Lecture Note
Local Fields and Hensel’s Lemmas
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University:
Massachusetts Institute of Technology -
Course:
18.785 | Number Theory I Academic year:
2021
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484
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8
Author:
chashitsuo2
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)) =
X
i
fi (a + (x − a)) =
i≥0
= f (a) +
X
i≥1
X
i≥0
X i
fi
aj (x − a)i−j
j
0≤j≤i
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
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