Lecture 25 Notes: Field Extensions

Lecture 25: Field Extensions 25 25.1 Field Extensions Primary Fields We have the following useful fact about fields: Fact 25.1 Every field is a (possibly infinite) extension of either Q, or Fp for a prime p. These are called the primary fields. Proof. In general, for any ring R, there is a unique ring homomorphism Z → R — we must have 1 7→ 1R , so then n 7→ 1R + · · · + 1R = nR for positive integers n, and −n 7→ −nR . {z } | n The image of the homomorphism is a quotient of Z — it’s either Z or Z/nZ. Now consider the kernel of this homomorphism. If R is an integral domain (note that all fields are domains), then either the homomorphism is one-to-one, or its kernel is (p) for a prime p — otherwise, the image would be Z/nZ for composite n, which is not a domain (as it has zero divisors). Now taking R = F to be a field, if the kernel is zero, then Z is a subring of F . But then Q = Frac(Z) must be inside F as well (since we can invert elements in a field) — in our original notation, the copy of Q in F is the fractions of the form nR /mR . On the other hand, if the kernel is (p), then we have a copy of Z/pZ in F , and we’re done. Definition 25.2 The generator of the kernel (as in the above proof) is called the characteristic of the field. So fields of characteristic 0 contain Q, and fields of characteristic p contain Z/pZ (and these are the only possible characteristics). 25.2 Algebraic Elements Last time, we defined algebraic elements in a field extension L/K: Definition 25.3 An element α ∈ L is algebraic over K if P (α) = 0 for some nonzero P ∈ K[x]. As stated last class, α is algebraic if and only if K(α)/K is finite (since a polynomial in α is the same as a linear relation between powers of α). We also looked at towers of extensions E/F/K — here E/K is called the composite extension, while E/F and F/K are called intermediate extensions. In particular, we saw the following theorem: Theorem 25.4 We have [E : K] = [E : F ] · [F : K]. In particular, E/K is finite if and only if both E/F and F/K are finite. This has some useful corollaries regarding algebraic elements. Corollary 25.5 If α, β ∈ L are algebraic over K, then α + β, αβ, and α β 111 are also algebraic. Lecture 25: Field Extensions Proof. If α and β are algebraic, then K(α)/K and K(α, β)/K(α) are both finite — since β satisfies a polynomial relation with coefficients in K, it satisfies the same polynomial relation with coefficients in K(α). So we can conclude that K(α, β)/K is finite, and therefore any element in it is algebraic. Corollary 25.6 Given an arbitrary extension, the set of elements in L which are algebraic over K form a subfield of L, called the algebraic closure of K in L. For example, the algebraic closure of Q in C is called the algebraic numbers. This is an abstract argument that doesn’t exactly tell us how to construct the polynomial; but it’s possible to come up with a procedure to write down an equation as well. Example √ 25.7 √ Let α = 2 and β = 3, and γ = α + β. How can we write down a polynomial equation for γ? 2 One possible method is that by Corollary √ 25.5,√we know √ that 1, γ, γ , . . . must be linearly dependent. In this case, they are all linear combinations of 1, 2, 3, and 6 with coefficients in Q — so they lie in a vector space of dimension at most 4. Then 1, γ, . . . , γ 4 are five elements in a four-dimensional vector space, so they must be linearly dependent; and using linear algebra, it’s possible to explicitly calculate this linear relation. There is another way to find the polynomial equation — right now we’ll present it as a guess, but later we’ll see that it’s part of a more general story. We’d like to find a polynomial P with γ as a root, so we can try to think about what the other roots of P should be. Suppose P factors as √ (x −√γ1 )(x − γ2 ) · · · , for γi ∈ C — it suffices to choose the γi such that P has rational coefficients, and γ1 = 2 + 3. √ √ √ √ We can guess that all of ± 2 ± 3 should be roots — from an algebraic perspective, if 2 + 3 shows up, we “should” be able to switch the sign of the square root (since there isn’t a difference between the two signs). So then we can take √ √ γ1 = 2 + 3 √ √ γ2 = 2 − 3 √ √ γ3 = − 2 + 3 √ √ γ4 = − 2 − 3. We can expand out the polynomial to see that it does indeed have rational coefficients (essentially, this involves using the equation a2 − b2 = (a − b)(a + b) twice). The main idea we used here is to exploit the symmetry between the roots (there is a group of symmetries acting on the roots, by replacing one of the square roots with its negative); we’ll later discuss ways to find these symmetries, using Galois theory. 25.3 Compass and Straightedge Construction Proposition 25.4 also relates to compass and straightedge constructions. It has the following corollary: Corollary 25.8 If E/F/K is a tower of finite extensions, then [F : K] | [E : K]. The problem of which regular n-gons can be constructed using a compass and straightedge can be rephrased algebraically in the following way (we won’t discuss the details here). Fact 25.9 A regular n-gon is constructible with compass and straightedge if and only if ζn = e2πi/n lies in an extension Q(α1 , αn ) such that αi2 ∈ Q(α1 , . . . , αn−1 ) for all i. 112 Lecture 25: Field Extensions This means we have a tower of quadratic extensions, where every step in this tower has degree 2 — more explicitly, we can define Fi = Q(α1 , . . . , αi ), with F0 = Q. Without loss of generality we can assume αi ̸∈ Fi−1 (or else adding it to the set of generators would be useless). Then we have the tower of extensions Fn /Fn−1 / · · · /F1 /F0 where [Fi : Fi−1 ] = 2 for all i. For convenience, we’ll assume n is prime. (The general case involves a few more details, but works very similarly.) Theorem 25.10 Let n = p be prime. Then a regular p-gon can be constructed if and only if p = 2k + 1. Primes p = 2k + 1 are called Fermat primes. There’s only 5 known Fermat primes (3, 17, 257, and 65537); it’s conjectured that there are no others, but we don’t even know whether there’s finitely or infinitely many. (Note that if 2k + 1 is prime, then k must be a power of 2 — otherwise, 2k + 1 can be factored.) We’ll only show one direction: that if ζp is constructible, then p is a Fermat prime. To prove this, the following proposition will be useful: Proposition 25.11 If p is prime, we have deg(ζn ) = p − 1, or equivalently [Q(ζp ) : Q] = p − 1. The extension Q(ζp ) is called a cyclotomic extension. Proof. We know ζp is a root of xp − 1. We can easily factor xp − 1 = (x − 1)(xp−1 + xp−2 + · · · + 1), so it suffices to show that the second factor, which we call P (x), is irreducible. Since the polynomial is primitive, it’s enough to show that it’s irreducible over Z. Now we can perform a trick — substitute t = x − 1. Then if we write P (x) = Q(t), we have tQ(t) = (t + 1)p − 1. But by expanding and using the Binomial Theorem, we then have  p−1  X p Q(t) = ti . i + 1 i=0 (For example, when p = 3, we have Q(t) = t2 + 3t + 3.) But the leading term is 1, and all other terms are divisible by p; and the free term is not divisible by p2 (in fact, none of the terms are divisible by p2 , but we only need to use the free term here). Now assume for contradiction that Q is reducible, so Q = Q1 Q2 for polynomials Q1 and Q2 of degree at least 1. Now consider the reduction mod p, where Q = Q1 Q2 . But Q is now tp−1 , and the only way to factor tp−1 in Fp [x] is as ti tp−1−i . But we have deg(Q1 ) = deg(Q1 ) > 1 (and the same is true for Q2 ), since the leading coefficients of Q1 and Q2 cannot be divisible by p (their product is the leading coefficient of Q, which is 1). So then we must have i ̸= 0, p − 1. But then since Q1 and Q2 are ti and tp−1−i for 0 < i < p − 1, their free terms must both be divisible by p. So the product of their free terms is divisible by p2 ; but this product is the free term of Q, which is not divisible by p2 . So this is a contradiction, and Q is irreducible. Proof of Necessity in Theorem 25.10. We’ve seen that deg(ζp ) = p − 1. So we have deg(ζp ) = p − 1. On the other hand, if ζp ∈ Fn for a field extension of the form described, then deg(ζp ) must divide [Fn : Q], which is a power of 2. So p − 1 must be a power of 2 as well. With our current tools, we can only show one direction — to show the other direction, we need a better extension of which fields can be obtained as the top floor of a tower of quadratic extensions. It’s necessary that the degree is a power of 2, but this may not be sufficient. In the case of Q(ζp ), the condition turns out to be sufficient as well (as we’ll see later). 113 Lecture 25: Field Extensions 25.4 Splitting Fields We’ve seen the construction where we start with an irreducible polynomial P ∈ F [x], and construct the field extension E = F [x]/(P ). This is an extension of F of degree n = deg(P ), and we can think of it as adjoining a root of the polynomial. But there’s another construction which also produces a finite extension from a polynomial, which is in some sense harder to control. Here, we do not require the polynomial to be irreducible. Definition 25.12 For a polynomial P ∈ F [x], a splitting field of P is an extension E/F such that: 1. P splits as a product of linear factors in E[x]; 2. E = F (α1 , . . . , αn ), where the αi are the roots of P . The first condition guarantees that P splits completely (so we can find all its roots) in E; the second prevents E from being too large (it only contains the elements which are necessary for P to split). Proposition 25.13 Given any polynomial P , its splitting field exists, and any two splitting fields of P are isomorphic. We’ll discuss the proof in more detail next time — the main idea is to add one root of P so that it splits partially, then add another root of any remaining irreducible factor, and so on. Example 25.14 √ √ The splitting field of P (x) = x3 − 2 over F = Q is E = Q( 3 2, ω 3 2), where ω is a primitive 3rd root of unity. We have [E : Q] = 6. On the other hand, we could start by adjoining ω: Example 25.15 √ The splitting field of P (x) = x3 − 2 over F = Q(ω) is E = F ( 3 2) — the polynomial x3 − 2 remains irreducible, but after adjoining one root, we already have all the others. Here [E : F ] = 3. Note that E is the same in both examples (even though F is not). Example 25.16 The splitting field of P (x) = xp−1 + · · · + 1 over F = Q is E = Q(ζp ) (since all roots are powers of ζp ), where [E : F ] = p − 1. 114