18.785 Number Theory I Full Lecture Notes (F2021)

18.785 Number theory I Lecture #1 1 1.1 Fall 2021 9/8/2021 Absolute values and discrete valuations Introduction At its core, number theory is the study of the integer ring Z. By the fundamental theorem of arithmetic, every element of Z can be written uniquely as a product of primes (up to multiplication by a unit ±1), so it is natural to focus on the prime elements of Z. If p is a prime, the ideal (p) := pZ it generates is a maximal ideal (Z has Krull dimension one), and the residue field Z/pZ is the finite field Fp with p elements (unique up to isomorphism). The fraction field of Z is the field Q of rational numbers. The field Q and the finite fields Fp together make up the prime fields: every field k contains exactly one of them, according to its characteristic: k has characteristic zero if and only if it contains Q, and k has characteristic p if and only if k contains Fp . The structure of the ring Z and the distribution of its primes are both intimately related to properties of the Riemann zeta function X Y ζ(s) = (1 − p−s )−1 . n−s = p As a function of the complex variable s, the Riemann zeta function is holomorphic and nonvanishing on Re(s) > 1 and admits an analytic continuation to the entire complex plane. It has a simple pole at s = 1, which implies that there are infinitely many primes (otherwise the product over primes on the RHS would be finite and converge). The distribution of its zeros in the critical strip 0 < s < 1 is directly related to the distribution of primes (via the explicit formula, which we will see later in the course). As you are probably aware, Riemann famously conjectured more than 150 years ago that the zeros of ζ(s) in the critical strip all lie on the critical line Re(s) = 12 ; this conjecture remains open. One can also consider finite extensions of Q, such as the field Q(i) := Q[x]/(x2 +1). These are called number fields, and each can be constructed as the quotient of the polynomial ring Q[x] by one of its maximal ideals; the ring Q[x] is a principal ideal domain and its maximal ideals can all be written as (f ) for some monic irreducible f ∈ Z[x]. Associated to each number field K is a zeta function ζK (s), and each of these has an associated conjecture regarding the location of its zeros (these conjectures all remain open). Number fields are one of two types of global fields that we will spend much of the course studying; the others are global function fields. Let Fq denote the field with q elements, where q is any prime power. The polynomial ring Fq [t] has much in common with the integer ring Z. Like Z, it is a principal ideal domain of dimension one, and the residue fields Fq [t]/(f ) one obtains by taking the quotient by a maximal ideal (f ), where f ∈ Fq [t] is any irreducible polynomial, are finite fields Fqd , where d is the degree of f . In contrast to the situation with Z, the residue fields of Fq [t] all have the same characteristic as its fraction field Fq (t), which plays a role analogous to Q. Global function fields are finite extensions of Fq (t) (this includes Fq (t) itself, an extension of degree 1). Associated to each global field k is an infinite collection of local fields corresponding to the completions of k with respect to its absolute values; when k = Q, these completions are the field of real numbers R and the p-adic fields Qp (as you will prove on Problem Set 1). The ring Z is a principal ideal domain (PID), as is Fq [t], and in such fields every nonzero prime ideal is maximal and thus has an associated residue field. For both Z and Fq [t] these residue fields are finite, but the characteristics of the residue fields of Z are all different (and distinct from the characteristic of its fraction field), while those of Fq [t] are all the same. We will spend the first part of this course fleshing out this picture, in which we are particularly interested in understanding the integral closure of the rings Z and Fq [t] in finite extensions of their fraction fields (such integral closures are known as rings of integers), and the prime ideals of these rings. Where possible we will treat number fields and function fields on an equal footing, but we will also note some key differences. Surprisingly, the apparently more complicated function field setting often turns out to be simpler than the number field setting; for example, the analog of the Riemann hypothesis in the function field setting (the Riemann hypothesis for curves), is not an open problem. It was proved by André Weil in the 1940s [5]; a further generalization to varieties of arbitrary dimension was proved by Pierre Deligne in the 1970s [3]. Zeta functions provide the tool we need to understand the distribution of primes, both in general, and within particular residue classes; the proofs of the prime number theorem and Dirichlet’s theorem on primes in arithmetic progressions both use zeta functions in an essential way. Dirichlet’s theorem states that for each integer m > 1 and each integer a coprime to m, there are infinitely many primes p ≡ a mod m. In fact, more is true: the Chebotarev density theorem tells us that for each modulus m the primes are equidistributed among the residue classes of the integers a coprime to m. We will see this and several other applications of the Chebotarev density theorem in the later part of the course. Before we begin, let us note the following. Remark 1.1. Our rings always have a multiplicative identity that is preserved by ring homomorphisms (so the zero ring in which 1 = 0 is not an initial object in the category of rings, but it is the terminal object in this category). Except where noted otherwise, the rings we shall consider are all commutative. 1.2 Absolute values We begin with the general notion of an absolute value on a field; a reference for much of this material is [4, Chapter 1]. Definition 1.2. An absolute value on a field k is a map | | : k → R≥0 such that for all x, y ∈ k the following hold: 1. |x| = 0 if and only if x = 0; 2. |xy| = |x||y|; 3. |x + y| ≤ |x| + |y|. If the stronger condition 4. |x + y| ≤ max(|x|, |y|) also holds, then the absolute value is nonarchimedean; otherwise it is archimedean. Example 1.3. The map | | : k → R≥0 defined by ( 1 if x 6= 0, |x| = 0 if x = 0, is the trivial absolute value on k. It is nonarchimedean. 18.785 Fall 2021, Lecture #1, Page 2 Lemma 1.4. An absolute value | | on a field k is nonarchimedean if and only if |1 + ··· + 1| ≤ 1 | {z } n for all n ≥ 1. Proof. See Problem Set 1. Corollary 1.5. In a field of positive characteristic every absolute value is nonarchimedean, and the only absolute value on a finite field is the trivial one. Definition 1.6. Two absolute values | | and | |0 on the same field k are equivalent if there exists an α ∈ R>0 for which |x|0 = |x|α for all x ∈ k. 1.3 Absolute values on Q To avoid confusion we will denote the usual absolute value on Q (inherited from R) by | |∞ ; it is an archimedean absolute value. ButQthere are are infinitely many others. Recall that any element of Q× may be written as ± q q eq , where the product ranges over primes and the exponents eq ∈ Z are uniquely determined (as is the sign). Definition 1.7. For a prime p the p-adic valuation vp : Q → Z is defined by ! Y eq := ep , vp ± q q and we define vp (0) := ∞. The p-adic absolute value on Q is defined by |x|p := p−vp (x) , where |0|p = p−∞ is understood to be 0. Theorem 1.8 (Ostrowski’s Theorem). Every nontrivial absolute value on Q is equivalent to | |p for some p ≤ ∞. Proof. See Problem Set 1. Theorem 1.9 (Product Formula). For every x ∈ Q× we have Y |x|p = 1. p≤∞ Proof. See Problem Set 1. 1.4 Discrete valuations Definition 1.10. A valuation on a field k is a group homomorphism k × → R such that for all x, y ∈ k we have
v(x + y) ≥ min v(x), v(y) . We may extend v to a map k → R ∪ {∞} by defining v(0) := ∞. For any any 0 < c < 1, defining |x|v := cv(x) yields a nonarchimedean absolute value. The image of v in R is the 18.785 Fall 2021, Lecture #1, Page 3 value group of v. We say that v is a discrete valuation if its value group is equal to Z (every discrete subgroup of R is isomorphic to Z, so we can always rescale a valuation with a discrete value group so that this holds). Given a field k with valuation v, the set A := {x ∈ k : v(x) ≥ 0}, is the valuation ring of k (with respect to v). A discrete valuation ring (DVR) is an integral domain that is the valuation ring of its fraction field with respect to a discrete valuation; such a ring A cannot be a field, since v(Frac A) = Z 6= Z≥0 = v(A). It is easy to verify that every valuation ring A is a in fact a ring, and even an integral domain (if x and y are nonzero then v(xy) = v(x) + v(y) 6= ∞, so xy 6= 0), with k as its fraction field. Notice that for any x ∈ k × we have v(1/x) = v(1) − v(x) = −v(x), so at least one of x and 1/x has nonnegative valuation and lies in A. It follows that x ∈ A is invertible (in A) if and only if v(x) = 0, hence the unit group of A is A× = {x ∈ k : v(x) = 0}, We can partition the nonzero elements of k according to the sign of their valuation. Elements with valuation zero are units in A, elements with positive valuation are non-units in A, and elements with negative valuation do not lie in A, but their multiplicative inverses are nonunits in A. This leads to a more general notion of a valuation ring. Definition 1.11. A valuation ring is an integral domain A with fraction field k with the property that for every x ∈ k, either x ∈ A or x−1 ∈ A. Let us now suppose that the integral domain A is the valuation ring of its fraction field with respect to some discrete valuation v (which we shall see is uniquely determined). Any element π ∈ A for which v(π) = 1 is called a uniformizer. Uniformizers exist, since v(A) = Z≥0 . If we fix a uniformizer π, every x ∈ k × can be written uniquely as x = uπ n where n = v(x) and u = x/π n ∈ A× and uniquely determined. It follows that A is a unique factorization domain (UFD), and in fact A is a principal ideal domain (PID). Indeed, every nonzero ideal of A is equal to (π n ) = {a ∈ A : v(a) ≥ n}, for some integer n ≥ 0. Moreover, the ideal (π n ) depends only on n, not the choice of uniformizer π: if π 0 is any other uniformizer its unique representation π 0 = uπ 1 differs from π only by a unit. The ideals of A are thus totally ordered, and the ideal m = (π) = {a ∈ A : v(a) > 0} is the unique maximal ideal of A (and also the only nonzero prime ideal of A). Definition 1.12. A local ring is a commutative ring with a unique maximal ideal. Definition 1.13. The residue field of a local ring A with maximal ideal m is the field A/m. 18.785 Fall 2021, Lecture #1, Page 4 We can now see how to determine the valuation v corresponding to a discrete valuation ring A. Given a discrete valuation ring A with unique maximal ideal m, we may define v : A → Z by letting v(a) be the unique integer n for which (a) = mn and v(0) := ∞. Extending v to the fraction field k of A via v(a/b) := v(a) − v(b) gives a discrete valuation v on k for which A = {x ∈ k : v(x) ≥ 0} is the corresponding valuation ring. Notice that any discrete valuation v on k with A as its valuation ring must satisfy v(π) = 1 for some π ∈ m (otherwise v(k) 6= Z), and we then have v(π) = 1 if and only if m = (π). Moreover, v must then coincide with the discrete valuation we just defined: for any DVR A, the discrete valuation on the fraction field of A that yields A as its valuation ring is uniquely determined. It follows that we could have defined a uniformizer to be any generator of the maximal ideal of A without reference to a valuation. Example 1.14. For the p-adic valuation vp : Q → Z ∪ {∞} we have the valuation ring na o Z(p) := : a, b ∈ Z, p6 | b , b with maximal ideal m = (p); this is the localization of the ring Z at the prime ideal (p). The residue field is Z(p) /pZ(p) ' Z/pZ ' Fp . Example 1.15. For any field k, the valuation v : k((t)) → Z ∪ {∞} on the field of Laurent series over k defined by   X v an tn  = n0 , n≥n0 where an0 6= 0, has valuation ring k[[t]], the power series ring over k. For f ∈ k((t))× , the valuation v(f ) ∈ Z is the order of vanishing of f at zero. For every α ∈ k one can similarly define a valuation vα on k as the order of vanishing of f at α by taking the Laurent series expansion of f about α. 1.5 Discrete Valuation Rings Discrete valuation rings are in many respects the nicest rings that are not fields. In addition to being an integral domain, every discrete valuation ring A enjoys the following properties: • noetherian: Every increasing sequence I1 ⊆ I2 ⊆ · · · of ideals eventually stabilizes; equivalently, every ideal is finitely generated. • principal ideal domain: Every ideal is principal (generated by a single element). • local : There is a unique maximal ideal m. • dimension one: The (Krull) dimension of a ring R is the supremum of the lengths n of all chains of prime ideals p0 ( p1 ( · · · ( pn (which need not be finite, in general). For DVRs, (0) ⊆ m is the longest chain of prime ideals, with length 1. • regular : The dimension of the A/m-vector space m/m2 is equal to the dimension of A. Non-local rings are regular if this holds for every localization at a prime ideal. • integrally closed (or normal ): Every element of the fraction field of A that is the root of a monic polynomial in A[x] lies in A. • maximal : There are no intermediate rings strictly between A and its fraction field. 18.785 Fall 2021, Lecture #1, Page 5 Various combinations of these properties can be used to uniquely characterize discrete valuation rings (and hence give alternative definitions). Theorem 1.16. For an integral domain A, the following are equivalent: • A is a DVR. • A is a noetherian valuation ring that is not a field. • A is a local PID that is not a field. • A is an integrally closed noetherian local ring of dimension one. • A is a regular noetherian local ring of dimension one. • A is a noetherian local ring whose maximal ideal is nonzero and principal. • A is a maximal noetherian ring of dimension one. Proof. See [1, §23] or [2, §9]. 1.6 Integral extensions Integrality plays a key role in number theory, so it is worth discussing it in more detail. Definition 1.17. Given a ring extension A ⊆ B, an element b ∈ B is integral over A if is a root of a monic polynomial in A[x]. The ring B is integral over A if all its elements are. Proposition 1.18. Let α, β ∈ B be integral over A ⊆ B. Then α + β and αβ are integral over A. Proof. Let f ∈ A[x] and g ∈ A[y] be such that f (α) = g(β) = 0, where f (x) = a0 + a1 x + · · · + am−1 xm−1 + xm , g(y) = b0 + b1 y + · · · + bn−1 y n−1 + y n . It suffices to consider the case A = Z[a0 , . . . , am−1 , b0 , . . . , bn−1 ], and B= A[x, y]
, f (x), g(y) with α and β equal to the images of x and y in B, respectively, since given any A0 ⊆ B 0 we have homomorphisms A → A0 defined by ai → ai and bi → bi and B → B 0 defined by x 7→ α and y 7→ β, and if x + y, xy ∈ B are integral over A then α + β, αβ ∈ B 0 must be integral over A0 . Let k be the algebraic closure of the fraction field of A, and let α1 , . . . , αm be the roots of f in k and let β1 , . . . , βn be the roots of g in k. The polynomial  Y h(z) = z − (αi + βj ) i,j has coefficients that may be expressed as polynomials in the symmetric functions of the αi and βj , equivalently, the coefficients ai and bj of f and g, respectively. Thus Q h ∈ A[z], and h(x+y) = 0, so x+y is integral over A. Applying the same argument to h(z) = i,j (z−αi βj ) shows that xy is also integral over A. 18.785 Fall 2021, Lecture #1, Page 6 Definition 1.19. Given a ring extension B/A, the ring à = {b ∈ B : b is integral over A} is the integral closure of A in B. When à = A we say that A is integrally closed in B. For a domain A, its integral closure (or normalization) is its integral closure in its fraction field, and A is integrally closed (or normal ) if it is integrally closed in its fraction field. Proposition 1.20. If C/B/A is a tower of ring extensions in which B is integral over A and C is integral over B then C is integral over A. Proof. See [1, Thm. 10.27] or [2, Cor. 5.4]. Corollary 1.21. If B/A is a ring extension, then the integral closure of A in B is integrally closed in B. Proposition 1.22. The ring Z is integrally closed. Proof. We apply the rational root test: suppose r/s ∈ Q is integral over Z, where r and s are coprime integers. Then  r n s + an−1  r n−1 s + · · · a1 r s + a0 = 0 for some a0 , . . . , an−1 ∈ Z. Clearing denominators yields rn + an−1 srn−1 + · · · a1 sn−1 r + a0 sn = 0, thus rn = −s(an−1 rn−1 + · · · a1 sn−2 r + a0 sn−1 ) is a multiple of s. But r and s are coprime, so s = ±1 and therefore r/s ∈ Z. Corollary 1.23. Every unique factorization domain is integrally closed. In particular, every PID is integrally closed. Proof. The proof of Proposition 1.22 works for any UFD. √ Example 1.24. The ring Z[ 5] is not √ √ a UFD (nor a PID) because it is not integrally closed: √ consider φ = (1 + 5)/2 ∈ Frac√Z[ 5], which is integral over Z (and hence over Z[ 5]), √ since φ2 − φ − 1 = 0. But φ 6∈ Z[ 5], so Z[ 5] is not integrally closed. The corollary implies that every discrete valuation ring is integrally closed. In fact, more is true. Proposition 1.25. Every valuation ring is integrally closed. Proof. Let A be a valuation ring with fraction field k and let α ∈ k be integral over A. Then αn + an−1 αn−1 + an−2 αn−2 + · · · + a1 α + a0 = 0 for some a0 , a1 , . . . , an−1 ∈ A. Suppose α 6∈ A. Then α−1 ∈ A, since A is a valuation ring. Multiplying the equation above by α−(n−1) ∈ A and moving all but the first term on the LHS to the RHS yields α = −an−1 − an−1 α−1 − · · · − a1 α2−n − a0 α1−n ∈ A, contradicting our assumption that α 6∈ A. It follows that A is integrally closed. 18.785 Fall 2021, Lecture #1, Page 7 Definition 1.26. A number field K is a finite extension of Q. The ring of integers OK is the integral closure of Z in K. Remark 1.27. The notation ZK is also sometimes used to denote the ring of integers of K. The symbol O emphasizes the fact that OK is an order in K; in any Q-algebra K of finite dimension r, an order is a subring of K that is also a free Z-module of rank r, equivalently, a Z-lattice in K that is also a ring. In fact, OK is the maximal order of K: it contains every order in K. Proposition 1.28. Let A be an integrally closed domain with fraction field K. Let α be an element of a finite extension L/K, and let f ∈ K[x] be its minimal polynomial over K. Then α is integral over A if and only if f ∈ A[x]. Proof. The reverse implication is immediate: if f ∈ A[x] then certainly α is integral over A. For the forward implication, suppose α is integral over A and let g ∈ A[x] be a monic polynomial for which g(α) = 0. In K[x] we may factor f (x) as Y f (x) = (x − αi ). i For each αi we have a field embedding K(α) → K that sends α to αi and fixes K. As elements of K we have g(αi ) = 0 (since f (αi ) = 0 and f must divide g), so each αi ∈ K is integral over A and lies in the integral closure à of A in K. Each coefficient of f ∈ K[x] can be expressed as a sum of products of the αi , and is therefore an element of the ring à that also lies in K. But A = à ∩ K, since A is integrally closed in its fraction field K. √ Example√1.29. We saw in Example 1.24 that (1 + 5)/2 is integral over Z. Now consider α = (1 + 7)/2. Its minimal polynomial x2 − x − 3/2 6∈ Z[x], so α is not integral over Z. References [1] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, 2013. [2] Michael Atiyah and Ian MacDonald, Introduction to commutative algebra, Addison– Wesley, 1969. [3] Pierre Deligne, La conjecture de Weil: I , Publications Mathématiques l‘I.H.É.S. 43 (1974), 273–307. [4] Jean-Pierre Serre, Local fields, Springer, 1979. [5] André Weil, Numbers of solutions of equations in finite fields, Bulletin of the American Mathematical Society, 55 (1949), 497–508. 18.785 Fall 2021, Lecture #1, Page 8 18.785 Number theory I Lecture #2 2 Fall 2021 9/13/2021 Localization and Dedekind domains After a brief review of some commutative algebra background on localizations, in this lecture we begin our study of Dedekind domains, which are commutative rings that play a key role in algebraic number theory and arithmetic geometry (named after Richard Dedekind). 2.1 Localization of rings Let A be a commutative ring (unital, as always), and let S be a multiplicative subset of A; this means S is closed under finite products (including the empty product, so 1 ∈ S), and S does not contain zero. The localization of A with respect to S is a ring S −1 A equipped with a ring homomorphism ι : A → S −1 A that maps S into (S −1 A)× and satisfies the following universal property: if ϕ : A → B is a ring homomorphism with ϕ(S) ⊆ B × then there is a unique ring homomorphism S −1 A → B that makes the following diagram commute: ϕ ← ← ← A ∃! ι → → B → S −1 A and one says that ϕ factors uniquely through S −1 A (via ι). As usual with universal properties, this guarantees that S −1 A is unique (hence well-defined), provided that it exists. To prove existence we construct S −1 A as the quotient of A × S modulo the equivalence relation (a, s) ∼ (b, t) ⇔ ∃u ∈ S such that (at − bs)u = 0. (1) We then use a/s to denote the equivalence class of (a, s) and define ι(a) := a/1; one can easily verify that S −1 A is a ring with additive identity 0/1 and multiplicative identity 1/1, and that ι : A → S −1 A is a ring homomorphism. If s is invertible in A we can view a/s either as the element as−1 of A or the equivalence class of (a, s) in S −1 A; we have (a, s) ∼ (a/s, 1), since (a · 1 − a/s · s) · 1 = 0, so this notation should not cause any confusion. For s ∈ S we have ι(s)−1 = 1/s, since (s/1)(1/s) = s/s = 1/1 = 1, thus ι(S) ⊆ (S −1 A)× . If ϕ : A → B is a ring homomorphism with ϕ(S) ⊆ B × , then ϕ = π ◦ ι, where π is defined by π(a/s) := ϕ(a)ϕ(s)−1 . If π : S −1 A → B is any ring homomorphism that satisfies ϕ = π ◦ ι, then ϕ(a)ϕ(s)−1 = π(ι(a))π(ι(s))−1 = π(ι(a)ι(s)−1 ) = π((a/1)(1/s)) = π(a/s), so π is unique. In the case of interest to us, A is actually an integral domain, in which case (a, s) ∼ (b, t) if and only if at − bs = 0 (we can always take u = 1 in the equivalence relation (1) above), and we can then identify S −1 A with a subring of the fraction field of A (which we note is the localization of A with respect to S = A6=0 ), and if T is a multiplicative subset A that contains S, then S −1 A ⊆ T −1 A. When A is an integral domain the map ι : A → S −1 A is injective, allowing us to identify A with its image ι(A) ⊆ S −1 A (in general, ι is injective if and only if S contains no zero divisors). When A is an integral domain we may thus view S −1 A as an intermediate ring that lies between A and its fraction field: A ⊆ S −1 A ⊆ Frac A. 2.2 Ideals in localizations of rings If ϕ : A → B is a ring homomorphism and b is a B-ideal, then ϕ−1 (b) is an A-ideal called the contraction of b to A and sometimes denoted bc ; when A is a subring of B and ϕ is the inclusion map we simply have bc = b ∩ A. If a is an A-ideal, in general ϕ(a) is not a B-ideal; but we can instead consider the B-ideal generated by ϕ(a), the extension of a to B, sometimes denoted ae . In the case of interest to us, A is an integral domain, B = S −1 A is the localization of A with respect to some multiplicative set S, and ϕ = ι is injective, so we view A as a subring of B. We then have ae = aB := (ab : a ∈ a, b ∈ B). (2) We clearly have a ⊆ ϕ−1 ((ϕ(a))) = aec and bce = (ϕ(ϕ−1 (b))) ⊆ b; one might ask whether these inclusions are equalities. In general the first is not: if B = S −1 A and a ∩ S 6= ∅ then ae = aB = B and aec = B ∩ A are both unit ideals, but we may still have a ( A. However when B = S −1 A the second inclusion is an equality; see [1, Prop. 11.19] or [2, Prop. 3.11] for a short proof. We also note the following theorem. Theorem 2.1. Let S be a multiplicative subset of an integral domain A. There is a oneto-one correspondence between the prime ideals of S −1 A and the primes ideals of A that do not intersect S given by the inverse maps q 7→ q ∩ A and p 7→ pS −1 A. Proof. See [1, Cor. 11.20] or [2, Prop. 3.11.iv]. Remark 2.2. An immediate consequence of (2) is that if a1 , . . . , an ∈ A generate a as an A-ideal, then they also generate ae = aB as a B-ideal. As noted above, when B = S −1 A we have b = bce , so every B-ideal is of the form ae (take a = bc ). It follows that if A is noetherian then so are all its localizations, and if A is a PID then so are all of its localizations. An important special case of localization occurs when p is a prime ideal in an integral domain A, and S = A − p (the complement of the set p in the set A). In this case it is customary to denote S −1 A by Ap := {a/b : a ∈ A, b 6∈ p}/ ∼, (3) and call it the localization of A at p. The prime ideals of Ap are then in bijection with the prime ideals of A that lie in p. It follows that pAp is the unique maximal ideal of Ap and Ap is therefore a local ring (whence the term localization). Warning 2.3. The notation in (3) makes it tempting to assume that if a/b is an element of Frac A, then a/b ∈ Ap if and only if b 6∈ p. This is not necessarily true! As an element of Frac A, the notation “a/b" represents an equivalence class; if a/b = a0 /b0 with b0 6∈ Ap , then in fact a/b = a0 /b0 ∈ Ap . As a trivial example, take A = Z, p = (3), a/b = 9/3 and a0 /b0 = 3/1. You may object that we should write a/b in lowest terms, but when A is not a unique factorization domain it is not clear what this means. Example 2.4. For a field k, let A = k[x] and p = (x − 2). Then Ap = {f ∈ k(x) : f is defined at 2}. The ring A is a PID, so Ap is a PID with a unique nonzero maximal ideal (the ideal pAp ), hence a DVR. Its maximal ideal is pAp = {f ∈ k(x) : f (2) = 0}. The valuation on the field k(x) = Frac A corresponding to the valuation ring Ap measures the order of vanishing of functions f ∈ k(x) at 2. The residue field is Ap /pAp ' k, and the quotient map Ap  Ap /pAp sends f to f (2). 18.785 Fall 2021, Lecture #2, Page 2 Example 2.5. Let p ∈ Z be a prime. Then Z(p) = {a/b : a, b ∈ Z, p - b}. As in the previous example, Z is a PID and Z(p) is a DVR; the valuation on Q is the p-adic valuation. The residue field is Z(p) /pZ(p) ' Fp and the quotient map Z(p)  Fp is reduction modulo p. 2.3 Localization of modules The concept of localization generalizes immediately to modules. As above, let A be a commutative ring, let S a multiplicative subset of A, and let M be an A-module. The localization S −1 M of M with respect to S is an S −1 A-module equipped with an A-module homomorphism ι : M → S −1 M with the universal property that if N is an S −1 A-module and ϕ : M → N is an A-module homomorphism, then ϕ factors uniquely through S −1 M (via ι). Note that in this definition we are viewing S −1 A-modules as A-modules via the canonical homomorphism A → S −1 A that is part of the definition of S −1 A. Our definition of S −1 M reduces to the definition of S −1 A in the case M = A. The explicit construction of S −1 M is exactly the same as S −1 A, one takes the quotient of the S −1 A-module M × S modulo the same equivalence relation as in (1): (a, s) ∼ (b, t) ⇔ ∃u ∈ S such that (at − bs)u = 0, where a and b now denote elements of M , and ι(a) := a/1 as before. Alternatively, one can define S −1 M := M ⊗A S −1 A (see [2, Prop. 3.5] for a proof that this is equivalent). In other words, S −1 M is the base change of M from A to S −1 A; we will discuss base change more generally in later lectures. ×s The map ι : M → S −1 M is injective if and only if the map M −→ M is injective for every s ∈ S. This is a strong condition that does not hold in general, even when A is an integral domain (the annihilator of M may be non-trivial), but it applies to all the cases we care about. In particular, if A lies in a field K (in which case A must be an integral domain whose fraction field lies in K) and M is an A-module that is contained in a K-vector space. In this setting multiplication by any nonzero s ∈ A is injective and we can view M as an A-submodule of any of its localizations S −1 M . We will mostly be interested in the case S = A − p, where p is a prime ideal of A, in which case we write Mp for S −1 M , just as we write Ap for S −1 A. Proposition 2.6. Let A be a subring of a field K, and let M be an A-module contained in a K-vector space V (equivalently, for which the map M → M ⊗A K is injective).1 Then \ \ M= Mm = Mp , m p where m ranges over the maximal ideals of A, p ranges over the prime ideals of A, and the intersections take place in V . T T Proof. The fact that M ⊆ m Mm is immediate. Now suppose x ∈ m Mm and consider the A-ideal a := {a ∈ A : ax ∈ M }. For each maximal ideal m we can write x = m/s for some m ∈ M and s ∈ A − m; we then have sx ∈ M and s ∈ a, but s 6∈ m, so a 6⊆ m. It follows that a must be the unit ideal, so 1 ∈ a and x = 1 · x ∈ M ; thus ∩m Mm ⊆ M . We now note that each Mp contains some Mm (since each p is contained in some m), and every maximal ideal is prime, so ∩m Mm = ∩p Mp . 1 The image is a tensor product of A-modules that is also a K-vector space. We need the natural map to be injective in order to embed M in it. Note that V necessarily contains a subspace isomorphic to M ⊗A K. 18.785 Fall 2021, Lecture #2, Page 3 An important special case of this proposition occurs when K = Frac A and V = K, in which case M is an A-submodule of K. Every ideal I of A is an A-submodule of K, and can thus be localized as above. The localization of I (as an A-module) at a prime ideal p of A is the same thing as the extension of I (as an A-ideal) to the localization of A at p. In other words, Ip = {i/s : i ∈ I, s ∈ A − p} = {ia/s : i ∈ I, a ∈ A, s ∈ A − p} = IAp . We also have the following corollary of Proposition 2.6. Corollary 2.7. Let A be an integral domain. Every ideal I of A (including I = A) is equal to the intersection of its localizations at the maximal ideals of A, and also to the intersection of its localizations at the prime ideals of A. T Example 2.8. If A = Z then Z = p Z(p) in Q. Proposition 2.6 and Corollary 2.7 are powerful tools, because they allow us work in local rings (rings with just one maximal ideal), which often simplifies matters considerably. For example, to prove that an ideal I in an integral domain A satisfies a certain property, it is enough to show that this property holds for all its localizations Ip at prime ideals p and is preserved under intersections. We now want to consider rings A that satisfy some further assumptions that make its localizations become even easier to work with. 2.4 Dedekind domains Proposition 2.9. Let A be a noetherian domain. The following are equivalent: (i) For every nonzero prime ideal p ⊂ A the local ring Ap is a DVR. (ii) The ring A is integrally closed and dim A ≤ 1. Proof. If A is a field then (i) and (ii) both hold, so let us assume that A is not a field, and put K := Frac A. We first show that (i) implies (ii). Recall that dim A is the supremum of the length of all chains of prime ideals. It follows from Theorem 2.1 that every chain of prime ideals (0) ( p1 ( · · · ( pn extends to a corresponding chain in Apn of the same length; conversely, every chain in Ap contracts to a chain in A of the same length. Thus dim A = sup{dim Ap : p ∈ Spec A} = 1, since every Ap is either a DVR (p 6= (0)), in which case dim Ap = 1, or a field (p = (0)), in which case dim Ap = 0. Any x ∈ K that is integral over A is integral over every Ap (since they T all contain A), and the Ap are integrally closed, since they are DVRs or fields. So x ∈ p Ap = A, and therefore A is integrally closed, which shows (ii). To show that (ii) implies (i), we first show that the following properties are all inherited by localizations of a ring: (1) no zero divisors, (2) noetherian, (3) dimension at most one, (4) integrally closed. (1) is obvious, (2) was noted in Remark 2.2, and (3) follows from Theorem 2.1 since, as argued above, we have dim Ap ≤ dim A. To show (4), suppose x ∈ K is integral over Ap . Then xn + an−1 n−1 a1 a0 x + ··· + x + =0 sn−1 s1 s0 18.785 Fall 2021, Lecture #2, Page 4 for some a0 , . . . , an−1 ∈ A and s0 , . . . , sn−1 ∈ A − p. Multiplying both sides by sn , where s = s0 · · · sn−1 ∈ S, shows that sx is integral over A, hence an element of A, since A is integrally closed. But then sx/s = x is an element of Ap , so Ap is integrally closed as claimed. Thus (ii) implies that every Ap is an integrally closed noetherian local domain of dimension at most 1, and for p 6= (0) we must have dim Ap = 1. Thus for every nonzero prime ideal p, the ring Ap is an integrally closed noetherian local domain of dimension 1, and therefore a DVR, by Theorem 1.16. Definition 2.10. A noetherian domain satisfying either of the equivalent properties of Proposition 2.9 is called a Dedekind domain. Corollary 2.11. Every PID is a Dedekind domain. In particular, Z is a Dedekind domain, as is k[x] for any field k. Remark 2.12. Every PID is both a UFD and a Dedekind domain. Not every UFD is a Dedekind domain √(consider k[x, y], for any √ field k),√and not every Dedekind domain is a UFD (consider Z[ −13], in which (1 + −13)(1 − −13) = 2 · 7 = 14). However (as we shall see), every ring that is both a UFD and a Dedekind domain is a PID. One of our first goals in this course is to prove that ring of integers of number fields and coordinate rings of global function fields are Dedekind domains. More precisely, we will prove that if A is a Dedekind domain and L is a finite separable extension of its fraction field, then the integral closure of A in L is a Dedekind domain. This includes the two main cases of interest to us, in which either A = Z and L is a number field, or A = Fq [t] and L is a global function field. Recall from Lecture 1 that number fields and global function fields are the two types of global fields (as we will prove in later lectures). 2.5 Fractional ideals Throughout this subsection, A is a noetherian domain (not necessarily a Dedekind domain) and K is its fraction field. Definition 2.13. A fractional ideal of a noetherian domain A is a finitely generated Asubmodule of its fraction field. Fractional ideals generalize the notion of an ideal: when A is noetherian the ideals of A are precisely the finitely generated A-submodules of A, and when A is also a domain we can extend this notion to its fraction field. Every ideal of A is also a fractional ideal of A, but fractional ideals are typically not ideals because they need not be contained in A. Some authors use the term integral ideal to distinguish the fractional ideals that lie in A (and are thus ideals) but we will not use this terminology. Lemma 2.14. Let A be a noetherian domain with fraction field K, and let I ⊆ K be an A-module. Then I is finitely generated if and only if aI ⊆ A for some nonzero a ∈ A. Proof. For the forward implication, if r1 /s1 , . . . , rn /sn generate I as an A-module, then aI ⊆ A for a = s1 · · · sn . Conversely, if aI ⊆ A, then aI is an ideal, hence finitely generated (since A is noetherian), and if a1 , . . . , an generate aI then a1 /a, . . . , an /a generate I. 18.785 Fall 2021, Lecture #2, Page 5 Remark 2.15. Lemma 2.14 gives an alternative definition of fractional ideals that can be extended to domains that are not necessarily noetherian; they are A-submodules I of K for which there exists a nonzero r ∈ A such that rI ⊆ A. When A is noetherian this coincides with our definition above. Corollary 2.16. Every fractional ideal of A can be written in the form a1 I, for some nonzero a ∈ A and ideal I. Definition 2.17. A fractional ideal of A is principal if it is generated by one element, that is, it has the form xA for some x ∈ K. We will also use the notation (x) := xA to denote the principal fractional ideal generated by x ∈ K. As with ideals, we can add and multiply fractional ideals: I + J := (i + j : i ∈ I, j ∈ J), IJ := (ij : i ∈ I, j ∈ J). Here the notation (S) means the A-module generated by S ⊆ K. As with ideals, we actually have I + J = {i + j : i ∈ I, j ∈ J}, but the ideal IJ is typically not the same as set {ij : i ∈ I, j ∈ J}, it consists of all finite sums of elements in this set. We also have a new operation, corresponding to division. For any fractional ideals I, J with J nonzero, the set (I : J) := {x ∈ K : xJ ⊆ I} is called a colon ideal. Some texts refer to (I : J) as the ideal quotient of I by J, but note that it is not a quotient of A-modules (for example, (Z : Z) = Z but Z/Z = {0}). We do not assume I ⊆ J (or J ⊆ I), the definition makes sense for any fractional ideals I and J with J nonzero.2 If I = (x) and J = (y) are principal fractional ideals then (I : J) = (x/y), so colon ideals can be viewed as a generalization of division in K × . Lemma 2.18. Let I and J be fractional ideals of a noetherian domain A with J nonzero. Then (I : J) is a fractional ideal of A. Proof. It is clear from the definition that (I : J) is closed under addition and multiplication by elements of A (since I is), so (I : J) is an A-module of the fraction field of A. To show that (I : J) is finitely generated, we first suppose that I, J ⊆ A are ideals. For any nonzero j ∈ J ⊆ A we have j(I : J) ⊆ I ⊆ A, so (I : J) is finitely generated, by Lemma 2.14. For the general case, choose a and b so that aI ⊆ A and bJ ⊆ A via Lemma 2.14. Then (I : J) = (abI : abJ) with abI, abJ ⊆ A, which we have already shown is finitely generated. Definition 2.19. A fractional ideal I is invertible if IJ = A for some fractional ideal J. Inverses are unique when they exist: if IJ = A = IJ 0 then J = JA = JIJ 0 = AJ 0 = J 0 . We may use I −1 to denote the inverse of a fractional ideal I when it exists. Lemma 2.20. A fractional ideal I of A is invertible if and only if I(A : I) = A (in which case (A : I) is its inverse). Before proving the lemma, note that I(A : I) ⊆ A always holds, since for y ∈ I and x ∈ (A : I) we have xy ∈ xI ⊆ A, by the definition of (A : I). The lemma states that this inclusion is an equality precisely when I is invertible. 2 The definition still makes sense when J is the zero ideal, but (I : (0)) = K will typically not be finitely generated as an A-module, hence not a fractional ideal. 18.785 Fall 2021, Lecture #2, Page 6 Proof. Suppose I is invertible, with IJ = A. Then jI ⊆ A for all j ∈ J, so J ⊆ (A : I), and A = IJ ⊆ I(A : I) ⊆ A, so I(A : I) = A. In the next lecture we will prove that in a Dedekind domain every nonzero fractional ideal is invertible, but let us first note that this is not true in general. Example 2.21. Consider the subring A := Z+2iZ of the Gaussian integers (with i2 = −1). The set I := 2Z[i] is a non-invertible A-ideal (even though it is an invertible Z[i]-ideal); indeed, we have (A : I) = Z[i] and I(A : I) = 2Z[i] ( A. 2.6 Invertible fractional ideals and the ideal class group In this section A is a noetherian domain (not necessarily a Dedekind domain) and K is its fraction field. Recall that a fractional ideal of A is a finitely generated A-submodule of K, and if I and J are fractional ideals, so is the colon ideal (I : J) := {x ∈ K : xJ ⊆ I}, and we say that a fractional ideal I is invertible if IJ = A for some fractional ideal J. The definition of (A : I) implies I(A : I) ⊆ A, and Lemma 2.20 implies that I is invertible precisely when this inclusion is an equality, in which case the inverse of I is (A : I). Ideal multiplication is commutative and associative, thus the set of nonzero fractional ideals of a noetherian domain form an abelian monoid under multiplication with A = (1) as the identity. It follows that the subset of invertible fractional ideals is an abelian group. Definition 2.22. The ideal group IA of a noetherian domain A is the group of invertible fractional ideals. Note that, despite the name, elements of IA need not be ideals. Every nonzero principal fractional ideal (x) is invertible (since (x)−1 = (x−1 )), and a product of principal fractional ideals is principal (since (x)(y) = (xy)), as is the unit ideal (1), thus the set of nonzero principal fractional ideals PA is a subgroup of IA . Definition 2.23. Let A be a noetherian domain. The quotient cl(A) := IA /PA is the ideal class group of A; it is also called the Picard group of A and denoted Pic(A).3 Example 2.24. If A is a DVR with uniformizer π then its nonzero fractional ideals are the principal fractional ideals (π n ) with n ∈ Z (including n ≤ 0). We have (π m )(π n ) = (π m+n ), thus the ideal group of A is isomorphic to Z (under addition). In this case PA = IA and the ideal class group cl(A) is trivial. Remark 2.25. A Dedekind domain is a UFD if and only if its ideal class group is trivial (we will prove this in the next lecture), thus cl(A) may be viewed as a measure of how far A is from being a UFD. More generally, the ideal class group of an integrally closed noetherian domain A is trivial when A is a UFD, and the converse holds if one replaces the ideal class group with the divisor class group. One defines a divisor as an equivalence class of fractional ideals modulo the equivalence relation I ∼ J ⇔ (A : I) = (A : J), and in an integrally closed noetherian domain A (or more generally, a Krull domain), the set 3 In general, the Picard group of a commutative ring A as the group of isomorphism classes of A-modules that are invertible under tensor product (equivalently, projective modules of rank one). When A is a noetherian domain, the Picard group of A is canonically isomorphic to the ideal class group of A and the two notions may be used interchangeably. 18.785 Fall 2021, Lecture #2, Page 7 of divisors forms a group that contains principal divisors as a subgroup; the divisor class group is defined as the quotient, and it is trivial if and only if A is a UFD (this holds more generally for any Krull domain, see [5, Thm. 8.34]). In a Dedekind domain, fractional ideals are always distinct as divisors and every nonzero fractional ideal is invertible, so the ideal class group and divisor class group coincide.4 References [1] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, 2013. [2] Michael Atiyah and Ian MacDonald, Introduction to commutative algebra, Addison– Wesley, 1969. [3] Pete L. Clark, Commutative algebra, 2015. [4] Anthony W. Knapp, Advanced Algebra, Digital Second Edition, 2016. [5] Max D. Larsen and Paul J. McCarthy, Multiplicative thoery of ideals, Academic Press, 1971. 4 In general, the divisor class group and the ideal class group (or Picard group) of an integrally closed noetherian domain A may differ when dim A > 1; see [3, Thm. 19.38] for a dimension 2 an example in which the ideal class group is trivial but the divisor class group is not (implying that A is not a UFD). 18.785 Fall 2021, Lecture #2, Page 8 18.785 Number theory I Lecture #3 3 Fall 2021 09/15/2021 Properties of Dedekind domains In the previous lecture we defined a Dedekind domain as a noetherian domain A that satisfies either of the following equivalent conditions: • the localizations of A at its nonzero prime ideals are all discrete valuation rings; • A is integrally closed and has dimension at most one. In this lecture we will establish several additional properties enjoyed by Dedekind domains, the most significant of which is unique factorization of ideals. As we noted last time, Dedekind domains are typically not unique factorization domains (the only exceptions are principal ideal domains), but ideals can be uniquely factored into prime ideals. 3.1 Invertible ideals in Dedekind domains Our first goal is prove that every nonzero fractional ideal in a Dedekind domain is invertible. We will use the fact that arithmetic of fractional ideals behaves well under localization. Lemma 3.1. Let I and J be fractional ideals of a noetherian domain A, and let p be a prime ideal of A. Then Ip and Jp are fractional ideals of Ap , as are (I + J)p = Ip + Jp , (IJ)p = Ip Jp , (I : J)p = (Ip : Jp ). The same applies if we localize with respect to any multiplicative subset S of A. Proof. Ip = IAp is a finitely generated Ap -module (since I is a finitely generated A-module; see Remark 2.2), hence a fractional ideal of Ap , and similarly for Jp . We have (I + J)p = (I + J)Ap = IAp + JAp = Ip + Jp , where we use the distributive law in K to get (I + J)Ap = IAp + JAp . We also have (IJ)p = (IJ)Ap = Ip Jp , since (IJ)Ap ⊆ Ip Jp obviously holds and by writing fractions over a common denominator we can see that Ip Jp ⊆ (IJ)Ap also holds. Finally (I : J)p = {x ∈ K : xJ ⊆ I}p = {x ∈ K : xJp ⊆ Ip } = (Ip : Jp ). For the last statement, note that no part of our proof depends on the fact that we localized with respect to a multiplicative set of the from A − p. Theorem 3.2. Let I be a fractional ideal of a noetherian domain A. Then I is invertible if and only if its localization at every maximal ideal of A is invertible, equivalently, if and only if its localization at every prime ideal of A is invertible. Proof. Suppose I is invertible. Then I(A : I) = A, and for any maximal ideal m we have Im (Am : Im ) = Am , by Lemma 3.1, so Im is also invertible. Now suppose Im is invertible for every maximal ideal m; then Im (Am : Im ) = Am for every maximal ideal m. Applying Lemma 3.1 and Proposition 2.6 yields \ \ \ I(A : I) = (I(A : I))m = Im (Am : Im ) = Am = A, m m so I is invertible. The same proof works for prime ideals. m Corollary 3.3. In a Dedekind domain every nonzero fractional ideal is invertible. Proof. If A is Dedekind then all of its localizations at maximal ideals are DVRs, hence PIDs, and in a PID every nonzero fractional ideal is invertible. It follows from Theorem 3.2 that every nonzero fractional ideal of A is invertible. An integral domain in which every nonzero ideal is invertible is a Dedekind domain (see Problem Set 2), so this gives another way to define Dedekind domains. Let us also note an equivalent condition that will be useful in later lectures. Lemma 3.4. A nonzero fractional ideal I in a noetherian local domain A is invertible if and only if it is principal. Proof. If I is principal then it is invertible, so we only need to show the converse. Let I be an the maximal ideal of A. We have II −1 = A, so Pninvertible fractional ideal, and let m be −1 , and each ai bi lies in II −1 = A. One of the i=1 ai bi = 1 for some ai ∈ I and bi ∈ I products ai bi , say a1 b1 , must be a unit, otherwise the sum would not be a unit (note that A = m t A× , since A is a local ring). For every x ∈ I we have a1 b1 x ∈ (a1 ), since b1 x ∈ A (because x ∈ I and b1 ∈ I −1 ). It follows that x = (a1 b1 )−1 a1 b1 x ∈ (a1 ), since (a1 b1 )−1 ∈ A, so we have I ⊆ (a1 ) ⊆ I, which shows that I = (a1 ) is principal. Corollary 3.5. A nonzero fractional ideal in a noetherian domain A is invertible if and only if it is locally principal, that is, its localization at every maximal ideal of A is principal. 3.2 Unique factorization of ideals in Dedekind domains We are now ready to prove the main result of this lecture, that every nonzero ideal in a Dedekind domain has a unique factorization into prime ideals. As a first step we need to show that every ideal is contained in only finitely many prime ideals. Lemma 3.6. Let A be a Dedekind domain and let a ∈ A be nonzero. The set of prime ideals that contain a is finite. Proof. Consider the following subsets S and T of the ideal group IA : S := {I ∈ IA : (a) ⊆ I ⊆ A}, T := {I ∈ IA : A ⊆ I ⊆ (a)−1 }. The sets S and T are both non-empty (they contain A) and partially ordered by inclusion. The elements of S are all ideals, and we have bijections ϕ1 : S → T I 7→ I −1 ϕ2 : T → S I 7→ aI with ϕ1 order-reversing and ϕ2 order-preserving. The composition ϕ := ϕ2 ◦ ϕ1 is thus an order-reversing permutation of S. Since A is noetherian, the set S satisfies the ascending chain condition: every chain I1 ⊆ I2 ⊆ I3 ⊆ · · · of ideals in S is eventually constant. By applying our order-reversing permutation ϕ we see that S also satisfies the descending chain condition: every chain I1 ⊇ I2 ⊇ I3 ⊇ · · · of ideals in S is eventually constant. Now if a lies in infinitely many distinct prime ideals p1 , p2 , p3 , . . ., then p1 ⊇ p1 ∩ p2 ⊇ p1 ∩ p2 ∩ p3 ⊇ · · · 18.785 Fall 2021, Lecture #3, Page 2 is a descending chain of ideals in S that must stabilize. Thus for n sufficiently large we have p1 · · · pn−1 ⊆ p1 ∩ · · · ∩ pn−1 = p1 ∩ · · · ∩ pn ⊆ pn . The prime ideal pn contains the product p1 · · · pn−1 , so it must contain one of the factors p1 , · · · , pn−1 (this is what it means for an ideal to be prime). But this contradicts dim A ≤ 1: we cannot have a chain of prime ideals (0) ( pi ( pn of length 2 in A. Corollary 3.7. Let I be a nonzero ideal of a Dedekind domain A. The number of prime ideals of A that contain I is finite. Proof. Apply Lemma 3.6 to any nonzero a ∈ I. Example 3.8. The Dedekind domain A = C[t] contains uncountably many nonzero prime ideals pr = (t − r), one for each r ∈ C. But any nonzero f ∈ C[t] lies in only finitely many of them, namely, the pr for which f (r) = 0; equivalently, f has finitely many roots. Let p be a nonzero prime ideal in a Dedekind domain A with fraction field K, let π be a uniformizer for the discrete valuation ring Ap , and let I be a nonzero fractional ideal of A. The localization Ip is a nonzero fractional ideal of Ap , hence of the form (π n ) for some n ∈ Z that does not depend on the choice of π (note that n may be negative). We now extend the valuation vp : K → Z ∪ {∞} to fractional ideals by defining vp (I) := n and vp ((0)) := ∞; for any x ∈ K we have vp ((x)) = vp (x). The map vp : IA → Z is a group homomorphism: if Ip = (π m ) and Jp = (π n ) then (IJ)p = Ip Jp = (π m )(π n ) = (π m+n ), so vp (IJ) = m + n = vp (I) + vp (J). It is order-reversing with respect to the partial ordering on IA by inclusion and the total order on Z: for any I, J ∈ IA , if I ⊆ J then vp (I) ≥ vp (J). Lemma 3.9. Let p be a nonzero prime ideal in a Dedekind domain A. If I is an ideal of A then vp (I) = 0 if and only if p does not contain I. In particular, if q is any nonzero prime ideal different from p then vq (p) = vp (q) = 0. Proof. If I ⊆ p then vp (I) ≥ vp (p) = 1 is nonzero. If I 6⊆ p then pick a ∈ I − p and note that 0 = vp (a) ≥ vp (I) ≥ vp (A) = 0, since (a) ⊆ I ⊆ A. The prime ideals p and q are nonzero, hence maximal (since dim A ≤ 1), so neither contains the other and vq (p) = vp (q) = 0. Corollary 3.10. Let A be a Dedekind domain with fraction field K. For each nonzero fractional ideal I we have vp (I) = 0 for all but finitely many prime ideals p. In particular, if x ∈ K × then vp (x) = 0 for all but finitely many p. Proof. For I ⊆ A this follows from Corollary 3.7 and Lemma 3.9. For I 6⊆ A let I = a1 J with a ∈ A and J ⊆ A. Then vp (I) = vp (J) − vp (a) = 0 − 0 = 0 for all but finitely many prime ideals p. This holds in particular for I = (x), for any x ∈ K × . We are now ready to prove our main theorem. Theorem 3.11. Let A be a Dedekind domain. The ideal group IA of A is the free abelian group generated by its nonzero prime ideals p. The isomorphism M IA ' Z p 18.785 Fall 2021, Lecture #3, Page 3 is given by the inverse maps Y p I 7→ (. . . , vp (I), . . .) pep ←[ (. . . , ep , . . .) Proof. Corollary 3.10 implies that the first map is well defined (the vector associated to I ∈ IA has only finitely many nonzero entries and is thus an element of the direct sum). For each nonzero prime ideal p, the maps I 7→ vp (I) and ep 7→ pep are group homomorphisms, and it follows that the maps in the theorem are both group homomorphisms. To see that the first map is injective, note that if vp (I) = vp (J) then Ip = Jp , and if this holds for every p then I = ∩p Ip = ∩p Jp = J, by Corollary 2.7. To see that it is surjective, note that Lemma 3.9 implies that for any vector (. . . , ep , . . .) in the image we have ! X Y ep vq (p) = eq , vq pep = p p Q which implies that p pep is the pre-image of (. . . , ep , . . .); it also shows that the second map is the inverse of the first map. Remark 3.12. When A is a DVR, the isomorphism given by Theorem 3.11 is just the ∼ discrete valuation map vp : IA −→ Z, where p is the unique maximal ideal of A. Corollary 3.13. Q In a Dedekind domain every nonzero fractional ideal I has a unique factorization I = p pvp (I) into nonzero prime ideals p.1 Remark 3.14. Every integral domain with unique ideal factorization is a Dedekind domain (see Problem Set 2). The isomorphism of Theorem 3.11 allowsQ us to reinterpret the operations we have defined Q on fractional ideals. If I = p pep and J = p pfp are nonzero fractional ideals then IJ = (I : J) = I +J = I ∩J = and for all I, J ∈ IA we have Y pep +fp , Y pmin(ep ,fp ) = gcd(I, J), Y Y pep −fp , pmax(ep ,fp ) = lcm(I, J), IJ = (I ∩ J)(I + J). A key consequence of unique factorization is that I ⊆ J if and only if ep ≥ fp for all p; this implies that J contains I if and only if J divides I. Recall that in any commutative ring, if J divides I (i.e. JH = I for some ideal H) then J contains I (the elements of I are H-linear, hence A-linear, combinations of elements of J and so lie in J), whence the slogan to divide is to contain. In a Dedekind domain the converse is also true: to contain is to divide. This leads to another characterization of Dedekind domains (see Problem Set 2). 1 We view A = Q p pvp (A) = Q p p0 = (1) as an (empty) product of prime ideals. 18.785 Fall 2021, Lecture #3, Page 4 Given that inclusion and divisibility are equivalent in a Dedekind domain, we may view I + J as the greatest common divisor of I and J (it is the smallest ideal that contains, hence divides, both I and J), and I ∩ J as the least common multiple of I and J (it is the largest ideal contained in, hence divisible by, both I and J).2 We also note that (where I = Q pp ep x ∈ I ⇐⇒ (x) ⊆ I ⇐⇒ vp (x) ≥ ep for all p, as above), and therefore I = {x ∈ K : vp (x) ≥ ep for all p}. We have I ⊆ A if and only if ep ≥ 0 for all p. Corollary 3.15. A Dedekind domain is a UFD if and only if it is a PID, equivalently, if and only if its class group is trivial. Proof. Every PID is a UFD, so we only need to prove the reverse implication. The fact that we have unique factorization of ideals implies that it is enough to show that every prime ideal is principal. Let p be a nonzero prime ideal in a Dedekind domain A that is also a UFD, let a ∈ p nonzero, and let a = p1 · · · pn be the unique factorization of a into irreducible elements. Now p contains and therefore divides (a) = (p1 ) · · · (pn ), so p divides (and therefore contains) some (pi ), which is necessarily a prime ideal (in a UFD, irreducible elements generate prime ideals). But A has dimension one, so we must have p = (pi ). 3.3 Representing ideals in a Dedekind domain Most Dedekind domains are not PIDs, so a typical Dedekind domain will contain ideals that require more than one generator. But it turns out that two generators always suffice, and we can even pick one of them arbitrarily. To prove this we need the following lemma. Recall that two A-ideals I and J are said to be relatively prime, or coprime, if I + J = A; equivalently, gcd(I, J) = (1). Lemma 3.16. Let A be a Dedekind domain and let I and I 0 be nonzero ideals. There exists an ideal J coprime to I 0 such that IJ is principal. Proof. Let p1 , . . . , pn be the nonzero prime ideals dividing I 0 (a finite list, by Corollary 3.7). For 1 ≤ i ≤ n define the ideal ai := p1 · · · pi−1 pi+1 · · · pn and choose ai ∈ I so that a i ∈ ai I and ai 6∈ pi I. Note that ai I ∩ pi I ( ai I because vpi (ai I ∩ pi I) = vpi (pi I) > vpi (I) = vpi (ai I), so such an ai exists. Each ai is necessarily nonzero, and satisfies vpi (ai ) = vpi (I) since vpi (ai ) ≥ vpi (ai I) = vpi (I) and vpi (ai ) < vpi (pi I) = vpi (I) + 1, and for j 6= i we have vpj (ai ) ≥ vpj (pj I) > vpj (I). We now define a := a1 + · · · + an , so that vpi (a) = vpi (ai ) = vpi (I) for 1 ≤ i ≤ n (by the nonarchimedean triangle equality; see Problem Set 1). We thus have vp (a) = vp (I) for all prime ideals p|I 0 . Now (a) is contained in I and therefore divisible by I (since A is a Dedekind domain), so (a) = IJ for some ideal J. For each prime ideal p|I 0 we have vp (J) = vp (a) − vp (I) = 0, so J is coprime to I 0 , and IJ = (a) is principal as desired. 2 It may seem strange at first glance that the greatest common divisor of I and J is the smallest ideal dividing I and J, but note that if A = Z then gcd((a), (b)) = (gcd(a, b)) for any a, b ∈ Z, so the terminology is consistent (note that bigger numbers generate smaller ideals). 18.785 Fall 2021, Lecture #3, Page 5 One can show that every integral domain satisfying Lemma 3.16 is a Dedekind domain (see Problem Set 2). Corollary 3.17 (Finite approximation). Let I be a nonzero fractional ideal in a Dedekind domain A and let p1 , . . . , pn be a finite set of nonzero prime ideals of A. Then I contains an element x for which vpi (x) = vpi (I) for 1 ≤ i ≤ n. Proof. Let I = 1s J with s ∈ A and J an ideal. As in the proof of Lemma 3.16, we can pick a ∈ J so that vpi (a) = vpi (J) for 1 ≤ i ≤ n. If we now let x = a/s then we have vpi (x) = vpi (a) − vpi (s) = vpi (J) − vpi (s) = vpi (I) for 1 ≤ i ≤ n as desired. Corollary 3.18. Let I be a nonzero ideal in a Dedekind domain A. The quotient ring A/I is a principal ideal ring (every ideal in A/I is principal). Proof. Let ϕ : A → A/I be the quotient map, let J¯ be an (A/I)-ideal and let J := ϕ−1 (J) be its inverse image in A; then I ⊆ J, and J¯ ' J/I as (A/I)-modules. By Corollary 3.17 we may choose a ∈ J so that vp (a) = vp (J) for all nonzero prime ideals p|I. For every nonzero prime ideal p we then have vp (J) ≤ vp (I) and ( min(vp (a), vp (I)) = vp (a) = vp (J) if p|I, vp ((a) + I) = min(vp (a), vp (I)) = vp (I) = 0 = vp (J) if p - I, so (a) + I = J (here we are using unique factorization of ideals; in a Dedekind domain two ideals with the same valuation at every nonzero prime ideal must be equal). If follows that J¯ ' J/I = ((a) + I)/I = ϕ((a)) = (ϕ(a)) is principal. The converse of Corollary 3.18 also holds; an integral domain whose quotients by nonzero ideals are principal ideal rings is a Dedekind domain (see Problem Set 2). Definition 3.19. A ring that has only finitely many maximal ideals is called semilocal. Example 3.20. The ring Z(3) ∩ Z(5) is semilocal, it has just two maximal ideals. Corollary 3.21. Every semilocal Dedekind domain is a principal ideal domain. Proof. If we let I 0 be the product of all the prime ideals in A and apply Lemma 3.16 to any ideal I we will necessarily have J = A and IJ = I principal. Theorem 3.22. Let I be a nonzero ideal in a Dedekind domain A and let a ∈ I be nonzero. Then I = (a, b) for some b ∈ I. Proof. We have (a) ⊆ I, so I divides (a) and we have II 0 = (a) for some nonzero ideal I 0 . By Lemma 3.16 there is an ideal J coprime to I 0 such that IJ is principal, so IJ = (b) for some b ∈ I. We have gcd((a), (b)) = gcd(II 0 , IJ) = I, since gcd(I 0 , J) = (1), and it follows that I = (a, b). Theorem 3.22 gives us a convenient way to represent ideals I in the ring of integers of a global field. We can always pick a ∈ Z or a ∈ Fq [t]; we will see in later lectures that there is a natural choice for a (the absolute norm of I). It also gives us yet another characterization of Dedekind domains: they are precisely the integral domains for which Theorem 3.22 holds. We end this section with a theorem that summarizes the various equivalent definitions of a Dedekind domain that we have seen. 18.785 Fall 2021, Lecture #3, Page 6 Theorem 3.23. Let A be an integral domain. The following are equivalent: • A is an integrally closed noetherian domain of dimension at most one. • A is noetherian and its localizations at nonzero prime ideals are DVRs. • Every nonzero ideal in A is invertible. • Every nonzero ideal in A is a (finite) product of prime ideals. • A is noetherian and “to contain is to divide" holds for ideals in A. • For every ideal I in A there is an ideal J in A such that IJ is principal. • Every quotient of A by a nonzero ideal is a principal ideal ring. • For every nonzero ideal I in A and nonzero a ∈ I we have I = (a, b) for some b ∈ I. Proof. See Problem Set 2. 18.785 Fall 2021, Lecture #3, Page 7 18.785 Number theory I Lecture #4 4 4.1 Fall 2021 09/20/2021 Étale algebras, norm and trace Separability In this section we briefly review some standard facts about separable and inseparable field extensions that we will use repeatedly throughout the course. Those familiar with this material should feel free to skim it. In this section K denotes any field, K is an algebraic closure that we will typically P choose to contain any extensions under consideration, P L/K and for any polynomial f = ai xi ∈ K[x] we use f 0 := iai xi−1 to denote the formal derivative of f (this definition also applies when K is an arbitrary ring). Definition 4.1. A polynomial f in K[x] is separable if (f, f 0 ) = (1), that is, gcd(f, f 0 ) is a unit in K[x]. Otherwise f is inseparable. If f is separable then it splits into distinct linear factors over over K, where it has deg f distinct roots; this is sometimes used as an alternative definition. Note that the property of separability is intrinsic to the polynomial f , it does not depend on the field we are working in; in particular, if L/K is any field extension the separability of a polynomial f ∈ K[x] ⊆ L[x] does not depend on whether we view f as an element of K[x] or L[x]. Warning 4.2. Older texts (such as Bourbaki) define a polynomial in K[x] to be separable if all of its irreducible factors are separable (under our definition); so (x−1)2 is separable under this older definition, but not under ours. This discrepancy does not change the definition of separable elements or field extensions. Definition 4.3. Let L/K be an algebraic field extension. An element α ∈ L is separable over K if it is the root of a separable polynomial in K[x] (in which case its minimal polynomial is necessarily separable). The extension L/K is separable if every α ∈ L is separable over K; otherwise it is inseparable. Lemma 4.4. An irreducible polynomial f ∈ K[x] is inseparable if and only if f 0 = 0. Proof. Let f ∈ K[x] be irreducible; then f is nonzero and not a unit, so deg f > 0. If f 0 = 0 then gcd(f, f 0 ) = f 6∈ K × and f is inseparable. If f is inseparable then g := gcd(f, f 0 ) is a nontrivial divisor of f and f 0 . This implies deg g = deg f , since f is irreducible, but then deg f 0 < deg f = deg g, so g cannot divide f 0 unless f 0 = 0. Corollary 4.5. Let f ∈ K[x] be irreducible and let p ≥ 0 be the characteristic of K. We n have f (x) = g(xp ) for some irreducible separable g ∈ K[x] and integer n ≥ 0 that are uniquely determined by f . Proof. If f is separable the theorem holds with g = f and n = 0; for uniqueness, note that n if p = 0 then pn 6= 0 if and only if n = 0, and if p > 0 and g(xp ) is inseparable unlessPn = 0 n n n because g(xp )0 = g 0 (xp )pnP xp −1 = 0 (by the previous lemma). Otherwise f (x) := fr xr 0 r−1 is inseparable and f (x) = rfr x = 0 (by the lemma), and this can occur only if p > 0 and fr = 0 for all r ≥ 0 not divisible by p. So f = g(xp ) for some (necessarily irreducible) g ∈ K[x]. If g is separable we are done; otherwise we proceed by induction. As above, the n uniqueness of g and n is guaranteed by the fact that g(xp )0 = 0 for all n > 0. Corollary 4.6. If char K = 0 then every algebraic extension of K is separable. Lemma 4.7. Let L = K(α) be an algebraic field extension contained in an algebraic closure K of K and let f ∈ K[x] be the minimal polynomial of α over K. Then # HomK (L, K) = #{β ∈ K : f (β) = 0} ≤ [L : K], with equality if and only if α is separable over K. Proof. Each element of HomK (L, K) is uniquely determined by the image of α, which must be a root β of f (x) in K. The number of these roots is equal to [L : K] = deg f precisely when f , and therefore α, is separable over K. Definition 4.8. Let L/K be a finite extension of fields. The separable degree of L/K is [L : K]s := # HomK (L, K). The inseparable degree of f is [L : K]i := [L : K]/[L : K]s We will see shortly that [L : K]s always divides [L : K], so [L : K]i is an integer (in fact a power of the characteristic of K), but it follows immediately from our definition that [L : K] = [L : K]s [L : K]i . holds regardless. Theorem 4.9. Let L/K be an algebraic field extension. and let φK : K → Ω be a homomorphism to an algebraically closed field Ω. Then φK extends to a homomorphism φL : L → Ω. Proof. We use Zorn’s lemma. Define a partial ordering on the set F of pairs (F, φF ) for which F/K is a subextension of L/K and φF : F → Ω extends φK by defining (F1 , φF1 ) ≤ (F2 , φF2 ) whenever F2 contains F1 and φF2 extends φF1 . Given any totally ordered subset C of F, let S E be the field {F : (F, φF ) ∈ C} and define φE : E → Ω by φE (x) = φF (x) for x ∈ F ⊆ E (this does not depend on the choice of F because C is totally ordered). Then (E, φE ) is a maximal element of C, and by Zorn’s lemma, F contains a maximal element (M, φM ). We claim that M = L. If not, then pick α ∈ L−M and consider the field F = M (α) ⊆ L properly containing M , and extend φM to ϕF : F → Ω be letting φF (α) be any root of αM (f ) in Ω, where f ∈ M [x] is the minimal polynomial of α over M and αM (f ) is the image of f in Ω[x] obtained by applying φM to each coefficient. Then (M, φM ) is strictly dominated by (F, φF ), contradicting its maximality. Lemma 4.10. Let L/F/K be a tower of finite extensions of fields and K be an algebraic closure of K that contains L. Then # HomK (L, K) = # HomK (F, K)# HomF (L, K). Proof. The result is immediate when F = K or L = F (the RHS is 1 times the LHS), so we assume K ( F ( L and decompose the extensions L/F and F/K into finite towers of non-trivial simple extensions K = K0 ( K1 ( · · · ⊆ Km = F = Km ( Km+1 ( · · · ( Kn = L, 18.785 Fall 2021, Lecture #4, Page 2 where Ki = Ki−1 (αi ) for 1 ≤ i ≤ n. To prove the lemma it suffices to show that # HomK0 (Kn , K) = n Y # HomKi−1 (Ki , K) i=1 for any tower of proper simple extensions K0 ( K1 ( · · · ( Kn . We now consider the map Φ : HomK (Kn , K) → K n ϕ 7→ (ϕ(α1 ), · · · , ϕ(αn )). This map Φ is injective, since ϕ : Kn = K0 (α1 , . . . , αn ) → K is uniquely determinedQby the images ϕ(α1 ), . . . ϕ(αn ), so it suffices to show that the image of Φ has cardinality ni=1 hi , where hi := # HomKi−1 (Ki , K). For any polynomial f ∈ K[x] let r(f ) := {β ∈ K : f (β) = 0} be the set of its K-roots. Each σ ∈ AutK (K) induces a bijection r(f ) → r(σ(f )) with #r(f ) = #r(σ(f )). Let fi be the minimal polynomial of αi over Ki−1 . Lemma 4.7 implies hi = #r(fi ), and for any ϕ ∈ HomK (Kn , K) we have hi = #r(ϕ(fi )), since Theorem 4.9 allows us to extend ϕ to an element of HomK (K, K) = AutK (K). Let ϕ0 : K → K be the inclusion map. For each ϕ ∈ HomK (Kn , K) we may define a compatible sequence of pairs (ϕi , βi ) with ϕi ∈ HomK (Ki , K) satisfying ϕi |K = ϕi−1 i−1 and βi ∈ r(ϕi−1 (fi )) by putting ϕi = ϕ|K and βi = ϕ(αi ); note that (β1 , . . . , βn ) uniquely i determine ϕ and the sequence of pairs (αi , βi ), so the set of such compatible sequences of pairs is in bijection with the image of Φ. We can construct such a compatible sequence by initially choosing β1 ∈ r(f1 ) and letting ϕ1 (α1 ) = β1 uniquely determine ϕ1 ∈ HomK (K1 , K); there are exactly h1 choices for the pair (α1 , β1 ). To extend this sequence we choose β2 ∈ r(ϕ1 (f2 )) and let ϕ2 (α2 ) = β2 and ϕ2 |K = ϕ1 uniquely determine ϕ2 ∈ HomK (K2 , K). There are 1 exactly h choices for the pair (α 2 2 , β2 ). Continuing in this fashion, we find there are exactly Qn i ) corresponding to some ϕ ∈ HomK (Kn , K). The image of i=1 hi sequences of pairs (αi , βQ the map Φ thus has cardinality ni=1 hi as desired. Corollary 4.11. Let L/F/K be a tower of finite extensions of fields. Then [L : K]s = [L : F ]s [F : K]s [L : K]i = [L : F ]i [F : K]i Proof. The first equality follows from the lemma and the second follows from the identities [L : K] = [L : F ][F : K] and [L : K] = [L : K]s [L : K]i . Theorem 4.12. Let L/K be a finite extension of fields. The following are equivalent: (a) L/K is separable; (b) [L : K]s = [L : K]; (c) L = K(α) for some α ∈ L separable over K; (d) L ' K[x]/(f ) for some monic irreducible separable polynomial f ∈ K[x]. Proof. The equivalence of (c) and (d) is immediate (let f be the minimal polynomial of α and let α be the image of x in K[x]/(f )), and the equivalence of (b) and (c) is given by Lemma 4.7. That (a) implies (c) is the Primitive Element Theorem, see [2, §15.8] or [3, §V.7.4] for a proof. It remains only to show that (c) implies (a). 18.785 Fall 2021, Lecture #4, Page 3 So let L = K(α) with α separable over K. For any β ∈ L we can write L = K(β)(α), and we note that α is separable over K(β), since its minimal polynomial over K(β) divides it minimal polynomial over K, which is separable. Lemma 4.7 implies [L : K]s = [L : K] and [L : K(β)]s = [L : K(β)] (since L = K(α) = K(β)(α)), and the equalities [L : K] = [L : K(β)][K(β) : K] [L : K]s = [L : K(β)]s [K(β) : K]s then imply [K(β) : K]s = [K(β) : K]. So β is separable over K (by Lemma 4.7). This applies to every β ∈ L, so L/K is separable and (a) holds. Corollary 4.13. Let L/K be a finite extension of fields. Then [L : K]s ≤ [L : K] with equality if and only if L/K is separable. Proof. We have already established this for simple extensions, and otherwise we my decompose L/K into a finite tower of simple extensions and proceed by induction on the number of extensions, using the previous two corollaries at each step. Corollary 4.14. Let L/F/K is a tower of finite extensions of fields. Then L/K is separable if and only if both L/F and F/K separable. Proof. The forward implication is immediate and the reverse implication follows from Corollaries 4.11 and 4.13. Corollary 4.15. Let L/F/K be a tower of algebraic field extensions. Then L/K is separable if and only if both L/F and F/K are separable. Proof. As in the previous corollary the forward implication is immediate. To prove the reverse implication, we assume L/F and F/K are separable and show that every β ∈ L is separable over K. If β ∈ F we are done, and if not we at least know that β is separable over F . Let M/K be the subextension of F/K generated by the coefficients of the minimal polynomial f ∈ F [x] of β over F . This is a finite separable extension of K, and M (β) is also a finite separable extension of M , since the minimal polynomial of β over M (β) is f , which is separable. By the previous corollary, M (β), and therefore β, is separable over K. Corollary 4.16. Let L/K be an algebraic field extension, and let F = {α ∈ L : α is separable over K}. Then F is a separable field extension of K. Proof. This is clearly a field, since if α and β are both separable over K then K(α) and K(α, β) are separable extensions of K (by the previous corollary), thus every element of K(α, β), including αβ and α + β, is separable over K and lies in F . The field F is then separable by construction. Definition 4.17. Let L/K be an algebraic field extension. The field F in Corollary 4.16 is the separable closure of K in L. When L is an algebraic closure of K it is simply called a separable closure of K and denoted K sep . When K has characteristic zero the notions of separable closure and algebraic closure necessarily coincide. This holds more generally whenever K is a perfect field. 18.785 Fall 2021, Lecture #4, Page 4 Definition 4.18. A field K is perfect if every algebraic extension of K is separable. All fields of characteristic zero are perfect. Perfect fields of positive characteristic are characterized by the following property. Theorem 4.19. A field K of characteristic p > 0 is perfect if and only if K = K p , that is, every element of K is a pth power, equivalently, the map x 7→ xp is an automorphism. Proof. If K 6= K p then for any α ∈ K − K p the polynomial xp − α is irreducible and the extension K[x]/(xp − α) is inseparable, implying that K is not perfect. Now suppose n K = K p and let f ∈ K[x] be irreducible. By Corollary 4.5, we havef (x) = g(xp ) for some separable g ∈ K[x] and n ≥ 0. If n > 0 then n f (x) = g(xp ) = g̃(xp n−1 )p , where g̃ is the polynomial obtained from g by replacing each coefficient with its pth root (thus g̃(x)p = g(xp ), since we are in characteristic p). But this contradicts the irreducibility of f . So n = 0 and f = g is separable. The fact that every irreducible polynomial in K[x] is separable implies that every algebraic extension of K is separable, so K is perfect. Corollary 4.20. Every finite field is a perfect field. n Proof. If a field K has cardinality pn then #K × = pn − 1, thus α = αp = (αp α ∈ K and every element of K is a pth power. n−1 )p for all Definition 4.21. A field K is separably closed if K has no nontrivial finite separable extensions. Equivalently, K is equal to its separable closure in any algebraic closure of K. Definition 4.22. An algebraic extension L/K is purely inseparable if [L : K]s = 1. Remark 4.23. The trivial extension K/K is both separable and purely inseparable (but not inseparable!); conversely, an extension that is separable and purely inseparable is trivial. Example 4.24. If K = Fp (t) and L = K[x]/(xp − t) = Fp (t1/p ), then L/K is a purely inseparable extension of degree p. Proposition 4.25. Let K be a field of characteristic p > 0. If L/K is purely inseparable of degree p then L = K(a1/p ) ' K[x]/(xp − a) for some a ∈ K − K p . Proof. Every α ∈ L − K is inseparable over K, and by Corollary 4.5 its minimal polynomial over K is of the form f (x) = g(xp ) with f monic. We have 1 < deg f ≤ [L : K] = p, so g(x) must be a monic polynomial of degree 1, which we can write as g(x) = x − a. Then f (x) = xp − a, and we must have a 6∈ K p since f is irreducible (a difference of pth powers can be factored). We have [L : K(α)] = 1, so L = K(α) ' K[x]/(xp − a) as claimed. Theorem 4.26. Let L/K be an algebraic extension and let F be the separable closure of K in L. Then L/F is purely inseparable. Proof. If L/K is separable then L = F the theorem holds, so we assume otherwise, in which case the characteristic p of K must be nonzero. Fix an algebraic closure K of K that contains L. Let α ∈ L − F have minimal polynomial f over F . Use Corollary 4.5 to write n f (x) = g(xp ) with g ∈ F [x] irreducible and separable, and n ≥ 0. We must have deg g = 1, since otherwise the roots of g would be separable over F , and therefore over K, but not lie 18.785 Fall 2021, Lecture #4, Page 5 n in in the separable closure F of K in L. Thus f (x) = xp − a for some a ∈ F (since f is monic and deg g = 1). Since we are in characteristic p > 0, we can factor f in F (α)[x] as n n n f (x) = xp − αp = (x − α)p . There is thus only one F -homomorphism from F (α) to K. The same statement applies to any extension of F obtained by adjoining any set of elements of L (even an infinite set). Therefore # HomF (L, K) = 1, so [L : F ]s = 1 and L/F is purely inseparable. Corollary 4.27. Every algebraic extension L/K can be uniquely decomposed into a tower of algebraic extensions L/F/K with F/K separable and L/F purely inseparable. Proof. By Theorem 4.26, we can take F to be the separable closure of K in L, and this is the only possible choice, since we must have [L : F ]s = 1. Corollary 4.28. The inseparable degree of any finite extension of fields is a power of the characteristic. Proof. This follows from the proof of Theorem 4.26. 4.2 Étale algebras We now want to generalize the notion of a separable field extension. By Theorem 4.12, every finite separable extension L/K can be explicitly represented as L = K[x]/(f ) for some separable irreducible f ∈ K[x]. If f is not irreducible then we no longer have a field, but we do have a ring K[x]/(f ) that is also a K vector space, in which the ring multiplication is compatible with scalar multiplication. In other words, L is a (unital) commutative Kalgebra whose elements are all separable over K. The notion of separability extends to elements of a K-algebra (even non-commutative ones): an element is separable over K if and only it is the root of some separable polynomial in K[x] (in which case its minimal polynomial must be separable). Recall that the minimal polynomial of an element α of a K-algebra A is the monic generator of the kernel of the K-algebra homomorphism K[x] → A defined by x 7→ α; note that if A is not a field, minimal polynomials need not be irreducible. It follows from the Chinese remainder theorem that if f is separable then the K-algebra K[x]/(f ) is isomorphic to a direct product of finite separable extensions of K. Indeed, if f = f1 · · · fn is the factorization of f into irreducibles in K[x] then K[x] K[x] K[x] K[x] = ' × ··· × , (f ) (f1 · · · fn ) (f1 ) (fn ) where the isomorphism is both a ring isomorphism and a K-algebra isomorphism. The separability of f implies that the fi are separable and the ideals (fi ) are pairwise coprime (this justifies our application of the Chinese remainder theorem). We thus obtain a Kalgebra that is isomorphic to finite product of separable field extensions K[x]/(fi ) of K. Algebras of this form are called étale algebras (or separable algebras). Definition 4.29. Let K be a field. An étale K-algebra is a K-algebra L that is isomorphic to a finite product of separable field extensions of K. The dimension of an étale K-algebra is its dimension as a K-vector space. When this dimension is finite we say that L is a finite étale K-algebra. A homomorphism of étale K-algebras is a homomorphism of K-algebras (which means a ring homomorphism that commutes with scalar multiplication). 18.785 Fall 2021, Lecture #4, Page 6 Remark 4.30. One can define the notion of an étale A-algebra for any noetherian domain A (we will consider this in a later lecture). Example 4.31. If K is a separably closed field then every étale K-algebra A is isomorphic to K n = K × · · · × K for some positive integer n (and therefore a finite étale K-algebra). Étale algebras are semisimple algebras. Recall that a (not necessarily commutative) ring R is simple if it is nonzero and has no nonzero proper (two-sided)Qideals, and R is semisimple if it is isomorphic to a nonempty finite product of simple rings Ri .1 A commutative ring is simple if and only if it is a field, and semisimple if and only if it is isomorphic to a finite product of fields; this applies in particular to commutative semisimple K-algebras. Every étale K-algebra is thus semisimple (but the converse does Q not hold). The ideals of a semisimple commutative ring R = ni=1 Ri are easy to describe; each corresponds to a subproduct. To see this, note that the projection maps R → Ri are surjective homomorphisms onto a simple ring, thus for any R-ideal I, its image in Ri is either the zero ideal or the whole ring (note that the image of an ideal under a surjective ring homomorphism is an ideal). In particular, for each index i, either every (r1 , . . . , rn ) ∈ I has ri = 0 or some (r1 , . . . , rn ) ∈ I has ri = 1; it follows that I is isomorphic to the product of the Ri for which I projects onto Ri . Q Proposition 4.32. Let A = Ki be a K-algebra written that is a product of field extensions Ki /K. Every surjective homomorphism ϕ : A → B of K-algebras corresponds to the projection of A on to a subproduct of its factors. Q Proof. The ideal ker ϕ is a subproduct of Ki , thus A ' ker ϕ × im ϕ and B = im ϕ is isomorphic to the complementary subproduct. Proposition 4.32 can be viewed as a generalization of the fact that every surjective homomorphism of fields is an isomorphism. Corollary 4.33. The decomposition of an étale algebra into field extensions is unique up to permutation and isomorphisms of factors. Proof. Let A be an étale K-algebra and suppose A is isomorphic (as a K-algebra) to two products of field extensions of K, say m Y i=1 Ki ' A ' n Y Lj . j=1 Q Composing with isomorphisms yields surjective K-algebra homomorphisms π : Lj → K i i Q and πj : Ki → Lj . Proposition 4.32 then implies that each Ki must be isomorphic to one of the Lj and each Lj must be isomorphic to one of the Ki (and m = n). Our main interest in étale algebras is that they naturally arise from (and are stable under) base change, a notion we now recall. Definition 4.34. Let ϕ : A → B be a homomorphism of rings (so B is an A-module), and let M be any A-module. The tensor product of A-modules M ⊗A B is a B-module (with multiplication defined by b(m ⊗ b0 ) := m ⊗ bb0 ) called the base change (or extension of scalars) of M from A to B. If M is an A-algebra then its base change to B is a B-algebra. 1 There are many equivalent (and a few inequivalent) definitions, but this is the simplest. 18.785 Fall 2021, Lecture #4, Page 7 We have already seen one example of base change: if M is an A-module and p is a prime ideal of A then Mp = M ⊗A Ap (this is another way to define the localization of a module). Remark 4.35. Each ϕ : A → B determines a functor from the category of A-modules to the category of B-modules via base change. It has an adjoint functor called restriction of scalars that converts a B-module M into an A-module by the rule am = ϕ(a)m (if ϕ is inclusion this amounts to restricting the scalar multiplication by B to the subring A). The ring homomorphism ϕ : A → B will often be an inclusion, in which case we have a ring extension B/A (we may also take this view whenever ϕ is injective, which is necessarily the case if A is a field). We are specifically interested in the case where B/A is a field extension and M is a finite étale A-algebra. Proposition 4.36. Suppose L is a finite étale K-algebra and K 0 /K is any field extension. Then L ⊗K K 0 is a finite étale K 0 -algebra of the same dimension as L. Proof. Without loss of generality we assume that L is actually a field; if not L is a product of fields and we can apply the following argument to each of its factors. By Theorem 4.12, L ' K[x]/(f ) for some separable f ∈ K[x], and if f = f1 f2 · · · fm is the factorization of f in K 0 [x], we have isomorphisms of K 0 -algebras Y L ⊗K K 0 ' K 0 [x]/(f ) ' K 0 [x]/(fi ), i in which each factor K 0 [x]/(fi ) is a finite separable extension of K 0 (as discussed above, this follows from the CRT because f is separable). Thus L ⊗K K 0 is a finite étale K 0 -algebra, and dimK L = deg f = dimK 0 K 0 [x]/(f ), so the dimension is preserved. Example 4.37. Any finite dimensional real vector space V is a finite étale R-algebra (with coordinate-wise multiplication with respect to some basis); the complex vector space V ⊗R C is then a finite étale C-algebra of the same dimension. Note that even when an étale K-algebra L is a field, the base change L ⊗K K 0 will often not be a field. For example, if K = Q and L 6= Q is a number field, then L ⊗K C will never be a field, it will be isomorphic to a C-vector space of dimension [L : K] > 1. Remark 4.38. In the proof of Proposition 4.36 we made essential use of the fact that the elements of an étale K-algebra are separable. Indeed, the proposition does not hold if L is a finite semisimple commutative K-algebra that contains an inseparable element. Corollary 4.39. Let L ' K[x]/(f ) be a finite separable extension of a field K defined by an irreducible separable polynomial f ∈ K[x]. Let K 0 /K be any field extension, and let f = f1 · · · fm be the factorization of f into distinct irreducible polynomials fi ∈ K 0 [x]. We have an isomorphism of finite étale K 0 -algebras Y L ⊗K K 0 ' K 0 [x]/(fi ) i where each K 0 [x]/(fi ) is a finite separable field extension of K 0 . Proof. This follows directly from the proof of Proposition 4.36. 18.785 Fall 2021, Lecture #4, Page 8 The following proposition gives several equivalent characterizations of finite étale algebras, including a converse to Corollary 4.39 (provided the field K is not too small). Recall that an element α of a ring is nilpotent if αn = 0 for some n, and a ring is reduced if it contains no nonzero nilpotents. Theorem 4.40. Let L be a commutative K-algebra of finite dimension and assume that the dimension of L is less than the cardinality of K. The following are equivalent: (a) L is a finite étale K-algebra. (b) Every element of L is separable over K. (c) L ⊗K K 0 is reduced for every extension K 0 /K. (d) L ⊗K K 0 is semisimple for every extension K 0 /K. (e) L = K[x]/(f ) for some separable f ∈ K[x]. The implications (a) ⇔ (b) ⇔ (c) ⇔ (d) ⇐ (e) hold regardless of the dimension of L. Qn Proof. To show (a) ⇒ Qn(b), let L = i=1 Ki with each Ki /K separable, and consider α = (α1 , . . . , αn ) ∈ L = i=1 Ki . Each αi ∈ Ki is separable over K with separable minimal polynomial fi ∈ K[x], and α is a root of f := lcm{f1 , . . . , fn }, which is separable (the LCM of a finite set of separable polynomials is separable), thus α is separable. To show (b) ⇒ (c), note that if α ∈ L is nonzero and separable over K it cannot be nilpotent (the minimal polynomial of a nonzero nilpotent is xn for some n > 1 and is therefore not separable), and separability is preserved under base change. The equivalence (c) ⇔ (d) follows from Lemma 4.42 below. To show (d) ⇒ (a), we first note we can assume L is semisimple (take K 0 = K), and it suffices to treat the case where L is a field. By base-changing to the separable closure of K in L, we can further reduce to the case that L/K is a purely inseparable field extension. If L = K we are done. Otherwise we may pick an inseparable α ∈ L, and, as in the proof of n Theorem 4.26, the minimal polynomial of α has the form f (x) = xp − a for some a ∈ K and n ≥ 1. Now consider γ := α ⊗ 1 − 1 ⊗ α ∈ L ⊗K L n n n We have γ 6= 0, since γ ∈ / K, but γ p = αp ⊗ 1 − 1 ⊗ αp = a ⊗ 1 − 1 ⊗ a = 0, so γ is a nonzero nilpotent and L ⊗K L is not reduced, contradicting (c) ⇔ (d). Q We have (e) ⇒ (a) form Corollary 4.39. For the converse, suppose L = ni=1 Li with each Li /K a finite separable extension of K. Pick a monic irreducible separable polynomial f1 (x) so that L1 ' K[x]/(f1 (x)), and then do the same for i = 2, . . . , n ensuring that each polynomial fj we pick is not equal to fi for any i < j. This can be achieved by replacing fj (x) with fj (x + a) for some a ∈ K × if necessary. Here we use the fact that there are at least n distinct choices for a, under our assumption that the dimension of L is less than the cardinality of K (note that if f (x) is irreducible then the polynomials f (x+a) are irreducible and pairwise coprime as a ranges over K). The polynomials f1 , . . . fn are then coprime and separable, so their product f is separable and L = K[x]/(f ), as desired. Remark 4.41. K-algebras of the form L = K[x]/(f (x)) are monogenic (generated by one element). Theorem 4.40 implies that finite étale K-algebras are monogenic whenever the base field K is big enough. This always holds if K is infinite, but if K is a finite field then not every finite étale K-algebra is monogenic. The recent preprint [5] gives exact bounds on the maximal number of generators needed for a finite étale K-algebra over a finite field. 18.785 Fall 2021, Lecture #4, Page 9 The following lemma is a standard exercise in commutative algebra that we include for the sake of completeness. Lemma 4.42. Let K be a field. A commutative K-algebra of finite dimension is semisimple if and only if it is reduced. Proof. If A is semisimple it is clearly reduced (otherwise we could project a nonzero nilpotent of A to a nonzero nilpotent in a field); we only need to prove the converse. Every ideal of a commutative K-algebra A is also a K-vector space; this implies that when dimK A is finite A satisfies both the ascending and descending chain conditions and is therefore noetherian and artinian. This implies that A has finitely many maximal ideals M1 , . . . Mn and that the intersection of these ideals (the radical of A) is equal to the set of nilpotent elements of A (the nilradical of A); see Exercises 19.12 and 19.13 in [1], for example. Taking the productL of the projection maps A  A/Mi yields a surjective ring hon momorphism ϕ : A  i=1 A/Mi from A to a product of fields. If A is reduced then ker ϕ = ∩Mi = {0} and ϕ is an isomorphism, implying that A is semismiple. Proposition 4.43. Suppose L is a finite étale K-algebra and Ω is a separably closed field extension of K. There is an isomorphism of finite étale Ω-algebras Y ∼ L ⊗K Ω −→ Ω σ∈HomK (L,Ω) that sends β ⊗ 1 to the vector (σ(β))σ for each β ∈ L. Proof. We may reduce to the case that L = K[x]/(f ) is a separable field extension, and we may then factor f (x) = (x − α1 ) · · · (x − αn ) over Ω, with the αi are distinct. We have a bijection between HomK (K[x]/(f ), Ω) and the set {αi }: each σ ∈ HomK (K[x]/(f ), Ω) is determined by σ(x) ∈ {αi }, and for each αi , the map x 7→ αi determines a K-algebra homomorphism σi ∈ HomK (K[x]/(f ), Ω). As in the proof of Proposition 4.36 we have Ω-algebra isomorphisms n n i=1 i=1 Y Ω[x] ∼ Y K[x] ∼ Ω[x] ∼ ⊗K Ω → → → Ω. (f ) (f ) (x − αi ) which map x ⊗ 1 7→ x 7→ (α1 , . . . , αn ) 7→ (σ1 (x), . . . , σn (x)). The element x ⊗ 1 generates L ⊗K Ω as an Ω-algebra, and it follows that β ⊗ 1 7→ (σ(β))σ for every β ∈ L. Remark 4.44. The proof of Proposition 4.43 does not require Ω to be separably closed. If L ' K[x]/(f ) as in Theorem 4.40 (with f not necessarily irreducible), we can take Ω to be any extension of K that contains the splitting field of f . Example 4.45. Let L/K = Q(i)/Q and Ω = C. We have Q(i) ' Q[x]/(x2 + 1) and Q(i) ⊗Q C ' C[x] C[x] C[x] Q[x] ⊗Q C ' 2 ' × ' C × C. 2 (x + 1) (x + 1) (x − i) (x + i) As C-algebra isomorphisms, the corresponding maps are determined by i ⊗ 1 7→ x ⊗ 1 7→ x 7→ (x, x) ≡ (i, −i) 7→ (i, −i). 18.785 Fall 2021, Lecture #4, Page 10 Taking the base change of Q(i) to C lets us see the two distinct embeddings of Q(i) in C, which are determined by the image of i. Note that Q(i) is canonically embedded in its base change Q(i) ⊗Q C to C via α 7→ α ⊗ 1. We have −1 = i2 = (i ⊗ 1)2 = i2 ⊗ 12 = −1 ⊗ 1 = −(1 ⊗ 1) Thus as an  isomorphism  of C-algebras, the basis (1 ⊗ 1, i ⊗ 1) for Q(i) ⊗Q C is mapped to the basis (1, 1), (i, −i) for C × C. For any (α, β) ∈ C × C, the inverse image of α+β α−β (1, 1) + (i, −i) 2 2i in Q(i) ⊗ C under this isomorphism is (α, β) = α+β α−β α+β α−β (1 ⊗ 1) + (i ⊗ 1) = 1 ⊗ +i⊗ . 2 2i 2 2i Now R/Q is an extension of rings, so we can also consider the base change of the Q-algebra Q(i) to R. But note that R is not separably closed and in particular, it does not contain a subfield isomorphic to Q(i), thus Proposition 4.43 does not apply. Indeed, as an R-module, we have Q(i) ⊗Q R ' R2 , but as an R-algebra, Q(i) ⊗Q R ' C 6' R2 . 4.3 Norms and traces We now introduce the norm and trace map associated to a finite free ring extension B/A. These are often defined only for field extensions, but in fact the same definition works without modification whenever B is a free A-module of finite rank. One can generalize further to projective modules (with some restrictions), but we will not need this. Definition 4.46. Let B/A be a (commutative) ring extension in which B is a free A-module of finite rank. The (relative) norm NB/A (b) and trace TB/A (b) of b (down to A) are the determinant and trace of the A-linear multiplication-by-b map B → B defined by x 7→ bx. As a special case, note that if A is a field and B is a finite A-algebra (a field extension, for example) then B is an A-vector space of finite dimension, hence a free A-module of finite rank. In practice one computes the norm and trace by picking a basis for B as an A-module and computing the matrix of the multiplication-by-b map with respect to this basis; this is an n × n matrix with entries in A whose determinant and trace are basis independent. It follows immediately from the definition that NB/A is multiplicative, TB/A is additive, we have group homomorphisms NB/A : B × → A× and TB/A : B → A, and if B1 /A and B2 /A are two ring extensions that are free A-modules of finite rank then NB1 ×B2 /A (x) = NB1 /A (x1 )NB2 /A (x2 ) for all x = (x1 , x2 ) ∈ B1 × B2 . and TB1 ×B2 /A = TB1 /A (x1 ) + TB2 /A (x2 ) Example 4.47. Consider A = R and B = C, which has the A-module basis (1, i). For
×b b = 2 + 3i the matrix of B → B with respect to this basis can be written as 23 −3 2 , thus   2 −3 NC/R (2 + 3i) = det = 13, 3 2   2 −3 TC/R (2 + 3i) = tr = 4. 3 2 18.785 Fall 2021, Lecture #4, Page 11 Warning 4.48. In order to write down the matrix of an A-linear transformation B → B with respect to basis for B as a free A-module of rank n, we not only need to pick a basis, we need to decide whether to represent elements of B ' An as row vectors with linear transformations acting via matrix multiplication on the right, or as column vectors with linear transformations acting via matrix multiplication on the left. The latter convention is often implicitly assumed in the literature (as in the example above), but the former is often used in computer algebra systems (such as Magma). We now verify that the norm and trace are well behaved under base change. Lemma 4.49. Let B/A be ring extension with B free of rank n over A, and let ϕ : A → A0 be a ring homomorphism. The base change B 0 = B ⊗A A0 of B to A0 is a free A0 -module of rank n, and for every b ∈ B we have ϕ(NB/A (b)) = NB 0 /A0 (b ⊗ 1) and ϕ(TB/A (b)) = TB 0 /A0 (b ⊗ 1). Proof. Let b ∈ B, let (b1 , . . . , bn ) be a basis for B as an A-module, and let M = (mij ) ∈ An×n ×b be the matrix of B → B with respect to this basis. Then (b1 ⊗ 1, . . . , bn ⊗ 1) is a basis for B 0 as an A0 -module (thus B 0 is free of rank n over A0 ) and M 0 = (ϕ(mij )) ∈ A0n×n is the ×b⊗1 matrix of B 0 → B 0 , and we have ϕ(NB/A (b)) = ϕ(det M ) = det M 0 = NB 0 /A0 (b ⊗ 1) ϕ(TB/A (b)) = ϕ(tr M ) = tr M 0 = NB 0 /A0 (b ⊗ 1) Theorem 4.50. Let K be a field with separable closure Ω and let L be a finite étale Kalgebra. For all α ∈ L we have Y X NL/K (α) = σ(α) and TL/K (α) = σ(α). σ∈HomK (L,Ω) σ∈HomK (L,Ω) Proof. Let n be the rank of L as a K-module. By the previous lemma and Proposition 4.43, NL/K (α) = N(L⊗K Ω)/Ω (α ⊗ 1) = NΩn /Ω (σ1 (α), . . . , σn (α)) = n Y σi (α). n X σi (α). i=1 Q The isomorphism L ⊗K Ω → σ Ω = Ωn of Prop. 4.43 sends α ⊗ 1 to (σ1 (α), . . . , σn (α)). Using the standard basis for Ωn , the matrix of multiplication-by-(σ1 (α), . . . , σn (α)) is just the diagonal matrix with σi (α) in the ith diagonal entry. Similarly, TL/K (α) = T(L⊗K Ω)/Ω (α ⊗ 1) = TΩn /Ω (σ1 (α), . . . , σn (α)) = i=1 The proof above demonstrates a useful trick: when working over a field that is not algebraically/separably closed, base change to an algebraic/separable closure. This often turns separable field extensions into étale algebras that are no longer fields. Proposition 4.51. Let L/K be a (not necessarily separable) finite extension, let K be an × algebraic closure of K Qdcontaining L. Let α ∈ L have minimal polynomial f ∈ K[x] with factorization f (x) = i=1 (x − αi ) in K[x], and let e = [L : K(α)]. We have NL/K (α) = d Y i=1 In particular, if f (x) = Pd i i=0 ai x , αie and TL/K (α) = e d X αi . i=1 then NL/K (α) = (−1)de ae0 and TL/K (α) = −ead−1 . 18.785 Fall 2021, Lecture #4, Page 12 Proof. See Problem Set 2. Corollary 4.52. Let A be an integrally closed domain with fraction field K and let L/K be a finite extension. if α ∈ L is integral over A then NL/K (α) ∈ A and TL/K (α) ∈ A. Proof. This follows immediately from Propositions 1.28 and 4.51. Theorem 4.53 (Transitivity of Norm and Trace). Let A ⊆ B ⊆ C be rings with C free of finite rank over B and B free of finite rank over A. Then C is free of finite rank over A and NC/A = NB/A ◦ NC/B and TC/A = TB/A ◦ TC/B . Proof. See [3, §III.9.4]. References [1] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, 2013. [2] Michael Artin, Algebra, 2nd edition, Pearson, 2010. [3] Nicolas Bourbaki, Algebra I: Chapters 1–3 , Springer, 1989. [4] Nicolas Bourbaki, Algebra II: Chapters 4–7 , Springer, 1989. [5] Uriya First, Zinovy Reichstein, Santiago Salazar, On the number of generators of a separable algebra over a finite field , arXiv:1709.06982, 2017. [6] Anthony W. Knapp, Advanced Algebra, Digital Second Edition, 2016. [7] Joseph J. Rotman, Advanced Modern Algebra, 2nd edition, Graduate Studies in Mathematics 114, AMS, 2010. 18.785 Fall 2021, Lecture #4, Page 13 18.785 Number theory I Lecture #5 5 Fall 2021 09/22/2021 Dedekind extensions In this lecture we prove that the integral closure of a Dedekind domain in a finite extension of its fraction field is also a Dedekind domain; this implies, in particular, that the ring of integers of a number field is a Dedekind domain. We then consider the factorization of prime ideals in Dedekind extensions. 5.1 Dual modules, pairings, and lattices In this section we work in a more general setting, where A is any commutative (unital) ring. Definition 5.1. Let A be a commutative ring and M an A-module. The dual module M ∨ is the A-module HomA (M, A) with scalar multiplication (af )(m) = af (m), where a ∈ A, f ∈ HomA (M, A), and m ∈ M . If ϕ : M → N is an A-module homomorphism, the dual homomorphism ϕ∨ : N ∨ → M ∨ is defined by ϕ∨ (g)(m) = g(ϕ(m)), for g ∈ N ∨ and m ∈ M . It is easy to check that taking duals preserves identity maps and is compatible with composition: if ϕ1 : M → N and ϕ2 : N → P are A-module homomorphisms, then ∨ (ϕ2 ϕ1 )∨ = ϕ∨ 1 ϕ2 . We thus have a contravariant functor from the category of A-modules to itself. This functor is compatible with (finite) direct sums, (M ⊕ N )∨ ' M ∨ ⊕ N ∨ . Lemma 5.2. Let A be a commutative ring. For all A-modules M and N the A-modules (M ⊕ N )∨ and M ∨ ⊕ N ∨ are canonically isomorphic. Proof. We have inverse A-module homomorphisms ϕ 7→ (m 7→ ϕ(m, 0), n 7→ ϕ(0, n)) and (φ, ψ) 7→ ((m, n) 7→ φ(m) + ψ(n)). If A is a field and M is finitely generated, then M is a vector space of finite dimension, is its dual space and we have M ∨∨ ' M . In general not every A-module is isomorphic to its double dual; those that are are said to be reflexive. We have already seen examples of reflexive modules: every invertible fractional ideal is isomorphic to the dual of its inverse, hence to its double dual, and is thus reflexive. M∨ Proposition 5.3. Let A be an integral domain with fraction field K and let M be a nonzero A-submodule of K. Then M ∨ ' (A : M ) := {x ∈ K : xM ⊆ A}; in particular, if M is an invertible fractional ideal then M ∨ ' M −1 and M ∨∨ ' M . Proof. For any x ∈ (A : M ) the map m 7→ xm is an A-linear map from M to A, hence an element of M ∨ , and this defines an A-module homomorphism ϕ : (A : M ) → M ∨ , since the map x 7→ (m 7→ xm) is itself A-linear. Since M ⊆ K is a nonzero A-module, it contains some nonzero a ∈ A (if a/b ∈ M , so is ba/b = a). If f ∈ M ∨ and m = b/c ∈ M then     b ac b b  ac  b f (a) f (m) = f = f = f = f (a) = m, c ac c ac c ac a where we have used the fact that a1 f (a2 /a3 ) = a2 f (a1 /a3 ) for any a1 , a2 , a3 ∈ A with a1 /a3 , a2 /a3 ∈ M , by the A-linearity of f . It follows that f corresponds to multiplication by x = f (a)/a, which lies in (A : M ) since xm = f (m) ∈ A for all m ∈ M . The map f 7→ f (a)/a defines an A-module homomorphism M ∨ → (A : M ) inverse to ϕ, so ϕ is an isomorphism. When M is an invertible fractional ideal we have M ∨ ' (A : M ) = M −1 , by Lemma 2.20, and M ∨∨ ' (M −1 )−1 = M follows. Example 5.4. As a Z-module, we have Q∨ = {0} because there are no non-trivial Z-linear homomorphisms from Q to Z; indeed, Q is a divisible group and Z contains no non-trivial divisible subgroups. It follows that Q∨∨ = {0} (but as Q-modules we have Q ' Q∨ ' Q∨∨ ). Similarly, the dual of any finite Z-module (any finite abelian group) is the zero module, as is the double dual. More generally, if A is an integral domain every dual (and double dual) A-module must be torsion free, but not all A-modules are torsion free. One situation where we can recover many of the standard results that hold for vector spaces of finite dimension (with essentially the same proofs), is when M is a free module of finite rank. In particular, not only is M reflexive, we have M ' M ∨ (non-canonically) and may explicitly construct a dual basis. Theorem 5.5. Let A be a commutative ring and let M be a free A-module of rank n. Then M ∨ is also a free A-module of rank n, and each basis (e1 , . . . , en ) of M uniquely determines ∨ ∨ a dual basis (e∨ 1 , . . . en ) of M with the property ( 1 i = j, e∨ i (ej ) = δij := 0 i 6= j. Proof. If n = 0 then M = M ∨ = {0} and the theorem holds. Now assume n ≥ 1 and fix an A-basis e := (e1 , . . . , en ) for M . For each a := (a1 , . . . , an ) ∈ An , define fa ∈ M ∨ by setting fa (ei ) = ai and extending A-linearly. The map a 7→ fa gives an A-module homomorphism An → M ∨ with inverse f 7→ (f (e1 ), . . . , f (en )) and is therefore an isomorphism. It follows that M ∨ ' An is a free A-module of rank n. n Now let e∨ i := fι̂ , where ι̂ := (0, . . . , 0, 1, 0, . . . , 0) ∈ A has a 1 in the ith position. Then ∨ ∨ n ∨ e∨ := (e∨ 1 , . . . , en ) is a basis for M , since (1̂, . . . , n̂) is a basis for A , and ei (ej ) = δij . ∨ The basis e is uniquely determined by e: it must be the image of (1̂, . . . , n̂) under the isomorphism a 7→ fa determined by e. Definition 5.6. Let A be a commutative ring and M an A-module. A (bilinear) pairing on M is an A-linear map h·, ·i : M × M → A. Explicitly, this means that for all u, v, w ∈ M and λ ∈ A we have hu + v, wi = hu, wi + hv, wi, hu, v + wi = hu, vi + hu, wi, hλu, vi = hu, λvi = λhu, vi. If hv, wi = hw, vi then h·, ·i is symmetric, if hv, wi = −hw, vi then h·, ·i is skew-symmetric, and if hv, vi = 0 then h·, ·i is alternating (the last two are equivalent provided char(A) 6= 2). The pairing h·, ·i induces an A-module homomorphism ϕ: M → M∨ m 7→ (n 7→ hm, ni) If ker ϕ = {0} then h·, ·i is nondegenerate, and if ϕ is an isomorphism then h·, ·i is perfect. Every perfect pairing is necessarily nondegenerate. If M is a vector space of finite dimension the converse holds, but this is not true in general, not even for free modules of finite rank: consider the pairing hx, yi := 2xy on Z, which is nondegenerate but not perfect. 18.785 Fall 2021, Lecture #5, Page 2 If M is a free A-module with basis (e1 , . . . , en ) and h·, ·i is a perfect pairing, we can ∼ apply the inverse of the isomorphism ϕ : M −→ M ∨ induced by the pairing to the dual ∨ 0 0 basis (e∨ 1 , . . . , en ) given by Theorem 5.5 to obtain a basis (e1 , . . . , en ) for M that satisfies he0i , ej i = δij . When h·, ·i is symmetric we can similarly recover (e1 , . . . , en ) from (e01 , . . . , e0n ) in the same way. We record this fact in the following proposition. Proposition 5.7. Let A be a commutative ring and let M be a free A-module of rank n with a perfect pairing h·, ·i. For each A-basis (e1 , . . . , en ) of M there is a unique basis (e01 , . . . , e0n ) for M such that he0i , ej i = δij . Proof. Existence follows from the discussion above: apply the inverse of the isomorphism ∨ ϕ : V → V ∨ induced by h·, ·i to the dual basis (e∨ 1 , . . . , en ) given by Theorem 5.5 to obtain 0 0 0 −1 ∨ 0 0 a basis (e1 , . . . , en ) for M with ei = ϕ (ei ). We then have e∨ i = ϕ(ei ) = m 7→ hei , mi and he0i , ej i = ϕ(e0i )(ej ) = e∨ i (ej ) = δij for 1 ≤ i, j ≤ n. If (f10 , . . . , fn0 ) is another basis for M with the same property then for each i we have he0i − fi0 , ej i = δij − δij = 0 for every ej , and therefore he0i − fi0 , mi = 0 for all m ∈ M , but then e0i − fi0 ∈ ker ϕ = {0}, since the perfect pairing h·, ·i is nondegenerate, and therefore fi0 = e0i for each i; uniqueness follows. Remark 5.8. In what follows the commutative ring A in Proposition 5.7 will typically be a field K and the free A-module M will be a K-vector space that we will denote V . We may then use A to denote a subring of K and M to denote an A-submodule of V . A perfect paring h·, ·i on the K-vector space V will typically not restrict to a perfect pairing on the A-module M . For example, the perfect pairing hx, yi = xy on Q does not restrict to a perfect pairing on the Z-module 2Z because the induced map ϕ : 2Z → 2Z∨ defined by ϕ(m) = (n 7→ mn) is not surjective: the map x 7→ x/2 lies in 2Z∨ = HomZ (2Z, Z) but it is not in the image of ϕ. We now introduce the notion of a lattice in a vector space. Definition 5.9. Let A be an integral domain with fraction field K and let V be a K-vector space of finite dimension. A (full) A-lattice in V is a finitely generated A-submodule M of V that spans V as a K-vector space. Remark 5.10. Some authors require A-lattices to be free A-modules. When A = Z (or any PID) this is not a restriction because M is necessarily torsion-free (it lies in a vector space) and any finitely generated torsion-free module over a PID is free (by the structure theorem for finitely generated modules over a PID). But when A is not a PID, finitely generated torsion-free A-modules will typically not be free. We do not want to exclude this case! In particular if L/K is an extension of number fields the ring of integers OL will typically not be a free OK -module (even though it is a free Z-module, as we shall shortly prove), but we still want to treat OL as an OK -lattice in L (this will be important in later lectures when we define the different ideal DL/K ). Definition 5.11. Let A be a noetherian domain with fraction field K, and let V be a K-vector space of finite dimension with a perfect pairing h·, ·i. If M is an A-lattice in V , its dual lattice (with respect to the perfect pairing h·, ·i on V ) is the A-module M ∗ := {x ∈ V : hx, mi ∈ A for all m ∈ M }. 18.785 Fall 2021, Lecture #5, Page 3 It is clear that M ∗ is an A-submodule of V , but it is not clear that it is an A-lattice in V (it must be finitely generated and span V ), nor is it obvious that it is isomorphic to the dual module M ∨ . In order to justify the term dual lattice, let us now prove both facts. We will need to use the hypothesis that A is noetherian, since in general the dual of a finitely generated A-module need not be finitely generated. Notice that h·, ·i is a perfect pairing on the K-module V that need not restrict to a perfect pairing on the A-module M . Theorem 5.12. Let A be a noetherian domain with fraction field K, let V be a K-vector space with a perfect pairing h·, ·i, and let M be an A-lattice in V . The dual lattice M ∗ is an A-lattice in V isomorphic to M ∨ . Proof. Let e := (e1 , . . . , en ) be a K-basis for V that lies in M , and let e0 := (e01 , . . . , e0n ) be the unique K-basis for V given by Proposition 5.7 that satisfies he0i , ej i = δij . To show that M ∗ spans V we write a finite set S of generators for M in terms of the basis e with coefficients in K and let d be the product of all denominators that P appear. We 0 ∗ 0 claim that de lies in M : for each ei and generator m ∈ S, if we put m = j mj ej then hde0i , mi = dhe0i , P j mj ej i = d P j mj he0i , ej i = d P j mj δij = dmi ∈ A, by our choice of d, and this implies de0i ∈ M ∗ . Thus M ∗ contains a basis de0 for V . We now show M ∗ is finitely generated. Let N := {a1 e1 + · · · + an en : a1 , . . . , an ∈ A} ' An be the free A-submodule of M spanned by e. The A-module N contains a basis for V and is finitely generated, so it is an A-lattice in V . The K-basis e0 for V lies in N ∗ , since he0i , ejP i = δij ∈ A, and we claim it is an A-basis for N ∗ . Given x ∈ N ∗ , if we write x = i xi e0i then hx, ei i = xi he0i , ei i = xi lies in A, since x ∈ N ∗ , so x lies in the A-span of e0 . It follows that N ∗ is a free A-module of rank n, and in particular, a finitely generated module over a noetherian ring and therefore a noetherian module (a module whose submodules are all finitely generated); see [1, Thm. 16.19]. From the definition of the dual lattice we have N ⊆ M ⇒ M ∗ ⊆ N ∗ , so M ∗ is a submodule of a noetherian module, hence finitely generated. We now show M ∗ ' M ∨ . We have an obvious A-module homomorphism ϕ : M ∗ → M ∨ givenP by x 7→ (m 7→ hx, mi), and the A-module homomorphism ψ : M ∨ → M ∗ defined by P f 7→ i f (ei )e0i is the inverse of ϕ. Indeed, for any x = i xi e0i ∈ M ∗ we have X X XX X ψ(ϕ(x)) = ϕ(x)(ei )e0i = hx, ei ie0i = xj he0j , ei ie0i = xi e0i = x, i i i j i P and for any f ∈ M ∨ and each generator m = mj ej for M we have
P P P P 0 0 0 ϕ(ψ(f ))(m) = ϕ i f (ei )ei (m) = i ϕ(f (ei )ei )(m) = i hf (ei )ei , j mj ej i = f (m), which implies ϕ(ψ(f )) = f and ϕ−1 = ψ; thus ϕ is an isomorphism from M ∗ to M ∨ . Corollary 5.13. Let A be a noetherian domain with fraction field K. If M1 , M2 are Alattices in K-vector spaces V1 , V2 with perfect pairings h·, ·i1 , h·, ·i2 (resp.), then h·, ·i1 +h·, ·i2 defines a perfect pairing on V1 ⊕ V2 and (M ⊕ N )∗ ' M ∗ ⊕ N ∗ . Proof. This follows from Lemma 5.2 and Theorem 5.12. 18.785 Fall 2021, Lecture #5, Page 4 Corollary 5.14. Let A be a noetherian domain with fraction field K, let V be a K-vector space with a perfect pairing h·, ·i, and let M be a free A-lattice in V with A-basis (e1 , . . . , en ). The dual lattice M ∗ is a free A-lattice in V that has a unique A-basis (e∗1 , . . . , e∗n ) that satisfies he∗i , ej i = δij . Proof. This follows from the proof of Theorem 5.12 with N = M and e∗i = e0i . You might wonder whether M ∗∗ = M for an A-lattice M in a vector space V . This is false in general, but it is true when A is a Dedekind domain and we have a symmetric perfect pairing on V . To prove this we first show that the dual lattice respects localization. Lemma 5.15. Let A be a noetherian domain with fraction field K, let V be a K-vector space of finite dimension with a perfect pairing h·, ·i, let M be an A-lattice in V , and let S be a multiplicative subset of A. Then S −1 M and S −1 M ∗ are (S −1 A)-lattices in V satisfying (S −1 M )∗ = S −1 M ∗ . Proof. It is clear that S −1 M are S −1 M ∗ are both S −1 A-lattices: each contains a basis for V (since M and M ∗ do), and both are finitely generated as S −1 A-modules (since M and M ∗ are finitely generated as A-modules). Let m1 , . . . mn be A-module generators for M (and therefore S −1 A-module generators for S −1 M ). If x is an element of (S −1 M )∗ then for each mi we have hx, mi i = ai /si for some ai ∈ A and si ∈ S, and if we put s = s1 · · · sn then hsx, mi i ∈ A for every mi , hence for all m ∈ M ; thus sx ∈ M ∗ and x ∈ S −1 M ∗ . Conversely, if x = y/s is an element of S −1 M ∗ with y ∈ M ∗ and s ∈ S, then hy, mi i ∈ A for every mi and hx, mi i = hy, mi i/s ∈ S −1 A for every mi , hence for all m ∈ S −1 M , and it follows that x ∈ (S −1 M )∗ . Proposition 5.16. Let A be a Dedekind domain with fraction field K, let V be a K-vector space of finite dimension with a symmetric perfect pairing h·, ·i, and let M be an A-lattice in V . Then M ∗∗ = M . Proof. By Proposition 2.6, it suffices to show (M ∗∗ )p = Mp for each maximal ideal p of A. By Lemma 5.15 we have (M ∗∗ )p = Mp∗∗ , so it is enough to show that the proposition holds when A is replaced by one of its localizations Ap (a DVR, since A is a Dedekind domain). So let us assume that A is a DVR. Then A is a PID and M and M ∗ are both torsion-free modules over a PID, hence free A-modules. So let us choose an A-basis (e1 , . . . , en ) for M , and let (e∗1 , . . . , e∗n ) be the unique dual A-basis for M ∗ that satisfies he∗i , ej i = δij (given ∗∗ ∗∗ that satisfies by Corollary 5.14). If we now let (e∗∗ 1 , . . . , en ) be the unique A-basis for M ∗ ∗∗ ∗ hei , ej i = δij and note that hei , ej i = δij (since h·, ·i is symmetric), by uniqueness, we must ∗∗ = M . have e∗∗ i = ei for all i, and therefore M 5.2 Extensions of Dedekind domains Let A be a Dedekind domain with fraction field K, let L/K be a finite extension, and let B be the integral closure of A in L. We wish to prove that B is a Dedekind domain, which we will do by showing that it is an A-lattice in L; this will imply, in particular, that B is finitely generated, which is really the only difficult thing to show. Let us first show that B spans L as a vector space (and in fact L is its fraction field). Proposition 5.17. Let A be a Dedekind domain with fraction field K, let L/K be a finite extension, and let B be the integral closure of A in L. Every element of L can be written as b/a with a ∈ A and b ∈ B. In particular, B spans L as a K-vector space and L is the fraction field of B. 18.785 Fall 2021, Lecture #5, Page 5 Proof. Let α ∈ L. By multiplying the minimal polynomial of α in K[x] by the product of the denominators of its coefficients, we obtain a polynomial in A[x]: g(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , with an 6= 0, that has α as a root. We can make this polynomial monic by replacing x with x/an and multiplying through by an−1 to obtain n n n−1 n−1 an−1 + an an−2 xn−2 · · · + an−2 n g(x/an ) = x + an−1 x n a1 x + an a0 . This is a monic polynomial with coefficients in A that has an α ∈ L as a root. Therefore an α ∈ B, since B is the integral closure of A in L, and α = b/an for some b ∈ B and an ∈ A as claimed. It follows that B generates L as a K-vector space (we have α = b · a1n with 1 an ∈ K), and B ⊆ L ⊆ Frac B implies L = Frac B (no smaller field can contain B). Proposition 5.18. Let A be a Dedekind domain with fraction field K, let L/K be a finite extension of fields, and let B be the integral closure of A in L. Then NL/K (b) ∈ A and TL/K (b) ∈ A for all b ∈ B. P Proof. The minimal polynomial f = di=0 ai xi ∈ K[x] of b has coefficients in A, by Proposition 1.28, and it then follows from Proposition 4.51 that NL/K (b) = (−1)de ae0 ∈ A and TL/K (b) = −ead−1 ∈ A (where e = [L : K(b)] ∈ Z). Definition 5.19. Let B/A be a ring extension with B a free A-module of finite rank. The trace pairing on B is the map B × B → A defined by hx, yiB/A := TB/A (xy). Theorem 5.20. Let L be a commutative K-algebra of finite dimension. The trace pairing h·, ·iL/K is a symmetric bilinear pairing. It is a perfect pairing if and only if L is a finite étale K-algebra. Proof. Bilinearity follows from the K-linearity of the trace map TL/K , and symmetry is immediate. The fact that L is a K-vector space implies that the trace pairing is perfect if and only if it is nondegenerate. If L is not reduced then the proposition holds, since it is not étale (by Theorem 4.40), and the trace pairing is degenerate: for any nonzero nilpotent x the map y 7→ TL/K (xy) must be the zero map, since every xy is also nilpotent and the trace of any nilpotent element z is zero (the matrix of the multiplication-by-z map is nilpotent, so its trace is zero). We now assume L is reduced, hence semisimple (by Lemma 4.42) and thus a product of fields. It suffices to consider the case that L is a field, since the trace pairing on a product of field extensions is nondegenerate if and only if the trace pairing on each factor is nondegenerate, and a product of field extensions is ètale if and only if each factor is ètale. As proved on Problem Set 2, TL/K is the zero map if and only if the field extension L/K is inseparable. If TL/K is the zero map then the trace pairing is clearly degenerate, and otherwise we may pick z ∈ L for which TL/K (z) 6= 0. Then for every x ∈ L× we have hx, z/xiL/K = TL/K (z) 6= 0, so x 7→ hx, yiL/K is not the zero map, and it follows that the trace pairing is nondegenerate. Remark 5.21. Theorem 5.20 gives another equivalent definition of a finite étale K-algebra in addition to the six listed in Theorem 4.40: a finite étale K-algebra is a commutative K-algebra of finite dimension for which the trace pairing is a perfect pairing. 18.785 Fall 2021, Lecture #5, Page 6 We now assume that L/K is separable. For the next several lectures we will be working in the following setting: A is a Dedekind domain with fraction field K, the extension L/K is finite separable, and B is the integral closure of A in L (which we will shortly prove is a Dedekind domain). As a convenient shorthand, we will write “assume AKLB" to indicate that we are using this setup. Proposition 5.22. Assume AKLB. Then B is an A-lattice in L, and in particular, B is finitely generated as an A-module. Proof. By Proposition 5.17, B spans L as a K-vector space, so it contains a basis (e1 , . . . , en ) for L as a K-vector space. Let M ⊆ B be the A-span of (e1 , . . . , en ). Then M is an A-lattice in L contained in B, and it has a dual lattice M ∗ that contains the A-module B ∗ := {x ∈ L : hx, biL/K ∈ A for all b ∈ B}. Proposition 5.18 implies that B ⊆ B ∗ , and we thus have inclusions M ⊆ B ⊆ B∗ ⊆ M ∗. By Theorem 5.12, M ∗ is an A-lattice in L, hence finitely generated, hence noetherian. It follows that its A-submodule B is finitely generated and thus an A-lattice in L. Remark 5.23. When L/K is inseparable, B need not be finitely generated as an A-module, not even when A is a PID; see [2, Ex. 11, p. 205]. We used the separability hypothesis in order to get a perfect pairing, which plays a crucial role in the proof of Theorem 5.12. Lemma 5.24. Let B/A be an extension of domains with B integral over A, and let q0 ( q1 be primes of B. Then q0 ∩ A ( q1 ∩ A and dim A ≥ dim B. Proof. We first replace B with B/q0 and replace A, q0 , and q1 with their images in B/q0 (the new B is integral over the new A, since the image of a monic polynomial in A[x] is a monic polynomial in (A/(q0 ∩ A))[x]). Then q0 = (0) and q1 is a nonzero prime ideal. Let α ∈ q1 be nonzero. Its minimal polynomial xn + an−1 xn−1 + · · · + a0 over K has coefficients in A (since α ∈ q1 ⊆ B is integral over A), with a0 6= 0 (otherwise divide by x). We have a0 = −a1 α − · · · − αn ∈ q1 , thus 0 6= a0 ∈ q1 ∩ A. So q1 ∩ A is not the zero ideal and therefore properly contains q0 ∩ A = {0}. We can apply this result repeatedly to any chain of distinct prime ideals in B to get a corresponding chain of distinct prime ideals in A. It follows that dim A ≥ dim B. Theorem 5.25. Let A be a Dedekind domain with fraction field K, let L/K be a finite separable extension, and let B be the integral closure of A in L. Then B is a Dedekind domain. Proof. Recall that we defined a Dedekind domain as an integrally closed noetherian domain of dimension at most one. Let us verify that each of these conditions holds: • B is an integrally closed domain (by definition); • B is finitely generated over the noetherian ring A (by Prop. 5.22), hence noetherian; • B has dimension at most 1, since dim B ≤ dim A ≤ 1, by Lemma 5.24. Thus B is a Dedekind domain. Remark 5.26. Theorem 5.25 holds without the assumption that L/K is separable. This follows from the Krull-Akizuki Theorem, see [4, Thm. 11.7] or [3, §VII.2.5], which is used to prove that B is noetherian even when it is not finitely generated as an A-module. Corollary 5.27. The ring of integers of a number field is a Dedekind domain. 18.785 Fall 2021, Lecture #5, Page 7 5.3 Splitting primes in Dedekind extensions We continue in the AKLB setup, in which A is a Dedekind domain, K is its fraction field, L/K is a finite separable1 extension, and B is the integral closure of A, which we now know is a Dedekind domain with fraction field L. As we proved in earlier lectures, every nonzero ideal in a Dedekind domain can be uniquely factored into prime ideals. Understanding the ideal structure of a Dedekind domain thus boils down to understanding its prime ideals. In order to simplify the language, whenever we have a Dedekind domain A, by a prime of A (or of its fraction field K), we always mean a nonzero prime ideal of A. If A has dimension zero then so does B, in which case there are no primes to consider, so we may as well assume dim A = 1, in which case dim B = 1 as well (if B is a field then so is B ∩ K = A). Henceforth our AKLB setup will include the assumption that A 6= K. Given a prime p of A, we can consider the ideal pB it generates in B (its extension to B under the inclusion map). The ideal pB need not be prime, but it can be uniquely factored into nonzero prime ideals in the Dedekind domain B. We thus have Y pB = qeq , q where q ranges over primes of B and the exponents eq ≥ 0 are zero for all but finitely many primes q. The primes q for which eq > 0 are said to lie over or above the prime ideal p. As an abuse of notation, we will often write q|p to indicate this relationship (there is little risk of confusion, the prime ideal p is maximal hence not divisible by any prime ideals of A other than itself). Lemma 5.28. Let A be a ring of dimension one contained in a Dedekind domain B. Let p be a prime of A and let q be a prime of B. Then q|p if and only if q ∩ A = p. Proof. If q divides pB then it contains pB (to divide is to contain), and therefore q ∩ A contains pB ∩ A which contains p; the ideal p is maximal and q ∩ A 6= A (since 1 6∈ q), so q ∩ A = p. Conversely, if q ∩ A = p then q = qB certainly contains (q ∩ A)B = pB, and B is a Dedekind domain, so q divides pB (in a Dedekind domain to contain is to divide). Lemma 5.28 implies that contraction gives us a surjective map Spec B → Spec A defined by q 7→ q ∩ A; to see why it is surjective, note that (0) ∩ A = (0), and if p is a nonzero element of Spec A then pB is nonzero and not the unit ideal, and therefore divisible by at least one q ∈ Spec B. The fibers of this map are finite; we use {q|p} to denote the fiber above a prime p of A. The primes p of A are all maximal ideals (since dim A = 1), so each has an associated residue field A/p, and similarly for primes q of B. If q lies above p then we may regard the residue field B/q as a field extension of A/p: the kernel of the map A ,→ B → B/q is p = A ∩ q, and the induced map A/p = A/(q ∩ A) → B/q is a ring homomorphism of fields, hence injective. Definition 5.29. Assume AKLB, and let p be a prime of A. The exponent eq in the Q e q factorization pB = q|p q is the ramification index of q, and the degree fq = [B/q : A/p] 1 Most of our proofs will not actually use the separability hypothesis (and even when they do, there may be another way to prove the same result, as with Theorem 5.25). In order to simplify the presentation we will use the separability assumption whenever it would be awkward not to. The cases we are most interested in (extensions of local and global fields) are going to be separable in any event. 18.785 Fall 2021, Lecture #5, Page 8 of the corresponding residue field extension is the residue degree (or inertia degree) of q. In situations where more than one extension of Dedekind domains is under consideration, we may write eq/p for eq and fq/p for fq . Lemma 5.30. Let A be a Dedekind domain with fraction field K, let M/L/K be a tower of finite separable extension, and let B and C be the integral closures of A in L and M respectively. Then C is the integral closure of B in M , and if r is a prime of M lying above a prime q of L lying above a prime p of K then er/p = er/q eq/p and fr/p = fr/q fq/p . Proof. It follows from Proposition 1.20 that the integral closure of B in M lies in C, and it contains C, since A ⊆ B. We thus have a tower of Dedekind extensions C/B/A. If r|q|p then the factorization of pC in C refines the factorization of pB in B, so er/p = er/q eq/p , and the residue field embedding A/p ,→ C/r factors as A/p ,→ B/q ,→ C/r, so fr/p = fr/q fq/p . Example 5.31. Let A := Z, with K := Frac A = Q, and let L := Q(i) with [L : K] = 2. The prime (5) factors in B = Z[i] into two distinct prime ideals: 5Z[i] = (2 + i)(2 − i). The prime (2 + i) has ramification index e(2+i) = 1, and e(2−i) = 1 as well. The residue field Z/(5) is isomorphic to the finite field F5 , and we also have Z[i]/(2 + i) ' F5 (this can be determined by counting the Z[i]-lattice points in a fundamental parallelogram of the sublattice (2 + i) in Z[i]), so f(2+i) = 1; we similarly have f(2−i) = 1. The prime (7) remains prime in B = Z[i]; its prime factorization is simply 7Z[i] = (7), where the (7) on the RHS denotes a principal ideal in B (this is clear from context). The ramification index of (7) is thus e(7) = 1, but its residue field degree is f(7) = 2, because Z/(7) ' F7 , but Z[i]/(7) ' F49 has dimension 2 has an F7 -vector space. The prime (2) factors as 2Z[i] = (1 + i)2 , since (1 + i)2 = (1 + 2i − 1) = (2i) = (2) (note that i is a unit). You might be thinking that (2) = (1 + i)(1 − i) factors into distinct primes, but note that (1 + i) = −i(1 + i) = (1 − i). Thus e(1+i) = 2, and f(1+i) = 1 because Z/(2) ' F2 ' Z[i]/(1 + i). P Let us now compute the sum q|p eq fq for each of the primes p we factored above: X q|(2) X q|(5) X q|(7) eq fq = e(1+i) f(1+i) = 2 · 1 = 2, eq fq = e(2+i) f(2+i) + e(2−i) f(2−i) = 1 · 1 + 1 · 1 = 2, eq fq = e(7) f(7) = 1 · 2 = 2. In all three cases we obtain 2 = [Q(i) : Q]; as we shall shortly prove, this is not an accident. √ Example 5.32. Let A := R[x], with K := Frac A = R(x), and let L := R( x3 + 3x). The integral closure of A in L is the Dedekind domain B = R[x, y]/(y 2 − x3 − 3x). Then [L : K] = 2. 18.785 Fall 2021, Lecture #5, Page 9 The prime (x − 1) factors in B into two distinct prime ideals: (x − 1) = (x − 1, y − 2)(x − 1, y + 2) (since y 2 − 4 = x3 + 3x − 4 ∈ (x − 1)). We thus have e(x−1,y−2) = 1, and f(x−1,y−2) = [B/(x − 1, y − 2) : A/(x − 1)] = [R : R] = 1. Similarly, e(x−1,y+2) = 1 and f(x−1,y+2) = 1. The prime (x + 1) remains prime in B (because y 2 = −1 has no solutions in R), thus e(x+1) = 1, and f(x+1) = [B/(x + 1) : A/(x + 1)] ' [C : R] = 2. The prime (x) factors in B as (x) = (x, y)2 , and we have e(x,y) = 2 and f(x,y)P = 1. As in the previous example, q|p eq fq = [L : K] in every case: X eq fq = e(x−1,y−2) f(x−1,y+2) + e(x−1,y+2) f(x−1,y+2) = 1 · 1 + 1 · 1 = 2, q|(x−1) X q|(x+1) X q|(x) eq fq = e(x+1) f(x+1) = 1 · 2 = 2. eq fq = e(x,y) f(x,y) = 2 · 1 = 2, P Before proving that q|p eq fq = [L : K] always holds, let us consider the quotient ring B/pB. The ring B/pB is typically not a field, so it is not a field extension of A/p, but it is an A/p-algebra. This follows from the fact that B contains A and pB contains p: given ā ∈ A/p and x̄ ∈ B/pB, if we choose lifts a ∈ A of ā and x ∈ B of x̄ then āx̄ = ax ∈ B/pB is the reduction of ax ∈ b and does not depend on the choice of a and x since any other choices would be congruent modulo pB. Lemma 5.33. Assume AKLB and let p be a prime of A. The dimension of B/pB as an A/p-vector space is equal to the dimension of L as a K-vector space. Proof. Let Ap := S −1 A and Bp := S −1 B be localizations of A and B (as A-modules), where S = A−p. Then Ap /pAp = S −1 A/(pS −1 A) ' A/p and Bp /pBp ' S −1 B/(pS −1 B) ' B/pB. It follows that if the lemma is true when A is a DVR then it is true in general, so we may assume that A is a DVR, and in particular, a PID. By Proposition 5.22, B is finitely generated as an A module, and as an integral domain containing A, it must be torsion free. It follows from the structure theorem for finitely generated modules over a PID that B is free of finite rank over A. By Proposition 5.17, B spans L as a K-vector space, so any A-basis for B is a K-basis for L. It follows that B has rank n := [L : K] as a free A-module, that is, B ' An . We then have pB ' pAn = (pA)n , so B/pB ' An /(pA)n ' (A/p)n is a free A/p-module of dimension n. Example 5.34. Let A = Z, B = Z[i], and consider p = (2). We have pB = 2Z[i] = (1 + i)2 , and B/pB = Z[i]/2Z[i] = Z[i]/(1 + i)2 is an F2 -algebra of dimension 2 = [Q(i) : Q]. It contains a nonzero nilpotent (the image of i + 1), so it is not a finite étale F2 -algebra. It is a ring of cardinality 4 and characteristic 2 isomorphic to F2 [x]/(x2 ). Theorem 5.35. Assume AKLB. For each prime p of A we have X eq fq = [L : K]. q|p 18.785 Fall 2021, Lecture #5, Page 10 Proof. We have B/pB ' Applying the previous proposition gives Y B/qeq q|p [L : K] = [B/pB : A/p] X [B/qeq : A/p] = q|p = X eq [B/q : A/p] q|p = X eq fq . q|p The second equality comes from the Chinese Remainder Theorem, and the third uses the fact that B/qeq has dimension eq as a B/q-vector space. Indeed, we have qeq = {x ∈ B : vq (x) ≥ eq }, and if π ∈ q is a uniformizer for Bq (a generator for qBq that we can force to lie in q by clearing denominators), the images of (π 0 , π 1 , . . . , π eq −1 ) in B/qeq are a B/q-basis for B/qeq ; to see that these elements are linearly independent, note that any non-zero B/q-linear combination can be lifted to an element of B with valuation strictly less than eq , which is therefore not an element of qe (hence nonzero in B/qe ). For each prime p of A, let gp := #{q|p} denote the cardinality of the fiber above p. Corollary 5.36. Assume AKLB and let p be a prime of A. Then gp is an integer in the interval [1, n], where n = [L : K], as are eq and fq for each q|p. We now define some standard terminology that we may use in the AKLB setting to describe how a prime p of K splits in L (that is, for a nonzero prime ideal p of A, how the ideal pB factors into nonzero prime ideals q of B). Definition 5.37. Assume AKLB, let p be a prime of A. • L/K is totally ramified at q if eq = [L : K] (equivalently, fq = 1 = gp = 1). • L/K is unramified at q if eq = 1 and B/q is a separable extension of A/p. • L/K is unramified above p if it is unramified at all q|p, equivalently, if B/pB is a finite étale algebra over A/p. When L/K is unramified above p we say that • p remains inert in L if q = pB is prime (equivalently, eq = gp = 1, and fq = [L : K]). • p splits completely in L if gp = [L : K] (equivalently, eq = fq = 1 for all q|p). In Example 5.34 above for the extension Q(i)/Q, the prime p = (2) is ramified and the quotient ring B/pB is not an étale A/p algebra, even though the residue field A/p ' F2 is a perfect field (note that B/pB is not a field). But when A/p is a finite field (or any perfect field), for any prime q|p the residue field B/q is necessarily a finite étale (A/p)-algebra, since it must be a separable field extension, and in this case q is unramified whenever eq = 1. This applies to our primary case of interest, where L/K is an extension of global fields. However, we will occasionally want to consider Dedekind domains A whose residue fields need not be perfect, in which case eq = 1 does not imply that q is unramified. 18.785 Fall 2021, Lecture #5, Page 11 References [1] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, 2013. [2] Zenon I. Borevich and Igor R. Shafarevich, Number theory, Academic Press, 1966. [3] Nicolas Bourbaki, Commutative Algebra: Chapters 1–7 , Springer, 1989. [4] Hideyuki Matsumura, Commutative ring theory, Cambridge University Press, 1986. 18.785 Fall 2021, Lecture #5, Page 12 18.785 Number theory I Lecture #6 6 Fall 2021 09/27/2021 Ideal norms and the Dedekind-Kummer theorem In order to better understand how ideals split in Dedekind extensions we want to extend our definition of the norm map to ideals. Recall that for a ring extension B/A in which B is a free A-module of finite rank, we defined the norm map NB/A : B → A as
×b NB/A (b) := det B −→ B , the determinant of the multiplication-by-b map with respect to an A-basis for B. If B is a free A-module we could define the norm of a B-ideal to be the A-ideal generated by the norms of its elements, but in the case we are most interested in (our “AKLB" setup) B is typically not a free A-module (even though it is finitely generated as an A-module). To get around this limitation, we introduce the notion of the module index, which we will use to define the norm of an ideal. In the special case where B is a free A-module, the norm of a B-ideal will be equal to the A-ideal generated by the norms of elements. 6.1 The module index Our strategy is to define the norm of a B-ideal as the intersection of the norms of its localizations at maximal ideals of A (note that B is an A-module, so we can view any ideal of B as an A-module). Recall that by Proposition 2.6 any A-module M in a K-vector space is equal to the intersection of its localizations at primes of A; this applies, in particular, to ideals (and fractional ideals) of A and B. In order to do this we first define the module index of two A-lattices, as originally introduced by Fröhlich [3]. Recall that an A-lattice M in a K-vector space V is a finitely generated A-submodule of V that spans V as a K-vector space (Definition 5.9). If M is a free A-module, then any A-basis for M is also a K-basis for V , and we must have M ' An , where n = dimK V . If A is a Dedekind domain, even when M is not free, its localization Mp at any prime p of A will be a free Ap -module. This follows from the following facts: (a) Ap is a DVR and therefore a PID, (b) Mp is a torsion-free Ap -module, since it lies in a K-vector space and Ap ⊆ K, and (c) any finitely generated torsion-free module over a PID is free. Definition 6.1. Let A be a Dedekind domain with fraction field K, let V be an ndimensional K-vector space, let M and N be A-lattices in V , and let p be a prime of A. Then Ap is a PID and we must have Mp ' Anp ' Np , as explained above. Choose an ∼ Ap -module isomorphism φp : Mp → Np and let φ̂p denote the unique K-linear map V → V extending φp . The linear map φ̂p is an isomorphism and therefore has nonzero determinant. The module index [Mp : Np ]Ap is the principal fractional Ap -ideal generated by det φ̂p :   [Mp : Np ]Ap := det φ̂p . This ideal does not depend on our choice of φp because any other choice can be written ∼ ∼ as φ1 φp φ2 for some Ap -module automorphisms φ1 : Mp −→ Mp and φ2 : Np −→ Np that necessarily have unit determinants. The module index [M : N ]A is the A-module \ [M : N ]A := [Mp : Np ]Ap , p where p ranges over primes of A and the intersection takes place in K. Each [Mp : Np ]Ap is an A-submodule of K (which need not be finitely generated), so their intersection is clearly an A-submodule of K, but it is not immediately clear that it finitely generated (or nonzero). We claim that in fact [M : N ]A is a nonzero fractional ideal of A whose localizations agree with all the local module indexes, that is for every prime p of A we have   [M : N ]A = [Mp : Np ]Ap . p This is obvious when M and N are free A-modules: fix a global A-module isomorphism ∼ φ : M → N so that (det φ̂)p = (det φ̂p ) for all primes p (where φp is just the Ap -module isomorphism induced by φ). To prove the general case we apply a standard “gluing" argument that will be familiar to those who have studied algebraic geometry. Proposition 6.2. Let A be a Dedekind domain with fraction field K and let M and N be A-lattices in a K-vector space of finite dimension. The module index [M : N ]A is a nonzero fractional ideal of A whose localization at each prime p of A is equal to the local module index [Mp : Np ]Ap . Proof. The finitely generated A-module M is locally free in the sense that the module Mp is a free Ap -module for every prime p. It follows from [2, Thm. 19.2] that there exist nonzero a1 , . . . , ar ∈ A generating the unit ideal such that each M [1/ai ] is a free A[1/ai ]-module (here M [1/ai ] denotes the localization of M with respect to the multiplicative set {ani : n ∈ Z≥0 }). We similarly have nonzero b1 , . . . , bs ∈ A generating the unit ideal such that each N [1/bj ] is a free A[1/bj ]-module. For any pair ai and bj , if we localize at the multiplicative set −1 −1 −1 n Sij := {am i bj : m, n ∈ Z≥0 } then Sij M and Sij N will both be free Sij A-modules and we will have   −1 −1 [Sij M : Sij N ]S −1 A = [Mp : Np ]Ap , ij p −1 for all primes p of A that do not contain either ai or bj , since we can fix a global Sij A-module −1 −1 isomorphism φ : Sij M → Sij N that induces Ap -module isomorphisms φp : Mp → Np with −1 (det φ̂)p = (det φ̂p ); note that if p contains either ai or bj then pSij A is the unit ideal (not −1 −1 −1 a prime ideal of Sij A), thus [Sij M : Sij N ]S −1 A is equal to the intersection ∩p [Mp : Np ]Ap ij over primes p that do not contain ai or bj . We now observe that since the sets {ai } and {bj } both generate the unit ideal, for every prime p there is a choice of ai and bj that do not lie in p. It follows that \ \ −1 −1 [M : N ]A = [Mp : Np ]Ap = [Sij M : Sij N ]S −1 A . p ij ij −1 −1 Moreover, [M : N ]A is a nonzero fractional ideal. To see this, let Iij := [Sij M : Sij N ]S −1 A . ij −1 Each Iij is a nonzero principal fractional Sij A-ideal, and we can choose a single α ∈ K × −1 −1 so that each αIij is an Sij A-ideal. The intersection of the αIij lies in ∩ij Sij A = A and is thus an A-submodule of A, hence an ideal, and finitely generated because A is noetherian. It follows that [M : N ]A is a fractional ideal of A, and it is nonzero, since it contains the product of the generators of the Iij , for example. The localization of the intersection of a finite set of A-modules is equal to the intersection of their localizations, thus     −1 −1 −1 −1 ([M : N ]A )p = ∩ij [Sij M : Sij N ]S −1 A = ∩ij [Sij M : Sij N ]S −1 A = [Mp : Np ]Ap ij p ij p as claimed. 18.785 Fall 2021, Lecture #6, Page 2 Proposition 6.2 implies that the module index [M : N ]A is an element of the ideal group IA . If M, N, P are A-lattices in V then [M : N ]A [N : P ]A = [M : P ]A , (1) ∼ since for each prime p we can write any isomorphism Mp → Pp as a composition of iso∼ ∼ morphisms Mp → Np → Pp ; we then note that the determinant map is multiplicative with respect to composition and multiplication of fractional ideals is compatible with localization. Taking P = M yields the identity [M : N ]A [N : M ]A = [M : M ]A = A, (2) thus [M : N ]A and [N : M ]A are inverses in the ideal group IA . We note that when N ⊆ M the module index [M : N ]A ⊆ A is actually an ideal (not just a fractional ideal), since in this case we can express a basis for Np as Ap -linear combinations of a basis for Mp , and the matrix for φ̂p will then have entries (and determinant) in Ap . Remark 6.3. In the special case V = K, an A-lattice in V is simply a fractional ideal of A. In this setting each module index [M : N ]A corresponds to a colon ideal (3) [M : N ]A = (N : M ). Note that the order of M and N is reversed. This unfortunate conflict of notation arises from the fact that the module index is generalizing the notion of an index (for example, [Z : 2Z]Z = ([Z : 2Z]) = (2)), whereas colon ideals are generalizing the notion of a ratio (for example, (Z : 2Z) = ((1) : (2)) = (1/2)). To see why (3) holds, let π be a uniformizer for Ap . Then Mp = (π m ) and Np = (π n ) for some m, n ∈ Z, and we may take φp to be the multiplication-by-π n−m map. We then have [Mp : Np ]Ap = (det φ̂p ) = (π n−m ) = (π n /π m ) = (Np : Mp ). It follows from the remark that if M and N are nonzero fractional ideals of A then M [M : N ]A = M (N : M ) = N. (note we are using the fact that A is a Dedekind domain; we always have M (N : M ) ⊆ N but equality does not hold in general), and if N ⊆ M then I := [M : N ]A ⊆ A is an ideal and we have M I = N = N A and therefore M/N ' A/I as quotients of A-modules. It follows that I = {a ∈ A : aM ⊆ N } is the annihilator of M/N , which is a cyclic A-module (has a single generator), since A/I is clearly cyclic (generated by the image of 1). Conversely, if we know that M/N ' A/I for nonzero fractional ideals N ⊆ M , then we necessarily have I = [M : N ]A . The following theorem generalizes this observation. Theorem 6.4. Let A be a Dedekind domain with fraction field K, and let N ⊆ M be Alattices in a K-vector space V of dimension r for which the quotient module M/N is a direct sum of cyclic A-modules: M/N ' A/I1 ⊕ · · · ⊕ A/In , where I1 , . . . , In are nonzero ideals of A. Then [M : N ]A = I1 · · · In . 18.785 Fall 2021, Lecture #6, Page 3 Proof. Let p be a prime of A, let π be a uniformizer for Ap , and let ej = vp (Ij ) for 1 ≤ j ≤ n. Pick a basis for Mp and an isomorphism φp : Mp → Np so that Mp /Np = coker φp . The matrix of φp is an r × r matrix over the PID Ap with nonzero determinant. It therefore has Smith normal form U DV , with U, V ∈ GLr (Ap ) and D = diag(π d1 , . . . , π dr ) for some uniquely determined nonnegative integers d1 ≤ · · · ≤ dr . We then have Ap /(π e1 ) ⊕ · · · ⊕ Ap /(π en ) ' Mp /Np = coker φ ' Ap /(π d1 ) ⊕ · · · ⊕ Ap /(π dr ). It follows from the structure theorem for modules over a PID that the non-trivial summands on each side are the invariant factors of Mp /Np , possibly in different orders. We Pprecisely P therefore have nj=1 ej = ri=1 di , and applying the definition of the module index yields [Mp : Np ]Ap = (det φp ) = (det D) = (π P di ) = (π P ej ) = (πpe1 ) · · · (πpen ) = (I1 · · · In )p . It follows that [M : N ]A = I1 · · · In , since the localizations ([M : N ]A )p = [Mp : Np ]Ap and (I1 · · · In )p coincide for every prime p. 6.2 The ideal norm In the AKLB setup the inclusion A ⊆ B induces a homomorphism of ideal groups: IA → IB I 7→ IB. We wish define a homomorphism NB/A : IB → IA in the reverse direction. As we proved in the previous lecture, every fractional B-ideal I is an A-lattice in L, so let us consider IB → IA I 7→ [B : I]A . Definition 6.5. Assume AKLB. The ideal norm NB/A : IB → IA is the map I 7→ [B : I]A . We extend NB/A to the zero ideal by defining NB/A ((0)) = (0). We now show that the ideal norm NB/A is compatible with the field norm NL/K .

Proposition 6.6. Assume AKLB and let α ∈ L. Then NB/A (α) = NL/K (α) . Proof. The case α = 0 is immediate, so assume α ∈ L× . We have   \

×α NB/A (α) = [B : αB]A = [Bp : αBp ]Ap = det(L −→ L) = NL/K (α) , p ×α since each Bp −→ αBp is an isomorphism of free Ap -modules that are Ap -lattices in L. Proposition 6.7. Assume AKLB. The map NB/A : IB → IA is a group homomorphism. Proof. Let p be a maximal ideal of A. Then Ap is a DVR and Bp is a semilocal Dedekind domain, hence a PID. Thus every element of IBp is a principal ideal (α) for some α ∈ L× , and the previous proposition implies that NBp /Ap : IBp → IAp is a group homomorphism, since NL/K is. For any I, J ∈ IB we then have \ \ NB/A (IJ) = NBp /Ap (Ip Jp ) = NBp /Ap (Ip )NBp /Ap (Jp ) = NB/A (I)NB/A (J). p p 18.785 Fall 2021, Lecture #6, Page 4 Corollary 6.8. Assume AKLB. For all I, J ∈ IB we have [I : J]A = NB/A (I −1 J) = NB/A ((J : I)) Proof. The second equality is immediate: (J : I) = I −1 J (because B is a Dedekind domain). The first follows from (1), (2), and the previous proposition. Indeed, we have −1 [I : J]A = [I : B]A [B : J]A = [B : I]−1 )NB/A (J) = NB/A (I −1 J). A [B : J]A = NB/A (I Corollary 6.9. Assume AKLB and let I be a fractional ideal of B. The ideal norm of I is the fractional ideal of A generated by the image of I under the field norm NL/K , that is,   NB/A (I) = NL/K (α) : α ∈ I . Proof. Let J denote the RHS. For any nonzero prime p of A, the localization of the ideal NB/A (I) = [B : I]A at p is [Bp : Ip ]Ap = NBp /Ap (Ip ). The fractional ideal NBp /Ap (Ip ) of Ap is principal, so NBp /Ap (Ip ) = Jp follows from the proposition, and NB/A (I) = \ NBp /Ap (Ip ) = p \ Jp = J. p The corollary gives us an alternative definition of the ideal norm in terms of the field norm. In view of this we extend our definition of the field norm NL/K to fractional ideals of B, and we may write NL/K (I) instead of NB/A (I). We have the following pair of commutative diagrams, in which the downward arrows map nonzero field elements to the principal fractional ideals they generate. We know that composing the maps K × → L× → K × along the top corresponds to exponentiation by n = [L : K] (see Problem Set 2); we now show that this is also true for the composition of the bottom maps. ← ← ← ← (x) ← IB → K× NB/A → → IB ← I7→IB NL/K (y) → IA (y) → → (x) L× ← K × -← → L× → IA Theorem 6.10. Assume AKLB and let q be a prime lying above p. Then NB/A (q) = pfq , where fq = [B/q : A/p] is the residue field degree of q. Proof. The (A/p)-vector space B/q has dimension fq (by definition); as a quotient of Amodules, we have B/q ' A/p ⊕ · · · ⊕ A/p, an fq -fold direct sum of cyclic A-modules A/p, and we may apply Theorem 6.4. Thus NB/A (q) = [B : q]A = p · · · p = pfq . Corollary 6.11. Assume AKLB. For I ∈ IA we have NB/A (IB) = I n , where n = [L : K]. Proof. Since NB/A and I 7→ IB are group homomorphisms, it suffices to consider the case were I = p is a nonzero prime ideal. We then have   P Y Y Y NB/A (pB) = NB/A  qeq  = NB/A (q)eq = peq fq = p q|p eq fq = pn . q|p q|p q|p 18.785 Fall 2021, Lecture #6, Page 5 6.3 The ideal norm in algebraic geometry The maps i : IA → IB and NB/A : IB → IA have a geometric interpretation that will be familiar to those who have studied algebraic geometry: they are the pushforward and pullback maps on divisors associated to the morphism of curves Y → X induced by the inclusion A ⊆ B, where X = Spec A and Y = Spec B. For the benefit of those who have not seen this before, let us briefly explain the connection (while glossing over some details). Dedekind domains naturally arise in algebraic geometry as coordinate rings of smooth curves (which for the sake of this discussion one can take to mean geometrically irreducible algebraic varieties of dimension one with no singularities). In order to make this explicit, let us fix a perfect field k and a polynomial f ∈ k[x, y] that we will assume is irreducible in k̄[x, y]. The ring A = k[x, y]/(f ) is a noetherian domain of dimension 1, and if we further assume that the algebraic variety X defined by f (x, y) = 0 has no singularities, then A is also integrally closed and therefore a Dedekind domain.1 We call A the coordinate ring of X, denoted k[X], and its fraction field is the function field of X, denoted k(X). Conversely, given a Dedekind domain A, we can regard X = Spec A as a smooth curve whose closed points are the maximal ideals of A (all of Spec A except the zero ideal, which is called the generic point). When the field of constants k is algebraically closed, Hilbert’s Nullstellensatz gives a one-to-one correspondence between maximal ideals (x−x0 , y−y0 ) and points (x0 , y0 ) in the affine plane, but in general closed points correspond to Gal(k̄/k)-orbits of k̄-points. Recall that the ideal group IA is isomorphic to the free abelian group generated by the nonzero prime ideals p of A. The corresponding object in algebraic geometry is the divisor group Div X, the free abelian group generated by the closed points P of X. The group P Div X is written additively, so its elements have the form D = nP P with all but finitely many of the integers nP equal to 0. A finite extension of Dedekind domains B/A induces a surjective morphism φ : Y → X of the corresponding curves X = Spec A and Y = Spec B. Primes q of B in the fiber above a prime p of A correspond to closed points Q of YQin the fiber of φ above a closed point P of X. The map IA → IB defined by p 7→ pB = q|p qeq corresponds to the pullback map φ∗ : Div X → Div Y induced by φ, which is defined by X φ∗ (P ) := eQ Q φ(Q)=P P P where eQ is the ramification index (one then extends Z-linearly: φ∗ ( nP P ) = nP φ∗ (P )). Geometrically we think of eQ as the “multiplicity" of Q in the fiber above P , although eQ is typically defined algebraically as the ramification index of the prime Q in the Dedekind extension B/A as we have done (alternatively, as we shall see in later lectures, it can be defined in terms of valuations on k(X) and k(Y ) associated to P and Q). In the other direction, the norm map NB/A : IB → IA , which sends q to NB/A (q) = pfq , corresponds to pushforward map φ∗ : Div Y → Div X induced by φ, which is defined by φ∗ (Q) := fQ φ(Q) = fQ P, 1 If A is not integrally closed, we can replace it by its integral closure, thereby obtaining the normalization of the curve X. One typically also takes the projective closure of X in order to obtain a complete curve; this corresponds to considering all absolute values (places) of the function field of X, not just those arising from primes. This distinction does not affect our discussion here but will become relevant in later lectures. 18.785 Fall 2021, Lecture #6, Page 6 where fQ counts the number of k̄-points in the Gal(k̄/k)-orbit corresponding to the closed point Q, equivalently, the degree of the field extension of k needed to split Q into fQ distinct closed points after base extension (here we are using our assumption that k is perfect). This is precisely the residue field degree of Q as a prime in the Dedekind extension B/A. Note that when k = k̄ we always have fQ = 1 (so over algebraically closed fields one typically omits fQ from the pushforward map and the degree formula below). If we compose the pushforward and pullback maps we obtain X eQ fQ P = deg(φ)P. φ∗ φ∗ (P ) = φ(Q)=P Here deg(φ) is the degree of the morphism φ : Y → X, which is typically defined as the degree of the function field extension [k(Y ) : k(X)], but one can take the above formula as an alternative definition (by Theorem 5.35). It is a weighted measure of the cardinality of the fibers of φ that reflects both the ramification and degree of each closed point in the fiber (and as a consequence, it is the same for every fiber and is an invariant of φ). 6.4 The ideal norm in number fields We now consider the special case A = Z, K = Q, where B = OL is the ring of integers of the number field L. In this situation we may simply write N in place of NB/A and call it the absolute norm. If q is a nonzero prime ideal of OL then Theorem 6.10 implies N(q) = (pf ), where p ∈ Z is the unique prime in q ∩ Z, and f is the degree of the finite field B/q as an extension of Fp ' Z/pZ. The absolute norm N(q) = [OL : q]Z = ([OL : q]) is the principal ideal generated by the (necessarily finite) index [OL : q] ∈ Z of q in OL as free Z-modules of equal rank; this is just the index of q in OL as additive groups. More generally, we have the following. Proposition 6.12. Let L be a number field with ring of integers OL . For any nonzero  OL -ideal a we have N(a) = [OL : a] . If b ⊆ a are nonzero fractional ideals of OL , then [a : b]Z = ([a : b]). Proof. The ring OL is a free Z module of rank n := [L : Q]. It is free because it is torsion-free and Z is a PID, and it has rank n because it contains a Q-basis for L, by Proposition 5.17. The same is true of any nonzero fractional ideal of OL : it is a torsion-free Z-module, hence free, and it has the same rank n as OL because it contains some nonzero principal fractional ideal αOL : the fact that OL spans L implies that αOL spans L, because the multiplication×α by-α map L → L is an invertible Q-linear transformation. Let us now fix Z-bases for OL and the nonzero OL -ideal a. Let Φ ∈ Zn×n be the matrix whose columns express each basis element for a in terms of our basis for OL . Multiplication by Φ defines a Z-module isomorphism from OL to a, since it maps our basis for OL to our basis for a. It follows that [OL : a]Z = (det Φ): for every prime p ∈ Z we can use the 18.785 Fall 2021, Lecture #6, Page 7 matrix Φ to define a Z(p) -module isomorphism φ(p) : (OL )(p) → a(p) with det φ̂(p) = det Φ (any Z-basis for a free Z-module M is also a Z(p) -basis for the free Z(p) -module M(p) ). We now observe that the absolute value of the determinant of Φ is equal to the index of a in OL : indeed, if we identify OL with Zn then | det Φ| is the volume of a fundamental parallelepiped for a, viewed as a sublattice of Zn . We thus have ([OL : a]) = (det Φ) = [OL : a]Z = N(a), which proves the first claim. For any α ∈ L× we have [a : b] = [αa : αb] and [a : b]Z = [αa : αb]Z , so we can assume without loss of generality that a and b are ideals in OL . We then have a tower of free Z-modules b ⊆ a ⊆ OL , and therefore [OL : a][a : b] = [OL : b]. Replacing both sides with the Z-ideals they generate, we have   N(a) [a : b] = N(b), and therefore ([a : b]) = N(a−1 b) = [a : b]Z , by Corollary 6.8, proving the second claim. Remark 6.13. Since Z is a principal ideal domain whose only units are ±1, we can unambiguously identify each fractional ideal with a positive rational number and view the absolute norm N : IOL → IZ as a homomorphism N : IOL → Q× >0 from ideal group of OL to the multiplicative group of positive rational numbers. If we write N(a) in contexts where an element of Z or Q (or R) is expected, it is always with this understanding. When a = (a) is a nonzero principal fractional ideal we may also write N(a) := N((a)) = |NL/Q (a)|; this is a positive rational number, and for a ∈ OL , a positive integer. 6.5 The Dedekind-Kummer theorem We now give a theorem that provides a practical method for factoring primes in Dedekind extensions. This result was proved by Dedekind for number fields, building on earlier work of Kummer, but we will give a version that works for arbitrary extensions of Dedekind domains B/A whose fraction fields are a finite separable extensions L/K (the AKLB setup). The primitive element theorem implies when L/K is a finite separable extension we can always write L = K(α) for some α ∈ L, and in the AKLB setup we can assume α ∈ B, by Proposition 5.17. This does not imply that B = A[α]; indeed, it may very will happen that there is no α ∈ B for which B = A[α]. Extensions L/K for which B = A[α] for some α ∈ B are said to be monogenic. This necessarily implies that B is a free A-module, hence it has an integral basis {β1 , . . . , βn } that is both an A-basis for B and a K-basis for L. But monogenicity is a much stronger condition: it implies that B has an integral power basis, one of the form {1, α, . . . , αn−1 }. When A = Z every B has an integral basis, but very few have an integral power basis. Examples of monogenic extensions include quadratic and cyclotomic number fields (as extensions of Q); see Problem Set 3 for proofs of these facts and some examples of non-monogenic number fields. We will first prove the Dedekind-Kummer theorem assuming we have a monogenic extension; in the next section we will address the general case. 18.785 Fall 2021, Lecture #6, Page 8 Theorem 6.14 (Dedekind-Kummer). Assume AKLB with L = K(α) and α ∈ B. Let f ∈ A[x] be the minimal polynomial of α, let p be a prime of A, and let f¯ = ḡ1e1 · · · ḡrer be its factorization into monic irreducibles in (A/p)[x]. Let qi := (p, gi (α)), where gi ∈ A[x] is any lift of ḡi in (A/p)[x] under the reduction map A[x] → (A/p)[x]. If B = A[α] then pB = qe11 · · · qerr , is the prime factorization of pB in B and the residue field degree of qi is deg ḡi . Before proving the theorem, let us give an example to illustrate its utility. Example 6.15. Let A = Z, K = Q, and L = Q(ζ5 ), where α = ζ5 is a primitive 5th root of unity with minimal polynomial f (x) = x4 + x3 + x2 + x + 1. Then B = OL = Z[ζ5 ] and we can use the theorem to factor any prime of Z in OL : • (2): f (x) is irreducible modulo 2, so 2Z[ζ5 ] is prime and (2) is inert in Q(ζ5 ). • (5): f (x) ≡ (x − 1)4 mod 5, so 5Z[ζ5 ] = (5, ζ5 − 1)4 and (5) is totally ramified in Q(ζ5 ). • (11): f (x) ≡ (x − 4)(x − 9)(x − 5)(x − 3) mod 11, so 11Z[ζ5 ] = (11, ζ5 − 4)(11, ζ5 − 9)(11, ζ5 − 5)(11, ζ5 − 3), and (11) splits completely in Q(ζ5 ). • (19): f (x) ≡ (x2 + 5x + 1)(x2 − 4x + 1) mod 19, so 19Z[ζ5 ] = (19, ζ52 + 5ζ5 + 1)(19, ζ52 − 4ζ5 + 1). The four cases above cover every possible prime factorization pattern in the cyclotomic extension Q(ζ5 )/Q (see Problem Set 3 for a proof). Proof of the Dedekind-Kummer theorem. We have B = A[α] ' A[x]/(f (x)) and therefore B A[α] A[x] (A/p)[x] (A/p)[x] = ' ' ¯ ' . qi (p, gi (α)) (f (x), p, gi (x)) (ḡi (x)) (f (x), ḡi (x)) The polynomial ḡi (x) is by assumption irreducible, thus (ḡi (x)) is a maximal ideal (because (A/p)[x] is a UFD of dimension 1), so the quotient (A/p)[x]/(ḡi (x)) is a field; indeed, it is an extension of the residue field A/p of degree deg gi . It follows that qi is a prime above p with residue field Q degreeQfqi = deg ḡi as claimed. Q The ideal i qei i = i (p, gi (α))ei = i (pB + (gi (α)))ei is divisible by pB, since if we expand the ideal product every term is clearly divisible by pB, including Y
gi (α)ei ≡ (f (α)) ≡ (0) mod pB. i The ḡi (x) are distinct as elements of (A/p)[x]/(f (x)) ' A[x]/(p, f (x)) ' A[α]/pA[α], and it follows that the gi (α) are distinct modulo pB. Therefore theQprime ideals qi are distinct, and we must then have ei ≥ eqi and {q|p} ⊆ {qi } in order for i qei i to be divisible by pB; we already showed that each qi is a prime above p, so we must have {qi } = {q|p}. Now ! Y e Y Y NB/A qi i = NB/A (qi )ei = (pfqi )ei = pei deg ḡi = pdeg f = p[L:K] , so P i ei fqi i = [L : K] = P i q|p eq fq . i We must have ei = eqi and the theorem follows. 18.785 Fall 2021, Lecture #6, Page 9 We now want to remove the monogenic hypothesis from Theorem 6.14. We can always write L = K(α) for some α ∈ B (since L/K is separable), but in general the ring A[α] may be a proper subring of B. The relationship between A[α] and B is characterized by the conductor of the extension B/A[α]. 6.6 The conductor of a ring We first give the general definition then specialize to subrings of Dedekind domains. Definition 6.16. Let S/R be an extension of commutative rings. The conductor of R in S is the largest S-ideal that is also an R-ideal; equivalently, it is the largest ideal of S contained in R. It can be written as c := {α ∈ S : αS ⊆ R} = {α ∈ R : αS ⊆ R}. If R is an integral domain, the conductor of R is the conductor of R in its integral closure. √ Example √ 6.17. The conductor of Z in Z[i] is (0). The √ conductor of Z[ −3] in Z[ζ3 ] is (2, 1 + −3)√(these may be viewed as generators over Z[ −3]√or Z[ζ3 ], or even just Z; note that (2, 1 + −3) = 2Z[ζ3 ] is principal in Z[ζ3 ] but not in Z[ −3]). We are interested in the case where R is a noetherian domain. Lemma 6.18. Let R be a noetherian domain. The conductor of R in its integral closure S is nonzero if and only if S is finitely generated as an R-module. Proof. This is a special case of Lemma 2.14. Recall that we defined a fractional ideal of a noetherian domain R as a finitely generated R-submodule of its fraction field. If R has nonzero conductor then its integral closure S is a fractional ideal of R that is also a ring. This means we can write S as 1r I for some r ∈ R and R-ideal I, and the conductor c is precisely the set of denominators r ∈ R for which S = 1r I for some R-ideal I (note that the representation 1r I is far from unique). 6.7 Orders in Dedekind domains We now introduce the notion of an order (in a Dedekind domain). This should not be confused with the notion of a reflexive, transitive, antisymmetric relation on a set, rather it is a literal translation of the German Ordnung, which refers to a ring of algebraic integers. Definition 6.19. An order O is a noetherian domain of dimension one whose conductor is nonzero, equivalently, whose integral closure is finitely generated as an O-module.2 Every Dedekind domain that is not a field is also an order. The integral closure of an order is always a Dedekind domain, but not every ring whose integral closure is a Dedekind domain is an order: as shown by Nagata [5, p. 212], one can construct noetherian domains of dimension one with zero conductor. But in the case of interest to us the conductor is automatically nonzero: in the AKLB setup B is finitely generated over A (by Proposition 5.22), hence over every intermediate ring between A and B, including all those whose integral 2 Not all authors require an order to have nonzero conductor (e.g. Neukirch [6, §I.12]), but nearly all of the interesting theorems about orders require this assumption, so we include it in the definition. 18.785 Fall 2021, Lecture #6, Page 10 closure is B. In particular, if A[α] and B have the same fraction field (so L = K(α)), then A[α] is an order in B (assuming B 6= L). There is an alternative definition of an order that coincides with our definition in the case of interest to us. Recall that an A-lattice in a K-vector space L is a finitely generated A-submodule of L that spans L as a K-vector space. Definition 6.20. Let A be a noetherian domain with fraction field K, and let L be a (not necessarily commutative) K-algebra of finite dimension. An A-order in L is an A-lattice that is also a ring. Remark 6.21. In general the K-algebra L (and the order O) in Definition 6.20 need not be commutative (even though A necessarily is). For example, the endomorphism ring of an elliptic curve is isomorphic to a Z-order in a Q-algebra L of dimension 1, 2, or 4. This Zorder is necessarily commutative in dimensions 1 and 2, where L is either Q or an imaginary quadratic field, but it is non-commutative in dimension 4, where L is a quaternion algebra; see Theorem 13.17 and Corollary 13.20 in [7]. Proposition 6.22. Assume AKLB and let O be a subring of L. Then O is an A-order in L if and only if it is an order with integral closure B. Proof. We first recall that under our AKLB assumption, dim A = 1, hence dim B = 1, since A = B ∩ K, and O ⊆ L is an A-module containing 1, so it contains A. Suppose O is an A-order in L. Then O is an A-lattice, hence finitely generated as an A-module, and therefore integral over A (see [1, Thm. 10.28], for example). Thus O lies in the integral closure B of A in L. The fraction field of O is a K-vector space spanning L, hence equal to L, so O and B have the same fraction field and B is the integral closure of O. Thus O is a domain of dimension 1 (since B is), and it is noetherian because it is a finitely generated over the noetherian ring A. The integral closure B of O is finitely generated over A, hence over O; therefore O is an order. Now suppose O is an order with integral closure B. It is an A-submodule of the noetherian A-module B, hence finitely generated over A. It contains a K-basis for L because L is its fraction field (take any K-basis for L written as fractions over O and clear denominators). Thus O is an A-lattice in L that is also a ring, hence it is an A-order in L. Remark 6.23. There may be subrings O of L that are orders but not A-orders in L, but these do not have B as their integral closure. Consider A = B = Z, K = L = Q, and O = Z(2) , for example. In this case O is a DVR, hence a Dedekind domain, hence an order, but it is not an A-order in L, because it is not finitely generated over A. But its integral closure is not B (indeed, O 6⊆ B). Remark 6.24. An A-order in L is a maximal order if it is not properly contained in any other A-order in L. When A is a Dedekind domain one can show that every A-order in L lies in a maximal order. Maximal orders are not unique in general, but in the AKLB setup B is the unique maximal order. As with Dedekind domains, we call a nonzero prime ideal p in an order O a prime of O, and if q is a prime of the integral closure B of O lying above p (dividing pB) then we may write q|p to indicate this. As in the AKLB setup, we have q|p if and only if q ∩ O = p, by Lemma 5.28. The fact that B is integrally closed ensures that every prime p of O has at 18.785 Fall 2021, Lecture #6, Page 11 least one prime q lying above it (this is a standard fact of commutative algebra). We thus have a surjective map Spec B  Spec O q 7→ q ∩ O If a prime q of B contains the conductor c, then so does p = q ∩ O (since c ⊆ O), and conversely. It follows that the map is Spec B → Spec O is still well-defined if we restrict to primes that do not contain c. In B we can factor c into a product of powers of finitely many primes q; it follows that only finitely many primes p of O contain c. Proposition 6.25. In any order O, only finitely many primes contain the conductor. We now show that when we restrict to primes that do not contain the conductor the map Spec B → Spec O becomes a bijection. Lemma 6.26. Let O be an order with integral closure B and conductor c and let p be a prime of O not containing c. Then pB is prime of B. Proof. Let q be a prime of B lying above p, so that p = q ∩ O, and pick an element s ∈ c not in p (and hence not in q). Claim: Op = Bq . To see that Op ⊆ Bq , note that if a/b ∈ Op with a ∈ O and b ∈ O − p, then b ∈ B − q, so a/b ∈ Bq . Conversely, if a/b ∈ Bq with a ∈ B and b ∈ B − q then sa ∈ O and sb ∈ O − p, so (sa)/(sb) = a/b ∈ Op ; here we have used that sB ⊆ O (since s ∈ c) and sb 6∈ q (since s, b 6∈ q), so sb 6∈ p. We now note that q0 |p ⇒ Bq0 = Op = Bq ⇒ q0 = q, so there is only one prime q lying above p. It follows that pB = qe for some e ≥ 1, and we claim that e = 1. Indeed, we must have pOp = qBq (this is the unique maximal ideal of the local ring Op = Bq written in two different ways), so qe Bq = qBq and therefore e = 1. Corollary 6.27. Let O be an order with integral closure B and conductor c. The restriction of the map Spec B → Spec O defined by q 7→ q ∩ O to prime ideals not containing c is a bijection with inverse p 7→ pB. We now note several conditions on primes of O that are equivalent to not containing the conductor; these notably include the property of being invertible. Theorem 6.28. Let O be an order with integral closure B and conductor c, and let p be a prime of O. The following are equivalent: (a) p does not contain c; (b) O = {x ∈ B : xp ⊆ p}; (c) p is invertible; (d) Op is a DVR; (e) pOp is principal. If any of these equivalent properties hold, then pB is a prime of B. Proof. See Problem Set 3. 18.785 Fall 2021, Lecture #6, Page 12 Remark 6.29. Orders in Dedekind domains also have a geometric interpretation. If O is an order, the curve X = Spec O will have a singularity at each closed point P corresponding to a maximal ideal of O that contains the conductor. Taking the integral closure B of O yields a smooth curve Y = Spec B with the same function field as X and a morphism Y → X that looks like a bijection above non-singular points (a dominant morphism of degree 1). The curve Y is called the normalization of X. Recall that two ideals I and J in a ring A are said to be relatively prime or coprime if I + J = A; we may also say that I is prime to J. When A is a noetherian domain this is equivalent to requiring that Ip + Jp = Ap for every prime ideal p of A; this follows from Proposition 2.6 and Lemma 3.1. For prime ideals p that do not contain J, we have Jp = Ap , in which case Ip + Jp = Ap certainly holds, so we only need to consider the case where p contains J. In this case Jp is contained in pAp and Ip + Jp = Ap if and only if Ip 6⊆ pAp , in which case Ip = Ap , equivalently, IAp = Ap . This leads to the following definition. Definition 6.30. Let A be a noetherian domain and let J be an ideal of A. A fractional ideal I of A is prime to J if IAp = Ap for all prime ideals p that contain J. The set of J ; it is a subgroup of the ideal group I . invertible fractional ideals prime to J is denoted IA A J is in fact a subgroup, we note that if p is any prime containing J then To check that IA (a) (1)Ap = Ap , (b) if IAp = Ap then I −1 Ap = I −1 IAp = Ap (c) if I1 Ap = Ap and I2 Ap = Ap then I1 I2 Ap = I2 Ap = Ap . Theorem 6.31. Let O be an order with integral closure B. Let c be any ideal of B contained c to I c in the conductor of O. The map q 7→ q ∩ O induces a group isomorphism from IB O and both groups are isomorphic to the free abelian group generated by their prime ideals. In particular, every ideal of O prime to the conductor a unique factorization into Q has Q efractional ei i prime ideals pi which matches the factorization IB = qi with pi = qi ∩ O. Proof. The B-ideal c lies in the conductor of O and is therefore also an O-ideal, so the c and I c are well defined and the map q → q ∩ O gives a bijection between the subgroups IB O sets of prime ideals contained in these subgroups, by Corollary 6.27; the theorem follows. We now return to the AKLB setup. Let O be an order in B with conductor c. For example, we could take O = A[α], where L = K(α) with α ∈ B, as in the DedekindKummer Theorem. Theorem 6.31 implies that we can determine how primes of A split in B by looking at their factorizations in O, provided we restrict to primes p that do not contain c ∩ A. This restriction ensures that the primes q of B and q0 = q ∩ O lying above p are all prime to c and hence to the conductor, so the factorizations of pB and pO will match up. In order to complete the picture, we now show that the residue field degrees of the primes in these factorizations also match. Proposition 6.32. Assume AKLB and let O be an order with integral closure B. Let c = (c0 ∩ A)B, where c0 is the conductor of O. Then O is an A-lattice in L and the c and I c commute with the isomorphism restrictions of the norm maps NB/A and NO/A to IB O c c IB → IO defined by q 7→ q ∩ O. If q is a prime of B that does not contain c and q0 = q ∩ O and p = q ∩ A, then NB/A (q) = NO/A (q0 ) = pfq and [B/q : A/p] = [O/q0 : A/p]. Proof. We first note that (c0 ∩ A)O ⊆ c0 , so c = (c0 ∩ A)B ⊆ c0 B = c0 , thus c is contained in the conductor of O. That O is an A-lattice in L follows from Proposition 6.22. Let q be a prime of B that does not contain c, and define q0 := q ∩ O and p := q ∩ A. If p0 is any prime 18.785 Fall 2021, Lecture #6, Page 13 of A other than p, then the localization of q at p0 contains B and the localization of q0 at p0 contains O (pick a ∈ p − p0 and note that a/a = 1 lies in both q and q0 ); we thus have NB/A (q)p0 = [Bp0 : qp0 ]Ap0 = [Bp0 : Bp0 ]Ap0 = Ap0 = [Op0 : Op0 ]Ap0 = [Op0 : q0p0 ]Ap0 = NO/A (q0 )p0 For the prime p we proceed as in the proof of Lemma 6.26 and pick s ∈ (c ∩ A) − p. We then find that Bp = Op and qp = q0p , and therefore NB/A (q)p = [Bp : qp ]Ap = [Op : q0p ]Ap = NO/A (q0 )p . Thus NB/A (q)p = NB/A (q0 )p for all primes p of A, and NB/A (q) = ∩p NB/A (q)p = ∩p NO/A (q0 )p = NO/A (q0 ). The proof that NB/A (q) = pfq in Theorem 6.10 does not depend on the fact that B is a Dedekind domain and applies equally to the order O. Thus NO/A (q0 ) = pfq0 , where fq0 := [O/q0 : A/p]. We therefore have fq0 = fq and [B/q : A/p] = [O/q 0 : A/p] as claimed. Corollary 6.33. The assumption B = A[α] in the Dedekind-Kummer theorem can be replaced with the assumption that pB is prime to the conductor of A[α] in B. Remark 6.34. In the special case where A = Z and L = Q(α) is a number field generated by an algebraic integer α, for any prime number p, the ideal pOL is prime to the conductor of A[α] if and only if p does not divide the index n of A[α] in OL , as we now explain. The conductor c is an OL -ideal with absolute norm [OL : c], and it is also an A[α]-ideal, hence contained in A[α], so [OL : c] = [OL : A[α]][A[α] : c] is divisible by n = [OL : A[α]]. If p|n then p|[OL : c] and pOL must have a prime of OL above p that divides c. Conversely if pOL is not prime to c then there is a prime q of OL above p that divides c, and it follows that p = [OL : q] divides [OL : c], hence p divides either OL : A[α]] or [A[α] : c]. The latter cannot hold because it would imply that q is an A[α]-ideal, hence divisible by the conductor c (and therefore equal to c), but then [OL : c] = [OL : q] and [OL : A[α]] = 1 which is impossible when A[α] has nontrivial conductor c = q. Remark 6.35. For number fields L = Q[x]/(xn + axm + b) with m|n, the article [4] gives a precise characterization of the primes p dividing [OL : A[α]] (equivalently, dividing the conductor of A[α], as argued above), including necessary and sufficient criteria for L to be monogenic. Remark 6.36. In Lecture 12 we will define the discriminant of an A-order, and for orders of the form A[α] this is just the principal A-ideal generated by the discriminant of the minimal polynomial f ∈ A[x] of α. In Problem Set 6 you will prove that this discriminant is equal to the product of the norm of the conductor of A[α] and the discriminant of the A-order B. An immediate practical consequence is that the Dedekind-Kummer theorem always holds for primes p of A that do not contain the discriminant of f , equivalently, primes for which the reduction of f modulo p is separable, which is useful because it is an easy condition to check. But we should note that this sufficient condition is not necessary. References [1] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, 2013. 18.785 Fall 2021, Lecture #6, Page 14 [2] David Eisenbud, Commutative algebra with a view toward algebraic geometry, Springer, 1995. [3] Albrecht Fröhlich, Ideals in an extension field as modules over the algebraic integers in a finite number field , Math. Z. 74 (1960), 29–38. [4] Anuj Jakhar, Sudesh K. Khanduja, Neraj Sangwan, On prime divisors of the index of an algebraic integer , J. Number Theory 2016 (166), 47–61. [5] Masayoshi Nagata, Local rings, John Wiley & Sons, 1962. [6] Jürgen Neukirch, Algebraic number theory, Springer, 1999. [7] Andrew V. Sutherland, 18.783 Elliptic curves, Lecture 13: Endomorphism algebras, Spring 2019, MIT OpenCourseWare. 18.785 Fall 2021, Lecture #6, Page 15 18.785 Number theory I Lecture #7 7 Fall 2021 9/29/2021 Galois extensions, Frobenius elements, and the Artin map In our standard AKLB setup, A is a Dedekind domain with fraction field K, and L/K is a finite separable extension of its fraction field (and B is the integral closure of A in L, also a Dedekind domain). We now consider the case where L/K is also normal, hence Galois, and let G := Gal(L/K) to denote the Galois group; we will use AKLBG to denote this setup. 7.1 Splitting primes in Galois extensions We begin by showing that the Galois group G acts on the ideal group IB (the invertible, equivalently, nonzero, fractional ideals of B) and that this action is compatible with the group structure of IB . More precisely, IB is a left G-module. Definition 7.1. Let G be a group. A left G-module is an abelian group M equipped with a left G-action that commutes with its group operation; in additive notation we have σ(a + b) = σ(a) + σ(b) for all σ ∈ G and a, b ∈ M . One similarly defines a right G-module as an abelian group with a right G-action that commutes with the group operation. Theorem 7.2. Assume AKLBG. For each fractional ideal I of B and σ ∈ G define σ(I) := {σ(x) : x ∈ I}. The set σ(I) is a fractional ideal of B, and this defines a group action on IB that makes it a left G-module. Moreover, the restriction of this action to Spec B makes it a G-set. Proof. We first show that σ(B) = B for all σ ∈ G. Each b ∈ B is integral over A, hence f (b) = 0 for some monic polynomial f ∈ A[x], and we have 0 = σ(0) = σ(f (b)) = f (σ(b)), so σ(b) is also integral over A, hence an element of B, since B is the integral closure of A in L. This proves σ(B) ⊆ B, and the same argument shows σ −1 (B) ⊆ B, hence B ⊆ σ(B) and therefore σ(B) = B as claimed. Each σ ∈ G = Gal(L/K) is a field automorphism of L and thus commutes with addition and multiplication. It follows that if I ⊆ L is a finitely generated B-module (a fractional ideal) then σ(I) is a finitely generated σ(B)-module, and σ(B) = B, so σ(I) is a finitely generated B-module, hence a fractional ideal as claimed. We clearly have σ((0)) = (0) for all σ ∈ G, so G permutes IB , the group of nonzero fractional ideals. We also have (στ )(I) = {(στ )(x) : x ∈ I} = {σ(τ (x)) : x ∈ I} = {σ(y) : y ∈ τ (I)} = σ(τ (I)), and the identity clearly acts trivially, so we have a left G-action on IB . Now let I, J ∈ IB and σ ∈ G. Each x ∈ IJ has the form x = a1 b1 +· · ·+an bn with ai ∈ I and bi ∈ J, and σ(x) = σ(a1 )σ(b1 ) + · · · + σ(an )σ(bn ) ∈ σ(I)σ(J). Thus σ(IJ) ⊆ σ(I)σ(J), and applying the same argument to σ(I), σ(J), and σ −1 implies σ −1 (σ(I)σ(J)) ⊆ IJ and therefore σ(I)σ(J) ⊆ σ(IJ). Thus σ(IJ) = σ(I)σ(J) for all I, J ∈ IB , implying that IB is a left G-module. Let p be a prime of B and let σ(p) = qe11 · · · qenn be the unique factorization of σ(p) in B. Applying σ −1 to both sides yields p = σ −1 (q1 )e1 · · · σ −1 (qn )en , and therefore n = 1 and e1 = 1, since p is prime, thus σ(p) = q1 is prime and the G-action on IB restricts to a G-action on MaxSpec B, and on Spec B, since G fixes {(0)} = Spec B − MaxSpec B. Recall that by a prime of A (or K) we mean a nonzero prime ideal of A, and similarly for B (and L), and for any prime p of A we use {q|p} to denote the set of primes q that lie above p (equivalently, for which p = A ∩ q); in other words, {q|p} is the fiber of the contraction map MaxSpec B → MaxSpec A above p. Corollary 7.3. Assume AKLBG. For each prime p of A the group G acts transitively on the set {q|p}; in other words, the orbits of the G-action on Spec B are the fibers of the contraction map Spec B → Spec A. Proof. Consider any σ ∈ G. For q|p we have pB ⊆ q and σ(pB) ⊆ σ(q), so σ(q)|p (note σ(pB) = pB and in a Dedekind domain, to contain is to divide). Thus {q|p} is closed under the action of G, we just need to show that it consists of a single orbit. Let {q|p} = {q1 , . . . , qn } and suppose that q1 and q2 lie in distinct G-orbits. The primes q1 , . . . , qn are maximal ideals, hence pairwise coprime, so by the CRT we have a ring isomorphism B B B ' × ··· × , q1 · · · qn q1 qn and we may choose b ∈ B such that b ≡ 0 mod q2 and b ≡ 1 mod σ −1 (q1 ) for all σ ∈ G (by hypothesis, σ(q2 ) 6= q1 for all σ ∈ G, since q1 , q2 lie in different G-orbits). Then b ∈ q2 and Y NL/K (b) = σ(b) ≡ 1 mod q1 , σ∈G hence NL/K (b) 6∈ A ∩ q1 = p. But NL/K (b) ∈ NL/K (q2 ) = pfq2 ⊆ p, a contradiction. As shown in the proof of Theorem 7.2, we have σ(B) = B for all σ ∈ G = Gal(L/K), thus each σ ∈ G restricts to a ring automorphism of B that fixes every element of the subring A = B ∩ K, and thus every element of any prime p of A. It follows that σ induces an isomorphism of residue field extensions σ̄ ∈ HomA/p (B/q, B/σ(q)) defined by σ̄(x + q) := σ(x) + σ(q) for x ∈ B, which we may more compactly write as σ̄(x̄) := σ(x) (but note that the x̄ and σ(x) are elements of different residue fields). Corollary 7.4. Assume AKLBG and let p be a prime of A. The residue field degrees fq := [B/q : A/p] are the same for every q|p, as are the ramification indices eq := vq (pB). Proof. For each σ ∈ G we have an isomorphism of the residue fields B/q and B/σ(q) that fixes A/p, so they clearly have the same degree fq = fσ(q) , and G acts transitively on {q|p}, by Corollary 7.3, so the function q 7→ fq must be constant on {q|p}. For each σ ∈ G we also have σ(p) = p and σ(B) = B, so σ(pB) = pB, and for each q|p,  Y  Y  Y  e eq = vq (pB) = vq (σ(pB)) = vq σ rer = vq σ(r)er = vq r σ−1 (r) = eσ−1 (q) . r|p r|p r|p The transitivity of the G-action on {q|p} again implies that q 7→ eq is constant on {q|p}. Corollary 7.4 implies that whenever L/K is Galois, we may unambiguously write ep and fp instead of eq and fq ; recall that we previously defined gp := #{q|p}. Corollary 7.5. Assume AKLBG. For each prime p of A we have ep fp gp = [L : K]. Proof. This follows immediately from Theorem 5.35 and Corollary 7.4. 18.785 Fall 2021, Lecture #7, Page 2 Example 7.6. Assume AKLBG. When n := [L : K] is prime there are just three ways a prime p of A can split in B: • ep = n and fp = gp = 1, in which case p is totally ramified in L; • fp = n and ep = gp = 1, in which case p remains inert in L if B/pB is finite étale; • gp = n and ep = fp = 1, in which case p splits completely in L if B/pB is finite étale. Recall from Definition 5.37 that we only defined the terms “remains inert” and “splits completely” for unramified primes, which includes the condition that all the residue field extensions B/q of A/p are separable, equivalently, that B/pB is finite ètale over A/p. This will automatically hold in the primary case of interest to us, where the residue field A/p is finite, hence perfect, and all residue field extensions are separable. 7.2 Decomposition and inertia groups Definition 7.7. Assume AKLBG. For each prime q of B the decomposition group Dq (also denoted Dq (L/K)) is the stabilizer of q in G. Lemma 7.8. Assume AKLBG and let p be a prime of A. The decomposition groups Dq for q|p are all conjugate in G, with #Dq = ep fp and [G : Dq ] = gp . Proof. Points in an orbit of group action have conjugate stabilizers, so the Dq for q|p are all conjugate, by Corollary 7.3. The orbit-stabilizer theorem implies [G : Dq ] = #{q|p} = gp . We have #G = [L : K] = ep fp gp , by Corollary 7.5, so #Dq = #G/[G : Dq ] = ep fp . Let us now consider a particular prime q|p of B (by writing q|p we define p as q ∩ A). As noted above, each σ ∈ G induces a residue field isomorphism σ̄ ∈ HomA/p (B/q, B/σ(q)). For σ ∈ Dq , we have σ(q) = q, in which case σ̄ ∈ AutA/p (B/q). Moreover, the map σ 7→ σ defines a group homomorphism πq : Dq → AutA/p (B/q), since for any x ∈ B we have στ (x̄) = στ (x) = σ(τ (x)) = σ(τ (x)) = σ(τ (x̄)). Note that B/q need not be a Galois extension of A/p even when L is a Galois extension of K, which is why we write AutA/p (B/q) and not Gal((B/q)/(A/p)). Proposition 7.9. Assume AKLBG and let q|p be a prime of B. The group homomorphism πq : Dq → AutA/p (B/q) defined by σ 7→ σ̄ is surjective and B/q is normal over A/p. Proof. Let F be the separable closure of A/p in B/q and for b̄ ∈ F , pick b ∈ B such that b ≡ b̄ mod q and b ≡ 0 mod σ −1 (q) (so σ(b) ≡ 0 mod q) for all σ ∈ G − Dq ; the CRT implies that such an b exists, since for σ ∈ G − Dq the ideals q and σ(q) are distinct and therefore coprime (since they are maximal ideals). Now define  Y g(x) := x − σ(b) ∈ A[x], σ∈G and let g denote the image of g in (A/p)[x]. Observe that b̄ is the root of a polynomial ḡ ∈ (A/p)[x] that splits completely in (B/q)[x], and our choice of b̄ was arbitrary, so this applies to every b̄ ∈ F × . It follows that F is a normal (hence Galois) extension of A/p, and we have Gal(F/(A/p)) ' AutA/p (B/q), since F is the separable closure of A/p in B/q. 18.785 Fall 2021, Lecture #7, Page 3 For each σ ∈ G − Dq we have σ(b̄) = 0 , so 0 is a root of g(x) with multiplicity at least m = #(G − Dq ), and the remaining roots are σ(b̄) for σ ∈ Dq , all of which are Gal(F/(A/p))-conjugates of b̄. It follows that g(x)/xm divides a power of the minimal polynomial f (x) of b̄, but f (x) is irreducible in (A/p)[x], so g(x)/xm is a power of f (x) and every Gal(F/(A/p))-conjugate of b̄ has the form σ(b̄) for some σ ∈ Dq . Applying this to b̄ chosen so that F = (A/p)(b̄) (by the primitive element theorem) shows that the map πq : Dq → AutA/p (B/q) ' Gal(F/(A/p)) is surjective. To show that B/q is a normal extension of A/p we proceed as we did for F : for each b ∈ B define g ∈ A[x] and g ∈ (A/p)[x] as above to show that every b ∈ B/q is the root of a polynomial in (A/p)[x] that splits completely in (B/q)[x]. Definition 7.10. Assume AKLBG, and let q|p be a prime of B. The kernel of the surjective homomorphism πq : Dq → AutA/p (B/q) is the inertia group Iq of q. Corollary 7.11. Assume AKLBG and let q|p be a prime of B. We have an exact sequence 1 −→ Iq −→ Dq −→ AutA/p (B/q) −→ 1, and #Iq = ep [B/q : A/p]i . We have shown that the residue field B/q is always a normal extension of the residue field A/p. Let us now suppose that it is also separable, hence Galois; this holds, for example, if A/p is a perfect field, and in particular, whenever A/p is a finite field. We then have Dq /Iq ' AutA/p (B/q) = Gal((B/q)/(A/p)). Proposition 7.12. Assume AKLBG, let q|p be a prime of B, and suppose B/q is a separable extension of A/p. We have a tower of field extensions K ⊆ LDq ⊆ LIq ⊆ L with ep = [L : LIq ] = #Iq ; fp = [LIq : LDq ] = #Dq /#Iq ; gp = [LDq : K] = #{q|p}. The fields LDq and LIq are the decomposition field and inertia field associated to q. Proof. The third equality follows immediately from Lemma 7.8. The second follows from Proposition 7.9 and the separability of (B/q)/(A/p), since Dq /Iq ' Gal((B/q)/(A/p)) has cardinality fp = [B/q : A/p]. We then have [L : LDq ] = #Dq = ep fp and #Dq /#Iq = fp , so #Iq = ep , so the first equality also holds. We now consider an intermediate field E lying between K and L. Let us fix a prime q|p of B, and let qE := q ∩ E, so that q|qE and qE |p, and let us use Gq (L/K) := AutA/p (B/q), Gq (L/E) := Aut(B∩E)/qE (B/q), GqE (E/K) := AutA/p ((B ∩ E)/qE ) to denote the automorphism groups of the residue field extensions associated to the tower K ⊆ E ⊆ L. We use the notation Dq (L/E) to denote the decomposition group of q relative to the extension L/E (note that L/E is Galois since L/K is), and similarly define Dq (L/K), as well as Iq (L/E) and Iq (L/K). In the case that E/K is also Galois, we similarly use DqE (E/K) and IqE (E/K) to denote the decomposition and inertia group of qE (subgroups of Gal(E/K)). 18.785 Fall 2021, Lecture #7, Page 4 Proposition 7.13. Assume AKLBG, let E be an intermediate field between K and L. Let q be a prime of B and let qE = q ∩ E and p = q ∩ K. Then Iq (L/E) = Iq (L/K) ∩ Gal(L/E) and Dq (L/E) = Dq (L/K) ∩ Gal(L/E). If E/K is Galois, then we have the following commutative diagram of exact sequences: ← ← ← ← ← ← ← → → ← ← ← ← → ← → GqE (E/K) 1 1 1 ← ← ← ← ← ← ← ← ← → DqE (E/K) → → Gq (L/K) → IqE (E/K) → → ← 1 → → → Gq (L/E) → Dq (L/K) → 1 → Dq (L/E) → Iq (L/K) → 1 → Iq (L/E) → 1 → ← 1 ← 1 → 1 → 1 → 1 Proof. Note that Dq (L/E) ⊆ Gal(L/E) ⊆ Gal(L/K). An element σ of Gal(L/K) lies in Dq (L/E) if and only if it fixes E (hence lies in Gal(L/E)) and satisfies σ(q) = q (hence lies in Dq (L/K)), which proves the first claim. For the second claim, the restriction of πq (L/K) : Dq (L/K) → Gq (L/K) to Dq (L/E) is the map πq (L/E) : Dq (L/E) → Gq (L/E), hence the kernels agree after intersecting with Gal(L/E). The exactness of the columns follows from Corollary 7.11; we now argue exactness of the rows. Each row corresponds to an inclusion followed by a restriction in which the inclusion is precisely the kernel of the restriction (for the first two rows this follows from the two claims proved above and for the third row it follows from the main theorem of Galois theory); exactness at the first two groups in each row follows. Surjectivity of the restriction maps follows from the bijection used in the proof of Lemma 4.10. We have a bijection HomK (L, Ω) → HomE (L, Ω) × HomK (E, Ω) whose second factor is restriction, and we may view this as a bijection φ : Gal(L/K) → Gal(L/E) × Gal(E/K). If σ ∈ Gal(E/K) stabilizes qE then φ−1 (1, σ) ∈ Gal(L/K) stabilizes q and restricts to σ; this implies surjectivity of the restriction maps in the first two rows, and for the third we replace L/E/K with the corresponding tower of residue field extensions (and forget about stabilizing qE ). We now argue commutativity of the four corner squares which suffices to prove the commutativity of the enitre diagram. The upper left square commutes because all the maps are inclusions. The upper right square commutes because inclusion and restriction commute. The lower left square commutes because the horizontal maps are inclusions and the vertical maps coincide on Dq (L/E). The lower right square commutes because the horizontal maps are restrictions and the vertical maps agree after restriction to E. Corollary 7.14. Assume AKLBG, let E be an intermediate field between K and L. Let q be a prime of B and let qE = q ∩ E and p = q ∩ K. Then • eqE /p = 1 if and only if E ⊆ LIq , and • eqE /p = fqE /p = 1 if and only if E ⊆ LDq , where Iq and Dq are the inertia and decomposition groups of q. 18.785 Fall 2021, Lecture #7, Page 5 Proof. Proposition 7.13 implies Iq (L/E) = Iq (L/K) ∩ Gal(L/E), and for F = LIq , we have Iq (L/F ) = Iq (L/K) = Gal(L/F ). We also have Gal(L/EF ) = Gal(L/E) ∩ Gal(L/F ), so Iq (L/E) = Iq (L/K) ∩ Gal(L/E) = Gal(L/F ) ∩ Gal(L/EF ) = Gal(L/EF ) = Iq (L/EF ). Now eq/qE = #Iq (L/E) = #Iq (L/EF ) = eq/qEF is equal to eq/qF = #Iq (L/F ) if and only if E ⊆ F . The first claim in the corollary follows, since Iq (L/F ) = Iq (L/K) implies eq/qF = eq/p which implies eqF /p = 1, since eq/qF eqF /p = eq/p , by Lemma 5.30. The proof of the second claim follows mutatis mutandis: replace Iq by Dq and eq/x by eq/x fq/x throughout. In our AKLBG setup, for any prime p of K we let Ip and Dp denote the subgroups of G = Gal(L/K) generated by the inertia groups Iq and the decomposition groups Dq of the primes q|p, respectively. These are the inertia and decomposition groups of p. The corresponding inertia field LIp and decomposition field LDp are Galois extensions of K that are characterized by the following corollary. Corollary 7.15. Assume AKLBG and let p be a prime of K. The fields LIp and LDp are Galois extensions of K, and for any intermediate field E we have eqE /p = 1 for all qE |p if and only if E ⊆ LIp , and eqE /p = fqE /p = 1 for all qE |p if and only if E ⊆ LDp . When A/p is a perfect field, the inertia field is the largest subfield of L in which p is unramified, and the decomposition field is the largest subfield in which p splits completely. Proof. The fact that G acts transitively on {q|p} means that Ip is generated by a complete set of conjugate subgroups Iq and is therefore stable under conjugation, hence normal, and similarly for Dp . It follows that the corresponding fixed fields are Galois extensions of K. The rest of the corollary follows immediately from Corollary 7.14. 7.3 Frobenius elements We now add the further assumption that the residue fields A/p (and therefore B/q) are finite for all primes p of K.1 This holds, for example, whenever K is a global field (a finite extension of Q or Fq (t)). In this situation B/q is necessarily a Galois extension of A/p (we don’t need Proposition 7.9 for this, finite extensions of finite fields are always Galois). Indeed, recall that every finite extension of a finite field F has a cyclic Galois group generated by the #F-power Frobenius automorphism x 7→ x#F . In order to simplify the notation, when working with finite residue fields we may write Fq := B/q and Fp := A/p; these are finite fields of p-power order, where p is the characteristic of Fp (and of Fq ). Note that the field K (and L) need not have characteristic p (consider the case of number fields), but if the characteristic of K is positive then it must be p (consider the homomorphism A → A/p from the integral domain A to the field A/p). Let q|p be a prime of B. Corollary 7.11 gives us an exact sequence
πq 1 −→ Iq −→ Dq −→ Gal Fq /Fp −→ 1. If p (equivalently, q) is unramified, then ep = eq = 1 and Iq is trivial. In this case we have an isomorphism ∼ πq : Dq −→ Gal(Fq /Fp ). 1 There exist Dedekind domains A (PIDs even) with a mixture of finite and infinite residue fields; see [1]. 18.785 Fall 2021, Lecture #7, Page 6 The Galois group Gal(Fq /Fp ) is the cyclic group of order fp = [Fq : Fp ] generated by the Frobenius automorphism x 7→ x#Fp . Note that the cardinality of the finite field Fp is necessarily a power of its characteristic p. If K = Q and p = (p) is a prime of Z, then Fp = Z/pZ is the field with p elements, but in general the field Fp need not be a prime field (consider K = Q(i) and p = (7)). Definition 7.16. Assume AKLBG with finite residue fields and q|p unramified. The inverse ∼ image of the Frobenius automorphism of Gal(Fq /Fp ) under πq : Dq −→ Gal(Fq /Fp ) is the Frobenius element σq ∈ Dq ⊆ G (also called the Frobenius substitution [2, §8]). Proposition 7.17. Assume AKLBG with finite residue fields and q|p unramified. The Frobenius element σq is the unique σ ∈ G such that for all x ∈ B we have σ(x) ≡ x#Fp mod q. Proof. Clearly σq has this property, we just need to show uniqueness. Suppose σ ∈ G has the desired property. For any x ∈ q we have x ≡ 0 mod q, and σ(x) ≡ x#Fp mod q implies σ(x) ≡ 0 mod q, so σ(x) ∈ q; it follows that σ(q) = q, and therefore σ ∈ Dq . The isomorphism πq : Dq → Gal(Fq /Fp ) maps both σ and σq to the Frobenius automorphism x 7→ x#Fp , so we must have σ = σq . Proposition 7.18. Assume AKLBG with finite residue fields and q|p unramified. For all q0 |p the Frobenius elements σq and σq0 are conjugate in G. Proof. By Corollary 7.3, G acts transitively on {q|p}, so let τ ∈ G be such that q0 = τ (q). For any x ∈ B we have σq (x) ≡ x#Fp mod q.   τ (σq (x)) ≡ τ x#Fp mod τ (q) (τ σq )(x) ≡ τ (x)#Fp mod q0 (τ σq )(τ −1 (x)) ≡ τ (τ −1 (x))#Fp mod q0 (τ σq τ −1 )(x) ≡ x#Fp mod q0 , where we applied τ to both sides in the second line and replaced x by τ −1 (x) in the fourth line. The uniqueness of σq0 given by Proposition 7.17 implies σq0 = τ σq τ −1 . Definition 7.19. Assume AKLBG with finite residue fields and q|p unramified. The conjugacy class of the Frobenius element σq ∈ G is the Frobenius class of p, denoted Frobp . It is common to abuse terminology and refer to Frobp as a Frobenius element σp ∈ G representing its conjugacy class (so σp = σq for some q|p); there is little risk of confusion so long as we remember that σp is only determined up to conjugacy (which usually governs all the properties we care about). There is, however, one situation where this terminology is entirely correct. If G is abelian then each conjugacy classes consists of a single element, in which we case Frobp = {σq : q|p} is a singleton set and there is a unique choice for σp (note that #{σq : q|p} = 1 does not imply #{q|p} = 1; the map q → σq is need not be injective). 18.785 Fall 2021, Lecture #7, Page 7 7.4 Artin symbols There is another notation commonly used to denote Frobenius elements that includes the field extension in the notation. Definition 7.20. Assume AKLBG with finite residue fields. For each unramified prime q of L we define the Artin symbol   L/K := σq . q Proposition 7.21. Assume AKLBG   with finite residue fields and q|p unramified. Then p L/K splits completely if and only if = 1. q Proof. This follows directly from the definitions: if p splits completely then ep fp = 1 and Dq = hσq i = {1}. Conversely, if Dq = hσq i = {1} then ep fp = 1 and p splits completely. We will see later in the course that the extension L/K is completely determined by the set of primes p that split completely in L. Thus in some sense the Artin symbol captures the essential structure of L/K. Proposition 7.22. Assume AKLBG with finite residue fields and let q|p be unramified. Let E be an intermediate field between K and L, and define qE := q ∩ E. Then     L/E L/K [FqE :Fp ] = , q q     L/K and if E/K is Galois then E/K is the restriction of to E. qE q Proof. For the first claim, note that #FqE = (#Fp )[FqE :Fp ] . The second claim follows from the commutativity of the lower right square in the commutative diagram of Proposition 7.13: the Frobenius automorphism x 7→ x#Fp of Gal(FqE /Fp ) is the restriction of the Frobenius automorphism x 7→ x#Fp of Gal(Fq /Fp ) to FqE . When L/K is abelian, the Artin symbol takes the same value for all q|p and we may instead write   L/K := σp . p In this setting we now view the Artin symbol as a function mapping unramified primes p to Frobenius elements σp ∈ G. We wish to extend this map to a multiplicative homomorphism from the ideal group IA to the Galois group G = Gal(L/K), but ramified primes q|p cause problems: the homomorphism πq : Dq → Gal(Fq /Fp ) is not a bijection when p is ramified (it has nontrivial kernel Iq of order eq = ep > 1). S denote the subgroup of I generated by the primes For any set S of primes of A, let IA A of A that do not lie in S (a free abelian group). Definition 7.23. Let A be a Dedekind domain with finite residue fields. Let L be a finite abelian extension of K = Frac A, and let S be the set of primes of A that ramify in L. The Artin map is the homomorphism   L/K S : IA → Gal(L/K) ·  m m  Y Y L/K ei pei i 7→ . pi i=1 i=1 18.785 Fall 2021, Lecture #7, Page 8 Remark 7.24. We will prove in later lectures that the set S of ramified primes is finite, but the definition makes sense in any case. One of the main results of class field theory is that the Artin map is surjective (this is part of what is known as Artin reciprocity). This is a deep theorem that we are not yet ready to prove, but we can verify that it holds in some simple examples. √ Example 7.25 (Quadratic fields). Let K = Q and L = Q( d) for some square-free integer d 6= 1. Then Gal(L/K) has order 2 and is certainly abelian. As you proved on Problem Set 2, the only ramified primes p = (p) of A = Z are those that divide the discriminant ( d if d ≡ 1 mod 4, D := disc(L/K) = 4d if d 6≡ 1 mod 4. If we identify Gal(L/K) with the multiplicative group {±1}, then !   √   Q( d)/Q L/K D = = ±1, = p (p) p where D p
is the Kronecker symbol. For odd primes p6 | D we have  D p  ( +1 if D is a nonzero square modulo p, = −1 if D is not a square modulo p, and for p = 2 not dividing D (in which case D = d ≡ 1 mod 4) we have   ( +1 if D ≡ 1 mod 8, D = 2 −1 if D ≡ 5 mod 8. The cyclotomic extensions Q(ζn )/Q provide another interesting example that you will have an opportunity to explore on Problem Set 4. References [1] Raymond C. Heitmann, PID’s with specified residue fields, Duke Math. J. 41 (1974), 565–582. [2] Jean-Pierre Serre, Local fields, Springer, 1979. 18.785 Fall 2021, Lecture #7, Page 9 18.785 Number theory I Lecture #8 8 Fall 2021 10/4/2021 Complete fields and valuation rings In order to make further progress in our investigation of how primes split in our AKLB setup, and in particular, to determine the primes of K that ramify in L, we introduce a new tool that allows us to “localize" fields. We have seen how useful it can be to localize the Dedekind domain A at a prime ideal p: this yields a discrete valuation ring Ap , a principal ideal domain with exactly one nonzero prime ideal, which is much easier to study than A, and from Proposition 2.6 we know that the localizations of A at prime ideals collectively determine the structure of A. Localizing A does not change its fraction field K. But there is an operation we can perform on K that is analogous to localizing A: we can construct the completion of K with respect to one of its absolute values. When K is a global field, this yields a local field, a term that we will define in the next lecture. At first glance taking completions might seem to make things more complicated, but as with localization, it simplifies matters by allowing us to focus on a single prime, and moreover, work in a complete field. We begin by briefly reviewing some standard background material on completions, topological rings, and inverse limits. 8.1 Completions Recall that an absolute value on a field K is a function | | : K → R≥0 for which 1. |x| = 0 if and only if x = 0; 2. |xy| = |x||y|; 3. |x + y| ≤ |x| + |y|. If in addition the stronger condition 4. |x + y| ≤ max(|x|, |y|) holds, then | | is nonarchimedean. This definition does not depend on the fact that K is a field, K could be any ring, but absolute values can exist only when K is an integral domains, since a, b 6= 0 ⇒ |a|, |b| 6= 0 ⇒ |ab| = |a||b| 6= 0 ⇒ ab 6= 0; of course an absolute value on an integral domain extends to an absolute value on its fraction field, but it will be convenient to consider absolute values on integral domains as well as fields. For a more general notion, we can instead consider a metric on a set X, which we recall is a function d : X × X → R≥0 that satisfies 1. d(x, y) = 0 if and only if x = y; 2. d(x, y) = d(y, x); 3. d(x, z) ≤ d(x, y) + d(y, z). A metric that also satisfies 4. d(x, z) ≤ max(d(x, y), d(y, z)) is an ultrametric and is said to be nonarchimedean. Every absolute value on a ring induces a metric d(x, y) := |x − y|, but not every metric on a ring is induced by an absolute value. The metric d defines a topology on X generated by open balls B0 and x ∈ X, and we call X a metric space. It is a Hausdorff space, since distinct x, y ∈ X have disjoint open neighborhoods B 0 there is an N ∈ Z>0 such that d(xn , x) <  for all n ≥ N ; the limit x is necessarily unique. The sequence (xn ) is Cauchy if for every  > 0 there is an N ∈ Z>0 such that d(xm , xn ) <  for all m, n ≥ N . Every convergent sequence is Cauchy, but the converse need not hold. A metric space in which every Cauchy sequence converges is said to be complete. When X is an integral domain with an absolute value | | that makes it a complete metric space we say that X is complete with respect to | |. Which sequences converge and which sequences are Cauchy depends very much on the absolute value | | that we use; for example, every integral domain is complete with respect to its trivial absolute value, since then every Cauchy sequence must be eventually constant and obviously converges. Equivalent absolute values necessarily agree on which sequences are convergent and which are Cauchy, so if an integral domain is complete with respect to an absolute value it is complete with respect to all equivalent absolute values. Definition 8.2. An abelian group G is a topological group if it is a topological space in which the map G × G → G defined by (g, h) 7→ g + h and the map G → G defined by g 7→ −g are both continuous (here G × G has the product topology). A commutative ring R is a topological ring if it is a topological space in which the maps R × R → R defined by (r, s) 7→ r + s and (r, s) 7→ rs are both continuous; the additive group of R is then a topological group, since (−1, s) 7→ −s is continuous, but the unit group R× need not be a topological group, in general. A field K is a topological field if it is a topological ring whose unit group is a topological group. If R is a ring with an absolute value then it is a topological ring under the induced topology, and its unit group is also a topological group ; in particular, if R is a field with an absolute value, then it is a topological field under the induced topology. These facts follow from the triangle inequality and the multiplicative property of an absolute value. Definition 8.3. Let X be a metric space. Cauchy sequences (xn ) and (yn ) are equivalent if d(xn , yn ) → 0 as n → ∞; this defines an equivalence relation on the set of Cauchy sequences in X and we use [(xn )] to denote the equivalence class of (xn ). The completion of X is the b whose elements are equivalence classes of Cauchy sequences with the metric metric space X d([(xn )], [(yn )]) := lim d(xn , yn ) n→∞ (this limit exists and depends only on the equivalence classes of (xn ) and (yn )). We may b via the map x 7→ x̂ = [(x, x, . . .)]. canonically embed X in its completion X b a ring by defining When X is a topological ring we extend the ring operations to X [(xn )] + [(yn )] := [(xn + yn )] and [(xn )][(yn )] := [(xn yn )]; 18.785 Fall 2021, Lecture #8, Page 2 the additive and multiplicative identities 0 := 0̂ and 1 := 1̂. When the metric on X is b via induced by an absolute value | |, we extend | | to an absolute value on X [(xn )] := lim |xn |. n→∞ This limit exists and depends only on the equivalence of (xn ), as one can show using the triangle inequality and the fact that (xn ) is Cauchy and R is complete. When X is a field b is also a field (this is false with a metric induced by an absolute value, the completion X in general, see Problem Set 4 for a counter example). Indeed, given [(xn )] 6= 0, we can choose (xn ) with xn 6= 0 for all n, and use the multiplicative property of the absolute value (combined with the triangle inequality), to show that (1/xn ) is Cauchy. We then have 1/[(xn )] = [(1/xn )], since [(xn )][(1/xn )] = [(1, 1, . . .)] = 1. If | | arises from a discrete valuation v on K (meaning |x| := cv(x) for some c ∈ (0, 1)), b by defining we extend v to a discrete valuation on X v([(xn )]) := lim v(xn ) ∈ Z, n→∞ for [(xn )] 6= 0̂ and v(0̂) := ∞. Note that for [(xn )] 6= 0̂ the sequence (v(xn )) is eventually constant (so the limit is an integer), and we have |[(xn )]| = cv([(xn )]) . 8.1.1 Topological fields with an absolute value Proposition 8.4. Let K be a field with an absolute value | | viewed as a topological field b be the completion K. The field K b is complete, and has under the induced topology, and let K the following universal property: every embedding of K as a topological field into a complete b into L which is an isomorphism field L can be uniquely extended to an embedding of K b is the unique topological field whenever K is dense in L. Up to a canonical isomorphism, K with this property. Proof. See Problem Set 4. b is (isomorphic to) itself, since we can The proposition implies that the completion of K b to the trivial embedding K b → K. b apply the universal property of the completion of K Completing a field that is already complete has no effect. In particular, the completion of K with respect to the trivial absolute value is K, since every field is complete with respect to the trivial absolute value. Two absolute values on the same field induce the same topology if and only if they are equivalent; this follows from the Weak Approximation Theorem. Theorem 8.5 (Weak Approximation). Let K be a field and let | · |1 , . . . , | · |n be pairwise inequivalent nontrivial absolute values on K. Let a1 , . . . , an ∈ K and let 1 , . . . , n be positive real numbers. Then there exists an x ∈ K such that |x − ai |i < i for 1 ≤ i ≤ n. Proof. See Problem Set 4. Corollary 8.6. Let K be a field with absolute values | |1 and | |2 . The induced topologies on K coincide if and only if | |1 and | |2 are equivalent. Proof. See Problem Set 4. 18.785 Fall 2021, Lecture #8, Page 3 The topology induced by a nonarchimedean absolute value has some features that may be counterintuitive to the uninitiated. In particular, every open ball is also closed, so the closure of B 1, which implies −1 |x | ≤ 1 and x−1 ∈ A, so rather than defining Av as the valuation ring of Kv we could define Av as the completion of A with respect to | |v and define Kv as its fraction field. We now give another characterization of Av as an inverse limit. Proposition 8.11. Let K be a field with absolute value | |v induced by a discrete valuation v, let A be the valuation ring of K, and let π be a uniformizer. The valuation ring of the completion Kv of K with respect to | |v is a complete discrete valuation ring Av with uniformizer π, and we have an isomorphism of topological rings A Av ' lim n . ←− π A n→∞ The first statement in the proposition is clear, since as noted above, Av is a complete DVR, and π is a uniformizer because v extends to a discrete valuation on Av with v(π) = 1. Before proving the non-trivial part of the proposition, let us check that we understand the topology of the inverse limit X := limn A/π n A. The valuation ring A is a closed ball ←− B≤1 (0) (and an open set) in the nonarchimedean metric space K, and this also applies to each of the sets π n A (they are closed balls of radius cn about 0). Each quotient A/π n A therefore has the discrete topology, since the inverse image of any point under the quotient map isQa coset of the open subgroup π n A. The inverse limit X is a subspace of the product space n A/π n A, whose basic open sets project onto A/π n A for all but finitely many factors (by definition of the product topology). It follows that every basic open U ⊆ X is the inverse image (under the canonical projection maps given by the inverse limit construction) of an open subset of A/π m A for some m ≥ 1; all open sets are unions of these basic open sets. And since A/π m A has the discrete m topology, its basic open Q sets are points, which are cosets a + πm A for some a ∈ A. Thus we can assume U = n Un where each Un is the image of a + π A under the quotient map A → A/π n A. We can alternatively describe the topology on X in terms of an absolute value: for nonzero x = (xn ) ∈ X = lim A/π n A, let v(x) be the least n ≥ 0 for which xn+1 6= 0, and ←− define |x|v := cv(x) . If we embed A in X in the obvious way, a 7→ (ā, ā, ā, . . .), the absolute value on X restricts to the absolute value | |v on A, and the subspace topology A inherits from X is the same as that induced by | |v . The open sets of X are unions of open balls B0 , then we say that w extends v with index e. Theorem 8.20. Assume AKLB and let p be a prime of A. For each prime q|p, the discrete valuation vq extends vp with index eq , and every discrete valuation on L that extends vp arises in this way. In other words, the map q 7→ vq gives a bijection from {q|p} to the set of discrete valuations of L that extend vp . Proof. For each prime q|p we have vq (pB) = eq (by definition of the ramification index eq ), 0 0 while vQ q (p B) = 0 for all primes p 6= p of A (since q lies above only the prime p = q ∩ A). np0 0 If I = p0 (p ) is any nonzero fractional ideal of A then   Y vq (IB) = vq  (p0 )np0 B  = vq (pnp B) = vq (pB)np = eq np = eq vp (I), p0 so vq (x) = vq (xB) = eq vp (xA) = eq vp (x) for all x ∈ K × ; thus vq extends vp with index eq . If q and q0 are two distinct primes above p, then neither contains the other and for any x ∈ q − q0 we have vq (x) > 0 ≥ vq0 (x), thus vq 6= vq0 and the map q 7→ vq is injective. Let w be a discrete valuation on L that extends vp with index e, let W = {x ∈ L : w(x) ≥ 0} be the associated DVR, and let m = {x ∈ L : w(x) > 0} be its maximal ideal. Since w|K = evp , the discrete valuation w is nonnegative on A = {x ∈ K : w(x) ≥ 0} therefore A ⊆ W , and elements of A with nonzero valuation are precisely the elements of p, thus p = m ∩ A. The discrete valuation ring W is integrally closed in its fraction field L, so B ⊆ W . Let q = m ∩ B. Then q is prime (since m is), and p = m ∩ A = q ∩ A, so q lies over p. The ring W contains Bq and is contained in Frac Bq = L. But there are no intermediate rings between a DVR and its fraction field, so W = Bq and w = vq (and e = eq ). 18.785 Fall 2021, Lecture #8, Page 9 References [1] Nicolas Bourbaki, General Topology: Chapters 1-4 , Springer, 1985. [2] Fernando Q. Gouvea, p-adic numbers, Springer, 1993. [3] Neal Koblitz, p-adic numbers, p-adic analysis, and zeta functions, Springer, 1984. [4] Alain M. Robert, A course in p-adic analysis, Springer, 2000. [5] Jean–Pierre Serre, A course in arithmetic, Springer, 1973. 18.785 Fall 2021, Lecture #8, Page 10 18.785 Number theory I Lecture #9 9 Fall 2021 10/06/2021 Local fields and Hensel’s lemmas In this lecture we introduce the notion of a local field ; these are precisely the fields that arise as completions of a global field (finite extensions of Q or Fq (t)), but they can be defined in a more intrinsic way. In later lectures we will see that global fields can also be defined in a more intrinsic way, as fields whose completions are local fields and which admit a suitable product formula. 9.1 Local fields Definition 9.1. A local field is a field with a nontrivial absolute value | | that is locally compact under the topology induced by | |. Recall that a topological space is locally compact if every point has a compact neighborhood.1 The topology induced by | | is given by the metric d(x, y) := |x − y|. A metric space is locally compact if and only if every point lies in a compact closed ball. Example 9.2. Under the standard archimedean absolute value both R and C are local fields but Q is not. Indeed no closed ball in Q is compact, since it is missing limit points (all irrational real numbers), and in a metric space a compact set must contains all its limit points. Finite fields are not local fields because they have no nontrivial absolute values. Our first goal is to classify local fields by showing that they are precisely the fields we get by completing a global field. As in the previous lecture, we use B0 about x, and B≤r (x) := {y : |y − x| ≤ r} to denote a closed ball. Open balls are always open sets and closed balls are always closed sets, but in a nonarchimedean metric space, open balls are both open and closed, as are closed balls. Remark 9.3. For nonarchimedean metric spaces whose metric is induced by a discrete valuation, every open ball of radius r is also a closed ball of some radius s ≤ r, but we need not have s = r; in particular, the closure of B 1 (such an α exists because | | is nontrivial). The map x 7→ αx is continuous and | | is multiplicative, so B≤|α|n s (0) is compact for every n ∈ Z>0 (recall that the continuous image of a compact set is compact). We thus have compact balls about 0 of arbitrarily large radii, implying that every closed ball B≤r (0) is a closed subset of a compact set, hence compact. For every z ∈ K the translation map x 7→ x + z is continuous, so every closed ball B≤r (z) is compact. This proves the forward implication, and the reverse implication follows immediately from the definition of local compactness. For the parenthetical, replace B≤s (0) in the argument above by any closed ball. Corollary 9.5. Let K be a local field with nontrivial absolute value | |. Then K is complete. 1 Weaker definitions of locally compact are sometimes used, but they all imply this one, and for Hausdorff spaces these weaker definitions are all equivalent to the one given here. Proof. Suppose not. Then there is a Cauchy sequence (xn ) in K that converges to a limit b − K. Pick N ∈ Z>0 so that |xn − x| < 1/2 for all n ≥ N (here we are using the x∈K b and consider the closed ball S := B≤1 (xN ) in K, which is compact extension of | | to K), by Lemma 9.4. The Cauchy sequence (xn )n≥N in S has a convergent subsequence whose limit lies in S ⊆ K, since S is compact and therefore sequentially compact (because K is a metric space). But this limit must be equal to x 6∈ K, a contradiction. Proposition 9.6. Let K be a field with absolute value | |v induced by a discrete valuation v with valuation ring A and uniformizer π. Then K is a local field if and only if K is complete and the residue field A/πA is finite. Proof. If K is a local field then K is complete, by Corollary 9.5, and the valuation ring A = {x ∈ K : v(x) ≥ 0} = {x ∈ K : |x|v ≤ 1} = B≤1 (0) is a closed ball, hence compact, by Lemma 9.4. The cosets x + πA of the subgroup πA ⊆ A are open balls B<1 (x), since y ∈ x + πA if and only if |x − y|v ≤ |π|v < 1. The collection {x + πA : x ∈ A} of cosets of πA is an open cover of A by disjoint sets which must be finite, since A is compact; thus A/πA is finite. Now suppose that K is complete and A/πA is finite. The valuation ring A ⊆ K is also complete, and Proposition 8.11 gives an isomorphism of topological rings A A = Â ' lim n . ←− π A n Each quotient A/π n A is finite, since A/πA is finite, and therefore compact; it follows that the inverse limit, and therefore A, is compact, by Proposition 8.10. Lemma 9.4 implies that K is a local field, since it contains a compact closed ball B≤1 (0) = A and | |v is nontrivial (recall that discrete valuations surject onto Z and are thus non-trivial by definition). Corollary 9.7. Let L be a global field with a nontrivial absolute value | |v . Then the completion Lv of L with respect to | |v is a local field. Proof. Let L/K be a finite extension with K = Q or K = Fq (t) and A = Z or A = Fq [t], so that K = Frac A. Then A is a Dedekind domain, as is its integral closure B in L, by Theorem 5.25 (and Remark 5.26 in the case that L/K is inseparable).2 If | |v is archimedean, then K = Q and the completion of L with respect to | |v must contain the completion of Q with respect to the restriction of | |v to Q, which must be isomorphic to R (as shown on Problem Set 1, every archimedean absolute value on Q is equivalent to the usual Euclidean absolute value). Thus Lv is a finite extension of R and must be isomorphic to either R or C (as a topological field), both of which are local fields. We now assume that | |v is nonarchimedean. We claim that in this case | |v is induced by a discrete valuation. Let C := {x ∈ L : |x|v ≤ 1} be the valuation ring of L with respect to | |v , and let m := {x ∈ L : |x|v < 1} be its maximal ideal, which is nonzero because | |v is nontrivial. The restriction of | |v to K is a nonarchimedean absolute value, and from the classification of absolute values on Q and Fq (t) proved on Problem Set 1, we can assume it is induced by a discrete valuation on A; in particular, |x|v ≤ 1 for all x ∈ A, and therefore 2 In fact, we can always choose K so that L/K is separable: if L has positive characteristic p, let Fq be the algebraic closure of Fp in L, choose a separating transcendental element t, and put K := Fq (t). Such a t exists because Fq is perfect and L/Fq is finitely generated, see [3, Thm. 7.20]. 18.785 Fall 2021, Lecture #9, Page 2 A ⊆ C. Like all valuation rings (discrete or not), C is integrally closed in its fraction field L, and C contains A, so C contains B, since B is the integral closure of A in L. The ideal q = m ∩ B is maximal, and the DVR Bq lies in C and must equal C, since there are no intermediate rings between a DVR and its fraction field (we cannot have C = L because C is not a field). It follows that the absolute value induced by vq is equivalent to | |v , since they have the same valuation rings. By choosing 0 < c < 1 appropriately, we can assume that | · |v = cvq (·) is induced by vq , which proves the claim. The residue field Bq /qBq ' B/q is finite, since B/q is a finite extension of the finite field A/p, where p = q ∩ A. If we now consider the completion Lv with valuation ring Bv , we can take any uniformizer π for q ⊆ B ⊆ Bv as a uniformizer for Bv , and we have Bq Bq B Bv = ' , ' q qBq πBq πBv so Bv /πBv is finite. Thus Lv is a complete field with an absolute value induced by a discrete valuation and finite residue field, and therefore a local field, by Proposition 9.6. In order to classify all local fields we require the following result from topology (here nondiscrete simply means that not every set is open). Proposition 9.8. A locally compact topological vector space over a nondiscrete locally compact field has finite dimension. Proof. See [4, Prop. 4-13.iv]. Theorem 9.9. Let L be a local field. If L is archimedean then it is isomorphic to R or C; otherwise, L is isomorphic to a finite extension of Qp or Fq ((t)). Proof. Let L be a local field with nontrivial absolute value | |; then L is complete, by Corollary 9.5. If L has characteristic zero then the prime field of L is Q, and L contains the completion of Q with respect to the restriction of | | to Q. By Ostrowski’s theorem, the restriction of | | to Q is equivalent to either the standard archimedean absolute value, in which case the completion is R, or it is equivalent to a p-adic absolute value, in which case the completion is Qp (which, by definition, is the completion of Q with respect to the p-adic absolute value). Thus L contains a subfield K isomorphic to R or to Qp for some prime p. If L has positive characteristic p then the prime field of L is Fp , and L must contain a transcendental element s, since no algebraic extension of Fp has a nontrivial absolute value (if |α| > 1 for some algebraic α ∈ L, then the restriction of | | to the finite field Fp (α) is nontrivial, but this is impossible). It follows that L contains Fp (s) and therefore contains the completion of Fp (s) with respect to | |. Every completion of Fp (s) is isomorphic to Fq ((t)) for some q a power of p and t transcendental over Fq (see Problem Set 5). Thus L contains a subfield K isomorphic to Fq ((t)). If K is archimedean then K = R is a local field, and if K is nonarchimedean then K = Qp or K = Fq ((t)) is a complete field with a discrete valuation and finite residue field, hence a local field by Proposition 9.6. The field K is therefore locally compact, and it is nondiscrete because its absolute value is nontrivial. Proposition 9.8 implies that L has finite degree over K. If K is archimedean then K = R, and L must be R or C; otherwise, L is a finite extension of Qp or Fq ((t)) as claimed. 18.785 Fall 2021, Lecture #9, Page 3 9.2 Hensel’s lemmas P Definition 9.10. Let R be a (commutative) ring, and let f (x) = Pfi xi ∈ R[x] be a polynomial. The (formal ) derivative f 0 of f is the polynomial f 0 (x) := ifi xi−1 ∈ R[x]. Note that the canonical ring homomorphism Z → R defined by 1 7→ 1 allows us to view the integers i = 1 + 1 + · · · + 1 as elements of R (the map Z → R will be injective only when R has characteristic zero, but it is well defined in any case). It is easy to verify that for all a, b ∈ R and f, g ∈ R[x] the formal derivative satisfies the usual identities: (af + bg)0 = af 0 + bg 0 , 0 0 (linearity) 0 (Leibniz rule) 0 (chain rule) (f g) = f g + f g , 0 0 (f ◦ g) = (f ◦ g)g , When the characteristic of R is positive, we may have deg f 0 < deg f − 1. Indeed, if R has characteristic p > 0 and g(x) = f (xp ) for some f ∈ R[x], then g 0 = f 0 (xp )pxp−1 = 0. P Lemma 9.11. Let R be a ring, let f = fi xi ∈ R[x] be a polynomial, and let a ∈ R. Then f (x) = f (a) + f 0 (a)(x − a) + g(x)(x − a)2 for a unique g ∈ R[x]. Proof. We have f (x) = f (a + (x − a)) = = f (a) + X i≥1 X i≥0 i fi (a + (x − a)) = X i fi aj (x − a)i−j j 0≤j 0 and f0 6= 0, since f is irreducible. Let mP:= min{vp (fi )}. Suppose for the sake of contradiction that m < 0, and let g := π −m f = ni=0 gi xi ∈ A[x]. Then g is an irreducible polynomial in A[x] with g0 , gn ∈ p, since m < 0 and f0 , fn ∈ A, and gi is a unit for some 0 < i < n, by the minimality of m. The reduction ḡ of g to k[x] has positive degree and constant term 0, and is therefore divisible by x. If we let ū := xd be the largest power of x dividing ḡ, then 0 < d ≤ deg ḡ < n and v̄ := ḡ/xd ∈ k[x] is coprime to ū (possibly deg v̄ = 0). Lemma 9.19 implies that g = uv for some u, v ∈ A[x] with 0 < deg u = deg ū < n. But this means g is not irreducible, a contradiction. Corollary 9.21. Let A be a complete DVR with fraction field K, and let L/K be a finite extension of degree n. Then α ∈ L is integral over A if and only if NL/K (α) ∈ A. P Proof. Let f = di=0 fi xi ∈ K[x] be the minimal polynomial of α. If α is integral over A then f ∈ A[x], by Proposition 1.28, and NL/K (α) = (−1)n f (0)e ∈ A, where e = [L : K(α)], by Proposition 4.51. Conversely, if NL/K (α) = (−1)n f (0)e ∈ A, then f (0) ∈ A, since f (0) ∈ K is a root of xe − (−1)n NL/K (α) ∈ A[x] and A is integrally closed. The constant coefficient of f thus lies in A, as does its leading coefficient (it is monic), so f ∈ A[x], by Lemma 9.20, and α is therefore integral over A. Theorem 9.22. Assume AKLB and that A is a complete DVR with maximal ideal p. Then B is a DVR whose maximal ideal q is necessarily the unique prime above p. Proof. We first show that #{q|p} = 1. At least one prime q of B lies above p, since the factorization of pB ( B is non-trivial. Now suppose for the sake of contradiction that q1 , q2 ∈ {q|p} with q1 6= q2 . Choose b ∈ q1 − q2 and consider the ring A[b] ⊆ B. The ideals q1 ∩ A[b] and q2 ∩ A[b] are distinct prime ideals of A[b] containing pA[b], and both are maximal, since they are nonzero and dim A[b] = dim A = 1 (note that A[b] is integral over A and therefore has the same dimension). The quotient ring A[b]/pA[b] thus has at least two 3 See [2, §II.6] for a proof of this. 18.785 Fall 2021, Lecture #9, Page 7 maximal ideals. Let f ∈ A[x] be the minimal polynomial of b over K, and let f¯ ∈ (A/p)[x] be its reduction to the residue field A/p. (A/p)[x] A[x] A[b] ' ' , ¯ (p, f ) pA[b] (f ) thus the ring (A/p)[x]/(f¯) has at least two maximal ideals, which implies that f¯ is divisible by two distinct irreducible polynomials (because (A/p)[x] is a PID). We can thus factor f¯ = ḡ h̄ with ḡ and h̄ coprime. By Hensel’s Lemma 9.19, we can lift this to a non-trivial factorization f = gh of f in A[x], contradicting the irreducibility of f . Every maximal ideal of B lies above a maximal ideal of A, but A has only the maximal ideal p and #{q|p} = 1, so B has a unique (nonzero) maximal ideal q. Thus B is a local Dedekind domain, hence a local PID, and not a field, so B is a DVR, by Theorem 1.16. Remark 9.23. The assumption that A is complete is necessary. For example, if A is the DVR Z(5) with fraction field K = Q and we take L = Q(i), then the integral closure of A in L is B = Z(5) [i], which is a PID but not a DVR: the ideals (1 + 2i) and (1 − 2i) are both maximal (and not equal). But if we take completions we get A = Z5 and K = Q5 , and now L = Q5 (i) = Q5 = K and B = Z5 [i] = Z5 = A is a DVR, since x2 + 1 has roots in F5 ' Z5 /5Z5 that we can lift to roots in Z5 via Hensel’s lemma. Remark 9.24. In the previous example you might wonder what happens to the factorization (5) = (1 + 2i)(1 − 2i) in B = Z(5) [i] if we replace A with its completion Z5 and consider B = Z5 [i] = Z5 . The two maximal ideals q1 = (1 + 2i) and q2 = (1 − 2i) in Z(5) [i] are coprime, thus vq1 (q2 ) = 0 and q2 Bq1 is the unit ideal, and conversely. No matter which maximal ideal we localize at, the RHS of the factorization (5) = (1 + 2i)(1 − 2i) is locally the product of the maximal ideal and the unit ideal. The same thing happens if we work in the completion of A. If we pick i ≡ 7 mod 25 as a root of x2 + 1 in Z5 we have (1 + 2i) = (5) and (1 − 2i) = (1), and the situation is reversed if we pick i ≡ 18 mod 25. References [1] Nicolas Bourbaki, General Topology: Chapters 1-4 , Springer, 1985. [2] Jürgen Neukirch, Algebraic number theory, Springer, 1999. [3] Anthony W. Knapp, Advanced algebra, Digital Second Edition, 2016. [4] Dinikar Ramakrishnan and Robert J. Valenza, Fourier analysis on number fields, Springer, 1999. [5] Stacks Project Authors, Stacks Project, http://stacks.math.columbia.edu. 18.785 Fall 2021, Lecture #9, Page 8 18.785 Number theory I Lecture #10 10 Fall 2021 10/13/2019 Extensions of complete DVRs Recall that in our AKLB setup, A is a Dedekind domain with fraction field K, the field L is a finite separable extension of K, and B is the integral closure of A in L; as we proved in Theorem 5.25, this implies that B is also a Dedekind domain (with L as its fraction field), and we proved in Theorem 9.22 that B is a DVR. We now want to show that B is complete. Definition 10.1. Let K be a field with absolute value | | and let V be a K-vector space. A norm on V is a function k k : V → R≥0 such that • kvk = 0 if and only if v = 0. • kλvk = |λ| kvk for all λ ∈ K and v ∈ V . • kv + wk ≤ kvk + kwk for all v, w ∈ V . Each norm k k induces a topology on V via the distance metric d(v, w) := kv − wk. Example 10.2. Let V be aPK-vector space with basis (ei ), and for v ∈ V let vi ∈ K denote the coefficient of ei in v = i vi ei . The sup-norm kvk∞ := sup{|vi |} is a norm on V (thus every vector space has at least one norm). If V is also a K-algebra, an absolute value k k on V (as a ring) is a norm on V (as a K-vector space) if and only if it extends the absolute value on K (fix v 6= 0 and note that kλk kvk = kλvk = |λ| kvk ⇔ kλk = |λ|). Proposition 10.3. Let V be a vector space of finite dimension over a complete field K. Every norm on V induces the same topology, in which V is a complete metric space. Proof. See Problem Set 5. Theorem 10.4. Let A be a complete DVR with fraction field K, maximal ideal p, discrete valuation vp , and absolute value |x|p := cvp (x) , with 0 < c < 1. Let L/K be a finite extension of degree n. The following hold. 1/n (i) There is a unique absolute value |x| := |NL/K (x)|p on L that extends | |p ; (ii) The field L is complete with respect to | |, and its valuation ring {x ∈ L : |x| ≤ 1} is equal to the integral closure B of A in L; (iii) If L/K is separable then B is a complete DVR whose maximal ideal q induces 1 |x| = |x|q := c eq vq (x) , where eq is the ramification index of q, that is, pB = qeq . Proof. Assuming for the moment that | | is actually an absolute value (which is not obvious!), for any x ∈ K we have 1/n 1/n |x| = |NL/K (x)|p = |xn |p = |x|p , so | | extends | |p and is therefore a norm on L. The fact that | |p is nontrivial means that |x|p 6= 1 for some x ∈ K × , and |x|a = |x|p = |x| only for a = 1, which implies that | | is the unique absolute value in its equivalence class extending | |p . Every norm on L induces the same topology (by Proposition 10.3), so | | is the only absolute value on L that extends | |p . We now show | | is an absolute value. Clearly |x| = 0 ⇔ x = 0 and | | is multiplicative; we only need to check the triangle inequality. It suffices to show |x| ≤ 1 ⇒ |x + 1| ≤ |x| + 1, since we always have |y + z| = |z||y/z + 1| and |y| + |z| = |z|(|y/z| + 1), and without loss of generality we assume |y| ≤ |z|. In fact the stronger implication |x| ≤ 1 ⇒ |x + 1| ≤ 1 holds: |x| ≤ 1 ⇐⇒ |NL/K (x)|p ≤ 1 ⇐⇒ NL/K (x) ∈ A ⇐⇒ x ∈ B ⇐⇒ x+1 ∈ B ⇐⇒ |x+1| ≤ 1. The first biconditional follows from the definition of | |, the second follows from the definition of | |p , the third is Corollary 9.21, the fourth is obvious, and the fifth follows from the first three after replacing x with x + 1. This completes the proof of (i), and also proves (ii). We now assume L/K is separable. Then B is a DVR, by Theorem 9.22, and it is complete because it is the valuation ring of L. Let q be the unique maximal ideal of B. The valuation vq extends vp with index eq , by Theorem 8.20, so vq (x) = eq vp (x) for x ∈ K × . We have 0 < c1/eq < 1, so |x|q := (c1/eq )vq (x) is an absolute value on L induced by vq . To show it is equal to | |, it suffices to show that it extends | |p , since we already know that | | is the unique absolute value on L with this property. For x ∈ K × we have 1 |x|q = c eq vq (x) 1 = c eq eq vp (x) = cvp (x) = |x|p , and the theorem follows. Remark 10.5. The transitivity of NL/K in towers (Corollary 4.53) implies that we can uniquely extend the absolute value on the fraction field K of a complete DVR to an algebraic closure K. In fact, this is another form of Hensel’s lemma in the following sense: one can show that a (not necessarily discrete) valuation ring A is Henselian if and only if the absolute value of its fraction field K can be uniquely extended to K; see [4, Theorem 6.6]. Corollary 10.6. Assume AKLB and that A is a complete DVR with maximal ideal p and let q|p. Then vq (x) = f1q vp (NL/K (x)) for all x ∈ L. Proof. vp (NL/K (x)) = vp (NL/K ((x))) = vp (NL/K (qvq (x) )) = vp (pfq vq (x) ) = fq vq (x). Remark 10.7. One can generalize the notion of a discrete valuation to a valuation, a surjective homomorphism v : K × → Γ, in which Γ is a (totally) ordered abelian group and v(x + y) ≥ min(v(x), v(y)); we extend v to K by defining v(0) = ∞ to be strictly greater than any element of Γ. In the AKLB setup with A a complete DVR, one can then define a valuation v(x) = e1q vq (x) with image e1q Z that restricts to the discrete valuation vp on K. The valuation v then extends to a valuation on K with Γ = Q. Some texts take this approach, but we will generally stick with discrete valuations (so our absolute value on L restricts to K, but our discrete valuations on L do not restrict to discrete valuations on K, they extend them with index eq ). Remark 10.8. Recall that a valuation ring is an integral domain A with fraction field K such that for every x ∈ K × either x ∈ A or x−1 ∈ A (possibly both). As you will show on Problem Set 6, if A is a valuation ring, then there exists a valuation v : K → Γ ∪ {∞} for some totally ordered abelian group Γ such that A = {x ∈ K : v(x) ≥ 0} is the valuation ring of K with respect to this valuation. 10.1 The Dedekind-Kummer theorem in a local setting Recall that the Dedekind-Kummer theorem (Theorem 6.14) allows us to factor primes in our AKLB setting by factoring polynomials over the residue field, provided that B is monogenic 18.785 Fall 2021, Lecture #10, Page 2 (of the form A[α] for some α ∈ B), or the prime of interest does not contain the conductor. We now show that in the special case where A and B are DVRs and the residue field extension is separable, B is always monogenic; this holds, for example, whenever K is a local field. To prove this, we first recall a form of Nakayama’s lemma. Lemma 10.9 (Nakayama’s Lemma). Let A be a local ring with maximal ideal p, and let M be a finitely generated A-module. If the images of x1 , . . . , xn ∈ M generate M/pM as an (A/p)-vector space then x1 , . . . , xn generate M as an A-module. Proof. See [1, Corollary 4.8b]. Before proving our theorem on local monogenicity, let us record some corollaries of Nakayama’s Lemma that will be useful to us later. Corollary 10.10. Let A be a local noetherian ring with maximal ideal p, let g ∈ A[x], and let B := A[x]/(g(x)). Every maximal ideal m of B contains the ideal pB. Proof. Suppose not. Then m+pB = B for some maximal ideal m of B. The ring B is finitely generated over the noetherian ring A, hence a noetherian A-module, so its A-submodules are all finitely generated. Let z1 , . . . , zn be A-module generators for m. Every coset of pB in B can be written as z + pB for some A-linear combination z of z1 , . . . , zn , so the images of z1 , . . . , zn generate B/pB as an (A/p)-vector space. By Nakayama’s lemma, z1 , . . . , zn generate B, in which case m = B, a contradiction. As a corollary, we immediately obtain a local version of the Dedekind-Kummer theorem that does not even require A and B to be Dedekind domains. Corollary 10.11. Let A be a local noetherian ring with maximal ideal p, let g ∈ A[x] be a polynomial with reduction ḡ ∈ (A/p)[x], and let α be the image of x in the ring B := A[x]/(g(x)) = A[α]. The maximal ideals of B are (p, gi (α)), where g1 , . . . , gm ∈ A[x] are lifts of the distinct irreducible polynomials ḡi ∈ (A/p)[x] that divide ḡ. Proof. By Corollary 10.10, the quotient map B → B/pB gives a one-to-one correspondence between maximal ideals of B and maximal ideals of B/pB, and we have B A[x] (A/p)[x] ' ' . pB (p, g(x)) (ḡ(x)) Each maximal ideal of (A/p)[x]/(ḡ(x)) is the reduction of an irreducible divisor of ḡ, hence one of the ḡi (because (A/p)[x] is a PID). The corollary follows. Theorem 10.12. Assume AKLB, with A and B DVRs with residue fields k := A/p and l := B/q. If l/k is separable then B = A[α] for some α ∈ B; if L/K is unramified this holds for every lift α of any generator ᾱ for l = k(ᾱ). Proof. Let pB = qe be the factorization of pB and let f = [l : k] be the residue field degree, so that ef = n := [L : K]. The extension l/k is separable, so we may apply the primitive element theorem to write l = k(ᾱ0 ) for some ᾱ0 ∈ l whose minimal polynomial ḡ is separable of degree equal to f . Let g ∈ A[x] be a monic lift of ḡ, and let α0 be any lift of ᾱ0 to B. If vq (g(α0 )) = 1 then let α := α0 . Otherwise, let π0 be any uniformizer for B and let α := α0 + π0 ∈ B (so α ≡ ᾱ0 mod q), and writing g(x + π0 ) = g(x) + π0 g 0 (x) + π02 h(x) for some h ∈ A[x] via Lemma 9.11, we have vq (g(α)) = vq (g(α0 + π0 )) = vq (g(α0 ) + π0 g 0 (α0 ) + π02 h(α0 )) = 1, 18.785 Fall 2021, Lecture #10, Page 3 so π := g(α) is also a uniformizer for B. We now claim B = A[α], equivalently, that 1, α, . . . , αn−1 generate B as an A-module. By Nakayama’s lemma, it suffices to show that the reductions of 1, α, . . . , αn−1 span B/pB as an k-vector space. We have p = qe , so pB = (π e ). We can represent each element of B/pB as a coset b + pB = b0 + b1 π + b2 π · · · + be−1 π e−1 + pB, where b0 , . . . , be−1 are determined up to equivalence modulo πB. Now 1, ᾱ, . . . , ᾱf −1 are a basis for B/πB = B/q as a k-vector space, and π = g(α), so we can rewrite this as b + pB = (a0 + a1 α + · · · af −1 αf −1 ) + (af + af +1 α + · · · a2f −1 αf −1 )g(α) + ··· + (aef −f +1 + aef −f +2 α + · · · aef −1 αf −1 )g(α)e−1 + pB. Since deg g = f , and n = ef , this expresses b + pB in the form b0 + pB with b0 in the A-span of 1, . . . , αn−1 . Thus B = A[α]. We now note that if L/K is unramified then l/k is separable (this is part of the definition of unramified), and e = 1, f = n, in which case there is no need to require g(α) to be a uniformizer and we can just take α = α0 to be any lift of any ᾱ0 that generates l over k. In our AKLB setup, if A is a complete DVR with maximal ideal p then B is a complete P DVR with maximal ideal q|p and the formula [L : K] = q|p eq fq given by Theorem 5.35 has only one term eq fq . We now simplify matters even further by reducing to the two extreme cases fq = 1 (a totally ramified extension) and eq = 1 (an unramified extension, provided that the residue field extension is separable).1 10.2 Unramified extensions of a complete DVR Let A be a complete DVR with fraction field K and residue field k. Associated to any finite unramified extension of L/K of degree n is a corresponding finite separable extension of residue fields l/k of the same degree n. Given that the extensions L/K and l/k are finite separable extensions of the same degree, we might wonder how they are related. More precisely, if we fix K with residue field k, what is the relationship between finite unramified extensions L/K of degree n and finite separable extensions l/k of degree n? Each L/K uniquely determines a corresponding l/k, but what about the converse? This question has a surprisingly nice answer. The finite unramified extensions L of K unr whose morphisms are K-algebra homomorphisms, and the finite sepaform a category CK rable extensions l of k form a category Cksep whose morphisms are k-algebra homomorphisms. These two categories are equivalent. Theorem 10.13. Let A be a complete DVR with fraction field K and residue field k := unr and C sep are equivalent via the functor F : C unr → C sep that sends A/p. The categories CK K k k each unramified extension L of K to its residue field l, and each K-algebra homomorphism ϕ : L1 → L2 to the k-algebra homomorphism ϕ̄ : l1 → l2 defined by ϕ̄(ᾱ) := ϕ(α), where α 1 Recall from Definition 5.37 that separability of the residue field extension is part of the definition of an unramified extension. If the residue field is perfect (as when K is a local field, for example), the residue field extension is automatically separable, but in general it need not be, even when L/K is unramified. 18.785 Fall 2021, Lecture #10, Page 4 is any lift of ᾱ ∈ l1 := B1 /q1 to B1 and ϕ(α) is the reduction of ϕ(α) ∈ B2 to l2 := B2 /q2 ; here q1 , q2 are the maximal ideals of the valuation rings B1 , B2 of L1 , L2 , respectively. unr and C sep , and In particular, F gives a bijection between the isomorphism classes in CK k if L1 , L2 and have residue fields l1 , l2 then F induces a bijection of finite sets ∼ HomK (L1 , L2 ) −→ Homk (l1 , l2 ). Proof. Let us first verify that F is well-defined. It is clear that it maps finite unramified extensions L/K to finite separable extensions l/k, but we should check that the map on morphisms does not depend on the lift α of ᾱ we pick. So let ϕ : L1 → L2 be a K-algebra homomorphism, and for ᾱ ∈ l1 , let α and α0 be two lifts of ᾱ to B1 . Then α − α0 ∈ q1 , and this implies that ϕ(α − α0 ) ∈ ϕ(q1 ) = ϕ(B1 ) ∩ q2 ⊆ q2 , and therefore ϕ(α) = ϕ(α0 ). The identity ϕ(q1 ) = ϕ(B1 ) ∩ q2 ⊆ q2 follows from the fact that ϕ restricts to an injective ring homomorphism B1 → B2 and B2 /ϕ(B1 ) is a finite extension of DVRs in which q2 lies over the prime ϕ(q1 ) of ϕ(B1 ). It’s easy to see that F sends identity morphisms to identity morphisms and that it is compatible with composition, so we have a well-defined functor. To show that F is an equivalence of categories we need to prove two things: • F is essentially surjective: each separable l/k is isomorphic to the residue field of some unramified L/K • F is full and faithful: the induced map HomK (L1 , L2 ) → Homk (l1 , l2 ) is a bijection. We first show that F is essentially surjective. Given a finite separable extension l/k, we may apply the primitive element theorem to write l ' k(ᾱ) = k[x] , (ḡ(x)) for some ᾱ ∈ l whose minimal polynomial ḡ ∈ k[x] is necessarily monic, irreducible, separable, and of degree n := [l : k]. Let g ∈ A[x] be any monic lift of ḡ; then g is also irreducible, separable, and of degree n. Now let L := K[x] = K(α), (g(x)) where α is the image of x in K[x]/g(x). Then L/K is a finite separable extension, and by Corollary 10.11, (p, g(α)) is the unique maximal ideal of A[α] (since ḡ is irreducible) and B A[α] A[x] (A/p)[x] ' ' ' ' l. q (p, g(α)) (p, g(x)) (ḡ(x)) We thus have [L : K] = deg g = [l : k] = n, and it follows that L/K is an unramified extension of degree n = f := [l : k]: the ramification index of q is necessarily e = n/f = 1, and the extension l/k is separable by assumption (so in fact B = A[α], by Theorem 10.12). We now show that the functor F is full and faithful. Given finite unramified extensions L1 , L2 with valuation rings B1 , B2 and residue fields l1 , l2 , we have induced maps ∼ HomK (L1 , L2 ) −→ HomA (B1 , B2 ) −→ Homk (l1 , l2 ). The first map is given by restriction from L1 to B1 , and since tensoring with K gives an inverse map in the other direction, it is a bijection. We need to show that the same is 18.785 Fall 2021, Lecture #10, Page 5 true of the second map, which sends ϕ : B1 → B2 to the k-homomorphism ϕ that sends α ∈ l1 = B1 /q1 to the reduction of ϕ(α) modulo q2 , where α is any lift of ᾱ. As above, use the primitive element theorem to write l1 = k(ᾱ) = k[x]/(ḡ(x)) for some ᾱ ∈ l1 . If we now lift ᾱ to α ∈ B1 , we must have L1 = K(α), since [L1 : K] = [l1 : k] is equal to the degree of the minimal polynomial ḡ of ᾱ which cannot be less than the degree of the minimal polynomial g of α (both are monic). Moreover, we also have B1 = A[α], since this is true of the valuation ring of every finite unramified extension in our category. Each A-module homomorphism in   A[x] HomA (B1 , B2 ) = HomA , B2 (g(x)) is uniquely determined by the image of x in B2 . This gives a bijection between HomA (B1 , B2 ) and the roots of g in B2 . Similarly, each k-algebra homomorphism in   k[x] , l2 Homk (l1 , l2 ) = Homk (ḡ(x)) is uniquely determined by the image of x in l2 , and there is a bijection between Homk (l1 , l2 ) and the roots of ḡ in l2 . Now ḡ is separable, so every root of ḡ in l2 = B2 /q2 lifts to a unique root of g in B2 , by Hensel’s Lemma 9.15. Thus the map HomA (B1 , B2 ) −→ Homk (l1 , l2 ) induced by F is a bijection. Remark 10.14. In the proof above we actually only used the fact that L1 /K is unramified. The map HomK (L1 , L2 ) → Homk (l1 , l2 ) is a bijection even if L2 /K is not unramified. Let us note the following corollary, which follows from our proof of Theorem 10.13. Corollary 10.15. Assume AKLB with A a complete DVR with residue field k. Then L/K is unramified if and only if B = A[α] for some α ∈ L whose minimal polynomial g ∈ A[x] has separable image ḡ in k[x]. Proof. The forward direction was proved in the proof of the theorem, and for the reverse direction note that ḡ must be irreducible, since otherwise we could use Hensel’s lemma to lift a non-trivial factorization of ḡ to a non-trivial factorization of g, so the residue field extension is separable and has the same degree as L/K, so L/K is unramified. Corollary 10.16. Let A be a complete DVR with fraction field K and residue field k, and let ζn be a primitive nth root of unity in some algebraic closure of K, with n prime to the characteristic of k. The extension K(ζn )/K is unramified. Proof. The field K(ζn ) is the splitting field of f (x) = xn − 1 over K. The image f¯ of f in k[x] is separable when p - n, since gcd(f¯, f¯0 ) 6= 1 only when f¯0 = nxn−1 is zero, equivalently, only when p|n. When f¯ is separable, so are all of its divisors, including the reduction of the minimal polynomial of ζn , which must be irreducible since otherwise we could obtain a contradiction by lifting a non-trivial factorization via Hensel’s lemma. It follows that the residue field of K(ζn ) is a separable extension of k, thus K(ζn )/K is unramified. When the residue field k is finite (always the case if K is a local field), we can give a precise description of the finite unramified extensions L/K. 18.785 Fall 2021, Lecture #10, Page 6 Corollary 10.17. Let A be a complete DVR with fraction field K and finite residue field Fq , and let L be a degree n extension of K. Then L/K is unramified if and only if L ' K(ζqn −1 ). When this holds, A[ζqn −1 ] is the integral closure of A in L and L/K is a Galois extension with Gal(L/K) ' Z/nZ. Proof. The reverse implication is implied by Corollary 10.16; note that K(ζqn −1 ) has den gree n over K because its residue field is the splitting field of xq −1 − 1 over Fq , which is an extension of degree n (indeed, one can take this as the definition of Fqn ). Suppose L/K is unramified. Then [l : k] = [L : K] = n and l ' Fqn has multiplicative group cyclic of order q n − 1 generated by some ᾱ. The minimal polynomial ḡ ∈ Fq [x] of ᾱ n n divides xq −1 − 1, and since ḡ is irreducible, it is coprime to the quotient (xq −1 − 1)/ḡ. By n Hensel’s Lemma 9.19, we can lift ḡ to a polynomial g ∈ A[x] that divides xq −1 − 1 ∈ A[x], and by Hensel’s Lemma 9.15 we can lift ᾱ to a root α of g, in which case α is also a root of n xq −1 − 1; it must be a primitive (q n − 1)-root of unity because its reduction ᾱ is. Let B be the integral closure of A in L. We have B ' A[ζqn −1 ] by Theorem 10.12, and n L is the splitting field of xq −1 − 1, since its residue field Fqn is (we can lift the factorization n of xq −1 − 1 from Fqn to L via Hensel’s lemma). It follows that L/K is Galois, and the bijection between (q n − 1)-roots of unity in L and Fqn induces an isomorphism Gal(L/K) ' Gal(l/k) = Gal(Fqn /Fq ) ' Z/nZ. Corollary 10.18. Let A be a complete DVR with fraction field K and finite residue field of characteristic p, and suppose that K does not contain a primitive pth root of unity. The extension K(ζm )/K is ramified if and only if p divides m. Proof. If p does not divide m then Corollary 10.16 implies that K(ζm )/K is unramified. If p divides m then K(ζm ) contains K(ζp ), which by Corollary 10.17 is unramified if and only if K(ζp ) ' K(ζpn −1 ) with n := [K(ζp ) : K], which occurs if and only if p divides pn − 1 (since ζp 6∈ K), which it does not; thus K(ζp ) and therefore K(ζm ) is ramified when p|m. Example 10.19. Consider A = Zp , K = Qp , k = Fp , and fix Fp and Qp . For each positive integer n, the finite field Fp has a unique extension of degree n in Fp , namely, Fpn . Thus for each positive integer n, the local field Qp has a unique unramified extension of degree n; it can be explicitly constructed by adjoining a primitive root of unity ζpn −1 to Qp . The n element ζpn −1 will necessarily have minimal polynomial of degree n dividing xp −1 − 1. Another useful consequence of Theorem 10.13 that applies when the residue field is finite is that the norm map NL/K restricts to a surjective map B × → A× on unit groups; in fact, this property characterizes unramified extensions. Theorem 10.20. Assume AKLB with A a complete DVR with finite residue field. Then L/K is unramified if and only if NL/K (B × ) = A× . Proof. See Problem Set 6. Definition 10.21. Let L/K be a separable extension. The maximal unramified extension of K in L is the subfield [ E⊆L K⊆E⊆L E/K fin. unram. where the union is over finite unramified subextensions E/K. When L = K sep is the separable closure of K, this is the maximal unramified extension of K, denoted K unr . 18.785 Fall 2021, Lecture #10, Page 7 Example 10.22. The field Qunr is an infinite extension of Qp with Galois group p Gal(Qunr lim Gal(Fpn /Fp ) ' lim Z/nZ =: Ẑ, p /Qp ) ' Gal(Fp /Fp ) = ← − ←− n n where the inverse limit is taken over positive integers n ordered by divisibility. The ring Ẑ is the profinite completion of Z. The field Qunr has value group Z and residue field Fp . p Theorem 10.23. Assume AKLB with A a complete DVR and separable residue field extension l/k. Let e and f be the ramification index and residue field degrees, respectively, and let q be the unique prime of B. The following hold: (i) There is a unique intermediate field extension E/K that contains every unramified extension of K in L and it has degree [E : K] = f . (ii) The extension L/E is totally ramified and has degree [L : E] = e. (iii) If L/K is Galois then Gal(L/K) is the decomposition group of Dq , Gal(L/E) is the inertia subgroup of Iq , and E/K is Galois with Gal(E/K) ' Dq /Iq ' Gal(l/k). Proof. (i) Let E/K be the finite unramified extension of K in L corresponding to the finite separable extension l/k given by Theorem 10.13; then [E : K] = [l : k] = f as desired. The maximal unramified extension E 0 of K in L has the same residue field l as L, which is also the residue field of E, and equivalence of categories given by Theorem 10.13 implies that the trivial isomorphism ` ' ` corresponds to an isomorphism E ' E 0 that allows us to view E as a subfield of L; the same applies to any unramified extension of K with residue field l, so E is unique up to isomorphism. (ii) Let n = [L : K]. Then [L : E] = [L : K]/[E : K] = n/f = ef /f = e. (iii) We have Dq ⊆ Gal(L/K) of order ef = [L : K], so this inclusion is an equality. If we put qE := q ∩ E then Proposition 7.13 implies IqE = Gal(L/E) ∩ Iq . These three groups all have order e and must coincide. The group Iq is a normal in Dq since it is the kernel of the surjective homomorphism πq : Dq → Gal(l/k)), so E/K is normal, hence Galois (it must be separable since L/K is), and it follows that Gal(E/K) ' Dq /Iq ' Gal(l/k). References [1] David Eisenbud, Commutative algebra with a view toward algebraic geometry, Springer, 1995. [2] Neal Koblitz, p-adic numbers, p-adic analysis, and zeta functions, Springer, 1984. [3] Serge Lang, Algebraic number theory, second edition, Springer, 1994. [4] Jürgen Neukirch, Algebraic number theory, Springer, 1999. 18.785 Fall 2021, Lecture #10, Page 8 18.785 Number theory I Lecture #11 11 Fall 2021 10/18/2021 Totally ramified extensions and Krasner’s lemma In the previous lecture we showed that in the AKLB setup, if A is a complete DVR with maximal ideal p then B is a complete DVR with maximal ideal q and [L : K] = n = eq fq ; see Theorem 10.4 (note that the AKLB setup includes the assumption that L/K is separable). In this setting we may unambiguously write eL/K for eq and fL/K for fq , since q is the unique prime of L. Provided the residue field extension is separable (always the case if K is a local field), we can decompose the extension L/K as a tower of field extensions L/E/K in which E/K is unramified (so eE/K = 1 and fE/K = fL/K ) and L/E is totally ramified (so eL/E = eL/K and fL/E = 1), by Theorem 10.23. In the previous lecture we classified unramified extensions of (fraction fields of) complete DVRs, and showed that when the residue field is finite (always true for local fields), unramified extensions are all cyclotomic extensions of the form K(ζn )/K for some n coprime to the residue field characteristic; see Corollary 10.17. In this lecture we will classify totally ramified extensions of complete DVRs. 11.1 Totally ramified extensions of a complete DVR Definition 11.1. Let A be a DVR with maximal ideal p. A monic polynomial f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 ∈ A[x] is Eisenstein (or an Eisenstein polynomial ) if ai ∈ p for 0 ≤ i < n and a0 6∈ p2 ; equivalently, vp (ai ) ≥ 1 for 0 ≤ i < n and vp (a0 ) = 1. Note that this means a0 is a uniformizer. Lemma 11.2 (Eisenstein irreducibility). Let A be a DVR with fraction field K and let f ∈ A[x] be Eisenstein. Then f is irreducible in both A[x] and K[x]. Proof. Suppose not. Then P P f = gh has P degree n ≥ 2 for some non-constant monic g, h ∈ A[x]. Put f = i fi xi , g = i gi xi , h = i hi xi . Then vp (f0 ) = vp (g0 h0 ) = vp (g0 ) + vp (h0 ) = 1, where p is the maximal ideal of A, and without loss of generality we may assume vp (g0 ) = 0 and vp (h0 ) = 1. Let i > 0 be the least i for which vp (hi ) = 0; such an i < n exists since h is monic and deg h < n. We have fi = g0 hi + g1 hi−1 + · · · + gi−1 h1 + gi h0 , with vp (fi ) ≥ 1 since f is Eisenstein and i < n, but the valuation of the RHS is zero, since vp (g0 hi ) = 0 and vp (gj hi−j ) ≥ 1 for 0 ≤ j < i, by the minimality of i, which is a contradiction. Thus f is irreducible in A[x], and since A is a DVR, and therefore a PID and thus a UFD, f is irreducible in K[x], by Gauss’s Lemma [1]. Remark 11.3. We can apply Lemma 11.2 to any polynomial f (x) over a Dedekind domain A that is Eisenstein over a localization Ap ; the rings Ap and A have the same fraction field K and f is then irreducible in K[x], hence in A[x]; this gives the Eisenstein criterion for irreducibility. Lemma 11.4. Let A be a DVR and let f ∈ A[x] be an Eisenstein polynomial. Then B = A[π] := A[x]/(f ) is a DVR with uniformizer π, where π is the image of x in A[x]/(f ). Proof. Let p be the maximal ideal of A. We have f ≡ xn mod p, so by 10.11 the P Corollary i ideal q = (p, x) = (p, π) is the only maximal ideal of B. Let f = fi x ; then p = (f0 ) and q = (f0 , π), and f0 = −f1 π − f2 π 2 − · · · − π n ∈ (π), so q = (π). The unique maximal ideal (π) of B is nonzero and principal, so B is a DVR with uniformizer π. Theorem 11.5. Assume AKLB with A a complete DVR and π a uniformizer for B. The extension L/K is totally ramified if and only if B = A[π] and the minimal polynomial of π is Eisenstein. Proof. Let n = [L : K], let p be the maximal ideal of A, let q be the maximal ideal of B (which we recall is a complete DVR, by Theorem 10.4), and let π be a uniformizer for B with minimal polynomial f . If B = A[π] and f is Eisenstein, then as in Lemma 11.4 we have p = qn , so vq extends vp with index eq = n and L/K is totally ramified. We now suppose L/K is totally ramified. Then vq extends vp with index n, which implies vq (K) = nZ. The set {π 0 , π 1 , π 2 , . . . , π n−1 } is linearly independent over K, since P i = 0 we must a π the valuations of π 0 , . . . π n−1 are distinct modulo vq (K) = nZ (if n−1 i i=0 have vq (ai π i ) = vq (aj π j ) for some nonzero ai and aj with i 6= j, which is impossible). Thus L = K(π). P Let f = ni=0 ai xi ∈ A[x] be the minimal polynomial of π. We have vq (f (π)) = ∞ and vq (ai π i ) ≡ i mod n for 0 ≤ i ≤ n. This is possible only if vq (a0 ) = vq (a0 π 0 ) = vq (an π n ) = vq (π n ) = n, and vq (ai ) ≥ n for 0 ≤ i < n. This implies that vp (a0 ) = 1, since vq extends vp with index n, and vp (ai ) ≥ 1 for 0 ≤ i < n. Thus f is Eisenstein and Lemma 11.4 implies that A[π] ⊆ B is a DVR, hence maximal, so B = A[π]. Example 11.6. Let K = Q3 . As shown √ in an √earlier problem √ set, there are just three √ distinct quadratic extensions of Q3 : Q3 ( 2), Q3 ( 3), and Q3 ( 6). The extension Q3 ( 2) is the unique unramified quadratic extension of Q3 , and we note that it can be written as a cyclotomic extension Q3 (ζ8 ). The other two are both ramified, and can be defined by the Eisenstein polynomials x2 − 3 and x2 − 6. Definition 11.7. Assume AKLB with A a complete DVR and separable residue field extension of characteristic p ≥ 0. The extension L/K is tamely ramified if p6 | eL/K (always true if p = 0); note that unramified extensions are tamely ramified. Otherwise L/K is wildly ramified if p|eL/K . A totally ramified extension L/K is totally tamely ramified if p6 | eL/K , and it is totally wildly ramified if eL/K is a power of p (a totally ramified extension that is wildly ramified need not be totally wildly ramified). Recall that ramification indices multiply in towers (Lemma 5.30), and separability is a transitive in towers (Corollary 4.14). This yields the following proposition, which we note applies to all nonarchimedean local fields. Proposition 11.8. The properties of being unramified, tamely ramified, wildly ramified, totally ramified, totally tamely ramified, and totally wildly ramified are all transitive in towers of extensions of fraction fields of complete DVRs with separable residue field extensions. Proof. This follows immediately from the transitivity of separability and the multiplicativity of ramification indices and degrees in towers. Remark 11.9. A compositum of totally ramified extensions need not be totally ramified. From √ Example 11.6 √ we see that the compositum of the totally ramified quadratic √ extensions Q3 ( 3) and Q3 ( 6) of Q3 contains the unramified quadratic extension Q3 ( 2) of Q3 . 18.785 Fall 2021, Lecture #11, Page 2 Theorem 11.10. Assume AKLB with A a complete DVR and separable residue field extension of characteristic p ≥ 0 not dividing n := [L : K]. The extension L/K is totally 1/n tamely ramified if and only if L = K(πA ) for some uniformizer πA of A. 1/n 1/n Proof. If L = K(πA ) then π = πA has minimal polynomial xn − πA , which is Eisenstein, so A[π] is a DVR by Lemma 11.4. This implies B = A[π], since DVRs are maximal, and Theorem 11.5 implies that L/K is totally tamely ramified, since p - n. Now assume L/K is totally tamely ramified and let p and q be the maximal ideals of A and B with uniformizers πA and πB respectively. Then vq extends vp with index eq = n n ) = n = v (π ). This implies that π n = uπ for some unit u ∈ B × . We have and vq (πB q A A B fq = 1, so B and A have the same residue field, and if we lift the image of u in B/q ' A/p to a unit uA in A and replace πA with u−1 A πA , we can assume that u ≡ 1 mod q. Now define g(x) := xn − u ∈ B[x] with reduction ḡ = xn − 1 in (B/q)[x]. We have ḡ 0 (1) = n 6= 0 (since p6 | n), so by Hensel’s Lemma 9.15 we can lift the root 1 of ḡ(x) in B/q to a root r of g(x) in B. Now let π := πB /r. Then π is a uniformizer for B and B = A[π] by Theorem 11.5, so n /r n = π n /u = π , so L = K(π 1/n ) as desired. L = K(π), and π n = πB A B A Proposition 11.11. Let L be a totally ramified extension of the fraction field K of a complete DV R. There is a unique intermediate field E such that E/K is totally tamely ramified and L/E is totally wildly ramified. Proof. Let e := eL/K be the ramification index and let p ≥ 0 be the characteristic of the residue field. If p 6 | e then the proposition holds with E = L, so we assume p|e, and put e = mpa with p - m (possibly m = 1). Let A be the valuation ring of K with maximal ideal p, and let B be the valuation ring of L (also a complete DVR) with maximal ideal q. As in the proof of Theorem 11.10, we can n = uπ with u ∈ B × and u ≡ 1 mod q. choose uniformizers πA of A and πB of B such that πB A m Let g(x) = x − u ∈ B[x]; as in the proof of the theorem we can construct a root r ∈ B of g(x) by Hensel lifting the root 1 of ḡ ∈ (B/q)[x]. Now consider the field extension pa n /r m = π n /u = π , so E = K(π 1/m ) with /r. We have π m = πB E := K(π), where π := πB A B A p - m. The polynomial xm − πA of π is Eisenstein, hence irreducible, and has π as a root, so E/K has degree m. By Theorem 11.10, the extension E/K is totally tamely ramified (the residue field extension is trivial, so it is certainly separable), and the extension L/E has degree pa and is thus totally wildly ramified. To see that E is unique, suppose E 0 ⊆ L is another totally tamely ramified extensions of 1/m K such that L/E 0 is totally wildly ramified. Then E 0 must also be of the form E 0 = K(πA ), n for intermediate field), by Theorem 11.10 and its proof (we can use the same πA = u−1 πB in other words, E and E 0 are both generated by (possibly different) roots of xm − πA . The ratio of these roots is a (not necessarily primitive) mth root of unity ζ ∈ L that must lie in K because L/K is totally ramified and the extension K(ζ)/K is necessarily unramified, by Corollary 10.16, since p - m. It follows that E 0 = E. Corollary 11.12. Let L be a finite separable extension of the fraction field field K of a complete DV R with separable residue field extension. There is a unique intermediate field E such that E/K is tamely ramified and L/E is totally wildly ramified. Proof. Let F be the maximal unramified extension of K in L. By Corollary 10.15 we can assume K = F (α) where α is an integral element whose minimal polynomial g has separable image in k[x], where k is the residue field of K. Applying the previous proposition to the 18.785 Fall 2021, Lecture #11, Page 3 totally ramified extension L/F yields a tamely ramified extension E/F with L/E totally wildly ramified. Unramified extensions are tamely ramified, so E/F/K is a tower of tamely ramified extensions, hence tamely ramified. Any field E 0 with L/E 0 totally wildly ramified must contain α, otherwise E 0 (α) would be a non-trivial unramified subextension L/E 0 (here we are again applying Corollary 10.15 and the fact that the image of the minimal polynomial of α over E 0 must divide g and thus has separable image in k[x] and in k 0 [x], where k 0 is the residue field of E 0 , since k 0 is an extension of k). Proposition 11.11 then implies E 0 = E. 11.2 Krasner’s lemma Let K be the fraction field of a complete DVR with absolute value | |. By Theorem 10.4 we can uniquely extend | | to any finite extension L/K by defining |x| := |NL/K (x)|1/n , where n = [L : K]; as noted in Remark 10.5, this induces a unique absolute value on K that restricts to the absolute value of K. Lemma 11.13. Let K be the fraction field of a complete DVR with algebraic closure K and absolute value | | extended to K. For all α ∈ K and σ ∈ AutK (K) we have |σ(α)| = |α|. Proof. The elements α and σ(α) must have the same minimal polynomial f ∈ K[x], since f (σ(α)) = σ(f (α)) = 0, so NK(α)/K (α) = f (0) = NK(σ(α))/K (σ(α)), by Proposition 4.51. It follows that |σ(α)| = |NK(σ(α))/K (σ(α))|1/n = |NK(α)/K (α)|1/n = |α|, where n = deg f . Definition 11.14. Let K be the fraction field of a complete DVR with absolute value | | extended to an algebraic closure K. For α, β ∈ K, we say β belongs to α if |β−α| < |β−σ(α)| for all σ ∈ AutK (K) with σ(α) 6= α, that is, β is strictly closer to α than it is to any of its conjugates. This is equivalent to requiring that |β − α| < |α − σ(α)| for all σ(α) 6= α, since every nonarchimedean triangle is isosceles (if one side is shorter than another, it is the shortest of all three sides). Lemma 11.15 (Krasner’s lemma). Let K be the fraction field of a complete DVR and let α, β ∈ K, with α separable over K. If β belongs to α then K(α) ⊆ K(β). Proof. Suppose not. Then β belongs to α but α 6∈ K(β). The extension K(α, β)/K(β) is separable and non-trivial, so there is an automorphism σ ∈ AutK(β) (K/K(β)) for which σ(α) 6= α (let σ send α to a different root of the minimal polynomial of α over K(β)). Applying Lemma 11.13 to β − α ∈ K, we have |β − α| = |σ(β − α)| = |σ(β) − σ(α)| = |β − σ(α)|, since σ fixes β. But this contradicts the hypothesis that β belongs to α, since σ(α) 6= α. Remark 11.16. Krasner’s lemma is another “Hensel’s lemma" in the sense that it characterizes Henselian fields (fraction fields of Henselian rings); although the lemma is named after Krasner [2], it was proved earlier by Ostrowski in [3]. P Definition 11.17. For a field K with absolute value | | the L1 -norm of f = fi xi ∈ K[x] is X kf k1 := |fi |, i It is easy to check that k k1 satisfies all the properties of Definition 10.1 and is thus a norm on the K-vector space K[x]. 18.785 Fall 2021, Lecture #11, Page 4 Q Lemma 11.18. Let K be a field with absolute value | | and let f := ni=1 (x − αi ) ∈ K[x] be a monic polynomial with roots α1 , . . . , αn ∈ L, where L/K is a field with an absolute value that extends | |. Then |α| < kf k1 for every root α of f . Proof. The lemma is clear for n ≤ 1, so assume n ≥ 2. If kf k1 = 1 then we must have f = xn and α = 0, in which case |α| = 0 < 1 = kf k1 and the lemma holds. Otherwise kf k1 > 1, and if |α| ≤ 1 the lemma holds, so let α is a root of f with |α| > 1. We have 0 = |f (α)| = αn + n−1 X i=0 fi αi ≥ |α|n − n−1 X i=0 |fi ||α|i ≥ |α|n − |α|n−1 n−1 X i=0 |fi | ≥ |α| − (kf k1 − 1), where we have used |a| = |a + b − b| ≤ |a + b| + | − b| = |a + b| + |b| to get the general inequality |a + b| ≥ |a| − |b| which we applied repeatedly to get the first inequality above, we used |α| > 1 to get the second (replacing |α|i with |α|n−1 in each term) and the third (dividing by |α|n−1 ≥ 1). Thus kf k1 − 1 ≥ |α|, and therefore kf k1 ≥ |α| + 1 > |α|. Theorem 11.19 (Continuity of roots). Let K be the fraction field of a complete DVR and f ∈ K[x] a monic irreducible separable polynomial. There exists δ = δ(f ) ∈ R>0 such that for every monic polynomial g ∈ K[x] with kf − gk1 < δ the following holds: Every root β of g belongs to a root α of f for which K(β) = K(α). In particular, every such g is separable, irreducible, and has the same splitting field as f . Proof. We first note that we can always pick δ < 1, in which case any monic g ∈ K[x] with kf − gk1 < δ must have the same degree as f , so we can assume deg g = deg f . Let us fix an algebraic closure K of K with absolute value | | extending the absolute value on K. Let α1 , . . . , αn be the roots of f in K, and write f (x) = n Y X (x − αi ) = fi xi . i i=0 Let  be the lesser of 1 and the minimum distance |αi − αj | between any two distinct roots of f . We now define n   > 0, δ := δ(f ) := 2(kf k1 + 1) P and note that δ < 1, since kf k1 ≥ 1 and  ≤ 1. Let g(x) = i gi xi be a monic polynomial of degree n with kf − gk1 < δ. We then have kgk1 ≤ kf k1 + kg − f k1 = kf k1 + kf − gk1 < kf k1 + δ, and for any root β ∈ K of g we have |f (β)| = |f (β) − g(β)| = |(f − g)(β)| = n X i=0 (fi − gi )β i ≤ n X i=0 |fi − gi ||β|i . We have |β| < kgk1 by Lemma 11.18, and kgk1 ≥ 1, so |β|i < kgki1 ≤ kgkn1 . Thus |f (β)| < kf − gk1 · kgkn1 < δ(kf k1 + δ)n < δ(kf k1 + 1)n ≤ (/2)n , and therefore n Y i=1 |β − αi | = |f (β)| < (/2)n . 18.785 Fall 2021, Lecture #11, Page 5 It follows that |β − αi | < /2 for at least one αi , and the triangle inequality implies that this αi must be unique since |αi − αj | ≥  for i 6= j. Therefore β belongs to α := αi . By Krasner’s lemma, K(α) ⊆ K(β), and we have n = [K(α) : K] ≤ [K(β) : K] ≤ n, so K(α) = K(β). It follows that g is the minimal polynomial of β, since deg(g) = [K(β) : K]. Thus g is irreducible, and it is also separable, since β ∈ K(β) = K(α) lies in a separable extension of K. We now observe that if a root β of g belongs to a root α of f , then for any τ ∈ AutK (K) and all σ ∈ AutK (K) such that σ(α) 6= α we have |τ (β) − τ (α)| = |τ (β − α)| = |β − α| < |α − σ(α)| = |τ (α − σ(α))| = |τ (α) − τ (σ(α))|. Noting that σ(α) 6= α ⇐⇒ τ (σ(α)) 6= τ (α), this implies that τ (β) belongs to τ (α). Now AutK (K) acts transitively on the roots of f and g, so every root β of g belongs to a distinct root α of f for which K(β) = K(α). Therefore g has the same splitting field as f . 11.3 Local extensions come from global extensions Let L̂ be a local field. From our classification of local fields (Theorem 9.9), we know that L̂ is (isomorphic to) a finite extension of K̂ = Qp (some p ≤ ∞) or K̂ = Fq ((t)) (some q). We also know that the completion of a global field at any of its nontrivial absolute values is a local field (Corollary 9.7). It thus reasonable to ask whether L̂ is the completion of a corresponding global field L that is a finite extension of K = Q or K = Fq (t). More generally, for any fixed global field K and local field K̂ that is the completion of K with respect to one of its nontrivial absolute values | |, we may ask whether every finite extension of local fields L̂/K̂ necessarily corresponds to an extension of global fields L/K, where L̂ is the completion of L with respect to one of its absolute values (whose restriction to K must be equivalent to | |). The answer is yes. In order to simplify matters we restrict our attention to the case where L̂/K̂ is separable, but this is true in general. Theorem 11.20. Let K be a global field with a nontrivial absolute value | |, and let K̂ be the completion of K with respect to | |. Every finite separable extension L̂ of K̂ is the completion of a finite separable extension L of K with respect to an absolute value that restricts to | |. One can choose L so that [L : K] = [L̂ : K̂], in which case L̂ = K̂ · L. Proof. Let L̂/K̂ be a separable extension of degree n. If | | is archimedean then K is a number field and K̂ is either R or √ C; the only nontrivial case is K̂ ' R and n = 2, and we may then assume that L̂ = K̂( d) ' C where d ∈ Z<0 is any nonsquare in K (such a d exists because K/Q is finite). We may assume without loss of generality that | | is the Euclidean √ absolute value on√K̂ ' R √ (it must be equivalent to it), and uniquely extend | | to L := K( d) by requiring | d| = −d. Then L̂ is the completion of L with respect to | |, and clearly [L : K] = [L̂ : K̂] = 2, and L̂ is the compositum of K̂ and L. We now suppose that | | is nonarchimedean, in which case the valuation ring of K̂ is a complete DVR and | | is induced by its discrete valuation. By the primitive element theorem (Theorem 4.12), we may assume L̂ = K̂[x]/(f ) where f ∈ K̂[x] is monic, irreducible, and separable. The field K is dense in its completion K̂, so we can find a monic g ∈ K[x] ⊆ K̂[x] such that kg − f k1 < δ for any δ > 0. It then follows from Theorem 11.19 that L̂ = K̂[x]/(g) (and that g is separable). The field L̂ is a finite separable extension of the fraction field of a complete DVR, so by Theorem 10.4 it is itself the fraction field of a complete DVR and has a unique absolute value that extends the absolute value | | on K̂. Now let L := K[x]/(g). The polynomial g is irreducible in K̂[x], hence in K[x], so [L : K] = deg g = [L̂ : K̂]. The field L̂ contains both K̂ and L, and it is clearly the smallest 18.785 Fall 2021, Lecture #11, Page 6 field that does (since g is irreducible in K̂[x]), so L̂ is the compositum of K̂ and L. The absolute value on L̂ restricts to an absolute value on L extending the absolute value | | on K, and L̂ is complete, so L̂ contains the completion of L with respect to | |. On the other hand, the completion of L with respect | | contains L and K̂, so it must be L̂. In the preceding theorem, when the local extension L̂/K̂ is Galois one might ask whether the corresponding global extension L/K is also Galois, and whether Gal(L̂/K̂) ' Gal(L/K). As shown by the following example, this need not be the case. Example 11.21. Let K = Q, K̂ = Q7 and L̂ = K̂[x]/(x3 − 2). The extension L̂/K̂ is Galois because K̂ = Q7 contains ζ3 (we can lift the root 2 of x2 + x + 1 ∈ F7 [x] to a root of x2 + x + 1 ∈ Q7 [x] via Hensel’s lemma), and this implies that x3 − 2 splits completely in L̂. But L = K[x]/(x3 − 2) is not a Galois extension of K because it contains only one root of x3 − 2. However, we can replace K with Q(ζ3 ) without changing K̂ (take the completion of K with respect to the absolute value induced by a prime above 7) or L̂, but now L = K[x]/(x3 − 2) is a Galois extension of K. In the example we were able to adjust our choice of the global field K without changing the local fields extension L̂/K̂ in a way that ensures that L̂/K̂ and L/K have the same automorphism group. Indeed, this is always possible. Corollary 11.22. For every finite Galois extension L̂/K̂ of local fields there is a finite Galois extension of global fields L/K and an absolute value | | on L such that L̂ is the completion of L with respect to | |, K̂ is the completion of K with respect to the restriction of | | to K, and Gal(L/K) ' Gal(L̂/K̂). Proof. The archimedean case is already covered by Theorem 11.20 (take K = Q), so we assume L̂ is nonarchimedean and note that we may take | | to be the absolute value on both K̂ and on L̂, by Theorem 10.4. The field K̂ is an extension of either Qp or Fq ((t)), and by applying Theorem 11.20 to this extension we may assume K̂ is the completion of a global field K with respect to the restriction of | |. As in the proof of the theorem, let g ∈ K[x] be a monic separable polynomial irreducible in K̂[x] such that L̂ = K̂[x]/(g) and define L := K[x]/(g) so that L̂ is the compositum of K̂ and L. Now let M be the splitting field of g over K, the minimal extension of K that contains all the roots of g (which are distinct because g is separable). The field L̂ also contains these roots (since L̂/K̂ is Galois) and L̂ contains K, so L̂ contains a subextension of K isomorphic to M (by the universal property of a splitting field), which we now identify with M ; note that L̂ is also the completion of M with respect to the restriction of | | to M . We have a group homomorphism ϕ : Gal(L̂/K̂) → Gal(M/K) induced by restriction, and ϕ is injective (each σ ∈ Gal(L̂/K̂) is determined by its action on any root of g in M ). If we now replace K by the fixed field of the image of ϕ and replace L with M , the completion of K with respect to the restriction of | | is still equal to K̂, and similarly for L and L̂, and now Gal(L/K) ' Gal(L̂/K̂) as desired. 11.4 Completing a separable extension of Dedekind domains We now return to our general AKLB setup: A is a Dedekind domain with fraction field K with a finite separable extension L/K, and B is the integral closure of A in L, which is also a Dedekind domain. Recall from Theorem 8.20 that if p is a prime of K (a nonzero prime ideal of A), each prime q|p induces a valuation vq of L that extends the valuation vp of K 18.785 Fall 2021, Lecture #11, Page 7 with index eq , meaning that vq |K = eq vp (and every valuation of L that extends vp arises in this way). We now want to look at what happens when we complete K with respect to the absolute value | |p induced by vp to obtain a complete field Kp , and similarly complete L with respect to | |q for some q|p to obtain Lq . This includes the case where L/K is an extension of global fields, in which case we get a corresponding extension Lq /Kp of local fields for each q|p; as proved below, the embedding K ,→ L induces an embedding Kp ,→ Lq of topological fields in which the absolute value | |p on Kp is equivalent to the restriction of | |q to Kp (if we define | |q as in Theorem 10.4, | |p will be the restriction of | |q ). In general the extension Lq /Kp may have smaller degree than L/K. If L ' K[x]/(f ), the irreducible polynomial f ∈ K[x] need not be irreducible over Kp . Indeed, this will necessarily be the case if there is more than one prime q lying above p; the Dedekind-Kummer theorem gives a one-to-one correspondence between irreducible factors of f in Kp [x] and primes q|p (via Hensel’s Lemma). The following theorem gives a complete description of the situation. Q Theorem 11.23. Assume AKLB, let p be a prime of A, and let pB = q|p qeq be the factorization of pB in B. Let Kp be the completion of K with respect to | |p , and let p̂ be the maximal ideal of its valuation ring. For each q|p, let Lq denote the completion of L with respect to | |q , and q̂ the maximal ideal of its valuation ring. The following hold: (1) Each Lq is a finite separable extension of Kp with [Lq : Kp ] ≤ [L : K]. (2) Each q̂ is the unique prime of Lq lying over p̂. (3) Each q̂ has ramification index eq̂ = eq and residue field degree fq̂ = fq . (4) [Lq : Kp ] = eq fq ; Q (5) The map L ⊗K Kp → q|p Lq defined by ` ⊗ x 7→ (`x, . . . , `x) is an isomorphism of finite étale Kp -algebras. (6) If L/K is Galois then each Lq /Kp is Galois and we have isomorphisms of decomposition groups Dq ' Dq̂ = Gal(Lq /Kp ) and inertia groups Iq ' Iq̂ . Proof. We first note that the Kp and the Lq are all fraction fields of complete DVRs; this follows from Proposition 8.11 (note that we are not assuming they are local fields). (1) For each q|p the embedding K ,→ L induces an embedding Kp ,→ Lq via the map [(xn )] 7→ [(xn )] on equivalence classes of Cauchy sequences; a sequence (xn ) that is Cauchy in K with respect to | |p , is also Cauchy in L with respect to | |q because vq extends vp . We may thus view Kp as a topological subfield of Lq , and it is clear that [Lq : Kp ] ≤ [L : K], since any K-basis b1 , . . . , bm for L ⊆ Lq spans Lq as a Kp -vector space: given a Cauchy sequence y := (yn ) of elements in L, if we write each yn as x1,n b1 +· · ·+xm,n bm with xi,n ∈ K we obtain Cauchy sequences x1 := (x1,n ), · · · , xm := (xm,n ) of elements in K (linear maps of finite dimensional normed spaces are uniformly continuous and thus preserves Cauchy sequences), and we can write [y] = [x1 ]b1 + · · · [xm ]bm as a Kp -linear combination of b1 , . . . , bm . The field L is a finite étale K-algebra, since L/K is a separable extension, so its base change L ⊗K Kp to Kp is a finite étale Kp -algebra, by Proposition 4.36. Let us now consider the Kp -algebra homomorphism φq : L ⊗K Kp → Lq defined by ` ⊗ x 7→ `x. We have φq (bi ⊗1) = bi for each of our K-basis elements bi ∈ L, and as noted above, b1 , . . . bm span Lq as Kp -vector space, thus φq is surjective. As a finite étale Kp -algebra, L⊗K Kp is by definition isomorphic to a finite product of finite separable extensions of Kp ; by Proposition 4.32, Lq is isomorphic to a subproduct and thus also a finite étale Kp -algebra; in particular, Lq /Kp is separable. 18.785 Fall 2021, Lecture #11, Page 8 (2) As noted above, the valuation rings of Kp and the Lq are complete DVRs, so this follows immediately from Theorem 9.22. (3) The valuation vq̂ extends vq with index 1, which in turn extends vp with index eq . The valuation vp̂ extends vp with index 1, and it follows that vq̂ extends vp̂ with index eq and therefore eq̂ = eq . The residue field of p̂ is the same as that of p: for any Cauchy sequence (an ) over K the an will eventually all have the same image in the residue field at p (since vp (an − am ) > 0 for all sufficiently large m and n). Similar comments apply to each q̂ and q, and it follows that fq̂ = fq . (4) It follows from (2) that [Lq : Kp ] = eq̂ fq̂ , since q̂ is the only prime above p̂, and (3) then implies [Lq :QKp ] = eq fq , by Theorem 5.35. (5) Let φ := q|p φq , where φq : L ⊗K Kp → Lq is theQsurjective Kp -algebra homomorphisms defined in the proof of (1). Then φ : L ⊗K Kp → q|p Lq is a Kp -algebra homomorphism. Applying (4) and the fact that taking the base change of a finite étale algebra does not change its dimension (see Proposition 4.36), we have X X Y dimKp (L ⊗K Kp ) = dimK L = [L : K] = eq fq = [Lq : Kp ] = dimKp Lq . q|p q|p q|p Q Pick a Kp -basis {βi } for q|p Lq , fix  > 0, and for each basis element βi = (βi,q )q|p use the weak approximation theorem proved in Problem Set 4 to construct αi ∈ L such that Q |αi − βi,q |q <  for all q|p. In the metric space q|p Lq (with the sup norm), each φ(αi ⊗ 1) is close to βi . The Kp -matrix whose jth column expresses φ(αj ⊗ 1) in terms of the basis {βi } is then close to the identity matrix (with respect to | |p ), and the determinant D of this matrix is close to 1 (the determinant is continuous). For sufficiently small  we must Q have D 6= 0, and then {φ(αi ⊗ 1)} is a basis for q|p Lq . It follows that φ is surjective and therefore an isomorphism, since its domain and codomain have the same dimension. (6) We now assume L/K is Galois. Each σ ∈ Dq acts on L and respects the valuation vq , since it fixes q (if x ∈ qn then σ(x) ∈ σ(qn ) = σ(q)n = qn ). It follows that if (xn ) is a Cauchy sequence in L, then so is (σ(xn )), thus σ is an automorphism of Lq , and it fixes Kp . We thus have a group homomorphism ϕ : Dq → AutKp (Lq ). If σ ∈ Dq acts trivially on Lq then it acts trivially on L ⊆ Lq , so ker ϕ is trivial. Also, eq fq = |Dq | ≤ #AutKp (Lq ) ≤ [Lq : Kp ] = eq fq , by Theorem 11.23, so #AutKp (Lq ) = [Lq : Kp ] and Lq /Kp is Galois, and this also shows that ϕ is surjective and therefore an isomorphism. There is only one prime q̂ of Lq , and it is necessarily fixed by every σ ∈ Gal(Lq /Kp ), so Gal(Lq /Kp ) ' Dq̂ . The inertia groups Iq and Iq̂ both have order eq = eq̂ , and ϕ restricts to a homomorphism Iq → Iq̂ , so the inertia groups are also isomorphic. Corollary 11.24. Assume AKLB and let p be a prime of A. For every α ∈ L we have X Y NL/K (α) = NLq /Kp (α) and TL/K (α) = TLq /Kq (α). q|p q|p where we view α as an element of Lq and NL/K (α) as an element of Kp via the canonical embeddings L ,→ Lq and K ,→ Kp . Proof. The norm and trace are defined as the determinant and trace of K-linear maps ×α L −→ L that are unchanged upon tensoring with Kp ; the corollary then follows from the isomorphism in part (5) of Theorem 11.23, which commutes with the norm and trace. 18.785 Fall 2021, Lecture #11, Page 9 Remark 11.25. Theorem 11.23 can be stated more generally in terms of equivalence classes of absolute values, or places. Rather than working with a prime p of K and primes q|p of L, one works with an absolute value | |v of K (for example, | |p ) and inequivalent absolute values | |w of L that extend | |v . Places will be discussed further in the next lecture. Q eq Corollary 11.26. Assume AKLB and let p be a prime of A. Let pB = q be the factorization of pB in B. Let Âp denote the completion of A with respect to | |p , Q and for each q|p, let B̂q denote the completion of B with respect to | |q . Then B ⊗A Âp ' q|p B̂q , as Âp -algebras Proof. After replacing A with Ap and B with Bp (localizing B as an A-module),Pwe may assume that A is a DVR and B/A is a free A module of rank n := [L : K] = q|p eq fq . Then B ⊗A Âp is a free Âp -module of rank n. Viewing Âp and Q the B̂q as valuation rings of KpP and Lq , it follows P from part (4) of Theorem 11.23 that B̂q is a free Âp -module of rank q|p [Lq : Kp ] = q|p eq fq Q = n. These isomorphic Âp -modules lie in isomorphic finite étale Kp -algebras L ⊗K Kp ' Lq , by part (5) of Theorem 11.23, and this Kp -algebra isomorphism restricts to an Âp -algebra isomorphism. Remark 11.27. Let A be a Dedekind domain with fraction field K. If we localize A at a prime p we obtain a DVR Ap with the same fraction field K. We can then complete Ap with respect to | |p to obtain a complete DVR Âp whose fraction field Kp is the completion of K with respect to | |p , and Âp is then the valuation ring of Kp . Alternatively, we could first complete A with respect to the absolute value | |p induced by p and then localize. But as explained in Lecture 8, completing A with respect to | |p is the same thing as taking the valuation ring of Kp , so the completion of A is already the complete DVR Âp we obtained by localizing and completing; there is no need to localize and nothing would change if we did. Completion not only commutes with localization, it makes localization unnecessary. References [1] Michael Artin, Algebra, 2nd edition, Pearson, 2010. [2] Marc Krasner, Théorie non abélienne des corps de classes pour les extensions finies et séparables des corps valués complets: principes fondamentaux; espaces de polynomes et transformation T ; lois d’unicité, d’ordination et d’existence, C. R. Acad. Sci. Paris 222 (1946), 626–628. [3] Alexander Ostrowski, Über sogenannte perfekte Körper , J. Reine Angew. Math. 147 (1917), 191–204 18.785 Fall 2021, Lecture #11, Page 10 18.785 Number theory I Lecture #12 12 12.1 Fall 2021 10/20/2021 The different and the discriminant The different We continue in our usual AKLB setup: A is a Dedekind domain, K is its fraction field, L/K is a finite separable extension, and B is the integral closure of A in L (a Dedekind domain with fraction field L). We would like to understand the primes that ramify in L/K. Recall that a prime q|p of L is unramified if and only if eq = 1 and B/q is a separable extension of A/p, equivalently, if and only if B/qeq is a finite étale A/p algebra (by Theorem 4.40).1 A prime p of K is unramified if and only if all the primes q|p lying above it are unramified, equivalently, if and only if the ring B/pB is a finite étale A/p algebra.2 Our main tools for studying ramification are the different DB/A and discriminant DB/A . The different is a B-ideal that is divisible by precisely the ramified primes q of L, and the discriminant is an A-ideal divisible by precisely the ramified primes p of K. Moreover, the valuation vq (DB/A ) will give us information about the ramification index eq (its exact value when q is tamely ramified). Recall from Lecture 5 the trace pairing L × L → K defined by (x, y) 7→ TL/K (xy); under our assumption that L/K is separable, it is a perfect pairing. An A-lattice M in L is a finitely generated A-module that spans L as a K-vector space (see Definition 5.9). Every A-lattice M in L has a dual lattice (see Definition 5.11) M ∗ := {x ∈ L : TL/K (xm) ∈ A ∀m ∈ M }, which is an A-lattice in L isomorphic to the dual A-module M ∨ := HomA (M, A) (see Theorem 5.12). In our AKLB setting we have M ∗∗ = M , by Proposition 5.16. Every fractional ideal I of B is finitely generated as a B-module, and therefore finitely generated as an A module (since B is finite over A). If I is nonzero, it necessarily spans L, since B does. It follows that every element of the group IB of nonzero fractional ideals of B is an A-lattice in L. We now show that IB is closed under the operation of taking duals. Lemma 12.1. Assume AKLB. If I ∈ IB then I ∗ ∈ IB . Proof. The dual lattice I ∗ is a finitely generated A-module, thus to show that it is a finitely generated B-module it is enough to show it is closed under multiplication by elements of B. So consider any b ∈ B and x ∈ I ∗ . For all m ∈ I we have TL/K ((bx)m) = TL/K (x(bm)) ∈ A, since x ∈ I ∗ and bm ∈ I, so bx ∈ I ∗ as desired. Definition 12.2. Assume AKLB. The different DL/K of L/K (and the different DB/A of B/A), is the inverse of B ∗ in IB . Explicitly, we have B ∗ := {x ∈ L : TL/K (xb) ∈ A for all b ∈ B}, and we define DL/K := DB/A := (B ∗ )−1 = (B : B ∗ ) = {x ∈ L : xB ∗ ⊆ B}. Note that B ⊆ B ∗ , since TL/K (ab) ∈ A for a, b ∈ B (by Corollary 4.52), and this implies DB/A = (B ∗ )−1 ⊆ B −1 = B. Thus the different is an ideal, not just a fractional ideal. 1 Note that B/qeq is reduced if and only if eq = 1; consider the image of a uniformizer in B/qeq . As usual, by a prime of A or K we mean a nonzero prime ideal of A, and similarly for B and L. The notation q|p means that q is a prime of B lying above p (so p = q ∩ A and q divides pB). 2 The different respects localization and completion. Proposition 12.3. Assume AKLB and let S be a multiplicative subset of A. Then S −1 DB/A = DS −1 B/S −1 A . Proof. This follows from the fact that inverses and duals are both compatible with localization, by Lemmas 3.1 and 5.15. Proposition 12.4. Assume AKLB and let q|p be a prime of B. Then DB̂q /Âp = DB/A B̂q , where Âp and B̂q are the completions of A and B at p and q, respectively. Proof. Let L̂ := L ⊗ Kp be the base Q change of the finite étale K-algebra L to Kp . By (5) of Theorem 11.23, we have L̂ ' q|p Lq . Note that even though L̂ need not be a field, in general, it is a free Kp -module of Pfinite rank, and is thus equipped with a trace map that necessarily satisfies TL̂/Kp (x) = q|p TL̂/Kp (x) that defines a trace pairing on L̂. Now B ⊗ Âp ; it is an Ap -lattice in L the Kp -vector space L̂. By Corollary 11.26, Q let B̂ :=L ∗ ∗ B̂ ' q|p B̂q ' q|p B̂q , by Corollary 5.13. It follows that q|p B̂q , and therefore B̂ ' B̂ ∗ ' B ∗ ⊗A Âp . In particular, B ∗ generates each fractional ideal B̂q∗ ∈ IB̂q . Taking inverses, DB/A = (B ∗ )−1 generates the B̂q -ideal (B̂q∗ )−1 = DB̂q /Âp . 12.2 The discriminant Definition 12.5. Let S/R be a ring extension in which S is a free R-module of rank n. For any x1 , . . . , xn ∈ S we define the discriminant disc(x1 , . . . , xn ) := discS/R (x1 , . . . , xn ) := det[TS/R (xi xj )]i,j ∈ R. Note that we do not require x1 , . . . , xn to be an R-basis for S, but if they satisfy a non-trivial R-linear relation then the discriminant will be zero (by linearity of the trace). In our AKLB setup, we have in mind the case where e1 , . . . , en ∈ B is a basis for L as a K-vector space, in which case disc(e1 , . . . , en ) = det[TL/K (ei ej )]ij ∈ A. Note that we do not need to assume that B is a free A-module; L is certainly a free K-module. The fact that the discriminant lies in A when e1 , . . . , en ∈ B follows immediately from Corollary 4.52. Proposition 12.6. Let L/K be a finite separable extension of degree n, and let Ω/K be a field extension for which there are distinct σ1 , . . . , σn ∈ HomK (L, Ω). For any e1 , . . . , en ∈ L we have disc(e1 , . . . , en ) = det[σi (ej )]2ij , and for any x ∈ L we have disc(1, x, x2 , . . . , xn−1 ) = Y i0 . Define the function χ : K × → R>0 by χ(x) := λx = µx (A)/µ(A). Then µx = χ(x)µ, and for all x, y ∈ K × we have χ(xy) = µxy (A) χ(x)µy (A) µx (yA) χ(x)χ(y)µ(A) = = = = χ(x)χ(y). µ(A) µ(A) µ(A) µ(A) Thus χ is multiplicative, and we claim that in fact χ(x) = |x|v for all x ∈ K × . Since both χ and | · |v are multiplicative, it suffices to consider x ∈ A − {0}. For any such x, the ideal xA is equal to pv(x) , since A is a DVR. The residue field k := A/p is finite, hence A/xA is also finite; indeed it is a k-vector space of dimension v(x) and has cardinality [A : xA] = (#k)v(x) . Writing A as a finite disjoint union of cosets of xA, we have µ(A) = [A : xA]µ(xA) = (#k)v(x) χ(x)µ(A), and therefore χ(x) = (#k)−v(x) = |x|v as claimed. It follows that µ(xS) = µx (S) = χ(x)µ(S) = |x|v µ(S), for all x ∈ K and measurable S ⊆ K. To prove uniqueness, if | | is an absolute value on K that induces the same topology as | |v then for some 0 < c ≤ 1 we have |x| = |x|cv for all x ∈ K × . Let us fix x ∈ K × with |x|v 6= 1 (take any x with v(x) 6= 0). If | | also satisfies µ(xS) = |x|µ(S) then   µ(xA) µ(xA) c c = |x| = |x|v = , µ(A) µ(A) which implies c = 1, meaning that | | and | |v are the same absolute value. 13.3 The product formula for global fields Definition 13.17. Let K be a global field. For each place v of K the normalized absolute value k kv : Kv → R≥0 on the completion of K at v is defined by kxkv := µ(xS) , µ(S) where µ is a Haar measure on Kv and S ⊆ Kv is a measurable set with finite nonzero measure (such as the set {x ∈ Kv : |x|v ≤ 1}, for example). This definition is independent of the choice of µ and S (by Theorem 13.14). If v is nonarchimedean then the normalized absolute value k kv is precisely the absolute value | |v defined in Proposition 13.16. If v is a real place then the normalized absolute value k kv is just the usual Euclidean absolute value | |R on R, since for the Euclidean Haar measure µR on R we have µR (xS) = |x|R µR (S) for every measurable set S. But when v is a complex place the normalized absolute value k kv is the square of the Euclidean absolute value | |C on C, since in C we have µC (xS) = |x|2C µC (S). Remark 13.18. When v is a complex place the normalized absolute value k kv is not an absolute value, because it does not satisfy the triangle inequality. For example, if K = Q(i) and v|∞ is the complex place of K then k1kv = |1|2C = 1 but k1 + 1kv = k2kv = |2|2C = 4 > 2 = k1kv + k1kv . Nevertheless, the normalized absolute value k kv is always multiplicative and compatible with the topology on Kv in the sense that the open balls B 1. If v is archimedean then Lw ' C and Kv ' R, in which case for any x ∈ Lw we have kxkw = µ(xS)/µ(S) = |x|2C = |xx|R = |NC/R (x)|R = kNLw /Kv (x)kv , where | |R and | |C are the Euclidean absolute values on R and C. We now assume v is nonarchimedean. Let πv and πw be uniformizers for the local fields Kv and Lw , respectively, and let f be the degree of the corresponding residue field extension w(x) kw /kv . Without loss of generality, we may assume x = πw , since kxkv = |x|v depends only on w(x). Theorem 6.10 and Proposition 13.16 imply kNLw /Kv (πw )kv = kπvf kv = (#kv )−f , so kNLw /Kv (x)kv = (#kv )−f w(x) . Proposition 13.16 then implies kxkw = (#kw )−w(x) = (#kv )−f w(x) = kNLw /Kv (x)kv . Remark 13.20. Note that if v is a nonarchimedean place of K extended by a place w|v of L/K, the absolute value k kw is not the unique absolute value on Lw that extends the absolute value on k kv on Kv given by Theorem 10.4, it differs by a power of n = [Lw : Kv ], but it is equivalent to it. It might seem strange to use a normalization here that does not agree with the one we used when considering extensions of local fields in Lecture 9. The difference is that here we are thinking about a single global field K that has many different completions Kv , and we want the normalized absolute values on the various Kv to be compatible (so that the product formula will hold). By contrast, in Lecture 9 we considered various extensions Lw of a single local field Kv and wanted to normalize the absolute values on the Lw compatibly so that we could work in Kv and any of its extensions (all the way up to K v ) using the same absolute value. These two objectives cannot be met simultaneously and it is better to use the “right" normalization in each setting. Theorem 13.21 (Product Formula). Let L be a global field. For all x ∈ L× we have Y kxkv = 1, v∈ML where k kv denotes the normalized absolute value for each place v ∈ ML . Proof. The global field L is a finite separable extension of K = Q or K = Fq (t).4 Let p be a place of K. By Theorem 13.5, any basis for L as a K-vector space is also a basis for Y L ⊗K Kp ' Lv v|p 4 Here we are using the fact that if Fq is the field of constants of L (the largest finite field in L), then L is a finite extension of Fq (z) and we can choose some t ∈ Fq (z) − Fq so that Fq (z) ' Fq (t) and L/Fq (t) is separable (such a t is called a separating element). 18.785 Fall 2021, Lecture #13, Page 7 as a Kv -vector space. Thus NL/K (x) = N(L⊗K Kp )/Kp (x) = Y NLv /Kp (x). v|p Taking normalized absolute values on both sides yields Y Y kxkv . kNLv /Kp (x)kp = NL/K (x) p = v|p v|p We now take the product of both sides over all places p ∈ MK to obtain Y Y Y Y kNL/K (x)kp = kxkv = kxkv . p∈MK p∈MK v|p v∈ML The LHS is equal to 1, by the product formula for K proved on Problem Set 1. With the product formula in hand, we can now give an axiomatic definition of a global field, which up to now we have simply defined as a finite extension of Q or Fq (t), due to Emil Artin and George Whaples [1]. Definition 13.22. A global field is a field K with at least one place whose completion at each of its places v ∈ MK is a local field Kv , and which has a product formula of the form Y kxkv = 1, v∈MK v for some where each normalized absolute value k kv : Kv → R≥0 satisfies k kv = | |m v absolute value | |v representing v and some fixed mv ∈ R>0 . Theorem 13.23 (Artin-Whaples). Every global field is a finite extension of Q or Fq (t). Proof. See Problem Set 7. References [1] Emil Artin and George Whaples, Axiomatic characterization of fields by the product formula for valuations, Bull. Amer. Math. Soc. 51 (1945), 469–492. [2] Alexander von Brill, Ueber die Discriminante, Math. Ann. 12 (1877), 87–89. [3] Joe Diestel and Angela Spalsbury, The Joys of Haar Measure, American Mathematical Society, 2014. 18.785 Fall 2021, Lecture #13, Page 8 18.785 Number theory I Lecture #14 14 14.1 Fall 2021 10/27/2021 The geometry of numbers Lattices in real vector spaces Recall that for an integral domain A with fraction field K, an A-lattice in a finite dimensional K-vector space V is a finitely generated A-submodule of V that contains a K-basis for V (see Definition 5.9). We now want to specialize to the case A = Z, but rather than working with the fraction field K = Q we will instead work with its completion R at the unique infinite place of Q. Remark 14.1. In this lecture we focus on number fields, but we will make remarks along the way about how to similarly treat global function fields (where one would take A = Fq [t] and work with its completion Fq (t)∞ ' Fq (( 1t )) at the unique infinite place of Fq (t)). You will be able to explore the function field case in more detail on Problem Set 7. Let V be an R-vector space of dimension n. Then V ' Rn is a locally compact group with a Haar measure that is unique up to scaling, by Theorem 13.14. Definition 14.2. A subgroup H of a topological group G is discrete if the subspace topology on H is the discrete topology (every point is open), and cocompact if H is a normal subgroup of G and the quotient G/H is compact (here G/H denotes the group G/H with the quotient topology given by identifying elements of G that lie in the same coset of H). Lemma 14.3. A subgroup G ⊆ V ' Rn is finitely generated if and only if it is discrete, in which case G ' Zm for some m ≤ n and G is then cocompact if and only if m = n. Proof. It follows from the structure theorem for finitely generated abelian group that if G is finitely generated it is a free Z-module, since it lies in the torsion free abelian group V , and the rank of G must be equal to the dimension m ≤ n of the subspace it spans. A subgroup G ' Zm of V ' Rn is discrete (0 is an isolated point), and cocompact if and only if m = n. Let v1 , . . . , vm be a basis for the subspace W of V spanned by G, and let Λ ' Zm be the subgroup of G generated by this basis. The quotient W/Λ ' Rm /Zm is compact (it is isomorphic to the closed unit cube in Rm ), so the image of the discrete group G under the quotient map π : W → W/Λ must be finite. We can thus choose a finite S ⊆ G such that π(S) = π(G), and then G is generated by v1 , . . . , vm and S. Definition 14.4. A (full) lattice in V ' Rn is a Z-submodule generated by an R-basis, equivalently, a discrete cocompact subgroup. Remark 14.5. A discrete subgroup of a Hausdorff topological group is always closed; see [1, III.2.1.5] for a proof. This implies that the quotient of a Hausdorff topological group by a normal discrete subgroup is Hausdorff (which is false for topological spaces in general); see [1, III.2.1.18]. It follows that the quotient of a Hausdorff topological group (including all locally compact groups) by a discrete cocompact subgroup is a compact group. These facts are easy to see in the case of lattices: Z is closed in R (as the complement of a union of open intervals), so Zn is closed in Rn . Given a lattice Λ in V , each Z-basis for Λ determines an isomorphism of topological groups Λ ' Zn and V ' Rn , and the quotient V /Λ ' Rn /Zn ' (R/Z)n (an n-torus), is compact Hausdorff and thus a compact group. Remark 14.6. You might ask why we are using the archimedean completion R = Q∞ rather than some other completion Qp . The reason is Z is not a discrete subgroup of Qp for any finite place p (elements of Z can be arbitrarily close to 0 under the p-adic metric). Similarly, Fq [t] is a discrete subgroup of Fq (t)∞ , but not of any other completion of Fq (t). Any basis v1 , . . . , vn for V determines a parallelepiped F (v1 , . . . , vn ) := {t1 v1 + · · · + tn vn : t1 , . . . , tn ∈ [0, 1)} ∼ that we may view as the unit cube by fixing an isomorphism ϕ : V −→ Rn that maps (v1 , . . . , vn ) to the standard basis of unit vectors for Rn . It then makes sense to normalize the Haar measure µ so that µ(F (v1 , . . . , vn )) = 1, and we then have µ(S) = µRn (ϕ(S)) for every measurable set S ⊆ V , where µRn denotes the standard Lebesgue measure on Rn . For any other P basis e1 , . . . , en of V , if we let E = [eij ]ij be the matrix whose jth column expresses ej = i eij vi , in terms of our normalized basis v1 , . . . , vn , then q √ p t t µ(F (e1 , . . . , en )) = | det E| = det E det E = det(E E) = det[hei , ej i]ij , (1) where hei , ej i is the canonical inner product (the dot product) on Rn . Here we have used the fact that the determinant of a matrix in Rn×n is the signed volume of the parallelepiped spanned by its columns (or rows). This is a consequence of the following more general result, which is independent of the choice of basis or the normalization of µ. Proposition 14.7. Let T : V → V be a linear transformation of V ' Rn . For any Haar measure µ on V and every measurable set S ⊆ V we have (2) µ(T (S)) = | det T | µ(S). Proof. See [11, Ex. 1.2.21]. If Λ is a lattice e1 Z + · · · + en Z in V , the quotient V /Λ is a compact group that we may identify with the parallelepiped F (e1 , . . . , en ) ⊆ V , which forms a set of unique coset representatives. More generally, we make the following definition. Definition 14.8. Let Λ be a lattice in V ' Rn . A fundamental domain for Λ is a measurable set F ⊆ V such that G V = (F + λ). λ∈Λ In other words, F is a measurable set of coset representatives for V /Λ. Fundamental domains exist: if Λ = e1 Z + · · · + en Z we may take the parallelepiped F (e1 , . . . , en ). Proposition 14.9. Let Λ be a lattice in V ' Rn and let µ be a Haar measure on V . Every fundamental domain for Λ has the same measure, and this measure is finite and nonzero. Proof. Let F and G be two fundamental domains for Λ. Using the translation invariance and countable additivity of µ (note that Λ ' Zn is a countable set) along with the fact that Λ is closed under negation, we obtain ! ! G G
µ(F ) = µ(F ∩ V ) = µ F ∩ (G + λ) = µ F ∩ (G + λ) = X λ∈Λ
λ∈Λ µ F ∩ (G + λ) = X λ∈Λ λ∈Λ
X
µ (F − λ) ∩ G = µ G ∩ (F + λ) = µ(G), λ∈Λ where the last equality follows from the first four (swap F and G). If we fix a Z-basis e1 , . . . , en for Λ, the parallelepiped F (e1 , . . . , en ) is a fundamental domain for Λ, and its 18.785 Fall 2021, Lecture #14, Page 2 closure is compact, so µ(F (e1 , . . . , en )) is finite, and it is nonzero because there is an isomorphism V ' Rn that maps the closure of F (e1 , . . . , en ) to the unit cube in Rn whose Lebesgue measure is nonzero (whether a set has zero measure or not does not depend on the normalization of the Haar measure and is therefore preserved by isomorphisms of locally compact groups). Definition 14.10. Let Λ be a lattice in V ' Rn and fix a Haar measure µ on V . The covolume covol(Λ) ∈ R>0 of Λ is the measure µ(F ) of any fundamental domain F for Λ. Note that covolumes depend on the normalization of µ, but ratios of covolumes do not. Proposition 14.11. If Λ0 ⊆ Λ are lattices in V ' Rn , then covol(Λ0 ) = [Λ : Λ0 ] covol(Λ). Proof. Fix a fundamental domain F for Λ and a set of coset representatives S for Λ/Λ0 . Then G F 0 := (F + λ) λ∈S is a fundamental domain for Λ0 , and #S = [Λ : Λ0 ] = µ(F 0 )/µ(F ) is finite. We then have X covol(Λ0 ) = µ(F 0 ) = µ(F + λ) = (#S)µ(F ) = [Λ : Λ0 ] covol(Λ), λ∈S since every translation F + λ of F is a fundamental domain for Λ. Definition 14.12. Let S be a subset of a real vector space. The set S is symmetric if it is closed under negation, and convex if for all x, y ∈ S we have {tx + (1 − t)y : t ∈ [0, 1]} ⊆ S. Theorem 14.13 (Minkowski’s Lattice Point Theorem). Let Λ be a lattice in V ' Rn and µ a Haar measure on V . If S ⊆ V is a symmetric convex measurable set that satisfies µ(S) > 2n covol(Λ), then S contains a nonzero element of Λ. Proof. See Problem Set 6. Note that the inequality in Theorem 14.13 bounds the ratio of the measures of two sets (S and a fundamental domain for Λ), and is thus independent of the choice of µ. Remark 14.14. In the function field analog of Theorem 14.13 the convexity assumption is not needed and the factor of 2n can be removed. 14.2 The canonical inner product Let K/Q be a number field of degree n with r real places and s complex places; then n = r + 2s, by Corollary 13.9. We now want to consider the base change of K to R and C: KR := K ⊗Q R ' Rr × Cs , KC := K ⊗Q C ' Cn . The isomorphism KR ' Rr × Cs follows from Theorem 13.5 and the isomorphism KC ' Cn follows from the fact that C is separably closed; see Example 4.31. We note that KR is an 18.785 Fall 2021, Lecture #14, Page 3 R-vector space of dimension n, thus KR ' Rn , but this is an isomorphism of R-vector spaces and is not an R-algebra isomorphism unless s = 0. We have a sequence of injective homomorphisms of topological rings (3) OK ,→ K ,→ KR ,→ KC , which are defined as follows: • the map OK ,→ K is inclusion; • the map K ,→ KR = K ⊗Q R is the canonical embedding α 7→ α ⊗ 1; • the map KR ' Rr × Cs ,→ Cr × C2s ' KC embeds each factor of Rr in a corresponding factor of Cr via inclusion and each C in Cs is mapped to C × C in C2s via z 7→ (z, z̄). To better understand the last map, note that each C in Cs arises as R[α] = R[x]/(f ) ' C for some monic irreducible f ∈ R[x] of degree 2, but when we base-change to C the field R[α] splits into the étale algebra C[x]/(x − α) × C[x]/(x − ᾱ) ' C × C. The composition K ,→ KR ,→ KC is given by the map x 7→ (σ1 (x), . . . , σn (x)), where HomQ (K, C) = {σ1 , . . . , σn }. If we put K = Q(α) := K[x]/(f ) and let α1 , . . . , αn ∈ C be the roots of f in C, each σi is the Q-algebra homomorphism K → C defined by α 7→ αi . If we fix a Z-basis for OK , its image under the maps in (3) is a Q-basis for K, an R-basis for KR , and a C-basis for KC , all of which are vector spaces of dimension n = [K : Q]. We may thus view the injections in (3) as inclusions of topological groups (but not rings!) Zn ,→ Qn ,→ Rn ,→ Cn . The ring of integers OK is a lattice in the real vector space KR ' Rn , which inherits an inner product from the canonical Hermitian inner product on KC ' Cn defined by 0 hz, z i := n X i=1 zi z̄i0 ∈ C. For elements x, y ∈ K ,→ KR ,→ KC the Hermitian inner product can be computed as X hx, yi := σ(x)σ(y) ∈ R, (4) σ∈HomQ (K,C) which is a real number because the non-real embeddings in HomQ (K, C) come in complex conjugate pairs. The inner product defined in (4) agrees with the restriction of the Hermitian inner product on KR ,→ KC . The metric space topology it induces on KR is the same as the Euclidean topology on KR ' Rn induced by the usual dot product on Rn , but the corresponding norm kxk := hx, xi has a different normalization, as we now explain. If we write elements z ∈ KC ' Cn as vectors (zσ ) indexed by the set σ ∈ HomQ (K, C) in some fixed order, we may identify KR with its image in KC as the set KR = {z ∈ KC : z̄σ = zσ̄ for all σ ∈ HomQ (K, C)}. 18.785 Fall 2021, Lecture #14, Page 4 For real embeddings σ = σ̄ we have zσ ∈ R ⊆ C, and for pairs of conjugate complex embeddings (σ, σ̄) we get the embedding z 7→ (zσ , zσ̄ ) = (zσ , z̄σ ) of C into C × C used to defined the map KR ,→ KC above. Each z ∈ KR can be uniquely written in the form (w1 , . . . , wr , x1 + iy1 , x1 − iy1 , . . . , xs + iys , xs − iys ), (5) with wi , xj , yj ∈ R. Each wi corresponds to a zσ with σ = σ̄, and each (xj + iyj , xj − iyj ) corresponds to a complex conjugate pair (zσ , zσ̄ ) with σ 6= σ̄. The canonical inner product on KR can then be written as 0 hz, z i = r X wi wi0 +2 i=1 s X (xj x0j + yj yj0 ). j=1 Thus if we take w1 , . . . , wr , x1 , y1 , . . . , xx , ys as coordinates for KR ' Rn (as R-vector spaces), in order to normalize the Haar measure µ on KR so that it is consistent with the Lebesgue measure µRn on Rn we define µ(S) := 2s µRn (S) (6) for any measurable set S ⊆ KR that we may view as a subset of Rn by expressing it in wi , xj , yj coordinates as above. With this normalization, the identity (1) still holds when we replace µRn with µ and the dot product on Rn with the Hermitian inner product on KR , that is, for any R-basis e1 , . . . , en of KR we still have q (7) µ(F (e1 , . . . , en )) = | det[hei , ej i]ij | Using the Hermitian inner product on KR ⊆ KC rather than the dot product on KR ' Rn multiplies 2s of the columns in the matrix [hei , ej i]ij by 2, and thus multiplies the RHS by √ 22s = 2s ; our normalization of µ = 2s µRn multiplies the LHS by 2s so that (7) still holds. Remark 14.15. In the function field case one replaces the separable closure C of R with a separable closure Fq (t)sep ∞ of Fq (t)∞ . The situation is slightly more complicated, since unlike C/R, the extension Fq (t)sep ∞ /Fq (t)∞ is not finite, but for any finite separable extension K/Fq (t) (a finite étale Fq (t)-algebra) one can base change K to Fq (t)∞ and Fq (t)sep ∞ ; these play the role of KR and KC . 14.3 Covolumes of fractional ideals Having fixed a normalized Haar measure µ for KR , we can now compute covolumes of lattices in KR ' Rn . This includes not only (the image of) the ring of integers OK , but also any nonzero fractional ideal I of OK : every such I contains a nonzero principal fraction ideal aOK , and if e1 , . . . , en is a Z-basis for OK then ae1 , . . . , aen is a Z-basis for aOK that is an R-basis for KR that lies in I. Recall from Remark 12.14 that the discriminant of a number field K is the integer DK := disc OK := disc(e1 , . . . , en ) ∈ Z. Proposition 14.16. Let K be a number field. Using the normalized Haar measure on KR defined in (6), p covol(OK ) = |DK |. 18.785 Fall 2021, Lecture #14, Page 5 Proof. Let e1 , . . . , en ∈ OK be a Z-basis for OK , let HomQ (K, C) = {σ1 , . . . , σn }, and define A := [σi (ej )]ij ∈ Cn×n . Then DK = disc(e1 , . . . , en ) = (det A)2 , by Proposition 12.6 Viewing OK ,→ KR as a lattice p in KR with basis e1 , . . . , en , we may use (7) to compute covol(OK ) = µ(F (e1 , . . . , en )) = | det[hei , ej i]ij |. Applying (4) yields hP i det[hei , ej i]ij = det = det(At A) = (det A)(det A). k σk (ei )σk (ej ) ij Noting that det A is the square root of an integer (hence either real or purely imaginary), we have covol(OK )2 = |(det A)2 | = |DK |, and the proposition follows. Recall from Remark 6.13 that for number fields K we view the absolute norm N : IOK → IZ I 7→ [OK : I]Z as having image in Q>0 by identifying N(I) ∈ IZ with a positive generator for N(I) (note that Z is a PID). Recall that [OK : I]Z is a module index of Z-lattices in the Q-vector space K (see Definitions 6.1 and 6.5), and for ideals I ⊆ OK this is just the positive integer [OK : I]Z = [OK : I]. When I = (a) is a principal fractional ideal with a ∈ K, we may simply write N(a) := N((a)) = |NK/Q (a)|. Corollary 14.17. Let K be a number field and let I be a nonzero fractional ideal of OK . Then p covol(I) = N(I) |DK | Proof. Let n = [K : Q]. Since covol(bI) = bn covol(I) and N(bI) = bn N(I) for any b ∈ Z>0 , without loss of generality we may assume I ⊆ OK (replace I with a suitable bI if not). Applying Propositions 14.11 and 14.16, we have p covol(I) = [OK : I] covol(OK ) = N(I) covol(OK ) = N(I) |DK | as claimed. 14.4 The Minkowski bound Theorem 14.18 (Minkowski bound). Let K be a number field of degree n with s complex places. Define the Minkowski constant mK for K as the positive real number   n! 4 s p |DK |. mK := n n π For every nonzero fractional ideal I of OK there is a nonzero a ∈ I for which N(a) ≤ mK N(I). To prove this theorem we need the following lemma. Lemma 14.19. Let K be a number field of degree n with r real and s complex places. For each t ∈ R>0 , the measure of the convex symmetric set n o X St := (zσ ) ∈ KR : |zσ | ≤ t ⊆ KR with respect to the normalized Haar measure µ on KR is µ(St ) = 2r π s tn . n! 18.785 Fall 2021, Lecture #14, Page 6 Proof. As in (5), we may uniquely write each z = (zσ ) ∈ KR in the form (w1 , . . . , wr , x1 + iy1 , x1 − iy1 . . . , xs + iys , xs − iys ) P with wi , xj , yj ∈ R. We will have σ |zσ | ≤ t if and only if r X i=1 s q X |wi | + 2 |xj |2 + |yj |2 ≤ t. (8) j=1 We now compute the volume of this region in Rn by relating it to the volume of the simplex n o Ut := (u1 , . . . , un ) ∈ Rn≥0 : u1 + · · · + un ≤ t ⊆ Rn , which is µRn (Ut ) = tn /n! (volume of the standard simplex in Rn scaled by a factor of t). If we view all the wi , xj , yj as fixed except the last pair (xs , ys ), then (xs , ys ) ranges over a disk of some radius d ∈ [0, t/2] determined by (8). If we replace (xs , ys ) with (un−1 , un ) ranging over the triangular region bounded by un−1 + un ≤ 2d and un−1 , un ≥ 0, we need to incorporate a factor of π/2 to account for the difference between (2d)2 /2 = 2d2 and πd2 ; repeat this s times. Similarly, if we hold everything but wr fixed and replace wr ranging over [−d, d] for some d ∈ [0, t] with ur ranging over [0, d], we need to incorporate a factor of 2 to account for this change of variable; repeat r times. We then have µ(St ) = 2s µRn (St ) = 2s  π s 2 2r µRn (U ) = 2r π s tn . n! Proof of Theorem 14.18. Let I be a nonzero fractional ideal of OK . By Theorem 14.13, if we choose t so that µ(St ) > 2n covol(I), then St will contain a nonzero a ∈ I. By Lemma 14.19 and Corollary 14.17, it suffices to choose t so that  n   t n!µ(St ) n!2n n! 4 s p = n r s > n r s covol(I) = n |DK |N(I) = mK N(I). n n 2 π n 2 π n π
n P Let us now pick t so that nt > mK N(I). Then St contains a ∈ I with σ |σ(a)| ≤ t Recalling that the geometric mean is bounded above by the arithmetic mean, we then have !n !n    n Y X 1 t n N(a) = N(a)1/n = |σ(a)|1/n ≤ |σ(a)| ≤ , n σ n σ Taking the limit as 14.5
t n n → mK N(I) from above yields N(a) ≤ mK N(I). Finiteness of the class group Recall that the ideal class group cl OK is the quotient of the ideal group IK of OK by its subgroup of principal fractional ideals. We now use the Minkowski bound to prove that every ideal class [I] ∈ cl OK can be represented by an ideal I ⊆ OK of small norm. It will then follow that the ideal class group is finite. Theorem 14.20. Let K be a number field. Every ideal class in cl OK contains an ideal I ⊆ OK of absolute norm N(I) ≤ mK , where mK is the Minkowski constant for K. 18.785 Fall 2021, Lecture #14, Page 7 Proof. Let [J] be an ideal class of OK represented by the nonzero fractional ideal J. By Theorem 14.18, the fractional ideal J −1 contains a nonzero element a for which N(a) ≤ mK N(J −1 ) = mK N(J)−1 , and therefore N(aJ) = N(a)N(J) ≤ mK . We have a ∈ J −1 , thus aJ ⊆ J −1 J = OK , so I = aJ is an OK -ideal in the ideal class [J] with N(I) ≤ mK as desired. Lemma 14.21. Let K be a number field of degree n and let M ∈ R>0 . The number of OK -ideals of norm N(I) ≤ M is bounded by (nM )log2 M (and in particular, finite). Proof. Let I be an ideal of absolute norm N(I) ≤ M and let I = p1 · · · pk be its factorization into (not necessarily distinct) prime ideals. We have N (pi ) ≥ 2 for each pi so k ≤ log2 M . There are less than M primes p ≤ M and at most n primes p of OK above each p ≤ M . It follows that the there are less than (nM )log2 M OK -ideals with norm N (I) ≤ M . Corollary 14.22. Let K be a number field. The ideal class group of OK is finite. Proof. By Theorem 14.20, each ideal class is represented by an ideal of norm at most mK , and by Lemma 14.21, the number of such ideals is finite. More generally, we have the following result, which can be applied to the analog of the ring of integers in any global field. Theorem 14.23. Let A be the ring Z or Fq [t] and let B be the integral closure of A in a finite separable extension of its fraction field. The ideal class group of B is finite. Proof. See Problem Set 7. Remark 14.24. The geometry of numbers is not a necessary ingredient to Corollary 14.22, there are purely algebraic proofs that apply to any global field; see [9] for an example. √ Remark 14.25. For imaginary quadratic fields K = Q( −d) it is known that the class number hK := # cl OK tends to infinity as d → ∞ ranges over square-free integers. This was conjectured by Gauss in his Disquisitiones Arithmeticae [3] and proved by Heilbronn [5] in 1934; the first fully explicit lower bound was obtained by Oesterlé in 1988 [7]. This implies that there are only a finite number of imaginary quadratic fields with any particular class number. It was conjectured by Gauss that there are exactly 9 imaginary quadratic fields with class number one, but this was not proved until the 20th century by Stark [10] and Heegner [4].1 Complete lists of imaginary quadratic fields for each class number hK ≤ 100 are now available [12]. By contrast, Gauss predicted that infinitely many real quadratic fields should have class number 1, however this question remains completely open.2 Corollary 14.26. Let K be a number field of degree n with s complex places. Then |DK | ≥ 1  nn n! 2    2 n π 2s 1 πe > 2 . 4 e n 4 Heegner’s 1952 result [4] was essentially correct but contained some gaps that prevented it from being generally accepted until 1967 when Stark gave a complete proof in [10]. 2 In fact it is conjectured that hK = 1 for approximately 75.446% of real quadratic fields with prime discriminant; this follows from the Cohen-Lenstra heuristics [2]. 18.785 Fall 2021, Lecture #14, Page 8 Proof. If I is an ideal and a ∈ I is nonzero, then N(a) ≥ N(I), so Theorem 14.20 implies   n! 4 s p mK = n |DK | ≥ 1, n π the first inequality follows. The second uses an explicit form of Stirling’s approximation, √  n n , n! ≤ e n e and the fact that 2s ≤ n. We note that πe2 /4 ≈ 5.8 > 1, so the minimum value of |DK | increases exponentially with n = [K : Q]. The lower bounds for n ∈ [2, 7] given by the corollary are listed below, along with the least value of |DK | that actually occurs. As can be seen in the table, |DK | appears to grow much faster than the corollary suggests. Better lower bounds can be proved using more advanced techniques, but a significant gap still remains. lower bound from Corollary 14.26 minimum value of |DK | n=2 3 3 n=3 13 23 n=4 44 275 n=5 259 4511 n=6 986 92799 n=7 6267 2306599 Corollary 14.27. If K is a number field other than Q then |DK | > 1; equivalently, there are no nontrivial unramified extensions of Q. Theorem 14.28. For every real M the set of number fields K with |DK | < M is finite. Proof. It follows from Corollary 14.26 that it suffices to prove this for fixed n := [K : Q], since for all sufficiently large n we will have |DK | > M for all number fields K of degree n. Case 1: Let K be a totally real field (so every place v|∞ is real) with |DK | < M . Then r = n and s = 0, so KR ' Rr × Cs = Rn . Consider the convex symmetric set √ S := {(x1 , . . . , xn ) ∈ KR ' Rn : |x1 | ≤ M and |xi | < 1 for i > 1} with measure p √ √ µ(S) = 2 M 2n−1 = 2n M > 2n |DK | = 2n covol(OK ). By Theorem 14.13, the set S contains a nonzero a ∈ OK ⊆ K ,→ KR that we may write as a = (a1 , . . . , an ) = (σ1 (a), . . . , σn (a)), where the σi are the n embeddings of K into C, all of which are real embeddings. We have N(a) = Y i σi (a) ≥ 1, since N(a) must be a positive integer, and |a2 |, . . . , |an | < 1, so |a1 | > 1 > |ai | for all i 6= 1. We claim that K = Q(a). If not, each ai = σi (a) would be repeated [K : Q(a)] > 1 times in the vector (a1 , . . . , an ), since there must be [K : Q(a)] elements of HomQ (K, C) that fix Q(a), namely, those lying in the kernel of the map HomQ (K, C) → HomQ (Q(a), C) induced by restriction. But this is impossible since ai 6= a1 for i 6= 1. The minimal polynomial f ∈ Z[x] of a is a monic irreducible polynomial of degree n.√The roots of f (x) in C are precisely the ai = σi (a) ∈ R, all of which are bounded by |ai | ≤ M . 18.785 Fall 2021, Lecture #14, Page 9 Each coefficient fi of f (x) is an elementary symmetric functions of its roots, hence also bounded in absolute value (certainly |fi | ≤ 2n M n/2 for all i). The fi are integers, so there are only finitely many possibilities for f (x), hence only finitely many totally real number fields K of degree n. Case 2: K has r real and s > 0 complex places, and KR ' Rr × Cs . Now let √ S := {(w1 , . . . , wr , z1 , . . . , zs ) ∈ KR : |z1 |2 < c M and |wi |, |zj | < 1 (j > 1)} with c chosen so that µ(S) > 2n covol(OK ) (the exact value of c depends on s and n). The argument now proceeds as in case 1: we get a nonzero a ∈ OK ∩ S for which K = Q(a), and only a finite number of possible minimal polynomials f ∈ Z[x] for a. Lemma 14.29. Let K be a number field of degree n. For each prime number p we have vp (DK ) ≤ nblogp nc + n − 1. In particular, vp (DK ) ≤ nblog2 nc + n − 1 for all p. Proof. We have vp (DK ) = vp (NK/Q (DK/Q )) = X fq vq (DK/Q ) q|p where DK/Q is the different ideal and fq is the residue degree of q|p. Using Theorem 12.27 to bound vq (DK/Q ) yields X X X vp (DK ) ≤ fq (eq − 1 + vq (eq )) = n − fq + fq eq vp (eq ) ≤ n − 1 + nblogp nc, q|p q|p q|p P where we have used −1 as an upper bound onP− q|p fq and blogp nc as an upper bound on each vp (eq ) (since eq ≤ n), and the fact that q|p eq fq = n (by Theorem 5.35). e Remark 14.30. The bound in Lemma 14.29 is tight; it is achieved by K = Q[x]/(xp − p), for example. Theorem 14.31 (Hermite). Let S be a finite set of places of Q, and let n be an integer. The number of extensions K/Q of degree n unramified outside of S is finite. Proof. By Lemma 14.29, since n is fixed, the valuation vp (DK ) is bounded for each p ∈ S and must be zero for p 6∈ S. Thus |DK | is bounded, and the theorem then follows from Proposition 14.28. Remark 14.32. In the function field analogs of Theorem 14.28 and Theorem 14.31 one requires K to be a separable extension of Fq (t) with constant field Fq (so K ∩ Fq = Fq ). This is not really a restriction in the sense that every global function field K contains a subfield Fq (t) for which this is true, but one needs to take q = #(K ∩ Fq ) and to choose t to be a separating element (such a t exists by [6, Thm. 7.20]). Unlike the number field setting where the embedding of the rational numbers Q in a number field K is unique, there are many ways to embed the rational function field Fq (t) in a global function field K. The notion of an absolute discriminant DK doesn’t really make sense in this setting, one can speak of the discriminant DK/Fq (t) only after fixing a suitable choice of Fq (t). As you showed on Problem Set 6, the valuation of the discriminant of an extension of global function fields is not bounded as a function of the degree, in general, and this means that the function field analog of Lemma 14.29 only holds when we use the discriminant of a separable extension. 18.785 Fall 2021, Lecture #14, Page 10 References [1] Nicolas Bourbaki, General Topology: Chapters 1-4 , Springer, 1995. [2] Henri Cohen and Hendrik W. Lenstra Jr., Heuristics on class groups of number fields, in Number Theory (Noordwijkerhout 1983), Lecture Notes in Mathematics 1068, Springer, 1984, 33–62. [3] Carl F. Gauss, Disquisitiones Arithmeticae, Göttingen (1801), English translation by Arthur A. Clark, revised by William C. Waterhouse, Spring-Verlag 1986 reprint of Yale University Press 1966 edition. [4] Kurt Heegner, Diophantische Analysis und Modulfunktionen, Math. Z. 56 (1952), 227– 253. [5] Hans Heilbronn, On the class number in imaginary quadratic fields, Quart. J. of Math. Oxford 5 (1934), 150–160. [6] Anthony W. Knapp, Advanced Algebra, Digital Second Edition, 2016. [7] Joseph Oesterlé, La probléme de Gauss sur le nombre de classes, Enseign. Math. 34 (1988), 43–67. [8] Michael Rosen, A geometric proof of Hermite’s theorem in function fields, J. Théor. Nombres Bordeaux 29 (2017), 799–813. [9] Alexander Stasinski, A uniform proof of the finiteness of the class group of a global field , American Mathematical Monthly 1228 (2021), 239–249. [10] Harold Stark, A complete determination of the complex quadratic fields of class–number one, Mich. Math. J. 14 (1967), 1–27. [11] Terence Tao, An introduction to measure theory, Graduate Studies in Mathematics 126, AMS, 2010. [12] Mark Watkins, Class numbers of imaginary quadratic fields, Math. Comp. 73 (2004), 907–938. 18.785 Fall 2021, Lecture #14, Page 11 18.785 Number theory I Lecture #15 15 Fall 2021 11/1/2021 Dirichlet’s unit theorem Let K be a number field. The two main theorems of classical algebraic number theory are: • The class group cl OK is finite. × • The unit group OK is finitely generated. We proved the first result in the previous lecture; in this lecture we will prove the second, which is due to Dirichlet. Dirichlet (1805–1859) died five years before Minkowski (1864– 1909) was born, so he did not have Minkowski’s lattice point theorem (Theorem 14.13) to work with. But we do, and this simplifies the proof considerably. 15.1 The group of Arakelov divisors of a global field Let K be a global field. As in previous lectures, we use MK to denote the set of places (equivalence classes of absolute values) of K. For each place v ∈ MK we use Kv to denote the completion of K with respect to v (a local field), and we have a normalized absolute value k kv : Kv → R≥0 defined by kxkv := µ(xS) , µ(S) where µ is a Haar measure on Kv and S is any measurable set of positive finite measure. This definition does not depend on the particular choice of µ or S; it is determined by the topology of Kv , which is an invariant of the place v (see Definition 13.17). When Kv is nonarchimedean its topology is induced by a discrete valuation that we also denote v, and we use kv to denote the residue field (the quotient of the valuation ring by its maximal ideal), which is a finite field (see Proposition 9.6). In Lecture 13 we showed that  −v(x) if v is nonarchimedean,  |x|v = (#kv ) kxkv = |x|R if Kv ' R,   2 |x|C if Kv ' C. While k kv is not always an absolute value (when Kv ' C it does not satisfy the triangle inequality), it is always multiplicative and defines a continuous homomorphism Kv× → R× >0 of locally compact groups that is surjective precisely when v is archimedean. Definition 15.1. Let K be a global field. A (multiplicative) Arakelov divisor is a sequence of positive real numbers c = (cv ) indexed by v ∈ MK with all but finitely many cv = 1 and cv ∈ kKv× k := {kxkv : x ∈ Kv× }.1 The set of Arakelov divisors Div K forms an abelian group under pointwise multiplication (cv )(dv ) := (cv dv ). The multiplicative group K × is canonically embedded in Div K via the map x 7→ (kxkv ), where it forms the subgroup Princ K of principal Arakelov divisors. Remark 15.2. Many authors define Div K as an additive group by taking logarithms (for nonarchimedean places v, one replaces cv = (#kv )−v(c) with the integer v(c)), as in [5] for example. The multiplicative convention we use here is due to Weil [6] and is better suited × 2 for our application to the multiplicative group OK . 1 When v is archimedean we have kKv× k = R>0 and this constraint is automatically satisfied. Weil calls them K-divisors [6, p. 422], while Lang uses MK -divisors [2, Ch.2 §5]. Neukirch works with additive Arakelov divisors that he also calls replete divisors [3, III.1.8]. 2 Definition 15.3. Let K be a global field. The size of an Arakelov divisor c is Y kck := cv ∈ R>0 . v∈MK The map Div K → R× >0 defined by c 7→ kck is a group homomorphism whose kernel contains Princ K (by the product formula, Theorem 13.21). Corresponding to each Arakelov divisor c is a subset L(c) of K defined by L(c) := {x ∈ K : kxkv ≤ cv for all v ∈ MK }. and a nonzero fractional ideal of OK defined by Y Ic := qv(c) v , v-∞ where qv := {a ∈ OK : v(a) > 0} is the prime ideal corresponding to the discrete valuation v that induces k kv , and v(c) := − log#kv (cv ) ∈ Z (so v(x) = v(c) if and only if kxkv = cv ). We have L(c) ⊆ Ic ⊆ K, and the map c 7→ Ic defines a group homomorphism Div K → IK . Observe that to specify an Arakelov divisor c it suffices to specify the fractional ideal Ic and the real numbers cv > 0 for v|∞ (a finite set). Remark 15.4. The quotient of Div K by the subgroup Princ K is denoted Pic K. The homomorphism Div K → IK sends principal Arakelov divisors to principal fractional ideals, thus the ideal class group cl OK is a quotient of Pic K. We have a commutative diagram → ← → Pic K → IK ← ← ← Div K → cl OK . The Arakelov divisors of size 1 form a subgroup of Div K denoted Div0 K that contains Princ K and surjects onto IK via the map Div K → IK (we are free to choose any Ic ∈ IK because we can always choose the cv at infinite places to ensure kck = 1). The quotient of Div0 K by Princ K is the Arakelov class group Pic0 K, which also admits the ideal class group cl OK as a quotient.3 As we will shall see, Pic0 K is a compact topological group that is finite when K is a global function field. See [5] for more background on Arakelov class groups and how to compute them. Remark 15.5. The set L(c) associated to an Arakelov divisor c is directly analogous to the Riemann-Roch space L(D) := {f ∈ k(X) : vP (f ) ≥ −nP for all closed points P ∈ X}, associated to a divisor D ∈ Div X of a smooth projective curve X/k, P which is a k-vector space of finite dimension. Recall that a divisor is a formal sum D = nP P over the closed points (Gal(k̄/k)-orbits) of the curve X with nP ∈ Z and all but finitely many nP zero. If k is a finite field then K = k(X) is a global field and there is a one-to-one correspondence between closed points of X and places of K, and a normalized absolute value k kP for 3 Neukirch uses CH1 (OK )0 to denote the (additive) Arakelov class group [3, III.1.10]. 18.785 Fall 2021, Lecture #15, Page 2 each closed point P (indeed, one can take this as a definition). The constraint vP (f ) ≥ −nP is equivalent to kf kP ≤ (#kP )nP , where kP is the residue field corresponding to P . If we put cP := (#kP )nP then c = (cP ) is an Arakelov divisor with L(c) = L(D). The Riemann-Roch space L(D) is finite (since k is finite), and we will prove below that L(c) is also finite (but note that when K is a number field the finite set L(c) is not a vector space). In §6.3 we described the divisor group Div X as the additive analog of the ideal group of the ring of integers A = OK , equivalently, the coordinate ring A = k[X], of the global function field K = k(X). When X is a smooth projective curve this is not a perfect analogy because divisors in Div X may include terms corresponding to “points at infinity” which do not correspond to a fractional ideal of A. The group of Arakelov divisors Div K takes these infinite places into account and is a better analog of Div X than IK when X is a smooth projective curve over a finite field and K is its function field. We now specialize to the case where K is a number field. Recall that the absolute norm N(I) of a fractional ideal of OK is the unique t ∈ Q>0 for which NOK /Z (I) = (t). We have N(Ic ) = Y v-∞ and therefore We also define N(qv )v(c) = Y (#kv )v(c) = v-∞ kck = N(Ic )−1 Y c−1 v , v-∞ Y (1) cv , v|∞ Rc := {x ∈ KR : |x|v ≤ cv for all v|∞}, which we note is a compact, convex, symmetric subset of the real vector space KR := K ⊗Q R ' Rr × Cs , where r is the number of real places of K, and s is the number of complex places. If we view Ic and L(c) as subsets of KR via the canonical embedding K ,→ KR , then L(c) = Ic ∩ Rc . Example 15.6. Let K = Q(i). The ideal (2 + i) lying above 5 is prime and corresponds to a finite place v1 , and there is a unique infinite place v2 |∞ which is complex. Let cv1 = 1/5, let cv2 = 10, and set cv = 1 for all other v ∈ MK . We then have Ic = (2 + i) and the image of L(c) = {x ∈ (2 + i) : |x|∞ ≤ 10} under the canonical embedding K ,→ KR ' C is the set of lattice points in the image of the ideal Ic that lie within the circle Rc ⊆ KR ' C of √ 2 radius 10. Note that k k√ v2 = | |C is the square of the usual absolute value on C, which is why the circle has radius 10 rather than 10. 18.785 Fall 2021, Lecture #15, Page 3 √ 10 The set L(c) is clearly finite; it contains exactly 9 points. Lemma 15.7. Let c be an Arakelov divisor of a global field K. The set L(c) is finite. Proof. We assume K is a number field; see Problem Set 7 for the function field case. The fractional ideal Ic is a lattice in KR (under the canonical embedding K ,→ KR ), and is thus a closed discrete subset of KR (recall from Remark 14.5 that lattices are closed). In KR we may view L(c) = Ic ∩ Rc as the intersection of a discrete closed set with a compact set, which is a compact discrete set and therefore finite. Corollary 15.8. Let K be a global field, and let µK denote the torsion subgroup of K × (equivalently, the roots of unity in K). The group µK is finite and equal to the kernel of the × . map K × → Div K defined by x 7→ (kxkv ); it is also the torsion subgroup of OK Proof. Each ζ ∈ µK satisfies ζ n = 1 for some positive integer n. For every place v ∈ MK we have kζ n kv = kζknv = 1, and therefore kζkv = 1. It follows that µK ⊆ ker(K × → Div K). Let c be the Arakelov divisor with cv = 1 for all v ∈ MK . Then ker(K × → Div K) ⊆ L(c) is a finite subgroup of K × and is therefore contained in the torsion subgroup µK . Every × element of µK is an algebraic integer (in fact a root of xn − 1), so µK ⊆ OK . It follows from Corollary 15.8 that for any global field K we have the following exact sequence of abelian groups 1 −→ µK −→ K × −→ Div K −→ PicK −→ 1. Proposition 15.9. Let K be a number field with s complex places, define  s 2 p BK := |DK |. π If c is an Arakelov divisor of size kck > BK then L(c) contains an element of K × . 18.785 Fall 2021, Lecture #15, Page 4 Proof. Our strategy is to apply Minkowski’s lattice point theorem (see Theorem 14.13) to the convex symmetric set Rc and the lattice Ic ⊆ K ⊆ KR ; we just need to show that if kck > BK then the ratio of the Haar measure of Rc to the covolume of Ic exceeds 2n , where n = r + 2s is the degree of K (which is the real dimension of KR ). As defined in §14.2, we normalize the Haar measure µ on the locally compact group KR ' Rr × Cs ' Rn so that µ(S) = 2s µRn (S) for measurable S ⊆ KR . For each real place v, the constraint kxkv = |x|R ≤ cv contributes a factor of 2cv to µ(Rc ), and for each complex place v the √ constraint kxkv = |x|2C ≤ cv contributes a factor of πcv (the area of a circle of radius cv ). We may then compute Q Q  s 2 2c πc s v v v real v complex µ(Rc ) 2 µRn (Rc ) = = covol(Ic ) covol(Ic ) covol(Ic ) Q r 2r (2π)s v|∞ cv 2 (2π)s kck n = p = p kck = 2 > 2n BK |DK |N(Ic ) |DK | where we have used Corollary 14.17 and (1) in the second line. Theorem 14.13 implies that L(c) = Rc ∩Ic contains a nonzero element (which lies in K × ⊆ KR , since Ic ⊆ K ⊆ KR ). Remark 15.10. The bound in Proposition 15.9 can be turned into an asymptotic, that is, for c ∈ Div K, as kck → ∞ we have ! 2r (2π)s #L(c) = p + o(1) kck. (2) |DK | This can be viewed as a multiplicative analog of the Riemann-Roch theorem for function P P fields, which states that for divisors D = nP P , as deg D := nP → ∞ we have dim L(D) = 1 − g + deg D. (3) The nonnegative integer g is the genus, an important invariant of a function field that is often defined by (3); one could similarly use (2) to define the nonnegative integer |DK |. For all sufficiently large kck the o(1) error term will be small enough so that (2) uniquely determines |DK |. Conversely, with a bit more work one can adapt the proofs of Lemma 15.7 and Proposition 15.9 to give a proof of the Riemann-Roch theorem for global function fields. 15.2 The unit group of a number field × Let K be a number field with ring of integers OK . The multiplicative group OK is the unit group of OK , and may also be called the unit group of K. Of course the unit group of the ring K is K × , but this is typically referred to as the multiplicative group of K. As a ring, the finite étale R-algebra KR = K ⊗Q R also has a unit group, and we have an isomorphism of topological groups4 Y Y Y KR× ' Kv× ' R× C× = (R× )r × (C× )s . v|∞ real v|∞ complex v|∞ 4 The additive group of KR is isomorphic to Rn as a topological group (and R-vector space), a fact we have used in our study of lattices in KR . But as topological rings KR ' Rr × Cs 6' Rn unless s = 0. 18.785 Fall 2021, Lecture #15, Page 5 Writing elements of KR× as vectors x = (xv ) indexed by the infinite places v of K, we now define a surjective homomorphism of locally compact groups Log : KR× → Rr+s (xv ) 7→ (log kxv kv ). It is surjective and continuous because each of the maps xv 7→ log kxv kv is, and it is a group homomorphism because Log(xy) = (log kxv yv kv ) = (log kxv kv + log kyv kv ) = (log kxv kv ) + (log kyv kv ) = Log x + Log y; here we have used the fact that the normalized absolute value k kv is multiplicative. Recall from Corollary 13.7 that there is a one-to-one correspondence between the infinite places of K and the Gal(C/R)-orbits of HomQ (K, C). For each v|∞ let us now pick a representative σv of its corresponding Gal(C/R)-orbit in HomQ (K, C); for real places v there is a unique choice for σv , while for complex places there are two choices, σv and its complex conjugate σ̄v . Regardless of our choices, we then have ( |σv (x)|R if v|∞ is real kxkv = |σv (x)σ̄v (x)|R if v|∞ is complex. The absolute norm N : K × → Q× >0 extends naturally to a continuous homomorphism of locally compact groups N : KR× → R× >0 Y (xv ) 7→ kxv kv v|∞ which is compatible with the canonical embedding K × ,→ KR× . Indeed, we have N(x) = NK/Q (x) = Y σ(x) σ Y = R v|∞ kxkv . We thus have a commutative diagram ← ← → Rr+s × Q× >0 -← → R>0 log → T ← ← Log N → → N ← K × -← → KR× → R, P where T : Rr+s → R is defined by T(x) = i xi . We may view Log as a map from K × to Rr+s via the embedding K × ,→ KR× , and similarly view N as a map from K × to R× >0 . We can succinctly summarize the commutativity of the above diagram by the identity T(Log x) = log N(x), which holds for all x ∈ K × , and all x ∈ KR× . The norm of a unit in OK must be a unit in × Z, hence have absolute value 1. Thus OK lies in the kernel of the map x 7→ log N(x) and 18.785 Fall 2021, Lecture #15, Page 6 × therefore also in the kernel of the map x 7→ T(Log x). It follows that Log(OK ) is a subgroup of the trace zero hyperplane Rr+s := {x ∈ Rr+s : T(x) = 0}, 0 which we note is both a subgroup of Rr+s , and an R-vector subspace of dimension r + s − 1. × The proof of Dirichlet’s unit theorem amounts to showing that Log(OK ) is a lattice in Rr+s 0 . Proposition 15.11. Let K be a number field with r real and s complex places, and let ΛK × be the image of the unit group OK in Rr+s under the Log map. The following hold: 0 (1) We have a split exact sequence of finitely generated abelian groups Log × 1 → µK → OK −→ ΛK → 0; (2) ΛK is a lattice in the trace zero hyperplane Rr+s 0 . Here µK is not a Haar measure, it denotes the group of roots of unity in K, all of which × × is clearly a root of unity. , and any torsion element of OK are clearly torsion elements of OK Log × Proof. (1) We first show exactness. Let Z be the kernel of OK −→ ΛK . Clearly µK ⊆ Z, r+s since ΛK ⊆ R0 is torsion free. Let c be the Arakelov divisor with Ic = OK and cv = 2 for v|∞, so that L(c) = {x ∈ OK : kxkv ≤ 2 for all v|∞}. × we have For x ∈ OK x ∈ L(c) ⇐⇒ Log(x) ∈ Log Rc = {z ∈ Rr+s : zi ≤ log 2}. The set on the RHS includes the zero vector, thus Z ⊆ L(c), which by Lemma 15.7 is a × , we must have Z ⊆ µK , so Z = µK and the sequence finite set. As a finite subgroup of OK × is exact (the map from OK to ΛK is surjective by the definition of ΛK ).
× ∩ L(c) is finite, We now show the sequence splits. Note that ΛK ∩ Log(Rc ) = Log OK since L(c) is finite. It follows that 0 is an isolated point of ΛK in Rr+s , and in Rr+s 0 , so ΛK r+s−1 and must be finitely generated, by Lemma 14.3. is a discrete subgroup of Rr+s ' R 0 × It follows that OK is finitely generated, since it lies in a short exact sequence whose left and right terms are finitely generated (recall that µK is finite, by Corollary 15.8). By the structure theorem for finitely generated abelian groups, the sequence must split, since µK × is the torsion subgroup of OK . (2) Having proved (1) it remains only to show that ΛK spans Rr+s 0 . Let V be the subspace r+s of R0 spanned by ΛK and suppose for the sake of contradiction that dim V < dim Rr+s 0 . The orthogonal subspace V ⊥ then contains a unit vector u, and for every λ ∈ R>0 the open ball B<λ (λu) does not intersect ΛK . Thus Rr+s contains points arbitrarily far away from 0 every point in ΛK (with respect to any norm on Rr+s ⊆ Rr+s ). To obtain a contradiction 0 it is enough to show that there is a constant M ∈ R>0 such that for every h ∈ Rr+s there 0 := is an ` ∈ ΛK for which kh − `k maxi |hi − `i | < M (here we are using k k to denote the sup norm on the R-vector space Rr+s ). Let us fix a real number B > BK , where BK is as in Proposition 15.9, so that for every c ∈ Div K with kck ≥ B the set L(c) contains a nonzero element, and fix a vector b ∈ Rr+s P with nonnegative components bi such that T(b) = i bi = log B. Let (α1 ), . . . , (αm ) be the 18.785 Fall 2021, Lecture #15, Page 7 list of all nonzero principal ideals with N(αj ) ≤ B (by Lemma 14.21 this is a finite list). Let M be twice the maximum of (r + s)B and maxj k Log(αj )k. Now let h ∈ Rr+s 0 , and define c ∈ Div K by Ic := OK and cv := exp(hi + bi ) for v|∞, where i is the coordinate in Rr+s corresponding to v under the Log map. We have X  Y cv = exp (hi + bi ) = exp T(h + b) = exp(T(h) + T(b)) = exp T(b) = B > BK , kck = v i thus L(c) contains a nonzero γ ∈ Ic ∩K = OK , and g = Log(γ) satisfies gi ≤ log cv = hi +bi . We also have T(g) = T(Log γ) = log P N(γ) ≥ 0, since N(γ) ≥ 1 for all nonzero γ ∈ OK . The vector w := g − h ∈ Rr+s satisfies i wi = T(v) = T(g) − T(h) = T(g) ≥ 0 and wi ≤ bi ≤ B which together imply |wi | ≤ (r + s)B, so kg − hk = kwk ≤ M/2. We also have log N(γ) = T(Log(γ)) ≤ T(h + b) = T(b) = log B, × so N(γ) ≤ B and (γ) = (αj ) for one of the αj fixed above. Thus γ/αj ∈ OK is a unit, and ` := Log(γ/αj ) = Log(γ) − Log(αj ) ∈ ΛK satisfies kg − `k = k Log(αj )k ≤ M/2. We then have kh − `k ≤ kh − gk + kg − `k ≤ M as desired (by the triangle inequality for the sup-norm). Dirichlet’s unit theorem follows immediately from Proposition 15.11. Theorem 15.12 (Dirichlet’s Unit Theorem). Let K be a number field with r real and s × ' µK × Zr+s−1 is a finitely generated abelian group. complex places. Then OK × under the Log map is the Proof. The image of the torsion-free part of the unit group OK r+s lattice ΛK in the trace-zero hyperplane R0 , which has dimension r + s − 1. We can restate this theorem in a more general form so that it applies to all global fields. As usual, when we consider global function fields we view them as extensions of Fq (t), with q chosen so that K ∩ Fq = Fq and t chosen so that K/Fq (t) is separable. Theorem 15.13 (Unit Theorem for Global Fields). Let K/F be a finite separable extension, with F = Q or F = Fq (t), let S ⊆ MK be the set of places of K lying above the × := {x ∈ K × : v(x) = 0 for all v ∈ MK − S}. unique infinite place of F , and define OK × Then OK ' µK × Z#S−1 is a finitely generated abelian group. Proof. For F = Q we have #S = r + s and this is Dirichlet’s unit theorem; for F = Fq (t), see [4, Prop. 14.1]. Remark 15.14. We should be careful how we interpret 15.13 in the case F = Fq (t). By applying an automorphism of Fq (t) (replace t by t−a for some a ∈ Fq , say) we can move any × degree-one place to infinity. This will change the group OK and may change the number of places of K above our new point at infinity. In contrast to the number field setting (where the place of Q at infinity is invariant because it is the only archimedean place) the ring OK and the set S are not intrinsic to K in the function field setting; they depend on the choice of the separating element t used to construct the separable extension K/Fq (t). 18.785 Fall 2021, Lecture #15, Page 8 √ Example 15.15. Let K = Q( d) be a quadratic field with d 6= 1 squarefree. If d < 0 then × × r = 0 and s = 1, in which √ case the unit group OK has rank 0 and OK = µK is finite. If d > 0 then K = Q( d) ⊆ R is a real quadratic field with r = 2 and s = 0, and the × × unit group OK has rank 1. The only torsion elements of OK ⊆ R are ±1, thus × OK = {±n : n ∈ Z}, × for some  ∈ OK of infinite order. We may assume  > 1: if  < 0 then replace  by −, and if  < 1 then replace  by −1 (we cannot have  = 1 ∈ µK ). The assumption  > 1 uniquely determines . This follows from the fact that for  > 1 we have |n | > || for all n > 1 and |n | ≤ 1 for all n ≤ 0. This unique  is the fundamental unit of OK (and of K). To explicitly determine , let D = disc OK (so D = d if d ≡ 1 mod 4 and D = 4d otherwise). Every element of OK can be uniquely written as √ x+y D , 2 where √x and Dy are integers of the same parity. In the case of a unit we must have N( x+y2 D ) = ±1, equivalently, x2 − Dy 2 = ±4. (4) Conversely, any solution (x, y) ∈ Z2 to the above equation has x and Dy with the same √ x+y D × parity and corresponds to an element of OK . The constraint  = > 1 forces x, y > 0. 2 √ √ |x−y D| This follows from the fact that −1 = < 1, so −2 < x − y D < 2, and adding and 2 √ subtracting x + y D > 2 shows x > 0 and y > 0 (respectively). Thus √we need only√consider positive integer solutions (x, y) to (4). Among such solutions, x1 + y1 D < x2 + y2 D implies x1 < x2 , so the solution that minimizes x will give us the fundamental unit . Equation (4) is a (generalized) Pell equation. Solving the Pell equation is a well-studied problem and there are a number of algorithms for doing so. The most well known uses continued fractions and is explored on Problem Set 7; this is not the most efficient method, but it is dramatically faster than an exhaustive search; see [1] for a comprehensive survey. A remarkable feature of this problem is that even when D is quite small, the smallest solution to (4) may be very large. For example, when D = d = 889 the fundamental unit is √ 26463949435607314430 + 887572376826907008 889 = . 2 15.3 The regulator of a number field Let K be a number field with r real places and s complex places, and let Rr+s be the 0 r+s r+s r+s−1 trace-zero hyperplane in R . Choose any coordinate projection π : R →R , and ∼ r+s−1 to endow Rr+s with a Euclidean measure. use the induced isomorphism Rr+s −→ R 0 0 × By Proposition 15.11, the image ΛK of the unit group OK is a lattice in Rr+s 0 , and we can r+s measure its covolume using the Euclidean measure on R0 . Definition 15.16. The regulator of a number field K is × RK := covol(π(Log(OK ))) ∈ R>0 , 18.785 Fall 2021, Lecture #15, Page 9 where π : Rr+s → Rr+s−1 is any coordinate projection; the value of RK does not depend on the choice of π, since we use π to normalize the Haar measure on Rr+s ' Rr+s−1 . If 0 × 1 , . . . , r+s−1 is a fundamental system of units (a Z-basis for the free part of OK ), then RK can be computed as the absolute value of the determinant of any (r + s − 1) × (r + s − 1) minor of the (r + s) × (r + s − 1) matrix whose columns are the vectors Log(i ) ∈ Rr+s . Example 15.17. If K √ is a real quadratic field with absolute discriminant D and fundax+y D mental unit  = , then r + s = 2 and the product of the two real embeddings 2 σ1 (), σ2 () ∈ R is N() = ±1. Thus log |σ2 ()| = − log |σ1 ()| and Log() = (log |σ1 ()|, log |σ2 ()|) = (log |σ1 ()|, − log |σ1 ()|). The 1×1 minors of the 2×1 transpose of Log() have determinant ± log |σ1 ()|; the absolute value of the determinant is the same in both cases, and since we have require the fundamental unit to satisfy  > 1 (which forces a choice of embedding), the regulator of K is simply RK = log . References [1] Michael J. Jacobson and Hugh C. Williams, Solving the Pell equation, Springer, 2009. [2] Serge Lang, Fundamentals of diophantine geometry, Springer, 1983. [3] Jürgen Neukirch, Algebraic number theory, Springer, 1999. [4] Michael Rosen, Number theory in function fields, Springer, 2002. [5] René Schoof, Computing Arakelov class groups, in Algorithmic Number Theory: lattices, number fields, curves, and cryptography. MSRI Publications 44 (2008), 447–495. [6] André Weil, Arithmetic on algebraic varieties, Annals of Mathematics (2) 53 (1951), 412–444. 18.785 Fall 2021, Lecture #15, Page 10 18.785 Number theory I Lecture #16 16 Fall 2021 11/3/2021 Riemann’s zeta function and the prime number theorem We now divert our attention from algebraic number theory to talk about zeta functions and L-functions. As we shall see, every global field has a zeta function that is intimately related to the distribution of its primes. We begin with the zeta function of the rational field Q, which we will use to prove the prime number theorem. We will need some basic results from complex analysis, all of which can be found in any introductory textbook (such as [1, 2, 3, 7, 12]). A short glossary of terms and a list of the basic theorems we will use can be found at the end of these notes.1 16.1 The Riemann zeta function Definition 16.1. The Riemann zeta function is the complex function defined by the series X ζ(s) := n−s , n≥1 for Re(s) > 1, where n varies over positive integers. It is easy to verify that this series converges absolutely and locally uniformly on Re(s) > 1 (use the integral test on an open ball strictly to the right of the line Re(s) = 1). By Theorem 16.17, it defines a holomorphic function on Re(s) > 1, since each term n−s = e−s log n is holomorphic. Theorem 16.2 (Euler product). For Re(s) > 1 we have X Y ζ(s) = (1 − p−s )−1 , n−s = p n≥1 where the product converges absolutely. In particular, ζ(s) 6= 0 for Re(s) > 1. The product in the theorem above ranges over primes p. This is a standard practice in analytic number theory that we will follow: the symbol p always denotes a prime, and any sum or product over p is understood to be over primes, even if this is not explicitly stated. Proof. We have X n−s = XY p−vp (n)s = n≥1 p n≥1 YX p e≥0 p−es = Y p (1 − p−s )−1 . To justify the second equality, consider the partial zeta function ζm (s), which restricts the summation in ζ(s) to the set Sm of m-smooth integers (those with no prime factors p > m). If p1 , . . . , pk are the primes up to m, absolute convergence implies X Y X X Y ei ζm (s) := (pe11 · · · pekk )−s = (p−s ) = (1 − p−s )−1 . n−s = i e1 ,...,ek ≥0 n∈Sm 1≤i≤k ei ≥0 p≤m For any δ > 0 the sequence of functions ζm (s) converges uniformly on Re(s) > 1 + δ to ζ(s); indeed, for any  > 0 and any such s we have |ζm (s) − ζ(s)| ≤ 1 X n≥m n−s ≤ X n≥m |n−s | = X n≥m n− Re(s) ≤ Z ∞ m 1 x−1−δ dx ≤ m−δ < , δ Those familiar with this material should still glance at §16.3.2 which touches on some convergence issues that are particularly relevant to number theoretic applications. for all sufficiently large m. It follows that the sequence ζQ m (s) converges locally uniformly to ζ(s) on Re(s) > 1. The sequence of functions Pm (s) := p≤m (1 − p−s )−1 clearly converges Q locally uniformly to (1 − p−s )−1 on any region in which the latter function is absolutely convergent (or even just convergent). For any s in Re(s) > 1 we have X p | log(1 − p−s )−1 | = X X1 X XX |p−s |e = (|ps | − 1)−1 < ∞, p−es ≤ e p p p e≥1 e≥1 P where we have used the identity log(1 − z) = − n≥1 n1 z n , valid for |z| < 1. It follows that Q −s −1 is absolutely convergent (and in particular, nonzero) on Re(s) > 1. p (1 − p ) Theorem 16.3 (Analytic continuation I). For Re(s) > 1 we have ζ(s) = 1 + φ(s), s−1 where φ(s) is a holomorphic function on Re(s) > 0. Thus ζ(s) extends to a meromorphic function on Re(s) > 0 that has a simple pole at s = 1 with residue 1 and no other poles. Proof. For Re(s) > 1 we have X 1 ζ(s) − = n−s − s−1 n≥1 Z 1 ∞ x −s Z X dx = n−s − n≥1 n+1 x −s n dx  = XZ n+1 n≥1 n
n−s − x−s dx. R n+1 For each n ≥ 1 the function φn (s) := n (n−s − x−s )dx is holomorphic on Re(s) > 0. For each fixed s in Re(s) > 0 and x ∈ [n, n + 1] we have Z x Z x Z x |s| |s| |s| −s −s −s−1 |n − x | = st dt ≤ dt = dt ≤ 1+Re(s) , s+1 1+Re(s) | n n n |t n t and therefore |φn (s)| ≤ Z n+1 n n−s − x−s dx ≤ |s| n1+Re(s) . For any s0 with Re(s0 ) > 0, if we put  := Re(s0 )/2 and U := B< (s0 ), then for each n ≥ 1, |s0 | +  =: Mn , n1+ s∈U P P and n Mn = (|s0 | + )ζ(1 + ) converges. The series n φn thus P converges locally normally on Re(s) > 0. By the Weierstrass M -test (Theorem 16.19), n φn converges to a function 1 that is holomorphic on Re(s) > 0. φ(s) = ζ(s) − s−1 sup |φn (s)| ≤ We now show that ζ(s) has no zeros on Re(s) = 1; this fact is crucial to the prime number theorem. For this we use the following ingenious lemma, attributed to Mertens.2 Lemma 16.4 (Mertens). For x, y ∈ R with x > 1 we have |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 1. 2 If this lemma strikes you as pulling a rabbit out of a hat, well, it is. For a slight variation, see [15, IV], which uses an alternative approach due to Hadamard. 18.785 Fall 2021, Lecture #16, Page 2 Proof. From the Euler product ζ(s) = log |ζ(s)| = − X p Q − p−s )−1 , we see that for Re(s) > 1 we have p (1 log |1 − p−s | = − since log |z| = Re log z and log(1 − z) = − X p P log |ζ(x + iy)| = Re log(1 − p−s ) = zn n≥1 n X X Re(p−ns ) p n≥1 n , for |z| < 1. Plugging in s = x + iy yields X X cos(ny log p) p n≥1 npnx , since Re(p−ns ) = p−nx Re(e−iny log p ) = p−nx cos(−ny log p) = p−nx cos(ny log p). Thus log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = X X 3 + 4 cos(ny log p) + cos(2ny log p) p n≥1 npnx . We now note that the trigonometric identity cos(2θ) = 2 cos2 θ − 1 implies 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ≥ 0. Taking θ = ny log p yields log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 0, which proves the lemma. Corollary 16.5. ζ(s) has no zeros on Re(s) ≥ 1. Proof. We know from Theorem 16.2 that ζ(s) has no zeros on Re(s) > 1, so suppose ζ(1 + iy) = 0 for some y ∈ R. Then y 6= 0, since ζ(s) has a pole at s = 1, and we know that ζ(s) does not have a pole at 1 + 2iy 6= 1, by Theorem 16.3. We therefore must have lim |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = 0, (1) x→1 since ζ(s) has a simple pole at s = 1, a zero at 1 + iy, and no pole at 1 + 2iy. But this contradicts Lemma 16.4. 16.2 The Prime Number Theorem The prime counting function π : R → Z≥0 is defined by X π(x) := 1; p≤x it counts the number of primes up to x. The prime number theorem (PNT) states that π(x) ∼ x . log x The notation f (x) ∼ g(x) means limx→∞ f (x)/g(x) = 1; one says that f is asymptotic to g. This conjectured growth rate for π(x) dates back to Gauss and Legendre in the late 18th century. In fact Gauss believed the asymptotically equivalent but more accurate statement3 Z x dt . π(x) ∼ Li(x) := 2 log t 3 More accurate in the sense that |π(x) − Li(x)| grows more slowly than |π(x) − x | log x as x → ∞. 18.785 Fall 2021, Lecture #16, Page 3 However it was not until a century later that the prime number theorem was independently proved by Hadamard [5] and de la Vallée Poussin [9] in 1896. Their proofs are both based on the work of Riemann [10], who in 1860 showed that there is a precise connection between the zeros of ζ(s) and the distribution of primes (we shall say more about this later), but was unable to prove the prime number theorem. The proof we will give is more recent and due to Newman [8], but it relies on the same properties of the Riemann zeta function that were exploited by both Hadamard and de la Vallée, the most essential of which is the fact that ζ(s) has no zeros on Re(s) ≥ 1 (Corollary 16.5). A concise version of Newman’s proof by Zagier can be found in [15]; we will follow Zagier’s outline but be slightly more expansive in our presentation. We should note that there are also “elementary" proofs of the prime number theorem independently obtained by Erdös [4] and Selberg [11] in the 1940s that do not use the Riemann zeta function, but they are elementary only in the sense that they do not use complex analysis; the details of these proofs are considerably more complicated than the one we will give. Rather than work directly with π(x), it is more convenient to work with the log-weighted prime-counting function defined by Chebyshev4 X ϑ(x) := log p, p≤x whose growth rate differs from that of π(x) by a logarithmic factor. Theorem 16.6 (Chebyshev). π(x) ∼ x log x if and only if ϑ(x) ∼ x. Proof. We clearly have 0 ≤ ϑ(x) ≤ π(x) log x, thus ϑ(x) π(x) log x ≤ . x x For every  ∈ (0, 1) we have ϑ(x) ≥ X   log p ≥ (1 − )(log x) π(x) − π(x1− ) x1− 1 and all  > 0 we have |F (λx) − F (x)| <  for all sufficiently large x. Fix λ > 1 and suppose there is an unbounded sequence (xn ) such that f (xn ) ≥ λxn for all n ≥ 1. For each xn we have Z λxn Z λxn Z λ f (t) − t λxn − t λ−t F (λxn ) − F (xn ) = dt ≥ dt = dt = c, 2 2 t t t2 xn xn 1 for some c > 0, where we used the fact that f is non-decreasing to get the middle inequality. Taking  < c, we have |F (λxn ) − F (xn )| = c >  for arbitrarily large xn , a contradiction. Thus f (x) < λx for all sufficiently large x. A similar argument shows that f (x) > λ1 x for all sufficiently large x. These inequalities hold for all λ > 1, so limx→∞ f (x)/x = 1. Equivalently, f (x) ∼ x. 5 The equality sign in the big-O notation f (x) = O(g(x)) is a standard abuse of notation; it simply means lim supx→∞ |f (x)|/|g(x)| < ∞ (and nothing more). In more complicated equalities a big-O expression should P be interpreted as a set of functions, one of which makes the equality true, for example, n≥1 n1 = log n+O(1). 18.785 Fall 2021, Lecture #16, Page 5 In order to show that the hypothesis of Lemma 16.8 is satisfied for f = ϑ, we will work with the function H(t) = ϑ(et )e−t − 1; the change of variables t = eu shows that Z ∞ Z ∞ ϑ(t) − t H(u)du converges . dt converges ⇐⇒ t2 1 0 We now recall the Laplace transform. Definition 16.9. Let h : R>0 → R be a piecewise continuous function. The Laplace transform Lh of h is the complex function defined by Z ∞ Lh(s) := e−st h(t)dt, 0 which is holomorphic on Re(s) > c for any c ∈ R for which h(t) = O(ect ). The following properties of the Laplace transform are easily verified. • L(g + h) = Lg + Lh, and for any a ∈ R we have L(ah) = aLh. • If h(t) = a ∈ R is constant then Lh(s) = as . • L(eat h(t))(s) = L(h)(s − a) for all a ∈ R. We now define the auxiliary function Φ(s) := X p−s log p, p which is related to ϑ(x) by the following lemma. Lemma 16.10. L(ϑ(et ))(s) = Φ(s) s is holomorphic on Re(s) > 1. Proof. By Lemma 16.7, ϑ(et ) = O(et ), so L(ϑ(et )) is holomorphic on Re(s) > 1. Let pn be the nth prime, and put p0 := 0. The function ϑ(et ) is constant on t ∈ (log pn , log pn+1 ), so Z log pn+1 Z log pn+1   1 −s e−st ϑ(et )dt = ϑ(pn ) e−st dt = ϑ(pn ) p−s − p n n+1 . s log pn log pn We then have (Lϑ(et ))(s) = Z ∞ e−st ϑ(et )dt = 0 ∞   1X −s ϑ(pn ) p−s − p n n+1 s n=1 ∞ ∞ 1X 1X = ϑ(pn )p−s ϑ(pn−1 )p−s n − n s s = = 1 s 1 s n=1 ∞  X n=1 ∞ X n=1  ϑ(pn ) − ϑ(pn−1 ) p−s n p−s n log pn = n=1 Φ(s) . s Let us now consider the function H(t) := ϑ(et )e−t − 1. It follows from the lemma and standard properties of the Laplace transform that on Re(s) > 0 we have LH(s) = L(ϑ(et )e−t )(s) − (L1)(s) = L(ϑ(et ))(s + 1) − 1 Φ(s + 1) 1 = − . s s+1 s 18.785 Fall 2021, Lecture #16, Page 6 Lemma 16.11. The function Φ(s) − that is holomorphic on Re(s) ≥ 1. 1 s−1 extends to a meromorphic function on Re(s) > 1 2 Proof. By Theorem 16.3, ζ(s) extends to a meromorphic function on Re(s) > 0, which we also denote ζ(s), that has only a simple pole at s = 1 and no zeros on Re(s) ≥ 1, by Corollary 16.5. It follows that the logarithmic derivative ζ 0 (s)/ζ(s) of ζ(s) is meromorphic on Re(s) > 0, with no zeros on Re(s) ≥ 1 and only a simple pole at s = 1 with residue −1 (see §16.3.1 for standard facts about the logarithmic derivative of a meromorphic function). In terms of the Euler product, for Re(s) > 1 we have6 !0 !0 Y X
0 ζ 0 (s) − = − log ζ(s) = − log (1 − p−s )−1 = log(1 − p−s ) ζ(s) p p  X log p X 1 X p−s log p 1 = + log p = = 1 − p−s ps − 1 ps ps (ps − 1) p p p X log p = Φ(s) + . ps (ps − 1) p The sum on the RHS converges absolutely and locally uniformly to a holomorphic function on Re(s) > 1/2. The LHS is meromorphic on Re(s) > 0, and on Re(s) ≥ 1 it has only a 1 simple pole at s = 1 with residue 1. It follows that Φ(s) − s−1 extends to a meromorphic 1 function on Re(s) > 2 that is holomorphic on Re(s) ≥ 1. 1 Corollary 16.12. The functions Φ(s + 1) − 1s and (LH)(s) = Φ(s+1) s+1 − s both extend to meromorphic functions on Re(s) > − 12 that are holomorphic on Re(s) ≥ 0. Proof. The first statement follows immediately from the lemma. For the second, note that   Φ(s + 1) 1 1 1 1 − = Φ(s + 1) − − s+1 s s+1 s s+1 is meromorphic on Re(s) > − 21 and holomorphic on Re(s) ≥ 0, since it is a sum of products of such functions. The final step of the proof relies on the following analytic result due to Newman [8]. Theorem 16.13. Let f : R≥0 → R be a bounded piecewise continuous function, and suppose its R ∞Laplace transform extends to a holomorphic function g(s) on Re(s) ≥ 0. Then the integral 0 f (t)dt converges and is equal to g(0). Proof. Without loss of generality weR assume f (t) ≤ 1 for all t ≥ 0. For τ ∈ R>0 , define Rτ ∞ gτ (s) := 0 f (t)e−st dt, By definition 0 f (t)dt = limτ →∞ gτ (0), thus it suffices to prove lim gτ (0) = g(0). τ →∞ For r > 0, let γr be the boundary of the region {s : |s| ≤ r and Re(s) ≥ −δr } with δr > 0 chosen so that g is holomorphic on γr ; such a δr exists because g is holomorphic on Re(s) ≥ 0, hence on some open ball B≤2δ(y) (iy) for each y ∈ [−r, r], and we may take 6 As is standard when computing logarithmic derivatives, we are taking the principal branch of the complex logarithm and can safely ignore the negative real axis where it is not defined since we are assuming Re(s) > 1. 18.785 Fall 2021, Lecture #16, Page 7 δr := inf{δ(y) : y ∈ [r, −r]}, which is positive because [−r, r] is compact. Each γr is a 2 simple closed curve, and for each τ > 0 the function h(s) := (g(s) − gτ (s))esτ (1 + rs2 ) is holomorphic on a region containing γr . Using Cauchy’s integral formula (Theorem 16.26) to evaluate h(0) yields  Z   1 1 s (2) g(0) − gτ (0) = h(0) = g(s) − gτ (s) esτ + 2 ds. 2πi γr s r We will show the LHS tends to 0 as τ → ∞ by showing that for any  > 0 we can set r = 3/ > 0 so that the absolute value of the RHS is less than  for all sufficiently large τ . Let γr+ denote the part of γr in Re(s) > 0, a semicircle of radius r. The integrand is absolutely bounded by 1/r on γr+ , since for |s| = r and Re(s) > 0 we have sτ g(s) − gτ (s) · e  1 s + 2 s r  = 1 2πi Z γr+  Z ∞ τ ∞ f (t)e−st dt · eRe(s)τ r s · + r s r eRe(s)τ 2 Re(s) · r r τ Re(s)τ − Re(s)τ e e 2 Re(s) = · · Re(s) r r 2 = 2/r . ≤ Therefore Z e− Re(s)t dt ·   1 s 2 1 1 sτ g(s) − gτ (s) e + 2 ds ≤ · πr · 2 = s r 2π r r (3) Now let γr− be the part of γr in Re(s) < 0, a truncated semi-circle. For any fixed r, the first term g(s)esτ (s−1 + sr−2 ) in the integrand of (2) tends to 0 as τ → ∞ for Re(s) < 0 and |s| ≤ r. For the second term we note that since gτ (s) is holomorphic on C, it makes no difference if we instead integrate over the semicircle of radius r in Re(s) < 0. For |s| = r and Re(s) < 0 we then have gτ (s)e sτ  1 s + 2 s r  = Z Z τ 0 τ f (t)e−st dt · eRe(s)τ r s · + r s r eRe(s)τ (−2 Re(s)) r r 0 ! − Re(s)τ Re(s)τ e e (−2 Re(s)) = 1− Re(s) r r ≤ e− Re(s)t dt · = 2/r2 · (1 − eRe(s)τ Re(s)), where the factor (1 − eRe(s)τ Re(s)) on the RHS tends to 1 as τ → ∞ since Re(s) < 0. We thus obtain the bound 1/r + o(1) when we replace γr+ with γr− in (3), and the RHS of (2) is bounded by 2/r + o(1) as τ → ∞. It follows that for any  > 0, for r = 3/ > 0 we have |g(0) − gτ (0)| < 3/r =  for all sufficiently large τ . Therefore limτ →∞ gτ (0) = g(0) as desired. 18.785 Fall 2021, Lecture #16, Page 8 Remark 16.14. Theorem 16.13 is an example of what is known as a Tauberian theorem. For a piecewise continuous function f : R≥0 → R, its Laplace transform Z ∞ Lf (s) := e−st f (t)dt, 0 is typically not defined on Re(s) ≤ c, where c is the least c for which f (t) = O(ect ). Now it may happen that the function Lf has an analytic continuation to a larger domain; for 1 example, if f (t) = et then (Lf )(s) = s−1 extends to a holomorphic function on C−{1}. But plugging values of s with Re(s) ≤ c into the integral usually does not work; in our f (t) = et example, the integral diverges on Re(s) ≤ 1. The theorem says that when Lf extends to a holomorphic function on the entire half-plane Re(s) ≥ 0, its value at s = 0 is exactly what we would get by simply plugging 0 into the integral defining Lf . More generally, Tauberian theorems refer to results related toRtransforms f → T (f ) that ∞ allow us to deduce properties of f (such as the convergence of 0 f (t)dt) from properties of T (f ) (such as analytic continuation to Re(s) ≥ 0). The term “Tauberian" was coined by Hardy and Littlewood and refers to Alfred Tauber, who proved a theorem of this type as a partial converse to a theorem of Abel. Theorem 16.15 (Prime Number Theorem). π(x) ∼ x log x . Proof. H(t) = ϑ(et )e−t − 1 is piecewise continuous and bounded, by Lemma 16.7, and its Laplace transform extends to a holomorphic function on Re(s) ≥ 0, by Corollary 16.12. Theorem 16.13 then implies that the integral Z ∞ Z ∞  H(t)dt = ϑ(et )e−t − 1 dt 0 0 converges. Replacing t with log x, we see that  Z ∞ Z ∞ 1 dx ϑ(x) − x ϑ(x) − 1 = dx x x x2 1 1 converges. Lemma 16.8 implies ϑ(x) ∼ x, equivalently, π(x) ∼ x log x , by Theorem 16.6. One disadvantage of our proof is that it does not give us an error term. Using more sophisticated methods, Korobov [6] and Vinogradov [14] independently obtained the bound ! x
, π(x) = Li(x) + O exp (log x)3/5+o(1) in which we note that the error term is bounded by O(x/(log x)n ) for all n but not by O(x1− ) for any  > 0. Assuming the Riemann Hypothesis, which states that the zeros of ζ(s) in the critical strip 0 < Re(s) < 1 all lie on the line Re(s) = 12 , one can prove π(x) = Li(x) + O(x1/2+o(1) ). More generally, if we knew that ζ(s) has no zeros in the critical strip with real part greater than c, for some c ≥ 1/2 strictly less than 1, we could prove π(x) = Li(x) + O(xc+o(1) ). There thus remains a large gap between what we can prove about the distribution of prime numbers and what we believe to be true. Remarkably, other than refinements to the o(1) term appearing in the Korobov-Vinogradov bound, essentially no progress has been made on this problem in the last 60 years. 18.785 Fall 2021, Lecture #16, Page 9 16.3 A quick recap of some basic complex analysis The complex numbers C are a topological field under the distance metric d(x, y) = |x − y| √ induced by the standard absolute value |z| := z z̄, which is also a norm on C as an Rvector space; all references to the topology on C (open, compact, convergence, limits, etc.) are made with this understanding. 16.3.1 Glossary of terms and standard theorems Let f and g denote complex functions defined on an open subset of C. • f is differentiable at z0 if limz→z0 f (z)−f (z0 ) z−z0 exists. • f is holomorphic at z0 if it is differentiable on an open neighborhood of z0 . • f is analytic at z0 if there of z0 in which f can be defined by P is an open neighborhood n a power series f (z) = n=0 an (z − z0 ) ; equivalently, f is infinitely differentiable and has a convergent Taylor series on an open neighborhood of z0 . • Theorem: f is holomorphic at z0 if and only if it is analytic at z0 . • Theorem: If C is a connected set containing a nonempty open set U and f and g are holomorphic on C with f|U = g|U , then f|C = g|C . • With U and C as above, if f is holomorphic on U and g is holomorphic on C with f|U = g|U , then g is the (unique) analytic continuation of f to C and f extends to g. • If f is holomorphic on a punctured open neighborhood of z0 and |f (z)| → ∞ as z → z0 then z0 is a pole of f ; note that the set of poles of f is necessarily a discrete set. • f is meromorphic at z0 if it is holomorphic at z0 or has z0 as a pole. • Theorem: at z0 then it can be defined by a Laurent series P If f is meromorphic n f (z) = n≥n0 an (z − z0 ) that converges on an open punctured neighborhood of z0 . • The order of vanishing ordz0 (f ) of a nonzero function f that is meromorphic at z0 is the least index n of the nonzero coefficients an in its Laurent series expansion at z0 . Thus z0 is a pole of f iff ordz0 (f ) < 0 and z0 is a zero of f iff ordz0 (f ) > 0. • If ordz0 (f ) = 1 then z0 is a simple zero of f , and if ordz0 (f ) = −1 it is a simple pole. • The residue resz0 (f ) of a function P f meromorphic at z0 is the coefficient a−1 in its Laurent series expansion f (z) = n≥n0 an (z − z0 )n at z0 . • Theorem: If z0 is a simple pole of f then resz0 (f ) = limz→z0 (z − z0 )f (z). • Theorem: If f is meromorphic on a set S then so is its logarithmic derivative f 0 /f , and f 0 /f has only simple poles in S and resz0 (f 0 /f ) = ordz0 (f ) for all z0 ∈ S. In particular the poles of f 0 /f are precisely the zeros and poles of f . 16.3.2 Convergence P P Recall that a series ∞ n=1 an of complex numbers converges absolutely if the series n |an | of nonnegative real numbers converges. An equivalent definition is that the function a(n) := an is integrable with respect to the counting measure µ on the set of positive integers N. Indeed, if the series is absolutely convergent then Z ∞ X an = a(n)µ, n=1 N 18.785 Fall 2021, Lecture #16, Page 10 and if the series is not absolutely convergent, the integral is not defined. Absolute convergence is effectively built-in to the definition of the Lebesgue integral, which requires that in order for the function a(n) = x(n) + iy(n) to be integrable, the positive real functions |x(n)| and |y(n)| must both be integrable (summable), and separately computes sums of the positive and negative subsequences of (x(n)) and (y(n)) as suprema over finite subsets. The measure-theoretic perspective has some distinct advantages. It makes it immediately clear that we may replace the index set N with any set of the same cardinality, since the counting measure depends only on the cardinality of N, not its ordering. We are thus free to sum over any countable index set, including Z, Q, any finite product of countable sets, and any countable coproduct of countable sets (such as countable direct sums of Z); such sums are ubiquitous in number theory and many cannot be meaningfully interpreted as limits of partial sums in the usual sense, since this assumes that the index set is well ordered (not the case with Q, for example). The measure-theoretic view makes P also makes it clear that we may convert any absolutely convergent sum• of the form X×Y into an iterated sum P P theorem. X Y (or vice versa), via Fubini’s Q We say that an infinite product is absolutely conn an of nonzero P Q complex numbers P vergent when the sum n log an is, in which case n an := exp( n log an ).7 This implies that an absolutely convergent product cannot converge to zero, and the sequence (an ) must converge to 1 (no matter how we order the an ). All of our remarks above about absolutely convergent series apply to absolutely convergent products as well. A series or product of complex functions fn (z) is absolutely convergent on S if the series or product of complex numbers fn (z0 ) is absolutely convergent for all z0 ∈ S. Definition 16.16. A sequence of complex functions (fn ) converges uniformly on S if there is a function f such that for every  > 0 there is an integer N for which supz∈S |fn (z)−f (z)| <  for all n ≥ N . The sequence (fn ) converges locally uniformly on S if every z0 ∈ S has an open neighborhood U for which (fn ) converges uniformly on U ∩S. When applied to a series of functions these terms refer to the sequence of partial sums. Because C is locally compact, locally uniform convergence is the same thing as compact convergence: a sequence of functions converges locally uniformly on S if and only if it converges uniformly on every compact subset of S. Theorem 16.17. A sequence or series of holomorphic functions fn that converges locally uniformly on an open set U converges to a holomorphic function f on U , and the sequence or series of derivatives fn0 then converges locally uniformly to f 0 (and if none of the fn has a zero in U and f 6= 0, then f has no zeros in U ). Proof. See [3, Thm. III.1.3] and [3, Thm. III.7.2]. P Definition 16.18. n (z) converges normally on a set S P P A series of complex functions n fP if n kfn k := n supz∈S |fn (z)| converges. The series n fnP (z) converges locally normally on S if every z0 ∈ S has an open neighborhood U on which n fn (z) converges normally. Theorem 16.19 (Weierstrass M-test). Every locally normally convergent series of P functions converges absolutely and locally uniformly. Moreover, a series n fn of holomorphic functions on converges locally normally converges to a holomorphic function f PS that 0 on S, and then n fn converges locally normally to f 0 . 7 In this definition we use the principal branch of log z := log |z| + i Arg z with Arg z ∈ (−π, π). 18.785 Fall 2021, Lecture #16, Page 11 Proof. See [3, Thm. III.1.6]. P Remark 16.20. To show a series n fn is locally normally convergent on a set S amounts to proving that for every z0 ∈ S there is an open neighborhood P U of z0 and a sequence of real numbers (Mn ) such that |fn (z)| ≤ Mn for z ∈ U ∩ S and n Mn < ∞, whence the term “M -test". 16.3.3 Contour integration We shall restrict our attention to integrals along contours defined by piecewise-smooth parameterized curves; this covers all the cases we shall need. Definition 16.21. A parameterized curve is a continuous function γ : [a, b] → C whose domain is a compact interval [a, b] ⊆ R. We say that γ is smooth if it has a continuous nonzero derivative on [a, b], and piecewise-smooth if [a, b] can be partitioned into finitely many subintervals on which the restriction of γ is smooth. We say that γ is closed if γ(a) = γ(b), and simple if it is injective on [a, b) and (a, b]. Henceforth we will use the term curve to refer to any piecewise-smooth parameterized curve γ, or to its oriented image of in the complex plane (directed from γ(a) to γ(b)), which we may also denote γ. Definition 16.22. Let f : Ω → C be a continuous function and let γ be a curve in Ω. We define the contour integral Z f (z)dz := γ Z b f (γ(t))γ 0 (t)dt, a whenever the integralR on the RHS (which is defined as a Riemann sum in the usual way) converges. Whether γ f (z)dz converges, and if so, to what value, does not depend on the parameterization of γ: ifR γ 0 is another parameterized curve with the same (oriented) image R as γ, then γ 0 f (z)dz = γ f (z)dz. We have the following analog of the fundamental theorem of calculus. Theorem 16.23. Let γ : [a, b] → C be a curve in an open set Ω and let f : Ω → C be a holomorphic function Then Z f 0 (z)dz = f (γ(b)) − f (γ(a)). γ Proof. See [2, Prop. 4.12]. Recall that the Jordan curve theorem implies that every simple closed curve γ partitions C into two components, one of which we may unambiguously designate as the interior (the one on the left as we travel along our oriented curve). We say that γ is contained in an open set U if both γ and its interior lie in U . The interior of γ is a simply connected set, and if an open set U contains γ then it contains a simply connected open set that contains γ. Theorem 16.24 (Cauchy’s Theorem). Let U be an open set containing a simple closed curve γ. For any function f that is holomorphic on U we have Z f (z)dz = 0. γ 18.785 Fall 2021, Lecture #16, Page 12 Proof. See [2, Thm. 8.6] (we can restrict U to a simply connected set). Cauchy’s theorem generalizes to meromorphic functions. Theorem 16.25 (Cauchy Residue Formula). Let U be an open set containing a simple closed curve γ. Let f be a function that is meromorphic on U , let z1 , . . . , zn be the poles of f that lie in the interior of γ, and suppose that no pole of f lies on γ. Then Z f (z)dz = 2πi γ n X reszi (f ). i=1 Proof. See [2, Thm. 10.5] (we can restrict U to a simply connected set). R it To see where the 2πi comes from, consider γ dz z with γ(t) = e for t ∈ [0, 2π]. In general one weights residues by a corresponding winding number, but the winding number of a simple closed curve about a point in its interior is always 1. Theorem 16.26 (Cauchy’s Integral Formula). Let U be an open set containing a simple closed curve γ. For any function f holomorphic on U and a in the interior of γ, Z 1 f (z) f (a) = dz. 2πi γ z − a Proof. Apply Cauchy’s residue formula to g(z) = f (z)/(z − a); the only poles of g in the interior of γ are a simple pole at z = a with resa (g) = f (a). Cauchy’s residue formula can also be used to recover the coefficients f (n) (a)/n! appearing in the Laurent series expansion of a meromorphic function at a (apply it to f (z)/(z −a)n+1 ). One of many useful consequences of this is Liouville’s theorem, which can be proved by showing that the Laurent series expansion of a bounded holomorphic function on C about any point has only one nonzero coefficient (the constant coefficient). Theorem 16.27 (Liouville’s theorem). Bounded entire functions are constant. Proof. See [2, Thm. 5.10]. We also have the following converse of Cauchy’s theorem. Theorem 16.28 (Morera’s Theorem). Let f be a continuous function and on an open set U , and suppose that for every simple closed curve γ contained in U we have Z f (z)dz = 0. γ Then f is holomorphic on U . Proof. See [3, Thm. II.3.5]. 18.785 Fall 2021, Lecture #16, Page 13 References [1] Lars V. Ahlfors, Complex analysis: an introduction to the theory of analytic functions of one complex variable, 3rd edition, McGraw-Hill, 1979. [2] Joseph Bak and Donald J. Newman, Complex analysis, Springer, 2010. [3] Rolf Busam and Eberhard Freitag, Complex analysis, 2nd edition, Springer 2009. [4] Paul Erdös, On a new method in elementary number theory which leads to an elementary proof of the prime number theorem, Proc. Nat. Acad. Scis. U.S.A. 35 (1949), 373–384. [5] Jacques Hadamard, Sur la distribution des zéros de la function ζ(s) et ses conséquences arithmétique, Bull. Soc. Math. France 24 (1896), 199–220. [6] Nikolai M. Korobov, Estimates for trigonometric sums and their applications, Uspechi Mat. Nauk 13 (1958), 185–192. [7] Serge Lange, Complex analysis, 4th edition, Springer, 1985. [8] David J. Newman, Simple analytic proof of the Prime Number Theorem, Amer. Math. Monthly 87 (1980), 693–696. [9] Charles Jean de la Vallée Poussin, Reserches analytiques sur la théorie des nombres premiers, Ann. Soc. Sci. Bruxelles 20 (1896), 183–256. [10] Bernhard Riemann, Über die Anzahl der Primzahlen unter einer gegebenen Grösse, Monatsberichte der Berliner Akademie, 1859. [11] Alte Selberg, An elementary proof of the Prime-Number Theorem, Ann. Math. 50 (1949), 305–313. [12] Elias M. Stein and Rami Shakarchi, Complex analysis, Princeton University Press, 2003. [13] Alfred Tauber, Ein Satz aus der Theorie der unendlichen Reihen, Monatsh f. Mathematik und Physik 8 (1897), 273–277. [14] Ivan M. Vinogradov, A new estimate of the function ζ(1 + it), Izv. Akad. Nauk SSSR. Ser. Mat. 22 (1958), 161–164. [15] Don Zagier, Newman’s short proof of the Prime Number Theorem, Amer. Math. Monthly 104 (1997), 705–708. 18.785 Fall 2021, Lecture #16, Page 14 18.785 Number theory I Lecture #16 16 Fall 2021 11/3/2021 Riemann’s zeta function and the prime number theorem We now divert our attention from algebraic number theory to talk about zeta functions and L-functions. As we shall see, every global field has a zeta function that is intimately related to the distribution of its primes. We begin with the zeta function of the rational field Q, which we will use to prove the prime number theorem. We will need some basic results from complex analysis, all of which can be found in any introductory textbook (such as [1, 2, 3, 7, 12]). A short glossary of terms and a list of the basic theorems we will use can be found at the end of these notes.1 16.1 The Riemann zeta function Definition 16.1. The Riemann zeta function is the complex function defined by the series X ζ(s) := n−s , n≥1 for Re(s) > 1, where n varies over positive integers. It is easy to verify that this series converges absolutely and locally uniformly on Re(s) > 1 (use the integral test on an open ball strictly to the right of the line Re(s) = 1). By Theorem 16.17, it defines a holomorphic function on Re(s) > 1, since each term n−s = e−s log n is holomorphic. Theorem 16.2 (Euler product). For Re(s) > 1 we have X Y ζ(s) = (1 − p−s )−1 , n−s = p n≥1 where the product converges absolutely. In particular, ζ(s) 6= 0 for Re(s) > 1. The product in the theorem above ranges over primes p. This is a standard practice in analytic number theory that we will follow: the symbol p always denotes a prime, and any sum or product over p is understood to be over primes, even if this is not explicitly stated. Proof. We have X n−s = XY p−vp (n)s = n≥1 p n≥1 YX p e≥0 p−es = Y p (1 − p−s )−1 . To justify the second equality, consider the partial zeta function ζm (s), which restricts the summation in ζ(s) to the set Sm of m-smooth integers (those with no prime factors p > m). If p1 , . . . , pk are the primes up to m, absolute convergence implies X Y X X Y ei ζm (s) := (pe11 · · · pekk )−s = (p−s ) = (1 − p−s )−1 . n−s = i e1 ,...,ek ≥0 n∈Sm 1≤i≤k ei ≥0 p≤m For any δ > 0 the sequence of functions ζm (s) converges uniformly on Re(s) > 1 + δ to ζ(s); indeed, for any  > 0 and any such s we have |ζm (s) − ζ(s)| ≤ 1 X n≥m n−s ≤ X n≥m |n−s | = X n≥m n− Re(s) ≤ Z ∞ m 1 x−1−δ dx ≤ m−δ < , δ Those familiar with this material should still glance at §16.3.2 which touches on some convergence issues that are particularly relevant to number theoretic applications. for all sufficiently large m. It follows that the sequence ζQ m (s) converges locally uniformly to ζ(s) on Re(s) > 1. The sequence of functions Pm (s) := p≤m (1 − p−s )−1 clearly converges Q locally uniformly to (1 − p−s )−1 on any region in which the latter function is absolutely convergent (or even just convergent). For any s in Re(s) > 1 we have X p | log(1 − p−s )−1 | = X X1 X XX |p−s |e = (|ps | − 1)−1 < ∞, p−es ≤ e p p p e≥1 e≥1 P where we have used the identity log(1 − z) = − n≥1 n1 z n , valid for |z| < 1. It follows that Q −s −1 is absolutely convergent (and in particular, nonzero) on Re(s) > 1. p (1 − p ) Theorem 16.3 (Analytic continuation I). For Re(s) > 1 we have ζ(s) = 1 + φ(s), s−1 where φ(s) is a holomorphic function on Re(s) > 0. Thus ζ(s) extends to a meromorphic function on Re(s) > 0 that has a simple pole at s = 1 with residue 1 and no other poles. Proof. For Re(s) > 1 we have X 1 ζ(s) − = n−s − s−1 n≥1 Z 1 ∞ x −s Z X dx = n−s − n≥1 n+1 x −s n dx  = XZ n+1 n≥1 n
n−s − x−s dx. R n+1 For each n ≥ 1 the function φn (s) := n (n−s − x−s )dx is holomorphic on Re(s) > 0. For each fixed s in Re(s) > 0 and x ∈ [n, n + 1] we have Z x Z x Z x |s| |s| |s| −s −s −s−1 |n − x | = st dt ≤ dt = dt ≤ 1+Re(s) , s+1 1+Re(s) | n n n |t n t and therefore |φn (s)| ≤ Z n+1 n n−s − x−s dx ≤ |s| n1+Re(s) . For any s0 with Re(s0 ) > 0, if we put  := Re(s0 )/2 and U := B< (s0 ), then for each n ≥ 1, |s0 | +  =: Mn , n1+ s∈U P P and n Mn = (|s0 | + )ζ(1 + ) converges. The series n φn thus P converges locally normally on Re(s) > 0. By the Weierstrass M -test (Theorem 16.19), n φn converges to a function 1 that is holomorphic on Re(s) > 0. φ(s) = ζ(s) − s−1 sup |φn (s)| ≤ We now show that ζ(s) has no zeros on Re(s) = 1; this fact is crucial to the prime number theorem. For this we use the following ingenious lemma, attributed to Mertens.2 Lemma 16.4 (Mertens). For x, y ∈ R with x > 1 we have |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 1. 2 If this lemma strikes you as pulling a rabbit out of a hat, well, it is. For a slight variation, see [15, IV], which uses an alternative approach due to Hadamard. 18.785 Fall 2021, Lecture #16, Page 2 Proof. From the Euler product ζ(s) = log |ζ(s)| = − X p Q − p−s )−1 , we see that for Re(s) > 1 we have p (1 log |1 − p−s | = − since log |z| = Re log z and log(1 − z) = − X p P log |ζ(x + iy)| = Re log(1 − p−s ) = zn n≥1 n X X Re(p−ns ) p n≥1 n , for |z| < 1. Plugging in s = x + iy yields X X cos(ny log p) p n≥1 npnx , since Re(p−ns ) = p−nx Re(e−iny log p ) = p−nx cos(−ny log p) = p−nx cos(ny log p). Thus log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = X X 3 + 4 cos(ny log p) + cos(2ny log p) p n≥1 npnx . We now note that the trigonometric identity cos(2θ) = 2 cos2 θ − 1 implies 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ≥ 0. Taking θ = ny log p yields log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 0, which proves the lemma. Corollary 16.5. ζ(s) has no zeros on Re(s) ≥ 1. Proof. We know from Theorem 16.2 that ζ(s) has no zeros on Re(s) > 1, so suppose ζ(1 + iy) = 0 for some y ∈ R. Then y 6= 0, since ζ(s) has a pole at s = 1, and we know that ζ(s) does not have a pole at 1 + 2iy 6= 1, by Theorem 16.3. We therefore must have lim |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = 0, (1) x→1 since ζ(s) has a simple pole at s = 1, a zero at 1 + iy, and no pole at 1 + 2iy. But this contradicts Lemma 16.4. 16.2 The Prime Number Theorem The prime counting function π : R → Z≥0 is defined by X π(x) := 1; p≤x it counts the number of primes up to x. The prime number theorem (PNT) states that π(x) ∼ x . log x The notation f (x) ∼ g(x) means limx→∞ f (x)/g(x) = 1; one says that f is asymptotic to g. This conjectured growth rate for π(x) dates back to Gauss and Legendre in the late 18th century. In fact Gauss believed the asymptotically equivalent but more accurate statement3 Z x dt . π(x) ∼ Li(x) := 2 log t 3 More accurate in the sense that |π(x) − Li(x)| grows more slowly than |π(x) − x | log x as x → ∞. 18.785 Fall 2021, Lecture #16, Page 3 However it was not until a century later that the prime number theorem was independently proved by Hadamard [5] and de la Vallée Poussin [9] in 1896. Their proofs are both based on the work of Riemann [10], who in 1860 showed that there is a precise connection between the zeros of ζ(s) and the distribution of primes (we shall say more about this later), but was unable to prove the prime number theorem. The proof we will give is more recent and due to Newman [8], but it relies on the same properties of the Riemann zeta function that were exploited by both Hadamard and de la Vallée, the most essential of which is the fact that ζ(s) has no zeros on Re(s) ≥ 1 (Corollary 16.5). A concise version of Newman’s proof by Zagier can be found in [15]; we will follow Zagier’s outline but be slightly more expansive in our presentation. We should note that there are also “elementary" proofs of the prime number theorem independently obtained by Erdös [4] and Selberg [11] in the 1940s that do not use the Riemann zeta function, but they are elementary only in the sense that they do not use complex analysis; the details of these proofs are considerably more complicated than the one we will give. Rather than work directly with π(x), it is more convenient to work with the log-weighted prime-counting function defined by Chebyshev4 X ϑ(x) := log p, p≤x whose growth rate differs from that of π(x) by a logarithmic factor. Theorem 16.6 (Chebyshev). π(x) ∼ x log x if and only if ϑ(x) ∼ x. Proof. We clearly have 0 ≤ ϑ(x) ≤ π(x) log x, thus ϑ(x) π(x) log x ≤ . x x For every  ∈ (0, 1) we have ϑ(x) ≥ X   log p ≥ (1 − )(log x) π(x) − π(x1− ) x1− 1 and all  > 0 we have |F (λx) − F (x)| <  for all sufficiently large x. Fix λ > 1 and suppose there is an unbounded sequence (xn ) such that f (xn ) ≥ λxn for all n ≥ 1. For each xn we have Z λxn Z λxn Z λ f (t) − t λxn − t λ−t F (λxn ) − F (xn ) = dt ≥ dt = dt = c, 2 2 t t t2 xn xn 1 for some c > 0, where we used the fact that f is non-decreasing to get the middle inequality. Taking  < c, we have |F (λxn ) − F (xn )| = c >  for arbitrarily large xn , a contradiction. Thus f (x) < λx for all sufficiently large x. A similar argument shows that f (x) > λ1 x for all sufficiently large x. These inequalities hold for all λ > 1, so limx→∞ f (x)/x = 1. Equivalently, f (x) ∼ x. 5 The equality sign in the big-O notation f (x) = O(g(x)) is a standard abuse of notation; it simply means lim supx→∞ |f (x)|/|g(x)| < ∞ (and nothing more). In more complicated equalities a big-O expression should P be interpreted as a set of functions, one of which makes the equality true, for example, n≥1 n1 = log n+O(1). 18.785 Fall 2021, Lecture #16, Page 5 In order to show that the hypothesis of Lemma 16.8 is satisfied for f = ϑ, we will work with the function H(t) = ϑ(et )e−t − 1; the change of variables t = eu shows that Z ∞ Z ∞ ϑ(t) − t H(u)du converges . dt converges ⇐⇒ t2 1 0 We now recall the Laplace transform. Definition 16.9. Let h : R>0 → R be a piecewise continuous function. The Laplace transform Lh of h is the complex function defined by Z ∞ Lh(s) := e−st h(t)dt, 0 which is holomorphic on Re(s) > c for any c ∈ R for which h(t) = O(ect ). The following properties of the Laplace transform are easily verified. • L(g + h) = Lg + Lh, and for any a ∈ R we have L(ah) = aLh. • If h(t) = a ∈ R is constant then Lh(s) = as . • L(eat h(t))(s) = L(h)(s − a) for all a ∈ R. We now define the auxiliary function Φ(s) := X p−s log p, p which is related to ϑ(x) by the following lemma. Lemma 16.10. L(ϑ(et ))(s) = Φ(s) s is holomorphic on Re(s) > 1. Proof. By Lemma 16.7, ϑ(et ) = O(et ), so L(ϑ(et )) is holomorphic on Re(s) > 1. Let pn be the nth prime, and put p0 := 0. The function ϑ(et ) is constant on t ∈ (log pn , log pn+1 ), so Z log pn+1 Z log pn+1   1 −s e−st ϑ(et )dt = ϑ(pn ) e−st dt = ϑ(pn ) p−s − p n n+1 . s log pn log pn We then have (Lϑ(et ))(s) = Z ∞ e−st ϑ(et )dt = 0 ∞   1X −s ϑ(pn ) p−s − p n n+1 s n=1 ∞ ∞ 1X 1X = ϑ(pn )p−s ϑ(pn−1 )p−s n − n s s = = 1 s 1 s n=1 ∞  X n=1 ∞ X n=1  ϑ(pn ) − ϑ(pn−1 ) p−s n p−s n log pn = n=1 Φ(s) . s Let us now consider the function H(t) := ϑ(et )e−t − 1. It follows from the lemma and standard properties of the Laplace transform that on Re(s) > 0 we have LH(s) = L(ϑ(et )e−t )(s) − (L1)(s) = L(ϑ(et ))(s + 1) − 1 Φ(s + 1) 1 = − . s s+1 s 18.785 Fall 2021, Lecture #16, Page 6 Lemma 16.11. The function Φ(s) − that is holomorphic on Re(s) ≥ 1. 1 s−1 extends to a meromorphic function on Re(s) > 1 2 Proof. By Theorem 16.3, ζ(s) extends to a meromorphic function on Re(s) > 0, which we also denote ζ(s), that has only a simple pole at s = 1 and no zeros on Re(s) ≥ 1, by Corollary 16.5. It follows that the logarithmic derivative ζ 0 (s)/ζ(s) of ζ(s) is meromorphic on Re(s) > 0, with no zeros on Re(s) ≥ 1 and only a simple pole at s = 1 with residue −1 (see §16.3.1 for standard facts about the logarithmic derivative of a meromorphic function). In terms of the Euler product, for Re(s) > 1 we have6 !0 !0 Y X
0 ζ 0 (s) − = − log ζ(s) = − log (1 − p−s )−1 = log(1 − p−s ) ζ(s) p p  X log p X 1 X p−s log p 1 = + log p = = 1 − p−s ps − 1 ps ps (ps − 1) p p p X log p = Φ(s) + . ps (ps − 1) p The sum on the RHS converges absolutely and locally uniformly to a holomorphic function on Re(s) > 1/2. The LHS is meromorphic on Re(s) > 0, and on Re(s) ≥ 1 it has only a 1 simple pole at s = 1 with residue 1. It follows that Φ(s) − s−1 extends to a meromorphic 1 function on Re(s) > 2 that is holomorphic on Re(s) ≥ 1. 1 Corollary 16.12. The functions Φ(s + 1) − 1s and (LH)(s) = Φ(s+1) s+1 − s both extend to meromorphic functions on Re(s) > − 12 that are holomorphic on Re(s) ≥ 0. Proof. The first statement follows immediately from the lemma. For the second, note that   Φ(s + 1) 1 1 1 1 − = Φ(s + 1) − − s+1 s s+1 s s+1 is meromorphic on Re(s) > − 21 and holomorphic on Re(s) ≥ 0, since it is a sum of products of such functions. The final step of the proof relies on the following analytic result due to Newman [8]. Theorem 16.13. Let f : R≥0 → R be a bounded piecewise continuous function, and suppose its R ∞Laplace transform extends to a holomorphic function g(s) on Re(s) ≥ 0. Then the integral 0 f (t)dt converges and is equal to g(0). Proof. Without loss of generality weR assume f (t) ≤ 1 for all t ≥ 0. For τ ∈ R>0 , define Rτ ∞ gτ (s) := 0 f (t)e−st dt, By definition 0 f (t)dt = limτ →∞ gτ (0), thus it suffices to prove lim gτ (0) = g(0). τ →∞ For r > 0, let γr be the boundary of the region {s : |s| ≤ r and Re(s) ≥ −δr } with δr > 0 chosen so that g is holomorphic on γr ; such a δr exists because g is holomorphic on Re(s) ≥ 0, hence on some open ball B≤2δ(y) (iy) for each y ∈ [−r, r], and we may take 6 As is standard when computing logarithmic derivatives, we are taking the principal branch of the complex logarithm and can safely ignore the negative real axis where it is not defined since we are assuming Re(s) > 1. 18.785 Fall 2021, Lecture #16, Page 7 δr := inf{δ(y) : y ∈ [r, −r]}, which is positive because [−r, r] is compact. Each γr is a 2 simple closed curve, and for each τ > 0 the function h(s) := (g(s) − gτ (s))esτ (1 + rs2 ) is holomorphic on a region containing γr . Using Cauchy’s integral formula (Theorem 16.26) to evaluate h(0) yields  Z   1 1 s (2) g(0) − gτ (0) = h(0) = g(s) − gτ (s) esτ + 2 ds. 2πi γr s r We will show the LHS tends to 0 as τ → ∞ by showing that for any  > 0 we can set r = 3/ > 0 so that the absolute value of the RHS is less than  for all sufficiently large τ . Let γr+ denote the part of γr in Re(s) > 0, a semicircle of radius r. The integrand is absolutely bounded by 1/r on γr+ , since for |s| = r and Re(s) > 0 we have sτ g(s) − gτ (s) · e  1 s + 2 s r  = 1 2πi Z γr+  Z ∞ τ ∞ f (t)e−st dt · eRe(s)τ r s · + r s r eRe(s)τ 2 Re(s) · r r τ Re(s)τ − Re(s)τ e e 2 Re(s) = · · Re(s) r r 2 = 2/r . ≤ Therefore Z e− Re(s)t dt ·   1 s 2 1 1 sτ g(s) − gτ (s) e + 2 ds ≤ · πr · 2 = s r 2π r r (3) Now let γr− be the part of γr in Re(s) < 0, a truncated semi-circle. For any fixed r, the first term g(s)esτ (s−1 + sr−2 ) in the integrand of (2) tends to 0 as τ → ∞ for Re(s) < 0 and |s| ≤ r. For the second term we note that since gτ (s) is holomorphic on C, it makes no difference if we instead integrate over the semicircle of radius r in Re(s) < 0. For |s| = r and Re(s) < 0 we then have gτ (s)e sτ  1 s + 2 s r  = Z Z τ 0 τ f (t)e−st dt · eRe(s)τ r s · + r s r eRe(s)τ (−2 Re(s)) r r 0 ! − Re(s)τ Re(s)τ e e (−2 Re(s)) = 1− Re(s) r r ≤ e− Re(s)t dt · = 2/r2 · (1 − eRe(s)τ Re(s)), where the factor (1 − eRe(s)τ Re(s)) on the RHS tends to 1 as τ → ∞ since Re(s) < 0. We thus obtain the bound 1/r + o(1) when we replace γr+ with γr− in (3), and the RHS of (2) is bounded by 2/r + o(1) as τ → ∞. It follows that for any  > 0, for r = 3/ > 0 we have |g(0) − gτ (0)| < 3/r =  for all sufficiently large τ . Therefore limτ →∞ gτ (0) = g(0) as desired. 18.785 Fall 2021, Lecture #16, Page 8 Remark 16.14. Theorem 16.13 is an example of what is known as a Tauberian theorem. For a piecewise continuous function f : R≥0 → R, its Laplace transform Z ∞ Lf (s) := e−st f (t)dt, 0 is typically not defined on Re(s) ≤ c, where c is the least c for which f (t) = O(ect ). Now it may happen that the function Lf has an analytic continuation to a larger domain; for 1 example, if f (t) = et then (Lf )(s) = s−1 extends to a holomorphic function on C−{1}. But plugging values of s with Re(s) ≤ c into the integral usually does not work; in our f (t) = et example, the integral diverges on Re(s) ≤ 1. The theorem says that when Lf extends to a holomorphic function on the entire half-plane Re(s) ≥ 0, its value at s = 0 is exactly what we would get by simply plugging 0 into the integral defining Lf . More generally, Tauberian theorems refer to results related toRtransforms f → T (f ) that ∞ allow us to deduce properties of f (such as the convergence of 0 f (t)dt) from properties of T (f ) (such as analytic continuation to Re(s) ≥ 0). The term “Tauberian" was coined by Hardy and Littlewood and refers to Alfred Tauber, who proved a theorem of this type as a partial converse to a theorem of Abel. Theorem 16.15 (Prime Number Theorem). π(x) ∼ x log x . Proof. H(t) = ϑ(et )e−t − 1 is piecewise continuous and bounded, by Lemma 16.7, and its Laplace transform extends to a holomorphic function on Re(s) ≥ 0, by Corollary 16.12. Theorem 16.13 then implies that the integral Z ∞ Z ∞  H(t)dt = ϑ(et )e−t − 1 dt 0 0 converges. Replacing t with log x, we see that  Z ∞ Z ∞ 1 dx ϑ(x) − x ϑ(x) − 1 = dx x x x2 1 1 converges. Lemma 16.8 implies ϑ(x) ∼ x, equivalently, π(x) ∼ x log x , by Theorem 16.6. One disadvantage of our proof is that it does not give us an error term. Using more sophisticated methods, Korobov [6] and Vinogradov [14] independently obtained the bound ! x
, π(x) = Li(x) + O exp (log x)3/5+o(1) in which we note that the error term is bounded by O(x/(log x)n ) for all n but not by O(x1− ) for any  > 0. Assuming the Riemann Hypothesis, which states that the zeros of ζ(s) in the critical strip 0 < Re(s) < 1 all lie on the line Re(s) = 12 , one can prove π(x) = Li(x) + O(x1/2+o(1) ). More generally, if we knew that ζ(s) has no zeros in the critical strip with real part greater than c, for some c ≥ 1/2 strictly less than 1, we could prove π(x) = Li(x) + O(xc+o(1) ). There thus remains a large gap between what we can prove about the distribution of prime numbers and what we believe to be true. Remarkably, other than refinements to the o(1) term appearing in the Korobov-Vinogradov bound, essentially no progress has been made on this problem in the last 60 years. 18.785 Fall 2021, Lecture #16, Page 9 16.3 A quick recap of some basic complex analysis The complex numbers C are a topological field under the distance metric d(x, y) = |x − y| √ induced by the standard absolute value |z| := z z̄, which is also a norm on C as an Rvector space; all references to the topology on C (open, compact, convergence, limits, etc.) are made with this understanding. 16.3.1 Glossary of terms and standard theorems Let f and g denote complex functions defined on an open subset of C. • f is differentiable at z0 if limz→z0 f (z)−f (z0 ) z−z0 exists. • f is holomorphic at z0 if it is differentiable on an open neighborhood of z0 . • f is analytic at z0 if there of z0 in which f can be defined by P is an open neighborhood n a power series f (z) = n=0 an (z − z0 ) ; equivalently, f is infinitely differentiable and has a convergent Taylor series on an open neighborhood of z0 . • Theorem: f is holomorphic at z0 if and only if it is analytic at z0 . • Theorem: If C is a connected set containing a nonempty open set U and f and g are holomorphic on C with f|U = g|U , then f|C = g|C . • With U and C as above, if f is holomorphic on U and g is holomorphic on C with f|U = g|U , then g is the (unique) analytic continuation of f to C and f extends to g. • If f is holomorphic on a punctured open neighborhood of z0 and |f (z)| → ∞ as z → z0 then z0 is a pole of f ; note that the set of poles of f is necessarily a discrete set. • f is meromorphic at z0 if it is holomorphic at z0 or has z0 as a pole. • Theorem: at z0 then it can be defined by a Laurent series P If f is meromorphic n f (z) = n≥n0 an (z − z0 ) that converges on an open punctured neighborhood of z0 . • The order of vanishing ordz0 (f ) of a nonzero function f that is meromorphic at z0 is the least index n of the nonzero coefficients an in its Laurent series expansion at z0 . Thus z0 is a pole of f iff ordz0 (f ) < 0 and z0 is a zero of f iff ordz0 (f ) > 0. • If ordz0 (f ) = 1 then z0 is a simple zero of f , and if ordz0 (f ) = −1 it is a simple pole. • The residue resz0 (f ) of a function P f meromorphic at z0 is the coefficient a−1 in its Laurent series expansion f (z) = n≥n0 an (z − z0 )n at z0 . • Theorem: If z0 is a simple pole of f then resz0 (f ) = limz→z0 (z − z0 )f (z). • Theorem: If f is meromorphic on a set S then so is its logarithmic derivative f 0 /f , and f 0 /f has only simple poles in S and resz0 (f 0 /f ) = ordz0 (f ) for all z0 ∈ S. In particular the poles of f 0 /f are precisely the zeros and poles of f . 16.3.2 Convergence P P Recall that a series ∞ n=1 an of complex numbers converges absolutely if the series n |an | of nonnegative real numbers converges. An equivalent definition is that the function a(n) := an is integrable with respect to the counting measure µ on the set of positive integers N. Indeed, if the series is absolutely convergent then Z ∞ X an = a(n)µ, n=1 N 18.785 Fall 2021, Lecture #16, Page 10 and if the series is not absolutely convergent, the integral is not defined. Absolute convergence is effectively built-in to the definition of the Lebesgue integral, which requires that in order for the function a(n) = x(n) + iy(n) to be integrable, the positive real functions |x(n)| and |y(n)| must both be integrable (summable), and separately computes sums of the positive and negative subsequences of (x(n)) and (y(n)) as suprema over finite subsets. The measure-theoretic perspective has some distinct advantages. It makes it immediately clear that we may replace the index set N with any set of the same cardinality, since the counting measure depends only on the cardinality of N, not its ordering. We are thus free to sum over any countable index set, including Z, Q, any finite product of countable sets, and any countable coproduct of countable sets (such as countable direct sums of Z); such sums are ubiquitous in number theory and many cannot be meaningfully interpreted as limits of partial sums in the usual sense, since this assumes that the index set is well ordered (not the case with Q, for example). The measure-theoretic view makes P also makes it clear that we may convert any absolutely convergent sum• of the form X×Y into an iterated sum P P theorem. X Y (or vice versa), via Fubini’s Q We say that an infinite product is absolutely conn an of nonzero P Q complex numbers P vergent when the sum n log an is, in which case n an := exp( n log an ).7 This implies that an absolutely convergent product cannot converge to zero, and the sequence (an ) must converge to 1 (no matter how we order the an ). All of our remarks above about absolutely convergent series apply to absolutely convergent products as well. A series or product of complex functions fn (z) is absolutely convergent on S if the series or product of complex numbers fn (z0 ) is absolutely convergent for all z0 ∈ S. Definition 16.16. A sequence of complex functions (fn ) converges uniformly on S if there is a function f such that for every  > 0 there is an integer N for which supz∈S |fn (z)−f (z)| <  for all n ≥ N . The sequence (fn ) converges locally uniformly on S if every z0 ∈ S has an open neighborhood U for which (fn ) converges uniformly on U ∩S. When applied to a series of functions these terms refer to the sequence of partial sums. Because C is locally compact, locally uniform convergence is the same thing as compact convergence: a sequence of functions converges locally uniformly on S if and only if it converges uniformly on every compact subset of S. Theorem 16.17. A sequence or series of holomorphic functions fn that converges locally uniformly on an open set U converges to a holomorphic function f on U , and the sequence or series of derivatives fn0 then converges locally uniformly to f 0 (and if none of the fn has a zero in U and f 6= 0, then f has no zeros in U ). Proof. See [3, Thm. III.1.3] and [3, Thm. III.7.2]. P Definition 16.18. n (z) converges normally on a set S P P A series of complex functions n fP if n kfn k := n supz∈S |fn (z)| converges. The series n fnP (z) converges locally normally on S if every z0 ∈ S has an open neighborhood U on which n fn (z) converges normally. Theorem 16.19 (Weierstrass M-test). Every locally normally convergent series of P functions converges absolutely and locally uniformly. Moreover, a series n fn of holomorphic functions on converges locally normally converges to a holomorphic function f PS that 0 on S, and then n fn converges locally normally to f 0 . 7 In this definition we use the principal branch of log z := log |z| + i Arg z with Arg z ∈ (−π, π). 18.785 Fall 2021, Lecture #16, Page 11 Proof. See [3, Thm. III.1.6]. P Remark 16.20. To show a series n fn is locally normally convergent on a set S amounts to proving that for every z0 ∈ S there is an open neighborhood P U of z0 and a sequence of real numbers (Mn ) such that |fn (z)| ≤ Mn for z ∈ U ∩ S and n Mn < ∞, whence the term “M -test". 16.3.3 Contour integration We shall restrict our attention to integrals along contours defined by piecewise-smooth parameterized curves; this covers all the cases we shall need. Definition 16.21. A parameterized curve is a continuous function γ : [a, b] → C whose domain is a compact interval [a, b] ⊆ R. We say that γ is smooth if it has a continuous nonzero derivative on [a, b], and piecewise-smooth if [a, b] can be partitioned into finitely many subintervals on which the restriction of γ is smooth. We say that γ is closed if γ(a) = γ(b), and simple if it is injective on [a, b) and (a, b]. Henceforth we will use the term curve to refer to any piecewise-smooth parameterized curve γ, or to its oriented image of in the complex plane (directed from γ(a) to γ(b)), which we may also denote γ. Definition 16.22. Let f : Ω → C be a continuous function and let γ be a curve in Ω. We define the contour integral Z f (z)dz := γ Z b f (γ(t))γ 0 (t)dt, a whenever the integralR on the RHS (which is defined as a Riemann sum in the usual way) converges. Whether γ f (z)dz converges, and if so, to what value, does not depend on the parameterization of γ: ifR γ 0 is another parameterized curve with the same (oriented) image R as γ, then γ 0 f (z)dz = γ f (z)dz. We have the following analog of the fundamental theorem of calculus. Theorem 16.23. Let γ : [a, b] → C be a curve in an open set Ω and let f : Ω → C be a holomorphic function Then Z f 0 (z)dz = f (γ(b)) − f (γ(a)). γ Proof. See [2, Prop. 4.12]. Recall that the Jordan curve theorem implies that every simple closed curve γ partitions C into two components, one of which we may unambiguously designate as the interior (the one on the left as we travel along our oriented curve). We say that γ is contained in an open set U if both γ and its interior lie in U . The interior of γ is a simply connected set, and if an open set U contains γ then it contains a simply connected open set that contains γ. Theorem 16.24 (Cauchy’s Theorem). Let U be an open set containing a simple closed curve γ. For any function f that is holomorphic on U we have Z f (z)dz = 0. γ 18.785 Fall 2021, Lecture #16, Page 12 Proof. See [2, Thm. 8.6] (we can restrict U to a simply connected set). Cauchy’s theorem generalizes to meromorphic functions. Theorem 16.25 (Cauchy Residue Formula). Let U be an open set containing a simple closed curve γ. Let f be a function that is meromorphic on U , let z1 , . . . , zn be the poles of f that lie in the interior of γ, and suppose that no pole of f lies on γ. Then Z f (z)dz = 2πi γ n X reszi (f ). i=1 Proof. See [2, Thm. 10.5] (we can restrict U to a simply connected set). R it To see where the 2πi comes from, consider γ dz z with γ(t) = e for t ∈ [0, 2π]. In general one weights residues by a corresponding winding number, but the winding number of a simple closed curve about a point in its interior is always 1. Theorem 16.26 (Cauchy’s Integral Formula). Let U be an open set containing a simple closed curve γ. For any function f holomorphic on U and a in the interior of γ, Z 1 f (z) f (a) = dz. 2πi γ z − a Proof. Apply Cauchy’s residue formula to g(z) = f (z)/(z − a); the only poles of g in the interior of γ are a simple pole at z = a with resa (g) = f (a). Cauchy’s residue formula can also be used to recover the coefficients f (n) (a)/n! appearing in the Laurent series expansion of a meromorphic function at a (apply it to f (z)/(z −a)n+1 ). One of many useful consequences of this is Liouville’s theorem, which can be proved by showing that the Laurent series expansion of a bounded holomorphic function on C about any point has only one nonzero coefficient (the constant coefficient). Theorem 16.27 (Liouville’s theorem). Bounded entire functions are constant. Proof. See [2, Thm. 5.10]. We also have the following converse of Cauchy’s theorem. Theorem 16.28 (Morera’s Theorem). Let f be a continuous function and on an open set U , and suppose that for every simple closed curve γ contained in U we have Z f (z)dz = 0. γ Then f is holomorphic on U . Proof. See [3, Thm. II.3.5]. 18.785 Fall 2021, Lecture #16, Page 13 References [1] Lars V. Ahlfors, Complex analysis: an introduction to the theory of analytic functions of one complex variable, 3rd edition, McGraw-Hill, 1979. [2] Joseph Bak and Donald J. Newman, Complex analysis, Springer, 2010. [3] Rolf Busam and Eberhard Freitag, Complex analysis, 2nd edition, Springer 2009. [4] Paul Erdös, On a new method in elementary number theory which leads to an elementary proof of the prime number theorem, Proc. Nat. Acad. Scis. U.S.A. 35 (1949), 373–384. [5] Jacques Hadamard, Sur la distribution des zéros de la function ζ(s) et ses conséquences arithmétique, Bull. Soc. Math. France 24 (1896), 199–220. [6] Nikolai M. Korobov, Estimates for trigonometric sums and their applications, Uspechi Mat. Nauk 13 (1958), 185–192. [7] Serge Lange, Complex analysis, 4th edition, Springer, 1985. [8] David J. Newman, Simple analytic proof of the Prime Number Theorem, Amer. Math. Monthly 87 (1980), 693–696. [9] Charles Jean de la Vallée Poussin, Reserches analytiques sur la théorie des nombres premiers, Ann. Soc. Sci. Bruxelles 20 (1896), 183–256. [10] Bernhard Riemann, Über die Anzahl der Primzahlen unter einer gegebenen Grösse, Monatsberichte der Berliner Akademie, 1859. [11] Alte Selberg, An elementary proof of the Prime-Number Theorem, Ann. Math. 50 (1949), 305–313. [12] Elias M. Stein and Rami Shakarchi, Complex analysis, Princeton University Press, 2003. [13] Alfred Tauber, Ein Satz aus der Theorie der unendlichen Reihen, Monatsh f. Mathematik und Physik 8 (1897), 273–277. [14] Ivan M. Vinogradov, A new estimate of the function ζ(1 + it), Izv. Akad. Nauk SSSR. Ser. Mat. 22 (1958), 161–164. [15] Don Zagier, Newman’s short proof of the Prime Number Theorem, Amer. Math. Monthly 104 (1997), 705–708. 18.785 Fall 2021, Lecture #16, Page 14 18.785 Number theory I Lecture #18 18 Fall 2021 11/10/2021 Dirichlet L-functions, primes in arithmetic progressions Having proved the prime number theorem, we would like to prove an analogous result for primes in arithmetic progressions. We begin with Dirichlet’s theorem on primes in arithmetic progressions, a result that predates the prime number theorem by sixty years. Theorem 18.1 (Dirichlet 1837). For all coprime integers a and m there are infinitely many primes p ≡ a mod m. In fact Dirichlet proved more than this. In a sense that we will make precise , he proved that for every fixed modulus m the primes are equidistributed among the residue classes in (Z/mZ)× . The equidistribution statement that Dirichlet was able to prove is a bit weaker than one might like, but it is more than enough to establish Theorem 18.1. Remark 18.2. Many of the standard tools of complex analysis we take for granted were not available to Dirichlet in 1837. Riemann was the first to seriously study ζ(s) as a function of a complex variable, some twenty years after Dirichlet proved Theorem 18.1. We will work in a more modern setting, but our approach follows the spirit of Dirichlet’s proof. 18.1 Infinitely many primes To motivate Dirichlet’s method of proof, let us consider the following (admittedly clumsy) proof that there are infinitely many primes. It is sufficient to show that the Euler product Y ζ(s) = (1 − p−s )−1 p diverges as s → 1+ . Of course we know ζ(s) has a pole at s = 1 (by Theorem 16.3), but let us suppose for the moment that we did not already know this. Taking logarithms yields X X log ζ(s) = − log(1 − p−s ) = p−s + O(1), (1) p p as s → 1+ , where we have used the asymptotic bounds X − log(1 − x) = x + O(x2 ) (as x → 0) and O(p−2s ) = O(1) (Re(s) > 1/2 + ). p We can estimate P 1 p≤x p via Mertens’ second theorem, one of three he proved in [4]. Theorem 18.3 (Mertens 1874). As x → ∞ we have X log p 1 (1) p = log x + R(x), where |R(x)| < 2. p≤x (2) X 1 p = log log x + B + O p≤x (3) X p≤x log  1− p1   1 log x  , where B =0.261497 . . . is Mertens’ constant; = −log log x − γ + O  1 log x  , where γ =0.577216 . . . is Euler’s constant. Proof. See Problem Set 9. 1 In fact, R(x) = −B3 + o(1) where B3 =1.332582 . . . is an explicit constant. P −s Thus not only does p diverge as s → 1+ , we can say with a fair degree of precision how quickly this happens. We should note, however, that Mertens’ estimate is not as strong as the prime number theorem. Indeed, as you will prove on Problem Set 9, the Prime Number Theorem is equivalent to the statement   X1 = log log x + B + o log1 x , p p≤x which is (ever so slightly) sharper than Mertens’ estimate.2 18.2 Dirichlet characters We now define the notion of a Dirichlet character. Historically, these preceded the notion of a group character; they were introduced by Dirichlet in 1831, well before the notion of an abstract group was in common use.3 In order to simplify the exposition we will occasionally invoke some standard facts about characters of finite abelian groups that we recall in §18.6. Definition 18.4. A function f : Z → C is called an arithmetic function.4 The function f is multiplicative if f (1) = 1 and f (mn) = f (m)f (n) for all coprime m, n ∈ Z; it is totally multiplicative (or completely multiplicative) if f (1) = 1 and f (mn) = f (m)f (n) for all m, n ∈ Z. For m ∈ Z>0 we say that f is m-periodic if f (n + m) = f (n) for all n ∈ Z, and we call m the period of f it is the least m > 0 for which this holds. Definition 18.5. A Dirichlet character is a periodic totally multiplicative arithmetic function χ : Z → C. The image of a Dirichlet character is a finite multiplicatively closed subset of C, hence the union of a finite subgroup of U(1) and a subset of {0}. The constant function 1(n) := 1 is the trivial Dirichlet character ; it is the unique Dirichlet character of period 1. Each mperiodic Dirichlet character χ restricts to a group character χ on (Z/mZ)× . Conversely, every group character χ of (Z/mZ)× can be extended to a Dirichlet character χ by defining χ(n) = 0 for n 6⊥ m; this is called extension by zero. Definition 18.6. A Dirichlet character of modulus m is an m-periodic Dirichlet character χ that is the extension by zero of a group character on (Z/mZ)× ; equivalently, an m-periodic Dirichlet character for which χ(n) 6= 0 ⇐⇒ m ⊥ n. Remark 18.7. Some authors only define Dirichlet characters of modulus m, thinking of them as extensions by zero of group characters on (Z/mZ)× , in which case every χ has an attached modulus m. But note that the function Z → C given by the extension by zero does not uniquely determine m (see Lemma 18.8 below). Indeed, the unique Dirichlet character of modulus 2 is a Dirichlet character of modulus 2k for all k ≥ 1. The Dirichlet characters of modulus m form a group under pointwise multiplication that is canonically isomorphic to the character group of (Z/mZ)× . Not every m-periodic Dirichlet 2 The error term in the PNT actually implies P 1 p≤x p = log log x + B + O   1 , but an o( log1 x ) bound is x already enough to show π(x) ∼ x/ log x. That the difference between a little-o and a big-O is the difference between proving the PNT and not proving it demonstrates how critical it is to understand error terms. 3 Galois’ seminal paper was rejected that same year; it wasn’t published until 12 years after his death. 4 Many authors restrict the domain of an arithmetic function to Z≥1 ; for the periodic arithmetic functions we are interested in here, this distinction is irrelevant, and it is slightly more natural to work with Z. 18.785 Fall 2021, Lecture #18, Page 2 character χ is a Dirichlet character of modulus m, since an m-periodic Dirichlet character need not vanish on n 6⊥ m, but if χ has period m then this holds. More generally, we have the following lemma. Lemma 18.8. Let χ be a Dirichlet character of period m. Then χ is a Dirichlet character of modulus m0 if and only if m|m0 |mk for some k (which holds in particular for m0 = m). Proof. To prove χ is a a Dirichlet character of modulus m we must show χ(n) 6= 0 ⇔ m ⊥ n. Suppose χ(n) 6= 0 with m 6⊥ n, and let p be a common divisor of m and n. Then χ(p) 6= 0, since χ(p)χ(n/p) = χ(n) 6= 0, and for any r ∈ Z we have χ(r)χ(p) = χ(rp) = χ(rp + m) = χ(r + m/p)χ(p), which implies χ(r) = χ(r + m/p), since χ(p) 6= 0. Thus χ is (m/p)-periodic, but this contradicts the minimality of the period m. Therefore χ(n) 6= 0 ⇒ m ⊥ n. For any n ⊥ m we can pick a = ne ≡ 1 mod m so that χ(1) = χ(a) = χ(ne ) = χ(n)e 6= 0, in which case χ(n) 6= 0. Thus n ⊥ m ⇒ χ(n) 6= 0, so χ is a Dirichlet character of modulus m. If m|m0 |mk , then the prime divisors of m0 coincide with those of m. It follows that n ⊥ m0 ⇐⇒ n ⊥ m ⇐⇒ χ(n) 6= 0, and χ is clearly m0 -periodic (since m|m0 ), so χ is a Dirichlet character of modulus m0 . Conversely, if χ is a Dirichlet character of modulus m0 , then χ is m0 -periodic, and therefore m|m0 , since m is the period of χ. And since χ is a Dirichlet character of modulus m and of modulus m0 , for each prime p we have p|m ⇐⇒ χ(p) = 0 ⇐⇒ p|m0 , thus the prime divisors of m and m0 coincide and m0 must divide some power mk of m. 18.2.1 Primitive Dirichlet characters Given a Dirichlet character χ1 of modulus m1 dividing m2 , we can always create a Dirichlet character χ2 of modulus m2 by taking the extension by zero of the restriction of χ1 to (Z/m2 Z)× ; in other words, let χ2 (n) := χ1 (n) for n ∈ (Z/m2 Z)× and χ2 (n) := 0 otherwise. If m2 is divisible by a prime p that does not divide m1 , the Dirichlet characters χ1 and χ2 will not be the same (χ2 (p) = 0 6= χ1 (p), for example), they will agree on n ∈ (Z/m2 Z)× but not on n ∈ (Z/m1 Z)× .5 We can create infinitely many new Dirichlet characters from χ1 in this way, but they will differ from χ1 only in a rather trivial sense. We would like to distinguish the Dirichlet characters that arise in this way from those that do not. Definition 18.9. Let χ1 and χ2 be Dirichlet characters of modulus m1 and m2 , respectively, with m1 |m2 . If χ2 (n) = χ1 (n) for n ∈ (Z/m2 Z)× then χ2 is induced by χ1 . A Dirichlet character that is not induced by any character other than itself is primitive. Lemma 18.10. A Dirichlet character χ2 of modulus m2 is induced by a Dirichlet character of modulus m1 |m2 if and only if χ2 is constant on residue classes in (Z/m2 Z)× that are congruent modulo m1 . When this holds, the Dirichlet character χ1 of modulus m1 that induces χ2 is uniquely determined. 5 Note that while #(Z/m1 Z)× ≤ #(Z/m2 Z)× , the set of integers n ∈ (Z/m1 Z)× (the n coprime to m1 ) contains the set of integers n ∈ (Z/m2 Z)× (the n coprime to m2 ) and is usually larger. 18.785 Fall 2021, Lecture #18, Page 3 Proof. If χ2 is induced by χ1 then it must be constant on residue classes in (Z/m2 Z)× that are congruent modulo m1 , since χ1 is. To prove the converse we first show that the surjective ring homomorphism Z/m2 Z → Z/m1 Z given by reduction modulo m1 induces a surjective homomorphism π : (Z/m2 Z)× → (Z/m1 Z)× of unit groups.6 Suppose u1 ∈ Z is a unit modulo m1 . Let a be the product of all primes dividing m2 /m1 but not u1 . Then u2 = u1 + m1 a is not divisible by any prime p|m1 (since u1 isn’t), nor is it divisible by any prime p|(m2 /m1 ): by construction, such a p divides exactly one of u1 and m1 a. Thus u2 is a unit modulo m2 that reduces to u1 modulo m1 and π is surjective. If χ2 is a Dirichlet character of modulus m2 constant on fibers of π we can define a Dirichlet character χ1 of modulus m1 via χ1 (n1 ) := χ2 (n2 ) for n1 ∈ (Z/m1 Z)× with n2 ∈ π −1 (n1 ) (any such n2 will do). Thus χ1 induces χ2 , and if χ01 also induces χ2 it must satisfy the same condition χ1 (n1 ) = χ2 (n2 ) that uniquely determines χ1 . Definition 18.11. A Dirichlet character χ induced by 1 is called principal (and is primitive if and only if χ = 1). For m ∈ Z>0 we use 1m to denote the principal Dirichlet character of modulus m; it corresponds to the trivial character of (Z/mZ)× . Lemma 18.12. Let χ be a Dirichlet character of modulus m. Then X χ(n) 6= 0 ⇐⇒ χ = 1m . n ∈ Z/mZ Proof. We have χ(n) = 0 for n 6∈ (Z/mZ)× , and the sum over (Z/mZ)× is nonzero if and only if χ restricts to the trivial character on (Z/mZ)× , by the orthogonality of characters; see Corollary 18.38. Note that the principal Dirichlet characters 1m and 1m0 necessarily coincide when m|m0 |mk ; for example the principal Dirichlet character of modulus 2 (the parity function) is the same as the principal Dirichlet character of modulus 4 (and every power of 2). Theorem 18.13. Every Dirichlet character χ is induced by a primitive Dirichlet character χ e that is uniquely determined by χ. Proof. Let us define a partial ordering  on the set of all Dirichlet characters by defining χ1  χ2 if χ1 induces χ2 . The relation  is clearly reflexive, and it follows from Lemma 18.10 that it is transitive. Let χ be a Dirichlet character of period m and consider the set X = {χ0 : χ0  χ}. Each χ0 ∈ X necessarily has period m0 dividing m and there is at most one χ0 of period m0 for each divisor m0 of m, by Lemma 18.10. Thus X is finite, and nonempty (since χ ∈ X). Suppose χ1 , χ2 ∈ X have periods m1 and m2 , respectively. Then m1 and m2 both divide m, as does m3 = gcd(m1 , m2 ). We have a commutative square of surjective unit group homomorphisms induced by reduction maps: ←  (Z/m3 Z)× .   (Z/m2 Z)× ←  (Z/m1 Z)× ← ← (Z/mZ)× 6 In fact, one can show that every surjective homomorphism of finite rings induces a surjective homomorphism of unit groups, but this does not hold in general (consider Z → Z/5Z, for example). 18.785 Fall 2021, Lecture #18, Page 4 From Lemma 18.10 we know that χ is constant on residue classes in (Z/mZ)× that are congruent modulo either m1 or m2 , and therefore χ is constant on residue classes in (Z/mZ)× that are congruent modulo m3 , as are χ1 and χ2 (which are determined by χ). It follows that there is a unique Dirichlet character χ3 of modulus m3 that induces χ, χ1 , and χ2 . Thus every pair χ1 , χ2 ∈ X has a lower bound χ3 under the partial ordering  that is compatible with the total ordering of X by period. This implies that X contains a unique element χ e that is minimal, both with respect to the partial ordering  and with respect to the total ordering by period; it must be primitive, by the transitivity of . Definition 18.14. The conductor of a Dirichlet character χ is the period of the unique primitive Dirichlet character χ e that induces χ. P Corollary 18.15. For a Dirichlet character χ of modulus m we have n∈Z/mZ χ(n) 6= 0 if and only if χ has conductor 1. Proof. This follows immediately from Lemma 18.12. Corollary 18.16. Let M (m) denote the set of Dirichlet characters of modulus m, let X(m) b denote the set of primitive Dirichlet characters of conductor dividing m, and let G(m) denote × the character group of (Z/mZ) . We have canonical bijections ∼ ∼ b M (m) −→ X(m) −→ G(m) χ 7−→ χ e 7−→ (n 7→ χ e(n)). Proof. By Theorem 18.13, the map χ → χ e is injective, and it is also surjective: each χ e ∈ X(m) induces the character χ ∈ M (m) by setting χ(n) := χ e(n) for n ∈ (Z/mZ)× and extending by zero. As previously noted, the map χ → (m 7→ χ(m)) defines a bijection b M → G(m) (a group isomorphism, in fact), and this bijection factors through the map χ 7→ χ e, since χ e(n) = χ(n) for n ∈ (Z/mZ)× . Remark 18.17. Corollary 18.16 implies that we can make X(m) a group by defining χ e1 χ e2 := χ ] ] e1 and χ e2 (which is typically 1 χ2 . Note that χ 1 χ2 is not the pointwise product of χ not primitive), it is the unique primitive character that induces the pointwise product. Example 18.18. 12-periodic Dirichlet characters, ordered by period m and conductor c. m 1 2 3 3 4 6 6 12 12 c 1 1 1 3 4 1 3 4 12 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 1 0 1 -1 0 0 0 0 0 3 1 1 0 0 -1 0 0 0 0 4 1 0 1 1 0 0 0 0 0 5 1 1 1 -1 1 1 -1 1 -1 6 1 0 0 0 0 0 0 0 0 7 1 1 1 1 -1 1 1 -1 -1 8 1 0 1 -1 0 0 0 0 0 9 1 1 0 0 1 0 0 0 0 10 1 0 1 1 0 0 0 0 0 11 1 1 1 -1 -1 1 -1 -1 1 mod-12 no no no no no yes yes yes yes principal yes yes yes no no yes no no no primitive yes no no yes yes no no no yes The fact that χ(n) ∈ {0, ±1} for all 12-periodic Dirichlet characters χ follows from the fact that the exponent of (Z/mZ)× is 2; thus (im χ) ∩ U(1) ⊆ µ2 = {±1}. 18.785 Fall 2021, Lecture #18, Page 5 18.3 Dirichlet L-functions Definition 18.19. The Dirichlet L-function associated to a Dirichlet character χ is Y X χ(n)n−s . L(s, χ) := (1 − χ(p)p−s )−1 = p n≥1 The sum and product converge absolutely for Re s > 1, since |χ(n)| ≤ 1, thus L(s, χ) is holomorphic on Re(s) > 1. For the trivial Dirichlet character 1 we have L(s, 1) = ζ(s). For the principal character 1m of modulus m induced by 1 we have ζ(s) = L(s, 1m ) Y p|m (1 − p−s )−1 . The product on the RHS is finite, hence bounded and nonzero as s → 1+ , so the L-function L(s, 1m ) has a simple pole at s = 1 with residue Y Y φ(m) ress=1 L(s, 1m ) = lim (s − 1)ζ(s) (1 − p−s ) = (1 − p−1 ) = . + m s→1 p|m p|m The L-functions of non-principal Dirichlet characters do not have a pole at s = 1. Proposition 18.20. Let χ be a non-principal Dirichlet character of modulus m. Then L(s, χ) extends to a holomorphic function on Re s > 0. Proof. Define the function T : R≥0 → C by T (x) := X χ(n). 0 0, since it is the limit of the uniformly converging sequence of functions φn (s) := s 0 T (x)x−s−1 dx (here we use the fact that T (x) is bounded), and is thus the analytic continuation of L(x, χ) to Re(s) > 0. Remark 18.21. In fact, L(s, χ) extends to a holomorphic function on C whenever χ is non-principal. 18.785 Fall 2021, Lecture #18, Page 6 18.4 Primes in arithmetic progressions We now return to our goal of proving Dirichlet’s theorem on primes in arithmetic progressions. It suffices to show that for any a ⊥ m the sum X p−s p ≡ a mod m is unbounded as s → 1+ . To convert this to a sum over all primes we use Proposition 18.37 to construct the indicator function ( X 1 if p ≡ a mod m, 1 χ(p/a) = φ(m) 0 otherwise χ∈X(m) where p/a is computed modulo m and χ ranges over primitive Dirichlet characters of conductor dividing m, which we identify with the character group of (Z/mZ)× via Corollary 18.16. As s → 1+ we have X X X 1 p−s = p−s χ(p/a) φ(m) p p ≡ a mod m χ∈X(m) = X χ∈X(m) = X χ∈X(m) = χ(1/a) X χ(p)p−s φ(m) p  χ(1/a)  log L(s, χ) + O(1) φ(m) X log ζ(s) + φ(m) χ∈X(m) χ6=1 χ(1/a) log L(s, χ) + O(1). φ(m) We now make the key claim that so long as χ is not principal, we have L(1, χ) 6= 0. This implies that log L(s, χ) = O(1) as s → 1+ and therefore X p−s = p ≡ a mod m log ζ(s) + O(1) φ(m) is unbounded as s → 1+ , since ζ(s) is. Moreover, Mertens’ second theorem implies X p≤x p ≡ a mod m 1 log log x ∼ . p φ(m) which proves that there are infinitely many primes p ≡ a mod m. We can makes this statement more precise by using the notion of Dirichlet density. Definition 18.22. The Dirichlet density of a set of primes S is given by P −s p∈S p d(S) := lim P −s , s→1+ pp 18.785 Fall 2021, Lecture #18, Page 7 defined whenever this limit exists (one can also define notions of lower and upper Dirichlet density using lim inf and lim sup which both agree with d(S) whenever it is defined). This definition differs from the more common notion of natural density #{p ≤ x : p ∈ S} . x→∞ #{p ≤ x} δ(S) := lim The Dirichlet density of the set S = {p ≡ a mod m} is P −s log ζ(s)/φ(m) 1 p∈S p d(S) = lim P −s = lim = . log ζ(s) φ(m) s→1+ s→1+ pp The primes are thus equidistributed modulo m in the sense that for all a ⊥ m we have X p≤x p ≡ a mod m 1 1 X1 1 ∼ ∼ log log x. p φ(m) p φ(m) p≤x We should note that this statement is weaker than the prime number theorem for arithmetic progressions, which states that π(x; m, a) ∼ 1 π(x), φ(m) where π(x; m, a) counts the primes p ≤ x for which p ≡ a mod m (see Problem Set 9). 18.5 Stieltjes integrals For the benefit of those who have not seen them before, we recall a few facts about Stieltjes integrals (also called Riemann-Stieltjes integrals), taken from [1, Ch. 7]. These generalize the Riemann integral but are less general than the Lebesgue integral; they provide a handy way for converting sums to integrals that is often used in analytic number theory. Definition 18.23. Let f and g be (real or complex valued) functions defined on a nonempty real interval [a, b]. For any partition P = (x0 , . . . , xn ) of [a, b] and sequence T = (t1 , . . . , tn ) with tk ∈ [xk−1 , xk ], we define the Riemann-Stieltjes sum S(P, T, f, g) := n X k=1   f (tk ) g(xk ) − g(xk−1 ) We say that f is Riemann-Stieltjes integrable with respect to g and write f ∈ S(g) if there is a (real or complex) number S such that for every  > 0 there is a partition P of [a, b] such that for every refinement P = (x0 , . . . , xn ) of P and every sequence T = (t1 , . . . , tn ) with tk ∈ [xk−1 , xk ] we have |S(P, T, f, g) − S| < .7 Rb When such an S exists it is necessarily unique and we denote it by a f dg, the RiemannStieltjes integral of f with respect to g. Improper Riemann-Stieltjes integrals are then defined as limits Z Z ∞ a b f dg := lim b→∞ a (and similarly for the lower limit), and we define 7 Ra b f dg f dg = − Rb a f dg and Ra a f dg = 0. This definition (due to Pollard) is more general than that originally given by Stieltjes but is now standard. 18.785 Fall 2021, Lecture #18, Page 8 Taking g(x) = x yields the Riemann integral. The Riemann-Stieltjes integral satisfies the usual properties of linearity, summability, and integration by parts. Proposition 18.24. Let f, g, and h be functions on [a, b] and let c1 and c2 be constants. The following hold: Rb Rb Rb • If f, g ∈ S(h) then a (c1 f + c2 g) dh = c1 a f dh + c2 a g dh. Rb Rb Rb • If f ∈ S(g), S(h) then a f d(c1 g + c2 h) = ci a f dg + c2 a f dh. Rb Rc Rb • If f ∈ S(g) then for any c ∈ [a, b] we have a f dg = a f dg + c f dg. Rb Rb • If f ∈ S(g) then g ∈ S(f ) and a f dg + a g df = f (b)g(b) − f (a)g(a). • If f = f1 + if2 and g = g1 + ig2 with f1 , f2 ∈ S(g1 ), S(g2 ) then Z b f dg = a Z b a f1 dg1 − Z b f2 dg2 a  +i Z a b f2 dg1 + Z a b  f1 dg2 . Proof. See [1, Thm. 7.2-7,7.50]. The last identity allows us to reduce complex-valued integrals to real-valued integrals. The following proposition allows us to reduce Stieltjes integrals to Riemann integrals. Proposition 18.25. Let f and g be real-valued functions on [a, b] and suppose g has a continuous derivative g 0 on [a, b]. Then Z b f dg = a Z b f (x)g 0 (x)dx. a Proof. See [1, Thm. 7.8]. Rb A key advantage of the Stieltjes integral a f dg is that neither the integrand f nor the integrator g is required to be continuous. It suffices for f and g to be of bounded variation and not share any discontinuities (and they can even share certain discontinuities, see Theorem 18.27). Definition 18.26. Let f be a (real or complex valued) function defined on a nonempty real interval [a, b]. Then f is of bounded variation if there exists a real number M such that n−1 X i=0 |f (xi+1 ) − f (xi )| < M for every partition P = (x0 , . . . , xn ) of [a, b]. If f has a continuous derivative f 0 on [a, b] Rb this is equivalent to requiring a |f 0 (x)|dx < ∞. Every piecewise monotone function is of bounded variation. In particular, any step function with finitely many discontinuities on [a, b] is of bounded variation. Theorem 18.27. Let f and g be functions on [a, b] of bounded variation such that for every c ∈ [a, b] the function f is continuous from the left at c and the function g is continuous Rb Rb from the right at c. Then a f dg and a g df both exist. Proof. See [2, Thm. 3.7]. 18.785 Fall 2021, Lecture #18, Page 9 Corollary 18.28. Let f and g be functions on [a, b] such that f and g are not Pboth discontinuous from the left or from the right at integers n ∈ [a, b], and let G(x) = a 1 since {p|p} ≤ n := [K : Q] and N(p) := [OK : p] ≥ p imply that X X log(1 − N(p)−s ) ≤ n log(1 − p−s ) , p p and the sum on the RHS converges on Re(s) > 1 since we know this holds for ζQ (s) = ζ(s). The following theorem is often attributed to Dirichlet, although he originally proved it only for quadratic fields (this is all he needed to prove his theorem on primes in arithmetic progressions, but we will use it in a stronger form). The formula for the limit in the theorem was proved by Dedekind [2, Supplement XI] (as a limit from the right, without an analytic continuation to a punctured neighborhood of z = 1), and analytic continuation was proved by Landau [3]. Hecke later showed that, like the Riemann zeta function, the Dedekind zeta function has an analytic continuation to all of C and satisfies a functional equation [1], but we won’t take the time to prove this; see Remark §19.13 for details. Theorem (Analytic Class Number Formula). Let K be a number field of degree n. The Dedekind zeta function ζK (z) extends to a meromorphic function on Re(z) > 1 − n1 that is holomorphic except for a simple pole at z = 1 with residue lim (z − 1)ζK (z) = z→1+ 2r (2π)s hK RK , wK |DK |1/2 where r and s are the number of real and complex places of K, respectively, hK := # cl OK is the class number, RK is the regulator, wK := #µK is the number of roots of unity, and DK := disc OK is the absolute discriminant. Recall that |DK |1/2 is the covolume of OK as a lattice in KR := K ⊗Q R ' Rr × Cs × (Proposition 14.16), and RK is the covolume of ΛK := Log(OK ) as a lattice in the trace-zero r+s hyperplane R0 (see Definition 15.16). The residue of ζK (z) at z = 1 thus reflects both the additive and multiplicative structure of the ring of integers OK . Remark 19.1. In practice the class number hK is usually the most difficult quantity in the analytic class number formula to compute. We can approximate the limit on the LHS to any desired precision using a finite truncation of either the sum or product defining ζK (s). Provided we can compute the other quantities to similar precision, this provides a method for computing (or at least bounding) the class number hK ; this explains the origin of the term “analytic class number formula”. You will have an opportunity to explore a computational application of this formula on Problem Set 9. Example 19.2. For K = Q we have n = 1, r = 1, s = 0, h = 1, w = #{±1} = 2, D = 1, and the regulator R is the covolume of a lattice in a zero-dimensional vector space, equivalently, the determinant of a 0 × 0 matrix, which is 1. In this case the theorem states that ζQ (z) = ζ(z) is holomorphic on Re z > 1 − 11 = 0 except for a simple pole at z = 1 with residue 21 (2π)0 · 1 · 1 = 1. lim (z − 1)ζQ (z) = z→1+ 2 · |1|1/2 19.1 Lipschitz parametrizability In order to prove the analytic class number formula we need an asymptotic estimate for the number of OK -ideals a with absolute norm N(a) bounded by a parameter t ∈ R>0 that we P will let tend to infinity; this is necessary for us to understand the behavior of ζK (z) = a N(a)−z as z → 1+ . Our strategy is to count points in Log(OK ∩ K × ) that lie inside a suitably chosen region S of Rr+s that we will than scale by t. In order to bound this count as a function of t we need a condition on S that ensures that the count grows smoothly with t; this requires S to have a “reasonable" shape. A sufficient condition for this is Lipschitz parametrizability. Definition 19.3. Let X and Y be metric spaces. A function f : X → Y is Lipschitz continuous if there exists c ∈ R>0 such that for all x1 , x2 ∈ X we have d(f (x1 ), f (x2 )) ≤ c · d(x1 , x2 ). Every Lipschitz continuous function is uniformly continuous, but the converse need not √ hold. For example,p the function f (x) = x on [0, 1] is uniformly continuous but not Lipschitz √ continuous, since | 1/n − 0|/|1/n − 0| = n is unbounded as 1/n → 0. Definition 19.4. A set B in a metric space X is d-Lipschitz parametrizable if it is the union of the images of a finite number of Lipschitz continuous functions fi : [0, 1]d → X. Before stating our next result, we recall the asymptotic notation f (t) = g(t) + O(h(t)) (as t → a), for real or complex valued functions f, g, h of a real variable t, which means lim sup t→a f (t) − g(t) < ∞. h(t) Typically a = ∞, and this is assumed if a is not specified. Lemma 19.5. Let S ⊆ Rn be a measurable set whose boundary ∂S := S − S 0 is (n − 1)Lipschitz parametrizable. Then #(tS ∩ Zn ) = µ(S)tn + O(tn−1 ), as t → ∞, where µ is the standard Lebesgue measure on Rn . Proof. It suffices to prove the lemma for positive integers, since #(tS ∩ Zn ) and µ(S)tn are both monotonically increasing functions of t and µ(S)(t + 1)n − µ(S)tn = O(tn−1 ). We can partition Rn as the disjoint union of half-open cubes of the form C(a1 , . . . , an ) = {(x1 , . . . , xn ) ∈ Rn : xi ∈ [ai , ai + 1)}, 18.785 Fall 2021, Lecture #19, Page 2 with a1 , . . . , an ∈ Z. Let C be the set of all such half-open cubes C. For each t > 0 define B0 (t) := #{C ∈ C : C ⊆ tS}, B1 (t) := #{C ∈ C : C ∩ tS 6= ∅}. For every t > 0 we have B0 (t) ≤ #(tS ∩ Zn ) ≤ B1 (t). We can bound B1 (t) − B0 (t) by noting that each C(a1 , . . . , an ) counted by this difference √ contains a point (a1 , . . . , an ) ∈ Zn within a distance n = O(1) of a point in ∂tS = t∂S. Let f1 , . . . , fm be Lipschitz functions [0, 1]n−1 → ∂S whose images cover ∂S, and let c1 , . . . cm be constants such that d(fi (x1 ), fi (x2 )) ≤ ci d(x1 , x2 ) for all x1 , x2 ∈ [0, 1]n−1 . For any y ∈ ∂S, we have y = fi (x1 , . . . , xn−1 ) for some i, and if we put rj = btxj c ∈ Z so that 0 ≤ xj − rj /t ≤ 1/t, then   √ rn−1 r1 d(y, fi ( rt1 , . . . , rn−1 )) ≤ c · d , . . . , ) (x , . . . , x ), ( < ci n/t ≤ c/t, i 1 n−1 t t t where c := √ n maxi ci . Thus every y ∈ ∂S lies within a distance c/t of a point in the set 
P = fi rt1 , . . . , rn−1 : 1 ≤ i ≤ m, 0 ≤ r1 , . . . , rn−1 ≤ t , t which has cardinality m(t + 1)n−1 = O(tn−1 ). It follows that every point of ∂tS is within a distance c of one of the O(tn−1 ) points in tP. The number of integer lattice points within a √ distance n of a point in t∂S is thus also O(tn−1 ), and therefore B1 (t) − B0 (t) = O(tn−1 ). We now note that B0 (t) ≤ µ(tS) ≤ B1 (t) and µ(tS) = tn µ(S); the lemma follows. Corollary 19.6. Let Λ be a lattice in an R-vector space V ' Rn and let S ⊆ V be a measurable set whose boundary is (n − 1)-Lipschitz parametrizable. Then #(tS ∩ Λ) = µ(S) n t + O(tn−1 ). covol(Λ) Proof. The case Λ ⊆ Zn is given by the lemma; note that the normalization of the Haar measure µ is irrelevant, since we are taking a ratio of volumes which is necessarily preserved under the isomorphism of topological vector spaces V ' Rn . We now note that if the corollary holds for sΛ, for some s > 0, then it also holds for Λ, since tS ∩ sΛ = (t/s)S ∩ Λ. For any lattice Λ, we can choose s > 0 so that sΛ is arbitrarily close to an integer lattice (for example, take s to be the LCM of all denominators appearing in rational approximations of the coordinates of a basis for Λ), which is necessarily a finite index subgroup of Zn . The corollary follows. Remark 19.7. Recall that covol(Λ) = µ(F ) for any fundamental region F for Λ, so the ratio µ(S)/ covol(Λ) = µ(S)/µ(F ) in Corollary 19.6 does not depend on the normalization of the Haar measure µ. p However, we plan to apply the corollary to Λ = OK and want to replace covol(OK ) with | disc(OK )| = |DK |1/2 via Proposition 14.16, which requires us to use the normalized Haar measure on KR defined in §14.2. 18.785 Fall 2021, Lecture #19, Page 3 19.1.1 Counting algebraic integers of bounded norm Recall from §15.2 that the unit group KR× of KR := K ⊗Q R is the locally compact group KR× ' Y v|∞ Kv× ' Y R× × real v|∞ Y C× . complex v|∞ We have a natural embedding K × ,→ KR× x 7→ (xv ), where v ranges over the r + s archimedean places of K; this allows us to view K × as a subgroup of KR× that contains the nonzero elements of OK . In Lecture 15 we defined the continuous homomorphism Log : KR× → Rr+s (xv ) 7→ (log kxv kv ), and proved that we have an exact sequence of abelian groups Log × 1 −→ µK −→ OK −→ ΛK → 0, := {x ∈ Rr+s : T(x) = 0} (where in which ΛK is a lattice in the trace-zero hyperplane Rr+s 0 T(x) is the sum of the coordinates of x). The regulator RK is the covolume of ΛK in Rr+s 0 with the Euclidean measure induced by any (see Definition 15.16), where we endow Rr+s 0 coordinate projection Rr+s → Rr+s−1 . By Dirichlet’s unit theorem (Theorem 15.12), we can write × = U × µK , OK × is free of rank r + s − 1 (the subgroup U is not uniquely determined, but let where U ⊆ OK us fix a choice). We want to estimate the quantity #{a : N(a) ≤ t}, where a ranges over nonzero ideals of OK , as t → ∞. As a first step, let us restrict our attention to principal ideals (α) ⊆ OK . We then want to estimate the cardinality of × {(α) : N(α) ≤ t}. For nonzero α, α0 ∈ K we have (α) = (α0 ) if and only if α/α0 ∈ OK , so this is equivalent to × {α ∈ K × ∩ OK : N(α) ≤ t}/OK , × where for any set S ⊆ KR× , the notation S/OK denotes the set of equivalence classes of S × 0 0 under the equivalence relation α ∼ α ⇔ α = uα for some u ∈ OK . If we now define × KR,≤t := {x ∈ KR× : N(x) ≤ t} ⊆ KR× ⊆ KR , then we want to estimate the cardinality of the finite set   × × KR,≤t ∩ OK /OK , 18.785 Fall 2021, Lecture #19, Page 4 where the intersection takes place in KR and produces a subset of KR× that we partition into × × equivalence classes modulo OK . To simplify matters, let us replace OK with the free group × U ⊆ OK ; we then have a wK –to–1 map   × × × (KR,≤t ∩ OK )/U −→ KR,≤t ∩ OK /OK . × It suffices to estimate the cardinality of (KR,≤t ∩ OK )/U and divide the result by wK . × Recall that for x = (xv ) ∈ KR , the norm map N : KR× → R× >0 is defined by N(x) := Y v|∞ kxv kv = Y |xv |R v real Y |xv |2C , v complex and satisfies T(Log x) = log N(x) for all x ∈ KR× . We now define a surjective homomorphism × ν : KR×  KR,1 x 7→ xN(x)−1/n . × The image of KR,1 under the Log map is precisely the trace zero hyperplane Rr+s in Rr+s 0 × ) = ΛK is a lattice. Let us fix a fundamental domain F for the in which Log(U ) = Log(OK r+s lattice ΛK in R0 so that
S := ν −1 Log−1 (F ) is a set of unique coset representatives for the quotient KR× /U . If we now define S≤t := {x ∈ S : N(x) ≤ t} ⊆ KR , we want to estimate the cardinality of the finite set S≤t ∩ OK . The set OK is a lattice in the R-vector space KR of dimension n. We have tS≤1 = S≤tn , so we can estimate the cardinality of S≤t = t1/n S≤1 via Corollary 19.6 with S = S≤1 and Λ = OK by replacing t with t1/n , provided that the boundary of S≤1 is (n − 1)-Lipschitz parametrizable, which we now argue. The kernel of the Log map is {±1}r × U(1)s , where U(1) := {z ∈ C : z z̄ = 1} is the unit circle in C. We thus have a continuous isomorphism of locally compact groups ∼ KR× = (R× )r × (C× )s −→ Rr+s × {±1}r × [0, 2π)s (1) x = (x1 , . . . , xr , z1 , . . . , zs ) 7−→ (Log x) × (sgn x1 , . . . , sgn xr ) × (arg z1 , . . . , arg zs ), where the map to Rr+s is the Log map, the map to {±1}r is the vector of signs of the r real components, and the map to [0, 2π)s is the vector of angles arg z such that z/|z| = ei arg z of the s complex components. The set S≤1 consists of 2r connected components, one for each element of {±1}r . We can parametrize each of these component using n real parameters as follows: • r + s − 1 parameters in [0, 1) that encode a point in F as an R-linear combination of Log(1 ), . . . , Log(r+s−1 ), where 1 , . . . , r+s−1 are a basis for U ; • s parameters in [0, 1) that encode an element of U(1)s ; 18.785 Fall 2021, Lecture #19, Page 5 • a parameter in (0, 1] that encodes the nth-root of the norm. These parameterizations define a continuously differentiable bijection from the set C = [0, 1)n−1 × (0, 1] ⊆ [0, 1]n to each of the 2r disjoint components of S≤1 ; it can be written out explicitly in terms of exponentials and the identity function. The boundary ∂C is the boundary of the unit ncube, which is clearly (n − 1)-Lipschitz parametrizable; thus each component of S≤1 , and therefore S≤1 itself, has a boundary that is (n − 1)-Lipschitz parametrizable. We now apply Corollary 19.6 to the lattice OK and the set S≤1 in the n-dimensional R-vector space KR with t replaced by t1/n , since S≤t = t1/n S≤1 . This yields      µ(S )  µ(S≤1 ) ≤1 1−1/n 1/n n 1/n n−1 #(S≤t ∩ OK ) = t + O t . (2) (t ) + O (t ) = covol(OK ) |DK |1/2 Our next task is compute µ(S≤1 ); as noted in Remark 19.7, we must use the normalized Haar measure µ on KR defined in §14.2 when doing so. We will use the isomorphism in (1) to make a change Q of coordinates, we just need to understand how this affects the Haar measure µ on KR = v|∞ Kv ' Rr × Cs . In terms of the standard Lebesgue measures dx and dA on R and C, we have µ = (dx)r (2dA)s , where the 2dA reflects the fact that the normalized absolute value k kv for each complex square of the Euclidean Q place v is the × × r absolute value on C. For each factor of KR = v|∞ Kv ' (R ) × (C× )s ⊆ Rr × Cs we define the maps R× → R × {±1} C× → R × [0, 2π) x 7→ (log |x|, sgn x) ` z 7→ (2 log |z|, arg z) `/2+iθ ±e ←[ (`, ±1) e dx 7→ e` d`µ{±1} ←[ (`, θ) 2dA 7→ 2e`/2 d(e`/2 )dθ = e` d`dθ, where d` is the Lebesgue measure on R, µ{±1} is the counting measure on {±1}, and dθ is the Lebesgue measure on [0, 2π). We thus have ∼ KR× −→ Rr+s × {±1}r × [0, 2π)s µ 7→ eT(·) µRr+s µr{±1} µs[0,2π) , where the trace function T(·) sums the coordinates of a vector in Rr+s . We now make one further change of coordinates: Rr+s → Rr+s−1 × R x = (x1 , . . . , xr+s ) 7→ (x1 , . . . , xr+s−1 , y := T(x)) eT(x) µRr+s 7→ ey µRr+s−1 dy. If we let π0 : Rr+s → Rr+s−1 denote the coordinate projection, then the Lebesgue measure of π0 (F ) in Rr+s−1 is, by definition, the regulator RK (see Definition 15.16). The Log map gives us a bijection   1 1 2 2 ∼ S≤1 −→ F + (−∞, 0] ,..., , ,..., , n n n n   1 1 2 2 1/n x = N(x) ν(x) 7→ Log ν(x) + log N(x) ,..., , ,..., . n n n n 18.785 Fall 2021, Lecture #19, Page 6 The coordinate y ∈ (−∞, 0] is given by y = T(Log x) = log N(x), so we can view S≤1 as an infinite union of cosets of Log−1 (F ) parameterized by ey = N(x) ∈ (0, 1]. Under our change of coordinates we thus have ∼ KR× −→ Rr+s−1 × R × {±1}r × [0, 2π)s S≤1 → π0 (F ) × (−∞, 0] × {±1}r × [0, 2π)s . Since RK = µRr+s−1 (π(F )), we have µ(S≤1 ) = Z 0 ey RK 2r (2π)s dy −∞ = 2r (2π)s RK . Plugging this into (2) yields #(S≤t ∩ OK ) = 19.2  2r (2π)s RK |DK |1/2    1 t + O t1− /n . (3) Proof of the analytic class number formula We are now ready to prove the analytic class number formula. Our main tool is the following theorem, which uses our analysis in the previous section to give a precise asymptotic estimate on the number of ideals of bounded norm. Theorem 19.8. Let K be a number field of degree n As t → ∞, the number of nonzero OK -ideals a of absolute norm N(a) ≤ t is  r    2 (2π)s hK RK 1−1/n , t + O t wk |DK |1/2 where r and s are the number of real and complex places of K, respectively, hK = # cl OK is the class number, RK is the regulator, wK := #µK is the number of roots of unity, and DK := disc OK is the absolute discriminant. Proof. In order to count the nonzero OK -ideals a of absolute norm N(a) ≤ t we group them by ideal class. For the trivial class, we just need to count nonzero principal ideals (α), × . equivalently, the number of nonzero α ∈ OK with N(α) ≤ t, modulo the unit group OK Dividing (3) by wK to account for the wK -to-1 map × × S≤t ∩ OK −→ (KR,≤t ∩ OK )/OK , we obtain #{(α) ⊆ OK : N(α) ≤ t} =  2r (2π)s RK wK |DK |1/2    1 t + O t1− /n . (4) To complete the proof we now show that we get the same answer for every ideal class; the nonzero ideals a of norm N(a) ≤ t are asymptotically equidistributed among ideal classes. Fix an ideal class [a], with a ⊆ OK nonzero (every ideal class contains an integral ideal, by Theorem 14.20). Multiplication by a gives a bijection ×a {ideals b ∈ [a−1 ] : N(b) ≤ t} −→ {nonzero principal ideals (α) ⊆ a : N(α) ≤ tN(a)} × −→ {nonzero α ∈ a : N(α) ≤ tN(a)}/OK . 18.785 Fall 2021, Lecture #19, Page 7 Let S[a],≤t denote the set on the RHS. The estimate in (4) derived from Corollary 19.6 applies to any lattice in KR , not just OK . Replacing OK with a in (4) we obtain  r    2 (2π)s RK 1 #S[a],≤t = t N(a) + O t1− /n wk covol(a)     2r (2π)s RK 1 = t N(a) + O t1− /n wk covol(OK )N(a)  r    2 (2π)s RK 1−1/n = t + O t , wk |DK |1/2 since covol(a) = N (a) covol(OK ), by Corollary 14.17. Note that the RHS does not depend on the ideal class [a]. Summing over ideal classes yields  r    X 2 (2π)s hK RK 1−1/n #{nonzero ideals b ⊆ OK : N(b) ≤ t} = #S[a],≤t = t+O t , wK |DK |1/2 [a]∈cl(OK ) as claimed. Lemma 19.9. Let a1 , a2 , . . . be a sequence of complex numbers and let σ be a real number. Suppose that a1 + · · · + at = O(tσ ) (as t → ∞). P Then the Dirichlet series an n−s defines a holomorphic function on Re s > σ. P Proof. Let A(x) := 0 σ we have ∞ X n=1 an n −s = Z ∞ 1− x−s dA(x) Z ∞ − A(x) dx−s − 1 1− Z ∞ = (0 − 0) − A(x)(−sx−s−1 ) dx 1− Z ∞ A(x) dx. =s s+1 1− x = A(x) xs ∞ Note that we used |A(x)| = O(xσ ) and Re(s) > σ to conclude limx→∞ A(x)/xs = 0. The integral on the RHS converges locally uniformly on Re(s) > σ and the lemma follows. Remark 19.10. Lemma 19.9 gives us an abscissa of convergence σ for the Dirichlet series P an n−s ; this is analogous to the radius of convergence of a power series. Lemma 19.11. Let a1 , a2 , . . . be a sequence of complex numbers that satisfies a1 + · · · + at = ρt + O(tσ ) (as t → ∞) P for some σ ∈ [0, 1) and ρ ∈ C× . The Dirichlet series an n−s converges on Re(s) > 1 and has a meromorphic continuation to Re(s) > σ that is holomorphic except for a simple pole at s = 1 with residue ρ. 18.785 Fall 2021, Lecture #19, Page 8 Proof. Define bn := an − ρ. Then b1 + · · · + bt = O(tσ ) and X X X X an n−s = ρ n−s + bn n−s = ρ ζ(s) + bn n−s . We have already proved that the Riemann zeta function ζ(s) is holomorphic on Re(s) > 1 and has a meromorphic continuation to Re(s) >P 0 that is holomorphic except for a simple pole at 1 with residue 1. By the previous lemma, bn n−s is holomorphic on Re(s) > σ, and since σ < 1, it is holomorphic at s = 1. So the entire RHS has a meromorphic continuation to Re(s) > σ that is holomorphic except for the simple pole at 1 coming from ζ(s), and the residue at s = 1 is ρ · 1 + 0 = ρ. We are now ready to prove the analytic class number formula. Theorem 19.12 (Analytic Class Number Formula). Let K be a number field of degree n. The Dedekind zeta function ζK (z) extends to a meromorphic function on Re(z) > 1 − n1 that is holomorphic except for a simple pole at z = 1 with residue lim (z − 1)ζK (z) = ρK := z→1+ 2r (2π)s hK RK , wK |DK |1/2 where r and s are the number of real and complex places of K, respectively, hK := # cl OK is the class number, RK is the regulator, wK := µK is the number of roots of unity, and DK := disc OK is the absolute discriminant. Proof. We have ζK (z) = X a N(a)−z = X at t−z , t≥1 where a ranges over nonzero ideals of OK , and at := #{a : N(a) = t} with t ∈ Z≥1 . If we now define 2r (2π)s hK RK , ρK := wK |DK |1/2 then by Theorem 19.8 we have a1 + · · · + at = #{a : N(a) ≤ t} = ρK t + O(t1− /n ) 1 (as t → ∞). P −z Applying Lemma 19.11 with σ = 1 − 1/n, we see that ζK (z) = at t extends to a meromorphic function on Re(z) > 1 − 1/n that is holomorphic except for a simple pole at z = 1 with residue ρK . Remark 19.13. As previously noted, Hecke proved that ζK (z) extends to a meromorphic function on C with no poles other than the simple pole at z = 1, and it satisfies a functional equation. If we define the gamma factors 1
ΓR (z) := π −z/2 Γ z2 , and ΓC (z) := ΓR (z)ΓR (z + 1) = 2(2π)−z Γ(z), and the completed zeta function ξK (z) := |DK |z/2 ΓR (z)r ΓC (z)s ζK (z), 1 The rightmost equality follows from the duplication formula for Γ(s). In older texts one may find ΓC (s) defined as (2π)−z Γ(z), which yields the same functional equation. 18.785 Fall 2021, Lecture #19, Page 9 where r and s are the number of real and complex places of K, respectively, then ξK (z) is holomorphic except for simple poles at z = 0, 1 and satisfies the functional equation ξK (z) = ξK (1 − z). In the case K = Q, we have r = 1 and s = 0, so ξQ (z) = ΓR (z)ζ(z) = π −z/2 Γ( z2 )ζQ (z), which is precisely the completed zeta function Z(z) we defined for the Riemann zeta function ζ(z) = ζQ (z) in Lecture 17 (without any extra factors to remove the zeros at z = 0, 1). 19.3 Cyclotomic zeta functions and Dirichlet L-functions Having proved the analytic class number formula, we now want to complete the proof of Dirichlet’s theorem on primes in arithmetic progressions that we began in the previous lecture. To do this we need to establish a connection between Dirichlet L-functions and Dedekind zeta functions of cyclotomic fields. ∼ Recall from Problem Set 4 that we have an isomorphism ϕ : Gal(Q(ζm )/Q) −→ (Z/mZ)× ϕ(σ) canonically defined by σ(ζm ) = ζm (independent of the choice of ζm ). The canonical bijection given by Corollary 18.16 allows us to identify the set X(m) of primitive Dirichlet characters of conductor dividing m with the character group of (Z/mZ)× ' Gal(Q(ζm )/Q).2 More generally, given any finite set of primitive Dirichlet characters, if we let m be the LCM of their conductors and consider the subgroup H of X(m) they generate, we may associate to H the subfield K := Q(ζm )φ(H) , where φ(H) := {σ ∈ Gal(Q(ζm )/Q) : χ(σ) = 1 for all χ ∈ H}; we may then regard H as the character group of Gal(K/Q) via Proposition 18.40. The same applies if we replace m with any multiple m0 , since H ⊆ X(m) ⊆ X(m0 ) for all m|m0 and we will get the same field K ⊆ Q(ζm ) ⊆ Q(ζm0 ). Conversely, for each subfield K of a cyclotomic field Q(ζm ) there is a corresponding subgroup H := {χ ∈ X(m) : χ(σ) = 1 for all σ ∈ Gal(Q(ζm )/K)}, for which K = Q(ζm )φ(H) . Note that K/Q is Galois, since Gal(Q(ζm )/Q) is abelian (every subgroup is normal), and we may view H as the character group of Gal(K/Q). We thus have a one-to-one correspondence between subgroups H ⊆ X(m) and subfields of K ⊆ Q(ζm ) in which H corresponds to the character group of Gal(K/Q) and K = Q(ζm )φ(H) . We will prove that under this correspondence, the Dedekind zeta function of ζK (s) is the product of the Dirichlet L-functions L(s, χ) for χ ∈ H. We first note the following. Proposition 19.14. Let p be a prime, let m be a positive integer, and let m0 = m/pvp (m) . Then Q(ζm0 ) is the maximal extension of Q in Q(ζm ) unramified at p. In particular, if p does not divide m then Q(ζm ) is unramified at p. Proof. By Corollary 10.18, the extension Qp (ζm0 )/Qp is unramified. It follows from Proposition 12.4 that Q(ζm0 )/Q is unramified at p. Applying the same argument to all primes q 6= p 2 As noted in Remark 18.17, the group operation on X(m) is not pointwise multiplication, one multiplies elements of X(m) by taking the unique primitive character that induces the pointwise product. 18.785 Fall 2021, Lecture #19, Page 10 dividing m shows that the extension Q(ζpvp (m) ) is ramified only at p. By Corollary 14.27, there are no nontrivial unramified extensions of Q, so every subfield of Q(ζpvp (m) ) that properly contains Q is ramified at p. Now Q(ζm ) is the compositum of Q(ζpvp (m) ) and Q(ζm0 ), 0 ) in Q(ζ ) contains a subfield of which intersect in Q, so any nontrivial extension of Q(ζm m Q(ζpvp (m) ) properly containing Q which must be ramified at p; the proposition follows. Theorem 19.15. Let H ⊆ X(m) be a group of primitive Dirichlet characters and let K = Q(ζm )φ(H) be the corresponding subfield of Q(ζm ), with φ(H) defined as above. Then Y ζK (s) = L(s, χ). χ∈H Proof. On the LHS we have Y
−1 Y Y
−1 ζK (s) = 1 − N(p)−s = 1 − N(p)−s , p p p|p and on the RHS we have Y YY
−1 Y Y
−1 L(s, χ) = 1 − χ(p)p−s = 1 − χ(p)p−s . χ∈H p χ∈H p χ∈H It thus suffices to prove Y p|p
? Y
1 − N(p)−s = 1 − χ(p)p−s (5) χ∈H for each prime p. Since K/Q is Galois, we have [K : Q] = ep fp gp , where ep is the ramification index, fp is the residue field degree, and gp = #{p|p}. On the LHS of (5) we have g p g p  Y
 , = 1 − (p−s )fp 1 − N(p)−s = 1 − (pfp )−s p|p which we note does not change if we replace K with the maximal subfield K 0 of K in which p is unramified (since K/K 0 is totally ramified at every prime of K 0 above p, only ep changes, not fp or gp ). On the RHS of (5), we have χ(p) = 0 for all χ ∈ H with conductor divisible by p, so we can replace H with the subgroup H 0 of Dirichlet characters with conductors 0 prime to p. It follows from Proposition 19.14 that K 0 = Q(ζm )φ(H ) (to see this, note that if we put m0 = m/pvp (m) then K 0 = K ∩ Q(ζm0 ) and H 0 = H ∩ X(m0 )). Thus without loss of generality we assume p6 | m, so K is unramified at p and we have #H = [K : Q] = fp gp . Since K/Q is abelian and unramified at p, the Artin map gives us a Frobenius element σp corresponding to the Frobenius automorphism x 7→ xp of the residue field, which by definition has order fp , so σp has order fp in Gal(K/Q). Viewing H as the character group of Gal(K/Q), the map χ 7→ χ(σp ) defines a surjective homomorphism from H to the group of fp -th roots of unity α ∈ U(1), and the kernel of this map has cardinality #H/fp = gp . Therefore  g p Y Y

g 1 − χ(p)p−s = 1 − αp−s p = 1 − (p−s )fp , χ∈H αfp =1 where the second equality follows from the identity Q αfp =1 (1 − αT ) = 1 − T fp ∈ C[T ]. 18.785 Fall 2021, Lecture #19, Page 11 19.4 Non-vanishing of Dirichlet L-functions with non-principal character We are now ready to prove the key claim needed to complete our proof of Dirichlet’s theorem on primes in arithmetic progressions. Theorem 19.16. Let ψ be any non-principal Dirichlet character. Then L(1, ψ) 6= 0. Proof. Let ψ be a non-principal Dirichlet character, say of modulus m. Then ψ is induced by a non-trivial primitive Dirichlet character ψe of conductor m e dividing m. The L-functions e of ψ and ψ differ at only finitely many Euler factors (1 − ψ(p)p−s )−1 (corresponding to primes p dividing m/m), e and these factors are clearly nonzero at s = 1, since p > 1. We thus assume without loss of generality that ψ = ψe is primitive. Let K be the mth cyclotomic field Q(ζm ). By Theorem 19.15 we have Y ζK (s) = L(s, χ), χ where χ ranges over the primitive Dirichlet characters of conductor dividing m, including ψ. By the analytic class number formula (Theorem 19.12), the LHS has a simple pole at s = 1, and the same must be true of the RHS. Thus Y ords=1 ζK (s) = ords=1 L(s, χ) χ −1 = ords=1 L(s, 1) −1 = ords=1 ζ(s) −1 = −1 + X Y L(s, χ) χ6=1 Y L(s, χ) χ6=1 ords=1 L(s, χ). χ6=1 Each χ 6= 1 in the sum is necessarily non-principal (since it is primitive). We proved in Proposition 18.20 that for non-principal χ the Dirichlet L-series L(s, χ) is holomorphic on Re(s) > 0, thus ords=1 L(s, χ) ≥ 0 for all χ appearing in the sum, which can therefore be zero if and only if every term ords=1 L(s, χ) is zero. So L(1, χ) 6= 0 for every non-trivial primitive Dirichlet character χ of conductor dividing m, including ψ. References [1] Erich Hecke, Über die Zetafunktion beliebiger algebraischer Zahlkörper , Nachr. Ges. Wiss. Göttingen (1917), 77–89. [2] P.G. Lejeune Dirichlet and Richard Dedekind, Vorlesungun über Zahlentheorie, Braunschweig F. Veiweg, 1894. [3] Edmund Landau, Neuer Beweis des Primzahlsatzes und Beweis des Primidealsatzes, Math. Ann. 56, 645–670. 18.785 Fall 2021, Lecture #19, Page 12 18.785 Number theory I Lecture #20 20 Fall 2021 11/17/2021 The Kronecker-Weber theorem In the previous lecture we established a relationship between finite groups of Dirichlet characters and subfields of cyclotomic fields. Specifically, we showed that there is a one-toone-correspondence between finite groups H of primitive Dirichlet characters of conductor dividing m and subfields K of Q(ζm ) under which H can be viewed as the character group of the finite abelian group Gal(K/Q) and the Dedekind zeta function of K factors as Y ζK (s) = L(s, χ). χ∈H Now suppose we are given an arbitrary finite abelian extension K/Q. Does the character group of Gal(K/Q) correspond to a group of Dirichlet characters, and can we then factor the Dedekind zeta function ζK (s) as a product of Dirichlet L-functions? The answer is yes! This is a consequence of the Kronecker-Weber theorem, which states that every finite abelian extension of Q lies in a cyclotomic field. This theorem was first stated in 1853 by Kronecker [2], who provided a partial proof for extensions of odd degree. Weber [7] published a proof 1886 that was believed to address the remaining cases; in fact Weber’s proof contains some gaps (as noted in [5]), but in any case an alternative proof was given a few years later by Hilbert [1]. The proof we present here is adapted from [6, Ch. 14] 20.1 Local and global Kronecker-Weber theorems We now state the (global) Kronecker-Weber theorem. Theorem 20.1. Every finite abelian extension of Q lies in a cyclotomic field Q(ζm ). There is also a local version. Theorem 20.2. Every finite abelian extension of Qp lies in a cyclotomic field Qp (ζm ). We first show that the local version implies the global one. Proposition 20.3. The local Kronecker-Weber theorem implies the global Kronecker-Weber theorem. Proof. Let K/Q be a finite abelian extension. For each ramified prime p of Q, pick a prime p|p and let Kp be the completion of K at p (the fact that K/Q is Galois means that every p|p is ramified with the same ramification index; it makes no difference which p we pick). We have Gal(Kp /Qp ) ' Dp ⊆ Gal(K/Q), by Theorem 11.23, so Kp is an abelian extension of Qp and the local Kronecker-Weber theorem implies that Kp ⊆ Qp (ζmp ) for some mp ∈ Z≥1 . Q Let np := vp (mp ), put m := p pnp (this is a finite product), and let L = K(ζm ). We will show L = Q(ζm ), which implies K ⊆ Q(ζm ). The field L = K · Q(ζm ) is a compositum of Galois extensions of Q, and is therefore Galois over Q with Gal(L/Q) isomorphic to a subgroup of Gal(K/Q)×Gal(Q(ζm )/Q), hence abelian (as recalled below, the Galois group of a compositum K1 · · · Kr of Galois extensions Ki /F is isomorphic to a subgroup of the direct product of the Gal(Ki /F )). Let q be a prime of L lying above a ramified prime p|p; as above, the completion Lq of L at q is a finite abelian extension of Qp , since L/Q is finite abelian, and we have Lq = Kp · Qp (ζm ). Let Fq be the maximal unramified extension of Qp in Lq . Then Lq /Fq is totally ramified and Gal(Lq /Fq ) is isomorphic to the inertia group Ip := Iq ⊆ Gal(L/Q), by Theorem 11.23 (the Iq all coincide because L/Q is abelian). It follows from Corollary 10.18 that Kp ⊆ Fq (ζpnp ), since Kp ⊆ Qp (ζmp ) and Qp (ζmp /pnp ) is unramified, and that Lq = Fq (ζpnp ), since Qp (ζm/pnp ) is unramified. Moreover, we have Fq ∩ Qp (ζpnp ) = Qp , since Qp (ζpnp )/Qp is totally ramified, and it follows that Ip ' Gal(Lq /Fq ) ' Gal(Qp (ζpnp )/Qp ) ' (Z/pnp Z)× . Now let I be the group generated by Q the union of the groups Ip ⊆ Gal(L/Q) for p|m. Since Gal(L/Q) is abelian, we have I ⊆ Ip , thus Y Y Y φ(pnp ) = φ(m) = [Q(ζm ) : Q]. #I ≤ #Ip = #(Z/pnp Z)× = p|m p|m p|m Each inertia field LIp is unramified at p (see Proposition 7.12), as is LI ⊆ LIp . So LI /Q is unramified, and therefore LI = Q, by Corollary 14.27. Thus [L : Q] = [L : LI ] = #I ≤ [Q(ζm ) : Q], and Q(ζm ) ⊆ L, so L = Q(ζm ) as claimed and K ⊆ L = Q(ζm ). To prove the local Kronecker-Weber theorem we first reduce to the case of cyclic extensions of prime-power degree. Recall that if L1 and L2 are two Galois extensions of a field K then their compositum L := L1 L2 is Galois over K with Galois group Gal(L/K) ' {(σ1 , σ2 ) : σ1 |L1 ∩L2 = σ2 |L1 ∩L2 } ⊆ Gal(L1 /K) × Gal(L2 /K). The inclusion on the RHS is an equality if and only if L1 ∩ L2 = K. Conversely, if Gal(L/K) ' H1 × H2 then by defining L2 := LH1 and L1 := LH2 we have L = L1 L2 with L1 ∩ L2 = K, and Gal(L1 /K) ' H1 and Gal(L2 /K) ' H2 . It follows from the structure theorem for finite abelian groups that we may decompose any finite abelian extension L/K into a compositum L = L1 · · · Ln of linearly disjoint cyclic extensions Li /K of prime-power degree. If each Li lies in a cyclotomic extension K(ζmi ), then so does L. Indeed, L ⊆ K(ζm1 ) · · · K(ζmn ) = K(ζm ), where m := m1 · · · mn . To prove the local Kronecker-Weber theorem it thus suffices to consider cyclic extensions K/Qp of prime power degree `r . There two distinct cases: ` 6= p and ` = p. 20.2 The local Kronecker-Weber theorem for ` 6= p Proposition 20.4. Let K/Qp be a cyclic extension of degree `r for some prime ` 6= p. Then K lies in a cyclotomic field Qp (ζm ). Proof. Let F be the maximal unramified extension of Qp in K; then F = Qp (ζn ) for some n ∈ Z≥1 , by Corollary 10.17. The extension K/F is totally ramified, and it must be tamely ramified, since the ramification index is a power of ` 6= p. By Theorem 11.10, we have K = F (π 1/e ) for some uniformizer π, with e = [K : F ]. We may assume that π = −pu for some u ∈ OF× , since F/Qp is unramified: if q|p is the maximal ideal of OF then the valuation vq extends vp with index eq = 1 (by Theorem 8.20), so vq (−pu) = vp (−p) = 1. The field K = F (π 1/e ) lies in the compositum of F ((−p)1/e ) and F (u1/e ), and we will show that both fields lie in a cyclotomic extension of Qp . 18.785 Fall 2021, Lecture #20, Page 2 The extension F (u1/e )/F is unramified, since vq (disc(xe −u)) = 0 for p - e, so F (u1/e )/Qp is unramified and F (u1/e ) = Qp (ζk ) for some k ∈ Z≥1 . The field K(u1/e ) = K · Qp (ζk ) is a compositum of abelian extensions, so K(u1/e )/Qp is abelian, and it contains the subextension Qp ((−p)1/e )/Qp , which must be Galois (since it lies in an abelian extension) and totally ramified (by Theorem 11.5, since it is an Eisenstein extension). The field Qp ((−p)1/e ) contains ζe (take ratios of roots of xe +p) and is totally ramified, but Qp (ζe )/Qp is unramified (since p6 | e), so we must have Qp (ζe ) = Qp . Thus e|(p − 1), and by Lemma 20.5 below, Qp ((−p)1/e ) ⊆ Qp ((−p)1/(p−1) ) = Qp (ζp ). It follows that F ((−p)1/e ) = F · Qp ((−p)1/e ) ⊆ Qp (ζn ) · Qp (ζp ) ⊆ Qp (ζnp ). We then have K ⊆ F (u1/e ) · F ((−p)1/e ) ⊆ Q(ζk ) · Q(ζnp ) ⊆ Q(ζknp ) and may take m = knp.   Lemma 20.5. For any prime p we have Qp (−p)1/(p−1) = Qp (ζp ). Proof. Let α = (−p)1/(p−1) . Then α is a root of the Eisenstein polynomial xp−1 + p, so the extension Qp ((−p)1/(p−1) ) = Qp (α) is totally ramified of degree p − 1, and α is a uniformizer (by Lemma 11.4 and Theorem 11.5). Let π = ζp − 1. The minimal polynomial of π is f (x) := (x + 1)p − 1 = xp−1 + pxp−2 + · · · + p, x which is Eisenstein, so Qp (π) = Qp (ζp ) is also totally ramified of degree p − 1, and π is a uniformizer. We have u := −π p−1 /p ≡ 1 mod π, so u is a unit in the ring of integers of Qp (ζp ). If we now put g(x) = xp−1 − u then g(1) ≡ 0 mod π and g 0 (1) = p − 1 6≡ 0 mod π, so by Hensel’s Lemma 9.15 we can lift 1 to a root β of g(x) in Qp (ζp ). We then have pβ p−1 = pu = −π p−1 , so (π/β)p−1 + p = 0, and therefore π/β ∈ Qp (ζp ) is a root of the minimal polynomial of α. Since Qp (ζp ) is Galois, this implies that α ∈ Qp (ζp ), and since Qp (α) and Qp (ζp ) both have degree p − 1, the two fields coincide. To complete the proof of the local Kronecker-Weber theorem, we need to address the case ` = p. Before doing so, we first recall some background on Kummer extensions. 20.3 The local Kronecker-Weber theorem for ` = p > 2 We are now ready to prove the local Kronecker-Weber theorem in the case ` = p > 2. Theorem 20.6. Let K/Qp be a cyclic extension of odd degree pr . Then K lies in a cyclotomic field Qp (ζm ). Proof. There are two obvious candidates for K, namely, the cyclotomic field Qp (ζppr −1 ), which by Corollary 10.17 is an unramified extension of degree pr , and the index p−1 subfield of the cyclotomic field Qp (ζpr+1 ), which by Corollary 10.18 is a totally ramified extension of degree pr (the pr+1 -cyclotomic polynomial Φpr+1 (x) has degree φ(pr+1 ) = pr (p − 1) and remains irreducible over Qp ). If K is contained in the compositum of these two fields then r K ⊆ Qp (ζm ), where m := (pp −1)(pr+1 ) and the theorem holds. Otherwise, the field K(ζm ) is a Galois extension of Qp with Gal(K(ζm )/Qp ) ' Z/pr Z × Z/pr Z × Z/(p − 1)Z × Z/ps Z, for some s > 0; the first factor comes from the Galois group of Qp (ζppr −1 ), the second two factors come from the Galois group of Qp (ζpr+1 ) (note Qp (ζpr+1 )∩Qp (ζppr −1 ) = Qp ), and the 18.785 Fall 2021, Lecture #20, Page 3 last factor comes from the fact that we are assuming K 6⊆ Qp (ζm ), so Gal(K(ζm )/Qp (ζm )) is nontrivial and must have order ps with 1 ≤ s ≤ r. It follows that the abelian group Gal(K(ζm )/Qp ) has a quotient isomorphic to (Z/pZ)3 , and the subfield of K(ζm ) corresponding to this quotient is an abelian extension of Qp with Galois group (Z/pZ)3 . By Proposition 20.7 below, no such field exists. Proposition 20.7. For odd p every totally wildly ramified Galois extension of Qp is cyclic. In particular, there is no abelian extension of Qp with Galois group (Z/pZ)3 when p is odd. Proof. See Problem Set 10 for the first statement. For the second, if Gal(K/Qp ) ' (Z/pZ)3 we can write G := Gal(K/Qp ) as the internal direct sum of the inertia subgroup I ≤ G and a cyclic subgroup H ≤ G, since LI is an unramified, hence cyclic extension of Qp with Galois group isomorphic to G/I ' H. But then LH is a totally wildly ramified abelian extension of Qp whose Galois group G/H is not cyclic. Remark 20.8. There is an alternative proof to Proposition 20.7 that is more explicit. One can show that for odd p the field Qp has exactly p ramified abelian extensions of degree p, namely, Qp [x]/(xp + pxp−1 + p(1 + ap)), for integers a ∈ [0, p − 1]; see [3, Prop. 2.3.1]. Any noncyclic totally wildly ramified abelian extension of Qp would contain at least p + 1 ramified abelian extensions of degree p, since (Z/pZ)2 has p + 1 quotients of order p. Remark 20.9. Another approach to Proposition 20.7 uses Kummer theory. One shows that for odd p the elementary abelian p-group Qp (ζp )× /Qp (ζp )×p has rank at most 2, and this rules out the existence of a (Z/pZ)3 extension; see [6, Lemma 14.8]. For p = 2 there is an extension of Q2 with Galois group isomorphic to (Z/2Z)3 : the cyclotomic field Q2 (ζ24 ) = Q2 (ζ3 ) · Q2 (ζ8 ). So the proof we used for p > 2 will not work. However we can apply a completely analogous argument. Theorem 20.10. Let K/Q2 be a cyclic extension of degree 2r . Then K lies in a cyclotomic field Q2 (ζm ). Proof. The unramified cyclotomic field Q2 (ζ22r −1 ) has Galois group Z/2r Z, and the totally ramified cyclotomic field Q2 (ζ2r+2 ) has Galois group Z/2Z × Z/2r Z (up to isomorphism). r Let m = (22 − 1)(2r+2 ). If K is not contained in Q2 (ζm ) then  r 2 s  Z/2Z × (Z/2 Z) × Z/2 Z with 1 ≤ s ≤ r Gal(K(ζm )/Q2 ) ' or   (Z/2r Z)2 × Z/2s Z with 2 ≤ s ≤ r and thus admits a quotient isomorphic to (Z/2Z)4 or (Z/4Z)3 . By Lemma 20.11 below, no extension of Q2 has either of these Galois groups, thus K must lie in Q2 (ζm ). Lemma 20.11. No extension of Q2 has Galois group isomorphic to (Z/2Z)4 or (Z/4Z)3 . Proof. As you proved on Problem Set 4, there are exactly 7 quadratic extensions of Q2 ; it follows that no extension of Q2 has Galois group (Z/2Z)4 , since this group has 15 subgroups of index 2 whose fixed fields would yield 15 distinct quadratic extension of Q2 . As you proved on Problem Set 5, there are only finitely many extensions of Q2 of any fixed degree d, and these can be enumerated by considering Eisenstein polynomials in Q2 [x] of degrees dividing d up to an equivalence relation implied by Krasner’s lemma. One finds that there are 59 quartic extensions of Q2 , of which 12 are cyclic; you can find a list of them here. It follows that no extension of Q2 has Galois group (Z/4Z)3 , since this group has 28 subgroups whose fixed fields would yield 28 distinct cyclic quartic extensions of Q2 . 18.785 Fall 2021, Lecture #20, Page 4 References [1] David Hilbert, Ein neuer Beweis des Kroneckerschen Fundamentalsatzes über Abelsche Zahlkörper , Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klass (1896), 29–39. [2] Leopold Kronecker, Uber die algebraisch auflösbaren Gleichungen I (1853), in Leopold Kronecker’s Werke, Part 4 (ed. K. Hensel), AMS Chelsea Publishing, 1968. [3] John W. Jones and David P. Roberts, A database of local fields, J. Symbolic Comput. 41 (2006), 80–97. [4] Serge Lang, Algebra, 3rd edition, Springer, 2002. [5] Olaf Neumann, Two proofs of the Kronecker-Weber theorem “according to Kronecker, and Weber", J. Reine Angew. Math. 323 (1981),105–126. [6] Lawrence C. Washington, Introduction to cyclotomic fields, 2nd edition, Springer, 1997. [7] Heinrich M. Weber, Theorie der Abel’schen Zahlkörper , Acta Mathematica 8 (1886), 193–263. 18.785 Fall 2021, Lecture #20, Page 5 18.785 Number theory I Lecture #21 21 Fall 2021 11/22/2021 Class field theory: ray class groups and ray class fields In the previous lecture we proved that every abelian extension L of Q is contained in a cyclotomic field Q(ζm ). The isomorphism Gal(Q(ζm )/Q) ' (Z/mZ)× then allows us to view Gal(L/Q) as a quotient of (Z/mZ)× . We would like to replace the base field Q with an arbitrary number field K, but we need analogs of the cyclotomic fields Q(ζm ) and the abelian Galois groups (Z/mZ)× . These analogs are ray class fields, and their Galois groups are isomorphic to ray class groups. Ray class fields are not, in general, cyclotomic extensions of K; their construction is rather more complicated. Before defining them, let us first recall some properties of the Artin map we defined in Lecture 7. 21.1 The Artin map Let L/K be a finite Galois extension of global fields, and let p be a prime of K. Recall that the Galois group Gal(L/K) acts on the set {q|p} (primes q of L lying above p) and the stabilizer of q|p is the decomposition group Dq ⊆ Gal(L/K). By Proposition 7.9, we have a surjective homomorphism πq : Dq → Gal(Fq /Fp ) σ 7→ σ := (α 7→ σ(α)), where α ∈ OL is any lift of α ∈ Fq := OL /q to OL and σ(α) is the reduction of σ(α) ∈ OL to Fq ; kernel of πq is the inertia group Iq . If q is unramified then Iq is trivial and πq is an isomorphism. The Artin symbol (Definition 7.20) is defined by   L/K := σq := πq−1 (x 7→ x#Fp ), q where (x 7→ x#Fp ) ∈ Gal(Fq /Fp ) is the Frobenius automorphism, a canonical generator for the cyclic group Gal(Fq /Fp ). Equivalently, σq is the unique element of Gal(L/K) for which σq (x) ≡ x#Fp mod q for all x ∈ OL . For q|p the Frobenius elements σq are all conjugate (they form the Frobenius class Frobp ), and when L/K is abelian they coincide, in which case we may write σp instead of σq (or use Frobp = {σp } to denote σp ), and we may write the Artin symbol as   L/K := σp . p Now assume L/K is abelian, let m be an OK -ideal divisible by every ramified prime m denote the subgroup of fractional ideals I ∈ I for which v (I) = 0 for all of K, and let IK p K p|m. The Artin map (Definition 7.23) is the homomorphism m m ψL/K : IK → Gal(L/K) Y Y  L/K np np p 7→ . p p6 |m p6 |m A key ingredient of class field theory that we will prove in this lecture is surjectivity of m . This allows us to identify Gal(L/K) with the quotient I m / ker ψ m . the Artin map ψL/K K L/K m Every p ∈ ker ψL/K is unramified and has the property that the Frobenius elements σq are trivial for all q|p, meaning that all the residue field extensions Fq /Fp are trivial. This implies that p splits completely in L (it is unramified and primes above it have residue degree one). m that splits completely in L lies in ker ψ m . Conversely, every prime p ∈ IK L/K Proposition 21.1. Let K ⊆ L ⊆ M be a tower of finite abelian extension of global fields and let m be an OK -ideal divisible by all primes p of K that ramify in M . We have a commutative diagram m ψM/K → Gal(M/K) res → Gal(L/K) → m ψL/K ← ← ← m IK where the vertical map is the homomorphism σ → σ|L induced by restriction. Proof. It suffices to check commutativity at primes p - m, which are necessarily unramified. The proposition then follows from Proposition 7.22. 21.2 Class field theory for Q We now specialize to K = Q. The Kronecker-Weber theorem tells us that every abelian extension L/K lies in a cyclotomic field Q(ζm ). Each σ ∈ Gal(Q(ζm )/Q) is determined by its action on ζm , and we have an isomorphism ∼ ω : Gal(Q(ζm )/Q) −→ (Z/mZ)× ω(σ) defined by σ(ζm ) = ζm . The primes p that ramify in Q(ζm ) are precisely those that divide m (by Corollary 10.18). For each prime p 6 | m the Frobenius element σp is the unique σ ∈ Gal(Q(ζm )/Q) for which σ(x) ≡ xp mod q for any (equivalently, all) q|(p). Thus ω(σp ) = p mod m, and it follows that the Artin map induces an inverse isomorphism (Z/mZ)× → Gal(Q(ζm )/Q): for every integer a coprime to m we have (a) ∈ IQm and ω −1 (ā) =  Q(ζm )/Q (a)  , where ā = a mod m. As you showed on Problem Set 4, the surjectivity of the Artin map follows immediately, since a ranges over all integers coprime to m. Now let L be a subfield of Q(ζm ). We cannot apply ω to Gal(L/Q), since Gal(L/Q) is a quotient of Gal(Q(ζm )/Q), not a subgroup, but the Artin map IQm → Gal(L/Q) is available; notice that the modulus m works for L as well as Q(ζm ), since any primes that ramify in L also ramify in Q(ζm ) and therefore divide m. By Proposition 21.1, the Artin map factors through the surjective homomorphism Gal(Q(ζm )/Q) → Gal(L/Q) induced by restriction and thus induces a surjective homomorphism (Z/mZ)× → Gal(L/Q). To sum up, we can now say the following about abelian extensions of Q: • Existence: for each integer m we have a ray class field Q(ζm ): an abelian extension ramified only at p|m with Galois group isomorphic to the ray class group (Z/mZ)× . • Completeness: every abelian extension of Q lies in a ray class field Q(ζm ). 18.785 Fall 2021, Lecture #21, Page 2 • Reciprocity: if L is an abelian extension of Q contained in the ray class field Q(ζm ), the Artin map IQm → Gal(L/Q) induces a surjective homomorphism from the ray class group (Z/mZ)× to Gal(L/Q), letting us view Gal(L/Q) as a quotient of (Z/mZ)× . All of these statements will be made more precise; in particular, we will refine the first two statements so that ray class fields are uniquely determined by the modulus m, and we will give an explicit description of the kernel of the Artin map that allows us to identify Gal(L/Q) with a quotient of (Z/mZ)× . But let us first consider how to generalize these statements to number fields other than Q and define the terms ray class field, and ray class group. In order to do so, we first need to make the role of the integer m more precise by introducing the notion of a modulus. 21.3 Moduli and ray class groups Recall that for a global field K we use MK to denote its set of places (equivalence classes of absolute values). We generically denote places by the symbol v, but for finite places, those arising from a discrete valuation associated to a prime p of K (a nonzero prime ideal of OK ), we may write p in place of v. We write v|∞ to indicate that v is an infinite place (one not arising from a prime of K); recall that when K is a number field all infinite places are archimedean, and they may be real (Kv ' R) or complex (Kv ' C). Definition 21.2. Let K be a number field. A modulus (or cycle) m for K is a function MK → Z≥0 with finite support such that for v|∞ have m(v) ≤ 1 with m(v) = 0 unless v Q we m(v) is a real place. We view m as a formal product v over MK , which we may factor as Y Y m = m0 m∞ , m0 := pm(p) , m∞ := v m(v) , p6 |∞ v|∞ where m0 is an OK -ideal and m∞ represents a subset of the real places of K; we use #m∞ to denote the number of real places in the support of m. If m and n are moduli for K we say that m divides n and write m|n if m(v) ≤ n(v) for all v ∈ MK . We define the product modulus mn by mn(v) := m(v) + n(v) for v - ∞ and mn(v) := max(m(v) + n(v), 1) for v | ∞; we also define gcd(m, n)(v) := min(m(v), n(v) and lcm(m, n)(v) := max(m(v), n(v)). The zero function is the trivial modulus, with m0 = (1) and #m∞ = 0. We use IK to denote the ideal class group of OK and define the following notation:1 • a fractional ideal I ∈ IK is coprime to m (or prime to m) if vp (I) = 0 for all p|m0 . m ⊆ I is the subgroup of fractional ideals coprime to m. • IK K m. • K m ⊆ K × is the subgroup of elements α ∈ K × for which (α) ∈ IK • K m,1 ⊆ K m is the subgroup of elements α ∈ K m with vp (α − 1) ≥ vp (m0 ) for all p|m0 and αv > 0 for v|m∞ (here αv is the image of α under K ,→ Kv ' R). m m m,1 . • Rm K ⊆ IK is the subgroup of principal fractional ideals (α) ∈ IK with α ∈ K The groups Rm K are called rays or ray groups. 1 This notation varies from author to author; there is no universally accepted notation for these objects (in particular, the modulus m may appear as a subscript rather than a superscript). Things will improve when we come to the adelic/idelic formulation of class field theory where there is more consistency. 18.785 Fall 2021, Lecture #21, Page 3 Definition 21.3. Let m be a modulus for a number field K. The ray class group for the modulus m is the quotient m m Clm K := IK /RK . A finite abelian extension L/K that is unramified at all places2 not in the support of m for m m → Gal(L/K) is equal to the ray group Rm is which the kernel of the Artin map ψL/K : IK K a ray class field for the modulus m. When m is the trivial modulus, the ray class group is the same as the usual class group ClK := cl(OK ), but in general the class group ClK is a quotient of the ray class group Clm K (as we will prove shortly). While not immediately apparent from the definition, we will see that ray class fields are uniquely determined by m, so it makes sense to speak of the ray class field for the modulus m (assuming existence). Remark 21.4. The definitions above make sense for any global field, but in our idealtheoretic treatment of class field theory we will mostly restrict our attention to number fields. Our adelic/idelic formulation of class field theory will address all global fields. Remark 21.5. If m(v) = 1 for every real place v of K then Clm K is a narrow ray class group. The narrow ray class group with m0 = (1) is the narrow class group; the usual class group ClK = cl OK is sometimes called the wide class group to distinguish the two. Note that the wide class group is a quotient of the narrow class group, thus smaller in general; this terminology can be confusing, but the thing to remember is that narrow equivalence is stronger than ordinary equivalence, so there are more narrow equivalence classes, in general. Of course for number fields with no real places (imaginary quadratic fields, in particular) there is no distinction. Example 21.6. For K = Q with the modulus m = (5) we have K m = {a/b : a, b 6≡ 0 mod 5} and K m,1 = {a/b : a ≡ b 6≡ 0 mod 5}. Thus m IK = {(1), (1/2), (2), (1/3), (2/3), (3/2), (3), (1/4), (3/4), (4/3), (4), (1/6), (6), . . .}, Rm K = {(1), (2/3), (3/2), (1/4), (4), (6), (1/6), (2/7), (7/2), . . .}. m,1 , but note that −2/3 ∈ K m,1 You might not have expected (2/3) ∈ Rm K , since 2/3 6∈ K and (−2/3) = (2/3). The ray class group is m m × Clm K = IK /RK = {[(1)], [(2)]} ' (Z/5Z) /{±1}, which is isomorphic to the Galois group of the totally real subfield Q(ζ5 )+ of Q(ζ5 ), which is the ray class field for this modulus. If we change the modulus to m = (5)∞ we instead m × get Rm K = {(1), (6), (1/6), (2/7), (7/2), . . .}, ClK ' (Z/5Z) , and the ray class field is Q(ζ5 ). Lemma 21.7. Let A be a Dedekind domain and let a be an A-ideal. Every ideal class in cl(A) contains an A-ideal coprime to a. Proof. Let I be a nonzero fractional ideal of A. For each prime p|a we can pick πp ∈ p such Q −v (I) that vq (πp ) = vq (p) for all q|a, by Corollary 3.17. If we then put α := p|a πp p , then vp (αI) = 0 for all p|a; thus αI is coprime to a and [αI] = [I]. 2 Archimedean places of K are unramified in L except for real places v with a complex place w of L above m m them. But if L is unramified at all p - m0 (necessary for ψL/K to be defined), and ker ψL/K = Rm K , then L will necessarily be unramified at all infinite places v - m∞ ; so in the definition of a ray class field it is enough for L to be unramified away from m0 . 18.785 Fall 2021, Lecture #21, Page 4 Now let S be the finite set of primes p for which vp (αI) < 0 and pick πp ∈ p such that vq (πp ) = vq (p) for all q ∈ S and q|a (again using Corollary 3.17). If we now put Q −v (αI) a := p∈S πp p ∈ A, then vp (aαI) ≥ 0 for all p and vp (aαI) = 0 for all p|a. Thus aαI is an A-ideal coprime to a and [aαI] = [I]. Theorem 21.8. Let m be a modulus for a number field K. We have an exact sequence × × 1 −→ OK ∩ K m,1 −→ OK −→ K m /K m,1 −→ Clm K −→ ClK −→ 1 and a canonical isomorphism K m /K m,1 ' {±1}#m∞ × (OK /m0 )× . Proof. Let us consider the composition of the maps K m,1 ⊆ K m and α 7→ (α): f g m K m,1 −→ K m −→ IK . × × ), ∩ K m,1 (since (α) = (1) ⇐⇒ α ∈ OK The kernel of f is trivial, the kernel of g ◦ f is OK × m m m,1 m m the kernel of g is OK , the cokernel of f is K /K , the cokernel of g ◦ f is ClK = IK /RK (by definition), and the cokernel of g is ClK (by Lemma 21.7). Applying the snake lemma (see [2, Lemma 5.13], for example) to the following commutative diagram with exact rows ← ← ← ← m → IK ← ∼ → ← → m → IK → K m /K m,1 → 1 π → g g◦f ← 1 ← f → K m,1 -← → K m ← 1 → 1 yields the exact sequence ker g ◦ f → ker g → ker π → coker g ◦ f → coker g → coker π: × × −→ K m /K m,1 −→ Clm 1 −→ OK ∩ K m,1 −→ OK K −→ ClK −→ 1, where the initial 1 follows from the fact that f is injective (and ker π = coker f ). We can write each α ∈ K m as α = a/b with a, b ∈ OK chosen so that (a) and (b) are both coprime to m0 . We now define the homomorphism   Y ϕ : Km →  {±1} × (OK /m0 )×  α 7→  v|m∞ Y v|m∞  sgn(αv ) × (ᾱ), where ᾱ = āb̄−1 ∈ (OK /m0 )× (here ā, b̄ are the images of a, b ∈ OK in OK /m0 , and they × both lie in (O (a) and (b) are coprime to m0 ). The ring (OK /m0 )× is isoQ K /m0 ) because morphic to p|m0 (OK /pm(p) )× , by the Chinese remainder theorem, and weak approximation (Theorem 8.5) implies that ϕ is surjective. The kernel of ϕ is clearly K m,1 , thus ϕ induces an isomorphism K m /K m,1 ' {±1}#m∞ × (OK /m0 )× . This isomorphism does not depend on the choices of a and b; every choice yields the same image ᾱ of α = a/b in (OK /m0 )× . 18.785 Fall 2021, Lecture #21, Page 5 Corollary 21.9. Let K be a number field and let m be a modulus for K. The ray class m m group Clm K is a finite abelian group whose cardinality hK := #ClK is given by hm K = × [OK φ(m)hK , × : OK ∩ K m,1 ] where hK := #ClK and φ(m) := #(K m /K m,1 ) = φ(m∞ )φ(m0 ), with Y (1 − N(p)−1 ). φ(m∞ ) = 2#m∞ , φ(m0 ) = #(OK /m0 )× = N(m0 ) p|m0 m In particular, hK divides hm K and hK divides hK φ(m). × × Proof. The exact sequence implies φ(m)/[OK : OK ∩ K m,1 ] = hm K /hK , and that both sides of this equality are integers. Computing the ray class number hm K is not a trivial problem, but there are algorithms for doing so; see [1], which considers this problem in detail. 21.4 Polar density We now want to prove the surjectivity of the Artin map for finite abelian extensions L/K of number fields (as noted in §21.2, we already know this for K = Q). In order to do so we first introduce a new way to measure the density of a set of primes that is defined in terms of a generalization of the Dedekind zeta function. Throughout this section and the next, all number fields are assumed to lie in some fixed algebraic closure of Q. Definition 21.10. Let K be a number field and let S be a set of primes of K. The partial Dedekind zeta function associated to S is the complex function Y ζK,S (s) := (1 − N(p)−s )−1 , p∈S which converges to a holomorphic function on Re(s) > 1 (by the same argument we used for ζK (s) in Lecture 18). If S is finite then ζK,S (s) is certainly holomorphic (and nonzero) on a neighborhood of 1. If S contains all but finitely many primes of K then it differs from ζK (s) by a holomorphic factor and therefore extends to a meromorphic function with a simple pole at s = 1, by Theorem 19.12. Between these two extremes the function ζK,S (s) may or may not extend to a function that is meromorphic on a neighborhood of 1, but if it does, or more generally, if some power of it does, then we can use the order of the pole at 1 (or the absence of a pole) to measure the density of S. n Definition 21.11. If for some integer n ≥ 1 the function ζK,S extends to a meromorphic function on a neighborhood of 1, the polar density of S is defined by m n m = −ords=1 ζK,S (s) ρ(S) := , n n1 n2 (so m is the order of the pole at s = 1, if one is present). Note that if ζK,S and ζK,s both extend to a meromorphic function on a neighborhood of 1 then we necessarily have n1 n1 n2 n2 n2 ords=1 ζK,S (s) = ords=1 ζK,S = n1 ords=1 ζK,S (s), 18.785 Fall 2021, Lecture #21, Page 6 which implies that ρ(S) does not depend on the choice of n. We will show below that (whenever it is defined) ρ(S) is a rational number in the interval [0, 1]. In Lecture 17 we encountered two other notions of density, the Dirichlet density P P −s −s p∈S N(p) p∈S N(p) d(S) := lim P = lim , 1 −s s→1+ s→1+ log s−1 p N(p) (the equality of the two expressions for d(S) follows from the fact that ζK (s) has a simple pole at s = 1, see Problem Set 9), and the natural density δ(S) := lim x→∞ #{p ∈ S : N(p) ≤ x} . #{p : N(p) ≤ x} On Problem Set 9 you proved that if S has a natural density then it has a Dirichlet density and the two coincide. We now show that the same is true of the polar density. Proposition 21.12. Let S be a set of primes of a number field K. If S has a polar density then it has a Dirichlet density and the two are equal. In particular, ρ(S) ∈ [0, 1] whenever it is defined. Proof. Suppose S has polar density ρ(S) = m/n. By taking the Laurent series expansion of n (s) at s = 1 and factoring out the leading nonzero term we can write ζK,S   X a 1 + ζK,S (s)n = ar (s − 1)r  , (s − 1)m r≥1 for some a ∈ C× . We must have a ∈ R>0 , since ζK,S (s) ∈ R>0 for s ∈ R>1 and therefore lims→1+ (s − 1)m ζK,S (s)n is a positive real number. Taking logs of both sides yields n X p∈S N(p)−s ∼ m log 1 s−1 (as s → 1+ ), which implies that S has Dirichlet density d(S) = m/n (note that log(a) = O(1) plays no role, since −m log(s − 1) → ∞ as s → 1+ ). Corollary 21.13. Let S be a set of primes of a number field K. If S has both a polar density and a natural density then the two coincide. We should note that not every set of primes with a natural density has a polar density, since the later is always a rational number while the former need not be. Recall that a degree-1 prime in a number field K is a prime with residue field degree 1 over Q, equivalently, a prime p whose absolute norm N(p) = [OK : p] = #Fp is prime. Proposition 21.14. Let S and T denote sets of primes in a number field K, let P be the set of all primes of K, and let P1 be the set of degree-1 primes of K. The following hold: (a) If S is finite then ρ(S) = 0; if P − S is finite then ρ(S) = 1. (b) If S ⊆ T both have polar densities, then ρ(S) ≤ ρ(T ). (c) If two sets S and T have finite intersection and any two of the sets S, T , and S ∪ T have polar densities then so does the third and ρ(S ∪ T ) = ρ(S) + ρ(T ). 18.785 Fall 2021, Lecture #21, Page 7 (d) We have ρ(P1 ) = 1, and ρ(S ∩ P1 ) = ρ(S) whenever S has a polar density. Proof. We first note that for any finite set S, the function ζK,S (s) is a finite product of nonvanishing entire functions and therefore holomorphic and nonzero everywhere (including at s = 1). If the symmetric difference of S and T is finite, then ζK,S (s)f (s) = ζK,T (s)g(s) for some nonvanishing functions f (s) and g(s) holomorphic on C. Thus if S and T differ by a finite set, then ρ(S) = ρ(T ) whenever either set has a polar density Part (a) follows, since ρ(∅) = 0 and ρ(P) = 1 (note that ζK,P (s) = ζK (s), and ords=1 ζK (s) = −1, by Theorem 19.12). Part (b) follows from the analogous statement for Dirichlet density proved on Problem Set 9. For (c) we may assume S and T are disjoint (by the argument above), in which case ζK,S∪T (s)n = ζK,S (s)n ζK,T (s)n for all n ≥ 1, and the claim follows. For (d), let P2 := P −P1 so that P = P1 tP2 . For each rational prime p there are at most := n [K : Q] (in fact n/2) primes p|p in P2 , each of which has absolute norm N(p) ≥ p2 . It follows by comparison with ζ(2s)n that the product defining ζK,P2 (s) converges absolutely to a holomorphic function on Re(s) > 1/2 and is therefore holomorphic (and nonvanishing, since it is an Euler product) on a neighborhood of 1; thus ρ(P2 ) = 0 and ρ(P1 ) = 1. We therefore have ρ(S ∩ P2 ) = 0, so ρ(S) = ρ(S ∩ P1 ) whenever ρ(S) exists, by (c). For a Galois extension of number fields L/K, let Spl(L/K) denote the set of primes of K that split completely in L. When K is clear from context we may just write Spl(L). Theorem 21.15. Let L/K be a Galois extension of number fields of degree n. Then ρ(Spl(L)) = 1/n. Proof. Let S be the set of degree-1 primes of K that split completely in L; it suffices to show ρ(S) = 1/n, by Proposition 21.14. Recall that p splits completely in L if and only if both the ramification index ep and residue field degree fp are equal to 1. Let T be the set of primes q of L that lie above some p ∈ S. For each q ∈ T lying above p ∈ S we have NL/K (q) = pfp = p, so N(q) = N(NL/K (q)) = N(p), thus q is a degree-1 prime, since p is. On the other hand, if q is any unramified degree-1 prime of L and p = q ∩ OK , then N(q) = N(NL/K (q)) = N(pfp ) is prime, so we must have fp = 1, and ep = 1 since q is unramified, which implies that p is a degree-1 prime that splits completely in L and is thus an element of S. Only finitely many primes ramify, so all but finitely many of the degree-1 primes in L lie in T , thus ρ(T ) = 1, by Proposition 21.14. Each p ∈ S has exactly n primes q ∈ T lying above it (since p splits completely), and we have Y Y Y (1 − N(q)−s )−1 = (1 − N(NL/K (q))−s )−1 = (1 − N(p)−s )−n = ζK,S (s)n . ζL,T (s) = q∈T It follows that ρ(S) = n1 ρ(T ) = q∈T 1 n p∈S as desired. Corollary 21.16. If L/K is a finite extension of number fields with Galois closure M/K of degree n, then ρ(Spl(L)) = ρ(Spl(M )) = 1/n. Proof. A prime p of K splits completely in L if and only if it splits completely in all the conjugates of L in M ; the Galois closure M is the compositum of the conjugates of L, so p splits completely in L if and only if it splits completely in M . 18.785 Fall 2021, Lecture #21, Page 8 Corollary 21.17. Let L/K be a finite Galois extension of number fields with Galois group G := Gal(L/K) and let H be a normal subgroup of G. The set S of primes for which Frobp ⊆ H has polar density ρ(S) = #H/#G. Proof. Let F = LH ; then F/K is Galois (since H is normal) and Gal(F/K) ' G/H. For each unramified prime p of K, the Frobenius class Frobp lies in H if and only if every σq ∈ Frobp acts trivially on LH = F , which occurs if and only if p splits completely in F . By Theorem 21.15, the density of this set of primes is 1/[F : K] = #H/#G. If S and T are sets of primes whose symmetric difference is finite, then either ρ(S) = ρ(T ) or neither set has a polar density. Let us write S ∼ T to indicate that two sets of primes have finite symmetric difference (this is clearly an equivalence relation), and partially order sets of primes by defining S - T ⇔ S ∼ S ∩ T (in other words, S − T is finite). If S and T have polar densities, then S - T implies ρ(S) ≤ ρ(T ), by Proposition 21.14. Theorem 21.18. If L/K and M/K are two finite Galois extensions of number fields then L ⊆ M ⇐⇒ Spl(M ) - Spl(L) ⇐⇒ Spl(M ) ⊆ Spl(L), L = M ⇐⇒ Spl(M ) ∼ Spl(L) ⇐⇒ Spl(M ) = Spl(L), and the map L 7→ Spl(L) is an injection from the set of finite Galois extensions of K (inside some fixed algebraic closure) to sets of primes of K that have a positive polar density. Proof. The implications L ⊆ M ⇒ Spl(M ) ⊆ Spl(L) ⇒ Spl(M ) - Spl(L) are clear, so it suffices to show that Spl(M ) - Spl(L) ⇒ L ⊆ M . A prime p of K splits completely in the compositum LM if and only if it splits completely in both L and M : the forward implication is clear and for the reverse, note that if p splits completely in both L and M then it certainly splits completely in L ∩ M , so we may assume K = L ∩ M ; we then have Gal(LM/K) ' Gal(L/K) × Gal(M/K), and if the decomposition subgroups of all primes above p are trivial in both Gal(L/K) and Gal(M/K) then the same applies in Gal(LM/K). Thus Spl(LM ) = Spl(L) ∩ Spl(M ). It follows that Spl(M ) - Spl(L) ⇒ Spl(LM ) ∼ Spl(M ). By Theorem 21.15, we have ρ(Spl(M )) = 1/[M : K] and ρ(Spl(LM )) = 1/[LM : K], thus Spl(LM ) ∼ Spl(M ) implies [LM : K] = ρ(Spl(LM ))−1 = ρ(Spl(M ))−1 = [M : K], in which case LM = M and L ⊆ M . This proves Spl(M ) - Spl(L) ⇒ L ⊆ M , so the three conditions in the first line of biconditionals are all equivalent, and this immediately implies the second line of biconditionals. The last statement of the theorem is clear, since Spl(L) has positive polar density, by Theorem 21.15. 21.5 Ray class fields and Artin reciprocity As a special case of Corollary 21.16, if F/K is a finite extension of number fields in which all but finitely many primes split completely, then [F : K] = 1 and therefore F = K. We will use this fact to prove that the Artin map is surjective. Theorem 21.19. Let L/K be an abelian extension of number fields and m a modulus divism m → Gal(L/K) is surjective. ible by all ramified primes. Then the Artin map ψL/K : IK 18.785 Fall 2021, Lecture #21, Page 9 m Proof. Let H ⊆ Gal(L/K) be the image of ψL/K and let F = LH be its fixed field, which we note is a Galois extension of K, since H is normal (because Gal(L/K) is abelian). For m the automorphism ψ m (p) ∈ H acts trivially on F = LH , therefore p each prime p ∈ IK L/K m contains all but finitely many primes p of K, so the splits completely in F . The group IK polar density of the set of primes of K that split completely in F is 1. Thus [F : K] = 1 and H = Gal(L/K), by Corollary 21.16. m We now show that the kernel of the Artin map ψL/K uniquely determines the field L. Theorem 21.20. Let m be a modulus for a number field K and let L and M be finite abelian m m extensions of K unramified at all primes not in the support of m. If ker ψL/K = ker ψM/K then L = M . In particular, ray class fields are unique whenever they exist. Proof. Let S be the set of primes of K that do not divide m. Each prime p in S is unramified in both L and M , and p splits completely in L (resp. M ) if and only if it lies in the kernel m m m m of ψL/K (resp. ψM/K ). If ker ψL/K = ker ψM/K then m m Spl(L) ∼ (S ∩ ker ψL/K ) = (S ∩ ker ψM/K ) ∼ Spl(M ), and therefore L = M , by Theorem 21.18. Theorem 21.19 implies that we have an exact sequence m m 1 → ker ψL/K → IK → Gal(L/K) → 1. One of the key results of class field theory is that for a suitable choice of the modulus m, m we have Rm K ⊆ ker ψL/K . This implies that the Artin map induces an isomorphism between m m Gal(L/K) and a quotient of the ray class group Clm K = IK /RK . When L is the ray class field for the modulus m, the Artin map allows us to relate subfields of L to quotients of the ray class group Clm K ' Gal(L/K) in a way that we will make more precise in the next lecture; this is known as Artin reciprocity. References [1] Henri Cohen, Advanced topics in computational number theory, Springer, 2000. [2] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, 2013. 18.785 Fall 2021, Lecture #21, Page 10 18.785 Number theory I Lecture #22 22 Fall 2021 11/24/2021 The main theorems of global class field theory In this lecture we refine the correspondence between quotients of ray class groups and subfields of ray class fields given by the Artin map so that we can more precisely state the main theorems of global class field theory (for number fields) in their ideal-theoretic form. Let us first recall the notational setup. We have a number field K and a modulus m : MK → Z≥0 that we as a formal Q view m(p) product over the places of K; we may write m is a product Q= m0 m∞ , where m0 := p over primes (finite places) of K and m∞ := v|∞ v m(v) defines a subset of the real places of K (recall that for v|∞ we have m(v) ≤ 1 with m(v) = 0 if v is not real). The moduli for K are partially ordered by the divisibility relation m|n, which holds if and only if m(v) ≤ n(v) for all v ∈ MK . We then define m ⊆ I , the subgroup of fractional ideals prime to m (equivalently, m ); • IK 0 K m; • K m ⊆ K × , the subgroup of α ∈ K × for which (α) ∈ IK • K m,1 ⊆ K m , the subgroup of α ∈ K m for which vp (α − 1) ≥ vp (m0 ) for p|m0 and αv > 0 for v|m∞ (here αv ∈ R is the image of α under the real-embedding v); m m m,1 (the ray group for m); • Rm K ⊆ IK the subgroup of ideals (α) ∈ IK with α ∈ K m m • Clm K := IK /RK (the ray class group for m); • Spl(L) := Spl(L/K), the set of primes of K that split completely in an extension L; m m → Gal(L/K), Artin map of an abelian extension L/K unramified at p - m. • ψL/K : IK In the previous lecture we defined the ray class field of K for the modulus m as a finite m abelian extension L/K unramified at all p - m such that the kernel of the Artin map ψL/K is equal to the ray group Rm K . We did not prove that such fields exist, but we did prove that there is at most one of them; see Theorem 21.20. Let K(m) denote this field. Assuming the ray class field K(m) exists, it follows from the surjectivity of the Artin m m → Gal(K(m)/K) proved in Theorem 21.19 that we have a canonical map ψK(m)/K : IK isomorphism m m Clm K = IK /RK ' Gal(K(m)/K) between the ray class group and the Galois group of the ray class field. More generally, if L is any intermediate field between K and K(m), the kernel of the Artin map is a subgroup m that contains the ray group C ⊆ IK m Rm K ⊆ C ⊆ IK , and we have an isomorphism m IK /C ' Clm K /C ' Gal(L/K) m m where C denotes the image of C in Clm K = IK /RK under the quotient map. m m containing Rm (a Thus if L is a subfield of K(m) then ker ψL/K is a subgroup of IK K congruence subgroup, as defined below). To prove that a given abelian extension L/K lies in a ray class field, it is enough to show that there exists a modulus m for K such that m Rm K ⊆ ker ψL/K , since we then have Spl(K(m)) - Spl(L) and L ⊆ K(m), by Theorem 21.18. In this lecture we want to better understand the structure of congruence subgroups, and to specify a minimal modulus m for which we should expect a given finite abelian extension L/K to lie in a subfield of the ray class field K(m); this minimal modulus is known as the conductor of the extension. So far we have not addressed this question even for K = Q (but see Problem Set 10); our proof of the Kronecker-Weber theorem showed that every abelian extension lies in some cyclotomic field Q(ζm ), but we made no attempt to determine such an integer m (or more precisely, a modulus m of the form m = (m)∞ or m = (m)). 22.1 Congruence subgroups Our presentation of congruence subgroups in this section follows [1, 3.3], but our notation differs slightly. Definition 22.1. Let K be a number field and let m be a modulus for K. A congruence m that contains Rm . We use C to denote subgroup for the modulus m is a subgroup C of IK K m m m the image of C in IK /RK = ClK under the quotient map. As explained above, congruence subgroups are precisely the groups we expect to arise as m m → Gal(L/K) associated to a finite abelian extension the kernel of an Artin map ψL/K : IK L/K, for a suitable choice of modulus m. The choice of m is critical; as can be seen in m Example 22.2 below, ker ψL/K need not be a congruence subgroup for the modulus m; there are constraints on the modulus m that must be satisfied beyond the basic requirement that m m must be divisible by all the primes of K that ramify in L (so that ψL/K is defined). Example 22.2. Let K = Q, and consider the cyclic cubic extension L := Q[x]/(x3 −3x−1), m which is ramified only at 3. The Artin map ψL/K is well-defined for any modulus m divisible by (3). The√ray class field for m = (3) is Q(ζ3 )+ = Q, and the ray class field for m = (3)∞ is m Q(ζ3 ) = Q( −3), neither of which contains L, so ker ψL/K does not contain Rm K for either of these moduli and is not a congruence subgroup. On the other hand, L is equal to Q(ζ9 )+ , m the ray class field for m = (9), so ker ψL/K contains (and is equal to) Rm K , and is thus a congruence subgroup for the modulus m = (9). m n If ker ψL/K is a congruence subgroup for the modulus m, then ker ψL/K is a congruence n n subgroup for each modulus n divisible by m. If m divides n then RK ⊆ Rm K and ψL/K is n m n the restriction of ψL/K to IK , which contains RK . If m and n are supported on the same m = I n and ψ m n m n primes, then IK K L/K = ψL/K , but the ray groups RK and RK may differ. To deal with these complications, we define an equivalence relation on congruence subgroups and show that each equivalence class has a canonical representative whose modulus divides the modulus of every equivalent congruence subgroup. Definition 22.3. Let K be a number field with moduli m1 and m2 . If C1 is a congruence subgroup for m1 and C2 is a congruence subgroup for m2 , then we say that (C1 , m1 ) and (C2 , m2 ) are equivalent and write (C1 , m1 ) ∼ (C2 , m2 ) whenever m1 m2 IK ∩ C 2 = IK ∩ C1 . Note that when m1 = m2 this reduces to C1 = C2 . Proposition 22.4. Let K be a number field. The relation (C1 , m1 ) ∼ (C2 , m2 ) is an equivm1 m2 alence relation. If (C1 , m1 ) ∼ (C2 , m2 ) then IK /C1 ' IK /C2 are related by a canonical isomorphism that preserves cosets of fractional ideals prime to both m1 and m2 . 18.785 Fall 2021, Lecture #22, Page 2 Proof. The relation ∼ is clearly symmetric, and reflexive. To show that it is transitive, let C1 , C2 , C3 be congruence subgroups for moduli m1 , m2 , m3 and suppose (C1 , m1 ) ∼ (C2 , m2 ) m3 m1 m2 m3 and (C2 , m2 ) ∼ (C3 , m3 ). Let I ∈ IK ∩ C1 and pick α ∈ K m1 m3 ,1 so that αI ∈ IK (this m1 m3 m1 is possible by Lemma 21.7 and Theorem 8.5). Then (α) ∈ RK ⊆ RK ⊆ C1 and I ⊆ C1 , m1 m2 m3 m2 so αI ∈ C1 , and we also have αI ∈ IK ⊆ IK , so m2 m1 αI ∈ IK ∩ C 1 = IK ∩ C2 ⊆ C2 , m1 m2 m3 m3 since C1 ∼ C2 , and αI ∈ IK ⊆ IK , so m3 m2 αI ∈ IK ∩ C 2 = IK ∩ C3 ⊆ C3 , −1 ∈ C , since C 1 m3 3 since C2 ∼ C3 . We have (α) ∈ Rm ⊆ Rm 3 3 K K , so (α) ∈ C3 and therefore (α) m1 m1 is a group. Thus α−1 αI = I ∈ C3 , and we also have I ∈ C1 ⊆ IK , so I ∈ IK ∩ C3 . Since m3 I ∈ IK ∩ C1 was chosen arbitrarily, this proves that m3 m1 IK ∩ C1 ⊆ IK ∩ C3 . The reverse inclusion follows by symmetry, so (C1 , m1 ) ∼ (C3 , m3 ) as desired. m1 For the last statement, for any fractional ideal I ∈ IK we can pick α ∈ K m1 ,1 so that m2 m2 αI ∈ IK (via Lemma 21.7 and Theorem 8.5). The image of αI in IK /C2 does not depend on m m2 0 m ,1 0 2 1 the choice of α, since for any α ∈ K with α I ∈ IK we have (αI)/(α0 I) = (α/α0 ) ∈ IK m1 m1 m2 m2 m2 0 1 and (α/α0 ) ∈ Rm K , so (α/α ) ∈ IK ∩ RK = IK ∩ RK ⊆ RK . This defines a group m1 m2 m2 m1 homomorphism ϕ : IK → IK /C2 . For I ∈ C1 , we have αI ∈ IK ∩ C1 = IK ∩ C2 ⊆ C2 , but m1 m2 for I ∈ IK − C1 we have αI ∈ IK − C1 and therefore αI 6∈ C2 , so ker ϕ = C1 . It follows m1 m2 that ϕ induces an injective homomorphism IK /C1 → IK /C2 , and by symmetry we have an m1 m2 injective homomorphism in the opposite direction, so IK /C1 ' IK /C2 as claimed. This isomorphism is independent of the choice of α used to define it (hence canonical), and for fractional ideals I coprime to both m1 and m2 we can choose α = 1, in which case m1 m2 the coset of I in IK /C1 will be identified with the coset of I in IK /C2 . We now observe that if C is a congruence subgroup for two moduli m1 and m2 , then (C, m1 ) ∼ (C, m2 ). In particular, each subgroup of IK lies in at most one equivalence class of congruence subgroups. We can thus view the equivalence relation (C1 , m1 ) ∼ (C2 , m2 ) as an equivalence relation on the congruence subgroups of IK and write C1 ∼ C2 without ambiguity. It follows from Proposition 22.4 that each equivalence class of congruence subgroups uniquely determines a finite abelian group that is the quotient of a ray class group. Within an equivalence class of congruence subgroups there can be at most one congruence subgroup for each modulus (since C1 ∼ C2 ⇔ C1 = C2 whenever C1 and C2 are congruence subgroups for the same modulus). The following lemma gives a criterion for determining when there exists a congruence subgroup of a given modulus within a given equivalence class. Lemma 22.5. Let C1 be a congruence subgroup of modulus m1 for a number field K. There exists a congruence subgroup C2 of modulus m2 |m1 equivalent to C1 if and only if m1 2 IK ∩ Rm K ⊆ C1 , 2 in which case C2 = C1 Rm K . 18.785 Fall 2021, Lecture #22, Page 3 m1 m2 m1 m2 Proof. Note that m2 |m1 implies IK ⊆ IK , so C1 ⊆ IK ⊆ IK . m1 m2 2 Suppose C2 ∼ C1 has modulus m2 . Then IK ∩ C2 = IK ∩ C1 = C1 , and Rm K ⊆ C2 , so m1 m2 m1 m2 m2 IK ∩ RK ⊆ C1 as claimed. Now suppose IK ∩ RK ⊆ C1 , and let C2 := C1 RK . Then C2 is a congruence subgroup of modulus m2 and we have m2 m1 m1 m2 m1 2 IK ∩ C1 = C1 = C1 (IK ∩ Rm K ) = IK ∩ C1 RK = IK ∩ C2 , so C1 ∼ C2 . The equivalence class of C1 contains at most one congruence subgroup of modulus 2 m2 , so if one exists it must be C2 = C1 Rm K . Proposition 22.6. Let C1 ∼ C2 be congruence subgroups of modulus m1 and m2 , respectively. There exists a congruence subgroup C ∼ C1 ∼ C2 with modulus n := gcd(m1 , m2 ). m2 m1 Proof. Put m := lcm(m1 , m2 ) and D := IK ∩ C 1 = IK ∩ C2 ; then m1 m2 m Rm K = RK ∩ RK ⊆ D ⊆ IK , so D is a congruence subgroup of modulus m, and we have m 1 IK ∩ Rm K ⊆D and m 2 IK ∩ Rm K ⊆ D, so D ∼ C1 ∼ C2 , by Lemma 22.5. To prove the existence of an equivalent congruence m ∩ Rn ⊆ D (again by Lemma 22.5). subgroup C of modulus n it suffices to show IK K n m So let a = (α) ∈ IK ∩ RK , and choose β ∈ K m ∩ K m2 ,1 so that αβ ∈ K m1 ,1 (this is possible by Theorem 8.5 because m = lcm(m1 , m2 ) and n = gcd(m1 , m2 )). Then (β) ∈ D m ∩ Rm1 ⊆ D, so β −1 βa = a ∈ D. Thus I m ∩ Rn ⊆ D and therefore C = DRn and βa ∈ IK K K K K is a congruence subgroup of modulus n equivalent to D ∼ C1 ∼ C2 . Corollary 22.7. Let C be a congruence subgroup of modulus m for a number field K. There is a unique congruence subgroup in the equivalence class of C whose modulus c divides the modulus of every congruence subgroup equivalent to C. Definition 22.8. Let C be a congruence subgroup for a number field K. The unique modulus c := c(C) given by Corollary 22.7 is the conductor of C, and we say that C is primitive if C = CRcK (the unique congruence subgroup of modulus c equivalent to C). Proposition 22.9. Let C be a primitive congruence subgroup of modulus m for a number field K. Then m is the conductor of every congruence subgroup of modulus m contained in C; in particular, m is the conductor of Rm K. Proof. Let C0 ⊆ C be a congruence subgroup of modulus m and let c be its conductor. Then m ∩ Rc ⊆ C ⊆ C, by Lemma 22.5, and this implies that there is a congruence c|m and IK 0 K subgroup of modulus c equivalent to C, and therefore m|c, so c = m. The proposition implies that a modulus m occurs as a conductor if and only if Rm K is primitive. This does not always hold: consider K = Q and m = (2), for example; the (2) (2) (2) (1) (1) (2) (2) (1) conductor of RQ = IQ is (1), since RQ ∩ IQ = IQ ∩ IQ implies RQ ∼ IQ . Thus (2) is not the conductor of any congruence subgroup for Q. 18.785 Fall 2021, Lecture #22, Page 4 22.2 Ray class characters We now want to prove a generalization of Dirichlet’s theorem on primes in arithmetic progressions. Given a congruence subgroup C for a modulus m we would like to compute the m that lie in C. We first Dirichlet density d(C) := d({p ∈ C}) of the set of prime ideals p ∈ IK need to generalize our notion of a Dirichlet character. Definition 22.10. Let K be a number field and let χ : IK → C be a totally multiplicative function with finite image; so χ(OK ) = 1, χ(IJ) = χ(I)χ(J) for all I, J ∈ IK , and χ restricts to a homomorphism from a subgroup of IK to a finite subgroup of U(1) whose m and Rm ⊆ ker χ, kernel we denote ker χ. If m is a modulus for K such that χ−1 (U(1)) = IK K then χ is a ray class character of modulus m and its kernel is a congruence subgroup of modulus m. Equivalently, χ is the extension by zero of a character of the finite abelian m m m group Clm K = IK /RK defined by setting χ(I) = 0 for I 6∈ IK . Example 22.11. For K = Q there is a one-to-one correspondence between Dirichlet characters χ : Z → C and ray class characters χ0 : IQ → C with χ(a) = χ0 ((a)) for all a ∈ Z≥1 . Each Dirichlet character χ of modulus m corresponds to a ray class character of modulus m = (m)∞ whose conductor divides (m) if and only if χ is an even Dirichlet character, meaning that χ(−1) = 1. Definition 22.12. Let χ1 , χ2 be ray class characters of moduli m1 , m2 of a number field K, m2 with m1 |m2 . If χ2 (I) = χ1 (I) for all I ∈ IK , then χ2 is induced by χ1 . A ray class character is primitive if it is not induced by any ray class character other than itself. Definition 22.13. The conductor of a ray class character χ is the conductor c(χ) := c(ker χ) of its kernel (as a congruence subgroup). Theorem 22.14. A ray class character is primitive if and only if its kernel is primitive. Every ray class character χ is induced by a unique primitive ray class character χ e. m /(ker χ) → U(1) be the Proof. Let χ be a ray class character of modulus m, let κ : IK group character induced by χ, and let C be the primitive congruence subgroup equivalent to ker χ with modulus c = c(χ) dividing m given by Corollary 22.7. By Proposition 22.4, ∼ c /C → m /(ker χ) that we can use to define a ray we have a canonical isomorphism ϕ : IK IK c /C. class character χ e of modulus c as the extension by zero of the character κ ◦ ϕ of IK c m e(I) = χ(I) for all The isomorphism ϕ preserves cosets of fractional ideals in IK ⊆ IK , so χ m and χ is induced by χ I ∈ IK e. If χ2 is a ray class character of conductor m2 induced by a ray class character χ1 of m2 m1 conductor m1 , then ker χ1 ∩ IK = ker χ2 = ker χ2 ∩ IK and ker χ1 ∼ ker χ2 , and we also m1 m2 note that if χ1 6= χ2 then IK 6= IK and m1 6= m2 . It follows that χ e is primitive, it is the unique primitive ray class character that induces χ. Thus χ is primitive if and only if it is equal to χ e, which holds if and only if ker χ = ker χ e is primitive. Theorem 22.14 is a direct generalization of Theorem 18.13 for Dirichlet characters. For a modulus m of K we use X(m) to denote the set of primitive ray class characters of conductor dividing m, which we note is in bijection with the character group of Clm K , and thus has a m group structure given by χ e1 χ e2 = χ ] 1 χ2 . Indeed, for each character of ClK , its extension by zero is a ray class character χ of modulus m induced by a primitive ray class character χ e whose conductor divides m, and each primitive ray class character χ e of conductor dividing m induces a ray class character χ of modulus m that determines a character of Clm K ; these two maps are inverses, hence bijections. This generalizes Corollary 18.16. 18.785 Fall 2021, Lecture #22, Page 5 Definition 22.15. A ray class character χ is principal if ker χ = χ−1 (U(1)). We use denote the unique primitive principal ray class character. 1 to Remark 22.16. For Dirichlet characters, 1 is the unique Dirichlet character of conductor 1, but for ray class characters this holds only when the class group ClK is trivial (as when K = Q). In general, the extension by zero of any character of ClK is a ray class character of conductor (1) and need not be principal (but is necessarily primitive). Like Dirichlet characters, each ray class character has an associated L-function. Definition 22.17. The Weber L-function L(s, χ) of a ray class character χ for a number field K is the complex function −1 X Y χ(a)N(a)−s , L(s, χ) := 1 − χ(p)N(p)−s = p a where the product is over prime ideals of OK and the sum is over nonzero OK -ideals; the product and sum both converge to a non-vanishing holomorphic function on Re(s) > 1 (this follows from comparison with the Dedekind zeta function ζK (s), since |χ(a)| ≤ 1). Example 22.18. For K = Q, Weber L-functions are Dirichlet L-functions. For any number field K, the Weber L-function for 1 is the Dedekind zeta function: L(s, 1) = ζK (s). More generally, we have the following theorem, which is analogous to Theorem 19.15 but avoids the need to assume the existence of a ray class field. Proposition 22.19. Let χ be a ray class character of modulus m for a number field K of degree n. Then L(s, χ) extends to a meromorphic function on Re(s) > 1 − n1 that has at most a simple pole at s = 1 and is holomorphic if χ is non-principal. Proof. Associated to each ray class γ ∈ Clm K we have a Dirichlet series X ζK,γ (s) := N(a)−s a∈γ that is holomorphic on Re(s) > 1. For the trivial modulus m, our proof of analytic class number formula (Theorem 19.12) implies that ζK,γ (s) has a meromorphic continuation to 1 − n1 with a simple pole at s = 1 and residue ρ = 2r (2π)2 RK /(ωK |DK |1/2 ), independent of γ. Recall that in our proof of Theorem 19.8 we treated each γ ∈ ClK = cl(OK ) separately and obtained the same value of ρ for each γ, leading to the residue ρK = hK ρ that appears in Theorem 19.12. The same proof works for Clm K , mutatis mutandi: replace covol(OK ) with covol(m0 ), × m := covol(π(Log(O × ∩ K m,1 ))), and replace the regulator RK = covol(π(Log(OK ))) with RK K m := #(µ ∩ K m,1 ). The exact value of ρ is not important to us replace wK = #µK with wK K here, the key point is that ζK,γ (s) has a meromorphic continuation to Re(s) > 1 − n1 with a simple pole at s = 1 whose residue ρ depends only on K and m (not γ). We then have X L(s, χ) = χ(γ)ζK,γ (s) γ∈Clm K = X γ∈Clm K   X χ(γ) ζK,γ (s) − ρ ζ(s) + χ(γ)ρ ζ(s), γ∈Clm K 18.785 Fall 2021, Lecture #22, Page 6 The first sum is a finite sum of functions holomorphic on Re(s) > 1 − n1 (since ζ(s) has a simple pole at s = 1 with residue 1), and the second sum vanishes whenever χ is nonprincipal (by Corollary 18.38). The proposition follows. We now prove a generalization of Dirichlet’s theorem on primes on arithmetic progressions for arbitrary number fields. We proved the nonvanishing of Dirichlet L-functions L(1, χ) for non-principal χ using the analytic class number formula for Q(ζm ), the ray class field Q((m)∞), by writing the Dedekind zeta function for Q(ζm ) as a product of Dirichlet L-functions (see Theorem 19.15). A similar approach works for Weber L-functions, assuming the existence of ray class fields K(m): the Dedekind zeta function of K(m) is equal to the product of the Weber L-functions for χ ∈ X(m). But we will prove the non-vanishing of L(1, χ) for non-principal χ without assuming the existence of ray class fields. For a congruence subgroup C, let X(C) denote the set of primitive ray class characters whose kernels contain C. If C is a congruence subgroup of modulus m then X(C) is a subgroup m /C and we may view X(C) as the character of X(m) isomorphic to the character group of IK m group of IK /C. Theorem 22.20. Let C be a congruence subgroup of modulus m for a number field K and m : C]. The set of primes {p ∈ C} has Dirichlet density let n := [IK ( 1 if L(1, χ) 6= 0 for all χ 6= 1 in X(C), d(C) = n 0 otherwise. In fact d(C) = n1 always holds, as we will prove in Corollary 22.22 below, but it is easier to prove the theorem as stated and then use this to derive the corollary. Proof. We proceed as in the proof of Dirichlet’s theorem on primes in arithmetic progressions (see §18.4). We first construct the indicator function for the set {p ∈ C}: ( 1 if p ∈ C, 1 X χ(p) = n 0 otherwise. χ∈X(C) Note that summing over χ ∈ X(C) is equivalent to summing over the character group of P m m /C is the IK /C, so Corollary 18.38 applies: therefore P χ(p) = 0 unless the image of p in IK identity, meaning that p ∈ C, in which case χ(p) = #X(C) = n. + As s → 1 we have X log L(s, χ) ∼ χ(p)N(p)−s , p and therefore X χ∈X(C) log L(s, χ) ∼ X X χ(p)N(p)−s χ∈X(C) p ∼n X N(p)−s . p∈C By Proposition 22.19, we may write L(s, χ) = (s − 1)e(χ) g(s) 18.785 Fall 2021, Lecture #22, Page 7 for some function g(s) that is holomorphic and nonvanishing on a neighborhood of 1, where e(χ) := ords=1 L(s, χ) is −1 when χ = 1, and e(χ) ≥ 0 otherwise. We have log X X 1 1 − e(χ) log ∼n N(p)−s . s−1 s−1 χ6=1 p∈C 1 Dividing both sides by n log s−1 yields 1− P χ6=1 e(χ) n thus d(C) = lim s→1+ ∼ P P p∈C p∈C N(p)−s 1 log s−1 N(p)−s 1 log s−1 = 1− (as s → 1+ ), P χ6=1 e(χ) n . The e(χ) are integers and the Dirichlet density is nonnegative, so either e(χ) = 0 for all χ 6= 1, in which case L(1, χ) 6= 0 for all χ 6= 1 and d(C) = n1 , or e(χ) = 1 for exactly one of the χ 6= 1 and d(C) = 0. (in fact this never happens, as noted above). Proposition 22.21. Let C be a congruence subgroup of modulus m for a number field K m : C]. For every I ∈ I m the set {p ∈ IC} has Dirichlet density and let n := [IK K ( 1 if L(1, χ) 6= 0 for all characters χ 6= 1 in X(C), d(IC) = n 0 otherwise. Proof. The proof is the same as in Theorem 22.20, except we now use the indicator function ( 1 if p ∈ IC, 1 X χ(I)−1 χ(p) = n 0 otherwise, χ∈X(C) and obtain X χ∈X(C) χ(I)−1 log L(s, χ) ∼ X X χ∈X(C) p χ(I)−1 χ(p)N(p)−s ∼ n X N(p)−s . p∈IC The rest of the proof is the same. Corollary 22.22. Let C be a congruence subgroup of modulus m for a number field K and m : C]. For every ideal I ∈ I m the set {p ∈ IC} has Dirichlet density 1/n, and for let n := [IK K every χ 6= 1 in X(C) we have L(1, χ) 6= 0. m be a complete set of coset representatives for C ⊆ I m . All Proof. Let I1 , . . . , In ∈ IK K m , hence in one of the cosets I C partitioning I m , but finitely many primes p of K lie in IK j K therefore d(I1 C) + · · · + d(In C) = 1. By Proposition 22.21, every term in this sum is either 0 or 1/n, and the equality implies they must all be equal to 1/n, which then implies L(1, χ) 6= 0 for all χ 6= 1 in X(C). 18.785 Fall 2021, Lecture #22, Page 8 Corollary 22.23. Let L/K be an abelian extension of number fields and let C be a congruence subgroup for a modulus m of K. If Spl(L) - {p ∈ C} then m [IK : C] ≤ [L : K], with equality whenever Spl(L) ∼ {p ∈ C}. Proof. We know from Theorem 21.15 that Spl(L) has polar density 1/[L : K], and this is also m : C], its Dirichlet density, by Proposition 21.12. The set {p ∈ C} has Dirichlet density 1/[IK by Theorem 22.22, and Spl(L) - {p ∈ C} (by assumption), so 1 1 = d(Spl(L)) ≤ d(C) = m . [L : K] [IK : C] 22.3 The conductor of an abelian extension We now introduce another notion of conductor, one attached to an abelian extension of number fields, which is defined as a product of local conductors attached to corresponding abelian extensions of the local field Kv for each place v ∈ MK . Definition 22.24. Let L/K be a finite abelian extension of local fields. The conductor c(L/K) is defined as follows.1 If K is archimedean then c(L/K) = 1 when K ' R and L ' C and c(L/K) = 0 otherwise. If K is nonarchimedean and p is the maximal ideal of its valuation ring OK , then c(L/K) := min{n : 1 + pn ⊆ NL/K (L× )} × × ). If L/K is a finite abelian extension , with 1 + p0 := OK (here 1 + pn is a subgroup of OK of global fields then its conductor is the modulus c(L/K) : MK → Z v 7→ c(Lw /Kv ) where Kv is the completion of K at v and Lw is the completion of L at a place w|v. (the fact that L/K is Galois ensures that c(Lw /Kv ) is the same for every w|v). As with any modulus, we may view the finite part of c(L/K) as an OK -ideal and the infinite part as a subset of ramified infinite places. It is not hard to show that conductor is supported on ramified places (in particular, it has finite support, as required for a modulus). More generally, we have the following. Proposition 22.25. Let L/K be a finite abelian extension of local or global fields. For each prime p of K we have   if and only if p is unramified, 0 vp (c(L/K)) = 1 if and only if p is ramified tamely,   ≥ 2 if and only if p is ramified wildly. Proof. See Problem Set 11. 1 Many authors use f(L/K) rather than c(L/K), we use c to avoid confusion with the residue field degree. 18.785 Fall 2021, Lecture #22, Page 9 The finite part of the conductor of an abelian extension divides the discriminant ideal and is divisible by the same set of primes, but the valuation of the conductor at these primes is typically smaller than that of the discriminant. For example, the discriminant of the extension Q(ζp )/Q is (p)p−2 , but its conductor is (p)∞. Lemma 22.26. Let L1 /K and L2 /K be two finite abelian extensions of a local or global field K. If L1 ⊆ L2 then c(L1 /K) divides c(L2 /K). Proof. If K ' R, C the result is clear, and for nonarchimedean local K we may apply × × NL2 /K (L× 2 ) = NL1 /K (NL2 /L1 (L2 )) ⊆ NL1 /K (L1 ). The global case follows. 22.4 Norm groups m m → Gal(L/K). We can now identify a candidate for the kernel of the Artin map ψL/K : IK Recall from Lecture 6 that the norm map NL/K : IL → IK can be defined by Y i qni i 7→ Y pini fi , i where pi := qi ∩ OK and fi := [Fqi : Fpi ] is the residue field degree. Definition 22.27. Let L/K be a finite abelian extension of number fields and let m be a modulus for K divisible by the conductor of L/K. The norm group (or Takagi group) associated to m is the congruence subgroup m m TL/K := Rm K NL/K (IL ), where ILm denotes the subgroup of fractional ideals in IL that are coprime to mOL . Proposition 22.28. Let L/K be a finite abelian extension of number fields and let m be a m m . modulus for K divisible by the conductor of L/K. Then ker ψL/K ⊆ TL/K m . Then p is coprime to m and splits Proof. Let p be a prime of K that lies in ker ψL/K completely in L, so ep = fp = 1. There is at least one prime q of L above p, and for this m . prime we have NL/K (q) = pfp = p (by Theorem 6.10), so p ∈ NL/K (ILm ) ⊆ TL/K We now record the following theorem, which is one of the two fundamental inequalities of class field theory (it was historically the “first" fundamental inequality of class field theory, proved by Weber, even though in modern terminology it is often called the second). Theorem 22.29. Let L/K be a finite abelian extension of number fields and let m be a modulus for K divisible by the conductor of L/K. Then m m [IK : TL/K ] ≤ [L : K]. m Proof. This follows from Corollary 22.23, since TL/K is a congruence subgroup that contains all the primes of K that split completely in L. Indeed, if p splits completely then fp = 1 and m m for any prime q|p we have NL/K (q) = pfp = p and therefore p ∈ TL/K = Rm K NL/K (IL ). m With Theorem 22.29 in hand, the proof of Artin reciprocity is reduced to showing TL/K ⊆ m ker ψL/K for any modulus m divisible by the conductor of L/K. We will prove this for the trivial modulus m over the course of the next two lectures. 18.785 Fall 2021, Lecture #22, Page 10 22.5 The main theorems of class field theory (ideal-theoretic version) We can give a more precise statement of the main theorems of class field theory. Let m be a modulus for a number field K. The three main theorems of class field theory state that: • Existence: The ray class field K(m) exists. • Completeness: If L/K is finite abelian then L ⊆ K(m) if and only c(L/K) | m. In particular, every finite abelian L/K lies in a ray class field. m m • Artin reciprocity: For each subextension L/K of K(m) we have ker ψL/K = TL/K m m with conductor c(L/K)|m and a canonical isomorphism IK /TL/K ' Gal(L/K). Artin reciprocity gives us a commutative diagram of canonical bijections: m L7→TL/K {quotients of Gal(K(m)/K)} 22.6 → ← → L7→Gal(L/K) m} → {congruence subgroups C ⊆ IK m /C C7→IK → ← ← {abelian L/K with c(L/K) | m} m ψL/K ← { quotients of Clm K} The Hilbert class field Definition 22.30. Let K be global field. The Hilbert class field of K is the maximal unramified abelian extension of K (the compositum of all finite unramified abelian extensions of K inside a fixed separable closure of K). While it is not obvious from the definition, it follows from the completeness theorem of class field theory that the Hilbert class field must be the ray class field for the trivial modulus, and in particular, that it is a finite extension of K. This is a remarkable result (which we will prove in a later lecture), since infinite unramified extensions of number fields do exist (they are necessarily nonabelian). Indeed, one way to construct such an extension is by considering a tower of Hilbert class fields. Starting with a number field K0 := K, for each integer n ≥ 0 define Kn+1 to be the be the Hilbert class field of Kn . This yields an infinite tower of finite abelian extensions K0 ⊆ K1 ⊆ K2 ⊆ · · · , S and we may then consider the field L := n Kn . There are two possibilities: either we eventually reach a field Kn with class number 1, in which case Km = Kn for all m ≥ n and L/K is a finite unramified extension of K, or this does not happen and L/K is an infinite unramified extension of K (which is necessarily nonabelian). It was a longstanding open question as to whether the latter could occur, but in 1964 Golod and Shafarevich proved that indeed it can; in particular, the field √ √ K0 = Q( −30030) = Q( −2 · 3 · 5 · 7 · 11 · 13) is the base of an infinite tower of Hilbert class field extensions. One might ask whether one can use an imaginary quadratic field of smaller discriminant than this. It is known that no imaginary quadratic field of discriminant |D| ≤ 420 has an infinite Hilbert class field tower [3]; they all stabilize at either K2 or K3 . 18.785 Fall 2021, Lecture #22, Page 11 Extensions arising from Hilbert class field towers are necessarily solvable, since they are towers of finite abelian extensions. One might ask whether infinite nonsolvable unramified extensions exist. As shown by Maire [2], they do, and this can happen even when the base field has class number one and the Hilbert class field tower is trivial. Indeed, the biquadratic extension √ √ Q( 17601097, 17380678572169893) has class number one and its maximal unramified extension is an infinite extension. References [1] Henri Cohen, Advanced topics in computational number theory, Springer, 2000. [2] Christian Maire, On infinite unramfieid extensions, Pacific J. Math. 192 (2000), 135– 142. [3] Ken Yamamura, Maximal unramified extensions of imaginary quadratic number fields of small conductors, Journal de Théorie des Nombres de Bordeaux 9 (1997) 405–448. 18.785 Fall 2021, Lecture #22, Page 12 18.785 Number theory I Lecture #23 23 Fall 2021 bonus lecture Tate cohomology In this lecture we introduce a variant of group cohomology known as Tate cohomology, and we define the Herbrand quotient (a ratio of cardinalities of two Tate cohomology groups), which will play a key role in our proof of Artin reciprocity. We begin with a brief review of group cohomology, restricting our attention to the minimum we need to define the Tate cohomology groups we will use. At a number of points we will need to appeal to some standard results from homological algebra whose proofs can be found in Section 23.6. For those seeking a more thorough introduction to group cohomology, see [1]; for general background on homological algebra, we recommend [7]. 23.1 Group cohomology Definition 23.1. Let G be a group. A G-module is an abelian group A equipped with a Gaction compatible with its group structure: g(a + b) = ga + gb for all g ∈ G, a, b ∈ A.1 This implies |ga| = |a| (where |a| := #hai is the order of a); in particular ga = 0 ⇔ a = 0. A trivial G-module is an abelian group with trivial G-action: ga = a for all g ∈ G, a ∈ A (so every abelian group can be viewed as a trivial G-module). A morphism of G-modules is a morphism of abelian groups α: A → B satisfying α(ga) = gα(a). Kernels, images, quotients, and direct sums of G-modules are also G-modules. Definition 23.2. Let A be a G-module. The G-invariants of A constitute the G-module AG := {a ∈ A : ga = a for all g ∈ G} consisting of elements fixed by G. It is the largest trivial G-submodule of A.Definition 23.3. Let A be a G-module and let n ∈ Z≥0 . The group of n-cochains is the abelian group C n (G, A) := Map(Gn , A) of maps of sets f : Gn → A under pointwise addition. We have C 0 (G, A) ' A, since G0 = {1} is a singleton set. The nth coboundary map dn : C n (G, A) → C n+1 (G, A) is the homomorphism of abelian groups defined by dn (f )(g0 , . . . , gn ) := g0 f (g1 , . . . , gn ) − f (g0 g1 , g2 , . . . , gn ) + f (g0 , g1 g2 , . . . , gn ) · · · + (−1)n f (g0 , , gn−2 , gn−1 gn ) + (−1)n+1 f (g0 , . . . , gn−1 ). Z (G, A) := ker dn and B n (G, A) := im dn−1 , n The group C (G, A) contains subgroups of n-cocycles and n-coboundaries defined by with B 0 (G, A) := {0}. n The coboundary map satisfies dn+1 ◦ dn = 0 for all n ≥ 0 (this can be verified directly, but we will prove it in the next section), thus B n (G, A) ⊆ Z n (G, A) for n ≥ 0 and the groups C n (G, A) with connecting maps dn form a cochain complex d0 d1 0 −→ C 0 (G, A) −→ C 1 (G, A) −→ C 2 (G, A) −→ · · · that we may denote CA . In general, a cochain complex (of abelian groups) is simply a sequence of homomorphisms dn that satisfy dn+1 ◦ dn = 0. Cochain complexes form a category whose morphisms are commutative diagrams with cochain complexes as rows. 1 Here we put the G-action on the left (one can also define right G-modules), and for the sake of readability we write A additively, even though we will be primarily interested in cases where A is a multiplicative group. Definition 23.4. Let A be a G-module. The nth cohomology group of G with coefficients in A is the abelian group H n (G, A) := Z n (G, A)/B n (G, A). Example 23.5. We can work out the first few cohomology groups explicitly by writing out the coboundary maps and computing kernels and images: • d0 : C 0 (G, A) → C 1 (G, A) is defined by d0 (a)(g) := ga − a (note C 0 (G, A) ' A). • H 0 (G, A) ' ker d0 = AG (note B 0 (G, A) = {0}). • im d0 = {f : G → A | ∃a ∈ A : f (g) = ga − a for all g ∈ G} (principal crossed homomorphisms). • d1 : C 1 (G, A) → C 2 (G, A) is defined by d1 (f )(g, h) := gf (h) − f (gh) + f (g). • ker d1 = {f : G → A | f (gh) = f (g) + gf (h) for all g, h ∈ G} (crossed homomorphisms). • H 1 (G, A) = crossed homomorphisms modulo principal crossed homomorphisms. • If A is a trivial G-module then H 1 (G, A) ' Hom(G, A). Lemma 23.6. Let α : A → B be a morphism of G-modules. We have induced group homomorphisms αn : C n (G, A) → C n (G, B) defined by f 7→ α ◦ f that commute with the coboundary maps. In particular, αn maps cocycles to cocycles and coboundaries to coboundaries and thus induces homomorphisms αn : H n (G, A) → H n (G, B) of cohomology groups, and we have a morphism of cochain complexes α : CA → CB : → ··· ← d0 → C 1 (G, B) → α2 → → C 0 (G, B) d2 ← ← → C 2 (G, A) α1 → α0 ← 0 d1 ← ← ← → C 1 (G, A) d1 → C 2 (G, B) ← ← d0 ← → C 0 (G, A) ← 0 d2 → ... Proof. Consider any n ≥ 0. For all f ∈ C n (G, A), and g0 , . . . , gn ∈ G we have  
αn+1 dn (f )(g0 , . . . , gn ) = αn+1 g0 f (g1 , . . . , gn ) − · · · + (−1)n+1 f (g0 , . . . , gn−1 ) = g0 (α ◦ f )(g1 , . . . , gn ) − · · · + (−1)n+1 (α ◦ f )(g0 , . . . , gn−1 ) = dn (α ◦ f )(g0 , . . . , gn ) = dn (αn (f ))(g0 , . . . , gn ), thus αn+1 ◦ dn = dn ◦ αn . The lemma follows. Lemma 23.6 implies that we have a family of functors H n (G, •) from the category of Gmodules to the category of abelian groups (note that id ◦f = f and (α ◦ β) ◦ f = α ◦ (β ◦ f )), and also a functor from the category of G-modules to the category of cochain complexes. Lemma 23.7. Suppose that we have a short exact sequence of G-modules α β 0 −→ A −→ B −→ C −→ 0. Then for every n ≥ 0 we have a corresponding exact sequence of n-cochains αn βn 0 −→ C n (G, A) −→ C n (G, B) −→ C n (G, C) −→ 0. 18.785 Fall 2021, Lecture #23, Page 2 Proof. The injectivity of αn follows from the injectivity of α. If f ∈ ker β n , then β ◦ f = 0 and im f ⊆ ker β = im α; via the bijection α−1 : im α → A we can define α−1 ◦f ∈ C n (G, A), and therefore ker β n ⊆ im αn . We also have im αn ⊆ ker β n , since β ◦ α ◦ f = 0 ◦ f = 0 for all f ∈ C n (G, A), and exactness at C n (G, B) follows. Every f ∈ C n (G, C) satisfies im f ⊆ C = im β, and we can define h ∈ C n (G, B) satisfying β ◦h = f : for each g0 , . . . , gn let h(g0 , . . . , gn ) be any element of β −1 (f (g0 , . . . , gn )). Thus f ∈ im β n and β n is surjective. Lemmas 23.6 and 23.7 together imply that we have an exact functor from the category of G-modules to the category of cochain complexes. Theorem 23.8. Every short exact sequence of G-modules β α 0 −→ A −→ B −→ C −→ 0 induces a long exact sequence of cohomology groups β0 α0 δ0 0 → H 0 (G, A) −→ H 0 (G, B) −→ H 0 (G, C) −→ H 1 (G, A) −→ · · · and commutative diagrams of short exact sequences of G-modules induce corresponding commutative diagrams of long exact sequences of cohomology groups. Proof. Lemmas 23.6 and 23.7 give us the commutative diagram → C n+1 (G, B) ← ← αn+1 → dn → → C n+1 (G, A) → 0 ← → C n (G, C) dn → ← dn ← 0 βn ← ← ← → C n (G, B) β n+1 → C n+1 (G, C) ← αn ← → C n (G, A) ← 0 → 0 dn We have B n (G, A) ⊆ Z n (G, A) ⊆ C n (G, A) −→ B n+1 (G, A) ⊆ Z n+1 (G, A) ⊆ C n+1 (G, A), thus dn induces a homomorphism dn : C n (G, A)/B n (G, A) → Z n+1 (G, A), and similarly for the G-modules B and C. The fact that αn maps coboundaries to coboundaries and cocycles to cocycles implies that we have induced maps C n (G, A)/B n (G, A) → C n (G, B)/B n (G, B) and Z n+1 (G, A) → Z n+1 (G, B); similar comments apply to β n . We thus have the following commutative diagram: → Z n+1 (G, B) → 0 dn ← ← αn+1 C n (G,C) B n (G,C) → → Z n+1 (G, A) → ← 0 → dn → dn βn ← C n (G,B) B n (G,B) ← ← ← → ← αn ← C n (G,A) B n (G,A) β n+1 → Z n+1 (G, C) The kernels of the vertical maps dn are (by definition) the cohomology groups H n (G, A), H n (G, B), H n (G, C), and the cokernels are H n+1 (G, A), H n+1 (G, B), H n+1 (G, C). Applying the snake lemma yields the exact sequence αn βn δn αn+1 β n+1 H n (G, A) −→ H n (G, B) −→ H n (G, C) −→ H n+1 (G, A) −→ H n+1 (G, B) −→ H n+1 (G, C), where αn and β n are the homomorphisms in cohomology induced by α and β (coming from αn and β n in the previous diagram via Lemma 23.6), and the connecting homomorphism δ n given by the snake lemma can be explicitly described as δ n : H n (G, C) → H n+1 (G, A) [f ] 7→ [α−1 ◦ dn (fˆ)] 18.785 Fall 2021, Lecture #23, Page 3 where [f ] denotes the cohomology class of a cocycle f ∈ C n (G, C) and fˆ ∈ C n (G, B) is a cochain satisfying β ◦ fˆ = f . Here α−1 denotes the inverse of the isomorphism A → α(A). The fact that δ n is well defined (independent of the choice of fˆ) is part of the snake lemma. The map H 0 (G, A) → H 0 (G, B) is the restriction of α : A → B to AG , and is thus injective (recall that H 0 (G, A) ' AG ). This completes the first part of the proof. For the second part, suppose we have the following commutative diagram of short exact sequences of G-modules ← ← ← ϕ → B0 ← ← ← → C β0 → ← → A0 β ψ α0 → → φ ← 0 → B → C0 ← α ← → A ← 0 → 0 → 0 By Lemma 23.6, to verify the commutativity of the corresponding diagram of long exact sequences in cohomology we only need to check commutativity at squares of the form ← δn → H n+1 (G, A) ← ← H n (G, C) ϕn ← H n (G, C 0 ) δ0 n → → φn+1 (1) → H n+1 (G, A0 ) Let f : Gn → C be a cocycle and choose fˆ ∈ C n (G, B) such that β ◦ fˆ = f . We have φn+1 (δ n ([f ])) = φn+1 ([α−1 ◦ dn (fˆ)]) = [φ ◦ α−1 ◦ dn (fˆ)]. Noting that ϕ ◦ f = ϕ ◦ β ◦ fˆ = β 0 ◦ ψ ◦ fˆ and φ ◦ α−1 = α0−1 ◦ ψ (as maps α(A) → A0 ) yields n n δ 0 (ϕn ([f ])) = δ 0 ([β 0 ◦ ψ ◦ f ]) = [α0 −1 ◦ dn (ψ ◦ fˆ)] = [α0−1 ◦ ψ ◦ dn (fˆ)] = [φ ◦ α−1 ◦ dn (fˆ)], thus diagram (1) commutes as desired. Definition 23.9. A family of functors F n from the category of G-modules to the category of abelian groups that associates to each short exact sequence of G-modules a long exact sequence of abelian groups such that commutative diagrams of short exact sequences yield commutative diagrams of long exact sequences is called a δ-functor. A δ-functor is said to be cohomological if the connecting homomorphisms in long exact sequences are of the form δ n : F n (G, C) → F n+1 (G, A). If we instead have δ n : F n+1 (G, C) → F n (G, A) then the δ-functor is homological. Theorem 23.54 implies that the family of functors H n (G, •) is a cohomological δ-functor. In fact it is the universal cohomological δ-functor (it satisfies a universal property that determines it up to a unique isomorphism of δ-functors), but we will not explore this further. 23.2 Cohomology via free resolutions P Recall that the group ring Z[G] consists of formal sums g ag g indexed by g ∈ G with coefficients ag ∈ Z, all but finitely many zero. Multiplication is given by Z-linearly extending the group operation in G; the ring Z[G] is commutative if and only if G is. As an abelian group under addition, Z[G] is the free Z-module with basis G, equivalently, the group of finitely supported functions G → Z under pointwise addition. 18.785 Fall 2021, Lecture #23, Page 4 The notion of a G-module defined in the previous section is equivalent to that of a (left) Z[G]-module: to define multiplication by Z[G] one must define a G-action, and the G-action on a G-module extends Z-linearly, since every G-module is also a Z-module. The multiplicative identity 1 of the ring Z[G] is the identity element of G; the additive identity 0 is the empty sum, which acts on A by sending a ∈ A to the identity element of A.2 For any n ≥ 0 we view Z[Gn ] as a G-module with G acting diagonally on the left: g · (g1 , . . . , gn ) := (gg1 , . . . , ggn ). This makes Z[G0 ] = Z a trivial G-module (here we are viewing the empty tuple as the identity element of the trivial group G0 ). Definition 23.10. Let G be a group. The standard resolution of Z by G-modules is the exact sequence of G-module homomorphisms d d d n 1 0 · · · −→ Z[Gn+1 ] −→ Z[Gn ] −→ · · · −→ Z[G] −→ Z −→ 0, where the boundary maps dn are defined by dn (g0 , . . . , gn ) := n X (−1)i (g0 , . . . , ĝi , . . . , gn ) i=0 and extended Z-linearly (the notation ĝi means from the tuple). The map d0 sends P omit gi P each g ∈ G to 1, and extends to the map g ag g 7→ g ag , which is also known as the augmentation map and may be denoted ε. Let us verify the exactness of the standard resolution. Lemma 23.11. The standard resolution of Z by G-modules is exact. Proof. The map d0 is clearly surjective. To check im dn+1 ⊆ ker dn it suffices to note that for any g0 , . . . , gn ∈ G we have dn (dn+1 (g0 , . . . , gn )) = X 0≤i≤n X (−1)i+j (g0 , . . . , ĝj , . . . , ĝi . . . , gn ) + 0≤j 0 Hn (G, A) for n > 0 Ĥ −n (G, A) := Ĥn−1 (G, A) Ĥ−n (G, A) := Ĥ n−1 (G, A). Note that Ĥ 0 (G, A) is a quotient of H 0 (G, A) ' AG (the largest trivial G-module in A) and Ĥ0 (G, A) is a submodule of H0 (G, A) ' AG (the largest trivial G-module quotient of A). Thus any morphism of G-modules induces natural morphisms of Tate cohomology and homology groups in degree n = 0 (and all other degrees, as we already know). We thus have functors Ĥ n (G, •) and Ĥn (G, •) from the category of G-modules to the category of abelian groups. Given that every Tate homology group is also a Tate cohomology group, in practice one usually refers only to the groups Ĥ n (G, A), but the notation Ĥn (G, A) can be helpful to highlight symmetry. Theorem 23.32. Let G be a finite group. Every short exact sequence of G-modules β α 0 −→ A −→ B −→ C −→ 0 induces a long exact sequence of Tate cohomology groups β̂ n α̂n δ̂ n · · · −→ Ĥ n (G, A) −→ Ĥ n (G, B) −→ Ĥ n (G, C) −→ Ĥ n+1 (G, A) −→ · · · , equivalently, a long exact sequence of Tate homology groups β̂n α̂ δ̂ n n · · · −→ Ĥn (G, A) −→ Ĥn (G, B) −→ Ĥn (G, C) −→ Ĥn−1 (G, A) −→ · · · . Commutative diagrams of short exact sequences of G-modules induce commutative diagrams of long exact sequences of Tate cohomology groups (equivalently, Tate homology groups). Proof. It follows from Theorems 23.8 and 23.21 that it is enough to prove exactness at the terms Ĥ 0 (G, •) = Ĥ−1 (G, •) and Ĥ0 (G, •) = Ĥ −1 (G, •). We thus consider the diagram ← α0 → BG ← → 0 N̂G ← → AG → CG N̂G → → ← N̂G ← 0 β0 ← → BG β0 → ← α0 → CG ← ← → AG ← δ0 ← H1 (C, G) δ0 → H 1 (A, G) 18.785 Fall 2021, Lecture #23, Page 11 whose top and bottom rows are the end and beginning of the long exact sequences in homology and cohomology given by Theorems 23.21 and 23.8, respectively; here we are using H0 (G, •) ' •G and H 0 (G, •) ' •G . For any [a] ∈ AG = A/IG A we have N̂G (α0 ([a])) = NG α(a) = α(NG a) = α0 (N̂G ([a])), so the first square commutes, as does the second (by the same argument). Applying the snake lemma yields an exact sequence of kernels and cokernels of N̂G β̂0 α̂ β̂ 0 α̂0 δ̂ Ĥ0 (G, A) →0 Ĥ0 (G, B) → Ĥ0 (G, C) → Ĥ 0 (G, A) → Ĥ 0 (G, B) → Ĥ 0 (G, C), where δ̂([c]) = [a] for any a ∈ A, b ∈ B, c ∈ C with α(a) = NG b and β(b) = c ∈ ker NG (that this uniquely defines the connecting homomorphism δ̂ is part of the snake lemma). Note that im δ0 = ker α0 = ker α̂0 ⊆ ker N̂G , since α0 is injective, so δ0 gives a well-defined map δ̂0 : Ĥ1 (G, C) → Ĥ0 (G, A) that makes the sequence is exact at Ĥ0 (G, A). Similarly, im N̂G ⊆ im β 0 = ker δ 0 , since β0 is surjective, so δ 0 induces a well-defined map δ̂ 0 : Ĥ 0 (G, C) → H 1 (A, G) that makes the sequence exact at Ĥ 0 (G, C). For the last statement of the theorem, suppose we have the following commutative diagram of exact sequences of G-modules ← ← ← ϕ → B0 ← ← ← → C β0 → ← → A0 β ψ α0 → → φ ← 0 → B → C0 ← α ← → A ← 0 → 0 → 0 By Theorems 23.21 and 23.8, we only need to verify the commutativity of the square ← δ̂ ← → Ĥ 0 (G, A) ← Ĥ0 (G, C) ← Ĥ0 (G, C 0 ) φ0 → → ϕ0 δ̂ 0 → Ĥ 0 (G, A0 ) Let a ∈ A, b ∈ B, c ∈ C satisfy α(a) = NG b and β(b) = c ∈ ker NG as in the definition of δ̂ above, so that δ̂([c]) = [a]. Now let a0 = φ(a), b0 = ψ(b), c = ϕ(c). Then α0 (a0 ) = α0 (φ(a)) = ψ(α(a)) = ψ(NG b) = NG ψ(b) = NG b0 β 0 (b0 ) = β 0 (ψ(b)) = ϕ(β(b)) = ϕ(c) = c0 ∈ ker NG , where we have used NG c0 = NG ϕ(c) = ϕ(NG c) = ϕ(0) = 0. Thus δˆ0 ([c0 ]) = [a0 ] and φ0 (δ̂([c])) = φ0 ([a]) = [φ(a)] = [a0 ] = δˆ0 ([c0 ]) = δˆ0 ([ϕ(c)]) = δˆ0 (ϕ0 ([c])), so φ0 ◦ δ̂ = δˆ0 ◦ ϕ0 as desired. Theorem 23.32 implies that the family Ĥ n (G, •) is a cohomological δ-functor, and that the family Ĥn (G, •) is a homological δ-functor. Corollary 23.33. Let G be a finite group. For any G-modules A and B we have Ĥ n (G, A ⊕ B) ' Ĥ n (G, A) ⊕ Ĥ n (G, B), for all n ∈ Z, and the isomorphisms commute with the natural inclusion and projection maps for the direct sums on both sides. 18.785 Fall 2021, Lecture #23, Page 12 Proof. For n 6= 0, −1 this follows from Corollaries 23.15 and 23.22. For n = 0, −1 it suffices to note that NG acts on A ⊕ B component-wise, and the induced morphism N̂G thus acts on (A ⊕ B)G = AG ⊕ BG component-wise. Theorem 23.34. Let G be a finite group and let B be an induced or co-induced G-module associated to some abelian group A. Then Ĥ n (G, B) = Ĥn (G, B) = 0 for all n ∈ Z. Proof. By Corollary 23.28, we only need to show Ĥ0 (G, B) = Ĥ 0 (G, B) = 0, and by Lemma 23.27 it suffices to consider the case B = IndG (A) = Z[G] ⊗Z A. Equivalently, we need to show that NG : B → B has kernel IG B and image B G . By definition, the Z[G]action on B = Z[G] ⊗Z A only affects the factor Z[G], so this amounts to showing that, as an endomorphism of Z[G], wePhave ker NG P = IG and im NG = Z[G]G . But this is clear: the action of NG on Z[G] is g∈G ag g 7→ ( g∈G ag )NG . The kernel of this action is the P augmentation ideal IG , and its image is Z[G]G = { g∈G ag g : all ag ∈ Z equal} = NG Z. Remark 23.35. Theorem 23.34 explains a major motivation for using Tate cohomology. It is the minimal modification needed to ensure that induced (and co-induced) G-modules have trivial homology and cohomology in all degrees. Corollary 23.36. Let G be a finite group and let A be a free Z[G]-module. Then we have Ĥn (G, A) = Ĥ n (G, A) = 0 for all n ∈ Z. Proof. Let S be a Z[G]-basis for A and let B be the free Z-module with basis S. Then A ' IndG (B) and the corollary follows from Theorem 23.34. 23.5 Tate cohomology of cyclic groups We now assume that G is a cyclic group hgi of finite order. In this case the augmentation ideal IG is principal, generated by g − 1 (as an ideal in the ring Z[G], not as a Z-module). For any G-module A we have a free resolution N g−1 g−1 N ε G G · · · −→ Z[G] −→ Z[G] −→ Z[G] −→ Z[G] −→ Z[G] −→ Z −→ 0. (2) The fact that augmentation ideal IG = (g − 1) is principal (because G is cyclic) ensures im NG = ker(g − 1), making the sequence exact. The group ring Z[G] is commutative, since G is abelian, so we need not distinguish left and right Z[G]-modules. For any G-module A we can view Z[G] ⊗Z[G] A as a G-module via g(h⊗a) = gh⊗a = h⊗ga and view HomZ[G] (Z[G], A) as a G-module via (gϕ)(h) := ϕ(gh).6 Theorem 23.37. Let G = hgi be a finite cyclic group and let A be a G-module. For all n ∈ Z we have Ĥ 2n (G, A) ' Ĥ2n−1 (G, A) ' Ĥ 0 (G, A) and Ĥ2n (G, A) ' Ĥ 2n−1 (G, A) ' Ĥ0 (G, A). Proof. We have canonical G-module isomorphisms HomZ[G] (Z[G], A) ' A ' Z[G] ⊗Z[G] A induced by ϕ 7→ ϕ(1) and a 7→ 1 ⊗ a, respectively, and these isomorphisms preserve the multiplication-by-g endomorphisms (that is, (gϕ)(1) = gϕ(1) and 1 ⊗ ga = g(1 ⊗ a)). Using the free resolution in (2), we can thus compute H n (G, A) using the cochain complex g−1 6 N g−1 N G G 0 −→ A −→ A −→ A −→ A −→ A··· , Note that we must have g1 g2 ϕ(h) = g1 (g2 ϕ)(h) = (g2 ϕ)(g1 h) = ϕ(g2 g1 h) = g2 g1 ϕ(h) in order for ϕ to be both a Z[G]-module morphism and an element of a Z[G]-module, so this will not work if G is not abelian. 18.785 Fall 2021, Lecture #23, Page 13 and we can compute Hn (G, A) using the chain complex N g−1 N g−1 G G · · · −→ A −→ A −→ A −→ A −→ A −→ 0. We now observe that AG = ker(g − 1), so for all n ≥ 1 we have H 2n (G, A) = H2n−1 (G, A) = ker(g − 1)/ im NG = coker N̂G = Ĥ 0 (G, A), so Ĥ 2n (G, A) = Ĥ2n−1 (G, A) = Ĥ 0 (G, A) for all n ∈ Z, since Ĥ −2n (G, A) = Ĥ2n−1 (G, A) and Ĥ−2n+1 = Ĥ 2n for all n ≥ 0. We also note that im(g − 1) = IG A, so for all n ≥ 1 we have H2n (G, A) = H 2n−1 (G, A) = ker NG / im(g − 1) = ker N̂G = Ĥ0 (G, A), so Ĥ2n (G, A) = Ĥ 2n−1 (G, A) = Ĥ0 (G, A) for all n ∈ Z, since Ĥ−2n (G, A) = Ĥ 2n−1 (G, A) and Ĥ −2n+1 = Ĥ2n for all n ≥ 0. It follows from Theorem 23.37 that when G is a finite cyclic group, all of the Tate homology/cohomology groups are determined by Ĥ0 (G, A) = ker N̂G = ker NG / im(g − 1) and Ĥ 0 (G, A) = coker N̂G = ker(g − 1)/ im NG . This motivates the following definition. Definition 23.38. Let G be a finite cyclic group and let A be a G-module. We define hn (A) := hn (G, A) := #Ĥ n (G, A) and hn (A) := hn (G, A) := #Ĥn (G, A). Whenever h0 (A) and h0 (A) are both finite, we also define the Herbrand quotient h(A) := h0 (A)/h0 (A) ∈ Q. Remark 23.39. Some authors define the Herbrand quotient via h(A) := h0 (A)/h1 (A) or h(A) := h0 (A)/h−1 (A) or h(A) := h2 (A)/h1 (A), but Theorem 23.37 implies that these definitions are all the same as ours. The notation q(A) is often used instead of h(A), and one occasionally sees the Herbrand quotient defined as the reciprocal of our definition (as in [2], for example), but this is less standard. Corollary 23.40. Let G be a finite cyclic group. Given an exact sequence of G-modules β α 0 −→ A −→ B −→ C −→ 0 we have a corresponding exact hexagon ← α̂0 → Ĥ 0 (G, B) ← δ̂0 Ĥ 0 (G, A) → ← Ĥ0 (G, C) Ĥ0 (G, B) → ← → → β̂0 β̂ 0 → δ̂0 Ĥ 0 (G, C) ← ← Ĥ0 (G, A) α̂ 0 Proof. This follows immediately from Theorems 23.32 and 23.37. Corollary 23.41. Let G be a finite cyclic group. For any exact sequence of G-modules α β 0 −→ A −→ B −→ C −→ 0, if any two of h(A), h(B), h(C) are defined then so is the third and h(B) = h(A)h(C). 18.785 Fall 2021, Lecture #23, Page 14 Proof. Using the exact hexagon given by Corollary 23.40 we can compute the cardinality h0 (A) = #Ĥ 0 (G, A) = # ker α̂0 # im α̂0 = # ker α0 # ker β 0 . Applying a similar calculation to Ĥ 0 (G, C) and Ĥ 1 (G, B) yields h0 (A)h0 (C)h0 (B) = # ker α̂0 # ker β̂ 0 # ker δ̂ 0 # ker α̂0 # ker β̂0 # ker δ̂0 . Doing the same for Ĥ 0 (G, B), Ĥ0 (G, A), Ĥ0 (G, C) yields h0 (B)h0 (A)h0 (C) = # ker β̂ 0 # ker δ̂ 0 # ker α̂0 # ker β̂0 # ker δ̂0 # ker α̂0 = h0 (A)h0 (C)h0 (B). If any two of h(A), h(B), h(C) are defined then at least four of the groups in the exact hexagon are finite, and the remaining two are non-adjacent, but these two must then also be finite. In this case we can rearrange the identity above to obtain h(B) = h(A)h(C). Corollary 23.42. Let G be a finite cyclic group, and let A and B be G-modules. If h(A) and h(B) are defined then so is h(A ⊕ B) = h(A)h(B). Proof. Apply Corollary 23.41 to the split exact sequence 0 → A → A ⊕ B → B → 0. Lemma 23.43. Let G = hgi be a finite cyclic group. If A is an induced or finite G-module then h(A) = 1. Proof. If A is an induced G-module then h0 (A) = h0 (A) = h(A) = 1, by Theorem 23.34. If A is finite, then the exact sequence g−1 0 −→ AG −→ A −→ A −→ AG −→ 0 implies #AG = # ker(g − 1) = # coker(g − 1) = #AG , and therefore h0 (A) = # ker N̂G = # coker N̂G = h0 (A), so h(A) = h0 (A)/h0 (A) = 1. Corollary 23.44. Let G be a finite cyclic group and let A be a G-module that is a finitely generated abelian group. Then h(A) = h(A/Ator ) whenever either is defined. Proof. Apply Corollary 23.41 and Lemma 23.43 to 0 → Ator → A → A/Ator → 0. Remark 23.45. The hypothesis of Corollary 23.44 actually guarantees that h(A) is defined, but we won’t prove this here. Corollary 23.46. Let G be a finite cyclic group and let A be a trivial G-module that is a finitely generated abelian group. Then h(A) = (#G)r , where r is the rank of A. Proof. We have A/Ator ' Zr , where Z is a trivial G-module. Then ZG = Z = ZG , and N̂G : ZG → ZG is multiplication by #G, so h(Z) = # coker N̂G /# ker N̂G = #G. Now apply Corollaries 23.42 and 23.44. Lemma 23.47. Let G be a finite cyclic group and let α : A → B be a morphism of G-modules with finite kernel and cokernel. If either h(A) or h(B) is defined then h(A) = h(B). 18.785 Fall 2021, Lecture #23, Page 15 Proof. Applying Corollary 23.41 to the exact sequences 0 → ker α → A → im α → 0 0 → im α → B → coker α → 0 yields h(A) = h(ker α)h(im α) = h(im α) = h(im α)h(coker α) = h(B), by Lemma 23.43, since ker α and coker α are finite. The lemma follows. Corollary 23.48. Let G be a finite cyclic group and let A be a G-module containing a sub-G-module B of finite index. Then h(A) = h(B) whenever either is defined. Proof. Apply Lemma 23.47 to the inclusion B → A. 23.6 A little homological algebra In an effort to keep these notes self-contained, in this final section we present proofs of the homological results that were used above. For the sake of concreteness we restrict our attention to modules, but everything in this section generalizes to suitable abelian categories. We use R to denote an arbitrary (not necessarily commutative) ring (in previous section R was the group ring Z[G]). Statements that use the term R-module without qualification are understood to apply in both the category of left R-modules and the category of right R-modules. 23.6.1 Complexes Definition 23.49. A chain complex C is a sequence of R-module morphisms d d d 2 1 0 · · · −→ C2 −→ C1 −→ C0 −→ 0, with dn ◦dn+1 = 0; the dn are boundary maps. The nth homology group of C is the R-module Hn (C) := Zn (C)/Bn (C), where Zn (C) := ker dn−1 and Bn (C) := im dn are the R-modules of cycles and boundaries, respectively; for n < 0 we define Cn = 0 and dn is the zero map. A morphism of chain complexes f : C → D is a sequence of R-module morphisms fn : Cn → Dn that commute with boundary maps (so fn ◦ dn = dn ◦ fn+1 ).7 Such a morphism necessarily maps cycles to cycles and boundaries to boundaries, yielding natural morphisms Hn (f ) : Hn (C) → Hn (D) of homology groups.8 We thus have a family of functors Hn (•) from the category of chain complexes to the category of abelian groups. The category of chain complexes has kernels and cokernels (and thus exact sequences). The set Hom(C, D) of morphisms of chain complexes C → D is an abelian group under addition: (f + g)n = fn + gn . The category of chain complexes of R-modules contains direct sums and direct products: if A and B are chain complexes of R-modules then (A ⊕ B)n := An ⊕ Bn and the boundary maps dn : (A ⊕ B)n+1 → (A ⊕ B)n are defined component-wise: dn (a ⊕ b) := dn (a) ⊕ dn (b). 7 We use the symbols dn to denote boundary maps of both C and D; in general, the domain and codomain of any boundary or coboundary map should be inferred from context. 8 In fact Hn (f ) : Hn (C) → Hn (D) is a morphism of R-modules, but in all the cases of interest to us, either the homology groups are all trivial (as occurs for exact chain complexes, such as the standard resolution of Z by Z[G]-modules), or R = Z (as in the chain complexes used to define the Ext and Tor groups below), so we will generally refer to homology groups rather than homology modules. 18.785 Fall 2021, Lecture #23, Page 16 Because the boundary maps are defined component-wise, the kernel of the boundary map of a direct sum is the direct sum of the kernels of the boundary maps on the components, and similarly for images. It follows that Hn (A ⊕ B) ' Hn (A) ⊕ Hn (B), and this isomorphism commutes with the natural inclusion and projection maps in to and out of the direct sums on both sides. In other words, Hn (•) is an additive functor (see Definition 23.16). This extends to arbitrary (possibly infinite) direct sums, and also to arbitrary direct products, although we will only be concerned with finite direct sums/products.9 Theorem 23.50. Associated to each short exact sequence of chain complexes β α 0 −→ A −→ B −→ C −→ 0 is a long exact sequence of homology groups · · · −→ Hn+1 (A) Hn+1 (α) −→ Hn+1 (B) Hn+1 (β) −→ Hn (α) δ Hn (β) n Hn (A) −→ Hn (B) −→ Hn (C) −→ · · · Hn+1 (C) −→ and this association maps morphisms of short exact sequences to morphisms of long exact sequences. In other words, the family of functors Hn (•) is a homological δ-functor. For n < 0 we have Hn (•) = 0, by definition, so this sequence ends at H0 (C) → 0. Proof. For any chain complex C, let Yn (C) := Cn /Bn (C). Applying the snake lemma to ← ← ← αn → Zn (B) → 0 dn ← ← → Zn (A) → Yn+1 (C) dn → → dn ← 0 βn+1 βn → ← → Yn+1 (B) ← αn+1 ← Yn+1 (A) → Zn (C) (where αn , βn , dn denote obviously induced maps) yields the exact sequence αn+1 βn+1 δ α βn n n Hn+1 (A) −→ Hn+1 (B) −→ Hn+1 (C) −→ Hn (B) −→ Hn (G). Hn (A) −→ The verification of the commutativity of diagrams of long exact sequences of homology groups associated to commutative diagrams of short exact sequences of chain complexes is as in the proof of Theorem 23.8, mutatis mutandi. Definition 23.51. Two morphisms f, g : C → D of chain complexes are homotopic if there exist morphisms hn : Cn → Dn+1 such that fn − gn = dn ◦ hn + hn−1 ◦ dn−1 for all n ≥ 0 (where h−1 := 0); this defines an equivalence relation f ∼ g, since (a) f ∼ f (take h = 0), (b) if f ∼ g via h then g ∼ f via −h, and (c) if f1 ∼ f2 via h1 and f2 ∼ f3 via h2 then f1 ∼ f3 via h1 + h2 . Lemma 23.52. Homotopic morphisms of chain complexes f, g : C → D induce they some morphisms of homology groups Hn (C) → Hn (D); we have Hn (f ) = Hn (g) for all n ≥ 0. Proof. Let [z] ∈ Hn (C) = Zn (C)/Bn (C) denote the homology class z ∈ Zn (C). We have fn (z) − gn (z) = dn (hn (z)) + hn−1 (dn−1 (z)) = dn (hn (z)) + 0 ∈ Bn (D), thus Hn (f )([z]) − Hn (g)([z]) = 0. It follows that Hn (f ) = Hn (g) for all n ≥ 0. 9 This does not imply that the Ext and Tor functors defined below commute with arbitrary direct sums and direct products; see Remarks 23.62 and 23.66. 18.785 Fall 2021, Lecture #23, Page 17 Definition 23.53. A cochain complex C is a sequence of R-module morphisms d0 d1 d2 0 −→ C 0 −→ C 1 −→ C 2 −→ · · · with dn+1 ◦dn = 0. The nth cohomology group of C is the R-module H n (C) := Z n (C)/B n (C), where Z n (C) := ker dn and B n (C) := im dn−1 are the R-modules of cocycles and coboundaries; for n < 0 we define C n = 0 and dn is the zero map. A morphism of cochain complexes f : C → D consists of R-module morphisms f n : C n → Dn that commute with coboundary maps, yielding natural morphisms H n (f ) : H n (C) → H n (D) and a functors H n (•) from the category of cochain complexes to the category of abelian groups. Cochain complexes form a category with kernels and cokernels, as well as direct sums and direct products (coboundary maps are defined component-wise). Like Hn (•), the functor H n (•) is additive and commutes with arbitrary direct sums and direct products. The set Hom(C, D) of morphisms of cochain complexes C → D forms an abelian group under addition: (f + g)n = f n + g n . Morphisms of cochain complexes f, g : C → D are homotopic if there are morphisms hn : C n+1 → Dn such that f n − g n = hn ◦ dn + dn−1 ◦ hn−1 for all n ≥ 0 (where h−1 := 0); this defines an equivalence relation f ∼ g.10 Theorem 23.54. Associated to every short exact sequence of cochain complexes β α 0 −→ A −→ B −→ C −→ 0 is a long exact sequence of homology groups H n (α) H n (β) δn · · · −→ H n (A) −→ H n (B) −→ H n (C) −→ H n+1 (A) H n+1 (α) −→ H n+1 (B) H n+1 (β) −→ H n+1 (C) −→ · · · and this association maps morphisms of short exact sequences of morphisms of long exact sequences, that is, the family of functors H n (•) is a cohomological δ-functor. For n < 0 we have Hn (•) = 0, by definition, so this sequence begins with 0 → H 0 (A). Proof. Adapt the proof of Theorem 23.50. Lemma 23.55. Homotopic morphisms of cochain complexes f, g : C → D induce the same morphisms of cohomology groups H n (C) → H n (D); we have H n (f ) = H n (g) for all n ≥ 0. Proof. Adapt the proof of Lemma 23.52. 23.6.2 Projective resolutions Recall that a projective R-module is an R-module P with the property that if π : M  N is a surjective morphism of R-modules, every R-module morphism ϕ : P → N factors through π: ← → π ϕ → M ← ← P ∃φ  N Free modules are projective, since we can then fix an R-basis {ei } for P and define φ(ei ) by picking any element of π −1 (ϕ(ei )) (note that the φ so constructed is in no way canonical). 10 Note the order of composition in the homotopy relations for morphisms of chain/cochain complexes. 18.785 Fall 2021, Lecture #23, Page 18 Definition 23.56. Let M be an R-module. A projective resolution of M is an exact chain complex P with P0 = M and Pn projective for all n > 0. Every R-module has a projective resolution, since (as noted earlier), every R-module M has a free resolution (we can always construct d0 : P1  M by taking P1 to be free module with basis M , then similarly construct d1 : P2  ker d0 , and so on). Proposition 23.57. Let M and N be R-modules with projective resolutions P and Q, respectively. Every R-module morphism α0 : M → N extends to a morphism α : P → Q of chain complexes that is unique up to homotopy. Proof. We inductively construct αn for n ≥ 1 (the base case is given). Suppose we have constructed a commutative diagram of exact sequences α0 α1 d1 → Q1 d0 → N → 0 ← → M ← ← ← ← → ··· d0 → ··· → P1 → ← ← ← d1 → → Pn−1 dn−2 → ← → Qn dn−1 → ← → Qn+1 → ← ← ··· dn → ··· αn−1 αn dn+1 dn−2 ← → Pn−1 ← dn−1 ← ← → Pn ← dn ← → Pn+1 ← dn+1 ← ··· → 0 Then dn−1 ◦ αn ◦ dn = αn−1 ◦ dn−1 ◦ dn = 0, so im(αn ◦ dn ) ⊆ ker dn−1 = im dn . We now define αn+1 : Pn+1 → Qn+1 as a pullback of the morphism αn ◦ dn : Pn+1 → im dn along the surjection dn : Qn+1 → im dn such that dn ◦ αn+1 = αn ◦ dn . Now suppose β : P → Q is another morphism of projective resolutions with β0 = α0 , and let γ = α − β. To show that α and β are homotopic it suffices to construct maps hn : Pn → Qn+1 such that dn ◦ hn = γn − hn−1 ◦ dn−1 (where h−1 = d−1 = 0). We have γ0 = α0 − β0 = 0, so let h0 := 0 and inductively assume dn ◦ hn = γn − hn−1 ◦ dn−1 . Then dn ◦ (γn+1 − hn ◦ dn ) = dn ◦ γn+1 − (dn ◦ hn ) ◦ dn = γn ◦ dn − (γn − hn−1 ◦ dn−1 ) ◦ dn = 0, so im(γn+1 − hn ◦ dn ) ⊆ Bn+1 (Q). The R-module Pn+1 is projective, so we can pullback the morphism (γn+1 − hn ◦ dn ) : Pn+1 → Bn+1 (Q) along the surjection dn+1 : Qn+1 → Bn+1 (Q) to obtain hn+1 satisfying dn+1 ◦ hn+1 = γn+1 − hn ◦ dn as desired. 23.6.3 Hom and Tensor If M and N are R-modules, the set HomR (M, N ) of R-module morphisms M → N forms an abelian group under pointwise addition (so (f + g)(m) := f (m) + g(m)) that we may view as a Z-module. For each R-module A we have a contravariant functor HomR (•, A) that sends each R-module M to the Z-module M ∗ := HomR (M, A) and each R-module morphism ϕ : M → N to the Z-module morphism ϕ∗ : HomR (N, A) → HomR (M, A) f 7→ f ◦ ϕ. To check this, note that ϕ∗ (f + g) = (f + g) ◦ ϕ = f ◦ ϕ + g ◦ ϕ = ϕ∗ (f ) + ϕ∗ (g), 18.785 Fall 2021, Lecture #23, Page 19 so ϕ∗ is a morphism of Z=modules (homomorphism of abelian groups), and id∗M = (f 7→ f ◦ idM ) = (f 7→ f ) = idM ∗ , (φ ◦ ϕ)∗ = (f 7→ f ◦ φ ◦ ϕ) = (f 7→ f ◦ ϕ) ◦ (f 7→ f ◦ φ) = ϕ∗ ◦ φ∗ , thus HomR (•, A) is a contravariant functor. Lemma 23.58. Let ϕ : M → N and φ : N → P be morphisms of R-modules. The sequence ϕ φ M −→ N −→ P −→ 0 is exact if and only if for every R-module A the sequence φ∗ ϕ∗ 0 −→ HomR (P, A) −→ HomR (N, A) −→ HomR (M, A) is exact. Proof. (⇒): If φ∗ (f ) = f ◦ φ = 0 then f = 0, since φ is surjective, so φ∗ is injective. We ∼ have ϕ∗ ◦φ∗ = (ϕ◦φ)∗ = 0∗ = 0, so im φ∗ ⊆ ker ϕ∗ . Let φ−1 : P → N/ ker φ. Each g ∈ ker ϕ∗ vanishes on im ϕ = ker φ inducing ḡ : N/ ker φ → A with g = ḡ ◦ φ−1 ◦ φ ∈ im φ∗ . (⇐): For A = P/ im φ and π : P → P/ im φ the projective map, we have φ∗ (π) = 0 and therefore π = 0, since φ∗ is injective, so P = im φ and φ is surjective. For A = P we have 0 = (ϕ∗ ◦ φ∗ )(idP ) = idP ◦φ ◦ ϕ = φ ◦ ϕ, so im ϕ ⊆ ker φ. For A = N/ im ϕ, and π : N → N/ im ϕ the projection map, we have π ∈ ker ϕ∗ = im φ∗ , thus π = φ∗ (σ) = σ ◦ φ for some σ ∈ Hom(P, A). Now π(ker φ) = σ(φ(ker φ)) = 0 implies ker φ ⊆ ker π = im ϕ. f g Definition 23.59. A sequence of morphisms 0 → A → B → C → 0 is left exact if it is exact at A and B (ker f = 0 and im f = ker g), and right exact if it is exact at B and C (im f = ker g and im g = C). A functor that takes exact sequences to left (resp. right) exact sequences is said to be left exact (resp. right exact). Corollary 23.60. For any R-module A the functor HomR (•, A) is left exact. Proof. This follows immediately from the forward implication in Lemma 23.58. Corollary 23.61. For any R-module A, the functor HomR (•, A) is an additive functor. Proof. See [6, Lemma 12.7.2] for a proof that this follows from left exactness; it is easy to check directly in any case. Remark 23.62. Corollary 23.61 implies that HomR (•, A) commutes with finite direct sums, but it does not commute with infinite direct sums (direct products are fine). Remark 23.63. The covariant functor HomR (A, •) that sends ϕ : M → N to (f 7→ ϕ ◦ f ) is also left exact. If M is a right R-module and A is a left R-module, the tensor product M ⊗R A is an abelian group consisting of sums of pure tensors m ⊗ a with m ∈ M and a ∈ A satisfying: • m ⊗ (a + b) = m ⊗ b + m ⊗ b; • (m + n) ⊗ a = m ⊗ a + m ⊗ a; • mr ⊗ a = m ⊗ ra. 18.785 Fall 2021, Lecture #23, Page 20 For each left R-module A we have a covariant functor • ⊗R A that sends each right Rmodule M to the Z-module M∗ := M ⊗R A, and each right R-module morphism ϕ : M → N to the Z-module morphism ϕ∗ : M ⊗R A → N ⊗R A m ⊗ a 7→ ϕ(m) ⊗ a For each left R-module A we also have a covariant functor HomZ (A, •) that sends each Z-module B to the right R-module HomZ (A, B) with ϕ(a)r := ϕ(ra) and each Z-module morphism ϕ : B → C to the right R-module morphism Hom(A, B) → Hom(A, C) defined by f 7→ ϕ ◦ f . Note that (ϕrs)(a) = ϕ(rsa) = (ϕr)(sa) = ((ϕr)s)(a), so HomZ (A, B) is indeed a right R-module. For any abelian group B there is a natural isomorphism of Z-modules ∼ HomZ (M ⊗R A, B) −→ HomR (M, HomZ (A, B)) (3) ϕ 7→ (m 7→ (a 7→ ϕ(m ⊗ a))) (m ⊗ a 7→ φ(m)(a)) ←[ φ The functors • ⊗R A and HomZ (A, •) are thus adjoint functors. One can view (3) as a universal property that determines M ⊗R A up to a unique isomorphism. Lemma 23.64. For any left R-module the functor • ⊗R A is right exact. Proof. Let ϕ φ 0 −→ M −→ N −→ P −→ 0, P be an exact sequence of right R-modules. For any i pi ⊗ ai ∈ P∗ we can pick ni ∈ N P P p ⊗ For any n ⊗ a) = such that φ(n ) = p and then φ( i i i i i P a, thus φ∗ is surjective. iP P P m ⊗ a ∈ M ⊗ A we have φ (ϕ ( m ⊗ a )) = φ(ϕ(m )) ⊗ a = 0 ⊗ a i i i = 0, so i ∗ ∗ i R i i i i i i im ϕ∗ ⊆ ker φ∗ . To prove im ϕ∗ = ker φ∗ it suffices to show that N∗ / im ϕ∗ ' P∗ , since the surjectivity of φ∗ implies N ∗ / ker ϕ∗ ' P∗ . For every abelian group B the sequence φ∗ φ∗ 0 −→ HomR (P, HomZ (A, B)) −→ HomR (N, HomZ (A, B)) −→ HomR (M, HomZ (A, B)) is left exact (by applying Corollary 23.60 to the right R-module HomZ (A, B); note that the corollary applies to both left and right R-modules). Equivalently, by (3), φ∗ ϕ∗ ∗ ∗ 0 −→ HomZ (P∗ , B) −→ HomZ (N∗ , B) −→ HomZ (M∗ , B), Applying Lemma 23.58 and the surjectivity of φ∗ yields the desired right exact sequence ϕ∗ φ M∗ −→ N∗ −→ P∗ −→ 0. Corollary 23.65. For any left R-module A, the functor • ⊗R A is an additive functor. Proof. See [6, Lemma 12.7.2] for a proof that this follows from right exactness; it is easy to check directly in any case. Remark 23.66. Corollary 23.65 implies that • ⊗R A commutes with finite direct sums, and in fact it commutes with arbitrary direct sums (but not direct products). 18.785 Fall 2021, Lecture #23, Page 21 Remark 23.67. For any right R-module A the functor A ⊗R • is also right exact. If A is an R-module and C is a chain complex of R-modules, applying the functor Hom(•, A) to the R-modules Cn and boundary maps dn : Cn+1 → Cn yields a cochain complex C ∗ of Z-modules C n := Cn∗ and coboundary maps dn := d∗n ,11 and morphisms f : C → D of chain complexes become morphisms f ∗ : C ∗ → D∗ of cochain complexes. We thus also have a contravariant left exact functor from the category of chain complexes to the category of cochain complexes. Proposition 23.68. Let A be an R-module and let •∗ denote the application of the functor Hom(•, A). Let f, g : C → D be homotopic morphisms of chain complexes of R-modules. Then f ∗ , g ∗ : D∗ → C ∗ are homotopic morphisms of cochain complexes of Z-modules. Proof. The morphisms f and g are homotopic, so their exist morphisms hn : Cn → Dn+1 such that fn − gn = dn ◦ hn + hn−1 ◦ dn−1 for all n ≥ 0. Applying the contravariant functor Hom(•, A) yields fn∗ − gn∗ = h∗n ◦ d∗n + d∗n−1 ◦ h∗n−1 , where h∗n : Dn+1 → Cn for all n ≥ 0, with h−1 = 0. Thus f ∗ and g ∗ are homotopic. Proposition 23.69. Let A be a left R-module and let •∗ denote the application of the functor • ⊗R A. Let f, g : C → D be homotopic morphisms of chain complexes of right R-modules. Then f∗ , g∗ : C∗ → D∗ are homotopic morphisms of chain complexes of Z-modules. Proof. The morphisms f and g are homotopic, so their exist morphisms hn : Cn → Dn+1 such that fn − gn = dn ◦ hn + hn−1 ◦ dn−1 for all n ≥ 0. Applying the covariant functor • ⊗R A yields fn ∗ − gn ∗ = dn∗ ◦ hn∗ + hn−1 ∗ ◦ dn−1 ∗ , where hn∗ : Cn+1 → Dn for all n ≥ 0, with h−1 = 0. Thus f∗ and g∗ are homotopic. 23.6.4 Ext and Tor functors Definition 23.70. Let P be a projective resolution of an R-module M . The truncation of P is the chain complex P with P 0 := P1 and P n := Pn+1 for all n > 0 (which need not be exact at P 0 ).12 Any morphism of projective resolutions f : P → Q induces a morphism f¯: P → Q of their truncations with f¯n := fn+1 . Theorem 23.71. Let P , Q be projective resolutions of an R-module M , let A be an R∗ ∗ module, and let •∗A denote application of HomR (•, A). Then H n (P A ) ' H n (QA ) for n ≥ 0. Proof. We will drop the subscript A in the proof to ease the notation. Let f : P → Q and g : Q → P be extensions of the identity morphism idM given by Proposition 23.57. The composition g ◦ f : P → P is an extension of idM , as is idP , so g ◦ f is homotopic to idP , by Proposition 23.57. We have (g ◦ f )0 = idM = (id P )0 , which implies that g ◦ f = ḡ ◦ f¯ and idP = idP are also homotopic (via the same homotopy; note h0 = 0 in the proof of Proposition 23.57). Similarly, f¯ ◦ ḡ and idQ are homotopic. ∗ ∗ ∗ ∗ Applying HomR (•, A) yields homotopic morphisms f¯∗ : Q → P and ḡ ∗ : P → Q , with f¯∗ ◦ ḡ ∗ homotopic to idP∗ = idP ∗ and ḡ ∗ ◦ f¯∗ homotopic to id∗Q = idQ∗ , by Proposition 23.68. ∗ ∗ By Lemma 23.55, f¯∗ and ḡ ∗ induce isomorphims H n (P ) ' H n (Q ) for all n ≥ 0. A 11 12 A This justifies our indexing the boundary maps dn : Cn+1 → Cn rather than dn : Cn → Cn−1 . The intuition is that the truncation of projective resolution of M can serve as a replacement for M . 18.785 Fall 2021, Lecture #23, Page 22 ∗ Definition 23.72. Let A and M be R-modules. ExtnR (M, A) is the abelian group H n (P A ) uniquely determined by Theorem 23.71 using any projective resolution P of M . If α : A → B is a morphism of R-modules the map ϕ 7→ α ◦ ϕ induces a morphism of cochain complexes ∗ ∗ P , A → P B and morphisms ExtnR (M, α) : ExtnR (M, A) → ExtnR (M, B) for each n ≥ 0. We thus have a family of functors ExtnR (M, •) from the category of R-modules to the category of abelian groups that is a cohomological δ-functor (by Theorem 23.54). Lemma 23.73. Let M be an R-module. The functors ExtnR (M, •) are additive functors and thus commute with finite direct sums and products. Proof. This follows from Corollary 23.61 and the fact H n (•) is an additive functor. Lemma 23.74. For any two R-modules M and A we have Ext0R (M, A) ' HomR (M, A). Proof. Let · · · → P2 → P1 → M → 0 be a projective resolution of M . Applying HomR (•, A) yields an exact sequence 0 → M ∗ → P1∗ → P2∗ → · · · , and we observe that ∗ ∗ ∗ Ext0R (M, A) = H 0 (P ) = Z 0 (P )/B 0 (P ) = ker(P1∗ → P2∗ )/ im(0 → P1∗ ) ' M ∗ . Theorem 23.75. Let P , Q be projective resolutions of a right R-module M . Let A be a left A A R-module, and let •A ∗ denote application of • ⊗R A. Then Hn (P ∗ ) ' Hn (Q∗ ) for n ≥ 0. Proof. We drop the superscript A in the proof to ease the notation. Let f : P → Q and g : Q → P be extensions of the identity morphism idM given by Proposition 23.57. As in the proof of Theorem 23.71, ḡ ◦ f¯ and idP are homotopic, as are f¯ ◦ ḡ and idQ . Applying • ⊗R A yields homotopic morphisms f¯∗ : P ∗ → Q∗ and ḡ∗ : Q∗ → P ∗ , with f¯∗ ◦ ḡ∗ homotopic to idP ∗ and f¯∗ ◦ ḡ∗ homotopic to idQ . By Lemma 23.52, f¯∗ and ḡ∗ induce ∗ isomorphisms Hn (P ∗ ) ' Hn (Q∗ ) for all n ≥ 0. Definition 23.76. Let A a left R-module and let M be a right R-module. TorR n (M, A) A is the abelian group Hn (P ∗ ) uniquely determined by Theorem 23.75 using any projective resolution P of M . If α : A → B is a morphism of left R-modules the map x ⊗ a 7→ x ⊗ ϕ(a) A B R R induces a morphism P ∗ → P ∗ and morphisms TorR n (M, α) : Torn (M, A) → Extn (M, B) for each n ≥ 0. This yields a family of functors TorR n (M, •) from the category of left R-modules to the category of abelian groups that is a homological δ-functor (by Theorem 23.50). Lemma 23.77. Let M be a right R-module. The functors TorR n (M, •) are additive functors and thus commute with finite direct sums and products. Proof. This follows from Corollary 23.65 and the fact Hn (•) is an additive functor. Lemma 23.78. For any two R-modules M and A we have TorR 0 (M, A) ' M ⊗R A. Proof. Let · · · → P2 → P1 → M → 0 be a projective resolution of M . Applying • ⊗R A yields the exact sequence · · · P2∗ → P1∗ → M∗ → 0, and we observe that TorR 0 (M, A) = H0 (P ∗ ) = Z0 (P ∗ )/B0 (P ∗ ) = ker(P1∗ → 0)/ im(P2∗ → P1∗ ) ' M∗ , Remark 23.79. One can also define ExtnR (M, A) and TorR n (M, A) using injective resolutions; see [7, §2.7] for a proof that this yields the same results. 18.785 Fall 2021, Lecture #23, Page 23 References [1] K. Brown, Cohomology of groups, Springer, 1982. [2] G. J. Janusz, Algebraic number fields, 2nd ed., AMS, 1992. [3] J. S. Milne, Class field theory, version 4.02, 2013. [4] J.-P. Serre, Local fields, Springer, 1979. [5] J.-P. Serre Galois cohomology, Springer, 1997. [6] Stacks Project Authors, Stacks Project, http://stacks.math.columbia.edu. [7] C. A. Weibel, An introduction to homological algebra, Cambridge Univ. Press, 1994. 18.785 Fall 2021, Lecture #23, Page 24 18.785 Number theory I Lecture #24 24 Fall 2021 bonus lecture Artin reciprocity in the unramified case Let L/K be an abelian extension of number fields. In Lecture 22 we defined the norm group m TL/K := NL/K (ILm )Rm K (see Definition 22.27) that we claim is equal to the kernel of the Artin m m map ψL/K : IK → Gal(L/K), provided that the modulus m is divisible by the conductor m m of L (see Definition 22.24). We showed that TL/K contains ker ψL/K (Proposition 22.28), and in Theorem 22.29 we proved the inequality m m m m [IK : TL/K ] ≤ [L : K] = [IK : ker ψL/K ] (1) (the equality follows from the surjectivity of the Artin map proved in Theorem 21.19). It only remains to prove the reverse inequality m m [IK : TL/K ] ≥ [L : K], (2) ∼ (3) which then yields an isomorphism m m /TL/K IK −→ Gal(L/K) induced by the Artin map. This result is known as the Artin reciprocity law. In this lecture we will prove (2) for cyclic extensions L/K when the modulus m is trivial (which forces L/K to be unramified), and then show that this implies the Artin reciprocity law for all unramified abelian extensions. 24.1 Some cohomological calculations If L/K is a finite Galois extension of global fields with Galois group G, then we can naturally × view any of the abelian groups L, L× , OL , OL , IL , PL as G-modules. When G = hσi is cyclic we can compute the Tate cohomology groups of any of these G-modules A, and their associated Herbrand quotients h(A). The Herbrand quotient is defined as the ratio of the cardinalities of A[σ − 1] , NG (A) A[NG ] Ĥ0 (A) := Ĥ0 (G, A) := ker N̂G = AG [N̂G ] = , (σ − 1)(A) Ĥ 0 (A) := Ĥ 0 (G, A) := coker N̂G = AG / im N̂G = if both are finite. We can also compute Ĥ0 (A) = Ĥ −1 (A) ' Ĥ 1 (A) = H 1 (A) as 1-cocycles modulo 1-coboundaries whenever it is convenient to do so. In the interest of simplifying the notation we omit G from our notation whenever it is clear from context. P × For the multiplicative groups OL , L× , IL , PL , the norm element NG := ni=1 σ i corresponds to the action of the field norm NL/K and ideal norm NL/K that we have previously defined, provided that we identify the codomain of the norm map with a subgroup of its × × domain. For the groups L× and OL this simply means identifying K × and OK as subgroups via inclusion. For the ideal group IK we have a natural extension map IK ,→ IL defined by I 7→ IOL that restricts to a map PK ,→ PL .1 Under this convention taking the norm of an 1 The induced map ClK → ClL need not be injective; extensions of non-principal ideals may be principal. Indeed, when L is the Hilbert class field every OK -ideal extends to a principal OL -ideal; this was conjectured by Hilbert and took over 30 years to prove. You will get a chance to prove it on Problem Set 10. element of IL that is (the extension of) an element of IK corresponds to the map I 7→ I #G , as it should, and IK is a subgroup of the G-invariants ILG .2 When A is multiplicative, the action of σ − 1 on a ∈ A is (σ − 1)(a) = σ(a)/a, but we will continue to use the notation (σ − 1)(A) and A[σ − 1] to denote the image and kernel of this action. Conversely, when A is additive, the action of NG corresponds to the trace map, not the norm map. In order to lighten the notation, in this lecture we use N to denote both the (relative) field norm NL/K and the ideal norm NL/K . We begin by recalling an elementary (but often omitted) lemma from Galois theory, originally due to Dedekind. Lemma 24.1. Let L/K be a finite extension of fields. The set AutK (L) is a linearly independent subset of the L-vector space of functions L → L. Proof. Suppose not. Let f := c1 σ1 + · · · + cr σr = 0 with ci ∈ L, σi ∈ AutK (L), and r minimal; we must have r > 1, the ci nonzero, and the σi distinct. Choose α ∈ L so σ1 (α) 6= σr (α) (possible since σ1 6= σr ). We have f (β) = 0 for all β ∈ L, and the same applies to f (αβ) − σ1 (α)f (β), which yields a shorter relation c02 σ2 + · · · + c0r σr = 0, where c0i = ci σi (α) − ci σ1 (α) with c01 = 0, which is nontrivial because c0r 6= 0, a contradiction. Theorem 24.2. Let L/K be a finite Galois extension with Galois group G := Gal(L/K), and for any G-module A, let Ĥ n (A) denote Ĥ n (G, A) and let N denote the norm map NL/K . (i) Ĥ 0 (L) and Ĥ 1 (L) are both trivial. (ii) Ĥ 0 (L× ) ' K × /N(L× ) and Ĥ 1 (L× ) is trivial. Proof. (i) We have LG = K (by definition). The trace map T : L → K is not identically zero (by Theorem 5.20, since L/K is separable), so it must be surjective, since it is K-linear. 0 Thus NG (L) = T(L) = K and Ĥ P(L) = K/K = 0. Now fix α ∈ L with T(α) = τ ∈G τ (α) P = 1, consider a 1-cocycle f : G → L (this means f (στ ) = f (σ) + σ(f (τ ))), and put β := τ ∈G f (τ )τ (α). For all σ ∈ G we have σ(β) = X σ(f (τ ))σ(τ (α)) = τ ∈G X (f (στ )−f (σ))(στ )(α) = τ ∈G X (f (τ )−f (σ))τ (α) = β−f (σ), τ ∈G so f (σ) = β − σ(β). This implies f is a 1-coboundary, so Ĥ 1 (L) = H 1 (L) is trivial. (ii) We have (L× )G = K × , so Ĥ 0 (L× ) = K × /NG L× = K × /N(L× ). Consider any × nonzero Lemma 24.1, P 1-cocycle f : G → L (now this means f (στ )P= f (σ)σ(f (τ ))). By × be a nonzero α 7→ f (τ )τ (α) is not the zero map. Let β = f (τ )τ (α) ∈ L τ ∈G τ ∈G element in its image. For all σ ∈ G we have X X X σ(β) = σ(f (τ ))σ(τ (α)) = f (στ )f (σ)−1 (στ )(x) = f (σ)−1 f (τ )τ (α) = f (σ)−1 β, τ ∈G τ ∈G τ ∈G so f (σ) = β/σ(β). This implies f is a coboundary, so Ĥ 1 (L× ) = H 1 (L× ) is trivial. Corollary 24.3 (Hilbert Theorem 90). Let L/K be a finite cyclic extension with Galois group Gal(L/K) = hσi. Then N(α) = 1 if and only if α = β/σ(β) for some β ∈ L× . Proof. By Theorem 23.37, Ĥ 1 (L× ) ' Ĥ −1 (L× ) = Ĥ0 (L× ) = L× [NG ]/(σ − 1)(L× ), and Theorem 24.2 implies L× [NG ] = (σ − 1)(L× ). The corollary follows. 2 Note that ILG = IK only when L/K is unramified; see Lemma 24.9 below. 18.785 Fall 2021, Lecture #24, Page 2 Remark 24.4. “Hilbert Theorem 90" refers to Hilbert’s text on algebraic number theory [1], although the result is due to Kummer. The result H 1 (Gal(L/K), L× ) = 0 implied by Theorem 24.2 is also often called Hilbert Theorem 90; it is due to Noether [2]. × Our next goal is to compute the Herbrand quotient of OL (in the case that L/K is a finite cyclic extension of number fields). For this we will apply a variant of Dirichlet’s unit theorem due to Herbrand, but first we need to discuss infinite places of number fields. If L/K is a Galois extension of global fields, the Galois group Gal(L/K) acts on the set of places w of L via the action w 7→ σ(w), where σ(w) is the equivalence class of the absolute value defined by kαkσ(w) := kσ(α)kw . This action permutes the places w lying above a given place v of K; if v is a finite place corresponding to a prime p of K, this is just the usual action of the Galois group on the set {q|p}. Definition 24.5. Let L/K be a Galois extension of global fields and let w be a place of L. The decomposition group of w is its stabilizer in Gal(L/K): Dw := {σ ∈ Gal(L/K) : σ(w) = w}. If w corresponds to a prime q of OL then Dw = Dq is also the decomposition group of q. Now let L/K be a Galois extension of number fields. If we write L ' Q[x]/(f ) then we have a one-to-one correspondence between embeddings of L into C and roots of f in C. Each embedding of L into C restricts to an embedding of K into C, and this induces a map that sends each infinite place w of L to the infinite place v of K that w extends. This map may send a complex place to a real place; this occurs when a pair of distinct complex conjugate embeddings of L restrict to the same embedding of K (which must be a real embedding). In this case we say that the place v (and w) is ramified in the extension L/K, and define the ramification index ev := 2 when this holds (and put ev := 1 otherwise). This notation is consistent with our notation ev := ep for finite places v corresponding to primes p of K. Let us also define fv := 1 for v|∞ and put gv := #{w|v} so that the following formula generalizing Corollary 7.5 holds for all places v of K: ev fv gv = [L : K]. Definition 24.6. For a Galois extension of number fields L/K we define the integers Y Y e0 (L/K) := ev , e∞ (L/K) := ev , e(L/K) := e0 (L/K)e∞ (L/K). v-∞ v|∞ Let us now write L ' K[x]/(g). Each embedding of K into C gives rise to [L : K] distinct embeddings of L into C that extend it, one for each root of g (use the embedding of K to view g as a polynomial in C[x], then pick a root of g in C). The transitive action of Gal(L/K) on the roots of g induces a transitive action on these embeddings and their corresponding places. Thus for each infinite place v of K the Galois group acts transitively on {w|v}, and either every place w above v is ramified (this can occur only when v is real and [L : K] is divisible by 2), or none are. It follows that each unramified place v of K has [L : K] places w lying above it, each with trivial decomposition group Dw , while each ramified (real) place v of K has [L : K]/2 (complex) places w lying above it, each with decomposition group Dw of order 2 (its non-trivial element corresponds to complex conjugation in the corresponding embeddings), and the Dw are all conjugate. 18.785 Fall 2021, Lecture #24, Page 3 Theorem 24.7 (Herbrand unit theorem). Let L/K be a Galois extension of number fields. Let w1 , . . . , wr be the real places of L, let wr+1 , . . . , wr+s be the complex places of L. × There exist ε1 , . . . , εr+s ∈ OL such that (i) σ(εi ) = εj if and only if σ(wi ) = wj , for all σ ∈ Gal(L/K); × (ii) ε1 , . . . , εr+s generate a finite index subgroup of OL ; (iii) ε1 ε2 · · · εr+s = 1, and every relation among the εi is generated by this one. × Proof. Pick 1 , . . . , r+s ∈ OL such that ki kwj < 1 for i 6= j; the existence of such i follows from the strong approximation theorem that we willQprove in the next lecture; the × product formula then implies ki kwi > 1. Now let αi := σ∈Dw σ(i ) ∈ OL . We have i Q Q kαi kwi = σ∈Dw ki kwi > 1 and kαi kwj = σ∈Dw ki kσ(wj ) < 1, since σ ∈ Dwi fixes wi i i and permutes the wj with j 6= i. Each αi is fixed by Dwi . Let G := Gal(L/K). For i = 1, . . . , r+s, let r(i) := min{j : σ(wi ) = wj for some σ ∈ G}, so that wr(i) is a distinguished representative of the G-orbit of wi . For i = 1, . . . , r + s let βi := σ(αr(i) ), where σ is any element of G such that σ(wr(i) ) = wi . The value of σ(αr(i) ) does not depend on the choice of σ because σ1 (wr(i) ) = σ2 (wr(i) ) if and only if σ2−1 σ1 ∈ Dwr(i) and αr(i) is fixed by Dwr(i) . The βi then satisfy Q (i). ni := The βi also satisfy (ii): a product γj i6=j βi cannot be trivial because kγj kwj < 1; in × isomorphic to Zr+s−1 which Q necessarily particular, β1 , . . . , βr+s−1 generate a subgroup of OL × r+s−1 × µL (see Theorem 15.12). But we must have i βini = 1 has finite index in OL ' Z for some tuple (n1 , . . . , nr+s ) ∈ Zr+s (with ni = nj whenever wi and wj lie in the same G-orbit, since every σ ∈ G fixes 1). The set of such tuples spans a rank-1 submodule of Zr+s from which we choose a generator (n1 , . . . , nr+s ) (by inverting some βi if necessary, we can make all the ni positive if we wish). Then εi := βini satisfy (i), (ii), (iii) as desired. Theorem 24.8. Let L/K be a cyclic extension of number fields with Galois group G = hσi. × The Herbrand quotient of the G-module OL is × )= h(OL e∞ (L/K) . [L : K] × × they be as in Theorem 24.7, and let A be the subgroup of OL Proof. Let ε1 , . . . , εr+s ∈ OL × generate, viewed as a G-module. By Corollary 23.48, h(A) = h(OL ) if either is defined, × , so we will compute h(A). since A has finite index in OL For each field embedding φ : K ,→ C, let Eφ be the free Z-module with basis {ϕ|φ} consisting of the n := [L : K] embeddings ϕ : L ,→ C with ϕ|K = φ, equipped with the G-action given by σ(ϕ) := ϕ ◦ σ. Let v be the infinite place of K corresponding to φ, and let Av be the free Z-module with basis {w|v} consisting of places of L that extend v, equipped with the G-action given by the action of G on {w|v}. Let π : Eφ → Av be the G-module morphism sending each embedding ϕ|φ to the corresponding place w|v. Let m := #{w|v} and define τ := σ m ; then τ is either trivial or has order 2, and in either case generates the decomposition group Dw for all w|v (since G is abelian). We have an exact sequence π 0 → ker π −→ Eφ −→ Av → 0, with ker π = (τ − 1)Eφ . If v is unramified then ker π = 0 and h(Av ) = h(Eφ ) = 1, since Eφ ' Z[G] ' IndG (Z), by Lemma 23.43. Otherwise, order {w|v} = {w0 , . . . , wm−1 } so    X  ker π = (τ − 1)Eφ = ai (wi − wm+i ) : ai ∈ Z ,   0≤i BK there exists a nonzero principal adele x ∈ K ⊆ AK for which kxkv ≤ kakv for all v ∈ MK . Proof. Let b0 := covol(K) be the measure of a fundamental region for K in AK under our normalized Haar measure µ on AK (by Theorem 25.12, K is cocompact, so b0 is finite). Now define 
b1 := µ z ∈ AK : kzkv ≤ 1 for all v and kzkv ≤ 41 if v is archimedean . Then b1 6= 0, since K has only finitely many archimedean places. Now let BK := b0 /b1 . Suppose a ∈ AK satisfies kak > BK . We know that kakv ≤ 1 for all almost all v, so kak 6= 0 implies that kakv = 1 for almost all v. Let us now consider the set  T := t ∈ AK : ktkv ≤ kakv for all v and ktkv ≤ 41 kakv if v is archimedean . From the definition of b1 we have µ(T ) = b1 kak > b1 BK = b0 ; this follows from the fact that the Haar measure on AK is the product of the normalized Haar measures µv on each of the Kv . Since µ(T ) > b0 , the set T is not contained in any fundamental region for K, so there must be distinct t1 , t2 ∈ T with the same image in AK /K, equivalently, whose difference x = t1 − t2 is a nonzero element of K ⊆ AK . We have  nonarch. v;  max(kt1 kv , kt2 kv ) ≤ kakv 1 kt1 − t2 kv ≤ kt1 kv + kt2 kv ≤ 2 · 4 kakv ≤ kakv real v;   1/2 2 1/2 1/2 2 1/2 2 1 (kt1 − t2 kv ) ≤ (kt1 kv + kt2 kv ) ≤ (2 · 2 kakv ) ≤ kakv complex v. Here we have used the fact that the normalized absolute value k kv satisfies the nonarchimedean triangle inequality when v is nonarchimedean, k kv satisfies the archimedean 1/2 triangle inequality when v is real, and k kv satisfies the archimedean triangle inequality when v is complex. Thus kxkv = kt1 − t2 kv ≤ kakv for all places v ∈ MK as desired. Remark 25.15. Lemma 25.14 should be viewed as an analog of Mikowski’s lattice point theorem (Thoerem 14.13) and a generalization of Proposition 15.9. In Theorem 14.13 we have a discrete cocompact subgroup Λ in a real vector space V ' Rn and a sufficiently large symmetric convex set S that must contain a nonzero element of Λ. In Lemma 25.14 the lattice Λ is replaced by K, the vector space V is replaced by AK , the symmetric convex set S is replaced by the set L(a) := {x ∈ AK : kxkv ≤ kakv for all v ∈ MK }, and sufficiently large means kak > BK , putting a lower bound on µ(L(a)). Proposition 15.9 is actually equivalent to Lemma 25.14 in the case that K is a number field: use the Arakelov divisor c := (kakv ) and note that L(c) = L(a) ∩ K. Theorem 25.16 (Strong Approximation). Let MK = S t T t {w} be a partition of the places of a global field K with S finite. Fix av ∈ K and v ∈ R>0 for each v ∈ S. There exists an x ∈ K for which kx − av kv ≤ v for all v ∈ S, kxkv ≤ 1 for all v ∈ T, (note that there is no constraint on kxkw ). 18.785 Fall 2021, Lecture #25, Page 9 Proof. Let W = {z ∈ AK : kzkv ≤ 1 for all v ∈ MK } as in the proof of Theorem 25.12. Then W contains a complete set of coset representatives for K ⊆ AK , so AK = K + W . For any nonzero u ∈ K ⊆ AK we also have AK = K + uW : given c ∈ AK write u−1 c ∈ AK as u−1 c = a + b with a ∈ K and b ∈ W and then c = ua + ub with ua ∈ K and ub ∈ uW . Now choose z ∈ AK such that Y 0 < kzkv ≤ v for v ∈ S, 0 < kzkv ≤ 1 for v ∈ T, kzkw > B kzk−1 v , v6=w where B is the constant in the Blichfeldt-Minkowski Lemma 25.14 (this is clearly possible: every z = (zv ) with kzv kv ≤ 1 is an element of AK ). We have kzk > B, so there is a nonzero u ∈ K ⊆ AK with kukv ≤ kzkv for all v ∈ MK . Now let a = (av ) ∈ AK be the adele with av given by the hypothesis of the theorem for v ∈ S and av = 0 for v 6∈ S. We have AK = K + uW , so a = x + y for some x ∈ K and y ∈ uW . Therefore ( v for v ∈ S, kx − av kv = kykv ≤ kukv ≤ kzkv ≤ 1 for v ∈ T, as desired. Corollary 25.17. LetQ̀K be a global field and let w be any place of K. Then K is dense in the restricted product v6=w (Kv , Ov ). Remark 25.18. Theorem 25.16 and Corollary 25.17 can be generalized to algebraic groups; see [1] for a survey. References [1] Andrei S. Rapinchuk, Strong approximation for algebraic groups, Thin groups and superstrong approximation, MSRI Publications 61, 2013. 18.785 Fall 2021, Lecture #25, Page 10 18.785 Number theory I Lecture #26 26 Fall 2021 12/01/2021 The idele group, profinite groups, infinite Galois theory 26.1 The idele group Let K be a global field. Having introduced the ring of adeles AK in the previous lecture, it is natural to ask about its unit group × × A× K = {(av ) ∈ AK : av ∈ Kv for all v ∈ MK , and av ∈ Ov for almost all v ∈ MK }. Here Ov× := Kv× ∩ Ov is the unit group of the valuation ring of Kv when v is nonarchimedean and isomorphic to R× or C× when v is archimedean. As noted in Lecture 25, the definition of AK does not actually depend on our choice of Ov at the finitely many archimedean places of K, but the choice we made ensures that every Ov× is a topological group. However, as a subspace of AK , the unit group A× K is not a topological group. Indeed, the inversion map a 7→ a−1 is not continuous. Example 26.1. Consider K = Q and for each prime p let a(p) = (1, . . . , 1, p, 1, . . .) ∈ AQ be the adele with a(p)p = p and a(p)q = 1 for q 6= p. Every basic open set U about 1 in AQ has the form Y Y U= Uv × Ov , v ∈S v 6∈S with S ⊆ MQ finite and 1v ∈ Uv , and it is clear that U contains a(p) for all sufficiently large p. It follows that limp→∞ a(p) = 1 in the topology of AQ . But notice that U does not contain a(p)−1 for any sufficiently large p, so limp→∞ a(p)−1 6= 1−1 in AQ . Thus the function a → a−1 is not continuous in the subspace topology for A× K. This problem is not specific to rings of adeles. For a topological ring R there is in general no reason to expect its unit group R× ⊆ R to be a topological group in the subspace topology. One notable exception is when R is a subring of a topological field (the definition × of which requires inversion to be continuous), as is the case for the unit group OK ; this explains why we have not encountered this problem before now. But the ring of adeles is not naturally contained in any topological field (note that it is not an integral domain). There is a standard solution to this problem: give the group R× the weakest topology that makes it a topological group. This is done by embedding R× in R × R via the map φ : R× → R × R r 7→ (r, r−1 ). We now declare φ : R× → φ(R× ) to be a homeomorphism; that is, we endow R× with the topology matching the subspace topology of φ(R× ) ⊂ R × R. The inversion map r 7→ r−1 is continuous in this topology because it is equal to composition of φ with the projection map R × R → R onto its second coordinate, both of which are continuous maps. × We now consider this construction in the case of A× K . The implied topology on AK has a basis of open sets of the form Y Y U0 = Uv × Ov× v ∈S v 6∈S where Uv ⊆ Kv× and S ⊆ MK is finite. To see this, note that in terms of the embedding −1 φ : A× K → AK × AK defined above, each φ(a) = (a, a ) lies in a product U × V of basic open sets U, V ⊆ AK , and this forces both a and a−1 to lie in Ov , hence in Ov×Q̀, for almost all v. The open sets U 0 are precisely the open sets in the restricted product (Kv× , Ov× ). This leads to the following definition. Definition 26.2. Let K be a global field. The idele group of K is the topological group Y a (Kv× , Ov× ) IK := v with multiplication defined component-wise, which we view as the subgroup A× K of AK endowed with the restricted product topology rather than the subspace topology. The canonical embedding K ,→ AK restricts to a canonical embedding K × ,→ IK , and we define the idele class group CK := IK /K × , a topological group. Remark 26.3. In the literature one finds the notations IK and A× K used interchangeably; they both denote the idele group defined above. But in this lecture we will temporarily use the notation A× K to denote the unit group of the ring AK in the subspace topology (which is not a topological group). Example 26.4. Let us again consider the sequence (a(p)) defined in Example 26.1. This × sequence lies in A× sequence Q and converges to 1 ∈ AQ under the subspace topology. But thisQ × does not converge to 1 in the topology of I . Indeed, consider the basic open set Q v Ov = Q × × p Zp × R of IQ . None of the a(p) = (1, . . . , 1, p, 1, . . .) lie in this open neighborhood of 1, so the sequence (a(p)) cannot converge to 1 in IQ (which means it cannot converge at all: if it converged to x 6= 1 in IQ it would converge to x 6= 1 in A× Q ⊆ AQ , which we know is not the case). The counterexample to the continuity of the inversion map x 7→ x−1 in A× Q is removed in IQ by adding more open sets to the topology; this makes it easier for maps to be continuous and harder for sequences to converge. We now define a surjective homomorphism IK → IK Y a 7→ pvp (a) where the product ranges over primes p of K and vp (a) := vp (av ), where v is the equivalence class of the p-adic absolute value k kp . The composition K × ,→ IK  IK has image PK , the subgroup of principal fractional ideals; we thus have a surjective homomorphism of the idele class group CK = IK /K × onto the ideal class group ClK = IK /PK and a commutative diagram of exact sequences: ← ← ← ← ← ← ← → CK → ClK ← ← → IK ← ←  → PK → IK  1 → K×  1 → 1 → 1 Proposition 26.5. Let K be a global field. The idele group IK is a locally compact group. Proof. It is clear that IK is Hausdorff, since its topology is finer than the topology of A× K ⊆ AK , which is Hausdorff by Proposition 25.9. For each nonarchimedean place v, the set Ov× = {x ∈ Kv× : kxkv = 1} is a closed subset of the compact set Ov , hence compact. This applies almost all v ∈ MK , and the Kv× are all locally compact, so the restricted Q̀ to product (Kv× , Ov× ) = IK is locally compact, by Proposition 25.6. 18.785 Fall 2021, Lecture #26, Page 2 Proposition 26.6. Let K be a global field. Then K × is a discrete subgroup of IK . Proof. We have K × ,→ K × K ⊆ AK × AK . By Theorem 25.12, K is a discrete subset of AK , and it follows that K × K is a discrete subset of AK × AK . The image of K × in AK × AK lies in the image of A× K ,→ AK × AK and in the discrete image of K ,→ AK × AK , and it follows that K × is discrete in A× K and therefore in IK , since having a finer topology only makes it easier for a set to be discrete. We proved last time that K is a discrete cocompact subgroup of AK , so it is natural to × is not a cocompact ask whether K × is a cocompact in A× K or IK . The answer is no, K subgroup of IK , thus the idele class group CK , while locally compact, is not compact. × Recall that for a number field K, the unit group OK is not a cocompact subgroup of × × r+s KR because Log(OK ) is not a (full) lattice in R ' Log(KR× ); it lies in the trace zero hyperplane Rr+s (see Proposition 15.11). In order to get a cocompact subgroup we need to 0 restrict IK to a subgroup that corresponds to the trace zero hyperplane. We have a continuous homomorphism of topological groups k k : IK → R× >0 a 7→ kak Q where kak := v kakv is the adelic norm defined in the previous lecture. We have kak > 0 × for a ∈ IK , since Q av ∈ Ov for almost all v: this implies that kakv = 1 for almost all v ×and the product v kakv is effectively a finite product, and it is nonzero because av ∈ Kv is nonzero for all v ∈ MK . Definition 26.7. Let K be a global field. The group of 1-ideles is the topological group I1K := ker k k = {a ∈ IK : kak = 1}, which we note contains K × , by the product formula (Theorem 13.21). A useful feature of the group of 1-ideles is that, unlike the group of ideles, its topology is the same as the subspace topology it inherits from AK . Lemma 26.8. The group of 1-ideles I1K is a closed subset of AK and IK , and the two subspace topologies on I1K coincide. Proof. We first show that I1K is closed in AK , and therefore also in IK , since it has a finer topology. Consider any x ∈ AK − I1K . We will construct an open neighborhood Ux of x that is disjoint from I1K . The union of the Ux is then the open complement of the closed set I1K . For each  > 0, finite S ⊆ MK , and x ∈ AK we define U (x, S) := {u ∈ AK : ku − xkv <  for v ∈ S and kukv ≤ 1 for v 6∈ S}, which is a basic open set of AK (a product of open sets Uv for v ∈ S and Ov for v 6∈ S). The case kxk < 1. Let S be a finite Q set containing the archimedean places v ∈ MK and all v for which kxkv > 1, such that v∈S kxkv < 1: such an S exists since kxk < 1 and kxkv ≤ 1 for almost all v. For all sufficiently small  > 0 the set Ux := U (x, S) is an open neighborhood of x disjoint from I1K because every y ∈ Ux must satisfy kyk < 1. The case kxk > 1. Let B be twice the product of all the kxkv greater than 1. Let S be the finite set containing the archimedean places v ∈ MK , all nonarchimedean v with 18.785 Fall 2021, Lecture #26, Page 3 residue field cardinality less than 2B, and all v for which kxkv > 1. For all sufficiently small  > 0 the set Ux := U (x, S) is an open neighborhood of x disjoint from I1K because for every y ∈ Ux , either kykv = 1 for all v 6∈ S, in which case kyk > 1, or kykv < 1 for some v 6∈ S, in which case kykv < 1/(2B) and kyk < 1. This proves that I1K is closed in AK , and therefore also in IK . To prove that the subspace topologies coincide, it suffices to show that for every x ∈ I1K and open U ⊆ IK containing x there exists open sets V ⊆ IK and W ⊆ AK such that x ∈ V ⊆ U and V ∩ I1K = W ∩ I1K ; this implies that every neighborhood basis in the subspace topology of I1K ⊆ IK is a neighborhood basis in the subspace topology of I1K ⊆ AK (the latter is a priori coarser than the former). So consider any x ∈ I1K and open neighborhood U ⊆ IK of x. Then U contains a basic open set V = {u ∈ AK : ku − xkv <  for v ∈ S and kukv ≤ 1 for v 6∈ S}, for some  > 0 and finite S ⊆ MK (take S = {v ∈ MK : kxk 6= 1 or πv (U ) 6= Ov } and  > 0 small enough). If we now put W := U (x, S) then x ∈ V ⊆ U and V ∩ I1K = W ∩ I1K as desired. Theorem 26.9 (Fujisaki’s Lemma). For any global field K, the principal ideles K × ⊆ IK are a discrete cocompact subgroup of the group of 1-ideles I1K . Proof. By Proposition 26.6, K × is discrete in IK , and therefore discrete in the subspace I1K . As in the proof of Theorem 25.12, to prove that K × is cocompact in I1K it suffices to exhibit a compact set W ⊆ AK for which W ∩ I1K surjects onto I1K /K × under the quotient map (here we are using Lemma 26.8: I1K is closed so W ∩ I1K is compact). To construct W we first choose a ∈ AK such that kak > BK , where BK is the BlichfeldtMinkowski constant in Lemma 25.14, and let W := L(a) = {x ∈ AK : kxkv ≤ kakv for all v ∈ MK }. Now consider any u ∈ I1K . We have kuk = 1, so k ua k = kak > BK , and by Lemma 25.14 there is a z ∈ K × for which kzkv ≤ ua v for all v ∈ MK . Therefore zu ∈ W . Thus every u ∈ I1K can be written as u = z −1 · zu with z −1 ∈ K × and zu ∈ W ∩ I1K . Thus W ∩ I1K surjects onto I1K /K × under the quotient map I1K → I1K /K × , which is continuous, and it follows that I1K /K × is compact. Remark 26.10. Fujisaki’s lemma appears in [7, Lemma 3.1.1] and can be used to prove many of the finiteness results we proved in Lecture 14, as well as Dirichlet’s unit theorem (Theorem 15.12). 1 := I1 /K × is the norm-1 Definition 26.11. For a global field K the compact group CK K idele class group. 1 is totally Remark 26.12. When K is a function field the norm-1 idele class group CK disconnected, in addition to being a compact group, and thus a profinite group. 26.2 Profinite groups In order to state the main theorems of class field theory in our adelic/idelic setup, rather than considering each finite abelian extension L of a global field K individually, we prefer to work in K ab , the compositum of all finite abelian extensions of K. This requires us to understand the infinite Galois group Gal(K ab /K), which is an example of a profinite group. 18.785 Fall 2021, Lecture #26, Page 4 Definition 26.13. A profinite group is a topological group that is an inverse limit of finite groups with the discrete topology. Given any topological group G, we can construct a profinite group by taking the profinite completion Y b := lim G/N ⊆ G G/N ←− N N where N ranges over finite index open normal subgroups, ordered by containment.1 If we are given a group G without a specified topology, we can make it a topological group by giving it the profinite topology. This is the weakest topology that makes every finite quotient discrete and is obtained by taking all cosets of finite-index normal subgroups as a basis. The profinite completion of G is (by construction) a profinite group, and it comes b that sends each g ∈ G to the sequence equipped with a natural homomorphism φ : G → G of Q its images (g N ) in the discrete finite quotients G/N , which we may view as an element of N G/N . The homomorphism φ is not necessarily injective; this occurs if and only if the intersection of all finite-index open normal subgroups of G is the trivial group (such a G is said to be residually finite), but we always have the following universal property for inverse limits. For every continuous homomorphism ϕ : G → H with H a profinite group, there is a unique continuous homomorphism that makes the following diagram commute b → G → ← ϕ φ ∃! → ← ← G H There is much one can say about profinite groups but we shall limit ourselves to a few remarks and statements of the main results we need, deferring most of the proofs to Problem Set 11. See [4] for a comprehensive treatment of profinite groups. Remark 26.14. Taking inverse limits in the category of topological groups is the same thing as taking the inverse limits in the categories of topological spaces and groups independently: the topology is the subspace topology in the product, and the group operation is the group operation in the product (defined component-wise). This might seem obvious, but the same statement does not apply to direct limits, where one must compute the limit in the category of topological groups, otherwise the group operation in the direct limit of the groups is not necessarily continuous under the direct limit topology; see [5].2 Remark 26.15. The profinite completion of G as a topological group is not necessarily the same thing as the profinite completion of G as a group if we forget its topology; this depends on whether the original topology on G contains the profinite topology or not. In particular, a profinite group need not equal to its profinite completion as a group; the group Gal(Q/Q) endowed with the Krull topology is an example (see below). Profinite groups that are isomorphic to their profinite completions as groups are said to be strongly complete; this is equivalent to requiring every finite index subgroup to be open (see Corollary 26.20 below). It is known that if G is finitely generated as a topological group (meaning it contains a finitely generated dense subgroup), then G is strongly complete [3]. This applies, for example, to Gal(F̄q /Fq ) for any finite field Fq , since the q-power Frobenius automorphism generates a dense subgroup (it is thus a topological generator ). 1 Recall that an inverse system has objects Xi and morphisms Xi ← Xj for i ≤ j. Here we have objects G/Ni and morphisms G/Ni ← G/Nj for i ≤ j; we want the indices ordered so that i ≤ j whenever Ni contains Nj ; containment induces a canonical morphism g + Ni ←[ g + Nj on the quotients. 2 For countable direct systems of locally compact groups this issue does not arise [5, Thm. 2.7]. 18.785 Fall 2021, Lecture #26, Page 5 Remark 26.16. For suitable restricted types of finite groups C (for example, all finite cyclic groups, or all finite p-groups for some fixed prime p), one can similarly define the notion of a pro-C group and the pro-C completion of a group by constraining the finite groups in the inverse system to lie in C. One can also define profinite rings or pro-C rings. Example 26.17. Here are a few examples of profinite completions: • The profinite completion of any finite group G is isomorphic to G with the discrete b is an isomorphism. topology; the natural map G → G b := lim Z/nZ = Q Zp , where the indices n are • The profinite completion of Z is Z ←−n b is injective but not surjective. ordered by divisibility; the natural map Z → Z • The profinite completion of Q is trivial because Q has no finite index subgroups other b = {1} is surjective but not injective. than itself. The natural map Q → Q b The image of G Lemma 26.18. Let G be a topological group with profinite completion G. b b under the natural map φ : G → G is dense in G. Proof. See Problem Set 11. We now give a topological characterization of profinite groups that can serve as an alternative definition. Theorem 26.19. A topological group is profinite if and only if it is a totally disconnected compact group. Proof. See Problem Set 11. Corollary 26.20. Let G be a profinite group. Then G is naturally isomorphic to its profinite completion. In fact, G ' lim G/U, ←− where U ranges over open normal subgroups (ordered by containment). However, G is isomorphic to its profinite completion as a group (in other words, strongly complete) if and only if every finite index subgroup of G is open. Proof. See Problem Set 11 for the first statement. For the second statement, if every finite index subgroup of G is open then every finite-index normal subgroup is open, meaning that the topology on G is finer than the profinite topology, and we get the same profinite completion under both topologies. Conversely, if G has a finite index subgroup H that is not open, then no subgroup of H is open (since H is the union of the cosets of any of its subgroups); in particular, the intersection of all the conjugates of H, which is a normal subgroup N , is not open in G, nor are any of its subgroups. If the topological group G is isomorphic to its profinite completion b as a group, then by the universal property of the profinite completion the natural map G b is an isomorphism, but the image of N under φ is an open subgroup of G b by φ: G → G construction, which is a contradiction. 18.785 Fall 2021, Lecture #26, Page 6 26.3 Infinite Galois theory The key issue that arises when studying Galois groups of infinite algebraic extensions (as opposed to finite ones) is that the Galois correspondence (the inclusion reversing bijection between subgroups and subextensions) fails spectacularly. As you proved on Problem Set 5 Q b in the case Gal(Fq /Fq ) ' Z ' p Zp , this happens for a simple reason: there are too many subgroups. Similarly, the absolute Galois group of Q has uncountably many subgroups of index 2 but Q has only countably many quadratic extensions, see [2, Aside 7.27]. Thus not all subgroups of an infinite Galois group Gal(L/K) correspond to subextensions of L/K. We are going to put a topology on Gal(L/K) that distinguishes those that do. Lemma 26.21. Let L/K be a Galois extension with Galois group G = Gal(L/K), If F/K is a normal subextension of L/K, then H = Gal(L/F ) is a normal subgroup of G with fixed field F , and we have an exact sequence 1 → Gal(L/F ) → Gal(L/K) → Gal(F/K) → 1, where the first map is inclusion, the second map is induced by restriction, and we have G/H ' Gal(F/K). This lemma is a list of things we already know to be true for finite Galois extensions, the point is simply to verify that they also hold for infinite Galois extensions; this seems prudent given the aforementioned failure of the Galois correspondence. Proof. If F/K is a normal subextension of L/K then the restriction map σ 7→ σ|F defines a homomorphism Gal(L/K) → Gal(F/K) whose kernel is a normal subgroup H = Gal(L/F ). The fixed field of H contains F by definition, and it must be equal to F : if we had α ∈ LH −F we could construct an element of H that sends α to a distinct root α0 6= α of its minimal polynomial f over F (this defines an element of Gal(E/F ), where E is the splitting field of f , which can be extended to Gal(L/F ) = H by embedding L in an algebraic closure and applying Theorem 4.9). The restriction map is surjective because any σ ∈ Gal(F/K) can be extended to Gal(L/K), by Theorem 4.9, thus the sequence in the lemma is exact, and G/H ' Gal(F/K) follows. Unlike the situation for finite Galois extensions, it can happen that a normal subgroup H of Gal(L/K) with fixed field F is not equal to Gal(L/F ); it must be contained in Gal(L/F ), but it could be a proper subgroup. This is exactly what happens for all but a countable number of the uncountably many index 2 subgroups H of G = Gal(Q/Q); the fixed field of H is Q but H ( G is not the Galois group of Q/Q, nor is the Galois group of Q/K for any subextension K/Q. It is thus necessary to distinguish the subgroups of Gal(L/K) that are actually Galois groups of a subextension. This is achieved by putting an appropriate topology on the Galois group. Definition 26.22. Let L/K be a Galois extension with Galois group G := Gal(L/K). The Krull topology on G has the basis consisting of all cosets of subgroups HF := Gal(L/F ), where F ranges over finite normal extensions of K in L. Under the Krull topology every open normal subgroup necessarily has finite index, but it is typically not the case that every normal subgroup of finite index is open. Thus the Krull topology on Gal(L/K) is strictly coarser than the profinite topology, in general (this holds for Gal(Q/Q), for example). However, the topological group we obtain by putting the Krull topology on Gal(L/K) is a profinite group. 18.785 Fall 2021, Lecture #26, Page 7 Theorem 26.23. Let L/K be a Galois extension. Under the Krull topology, the restriction maps induce a natural isomorphism of topological groups φ : Gal(L/K) → lim Gal(F/K), ←− where F ranges over finite Galois extensions of K in L. In particular, Gal(L/K) is a profinite group whose open normal subgroups are precisely those of the form Gal(L/F ) for some finite normal extension F/K. Proof. Every α ∈ L is algebraic over K, hence lies in some finite normal subextension F/K (take the normal closure of K(α)). Every automorphism in Gal(L/K) is thus uniquely determined by its restrictions to finite normal F/K, which implies that φ is injective. Given an element (σF ) ∈ lim Gal(F/K), we can define an automorphism σ ∈ Gal(L/K) by simply ←− putting σ(α) = σF (α), where F is the normal closure of K(α) (the fact that this actually gives an automorphism is guaranteed by the inverse system of restriction maps used to define lim Gal(F/K)). Thus φ is surjective. ←− By Lemma 26.21, if we put G := Gal(L/K) and HF := Gal(L/F ), then we can view φ as the natural map φ : G → lim G/HF , ←− which is continuous, and we have shown it is a bijection. To prove that φ is an isomorphism of topological groups it remains only to show that it is an open map. For this it suffices to show that φ maps open subgroups H ⊆ G to open sets in lim G/HF , since every open set ←− in G is a union of cosets of open subgroups. If H = Gal(L/F ) then φ(H) = {(σE ) : σE |E∩F = id|E∩F } = πF−1 (id|F ), where E/K ranges over finite normal subextensions of L/K and πF is the projection map from the inverse limit to Gal(F/K). The singleton set {id|F } is open in the discrete group Gal(F/K), so its inverse image under the continuous projection πF is open in G. The last statement follows from Corollary 26.20 and Lemma 26.21. Theorem 26.24 (Fundamental theorem of Galois theory). Let L/K be a Galois extension and let G := Gal(L/K) be endowed with the Krull topology. The maps F 7→ Gal(L/F ) and H 7→ LH define an inclusion reversing bijection between subextensions F/K of L/K and closed subgroups H of G. Under this correspondence, subextensions of finite degree n correspond to subgroups of finite index n, and normal subextensions F/K correspond to normal subgroups H ⊆ G such that Gal(F/K) ' G/H as topological groups. Proof. We first note that every open subgroup of G is closed, since it is the complement of the union of its non-trivial cosets, all of which are open, and closed subgroups of finite index are open by the same argument. The correspondence between finite Galois subextensions F/K and finite index closed normal subgroups H then follows the previous theorem, and we have [F : K] = [G : H] because G/H ' Gal(F/K), by Lemma 26.21. If F/K is any finite subextension with normal closure E, then H = Gal(L/F ) contains the normal subgroup N = Gal(L/E) with finite index. The subgroup N is open and therefore closed, thus H is closed since it is a finite union of cosets of N . The fixed field of H is F (by the same argument as in the proof of Lemma 26.21), thus finite subextensions correspond to closed subgroups of finite index. Conversely, every closed subgroup H of 18.785 Fall 2021, Lecture #26, Page 8 finite index has a fixed field F of finite degree, since the intersection of its conjugates is a normal closed subgroup N = Gal(L/E) of finite index whose fixed field E contains F and has finite degree. The degrees and indices match because [G : N ] = [G : H][H : N ] and [E : K] = [F : K][E : F ]; by the previous argument for finite normal subextensions, [E : K] = [G : N ] and [E : F ] = [H : N ] (for the second equality, replace L/K with L/F and G with H). Any subextension F/K is the union of its finite subextensions E/K. The intersection of the corresponding closed finite index subgroups Gal(L/E) is equal to Gal(L/F ), which is therefore closed. Conversely, every closed subgroup H of G is an intersection of basic closed subgroups, all of which have the form Gal(L/E) for some finite subextension E/K, thus H = Gal(L/F ), where F is the union of the E. The isomorphism Gal(F/K) ' G/H for normal subextensions/subgroups follows directly from Lemma 26.21. Corollary 26.25. Let L/K be a Galois extension and let H be a subgroup of Gal(L/K) with fixed field F . The closure H of H in the Krull topology is Gal(L/F ). Proof. The Galois group Gal(L/F ) contains H, since it contains every σ ∈ Gal(L/K) that fixes F (by definition), and Gal(L/F ) is a closed subgroup of Gal(L/K) with LGal(L/F ) = F , by Theorem 26.24. We thus have H ⊆ H ⊆ Gal(L/F ) with the same fixed field F . The last two groups are closed and therefore equal under the bijection given by Theorem 26.24. We conclude this section with the following theorem due to Waterhouse [6]. Theorem 26.26 (Waterhouse 1973). Every profinite group G is isomorphic to the Galois group of some Galois extension L/K. Proof sketch. Let X be the disjoint union of the finite discrete quotients of G equipped with the G-action induced by multiplication. Now let k be any field and define L = k(X) as a purely transcendental extension of k with indeterminates for each element of X. We can view each σ ∈ G as an automorphism of L that fixes k and sends each x ∈ X to σ(x), and since G acts faithfully on X, we can view G as a subgroup of Autk (L). Now let K = LG . Then L/K is a Galois extension with G ' Gal(L/K), by [6, Thm. 1]. Remark 26.27. Although this proof lets us choose any field k we like, we have no way to control K. In particular, it is not known whether every profinite group G is isomorphic to a Galois group over K = Q; indeed, this is not even known for all finite groups G. References [1] Nicolas Bourbaki, General Topology: Chapters 1-4 , Springer, 1995. [2] J.S. Milne, Fields and Galois theory, version 4.51, 2015. [3] Nikolay Nikolov and Dan Segal, On finitely generated profinite groups I: strong completeness and uniform bounds, Annals of Mathematics 165 (2007), 171–238. [4] Luis Ribes and Pavel Zalesskii, Profinite groups, second edition, Springer, 2010. [5] N. Tatsuuma, H. Shimomura, and T. Hirai, On group topologies and unitary representations of inductive limits of topological groups and the case of the group of diffeomorphisms, J. Math. Kyoto Univ. 38 (1998), 551–578. 18.785 Fall 2021, Lecture #26, Page 9 [6] William C. Waterhouse, Profinite groups are Galois groups, Proceedings of the American Mathematical Society 42 (1974). [7] André Weil, Adeles and algebraic groups, Springer, 1982. 18.785 Fall 2021, Lecture #26, Page 10 18.785 Number theory I Lecture #27 27 Fall 2021 12/6/2021 Local class field theory In this lecture we give a brief overview of local class field theory. Recall that a local field is a locally compact field whose topology is induced by a nontrivial absolute value (Definition 9.1). As we proved in Theorem 9.9, every local field is isomorphic to one of the following: • R or C (archimedean, characteristic 0); • finite extension of Qp (nonarchimedean, characteristic 0); • finite extension of Fq ((t)) (nonarchimedean, characteristic p > 0). In the nonarchimedean cases, the ring of integers of a local field is a complete DVR with finite residue field. The goal of local class field theory is to classify all finite abelian extensions of a given local field K. Rather than considering each finite abelian extension L/K individually, we will treat them all at once, by working in the maximal abelian extension of K inside a fixed separable closure K sep . Definition 27.1. Let K be field with separable closure K sep . The field [ K ab := L L ⊆ K sep L/K finite abelian is the maximal abelian extension of K (in K sep ). We also define [ L, K unr := L ⊆ K sep L/K finite unramified the maximal unramified extension of K (in K sep ). The field K ab contains the field K unr ; this is obvious in the archimedean case, where we have K = K unr is R or C and K ab = K sep = C (note that the extension C/R is ramified). In the nonarchimedean case the inclusion K unr ⊆ K ab follows from Theorem 10.13, which implies that K unr is isomorphic to the algebraic closure of the residue field of K, which is an abelian extension because it is pro-cyclic (every finite extension of the residue field is cyclic because the residue field is finite). We thus have a tower of field extensions K ⊆ K unr ⊆ K ab ⊆ K sep . By Theorem 26.23, the Galois group Gal(K ab /K) is the profinite group Gal(K ab /K) ' lim Gal(L/K), ←− L where L ranges over the finite extensions of K in K ab , ordered by inclusion (note that every finite extension of K in K ab is normal because every open subgroup of the abelian group Gal(K ab /K) is a normal subgroup). Like all Galois groups, the profinite group Gal(K ab /K) is a totally disconnected compact group; see Problem Set 11. By Theorem 26.24, we have the Galois correspondence { extensions of K in K ab } ←→ { closed subgroups of Gal(K ab /K) } L 7−→ Gal(K ab /L) (K ab )H ←−[ H. Finite abelian extensions L/K correspond to open subgroups of Gal(K ab /K) (which must have finite index since Gal(K ab /K) is compact). When K is an archimedean local field its abelian extensions are easy to understand; either K = R, in which case C is the unique nontrivial abelian extension, or K = C and there are no nontrivial abelian extensions. Now suppose K is a nonarchimedean local field with ring of integers OK , maximal ideal p, and residue field Fp := OK /p. If L/K is a finite unramified extension with residue field Fq := OL /q, Theorem 10.13 gives us a canonical isomorphism Gal(L/K) ' Gal(Fq /Fp ) = hx 7→ x#Fp i, between the Galois group of L/K and the Galois group of the residue field extension Fq /Fp . The group Gal(Fq /Fp ) is generated by the Frobenius automorphism x → x#Fp , and we use FrobL/K ∈ Gal(L/K) to denote the corresponding element of Gal(L/K); note that FrobL/K is an element, not just a conjugacy class, because Gal(L/K) is abelian. Every finite unramified extension of local fields L/K thus comes equipped with a canonical generator FrobL/K for its Galois group (which is necessarily cyclic). In this local unramified setting, the Artin map is very easy to understand. The ideal group IK is the infinite cyclic group generated by the prime ideal p, and the Artin map ψL/K : IK → Gal(L/K) p 7→ FrobL/K , corresponds to the quotient map Z → Z/nZ, where n := [L : K]. We can extend the Artin map to K × by defining ψL/K (x) := ψL/K ((x)); this map sends every uniformizer π to the Frobenius element FrobL/K ; note that since OK is a DVR, hence a PID, every ideal in I is of the form (x) for some x ∈ K × , so defining the Artin map on K × rather than IK does not lose any information when K is a local field. 27.1 Local Artin reciprocity Local class field theory is based on the existence of a continuous homomorphism θK : K × → Gal(K ab /K) known as the local Artin homomorphism (or local reciprocity map), which is described by the following theorem. Theorem 27.2 (Local Artin Reciprocity). Let K be a local field. There is a unique continuous homomorphism θK : K × → Gal(K ab /K) with the property that for each finite extension L/K in K ab , the homomorphism θL/K : K × → Gal(L/K) given by composing θK with the natural map resL/K : Gal(K ab /K)  Gal(L/K) satisfies: 18.785 Fall 2021, Lecture #27, Page 2 • if K is nonarchimedean and L/K is unramified then θL/K (π) = FrobL/K for every uniformizer π of OK ; • θL/K is surjective with kernel NL/K (L× ), inducing K × /NL/K (L× ) ' Gal(L/K). The natural map resL/K : Gal(K ab /K)  Gal(L/K) can be viewed as any of • the map induced by restriction σ 7→ σ|L (note that σ(L) = L because L/K is Galois); • the quotient map Gal(K ab /K)  Gal(K ab /K)/Gal(K ab /L); • the projection coming from Gal(K ab /K) = limL Gal(L/K) ⊆ ←− L ranges over finite extensions of K in K ab ). Q L Gal(L/K) (where These are equivalent descriptions of the same surjective homomorphism of topological groups (where the finite group Gal(L/K) has the discrete topology). We will not have time to prove this theorem, but we would like to understand exactly what it says. The homomorphisms θL/K form a compatible system, in the sense that if L1 ⊆ L2 then θL1 /K = resL2 /L1 ◦θL2 /K , where resL2 /L1 is the natural map from Gal(L2 /K) to Gal(L1 /K) = Gal(L2 /K)/Gal(L2 /L1 ). Indeed, the maps resL2 /L1 are precisely the maps that appear in the inverse system defining limL Gal(L/K) ' Gal(K ab /K). ←− It is first worth contrasting local Artin reciprocity with the more complicated global version of Artin reciprocity that we saw in Lecture 21: • There is no modulus m; working in K ab addresses all abelian extensions of K at once. × • The ray class groups Clm K are replaced by quotients of K . × × m • The Takagi group NL/K (ILm )Rm K ⊆ IK is replaced by NL/K (L ) ⊆ K . 27.2 Norm groups Definition 27.3. A norm group of a local field K is a subgroup of the form N(L× ) := NL/K (L× ) ⊆ K × , for some finite abelian extension L/K. Remark 27.4. Removing the word abelian does not change the definition above. If L/K is any finite extension (not necessarily Galois), then N(L× ) = N(F × ), where F is the maximal abelian extension of K in L; this result is known as the Norm Limitation Theorem (see [1, Theorem III.3.5]). So we could have defined norm groups more generally. This is not relevant to classifying the abelian extension of K, but it demonstrates a key limitation of local class field theory (which extends to global class field theory): norm groups tell us nothing about nonabelian extensions of K. Theorem 27.2 implies that the Galois group of any finite abelian extension L/K of a local fields is canonically isomorphic to the quotient K × /NL/K (L× ). In order to understand the finite abelian extensions of a local field K, we just need to understand its norm groups. Corollary 27.5. The map L 7→ N(L× ) defines an inclusion reversing bijection between the finite abelian extensions L/K in K ab and the norm groups in K × which satisfies × (a) N((L1 L2 )× ) = N(L× 1 ) ∩ N(L2 ) and × (b) N((L1 ∩ L2 )× ) = N(L× 1 )N(L2 ). In particular, every norm group of K has finite index in K × , and every subgroup of K × that contains a norm group is a norm group. 18.785 Fall 2021, Lecture #27, Page 3 Here we write L1 L2 for the compositum of L1 and L2 inside K ab (the intersection of all subfields of K ab that contain both L1 and L2 ). Proof. We first note that if L1 ⊆ L2 are two extensions of K then transitivity of the field norm (Corollary 4.53) implies NL2 /K = NL1 /K ◦ NL2 /L1 , × × and therefore N(L× 2 ) ⊆ N(L1 ); the map L 7→ N(L ) thus reverses inclusions. × This immediately implies N((L1 L2 )× ) ⊆ N(L1 ) ∩ N(L× 2 ), since L1 , L2 ⊆ L1 L2 . For the reverse inclusion, let us consider the commutative diagram ← ← K× θL1 L2 /K -←→ θL1 /K ×θL2 /K → Gal(L1 L2 /K) res×res → Gal(L1 /K) × Gal(L2 /K) × × lies in the kernel of θ By Theorem 27.2, each x ∈ N(L× L1 /K and θL2 /K , 1 ) ∩ N(L2 ) ⊆ K hence in the kernel of θL1 L2 /K (by the diagram), and therefore in N((L1 L2 )× ), again by Theorem 27.2. This proves (a). We now show that L 7→ N(L× ) is a bijection; it is surjective by definition, so we just × need to show it is injective. If N(L× 2 ) = N(L1 ) then by (a) we have × × × N((L1 L2 )× ) = N(L× 1 ) ∩ N(L2 ) = N(L1 ) = N(L2 ), and Theorem 27.2 implies Gal(L1 L2 /K) ' Gal(L1 /K) ' Gal(L2 /K), which forces L1 = L2 ; thus L 7→ N(L× ) is injective. We now prove (b). The field L1 ∩ L2 is the largest extension of K that lies in both × × × L1 and L2 , while N(L× 1 )N(L2 ) is the smallest subgroup of K containing both N(L1 ) and × × N(L2 ); they therefore correspond under the inclusion reversing bijection L 7→ N(L ) and × we have N((L1 ∩ L2 )× ) = N(L× 1 )N(L2 ) as desired. The fact that every norm group has finite index in K × follows immediately from the isomorphism Gal(L/K) ' K × /NL/K (L× ) given by Theorem 27.2, since Gal(L/K) is finite. Finally, let us prove that every subgroup of K × that contains a norm group is a norm group. Suppose N(L× ) ⊆ H ⊆ K × , for some finite abelian L/K, and subgroup H of K × , and put F := LθL/K (H) . We have a commutative diagram ← ← θL/K res → Gal(F/K) → θF/K → Gal(L/K) ← K× in which Gal(L/F ) = θL/K (H) is precisely the kernel of the map Gal(L/K) → Gal(F/K) induced by restriction. It follows from Theorem 27.2 that H = ker θF/K = N(F × ) is a norm group as claimed. Lemma 27.6. Let L/K be any extension of local fields. If N(L× ) has finite index in K × then it is open. 18.785 Fall 2021, Lecture #27, Page 4 Proof. The lemma is clear if K is archimedean (either L = K and N(L× ) = K × , or L ' C, K ' R, and [K × : N(L× )] = [R× : R>0 ] = 2), so assume K is nonarchimedean. Suppose × × [K × : N(L× )] < ∞. The unit group OL is compact, so N(OL ) is compact (since N : L× → × × K is continuous), thus closed in the Hausdorff space K . For any α ∈ L, × × α ∈ OL ⇐⇒ |α| = 1 ⇐⇒ |NL/K (α)| = 1 ⇐⇒ NL/K (α) ∈ OK , and therefore × × N(OL ) = N(L× ) ∩ OK . × × It follows that N(OL ) is the kernel of the homomorphism OK ,→ K ×  K × /N(L× ) and × × × × × therefore [OK : N(OL )] ≤ [K : N(L )] < ∞. Thus N(OL ) is a closed subgroup of finite × index in OK , hence open (its complement is a finite union of closed cosets, hence closed), × × and OK is open1 in K × , so N(OL ) is open in K × , and therefore N(L× ) is open in K × , since × N(L× ) is a union of cosets of the open subgroup N(OL ). Remark 27.7. If K is a local field of characteristic zero then one can show that in fact every finite index subgroup of K × is open (whether it is a norm group or not), but this is not true in positive characteristic. 27.3 The main theorems of local class field theory Corollary 27.5 implies that all norm groups of K have finite index in K × , and Lemma 27.6 then implies that all norm groups are finite index open subgroups of K × . The existence theorem of local class field theory states that the converse also holds. Theorem 27.8 (Local Existence Theorem). Let K be a local field and let H be a finite index open subgroup of K × . There is a unique extension L/K in K ab with NL/K (L× ) = H. The local Artin homomorphism θK : K × → Gal(K ab /K) is not an isomorphism; indeed, it cannot be, because Gal(K ab /K) is compact and K × is not. However, the local existence theorem implies that after taking profinite completions the local Artin homomorphism becomes an isomorphism. Theorem 27.9 (Main Theorem of Local Class Field Theory). Let K be a local field. The local Artin homomorphism induces a canonical isomorphism of profinite groups. ∼ d × −→ θbK : K Gal(K ab /K) Proof. The Galois group Gal(K ab /K) is a profinite group, isomorphic to the inverse limit Gal(K ab /K) ' lim Gal(L/K), ←− (1) L where L ranges over the finite extensions of K in K ab ordered by inclusion; see Theorem 26.23. It follows from Lemma 27.6, Theorem 27.8, and the definition of the profinite completion, that d × ' lim K × /N(L× ), K (2) ←− L 1 Recall that in a nonarchimedean local field, |K × | is discrete in R>0 and we can always pick  > 0 so × that OK = {x ∈ K × : 1 −  < |x| < 1 + }, which is clearly open in the metric topology induced by | |. 18.785 Fall 2021, Lecture #27, Page 5 where L ranges over finite abelian extensions of K (in K sep ). By local Artin reciprocity (Theorem 27.2), for each finite abelian extension L/K we have an isomorphism ∼ θL/K : K × /N(L× ) −→ Gal(L/K), and these isomorphisms commute with inclusion maps between finite abelian extensions of K. We thus have an isomorphism of the inverse systems appearing in (1) and (2). The isomorphism is canonical because the Artin homomorphism θK is unique and the isomorphisms in (1) and (2) are both canonical. d ×. In view of Theorem 27.9, we would like to better understand the profinite group K d × is either trivial or the cyclic group of order 2, so let us assume If K is archimedean then K that K is nonarchimedean. If we pick a uniformizer π for the maximal ideal p of OK , then × we can uniquely write each x ∈ K × in the form uπ v(x) , with u ∈ OK and v(x) ∈ Z. This defines an isomorphism ∼ × ×Z K × −→ OK
x 7−→ x/π v(x) , v(x) . Taking profinite completions (which commutes with products), we obtain an isomorphism d × ' O × × Z, b K K since the unit group × × O /(1 + pn ) OK ' F× p × (1 + p) ' Fp × lim ←− K n is already profinite (hence isomorphic to its profinite completion, by Corollary 26.20). Note d × ' O× × Z b is far from canonical; it depends on our choice of π, that the isomorphism K K and there are uncountably many π to choose from. We have a commutative diagram of exact sequences of topological groups → Z ← ← ← φ ← → Gal(K ab /K) → 0 res → Gal(K unr /K) ← ← ← → Gal(K ab /K unr ) v θK → → ← 1 o ← → K× -← → × → OK ← 1 → 1 in which the bottom row is the profinite completion of the top row. The map φ on the right is given by b ' Gal(Fp /Fp ) ' Gal(K unr /K), Z ,→ Z and sends 1 to the sequence of Frobenius elements (FrobL/K ) in the profinite group Gal(K unr /K) ' lim Gal(L/K) ⊆ ←− L Y Gal(L/K), L where L ranges over finite unramified extensions of K; here we are using the canonical isomorphisms Gal(L/K) ' Gal(Fq /Fp ) given by Theorem 10.13. The Frobenius element φ(1) is a topological generator for Gal(K unr /K), meaning that it generates a dense subset. 18.785 Fall 2021, Lecture #27, Page 6 Remark 27.10. The Frobenius element φ(1) ∈ Gal(K unr /K) corresponds to the Frobenius automorphism x 7→ x#Fp of Gal(Fp /Fp ); both are canonical topological generators of the Galois groups in which they reside, and both are sometimes referred to as the arithmetic Frobenius. There is another obvious generator for Gal(K unr /K) ' Gal(Fp /Fp ), namely φ(−1), which is called the geometric Frobenius (for reasons we won’t explain here). × The group Gal(K ab /K unr ) ' OK corresponds to the inertia subgroup of Gal(K ab /K). The top sequence splits (but not canonically), hence so does the bottom, and we have × b Gal(K ab /K) ' Gal(K ab /K unr ) × Gal(K unr /K) ' OK × Z. For each choice of a uniformizer π ∈ OK we get a decomposition K ab = Kπ K unr correspond× Z ing to K × = OK π . The field Kπ is the subfield of K ab fixed by θK (π) ∈ Gal(K ab /K). Equivalently, Kπ is the compositum of all the totally ramified finite extensions L/K in K ab for which π ∈ N(L× ). Example 27.11. Let K = Qp and pick π = p. The decomposition K ab = Kπ K unr is [ [ n) · Qab = Q (ζ Qp (ζm ), p p p n m⊥p × ). where the first union on the RHS is fixed by θK (p) and the second is fixed by θK (OK Constructing the local Artin homomorphism is the difficult part of local class field theory. However, assuming the local existence theorem, it is easy to show that the local Artin homomorphism is unique if it exists. Proposition 27.12. Let K be a local field and assume every finite index open subgroup of K × is a norm group. There is at most one homomorphism θ : K × → Gal(K ab /K) of topological groups that has the properties given in Theorem 27.2. Proof. The proposition is clear when K is archimedean, so assume it is nonarchimedean. Let p = (π) be the maximal ideal of OK , and for each integer n ≥ 0 let Kπ,n /K be the finite abelian extension given by Theorem 27.8 corresponding to the finite index subgroup × , and (1 + pn )hπi of K × ; here 1 + pn and hπi denote subgroups of K × , with 1 + p0 := OK × × we note that K ' OK hπi. ab /K) is a continuous homomorphism as in Theorem 27.2. Suppose θ : K × → Gal(K S Then θ(π) fixes Kπ := n Kπ,n , since π ∈ N(Kπ,n ) = ker θKπ,n /K . We also know that θL/K (π) = FrobL/K for all finite unramified extensions L/K, which uniquely determines the action of θ(π) on K unr , and hence on K ab = Kπ K unr . Now suppose θ0 : K × → Gal(K ab /K) is another continuous homomorphism as in Theorem 27.2. By the argument above we must have θ0 (π) = θ(π) for every uniformizer π of OK , and K × is generated by its subset of uniformizers: if we fix one uniformizer π, every × x ∈ K × can be written as uπ n = (uπ)π n−1 for some u ∈ OK and n ∈ Z, and uπ is another 0 × uniformizer). It follows that θ(x) = θ (x) for all x ∈ K and therefore θ = θ0 is unique. Remark 27.13. One approach to proving local class field theory uses S the theory of formal groups due to Lubin and Tate to explicitly construct the fields Kπ = n Kπ,n used in the × proof of Proposition 27.12, along with a continuous homomorphism θπ : OK → Gal(Kπ /K) × unr that extends uniquely to a continuous homomorphism θ : K → Gal(Kπ K /K). One then shows that K ab = Kπ K unr (using the Hasse-Arf Theorem), and that θ does not depend on the choice of π; see [1, §I.2-4] for details. 18.785 Fall 2021, Lecture #27, Page 7 27.4 Finite abelian extensions Local class field theory gives us canonical bijections between the following sets: (1) finite-index open subgroups of K × (which are necessarily normal); (2) open subgroups of Gal(K ab /K) (which are necessarily normal and of finite index); (3) finite extensions of K in K ab (which are necessarily normal). d × ' Gal(K ab /K) given by The bijection from (1) to (2) is induced by the isomorphism K Theorem 27.9 and is inclusion preserving. The bijection from (2) to (3) follows from Galois theory (for infinite extensions), and is inclusion reversing, while the bijection from (3) to (1) is via the map L 7→ N(L× ), which is also inclusion reversing. References [1] J.S. Milne, Class field theory, version 4.02, 2013. [2] Jean-Pierre Serre, Local fields, Springer, 1979. 18.785 Fall 2021, Lecture #27, Page 8 18.785 Number theory I Lecture #28 28 Fall 2021 12/8/2021 Global class field theory, the Chebotarev density theorem Recall that a global field is a field with a product formula whose completions at nontrivial absolute values are local fields. By the Artin-Whaples theorem (see Problem Set 7), every such field is either • a number field : finite extension of Q (characteristic zero); • a global function field : finite extension of Fq (t) (positive characteristic). In Lecture 25 we defined the adele ring AK of a global field K as the restricted product n o Y a Y AK := (Kv , Ov ) = (av ) ∈ Kv : av ∈ Ov for almost all v , v where v ranges over the places of K (equivalence classes of absolute values), Kv denotes the completion of K at v, and Ov is the valuation ring of Kv if v is nonarchimedean, and equal to Kv otherwise. As a topological ring, AK is locally compact and Hausdorff. The field K is canonically embedded in AK via the diagonal map x 7→ (x, x, x, . . .) whose image is discrete, closed, and cocompact; see Theorem 25.12. In Lecture 26 we defined the idele group n o Y a Y IK := (Kv× , Ov× ) = (av ) ∈ Kv× : av ∈ Ov× for almost all v , which coincides with the unit group of AK but has a finer topology (using the restricted product topology ensures that a 7→ a−1 is continuous, which is not true of the subspace topology). As a topological group, IK is locally compact and Hausdorff. The multiplicative group K × is canonically embedded as a discrete subgroup of IK via the diagonal map x 7→ (x, x, x, . . .), and the idele class group is the quotient CK := IK /K × , which is locally compact but not compact. 28.1 The idele norm The idele group IK surjects onto the ideal group IK of invertible fractional ideals of OK via the surjective homomorphism ϕ : IK → IK Y a 7→ pvp (a) , where vp (a) is the p-adic valuation of the component av ∈ Kv× of a = (av ) ∈ IK at the finite place v corresponding to the absolute value k kp . We have the following commutative diagram of exact sequences: ← ← ← ← ← ← → ClK ← → IK → CK  ← → PK ϕ   ← 1 x7→(x) → IK ← → K× ← 1 → 1 → 1 where PK is the subgroup of principal ideals and ClK := IK /PK is the ideal class group. Definition 28.1. Let L/K is a finite separable extension of global fields. The idele norm NL/K : IL → IK is defined by NL/K (bw ) = (av ), where each av := Y NLw /Kv (bw ) w|v is a product over places w of L that extend the place v of K and NLw /Kv : Lw → Kv is the field norm of the corresponding finite separable extension of local fields Lw /Kv . It follows from Corollary 11.24 and Remark 11.25 that the idele norm NL/K : IL → IK agrees with the field norm NL/K : L× → K × on the subgroup of principal ideles L× ⊆ IL . The field norm is also compatible with the ideal norm NL/K : IL → IK (see Proposition 6.6), and we have the following commutative diagram: ← ← ← ← → IK NL/K → K× NL/K → → NL/K → IL ← ← → IL ← L× → IK The image of L× in IL under the composition of the maps on the top row is precisely the group PL of principal ideals, and the image of K × in IK is similarly PK . Taking quotients yields induced norm maps on the idele and ideal class groups, both of which we also denote NL/K , and we have a commutative square CK 28.2 ← ← → NL/K  ClL NL/K → ← ← CL  ClK The Artin homomorphism We now construct the global Artin homomorphism using the local Artin homomorphisms we defined in the previous lecture. Let us first fix once and for all a separable closure K sep of our global field K, and for each place v of K, a separable closure Kvsep of the local field Kv . Let K ab and Kvab denote maximal abelian extensions within these separable closures; henceforth all abelian extensions of K and the Kv are assumed to lie in these maximal abelian extensions. By Theorem 27.2, each local field Kv is equipped with a local Artin homomorphism θKv : Kv× → Gal(Kvab /Kv ). For each finite abelian extension L/K and each place w|v of L, composing θKv with the natural map Gal(Kvab /Kv ) → Gal(Lw /Kv ) yields a surjective homomorphism θLw /Kv : Kv× → Gal(Lw /Kv ) with kernel NLw /Kv (L× w ). When Kv is nonarchimedean and Lw /Kv is unramified we have θLw /Kv (πv ) = FrobLw /Kv for all uniformizers πv of Kv . Note that by Theorem 11.20, every finite separable extension of Kv is of the form Lw for some place w|v. 18.785 Fall 2021, Lecture #28, Page 2 We now define an embedding of Galois groups ϕw : Gal(Lw /Kv ) ,→ Gal(L/K) σ 7→ σ|L The map ϕw is well defined and injective because every element of Lw can be written as `x for some ` ∈ L and x ∈ Kv (any K-basis for L spans Lw as a Kv vector space), so each σ ∈ Gal(Lw /Kv ) is uniquely determined by its action on L, which fixes K ⊆ Kv . If v is archimedean then ϕw (Gal(Lw /Kv )) is either trivial or generated by the involution corresponding to complex conjugation in Lw ' C. If v is a finite place and q is the prime of L corresponding to w|v, then ϕw (Gal(Lw /Kv )) is the decomposition group Dq ⊆ Gal(L/K); this follows from parts (5) and (6) of Theorem 11.23. More generally, for any place v of K, the Galois group Gal(L/K) acts on the set {w|v}, via |α|σ(w) := |σ(α)|w , and ϕw (Gal(Lw /Kv )) is the stabilizer of w under this action. It thus makes sense to call ϕw (Gal(Lw /Kv )) the decomposition group of the place w. For w|v the groups ϕw (Gal(Lw /Kv )) are necessarily conjugate, and in our abelian setting, equal. Moreover, the composition ϕw ◦ θLw /Kv defines a map Kv× → Gal(L/K) that is independent of the choice of w|v: this is easy to see when v is an unramified nonarchimedean place, since then ϕw (θLw /Kv (πv )) = Frobv for every uniformizer πv of Kv , and this determines ϕw ◦ θLw /Kv since the πv generate Kv× . For each place v of K we now embed Kv× into the idele group IK via the map ιv : Kv× ,→ IK α 7→ (1, 1, . . . , 1, α, 1, 1, . . .), whose image intersects K × ⊆ IK trivially. This embedding is compatible with the idele norm in the following sense: if L/K is any finite separable extension and w is a place of L that extends the place v of K then the diagram ← L× w NLw /Kv ιw ιv ← IL -←→ -←→ → Kv× NL/K → IK commutes. Now let L/K be a finite abelian extension. For each place v of K, let us pick a place w of L extending v and define θL/K : IK → Gal(L/K) Y (av ) 7→ ϕw (θLw /Kv (av )), v where the product takes place in Gal(L/K). The value of ϕw (θLw /Kv (av )) is independent of our choice of w|v, as noted above. The product is well defined because av ∈ Ov× and v is unramified in L for almost all v, in which case v) ϕw (θLw /Kv (av )) = Frobv(a = 1, v It is clear that θL/K is a homomorphism, since each ϕw ◦ θLw /Kv is, and θL/K is continuous because its kernel is a union of open sets: each a := (av ) ∈ ker θL/K lies in an open set 18.785 Fall 2021, Lecture #28, Page 3 Q Ua := US × v6∈S Ov× ⊆ ker θL/K , where S contains all ramified v and all v for which Q × a Qv 6∈ Ov× , and US is the kernel of (av )v∈S 7→ v∈S ϕw (θLw /Kv (av ))), which is open in v∈S Kv . If L1 ⊆ L2 are two finite abelian extensions of K, then θL1 /K (a) = θL2 /K (a)|L1 for all a ∈ IK . The θL/K form a compatible system of homomorphisms from IK to the inverse limit limL Gal(L/K) ' Gal(K ab /K), where L ranges over finite abelian extensions of K in K ab ←− ordered by inclusion. By the universal property of the profinite completion, they uniquely determine a continuous homomorphism. Definition 28.2. Let K be a global field. The global Artin homomorphism is the continuous homomorphism θK : IK → lim Gal(L/K) ' Gal(K ab /K) ←− L defined by the compatible system of homomorphisms θL/K : IK → Gal(L/K), where L ranges over finite abelian extensions of K in K ab . The isomorphism Gal(K ab /K) ' lim Gal(L/K) is the natural isomorphism between a ←− Galois group and its profinite completion with respect to the Krull topology (Theorem 26.23) and is thus canonical, as is the global Artin homomorphism θK : IK → Gal(K ab /K). Proposition 28.3. Let K be global field. The global Artin homomorphism θK is the unique continuous homomorphism IK → Gal(K ab /K) with the property that for every finite abelian extension L/K in K ab and every place w of L lying over a place v of K the diagram ← Kv× θLw /Kv ϕw ιv ← IK -←→ -←→ → Gal(Lw /Kv ) θL/K → Gal(L/K) commutes, where the homomorphism θL/K is defined by θL/K (a) := θK (a)|L . Proof. That θK has this property follows from its construction. Now suppose that there 0 : I → Gal(K ab /K) with the same property. We is another continuous homomorphism θK K Q ab may view elements of Gal(K /K) ' lim Gal(L/K) as elements of L/K Gal(L/K), where ←− 0 are not identical, then L varies over finite abelian extensions of K in K ab . If θK and θK 0 there must be an a ∈ IK and a finite abelian extension L/K for which θL/K (a) 6= θL/K (a). Let S be a finite set of places of K that includes all places v for which av 6∈ Ov× := 1 for v ∈ S and bv := av for and all ramified places Q of L/K. Define b ∈ IK by bv v 6∈ S, so that a = b v∈S ιv (av ). Then θLw /Kv (bv ) = 1 for all places v, so we must have 0 θL/K (b) = 1 = θL/K (b), and for v ∈ S we have 0 θL/K (ιv (av )) = ϕw (θLw /Kv (av )) = θL/K (ιv (av )), by the commutativity of the diagram in the proposition. But then Y Y 0 0 0 θL/K (a) = θL/K (b) θL/K (ιv (av )) = θL/K (b) θL/K (ιv (av )) = θL/K (a), v∈S v∈S 0 = θ as claimed. which is a contradiction. So θK K 18.785 Fall 2021, Lecture #28, Page 4 28.3 The main theorems of global class field theory In the global version of Artin reciprocity, the idele class group CK := IK /K × plays the role that the multiplicative group Kv× plays in local Artin reciprocity (Theorem 27.2). Theorem 28.4 (Global Artin Reciprocity). Let K be a global field. The kernel of the global Artin homomorphism θK contains K × , and we thus have a continuous homomorphism θK : CK → Gal(K ab /K), with the property that for every finite abelian extension L/K in K ab the homomorphism θL/K : CK → Gal(L/K) obtained by composing θK with the natural map Gal(K ab /K)  Gal(L/K) is surjective with kernel NL/K (CL ), inducing an isomorphism CK /NL/K (CL ) ' Gal(L/K). Remark 28.5. When K is a number field, θK is surjective butQnot injective; Q its kernel is the connected component of the identity, including the image of v|∞ R>0 × v<∞ 1 ⊆ IK , which injects into CK . When K is a global function field, θK is injective but not surjective; its image is dense in Gal(K ab /K). We also have a global existence theorem. Theorem 28.6 (Global Existence Theorem). Let K be a global field. For every finite index open subgroup H of CK there is a unique finite abelian extension L/K in K ab for which NL/K (CL ) = H. As with the local Artin homomorphism, taking profinite completions yields an isomorphism that allows us to summarize global class field theory in one statement. Theorem 28.7 (Main theorem of global class field theory). Let K be a global field. The global Artin homomorphism θK induces a canonical isomorphism of profinite groups. ∼ ab d θbK : C K −→ Gal(K /K) We then have an inclusion reversing bijection { finite index open subgroups H of CK } ←→ { finite abelian extensions L/K in K ab } H 7→ (K ab )θK (H) NL/K (CL ) ←[ L and corresponding isomorphisms CK /H ' Gal(L/K), where H = NL/K (CL ). We also note that the global Artin homomorphism is functorial in the following sense. Theorem 28.8 (Functoriality). Let K be a global field and let L/K be any finite separable extension (not necessarily abelian). Then the following diagram commutes θL res → ← → NL/K CK → Gal(Lab /L) ← ← ← CL θK → Gal(K ab /K). 18.785 Fall 2021, Lecture #28, Page 5 28.4 Relation to ideal-theoretic version of global class field theory Let K be a number field let m : MK → Z≥0 be a modulus for K, which we view as Q and e v a formal product m = v v over the places v of K with ev ≤ 1 when v is archimedean and ev = 0 when v is complex (see Definition 21.2). For each place v we define the open subgroup  ×  if v6 | m, where Ov× := Kv× when v is infinite), Ov m UK (v) := R>0 if v|m is real, where R>0 ⊆ R× ' Ov× := Kv× ,   1 + pev if v|m is finite, where p = {x ∈ Ov : |x|v < 1}, Q m m := and let UK v UK (v) ⊆ IK denote the corresponding open subgroup of IK . The image m m U K of UK in the idele class group CK = IK /K × is a finite index open subgroup. The idelic version of a ray class group is the quotient m m m × CK := IK /(UK K ) = CK /U K , and we have isomorphisms m CK ' Clm K ' Gal(K(m)/K), where Clm K is the ray class group for the modulus m (see Definition 21.3), and K(m) is the corresponding ray class field, which we can now define as the finite abelian extension L/K m for which NL/K (CL ) = U K , whose existence is guaranteed by Theorem 28.6. m If L/K is any finite abelian extension, then NL/K (CL ) contains U K for some modulus m; m this follows from the fact that the groups U K form a fundamental system of open neighborhoods of the identity. Indeed, the conductor of the extension L/K (see Definition 22.24) is precisely the minimal modulus m for which this is true. It follows that every finite abelian extension L/K lies in a ray class field K(m), with Gal(L/K) isomorphic to a quotient of a m. ray class group CK 28.5 The Chebotarev density theorem We conclude this lecture with a proof of the Chebotarev density theorem, a generalization of the Frobenius density theorem you proved on Problem Set 10. Recall from Lecture 18 and Problem Set 9 that if S is a set of primes of a number field K, the Dirichlet density of S is defined by P P −s −s p∈S N(p) p∈S N(p) d(S) := lim P , = lim 1 −s s→1+ s→1+ log s−1 p N(p) whenever this limit exists. As you proved on Problem Set 9, if S has a natural density then it has a Dirichlet density and the two coincide (and similarly for polar density). In order to state Chebotarev’s density theorem we need one more definition: a subset C of a group G is said to be stable under conjugation if στ σ −1 ∈ C for all σ ∈ G and τ ∈ C. Equivalently, C is a union of conjugacy classes of G. Theorem 28.9 (Chebotarev density theorem). Let L/K be a finite Galois extension of number fields with Galois group G := Gal(L/K). Let C ⊆ G be stable under conjugation, and let S be the set of primes p of K unramified in L with Frobp ⊆ C. Then d(S) = #C/#G. 18.785 Fall 2021, Lecture #28, Page 6 Note that G is not assumed to be abelian, so Frobp is a conjugacy class, not an element. However, the main difficulty in proving the Chebotarev density theorem (and the only place where class field theory is used) occurs when G is abelian, in which case Frobp contains a single element. The main result we need is a corollary of the generalization of Dirichlet’s theorem on primes in arithmetic progressions to number fields that we proved in Lecture 22, a special case of which we record below. Proposition 28.10. Let m be a modulus for a number field K and let Clm K be the corresponding ray class group. For every ray class c ∈ Clm the Dirichlet density of the set of K m primes p of K that lie in c is 1/#ClK . Proof. Apply Corollary 22.22 to the congruence subgroup C = Rm K. The Chebotarev density theorem for abelian extensions follows from Proposition 28.10 and the existence of ray class fields, which we now assume.1 Corollary 28.11. Let L/K be a finite abelian extension of number fields with Galois group G. For every σ ∈ G the Dirichlet density of the set S of primes p of K unramified in L for which Frobp = {σ} is 1/#G. Proof. Let m = cond(L/K) be the conductor of the extension L/K; then L is a subfield of the ray class field K(m) and Gal(L/K) ' Clm K /H for some subgroup H of the ray class group. For each unramified prime p of K we have Frobp = {σ} if and only if p lies in one of the ray classes contained in the coset of H in Clm K /H corresponding to σ. The Dirichlet density of the set of primes in each ray class is 1/#Clm K , by Proposition 28.10, and there are #H ray classes in each coset of H; thus d(S) = #H/#Clm K = 1/#G. We now derive the general case from the abelian case. Proof of the Chebotarev density theorem. It suffices to consider the case where C is a single conjugacy class, which we now assume; we can reduce to this case by partitioning C into conjugacy classes and summing Dirichlet densities (as proved on Problem Set 9). Let S be the set of primes p of K unramified in L for which Frobp is the conjugacy class C. Let σ ∈ G be a representative of the conjugacy class C, let Hσ := hσi ⊆ G be the subgroup it generates, and let Fσ := LHσ be the corresponding fixed field. Let Tσ be the set of primes q of Fσ unramified in L for which Frobq = {σ} ⊆ Gal(L/Fσ ) ⊆ Gal(L/K) (note that the Frobenius class Frobq is a singleton because Gal(L/Fσ ) = Hσ is abelian). We have d(Tσ ) = 1/#Hσ , since L/Fσ is abelian, by Corollary 28.11.2 As you proved on Problem Set 9, restricting to degree-1 primes (primes whose residue field has prime order) does not change Dirichlet densities, so let us replace S and Tσ by their subsets of degree-1 primes, and define Tσ (p) := {q ∈ Tσ : q|p} for each p ∈ S. Claim: For each prime p ∈ S we have #Tσ (p) = [G : Hσ ]. Proof of claim: Let r be a prime of L lying above q ∈ Tσ (p). Such an r is unramified, since p is, and we have Frobr = σ, since Frobq = {σ}. It follows that Gal(Fr /Fq ) = hσ̄i ' Hσ . 1 This assumption is not necessary; indeed Chebotarev proved his density theorem in 1923 without it. With slightly more work one can derive the general case from the cyclotomic case L = K(ζ), where ζ is a primitive root of unity, which removes the need to assume the existence of ray class fields; see [4] for details. 2 Note that the integers #Hσ and [G : Hσ ] do not depend on the choice of σ (the Hσ are all conjugate). 18.785 Fall 2021, Lecture #28, Page 7 P Therefore fr/q = #Hσ and #{r|q} = 1, since #Hσ = [L : Fσ ] = r|q er/q fr/q . We have fr/p = fr/q fq/p = fr/q = #Hσ , since fq/p = 1 for degree-1 primes q|p, and er/p = 1, thus #G = [L : K] = X er/p fr/p = #{r|p}#Hσ = #Tσ (p)#Hσ , r|p so #Tσ (p) = #G/#Hσ = [G : Hσ ] as claimed. We now observe that X N(p)−s = p∈S XX σ∈C p∈S X #C X 1 N(q)−s = N(q)−s [G : Hσ ] [G : Hσ ] q∈Tσ (p) q∈Tσ since N(q) = N(p) for each degree-1 prime q lying above a degree-1 prime p, and therefore d(S) = #C #C #C d(Tσ ) = = . [G : Hσ ] [G : Hσ ]#Hσ #G Remark 28.12. The Chebotarev density theorem holds for any global field; the generalization to function fields was originally proved by Reichardt [3]; see [2] for a modern proof (and in fact a stronger result). In the case of number fields (but not function fields!) Chebotarev’s theorem also holds for natural density. This follows from results of Hecke [1] that actually predate Chebotarev’s work; Hecke showed that the primes lying in any particular ray class have a natural density. References [1] Erich Hecke, Über die L-Funktionen und den Dirichletschen Primzahlsatz für einen beliebigen Zahlkörper , Nachrichten von der Königlichen Gesellschaft der Wissenschaften zu Göttingen, Mathematisch–Physikalische Klasse (1917) 299–318. [2] Michiel Kosters, A short proof of a Chebotarev density theorem for function fields, arXiv:1404.6345. [3] Hans Reichardt, Der Primdivisorsatz für algebraische Funktionenkörper über einem endlichen Konstantenkörper , Mathematische Zeitschrift 40 (1936) 713–719. [4] Peter Stevenhagen and H.W. Lenstra Jr., Chebotarev and his density theorem, Math. Intelligencer 18 (1996), 26–37. 18.785 Fall 2021, Lecture #28, Page 8