Algebraic Number Theory

Andrew Kobin Spring 2016 Contents Contents

Contents

0 Introduction 1 0.1 Attempting Fermat’s Last Theorem ...... 1

1 Algebraic Number Fields 4 1.1 Integral Extensions of Rings ...... 4 1.2 Norm and Trace ...... 5 1.3 The Discriminant ...... 6 1.4 Factorization of Ideals ...... 11 1.5 Ramification ...... 15 1.6 Cyclotomic Fields and Quadratic Reciprocity ...... 21 1.7 Lattices ...... 24 1.8 The Class Group ...... 26 1.9 The Unit Theorem ...... 31

2 Local Fields 34 2.1 Discrete Valuation Rings ...... 34 2.2 The p-adic Numbers ...... 37 2.3 Absolute Values ...... 42 2.4 Local Fields ...... 49 2.5 Henselian Fields ...... 52 2.6 Ramification Theory ...... 54 2.7 Extensions of Valuations ...... 61 2.8 Galois Theory of Valuations ...... 65 2.9 Higher Ramification Groups ...... 69 2.10 Discriminant and Different ...... 75

i 0 Introduction

0 Introduction

These notes follow a course on algebraic number theory taught by Dr. Andrew Obus at the University of Virginia in Spring 2016. The main topics covered are:

ˆ Algebraic number fields (the global case)

ˆ The ideal class group

ˆ Structure of the unit group

ˆ The p-adic numbers (the local case)

ˆ Hensel’s Lemma

ˆ Ramification theory

ˆ Further topics, including adeles and ideles

The main companion for the course is Neukirch’s Algebraic Number Theory. Other great ref- erences include Cassels and Frohlich’s Algebraic Number Theory, Janusz’s Algebraic Number Fields, Lang’s Algebraic Number Theory, Marcus’s Number Fields and Weil’s Basic Number Theory.

0.1 Attempting Fermat’s Last Theorem

Algebraic number theory was developed primarily as a set of tools for proving Fermat’s Last Theorem. We recall the famous (infamous?) theorem here.

Fermat’s Last Theorem. The equation xn + yn = zn has no solutions in positive integers for n ≥ 3.

In attempting to prove the theorem, we first remark that the n = 4 case is elementary; it’s just a matter of parametrizing the Pythagorean triples (x, y, z) that solve x2 + y2 = z2 and noticing that not all three can be perfect squares. With this, we can reduce to the case when n = p, an odd prime. There are two cases:

ˆ Case 1: x, y, z are all relatively prime to p.

ˆ Case 2: p divides exactly one of x, y, z.

We will show a proof for the first few primes in Case 1; the other case uses similar tech- niques. Let ζ be a primitive pth root of unity (e.g. ζ = e2πi/p) and assume Z[ζ] is a unique factorization domain (UFD). This was the classical approach, but number theorists quickly realized that Z[ζ] is not always a UFD. In fact, it is an open question whether there are an infinite number of primes p for which Z[e2πi/p] is a UFD. In any case, the assumption that Z[ζ] is a UFD holds for p < 23 so we will have proven a number of cases of Fermat’s Last Theorem with the following proof.

1 0.1 Attempting Fermat’s Last Theorem 0 Introduction

Proof. Suppose x, y, z are positive integers satisfying xp + yp = zp. We may assume x, y, z are relatively prime in Z. The equation above may be factored as

p Y (x + ζiy) = zp (∗) i=1

For p = 3, the only cubes mod 9 are ±1 and 0 so there are no solutions for (*) where 3 - xyz. So we may assume p ≥ 5. We need the following lemmas: p−1 Y Lemma 0.1.1. p = (1 − ζi). i=1

tp−1 Proof. Consider expanding t−1 in two ways: tp − 1 (t − ζ) · (t − ζp−1) = = tp−1 + ... + t + 1. t − 1 Then plugging in t = 1 gives the result. Lemma 0.1.2. For any 0 ≤ i < j ≤ p − 1, the elements x + ζiy and x + ζjy are coprime in Z[ζ]. Proof. Suppose that π ∈ Z[ζ] is a prime which divides x + ζiy and x + ζjy. Then π divides ζiy(1 − ζj−i). Notice that ζi is a unit and p - y by assumption, but 1 − ζj−i | p. So in particular, π | y and thus π | yp. Since π is a prime, π | y or π | p. Repeating the argument for x shows that π | x or π | p. Since x and y are coprime in Z, we cannot have π | x and π | y simultaneously, so π | p. By assumption we have that π divides xp + yp and therefore also zp in Z, but (p, z) = 1 so the Euclidean algorithm implies that π | 1. Therefore x + ζiy and x + ζjy are relatively prime in Z[ζ]. Now, each factor x + ζiy must be a pth power in Z[ζ], possibly multiplied by a unit. Write x + ζy = utp for u ∈ Z[ζ]∗ and t ∈ Z[ζ]. Lemma 0.1.3. u/u¯ is a pth root of unity.

Proof. It is simple to show that u/u¯ and all of its Galois conjugates have modulus 1 in C; this is then true for all powers of u/u¯ as well. Then the degree of u/u¯ and all of its powers is bounded. Since all of these are algebraic integers, there are only finitely many possible choices for their minimal polynomials. Hence the set {(u/u¯)k : k ∈ N} is finite. This proves u/u¯ is a root of unity in Z[ζ]. In particular, (u/u¯)2p = 1 but we want to show it is a pth root of unity. Suppose (u/u¯)p = −1. Then up = −u¯p. Since u ∈ Z[ζ] we may write

2 p−2 u = a0 + a1ζ + a2ζ + ... + ap−2ζ

for unique ai ∈ Z; this follows from unique factorization in Z[ζ]. Now

p p p p u ≡ a0 + a1 + ... + ap−2 (mod p)

≡ a0 + a1 + ... + ap−2 (mod p) by Fermat’s Little Theorem.

2 0.1 Attempting Fermat’s Last Theorem 0 Introduction

In particular, up is conjugate to a real number mod p. Likewise, we can write −u¯ as −u¯ = p−1 2 −(a0 + a1ζ + ... + ap−2ζ ) so

p −u¯ ≡ −a0 − a1 − ... − ap−2 (mod p).

p This implies a0 + a1 + ... + ap−2 ≡ 0 (mod p) so p | u . However, this is impossible if u is a unit. Therefore (u/u¯)p = 1. Putting these results together, we can now write

x + ζy = ζjut¯ p ≡ ζju¯t¯p ≡ ζj(x + ζy¯ ) (mod p).

Expanding this out gives us

x + ζy − ζjx − ζj−1y ≡ 0 (mod p). (∗∗)

∼ p−1 ∼ p−1 Now Z[ζ]/(p) = Z[x]/(p, x + ... + x + 1) = Fp[x]/(x + ... + x + 1). Thus the images p−2 p−1 of 1, x, . . . , x are Fp-linearly independent in this ring. This implies 1, ζ, . . . , ζ are Z- linearly independent in Z[ζ]/(p). Since x, y ∈ Z, the only possibilities in (**) for j are j = 0, 1, 2, p − 1. If p = 0, 2, p − 1, it is easy to simplify (**) and produce a nontrivial ζ2 term, which is impossible. If j = 1, (**) becomes

(x − y)(1 − ζ) ≡ 0 (mod p).

Qp−1 i Thus i=2 (1 − ζ ) divides x − y but since x − y ∈ Z, it must be that p | (x − y). Rearranging the equation xp + yp = zp to read xp + (−z)p = yp and repeating the argument so far shows that p | (x + z) as well. Thus y ≡ x ≡ −z (mod p). But then

0 = xp + yp − zp ≡ 3xp (mod p)

which implies p | x, contradicting the assumption that p 6= 3. Therefore no solutions exist to xp + yp = zp for p > 5 such that p - xyz. This proof fails for general primes p in two places: as we mentioned, not every ring Z[e2πi/p] is a UFD; moreover, there can be many more units than just the roots of unity in Z[e2πi/p]. This motivates the study of ideal class groups – which measure how far from being a PID (and a UFD) a ring of integers is – and unit groups in algebraic number theory.

3 1 Algebraic Number Fields

1 Algebraic Number Fields

1.1 Integral Extensions of Rings

Let A ⊆ B be rings.

Definition. An element x ∈ B is integral over A if it is a root of a monic polynomial with coefficients in A. We say B is integral over A if every element of B is integral over A.

Definition. The integral closure of A in B is the set of all x ∈ B which are integral over A. If A is equal to its integral closure in B then we say A is integrally closed in B. In particular, if A is a domain and B is the fraction field of A then we simply say that A is integrally closed.

Lemma 1.1.1. x ∈ B is integral over A if and only if A[x] is a finitely generated A-module.

n n−1 n Pn−1 i Proof. ( =⇒ ) If x + an−1x + ... + a0 for ai ∈ A then x ∈ M := i=1 Ax which is a finitely generated A-module. By induction, for all m ≥ n, xm ∈ M. This implies A[x] = M, so in particular A[x] is finitely generated. ( ⇒ = ) Suppose A[x] is generated by f1(x), . . . , fn(x) where fi are polynomials in a single n variable over A. Let d ≥ max{deg fi}i=1. Then

n d X x = aifi(x) i=1

d Pn for some choice of ai ∈ A. This shows that x is a root of the polynomial t − i=1 aifi(t) so x is integral over A.

Theorem 1.1.2. The integral closure of A in B is a ring.

Proof. It suffices to prove that the integral closure A¯ is closed under the addition and mul- tiplication of B. If x, y ∈ A¯, Lemma 1.1.1 shows A[x, y] is finitely generated. This implies that the submodules A[x + y] and A[xy] are also finitely generated, so x + y, xy ∈ A¯. Hence A¯ is a ring. Let A ⊂ B be a subring. We will make use of the following facts about integral extensions of rings:

ˆ Every UFD is integrally closed.

ˆ If A is a domain, B is finite over A if and only if B is integral over A and B is finitely generated as an A-module.

ˆ Suppose C ⊇ B ⊇ A are all rings. If C is integral over B and B is integral over A then C is integral over A.

ˆ If B is integral over A then S−1B is integral over S−1A for any multiplicatively closed subset S ⊂ A.

4 1.2 Norm and Trace 1 Algebraic Number Fields

The two most important objects in global algebraic number theory are defined next.

Definition. K is a number field if K is a finite field extension of Q. Definition. For a number field K ⊃ Q, the integral closure of Z in K is called the ring of integers of K, written OK . Examples.

1 The ring of integers of Q is Z. √ √ h 1+ −3 i 2 For K = Q( −3), the ring of integers is OK = Z 2 .

2πi/p 3 For a prime p, the cyclotomic field K = Q(ζp) = Q(e ) has ring of integers OK = Z[ζp].

It turns out that OK is always a free Z-module of rank [K : Q]. Thus we can think of OK as a lattice embedded in the vector space K.

1.2 Norm and Trace

Two important maps for understanding number fields are introduced in this section. Let L/K be a finite field extension and fix x ∈ L.

Definition. The norm of x is the element NL/K (x) = det Tx ∈ K, where Tx : L → L is the K-linear map Tx(`) = x`.

Definition. The trace of x is TrL/K (x) = tr Tx, where tr denotes the trace. Note that the norm and trace are defined for any finite extension L/K, not just number fields. We will often drop the subscript and write N(x) and Tr(x) when the extension is understood.

× × Lemma 1.2.1. The norm map NL/K : L → K is a homomorphism of multiplicative groups, and the trace map TrL/K :(L, +) → (K, +) is a homomorphism of abelian groups.

Theorem 1.2.2. Suppose L/K is a finite, separable extension of fields. Let σ1, . . . , σn be the distinct embeddings L,→ K where K is the algebraic closure of K. Then for all x ∈ L,

n n Y X NL/K (x) = σi(x) and TrL/K (x) = σi(x). i=1 i=1

n−1 Proof. Assume σi(x) 6= σj(x) when i 6= j. A basis of L/K is 1, x, . . . , x and the matrix for Tx in this basis is   0 0 ··· 0 −a0 1 0 ··· 0 −a1   .   ..  0 1 0 −a2  . . . . .  ......  0 0 ··· 1 −an−1

5 1.3 The Discriminant 1 Algebraic Number Fields

n where f(x) = a0 + a1x + ... + anx is the minimal polynomial of x over K. In this case f is also the characteristic polynomial of x, so by linear algebra, Tr(x) is equal to the sum of the roots of f and N(x) is equal to the product of the roots of f. This implies the result. √ Example 1.2.3. Let K = Q( d) for d a squarefree√ integer√ (this means d = ±p1p2 ··· pr in its prime factorization). Then an element x = a + b d ∈ Q( d) has norm N(x) = a2 − b2d and trace Tr(x) = 2a.

1.3 The Discriminant

In this section let L/K be a finite, separable extension of fields and let {α1, . . . , αn} be a K-basis of L, so that [L : K] = n. Also denote by σ1, . . . , σn : L,→ K the n distinct K-embeddings of L into the algebraic closure of K.

Definition. The discriminant of the basis {α1, . . . , αn} is

2 dL/K (α1, . . . , αn) = [det(σi(αj))] .

Proposition 1.3.1. Let A = [TrL/K (αiαj)]. Then dL/K (α1, . . . , αn) = det A. In particular, dL/K (α1, . . . , αn) lies in K. Pn Proof. By Theorem 1.2.2, TrL/K (αiαj) = k=1 σk(αi)σk(αj). Thus A = BC, where

T B = (σk(αi)) and C = (σk(αj)).

2 Taking the determinant gives us det A = (det B)(det C) = (det C) = dL/K (α1, . . . , αn). One case of interest is when L = K(α) is a simple extension and {1, α, α2, . . . , αn−1} is a basis for L as a K-vector space. Then the discriminant of α is defined to be

2 n−1 dL/K (α) := dL/K (1, α, α , . . . , α ).

Lemma 1.3.2. For any algebraic element α over K, dL/K (α) equals the discriminant of the minimal polynomial of α.

Proof. Set L = K(α) and let αi = σi(α) for each embedding σi : L,→ K. Then

 n−1 1 α1 ··· α1 1 α ··· αn−1  2 2  dL/K (α) = det . . .. .  . . . .  n−1 1 αn ··· αn This is a Vandermonde determinant, which evaluates to

Y Y 2 dL/K (α) = (αi − αj) = (αi − αj) . 1≤i,j≤n 1≤i

Since K(α)/K is separable, dL/K (α) 6= 0. In fact, the product formula above is precisely the discriminant of f, the minimal polynomial of α over K.

6 1.3 The Discriminant 1 Algebraic Number Fields

Proposition 1.3.3. For any K-basis {α1, . . . , αn} of L, dL/K (α1, . . . , αn) 6= 0. Proof. Since L/K is finite and separable, L = K(θ) for some θ ∈ L by the primitive element n−1 theorem. Then by Lemma 1.3.2, dL/K (1, θ, . . . , θ ) 6= 0. Let A ∈ GLn(K) be the change n−1 of basis matrix from {α1, . . . , αn} to {1, θ, . . . , θ }. Then for each 1 ≤ i, j ≤ n,

j−1 det(σi(αj)) = (det A)(det(σi(θ )).

Both determinants on the right are nonzero, so det(σi(αj)) 6= 0 which implies finally that dL/K (α1, . . . , αn) 6= 0 by the definition of disciminant. The proof of Proposition 1.3.3 gives us the following useful formula: If A, B are two K-bases for L with change of basis matrix A, then

2 dL/K (A) = (det A) dL/K (B). √ Example 1.3.4. √Take our favourite example, K = Q( d) over Q, where d is a squarefree integer. Then {1, d} is a basis for L, and its discriminant is

√ 2 √  1 d  √ d (1, d) = det √ = (−2 d)2 = 4d. K/Q 1 − d

This matches the fact that the discriminant of x2 − d is 4d. Suppose A ⊆ K is integrally closed with fraction field K. Let B be the integral closure of A in L. Observe that if x ∈ B then all conjugates of x in K are integral over K. Thus NL/K (x) ∈ A and TrL/K (x) ∈ A since A is integrally closed.

× × Lemma 1.3.5. If x ∈ B then NL/K (x) ∈ A .

Proof. By Lemma 1.2.1, NL/K is a homomorphism of groups.

Lemma 1.3.6. Suppose α1, . . . , αn ∈ B form a K-basis of L. Let d = dL/K (α1, . . . , αn). Then dB ⊆ Aα1 + ... + Aαn. Pn Proof. Let a1, . . . , an ∈ K such that α := i=1 aiαi ∈ B. Then (a1, . . . , an) is a solution to the system of linear equations

n X TrL/K (αiα) = TrL/K (αiαj)xj, 1 ≤ i ≤ n. j=1 The matrix corresponding to this system has determinant d by Proposition 1.3.1. Thus 1 each aj can be written as d times an A-linear combination of Tr(αiα). Since αi, α ∈ B, Tr(αiα) ∈ A so dαj ∈ A for each j. Thus

n X dα = dajαj ∈ Aα1 + ... + Aαn. j=1

Since α ∈ B was arbitrary, we have shown that dB ⊆ Aα1 + ... + Aαn.

7 1.3 The Discriminant 1 Algebraic Number Fields

Proposition 1.3.7. If A is a PID and M ⊆ L is a finitely generated B-module, then M is free of rank n = [L : K] as an A-module. In particular, B is free of rank n as an A-module.

Proof. Let {α1, . . . , αn} ⊂ B be a basis for L/K. We know the rank of B, which is well- defined over a PID, is at most n. On the other hand, since the αi are linearly independent, the rank of B is at least n. Thus the rank of B equals n. Now suppose M is finitely generated as a B-module, say by elements µ1, . . . , µr. Then there exists an a ∈ A such that aµi ∈ B for each i (by a homework problem). Thus daµi ∈ M0 := Aα1 + ... + Aαn by Lemma 1.3.6. So daM ⊆ M0. By the structure theory of modules over a PID, since M0 is free, daM is also free, so M is free of rank at most n. On the other hand, rank M ≥ rank B = n by assumption so rank M = n. The second statement follows from taking M = B.

Definition. In the situation above, an A-basis for B is called an integral basis.

The most important case of Proposition 1.3.7 is when K = Q, A = Z and B ⊂ L is the integral closure of A in some number field L.

Proposition 1.3.8. Let {α1, . . . , αn} and {β1, . . . , βn} be two integral bases for B/Z. Then

dL/Q(α1, . . . , αn) = dL/Q(β1, . . . , βn).

0 2 0 Proof. Let d = dL/Q(α1, . . . , αn) and d = dL/Q(β1, . . . , βn). Then d = (det M) d for some 0 M ∈ GLn(Z). Thus det M = ±1 so d = d . This allows us to define:

Definition. The discriminant of a number field K/Q is

dK := dK/Q(α1, . . . , αn)

for any Z-basis {α1, . . . , αn} of OK . √ √ Example 1.3.9. The quadratic field K = Q(√ 2) has integral basis {1, 2}. Then dK = 8. In general, for a quadratic extension K = Q( d), the discriminant is given by ( 4d, d ≡ 2, 3 (mod 4) dK = d, d ≡ 1 (mod 4)

Example 1.3.10. Let ζ be a primitive prth root of unity and let K = Q(ζ). We know that r r−1 r [K : Q] = ϕ(p ) = p (p − 1). Set n = ϕ(p ). We will show that OK = Z[ζ] for every r prime power p . First, we compute the discriminant dK = dK/Q(ζ). To do so, we need the following lemma.

Lemma 1.3.11. If f is the minimal polynomial of α and K = Q(α),

0 dK/Q(α) = ±NK/Q(f (α)).

8 1.3 The Discriminant 1 Algebraic Number Fields

Proof. Let σ1, . . . , σn be the Q-embeddings of K into Q. We may assume σ1 = id : K,→ Q. Then by Lemma 1.3.2, Y 2 dK/Q(α) = (σi(α) − σj(α)) . 1≤i

n 0 Y f (α) = (α − σi(α)). i=2 Finally, compute the norm of f 0(α):

n ! 0 Y Y NK/Q(f (α)) = σj (α − σi(α)) 1≤j≤n i=2 Y = (σi(α) − σj(α)) 1≤i6=j≤n Y 2 = ± (σi(α) − σj(α)) = dK/Q(α). 1≤i

(Note: in fact, Lemma 1.3.11 holds for any finite separable extension L/K.) Now, to

compute dK/Q(ζ) in our example, let f(x) be the minimal polynomial of ζ over Q. We may write this in two ways:

pr x − 1 r−1 r f(x) = or (xp − 1)f(x) = xp − 1. xpr−1 − 1

Differentiating the second expression gives us f 0(x)(xpr−1 − 1) + f(x)(pr−1xpr−1−1) = prxpr−1. Then plugging in ζ and solving for f 0(ζ) produces

prζr−1 f 0(ζ) = . ζpr−1 − 1 Take the norm of this expression:

N(pr) N(f 0(ζ)) = ± = ±pa for some a ∈ . N(ζpr−1 − 1) Z

a r r Thus by Lemma 1.3.11, dK/Q(ζ) = ±p for some a ∈ Z, where a ≤ ϕ(p )p . It turns out that it’s easier to work with 1 − ζ in this example. In general this creates no obstacles, since dK/Q(1 − ζ) = dK/Q(ζ). In our case, we observe that

Y 2 Y 2 dK/Q(1 − ζ) = (1 − σi(ζ) − (1 − σj(ζ))) = (σi(ζ) − σj(ζ)) = dK/Q(ζ). 1≤i

a Thus dK/Q(1 − ζ) = ±p . To proceed, we need the following generalization of Lemma 0.1.1.

9 1.3 The Discriminant 1 Algebraic Number Fields

Y Lemma 1.3.12. (1 − ζk) = p. p-k 1≤k≤pr Proof. Consider

pr x − 1 r−1 r−1 r−1 f(x) = = 1 + xp + x2p + ... + x(p−1)p . xpr−1 − 1 Plugging in x = 1 gives the result.

Observe that for any two k1, k2 ∈ N not divisible by p,

1 − ζk1 ∈ Z[ζ]. 1 − ζk2

k Then by symmetry, 1−ζ 1 is a unit in [ζ] for all such k , k . We will now show O = [1−ζ]. 1−ζk2 Z 1 2 K Z 2 n−1 Consider the basis {1, 1 − ζ, (1 − ζ) ,..., (1 − ζ) } for K/Q. If x ∈ OK , we can write x in the following manner by Lemma 1.3.6:

n−1 X bi x = (1 − ζ)i for b ∈ , pa i Z i=0

a bi using the fact that dK/Q(1 − ζ) = ±p . If pa ∈ Z for each i, then we’re done. If not, multiply c bi c 1 c by some p so that all pa p ∈ p Z but not all of them lie in Z. Note that p x ∈ OK , so we may replace x with pcx and write

n−1 X bi x = (1 − ζ)i, b ∈ . p i Z i=0

Suppose x 6∈ Z[1 − ζ]. Subtracting off the terms where p | bi if necessary, we may assume bi = 0 whenever p | bi. Let j be the smallest index with p - bj. Then

n−1 X bi x = (1 − ζ)j, p b . p - j i=j

p n The element (1−ζ)j+1 lies in Z[1 − ζ] since j + 1 ≤ n and (1 − ζ) | p by Lemma 1.3.12. p Therefore we may multiply the expression for x by (1−ζ)j+1 to obtain

b x = j + (terms in [1 − ζ]). 1 − ζ Z

n n  bj  bj bj  bj  Note that N 1−ζ = N(1−ζ) = p is not divisible by p, Thus N 1−ζ 6∈ Z but this con- bj tradicts the fact that 1−ζ ∈ OK . Hence x ∈ Z[1 − ζ] which finally proves the claim that OK = Z[1 − ζ] = Z[ζ].

10 1.4 Factorization of Ideals 1 Algebraic Number Fields

The following theorem allows us to generalize Example 1.3.10 to all Q(ζ) where ζ is a primitive nth root of unity. Theorem 1.3.13. Let A be an integrally closed integral domain with field of fractions K and suppose L/K and M/K are finite separable extensions with ω1, . . . , ωn an integral basis for L with respect to A and α1, . . . , αm an integral basis for M with respect to A. Further suppose dL/K (ω1, . . . , ωn) and dM/K (α1, . . . , αm) are relatively prime in A. Then {ωiαj} is an integral basis for the compositum LM over A and

m n dLM/K (ωiαj) = dL/K (ωi) dM/K (αj) .

Corollary 1.3.14. If ζm is an mth root of unity then OQ(ζm) = Z[ζm].

Q a1 ar Proof. Factor m = p ··· p . Then (ζm) = (ζ a1 ) ··· (ζ ar ). Moreover, for distinct 1 r Q Q p1 Q pr primes p 6= q, dQ(ζp)/Q and dQ(ζq)/Q are relatively prime. Therefore by Theorem 1.3.13, the ring of integers of Q(ζm) is

O (ζ ) = [ζ a1 , . . . , ζ ar ] = [ζm]. Q m Z p1 pr Z

1.4 Factorization of Ideals

Let K be a number field. We have seen that unique factorization may fail in OK , as we recall in the example below. √ √ Example 1.4.1. The quadratic field K = Q( −5) has ring of integers OK = Z[ −5]. In this ring, 6 has two different factorizations: √ √ 6 = 2 · 3 = (1 + −5)(1 − −5). √ Therefore unique factorization fails√ in Z[ −5]. To√ see that these are the only two factor- izations of 6, observe that N(1 +√ −5) = N(1 − −5) = 6, but there are no solutions in integers to the equation N(a + b −5) = a2 + 5b2 = 2, 3. It is our goal in this section to in some fashion repair the failure of unique factorization in OK , and in an arbitrary Dedekind domain A (to be defined below). Then we will further study the problem of determining all factorizations of an element in an integral extension. For the unique factorization√ problem, it would be nice (even ‘ideal’) if there were some objects p1, p2, p3, p4 ∈ Z[ −5] such that √ 2 = p1p2 1 + −5 = p1p3 √ 3 = p3p4 1 − −5 = p2p4.

In fact, the exact objects we are looking for are prime ideals in OK . In order to describe a unique factorization into prime ideals, recall that for ideals I,J ⊂ A, their ideal product is

( n ) X IJ = xiyi : xi ∈ I, yi ∈ J . i=1

11 1.4 Factorization of Ideals 1 Algebraic Number Fields

Definition. An integral domain A is a Dedekind domain if

(1) A is integrally closed.

(2) A is Noetherian.

(3) All nonzero prime ideals of A are maximal.

The main theorem we will prove is:

Theorem 1.4.2. If A is a Dedekind domain, then every nonzero ideal I ⊂ A has a factor- Qn ai ization I = i=1 pi for distinct prime ideals pi ⊂ A which are unique up to ordering.

Theorem 1.4.3. For every number field K, OK is a Dedekind domain.

Proof. (1) OK is integrally closed since by definition it is the integral closure of Z in K. (2) We have seen (Prop. 1.3.7) that OK lies inside a finitely generated Z-module. By commutative algebra, this is sufficient to conclude that OK is Noetherian. (3) The property of nonzero prime ideals being maximal is alternatively known as Krull dimension 1. It is known that finite integral extensions which are integrally closed preserve Krull dimension, e.g. by the going up theorem.

Since Z is Dedekind, integer unique factorization can be captured by Theorem 1.4.2 by associating a prime√ integer p ∈ Z with the principal prime ideal it generates: (p) ⊂ Z. For example, in Z[ −5] we have √ √ √ √ (6) = (2, 1 + −5)(2, 1 − −5)(3, 1 + −5)(3, 1 − −5).

Lemma 1.4.4. If A is Dedekind, every nonzero ideal I ⊂ A contains a finite product of prime ideals.

Proof. Let M be the set of nonzero ideals of A not divisible by a finite product of primes. Since A is Noetherian, there exists a maximal element a ∈ M. Then a must not be prime, so there exist elements b1, b2 ∈ A r a such that b1b2 ∈ a. Consider the ideals a + (b1) and a + (b2). Since a is maximal in M, each of these contains a finite product of prime ideals. Then (a + (b1))(a + (b2)) ⊆ a contains a product of primes, a contradiction. Hence M is empty. The classic proof of unique factorization of integers relies on being able to cancel out primes (by dividing), so to mimic this in our proof of Theorem 1.4.2, we define an analogy of inverses for ideals.

Definition. If J ⊂ A is an ideal, the fractional ideal generated by J is the A-module

J −1 := {x ∈ K | xJ ⊆ A}.

Lemma 1.4.5. For every ideal J ⊂ A, J −1 is an A-submodule of K.

Notice that J −1 ⊇ A so for any proper ideal J ( A, J −1 is not an ideal of A.

12 1.4 Factorization of Ideals 1 Algebraic Number Fields

Lemma 1.4.6. If p ⊂ A is a prime ideal, p−1 6= A.

Proof. Let x ∈ p. Then By Lemma 1.4.4, (x) ⊇ p1 ··· pr for prime ideals pi ⊂ A. Assume r is minimal among such products of primes contained in (x). We claim that p = pi for some 1 ≤ i ≤ r. If not, there exists an ai ∈ pi r p for each i, by maximality of prime ideals. Then a1 ··· ar ∈ p1 ··· pr ⊆ (x) ⊆ p, a contradiction. Thus p = pi for some i. Assume p = p1. −1 Then by minimality of r, we know (x) ) p2 ··· pr. Let b ∈ p2 ··· pr r (x). Then x b 6∈ A, but x−1bp ⊆ A. So x−1b ∈ p−1 r A. Lemma 1.4.7. If a ⊂ A is an ideal and p ⊂ A is a prime ideal, then ap−1 ) a. Proof. Certainly ap−1 ⊇ a since p−1 ⊃ A. Suppose ap−1 = a and let x ∈ p−1. Then xa ⊆ a, so in particular, left multiplication by x is an element of the A-algebra EndA(a). Since A is Noetherian, EndA(a) is finitely generated. Clearly EndA(a) is also a faithful A-module, so by a well-known characterization of integrality (cf. Atiyah-Macdonald), x is integral over A. Then since A is integrally closed, x ∈ A. We have shown p−1 = A, but this contradicts Lemma 1.4.6. Therefore ap−1 ) a. Corollary 1.4.8. For any prime ideal p ⊂ A, pp−1 = A.

Proof. By Lemma 1.4.7, we have p ( pp−1 ⊆ A, but primes are maximal in a Dedekind domain, so pp−1 = A.

Corollary 1.4.9. For an ideal a ⊂ A and a prime ideal p ) a, ap−1 ( A. Proof. If ap−1 = A then p = app−1 = a, a contradiction. We are now prepared to prove the unique factorization theorem for nonzero ideals in a Dedekind domain. Proof. (of Theorem 1.4.2) Let M be the collection of nonzero, non-unital ideals in A that do not have a factorization into prime ideals. Since A is Noetherian, M has a maximal element a. As before, a cannot be prime so it is contained in a prime ideal p. By Lemma 1.4.7, ap−1 ) a so ap−1 6∈ M. On the other hand, by Corollary 1.4.9 we have ap−1 6= (1) so −1 −1 ap = p1 ··· pr. Multiplying by p gives us a = app = pp1 ··· pr which shows a has a prime factorization. Thus M must be empty. This proves the existence part of the theorem. For uniqueness, suppose a = p1 ··· pr = q1 ··· qs for prime ideals pi, qj ⊂ A. Then by the Q proof of Lemma 1.4.6, p1 ⊃ j qj implies p1 = qj for some 1 ≤ j ≤ s. Multiplying both −1 sides by p1 cancels out terms, yielding shorter prime factorizations of a which are equal by induction. The base case of this induction is easy: if a is prime then it only has the trivial factorization a = a. This finishes the proof of unique factorization of ideals in a Dedekind domain. Remark. For an ideal a ⊂ A and a prime ideal p ⊂ A, we will use the expressions p ⊇ a (‘p contains a’) and p | a (‘p divides a’) to mean the same thing: p appears in the prime factorization of a. If I,J ⊂ A are ideals, we will write (I,J) = 1 if I + J = (1), that is, if I and J are relatively prime in A.

13 1.4 Factorization of Ideals 1 Algebraic Number Fields

Definition. A fractional ideal of A is any finitely generated A-submodule of K. Example 1.4.10. For any ideal J ⊂ A, J −1 is a fractional ideal. Proposition 1.4.11. The nonzero fractional ideals of A form a group under ideal multipli- cation, with identity (1).

Q ai Proof. By Theorem 1.4.2, fractional ideals of the form pi , with ai ∈ Z and pi ⊂ A prime, form a group which is isomorphic to a direct sum of copies of Z. Let M be any fractional ideal. Then M is finitely generated, so there exists an element x ∈ K such that xM ⊂ A is an ideal. Since (x) and xM have prime factorizations, so does M = (x)−1xM. Hence all fractional ideals form a group under multiplication. Corollary 1.4.8 shows that (1) is the identity element in this group. Proposition 1.4.12. If I ⊂ A is an ideal then II−1 = (1).

−1 −1 Proof. Suppose I = p1 ··· pr is the prime factorization of I. Then J = p1 ··· pr is a fractional ideal, and by Corollary 1.4.8, IJ = (1). It remains to show J = I−1. First, since IJ = (1) we have J ⊆ I−1. If x ∈ I−1 then xI ⊂ A so

−1 −1 −1 −1 xIp1 ··· pr ⊆ p1 ··· pr = J. Thus Ax ⊆ J so x ∈ J. This proves J = I−1 as required. Corollary 1.4.13. If A is a Dedekind domain and a unique factorization domain, then A is also a PID.

Definition. For a Dedekind domain A, let JA denote the group of fractional ideals of A. The ideal class group of A is defined as the quotient group

CA = JA/PA where PA is the subgroup of JA consisting of the principal fractional ideals of A.

Clearly |CA| = 1 if and only if A is a PID (and therefore a UFD), so the ideal class group is a direct measure of the failure of unique factorization in A. Moreover, the ideal class group corresponds to an exact sequence of groups

× × 1 → A → K → JA → CA → 1.

We will study this further when we characterize the unit group K× in Section 1.9.

Lemma 1.4.14. Every class in CA can be represented by an ideal I ⊂ A.

Example 1.4.15. The ring A = C[x, y]/(y2 − x3 − x) is a Dedekind domain. It turns out that the ideal class group CA has cardinality equal to |C|, so this example shows that ideal class groups can be particularly bad. In particular, unique factorization fails in A: x3 − x = y2 = x(x − 1)(x + 1). One of the most important results in algebraic number theory is the following theorem, which we will prove in Section 1.8.

14 1.5 Ramification 1 Algebraic Number Fields

Theorem. For a number field K, the class group CK := COK is finite. √ √ √ Example 1.4.16. Let K = Q( −5) and√ recall that 6 = 2 · 3 = (1 + −5)(1 − −5). What do 2 and 3 split into as ideals in OK = Z[ −5]? It turns out that √ √ √ √ 2OK = (2, 1 + −5)(2, 1 − −5) and 3OK = (3, 1 + −5)(3, 1 − −5).

The underlying principle√ governing this splitting behavior is the fact that the minimal poly- nomial x2 + 5 of −5 splits differently mod 2 and 3:

x2 + 5 ≡ (x + 1)2 (mod 2) and x2 + 5 ≡ (x + 1)(x − 1) (mod 3).

1.5 Ramification

In this section let L/K be a finite separable field extension, let OK be a Dedekind domain with field of fractions K and let OL be the integral closure of OK in L. Put n = [L : K].

Lemma 1.5.1. OL is a Dedekind domain. Proof. This is the same proof as for Theorem 1.4.3.

Lemma 1.5.2. If p ⊂ OK is a prime ideal then pOL 6= OL.

−1 Proof. Take x ∈ p r OK , which exists by Lemma 1.4.6. Then xp ⊆ OK so xpOL ⊆ OL. If pOL = OL then we have xpOL = xOL ( OL, a contradiction. Therefore pOL 6= OL.

Now fix a nonzero prime ideal p ⊂ OK . By Theorem 1.4.2, p considered as an ideal of OL has a unique factorization e1 eg pOL = P1 ··· Pg

where the Pi ⊂ OL are distinct primes and each ei > 1. Note that for each i, OL/Pi is a finite dimensional OK /p-vector space. (This follows from the fact that Pi ∩ OK = p.) We say the Pi are the primes of OL lying over p. By unique factorization, these are the only primes lying over p.

Definition. For a prime Pi in the factorization of pOL, the index fi = [OL/Pi : OK /p] is called the inertial degree of Pi (over p) and the exponent ei is called the ramification index of Pi (over p). We say the prime p is totally split if ei = fi = 1 for all 1 ≤ i ≤ g; p is totally ramified if g = 1 and f1 = 1; and p is inert if g = 1 and e1 = 1.

Definition. If any ei > 1 or (OL/Pi)/(OK /p) is inseparable, we say the prime p is ramified (in OL). Otherwise p is unramified.

Qg ei Theorem 1.5.3. For any prime p ⊂ OK with prime factorization pOL = i=1 Pi , we have Pg i=1 eifi = n = [L : K].

15 1.5 Ramification 1 Algebraic Number Fields

Proof. By the Chinese remainder theorem, we can write g g Y ei ∼ M ei OL/pOL = OL/ Pi = OL/Pi . i=1 i=1

ei To prove the theorem, we show that [OL/pOL : OK /p] = n and [OL/Pi : OK /p] = eifi for each 1 ≤ i ≤ g. For the first equality, take {ω1,..., ωm} to be a basis for OL/pOL as an OK /p-vector space. Lift these elements to ω1, . . . , ωm ∈ OL. Suppose

a1ω1 + ... + amωm = 0 for coefficients ai ∈ OK .

−1 Let a = (a1, . . . , am) ⊂ OK and let x ∈ a r ap; such an element exists by Lemma 1.4.7. Then xai ∈ OK for all i, but xai 6∈ p for some i. Replacing ai with xai and reducing mod p gives us a linear dependence, contradicting the assumption that ω1,..., ωm are a basis of OL/pOL. Hence ω1, . . . , ωm must be linearly independent in OK . To show they span OK , let M = ω1OK + ... + ωmOK ⊆ OL. Since the ωi generated OL/pOL, we get M + pOL = OL. In other words, p(OL/M) = OL/M. By Nakayama’s Lemma, this means OL/M is killed by some x ∈ 1 + p. In particular, such an x is necessarily ωm ω1 ωm nonzero so xOL ⊆ M. Thus ω1xOK +...+ x OK ⊇ OL. This implies that x K+...+ x K = L so the ωi span OL as an OK -vector space. This of course is only possible if m = n, so we have the first equality. Now consider the sequence

ei ei 2 ei ei−1 ei OL/Pi ⊇ Pi/Pi ⊇ Pi /Pi ⊇ · · · ⊇ Pi /Pi ⊇ 0. ν ν+1 Taking each quotient in the chain yields something of the form Pi /Pi , and by unique ν ν+1 factorization, each of these quotients is nontrivial. Thus we can choose x ∈ Pi r Pi . Consider the map

ν ν+1 ϕ : OL −→ Pi /Pi α 7−→ xα.

Certainly ker ϕ = Pi since Pi ⊆ ker ϕ and primes are maximal in OL. Also, ϕ is surjective ν+1 ν+1 ν ν+1 ν ∼ ν ν+1 since Pi ( (x)+Pi ⊆ Pi which implies (x)+Pi = Pi . Therefore OL/Pi = Pi /Pi as OK /p-vector spaces. Adding these up gives us

ei ei X dimOK /p OL/Pi = dimOK /p OL/Pi = eifi. j=1 Pg This proves both claims, and this is of course enough to conclude that n = i=1 eifi.

Let θ ∈ OL be a primitive element of L/K, that is, L = K(θ). It is not always guaranteed that OK [θ] = OL. However, we have a way of measuring how far off from the whole ring OL the submodule OK [θ] really is. Definition. The conductor of the extension L/K is the ideal

f := {α ∈ OK | αOK ⊆ OK [θ]} ⊂ OK where L = K(θ).

16 1.5 Ramification 1 Algebraic Number Fields

Example 1.5.4. If OK [θ] = OL, then f = (1). √ √ Example 1.5.5. For K = Q and L = Q( −3), the conductor is f = (2, 1 + −3). Note that f is always nonzero.

Theorem 1.5.6. Let L/K be a finite separable extension with L = K(θ). Suppose p ⊂ OK is prime and pOL + f = (1), where f is the conductor of the extension L/K. Let ϕ(x) be the minimal polynomial of θ over K. If ϕ(x) factors completely in (OK /p)[x] as

e1 eg ϕ(x) = ϕ1(x) ··· ϕg(x) mod p

Qg ei with deg ϕi = fi, then the factorization of p in OL is pOL = i=1 Pi where for each i, Pi is a prime ideal with ramification index ei and inertia degree fi, given explicitly by Pi = ϕi(θ)OL + pOL for any lift ϕi(x) of ϕi(x) in OK [x].

0 Proof. Set O = OK [θ]. We will prove the following isomorphisms:

∼ 0 0 ∼ OL/pOL = O /pO = (OK /p)[x]/ϕ(x)

0 where again ϕ(x) is the minimal polynomial of θ over K. Clearly O ⊆ OL so we have a map 0 0 0 O /pO → OL/pOL. By assumption, pOL + f = OL but f ⊆ OL so we have pOL + O = OL. 0 0 Hence the map is surjective. On the other hand, pO ⊆ pOL ∩ O and

0 0 0 0 pOL ∩ O = (pOL ∩ O )(pOL + f) ⊆ pO + fpOL ⊆ pO .

This proves injectivity, so the first isomorphism is proven. The second isomorphism is im- mediate from the fact that

0 0 ∼ ∼ O /pO = OK [x]/(ϕ(x), p) = (OK /p)[x]/ϕ(x).

Now by the Chinese remainder theorem, we may write

g ∼ ∼ M ei OL/pOLL = (OK /p)[x] = (OK /p)[x]/ϕ(x) . i=1

The prime ideals on the right are just the ideals (ϕi(x)). Set R = (OK /p)[x]/ϕ(x) and Qg ei notice that [R/(ϕi(x)) : OK /p] = fi = deg ϕi and i=1 ϕi(x) = 0. The primes in OL/pOL corresponding under the above isomorphism to the ϕi(θ) are Pi := ϕi(θ)OL + pOL. Notice Qg ei that i=1 Pi ⊆ pOL, but since

g g X Y ei dimOK /p OL/pOL = eifi = dimOK /p OL/ Pi , i=1 i=1

Qg ei we have i=1 Pi = pOL. This proves the theorem.

17 1.5 Ramification 1 Algebraic Number Fields

Example 1.5.7. Let OK = Z[i] be the Gaussian integers. Here the conductor is f = (1). Consider how x2 + 1 splits mod 13:

x2 + 1 ≡ (x − 5)(x + 5) (mod 13).

Then by Theorem 1.5.6, the ideal (13) splits in Z[i] in the following way:

13Z[i] = (13, 5 + i)(13, −5 + i) = (3 + 2i)(3 − 2i).

For a prime p ⊂ OK and a prime P ⊂ OL lying over p, write kp = OK /p and `P = OL/P.

Proposition 1.5.8. If p ⊂ OK is a nonzero prime such that p + f = (1) in OL, then p is unramified if and only if p does not divide the principal ideal (dL/K (θ)) generated by the discriminant of L/K.

Q 2 Proof. We know dL/K (θ) = i

⇐⇒ p is relatively prime to each θi − θj in OM for some normal closure M of L/K

⇐⇒ p - (dL/K (θ)) in OK .

Hence p is unramified precisely when p - (dL/K (θ)). We now discuss Hilbert’s program for ramification theory. Assume that L/K is Galois and let G = Gal(L/K). Note that σ(OL) = OL for all σ ∈ G. If p ⊂ OK is a prime and Qg ei pOL = i=1 Pi , then each σ ∈ G acts on the primes lying over p: σ(Pi) = Pj for some 1 ≤ j ≤ g. The key observation is that this action is transitive.

Proposition 1.5.9. For any prime p ⊂ OK , G = Gal(L/K) acts transitively on the set of primes of OL lying over p.

Proof. Suppose not. Then there is some pair of primes Pi, Pj lying over p such that σPj 6= Pi for all σ ∈ G. By the Chinese remainder theorem, pick x ∈ Pj such that x ≡ 1 (mod σPi) Q for all σ ∈ G. Then NL/K (x) ∈ Pj ∩ OK = p. On the other hand, NL/K (x) = σ∈G σ(x) but σ(x) 6∈ Pi for any σ, so NL/K (x) 6∈ p. This is impossible, so there is some σ ∈ G such that σPj = Pi.

Corollary 1.5.10. When L/K is Galois, for any prime p ⊂ OK , all ramification indices ei and all inertia degrees fi for primes over p are equal, and therefore [L : K] = efg, where e = ei and f = fi for any prime Pi | p.

18 1.5 Ramification 1 Algebraic Number Fields

ν ν Proof. An ideal Pi divides pOL if and only if σPi divides pOL for all σ ∈ G, which by ν Proposition 1.5.9 is equivalent to Pj dividing pOL for all 1 ≤ j ≤ g. Therefore the ramifi- cation indices are all equal; let e denote any one of them. Now given 1 ≤ i, j ≤ g, suppose σ ∈ G is a permutation taking Pj to Pi, that is, Pi = σPj. Then σ determines an isomor- phism OL/Pj → OL/Pi. Therefore fi = fj. Let f denote any of the inertial degrees. Then Pg finally, by Theorem 1.5.3 we have [L : K] = i=1 ef = efg.

Fix a prime P ⊂ OL lying over p.

Definition. The subgroup DP = {σ ∈ G | σ(P) = P} of G is called the decomposition group of P.

Clearly by the orbit-stabilizer theorem, |DP| = ef where e and f are the ramification index and inertia degree of p, respectively. By Galois theory, there is a field extension ZP/K DP corresponding to the subgroup DP ≤ G, which is explicitly the fixed field ZP = L .

Definition. For a prime P | p, the field ZP is called the decomposition field of P.

L

DP

ZP G

K

0 0 −1 Lemma 1.5.11. If σP = P for two primes P, P lying over p, then DP0 = σDPσ for some σ ∈ G. Proof. This is a more general fact about the stabilizers of a transitive group action. Note that for σ, τ ∈ Gal(L/K),

−1 −1 τ στ ∈ DP ⇐⇒ τ στP = P ⇐⇒ στP = τP ⇐⇒ σ ∈ DτP

−1 −1 which implies that σ ∈ DP ⇐⇒ τστ ∈ DP. Hence τDPτ = DτP. The ramification index and inertia degree are transitive in any tower of Galois field extensions:

Lemma 1.5.12. For a Galois tower of number fields M ⊃ L ⊃ K and a prime p ⊂ OK , let Q ⊂ OM be a prime lying over p and set P = Q ∩ L. Then e(Q | p) = e(Q | P)e(P | p) and f(Q | p) = f(Q | P)f(P | p).

Proof. Clearly P is a prime lying over p in OL, so e(P | p) and f(P | p) are defined. Then e(Q | p) = e(Q | P)e(P | p) is immediate by unique factorization in Dedekind domains, and f(Q | p) = f(Q | P)f(P | p) follows from Corollary 1.5.10 and the fact that [M : K] = [M : L][L : K]. The decomposition field is characterized by the following proposition.

19 1.5 Ramification 1 Algebraic Number Fields

Proposition 1.5.13. Let a = P ∩ ZP be a prime below P in ZP. Then

(1) P is the only prime in OL lying over a. (2) If e = e(P | p) and f = f(P | p) then e(P | a) = e, f(P | a) = f and e(a | p) = 1 = f(a | p).

L P e f

ZP a 1 1 K p

Proof. (1) For all σ ∈ DP = Gal(L/ZP), σP = P. By Proposition 1.5.9, DP acts transitively on the primes over a, so P must be the unique one. (2) Since |DP| = ef, e(P | a)f(P | a) = ef but by Lemma 1.5.12, e(P | a) divides e and f(P | a) divides f. Therefore e(P | a) = e and f(P | a) = f, and the others are 1 by Lemma 1.5.12.

Remark. Every σ ∈ DP induces an automorphism ϕσ : OL/P → OL/P which fixes kp = OK /p ⊆ `P = OL/P. Thus we get a map

ϕ : DP −→ Aut(`P/kp)

σ 7−→ ϕσ.

Proposition 1.5.14. ϕ : DP → Aut(`P/kp) is surjective and `P/kp is a normal extension.

Proof. By Proposition 1.5.13, ka = kp for any prime ideal a in the ring of integers of the decomposition field, so we can replace K with Z = ZP and G with DP. Thus P is the ¯ only prime lying over p. Take θ ∈ `P and let θ ∈ OL be any lift, with minimal polynomials ¯ ¯ ¯ g¯(x) ∈ kp[x] and f(x) ∈ K[x], respectively. Certainly f(θ) = 0 mod p sog ¯ | f in kp[x]. ¯ Since L/K is normal (it is a Galois extension), f splits over L. This implies f splits over `P, sog ¯ splits as well. This proves `P/kp is a normal extension. ¯ Now choose θ generating the separable closure of kp in `P. Letσ ¯ ∈ Aut(`P/kp). Then σ¯θ¯ is a root ofg ¯ and thus of f¯. Since f splits in L, there exists a root α ∈ L of f such ¯ thatα ¯ =σ ¯θ in `P. Choose σ ∈ G = DP such that σθ = α, which is possible since L/K is ¯ normal. Then ϕ(σ) =σ ¯ because θ generates the separable closure of kp in `P. This proves ϕ is surjective.

Definition. The kernel IP = ker ϕ ≤ DP is called the inertia group of P. Explicitly,

IP = {σ ∈ G | σ(x) ≡ x mod P for all x ∈ OL}.

IP Definition. The fixed field TP = L is called the inertia field of P (over p).

20 1.6 Cyclotomic Fields and Quadratic Reciprocity 1 Algebraic Number Fields

By Proposition 1.5.14, we have an exact sequence

1 → IP → DP → Gal(`P/kp) → 1.

Proposition 1.5.15. Let b = P ∩ TP and a = P ∩ ZP = b ∩ ZP be prime ideals in the inertia and decomposition fields, respectively. Set e = e(P | p) and f = f(P | p). Then e(P | b) = e, f(b | a) = f and e(b | a) = f(P | b) = 1.

L P e 1

TP b 1 f

ZP a 1 1 K p

Proof. Let Z = ZP and T = TP. In light of Proposition 1.5.13, it’s enough to show `P = OT /b and |DP/IP| = f. By the exact sequence 1 → IP → DP → Gal(`P/kp) → 1, ∼ DP/IP = Gal(`P/kp) so if `P = OT /b, then

|DP/IP| = | Gal((OT /b)/kp)| = | Gal((OT /b)/(OZ /a))| = f.

Therefore it suffices to prove the former statement, that is, `P = OT /b. The decomposi- tion/inertia group exact sequence for the extension L/T is

1 → IP → IP → Gal(`P/(OT /b)) → 1

which implies `P = OT /b as claimed.

1.6 Cyclotomic Fields and Quadratic Reciprocity

Recall that when ζm is a primitive mth root of unity and K = Q(ζm), the ring of integers of this cyclotomic number field is OK = Z[ζm]. This was proven in Corollary 1.3.14. In this section, we further elaborate on the properties of Q(ζm) and Z[ζm] and use algebraic number theory to prove Gauss’s celebrated quadratic reciprocity law. Recall the following definition from elementary number theory.

Definition. If p is an odd prime and a ∈ Z, the Legendre symbol of a mod p is  1 p - a and a is a square (mod p) a  = −1 a is not a square (mod p) p 0 p | a.

21 1.6 Cyclotomic Fields and Quadratic Reciprocity 1 Algebraic Number Fields

 a  Proposition 1.6.1. Suppose a, p ∈ Z with p prime and (2a, p) = 1. Then p = 1 if and √ only if the prime ideal (p) splits completely in OQ( a). Proof. The conductor f√ divides 2, so (p) splitting in O √ is equivalent to (p) splitting in √ a Q( a) Z[ a]. This, in turn, is equivalent to x2 − a splitting mod p, by Theorem 1.5.6, i.e. a ≡ x2 √ (mod p) for some x ∈ Z. Hence (p) splits in OQ( a) if and only if a is a square mod p.

×  a  (p−1)/2 Remark. Since Fp is cyclic, it’s easy to show that p ≡ a (mod p) for any a ∈ Z — this is called Euler’s criterion. In particular, for a = −1 we have ( −1 1, p ≡ 1 (mod 4) = p −1, p ≡ −1 (mod 4).

Most elementary proofs of quadratic reciprocity exploit this characterization of the Legendre symbol in some fashion. Here we prove the reciprocity law by considering the factorization of (p) in the ring Z[ζq]. Theorem 1.6.2 (Quadratic Reciprocity). Let p and q be distinct, odd primes. Then

p q  = (−1)(p−1)(q−1)/4. q p

∗ (q−1)/2  q∗   q  (p−1)(q−1)/4 First, set q = (−1) q so that p = p (−1) by Euler’s criterion. The  p   q∗  statement we must then prove is that q = p . Example 1.6.3. The beauty of the quadratic reciprocity law is that it allows for fast com- putations of the Legendre symbol. For example, is 15 a square mod 37? Rather than trying 2 to compute all squares mod 37, or trying to factor x − 37 in F37, we can use reciprocity. Since 37 ≡ 1 (mod 4), we have:

15  5   3  37 37 2 1 = = = = (−1)(1) = −1. 37 37 37 5 3 5 3

So 15 is not a square mod 37.

Lemma 1.6.4. Suppose n ≥ 2 is an integer with prime factorization n = Q pν(p), where the product is over all primes p and ν(p) ≥ 0 for all p. For each prime p, let fp be the ν(p) ϕ(pν(p)) multiplicative order of p mod n/p . Then in R = Z[ζn] we have pR = (p1 ··· pr) for distinct prime ideals p1,..., pr ⊂ R such that f(pi | p) = fp for each 1 ≤ i ≤ r. Proof. Fix a prime p and set m = n/pν(p) so that n = pν(p)m. Consider the number field K = Q(ζn). We know the conductor of ζn in OK is f = 1. Let γn be the nth cyclotomic ν(p) polynomial and let {αi} be the primitive p th roots of unity and {βj} be the primitive mth roots of unity. Then by the Chinese remainder theorem,

× ∼ ν(p) × × (Z/nZ) = (Z/p Z) × (Z/mZ)

22 1.6 Cyclotomic Fields and Quadratic Reciprocity 1 Algebraic Number Fields

so we can write Y γn(x) = (x − αiβj). i,j

Note that all the αi are 1 in any field of characteristic p. Thus, modulo p,

Y ϕ(pν(p)) ϕ(pν(p)) γn(x) ≡ (x − βj) = γm(x) . j

ν(p) This allows us to reduce to the case when m = n, that is, the case when p = 1. Letγ ¯m(x) m denote the factorization of γm(x) mod p. Since x − 1 is separable over Fp (m is relatively m prime to p) andγ ¯m(x) | x − 1, we have thatγ ¯m(x) is also separable over Fp. The smallest extension of Fp containing a primitive mth root of unity (and thus all of them) is Fpfp . Thus γ¯m splits over Fpfp and each irreducible factor ofγ ¯m over Fp is the minimal polynomial of some primitive mth root of unity, each of which having degree fp. This impliesγ ¯m is a product of degree fp irreducible polynomials over Fp. By Theorem 1.5.6, we have

ϕ(pν(p)) pR = (p1 ··· pr) .

ϕ(n) Remark. In general, Theorem 1.5.10 implies that r = ν(p) . ϕ(p )fp

Corollary 1.6.5. An odd prime integer p is ramified in Q(ζn)/Q if and only if p | n, and p = 2 is ramified if and only if 4 | n.

p (q−1)/2 Lemma 1.6.6. If q is an odd prime integer, then (−1) q ∈ Q(ζq). Proof. Set X a τ = ζa. q q a∈(Z/qZ)× 2 (q−1)/2 Then τ ∈ Q(ζq) and τ = (−1) q. We are now able to prove quadratic reciprocity (Theorem 1.6.2). Proof. Let p and q be distinct odd primes and set q∗ = (−1)(q−1)/2q. Consider the tower of √ ∗ number fields Q(ζq) ⊃ Q( q ) ⊃ Q, with Galois groups as shown:

Q(ζq) Z/((q − 1)/2)Z √ Z/(q − 1)Z Q( q∗) Z/2Z Q

23 1.7 Lattices 1 Algebraic Number Fields

 q∗  Then we determine the reciprocity law for p as follows:

q∗  √ = 1 ⇐⇒ (p) splits in ( q∗) by Proposition 1.6.1 p Q √ ∗ ⇐⇒ Q( q ) ⊆ ZP, the decomposition field for any prime P over (p) ⇐⇒ there exist an even number of primes in Z[ζq] lying over (p) q − 1 ⇐⇒ is even, where fp is the multiplicative order of p mod q fp q − 1 ⇐⇒ f divides p 2 q−1 ⇐⇒ p 2 ≡ 1 (mod q) p ⇐⇒ = 1. q

Thus quadratic reciprocity is proven.

Corollary 1.6.7. If q is an odd prime, then ( 2 1, q ≡ 1, 7 (mod 8) = q −1, q ≡ 3, 5 (mod 8).

∗ q−1 ∗ Proof. Set q = (−1) 2 q, so that q ≡ 1 (mod 4). Then √    ∗  2 √ 1 + q = 1 ⇐⇒ (2) splits in O ∗ = q Q( q ) Z 2 1 − q∗ ⇐⇒ f(x) = x2 − x + splits mod 2 4 ⇐⇒ q∗ ≡ 1 (mod 8) ⇐⇒ q = 1, 7 (mod 8).

1.7 Lattices

One perspective on rings of algebraic integers is to view them as lattices. For example, Z[i] is very clearly a lattice in C spanned by the vectors 1 and i. We will show that any ring n of integers OK in a number field K/Q is a lattice in some R . This is the beginning of Minkowski’s so-called theory of geometry of numbers.

n Definition. A Z-module Γ ⊆ R is a lattice of rank m if Γ = Zv1 + ldots + Zvm for R-linearly independent vectors v1, . . . , vm. If m = n then we say Γ is a complete lattice, or has full rank in Rn.

24 1.7 Lattices 1 Algebraic Number Fields

Definition. For a lattice Γ ⊆ Rn, the set

Φ = {x1v1 + ... + xmvm | 0 ≤ xi < 1} is called the fundamental domain of Γ, also sometimes called the fundamental paral- lelopiped.

Observe that Γ is a complete lattice in Rn if and only if Γ + Φ = Rn.

Definition. A subgroup W ⊆ Rn is said to be discrete if every point in W is open in the topology on Rn, that is, if every point x ∈ W has a neighborhood U in Rn such that U ∩ W = {x}.

Proposition 1.7.1. If Γ ⊆ Rn is a subgroup, then Γ is discrete if and only if Γ is a lattice.

Proposition 1.7.2. If Γ ⊆ Rn is a lattice, then Γ is complete if and only if there exists a bounded set M such that Γ + M = Rn. Proof. ( =⇒ ) When Γ is complete, M = Φ works. ( ⇒ = ) If Γ is not complete, let V (Rn be the R-span of Γ. Then V lies in some hyperplane H in Rn. Choose d > 0. Then for any bounded set of diameter diam(M) < d, all points further than d from H do not lie in Γ + M ⊆ H + M. Hence Γ + M 6= Rn.

n Definition. If Γ = Zv1 + ... Zvn is a complete lattice in R , we define the volume of Γ to be the volume of the parallelopiped spanned by v1, . . . , vn:

 | |  vol(Γ) := vol(Φ) = | det A| where A = v1 ··· vn . | |

Note that since det(AT A) = (det A)2, we can write the volume formula as q vol(Γ) = det(vivj).

Definition. A region Ω ⊆ Rn is centrally-symmetric if x ∈ Ω implies −x ∈ Ω. Minkowski’s theorem is the key result in the geometry of numbers which allows us to describe lattices like OK and UK , the ring of integers and unit group, respectively, in a number field K/Q.

Theorem 1.7.3 (Minkowski). If Γ is a complete lattice in Rn and X is a centrally-symmetric, convex region of Rn such that vol(X) > 2n vol(Γ), then X contains a nonzero point of Γ.

Proof. By a linear change of variables, we may assume Γ = Zn. Then vol(Γ) = det(I) = 1. n 1
Suppose X is as described, with vol(X) > 2 . Then vol( 2 > 1. We claim that there exist

25 1.8 The Class Group 1 Algebraic Number Fields

1
1
1
lattice points γ1 6= γ2 in Γ such that 2 X + γ1 ∩ 2 X + γ2 6= ∅. If not, 2 X + γ1 ∩ Φ is 1
disjoint from 2 X + γ2 ∩ Φ for all distinct γ1, γ2 ∈ Γ. Thus 1 = vol(Γ) X 1

≥ vol 2 X + γ ∩ Φ γ∈Γ X 1
= vol (Φ − γ) ∩ 2 X γ∈Γ 1
n = vol 2 X since Φ + Γ = R > 1,

1
1
a contradiction. Therefore there exist such γ1, γ2 ∈ Γ. Now take x ∈ 2 X + γ1 ∩ 2 X + γ2 . Then for some x1, x2 ∈ X, we have

1 1 1 x = 2 x1 + γ1 = 2 x2 + γ2 =⇒ γ1 − γ2 = 2 (x2 − x1),

which is just the midpoint of the line between x2 and −x1. By convexity and central- symmetry, this implies γ1 − γ2 ∈ X, but since γ1 6= γ2 we have found a nonzero lattice point in X.

Remark. Note that the inequality in Minkowski’s theorem must be sharp, for if Γ = Zn, then vol(Γ) = 1, whereas the centrally-symmetric, convex set

X = {(x1, . . . , xn) | −1 < xi < 1}

has volume 2n but contains no nonzero lattice points.

1.8 The Class Group

Let K be a number field of degree n = [K : Q] and let T be the set of all field embeddings τ : K,→ C. Define subsets TR ⊆ T , consisting of all real embeddings of K, and TC ⊆ T , consisting of all complex embeddings of K, and set r = |TR| and 2s = |TC|. Since the complex embeddings come in pairs τ, τ¯ ∈ TC, the 2s makes sense. There is an isomorphism of vector spaces Y K ⊗Q C −→ C =: KC τ∈T

x ⊗ y 7−→ (τ(x)y)τ .

Further, there is a canonical embedding

∼= j : K K ⊗Q C KC

x x ⊗ 1 (τ(x))τ .

26 1.8 The Class Group 1 Algebraic Number Fields

There is an involution F on K ⊗Q C given by F (x⊗y) = x⊗y¯, which corresponds to complex conjugation F ((xτ )τ ) = (¯xτ¯)τ in KC. Therefore the following diagram commutes: ∼= K ⊗Q C KC

F F ∼= K ⊗Q C KC

The fixed points under the involution F are the subset

KR = {(xτ )τ | xτ ∈ R for τ ∈ TR and xτ =x ¯τ¯ for τ ∈ TC}. ∼ This subset KR ⊆ KC corresponds to the field K ⊗Q R = KR. Note that j(K) ⊆ KR. The trace map also respects the inclusion KR ⊆ KC:

Tr : KC −→ C

KR −→ R X (xτ )τ 7−→ xτ . τ∈T

Observe that Tr ◦j : K → R is just equal to the field trace, TrK/Q, as defined in Section 1.2. Likewise, the norm map

N : KC −→ C

KR −→ R Y (xτ )τ 7−→ xτ τ∈T respects KR ⊆ KC and satisfies N ◦ j = NK/Q : K → R. Recall that r = |TR| and 2s = |TC|, so that r + 2s = |T | = n = [K : Q]. There is an isomorphism

r+2s n f : KR −→ R = R (x1, . . . , xr, y1, y¯1, . . . , ys, y¯s) 7−→ (x1, . . . , xr, Re(y1), Im(y1),..., Re(ys), Im(ys)).

r+2s It is sometimes more useful to think of KR as R in this way. There is a standard Hermitian inner product on KC, which restricts to an inner product on KR called the Minkowski inner product. In Rn, this corresponds to the canonical real inner product: 0 0 if ~u = (u1, . . . , ur, z1, z1, . . . , zs, zs) 0 0 and ~v = (v1, . . . , vr, w1, w1, . . . , ws, ws) r s X X 0 0 then h~u,~vi = uivi + 2 (wizi + wizi). i=1 i=1

For K a number field with ring of integers OK , let JK = JOK be the group of fractional

ideals, PK = POK the subgroup of principal fractional ideals and let CK = JK /PK be the class group. Our goal is to prove that CK is a finite group.

27 1.8 The Class Group 1 Algebraic Number Fields

Definition. For a nonzero ideal a ⊂ OK , the ideal norm of a is N (a) = [OK : a]. The following lemma explains the terminology ‘ideal norm’.

Lemma 1.8.1. For any principal ideal (α) ⊂ OK , N ((α)) = |NK/Q(α)|.

Lemma 1.8.2. Ideal norm is multiplicative. That is, for any nonzero ideals a, b ⊂ OK , N (ab) = N (a)N (b). Proof. If a and b are relatively prime, this follows from the Chinese remainder theorem. a a Thus it suffices to show that N (p ) = N (p) for every prime p ⊂ OK and exponent a ≥ 0. By considering the filtration of OK by powers of p, we have a 2 a−1 a [OK : p ] = [OK : p][p : p ] ··· [p : p ].

j j+1 For each 0 ≤ j ≤ a − 1, p /p is a simple OK /p-module and thus a 1-dimensional vector j j+1 a a space, so [p : p ] = |OK /p|. It follows that [OK : p ] = [OK : p] . By Theorem 1.4.2, we can extend the ideal norm N to fractional ideals of K by:

−1 −1 N (a ) = [OK : a] .

× This determines a homomorphism N : JK → Q .

Lemma 1.8.3. Given any constant c > 0, there exist only finitely many ideals a ⊂ OK with norm N (a) < c. Proof. By Lemma 1.8.2, it suffices to prove this statement for prime ideals. For each prime integer p ∈ Z, Theorem 1.4.2 implies that there are only finitely many prime ideals p ⊂ OK lying over (p). For each of these p, we have N (p) = pf for some f – in fact, this f is the residue degree of p/(p) as defined in Section 1.5. Therefore any prime ideal p with N (p) < c must lie above a prime p ∈ Z such that pf < c. There are only finitely many of these, so we are done. ∼ n Proposition 1.8.4. If a ⊂ OK is a nonzero ideal, then Γ = j(a) ⊆ K = R is a complete p R lattice with volume vol(Γ) = |dK |N (a).

Proof. It is routine to prove that j(a) is a lattice – in fact, it suffices to show OK is a lattice since a is a discrete subgroup. Now if α1, . . . , αn is a Z-basis for a and T = {τ1, . . . , τn} is the set of embeddings K,→ C, then 2 2 |dK |N (a) = |dK/Q(α1, . . . , αn)| = | det(τi(αk))| . On the other hand,

2 vol(Γ) = | det(hj(αi), j(αk)iik)| n !

X = det τ`(αi)τ`(αk) `=1 ik ∗ = | det(AA )| where A = (τi(αk))ik = | det A|2. This implies the formula and in particular vol(Γ) > 0 so j(a) must be a complete lattice.

28 1.8 The Class Group 1 Algebraic Number Fields

Lemma 1.8.5. For any nonzero ideal a ⊂ OK , let cτ > 0 for each τ ∈ T be such that s Y  2  c > p|d |N (a). τ π K τ∈T

Then there exists some α ∈ a r {0} such that |τ(α)| < cτ for all τ ∈ T .

Proof. Define X = {(zτ )τ ∈ KR : |zτ | < cτ for each τ ∈ T }. Then it is easy to verify that X n n is centrally-symmetric and convex. Viewing X in R via the isomorphism f : KR → R , we see that its image is

n 2 2 2 f(X) = {(xτ )τ ∈ R : |xτ | < cτ for τ ∈ TR and xτ + xτ¯ < cτ for τ ∈ TC} which has volume ! s ! Y Y vol(f(X)) = 2r c 2πc2 τ τi i=1 τ∈TR r+s s Y = 2 π cτ τ∈T r+2sp > 2 |dK |N (a) = 2n vol(j(a)). Therefore by Minkowski’s theorem (1.7.3), f(X) contains a nonzero lattice point of j(a). Let α be the corresponding nonzero point in a. Then it is clear α satisfies the desired condition.

Theorem 1.8.6. For any nonzero ideal a ⊂ OK , there exists a nonzero element α ∈ a such that  2 s |N (α)| ≤ p|d |N (a). K/Q π K

Proof. By Theorem 1.2.2, for any α ∈ OK we have Y |NK/Q(α)| = |τ(α)|. τ∈T

For ε > 0, if τ ∈ T such that cτ > 0 and s Y  2  c = p|d |N (a) + ε, τ π K τ∈T then by Lemma 1.8.5, there exists a nonzero α ∈ a such that |τ(α)| < cτ for all τ. That is, s Y  2  |τ(α)| < p|d |N (a) + ε. π K τ∈T

Letting ε → 0, the fact that |NK/Q(α)| ∈ N0 implies that α ∈ a may be chosen such that  2 s |N (α)| ≤ p|d |N (a). K/Q π K

29 1.8 The Class Group 1 Algebraic Number Fields

Corollary 1.8.7. For any number field K/Q, the class group CK is finite. 2
s p Proof. It suffices to show every ideal class in CK contains an ideal of norm at most π |dK |, since then Lemma 1.8.3 says there are a finite number of these. Fix a class C ∈ CK and −1 pick fractional ideal a ∈ C such that a ⊂ OK is an ideal. By Theorem 1.8.6, there exists α ∈ a−1 such that  2 s N ((α)) = |N (α)| < p|d |N (a−1). K/Q π K −1 Note that αa ⊆ OK . Since norm is multiplicative (Lemma 1.8.2), we have

N (αa) = N ((α))N (a)  2 s < p|d |N (a−1)N (a) π K  2 s = p|d |. π K

Then the ideal αa is in C and satisfies the desired norm bound. This completes the proof.

Definition. For a number field K, the finite number hK = |CK | is called the class number of K.

The preferred setting for algebraic number theory is obviously when the class number is 1, since then OK is a PID and thus a UFD. However, having class number 1 is a substantial restriction on number fields. For example,√ Heegner (and others later) proved that the only imaginary quadratic number fields Q( d), where d < 0 is squarefree, with class number 1 are for d = −1, −2, −3, −7, −11, −19, −43, −67, −163. For real quadratic number fields, the situation is wide open. It is conjectured that there are infinitely many real quadratic fields of class number 1, but this remains unsolved. √ Example 1.8.8. We will compute the class group of K = Q( −5). Here, dK = −20 since −5 ≡ −1 (mod 4) so the Minkowski bound in Corollary 1.8.7 becomes

 2 1 √ N (a) ≤ 20 ≈ 2.84 < 3. π

In particular every ideal class in CK has an ideal with norm 1 or 2. Thus any nonprincipal class contains some ideal lying over (2). Notice that x2 + 5 ≡ (x + 1)2 (mod 2), so by Theorem 1.5.6, √ (2) = (2, 1 + −5)2 = p2. √ ∼ One can check that p = (2, 1 + −5) is not principal, so we deduce that CK = h[p]i = Z/2Z.

30 1.9 The Unit Theorem 1 Algebraic Number Fields

1.9 The Unit Theorem

Let K be a number field of degree n = [K : Q] with ring of integers OK . As in Section 1.8, let

T = Hom(K, C) = {τ1, . . . , τr, σ1, σ¯1, . . . , σs, σ¯s} Y F KC = C,KR ⊆ KC and j : K,→ KR. τ∈T

Also set K× = Q × and K× = K× ∩ K . In fact we have an embedding j : K× ,→ K×. C τ∈T C R C R R Let µ(K) be the set of roots of unity in K, i.e. µ(K) = {x ∈ K | xa = 1 for some a ∈ N}. Define the map

L : K× −→ r+s R R 2 2 (xτ1 , . . . , xτr , xσ1 , x¯σ1 , . . . , xσs , x¯σs ) 7−→ (log |xτ1 |,..., log |xτr |, log |xσ1 | ,..., log |xσs | ).

Then L is a homomorphism of groups which takes multiplication in K× to addition in r+s. R R Lemma 1.9.1. The diagram

j L × K× r+s K R R

NK/Q N Tr log | · | × Q R× R

commutes.

Proof. This follows from the definitions of the norm and trace maps in Section 1.2 and their

extensions to KC (and KR) in Section 1.8. We will prove:

Theorem 1.9.2 (Dirichlet’s Unit Theorem). Let K be a number field of degree n = r + 2s. × ∼ r+s−1 Then OK = Z × µ(K). To start, define the sets

S = {x ∈ K× | N(x) = ±1} R r+s H = L(S) = {x ∈ R | Tr(x) = 0} × Γ = L ◦ j(OK ) ⊆ H.

Our strategy for proving the unit theorem is to show that Γ is a complete lattice in the hyperplane H with ker(L ◦ j) = µ(K). The unit theorem will then follow from the theory of finitely generated modules over Z.

31 1.9 The Unit Theorem 1 Algebraic Number Fields

Proposition 1.9.3. There is a short exact sequence of groups

× L◦j 1 → µ(K) → OK −−→ Γ → 1.

Proof. Clearly µ(K) ⊆ ker(L ◦ j). Thus it suffices to show that if |τ(x)| = 1 for all τ ∈ T , then x ∈ µ(K). First, there exists a bounded domain in K× containing all the j(x) for R x ∈ OK for which |τ(x)| = 1 for all τ ∈ T . From Proposition 1.8.4, we know that j(OK ) is a lattice in KR so there can only be finitely many x ∈ OK with |τ(x)| = 1 for all τ. Further, 2 3 since for any such x ∈ OK , x, x , x ,... all have this property as well, there must be some m ∈ N such that xm = 1. Therefore x ∈ µ(K). ∼ r+s−1 The proof of Dirichlet’s unit theorem now comes down to showing that Γ = Z . To ∼ r+s−1 do this, we show that Γ is a complete lattice inside H = R .

× Lemma 1.9.4. Given a ∈ Z, up to multiplication by elements of OK , there are only finitely many α ∈ OK with NK/Q(α) = a.

Proof. An equivalent statement is that each coset of OK /aOK has at most one element of of norm a, up to a unit. Suppose α, β ∈ OK are two such elements; that is, β = α + aγ for some γ ∈ OK . Then β a N(α) = 1 + γ = 1 + γ ∈ O α α α K α α × × since N(α)/α ∈ OK . Similarly, β ∈ OK so β ∈ OK . Thus for some u ∈ OK , α = uβ, proving the lemma. Now we prove Theorem 1.9.2. Proof. We first demonstrate that Γ is a lattice. By Proposition 1.7.1, it’s equivalent to show that Γ is discrete and to do this, we show the point 0 ∈ Γ is an isolated point, i.e. every bounded set in H containing 0 contains only finitely many points in Γ. Let X ⊆ H be such a −1 −1 R bounded set. Then L (X) ⊆ S is also bounded, so L (X) is bounded in K . Since j(OK ) −1 is a lattice in KR (follows from Proposition 1.8.4), j(OK ) ∩ L (X) is finite. Applying L, we get that Γ ∩ X is finite, which implies 0 is isolated and hence Γ is a discrete subgroup. S To prove Γ is complete, we exhibit a bounded set M ⊆ H such that H = γ∈Γ(M + γ) and apply Proposition 1.7.2. Since L : S → H is surjective, it will be enough to construct S a bounded set B ⊆ S such that S = × Bj(ε), where Bj(ε) the translate of B by ε∈OK j(ε). There is a subtlety here: if B ⊆ S is bounded, so is L(B) ⊆ H but only because the

logarithms of the elements in B stay away from 0. Now S ⊆ KR, so for all τ ∈ T , pick cτ > 0 such that cτ¯ = cτ and s Y  2  C := c > sqrt|d |, τ π K τ∈T Q Note that for all y ∈ S, τ∈T |τ(y)|cτ = C by definition of S. This means that if y = (yτ ) ∈ S and

Xy = {xy | x ∈ X} = {(zτ )τ ∈ KR : |zτ | < cτ |yτ |}

32 1.9 The Unit Theorem 1 Algebraic Number Fields

then Xy contains some j(α) for α ∈ OK r {0} by Lemma 1.8.5. Now by Lemma 1.9.4, there exist elements α1, . . . , αN ∈ OK such that any α ∈ OK with j(α) ∈ Xy is of the form αiε × for some 1 ≤ i ≤ N, ε ∈ OK . Define

N [ −1 B = S ∩ Xj(αi) . i=1

It is immediate from the definition of the αi that |NK/Q(αi)| < C, and since X is bounded, we get that B is bounded. Moreover, if y ∈ S the above shows that Xy−1 contains some j(α) −1 for α ∈ OK such that |NK/Q(α)| < C. Thus there exists x ∈ X such that xy = j(α), and hence y = xj(α)−1 so S is covered by these bounded sets B. Hence by the initial comments, Proposition 1.7.2 implies Γ is a complete lattice. ∼ r+s−1 Finally, by the theory of finitely generated modules over Z, we have OK = Z × (OK )tors, but it is clear (see Proposition 1.9.3) that the torsion part of OK is precisely µ(K). ∼ r+s−1 Hence OK = Z × µ(K) as required. √ √ Example 1.9.5. Let d > 0 be a squarefree integer, K = Q( d) and take α = a+b d ∈ OK . 1 That is, a, b ∈ Z when d 6≡ 1 (mod 4) and a, b ∈ 2 Z when d ≡ 1 (mod 4). Then √ √ × 2 2 a + b d ∈ OK ⇐⇒ NK/Q(a + b d) = ±1 ⇐⇒ a − b d = ±1. In a real quadratic number field, r = 2, s = 0 and µ(K) = {±1} so Theorem 1.9.2 gives us

× m OK = {±ε | m ∈ Z}

× 2 2 for some εOK . (Such an ε is called a fundamental unit of K.) The equation a − b = ±1 is known as Pell’s equation, so the unit theorem says that the solutions to Pell’s equation over Z form a rank 1 abelian group. √ √ √ × For example, when d = 6 and OK = Z[ 6], 5+2 6 is a unit√ in OK with inverse 5−2 6. 2 2 Notice√ that 5 − 6 · 2 · 6 = 1 and one can check that 5 + 2 6 is a√ fundamental unit for Q( 6). Therefore all solutions to a2 − 6b2 = 1 are of the form (5 + 2 6)k for k ∈ Z.

× Definition. A set of units ε1, . . . , εr+s−1 ∈ OK such that all units in OK are of the form ν1 νr+s−1 ζε1 ··· εr+s−1 for ζ ∈ µ(K) and νi ∈ Z is called a system of fundamental units in K.

× Definition. For Γ ⊆ H, the complete lattice image of OK under L ◦ j, the volume vol(Γ) is called the regulator of K.

× Corollary 1.9.6. If ε1, . . . , εr+s−1 is a system of fundamental units in OK , then the regulator of K is √ vol(Γ) = r + s det((L ◦ j(εi))k)ik.

33 2 Local Fields

2 Local Fields

2.1 Discrete Valuation Rings

Definition. A local Dedekind domain A is called a discrete valuation ring (DVR for short). Its residue field is, as with any local ring, the quotient k = A/m where m is the unique maximal ideal of A. The following definition and proposition explain the where the term discrete valuation ring comes from.

Definition. Let A be a ring. Then a valuation on A is a function v : A r {0} → Z≥0 satisfying:

(i) v(xy) = v(x) + v(y) for all x, y ∈ A r {0}. (ii) v(x + y) ≥ min{v(x), v(y)} for all x, y ∈ A r {0}. (iii) v(x) = 0 if and only if x ∈ A×. A valuation v is a discrete valuation if it is surjective. Proposition 2.1.1. For an integral domain A, the following are equivalent: (1) A is a DVR. (2) There is a discrete valuation v on A. Proof. (i) =⇒ (ii) Since A is a DVR, it is a PID by commutative algebra so each x ∈ A can be written uniquely as x = uπn for π generating the maximal ideal m ⊂ A. Define v(x) = n. Then one verifies v is a discrete valuation on A. (ii) =⇒ (i) The maximal ideal is m = {x ∈ A | v(x) > 0}. It’s easy to check that A is local, integrally closed and therefore a DVR. It is common to extend a valuation v on A to the field of fractions K of A by setting a
v(0) = ∞ and v b = v(a) − v(b) to get a function v : K → Z ∪ {∞}. Example 2.1.2. Let p be a prime and consider the localization of Z at the prime ideal (p):  a Z(p) = b ∈ Q : a, b ∈ Z, p - b .

a
a r a0 0 0 Then Z(p) is a DVR with valuation v b = r if we can write b = p b0 for integers a , b not divisible by p. Example 2.1.3. Let k be a field and consider the polynomial ring k[t]. Localizing at the maximal ideal (t), we get a discrete valuation ring

n p o C[t](t) = q ∈ k(t): p, q ∈ k[t], t - q ,

 p  p r p0 where, much like Example 2.1.2, the valuation is v q = r if we can write q = t q0 for polynomials p0, q0 ∈ k[t] not divisible by t.

34 2.1 Discrete Valuation Rings 2 Local Fields

Example 2.1.4. Let k be a field and consider the power series ring k[[t]] with maximal ideal (t). Then the local ring k[[t]](t) is a DVR with valuation

∞ ! X i v ait = min{i ≥ 0 | ai 6= 0}. i=0

Example 2.1.5. Let Fq be a finite field with q elements and consider the function field k = Fq(t) in one variable. Then the discrete valuations on k are parametrized by the set of irreducible monic polynomials f ∈ Fq[t], together with a point at ∞ which corresponds to the degree valuation: g
v∞ h = deg h − deg g. Lemma 2.1.6. Let A be a Dedekind domain and take a nonzero element α ∈ A with fac- Qr vi torization (α) = i=1 pi , with pj prime ideals and vi ≥ 1. Then for any pj,

vj xApj = pj Apj while for any prime ideal p not dividing (x), xAp = Ap.

Proof. This just comes from the commutative algebra correspondence between ideals in Ap and ideals in A contained in p.

Theorem 2.1.7. Let A be a Dedekind domain with field of fractions K. Then there are bijective correspondences

nonzero prime ideals discrete valuation rings discrete valuations ←→ ←→ . p ⊂ A R ⊂ K v : K → Z ∪ {∞}

Proof. A prime ideal p determines a local ring Ap which is a discrete valuation ring with valuation ( v , if p = p for (x) = Qr pvi v(x) = j j i=1 i 0, if p - (x). It follows from Lemma 2.1.6 that v is a discrete valuation. Proposition 2.1.1 shows that DVRs and discrete valuations are in bijection. Finally, if s : A → K is the canonical embedding −1 and R ⊂ K is a DVR with maximal ideal mR, then s (mR) is a nonzero prime of A. Definition. Let A be a Dedekind domain with field of fractions K and suppose S ⊆ Spec A S contains all but finitely many prime ideals of A. Set U = p∈S p and define the “semi- localization” −1 n f o AS = U A = g ∈ K : f, g ∈ A, g 6∈ p for any p ∈ S .

Lemma 2.1.8. AS is a Dedekind domain. Proof. It is a standard fact from commutative algebra that the localization of a Dedekind domain at any multiplicative set is also Dedekind.

Let CA and CAS denote the class groups of the Dedekind domains A and AS, respectively.

35 2.1 Discrete Valuation Rings 2 Local Fields

Proposition 2.1.9. Let A be a Dedekind domain and S ⊆ Spec A a set of primes excluding only finitely many of the primes of A. Then there is an exact sequence

× × M × × 1 → A → AS → K /Ap → CA → CAS → 1. p6∈S

× × Proof. First, A ,→ AS is a natural inclusion (by the universal property of localization), × × × while the direct sum of the natural inclusions AS ,→ K /Ap for p 6∈ S give the map

× M × × AS → K /Ap . p6∈S

× × For each DVR Ap ⊂ K, the associated valuation vp : K → Z is surjective with kernel Ap so we get an isomorphism M × × ∼ M K /Ap = Z. p6∈S p6∈S

The map CA → CAS is given by [I] 7→ [IAS], and the middle map comes from M Z −→ CA p6∈S " # Y ap (ap)p6∈S 7−→ p . p6∈S One can check that all of these maps are well-defined. Now exactness at A× is trivial: this map is an embedding by the universal property of localization. For CA → CAS , it is a commutative algebra fact again that every ideal of AS is an extended ideal of A so we have surjectivity. × × L × × × × For exactness at AS , let f : AS → p6∈S K /Ap . Then clearly im(A ,→ AS ) ⊆ ker f. × On the other hand, if x ∈ ker f, consider the prime factorization of Ax. Since x ∈ AS , no p f outside S appears in the prime factorization of xA, while if p ∈ S, then x = g with g 6∈ p so −1 g p occurs in the factorization of xA with nonnegative exponent. The same holds for x = f : if p occurs in the factorization of x−1A, it occurs with nonnegative exponent. Hence the exponent must be zero, so x has trivial prime factorization and thus x ∈ A×. L × × × For exactness at p6∈S K /Ap , take x ∈ AS and suppose that Y xA = pvq p∈Spec A L for vq ≥ 0. Then x maps to (vp)p6∈S in p6∈S Z, and all primes p ⊂ A with vp 6= 0 lie outside S by the previous paragraph. Thus Y xA = pvq p6∈S

Q vp so (vp)p6∈S maps to [ p ] = [xA] = 1 in the class group of A. This proves the sequence is a L × × complex at p6∈S K /Ap .

36 2.2 The p-adic Numbers 2 Local Fields

Q vp × Conversely, if p6∈S p = xA is principal, we just need to show that x ∈ AS . We know that x ∈ K×. Further, observe that for any p 6∈ S, all elements of p and of p−1 lie in −1 −1 S AS: if y ∈ p then pp = A allows us to write xy ∈ A for some z ∈ p r q∈S q so that zy Q vp × y = z ∈ AS. Therefore any element of p6∈S p lies in AS, so x ∈ AS . This proves exactness at the middle term. Finally, the sequence is a complex at CA because for any p 6∈ S,[pAS] = 1. On the other × hand, suppose I is a fractional ideal of A such that IAS = xAS for some x ∈ K ; without Q vp loss of generality we may assume IAS = AS. Write I = p p . Notice that if vq > 0 for any q ∈ S, then Y vp IAS = (pAS) ⊆ qAS 6= AS, p contradicting IAS = AS. Therefore none of the p in the factorization of IAS lie in S, so IAS L lies in the image of p6∈S Z. Hence the entire sequence is exact. Corollary 2.1.10. Let A be a Dedekind domain and S ⊆ Spec A a set of primes excluding

finitely many primes of A. Then if the class number |CA| is finite, so is |CAS |. × Definition. For a number field K and a cofinite set of primes S of OK , OK,S is called the

group of S-units of K and CK,S = COK,S the S-class group of K.

Corollary 2.1.11 (Dirichlet’s S-Unit Theorem). If A = OK is the ring of integers in an algebraic number field of degree n = r + 2s, then × ∼ r+s−1+N OK,S = Z × µ(K) where N is the finite number of primes excluded from S.

× Proof. By the ordinary unit theorem (1.9.2), it is enough to show that the rank of OK,S is r + s − 1 + N. By Corollaries 1.8.7 and 2.1.10, CK and CK,S are finite so taking the × alternating sum of ranks on the exact sequence in Proposition 2.1.9, we get rank(OK,S) = × rank(OK ) + N. Corollary 2.1.12. For any number field K, there exists a cofinite set S of prime ideals of OK such that CK,S = 1.

Proof. Let a1,..., am be representatives of the class group CK and take T to be the set of all L prime divisors of any aj. Then S = Spec(OK )rT is the desired set: the map p6∈S Z → CK in Proposition 2.1.9 is surjective, so by exactness, CK,S = 1.

2.2 The p-adic Numbers

In this section we define and explore some basic properties of the p-adic numbers, first dis- covered by Kurt Hensel. His original inspiration for defining such numbers was the ubiquity of power series expansions in analysis and their potential utility in number theory. Let K be a field and take some polynomial f(x) ∈ K[x]. Given a ∈ K, we can write

n X i f(x) = ai(x − a) for some ai ∈ K. i=0

37 2.2 The p-adic Numbers 2 Local Fields

(i) Observe that the coefficients ai are related to derivatives f (a), as in Taylor’s theorem. If g(x) instead we have a rational function f(x) = h(x) ∈ K[x](x−a) for g, h ∈ K[x] where h(a) 6= 0, then we can still write a formal power series expansion of f(x) about x = a:

∞ f(x) X ≈ a (x − a)i for a ∈ K. g(x) i i i=0

This is the beginning of a fruitful dictionary between the integers Z and polynomial rings over a field: Z K[x] prime ideal (p) maximal ideal (x − a) reduction of a mod p evaluation f(a) reduction of a mod pn+1 nth derivative f (n)(a)

Running with this idea, given a positive integer x ∈ Z, we can write

n X i x = aip for ai ∈ {0, 1, . . . , p − 1}. i=0

If x ∈ Z(p), the localization at (p) (see Example 2.1.2), then we would like to write a formal P∞ i power series i=0 aip with ai ∈ {0, 1, . . . , p − 1} that represents x. Example 2.2.1. Take p = 5 and x = 233. Then the 5-adic expansion gives a “power series” for 233: 233 = 3 · 1 + 1 · 5 + 4 · 52 + 1 · 53 + 0 · 54 + ...

P∞ i Definition. For a prime p, a p-adic integer is a formal infinite sum i=0 aip for ai ∈ {0, 1, . . . , p − 1}. The set of all p-adic integers is denoted Zp. Notice that every p-adic integer has a well-defined residue class modulo pn for each n ≥ 0. On the other hand, every element of the local ring Z(p) has a well-defined residue class mod n P∞ i p . For x ∈ Z(p), we will write x = i=0 aip if both of these objects have the same residue n mod p for all n ≥ 0. In other words, we have a map Z(p) → Zp. To see that the map is P∞ i n injective, suppose x, y ∈ Z(p) with x = i=0 aip = y. Then x − y ≡ 0 (mod p ) for all n ≥ 0, so we must have x = y.

Example 2.2.2. Beware that these “p-power series” expansions do not always behave as they do in the analytic case. For example, take x = −1. Then for each n ≥ 0,

n−1 X (p − 1)pi = pn − 1 ≡ −1 (mod pn). i=0

P∞ i Thus −1 has p-adic expansion i=0(p − 1)p for any prime p. When p = 2, this gives the famous “identity” −1 = 1 + 2 + 4 + 8 + 16 + ...

38 2.2 The p-adic Numbers 2 Local Fields

In ordinary integers, such a sum does not converge, but in 2-adic land it does! Alternatively, the power series 1 = 1 + x + x2 + x3 + ... 1 − x does not converge for x = 2, but it does converge in 2-adic numbers! In general, the above shows that 1 = 1 + p + p2 + p3 + ... 1 − p

is valid in Zp.

In the polynomial ring case, we have strict containments of rings K[x] ( K[x](x−a) ( K[[x − a]]. Similarly, we have containments of sets Z(Z(p) (Zp for any prime p. Our next goal is to give Zp the structure of a ring. Informally, we can think of a p-adic integer as a sequence of residue classes in Z/pZ, Z/p2Z, Z/p3Z,... which are compatible with the sequence of homomorphisms

λ3 3 λ2 2 λ1 ··· −→ Z/p Z −→ Z/p Z −→ Z/pZ. (In commutative algebra, this system of abelian groups and homomorphisms is called an inverse system and such a sequence of residue classes is called a coherent sequence.) Then 2 3 we can view Zp as a (proper) subset of Z/pZ × Z/p Z × Z/p Z × · · · :

∞ i Zp = {x = (xi)i=1 | xi ∈ Z/p Z and λi(xi+1) = xi for all i ∈ N}.

i In other words, p is an inverse limit, p = lim /p . Z Z ←− Z Z

Lemma 2.2.3. If x = (xi) and y = (yi) are coherent sequences of residue classes in i (Z/p Z)i∈N then so are x + y = (xi + yi) and xy = (xiyi). That is, Zp is a subring of Q∞ i i=1 Z/p Z.

Further, Zp is the completion of the DVR Z(p) with respect to a certain metric topology called the p-adic topology, which we will discuss further in Section 2.3. One important fact is that Z(p) is a dense subring of Zp. Lemma 2.2.4. Let p be prime. Then

(1) The image of (p)Z(p) in Zp is a maximal ideal, also denoted by (p).

n n+1 (2) Zp is a DVR with discrete valuation vp(x) = n if x ∈ (p ) but x 6∈ (p ).

h 1 i (3) The field of fractions of Zp is Zp p .

Definition. The field of fractions of Zp is called the field of p-adic numbers, written Qp.

−m By definition any element of Qp can be written as p x for some x ∈ Zp and m ≥ 0:

m m m X 1 X X b = b pm−ip−m = p−m b pm−i. i pi i i i=0 i=1 i=0

39 2.2 The p-adic Numbers 2 Local Fields

−m −r −m m−r Addition in Qp is given by p x + p y = p (x + p y) if m ≥ r, while multiplication is −m −r −(m+r) simply (p x)(p y) = p xy. Note that Qp is a field of characteristic 0, so it contains Q as a subfield. More formally, there is a canonical embedding Q ,→ Qp making the following diagram commute:

Q Qp

Z(p) Zp

P∞ i Concretely, elements of Qp may be thought of as p-adic Laurent series i=−m aip with ai ∈ {0, 1, . . . , p − 1}. By analogy, the field of fractions of K[[x − a]] is K((x − a)), the field of Laurent series over K.

Definition. For every prime integer p ∈ Z, the p-adic valuation on Q is the valuation m a vp : Q → Z ∪ {∞} defined by vp(x) = m if x = p b for a, b ∈ Z with p - ab, and vp(0) = ∞. Definition. A valuation v on a ring A is called nonarchimedean if for every x, y ∈ A, v(x + y) ≥ min{v(x), v(y)} with equality if and only if v(x) 6= v(y).

Lemma 2.2.5. Every p-adic valuation on Q is nonarchimedean.

Definition. For a prime p, the (normalized) p-adic absolute value on Q is defined by −vp(x) |x|p = p for x 6= 0 and |0|p = 0.

Lemma 2.2.6. The p-adic absolute value is a norm on Q for all primes p.

Thus every p-adic valuation gives rise to a metric topology on Q: dp(x, y) = |x − y|p. This topology is called the p-adic topology on Q. For the standard absolute value inducing the (Euclidean) metric topology on Q, we will write | · |∞.

Lemma 2.2.7 (Product Formula). Let x ∈ Q be nonzero. Then Y |x|p = 1, p

where the product is over all primes p plus the “infinite prime” p = ∞.

Proof. Since norms are multiplicative, it’s enough to check the product formula when x is prime and x = −1. When x = −1, | − 1|p = 1 for all primes p and | − 1|∞ = 1 so the product formula holds trivially. If x = q is prime, we have  q, p = ∞  1 |q|p = q , p = q 1, p 6= q, ∞.

Thus the product formula holds in this case as well.

40 2.2 The p-adic Numbers 2 Local Fields

The following lemma demonstrates one of the curious aspects of topologies defined by nonarchimedean absolute values.

Lemma 2.2.8. For p prime and a ∈ Q, define the p-adic ball around a of radius r:

Bp(a, r) = {c ∈ Q : |c − a|p < r}.

Then every point b ∈ Bp(a, r) is in fact the center of the ball. The same holds for any closed ball Bp(a, r).

Proof. Suppose c ∈ Bp(a, r) is any other point in the ball, so that |a − c|p < r. Since b ∈ Bp(a, r), we have

|b − c|p = |b − a + a − c|p ≤ max{|b − a|p, |a − c|p} < r.

Hence c ∈ Bp(b, r), so Bp(a, r) ⊆ Bp(b, r). Reversing the roles of a and b gives Bp(a, r) = Bp(b, r).

It is not hard to show Q is not complete with respect to | · |p for any prime p, and we know from real analysis that | · |∞ does not define a complete topology on Q either. Thus we can complete Q with respect to any of these topologies by constructing the ring of Cauchy sequences and taking the quotient by the ideal of sequences whose limit is 0.

Lemma 2.2.9. The completion of Q with respect to any valuation | · |p, for p prime or p = ∞, is a topological field. Moreover, this completion is precisely Qp if p is prime and R if p = ∞. Finally, when p is prime, Zp = {x ∈ Qp : |x|p ≤ 1}. Proof. (Sketch) The p = ∞ case is dealt with in a basic real analysis course, so assume p is P∞ i a finite prime. We may identify any p-adic number i=−m aip with the Cauchy sequence (sn) defined by n X i sn = aip ∈ Q. i=−m On the other hand, for any n, any Cauchy sequence is eventually constant mod pn. Thus we may associate such a sequence (sn) to a sum

n−1 X i aip i=−m

P∞ i for each n ∈ N. Given this identification, we can treat i=−m aip as a convergent power series in Qp. We know that

∞ X i m aip = p i=−m p by the ultrametric property, so ∞ X i y = aip ∈ {x ∈ Qp : |x|p ≤ 1} ⇐⇒ m ≤ 0 ⇐⇒ y ∈ Zp. i=−m Therefore the p-adic integers are as described.

41 2.3 Absolute Values 2 Local Fields

We now have three different interpretations of the field of p-adic numbers Qp: ˆ Formal power series (an analytic interpretation); ˆ The fraction field of Zp (an algebraic interpretation); ˆ The completion of Q with respect to a norm | · |p (a topological interpretation).

Proposition 2.2.10. For any prime p, Zp is the closure of Z in Qp. P∞ i Proof. If x ∈ Zp, write x = i=0 aip . Then x is the convergent limit of the sequence Pn i sn = i=0 ai ∈ Z. On the other hand, if x 6∈ Zp then |x|p > 1 but no sequence (yn) ⊆ Z can converge to x because |yn|p ≤ 1 for all n. Therefore Zp = Z. × Notice that Zp = {x ∈ Zp : |x|p = 1}. This description of units will become useful in later results. ∼ Theorem 2.2.11. For any prime p, Zp = Z[[x]]/(x − p) as rings. Proof. Consider the map

ϕ : Z[[x]] −→ Zp ∞ ∞ X i X i aix 7−→ aip , i=0 i=0 where the power series on the right is treated as a convergent power series per previous remarks. Clearly ϕ is surjective by the definition of Zp. Moreover, it is a ring homomorphism P∞ i Pn i by construction and (x−p) ⊆ ker ϕ. If y ∈ ker ϕ, then y = i=0 aix such that i=0 aip ≡ 0 n+1 1 n (mod p ) for all n ≥ 0. For each n, let bn = − pn+1 (a0 + a1p + ... + anp ). Then

2 2 (b0 + b1x + b2x + ...)(x − p) = (a0 + a1p + a2p + ...)

so y ∈ (x − p) and hence ker ϕ = (x − p). Now apply the first isomorphism theorem.

2.3 Absolute Values

In this section we generalize the notion of the p-adic valuation, absolute value and metric topology to any field K.

Definition. Let K be a field. An absolute value on K is a function | · | : K → R such that (1) |x| ≥ 0 for all x ∈ K, with |x| = 0 if and only if x = 0.

(2) |xy| = |x| |y| for all x, y ∈ K.

(3) |x + y| ≤ |x| + |y| for all x, y ∈ K.

Remark. Axiom (3) implies that |ζ| = 1 for any root of unity ζ ∈ K such that ζn = 1.

42 2.3 Absolute Values 2 Local Fields

Definition. An absolute value | · | : K → R≥0 is called nonarchimedean if |x + y| ≤ max{|x|, |y|} for any x, y ∈ K. Otherwise | · | is called archimedean. Example 2.3.1. The trivial absolute value is defined for any field K: ( 1, x 6= 0 |x|0 = 0, x = 0.

Example 2.3.2. The standard absolute value ( x, x ≥ 0 |x| = −x, x < 0 is an archimedean absolute value on Q. Example 2.3.3. For any prime number p ∈ Z, the p-adic absolute value defined in Sec- tion 2.2 is a nonarchimedean absolute value on Q. The following result establishes an easy condition to check for when an absolute value is nonarchimedean.

Lemma 2.3.4. An absolute value | · | : K → R≥0 is nonarchimedean if and only if |x| ≤ 1 for all x ∈ {n1K : n ∈ Z}. Proof. ( =⇒ ) is immediate from the definition of nonarchimedean. ( ⇒ = ) Suppose |x| ≥ |y| for x, y ∈ K. Then |x|ν|y|n−ν ≤ |x|n for any 0 ≤ ν ≤ n so we have

|x + y|n = |(x + y)n| n X n = xνyn−ν by the binomial theorem ν ν=0 n   X n ν n−ν ≤ |x| |y| by the triangle inequality ν ν=0 n X n ≤ |x|n since ∈ ν Z ν=0 = (n + 1)|x|n. √ So |x + y| ≤ n n + 1|x|. Taking n → ∞,(n + 1)1/n approaches 1 so we get |x + y| ≤ |x|. Hence | · | is nonarchimedean. Corollary 2.3.5. If char K = p > 0, then every absolute value on K is nonarchimedean.

Definition. Two absolute values | · |1 and | · |2 on K are said to be equivalent, written | · |1 ∼ | · |2, if they induce the same metric topology on K, i.e. if there are constants r, s > 0 such that for every x, y ∈ K,

r s |x − y|2 ≤ |x − y|1 and |x − y|1 ≤ |x − y|2.

43 2.3 Absolute Values 2 Local Fields

Proposition 2.3.6. If | · |1 and | · |2 are two nontrivial, equivalent absolute values on K then s there exists a constant s > 0 such that |x|1 = |x|2 for all x ∈ K.

n n Proof. Notice that if | · |1 ∼ | · |2 then x → 0 in | · |1 if and only if x → 0 in | · |2. This × implies that |x|1 < 1 if and only if |x|2 < 1. Now let y ∈ K satisfy |y|1 > 1 and take x ∈ K α so that |x|1 = |y|1 for α ∈ R. If mi, ni ∈ Z are sequences of integers such that each ni > 0 mi mi α mi/ni and converges from above to α but 6= α for any i, then |x|1 = |y| < |y| for all i. ni ni 1 1 Thus ni ni x x mi/ni < 1 =⇒ < 1 =⇒ |x|2 < |y| . mi mi 2 y 1 y 2

mi α mi Taking i → ∞ so that → α, we get |x|2 ≤ |y| . If we take such a sequence converging ni 2 ni α α to α from below, we get |x|2 ≥ |y|2 , so |x|2 = |y|2 . Thus log |x| log |y| 1 = 1 for all x ∈ K×. log |x|2 log |y|2

log |x|1 This shows that the function s = is a constant function. Hence it follows that |x|1 = log |x|2 s |x|2 for all x ∈ K. Corollary 2.3.7. Each equivalent class of absolute values on a field K is characterized uniquely by the set {x ∈ K : |x| < 1} for any | · | in the class.

Theorem 2.3.8 (Ostrowski). Every nontrivial absolute value | · | on Q is equivalent to | · |p for some prime p if | · | is nonarchimedean and | · |∞ if | · | is archimedean.

Proof. First suppose | · | : Q → R≥0 is nonarchimedean. Let p ∈ N be minimal such that |p| < 1, which exists since | · | is nontrivial and multiplicative; the latter even implies p can be chosen prime. Set I = {x ∈ Z : |x| < 1}. Then I is an ideal of Z by the nonarchimedean property and Lemma 2.3.4. We certainly have I ⊇ (p) but since (p) is a maximal ideal, we must have I = (p). Thus if a ∈ Z and p - a, |a| = 1. So for any m ∈ Z such that p - m, we have |pnm| = |p|n|m| = |p|n.

s s  1  This shows that |·| = |·|p where s is the unique positive number satisfying |p| = p . Thus all nonarchimedean absolute values on Q are equivalent to a p-adic absolute value. (We call the absolute value with s = 1 above the normalized p-adic absolute value, as in Section 2.2.) Now assume |·| is archimedean. Suppose that for all m, n ∈ Z with m, n > 1, the absolute value satisfies the following property: |m|1/ log m = |n|1/ log n (∗). Then for s > 0 such that es = |n|1/ log n (for any n > 1), we have

log m |m| = |n|1/ log n
= es log m = ms = |m|s.

s Therefore |m| = |m|∞ and this holds for all m ∈ Q by multiplicativity. Thus it suffices to check that any archimedean absolute value satisfies property (∗).

44 2.3 Absolute Values 2 Local Fields

Fix m, n ∈ Z with m, n > 1 and write m in base n: r m = a0 + a1n + ... + arn for 0 ≤ ai < n. log m Note that r ≤ log n . Then r |m| = |a0 + a1n + ... + arn | r X i ≤ |ai| |n| by the triangle inequality i=0  log m ≤ 1 + |n| · |n|log m/ log n log n  log m = 1 + |n|1+log m/ log n. log n Replacing m with mk for k > 1, we get  k log m |m|k ≤ 1 + n1+k log m/ log n log n  k log m1/k =⇒ |m| ≤ 1 + |n|1/k+log m/ log n. log n Letting k → ∞, we then obtain |m| ≤ |n|log m/ log n, or |m|1 log m ≤ |n|1/ log n. Reversing the roles of m and n gives the other inequality, establishing property (∗) and completing the proof. The following theorem may be seen as a certain generalization of the Chinese remainder theorem.

Theorem 2.3.9 (Weak Approximation). Suppose | · |1,..., | · |n are inequivalent absolute values on K and choose a1, . . . , an ∈ K. Then for all ε > 0, there exists an x ∈ K such that |x − ai|i < ε.

Proof. For n = 1 this is trivial, so assume n ≥ 2. Since | · |1 and | · |n are not equivalent, we know there exists α ∈ K such that |α|1 < 1 but |α|n ≥ 1. Likewise, there exists β ∈ K β such that |β|1 ≥ 1 and |β|n < 1. Let y = α so that |y|1 > 1 and |y|n < 1. We will show that there exists some z ∈ K such that |z|1 > 1 but |z|j < 1 for all 2 ≤ j ≤ n. The base case of this statement was just proven, so to induct, pick z ∈ K such that |z|1 > 1 and |z|j < 1 m for 2 ≤ j ≤ n − 1. If |z|n < 1 then we are done. If |z|n = 1 then z y will work for some zm sufficiently large m. Finally, if |z|n > 1 then let tm = 1+zm so that as m → ∞, |tm|1 → 1, |tm|n → 1 and |tm|j → 0 for all 2 ≤ j ≤ n − 1. Then tmy will work for sufficiently large m. Now given z ∈ K such that |z|1 > 1 and |z|j < 1 for 2 ≤ j ≤ n, consider the same zm sequence tm = 1+zm . As m → ∞, we have m z 1 |tm|1 = m = 1 − m −→ 1 1 + z 1 1 + z 1 m z m |tm|j = m ≤ |z|j −→ 0 for all 2 ≤ j ≤ n. 1 + z j

45 2.3 Absolute Values 2 Local Fields

Therefore one can find z1 such that |z1 − 1|1 < ε and |z1|j < ε for 2 ≤ j ≤ n. Repeat the process to pick z2, . . . , zn with |zj − 1|j < ε and |zj|` < ε for ` 6= j. Then setting x = a1z1 + . . . anzn gives an element satisfying the desired norm conditions. There exists a generalization, naturally called the strong approximation theorem, which we won’t prove. Theorem 2.3.10 (Strong Approximation). Let S be a set of equivalence classes of absolute valuations on a field K such that S does not contain at least one absolute value on K. Then for any nonequivalent | · |1,..., | · |n ∈ S, elements a1, . . . , an ∈ K and ε > 0, there exists an x ∈ K such that |x − ai|i < ε for each 1 ≤ i ≤ n and |x| < 1 for all | · | ∈ S r {| · |1,..., | · |n}. Proposition 2.3.11. The only fields that are complete with respect to an archimedean ab- solute value are (R, | · |∞) and (C, | · |∞). We now connect the theory of nonarchimedean absolute values with discrete valuations on K (Section 2.1). Proposition 2.3.12. Given a nonarchimedean absolute value | · | on K, setting v(x) = − log |x| for all x ∈ K× and v(0) = ∞ defines a discrete valuation v : K → R ∪ {∞}. Proof. For all x, y ∈ K, we have |xy| = |x| |y| which implies v(xy) = v(x) + v(y). Likewise, |x + y| = max{|x|, |y|} implies v(x + y) ≥ min{v(x), v(y)}. Definition. For a nonarchimedean absolute value | · | on a field K, define O := {x ∈ K× | v(x) ≥ 0} ∪ {0} = {x ∈ K× : |x| ≤ 1} ∪ {0} O× := {x ∈ K | v(x) = 0} = {x ∈ K : |x| = 1} m := {x ∈ K | v(x) > 0} = {x ∈ K : |x| < 1} κ := O/m, called respectively the valuation ring, group of units, valuation ideal and residue field of | · |. Example 2.3.13. The analogy between p-adic numbers and power series is borne out by these concepts:

(K, | · |) (Qp, | · |p) (C((t)), | · |t O Zp C[[t]] × × × O Zp C[[t]] m pZp (t) κ Fp C Definition. If K is a field with a nonarchimedean absolute value and associated discrete valuation, we will call the triple (K, | · |, v) a discretely valued field. If (K, | · |, v) is a discretely valued field, then we have filtrations O ⊃ m ⊇ m2 ⊇ m3 ⊇ · · · (of ideals) O× ⊇ U (1) ⊇ U (2) ⊇ U (3) ⊇ · · · (of subgroups) where U (n) = {x ∈ O× | x ≡ 1 mod mn} = {x ∈ O× | v(x) ≥ n}.

46 2.3 Absolute Values 2 Local Fields

Proposition 2.3.14. Let (K, | · |, v) be discretely valued. Then for any n, (1) O×/U (n) ∼= (O/mn)×. (2) U (n)/U (n+1) ∼= O/m = κ. Proof. (1) It is clear that the natural map O× → (O/mn)× is surjective with kernel U (n). (2) Pick a generator π of m. Then the map U (n) −→ O/m 1 + πna 7−→ a mod m

is surjective with kernel U (n+1).

If v is a discrete valuation on K, we can form the completion Kb of K with respect to the absolute value | · | = | · |v. Similar to Lemma 2.2.9, we have: Lemma 2.3.15. For any valuation v on K,

(a) The completion Kb with respect to | · | is a field.

(b) | · | extends uniquely to an absolute value on Kb.

(c) K embeds as a dense subset of Kb. We will also denote by | · | the unique extension of | · | to Kb. Define the completions of the valuation ring and valuation ideal of | · | in Kb:

Ob = {x ∈ Kb × : |x| ≤ 1} ∪ {0} × mb = {x ∈ Kb : |x| < 1}.

Lemma 2.3.16. For any absolute value | · | on K, Ob/mb = O/m. Let R ⊆ O be a system of representatives of O/m such that 0 is one of the representatives. Then all elements of Kb can be written uniquely in the form

m 2 π (a0 + a1π + a2π + ...)

with all ai ∈ R and m ≤ 0. This generalizes the construction of Qp in Section 2.2. Example 2.3.17. If K = k(t), there is an absolute value | · | on K defined by |f| = e−m m a where f = t b for t - a, b. Then Kb = k((t)) and Ob = k[[t]]. Thus it is natural to view completions of discretely valued fields as “power series in π”, justifying in particular the analogy in Example 2.3.13. Proposition 2.3.18. For any discretely valued field (K, | · |, v), the completions of the val- uation ring and group of units are inverse limits:

Ob = lim O/mn ←− Ob× = lim(O/mn)× = lim O×/U (n). ←− ←−

47 2.3 Absolute Values 2 Local Fields

For the rest of the section, assume K is a field which is complete with respect to a discrete, nonarchimedean absolute value | · |.

Theorem 2.3.19 (Hensel’s Lemma). Suppose f(x) ∈ O[x] is a monic polynomial of degree n and f¯(x) ∈ κ[x] admits a factorization

f¯(x) =g ¯(x)h¯(x)

for g,¯ h¯ relatively prime, monic polynomials over κ of degrees r and n − r, respectively. Then

f(x) = g(x)h(x)

for g(x), h(x) ∈ O[x] with deg g = r, deg h = n − r, g¯(x) = g(x) mod m and h¯(x) = h(x) mod m.

k Proof. The idea is to find gk, hk ∈ O[x] inductively such that gkhk − f ∈ m for all k ∈ N, ¯ satisfying the conditions deg gk = r, deg hk = n − r, g¯ ≡ gk mod m and h ≡ hk mod m. ¯ For k = 1, let g1 and h1 be any monic lifts ofg, ¯ h to O[x] with the correct degrees. To ¯ induct, assume gk, hk have been constructed. By hypothesis, (¯g) + (h) = (1) in κ[x] so for allq ¯ ∈ κ[x], there exista, ¯ ¯b ∈ κ[x] such thata ¯g¯ + ¯bh¯ =q ¯. If degq ¯ < n, then we can take ¯ k dega ¯ < n − r and deg b < r. Let m = (π) and write gkhk − f = qπ for some q ∈ O[x] with deg q < n. Now leta, ¯ ¯b ∈ κ[x] be as above forq ¯, the reduction of this q mod m. Let a, b ∈ O[x] be lifts ofa, ¯ ¯b with the same degrees and set

k k gk+1 = gk − π b and hk+1 = hk − π a.

Then we have

k k gk+1hk+1 = (gk − π b)(hk − π a) k k 2k = gkhk − π bhk − π agk + π ab k k+1 ≡ gkhk − π (agk + bhk) (mod π ) k k+1 ≡ gkhk − π q (mod π ) ≡ f (mod πk+1) by induction.

Therefore gk+1, hk+1 are constructed. Now note that the coefficients of the sequences (gk) and (hk) form Cauchy sequences in K. Since K is assume to be complete, each sequence of coefficients converges so we can define the pointwise limits g = limk→∞ gk and h = limk→∞ hk which exist in O[x]. It is routine to verify that these g, h are the functions we seek. We recover the following result, which is sometimes known as Hensel’s lemma but is really only a special case.

Corollary 2.3.20. If f(x) ∈ O[x] such that f¯(x) ∈ κ[x] has a simple root in κ then f(x) has a simple root in O.

Proof. Apply Theorem 2.3.19 with r = 1.

48 2.4 Local Fields 2 Local Fields

2 Example 2.3.21. Consider f(x) = x − 14 in Z5. Then the residue field is F5 and x2 − 14 = (x − 2)(x + 2) (mod 5). √ 2 Thus by Hensel’s Lemma, x −14 = (x−α)(x+α) for some α ∈ Z5. In particular, 14 ∈ Z5.

Corollary 2.3.22. For each prime p, all (p − 1)st roots of unity lie in Zp. p−1 Proof. Consider the polynomial f(x) = x − 1. Then f(x) splits completely in Fp and in p−1 particular there are no multiple roots of f. Thus, x −1 splits completely in Zp by Hensel’s Lemma so all (p − 1)st roots are in Zp. Definition. A function f ∈ O[x] is said to be primitive if some coefficient of f is a unit in O. The following version of Hensel’s Lemma will be useful. Theorem 2.3.23 (Hensel’s Lemma II). Suppose f(x) ∈ O[x] is a primitive polynomial such that f¯(x) =g ¯(x)h¯(x) in κ[x], with g,¯ h¯ coprime. Then f(x) = g(x)h(x) in O[x] for g, h ∈ O[x] such that deg g = degg, ¯ deg h = deg h,¯ g ≡ g¯ mod m and h ≡ h¯ mod m.

2 Example 2.3.24. Let K = Q5 and consider the polynomial f(x) = 5x + 8x + 5. Then ¯ f(x) = 8x is a coprime factorization in F5 so there exist g, h ∈ Z5[x], each of degree 1, such that f(x) = g(x)h(x).

Pn i Corollary 2.3.25. If K is a complete nonarchimedean field and f(x) = i=0 aix ∈ K[x] is an irreducible, monic polynomial with a0 ∈ O, then every ai ∈ O.

× Proof. Scale f so that it is primitive in O[x]. Let r be the minimal integer such that ar ∈ O . Then ¯ r n−r f(x) ≡ x (ar + ... + x ) mod m.

If 0 < r < n, this contradicts Theorem 2.3.23 and the irreducibility of f. If r = 0, then a0 is a unit after scaling, or in other words, no scaling took place. Likewise, if r = n, no scaling took place. In all cases, f must be primitive to begin with, so all coefficients lie in O.

2.4 Local Fields

Definition. A local field is a complete, discretely valued field with finite residue field.

Example 2.4.1. For any prime integer p, the p-adic field Qp and the field of Laurent series Fp((t)) are both local fields. Remark. Elsewhere in the literature, it is sometimes required that a discretely valued field has a perfect residue field to be local. Other times, the residue field is allowed to be arbitrary. Many times R and C are included in the definition of local field, as they bear similarities to the prototypical examples of local fields Qp and Fp((t)). Lemma 2.4.2. A field K is a local field if and only if K admits a discrete, nonarchimedean valuation with respect to which K is locally compact.

49 2.4 Local Fields 2 Local Fields

Proof. ( =⇒ ) Since K is a topological field, it’s enough to show that K has a compact open neighborhood of 0. Notice that OK is an open neighborhood of 0. If mK is the maximal ideal of OK , then Proposition 2.3.18 gives us

∼ n OK = lim OK /m ←−

Q n Q n which is closed in the product OK /m . By Tychonoff’s theorem, OK /m is compact Q n and therefore OK ⊂ OK /m is compact. ( ⇒ = ) For exercise. Note that K itself is not compact, as

−1 −2 K = OK ∪ mK OK ∪ mK OK ∪ · · · is an open cover with no finite subcover.

Theorem 2.4.3. Every local field is a finite extension of Qp or Fp((t)) for some prime integer p.

Proof. Let K, OK , mK , πK , κ, v be as usual and let char κ = p. If char K = 0, certainly

K ⊇ Q so p ∈ mK , which means that v|Q must be equal to the p-adic valuation vp on Q. Since K is complete, we have K ⊇ Qp. It will follow from the fundamental equality × × (Proposition 2.6.1) that [K : Qp] = ef, where e = [v(K ): vp(Qp )] < ∞ because K is discretely valued and f = [κ : Fp] < ∞ because κ is a finite field of characteristic p. Therefore K/Qp is a finite extension. On the other hand, suppose char K = p. Then κ = Fp(α) for some α algebraic over Fp which has minimal polynomial f ∈ Fp[t]. Then f is separable because Fp is perfect, so by Hensel’s Lemma (Theorem 2.3.19), f splits completely over K (viewed as a polynomial with coefficients in K ⊇ Fp). Thus κ is isomorphic to a subfield of K; assume κ ⊆ K. Since K P∞ i is complete and discretely valued, all elements of K are of the form i=−N aiπK for ai ∈ κ. ∼ This implies K = κ((πK )) = κ((t)). Finally, since κ/Fp is a finite extension, κ((t))/Fp((t)) is a finite extension and thus so is K/Fp((t)). This completes the proof. Let K be a local field with residue field κ. Then char κ = p > 0 for some prime p. When char K = 0, we call this the mixed characteristic case, whereas char K = p is called the equal characteristic case.

Corollary 2.4.4. The only locally compact fields are R, C and finite extensions of Qp and Fp((t)) for p prime.

Let K be a local field, with OK , mK , πK , κ and v as usual and set q = |OK /mK | = |κ|. We now describe the group structure of K×.

Proposition 2.4.5. For any local field K,

× ∼ (1) K = Z × Z/(q − 1)Z × U

(1) n where U = {1 + x ∈ OK | x ∈ mK }.

50 2.4 Local Fields 2 Local Fields

× n × Proof. If α ∈ K then α = πK u for a unique unit u ∈ OK and n ∈ Z. Now by Corol- × ∼ lary 2.3.22, µq−1 ⊆ K and it is easy then to see that Fq = µq−1. So u factors uniquely as n ∼ u = xv, whereu ¯ = x ∈ µq−1 andv ¯ = 1. Thus α = πK xv uniquely. Identifying hπK i = Z and µq−1 ∈ Z/(q − 1)Z, we get the desired isomorphism. Let K be a characteristic 0 local field, with char κ = p. We next define analogues of the logarithmic and exponential functions for K. Proposition 2.4.6. There exists a unique homomorphism log : K× → K satisfying (1) log(p) = 0. x2 x3 (2) For all 1 + x ∈ U (1), log(1 + x) = x − + − ... 2 3

x2 x3 Proof. If v(x) > 0 then the infinite sum x − 2 + 3 − ... converges so this power series is well-defined on U (1). Note also that if such a log function is defined, it must necessarily satisfy log(ω) = 0 for any root of unity ω, since

0 = log(1) = log(ωn) = n log(ω) =⇒ log(ω) = 0.

e By Proposition 2.4.6, we may write p = πK ω(p)u(p) for unique e ∈ Z, ω(p) ∈ µq−1 and (1) 1 u(p) ∈ U . Define log(πK ) = − e log(u(p)). This is well-defined since the decomposition of p × n is unique. Now for any α ∈ K , use Proposition 2.4.6 to write α = πK ωu for n ∈ Z, ω ∈ µp−1 (1) × and u ∈ U . Extend the definition of log to K by log(x) = n log(πK ) + log(u). This converges and log(p) = 0 by construction. Further, it’s immediate by the definition that log is a homomorphism on hπK i and µq−1. One can check that the power series expansion converges on U (1) by computing valuations. Moreover, by the power series identity,

log((1 + x)(1 + y)) = log(1 + x + y + xy)

for all 1 + x, 1 + y ∈ U (1), so log is indeed a homomorphism. This defines the formal logarithm on K. Next, define the exponential function

x2 x3 exp(x) = 1 + x + + + ... 2! 3!

n e Lemma 2.4.7. exp(x) converges on mK whenever n > p−1 , where e = eK/Qp = v(p).

22 23 Example 2.4.8. In K = Q2, exp(2) = 1 + 2 + 2! + 2! + ... does not converge. This is 1 reflected by the fact that v(2) = 1 6> 2−1 = 1. Proposition 2.4.9. For any local field K of characteristic 0, exp : mn → K× is a homo- (n) e morphism with images in U whenever n > p−1 . Lemma 2.4.10. The functions exp and log are continuous on their domains.

e n (n) (n) n Theorem 2.4.11. When n > p−1 , exp : m → U and log : U → m are inverse isomorphisms of topological groups.

51 2.5 Henselian Fields 2 Local Fields

× ∼ (1) (1) Now K = Z × Z/(q − 1)Z × U (Proposition 2.4.6) and one can show that U is a x (1) Zp-module via the action x · u = u for all u ∈ U and x ∈ Zp. One also computes the (1) (1) torsion part of U to be U ∩ µ∞, where µ∞ is the set of all roots of unity in K. For any (n) (n) n n ≥ 1, the rank of the Zp-submodule U is rankZp U = rankZp m = rankZp OK . Putting everything together, we get:

× ∼ Theorem 2.4.12. If K is a characteristic 0 local field of degree d = [K : Qp], then K = d Z × (K ∩ µ∞) × Zp.

2.5 Henselian Fields

Many useful number theoretic properties of a field may be derived solely from the lifting property in Hensel’s Lemma, so we may weaken the completeness assumptions at the end of Section 2.3 as follows.

Definition. A field K is Henselian if there exists a nonarchimedean absolute value | · | on K with valuation ring O such that Hensel’s Lemma (either Theorem 2.3.19 or 2.3.23) holds for irreducible polynomials in O[x].

Example 2.5.1. By Hensel’s Lemma, complete, discretely valued fields are Henselian.

Suppose (K, | · |, v) is a nonarchimedean field. Taking its completion Kb, we can consider the subextension K ⊆ Kh ⊆ Kb defined by

Kh = {α ∈ Kb | α is separable over K}.

Then v and | · | extend uniquely to Kb (Lemma 2.3.15); denote their restrictions to Kh ⊆ Kb h h also by v and |·|. This makes K into a nonarchimedean field with valuation ring O := OKh . Note that O ⊆ Oh ⊆ Ob. Since the value groups and residue fields of K and Kb are the same (Lemma 2.3.16), the value group and residue field of Oh must coincide with these as well.

Lemma 2.5.2. Kh is Henselian.

Proof. Factoring a monic polynomial f(x) ∈ K[x] can be done over the algebraic closure K of K if it can be done over any extension of K. Thus Hensel’s Lemma holds for K ∩ Kb = Ksep ∩ Kb = Kh.

Definition. For a nonarchimedean field (K, |·|, v), the field Kh ⊆ Kb is called the Henseliza- tion of K.

Theorem 2.5.3. If (K, | · |) is a Henselian field and L/K is an algebraic extension, then there is a unique absolute value | · |L on L extending | · |. Further, if L/K is finite of degree n then q n |x|L = |NL/K (x)|

and L is complete with respect to | · |L if K is complete with respect to | · |.

52 2.5 Henselian Fields 2 Local Fields

np0 Proof. (Sketch) Let |x|L = |NL0/K (x)| for some finite extension L0/K containng x, where n0 = [L0 : K]. One can show that |x|L is independent of the choice of L0, so it’s enough to prove the theorem when L/K itself is finite. We now demonstrate that | · |L is a nonar- chimedean absolute value on L. For any x, y ∈ L, |xy|L = |x|L|y|L follows from multiplicativity of the norm (Lemma 1.2.1). Moreover, |x|L = 0 if and only if NL/K (x) = 0 if and only if x = 0. Finally, for α, β ∈ L with |α| ≤ |β|, we have   α α + 1 ≤ max , 1 = 1 if and only if |x| ≤ 1 implies |x + 1| ≤ 1 for all x ∈ L. β β

Thus it’s enough to show that OL = {x ∈ L : |x|L ≤ 1} is a ring and is the integral closure of O in L. For x ∈ L, we have that

d x is integral over O ⇐⇒ x + ... + a1x + a0 = 0 for an irred. polynomial with ai ∈ O d ⇐⇒ x + ... + a1x + a0 = 0 irred., with ai ∈ K, a0 ∈ O, by Cor. 2.3.25

⇐⇒ NL/K (x) ∈ O

⇐⇒ |NL/K (x)| ≤ 1

⇐⇒ |x|L ≤ 1.

It follows that OL is the integral closure of O in L. Now |x| ≤ 1 ⇐⇒ |x + 1| ≤ 1 for all x ∈ L follows immediately, so | · |L is an absolute value on L. 0 0 0 To prove uniqueness, suppose | · |L also extends | · | to L. Let OL = {x ∈ L : |x|L ≤ 1}. If d x ∈ OL, x 6= 0, then f(x) = 0 for some irreducible, monic polynomial f(t) = t +...+a1t+a0 d 1−d −d with coefficients ai ∈ O. Dividing out by x , we get 1 + ... + a1x + a0x = 0, which can in turn be written −1 1−d −d 1 = −ad−1x − ... − a1x − a0x . 0 0 By the nonarchimedean property, |ai|L ≤ 1 for all i, so if |x|L > 1 then we would have −1 0 0 |x |L < 1 and therefore the above equation would imply |1|L < 1, a contradiction. Thus 0 0 0 |x|L ≤ 1 which means x ∈ OL. It follows that | · |L and | · |L are in fact equivalent, for if not, 0 the weak approximation theorem (2.3.9) would give an element y ∈ L such that |y|L > 1 but |y|L < 1, which we just showed was impossible. Finally, the two absolute values are in fact equal since they agree on K. For the statement about completeness, see Neukirch II.4.9.

Example 2.5.4. Theorem 2.5.3 need not hold if K is not Henselian. For instance, K = Q with the 5-adic absolute value | · | = | · |5 is not Henselian. If L = Q(i) then one can define two distinct absolute values on L: a a |x| = 5−m if x = (1 + 2i)m and |x| = 5−m if x = (1 − 2i)m . 1 b 2 b

Both of these extend | · |5 to L, but they are clearly inequivalent. The converse of Theorem 2.5.3 is true, that is, the property of unique extension of absolute values characterizes Henselian fields. Theorem 2.5.5. Suppose (K, |·|, v) is a nonarchimedean field such that |·| extends uniquely to any algebraic extension L/K. Then K is Henselian.

53 2.6 Ramification Theory 2 Local Fields

Proof. We will prove that K satisfies the first version of Hensel’s Lemma (Theorem 2.3.19) for monic polynomials. Let f ∈ O[x] be monic with nonzero constant term, i.e. f(x) = n a0 + a1x + ... + x . (If a0 = 0, we may divide out by x and apply the proof to the remaining factor.) First, if f is irreducible, let L/K be a splitting field of f. By hypothesis, | · | extends uniquely to L so OL, mL, πL and λ := OL/mL are all defined for this field. Observe that any σ ∈ Gal(L/K) preserve | · |, since otherwise |x|0 = |σ(x)| is a distinct absolute value on L extending | · |. So Gal(L/K) acts on OL, mL and λ. If α ∈ L is a root of f(x), then a0 is a Q power of σ∈Gal(L/K) σ(α) and so

Y µ µ |α0| = |σ(α)| = |α| σ∈Gal(L/K) for some µ. Since |a0| ≤ 1, we must also have |α| ≤ 1, so α ∈ OL. Thus α has an imageα ¯ in λ = OL/mL. Since each σ(α) lies in OL and as σ ranges over Gal(L/K) these constitute all roots of f, all roots of f¯ in λ must be of the formσ ¯(¯α) where σ ∈ Gal(L/K) andσ ¯ is the automorphism in Gal(λ/κ) induced by σ (as in Proposition 1.5.14). Then all roots of f¯ in λ are Galois conjugate in λ/κ. The only possibility is that f¯(x) = ϕ(x)m for some m ∈ N and some irreducible polynomial ϕ ∈ κ[x]. (In fact, it’s not too hard to see that ϕ must be equal to the minimal polynomial ofα ¯ over κ.) Now let f ∈ O[x] be monic but not necessarily irreducible. Write f = f1 ··· fr for monic, ¯ ¯ ¯ irreducible polynomials fj ∈ O[x]. Then f = f1 ··· fr in κ[x] so by the irreducible case above, ¯ ¯ ¯ each fj is a power of an irreducible polynomial. If f =g ¯h is a coprime, monic factorization in κ[x], then Y ¯ ¯ Y ¯ g¯ = fj and h = fj j∈J j6∈J Q Q for some subset J ⊆ {1, . . . , r}. Letting g = j∈J fj and h = j6∈J fj, we get that f = gh in O. So K is Henselian. Corollary 2.5.6. Every algebraic extension of a Henselian field is Henselian. In particular, every finite extension of a Henselian field is also Henselian. Corollary 2.5.7. Let (K, | · |) be a complete nonarchimedean field and L/K an algebraic extension. Then there is a unique absolute value | · |L on L which extends | · | and is of the pn form |x|L = |NL/K (x)| if L/K is finite of degree [L : K] = n. Moreover, L is complete with respect to this | · |L.

2.6 Ramification Theory

Let (K, |·|, v) be a nonarchimedean field and L/K an algebraic extension. Then the extension of absolute values to L induces an extended valuation

× w : L −→ R α 7−→ v(NL/K (α)). Moreover, by Theorem 2.5.3, if K is Henselian then w is the unique such valuation on L extending v.

54 2.6 Ramification Theory 2 Local Fields

Definition. For a Henselian field (K, | · |, v) and an algebraic extension (L, | · |L, w), the × × ramification index is e = eL/K = [w(L ): v(K )] and the inertial degree is f = fL/K = [λ : κ]. Notice that if v is a discrete valuation and w is its extension to L/K, we have

e w(πL) = ew(πL) = v(πK ) = w(πK ), e e so (πL) = (πK ) in OL, i.e. mL = mK OL. In particular, this is consistent with the ramification theory in the global case (`ala Section 1.5; after all, a DVR is a Dedekind domain). In fact, in the local case, it turns out that ramification behavior is much nicer: a prime only ramifies or remains inert, never splits.

Proposition 2.6.1. Let K be Henselian, L/K a finite extension and e = eL/K and f = fL/K the ramification index and inertial degree, respectively. Then [L : K] ≥ ef with equality if and only if v is a discrete valuation and L/K is separable.

Proof. Pick elements ω1, . . . , ωf ∈ OL which reduce modulo mK to a basis of λ/κ. Also pick × × × π0, π1, . . . , πe−1 ∈ L such that w(π0), w(π1), . . . , w(πe−1) are representatives of w(L )/v(K ). It then suffices to prove the products ωiπj are linearly independent over K. Suppose P i,j aijωiπj = 0 where aij ∈ K are not all 0. Collecting the terms of minimal valuation in this sum, it will be enough to show that the sum of these lowest-valuation terms has the same valuation as each individually. Observe that all these terms must share the same index j, because × w(aijωiπj) = w(aij) + w(πj) ≡ w(πj) mod w(K ), so different j correspond to different valuations. Fix this j and consider X aijωjπj i∈I

where I ⊆ {1, . . . , f} corresponds to the subset of terms of minimal valuation. Then w(aij) × is constant over i ∈ I, say w(aij) = a, so aij = εbij for some ε ∈ K and bij satisfying w(bij) = 0. Thus X επj bijωj 6≡ 0 mod mL i∈I sinceω ¯1,..., ω¯f are a basis for λ/κ. So ! X w aijωiπj = w(επj) = w(aij) = a i∈I and the linear independence is proved. j Now assume v is discrete and L/K is separable. Then each πj = πL. Define the OL- submodules

X X j M = OK ωiπj = OK ωiπL i,j i,j X N = OK ωi. i

55 2.6 Ramification Theory 2 Local Fields

e−1 Then M = N + πLN + ... + πL N. We will show M = OL. Write

OL = N + πLOL

= N + πL(N + πLOL)

= N + πL(N + πL(N + πLOL)) 2 e−1 e = N + πLN + πLN + ... + πL N + πLOL after e expansions e = M + πLOL = M + πK OL.

Now OK is a local ring (it’s a DVR) and since L/K is separable, OL is a finitely generated OK -module. Therefore by Nakayama’s Lemma, OL = M. Hence [L : K] = ef. Remark. For complete fields with discrete valuations, the ‘fundamental equality’ in Propo- sition 2.6.1 holds even without the separable assumption.

Let K be a Henselian field with OK , mK , κ and v as usual, and let L/K be an algebraic extension with extensions OL, mL, λ and w of the objects for the corresponding objects for K.

Definition. We say a finite extension L/K is unramified if fL/K = [L : K] and λ/κ is separable. If L/K is infinite, we say the extension is unramified if it is the union of finite unramified extensions. In all other cases L/K is ramified.

Notice that for a finite extension, fL/K = [L : K] implies eL/K = 1. Proposition 2.6.2. Suppose L/K is an unramified extension, K0/K is an algebraic exten- sion and L0 = LK0 is the compositum inside a fixed algebraic closure K/K. Then L0/K0 is an unramified extension.

L L0

ur

K K0 alg.

Proof. We may assume L/K and K0/K are finite. By hypothesis, λ/κ is separable so λ = κ(¯α) for someα ¯ ∈ λ by the primitive element theorem. Liftα ¯ to some α ∈ L. Then

[L : K] = fL/K = [λ : κ] = deg(¯α) ≤ deg(α) ≤ [L : K] implies deg(α) = [L : K], so L = K(α). This means L0 = K0(α). Let g be the minimal polynomial of α over K0 and f be the minimal polynomial of α over K. Since f¯ is separable and g divides f,g ¯ is also separable. Ifg ¯ were reducible, g would be reducible by Hensel’s Lemma (Theorem 2.3.19), but this is impossible since g is a minimal polynomial. Thusg ¯ is 0 0 0 irreducible over κ = OK0 /mK0 and separable. If λ is the residue field of L , then [λ0 : κ0] ≥ degg ¯ = deg g = [L0 : K0]. On the other hand, Proposition 2.6.1 gives us [λ0 : κ0] ≤ [L0 : K0] so we have equality. Further, λ0 is the splitting field over κ0 ofg ¯, so λ0/κ0 is separable and hence L0/K0 is unramified.

56 2.6 Ramification Theory 2 Local Fields

Corollary 2.6.3. Let K be a local field, L, L0 unramified, algebraic extensions of K and LL0 ⊆ K their compositum inside an algebraic closure K. Then LL0/K is unramified.

K

LL0

L ur L0 ur ur K

Proof. Assume all extensions are finite. By Proposition 2.6.2, LL0/L and LL0/L0 are unram- ified. Further, towers of separable extensions are separable and f is multiplicative in towers (Lemma 1.5.12), so it follows that

0 0 fLL0/K = fL/K fLL0/L = [L : K][LL : L] = [LL : K]. Therefore LL0/K is unramified. Corollary 2.6.4. If L/K is an algebraic extension, there exists a maximal unramified sub- field K ⊆ T ⊆ L. Proof. By Corollary 2.6.3, we may take T to be the compositum inside an algebraic closure K/K of all unramified extensions L/K. Definition. The maximal unramified extension of a Henselian field K is the maximal unramified intermediate extension of K/K, denoted Kur. Lemma 2.6.5. For an algebraic extension L/K with maximal unramified subextension K ⊆ T ⊆ L, the residue field τ of T is equal to the separable closure of κ in λ. Proof. Let κsep be the separable closure of κ in λ and let τ be the residue field of T . Clearly τ ⊆ κsep ∩ λ. On the other hand, givenα ¯ ∈ κsep ∩ λ with minimal polynomial f¯ over κ, we know f¯ is separable. Lift f¯ to a monic polynomial f in L[x]. By Hensel’s Lemma (Theorem 2.3.19), f has a root α ∈ L liftingα ¯. Then K(α)/K is unramified since

[K(α): K] ≤ deg f = deg f¯ = [κ(¯α): κ]

and κ(¯α)/κ is separable. Hence K(α) ⊆ T , soα ¯ ∈ τ. Corollary 2.6.6. For any Henselian field K with residue field κ, Kur ∼= κsep. Definition. Let K be Henselian, char κ = p and L/K an algebraic extension. If L/K is finite, the extension is called tamely ramified if λ/κ is separable and p - [L : T ], where T is the maximal unramified subextension of L/K. If L/K is infinite, we say it is tamely ramified if every finite subextension T ⊆ M ⊆ L is tamely ramified.

57 2.6 Ramification Theory 2 Local Fields

If K is any discretely valued field of characteristic 0 with perfect residue field κ of char- acteristic 0, then saying L/K is tamely ramified is equivalent to saying p - eL/K .

Lemma 2.6.7. If L/K is a tame extension and eL/K = fL/K = 1, then L = K.

Proof. Suppose α ∈ L r K. Let m = deg(α) and note that p - m because L/K is tame. Set 1 β = α − m TrL/K (α). Then 1 Tr(β) = Tr(α) − m Tr(α) = 0. m

× Since eL/K = 1, there exists b ∈ K with v(b) = w(β). Set ε = β/b. Thus Tr(ε) = 0 = w(ε).

Further, fL/K = 1 implies TrL/K (ε) = mε¯ because all conjugates of ε in a normal closure of L/K have the same image in λ = κ. But Tr(ε) = 0 implies mε¯ = 0, but this contradicts w(ε) = 0. Hence L = K as claimed. We have the following characterization of tame extensions (tamely ramified extensions) of a Henselian field.

Theorem 2.6.8. Suppose L/K is a finite extension, with maximal unramified subfield T . Then L/K is tame if and only if L/T is generated by prime-to-p roots of elements of T .

Proof. (Sketch) By definition of T , L/K is tamely ramified if and only if L/T is tamely ramified so we may assume K = T . ( ⇒ √= ) Adjoining one prime-to-p root at a time and applying induction, we may assume m × L = K( a) for a ∈ K and p - m. If m - v(a) in v(K ), then eL/K = m so [L : K] = m. Since p - m, this means fL/K = 1 so L/K is tame. On the other hand, if m | v(a) then we can multiply a by√ an mth power of an element of K to get v(a) = 0. Thena ¯ is an mth power m × m in κ, or else κ( a¯) is an inseparable extension of κ, contradicting K = √T . Buta ¯ ∈ (κ ) implies a ∈ (κ×)m by Hensel’s Lemma (Corollary 2.3.20). Hence L = K( m a) = K, so in all cases L/K is tame. ( =⇒ ) Suppose L/K is tame and set n = [L : K]. Then p - n. Since for any α ∈ L, 1 × × w(α) = n v(NL/K (α)) by Theorem 2.5.3, we have p - [w(L ): v(K )] = eL/K . Pick γ ∈ L such that w(γ) 6∈ v(K×). (If w(L×) = v(K×), skip this step.) Let m be the order of w(γ in w(L×)/v(K×). Then p - m so we can write γm = cε for c ∈ K and ε ∈ L such that w(ε) = 0. Since λ = κ, we can assumeε ¯ = 1 in λ. By Hensel’s Lemma (Theorem 2.3.19), ε is then an mth power in L; write ε = (ε0)m for ε0 ∈ L. Hence γ
m = c ∈ K×. Now √ ε0 γ
m × × replace K with K ε0 = K( c) and repeat the procedure until w(L ) = v(K ). This shows eL/K = 1 = fL/K so L = K by Lemma 2.6.7 and we are done. Corollary 2.6.9. The fundamental equality [L : K] = ef holds for all finite tame extensions L/K.

Corollary 2.6.10. Given a tame extension L/K and algebraic extension K0/K and their compositum L0 = LK0 ⊆ K, L0/K is also tame.

58 2.6 Ramification Theory 2 Local Fields

L L0

tame

K K0 alg.

Proof. By Corollary 2.6.4, there is a maximal unramified subfield K ⊆ T ⊆ L. Then by Proposition 2.6.2, TK0/K0 is also unramified. Let T 0 be the maximal unramified subfield of the extension L0/K0, so that we have the following diagram of fields L L0

T 0

T TK0

ur ur

K K0 alg.

By Theorem 2.6.8, L/T is generated by mth roots, so L0/T K0 is generated by mth roots and in turn L0/T 0 is generated by mth roots. This proves, once again by Theorem 2.6.8, that L0/K0 is tame.

Corollary 2.6.11. Let L, L0 be two tamely ramified, algebraic extensions of K. Then their compositum LL0 ⊆ K is tamely ramified.

Proof. Same as the proof of Corollary 2.6.3.

Corollary 2.6.12. If L/K is an algebraic extension, there exists a maximal tamely ramified subfield K ⊆ V ⊆ L.

Definition. The maximal tame extension of a Henselian field K is the maximal tamely ramified extension of K/K, denoted Ktame.

In analogy with the decomposition/inertia field tower in the global case (Proposition 1.5.15), we have the following tower of Henselian fields, along with corresponding residue fields and value groups.

59 2.6 Ramification Theory 2 Local Fields

L λ w(L×)

V ν = κsep ∩ λ w(V ×) = w(L×)(p)

T τ = κsep ∩ λ w(T ×)

K κ v(K×)

Definition. Let L/K be an algebraic extension of Henselian fields with maximal unramified and maximal tame extensions K ⊆ T ⊆ V ⊆ L. We say L/K is totally ramified if T = K and wildly ramified if V 6= L.

a Remark. When L/K is a finite extension, we can write eL/K = p e for some p - e, which is in fact the ramification indices of V/K and V/T : eV/K = e = eV/T . Therefore [V : T ] = e.

Example 2.6.13. Let K be a local field and consider the cyclotomic extension K(ζn)/K for ζn a primitive nth root of unity. By Theorem 2.4.3, K is a finite extension of either Qp or Fp((t)) for some prime p. Suppose that p - n; set κ = Fq where p | q. If f = ordn q, i.e. f q ≡ 1 (mod n), then we will show K(ζn)/K is uramified of degree f. Note that Fqf /Fq is the smallest extension of Fq containing an nth root of unity. Let g(x) be the minimal polynomial of ζn over K. Then g is separable andg ¯ is irreducible in Fq[x] – if not, g has multiple roots, but all nth roots of unity have distinct reductions in Fqf , so this is impossible. Thus degg ¯ = f so deg g = f and hence K(ζn)/K is unramified of degree f.

Lemma 2.6.14. For any n ≥ 1, OK(ζn) = OK [ζn].

Proof. Let L = K(ζn). Then OL = OK [ζn] + mLOL but since OL and OK are local rings, Nakayama’s Lemma implies OL = OK [ζn]. (Compare this to the global case in Corollary 1.3.14.) m Now suppose p | n. To simplify things, we will assume now that K = Qp and n = p for some m ≥ 1.

∼ m × Lemma 2.6.15. The extension Qp(ζn)/Qp is totally ramified, with Gal(Qp(ζn)/Qp) = (Z/p Z) ,

OQp(ζn) = Zp[ζn] and mQp(ζn) = (1 − ζn), where |N(1 − ζn)| = p. Proof. Let

(x + 1)n − 1 (x + 1)pm − 1 h(x) = = (x + 1)n/p − 1 (x + 1)pm−1 − 1 = 1 + (x + 1)pm−1 + ... + (x + 1)(p−1)pm−1

60 2.7 Extensions of Valuations 2 Local Fields

be the minimal polynomial of 1 − ζn over Qp. Then h(x) is an Eisenstein polynomial whose constant coefficient is p. Thus h(x) is irreducible, so h(x) = 1 + (x + 1)pm−1 + ... + (x + 1)(p−1)pm−1 = 1 + (xpm−1 + 1) + (xpm−1 + 1)2 + ... + (xpm−1 + 1)p−1 + A where A is divisible by p = x(p−1)pm−1 + p + A0 where A0 is divisible by p.

m × m m−1 This implies Gal(Qp(ζn)/Qp) ,→ (Z/p Z) but both groups have order ϕ(p ) = (p−1)p , so the map is an isomorphism. Next, 1−ζn is a prime element of Qp(ζn), so it is a uniformizer. Moreover, Y N(1 − ζn) = (1 − σ(ζn)) = h(1) = ±p. σ∈(Z/pmZ)×

Let w be the unique extension of v = vp from Qp to Qp(ζn). Then 1 1 1 1 w(1 − ζn) = v(N(1 − ζn)) = · v(p) = = . ϕ(n) ϕ(n) ϕ(n) [Qp(ζn): Qp]

It follows that eQp(ζn)/Qp = [Qp(ζn): Qp] so this extension is totally ramified. m 0 0 For the general case, let n = p n where p - n . Then we still have OQp(ζn) = Zp[ζn] by Lemma 2.6.14, and the following tower gives the full ramification theory for Qp(ζn)/Qp:

L = Qp(ζn)

V = Qp(ζpn0 ) = T (ζp)

T = Qp(ζn0 )

K = Qp

2.7 Extensions of Valuations

Let K be any field with an absolute value | · |v and fix an algebraic extension L/K. We will see that there is a correspondence between extensions of | · |v to L and embeddings of L into the completion Kv. From one perspective, this will generalize and simplify Galois theory for fields with an absolute value, completely subsuming the ramification theory of Section 1.5. Let Kv denote the completion of K with respect to |·|v. There exist embeddings L,→ Kv since L embeds into K be classic Galois theory. Given such an embedding τ : L,→ Kv, we

61 2.7 Extensions of Valuations 2 Local Fields

know by Theorem 2.5.3 that | · |v on Kv extends uniquely to a valuation | · |v¯ on Kv such that for any finite extension Kv ⊆ M ⊆ Kv, the valuation is given by

1/[M:Kv] |x|v¯ = |NM/Kv (x)|v .

Define w on L by |x|w = |τ(x)|v¯ for this fixed embedding τ. We will write w | v, read “w extends v”. Now let Lw be the closure of τ(L) in Kv with respect to the topology induced by w. Abstractly, assuming L/K is finite, Lw = Lw, the completion of L with respect to |·|w in Kv. If L/K is infinite, then Lw is the union of the completions of all finite intermediate extensions of L/K with respect to | · |w. Note that | · |w extends to Lw by restricting | · |v¯ to Lw ⊆ Kv.

Lemma 2.7.1. For L/K and w | v as above, Lw = τ(L)Kv ⊆ Kv.

Proof. Suppose L/K is finite. Then τ(L)Kv ⊆ Lw. On the other hand, Theorem 2.5.3 implies τ(L)Kv is complete with respect to | · |w and therefore Lw ⊆ τ(L)Kv. Generalizing to the infinite case is straightforward.

From now on we will write Lw = LKv = τ(L)Kv. There is a diagram of field extensions in Kv

L Lw

K Kv

sometimes called the “local-to-global principle” for algebraic extensions. This terminology is reflected in the example of a function field K = k(t): one may pass from extensions L/k(t) of function fields to extensions Lw/k((t)) of fields of power series, that is, from global functions to local functions.

Lemma 2.7.2. Every extension of valuations w | v on L arises from an embedding τ : L,→ Kv as w =v ¯ ◦ τ.

Proof. Define Lw ⊆ Kv as above. Then Lw/Kv is algebraic and w is the unique extension of v on Kv to Lw. Thus for any embeddingτ ¯ : Lw ,→ Kv, we must havev ¯ ◦ τ¯ = w. Restricting τ¯ to L defines an embedding τ : L,→ Kv satisfyingv ¯ ◦ τ = w.

Lemma 2.7.3. Two embeddings τ1, τ2 : L,→ Kv give rise to the same absolute value on L if and only if τ2 = σ ◦ τ1 for some σ ∈ Aut(L/K).

Proof. ( ⇒ = ) is clear by the uniqueness of | · |v¯ on Kv. 00 00 ( =⇒ ) Suppose |τ1(x)|v¯ = |τ2(x)|v¯ for all x ∈ L. Define σ : τ1(L) → τ2(L) by σ = −1 0 00 τ2 ◦ τ1 and use continuity to extend to a map σ : τ1(L)Kv → τ2(L)Kv. (Note that σ is 0 continuous on τ1(L) precisely because |τ1(x)|v¯ = |τ2(x)|v¯.) Then σ is a Kv-isomorphism of 0 algebraic extensions of Kv, so by classic Galois theory, σ extends to a Kv¯-automorphism σ which necessarily satisfies τ2 = σ ◦ τ1.

62 2.7 Extensions of Valuations 2 Local Fields

Theorem 2.7.4. For any absolutely valued field (K, |·|v, v), there is a one-to-one correspon- dence extensions of valuations Galois orbits of embeddings ←→ . w | v to L L,→ Kv

Proof. An extension of valuations w | v determines an embedding τ : L,→ Kv by Lemma 2.7.2. The correspondence is bijective up to Galois conjugacy by Lemma 2.7.3. Now let L/K be finite, L = K(α) for some α ∈ L and let f be the minimal polynomial m1 mr of α over K. Factor f into irreducible polynomials f = f1 ··· fr over Kv. Then the K-embeddings L,→ Kv are precisely determined by which root of some fi is the image of α. Two embeddings are conjugate if and only if they take α to two roots of the same fi. Therefore Theorem 2.7.4 implies:

Corollary 2.7.5. For a simple extension L = K(α) with minimal polynomial f ∈ K[x], the embeddings L,→ Kv are in one-to-one correspondence with the irreducible factors of f.

Explicitly, an irreducible factor fi | f determines a valuation wi | v by |x|wi = |τi(x)|v¯, where τi; L,→ Kv is the embedding where τi(α) = αi is a root of fi. √ 2 Example 2.7.6. Let K = Q, L = Q( 14), f(x) = x − 14 and v = v5 the 5-adic valuation. Then over Q5, f splits as f(x) = x2 − 14 = (x − b)(x + b) for some b ≡ 2 (mod 5) such that b2 = 14. There are two embeddings of this quadratic number field into the 5-adic number field: √ Q( 14) −→ Q5 √ τ1 : 14 7−→ b √ τ2 : 14 7−→ −b. √ √ These give√ rise to two different extensions of v to Q( 14), say w1 and w2, with w1( 14−2) > 0 and w2( 14 + 2) > 0 for example. So they are indeed distinct. Notice that √ √ 5O √ = (5, 14 − 2)(5, 14 + 2) Q( 14) so the valuation theory completely captures the ramification theory in Section 1.5.

More generally, suppose L/K is a finite extension of number fields and fix a prime ideal p ⊂ OK with factorization e1 er pOL = P1 ··· Pr for distinct prime ideals Pi ⊂ OL and ei > 0. Let v be the p-adic valuation on K, i.e. v(x) = n if and only if x ∈ pn r pn+1. In this case, we get r different extensions of v to 1 L: v1, . . . , vr, where vi = vP , the normalization of the Pi-adic valuation on L by the ei i ramification index ei. To see this, assume OL = OK [α] and p is unramified in OL (there

63 2.7 Extensions of Valuations 2 Local Fields

are only finitely many ramified primes anyway). Then each ei = 1, so we have the following equivalences:

prime factors of pOL ←→ irreducible factors of f(x) mod p by Theorem 1.5.6

←→ irreducible factors of f(x) in Kv by Hensel’s Lemma

←→ embeddings L,→ Kv by Corollary 2.7.5 ←→ extensions of valuations w | v to L by Theorem 2.7.4.

Assume L/K is finite and consider the map Y ϕ : L ⊗K Kv −→ Lw w|v

a ⊗ b 7−→ (ab)w ∼ where ab is viewed in LKv = Lw. Proposition 2.7.7. If L/K is separable, then ϕ is an isomorphism.

Proof. Write L = K(α) and let f be the minimal polynomial of α over K. Then f factors over K as v Y f = fw w|v with no repeated factors since f is separable. For each w | v, view Lw inside Kv and let αw be the image of α in Kv under an embedding corresponding to w. Then Lw = Kv(αw) and fw is the minimal polynomial of αw over Kv. This corresponds to the commutative diagram Y Kv[x]/fw Kv[x]/f w|v

∼= ∼=

ϕ Y Lw L ⊗K Kv w|v

where the top row is by the Chinese remainder theorem, the left isomorphism is x 7→ α ⊗ 1 and the right isomorphism is x 7→ (αw)w. Therefore ϕ is an isomorphism. Corollary 2.7.8. If L/K is separable, then X [L : K] = e(w | v)f(w | v) w|v

× × where e(w | v) = [w(L ): v(K )] and f(w | v) = [λw : κv].

64 2.8 Galois Theory of Valuations 2 Local Fields

Proof. First note that [L : K] = [L ⊗K Kv : Kv] by basic algebra. Then X [L ⊗K Kv : Kv] = [Lw : Kv] by Proposition 2.7.7 w|v X = e(w | v)f(w | v) by Corollary 2.6.9. w|v P Therefore [L : K] = w|v e(w | v)f(w | v) as claimed. Definition. For L/K a separable extension with extension of valuations w | v, e(w | v) = × × [w(L ): v(K )] is called the ramification index of w | v and f(w | v) = [λw : κv] is called the inertial degree of w | v.

Example 2.7.9. Let K = Q and let L be any number field. Then the archimedean absolute value | · |∞ completes to the reals: Q∞ = R, and the corresponding base change from Proposition 2.7.7 is ∼ Y L ⊗Q R = Lw w|∞ ∼ ∼ where Lw = R or C. For example, if L is imaginary quadratic, L ⊗Q R = C, whereas if L is ∼ ∼ r s real quadratic, L ⊗Q R = R × R. In general, L ⊗Q R = R ⊗ C , where [L : Q] = r + 2s as in Section 1.8.

2.8 Galois Theory of Valuations

Assume L/K is a Galois extension with Galois group G = Gal(L/K). Then G acts on the set of extensions | · |w of | · |v to L by σ(| · |w)(x) = |σ(x)|w for all x ∈ L. Proposition 2.8.1. For L/K finite Galois, G acts transitively on the set of extensions of | · |v to L. Proof. If not, there exist disjoint G-orbits of absolute value extensions. Since all extensions of | · |v agree on K, any nonequivalent extensions must be distinct. Thus there exists some 0 x ∈ L with |σ(x)|w < 1 but |σ(x)|w0 > 1 for some w, w from distinct G-orbits and for all σ ∈ G, by the weak approximation theorem (2.3.9). Let Y α = σ(x). σ∈G

Then α ∈ K but |α|v < 1 and |α|v > 1 simultaneously, a contradiction. Hence G acts transitively. Let L/K be a Galois extension, w | v an extension of valuations and set

OL,w = {x ∈ L : |x|w ≤ 1} (the valuation ring for w)

PL,w = {x ∈ L : |x|w < 1} (the valuation ideal for w).

65 2.8 Galois Theory of Valuations 2 Local Fields

Definition. For an arbitrary extension of valuations w | v, we define the decomposition group for w by Gw = {σ ∈ G : |σ(x)|w = |x|w for all x ∈ L}. If w and v are nonarchimedean valuations, we also define the inertia group and ramifi- cation group for w respectively by

Iw = {σ ∈ Gw : σ(x) ≡ x mod PL,w for all x ∈ OL,w}  σ(x)  R = σ ∈ G : ≡ 1 mod P for all x ∈ L× . w w x L,w

Notice that for any w | v, we have Rw ≤ Iw ≤ Gw ≤ G. If the extension is to be emphasized, we will write Gw(L/K),Iw(L/K) and Rw(L/K).

Lemma 2.8.2. The subgroups Gw,Iw and Rw are closed subgroups of G = Gal(L/K).

Proof. We prove Gw ≤ G is closed and remark that the proofs for Iw and Rw are similar. Let σ ∈ G be in the closure of Gw and let K ⊆ M ⊆ L such that M/K is finite Galois. Then there exists σM ∈ Gw ∩ σ Gal(L/M), so σM |M = σ|M . Further, σM ∈ Gw implies w ◦ σM = w and so w ◦ σ|M = w ◦ σM |M = w, or σ ∈ Gw. Therefore Gw is closed in G. Suppose L/K and K0/K are Galois extensions and set L0 = LK0 ⊆ K: τ L L0

τ K K0 Set G = Gal(L/K) and G0 = Gal(L0/K0). Then any embedding τ : K,→ K0 induces a homomorphism τ ∗ : G0 −→ G σ 7−→ τ ∗(σ)(x) := τ −1στ(x). 0 0 0 0 0 Now let w a valuation on L , v = w |K0 , w = w ◦ τ and v = w|K . Proposition 2.8.3. The induced map τ ∗ : G0 → G induces homomorphisms 0 0 Gw0 (L /K ) −→ Gw(L/K) 0 0 Iw0 (L /K ) −→ Iw(L/K) 0 0 Rw0 (L /K ) −→ Rw(L/K). 0 0 0 ∗ 0 Proof. Suppose σ ∈ Gw0 = Gw0 (L /K ) and σ = τ (σ ) ∈ G. Then w(σ(x)) = w(τ ∗(σ0)(x)) = w(τ −1σ0τ(x)) = w0(σ0(τ)(x)) 0 0 0 = w (σ (x)) since σ ∈ Gw0 = w(x). ∗ 0 Therefore τ (σ ) = σ ∈ Gw. The proof is similar for the maps on inertia and ramification groups.

66 2.8 Galois Theory of Valuations 2 Local Fields

The most important case of this proposition is for the “local-to-global principle” of Sec- 0 0 tion 2.7, i.e. when K = Kv is the completion of K at v and L = Lw = LKv by Lemma 2.7.1. τ L Lw

K Kv

Lemma 2.8.4. Let σ ∈ G. Then σ ∈ Gw if and only if σ is continuous with respect to | · |w

Proof. ( =⇒ ) is clear since |x|w = |σ(x)|w for all x ∈ L implies continuity. ( ⇒ = ) If σ is continuous, then |x|w < 1 if and only if |σ(x)|w < 1, but then Corollary 2.3.7 implies | · |w and σ(| · |w) are equivalent. Hence σ ∈ Gw.

Proposition 2.8.5. If τ : L,→ Lw is an embedding, then the maps

=∼ Gw(L/K) −→ G(Lw/Kv) =∼ Iw(L/K) −→ I(Lw/Kv) =∼ Rw(L/K) −→ R(Lw/Kv) induced by τ are isomorphisms.

Proof. Note that τ(L) is dense in Lw with respect to | · |w, so there can’t be two different ∗ elements of Aut(Lw) with the same restriction to τ(L). This implies τ is injective. On the other hand, if σ ∈ Gw then σ is continuous with respect to | · |w by Lemma 2.8.4, so σ ∗ extends to an automorphism of Lw respecting the topology generated by | · |w. Hence τ is also surjective. So up to restriction to a decomposition group, the Galois theory of L/K is the same in the global case as it is in the local case. Definition. For a Galois extension L/K and a fixed extension w | v of valuations, define Gw Iw the decomposition field Zw = L , the inertia field Tw = L and the ramification Rw field Vw = L . We have a tower of fields and valuations:

K Zw Tw Vw L

v wZ wT wV w

Proposition 2.8.6. Let L/K be a Galois extension and fix w | v. Then

(1) w is the only extension of wZ to L.

(2) Zw = L ∩ Kv.

67 2.8 Galois Theory of Valuations 2 Local Fields

(3) e(wZ | v) = f(wZ | v) = 1. (4) There is a short exact sequence

1 → Iw → Gw → Gal(λ/κ) → 1

where κ and λ are the residue fields of Kv and Lw, respectively.

(5) Tw is the maximal unramified extension of Zw in L.

Proof. (1) Gw = Gal(L/Zw) acts transitively on such extensions, but by definition Zw is the subfield of L/K fixed by this group. ∼ Gw (2) By Proposition 2.8.5, Gw = Gal(Lw/Kv) and Zw = L , so we must have Zw ⊆ Kv. It follows that Zw = L ∩ Kv. (Really, this is all taking place in Kv after applying some embedding τ : L,→ Lw.) (3) follows from (2). (4) Exactly the same as Proposition 1.5.14. (5) We may assume K = Zw. Further, Proposition 2.8.5 allows us to assume K = Kv is complete. Let λs be the separable closure of κ in λ. Then certainly λs/κ is Galois. Let T/K be the maximal unramified subextension of L/K; by Lemma 2.6.5, we know T/K is Galois with residue field λs. Thus there is a homomorphism ϕ : Gal(T/K) → Gal(λs/κ) which is surjective by (4). Further, since T/K is unramified, [T : K] = [λs : κ] which implies ϕ is injective and hence an isomorphism. This means any σ ∈ Gw acts trivially on λs if and only if σ ∈ Gal(L/T ). In other words, Iw = Gal(L/T ) so by Galois theory, Tw = T .

The inertia subgroup Iw ≤ Gw is characterized as the kernel of the map Gw → Gal(λ/κ). We now describe a similar characterization for the ramification subgroup Rw ≤ Iw. Write × × × × × χ(L/K) = Hom(w(L )/v(K ), λ ). Given σ ∈ Iw and δ ∈ w(L )/v(K ), choose x ∈ L such that w(x) = δ. This defines a map

ψ : Iw −→ χ(L/K)  σ(x)  σ 7−→ δ 7→ mod P . x L,w

σ(x) |σ(x)|w 0 Note that x = |x| = 1 so indeed δ ∈ χ(L/K). Also, if x = xau for |u|w = 1 and w w a ∈ K, then σ(xau) σ(x) σ(u) σ(x) ψ(σ)(x0) = = · ≡ mod P xau x u x L,w

since σ ∈ Iw. Thus the homomorphism ψ is well-defined. It is now clear that Rw = ker ψ by the definition of the ramification group.

Proposition 2.8.7. Let char κ = p. If p > 0 then Rw is the unique Sylow p-subgroup of Iw, and if p = 0, then Rw = 1.

Proof. As before, we may assume K = Tw and K = Kv is complete. Also assume L/K is finite (the infinite case follows from taking limits). Let char κ = p > 0. We first show Rw × × contains all Sylow p-subgroups of Iw. Since w(L )/v(K ) is finite, any homomorphism into

68 2.9 Higher Ramification Groups 2 Local Fields

λ× takes values in the roots of unity of λ×, none of which have p-power order, so p does not divide |χ(L/K)|. Thus Iw/Rw has no elements of p-power order, so Rw must contain all Sylow p-subgroups of Iw as claimed. Next, we show every element of Rw has p-power order. Suppose to the contrary that 0 hσi 0 there exists a σ ∈ Rw with prime order `, for p 6= `. Take K = G with residue field κ . Then (5) of Proposition 2.8.6, together with Lemma 2.6.5, implies λ/κ is purely inseparable 0 0 0 (assuming K = Tw), so λ/κ is purely inseparable. Suppose L/K is not tame. Then λ/κ is not separable. Takeα ¯ ∈ λ r κ0 and lift to some α ∈ L. Then L = K0(α) and α has a minimal polynomial f(x) over K0. By Hensel’s Lemma, f¯(x) =g ¯(x)m for someg ¯(x) ∈ κ0[x] ¯ so we must haveg ¯(¯α) = 0. Hence degg ¯ | deg f | `, contradicting pure√ inseparability. Hence L/K0 is a tame extension. This implies by Theorem 2.6.8 that L = K0( ` a) for some a ∈ K0. 0 Since L/K is Galois, we have √ √ σ( ` a) = ζ ` a for an `th root of unity ζ ∈ L not equal to 1. This means √ σ( ` a) √ = ζ 6≡ 1 mod PL,w. ` a

This contradicts σ ∈ Rw, so every element in Rw has p-power order. Combined with the first paragraph, this says that Rw is itself a Sylow p-subgroup and since it is the kernel of ψ and thus normal, Rw is the unique one.

Corollary 2.8.8. Vw is the maximal tamely ramified extension of Zw in L. Corollary 2.8.9. There is an exact sequence

1 → Rw → Iw → χ(L/K) → 1.

2.9 Higher Ramification Groups

In Section 2.8, we constructed a sequence of subgroups Rw ≤ Iw ≤ Gw ≤ G. This is really the beginning of a filtration of subgroups for G = Gal(L/K), which we construct in this section. Assume (K, v) is Henselian, where v is a discrete, normalized valuation. Let OK , mK , πK and κ be as usual. For a finite Galois extension L/K with Galois group G = Gal(L/K), let w be the extension of v to L (unique by Theorem 2.5.3) and define the normalized extension of v to L by vL = eL/K w. Let OL, mL, πL and λ be as usual. Finally, assume λ/κ is separable and char κ = p.

Definition. For each s ∈ [−1, ∞), define the sth higher ramification group

Gs = {σ ∈ G | vL(σ(a) − a) ≥ s + 1 for all a ∈ OL}.

(These may also be referred to as the ramification groups of G for the lower numbering.)

69 2.9 Higher Ramification Groups 2 Local Fields

Example 2.9.1. Clearly G−1 = G and G0 = I = IvL is the inertia group. Moreover, if

R = RvL is the ramification group of G, we have σ(a)  σ ∈ R ⇐⇒ v − 1 ≥ 1 for all a ∈ O L a L σ(a) − a ⇐⇒ v ≥ 1 for all a ∈ O . L a L

 σ(a)−a  If a ∈ mL, then vL a = vL(σ(a)−a)−vL(a) so vL(σ(a)−a) ≥ vL(a)+1 ≥ 2. Likewise × for a ∈ OL , so G1 = R is the ramification group.

Lemma 2.9.2. Gs is a normal subgroup of G for all s ≥ 0.

Proof. Take τ ∈ Gs, σ ∈ G and a ∈ L. Then −1 −1 −1 vL(στσ (a) − a) = vL(τ(σ (a)) − σ (a)) −1 so if vL(τ(x) − x) ≥ s + 1 for all x ∈ OL, then vL(στσ (x) − x) ≥ s + 1 for all x ∈ OL and vice verse, since σ acts on G by automorphisms.

The higher ramification groups Gs form a filtration of G:

G = G−1 ⊇ G0 ⊇ G1 ⊇ G2 ⊇ · · · Moreover, the quotients in this filtration are described by the following proposition. For each (s) × s ≥ 0, let UL = {x ∈ OL : vL(x − 1) ≥ s}. Proposition 2.9.3. For all s ≥ 0, the map (s) (s+1) Gs/Gs+1 −→ UL /UL σ(π ) σ 7−→ L πL is an injective homomorphism of groups.

 σ(πL)  Proof. If σ ∈ Gs+1 then vL(σ(πL) − πL) ≥ s + 2 which implies vL − 1 ≥ s + 1, i.e. πL σ(πL) ∈ U (s+1). Therefore the map is well-defined. To see that it is a homomorphism, take πL L σ, τ ∈ Gs and consider: στ(π ) στ(π ) τ(π ) L = L · L πL τ(πL) πL

σ(uπL) τ(πL) × = · for some u ∈ OL uπL πL σ(u) σ(π ) τ(π ) = · L cdot L . u πL πL  σ(u)  σ(u) (s+1) Since σ ∈ Gs, vL(σ(u) − u) ≥ s + 1, so vL u − 1 ≥ s + 1 and thus u ≡ 1 in UL . Hence στ(πL) = σ(πL) · τ(πL) in U (s)/U (s+1). πL πL πL L L  σ(πL)  Finally, suppose σ ∈ Gs+1. Then vL(σ(πL) − πL) = s + 1 so vL − 1 = s and in πL particular σ(πL) 6= 1 in U (s)/U (s+1). Hence the map is injective. πL L L

70 2.9 Higher Ramification Groups 2 Local Fields

Corollary 2.9.4. For any L/K with Galois group G,

× ∼ (1) There is an embedding G0/G1 ,→ λ . In particular, G0/G1 = µ`, the group of `th roots of unity in λ, for some p - `. ∼ (2) For each s ≥ 1, there is an embedding Gs/Gs+1 ,→ (λ, +). In particular, Gs/Gs+1 = (Z/pZ)a for some a. Proof. Apply Proposition 2.3.14.

Example 2.9.5. The corollary implies G1 is the unique Sylow p-subgroup of G0 = I, so by Proposition 2.8.7, G1 = R, the ramification group. This confirms Example 2.9.1. Higher ramification groups give us an idea about the general shape of the Galois group of an extension L/K.

Lemma 2.9.6. G0 is isomorphic to a semidirect product P o Z/mZ where P is a p-group and m ∈ Z, p - m. Proof. Apply the Schur-Zassenhaus theorem.

Corollary 2.9.7. G0 is solvable. Corollary 2.9.8. If L/K is totally ramified and Galois, then Gal(L/K) is solvable.

Example 2.9.9. Consider the local function field K = Fp((t)). Then any finite Galois extension L/K is totally ramified and hence has solvable Galois group. In particular, the inverse Galois problem does not hold for K.

Example 2.9.10. Let K = C((t)) be the global function field over k = C. Then one can ∼ prove GK := Gal(C((t))/C((t))) = Zb, the profinite completion of the integers. Since C is algebraically closed of characteristic zero, for any finite Galois extension L/C((t)) we get G0 = G and G1 = {1}. Fix a tower of Galois field extensions L ⊃ L0 ⊃ K with G = Gal(L/K) and H = Gal(L/L0). Compare the filtrations

G−1 ⊇ G0 ⊇ G1 ⊇ G2 ⊇ · · ·

and H−1 ⊇ H0 ⊇ H1 ⊇ H2 ⊇ · · ·

One can see that by the definitions of these higher ramification groups, for each s ≥ −1, 0 0 ∼ Hs = Gs ∩ H. On the other hand, if G = Gal(L /K) = G/H, it is not clear that the filtrations

G−1 ⊇ G0 ⊇ G1 ⊇ G2 ⊇ · · · 0 0 0 0 and G−1 ⊇ G0 ⊇ G1 ⊇ G2 ⊇ · · · are even related at all.

71 2.9 Higher Ramification Groups 2 Local Fields

Lemma 2.9.11. If L/K is Galois and the residue extension λ/κ is separable, there exists x ∈ OL such that OL = OK [x].

Proof. By the fundamental equality (Proposition 2.6.1), eL/K fL/K = [L : K]. Since we are assuming λ/κ is separable, we may choosex ¯ ∈ λ such that λ = κ(¯x). Let f¯(t) be the minimal ¯ polynomial ofx ¯ over κ. Then by Hensel’s Lemma, there is a lift f(t) ∈ OK [t] of f(t). Lift i j x¯ to an element x ∈ OL. We know vL(f(x)) > 0. If vL(f(x)) = 1, the elements f(x) x for 0 ≤ i < eL/K and 0 ≤ j < fL/K generate OL as an OK -module since the number of these is [L : K]. In this case, it is clear that OK [x] = OL. On the other hand, if vL(f(x)) > 1, replace x with x + πL, so that

0 2 f(x + πL) = f(x) + πLf (x) + O(πL).

0 × ¯ ¯0 Then f (x) ∈ OL since f is separable and f (¯x) 6= 0. Also, vL(f(x)) > 1 implies that vL(f(x + πL)) = 1. So in all cases, OL is generated by some x ∈ OL as an OK -module.

Let x ∈ OL such that OL = OK [x]. For each nontrivial σ ∈ G = Gal(L/K), write

iL/K (σ) = vL(σ(x) − x) and also set iL/K (1) = ∞. In fact, iL/K (σ) = miny∈OL {vL(σ(y) − y)} since for any y ∈ OL, we may write

n y = a0 + a1x + ... + anx

n n for n ∈ N, ai ∈ OK and have σ(y) − y = a1(σ(x) − x) + ... + an(σ(x ) − x ). By a k k binomial expansion, each σ(x ) − x is divisible by σ(x) − x so it follows that vL(σ(y) − y) ≥ vL(σ(x)−x). In particular, this implies usefully that the definition of iL/K (σ) is independent of any generator chosen for OL. The higher ramification groups can thus be written

Gs(L/K) = {σ ∈ G | iL/K (σ) ≥ s + 1}.

Now return to the situation where L ⊃ L0 ⊃ K and L0/K is Galois.

Lemma 2.9.12 (Tate). For any σ0 ∈ G0 = Gal(L0/K),

0 1 X iL0/K (σ ) = iL/K (σ). e 0 L/L σ∈G 0 σ|L0 =σ

Proof. If σ0 = 1 then both sides are infinite so the equality holds. Assume σ0 6= 1. By Lemma 2.9.11, OL0 = OK [y] for some y ∈ OL0 ; as above, let OL = OK [x]. Then

0 0 1 0 iL0/K (σ ) = vL0 (σ (y) − y) = vL(σ (y) − y) eL/L0

0 0 which we will rewrite as eL0/LiL0/K (σ ) = vL(σ (y) − y). It therefore suffices to show

0 X vL(σ (y) − y) = iL/K (σ). 0 σ|L0 =σ

72 2.9 Higher Ramification Groups 2 Local Fields

Immediately, we have that X Y iL/K (σ) = vL(στ(x) − x) 0 σ|L0 =σ τ∈H

0 Q Set a = σ (y) − y and b = τ∈H (στ(x) − x). If f(t) ∈ OK [t] is the minimal polynomial of x over K, then Y Y f(t) = (t − τx) =⇒ (σf)(t) = (t − στx) τ∈H τ∈H Y =⇒ (σf)(x) = (x − στx) τ∈H =⇒ (σf)(x) − f(x) = (−1)|H|b since f(x) = 0.

0 But the coefficients of σf −f lie in OL0 , so they are all divisible by σ (y)−y = a. This shows a | b. On the other hand, let g(t) ∈ OK [t] be any polynomial and set y = g(x). Then x is ˜ ˜ a root of the polynomial g(t) − y ∈ OL[t] so g(t) − y = f(t)h(t) where f(t) is the minimal polynomial of x over L0. Then

a = σ0(y) − y = σ(g(t) − y) − (g(t) − y) = (σf˜)(t)(σh)(t) − f˜(t)h(t).

Evaluating this at t = x, we get

a = (σf˜)(x)(σh)(x) = (−1)|H|b(σh)(x)

as above. Thus b divides a, so we have a = b and thus vL(a) = vL(b) as required. Define the function

ϕL/K :[−1, ∞) −→ [−1, ∞) Z s dx s 7−→ 0 [G0 : Gs]

−1 where formally we set [G0 : G−1] = [G : G0] . Then ϕL/K is piecewise-linear, nondecreasing and if gs = |Gs|, then we can explicitly write 1 ϕL/K (s) = (g1 + ... + gm + (s − m)gm+1) g0

for any m ∈ N such that 0 < m ≤ s ≤ m + 1. Also, ϕL/K (s) = s for −1 ≤ s ≤ 0. By this reformulation, we can see that the slope of ϕ (s) is gm+1 for all s, where m < s < m + 1, L/K g0 but when s ∈ , the slope is gs−1 . This implies: Z g0 1 X Lemma 2.9.13. For any s ≥ −1, ϕL/K (s) = min{iL/K (σ), s + 1} − 1. g0 σ∈G

73 2.9 Higher Ramification Groups 2 Local Fields

Theorem 2.9.14 (Herbrand). Let L0/K be a Galois extension and H = Gal(L/L0) and G0 = Gal(L0/K).Then for any s ≥ −1,

0 Gs(L/K)H/H = Gt(L /K)

where t = ϕL/L0 (s).

0 0 0 Proof. Fix σ ∈ G and pick σ ∈ G such that σ|L0 = σ and iL/K (σ) is maximal among all such 0 0 0 σ ∈ G restricting to σ on L . We claim iL0/K (σ ) − 1 = ϕL/L0 (iL/K (σ) − 1). Set m = iL/K (σ) and fix τ ∈ H. Then if τ ∈ Hm−1, we have iL/K (τ) ≥ m by the above description of the higher ramification groups, as well as

vL(στ(x) − x) = vL(στ(x) − τ(x) + τ(x) − x)

≥ max{vL(στ(x) − τ(x)), vL(τ(x) − x)} = max{m, m} = m.

But by maximality, this implies vL(στ(x) − x) = m. On the other hand, if τ ∈ Hm−1, then iL/K (τ) < m so vL(στ(x)−x) = iL/K (τ). Thus iL/K (στ) = vL(στ(x)−x) = min{m, iL/K (τ)}. By Lemma 2.9.12,

0 1 X iL0/K (σ ) = iL/K (σ) e 0 L/L τ∈H 1 X = min{m, iL/K (τ)} h0 τ∈H

= ϕL/L0 (iL/K (σ) − 1) + 1 by Lemma 2.9.13.

So the claim holds. Now for σ0 ∈ G0 = G/H,

0 σ ∈ Gs(L/K)H/H ⇐⇒ iL/K (σ) − 1 ≥ s

⇐⇒ ϕL/L0 (iL/K (σ) − 1) ≥ ϕL/L0 (s) = t 0 ⇐⇒ iL0/K (σ ) − 1 ≥ t 0 0 ⇐⇒ σ ∈ Gt(L /K).

0 Hence Gs(L/K)H/H = Gt(L /K).

t Definition. Let L/K be a Galois extension. Then the subgroups G := Gs for t = ϕL/K (s) are called the higher ramification groups for the upper numbering of G.

Since ϕL/K (s) is monotone in s, it has an inverse function ψL/K :[−1, ∞) → [−1, ∞). Lemma 2.9.15. For a tower L ⊃ L0 ⊃ K of Galois extensions,

ϕL/K = ϕL0/K ◦ ϕL/L0 and ψL/K = ψL/L0 ◦ ψL0/K .

Proof. We prove the statement for the ϕ maps; the other statement follows from the fact −1 that each ψ = ϕ . By Theorem 2.9.14, we know that if t = ϕL/L0 (s) then Gs(L/K)/Hs =

74 2.10 Discriminant and Different 2 Local Fields

∼ GsH/H = (G/H)t. Thus |Gs| = |Hs| |(G/H)t| and comparing the derivatives of ϕL/K (s) and ϕL0/K ◦ ϕL/L0 (s), we see that

0 1 ϕL/K (s) = |Gs| eL/K 1 = |Hs| |(G/H)t| as in Lemma 1.5.12 eL/L0 eL0/K 1 1 = |Hs| |(G/H)t| eL/L0 eL0/K 0 0 = ϕL/L0 (s)ϕL0/K (t) for s 6∈ Z 0 = (ϕL0/K ◦ ϕL/L0 ) (s) by the chain rule.

Thus ϕL/K (s) and ϕL0/K ◦ ϕL/L0 (s) differ by a constant away from s ∈ Z, but since both are continuous and equal to 0 at s = 0, they must be equal. Theorem 2.9.16. For all t ≥ −1, Gt(L/K)H/H = Gt(L0/K). Proof. Let t ≥ −1. Then

t G (L/K)H/H = GψL/K (t)(L/K)H/H by definition of the upper numbering = G (L0/K) ϕL/L0 ◦ψL/K (t) = G (L0/K) by Lemma 2.9.15 ϕL/L0 ◦ψL/L0 ◦ψL0/K (t) = G (L0/K) ψL0/K (t) = Gt(L0/K).

This shows the advantage of the ramification groups of upper numbering: they are in- variant under passage to a Galois subextension L0/K of L/K. By construction, the “jumps” in the filtration Gs can only occur at integers. However, this is not necessarily true of the ramification groups of upper numbering Gt. However, we have: Theorem 2.9.17 (Hasse-Arf). If L/K is an abelian extension and Gt is a jump in the upper filtration of G = Gal(L/K), then t ∈ Z.

2.10 Discriminant and Different

We conclude the chapter by giving an application to the ramification theory of number fields, generalizing the criterion for ramification given in Proposition 1.5.8. The first few results apply to general Dedekind domains, so let A be a Dedekind domain with field of fractions K, take a finite separable extension L/K and let B be the integral closure of A in L. We will assume all residue field extensions are separable. The trace form of L/K is the K-bilinear map T : L × L −→ K

(x, y) 7−→ TrL/K (xy).

75 2.10 Discriminant and Different 2 Local Fields

Definition. Let J be a fractional ideal of A. Then the dual of J is

J ∗ = {x ∈ L | T (x, y) ∈ A for all y ∈ J}.

Lemma 2.10.1. For any fractional ideal J of A, J ∗ is a fractional ideal.

Example 2.10.2. B is a fractional ideal of A, so the dual B∗ is defined. It is clear that B∗ ⊇ B.

Definition. The different of the ring extension B/A is defined as the inverse of the dual of B: ∗ −1 DB/A = (B ) .

Notice that the different DB/A is an actual ideal of B. Proposition 2.10.3. Let A be a Dedekind domain, K its field of fractions, L/K a finite separable extension and B the integral closure of A in L. Then

(i) If K ⊆ L ⊆ M with C the integral closure of A in M, then DC/A = DC/BDB/A.

−1 (ii) If S ⊆ A is any multiplicatively closed subset, then DS−1B/S−1A = S DB/A. (iii) If p ⊂ A is a prime ideal and q ⊂ B is any prime lying over p, then

D B = D B/A bq Bbq/Abp

where Bbq (resp. Abp) is the valuation ring of the completion of L (resp. K) at the place | · |q (resp. | · |p). Proof. (i) Suppose I is a fractional ideal of M. Then

−1 I ⊆ DC/B ⇐⇒ TrM/L(I) ⊆ B −1 −1 −1 ⇐⇒ DB/A TrM/L(I) ⊆ DB/AB = DB/A −1 ⇐⇒ TrL/K (DB/A TrM/L(I)) ⊆ A −1 −1 ⇐⇒ TrL/K (TrM/L(DB/AI)) ⊆ A since DB/A ⊆ B −1 ⇐⇒ TrM/K (DB/AI) ⊆ OK by transitivity of trace −1 −1 ⇐⇒ DB/AI ⊆ DC/A −1 ⇐⇒ I ⊆ DC/ADB/A.

−1 −1 Therefore by unique factorization of fractional ideals (Theorem 1.4.2), DC/B = DC/ADB/A so by inverting, we get DC/A = DC/BDB/A. (ii) is easy. (iii) We may assume A is in fact a DVR. Then the property is shown by proving that B∗ ∗ ∗ is dense in Bbq , where Bbq is the dual of the fractional ideal Bbq. The following is an example of a so-called ‘local-to-global principle’ in number theory.

76 2.10 Discriminant and Different 2 Local Fields

Corollary 2.10.4. For any A, K, L, B, p, q as above, the different may be computed locally: Y D = (D ∩ B) B/A Bbq/Abp q|p where the product is taken over all primes p ⊂ A and all q ⊂ B lying over p.

Let L/K be an extension of number fields, with rings of integers OK and OL. We will

write DL/K to denote the different DOL/OK . We may assume OL = OK [α] for α ∈ L with minimal polynomial f(x) over K.

0 Lemma 2.10.5. DL/K = (f (α)).

n−1 n Proof. Write f(x) = a0 + a1x + ... + an−1x + x ∈ OK [x]. Then f(x) = b + b x + ... + b xn−1 x − α 0 1 n−1

2 n−1 for bi ∈ OK . We show the dual basis of {1, α, α , . . . , α } with respect to the trace form n o b0 bn−1 is precisely f 0(α) ,..., f 0(α) . To see this, let α1, . . . , αn be the distinct roots of f(x). Then the polynomial r X f(x) αr g(x) = xr − · i x − α f 0(α ) i=1 i i

is monic of degree strictly less than n, but α1, . . . , αn are all roots of g. This implies g = 0, so n X f(x) αr · i = xr x − α f 0(α ) i=1 i i  r  f(x) αi r f(x) for each 0 ≤ r ≤ n − 1. Thus TrL/K · 0 = x for 0 ≤ r ≤ n − 1, but = x−αi f (αi) x−α n−1 b0 + b1x + ... + bn−1x so comparing degrees, we get

 b αj  Tr i = δ . L/K f 0(α) ij

Thus the dual basis is as claimed. Now notice that the bi satisfy recursive equations: bn−1 = 1, bn−2 − αbn−1 = an−1, and so on. Solving this yields the identity

i−1 i−2 bn−i = α + an−1α + ... + an−i+1

0 which shows that b0, . . . , bn−1 generate OL. This implies DL/K = (f (α)). The different has an important relationship with the discriminant of a field extension, which further relates it to ramification theory.

Theorem 2.10.6. Let L/K be an extension of discretely valued fields and q ⊂ OL a prime ideal. Then

77 2.10 Discriminant and Different 2 Local Fields

(i) q is ramified in OL if and only if q divides the different DL/K .

s (ii) If s is the maximal exponent such that q | DL/K , p = q ∩ OK and e = e(q | p), then s = e − 1 when q | p is tamely ramified and e ≤ s ≤ vq(e) + e − 1 when q | p is wildly ramified.

(iii) If L/K is Galois with Galois group G = Gal(L/K), then

∞ X s = (|Hi| − 1) i=0

where H = Dq is the decomposition group of q and Hi are the higher ramification groups.

Proof. By Proposition 2.10.3(iii), we may assume OL and OK are complete DVRs. Write OL = OK [α] and let f be the minimal polynomial of α over K. Then by Lemma 2.10.5, 0 DL/K = (f (α)). Under the assumption of completeness, we have unique prime ideals p = mK ⊂ OK and q = mL ⊂ OL. (i) If L/K is unramified, thenα ¯ is a simple root of f¯ = f mod q becauseα ¯ must generate a separable extension of residue fields of degree deg f. Thus f¯0(¯α) 6= 0 and thus 0 DL/K = (f (α)) = (1). The converse will follow directly from (ii). (ii) By Proposition 2.10.3(i), we may assume L/K is totally ramified. Write

e e−1 f(x) = x + a1x + ... + ae−1x + ae where ai ∈ OK and e = eL/K . Then f(x) is Eisenstein since α may be taken to be a uniformizer of OL. In particular,

0 e−1 e−2 f (α) = eα + (e − 1)a1α + ... + ae−1.

0 Since all ai ∈ OK , e | vL(ai) for each ai and vL(α) = 1, so each term in f (α) has a different 0 0 valuation. Thus vL(f (α)) = e − 1 when p - e (the tame case) and vL(f (α)) ≤ vL(e) + e − 1 (the wild case) since OL is a DVR. (iii) Now suppose L/K is Galois. Then Y f 0(α) = (α − σ(α)). σ∈Gr{1}

By Proposition 2.8.5, H = Dq = Gal(L/K) = G and by the above,

0 X s = vL(f (α)) = iL/K (σ) σ∈Gr{1} = #{(σ, i) | σ ∈ Gi r {1}, i ≥ 0} ∞ X = (|Gi| − 1). i=0

78 2.10 Discriminant and Different 2 Local Fields

Let L/K be an extension of number fields. Recall from Section 1.3 the definition of the discriminant dL/K (α1, . . . , αn) for a K-basis {α1, . . . , αn} of L:

2 dL/K (α1, . . . , αn) = [det(σi(αj))] .

As in Proposition 1.5.8, define the discriminant ideal DL/K = (dL/K (α1, . . . , αn)) for any such basis. Theorem 2.10.7. For an extension L/K, the discriminant ideal is the ideal norm of the different: DL/K = NL/K (DL/K ).

Proof. Again, we may assume OK and OL are DVRs by Proposition 2.10.3(iii). In particu- lar, OK is a PID (Proposition 2.1.1) so OL admits an integral basis α1, . . . , αn by Proposi- tion 1.3.7. Then DL/K = (dL/K (α1, . . . , αn)) by definition. On the other hand, OL is also a −1 −1 ∗ ∗ ∗ ∗ PID so DL/K = βOL for some β ∈ L. By definition, DL/K = (α1, . . . , αn) where {α1, . . . , αn} −1 is the dual basis to {α1, . . . , αn} with respect to the trace form. Then DL/K = (α1β, . . . , αnβ), so we have

∗ ∗ 2 dL/K (α1, . . . , αn) = dL/K (βα1, . . . , βαn) = NL/K (β) dL/K (α1, . . . , αn).

∗ ∗ −2 ∗ This implies (dL/K (α1, . . . , αn)) = NL/K (DL/K )DL/K . Now using the pairing Tr(αiαj ) = δij, T ∗ ∗ ∗ −1 we obtain [σi(αj)] [σi(αj )] = In so dL/K (α1, . . . , αn) = dL/K (α1, . . . , αn) . It follows that 2 2 DL/K = NL/K (DL/K ) but since the norm is multiplicative, we obtain the desired expression.

Corollary 2.10.8. For a finite separable extension of discretely valued fields, Y DL/K = (DLq/Kp ∩ OK ). q|p

where the product is taken over all primes p ⊂ OK and all q ⊂ OL lying over p. We also obtain a strengthening of Proposition 1.5.8:

Corollary 2.10.9. Let L/K be a finite separable extension. Then a prime p ⊂ OK is ramified in OL if and only if p divides the discriminant DL/K . Proof. This is immediate from Theorems 2.10.6(i) and 2.10.7. Let us now derive an important consequence for number fields.

Theorem 2.10.10. For any fixed d, N > 0, there exist only finite many number fields K/Q with discriminant dK and degree n = [K : Q] satisfying |dK | ≤ d and n ≤ N.

Proof. First note that if K/Q has discriminant dK satisfying |dK | ≤ d and n = [K : Q] ≤ N, n then K(i)/Q has discriminant |dK(i)| ≤ (4d) and [K(i): Q] ≤ 2N, so we are free to assume i ∈ K. In particular, we may assume all embeddings of K into C are complex. Fix one of these, τ0 : K,→ C. Let X ⊆ KR be the set of all (zτ ) ∈ KR satisfying the following conditions:

79 2.10 Discriminant and Different 2 Local Fields

√ ˆ Im(zτ0 ) < C d for some constant C; ˆ Re(zτ0 ) < 1;

ˆ |zτ | < 1 for all τ 6∈ {τ0, τ¯0}.

It is clear that X is centrally-symmetric√ and convex. If C is chosen large enough, we can n guarantee that vol(X) > 2 d. Then by Minkowski’s theorem√ (1.7.3), X contains a lattice ∗ point j(α) for some α ∈ OK . In particular, Im(τ0(α)) < C d, Re(τ0(α)) < 1 and for any τ 6= τ0, τ¯0, |τ(α)| < 1. It now suffices to show K = Q(α) since these conditions impose a bound on the degree of the minimal polynomial of α over Q, and hence on the number of such K. On one hand, |NK/Q(α)| ≥ 1, but |τ(α)| < 1 for all τ 6= τ0, τ¯0, so we must have |τ0(α)| > 1. Thus Im(τ0(α)) > 0 so τ0(α) 6=τ ¯0(α). Also, τ0(α) 6= τ(α) for all τ 6= τ0, τ¯0 so α has distinct images under all embeddings K,→ C. This implies K = Q(α) so we are done.

Proposition 2.10.11. If K is a number field with discriminant dK and degree n = [K : Q], then nn π n/2 p|d | ≥ . K n! 4

Proof. By Theorem 1.8.6, there is some α ∈ OK with n!  4 s 1 ≤ |N (α)| ≤ p|d | K/Q nn π K

where s is the number of pairs of complex embeddings K,→ C. Rearranging this, we get nn π s nn π n/2 p|d | ≥ ≥ K n! 4 n! 4 since 2s ≤ n.

Corollary 2.10.12. For any d > 0, there are finitely many number fields K/Q of discrimi- nant |dK | ≤ d.

nn π
n/2 Proof. Define the sequence an = n! 4 . Then

a π 1/2  1 n π 1/2 n+1 = 1 + −→ e > 1 as n → ∞ an 4 n 4

so the sequence (an) increases geometrically. But by Proposition 2.10.11, |dK | ≥ an so there can only be finitely many number fields K of bounded discriminant.

Corollary 2.10.13. The only number field K with discriminant dK = ±1 is K = Q.

Proof. Let (an) be the sequence defined in the proof of Corollary 2.10.12. For all n ≥ 2, an > 1 so |dK | > 1 by Proposition 2.10.11.

Corollary 2.10.14. There are no unramified extensions of Q.

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