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Local Class Field Theory

KENKICHI IWASAWA Princeton University

OXFORD UNIVERSITY PRESS • New York CLARENDON PRESS - Oxford

1986

Oxford University Press

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Copyright C) 1986 by Kenkichi Iwasawa

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Library of Congress Cataloging-in-Publication Data Iwasawa, Kenkichi, 1917–

Local class field theory. (Oxford mathematical monographs) Bibliography: p. Includes index. 1. Class field theory. I. Title. II. Series.

QA247.195413 1986 512'.74 85-28462 ISBN 0-19-504030-9

British Library Cataloguing in Publication Data Iwasawa, Kenkichi

Local class field theory.—(Oxford mathematical monographs) 1. Fields, Algebraic I. Title 512'.3 QA247 ISBN 0-19-504030-9

2 4 6 8 9 7 5 3 1

Printed in the United States of America on acid-free paper

Preface

Local class field theory is a theory of abelian extensions of so-called local fields, typical examples of which are the p-adic number fields. This book is an introduction to that theory.

Historically, local class field theory branched off from global, or classical, class field theory, which studies abelian extensions of global fields—that is, algebraic number fields and algebraic function fields with finite fields of constants. So, in earlier days, some of the main results of local class field theory were derived from those of the global theory. Soon after, however, in the 1930s, F. K. Schmidt and Chevalley discovered that local class field theory can be constructed independently of the global theory; in fact, the former provides us essential devices for the proofs in the latter. Around 1950, Hochschild and Nakayama brought much generality and clarity into local class field theory by introducing the cohomology theory of groups. Classical books such as Artin [1] and Serre [21] follow this cohomological method. Later, different approaches were proposed by others—for ex-ample, the method of Hazewinkel [11], which forgoes cohomology groups, and that of Kato [14], based on algebraic K-theory. More recently, Neukirch also introduced a new idea to local class field theory, which applies as well to global fields.

Meanwhile, motivated by the analogy with the theory of complex multiplication on elliptic curves, Lubin and Tate showed in their paper [19] of 1965 how formal groups over local fields can be applied to deduce important results in local class field theory. In recent years, this idea has been further pursued by several mathematicians, in particular by Coleman. Following this trend, we shall try in this book to build up local class field theory entirely by means of the theory of formal groups. This approach, though not the shortest, seems particularly well suited to prove some deeper theorems on local fields.

In Chapters I and II, we discuss in the standard manner some basic definitions and properties of local fields. In Chapter III, we consider certain infinite extensions of local fields and study formal power series with coefficients in the valuation rings of those fields. These results are used in Chapters IV and V, where we introduce a generalization of Lubin–Tate formal groups and construct similarly as in [19] abelian extensions of local fields by means of division points of such formal groups. In Chapter VI, the main theorems of local classfield theory are proved: we first show that the abelian extensions constructed in Chapter V in fact give us all abelian extensions of local fields, and then define the so-called norm residue maps and prove important functorial properties of such maps. In Chapter VII, the classical results on finite abelian extensions of local fields are deduced from the main theorems of Chapter VI. In the last chapter, an explicit reciprocity theorem of Wiles [25] is proved, which generalizes a beautiful formula of Artin–Hasse [2] on norm residue symbols.

vi Preface

The book is almost self-contained and the author tried to make the exposition as readable as possible, requiring only some basic background in algebra and topological groups on the part of the reader. The contents of this book are essentially the same as the lectures given by the author at Princeton University in the Spring term of 1983. However, the original exposition in the lectures has been much improved at places, thanks to the idea of de Shalit [6].

In 1980, the author published a book [13] on local class field theory in Japanese from Iwanami-Shoten, Tokyo, which mainly followed the idea of Hazewinkel fil]. When the matter of translating this text into English arose, the author decided to rewrite the whole book in the manner just described. In order to give the reader some idea of other approaches in local class field theory, a brief account of cohomological method and Hazewinkel's method are included in an Appendix. At the end of the book, a short list of references is attached, containing only those items in the literature mentioned in the text; for a more complete bibliography on local fields and local class field theory, the reader is referred to Serre [21]. An index and a table of notations are also appended for the convenience of the reader.

The author expresses here his hearty gratitude to D. Dummit and E. Friedman who carefully read the book in manuscript and offered many valuable suggestions for its improvement. He also thanks D. W. Degen-hardt of Oxford University Press for his help in publishing this book.

Princeton September 1985

K.I.

Contents

Chapter I. Valuations 3

1.1. Some Basic Definitions 3 1.2. Complete Fields 7 1.3. Finite Extensions of Complete Fields 12

Chapter II. Local Fields 18

2.1. General Properties 18 2.2. The Multiplicative Group k x 22 2.3. Finite Extensions 25 2.4. The Different and the Discriminant 29 2.5. Finite Galois Extensions 32

Chapter III. Infinite Extensions of Local Fields 35

3.1. Algebraic Extensions and Their Completions 35 3.2. Unramified Extensions and Totally Ramified Extensions 36 3.3. The Norm Groups 40 3.4. Formal Power Series 43 3.5. Power Series over ok 45

Chapter IV. Formal Groups FAX, Y) 50

4.1. Formal Groups in General 50 4.2. Formal Groups Ff(X, Y) 53 4.3. The o-Modules W'fl 57 4.4. Extensions Lnig 61

Chapter V. Abelian Extensions Defined by Formal Groups 65

5.1. Abelian Extensions Ln and lc" 65 5.2. The Norm Operator of Coleman 69 5.3. Abelian Extensions L and lc, 75

Chapter VI. Fundamental Theorems 80

6.1. The Homomorphism Pk 80 6.2. Proof of L k = lcab 84 6.3. The Norm Residue Map 88

viii Contents

Chapter VII. Finite Abelian Extensions 98

7.1. Norm Groups of Finite Abelian Extensions 98 7.2. Ramification Groups in the Upper Numbering 101 7.3. The Special Case k7'nlk 107 7.4. Some Applications 110

Chapter VIII. Explicit Formulas 116

8.1. a-Sequences 116 8.2. The Pairing (a, 13 ) 120 8.3. The Pairing [a, O], 123 8.4. The Main Theorem 127 8.5. The Special Case for k =Qp 133

Appendix 137

A.1. Galois Cohomology Groups 137 A.2. The Brauer Group of a Local Field 141 A.3. The Method of Hazewinkel 146

Bibliography 151

Table of Notations 153

Index 155

LOCAL CLASS FIELD THEORY

Chapter I

Valuations

In this chapter, we shall briefly discuss some basic facts on valuations of fields which will be used throughout the subsequent chapters. We shall follow the classical approach in the theory of valuations, but omit the proofs of some elementary results in Section 1.1, which can be found in many standard textbooks on algebra.t For further results on valuations, we refer the reader to Artin [1] and Serre [21].

1.1. Some Basic Definitions

Let k be a field. A function v(x) on k, x E k, is called a valuation of k if it satisfies the following conditions:

(i) v(x) is a real number for x *0; and v(0) = + GC.

(ii) For any x, y in k,

min(v(x), v(y)) Is v(x + y).

(iii) Similarly, v(x) + v(y)= v(xy).

EXAMPLE 1. Let

v0(0) = + 00, vo(x) = 0 for all x *0 in k.

Then vo is a valuation of k. It is called the trivial valuation of k.

EXAMPLE 2. Let p be a prime number. Each non-zero rational number x can be uniquely written in the form x = pey, where e is an integer and y is a rational number whose numerator and denominator are not divisible by p. We define a function vp on the rational field Q by

v(0) = + 00; vp (x) = e, if x *0 and x = pey as above.

Then vp is a valuation of Q; it is the well-known p-adic valuation of the rational field. vp is the unique valuation v on Q satisfying v(p)= 1.tt EXAMPLE 3. Let F be a field and let F(T) denote the field of all rational functions in an indeterminate T with coefficients in F. Then each x *0 in F(T) can be uniquely written in the form x = Tey, where e is an integer and y is a quotient of polynomials in F[T], not divisible by T. Putting v(0) = + 00 and v(x) = e for x *0 as above, we obtain a valuation v of the field k = F(T) such that v(T)=1. Instead of T, we can choose any irreducible

t See, for example, Lang [16] and van der Waerden [23]. tt Compare van der Waerden [23].

4 Local Class Field Theory

polynomial f(T) in F(T) and define similarly a valuation v on k = F(T) with v(f) = 1.

Let v be a valuation of a field k. Then it follows immediately from the definition that

v(±1)= 0, v(–x)= v(x), v(x) < v(y) v(x + y)= v(x).

Here 1 (= 1 k ) denotes the identity element of the field k. Let

o {x I x E k, v(x) 0},

p = I X E k, v(x)> O}.

Then o is a subring of k continuing 1 and p is a maximal ideal of o so that

is a field. o, p, and f are called the valuation ring, the maximal ideal, and the residue field of v, respectively. By (iii) above, the valuation v defines a homomorphism

v:k x –*R±

from the multiplicative group le of the field k into the additive group R± of real numbers. Hence the image v(k x ) is a subgroup of R± , and we have

le 1 U v(kx) where

U = Ker(v) = {x E , v(x) = 0}.

U is called the unit group of the valuation v. Let v be a valuation of k. For any real number a> 0, define a function

,u(x) on k by

,u(x) = crv(x), for all x E k.

Then /.2 is again a valuation of k. When two valuations v and p, on k are related in this way—namely, when one is a positive real number times the other—we write

v

and say that v and p. are equivalent valuations of k. Equivalent valuations have the same valuation ring, the same maximal ideal, the same residue field, the same unit group, and they share many other important properties.

Let v be a valuation of a field k. For each x E k and a E R, let

N(x, a) = {y I y E k, v(y – x)> a} .

This is a subset of k containing x. Taking the family of subsets N(x, a) for all a E R as a base of neighbourhoods of x in k, we can define a Hausdorff topology on k, which we call the v-topology of k. k is then a topological field in that topology; the valuation ring o is closed in k and the maximal ideal p is open in k. A sequence of points, x 1 , x2 , x3 , . . . , in k converges to

Valuations 5

x E k in the v-topology—that is,

lim xn = x, n->00

if and only if lim v(xn — x) = + X.

When this is so, then bin v(x) = V(X).

In fact, if x * 0, then v(x) = v(x) for all sufficiently large n. A sequence x l , x2 , x 3 , . . . in k is called a Cauchy sequence in the v-topology when

v(x,n — xn )—* + 00, as m, n---> + 00 •

A convergent sequence is of course a Cauchy sequence, but the converse is not necessarily true. The valuation v is called complete if every Cauchy sequence in the v-topology converges to a point in k. If v is complete then the infinite sum

00

E xn = hill E X n n=1

converges in k if and only if

v(xn )--> + 00, as n --> + oo.

A valuation v of k is called discrete if v(k x ) is a discrete subgroup of It—that is, if

v(k x) = Zi3 = Inf3 I n = 0, ±1, ±2,. . .1

for some real number f3 a 0. If p = 0, then v is the trivial valuation vc, in Example 1. When )3 = 1—that is, when

v(V)= Z = {0, ±1, ±2, .. .},

v is called a normalized, or normal, valuation of k. It is clear that a valuation v of k is discrete but non-trivial if and only. if v is equivalent to a normalized valuation of k.

Let k' be an extension field of k, and 3/' a valuation of k'. Let y' 1 k denote the function on k, obtained from y' by restricting its domain to the subfield k. Then y' 1 k is a valuation of k, and we call it the restriction of y' to the subfield k. On the other hand, if v is a valuation of k, any valuation y' on k' such that

I, ' 1 k = y

is called an extension of v to k'. When y' on k' is given, its restriction I, ' 1 k is always a well-determined valuation of k. However, given a valuation v on k, it is not known a priori whether I/ can be extended to a valuation y' of k'. The study of such extensions is one of the main topics in the theory of valuations.


Let I, ' I k = v as stated above. Then the v'-topology on k' induces the v-topology on the subfield k so that k is a topological subfield of k'. Let o',

and f' denote the valuation ring, the maximal ideal, and the quotient field of v', respectively:

= {.7C' E k' I v'(x')a-0},

p' = k' I v' (x')> O}, = 07p ,.

Then

so that o=o'nk, p=p'nk=p'no

f = o/p = o/(p' n 0) = (0 + p')Ip' ç o'ip' = f'.

Thus the residue field f of v is naturally imbedded in the residue field f' of v'. On the other hand, if' I k = v also implies

v(k x) v'(k'') R+ . Let

e = e(v' I v)= [v'(le x ): v(0)], f = f(v' I v)= [f' :

where [vi(le x ):(k x )] is the group index and [f' : f] is the degree of the extension f1f. e and f are called the ramification index and the residue degree of v'/v, respectively. They are either natural numbers 1, 2, 3, . . . , or +00.

The following proposition is a fundamental result on the extension of valuations.

PRoposmoN 1.1. Let v be a complete valuation of k and let k' be an algebraic extension of k. Then v can be uniquely extended to a valuation y' of k': 3/' I k = v. If, in particular, k' I k is a finite extension, then 1/' is also complete, and

1 v'(x')=–

nv(N vik (x)), for all x' E k',

where n =[k' :k] is the degree and N kvk is the norm of the extension k' I k. Proof. We refer the reader to van der Waerden [23]. •

COROLLARY. Let k' I k, v, and y' be as stated above and let a be an automorphism of k' over k. Then v'o a = v'—that is,

vi (o-(x'))= v'(x'), for all x' E k', so that

v(o')= o', cr(p')= p'.

Hence a is a topological automorphism of k' in the v'-topology, and it induces an automorphism a' of f' over f:

J' . 1 2->

Proof. The proof can easily be reduced to the special case where k' I k is

Valuations 7

a finite extension. The first part then follows from Nkyk (a(x 1 ))= N kik(x 1 )- The second part is obvious. •

Let v be a valuation of k, not necessarily complete. It is well known that there exists an extension field k' of k and an extension V of v on k' such that V is complete and k is dense in k' in the v'-topology of k'. Such a field k' is called a completion of k with respect to the valuation v. More precisely, we also say that the pair (k', v') is a completion of the pair (k, v).

Let (k", v") be another completion of (k, v). Then there exists a k-isomorphism u: k' 2-, k" such that V = v" 0 a. Thus a completion is essentially unique, and hence (k', v') is often called the completion of (k, v). By the definition, each x' in k' is the limit of a sequence of points, x i , x2 , . . . , in k in the 'V-topology:

x' = lim xn . n--.00

Then

V f (X 1 ) = liM V f (X n ) = liM V (X n)* rt--.00

Hence if x' *0, then v'(x ') = v(x) for all sufficiently large n. It follows that

3/ 1 (lex)= v(e), fi = f, so that

e(v' I v)= f(v' I v) = 1

in this case. It is also clear that if (k', v') is a completion of (k, v) and if ti = cry, a >0, then (k', te), with if = cry', is a completion of (k, it).

EXAMPLE 4. Let K be an extension of k and let /2 be an extension of v on k to the extension field K:ttIk= v. Suppose that tt is complete. Let k' denote the closure of k in K in the s-topology. Then k' is a subfield of K, and (k', v'), with V = ti I k', is a completion of (k, v).

To study a valuation v on a field k, we often imbed (k, v) in its completion (k', v'), investigate the complete valuation V, and then deduce from it the desired properties of v. For example, in this manner we can deduce from Proposition 1.1 that if K is an algebraic extension of k, then every valuation on k has at least one extension on K.

1.2. Complete Fields Let v be a valuation of a field k. We say that k is a complete field with respect to v, or, simply, that (k, v) is a complete field, if v is a complete, normalized valuation of k. t

Let v be a normalized valuation of a field k, not necessarily complete, and let (k', v') be the completion of (k, v). Then v 1 (k")= v(kx)= Z by

t Some authors call a field a complete field if it is associated with a complete valuation, not necessarily normalized.


Section 1.1. Since 3/' is complete, (k', V) is a complete field. Many natural examples of complete fields are obtained in this manner.

EXAMPLE 5. Let p be a prime number and let vp be the p-adic valuation of the rational field Q in Example 2, Section 1.1. Since vp is a normalized valuation, the completion (k', v') of (Q, vp ) is a complete field. k' is nothing but the classical p-adic number field Qp , and vi, often denoted again by vp , is the standard p-adic valuation of Qp . For (Qp , vp ) , the valuation ring is the ring Zp of p-adic integers and the maximal ideal is pZp

so that the residue field is Zp /pZp = Fp , the prime field with p elements. Note that V(p) = 1.

EXAMPLE 6. Let F be a field, T an indeterminate, and F((T)) the set of all formal Laurent series of the form

E an Tn, an E F, — 00 <<n

where — 00 n indicates that there are only a finite number of terms an Tn with n <O, an O. Then k = F((T)) is an extension field of F in the usual addition and multiplication of Laurent series. Let v(0) = + 00 and let v(x) = i if

00

x o, x = E an Tn, with ai O. n=i

Then one checks easily that (k, v) is a complete field. The valuation ring is the ring F[[T]] of all (integral) power series in T over F, the maximal ideal is TF[[x]], the residue field is F[[T]]ITF[[T]]--= F, and v(T) = 1. The field k contains the subfield F(T) of all rational functions of T with coefficients in F, and the restriction v F(T) is the normalized valuation of F(T) in Example 3, Section 1.1. Furthermore, (k, v) is the completion of (F(T), yI F(T)).

Now, let (k, v) be any complete field and let o, p, and = oip denote, respectively, the valuation ring, the maximal ideal, and the residue field of the valuation V. They are also called the valuation ring, and so on, of the complete field (k, v). Since v(k x ) = Z, there exists an element r in k such that

v(z)= 1.

Any such element r is called a prime element of (k, v). Fix jr. Since p = {x E k v(x) % 1} in this case,

= (z) = oz. Hence, for any integer n 0,

pn = (.70) = 0.70 = {x E k I v(x) n}.

In general, let a be an o-submodule of k, different from {0}, k. As a* k, the set {v(x) I x E a, x *0 } is bounded below in Z, and if n denotes the

Valuations 9

minimum of the integers in this set, then

a = {X E k I v(x) n} = Og n .

Such an o-submodule a of k, a* {0}, k, is called an ideal of (k, v). The set of all ideals of (k, v) forms an abelian group with respect to the usual multiplication of o-submodules of k. By the above, it is an infinite cyclic group generated by p.

An ideal of (k, v), contained in o, is nothing but a non-zero ideal of the ring o in the usual sense. Hence the sequence

{0 } p n p2 p p0 = 0 (1.1)

gives us all ideals of the ring o. Since prz = (Tun), o is a principal ideal domain. Furthermore, in this case v(x) n<#.v(x)> n —1 for any n E Z. Hence all the ideals If, 0, in (1.1) are at the same time open and closed in k, and they form a base of open neighbourhoods of 0 in the v-topology of k. Since pn = (Tun)* {0}, the field k is totally disconnected and non-discrete as a topological space.

Next, we consider the multiplicative group k x of k. Let (g) denote the cyclic subgroup of k x , generated by a prime element Tr. Then the isomorphism kx I U v(k x )= Z in Section 1.1 implies

k = (it) x U, (n) Z. Let

Uo = U, Un = 1 + pn = {x Eolx 1 mod P n

for n

and let f ± and f x denote the additive group and the multiplicative group of f = o/p, respectively. Then we have a sequence of subgroups of k x :

{ 1 } • - • Un c • • • c Uo = U k x (1.2) such that

Uo/ Ui 4 X f[ Un/ Un +1 4f± for n (1.3)

In fact, the canonical ring homomorphism f = o/p induces a homomorphism of multiplicative groups, U—* fx, which in turn induces U0/ U1 2-> fx. On the other hand, for r 1, if we write an element of Un = 1 + pn = 1 + ogn in the form 1 + xgn with x E o, then the map

1 + xgn mod U„ 1 1--->x mod p

defines an isomorphism Unl U1 ft The multiplicative group k x is a topological abelian group in the topology induced by the v-topology of k. The groups tin , 0, are open and closed subgroups of 10 and they form a base of open neighbourhoods of 1 in the topological group k x. Hence k x is again totally disconnected and non-discrete as a topological space.

So far we have not yet used the fact that v is a complete valuation of k. But, now, this will be used in an essential manner. Let A be a complete set of representatives of the residue field f = o/p in o—that is, a subset of o such that each residue class of o mod p contains a unique element in A. We assume that A contains 0, namely, that 0 is the representative of p in A. For


each n E Z, fix an element sr,, in k such that

v(71-0= n,

and consider an infinite sum of the form

anZn) — 00<<n

where the an 's are elements of A and the sum over — œ « n means the same as in Example 6 above. Since v is complete and since v(angn ) = v(an ) + v(Jrn ) n so that v(an,x,)--* + co as n—* +00, such an infinite sum always converges to an element in k.

PROPOSITION 1.2. (i) Each x in k can be uniquely expressed in the form

x = E anz n , with an E A. --0.0<<n

If x * 0 and if ai * 0, an = 0 for all n <i, then

v(x)= i. (ii) Let

x = E angn , Y = E bng„, an , b E A.

Then, for any integer i,

v(x — y) i <=> an = b„ for all n < j.

Proof. We first prove (ii). If an = bn for all n, then x = y, v(x — y) = + co, and statement (ii) is trivial. Hence, assume that there is an integer m such that am bm , an = b„ for all n<m. Then

CO

X — y = E (an — bn) 7rn• n=m

Since am , b,n E A, am bm implies am 0 bm mod p so that

v(am — bm ) = 0, v((a,n — bm ),Irm )= m.

Therefore, for n>m, v((an — bn). 7rn)a. v(zn)= n> m = v((a,n — bm)Irm) , and it follows that

v(x — y) = v((a m — bm ),Irm )= m.

Statement (ii) is then obvious. The argument also proves the uniqueness and the formula v(x) = i in (i). Therefore, it only remains to show that each X E k can be expanded into an infinite series as in (i). We may assume that x *0, v(x) = i< +00. Now, by the definition of A,

o=A +p= fa+p la eAl.

Since V = {x E k I v(x) n} = or,, for n E Z, it follows that p n =Ann + =A

'7U + 147r n4-1+ • • • + Agm Pm±1, for all m a- n.

Valuations 11

As X E pi, we see that there exists a sequence of elements in A, ai , ai+2, . . . , such that

x = mod pi± 1 , for any j n=i

It then follows that cc

x = lim E angn = E an7rn . •

n=i n=i

Let A denote the set of all sequences (a0 , a l , a2 , .), where an are taken arbitrarily from the set A defined above. Thus A' is the set- theoretical direct product of the sets A n = A for all n 0:

CC

A' = A n , A n = A. n=0

Introduce a topology on A" as the direct product of discrete spaces A n ,

COROLLARY OF PROPOSITION 1.2. The map CC

(a0 , al , a2 , E a ,rr n n n=0

defines a homeomorphism of A' onto the valuation ring o of (k, v). Proof. (i) shows that the map is bijective and (ii) implies that it is a

homeomorphism. •

Let 7r be a prime element of k:v(g)= 1. Then we may choose irn as irn in the above proposition. Hence we see that each X E k can be uniquely expressed in the form

X = E ane, with an E A, -oe«n

and that if x * 0, ai * 0, and an = 0 for n <i, then

v(x) = v(E a nzn)= j. n=i

It follows in particular that

00

(1.4)

O = angn an E = ane I an E n=0 n=1

EXAMPLE 7. Let (k, v)= (Qp , vp ) in Example 5. Then we may set

A = {0, 1, , p — 1 } , ir =p

so that each X E o = Zp ; that is, each p-adic integer x can be uniquely


expressed in the form

X = E anpn, an e Z, 0 an <p. n=0

This is of course the well-known p-adic expansion of x. Next, let (k, v) be the complete field in Example 6: k = F((T)). In this case, we may set

A = F, yr = T.

For x E o, the expansion CO

x = E an T n , an E A = F n=0

mentioned above, is then nothing but the formal expression for x as an element of the power series ring o = F[M].

REMARK. Let k be a field with a non-trivial, discrete, complete valuation y on it. Then y is equivalent to a unique normalized valuation if of k: y — v, and y is again complete so that (k, II) is a complete field. Since the valuation ring, the maximal ideal, the residue field, and so on, are the same for y and v, to study a complete field is essentially the same as investigating a field k with such a valuation y.

1.3. Finite Extensions of Complete Fields

A complete field (k', v') is called an extension of a complete field (k, v) if k' is an extension field of k and if the restriction of y' on k is equivalent to V:

k c k' , v' I k — v.

In such a case, we shall also say that (k', v') is a complete extension of (k, v), or, in short, that le I k is an extension of complete fields.

Let (k, v) and (k', v') be as above and let

e = e(v 1 1,u), f = f(v 1 1,u).

e and f are then denoted also by e(k' lk) and f(k' lk), respectively, and they are called the ramification index and the residue degree of the extension k' 1 k of complete fields:

e = e(k' lk)= e(v 1 I,u), f = f(k' lk)= f(v' I ,u).

Let ,u = ay, œ> O. Then ,u(kx) = œv(kx)= crZ so that

e = [v1 (k"): ,u(kx)]= [Z: aZ] = a. Therefore

3/' I k = ev. (1.5)

This characterizes e = e(k' lk) and it also proves that

e(k' I k)< +cc.

Valuations 13

Let f = o/p and f' = o'/p' be the residue fields of (k, v) and (k', v'), respectively. Since y v, f is also the residue field of It so that

f = [f' :f]. (1.6)

In general, f(k' I k) is not finite. However, it is clear from (1.5) and (1.6) that if (k", v") is a complete extension of (k', v'), then

e(k" I k) = e(k"Ik')e(k' I k), f(k"I k)= f(k"I k')f(k' I k). (1.7)

Now, with the notation introduced above, let co i , . . . , cos be any finite number of elements in f', which are linearly independent over f, and for each i, choose an element in o' that belongs to the residue class coi in f' = o1p 1 . Fix a prime element g' of k' and let

= 0j<e=e(kilk).

LEMMA 1.3. (i) Let

= E y,, with yi E k. i = 1

Then v'(y 1 )= min(ey(yi), 1 s i

and . . . , are linearly independent over k. (ii) Let

= xim,, with Xij E k, lsiss, j<e.

Then v'(x') = min(ey(x 4)+ j, 1 i s, O 5f < e)

and the elements TN, 1 i s s, o s j < e, are linearly independent over k. Proof. (i) If y1 = • = y5 = 0, then y' = 0, and the statement is trivial.

Hence, renumbering if necessary, let

v(yi ) = • • • = v(y,.) < v(yi), for r

Put z. = Then z1 — 1, Zi E o for all i, and

yvyi = E E . i = 1

If y "Yi E p', it would follow from the above that co i is a linear combination of w2 ,. . . , cos with coefficients in f = alp, contradicting the assumption that (0 1, , ws are linearly independent over f. Therefore, y' 4 p'—that is,

'(y ' ly 1 )= 0—and we have

vi(y 1 )= vi(y)= ev(y i ) = min(ey(yi ), 1 s j s).

Assume now that Esi = i yii = 0 for some y e k. Then it follows from v'(0) = + 00 that v(y) = + 0c—that is, yi = 0 for all i. Hence , are linearly independent over k.


(ii) Let

y1 =x, o < e i= 1

so that e-1

X I = E ;=.

By (i), if yj *0, then v'(yi) is a multiple of e by (1.5) so that v'(ygr;) j mod e. Hence the finite values in v'(yo), . , v'(y e _ i ) are distinct, and it follows that

v'(x') = min(vi(yiz"), 0 < e).

As v'(yi) = min(ev(x 4), 1 i -s), we obtain

v'(x') = min(ev(x,;)+ j, 0 j < e).

The linear independence of tio , 1 j s , O sf <e, can be proved similarly as that of , in (i). •

LEMMA 1.4. Let (k', v') be a complete extension of a complete field (k, v). Assume that f(k' lk) is finite. Then k' lk is a finite extension and

[k' :1(]= ef, e = e(k' lk), f = f(k' lk).

Proof. Let w 1 ,. , wf be a basis of f' over f and let E o' be an element, taken from the residue class w i in f' = As in Proposition 1.2, let A be a complete set of representatives of f = o/p in o, containing 0, and let

, af E A}. i=1

Then A' is a complete set of representatives of f' = o'/p' in o'. Fix a prime element a of k and a prime element 7r. ' of k': v(z)= v'(2e)= 1. Writing each integer m in the form

m =te+j, t =0, ±1, ±2, , j= 0, 1, , e —1, we put

= Since y' k = ev, we then have

v'(7r,n' ) = et +1 = m.

Hence we may apply Proposition 1.2 for (k', v'), A', and n-„,' , and see that each x' E k' can be uniquely written in the form

x' = E d„rem , with a',n e A '. --.«m

Let

am = E aj , mj , a m E A. i= 1

Valuations 15

Then

where

X' = E E aj , m jr , = E xxei, i=1

= E ai,te +Jai E k. -00«t

0j<e,

Thus every x' in k' is a linear combination of 7/4 = , 1 j f, 0 j <e, with coefficients in k. However, the elements ni; are linearly independent over k by Lemma 1.3. Therefore, [k' : k] = ef < +00. •

With these lemmas, we are now able to prove the following

PROPOSITION 1.5. Let (k, y) be a complete field, and k' a finite extension of k. Then there exists a unique normalized valuation y' on k' such that

y' k y.

(k', y') is then a complete extension of (k, y) and

:k] = ef, for e = e(k' lk), f = f(k' lk),

y'(x 1 ) = -1

y(N k ,,k(f)), for x' E k'.

Proof. By Proposition 1.1, there is a unique valuation on k' such that II' k = v. Furthermore, such a valuation is complete and it satisfies

1 for x' E k',

with n =[k' :k]. Since v(kx) = Z, it follows that

so that 1.1' is discrete. Hence, there exists a unique normalized valuation y' on k' such that y' — ice. It then follows that y' is the unique normalized valuation on k' such that y' I k v. As tir is complete, y' is also complete, and (k', y') is a complete extension of (k, y). Let co l , , cos be any finite number of elements in f' that are linearly independent over f and let be an element in the residue class co„ 1 i s, in f' = oi/pi. Then, by Lemma 1.3, . . , are linearly independent over k so that s =[k' :k]. This implies that f = f(k' lk)= [f' : f] n < +00. Therefore, by Lemma 1.4, n = ef. Now, since I, ' — 1f—that is, I, ' = , a> 0—we have

vi(x) = Pv(N kyk (f)), for x' E k' ,

with p = aln> O. For x E k, this implies

v'(x) = Pv(f) = Pny(x).


However, vi(x) = ev(x) by (1.5). As n = ef, it follows that fi = 1/f so that

1 v (i(xi) = - v Nkyk (xi)), for x' ek'. •

Now, let { . . . , an } denote the elements

7/ 0 = 1 i f, 05_j<e

in Lemma 1.3, arranged in some order. In the proof of Lemma 1.4, we saw that ef =[k' :k]= n and that {a l , . . . , ci} is a basis of k' over k. Let

x, E k.

Then it follows from Lemma 1.3 (ii) that for any integer r,

re.(=>v(x,) r for i = 1, . . . , n.

For r = 0, this means that {a l , , an } is a free basis of the o-module : o' = oai Ef) oœn . Let kn (resp. O n ) denote the direct sum of n copies

of k (resp. o) and let k n be given the product topology of the v-topology on k. The above equivalence then shows that the k-linear map

kn ,

(x1, = E

is a topological isomorphism and it induces a topological o-isomorphism

On 4 = Dal e • •e oa'n.

(1.8)

PROPOSITION 1.6. Let (k, v) and (k', v') be the same as in Proposition 1.5, and let o and o' be their valuation rings. Then the trace map and the norm map of the finite extension k' /k:

are continuous in the v-topology of k and the v'-topology of k', and

g 0, N k'lk(° ' ) g 0 .

Proof. Let x' E k' and let

X'ci= >Xcij , x i, G k,

for the basis {al , . . . , an } stated above. Then x0 depends continuously on x' E k'. Since T k ,,k (x') and N k ,,k (f) are the trace and the determinant of the n X n matrix (x0), respectively, the first half is proved. If x' E 0', then x0 E o for all i, j by (1.8). Therefore, the second half is also clear. •

Now, let a' be an ideal of (k', v')—that is, an o'-submodule of k' different from {0 } , k' (cf. Section 1.2). We define the norm N k ,,k (a') of

Valuations 17

a' to be the o-submodule of k generated by Nkyk (x') for all x' E a'. Then Nkyk (a') is an ideal of (k, y) and it follows from the formula for yi(x') in Proposition 1.5 that if a' = rn, m E Z, then

Nvik(a 1 )=Pmf, with f = f(k'lk).

Hence, in particular,

Nk ',k(o' ) = 0, Nvik(P 1 ) = V.

Chapter II

Local Fields

A complete field, defined in Section 1.2, is called a local field when its residue field is a finite field. In this chapter, we shall discuss some preliminary basic results on such local fields.

2.1. General Properties Let (k, v) be a local field. By the definition, (k,v) is a complete field and its residue field f = o/p is a finite field. Let q be the number of elements in f—that is,

f = Fq ,

Fq denoting, in general, a finite field with q elements. When f is a field of characteristic p—namely, when q is a power of a prime number p—the local field (k, v) is called a p-field. This happens if and only if

po ç p, that is, v(p • l k ) > 0

where 1 k denotes the identity element of the field k. Hence it follows that if (k, v) is a p-field, then the characteristic of the field k is either 0 or p.

EXAMPLE 1. Let Qp be the p-adic number field, and vp the p-adic valuation on Qp (cf. Example 5, Section 1.2). Then (Qp , vp ) is a complete field and its residue field is

f= o/p=Zp /pZp =Fp .

Hence (Qp , vp ) is a p-field of characteristic O. On the other hand, let F be any finite field of characteristic p; for example, F = Fp . Let k = F((T)) be the field of formal Laurent series in T with coefficients in F, and let VT be the standard valuation of k (cf. Example 6, Section 1.2). Then (k, V T ) is a complete field and its residue field is

f = o/p = FRTfilTF[[T]]----- F.

Therefore (k, V T ) is a p-field of characteristic p.

EXAMPLE 2. Let k be a finite algebraic number field—that is, a finite extension of the rational field Q—and let o denote the ring of all algebraic integers in k. It is known that each maximal ideal m of o defines a normalized valuation v of k with finite residue field. Hence the completion (k', v') of (k, v) is a local field. In fact, if p is the prime number contained in m, then (k', v') is a p-field. (Qp , vp ) in Example 1 is a special case of such (k', v') for k = Q, m =pZ. Application of the theory of local fields for algebraic number fields is based on this fact. Compare Lang [17].

Local Fields 19

PROPOSITION 2.1. Let (k, y) be a local field. Then k is a non-discrete, totally disconnected, locally compact field in its v-topology. The valuation ring o(= p°) and the ideals pn of o, n al, are open, compact subgroups of the additive group of the field k, and they form a base of open neighborhoods of 0 in k. Furthermore, o is the unique maximal compact subring of k.

Proof. Let A be a complete set of representatives of f = o/p in o, containing 0 (cf. Section 1.2). Since f is a finite field, A is a finite set. Hence the set A = fr= c, An , An = A, in the Corollary of Proposition 1.2 is a compact set as a direct product of finite sets A n , n a O. Therefore, by the same Corollary, o is compact in the v-topology. We already stated in Section 1.2 that each pn, n a 0, is at the same time open and closed in k and that the pn's in (1.1) gives us a base of open neighbourhoods of 0 in k so that k is a non-discrete, totally disconnected topological field. Since o is compact, k is locally compact, and since pn is closed in o, pn is also compact. Let R be any compact subring of k. Then the compactness implies that the set { v(x) IxER } is bounded below in the real field R. If x E R, then xn E R and v(f) = nv(x) for all n al. Hence, by the above remark, v(x) _.! 0—that is, X E O. This proves that R c 0 so that o is the unique maximal compact subring of the field k. •

It is known that if k is a non-discrete locally compact field, then k is either the real field R, the complex field C, or a local field in its v-topology.t Thus, as a topological field, a local field is characterized by the property that it is locally compact, but is neither discrete nor connected.

Now, let a be a prime element of a local field (k, v):y(n) = 1. Since pn = on", na 0, the map xi-->xan, x E o, induces an isomorphism of o-modules:

oip 4 priir+1, for n a O.

Since f =Fq , [o: 0 = q, it follows that

[o : pi = qn, for n a O.

Thus o/pn is a finite ring with qn elements. Since o is compact and the intersection of all pn, n a 0, is 0, we see that

U = lim o/pn, 4---

where the inverse limit is taken with respect to the canonical maps o/pm ---> o/pn for all m _.__ n a O. Therefore, if (k, v) is a p-field so that q is a power of p, then the additive group of o is an abelian pro-p-group--that is, an inverse limit of a family of finite abelian p-groups (cf. Section 2.2).

For the proof of the next proposition, we first prove the following

LEMMA 2.2. Let f = o/p = CI for a p-field (k, v). Then, for each x E o, the

t Compare Weil [24]. Note that a local field in that book is, by definition, a non-discrete locally compact field.


limit

(0(x) = Ern xqn n-0.00

exists in o, and the map w:o--> o has the following properties:

w(x) x mod p, w(x)q = w(x), w(xy)= w(x)w(y).

Proof. By induction, we shall prove the congruences

xq n xqn-1 mod pn

for all n 1. For n =1, x'1 x mod p follows from the fact that f= o/p is a finite field with q elements. Assume the congruence for n :1 so that

n-1 X qn = X q + y with y E pn . Then

n+1 ■—• ( Xq —

) 2:j

inn X

-1 y- .

i=o

For 0 < i < q, the integer

(q) -q (q - 1) i

is divisible by p so that (flyq -i is contained in p n±1 . Since the same is

obviously true for i = 0, we obtain nn+1 n ±1

inou .

Now, we see from these congruences that fen l,„ 1 is a Cauchy sequence in in the y-topology. As y is complete and U is closed in k, the sequence

converges to an element w(x) in U. It is clear that the congruences yield xq n x mod p for all n 1. Hence w(x)L----- x mod p. We also see

co(x)l = lim xq n±1 = w(x), w(xy) = lim xq nyq n = w(x)w(y). • n—>cc n—*oc

PROPOSITION 2.3. Let (k, y) be as stated above and let

V = {x e k I xq -1 = 1}, A = V U {0} = {x e k x`i = x} .

Then A is a complete set of representatives of f = o/p in o, containing 0; V is the set of all (q — 1)st roots of unity in lc; and the canonical ring homomorphism o-->f=o/p induces an isomorphism of multiplicative groups:

L'; f

In particular, V is a cyclic group of order q —1.

Proof. Let A' = {w(x)i xED}. As w(x) x mod p, each residue class of )3 mod p contains at least one element in A', and as w(x)q = w(x), A' is a subset of A. However, the number of elements x in k satisfying x'1 — x = 0 is at most q, while the number of elements in f = o/p is q. Hence A = A' and

Local Fields 21

A is a complete set of representatives of f = o/p in o. Obviously, 0= w(0) E A. Since co(xy)= co(x)co(y), the other statements on V are clear. •

REMARK. This proposition is also an easy consequence of the well-known Hensel's Lemma.t

PROPOSITION 2.4. Let (k, v) be a p-field of characteristic 0:Q c k. Let e = v(p), where p = p • 1k and let f = Fq , q = pf, for the residue field f = o/p. Then (k, v) is a complete extension of the p-adic number field (Qp , vi,), and

[k:Qp ] = ef < +00.

Furthermore, the valuation ring o of (k, v) is a free Zr-module of rank ef =[k: Qr ].

Proof. Let A = y 1 Q. Since (k, v) is a p-field and p =p • l k # 0,

O< )(p)= v(p)< +00.

However, it is known that such a valuation A on Q is equivalent to the p-adic valuation yr, of Q (cf. the remark in Example 2, Section 1.1.). Let k' denote the closure of Q in k. As (k, v) is complete, (k', v I k') is a completion of (Q, A,) (cf. Example 4, Section 1.1). It then follows from A — yr that k' = Qp and y I Qp — yr . Hence (k, v) is a complete extension of (Qp , vp ) in the sense of Section 1.3. Since yr (p) = 1, y(p) = e, and y 1 Qp = e(klQr)vp , we obtain e(klQr)= e. On the other hand, f = Fq , q = pf, and Zp /pZp = Fr

imply f (k I Qr) =f < +00. Therefore, [k :Qr] = ef by Proposition 1.4. That o is a free Zr-module of rank ef is a special case of (1.8) in Section 1.4. •

PROPOSITION 2.5. Let (k, v) be a p-field of characteristic p with f = o/p = Fg and let (Fg ((T)), V T ) denote the p-field of Laurent series in T stated in Example 1. Then k contains Fg as a subfield and there exists an Fq -isomorphism

(k, v):_'; (F g ((T)), V T ),

namely, an Fg-isomorphism k 2;Fg ((T)), which transfers the valuation v of k to the valuation V T of Fg ((T)).

Proof. Since k is a field of characteristic p, the set A = {x E k 1 xq = x} in Proposition 2.3 is a subfield of k with q elements. Then Fq = A c k. Now, fix a prime element it of k and apply Proposition 1.2 for A = Fq and

= .7r", n E Z. We then see that the map

E anz" F-0 E a,., T", an E Fq , n n

defines an Fg -isomorphism (k, v):_•;(Fg ((T)), VT). •

t Compare van der Waerden [23].


2.2. The Multiplicative Group le

Let (k, y) be again a p-field with residue field f = o/p =Fq . We shall next study the multiplicative group le of the field k.

PROPOSMON 2.6. The group k x is a non-discrete, totally disconnected, locally compact abelian group in the topology induced by the y-topology of k. The unit group U (=U0) and its subgroups Un =1+ , n are open, compact subgroups of le , and they form a base of open neighbourhoods of 1 in le. Furthermore, U is the unique maximal compact subgroup of le.

Proof. By Proposition 2.1, Un = 1 + pn, n 1, are open, compact subgroups of kx. Since U/1/1 fx by (1.3), U/1/1 is a finite group. Hence U is also an open, compact subgroup of le . That U is the unique maximal compact subgroup of le can be proved similarly as the fact that o is the unique maximal compact subring of k. The rest of the proposition is clear from what we mentioned in Section 1.2. •

Now, let r be a prime element of (k, v). By Section 1.2, we have

= (a) x U, (7r) Z,

U I , Un /U„ ±i f± , for n 1.

Since f = Fq , it follows that

[U: = q — 1, [Un :U„i]= q

so that

[U: Un] = (q —1)qn-1, [U1 :Un]= q" 1 , for n 1. (2.1)

We also know that the canonical ring homomorphism f = o/p induces the isomorphism U/Ui f X . However, by Proposition 2.3, the same ring homomorphism induces V fx , where V = E k xq' = 11 is a subgroup of U = {x E k v(x) = 01. Hence we obtain

U= V x = (a) x V x (2.2)

where (a) Z, V Z/(q — 1)Z. Therefore, the structure of the abelian group k x will be completely known if we can determine the structure of U1 .

In general, let G be a finite abelian p-group. Then G can be regarded as a module over Z/pnZ whenever n is sufficiently large. As Z/pnZ Zr /pnZr , we see that we can canonically define a structure of 4-modu 1e on such a group G. Let G, now, be an abelian pro-p-group--that is,

G = lim G,

where {Gi } is a family of finite abelian p-groups. Since every G. is a Zr-module in the natural manner, the inverse limit G can also be made into a Zr -module, and one checks easily that the structure of Zr -module on G thus defined is independent of the way G is expressed as an inverse limit of

Local Fields

23

finite abelian p-groups. In short, each abelian pro-p-group G is a Zr-module in a canonical manner.

Now, just as o is the inverse limit of o/pn, n _-__ 0, the compact group U1 is the inverse limit of finite abelian p-groups Ut/Un with respect to the canonical maps Ui /U,n --> U1 /U,., for m ___- n _.! 0:

Ui = lim U1 / U,,. 4.---

Thus U1 is a Zr -module by the above remark, and one also sees that the Un , n _-__ 1, are Zr-submodules of U1 . We shall next study the structure of the Zr -module U1 .

Let (k, y) be a p-field of characteristic O. By Proposition 2.4, k is a finite extension of the p-adic number field Qp and its degree d =[k: Qp ] is given by

d = ef,

where e = e(k I Qp) = y(p), and where f = f(k1Q p ) is the exponent of q = pf. Let

W = the set of all roots of unity with p-power orders in k.

Since (2.2) induces kx/Ui ------- Z el Z/(q — 1)Z, we see that W is a subgroup of U1 ; in fact, W is the torsion submodule of the Zr -module Up

PROPOSITION 2.7. Let (k, y) be a p-field of characteristic 0 and let W be the group of all p-power roots of unity in k. Then W is a finite cyclic subgroup of U1 =1+ p and Ui lW is a free Zr-module of rank d =[k: Qp ]:

Ui = w e zpd, w — zipaz, a __ O.

Proof. Let n> e = y(p) and let Ui,', = {x" Ix e Un }. (tg is not the direct sum of p-copies of Un .) U1', is the image of the compact group Un under the continuous endomorphism xi--- > x" so that it is a compact subgroup of Un . Now, since n> e,

Uft --= (1 ± P n r ---= 1 ±pPn MOd P 2n ,

pv = vi+e, 13 2,n = vz +e +1 . where Hence we have

Un -F e — tgUn+e+1.

As this holds for any n> e, we also obtain

Un, = U1',U,,, for all m __-_ n + e + 1.

This shows that Un , is the closure of the compact subgroup U1', of Un so that

Un , = (gt , for n _-__e,

and it follows that

[Un :U ]=[U,,:Un±e]= qe = pef = pd.


In general, it is easy to see (the proof is left to the reader as an exercise) that if A is a compact 4-modu 1e (in additive notation) such that AlpA is finite, then A is finitely generated over Zr,. Since

+co,

U1 is finitely generated over Zr,. Hence its torsion submodule W is finite. Furthermore, as a finite subgroup of the multiplicative group kx of the field k, W is also cyclic. Now, choose n> e large enough so that W n un = 1. By the structure theorem for finitely generated Zr-modules, we see from [Un : Ufd =pc', [Ui :Un ]< +co that Ui/W Zpd, W Zpd. •

COROLLARY. For a p-field (k, y) of characteristic 0,

kx= (.70xU=<.7r)xVxU1 = (.7)><VxWxU'

Z Z/(q — 1)Z Es Z/paZp Zpd, a d =[k:Qp ].

Hence V x w is the group of all roots of unity in k and it is a finite cyclic group of order (q —1)pa.

REMARK. The proposition can be proved also by using the p-adic logarithm for k:f

X2 x 3

1 •og(l+x)=x--+-- • •, x 3

x e p.

We now consider the Zr -module U1 = U(k) for a p-field (k, y) of characteristic p. By Proposition 2.5, we may assume that

k =Fq ((T)), U1 =1+ p =1+ TFJ[T]]

where q=pf, f=[Fq :Fp ] 1. Let {w 1 , . . . , cod be a basis of Fq over the prime field Fr,. For each integer n 1, prime to p, let A n denote the direct product of f copies of Zp . Take any element a, in A n : = (a1 , . . , af ), ai E Zp . We define an element g(œ) in Un = 1 + If by

gn (a) = ± coi Tn)az.

Let bi E Z, mod pZp for 1 i f and let w = EL biw i E Fq . Then

g(a)=- II (1 + i Tn) b, 1 + cor mod pn+ 1 .

Let m = nps, s E Z, s O. Since k is a field of characteristic p, we have

gn (psa)= gn (a)Ps coPT'a mod r4-1 .

When a ranges over all elements of A n , w and, hence, at run over all elements of Fq- Therefore, we see from the above that

U,n = gn (psA n )U,n±i , for m = nps. (2.3)

t See Lang [17], Chapter IX.

Local Fields 25

Note also that gn (psa)----1 mod pm+ 1 <=> 0 )=0<=>b i -- 0 mod p, 1-i-f.t:4 ai *OmodpZr , 1 - if<=',>aepAn . Thus

a' f PA n <=> gn(Ps CO f U„,±1, for m = nps. (2.4)

PROPOSITION 2.8. Let (k, y) be a p-field of characteristic p. Then the Zr-module Ul = U1 (k) is topologically isomorphic to the direct product of countably infinite copies of Zr.

Proof. Keeping the notation introduced above, let A be the direct product of the 4-modu 1es A n for all integers n -_1, prime to p:

A = n A n . n

A is a compact Zr-module in an obvious manner. We define a map g:A--->U1 by

g()= n gn (an ), for = (. .. , an , ...) E A, an E A n . n

Since gn (œn )E un , the product on the right converges in U1 and g defines a continuous Zr-homomorphism. Let m be any positive integer and let m = nps, (n, p)= 1, s -.. O. Since gn (A n ) ç g(A), it follows from (2.3) that each coset of Um / U,n±i is represented by an element of g(A). Hence g(A) is dense in U1 . However, as A is compact and g is continuous, g(A) is compact, hence closed, in U1 . Therefore, g(A)= U1—that is, g is surjective.

Next, let = (• • • , a'n, • • .) EA, *O. Then an 00 for some n. Such an an can be uniquely written as an = ps 13n with s = s(a):- 0 and fin E A n ,

fin f PA n . It then follows from (2.4) that

gn (CYn ) E U,n , gn(ain) f U.-1-1, for m = m(œn )=nps.

Since the n's are prime to p, the integers m(an ) are distinct for all an 's, an 0 O. Hence, let n denote the integer, prime to p, such that an *0 and m(an ) < m(an ) for all n' * n with an , O. Then, for all n' * n,

gn ,(an ,) E U,n± i, with m = m(an)<m(an ,).

Therefore,

gM --'-'gn(crn)* 1 mod Uni+i, g() o 1,

and this proves that g is injective. Thus g :A 4 U1 . Since A is obviously the direct product of countably infinite copies of Zr , the proposition is proved. •

By (2.2) and Propositions 2.7 and 2.8, the structure of the multiplicative group le of a local field k is completely determined.

2.3. Finite Extensions

Let (k, y) be a local field with residue field f = o/p =Fq . Let k' be any finite extension over k. By Proposition 1.5, there is a unique normalized valuation


V on k' such that y' 1 k — y, and (k', y') is then a complete extension of (k, y) with

[k' :k]= ef, e = e(le lk), f = f(k' lk).

Let f' = ovp' be the residue field of (k', v'). Then

[f' : f] = f(k'lk) [1c' :k]< +00

so that f' is a finite field with qf elements: f' = Fq ,, q' = qf. Therefore, (k', y') is a local field; it is a p-field if (k, y) is a p-field. In this manner, a finite extension k' of k always provides us a local field (k', v'). In such a circumstance, we shall often say, without specifying the valuations, that le lk is a finite extension of local fields.

Taking account of the above remark, we can get from Propositions 2.4 and 2.5, the following simple description of the family of all p-fields for a given prime number p, although the statement here is not as precise as in those propositions:

THEOREM 2.9. A field k is a p-field of characteristic 0 if and only if it is a finite extension of the p-adic number field Qp , and k is a p-field of characteristic p if and only if it is a finite extension of the Laurent series field F p ((T)).

Now, let le lk be a finite extension of local fields and let

n =[k' :k]= ef, e = e(k' lk), f = f(k' lk).

The extension k' lk is called an unramified extension if

e =1, f = n,

and it is called a totally ramified extension if

e = n, f =1.

In general, let z and Jr' denote prime elements of k and k', respectively: v(z) = v 1 (7r 1 ) = 1. Then it follows from (1.5) and Proposition 1.5 that

e = vi(z), f = y(Nkyk (a- ')).

Therefore, the extension lelk is unramified if and only if a prime element z is also a prime element of k', and /elk is totally ramified if and only if the norm Nkyk (z) of a prime element z' of k' is a prime element of k. Let k" be a finite extension of k':

k c k' c k".

Then it follows from (1.7) that k"lk is unramified (resp. totally ramified) if and only if both k"Ik' and lelk are unramified (resp. totally ramified).

Let lelk be again an arbitrary finite extension of local fields and let, as before, n =[k' :k]= ef and f = Fq , f' = Fq ,, for the residue fields with q' = qf. Further, let

ko =k(A'), k ckock'.

Local Fields 27

LEMMA 2.10. ko is a splitting field of the polynomial xq' — x over k and k olk is an unramified cyclic extension with degree [ko :k]= f.

Proof. By Proposition 2.3, A' is a complete set of representatives of f' = oVp' = Fq , in o' so that it contains q' elements—that is, all roots of

— X. Hence /co = k(A') is a splitting field of X"' — X over k. As X"' — X has q' distinct roots, k o/k is a separable, hence Galois, extension. Let fo denote the residue field of ko : f fo f'. Then A' c ko implies fo = f' = Fq ,, q' =qf. As f is a finite field, fo ( = f') is a cyclic extension of degree f over f. Now, by the Corollary of Proposition 1.1, each a in the Galois group Gal(ko/k) induces an automorphism a' in Gal(fo/f), and the map a 1--> a' defines a homomorphism

Gal(ko/k)—> Gal(fo/f).

Assume that a' = 1—that is, a'— 1. Since the set A' of all roots of X"' — X is mapped onto itself by a and since A' is also a complete set of representatives of fo = ocipo in the valuation ring oo of 1(0 , a' =1 implies that a fixes every element of A'. As ko = k(A'), it follows that a = 1. Thus the above homomorphism Gal(k o/k)—> Gal(fo/f) is injective, and we obtain

[ko : k] = [Gal(ko/k):1] [Gal(f0/f):1]= [fo : f].

On the other hand, by Proposition 1.5,

[fo : f] = f (k olk) [k 0 : k].

Therefore,

[ko :k]= f(kol k)= [to : r] = [f' :f] =f

and k o/k is an unramified Galois extension of degree f. It also follows that the above homomorphism of Galois groups is an isomorphism:

Gal(ko/k) Gal(fo/f)

so that Gal(ko/k) is a cyclic group of order f. •

PROPOSMON 2.11. Let (k, v) be a local field with residue field f = o/p = Fir Then, for each integer n there exists an unramified extension k' /k with degree [k' :k]= n, and k' is unique up to an isomorphism over k. The field k' is a splitting field of the polynomial Xq n — X over k, and it is a cyclic extension of degree n over k. Let f' be the residue field of the local field (k', v'). Then each element a of Gal(k'/k) induces an automorphism a' of fig, and the map a'—* a' defines an isomorphism

Gal(k'/k) Gal(f'/f).

Proof. Since the finite field f has a separable extension of degree n over it, there exists a monic irreducible polynomial g(X) with degree n in f[X]. Let f(X) be a monic polynomial of degree n in o[X] such that g(X) is the reduction of f(X) mod p. Let w be a root of f(X):f(w)= 0, and let k' = k(w). If f(X) = Xn + 7:01 aiX i, ai E o, then w" = ai w i, ai E o,

28


and this implies that v'(w) 0—that is, W E 0', 0' being the valuation ring of (k', v'). Hence, let co denote the residue class of w in f' = o'/13' for (k', v'). Then clearly g(c0= O. As g(X) is irreducible in f[X], it follows that [f(w): f] = deg g(X)= n. On the other hand, f(w)= 0 implies [k':k]= [k(w):k] deg f(X)= n. Hence, by Proposition 1.5,

n

and it follows that [k':k]=f(k'lk)=n so that k'lk is an unramified extension of degree n. This proves the existence of k'.

Now, let k' be any unramified extension of degree n over k. Apply the above lemma for k' I k. Since n= f in this case, we see that k' = ko so that the field k' has the properties stated in the proposition. Furthermore, as a splitting field of JO" – X over k, such a field k' is unique for n 1 up to an isomorphism over k. Hence the proposition is proved. •

Now, since f = Fq , the Galois group Gal(f'/f) in the above proposition is generated by the automorphism

wq, for CO E f'.

Let q7 denote the automorphism of k'lk, corresponding to the above automorpliism of Vif under the isomorphism Gal(k7k) Gal(r/f) in Proposition 2.11. Then q9 is a generator of the cyclic group Gal(k'/k) and it is uniquely characterized by the property that

cp(y)--=- yq mod 13', for all y E 0'.

This cp is called the Frobenius automorphism of the unramified extension k'/k; it will play an essential role throughout the following.

Next, let k'lk be again an arbitrary, not necessarily unramified, finite extension of local fields and let e = e(k' lk), f = f(k' lk), n =[k' :k]= ef,

=

PROPOSMON 2.12. The splitting field 1( 0 of Xq' – X over k in Lemma 2.10 is the unique maximal unramified extension over k, contained in k'. k' /k 0 is a totally ramified extension and

[k':k0]= e, [ko :k]=f.

Proof. We already know that ko/k is an unramified extension with ['co : k]= f. Let k 1 be any unramified extension of k, contained in k', and let fi =f(k1/k)=[k1:k], q"= qfl. By Proposition 2.11, k 1 is the splitting field, in k', of the polynomial Xe?" – X over k. As f =f(k'lk)=f(k'lk i )f(k i lk), fi is a factor of f so that q" – 1 divides q' – 1, and Xq"– X divides Xq' – X in k[X]. Therefore, k 1 is contained in the splitting field ko , in k', of X"' – X over k. This proves that ko is the unique maximal unramified extension of k, contained in k'. Now, f (k ol = [ko : 1(1 = f = (k' I k) implies f(k' I k o) = 1— namely, that k'/k 0 is totally ramified. As ef = n =[k':k], f = [ko :k], we also have [k' :k0]= e. •

Local Fields 29

The field 1(0 in the above proposition is called the inertia field of the extension k'/k.

2.4. The Different and the Discriminant

Let k'lk still be a finite extension of local fields and let f = o/p, f' = and so on, be defined as above. Throughout this section, we assume that k' /k is a separable extension.

By Proposition 1.6, the trace map Tkvk:k ' --> k is continuous and Tk ,,k (o') OE o. Let T = Tkvk for simplicity, and define

m = {x' E k'IT(x'o') _g ol.

Since k'lk is separable, there exists an element x' E k' such that T(x') * O. Hence T(k')= k, and it follows from the definition that m is an o'- submodule of k', different from k'. Furthermore, T(o') o implies that o' m, m {0}. Therefore m is an ideal of (k', y') (cf. Section 1.2), containing o' so that its inverse 2 = rrt -1 is a non-zero ideal of the ring o'. We call 2 the different of the extension k' /k and denote it by 2(k' lk):

Let D(k'lk)= Nk7k( 2(le lk)).

This is a non-zero ideal of o and it is called the discriminant of k' /k. To find a simple description of the different 2(lelic), we need the

following

LEMMA 2.13. There exists an element w in o' such that 1, w, wn -1 form a free basis of the o-module o':

o' = ED ow ED • • •

In particular, we have

o' = o[w],

where o[w] denotes the ring of all elements of the form h(w) with h(X) E o[X].

Proof. We first note that f' = f(w) for some co E f', because f'/f is a separable extension. Let g(X) be the minimal polynomial of co over f; g(X) is a monic polynomial in f[X] with deg g(X) = [f' : f] = f(k'lk) =f. Let f(X) be a monic polynomial of degree f in o[X] such that g(X) is the reduction of f(X) mod p, and let w be an element of o', belonging to the residue class co in f' = o'/p'. Clearly, g(w) = 0 implies f(w)=----- 0 mod p'. Let w' = w + where is a prime element of k'. Then w' again belongs to the same residue class co, and

f(w') f(w)+ f' (w)a' mod p' 2 ,

where f' = df I dX. However, as f' = f(co) is a separable extension of f, we

30


have g'((o) 0 for g' = dgIdX, and this implies f'(w) 0 mod There- fore, it follows from the above congruence that either f(w)0 mod p'2 or f(w')*0 mod p' 2 . Replacing w by w' if necessary, we see that the residue class co contains an element w such that f(w)--=-- 0 mod p', f(w)+0 mod p' 2 . Thus f(w) is a prime element of k'. Since 1, co, . . . , co-f-1 form a basis of f' over f, the proof of (1.8) shows that the ef elements

lb; = 1 <e

form a basis of the o-module o'. Since ef = n, deg f(X) =f, it follos that 1, w, wn-i generate the o-module o'. Then those n elements also generate the k-module k' so that they form a basis of k' over k. Therefore,

o' = o ow ED • • • e own -1 = o[w]. •

COROLLARY. Let k' /k be totally ramified and let z' be any prime element of k'. Then

= o e e • • • e 07t n = ok'].

Proof. In this case, f(le lk)= 1 so that f' = f. Hence, in the above proof, we may set

= 0, g(X)=f(X)=X, w = . •

PROPOSITION 2.14. Let w be an element of o' as in Lemma 2.13 and let f(X) denote the minimal polynomial of w over k. Then

a (k' lk)= f' (w)0'

where f'(X)= dfldX.

Proof. Let f(X)= E7i01 • , ai E k. Since o' = o ow e • • • e

ow', wn is a linear combination of 1, w, . . . , wn -1 with coefficients in o. Hence ai E o for 0 i <n. Let w1 , , wn be the roots of f(X) in an algebraic closure of k'. Since elk is separable, these roots are distinct and they are the conjugate of w over k. By a classical formula of Euler,

1 n 1 1

f(X) =

f' (wi ) X —

Putting Y = X-1 , we obtain as formal power series

yn xn, 1 Y wm-1) Tn. = 2.4 = E

1 +a 1 Y+ • • + ao Yn i =1 (wi) 1 — w T

iY m=i (f(w)

Since ai E o for 0 i < n, it follows that T(w if (w) -1 ) E o for all i 0 and that

ir (wn-if , (w) -1 ) _ 1, T(wr(w) -1) = 0, for 0 i <n — 1. (2.5)

Now, let y E k' and let n —1

y = bi wi, i=0

bi e k.

(T(yiyi)) = ( w7-1

Local Fields 31

Then

yf' (w) -1 E m = <=> T(yf' (w) -1 14,') E

n-1

<4. E bi T(w 1-3f'(w)) E o, for 0 < n. i=o

By (2.5), this condition for j = 0, 1, . . . , n —1 yields successively bn-i o, , bo E C). Hence

yr(w) -1 E <>.y E o' = o e ow e • -

so that 2 -1 = of(w) -1—that is, 2 =f1 (w)o' .

PROPOSITION 2.15. Let yn be a basis of k' over k such O' = oyi e • • • e oyn (e.g., 1, w,... , 10 -1 in Lemma 2.13). Then

D(k' I k)= det(Tkyk(YiY;))o.

for 0-j<n

ED own- 1

•

that

Proof. It is easy to see that the independent of the choice of {y',. Hence we may assume that {Yi,. proof of Proposition 2.14, let w1 , . Then

so that

right-hand side of the above equality is , yn } such that o' = °he • • • e oy n • y yn } = {1, w w n-1

}

As in the wn denote the conjugates of w over k.

• 1 Wi, . y w7-1

Wn •

y w nn - 1 1 , wny yyyy w nn - 1

det(T (YiYi)) = 11 (wi - 4 )2 = ± f'() = ±Nkyaf'(w)).

i=1

Hence, by Proposition 2.14,

det(T(YiY1)) 0 = Nkva r (w))0 = Nkva r (w)o r )=- Nkvk( 2 (1C/k)) = D(kr I k).

•

PROPOSITION 2.16. Let k"I k be a finite separable extension of local fields and let k k' k". Then

2(k"lk)= 2(k"1102(kr 1 k).

Proof. Let m = 2(kr 1 k)', mr = 2(k"110 -1 , m" = 2(k"lk) -1 . Let o" denote the valuation ring of k" and let m = pra = :Cap', where z' is a prime element of k'. Since o"

Teik(z o") = Tk 7k(Tvik , (z o")0 ')

32


for any Z E k". Hence

Z E Ill" <=> Tvik(Z 0") ç o<=> Tkvk (Tkve (zo")o) ç o

<=> Telle (zo") g m = ;Cap' <=> Tenc ,(7r 1-azo") c o'

<=> Jr ' aZ E III' <7> Z E 7r a lll = Tne .

Therefore, m" = mm'—namely, 2(k"lk)=2(k"11( 1 )2(1‘ 1 1k). • We now consider the different 2(k 1 lk) in the case where k' /k is either

unramified or totally ramified. Suppose first that lelk is unramified. Let w and f(X) be as in Proposition 2.14 and let co denote the residue class of w mod p' : 0) E f' = 0 1 /p'. Since o' = o @ ow @ • • • S own-1 and [f' : f] = [k':I1=n, we see that f' = f(w) and the minimal polynomial g(X) of co over f is the reduction of f(X) mod p. As fi/f is separable, Ow) O. Hence f'(w)* 0 mod p', and it follows from Proposition 2.14 that

2(k 1 lk)= 0'.

Next, let 'elk be totally ramified and let e=n=[k':k]>1. By the Corollary of Lemma 2.13, we have then o' = o ED oz' ED - • - ED ozin -1 for any prime element 7r' of (k', v'). Let

n-1

f(X)=X' 1 i=0

ai E 0

be the minimal polynomial of a' over k. From f(70= 0 and I, ' 1 k = ev = nv, we then see that v(ai) > 0—that is, ai E p, for 0 s jS n-1. It then follows that

n-1

f' (70= nnin-1 + E iain"-1 7=0 mod p' i=0

so that

2(kilk)=T(n 1 )0' * 0 ' .

PROPOSITION 2.17. Let le lk be a finite separable extension of local fields. Then 'elk is unramified if and only if 2(kilk)=0'— that is, if and only if D(k'lk)= 0.

Proof. Let 1(0 denote the inertia field of the extension k' lk:kc k 0 c k' (cf. Proposition 2.12). Then

g(lelk)= 2 (klk 0)2(k 0lk)=2(kilko)

because 2(k 0Ik)= 00 by the above remark. By the same remark, (k' / k0)= 0' if and only if e=[k':k0]=1. Hence the proposition is proved. •

2.5. Finite Galois Extensions Let le lk now be a finite Galois extension of local fields and let

G = Gal(k'/k).

Local Fields 33

Let a be any element of G. By the Corollary of Proposition 1.1,

=

for n -_1,

o' and p' being the valuation ring and the maximal ideal of k'. Hence, for each n-_1, a induces a ring automorphism

0.n:0 , 4 in-1-1 4 o vp ,n+1..

The map a i--> a, then defines a homomorphism of G into the group of automorphisms of the ring o'ip'n+i. Let G„ denote the kernel of this homomorphism:

Gn = {GrEGI a(y) --= y mod pin+ 1 for all y E O i } .

Gn is a normal subgroup of G and Gn .+. 1 c Gn for n _-_ O. Furthermore, if a * 1, then there exists y in o' such that a(y)* y, so a(y) * y mod pm 4-1 for sufficiently large n. Therefore, Gm = 1 for all large m, and we have a sequence of normal subgroups of G:

1= • • • = G,n c G,n _ i c • - • c Gi c Go c G.

These groups Gn are called the ramification groups (in the lower numbering) of the Galois extension k' lk.

PROPOSITION 2.18. Let 1(0 be the inertia field of the extension k' I k (cf. Proposition 2.12). Then

Go = Gal(k7k0), G/Go = Gal(k0/k) '4 Gal(f7f)

so that G I Go is a cyclic group and

[G0 :1] = e , [G : Go] = f

Proof. Each a in G induces an automorphism a' (= a()) of f' = and a' obviously fixes elements of f. Hence a 1--> a' defines a natural homomorphism G = Gal(k7k)--> Gal(f' If), and by definition, Go is the kernel of this homomorphism. For the Galois extension k0/k, we have a similar natural homomorphism Gal(ko/k)---> Gal(f o/f), where fo denotes the residue field of ko . Since k c k o c k', f c fo c f', there also exist canonical homomorphisms Ga1(k7k)---> Gal(k o/k) and Ga1(f1f)---> Gal(f o/f), and those maps define a commutative diagram

Gal(V/k) ---1--> Gal(f 1 /0

i Gal(ko /k) --> Gal(fo /f).

However, by Proposition 2.12, f (le I ko) = 1, f' = fo and by Proposition 2.11, Gal(k0/k) '4 Gal(fo/f). Hence

Go = Ker(œ) = Ker(f3) = Gal(k11k0).


The rest of the proposition then follows immediately from Propositions 2.11 and 2.12. •

PROPOSITION 2.19. For each n 0, there exists an injective homomorphism

Gn IG„±i ----> Ufn IU'n+i,

where U(') = U' is the unit group of k' and U'i =1+ p" for i- 1.

Proof. Let z' be any prime element of k'. Then Cr E G„ implies a(z')=-- z' mod p'n± 1 so that a(z)z'' ---= 1 mod p '—that is, a(r)r' 1 E 1.1:i . If z" is another prime element of k', then ;0= zfu with u E U', and a(u) - u mod pfn+i implies a(u)u 1 E--- 1 mod pfn ±1 so that

a(z")a"-1 mod Un' +1 .

Hence the map An : Gn ---> U/14+1,

al- a(7r)af-1 mod Urn±1

is independent of the choice of the prime element :C. Let r e Gn . Then = r(a- f) is another prime element of IC. Hence, it follows from

ar(1 .011.,-i = (a(7Vl7V ff-1)(r(a., )a ,-1 )

that An(ar)=

namely, that A n :Gn ---> Unf/Unf ±i is a homomorphism. Let A n (a)= 1 for a E Gn . This means cr(af)irf -1 ----- 1 mod p'n +1 so that a(z)=- z' mod pfn+ 2 . Let /cc, be as in Proposition 2.18 and let 00 be the valuation ring of /c c,. Since k'/k0 is totally ramified by Proposition 2.12, it follows from the Corollary of Lemma 2.13 that of = 00[34 Hence, for a E G„ c Go = Gal(leko), a(z')-7---- z' mod pul± 2 holds if and only if a(y)---- y mod p"1 ± 2 for every y E o'—that is, a E Gn±1 . Thus Ker(A n ) = Gpi-F1, and An induces an injective homomorphism GnIGn+i---->Unf lUnr +1. •

COROLLARY. A finite Galois extension le lk of local fields is always a solvable extension—that is, Gal(le lk) is a solvable group.

Proof. GIG() is cyclic by Proposition 2.18 and Gn /Gn±i is abelian for n- 1 by Proposition 2.19. Since Gm = 1 for sufficiently large m, G = Gal(lcf/ k) is a solvable group. •

Now, suppose that k is a p-field so that both q and q' are powers of p. By (1.3) for (k', y'),

U fn/Vi+i --- ff ±, for n- 1.

Hence, it follows from Proposition 2.19 that

Go/G1 = a cyclic group, [Go: Gd I (qf —1),

Gn iGn+i = an abelian group of type (p,. . . , p), [G,,: G,,+1] q', for n- I.

In particular, [Go : G1 ] is prime to p and [G1 :1] is a power of p.

Chapter III

Infinite Extensions of Local Fields

This chapter consists of preliminary results for the remaining chapters. In the first part, some infinite extensions of local fields are discussed and then, in the second part, power series with coefficients in the valuation rings of those infinite extensions are studied. Here and in the following chapters we need some fundamental facts on infinite Galois extensions and their Galois groups—namely, profinite groups. A brief account of these can be found in Cassels–Frbhlich [3], Chapter V.

3.1. Algebraic Extensions and Their Completions

Let (k, y) be a local field with residue field f = o/p = Fq . Let Q be a fixed algebraic closure of k, and ti the unique extension of y on Q (cf. Proposition 1.1). We denote by (Q, ft) the completion of (Q, p). Let F be any intermediate field of k and Q:

kcFc S2 c S-72 .

The closure F of F in .s.-.2 in the ft-topology is a subfield of -0. Let

PF = P I F1

Then piF is the unique extension of y on the algebraic extension F over k, and (F, t t) is the completion of (F, pF). Now, any algebraic extension over k is k-isomorphic to a field F such as mentioned above. Hence, in order to study algebraic extensions over _k and their completions, it is sufficient to consider the pairs (F, fiF) and (F, tif) as stated above. Let

o F = the valuation ring of /IF,

p F = the maximal ideal of !A F,

ff = oFlpF = the residue field of tiF ,

and let OA PP, and ff, be defined similarly for pp. Since Ali flp F) = 1, the injection 0 E ---> op identifies f F with fp:

fF =

Let a be any automorphism of F over k. Then, by the Corollary of Proposition 1.1, a is a topological automorphism of F in the HF-topology, and by continuity, it can be uniquely extended to a topological automorph-


ism a of F in the pp-topology. We then have

/IF ° a = /IF, a(0 F) = oF, a(PnF) = PnF,

tip° a = tip, a(oP)= oP, No) = VI, nl,

and a and a induce the same automorphism of fF = fi, over f.

LEMMA 3.1. Let E be a finite extension of F in Q:k cFcEcS2 S72, and let E denote the closure of E in Q. Then

EF = -E.

Furthermore, if EIF is a separable extension, then

E n P = F.

Proof. Clearly F ç EP ç E and EF/F is a finite extension. Since pp is complete, it follows from Proposition 1.1 that ft I EF is a complete valuation on EP so that EF is closed in C2 in the ft-topology. Hence a = E. Assume now that EIF is separable. Since EIF is separable, there is a finite Galois extension E' over F, containing E:F cE cE', and in order to prove E n F = F, it is sufficient to show that E' n F = F. Hence, replacing E by E', we may suppose that EIF itself is a finite Galois extension. Then E = EP is a finite Galois extension over F and

[E:F]=[EF :F]= [E:E n F].

Now, by continuity, each a in Gal(E/F) can be uniquely extended to an automorphism a in Gal(E/F), and a'- a. defines a monomorphism Gal(E/ F)----> Gal(E/F). Hence

[E : F] [E : P]= [E : E n F].

Since FErlf-. = E, it follows that EnF=F. •

REMARK. Obviously, the second part of the lemma can be generalized for any separable algebraic extension EIF, not necessarily of finite degree.

3.2. Unramified Extensions and Totally Ramified Extensions

Let kcFcQ be as above. We call F lk an unramtfied extension if every finite extension k' over k in F, k c k' c F, is unramified in the sense of Section 2.3—that is, e(k' lk)=1. It is clear that if Flk is unramified and k c F' cF, then F'lk is also unramified.

LEMMA 3.2. Let kcFc Q. Then F I k is unramtfied if and only if PI F) is a normalized valuation on F: p(Fx)= Z.

Proof. In general, let k' be any finite extension over k in Q :k c k' c Q. By Section 2.3, k' is a local field with respect to a unique normalized valuation y' such that y' I k — v, and y' f k = ev with e = e(k' lk). Since Pk' 1 k = v for pk , = p f k', it follows from the uniqueness that I, ' = epic so that pk ,(k'')= (11e)v'(k")= (1/e)Z. Hence k' /k is unramified—that is,

Infinite Extensions of Local Fields 37

e =1—if and only if 11(k') = tik ,(k)= Z. This proves the lemma for a finite extension k'/k. The case for an arbitrary algebraic extension Pk then follows immediately. •

For each integer n -?--1, there exists a unique unramified extension k nur

over k in Q with degree [kn,: k] = n, namely, the splitting field of the polynomial Xq n - X over k in Q (cf. Proposition 2.11). Since k:r/k is a cyclic extension, one sees immediately that

k:r c k tTr <=>n I m, for n, t n ?-_ 1.

Hence the union k ur of all n _?-_1, is a subfield of Q:

kur = U k:r, k c k, c Q. n 1

For simplicity, we shall often write K for kur:

K = k,.

It is clear that k,lk is an unramified extension. On the other hand, if F is an unramified extension over k in Q and if a E F, then k' = k(cr) is a finite extension of k in F so that k' I k is unramified. Hence k' = k nur for n =[k' :k] and a E k:r c kur. Thus

k c F kur.

Therefore k, is the unique maximal unramified extension over k in Q. For n ?-- 1, let on and r denote the valuation ring and the residue field of

k unr, respectively, and let fic =--- oic/Pic be the residue field of K = kur.

PROPOSITION 3.3. fK is an algebraic closure of the residue field f (= fk ) of k. Each a in Gal(knr /k) induces an autotnorphistn a' of fK lf, and the map al-4 a' defines a natural isomorphism

Gal(k,/k) L.-, Gal(f Klf).

Proof. If m I n, then k lenur kunr k, so that fcr c fn C f lc . Since ox is clearly the union of on for all n ?:1, t ic is the union of r for all n 1. Since [fn : f] = [knur :k]= n, and since the finite field f has a unique extension with degree n in any algebraic closure, it follows that f K is an algebraic closure of f. Now, it is clear that

Gal(kur /k) = lim Gal(k r/k),

Gal(fK/f) = lim Gal(f n /f),

where the inverse limits are taken with respect to the canonical maps Gal(Cr/k)---> Gal(k: r/k), Gal(fm/f)---> Gal(fn/f) for nlm, m, n 1. As ex-plained in general in Section 3.1, each a in Gal(k,/k) induces an automorphism a' in Gal(fK/f). However, by Proposition 2.11, the map a- a' induces an isomorphism Gal(k'nz r/k) :-., Gal(fn/f) for each n a- 1.

38


Hence it follows that a'– a' defines an isomorphism Gal(kur /k) 2; Gal(fK / f). •

Since f = Fq , the map co 1–> wq, co E fK , defines an automorphism of fic over f. Let cp denote the corresponding element in Gal(kur /k) under Gal(kur /k) 2; Gal(fK/f)—namely, the unique element in Gal(kur /k) satisfying

cp(a)=- e mod p K, for all a E O K , K = kw..

As cp is uniquely associated with the local field k (with Q fixed) in this manner, it is called the Frobenius automorphism of k,I k, or, of k, and is denoted by Cpk • It is clear that cpk induces on each k:r, n 1, the Frobenius automorphism Tin of kn,lk (cf. Section 2.3). Since Gal(kn,/k) is the cyclic group of order n generated by cpn , the map a mod n 1– cpan , a E Z, defines an isomorphism

Z/nZ 2; Gal(kl/k).

For n I m, let Z/na ---> Z/nZ be the natural homomorphism defined by a mod m 1– a mod n, a E Z, and let

Z = lim Z/nZ <---

with respect to those maps for n 1 m. Since Tm 1 kn = Pk I kn = ci9n for n I m, the diagram

Z/mZ"'--->" Gal(k unr /k)

Z/nZ --'--> Gal(leuir /k)

is commutative. Hence we obtain an isomorphism of profinite (totally disconnected, compact) abelian groups:

Z 2; Gal(k,/k) = lim Gal(knur/k). (3.1)

Now, the natural homomorphisms Z---> Z/nZ, n 1, induce a monomorph-ism Z-4 i so that Z may be regarded as a dense subgroup of Z. In the isomorphism (3.1), 1 in Z is then mapped to the Frobenius automorphism Pk of k so that (3.1) induces

Z4(990, n,-4 q

between the subgroups. Since Z is dense in i, the cyclic group <cpk > is dense in Gal(k,/k). Hence k is the fixed field of Cpk in k, and the topological isomorphism (3.1) is uniquely characterized by the fact that

REMARK. For each prime number p, let Zp± denote the additive group of


all p-adic integers. It is easy to see that i is topologically isomorphic to the direct product of the compact groups Zp+ for all prime numbers p.

Let k be a p-field so that q is a power of p, and let Vœ be the multiplicative group of all roots of unity in Q with order prime to p. For n _-_1, let 17n denote the subgroup of all (q n — 1)st roots of unity in Q. Then

V. = U vn . rt.1

Since kl. is the splitting field of Xq n — X over k in Q, we have

kriiir = k(Vu ), k ur = k(Vco). (3.2)

On the other hand, we see from Proposition 2.3 that the canonical ring homomorphism O irc -->fic = C 'KIP K induces an isomorphism

V. -4 fk.

Hence it follows from cpk (n):----- rig mod p K that

cpk (n) = nq, for n E V.. (3.4)

The above equality also uniquely characterizes the Frobenius automorphism 99k of k.

Let k' be any finite extension of k in Q so that k' is again a local field. Then it follows from (3.2) that k'k u,.= k'(V,c). Hence k'k ur is the maximal unramified extension lc'u,. over k' in Q:

Cr = k'k ur .

Let (pie be the Frobenius automorphism of k' and let f = f(k7k). Then f' = Fq , with q' = qf for the residue field f' of k'. Hence q9k ,(q) = nq' = nqf for all n E Vc , and it follows that

CPk' I kur = 99{c, with f = f(k' lk).

Now, let F be an algebraic extension of k in Q: k = F = Q. Similarly as for unramified extensions, we define Flk to be a totally ramified extension if every k' such that k ç k' ç F, [k' :k]< +00, is a totally ramified extension—that is, f(kilk)= 1. Clearly, if k = F' =F and FIk is totally ramified, then Filk is also totally ramified. Let k' be any finite extension of k in Q and let 1(0 denote the inertia field of the extension le lk (cf. Proposition 2.12). Then /co = k' n k,, and [lco :k]= f(k' I k) by Proposition 2.12. Therefore k' /k is totally ramified if and only if k' n kur = k. It follows that in general F lk is totally ramified if and only if

F n kur = k.

Let L be an algebraic extension of k such that

L = Fku„ k = F n kur .

Then LI F is a Galois extension, and the restriction map a +— a I kir defines


an isomorphism

Gal(L/F) 2; Gal(k,/k).

Hence there exists a unique element tp in Gal(L/F) such that lp kur= Pk,

the Frobenius automorphism of k. In other words, Pk has a unique extension lp in Gal(L/F). Let F F' c L. Then it follows from the above isomorphism that

[F' n k ur :k]=[F' :F].

Hence Fill( is totally ramified only when F' =F, and we see that F is a maximal totally ramified extension over k contained in L.

LEMMA 3.4. Let E be a Galois extension over k, containing k u,.. Let lp be an element of Gal(E/k) such that lp I kur = (Pk and let F be the fixed field of tp in E. Then

Fku,.= E, F n k,= k, Gal(E/F) Gal(k,/k).

In particular, F is a maximal totally ramified extension over k in E. Proof. Clearly F n k„ is the fixed field of CPk = I kur in kur. Hence

F n k,= k. Let M be any field such that

FcMcE, [M:F]=n<+00,

and let F' = FkZr, F" = F' M. Since F n k,= k, we have [F' :F]= n. Hence F" is a finite extension over F, containing both M and F'. As F is the fixed field of lp in E, (V) is dense in Gal(E/F) so that Gal(F"/F) is a finite cyclic group, generated by lp I F". Therefore it follows from [M:F]=[F' :F]= n that M = F' = Fk ,. Fk,. Since this holds for any M, we obtain Fkur = E. •

3.3. The Norm Groups

For each algebraic extension Flk, kcF Q, let U(F) denote the unit group of F:

U(F)= Ker(p F

LEMMA 3.5. Let le lk be a finite extension: k k' Q. Then N kvic (U(V)) is a compact subgroup of U = U(k), and Nkyk (e x ) is a closed subgroup of k x .

Proof. By Proposition 1.6, the norm map Nkyk:le k is continuous, and by Proposition 2.6, U(k') is compact. Hence Nkyk (U(V)) is a compact subgroup of k x and is closed in k x . Furthermore, the formula for v'(f) in Proposition 1.5 shows

Nvik(U(V))N (k = - 101c\ n U(k) U= U(k).

Since U is open in k x , it follows from the above that Nkvk (le x )/N kwc (U(ki)) is a discrete subgroup of k x /Nkyk (U(k 1 )). Hence Nk7k (k x) is closed in k x . •


For any algebraic extension Flk, we now define

N(Flk)= fl N kvk (le x ), NU(F lk)= fl N kyk(U(le)),

where the intersections are taken over all fields k' such that

k ck' F, :kJ< +00.

We call N(Flk), and NU(Flk) the norm group and the unit norm group of the extension Flk, respectively. Clearly, if Flk is finite, then N(Flk)= N FIk(Fx ), NU(F lk)= NFIk(U(F)). In general, it follows from Lemma 3.5 that N(F lk) is a closed subgroup of le and NU(Flk) is a compact subgroup of U = U(k). In fact,

NU(Flk)= N(F lk) n U(k).

It is clear from the definition that if kcFcF' c Q, then

N(FrIk)cN(Flk), NU(F' lk)cNU(Flk).

Furthermore, if fk i l is a family of fields such that

k ck i c F, [ki :k ] < +00, F = the union of all ki ,

then

N(Flk)= n N ki ,k (k i ), NU(Flk)= n Nk i,k(u(ki)).

We shall next consider such norm groups for unramified extensions and totally ramified extensions.

LEMMA 3.6. Let le lk be a finite unramified extension. Then

NU(k' lk)= U(k).

Proof. As before, let f and f' denote the residue field of k and k', respectively. By (1.3), a prime element 7r of k defines isomorphisms

U/Ui 2-> V', WU, f±, for i 1,

where U = U0 = U(k) and Ui = 1 + pi, i 1. Now, as k' lk is unramified, Jr is also a prime element of k' (cf. Section 2.3). Hence it also defines similar isomorphisms for k':

LI L1_), 1 IJ:11.1:±1 f 1 +,

where U' =IA= U(k'). U; =1+ i 1. By Proposition 2.11, there is a natural isomorphism Gal(k'/k) Gal('/f). Therefore, we obtain the follo-wing commutative diagrams:

U'/ f'' U; U;+ f

IN IT U/Ui

42


where N = Nkyk, and T', N' are the trace map and the norm map of the extension f' if, respectively. However, since ra is an extension of finite fields, both T' and N' are surjective maps. Hence it follows from the above diagrams that the maps N are also surjective so that

Nrok(U')Ui = U, for all i 1.

Since N k ,,k(U') is compact, hence, closed, in U, we obtain Nk , ,k(U') = U. •

PROPOSITION 3.7. Let Flk be an unramified algebraic extension. then

NU(F lk)= U(k).

If, furthermore, F 1 k is an infinite extension, then

N(F lk)= U(k).

In particular,

N(k,1k)= NU(k,1k)= U(k).

Proof. The first part is an immediate consequence of Lemma 3.6. Let k'lk be the unramified extension of degree n_?_1: k' = let:, Then a prime element jr of k is also a prime element of k' so that k' x = <70 x U(k'). Hence

N kvk (k' x )= (an) x N kvk (U(k'))= (an) x U(k)

_ç k x = (30 X U(k).

If Flk is an infinite extension, there exists k' such that k c k' F with arbitrarily large n = [k' : k]. Hence the second part is proved. •

PROPOSITION 3.8. An algebraic extension Flk is totally ramified if and only

if N(F lk) contains a prime element of k. Proof. Suppose first that F is not totally ramified—that is, F n kur k.

Then k ç knur ç F for some n 2, and the proof of Proposition 3.6 shows that

N(F Ik) ç N(k11 k)= (Zn) X U(k).

Since n 2, N(Flk) contains no prime element of k. Suppose next that Flk is totally ramified. For each finite extension lelk

such that k ç k' ç F, let S(k ') denote the set of all prime elements of k, contained in N(le I k). Let z' be a prime element of k'. Since k'lk is totally ramified, N kvk (a) is then a prime element of k (cf. Section 2.3), and we see

S(k') = N kvk (z' U(k')) = N ielk(a')NU(k' lk).

Hence S(1e) is a non-empty compact subset of the compact set S(k)= gU(k), 7V being a prime element of k. Furthermore, if both Clk and 16,1k are finite extensions in F, then clearly S(kik)c S(16) n S(C). Therefore, it follows from the compactness of S(k) that the intersection of S(k') for all k'


is non-empty, and any element in that intersection is a prime element of k, contained in N(F lk). •

3.4. Formal Power Series

In general, for any commutative ring R with identity 1*0, let

S = R[[Xi ,

denote the commutative ring of all formal power series

f(X i , . . , Xn ) = E a • • X , a 1 E R,

where X1 ,. , X,, are indeterminates and i = (i 1 , , in ) ranges over all n-tuples of integers For f, g E S and for any integer d 0, we write

f g mod deg d

if the power series f -g contains no terms of total degree less than d. Let Yi ,.. ., Ym be another set of indeterminates that let g1 ,. . . , gn be power series in R[[171 , . . . , Ym ]] such that gi = 0 mod deg 1 for 1 i n. For any f(X i , . . , Xn ) in S, one can then substitute g • for Xi , 1 i n, and obtain a well-defined power series

f(g i (Y i , • • • , Ym), • • • , gn(Yi, • • • Ym))

in R[[171 , . . , Ym ]]. We shall often denote the above power series by

fqgi, • • • ,g,,)•

Let X be.an indeterminate and let M be the ideal of R[[X]], generated by X: M = (X) = X • R[[X]]. M is the set of all f (X) in R[[X]] such that f 0 mod deg 1. Let f, g E M. As a special case of the above (n = m = 1), the power series f o g is defined and it again belongs to M. Thus M becomes a semi-group with respect to the multiplication f og and the power series X is the identity element of this semi-group: Xof = foX=f. Hence, if fog=g0f=Xforf,geM, we writef=g - 1 , g =f'. Let

00

f(x)=-- E anXn, an E R. n=1

Then f is invertible (i.e., it has the inverse f-1 ) in M if and only if a l is invertible in the ring R—that is, al b = 1 for some b in R.

In the following, we consider the case where R is a topological ring. In such a case, we introduce a topology on S = R[[Xi , , Xn ]] so that the map

f = E • • • X l,71-

defines a homeomorphism of S onto the direct product of infinitely many copies of the topological space R, indexed by i = (i 1 , , in ). S then becomes a topological ring.


As in Section 3.1, let

= 00/P0

be the residue field of 0. We apply the above for the topological ring oo. Let

f = E • • X, a 1 E 00,

and let g1 ,. . , g, be power series in oti[Yi , . . . , Y,n ]] such that the constant terms of g l , , g, are contained in pi). Then we can see easily that

converges 00[Pi, f 0 •

f(xi, • • • , defined in

E WI • • gn( 1711 • Yrn) in

in o[[171 , . . , Ym ]] in the above-mentioned topology on . , Ki ]]. The resulting power series will be denoted again by , g„). In particular, for any a,..., a, in pc,, the value of Xn ) at X1 = ai , , Xn = an—namely, f(œ,.. , an )—is well oc-2 .

LEMMA 3.9. Let a E pc, and suppose f(a) = 0 for a power series f(X) in oa[X]]. Then f(X) is divisible by X – a in o[[X]]—namely,

f(X)= (X – a)g(X), with g(X) E

Proof. Let f(X)= E7=0 aiXt, cri E Do, and let CO

fi i= E i=0

Since a E Po, such sums converge in ot--2 and OC

g(X)= i=0

satisfies f(X)= (X – a)g(X). •

In the above, let a be algebraic over k —that is, a E p„, (ci E Q, ,u (ci)> 0)—and let f(X) be a power series in the subring o[[X]] of o[[X]], o being the valuation ring of k. Take a finite extension k' over k such that a E k' Q. Then k' is a local field and the infinite sum (3.5) converges in the valuation ring o' of k'. Hence f(X)= (X – a)g(X) holds with g(X) E ol[X]].

LEMMA 3.10. Let h(X) be a monic polynomial in o[X] such that

for 0. (3.5)

h(X)=H(x - cri )

with distinct a i , . . . , am in p„,. Let f(X) be a power series in o[[X]] such that

0 > 0 k Ok


f(cri)= 0 for 1isi ism. Then

f(X)= h(X)g(X)

with a power series g(X) in °PCB Proof. Let k' = k(ai , , an ). Then k' is a finite Galois extension

over k in Q, and by the above remark, f(X) is divisible by h(X)= frin=1 (X - cri) in n'[[X]]—namely, f(X)= h(X)g(X) with g(X) in o'RX]]. For each a in Gal(k'/k), let f(X) denote the power series in o'[[X]] obtained from f(X) by replacing each coefficient ci of f(X) by a(a). Then we have f "(X) = ha (X)g"(X), where g"(X) and h"(X) are defined similarly as f"(X). Here, r =f, h' = h because f, g E 0[[,(]1 k[[X]]. Hence it follows from f = hg = he that g = gcr. As this holds for every a in Gal(k'/k), g(X) is a power series in oRn. •

3.5. Power Series over ok-

As in Section 3.2, let K denote the maximal unramified extension of the local field k in Q:K=k„,.. Let CPk be the Frobenius automorphism of k: cpk E Gal(K/k), and let Cpk be the natural extension of Pk on the closure (completion) k of K in Q (cf. Section 3.1). For simplicity, we shall denote both Pk and CPk by cp. By definition, cp induces the automorphism coq on the residue field fK = rk = ok/pk so that

aT(=q)(a))----- aq mod pk , for all a E Ok.

Now, let OR also denote the additive group of the valuation ring oR and let U(k) be the multiplicative group of units in the complete field K. We consider the endomorphisms

1-4 (99 1)(œ) = 99 (œ) Cr)

cp - 1: U(K)--> U(k),

=

LEMMA 3.11. The following sequences are exact:

1 --> U(k) --> U(k) 9-1-L>1 U(k) --> 1.

Proof. Since the proofs are similar in both cases, we shall prove here only the exactness of the second sequence. By Proposition 3.3, fR (= fK ) is algebraically closed. Hence the maps fR---> fR defined by co 1--> cog - w and col--> cog' are both surjective. This implies that

( 49 — 1 ) 0 k ± Pk= OR U(K)T -1 (1+ 1)0= U(K). (3.6)


Let E U(k). We shall define, by induction, a sequence of elements frinIn-0 in U(k) satisfying

tir i mod pV-1 , tin Tin +1 mod pr, for all n 0. (3.7)

By (3.6), there exists no E U(k) such that 09 -1 mod p K . Hence, suppose that we have found n o , n i , , ij,, n 0, which satisfy the required conditions. Let a- be a prime element of k. Since Klk is unramified, z is then also a prime element of K and of k: Pk= zok. Hence, by (3.7):

= 1 ± a7rn+1 with a E Ok.

By (3.6), there exists f3 E 0k such that a (cp - 1)fi mod p k . Let

nn±i = nn(1 + fizn+ 1 ).

Then, clearly, 71+1 E ?i U(k) and n -= n +1 mod VI». SInce q9(z) = for ,n

E k, we also have

071= nf, -1 (1 + (9(0) — 0)r' 1 ) =nfi -1 (1 + cYzn+') = mod pnk±2.

Thus the existence of the sequence {rin },0 is proved. Since k is complete, it then follows that ri = lirnn . tin exists in U(k) and satisfies re9-1 = Hence the exactness of U(K)---> U(K)--> 1 is verified.

It is obvious that U(k) is contained in the kernel of U(k)-> U(k). Suppose that 99-1 = 1—that is, cp() = —for an element in U(K). It follows from (3.3) that the set A = {0} U V. is a complete set of repre-sentatives for fk (= fK) in ok• With n- as above, it follows from Proposition 1.2 that the element in U(k) ok can be uniquely expressed in the form

OC E anzn' n=0

Applying q9, we obtain

with an EA = {0} U V..

OC

= q2()= E 99 (an)JE' - n=0

However, by (3.4), cp(a) = e A for every a E A. Hence, by the uniqueness,

an = q9(an ) = dn', for all n O.

Therefore, either an = 0 or an is an element of the cyclic group V of order q — 1 in Proposition 2.3. Consequently, Eknu(k)= U(k), and the exactness of 1-> U(k)--> U(k)-> U(k) is also proved. •

Now, for each power series

f(Xi, Xn)= E • • a, i ,...,,„ E 0k,

in ok[[Xi , . . . , Xn ]], let fg' denote the power series of od[Xi , . . , Xn]]


defined by

, Xn ) = E q9(ai,,..., in)Z • • •

By Lemma 3.2, /I K (= 1.2 K) is a normalized valuation on K. Hence ,uk (= ft I k) is a normalized valuation on the completion k of K.

PROPOSITION 3.12. Let A- 1 and 7r2 be prime elements of k: k(a" 1) = =1, and let fi and f2 be power series in ok[[X]] such that

fi (X) i (X), 7r2X mod deg 2, fi (X) f 2 (X) X(' mod k ,

q being the number of elements in the residue field f of the ground field k. Let

L(X i , . . , Xm ) = cri Xi + • + am Xm , E OR

be a linear form in X1 ,. . . , Xm over ok such that

, Xm )= .7r2 L`P(X 1 , , Xm ).

Then there exists a unique power series F = F(Xi , . . , Xm ) in ok[[Xi , . . . , Xm ]] such that

F L mod deg 2, fi 0 F = F9' 0 f2 ,

namely,

Xm))= F T V2( )(1), • • • f2(Xm)).

Proof. Let F1 = L. Then the assumptions on f2 , and L imply

f1 oF1 Fof2 moddeg 2 .

Let n 1 and suppose that we have found a polynomial Fn of total degree in ok[Xi , . . , X„,] such that

fi 0 Fn —= ° f2 mod deg n + 1. (3.8)

Let H,, 1 be a homogeneous polynomial of degree n + 1 in ok[[Xi , , Xm ]] and let

Fn+i = Fn Hn+1•

Then F,, ±i is a polynomial of total degree n +1 in ok[Xi , , Xm ] and it satisfies

Fn mod deg n +1.

We shall next show that there exists a unique H„ ±1 such that the above F,, ±1 satisfies

Fn+i —= FL-112 mod deg n +2. (3.9)

To see this, let

G,, ±1 = 0 Fn —Fo f2.

48


Then, by (3.8), G,, 1 0 mod deg n + 1. Since f r1X, f2 z2X mod deg 2,

Fn +1 = fl(Fn Hn+1) == fl ° Fn mod deg n + 2,

F/±1 0f2 = F2'.f2 +1-1,f+1 ,12 = nof2 + 7r3 +1H,',°+1 mod deg n + 2.

Hence (3.9) is equivalent to the congruence

G,, +1 + — .7r2H-11a+1 0 mod deg n + 2. (3.10)

On the other hand, since fi f2 = X" mod Ilk and ag9 = aq mod Il k for a E Ok, we see from the definition of G„ +1 that

Gn +1 Fn(X11 • • • Xm ri F ,9,9(XY, . . . , 0 mod pk.

Now, take a monomial Xi = • • X of degree n + 1 in ok[Xi , . . . , X„ ] . By the above, the coefficient of Xi in Gn+ 1 is an element of the form —z ip with fi E ok. Let a be the coefficient of X in H,, ± 1. Then the coefficient of the same Xi in .7r 1 Hn +1 — +1m+1 is _ avicrcp. Since G,, 1 = 0 mod deg n + 1, we see that (3.10) holds if and only if the coefficient a of Xi in Hn±i satisfies

r 1)6 + — = 0

for every monomial Xi of degree n + 1. Let y = AITI.jrn +1 so that pk(y) = 1. Then the above equation for a can be written as

(3.11)

where fi and y are known quantities. However, since lik(y) 1,

a' = f3+ 7042 71+42 0 °22 ± • • (3.12)

converges in ok and it obviously satisfies (3.11). Furthermore, if both al and a2 are solutions of (3.11), then

= 7(a'' -

a'2) = k(c19 (oei — cr2)) = Act'T k(7) 1

so that ,uk(ai — a2) = +00, al = a2 . Therefore, for each Xi, the equation (3.11) has a unique solution a in ok, and it follows that there exists a unique H,, 1 satisfying (3.10), and so Fn +i satisfying (3.9). Now, starting from F1 = L, we can define successively a sequence of polynomials F„, n 1, in o k[Xi , . .. , Xm ] such that deg Fn n and

° Fn Fi ° f21 Fn ±i Fn mod deg n + 1, for all n 1.

The second congruence shows that there exists a power series F in

ok[[Xi , . . . , Xn ]] such that F Fn mod deg n + 1 for all n 1. It is then clear that this power series F satisfies

F = L mod deg 2, fi 0 F = Fg'of2 .

Finally, to prove the uniqueness, let F' be any power series in

ok[[Xi , . . . , Xm ]], satisfying the conditions for F as above. For 1, let F


denote the sum of the terms of degree isn in the power series F' and let F,,' +1 = F, + II ±i. Then F' -=-- L mod deg 2 implies Pi = L= F1 , and the uniqueness of T4 +1 in the above proof yields successively F' = F all n-. 1. Hence F' = F. •

Now, let k' be the unramified extension of degree n over k in Q :k' = knur, k ck' c K c k, and let o' denote the valuation ring of k'. Let .7r 1 and A-2 be prime elements of k'. Since KIk' is unramified, z 1 and z2 are also prime elements of K and K. In the above proposition, suppose further that

fi, f2 E 01[X]] and L E 01)(1, . . . , X,]. Since k' is complete and p(k')= k', the element a in (3.11) belongs to o', and it follows that the power series F in the proposition has coefficients in o':

F(Xi , . . . , Xm ) E O'[[Xi, . . . , Xm ]] . (3.13)

This remark will be used often later.

Chapter IV

Formal Groups FAX, Y)

In this chapter, we shall first briefly explain the general notion of formal groups, and then study a certain family of formal groups over the ring R = ok, which will play an essential role in the subsequent chapters. Notations introduced in Chapter III will be retained.

4.1. Formal Groups in General

We shall first briefly discuss some fundamental facts on formal groups in general.t Let R be a commutative ring with 1 0 0 and let X, Y, and Z be indeterminates. A power series F(X, Y) in R[[X, Y] l is called a formal group over R if it satisfies the following conditions:

(i) F(X, Y) X + Y mod deg 2, (ii) F(F(X, Y), Z)= F(X, F(Y, Z)),

(iii) F(X, Y) = F(Y, X).

Note that (i) implies F(0, 0) = 0 so that both sides of (ii) are well-defined power series in R[[X, Y, Z1]. In the general theory of formal groups, a power series such as F(X, Y) above is called, more specifically, a one-dimensional, commutative formal group (or, formal group law) over the ring R. However, we shall call it here simply a formal group, because no other type of formal groups will appear in the following.

Let Y = Z = 0 in (i) and (ii). Then we obtain

F(X, 0) X mod deg 2, F(F(X, 0), 0) = F(X, 0).

From the first congruence, we see that f(X)= F(X, 0) has an inverse f -1 in M = XR[[X]] in the sense of Section 3.4. It then follows from the second equality that F(X, 0) = X. Similarly, or by (iii), F(0, Y) = Y. Therefore,

00

F(X, Y) = X + Y + E ciiXEY', Cii E R, (4.1) i,J=1

namely, F contains no terms like X2 . We then see easily that the equation F(X, Y) = 0 can be uniquely solved for Y in M—namely, that there is a unique power series

00

F (X) -x + biX i, bi E R,

such that

i =2

F(X, iF(X))= 0. (4.2)

t For the theory of formal groups, see FrOhlich [8].

Formal Groups Ff (X, Y) 51

For f, g E M = XR[[X]], let

f 4F- g = F(f(X), g(X)).

Then f g again belongs to M, and it follows from (ii), (iii), and (4.2), that

the set M forms an abelian group with respect to the addition f 4F- g, with inverse 'F(f) for f. We denote it by MF.

Now, let G(X, Y) be another formal group over R and let f(X) be a power series in M = XR[[X]] such that

f(F(X, Y)) = G(f(X), f(Y)). (4.3)

We call such f a morphism from F to G, and write

If, in particular, f has the inverse f-1 in M, then r is a morphism from G to F. In such a case, we call f an isomorphism and write

As before, when there is no risk of confusion, an equality like (4.3) will be simply written as

fo F= Gof.

In general, if F(Xi , . . . , X,i ) is any power series in R[[Xi , . . . , X,,]] and if f c M = XR[[X]] is invertible in M :f' E M, then we define a power series

X„,) fl R[[Xi , , X„, ]] by

Ff (Xi , , X„,) = fo Fo f-1

=f(FV-1 (X1) , • • •

One checks easily that if F(X, Y) is a formal group over R, then G = Ff is again a formal group over R and

Let Hom R (F, G) = the set of all morphisms f :F--3 G,

EndR (F) =HomR (F, F).

For simplicity, these will also be denoted by Hom(F, G) and End(F), respectively.

LEMMA 4.1. Hom(F, G) is a subgroup of the abelian group MG, and End(F) is a ring with respect to the addition f g and the multiplication fo g.

Proof. Let f, g c Hom(F, G) and let h =f AG- g. Then

hoF=f0F4G-g0F=Gof-iG-Gog=G(G0f,Gog),

where Gof= G(f(X), f(Y)), Gog = G(g(X), g(Y)). Using (ii), (iii), for G,


we obtain

G(G 0 f, Gog) = G(G(f(X), g(X)), G(f(Y), g(Y)))

= G((f -.- g)(X), (f -- g)(Y))= Go h.

Hence h = f 4G- g belongs to Hom(F, G). As stated earlier, 'G(f) = ic of is the

inverse of f in the abelian group MG. Again by (ii), (iii), for G, we obtain

G(G(X, Y), G(i G (X), i G (Y))= G(G(X, i G (X)), G(Y, i G (Y))= G(0, 0) = 0,

that is, G 0 iG = iG 0 G. Hence

iG(f)01' = iG ot i F = iG oG 0 f = GoiG of=GoiG (f)

so that 1G(f) E Hom(F, G). Since clearly 0 E Hom(F, G), this proves that Hom(F, G) is a subgroup of MG. To see that End(F) is a ring, we have only to note that if f E End(F), g, h E M, then

fo (g 4F- h) = f(F(g(X), h (X))) = F(f(g(X)), f(h(X)))

=fog 4F- foh.

Note also that X is the identity element of the ring End(F). • Now, let 0 be an isomorphism from F to G:

0 : F 4G, that is, F8 = G.

Let f E End(F):foF =Fof. Then we have

feo Fe . Fe 0 fe, f 6 E End(G). Since

F(f, g) ° = Fe(f e, g °)= G(fe, g 6), (fog) ° = fe cig e

for f, g E End(F), we see that the map f 1--> f ° defines a ring isomorphism

0 : End(F) 4 End(G). Let

Ga(X, Y) = X + Y.

Obviously Ga is a formal group over R. It is called the additive (formal) group over R.

LEMMA 4.2. Suppose that R is a commutative algebra over the rational field Q. Then, for each formal group F(X, Y) over R, there exists a unique isomorphism

L-, Ga

such that il,(X)= X mod deg 2. Proof. Let F1 = 3F/3Y. Differentiating, with respect to Z, both sides of

Formal Groups FAX, Y) 53

(ii) in the definition of F(X, Y), we obtain

Fi(F(X, Y), Z)= Fi (X, F(Y, Z))Fi (Y, Z).

Since F(X, 0) = X, it follows that

Fi(F(X, Y), 0) = Fi (X, Y)Fi (Y, 0). (4.4)

Now, as F(X, Y)=== X + Y mod deg 2 implies FAX, 0)------ 1 mod deg 1, there is a power series /p(X) in R[[X]] such that

00

an E R. n=1

Let œ an Â(x)=x+

n=1 n a

with –2 E R, n

so that clAldX = lp. Then (4.4) can be written as

a a v(F(X, Y))Fi (X, Y) = p(Y), that is, —

3Y 4F(X, Y))= — A(Y). a Y

Therefore, A(F(X, Y)) = 0(X) + A(Y) with a power series 0(X) in R[[X]]. But, putting Y = 0 in this equality, we find that A(X) = 0(X). Thus

A(F(X, Y))= A(X) -I- A(Y), A: F 4 Ga.

To see the uniqueness of such A, it is sufficient to consider the case F = Ga . Hence, let A: Ga 4 Ga be any automorphism of Ga, satisfying A(X)=- X mod deg 2. Then A ° Ga = Ga ° A—that is,

Â(X + Y) = A(X) + A(Y), A(X)===.- X mod deg 2.

For A'(X) = dAldX, we then have

A' (X + Y) = A' (X), A' (X) --=. 1 mod deg 1.

Therefore, A'(Y) = A'(0) = 1. Since R is a Q-algebra, it follows that

A(X)= X.

Thus A is unique. •

REMARK. The lemma can be applied for formal groups over any field of characteristic O. Note also that the argument in the second half yields that End(Ga ) consists of precisely aX for a e R.

4.2. Formal Groups FAX, Y)

Let (k, V) be a local field with residue field f = o/p --= Fq and let Q, 0, K = lc, k, fk= ok/Pk, T = g9k = Pk, and so on, be the same as in Section 3.5. We shall next define a special type of formal group Ff (X, Y) over

R = OK.


For each prime element .7r of k, ,i(r)= 1, let 97, denote the family of all power series f(X) in R[[X]] such that

f (X) JrX mod deg 2, f(X) --=- Xq mod p,.

Forexample, the polynomial :rX + Xq belongs to .97,. It is also clear that if f E , then p9 E 97 . The union of the sets 37, , for all prime elements Jr of k, will be denoted by 37..

PROPOSITION 4.2. For each f E there exists a unique formal group FAX, Y) over R such that f E HOMR(Ff, FD—that is,

fo Ff = Fr ° f.

Proof. Apply Proposition 3.12 for zi = z2 = Jr , f , =f2 =f, and L(X, Y) = X + Y, m = 2. Then there exists a unique power series F(X, Y) in R[[X, Y]] such that

F(X, Y) —= X+ Y mod deg 2, f 0 F = Fo f. (4.5) Let

Fi(X, Y, Z)= F(F(X, Y), Z),

F2(X, Y, Z)= F(X, F(Y, Z)).

It follows from (4.5) that

Fi (X, Y, Z) F(X, Y)+ZE----X+ Y + Z mod deg 2,

foF1 = F T (f(RX, Y)) , f(Z)) = F 97(F 97(f(X), f(Y)), f(Z))= FT 0 f.

Similarly,

F2(X, Y, Z) —= X + Y + Z mod deg 2, fo F2 = F of.

Hence the uniqueness of Proposition 3.12 with L(X, Y, Z)= X + Y + Z, m = 3, implies F1 = F2—namely,

F(F(X, Y), Z)= F(X, F(Y, Z)).

Next, let G(X, Y) = F(Y, X). Then

G(X, Y) —= X + Y mod deg 2, fo G=G990f.

Since F(X, Y) is the only power series in R[[X, Y]] satisfying (4.5), we obtain F = G—namely , ,

F(X, Y) = F(Y, X).

Thus F(X, Y) is a formal group over R. It is then clear that F 97 is also a formal group over R. Writing Ff for F, we see that Ff is the unique formal group over R such that fo Ff = FfT f—that is, f E HOM(Ff, FcfP). •

Now, it follows from fo Ff = FT of that

fi99 ° = (FM op',

Formal Groups FAX, Y) 55

and this implies

Fr = Ffv, where f cP E

Let f be again a power series in 3.- and let a be any element in the valuation ring o of k. Applying Proposition 3.12 for L(X)= aX, m = 1, we see that there exists a unique power series cp(X) in R[[X]] such that cp(X). -------- aX mod deg 2, f o 4) = 4) 99 of. We denote this power series 4)(X) by [a]f : 0 =[a]f . Thus [a]f is the unique power series in R[[X]] satisfying

[a]f --=- aX mod deg 2, f 0[4 = [a]r 0 f. (4.7)

PROPOSITION 4.4. For each a E 0, [a]f belongs to EndR (Ff ), and the map a 1---> [a]f defines an injective ring homomorphism

o --0 EndR(Ff).

Proof. Let 4) = [a]f . Then

f000ff =0920foff =0920F7of .-(00Ff )Pof,

foFf o 0= Frofoo = FrooTof =(Ff oo)Pof,

4) ° Ff==--- Ff ° (1) --=-- a(X + Y) mod deg 2.

Hence, by the uniqueness of Proposition 3.12,

4)0Ff = Ff 0 4), that is, 4) = [a]f 6 EndR (Ff).

Let a, b E o, then

fo ([4 ±. [b]f ) —f° Ff ([4, [b]f) — FTV o [a]f , f °[b]f)

= P9([a]7 'f, [b]7 1) = ([4 ± [MX ° f,

[a]f ±,[b]f ---=-- [a]f + [b]f =----- (a + b)X mod deg 2.

By the Definition (4.7) for [a + b]f , we then see

[a]f -i; [b]f = [a + b]f .

Similarly, [a]f 0[6].f — [ab]f .

Hence a 1—> [a]f defines a ring homomorphism o --> EndR (Ff). Since [a]f -.--=--- aX mod deg 2, the homomorphism is injective. •

Let z and a' be prime elements of k and let f E ,97,, f' E 37, We shall next compare the formal groups Ff and Fr over R = °K• Since ,uk(z) = itt Az') = 1,

a' = :r, with E U(k).


By Lemma 3.11, there exists an element ri of U(K) such that = Tr-1 .

Let L(X)= 71X. Then a' L(X)= aLT(X).

Hence, applying Proposition 3.12 for fi = f', f2 = f, al = a', a2 = a, L(X)= 71X, m =1, we see that there exists a unique power series 0(X) in R[[X]] such that

0 (X) ==- 71 X mod deg 2, f ' o 0 = ()To!

(4.8)

Since n E U(K), O(X) is invertible in M = XR[[X]]: 0 -1- E M.

PROPOSITION 4.5. The above 0(X) has the following properties:

0 :Ff. -4 FT , that is, 1 7 = Fr,

[a]7 = [a]r, for a E I/

Proof.

roe° ft. = 0 99 of o Ff = 0 q' o Fiof = (0 o Ff)99 of,

f f OFf 00=FrOf t 00 = FrOOTOf =(F .f.,00) 99 0f,

00 Ff == Ff 0 0 71(X + Y) mod deg 2.

Hence, by the uniqueness of Proposition 3.12 with L(X, Y) = 71(X + Y),

0 ° Ff = F f ° 0, that is, fl = F f.

The proof for [a]7 = [a]r is similar. • It follows in particular from the above proposition that the formal groups

Ff over R are isomorphic to each other for all power series f in the family 37. The isomorphism O(X) in this proposition will be used quite often in the sequel.

EXAMPLE. Let k be the p-adic number field: k = Qp , and let .7r = p. Then

f (X) = (1+ xy — 1 = pX + (132 )X 2 + • • - + XP

belongs to the family 37p . Let

F(X, Y)= (1+ X)(I+ Y) — 1= X + Y + XY.

One checks immediately that F is a formal group over Zp , hence over R = ok, satisfying f 0 F = F99 0f. Therefore,

Ff =F=X+Y+XY.

For each a E Zp , define

(1 + XY = i (a )r n=13 n /

Formal Groups Ff (X, Y)

57

where ( a) =_a(a — 1) • • • (a — n +1) E Z for n

\ni n!

Then it follows from the definition of [cdf in (4.8) that

[a]f = (1+ X)a —1= E ( a)Xn , for a E Zp .

n=1 n

For n 0, let (cf. Section 4.3 below)

{oe E Pc2 I [Pn lf(œ) 0 } •

Since [pn±l ]f (X) = (1 + X)P n+1 — 1,

= —11 E Q, 11.

Hence Q(W) is the cyclotomic field of pn ±lth roots of unity over Qp in Q. It is a well-known classical fact that such a cyclotomic field Q(W) is an abelian extension over Q ,. and that the Galois group of Qp (14/7)/Qp is naturally isomorphic to the factor group 41(1 + np +izp ‘ . ) In the following sections, we shall show in general that similar results can be proved for an arbitrary local field (k, y) and for the extension k(W2), defined by means of the formal group FAX, Y) introduced above.

4.3. The o-Modules W.7

As before, let oo/p o be the residue field of the completion 572 of a fixed algebraic closure Q of k. For simplicity, let m denote the maximal ideal po of o c-2 :

Let k be the completion of K = /cur and let FAX, Y) be the formal group over R = ok in Section 4.2, associated with a power series f in the family g. For a, p in m and for any a in the valuation ring o of the ground field k, let

= Ff(cr, 0), aloe = [a]f(œ), (4.9)

[a]f (X) being the power series in (4.7). By the general remark in Section 3.4, both Ff (a, p) and [a]f (a) are well defined in oo. Furthermore, it follows from FAX, 0, [a]f (X) = 0 mod deg 1 that those elements again belong to m

a + 13, a ci E m, for a, /3 E m, a E O.

By (ii), (iii), and (4.2) in Section 4.1, we see immediately that m is an abelian group with respect to the addition a ±. 0, the inverse of a being

given by iF(a) for F = Ff in (4.2). By Proposition 4.4—that is, [cdf o Ff =


F1 0 [a]f, and so on— we then also see that the map

0 x m---> m,

(a, a)1—> a 1 a

defines an o-module structure on m. We shall denote m by mf when it is viewed as an o-module in this manner. For integers n —1, define

vin+1 a={a-alaEpn+1} , f f

for a E Tilf,

= foe € trif = 01.

Then we obtain a sequence of o-submodules of mf :

{0} = W7 1 C • • • 1447 • • • Wf,

where Wf denotes the union of W7 for all n —1. Now, let f' be another power series in the family Y. Let 0(X) denote the

power series in XR[[X]], satisfying (4.8). Then 0(X) has an inverse 0 -1(x) in XR[[X]] and F7. = Fr, [a]; =[a]f., by Proposition 4.5. Hence it follows from (4.9) that 0 defines an isomorphism of o-modules:

0 : trif '4 air ,

al—) 0(a),

which induces o-isomorphisms

0 : W7 -4 W7,, n —1. (4.10)

LEMMA 4.6. For f E g and i 0, let

Then

gi = ofi _ i o • • • ofo, g_ 1 (X)= X.

W7 = {a E gn (a) = 0}, for n —1.

Proof. Since W.F' = {0}, g_ i (X)= X, the statement is clear for n = —1. Let O. Let f' and 0 be as stated above so that f 00= O'Pof. Then we have

o 092` = 0 921+1 of, f: = 0 92"-I ofi 0 -921 , i 0, and

=o • • • of;) = 60 92n+1 ogn o n O.

Since 0(m) = m, e ( r.)=W7,, we see that if the lemma holds for f, then it also holds for f'. Now, let .7v be a prime element of (k, v). Since Klk is unramified, .7v is then also a prime element of K and K. Hence

f (X) = .7rX + X('

belongs to the family g. Since .7v e o, f = .7rX mod deg 2 and f of =r2 of, it follows from (4.7) that

f = [z]p J = f,

gn = f . . .of = [jrn

Formal Groups FAX, Y)

59

As pn +1 = wrn+1, we obtain

W7 = { cr E rilf 1 Pn±1 }- a = 0) = fa E ni 1 gn (a)= 0).

By the earlier remark, the lemma then holds for an arbitrary f in g. •

Fix an integer m 1 and let k' denote the unique unramified extension of degree m over k in Q:

k' _m r.

Let f' = o'/W be the residue field of k' and let .7r be any prime element of k'. Then

f(X)= .7i-X + Xq

is a polynomial in o'[X], belonging to the family g. Hence, checking the proofs of Proposition 4.3 and of (4.7), we see from the Remark (3.13) that

F(X, Y) E 0 1 [[X, Y]], [a]f (X) E 0 1 [[X]], for a E O. (4.11) Let

hn (X)= .702" + gn _ 1 (X)q -1 , for n O. Then

n 0, so that

gn (X)= hn (X)h n _ i(X) • • • ho(X)X.

LEMMA 4.7. (i) h(X) is a monic separable polynomial of degree (q —1)qn in o'[X] and it is irreducible even in the polynomial ring k[X],

(ii) g(X) is a monic separable polynomial of degree qn +1 in o'[X] and W7 is the set of all roots of g(X) in Q. Hence the order of W7 is qn +1 and ki(W7) is a finite Galois extension over k'.

(iii) Let h(a0) = 0 for n 0, cro E Q. Then

ao E 14/7, cro Wr 1 , (q — 1)qn = [k"(a0):k'] 5- [k' (W7): k']

and

cpn(.7r)(= 7 n )= N(—a0) E N(le(a0)11e)

where N denotes the norm of the extension kf(a 0)11e . Proof. It follows from gn = fn o • • • 0 fo that

g(X) = anX + • • - + Xq n+ ' =--- Xq n+i mod p ' , an = zi-i-cp-f----i-cp- , (4.12) hn (X) = .702n + (a n _ IX + • • • + r n )q-1 =-=- X(q-1)qn mod p '

so that gn (X) and hn (X) are monic polynomials in (IX] with degrees q`H-1 and (q —1)q", respectively. Since be n (= pn(r)) is a prime element in k, we also see from the above that hn (X) is a so-called Eisenstein polynomial in R[X], R = o. Hence, it is irreducible even in k[X].t Suppose that (k, y)

t Compare Lang [16].

60


is a p-field so that q is a power of p. Then

(q

dh„ -1)4:1r-2 + • • • , an_ 1 = dX

where (q - 1)4:1 0 even when the characteristic of k (and of k) is p. Hence hn (X) is a separable irreducible polynomial of degree (q - 1)qn in k[X]. Therefore, hi (X)* h i (X) for i j so that gn = hn • • hoX is again a separable polynomial—that is a polynomial without multiple roots. Let a E W7. By Lemma 4.6, a is an element of mf = pn such that gn (a)= 0. But, as gn (X) is a polynomial in o'[X] c le[X], a is a root of gn in the algebraic closure Q of k. Conversely, if gn (a)= 0 for a E Q, then it follows from (4.12) that p(œ) >0, a E mf. Therefore, W7 is the set of all roots of the separable polynomial gn (X) of degree qn +1 , contained in Q. This of course implies that the order of W7 is qn+1 : [1477 :01= q' and that k'(14/7)/k ' is a finite Galois extension. Thus (i) and (ii) are proved. The first part of (iii) follows from (i) and from hi (X)*It i(X) for i *j. The second part is clear from (4.12) for hn(X). •

REMARK. Actually equality holds in (iii) above. Compare Proposition 5.2(ii) below.

LEMMA 4.8. Let f be any power series in the family Pie in R[[X11. Then:

(i) The order of the o-module W7 is qn': [W7: 0] = qn', -1. (ii) Fix an element a() E W7, a() o W7 -1 , 0. Then W7 = 0 iao and the

map a 1-> a ao induces an isomorphism

(iii) W7 = Wri for05-i5_n. Proof. (i) This is clear from (4.10) and Lemma 4.7. (ii) Note first that since [W7 :0] = [W7 -1 :0]= n, q there exists

ao E 1477, cro W7 -1 . Clearly ai--> a ac, defines an 0-homomorphism o—> W7. Since pn 4-1 3. cro = 0, Pn ac, *0, the kernel of the homomorphism is an ideal of o, containing pn+ 1 but not pn. Hence the kernel is pn+ 1 • It then follows from [o :1) -4-1 ] = N/7:01= qn' that o—> W7 induces an isomorphism

2.,* urn VY f SO that W7 = •oi a 0 •

(iii) Let ir be a prime element of k : = zo. Then ao E W r), a0 0 W7 -1 , implies ai l ce0 E Wri, 71-5a0 0Wr i-1 . Therefore, by (ii), W7 = z' • 0 • a.= 0 • zi • an= Wnf -i ff ,, f f ._. ,• •

Let f again be any power series in 3.7 and let

End(W7) = the ring of all endomorphisms of the o-module W7,

Aut(W7) = the group of all automorphisms of the o-module W7, = the multiplicative group of all invertible elements in the

ring End(W7).

Formal Groups Ff (X, Y)

61

For each a E 0, let Ea :W.7 WY,

f3I— a 13 = [4(0).

Then Ea is an endomorphism of 1477: Ea E End(W.7). If, in particular, a E U = U(k), the unit group of k, then [alf =[a]1' implies E a -i =E -a-1 SO

that Ea E Allt(W.D. It is then clear that the map al—) Ea defines homomorph-isms of rings and groups:

0-->End(W.7), AUI(W. ).

PROPOSITION 4.9. The above homomorphisms induce a ring isomorphism

o/pn +1 4End(147.7)

and a group isomorphism n 0

where Un+1= 1+ Proof. Fix ao E 1477, o40 1$ so that W7 = 0 1 ceo by Lemma 4.7(ii).

Let E E End(W7). Then E(a0) E W.7 = 01 œo so that E(œ 0) = a frao for some a E D. Since Wfi = 0). cro , it follows that E(a) = a a for all a E 1477--that is,

E = Ea . Hence o-->End(W.7) is surjective. As on ±l i W.7 = 0, pn;.W.7= 0 (cf. Lemma 4.8(iii)), the kernel of o 1—End(W) is pn+ 1 so that

0/r ±i End(W7). Since U/Un+i is the multiplicative group of the ring 0/pn +1 , the group isomorphism U/Un+i Aut(W7) follows from this ring isomorphism just proved. •

REMARK. Although the isomorphism o/pn +1 :...-.> W.7 in Lemma 4.8 is defined by means of a choice of ceo E W, ceo l$ WI', both isomorphisms in the above proposition are canonical.

4.4. Extensions Ln/k

Let 0 and let W.7 be the 0-module of the last section, associated with a power series f in W Inf = m Q. We shall next consider the subfield 1( /177) of Q. LEMMA 4.10. 1Z(W.7) is a finite Galois extension over k and it does not depend on the choice off in the family Si.

Proof. Let f(X)= 7rX + X' as in Lemma 4.7, a- being a prime element of k' = k. By (ii) of the same lemma, le(W .7)/k 1 is a finite Galois extension. Since k' is a subfield of k, k(w . )IR is then also a finite Galois extension. Since k is complete in the pk-topology, it follows from Proposition 1.1 that _is is also complete and is a closed subfield of n in the A-topology. Now, let f' be any power series in ge and let 0(X) be the power series in (4.8) so that

0(14/7)= W7, 0-1(1477)=


by (4.10). Since 0 E R[[X]], R = ok, and since k(W7) is closed in 0, we see that

W7, = 0(W7) ç k(W7), k ç k(W7,) ç k(W).

Therefore, k(W7,) is a finite extension of k and, hence, is closed in S-2. It then follows similarly that

W7 = 0 -1 ( 4/7,) ç k(W7,) so that k(14/7) = k(W7,). •

In the following, we shall denote the field k(W7) by Ln:

Ln = k(WD, for every f E 37, n- —1.

As the notation indicates, Ln is in fact the closure in 0 of a certain subfield L n of Q. Choosing a special f in 37, this will be proved in Proposition 5.2(i). Note that L -1 = k because W7 1 = {0}.

PROPOSITION 4.11. For n- 0, there exists a natural homomorphism

(5 " : U = U(k)—> Gal(Ln I k)

such that for each u in U,

f (ôn(u)(a)=

U - a =[u]f (a))

for every f in 3"; and for every a in W7. The homomorphism (5 " induces an isomorphism

U/Un+1 -4 Gal( L n / k)

so that Ln I k is an abelian extension with degree [Ln :k] = (q —1)qn. Proof. Again, let f(X)= zX + Xq and hn (a0)= 0 as in Lemma 4.7. By

(i) of the same lemma, h(X) is an irreducible polynomial of degree (q —1)qn in MX]. Hence

(q — 1)q " = [k(œ o) : k] 5- [k- (1477) : k] = [L n : k]. (4.13)

Let a be any element of Gal(Ln/k). Since a is continuous and since Ff (X, Y) and [a]f (X), for a E o, are power series with coefficients in R = ok c k, one has

(a). /6 ) G = Ff(cr, /6 ) G = Ff(ce , g cr)= a' ci t V,

(a i. fir = [a]f (a)° = [a]f (e) = a i. a'

for a, fi E Illf and a E O. Thus a defines an automorphism of the o-module mf and it induces an automorphism a' of the o-submodule W7 = {a E

rtlf 1 Pn+1 ). a = 0} in mf. Clearly the map a'- a' defines a homomorphism

Gal(Ln/k)—> Aut(W7),

Formal Groups Ff (X, Y) 63

and since Ln = k(W7), this homomorphism is injective. As [Aut(W7) :1] = [U: Un+1] = (q —1)qn by Proposition 4.9, it follows that

—1)qn.

Comparing this with (4.13), we see that

L n = k(cro), [Ln : K. ] = (q —1)e, Gal(Ln I 10:4 Aut(W7). (4.14)

Let b n denote the product of the sequence of maps

U—> U I Un+i -4 Aut(I477) .4 Gal(Ln I 10,

where U—>U1Un±i is the canonical homomorphism, U/Un+1'4 Aut(W7) is the isomorphism of Proposition 4.9, and Aut(W7).2-, Gal(Ln /10 is the inverse of the isomorphism in (4.14). Then (5 " : U —* G al(L n I k) obviously induces U/Un+i 4 Gal(Ln/k), and checking the definitions of the maps in the above sequence, we find easily that (5n (u)(a) = [u]f (a)= u1 a for u E U, œ E W7.

Now, let f' be any power series in the family g and let 0(X) be the power series in (4.8) so that W7, = 0(W7) by (4.10). Let a' E 14/7, and let a' = 0(a) with a E W7. Then, for u E U,

u i, a' =[u]r(e)=[ur(0(a))= 0([14-(a))=

However, since 0(X) E R[EXN, where R = 0k and since 6n(u) E Gal(Ln / K), we have

0(6n(u)(a))= 6a(u)(0(a))= 6n(u)(a') so that

= u1 œ', for f' E Ye, a' E W7 , .

This completes the proof of the proposition. •

Since W7 OE W7 ±1 and Wf is the union of all W7, n —1, we have a sequence of subfields of n:

k=r-i.L0.•,Ln.....r.n, where

L = the union of all L " , n _._ —1

= K(VVf ), for any f E .

PROPOSITION 4.12. L/k is an abelian extension and there exists a topological isomorphism

6:U = U(k):_-, Gal(L/k)

which induces the homomorphism 6" :U—> Gal(Ln I k) in Proposition 4.11 for each n _-_ 0.

Proof. It is clear from Proposition 4.11 that (5 n is the product of b n±' and the canonical homomorphism Gal(Ln +1 /10—> Gal(Ln/K). Hence we


have

(5 : U = lim U/U„1 2-.> Gal(L/k) = urn Gal(Ln/k). <-- 4— •

Since the structure of the compact group U = U (k) is explicitly described in Section 2.2, the structure of the Galois group Gal(L/k) is now completely determined.

Chapter V

Abelian Extensions Defined by Formal Groups

Still keeping the notation introduced earlier, let k denote the completion of the maximal unramified extension K = k„ of a local field (k, v). In the last chapter, we constructed certain abelian extensions of k by means of the formal groups FAX, Y) over R = ok • In the present chapter, we shall similarly construct abelian extensions of the ground field k, by choosing suitable power series f in the family Y. We shall then study the properties of such abelian extensions.

5.1. Abelian Extensions L n and le77' n For each integer m 1, let om again denote the valuation ring of the unique unramified extensions k: of degree m over k in Q. Let a- be a prime element of km, so that it is also a prime element of K = k, and of K. Let 37, be the family of power series in RRX]], R = ok, as defined in Section 4.2, and let

= g, n omuxj].

For example, f(X)= :IX + X" in Lemma 4.7 belongs to 377. We put

gm = the union of 377: for all prime elements it of k:,

3,— = the union of gm for all integers m 1.

LEMMA 5.1. Let m 1, n O.

(i) Let f E 377. Then the field k(W) depends only on m, n, and a-, and is independent of the choice off in the family

(ii) Let f E 37'. Then the field K(1477) depends only on n and is independent of the choice off in 3:'.

Proof. (i) The proof is similar to that of Lemma 4.10. Namely, let f(X)= :IX + X" and let f'(X) be any other power series in g'77. Since we may put it = a', = ?I = 1 in the proof of (4.8), it follows from the remark (3.13) that in this case, the power series 0(X) in (4.8) belongs to om[[X]]. By Lemma 4.7(ii), k(W7) is a finite Galois extension over k. Hence it is a local field and is closed in O. Since 0(X) E 01[X]1, we see from (4.10) that

Wri, = 0(1417) ç k74,.(1417), k ç k(147) ç k(1477).

It then follows that k(W) is again a local field so that k i:(1477) ç k:(1477) since also 0 -1 (X) E om[[X]]. Hence k:(1477) = k iT,.(1477) for any f' E 377.


(ii) Let f E .cYP . Then f belongs to 3,--7T for some m 1 and 7E E k. Let

E = K(W7)= K k(W7)

and let E denote the closure of E in C-2. By (i) above and by Lemma 4.7, kumr(W7)/k i: is a finite Galois extension for any f E ge,m. Hence El K is also a finite Galois extension, and by Lemma 3.1 and the definition of Ln,

= Ek = k(wr.o= Ln, K=E fl K.

Let f' be another power series in `,Y, and let E' = K(W7). Then E'IK is again a finite Galois extension and E' = Ln. Let M=EE'. Since both MIE and MIE' are finite Galois extensions, it follows from Lemma 3.1 that

E=mnE=mnLn=mnEi=E",

namely, K(14,7) = K(147). •

In view of the above lemma, we put

kn = k (W , for any f E -1,

Ln = K(W 7), for any f E -1.

Note that

= k, L -1 = kur (= K)

because W7 1 = {0 } .

PROPOSMON 5.2. Let 0, and let a- be a prime element of k mur. (i) The field Ln in Section 4.4 is the closure of Ln in Q, and

Ln = kLn , K = kn Ln , Gal(Ln I k) Gal(Ln I K).

(ii) Ln= Ken,' n , k = K n km,'n so that km,'n is a maximal totally ramified extension over kmi„. in L" and

Gal(Ln I k Tr) = Gal(Ln I k 77n) X Gal(Ln/K)

Gal(K/k:) X Gal(kr/k:).

(iii) Lnlk, LnIK, k'77' 21k, and k 7T'nlk: are abelian extensions and

[Ln : K] = [k77 : k ] = (q - 1)qn , [k 7T'n : k] = m(q - 1)qn

Proof. It was already proved above that both Ln I K and k"/k: are finite Galois extensions and that È = Ek, K=Enk for E = K(1477)= L' 2 . Hence Gal(Ln/k) Gal(Ln/K), [Ln : K] = [Ln : k] = (q - 1)qn by Proposi-tion 4.11, proving (i). It is clear that

Ln = Kk jnri'n, K n k 7T'n k'n 7T,

so that

(q -1)qn = [Ln : K] = [k 7T'n : K n k'] [krjrn'n : Cr]. (5.1)

Let f(X)= irX + Xq,ir E k:, and let hn (cro) = 0 as in Lemma 4.7(iii). Then

= cro = f[a]f (a0) la E 0)

Abelian Extensions Defined by Formal Groups 67

by Lemma 4.8(ii). However, since f E 01[X]], it follows from (4.7) and the remark (3.13) that [a]f belongs to ol[X]]. Hence

1477 g kiTr(a'0), km , n = k in,zr(W 7.) = k ,

and, by Lemma 4.7(iii),

[141'n :1(:]=[k:(a0):k:]= (q —1)qn.

Comparing this with (5.1), we obtain k= K n k". This proves (ii). The statement on Ln I K and kr/C: in (iii) are consequences of (i), (ii), and Proposition 4.11. To see that Ln I k is abelian, take a prime element a- of the ground field k. By (i), (ii),

Gal(Ln/K) Gal(Id n/k lur), where kil„.= k. Hence ldn I k is abelian by Proposition 4.11. Since Ln = Kid" and KI k is abelian, L n I k is also an abelian extension. The rest of (iii) follows from this and from (i) and (ii). •

The following proposition is an immediate consequence of Proposition 4.11 and Proposition 5.2 above:

PROPOSITION 5.3. (i) For each 0, there exists a homomorphism

o n : U = U(k)—> Gal(Ln/K)

such that for each f E 37° , each u E U, and each a E 14/7,

6n(u)(a) = u a(=[u]f (a)),

and it induces an isomorphism

U I Un+1.4Gal(Ln

(ii) Let m 1 and let r be a prime element of k:. Then ô ' induces a homomorphism

(57r : U = U(k)—> Gal(kT'nlk'unr)

such that for each f E 3777, each u E U, and each a E W7,

67r(u)(a) = u a( = [u]f (a)),

and •57, in turn induces an isomorphism

U/ 144-14 Gal(k7n/kiT)-

COROLLARY. Let 0 i n. Then the isomorphism UlUn+i induces

Gal(kr:' n 1 k,Tr)

Un+i2-> (57r(Ui+i) = Gal(k7'n1

Proof. It is clear from the above proposition that (57r(u)lk7r" = for u E U so that

u E <4> to (U) = 1 <=> 67,(u)11(7.:'i =1<=> (57,(u) E Gal(kT'n/kTr ' i).

Hence Ui+i/Un+i2-› (57r(Ui+i) = Ga1(k7 n/k;71. •

68


PROPOSITION 5.4. Let n 0 and let a- be a prime element of k mur. Then:

(i) kTr 'nIkrun,. is a totally ramified finite abelian extension and .71- is contained in the norm group of k7'n1krunr : 7t E N(k7'"Ikrunr).

(ii) Let f E 377 and let a E W7, a 0 W7 -1 . Then a is a prime element of the local field k" so that

k" = kiTr(a), o'n'n =

O m ' n and om being the valuation rings of 1(7' n and krunr, respectively. Proof. (i) Since K n ic;;" = cr, the extension k7'n/k ul. is totally rami-

fied. By Lemma 4.7(iii), q9n(z) is contained in N(k"Ik:). Since k"lk is abelian by Proposition 5.2, the automorphism cp of krunrIk can be extended to an automorphism of k7'n/k such that cp(N(k 7T'Ikrunr))= N(k7mIkZ.). Hence

E N(kr I krunr). (ii) Let f(X)= irX + r. Let f' be any power series in 3-, 7 1, and let

a' E a' 0 W7,-1 . Put oto = 0 -1 (a'), where 0(X) is the power series in (4.8). Then, by (4.10), ceo E W, cro W7 -1 . Hence h(œ0) = 0 and cpn(z)=

N(—œ 0) by Lemma 4.7. Since cpn(z) is a prime element of Cr, it follows that —ceo and, hence, ceo are prime elements of 14". The proof of Lemma 5.1(i) shows that 0(X) E O m [[X]1 and 0(X) = X mod deg 2. Therefore, a' = 0(a) is also a prime element of 14" and it follows from the Corollary of Lemma 2.13 that o'n = e[œ], = krunr(a1 ). Changing the notation from f' to f, we see that (ii) is proved.

COROLLARY. Let f E 3-1 and let a E W7, ce W7 -1 so that krjr = k(œ).

The complete set of conjugates of a over the subfield km„ is given by the

set of all f3 EW , /3WÏ 1 , For 0:5_ i 5_.n, the complete set of conjugates of a over this subfield k"d' i

is given by a 4_ vr -i-i = la 4_ y i y

f f I

Proof. (i) Since k rna ' n = krun,.(a), it follows from Proposition 5.3(ii) that the complete set of conjugates of ce over km, is given by the elements ce for all u E U. By Lemma 4.8(ii), (iii), this is the set of all 16 E W7 = o ce,

et w7-1 =

(ii) By the Corollary of Proposition 5.3, the complete set of conjugates of a over kr;r!' i is given by u;. a for all u E Ui + i = 1 ± 1) 7 + 1—namely, by the set

(1 + 1)')). ce

where p 1 + 1 ). = W7 -1 by Lemma 4.8(iii). •

EXAMPLE 1. Let us consider the case n = 0 in the above. Let f(X)=

+ r, E WY, ce W7 = {0}—that is, a *0. Then

cfre = = krur,.(q-V-7r).


As in Section 2.1, let U = V x Ul , where V is a cyclic group of order q - 1, and let w(u), for U E U, denote the unique element of V such that u --- co(u) mod p. Since [u]f = uX mod deg 2, we then have

=[u]f (a)-=--- ua = co(u)a mod pL.

However, since a =q-k-lz, its conjugate (5 °(u)(a) over lenur must be an element of the form ?la with 77 E V. It then follows from V .1-_-, r that

co(u) = ?I, (5°(u)(co= co(u)a, for u E U.

EXAMPLE 2. Let us consider the example stated at the end of Section 4.2—namely, the case:

k = Qp , a- =p, f(X) = (1+ x)" —1E g;p1 ,

FAX, Y) = (1 + X)(1 + Y) - 1.

In this case, we know that

W2={ - 11eQ, V n+1 = 11, forn0

so that kA '" = Q(W7) is the cyclotomic field of pn± ith roots of unity over Qp = k ui r. Let u E U, a= -1 E VVf . Then ô(u) in Proposition 5.4 satisfies

[u]f (a) = (1 + a )" - 1, that is, 67,(u)( ) = V `,

and (5 pn induces an isomorphism

U/Un+i = zp(1+ pn±1zp ):_-_, Gal(Qp (W2)/Q p ).

Furthermore, by Proposition 5.5, we also know that Qp (W2)/Qp is a totally ramified abelian extension of degree (p - 1)/3" and that p is a norm from the extension Q(W2)/Q.

5.2. The Norm Operator of Colemant

In this section, we shall study the field ley'r" for m = 1, 7r E k = kir, and denote it simply by le;r :

1c7r = lcin, n -1.

These kn,, n -1, are the abelian extensions of k, originally introduced by Lubin-Tate [19]. Let S denote the power series ring o[[X]], and Sx the multiplicative group of the ring S:

S = D[[X]]

Sx = {g(X) E S 1 g(0) 0 0 mod p, that is, g(0) E U}.

Since the valuation ring o of the local field (k, y) is compact, S is a compact ring in the topology introduced in Section 4.1. Let 7r be a prime element of k and let f(X) be a power series in the family S7-e!r = 3--...„ n S.

t Compare Coleman [5].


LEMMA 5.5. Let g(X) be a power series in S such that

g(141) = 0, that is, g(y) = 0, for all y E W.

Then g(X) is divisible by [z ]t. in S:

g(X)=[z]fh(X), h(X) E S.

Proof. First, consider the case where f(X)= 7rX + Xq. Since a E 0, it follows from (4.7) that

f(X)= [n]p

By Lemma 4.7(ii), W.ri)- is the set of all roots, in Q, of the separable polynomial f (= go) in o[X]. Hence, if g(141) = 0, then g is divisible by f = [nit. in S by Lemma 3.10. In general, let f'(X) be any power series in 3,1 and let 0(X) be the power series in (4.8). As stated in the proof of Lemma 5.1(i), 0(X) then belongs to S = q[X]]. Now, let g' (X) be a power series in S such that g'(W (f),)= 0. Since WY, = 0(WY) by (4.10), we then have (g' o 0)(WY) = 0 with g' o 0 E S. Therefore, by the above,

g' 0 0 =[z]fh, g' = ([zi] 0 0-1)(h 0 0"), with h, h o 0 -1 E S.

However, [n]f 0 0 = 0-1 o [n]f ,, by Proposition 4.5, and 0 -1 (X) is divis-

ible by X in S. Hence g'(X) is divisible by [a]p in S. •

REMARK. By a similar argument, we can also prove that if g(W7) =0 for g E S, then g(X) is divisible by krn ± lf in S.

In the following, we shall fix a power series f in 3;1= .97„ n S and omit, for simplicity, the suffix f in [a]f , 14 f 7, and so on.

Let

trz = On/Prz, n —1,

be the residue field of lc,', = k(W) and let a E Wn c m = po . Since a E P n = at n kn,, it follows from the general remark in Section 3.4 that Ff(X, a) is a well-defined power series in on [[X]]. Let

X 4- a ( = X -if- a) = FAX, a).

Then X -î- a --- a mod deg 1 by (4.1). Hence, if g(X) is any power series in S = o[[X]], it again follows from the same remark that g(X + a) is a well-defined power series in on [[X]].

LEMMA 5.6. Let g(X) be a power series in S satisfying

g(X + y)= g(X), for all y E W ° .

Then there exists h(X) in S such that

g = h 0 [n].

Proof. Let ao = g(0) E 0 and let h 1 = g — ao . Then h l (y) = g(y) — g(0) =

Abelian Extensions Defined by Formal Groups

71

0 for all 7 E W° . Hence, by Lemma 5.5, h l is divisible by [n] in S:

h i = g i [a], g = ao+kigi, g i E S.

Since

[zi(X 4- 7) = [7r1(X) 4- [7r1(Y) = kl(X),

it follows that g i (X -i- 7) = g 1 (X) for all 7 E W°. Therefore, by the same argument,

g1 = al ± [zig2, a l = g i (0) E 0, g2 E S

so that

g = (10 + ai [n] + [JO,

where g2(X) again satisfies g2(X -1- y)= g 2(X) for all y E W° . Thus we can find a sequence, ao , al , a2 , . . . in o and a sequence g l , g2 , . . . in S such that

g = a0 + adz] .4_ ... + a jar +prrign+i, for all n _- O.

Hence g = h 0 [a] with h(X)= (10 + ai X + - - • + anXn + • • • in S. •

REMARK. Again the lemma can be generalized for IV" and [a.n+1].

The power series h in Lemma 5.6 is unique for g. This is a consequence of the following lemma:

LEMMA 5.7. Let g = h 0 [sr] for g, h E S and let p be as before the maximal ideal of k. Then, for each n _- 0,

g ------- 0 mod pn.(=> h ------- 0 mod pn.

Consequently,

g =0•(=>h =O.

Proof. 4= is obvious. We prove by induction on n. For n = 0, this is trivial. Hence, let g -7-- - - - 0 mod pn, n _- 1. Then g 1------ 0 mod pn -1 , g = an-lg i , with g 1 E S. By the induction assumption, we have h --.÷-- 0 mod V-1 , h = zn'h i with h l E S, and it follows from g = h 0 [a] that g 1 = h l 0 [a]. However, g --=, 0 mod pn implies g 1 -=--- 0 mod p. Since [a] = f(X)- -- Xq mod p, we see

h i (Xq)-7--- h i 0 [z]= gi --=:- 0 mod p

so that h i (X)----- 0 mod p. Hence h = n'lh i --- 0 mod r. •

REMARK. The proof shows that the lemma holds also for power series g, h in R = ok[[X]] and for the maximal ideal pk of K.

Now, let h(X) be any power series in S = o[[X]] and let

h 1 (X)=1-1h(X -i- y), y EW° . Y

Then h i (X) is a power series in oo[[X[], 00 being the valuation ring of


k °,T = k(W °). By the Corollary 1 of Proposition 5.4 for m = 1, all y E W°, y *0, are conjugate to each other over k. Hence

lei' = h1 , for all a E Gal(kVic)

so that h i (X) belongs to S = o[[X]]. Furthermore, since

(X 4- y) 4- y' = X 4- (y 4- y'), for y, y', y 4- y' E W °,

h i (X) satisfies h i (X+ y) = h i(X) for all y E W °. Therefore, by Lemmas 5.6 and 5.7,

h l = h2 ° [z]

with a unique power series h 2(X) in S. We shall denote this h 2 by Nf(h), or simply, by N(h). Thus N(h) (= Nf (h)) is the unique power series in S such that

N(h) 0 [n] =II h(X 4- y), 7

yEW°. (5.2)

The map

N (= Nf ): S —> S

is called the norm operator on S, associated with f E .9;1. We shall next prove some basic properties of the operator N.

LEMMA 5.8.

(i) N(h 1 h 2) = N(h 1)N(h 2), for h 1 , h2 E S. (ii) N(h)=- h mod I), for h E S.

(iii) h E XES' for i -- 0 N(h) E X iS' , S' being the multiplicative group of the ring S.

(iv) h =-- 1 mod p', i _- 1 N(h )7-7. 1 mod p'+ 1 . Proof. (i) is obvious from the uniqueness in (5.2). (ii) [n]= f (X) -=-- Xq mod p implies

N(h) 0 [sr] ----- N(h)(r) mod p.

On the other hand, y E W° po and [W° :0] = q, f = oip = Fq imply

X + y == X mod po , n h(X-i- y)=----h(X)q -=-- h(r) mod po . ,,

Hence, it follows from (5.2) that (Nh)(r)-= h(r)mod so that (Nh)(X)- --= h(X) mod p.

(iii) Let h E S x—that is, h(0»0 mod p. Then N(h)(0)= h (0) 0 0 mod p by (ii) so that N(h) E S x . For h(X)= X, X 4-0 = X implies

N(X)= Xh i (X), with h l E S. (5.3)

Hence, by (5.2),

[n](hi 0 [a ] ) = x 11 (x4- 7), yEW°. y*C1


Dividing the both sides by X and putting X = 0, we obtain

nh 1 (0) = 11 7, y e W°. )1C1

By the Corollary (i) of Proposition 5.4, the product on the right is the norm of y *0 in W° for the extension k °,Ik. By Proposition 4.2(ii) and Proposition 5.4(ii), such a y is a prime element of the totally ramified extension k (7), over k. Hence the norm of y—namely, .nh 1 (0)—is a prime element of k. Therefore, h 1 (0) is a unit of k and it follows from (5.3) that N(X) E XS x . By (i), we now see that h E XiS x implies N(h) E XiS x .

(iv) Write h in the form h =1+ nih i , i _- 1, h l E S. Then

N(h) 0 [a] = fi (1 + zth i(x 4- y))7---- (1 + nih i (X))q mod Zi P0 7

=1+ wrzhi(X)+ • • • + azghi (X)q =1 mod zzpo .

Let N(h) = 1 + h2 , h2 E S. Then it follows that

h2 0 [sr] =. 0 mod n i Po , hence, mod rl •

Therefore, h 2 =- 0 mod p t—namely, N(h)=- 1 mod p' +1—by Lemma 5.7. •

We now define the iteration Nn of the norm operator N on S by

N°(h)= h, Nn(h)= N(Nn -1(h))= N(N(. . .(N(h)). . .)), for n _- 1.

LEMMA 5.9. (i) Nn(h) 0 [zn] = flu, h(X --i- a), for a E Wn -1 , n _- 0. (ii) If h E XiS x , i -' 0, then Nn(h)IN(h) E Sx and

Nn'(h) . Nn(h) mod pn +1, n _. O.

Proof. (i) For n = 0, 1, this is trivial. Let n _- 2 and assume that the equality in (i) holds for n –1. Let A be a set of representatives for W'1/W° in W': Wn-1 =A 4_ W°= fa, 4_ 7 1 a E A, y E WI. Then

H h(X -1- a) = n n h(X -4- a' 4- y) = H N(h)([1r](X4- a)), a, E wn-i. crEA YE W° crEA

where [z](X -i- a)=Pri(X)--i-Pri(a)=Pri(X) 4- a • a, a E A.

Since Wn-2 = a • Wn -1 = .7r • A, it follows that

{I h(X -1- a)= n N(h)(Pri(X) 4-1), crEwn-i p E Wn -2

and by the induction assumption, the last term is equal to

Nn -1(N(h)) 0 Prn-lli[z](X)) = Nn (h) 0 [Jrn].

(ii) By Lemma 5.8(iii), h E X'S x , i -' 0, implies Nn(h)E Xi S x for all n -- 0. Hence Nn+ 1 (h)INn(h) E Sx. By Lemma 5.8(ii), we have

h i –= 1 mod p, for h i = N(h )Ih E Sx.


Hence, by (iv) of the same lemma, we obtain successively

N(h i) = 1 mod p2 ,. . . , Nn (h i ) 1 mod

Since Nn(h i) = Nn+ 1 (h)INn(h), the assertion in (ii) is proved.

We still keep fixed a power series f(X) in 3-,,r1 = 977, n S and choose an element a such that a E Wn, cy Wn-1, 0. Let

ai =7vfl = [7v - ](a), for 0

Then ai E W k , ci$W i-1 so that by Proposition 5.4(ii) for m =1, a is a prime element of le,- , and

0 i = o[ai], pi = o•ai

for the valuation ring and the maximal ideal of k.

LEMMA 5.10. Let Oi E Zn-i 1)00i for 0 S j s n. Then there exists a power series h(X) in S = 0[[X]] such that

h(œ1)=f3 , for 0 i n.

Proof. Let gi(x) =

Since PI 7viX mod deg 2 for j 0,

= 0 pril zn-i [z i ± l j mod deg 2 in [7v i± 1 ].

Hence

gi(x) = (7vn-i a2[7v1+12 ai , a2 , E

so that g• E S. Now, it follows from pt iliai)= ao , [7v 1 +1 1(ai )= 0 that gi(co = jrn-i cyo.

If 0 Si < i, then Pr ilia;) = 0 implies g1 (cri) = 0; and if i <j s n, then Pri±llicri ) = œfi 0 and kn'licx;) = 0 imply again gi (cri) = 0. Since

Pi E 7vn-iP00i, where po i = ao[ai], we may write in the form

= i crohi(cri), with hi (X) E o[X]. Then

h = gi hi i=o

belongs to S and it clearly satisfies h(ai)= f3. for 0 s is n. •

We now prove a main result of Coleman [5].

PROPOSITION 5.11. Let 7r be a prime element of k (= kul r) and let f E gni = 37, n 0[[X]]. Fix an element a E W7, a fit W7 -1 , n•._ 0, and let ai = ci for Os i n. Let be any element of U(k) and let

O s iS n,


where Nn , i denotes the norm from kq to k i,r . Then there exists a power series h(X) in o[[X]] such that

h(ai), for 0 i n.

Proof. Since U(k) c o n = o[a], we have

= h(a), with hi(X) E 0[X].

Furthermore, E U(14) implies that h 1 (0) 0 0 mod p so that h i (X) E 5' in S = 0[[X]]. By Lemma 5.9(i),

N'2 (h 1 ) 0 [7r'l = fl h i (X -i- y), y E Wn—i-1 , i s n. Y

Put X = a in the above. Then, by the Corollary of Proposition 5.4,

Nn — i(hi)(cyi) = fl h( - i- = Nn , i (h i (a)) = Nn,i()=

0 i n. Y

Let h 2 = Nn(h i ) E S x . Then, by Lemma 5.9(ii),

N'(h i ) Nn-14-1 (121) • • •

Putting X = ai , we obtain

= N'(h i )(ai)= h2(ai ) mod pn'

so that

Pi = h2(oe1) E Zn-i±10 i

Therefore, by Lemma 5.10, pi = h 3 (œ), O sis n , with some h 3 E S. Consequently,

for h = h2 + h3 in S = 0[[X]]• •

5.3. Abelian Extensions L and k,

Let r be a prime element of ktun,., m Since Wr' c 14/7 for f E it follows from the definition of Crn'n and Ln in Section 5.1 that

k = ktu;.= k"nn' ° c • • - Q,

k ur KL 1 L 2 c • c L n • • • CS-2.

Let C'oe = the union of 14" for all n —1,

L = the union of L n for all —1.

Clearly bm,œ _ k in,',.(Wf ), for any f E YPInn ,

L = ki„.(Wf ), for any f E


PROPOSITION 5.12. L is an abelian extension over k and

L= ki,k,',7' - , ka.= ki, n kqi'`°.

Hence k' is a maximal totally ramified extension over ka in L and

Gal(LI k)= Gal(L/km1 x Gal(L/k ur ) 4 Gal(kur /k) x Gal(k7''/Ic iT)-

Proof. By Proposition 5.2(iii), each L n lk is an abelian extension. Hence Llk is also an abelian extension. The rest is a consequence of Proposition 5.2(ii). •

PROPOSITION 5.13. There exist topological isomorphisms

45:UGal(LIki,),

6,:U 4Gal(k';'/k,7.),

which induce, for each n a- 0, the homomorphisms 45 n and ô nn , respectively, of Proposition 5.3.

Proof. Similar to the proof of Proposition 4.12. •

Note that by Proposition 5.3, e• has the property:

6(u)(a) = u -if- CY, U E U, CV E 14/7

for any n 0 and for any f in gi'. Similarly for 45„. For m =1 and z E k = kui „ let k l„!" be denoted by k„:

k, = k l.„!' = the union of kn.„( =k/) for all n -. —1.

As a special case of the above proposition, we see that

L= kurk„, k = k„,. n lc,

so that k, is a maximal totally ramified extension over k in L and

Gal(L/k) = Gal(L/k,) x Gal(L/k,)

Gal(k,/k) x Gal(k,/k), (5.4)

6,:U 4Gal(k,lk).

Now, fix a prime element ir of k and an integer m 1. Let

J, = ka,

cp' = cpk , = the Frobenius automorphism of k'.

Clearly cp' = cpm , cp = 99k being the Frobenius automorphism of k. Let .7ri be any prime element of k'. Since .7r is also a prime element of k', we have

.7ri = .7r, with E U(k'). Let

U = Nvik()E U(k).


The following lemma relates the action of the Frobenius automorphism on the formal groups for r and z' to the action of the endomorphism [4. LEMMA 5.14. Let f E ,97;„ f' E ,977, and let 0(X) be the power series in (4.8). Then

= O 0[4.

Proof. Let gm _ i and g_ defined for f and f' as in Lemma 4.6. The proof of the same lemma shows that

em _ 1 0 O = eqf 0 gm _ i • (5.5)

Let x = Nkyk (7C)= zmu.

It is clear from the definition of g,n' _ 1 that gm' _ 1(X) xX mod deg 2. On the other hand, since f' E f"P' = f', we see from the definition of gm' _ 1 that

o gm' _1= g,','_1 o f'. Hence, by the uniqueness in (4.8) and Proposition 4.5,

= = , g_i 0 0 = 0 0[4 = 0 0[4 °WI.

However, since f belongs to ST,r1 , we have f = g 1 = [e ]f . Therefore, it follows from (5.5) that

0 0 [4. o k nz lf = (13' o [elf

As [zm]f = [.7df 0 • • • 0 [z]f, the equality of the lemma follows from the remark after Lemma 5.7. •

LEMMA 5.15. Let z' be another prime element of k and let z' = zu with u E Un±1.• Then

k7r = (z) x Un+i N(kriztlk), —1.

Proof. Apply the preceding lemma for m =1, cp = cp' , = ru. Let a' E 14/7. By (4.9), W. = 0(W .7) so that a' = 6(a) with a E W. By Lemma 5.14,

0g2 (a) =

But, as u E E W7, it follows from Proposition 5.4(ii) that [u]f (a)= a so that 0"(a) = 0(a). On the other hand, since knr nk",= k by Proposition 5.2(ii) for m = 1, the Frobenius automorphism cp of k,lk can be extended to the Frobenius automorphism of L n = knrktin over knn and, hence, to an automorphism of its closure L n over k7r. As cr e 14/7 c 14, we then obtain

0(ar = 09) (a)= 0(a), for a" = 0(a) E L" = k,(147).

Therefore, a" = 0(a) e k nn because cp is the Frobenius automorphism of L n I knn. Thus

k7r , = k(W7) k7 r.

Since r =14-1 .7C, u E Un4-1, we obtain similarly k Hence /On = kn„,

78


for sr' = 7ru, u E U„1. By Proposition 5.4(i) both r and .7ri are contained in N(141k) = N(klk). Consequently, u = .7e1.7r E N(Cirlk) for every U E

U1—namely, U„ 1 N(k71k). Hence (sr) x U„i ç N(141k). •

PROPOSITION 5.16. For any prime element r of k,

N(k7,1k)= (z) X Un+1) for n —1.

Proof. Since we know

(7r) x Un+1 N(Cirlk) ç kx (sr) x U,

it is sufficient to show that

NU(k7r1k)= U n N(141k)c Un+1.

Let u E NU(141k)—namely, u= N'(), where E U(14) and N' denotes the norm of the extension 1c7r/k. Fix an element a E 14/7, a Et W. As in the proof of Proposition 5.11, we then have

= h(a), with hi(X) E Sx.

By Lemma 5.9(ii),

Nn(h i ), Nn (h 1 ) E S x , Nn+1(h1) Nn (hi) mod pn± 1•

Let u 1 = Nn(k)(0), u 2 = Nn +1 (h 1)(0). Then it follows from the above that

141, U2 E U = U(k), u 1 mod prz

However, by Lemma 5.9(i)

u i = fl hi(P), u2= JJ hi (P) O E Wn-1 Wn

so that

u2/u 1 = hag), where 0 E 1417, 3 W7 .

Therefore, by the Corollary (i) of Proposition 5.4 for m =1,

u 2lu 1 = N'(hi (a))= N'()= u.

Hence u2 u1 mod pn+ 1 implies u 1 mod f 1—namely, u E Un+1. Thus NU(kn7 lk)cUn±1 . •

PROPOSITION 5.17. For any prime element r of k,

N(k„lk)= (z).

More generally, if F is a totally ramified extension over k in Q, containing k„:k c k„ F Q, then

N(F lk)= (z).

Proof. Since the intersection of Un+1 = 1 + pn+ 1 for all 0 is 1, the first part follows immediately from Proposition 5.16. Let F be as stated


above. Then

N(F I k) c N(k„r I k) = (z).

However, since Flk is totally ramified, N(Flk) contains, by Proposition 3.8, a prime element of k. Since .7r is the only prime element of k in (.7r), it follows that .7r E N(F I k) so that N(F I k)= (z). •

REMARK. The fields km„r 'n in Section 5.1 were introduced by de Shalit [61 as generalizations of the abelian extensions kn„r in Lubin-Tate [19]. He also generalized the result of Sections 5.2 and 5.3 and proved, for example, that

N(k7;'° c I k) = (Nk7k (z))

for any prime element .7r of k' = k. However, since Proposition 5.17 (i.e., the case m = 1 in the above equality) is sufficient for our applications in Chapter VI, we discussed here only the classical case of kn,r for the sake of simplicity.

Chapter VI

Fundamental Theorems

Let kab denote the maximal abelian extension, in Q, of the local field (k, II).

Using the results on the abelian extensions L" , km,' , and lc", over k obtained in the last chapter, we are now going to prove fundamental theorems in local class field theory. Namely, we shall define the so-called norm residue map

pk :kx -4 Gal(k ab l k)

and then discuss the functorial properties of Pk with respect to a change of the ground field k.

6.1. The Homomorphism Pk

In Section 5.3, we introduced an abelian extension L of the local field (k, v). L will be denoted also by Lk when the ground field k is varied. Clearly

kcLck ab ,

where kab is, as stated above, the maximal abelian extension over k in Q. For each prime element z of (k, v), we also defined in Section 5.3 an abelian extension k, of k, contained in L. Let cpk denote, as before, the Frobenius automorphism of k: Cpk E Gal(ku,./k). Since

Gal(L/k,) -2.;. Gal(ku,./k)

by (5.4), there exists a unique element /pit in Gal(L/k) such that

Vnlkur= 991c , ip, 1 k, = 1. (6.1)

As (cpk ) is dense in Gal(ku,./k), (v, ) is a dense subgroup of Gal(L/k7r ) so that k the fixed field of ip, in L.

Now, fix a prime element zo of (k, y) and, for simplicity, write /p for IP A° . Since kx = (z o ) x U, each x in k > can be written uniquely in the form

x = z'onu, m E Z, U E U.

In fact, here m = v(x). For such an x, let

p(x) = 111"45(u -1 ), (6.2) where

45 : U Gal(L/k,)

is the isomorphism in Proposition 5.13. It is then clear that the map xi--> p(x) defines a homomorphism of abelian groups

p:kx -4 Gal(L/k)

Fundamental Theorems 81

satisfying

P(x) I kur = kur = 4197: , with m = v(x).

LEMMA 6.1. Let z' be a prime element of k' = km„, m_1, and let

x = Nelk (z 1 ).

Then p(x) is the unique element a of Gal(L/k) such that

a I k„ = CP rkn , a = 1.

(6.3)

Proof. Since cp /T is the Frobenius automorphism Tie of k': cp'icn = p E Gal(ki,/k 1 ), it follows from Proposition 5.12 that there is a unique element a in Gal(L/k) with the properties stated above. As p(x) satisfies (6.3), it is sufficient to prove that p(x)1 k7rzic° = 1—namely, that

p(x) = 1, for all 1.

Now, the prime element :t o of k is also a prime element of the unramified extension k' over k. Hence

= 7ro with some in U(k'), so that

x = Nkyk (z 1 ) = aVu, u = Af ietk() E U = U(k).

Let f E ,97;r0, f' E 3777, , and let 0(X) be the power series in (4.8) for f, f'. Then, by Lemma 5.14,

0 ' = 0 ° [U ]f,

where q9' denotes the extension of 99k , =9972, on the completion k of K = k,. Let a' E 1477, = 0(W7) and let a" = 0(a), Ci E W7. By (6.2) and Proposition 5.3,

p(x)(a)= b(u')(a)=[ulf (a).

Using the fact that 0(X) E ok[[X]] and that p(x) extends to cp' on k, we see from ci' = 0(a) that

p(x)(e) = 0q9 ' (p(x)(œ)) = 0 0 [u]f 0 [ulf (a) = 0(a) = a".

Since k7rzin = k' (1477), we obtain p(x)1 4Y' = 1. •

REMARK. The above proof shows why we define p(x) by (6.2) instead of p(x)= vmb(u).

PROPOSITION 6.2. There exists a unique homomorphism

Pk: k x Gal(L/k) such that

Pk(z)= 1P7r

for every prime element .7r of k. Proof. Applying Lemma 6.1 for m =1, = n, we see that p(z) is the

82


unique element of Gal(L/k) such that

p(7r) k i, = cpk , p(z) 1c7r = 1.

Hence p(z) = 7r by (6.1). Thus the homomorphism p : kx Gal(L/k), defined above by means of a fixed prime element irc, of k, has the property mentioned in the proposition. Let u E U and let z' be a prime element of k. Then ir" = yru is again a prime element of k and u= el.e. Hence the multiplicative group k x is generated by the prime elements of k. This implies the uniqueness of pk. •

Since pk = p, it follows from (6.2) that Pk induces the topological isomorphism

U

45(u -1 )

on the subgroup U of k>. Let .7r. be a prime element of k and let

x zmu, m V(X) E Z, U E U,

for an element x of k>. Then

Pk(X) = pk (z)m Pk(u ) = 1P7:45 04 -1 ),

Pk(x) I kur= ir kur (PT, m = v(x).

(6.4)

(6.5)

Thus (6.2) and (6.3) hold not only for the particular prime element .7r0 but also for any prime element r of k. (This is also clear from the fact that P = Pk is independent of the choice of 7r0.)

PROPOSITION 6.3. (i) Pk is injective and is continuous in the v-topology of kx and Krull topology of the Galois group Gal(L/k).

(ii) The image of Pk is a dense subgroup of Gal(L/k) and consists of all elements a in Gal(L/K) such that a kur = 99tkn for some integer m. In particular, if a I kur = Pk, then there is a unique prime element r of k such that a =

Proof. (i) Let pk (x)= 1 for x = zniu in (6.5). Then CP tkn = pk (x)1 k, = 1, and since (cpk it follows that m = 0, x = u, 6(u')= pk (u)= pk (x)= 1. As 6 is an isomorphism, we then see that u =1—namely, x = 1. Thus Pk is injective. The continuity of Pk follows from the fact that it induces a topological isomorphism U Gal(L/k,) on the open subgroup U of k x.

(ii) By (5.4), Gal(L/K) = Gal(L/k,) x Gal(L/k„). Hence the first part of (ii) is a consequence of (6.5), p(U)= Gal(L/k,), and the fact that ( ip,) is a dense subgroup of Gal(L/k,). The second part is also clear from (6.5) for

= 1. •

Now, let k' I k be a finite extension of local fields and let

Pk k x Gal(Lk /k), Gal(Lk ik ')

be the associated homomorphisms given by Proposition 6.2. We shall next prove an important preliminary result on the relation between Pk and pk,.


For simplicity, let L= Lk, L' = Lie, and let

E=LnLi.

Since keEeL and Llk is abelian, pk (x), for x E kx , induces an auto-morphism of Elk. Let x' E k''. Then pk ,(xi) is an automorphism of L' over k', and hence, over k. Therefore, pk ,(xi) also induces an automorphism of Elk.

LEMMA 6.4. Suppose that le I k is a totally ramified finite extension of local fields. Then

p k ,(x')1 E = pk (N kvk (x'))1E, for all z' e /ex .

Proof. Since k'x is generated by the prime elements of k', it is sufficient to prove the above equality for a prime element z' of k'. Extend p k ,(z) in Gal(L'/k') to an automorphism r of the algebraic closure Q over k and let F denote the fixed field of r in Q. Since the fixed field of

=pk ,(7C) in L' is Cr . by (6.1),

F n L' = Cc, 1 m le,,,.= Cr ,n Cu.= k'

so that Flk' is totally ramified and k' eV, e F. Hence, by Proposition 5.17,

N(F 110= (z'). (6.6)

On the other hand, t I le, = pk ,(JT')1k:‘,.= Cpk , by (6.5) and cp k , I k,= 'Pk because k'lk is totally ramified so that Gal(V„./k'):_-, Gal(k,/k). Hence

r 1 kur= ( r I Kr) 1 kur = Pk' 1 kur = cPk-

Let a= r 1 L. Then a 1 k,= cpk by the above. Therefore, by Proposition 6.3(ii),

a = Pk(z)

for some prime element .7r of k. Since k the fixed field of pk (z) in L by (6.1),

kek, r eF= the fixed field of r in Q,

and since both Flk' and kilk are totally ramified, Flk is also totally ramified. Hence, again by Proposition 5.17,

N(F lk)= ( z ) .

It then follows from (6.6) that

Nk 747C) E N(Flk)= (z).

However, as le lk is totally ramified, Nkyk (z) is a prime element of k. Therefore, we see from the above that

Nkyk(7c)=.7r

so that

Ple(z i )1E=TIE=GIE=Pk(z)1E=Pk(Nkvk(7C))1E. •

84


COROLLARY. Let le lk be a totally ramified finite extension of local fields such that k c k' c Lk. Then

Pk(N(le I k)) I k' = 1.

Proof. Clearly k c k' E---lini,'. Hence

pk (N(k' I k))1 k' = p k ,(k' x)1 k' =1. •

6.2. Proof of Lk = kab

In this section, we shall prove that the abelian extension L (= L k ) over k introduced in Section 5.3 is actually the maximal abelian extension kab of k.

We first prove the following key lemma:

LEMMA 6.5. Let (k, y) be a p-field and let (k', y') be a cyclic extension of degree p over k—that is, [k' :k]= p. Then

N(kilk)0 le.

Proof. Let G = Gal(k7k) and let G,,, n_-_. 0, be the ramification groups of /elk defined in Section 2.5. Since G is a cyclic group of order p, each G„ is either G or 1. Suppose first that Go = 1. By Proposition 2.18, k' I k is then unramified: e(k' lk)=1, f(k' lk)= p, and it follows from the formula in Proposition 1.5 that v(N kyk (f)) is divisible by p for every x' E le . Hence a prime element Jr of k is not contained in N(k' lk), and consequently N (le I k) 0 le .

Next, suppose that Go = G so that k'lk is totally ramified by Proposition 2.18. By the remark after the Corollary of Proposition 2.19, the index [Go : G1 ] is prime to p. Hence there is an integer s.-..1 such that

G = Go = G1 = • • - = Gs, G s+1 = G+2 — • ' • = 1.

Fix a prime element .7r' of (k', v') and let f(X) denote the minimal polynomial of Jr' over k. By the corollary of Proposition 2.13 and Proposition 2.14, the different g = g(k'lk) of k'lk is given by g = f'(r')o', where f' = dfldX and o' is the valuation ring of (k', v'). Clearly f' (Jr') = IL (z' — cr(70) with a ranging over all elements *1 in G. Since G =G„ Gs+1 =1 implies v '(z' – o-(z')) = s +1 for all a01, we obtain

vy(7e))= (p – 1) (s ± 1), g =

= 1 + p's+ 1 and let pi being the maximal ideal of (k', v'). Now, let x' E U+1

X ' = 1+y', y' ey's+1. Then

Nkyk (f) = ri (1 + a(Y i )) , a E G,

= 1 + Ey' A +Nky k(yi), A

(6.7)

where A ranges over all elements of the form A = a l + • • • + a„ 1 -. t -.

Fundamental Theorems

85

p —1, in the group ring Z[G], with distinct al , , at in G. Since p is a prime, if a 01, then o-A 0 A for any such A. Hence A, o-A„ o-P -1A are distinct elements of Z[G], and

E Y fm = Teik(Y

for the partial sum on i = A, . . . , aP-1 A. However, since le I k is totally ramified, y' E p's ±1 implies

r E p qp-1)(s+1)+(s+1) = p rp(s-1-1) = ps+1

for the maximal ideal p of (k, y). Therefore,

Tkyk(Y E TV/Ic(Ps+1-1) = ps+1T kvic(g —1) ps+1

by the definition of g = gi(le I k). On the other hand, by Proposition 1.5, y' E P fs+1 also implies Nkv jyf) E Ps±1 . Hence it follows from (6.7) that

Nkyk (x') 1 mod ps+ 1 , for x' =1+ y' E U fs-1-1.

Thus Nrok(U.:+i) Us+i , and the norm map Nkvk induces a homomorphism

W/U.: +1 —> U/Us+1. (6.8)

Let = o- (7e)1A . ' for the prime element A . ' of (k', y') and for a 01 in G. Then it follows from y'(o- (.7e) — z') = s + 1 that E Usf , V ± i. Since Nk ,,k() = 1, we see that the above homomorphism is not injective. However, since le I k is totally ramified and oi/pf = o/p = Fq , it follows from (2.1) that

[(/':U.: +1]= [U:Us+1] = (q —1)qs.

Hence (6.8) is not surjective. This implies Nkyk (W) 0 U and consequently N(k' I k)= Nkyk(lex)* kx. •

REMARK. By the so-called fundamental equality in local class field theory (cf. Section 7.1 below),

:N(k' lk)] = [k' :Id= p.

Hence the lemma is trivial if one is allowed to use the above equality. For a p-field (k, y) of characteristic 0, it may also be proved by using Herbrand quotients. Compare Lang [17], Chap. IX, §3.

LEMMA 6.6. Let (k, y) be a p-field and let k' I k be a cyclic extension of degree p: k c k' c k ab . Then k' is contained in L: k c k' c L c k ab .

Proof. Assume that k' is not contained in L. Since [k' :11= p, it follows that

k' n L = k, Gal(k' LI k) Gal(k1k) x Gal(L/k).

Let Jr be any prime element of k. Then Gal(L/k) Gal(k„ r /k) x Gal(k„/k) by (5.4). Hence we see that k' k„ n kar = k so that k'k„ is a totally ramified extension over k, containing k„. Therefore, by Proposition 5.17, N(k'k„lk)


= (n), and consequently

E N(k'k,1k) N(le lk).

Since le is generated by the prime elements of k, it would then follow N(k' lk)= le, which contradicts Lemma 6.5. Hence k' c L. •

LEMMA 6.7. Let p be a prime number and let k be a field (not necessarily a local field), containing a primitive pth root of unity 4 • Let k' be a cyclic extension of degree ps, s 0, over k. Suppose that k' is contained in a cyclic extension k" of degree ps ±1 over k: k c k' c k". Then 4 is the norm of an element of k'.

Proof. Let a be a generator of Gal(k"/k) and let T = , so that r is a generator of Gal(k"/V). Since 4 E k' c k", it follows from Kummer theory that k" contains an element a such that

k" = k' (a), ar-1 = p .

Let fi = acr-1 . Then pr-1 = ( ar-1)a-1 •pa-1 = 1.

Hence ig is an element of k'. Since

<ps,

satisfies Nk7k(f)= œt - l= •

REMARK. The converse of the lemma is also true if s > 1. See Exercise 7 in Bourbaki, Algebra, Chap. V, §11.

Let (k, y) be a p-field. We now prove kab = L in several steps.

(i) Since k k ur c kab , extend the Frobenius automorphism 'Pk of k to an automorphism ip of kb over k. Let F denote the fixed field of ip in kab . Then

Fk„= kab , F n k ur = k, Gal(kab l Gal(k,/k)

by Lemma 3.4. Let a = I L. Then a kur = 11) kur = To and it follows from Proposition 6.3(ii) that a = Pk(7r) for some prime element a- of k. Since a = if) I L, L n F is the fixed field of a in L and since a = Pk(iv) = we see from the remark after (6.1) that

kck„=LnF=F. We shall prove that k7, = F.

(ii) Let k' be a finite extension of k such that

kck („) ck'c F.

Since k,nF =k, k' lk is a totally ramified finite abelian extension and f = f' = Fq for the residue fields of k and k'. Therefore, by the remark at the end of Section 2.5, the degree [k' :lc] is the product of a factor of q —1 and a power of p. However, [k (7),: k] = q — 1 by Proposition 5.2(iii) for m = 1. Hence [k' :14) ] is a power of p for any k' as above.


(iii) Assume now that k,* F.

Then there exists a finite cyclic extension E'lk such that

k E' F, E' k.

By (ii), [E' : E' n k°,] = [E' le; :14] is a power of p, while [E' n : k] is prime to p. Hence, there is a cyclic extension Elk of p-power degree such that

E(E' n C ) = E', E n (E' ne)= k.

Since E' k , E is not contained in k„. Replacing E by a subfield if necessary, we see that there exists a finite cyclic extension Elk with p-power degree such that

[E:Enk„]=p.

Let

k' = E n k„, [k' :k]= ps, [E:k]_ ps+i, s , 0.

By (6.4) and Gal(L/kur) a-, Gal(k,/k), Pk induces an isomorphism

U Gal(k„ /k).

Let U' be the subgroup of U such that

Gal(k,/k ')

under the above isomorphism. Since U/U' Ga1(k7k), U/U' is a cyclic group of order ps• Hence there is a character x of U/U' with order To , which we view as a continuous character of the compact group U with Ker(x) = U'. We shall next show that there is a continuous character A of U with order p'+' such that

X =

Consider first the case where k contains no primitive pth root of unity. By Proposition 2.7 (for a = 0) or Proposition 2.8, U1 = 1 + p is isomorphic to the direct product of finitely or infinitely many copies of Z„. Since U = V x U1 by (2.2) and since x(V) = 1 for the group V of order q —1, the existence of A, is clear because the character group of Z„ is isomorphic to Qp /Z„, a divisible group. Suppose next that k contains a primitive pth root of unity 4. In this case, the p-field k has characteristic O. Since k' is contained in a cyclic extension E of degree ps' over k, it follows from Lemma 6.7 that E N(k 1 lk). Hence, by the Corollary of Proposition 6.4, Pk() I k' =1, so that 4 E U', x(4)=1. It then follows again from Proposition 2.7 and (2.2) that there is a character A, of U satisfying x = .

(iv) Let x and A, be as above. Let U"= Ker(A) and let k" be the subfield of k,, corresponding to U":

U/ U" 2-, Gal(k"ik).


Then k c k' c k" c k„, E n k" = k' ,

and both Elk and k"lk are cyclic extensions of degree ps±1 with [E:le]= [k" : k'] =p. Hence there exists a cyclic extension k* I k with [k* : k] =p such that

Ek" = k*k".

However, by Lemma 6.6, k* is contained in k, = L n F. Hence E c k*k" c k„, and this contradicts k' = E n k,* E. Thus the assumption F *k, in (iii) leads to a contradiction and it is proved that

F = k„, kab = kurF = kurk„ = L.

THEOREM 6.8. For each local field (k, y),

kab = Lk, namely,

kab = kurk, for any prime element Jr of k.

EXAMPLE. Let (k, y) = (Qp , vp ). In Example 2 of Section 5.1, we have seen that in this case

kpn = Qp (W p.+1), n.-_. —1,

with Wpn+i the group of all pn'th roots of unity in Q. Hence

kp = U1(7,= Qp (Wp..), n

Wp. being the group of all p-power roots of unity in Q. On the other hand, by (3.2),

k ur = Qp (V.),

where V. is the group of all roots of unity in Q with order prime to p. Therefore, for (k, v)= (Qp , vu ), Theorem 6.8—that is, kab = kurkp—states that the maximal abelian extension (Qp)ab of Qp is generated over Qp by all roots of unity in Q. Obviously, this is the analogue for Qp of the classical theorem of Kronecker, which states that the maximal abelian extension of the rational field Q is generated over Q by all roots of unity in the algebraic closure of Q.

6.3. The Norm Residue Map

By Proposition 6.2 and Theorem 6.8, we now see that with each local field (k, y) is associated a homomorphism

pk :k x —>Gal(kab lk).

Pk is called the norm residue map, or, Artin map, of the local field (k, y). Some of the properties of Pk were already given by Proposition 6.3 and


(6.4). For convenience, we shall restate such properties of pk :k x -> Gal(kab /k) below:

(i) Pk is injective and is continuous in the v-topology of k x and Krull topology of Gal(kab /k).

(ii) The image of Pk is a dense subgroup of Gal(kab /k) and consists of all elements a in Gal(kab /k) such that a 1 k,= Vk'l for some integer m. In particular, if a ku r = then there is a unique prime 1 (Pk, element g of k such that pk (z)= a.

(iii) Pk induces a topological isomorphism

U2', Gal(kab /k,),

ul--> (5(u')

on the subgroup U of k > , where (5 : U :1-* Gal(kab ik,) is the isomorphism in Proposition 5.13. For any prime element g of k, it also induces

U :-_-, Gal(k,/k), ul->

where (5, is the isomorphism in (5.4).

We shall next study how Pk depends upon the ground field (k, v). Let k' I k be a finite extension of local fields and let

pk : le -> Gal(kab l k),

P k': IC " --> Gal(eab/k f)

be their norm residue maps. Since /cable I k' is an abelian extension, we have

k c k ab = /cable = k b

so that the map a-* a I kb , a e Gal(kai b/k f), defines a homomorphism

res: Gal(k fabile)--> Gal(kab /k).

THEOREM 6.9. Let N jell,: k"--> k x be the norm map of k' I k. Then the following diagram is commutative:

k'x ---> Gal(k b 1 le)

IN Ires

kx --> Gal(kab I k).

In other words,

Ple(x f )lkab = Pk(N kyk(X)), for all X f E k fx .

Proof. Let 1(0 be the inertia field of the finite extension le I k so that k o/k is unramified and k'/k0 is totall ramified (cf. Section 2.3). It is clear that if the theorem holds for the extensions k o/k and k f/ko , then it also holds for

90


/elk. Hence it is sufficient to prove the theorem in the cases where k'lk is either unramified or totally ramified. In the latter case, the theorem is an immediate consequence of Lemma 6.4 because we now know by Theorem 6.8 that

Lk = k ab y Lk' = kb, E = L k n Lk , --- kab•

Hence, suppose that le I k is unramified and let k' = k, m = :kJ. As usual, it is enough to show that for any prime element z' of k',

P le(z i ) I kab = PAN k'lk(Z 1 ))•

By Proposition 5.4(i), k72;n1kta';. is a totally ramified finite abelian extension and 7rt is contained in N(142:'' I k). Therefore, applying the Corollary of Lemma 6.4 for k' = ktunr c kthiin c Lk , = kb, we see that p k , (zi) lejrn/ n = 1 for all n 0. Noting that k c c k ab by Proposition 5.2(iii), we find

a 114Y° = P e 1, for a = (e) I kab On the other hand,

.

f kur Ple(.701 kur = 99 k' = CP rkn

by (6.5). Hence, by Lemma 6.1,

P/e(b7r') kab = CT = Pk(Nk'lk(• 70).

COROLLARY. Let k' I k be a finite extension of local fields and let x E . Then

Pk(x)1(k' n kab) = 1 47>x E N(k' I k).

Proof. Let x = N kyk (f), x' E k'x . Then

Pk(x) f (k' n kab) = Pk,(x') (k' n kab ) = 1

because pk , (x) E Gai(kablle). Conversely, suppose that Pk(x) I (k' n kab) = 1. Since Gal(kabk' I k') Gal(kab 1 k' n kab), pk (x) can be extended to an automorphism in Gal(kable I ki) and, then, to an automorphism a of Gal(kL b l k') : a f kab = Pk(x). Let

k tun, = k' n kur = (k' n kab)n kar, m 1.

Then p k (x) f ktunr = 1 so that p k (x) f k, is, by (6.4), a power of the Frobenius automorphism CP tkn over 1(. Since Gal(ka' ,.1 Gal(kur /k:), a k is then a power of the Frobenius automorphism Tie over k', and it follows again from Proposition 6.3(ii) that

a =Pk'(x), for some x' E k' > .

Therefore, by the theorem above,

Pk(x) = a f kab = Pk(Nk'lk(X t ))•

Since Pk is injective by Proposition 6.3(i), it follows

x = N kyk(X ' ) E N(k' I k). •

•


In Proposition 6.2, we defined the map Pk by means of ip, in (6.1), which in turn depends on the field k, constructed by the formal group FAX, Y) for f e 37,1 (cf. Section 5.3). However, we are now able to give a simple description of the norm residue map Pk as follows:

THEOREM 6.10. Pk is uniquely characterized as a homomorphism

p: le —>Gal(kablk)

with the following two properties:

(i) For each prime element g of k, P(101 kur = (Plcy (Pk being the Frobenius automorphism of k.

(ii) For each finite abelian extension k' over k, p(N(k' lk))1k' =1. Proof. By (6.5) and the Corollary of Theorem 6.9, Pk has properties (i)

and (ii). Let p :V' —> Gal(kablk) be any homomorphism satisfying (i) and (ii), and let it be any prime element of k. Since it E N(141k) by Proposition 5.4(i) for m = 1, it follows from (ii) that p(g) I kr,z, = 1 for all n.-_. 0. Hence P(.701 kr = 1. As P(.701 kur = cpk, we see from (6.1) that p(z)= ip, = pk (.7r). Therefore, p(x)= pk (x) for every x E le—that is, p = Pk . •

We can see now that Pk is naturally associated with the local field (k, y) in the following sense. Namely, let

o- :(k, y).4(le , y')

be an isomorphism of local fields—that is, an isomorphism a:V:4k' such that y = y' 0 a. We can extend a to an isomorphism of fields

a:kab .4 Vab

and define an isomorphism of Galois groups:

a*: Gal(kab /k) .4 Gal(k ab/V), -1 al—> OT a .

We then have the following theorem:

THEOREM 6.11. The following diagram is commutative:

V' ')L ->c Gal(kab I k)

16 16* k''' 9`—>`' Gal(kiablk').

Proof. It is clear that a :kab k'ab and a* :Gal(kablk)_-_-> Gal(leablk) induce

a: k u,..:_-> k'u„ a* : Gal(k,/k) :4 Gal(k,/k ') and

a* (Tk) = acPka-1 = (Pk'

92


on Co.. Let p' = o- * 0 pk 0 o- ' : --> Gal(leablk),

and let a' be any prime element of k'. Since y = y' o a, a = o-'(a') is a prime element of k. Hence

(Pk ° (1-1 )(z i )1 kw. = Pk(z)1kur = q9k,

P Kr= a* (99k)=

Let E' E' be any finite abelian extension over k' and let E = o- '(E'). Then Elk is a finite abelian extension over k and

a: N(Elk)2-,N(E' Ile), a* : Gal(Elk) Gal(E'/k').

Hence (Pk° a 1 )(N(E' lle))1E = pk (N(Elk))1 E =1,

(N(E' 1 10)i E' = a*(pk (N(Elk))1E' = 1.

Therefore, by Theorem 6.10, p' = Pk' • • COROLLARY. Let a be an automorphism of a local field (k, y). Then

Pk(a(x))= apk (x)o--1 , for all x E k x ,

where a on the right denotes any automorphism of k ab , which extends the given a:kk.

Proof. Apply the theorem for k = k'. •

REMARK. Theorem 6.11 can also be proved directly by going back to the definition of Pk in Proposition 6.2.

To explain the next result on Pk, we need some preparations. Let Q., denote the maximal Galois extension over the local field (k, y) and let k' be a finite separable extension over k so that

kclecQs cQ, n =[1(' :k]< + Let

G = Gal(Qs /k), G' = Gal(Qs/kab),

H= Gal(Qs /k 1 ), H' = Gal(Qs/k Lb).

Then H is an open subgroup of the compact group G with index [G :H] = :k]= n, and G' and H' are the topological commutator subgroups of G and H, respectively. Let {x 1 ,. , TO be a set of representatives for the left cosets HT, t E G:

G=LJ Hr. i=1

Then, for each a E G and each index j, 1 i n, there exist a unique element hi E H and a unique index such that

'ria =


Denoting the above hi by hi (a), we define an element t G ,H(o-) in HI H' = Gal(klabl le) by

t G IH(a) = hi(0-)H, for a E G. (6.9)

It is known from group theory that t G ,H(o-) depends only upon CIE G, namely, that it is independent of the choice of the representatives

. , rn for the left cosets HT, and that a 1-tGIH(a) defines a homomorphism

which is called the transfer map from G to H.t In the following, we shall denote tGai also by tkvk and call it the transfer map from k to k'. Since HIH' is an abelian group, the homomorphism tkvk tGiH) can be factored as

G-->GIG' --> HITE.

For convenience, the induced map GIG'--> HIH' will also be denoted by tkvk. Thus for each finite separable extension k' /k, t kw, denotes homo-morphisms of Galois groups

Gal(S2s /k)--> Gal(Kt,/k 1 ), Gal(kab /k)--> Gal(Kb/k 1 ).

We shall next describe some properties of the transfer maps which follow from the Definition (6.9) by purely group-theoretical arguments:

(1) If k c k' k", then tk"lk =

(2) For a E G = Gal(52,/k), n =[k' : k]= [G : H],

tic7k (o-)G 1 = o-"G' , that is, tkvk(a) k ab = an I kab•

(3) Let k' I k be a Galois extension and let a E H = Gal(Qs /k 1 ). Then

\ tkvk(a)= Tar -11/1 = (II Tar

-1 )

where r ranges over a set of representatives for the factor group G I H = Gal(k'/k).

(4) Let k' /k be a cyclic extension and let aH be a generator of GIH= Gal(k'/k). Then t k ,,k (o-) = o-"H'—that is,

tvik(a)= an Kb, n =[k' : k]= [G :H].

Now, for a finite separable extension k' I k of local fields, consider the

t For group-theoretical properties of transfer maps, see, for example, Hall [14

kb,

94

diagram

(k' I k)

le — P2--.> Gal(kab /k)

I" Gal(kL b lk')


where the vertical map on the left is the natural injection of le into k''. We shall prove below that the above diagram (k' I k) is commutative. We first establish some preliminary results.

LEMMA 6.12. Suppose that (k' lk) is commutative. Then tkyk:Gal(kabl

k)--->Gal(kL blk') is injective. Proof. Let tkyk (a) = 1 for a E Gal(kab /k). Extend a to an automorph-

ism of Gal(52s /k) and call it again a. Then, by (2) above,

an (= an 1 k ab ) = t k'/k 7)1 kab =1, 11 =7- fk' : k].

Since Gal(k,/k):-_-,i and i is a torsion-free abelian group (cf. Section 3.2), it follows that a E Gal(kab /k„). Hence, by (6.4), a = pk (u) for some u E U. As (k' I k) is commutative, we then have

Pic(u) = tk , /k(Pk(u)) = tkyk(a)= 1.

Since Pk , defines U':-_--,Gal(leabl Co.), it follows that u =1, a = pk (u) = 1. •

LEMMA 6.13. Let k k' k".

(i) If both (k"Ik') and (k' /k) are commutative, then so is (k"lk), (ii) If both (k"Ik') and (k"lk) are commutative, then so is (k' /k). Proof Consider the diagram

(k' I k)

(k"Ik')

kx --> Gal(kab /k)

I I k' x --> Gal(leab /k 1 ).

I 1

k"' --> Gal(Kb I k")

By (1) above, the outside rectangle of the above diagram is (k"lk). Hence (i) is obvious. If (k"Ik') is commutative, then

t k7k , : Ga1(k ai t,/k 1 )---> Gal(K b /k")

is injective by Lemma 6.12. Hence (ii) follows.

LEMMA 6.14. Let le lk be a finite Galois extension and let x E N(k' lk). Then

P v(x) = tk7k(Pk(x)).


Proof. Let x = Nk7k (f), x' E k' x , namely,

x =

where T ranges over all elements of Gal(k'/k) = GIH. Extend each r to an automorphism in Gal(Q s /k) = G and denote it again by T. Then it follows from (3) above that for a = pk ,(x') E Ga1(kai b /k 1 ) and for tkyk:Gal(kabi

k)--* Gal(kai b /k 1 ),

tk7k(a I kab) = Tar -1

on k a' I,. Here

kab = Pk'(-) I kab = pk(Nkyk(x 1 ))— Pk(x)

by Theorem 6.9. On the other hand, on eat),

pk , (x) =LI pk,(r(f))= FI TPk , (x')T - ' = TGT-1

by the Corollary of Theorem 6.11. Hence p le (x) = t ( ok. (x))

LEMMA 6.15. If k' I k is unramified, then the diagram (k' I k) is commutative. Proof. Since le is generated by the prime elements of k, it is sufficient

to prove

Pk'() = tkvk(Pk(z))

for any prime element .7r of k. In this case, ka' b /k is a Galois extension. Hence, extend zp,r = Pk (z) to an automorphism zp of k a' b over k and let F denote the fixed field of lp in k a' b . Since zp I k,=Pir kur = cpk , it follows from Lemma 3.4 that

Fk„=- Vab, F n k„ = k, Gal(k'abl Gal(kur /k).

Furthermore,

Fnkab =k,, kck,cF

because F n ka, is the fixed field of 1p7r = kab in kab• Hence, by Proposition 5.17,

N(F I k) = (.7r).

Let k' = k r, n =[k' :k], and let F 'Fk r, =

Then F' is the fixed fi eld of zp' in CI, and

k'ar = I kur = cPrki = 'Pk ' •

Therefore, by Proposition 6.3, there exists a prime element .7r' of k' such that

= F' =


and by Proposition 5.17,

N(F'Ik')=N(klk')= (z').

Now, let E be any finite extension of k in F:kc Ec F, [E:k]< +00. Since F nk' = k, we then have

[Ek' :k']=[E:k], N(Elk) ç N(Ek' Ik').

As F' (= Fk') is the union of all such fields Ek' , it follows that

(z)=N(Flk)gN(F'110= (JO.

Hence Jr = z' because both Jr and Jr' are prime elements of k'. Now extend ip to an automorphism ci of Qs over k, so that a I kab ---7 li) I kab = Pk(). Then it follows from (4) above that

tk7k(Pk(2t)) = tk7k(a) = an I kb =1P' = Pk , (7')= Ple(z). •

We are now ready to prove the following

THEOREM 6.16 Let k' /k be a finite separable extension of local fields. Then the diagram

le Pk -40

(k' lk) I

k' x ‘- ' ->e

Gal(kab lk)

fic'lk

Gal(k'ab lk')

is commutative. In other words,

Pe(x)=- telk(pk(x)), for all x E kx.

Proof. Since le lk is separable, k' can be imbedded in a finite Galois extension k" over k: k c k' c k". By Lemma 6.13(ii), we see that it suffices to prove the theorem in the case where k'lk is a finite Galois extension. But, by the Corollary of Proposition 2.19, Gal(k'/k) is then a solvable group. Hence, by Lemma 6.13(i), we may even assume that k'lk is an abelian extension. Let

n = [lc' :k], E = k:14r, E' = Ek',

and let z be a prime element of k. Then pk (z) I E (= (NI krai r) is a generator of Gal(E/k) and pE(z) = tok(Pk(z)) by Lemma 6.15. Since k ç k' g kab g Eab, [k' : k] = n, it follows from (4) above applied for Elk that

PA:0 1 k' = t ok(Pk(z))1 le = Pk(z)n 1k' = 1.

Hence pE(z)1E' =1 for the finite abelian extension E' over E, and it follows from the Corollary of Theorem 6.9 that z E N(E1E). Therefore, by Lemma 6.14,

PE-(z)= tEvE(PE(z))= tE7E(tok(Pk(z)))= tE7k(Pk(z))•


Since le is generated by prime elements, this shows that the diagram (E' I k) is commutative. Since (E' I le) is commutative by Lemma 6.15, it follows from Lemma 6.13(ii) that (k' I k) is also commutative. •

COROLLARY. The transfer map

tkyk:Gal(k ablk) ---> Gal(k'abl lc') is injective.

Proof. Since (k' /k) is commutative, this follows from Lemma 6.12. •

Chapter VII

Finite Abelian Extensions

In this chapter we shall first prove some important results on finite abelian extensions of local fields that constitute the main theorems of local class field theory in the classical sense. We shall then discuss the ramification groups, in the upper numbering, for such finite abelian extensions.

7.1. Norm Groups of Finite Abelian Extensions

Keeping the notation introduced in the last chapter, let

Pk:k x—> Gal(kab/k)

denote the norm residue map of a local field (k, v). Let k' be any finite abelian extension over k: k c k' c k ab , [le :k] < + co. We denote by p the ....e product of Pk and the canonical restriction homomorphism

Gal(kab /k)---> Gal(k'/k) = Gal(k ab /k)/Gal(k ab /k ').

Thus

Pk'lk:k x— *Gal(k ' /k).

THEOREM 7.1. Let k' lk be a finite abelian extension of local fields. Then the above homomorphism pkvk induces an isomorphism

kx1N(k7k) 4' Gal(k'/k).

Furthermore,

N(k'lk)= piV(Gal(k ablV))

and Gal(kab lk') is the closure of pk (N(k' I k)) in Gal(kab lk). Proof. By the Corollary of Theorem 6.9, the kernel of pkvk is N(k' lk).

By (ii) at the beginning of Section 6.3, the image of Pk is a dense subgroup of Gal(kab /k). Since Gal(kab /k') is an open normal subgroup of Gal(kab /k), Pk'lk is surjective. Hence k x 1N(V1k): -_-> Gal(k'/k) and N(k'lk)=

Pi-c l (Gal(kabl le)). The isomorphism shows that Pk maps each coset of le mod N(k' lk) into a coset of Gal(kab /k) mod Gal(kab /k '). Since the cosets are closed and since the image of Pk is dense in Gal(kab /k) and pk (N(k'lk))c Gal(k ablk'), it follows that Gal(kab /le) is precisely the closure of p(N(k' lk)) in Gal(kab/k). •

COROLLARY. For a finite abelian extension k' lk of local fields,

[k x :N(k' I k)]= [k' : k].

The above equality is called the fundamental equality in local class field

Finite Abelian Extensions 99

theory because in the classical approach, the proof of this equality is one of the first important steps in building up local class field theory. We also note that the norm residue map Pk is so named because it induces an isomorphism of the residue class group of le modulo the norm group N(k'lk) onto the Galois group Gal(k'/k).

PROPOSITION 7.2. (i) Let k' lk be any finite extension of local fields. Then

N(k' I k)= N((k' Clk ab )1k), [le :N(le lk)]-[k' :k].

Equality holds in the second part if and only if k' 1 k is an abelian extension. (ii) Let le lk be a finite extension and k"lk a finite abelian extension of

local fields. Then

N(k'lk) ç N(k"lk)<=>kc k" c k'.

Proof. (i) Applying the Corollary of Theorem 6.9 for k' lk and (k' n kab )Ik, we obtain

x E N(k' I k)<=> pk (x)1 k' n kab = 1 <=> x E N((k' n k ab /k)

for x E le . Hence N(k' I k)= N((k' n k ab )Ik). It then follows from the Corollary of Theorem 7.1 for (k' n kab )Ik that

[le :N(k1k)]= [le :N((k' n kab )/k)] = [k' n kab : k] _[k' :k].

Equality holds if and only if k' n k ab = k'—namely, when k' lk is abelian. (ii) Since k"lk is abelian,

kc k"cle.(=>kc k"ck i nkab=kab

<=>Gal(kab l (le n kab)) g Gal(kab lk").

By Theorem 7.1, the last inclusion is equivalent to

N(k' I k)= N((k' n k ab )Ik) ç N(k"lk). •

Now, let k'lk be a finite abelian extension of local fields and let

k c k" c k'.

Consider the diagram

VIN(lelk) Gal(k'/k)

I I

le IN(k"lk) '--=–> Gal(k"/k),

where the horizontal isomorphisms are those in Theorem 7.1, the vertical map on the left is the one defined by the inclusions N(k'lk)c N(k"lk)c le (cf. Proposition 7.2(ii)), and the vertical map on the right is the canonical (restriction) homomorphism.

100


PROPOSITION 7.3. The above diagram is commutative. Hence the isomorph-ism kx1N(k1k) Gal(k1k) induces

N(k"I k)IN(k' I k) Gal(k1k").

In other words,

Pkyk(N(k"lk))= Gal(k'/k").

Proof. The commutativity of the diagram follows from the fact that both horizontal isomorphisms are induced by the same pk :k x —> Gal(kab /k). Hence the diagram yields an isomorphism between the kernels of the vertical maps—namely, N(k"lk)IN(k' I Gal(k1k"). •

LEMMA 7.4. Let m 1, n 0 and let 7t be a prime element of a local field (k, v). Let

E = kiTrk;ir

with the abelian extensions kmur and kn„ over k, defined in Sections 3.2 and 5.2, respectively. Then

N(Elk)= (:rm) x

Proof. By Proposition 5.4(i) for m =1, kn„ is a totally ramified abelian extension over k and r is the norm of a prime element jr' of kn„. E = rure, is then an unramified extension of degree m over kn, so that

N(E114)= (It'm X U(k),

as shown in the proof of Proposition 3.7. However, NU(141k)= U„ i by Proposition 5.16. Hence, taking the norm from Cir to k, we obtain from the above that

N(Elk)= (zrn) X Un+1.

We note in passing that the field E = k unrk;ir above is actually 14" of Section 5.1 for Jr in k. •

THEOREM 7.5. For each closed subgroup H of kx with finite index, there exists a unique finite abelian extension k' I k such that

H = N(k' I k).

Proof. Let m = [kx :Hl< +00. Then n'n E H for a prime element r of k. Since H n v is a closed subgroup of U= U(k) with [U : H n v] m < +00, there is an integer n 0 such that v„ ± , H n Uc H. Hence, by Lemma 7.4,

N(Elk)= (am ) x Un .“ H kx

for E = krunrk7r. Now, by Theorem 7 .1, PElk induces

kx IN(Elk) Gal(E/k). (7.1)

Therefore, there exists a field k' such that

k' E, Gal(E/k')


under the above isomorphism (7.1). However, by Proposition 7.3,

N(k' I k)IN(EI k):-., Gal(E/k')

also under (7.1). Hence H = N(k' I k). The uniqueness of k' follows from Proposition 7.2(ii). M

Classically, the above result is referred to as the existence theorem and uniqueness theorem in local class field theory.

Let k' I k be a finite extension of local fields. By Proposition 3.5 and Proposition 7.2(i), the norm group N(k' lk) is a closed subgroup of finite index in le. Therefore, we see from Theorem 7.5 that the map k'i--> H = N(k' I k) defines a one—one correspondence between

the family of all finite abelian extensions k' over k (in Q)

and

the family of all closed subgroups H with finite indices in k' .

Furthermore, by Proposition 7.2(ii), this correspondence reverses inclusion, namely, if k i i--> 111 , k 2 1--> H2, then

k 1 c k2 <=> Hi c H2.

Hence

N(k'k"lk)= N(k' I k) n N(k"lk), N((k' n k")I k)= N(k' lk)N(k"lk)

for any finite abelian extensions k'lk and k"lk.

REMARK. The one—one correspondence mentioned above can be extended to a similar one—one correspondence between the family of all abelian extensions of k (in Q) and the family of all closed subgroups of le. See Artin [1].

THEOREM 7.6. Let le lk be a finite extension of local fields. Let E be any finite abelian extension of k and let E' = Ek'. Then

N(E' I k') = {x' E k'IN kyk (x') E N(Elk)}.

Proof. Note first that E' I k' is a finite abelian extension and that kcEc E' c k a' b . Let X' E k", x = Nkyk(f). Then, by Theorem 6.9,

pk ,(x') I E = pk (x) I E.

Since E' = Ek', it follows from Theorem 7.1 that

X' E N(E'lk 1 )<=>pk ,(x') I E' = l<=> pk ,(x 1 )1E =1

<=>

Pk (x)1E =1 <=>x E N(E I k). •

7.2. Ramification Groups in the Upper Numbering

Let k' I k be a finite Galois extension of local fields and let

G = Gal(k'/k).

102


In Section 2.5, we defined a sequence of normal subgroups G„, n ___- 0, of G:

lc • • • c G, c Gn _ i c • • - c Go c G,

called the ramification groups (in the lower numbering) for the extension k'/k. Let

k c k" c k' , H = Gal(k' /k") e G

and let 11,,, n_._.- 0, be the ramification groups for k'/k". Then it follows immediately from the definition of G„ and H„ that

Hn = Gn n H, for n __-. O.

Suppose now that k"I k is a Galois extension so that H is a normal subgroup of G and

G I H = Gal(k"/k).

Let (G I H),, n ___- 0, be the ramification groups for the extension k"/ k. A natural relation one may expect between G,, and (G/ H),, might be that (G I Inn is the image of G,, under the canonical (restriction) homomorphism

G = Gal(k ' /k)--> GI H = Gal(k"/k).

However, this is not true in general, and in order to obtain a simple relation between G„ and (G/H),, we have to change the "numbering" of the ramification groups, as discussed below.

Let G „ , n .---- 0, be as above. For each real number r -..- -1, we define a subgroup G,. of G as follows:

Gr = Gn, ifn-l<r_n,neZ,n-- 0.

Let gr = [Gr: 1], for r _-_ -1.

Then g,. = g,, for n - 1 < r_.- n, n ...- 0, and

g0 = [G0 : 1] = e(k' I k)

by Proposition 2.18. We consider a real-valued function (1)(r) for r -.-- -1 with the following two properties:

(i) 4)(r) is a continuous, piecewise linear function of r ___. -1 with 0(0) = 0,

(ii) The derivative 0' (r) of cp at r (t Z is equal to grigo.

Drawing the graph of cp(r), we see easily that there exists a unique such function OW and that in fact

0(0= r, for -1 r 0,

1 -'------ (gi + • • • +gn-i + (r - n +1)g,), forn-l<r_n,n..._ 1, n EZ.

go (7.2)


Since g; is a factor of gi for integers i, j, 0 i j, it follows from the above that if (j(r) is an integer, then so is r.

For each a in G = Gal(k'/k), let

i(a) (=ic(cr)) =min(v'(a(Y) — Y) I Y E 0 '),

where o' is the valuation ring of the normalized valuation y' of k'. Since a(y) — y E 0', we always have i(a)__ 0, and, in particular, a(1) = +00. On the other hand, if a *1, then a(y)* y for some y E O'. Hence

0 i(a)< +x, for all a * 1 in G.

By Lemma 2.13, there exists an element w in o' such that o' = o[w]. If y is an element of o' = o[w], then a(y) —y is divisible by a(w) — w in o' = o[w]. Hence

v'(a(y)— y)a- v'(a(w)— w), for all y E O',

and it follows that

i(a) = v'(a(w)— w), for a E G.

It is then clear from the definition of Gr that

Gr = {a E G I i(a) r+ 1}

= {a E G v 1 (a(w)— w) r+1}, r —1.

Consider now a function AO, r —1, defined by

AO= —1+ —1 E min(i(a), r +1).

go aEG

(7.3)

Clearly )L(r) is continuous and piecewise linear because each min(i(a), r + 1) is such a function. Furthermore A(0) = —1 + (go lgo)= 0, and for r in the open interval i —1<r <i, i E Z, j a' 0, A(r) is given by

A(r)= a + II go

where a and b are constants and b is the number of a's in G such that i(a) r +1—namely, b =[G,.:1]= gr . Hence )'(r) = g,.1 g0 , and it follows from the definition of (p(r) that (p(r) = A(r)—that is,

0(0= —1+ —1 E min(i(a), r + 1), for r —1.

go aEG (7.4)

Since 4 (r) is, by (7.2), a strictly increasing function of r —1, let v(r) denote the inverse function of cp(r):

(POP(r))= 110(r)) = r, for r —1.

We then define subgroups G r of G by

G r = G v(0, G q9(s) = Gs , for real r, s —1.


For example,

G -1 = G_ i = G, Gr = G r = Go , for —1 < r 0.

These subgroups Gr, r —1, are called the ramification groups, in the upper numbering, for the finite Galois extension k' /k.

Now, as mentioned above, let k" be a Galois extension of k, contained in k', and let

H = Gal(k7k"), GIH = Gal(k"/k).

Then we have the following theorem:

THEOREM 7 .7 . Let Gr and (G I H)r, r —1, be the ramification groups in the upper numbering for k' lk and k"lk, respectively. Then (GIH)r is the image of Gr under the canonical homomorphism GIH:

(GIH)r=GrHIH, for —1.

We shall next prove this theorem in several steps. (i) For each a' E GIH,

1

ical(01 E ic(a), e

where i G/H denotes the i-function for k"lk, e' =e(k'lk"), and the sum on the right is taken over all elements a E G such that a 1--> a' under

G/H—that is, a' = aH.

Proof. If a' =1, then both sides are +0c. Hence assume a' 01. Let o" be the valuation ring of the normalized valuation le on k". By Lemma 2.13, there exist elements w and z such that

o' = o[w], o"= o[z].

Since y' k" = e' v", it follows from (7.3) that

iG (a) = v r(a(w) — w), 1

v"(a'(z)— z)=—e'

v'(a(z)— z).

Fix an element a E G such that a' = aH. Since

1 1 1 E iG (") —er TE EH 1?("(w) w) =

f

("(w) w))' x e tE rEH

the equality to be proved is equivalent to the following:

11 ' (CY(Z ) — z)= v'( fl (or(w) — w)). -L-EH

Let s = [H:1 ] = [k' :k"1 and let g(X) denote the minimal polynomial of w over k". Since k' = k"(w) and v'(r(w)) = v'(w) 0, we see that

g(X)= fl (X — r(w)) TE H

=Xs +ai Xs -1 + • • • + as , aiEo",


Let

ga(X)= Xs + a(ai )Xs -1 + • • • + cr(as), Gr(ai) E 10"

= fl ot(w)). TEH

Then

Ow) - g(w)= Ow) = H (w - arm). rEH

However, the coefficients a(a)- ai of e(X)- g(X) are divisible by a(z) - z in o" = o[z]. Hence we obtain

y' (w - v 1 (a(z)- z). rEff

On the other hand, it follows from z E 0" = o[w] that there is a polynomial h(X) in o[X] such that

z = h(w).

As h(X)- z is a polynomial of o"[X] that vanishes for X = w, it follows that

h(X)- z = g(X)g i (X), with g(X) E 0"[X].

Applying a to the coefficients of both sides and using h(X) E 0[X] g k[X],

we obtain

h(X)- a(z) = ga(X)0X),

where g(X) is a polynomial of o"[X], defined similarly as e(X). Put X = w in the above. Then

z - a(z) = e(w)e(w)= (w- ax(w))e(w), rEH

where gaw) E 0'. Hence

v'(z - v 1 ( fl (w - TEH

This completes the proof of (i). • (ii) For a' E G/H, let

j(cr') = max(i G (cr) a E G, aH = a').

Then

iG/H( 0 ) - 1 = Ceik-(j(ai ) - 1 ),

where 0 k ,,k ” denotes the 0-function for the extension k'llc". Proof. We may again assume a' *1. Fix an element a E G such that

aH = a', Au') = iG (a). Let n = Au') 0. By (7.4) for le /k",

(Pkvv{i(cr') - 1 ) =- (Pkvie(n - 1) = -1+ —1 E min(iH (T), n). e' rEH


Suppose 1H(r) _-_ n so that min(iH(r), n)= n. Then r E H,-i= H n Gn_i. Since iG (a)= n, so that a E Gn - i , it follows that CT E Gn _ i , iG (a-t- ).-_ n. By the choice of a, we then obtain iG (a-t- )= n = min(iH (r), n). Next, suppose iH(r) < n. Then rotz_i=iinGn_1. Since a e Gn _ i , it follows that iG (ar)= iH(T)= min(iH(r), n) also in this case. Therefore, by (i),

(Preik"U(cr 1 ) - 1 ) = -1 +4 E ic (ITO = -1 + icul(a1 )• e t-EH

(iii) (Herbrand's Theorem). Let s = Okve (r), r .-_ -1. Then

Gr HIH =(G1H) s .

Proof. a' E Gr H I H <=> iG (a).-_r +1 for some o c G such that a' = aH

_-_ r +1

4 ' (I) k , ik-(J(G') — 1 ) --' Cc vic-(0, because cp kyk ” is an increasing function

.(=>icin(cri ) - 1 _-- s, by (ii)

.(=> a' E (G I H)s . •

(iv) (Pk'/k = Ok"lk ° (Pk'/k", Pk'/k = Vk'lk" 0111 1c"Ik•

Proof. Since lp is the inverse function of ofp, it is sufficient to prove the first equality. Let A = do . k"lk ° (Pk'/k». It is clear that A is continuous, piecewise linear, and A(0) = 0. Let r _-_ -1, r f Z. Then ofp kvv (r) Et Z by the remark after (7.2), and the derivatives satisfy

A'(r)= Ok' ”Ik (s)(p k' 7ku(r), with s

where 1 1

q../k (s) = — [(GI H), :1], Cc'/Ic"(r) = "i., [Hr :1] e"

with e' = e(k' I k"), e"= e(k"lk). However, by (iii) above,

(GI H), = Gr HIH 2 Gr /(Gr n H) = Gr l Hr .

Since e'e"= e = e(k1 lk), it follows that

1 A'(r)= [Gr :1] = 4);, 7k (r).

Therefore, by the uniqueness of the function cp(r), (pkvic (r)= A(r) for r .-__ -1. •

We are now ready to complete the proof of Theorem 7.7. Let s = 1P kvk(r), r --- -1, so that Gr = G. By (iii) above,

GrHIH=QH1H=(G1H)„ with t = Ok7e(5).

By (iv),

•

= (Pkvie(r),

4) k"/k(t) = (I; 0 k"ik(4) kyk-(s)) = 0 le ik(s) = (i) kyk(1 I 1 k7k(r))= 7'.


Hence

GrHIH=(GIH),=(GIH)r. •

COROLLARY OF THEOREM 7.7. Let both k i lk and k2lk be finite Galois extensions of local fields (in Q) and let k' = k1 k 2 . Then

Gal(k i/k)r = 1, Gal(k 2/kr = 1.(=> Gal(k' I Or =1.

Proof. By the theorem, the left hand side yields

Gal(ki lk)r Gal(k1k 1 ), Gal(k'/k)r c Gal(V/k2).

However, since k' = k 1 k2 , the intersection of Gal(ki/k i) and Gal(k i/k 2) is 1. Hence Gal(k'/k)r = 1. The converse is clear. •

LEMMA 7.8. Let ki I k be a finite Galois extension of local fields. Let p' denote the maximal ideal of k' and let

g(ki lk)= pia, a .-_0,

for the different g(ki I k) of the extension k' /k. Then OC

a = E i(a)= E (gn — 1), a E G = Gal(k ilk), a*1 n=0

where gn =[Gn :1], n = 0, 1, 2, . . . . Proof. By Proposition 2.14,

(ki I k)= fi(w)oi ,

where w is an element of o' as in Lemma 2.13 and fi(X) is the derivative of the minimal polynomial f(X) of w over k. Since

f(X)=Fi(X — a(w)), a E G, a

it follows from (7.3) that

a = vi(fi(w))= E v'(w — cr(w))= E i(a). a*1 a*1

Let gni =gn — 1. Since i(a)= n if and only if Gr E Gn - i , a If Gn , we see that =

E i(a) = E n(g'n _ i — gi,„)= (g(i) — g;) + 2(g; — a + • • • a*1 n=0

= .g-') ± el. ± g;. ± . • 7.3. The Special Case kTm/k

Let m .-_ 1, n.-.. 0 and let 7r be a prime element of Cr. Let

G = Gal(kT . n/k)

for the finite abelian extension k" over k, defined in Section 5.1. We shall next determine the ramification groups G,. and Gr, r .-_1, for the extension

108


1c 77n/k. Let

H = Gal(k 77n/kZir).

By Proposition 5.2, k ur n k7, n = Ica—namely, ka is the inertia field of the extension Kr/k. Hence Go = H so that

= Hi , for all integers

Therefore, we also have

Gr Gr = H', for all real

Since G = G -1 = G, Gr = G r r

11 for –1 < r 0,

we see that it is sufficient to determine Hr and Hr for the totally ramified extension k"/k.

LEMMA 7.9. Let (5:U-- H be the surjective homomorphism in Proposition 5.3(ii) and let a = (5n,(u) H with u E U = U(k). If u Ui , u U1 1, 0 n, then

where iH is the function on H defined in Section 7.2 and q is the number of elements in the residue field = o/p of k.

Proof. Let f = and let a E W7, a WI'. By Proposition 5.4(ii), is a prime element of k 7"ri'n and o'n'n = orn[a]. Hence, by ( .3),

iH(a)= v'(a(a)– a),

where y' denotes the normalized valuation of the local field k,Pt Let u E U, u Ui—that is, u* 1 mod p. Since v'($x) = 1, v'(u – 1) =0, we then have

a(a)=[u]f (a)-= ua mod a2, v'(a(a) – = 1.

Therefore, the lemma is proved for i = 0. Let now u E 14, u Ui+1 for 1 j n. Write u in the form u = 1 4- y with u E pi± 1 • Then

a(a)=[u]f (a)= u a =1.1.a -if- v I. = 4- 13,

where 13 = E = Wr i, fi pi4-1 .px = Wri-1 by Lemma 4.8. There- fore, by Proposition 5.4(ii), )3 is a prime element of Since len,r ' 2 1kmi, and, hence, are totally ramified, we see from Proposition 5.2(iii) that

'v'(/3) = [k7'n :k 7'n -1 ] = q i.

On the other hand, it follows from (4.1) that

cy 4- fi = Pf (a, fi) + fi mod 0.


Therefore,

iH(a)= v'(a(cY) cr)=I'Vcr 13 ) – cv)= v%(.3 )= q •

We now consider Hr and Hr for real numbers r 0. Let j be an integer such that qa j < qa+1 for a E Z, 05_a5_n. Then, for a=6,n(u) in Lemma 7.9,

-Fifix(a)a'j + 1•(=>q' j + 1•(=>i a + 1•(=>u E Ua+1.

Hence, by the Corollary of Proposition 5.3,

Hi = 6(11, 44) = Gal(k7'n/kT'a).

Let [H1 : 1] = hj . Noting that 141 'n I leunr is totally ramified, we obtain from the above the following table for H, and hi :

H_ 1 = Ho = Gal(k7mIka), ho = (q –1)qn,

= • • • = Hq _ 1 = Gal(kT'n

h1 =q'2 ,

Hqa =• • • = H q a+1_ 1 = Gal(k7 m/k7rl ' a ), h1 = qn- a,

Hqn-i= • • = lign_ i = Gal(k7'ni

h,= q,

He = Hq n ±i = • • • = 1, h; =1.

Now, by the definition of Hr for real –1, the map ri-->Hr is continuous on the left—that is, Hr = Hr _, for small E > 0. The above table shows that r'-->Hr is discontinuous on the right exactly at r = 0, q –1, . , qn –1. Since Hr = 114)( r ) with a continuous function cp(r), we see that the map ri-->Hr is continuous on the left and the discontinuity on the right occurs exactly at

r = 0(0), 11)(q –1), . , (1)(qn –1).

However, by (7.2),

1 P = + • • + h i )

ho

for integers 0. Hence it follows from the above table that

cp(qa – 1) = —1

a(q –1)qn = a, for 0 5_ a 5_ n. ho

By the remark on the relation between Gr , Gr and Hr , Hr mentioned earlier, we now obtain the following result:

PROPOSITION 7.10. For the finite abelian extension km,'n1k, the ramification


groups in the upper numbering, Gr are given as follows:

G ' = Gal(kTr 'n/k), for r = —1.

= Gal(KT . n/k unr), for —1 < r 0.

= Ga1(141 'n74"), for i-1<r5_i,iEZ,15_i5_n.

= 1 for r> n.

Hence

1 =G±J c G2 c. G1 c GO G-1 = G.

[G : G°] = m, [G° : = q — 1, [G : G'+'] = q, for 1 5 i n.

PROPOSITION 7.11. Let o' denote the valuation ring of le", and po the maximal ideal of k 77,1 ' ° . Then

2(142 ' 2 I k) = (k 77,1 ' 2 I = Pn±i P cT io',

p being, as usual, the maximal ideal of k. Proof. Since k iT.Ik is unramified, the first equality follows from Propo-

sitions 2.16, 2.17. Let p' be the maximal ideal of k 77'n and let

(Knr 'n/Kir) = p'a, a 0.

Using the values of hi =[Hi : 1] in the table obtained above, we see from Lemma 7.8 that

a = (hi -1) i=0

= (q — 1)qn — 1 + (q — 1)(qn — 1) + (q — 1)q(qn -1 — 1) + • •

+ (q —1)qn -1 (q —1)

= (n +1)(q —1)qn — qn.

Since k'n/kZ. is totally ramified and [142 ' 2 = (q —1)qn, [141 ' 2 :1‘71= qn, po', and Poo' are the (q — 1)qnth and qnth powers of p', respectively. Therefore,

2 (k`77,1'nlkiT)= p a = pn -f-1 piT1 0 •

7.4. Some Applications

Let k'lk be a finite abelian extension of local fields and let

P k'lk:kx G = Gal(k'/k)

be the homomorphism defined in Section 7.1.

THEOREM 7.12. Let G', r —1, denote the ramification groups in the upper numbering for k' I k. Then

G r = pkvk(Ui),


Proof. Fix a prime element ir of k so that

k k' k ab = kurk„

by Theorem 6.8. Let f E gr'!„ Then, by the definitions in Sections 5.1 and 5.2,

= = k(W), le;r1 " 1 = k(TV'f) = k iTrkT , Ln = kur(WD = k ur 14.

As k i, is the union of k for all m 1 and k„ is the union of 14 for all n 0, it follows that k i,k„ is the union of the fields k" 1 = kk7, for all m 1, n 0. Hence

k c k' c E = k" = krunrk;

for sufficiently large m 1 and n 0. By definition, Pk'ik is then the product of PElk:kx Gal(E/k) and the canonical (restriction) homomorphism Gal(Elk)-0 Gal(k'/k). However, by Theorem 7 .7 , Gr (= Gal(k' I k)r) is the image of Gal(Elk)r, r —1, under the same homomorphism Gal(E/

Gal(k7k). Therefore, it is sufficient to prove the equalities of the theorem for the extension Elk instead of k'lk. Now, by Lemma 7.4 and Proposition 7.10,

N(k,T' i-1 Ik)= N(kki„-11k)= (am) X U„

Gal(E/k)' = Gal(E/k7' 1 ), for i — l< r 5_ i, 0,

with 141 ' -1 = k;1 = k. On the other hand, by Proposition 7.3, _ Gal(E/14" -1). PE/k(N(k7' ))

Since (7rm) is contained in the kernel N(Elk) = (irm) X U n+i of D : Elk

Ex -4 Gal(E/k), we obtain

PE/k(Ui) = PE/k((irm ) X (4) = Gal(E/k7' i-1) = Gal(E/k)r. •

Let k' I k now be any finite Galois extension of local fields and let G = Gal(k'/k). The definition of the subgroups G,., r —1, of G in Section 7.2 states that Gr = G, if i is the integer satisfying i — l< r 5_ i, 0. Hence it is natural to ask whether similar equalities:

Gr = G i , fori-1<r5_0:0, iEZ (7.5)

hold also for the ramification groups Gr in the upper numbering. This is not true in general.t However, for an abelian extension k'lk, we have the following theorem of Hasse-Arf, which is an immediate consequence of the above Theorem 7.12:

THEOREM 7.13. Let k' I k be a finite abelian extension of local fields and let r be a real number —1. Then

Gr = G i ,

Let k' I k still be a finite abelian extension. It is clear from Theorem 7.12

t See Serre [21], p. 84.

112


that

II, Œ N(le I k)<=> Pkyk (Un )= l<=>Gal(le 1 k)'2 = 1, n .-_ O.

Since any of these equivalent statements holds whenever the integer n is sufficiently large, let c(k' lk) denote the minimal integer n _-_ 0 for which the above conditions are satisfied, and define

f(k7k) = pC(k '4) ,

p being, as usual, the maximal ideal of k. The ideal f(le lk) of k is called the conductor of the abelian extension k' lk. It follows immediately from the definition that

f(k1k)=o<=>c(kilk)=0<=>Gal(10k)° = 1.(=>k ilk is unramified,

namely, by Proposition 2.16,

f(kilk)=o<=>g(kilk)= o'<=>D(lelk)= 0,

o and o' being the valuation rings of k and k', respectively. If k' lk is ramified, then f(k 1 lk) OE p and it is the largest ideal of o such that

Pk' /k(1 ± f(le lk))= 1 .

We shall next discuss the relation between f(le lk) and D (le I k) for finite abelian extensions k' lk .

Let Gn again denote the ramification groups, in the lower numbering, for the abelian extension k'lk. Let gn = [Ga : 1] and let 0. be the increasing function associated with kilk (cf. Section 7.2). Let

kOEk"OE k', H= Gal(k'/k"), GIH= Gal(k"/k).

LEMMA 7.14. Suppose Gn i H, Gn±1 OE H for an integer n._- —1. Then

c(k"lk)= 1 + 0(n)

1 = — (go+ gi+ • • • +g),

go

where the last sum is meant to be zero if n = —1.

Proof. By Theorem 7.7,

GrH I H = G(1)(r)H I H = (G I H) (r) .

Since Gr = G,±1 for n < r _._ n + 1 and since 0 is continuous, it follows from the assumption that

(G/H) ( n ) 0 1, (G/H) (n )+ E = 1, for small E > O.

Hence, 'by Theorem 7.13, 11)(n) is an integer, and

(G/H)(1' (" ) 0 1, (G/H) (n )±1 = 1.

Therefore, by (7.2) and by the definition of c(k"lk),

c(k"lk) = 1 + cp(n) = —1

(go + gi + • • • + gn ). • go


COROLLARY. Let Gn *1, G 1 =1 for an integer n .__ —1. Then

c(k' lk)= 1 + 0(n)

1 =— (go+ gi+ • • • + gn).

go

REMARK. ifp(n)= c(k' lk) — 1 motivated the definition of the function 0(4

Now, for each character x of the finite abelian group G = Gal(k'/k), let

Hx = the kernel of the homomorphism x : G—*Cx ,

kx = the fixed field of Hx in k',

f(x) = f(k x 1k), the conductor of kx lk.

THEOREM 7.15. Let D(k' lk) be the discriminant of the finite abelian extension k' lk. Then

D(k'lk)= n f(x), x

where the product is taken over all characters x of Gal(k'/k). Proof. Let 2(k' 1 k) be the different of k' 1 k and let g(le lk)= p'a

,

a .._ O. Then, by Sections 1.3 and 2.4,

D(k' 1 k) = Neik( 2 (k 1k))= VI.,

where f = f(k' lk)= glgo with g =[G :1]= [k' :k], go = [G0 :1]= e(k' 1 k). By Lemma 7.8,

00

a = E (gi — 1), for gi =[G,:1], i .__0. i=1

Hence the equality of the theorem is equivalent to

g0=0 x

where c(kx 1k) is defined as above. Now, fix a character x, and for each i --- 0, let

, , , X(t-7,) = — L AO, with a E G.

go a

Then

x(Gi) = 1, if x I

if X 1

Gi = 1, that is,

=0,

G,* 1, that is, Gi &Hx .

Let x * 1 so that Hx * G and G„ ck Hx , G„,1 Œ Hx for some n ._- —1. By Lemma 7.14, we then see

1 ' c(kx 1k)=— E g,(1 — x(Gi)).

g0 i=0


If x = 1, IL= G, then k= k, c(kx 1k). 0, while x(G,)= 1 for all i O. Hence the above equality holds also in this case. When i is fixed and X ranges over all characters of G,

E x(G) = the number of characters x of G such that Gi c Hx

= the number of characters of GIG,

= g I gi .

Therefore, ,

• go i=o gi g01 =0

This theorem is called the conductor-discriminant theorem for finite abelian extensions of local fields.

EXAMPLE. Let lelk be a cyclic extension of degree 1, a prime number. Then

X= 111x = G, k= k, f(X)= f(k1k)= 0,

X 1 11x =1, k= k', f(X)=f(k1k).

Hence

D(k' lk)= f(k'10 1-1 .

Theorem 7.15 is a generalization of the above equality, originally obtained by Takagi.

In the above discussions, we deduced Theorem 7.12 and, hence, Theorem 7.13 from Theorems 6.8 and 7.7 and Proposition 7.10 on the ramification groups of k mn ' n 1k. We shall next show that, conversely, Theorem 6.8 is a consequence of Theorems 7.7 and 7.13 and Proposition 7.10.

LEMMA 7.16. Let le lk be a totally ramified, finite, Galois extension of local fields and let q be the number of elements in the residue field of k: = Fq . Suppose that (7.5) holds for the ramification groups G r of k' lk for all real r —1. Then

[G° : G11 (q — 1), [Gn : Gn+11 q, for n 1

so that

[G : G n± 11 (q —1)qn, for every n 0.

Proof. Since k' lk is totally ramified, G = Go = G° and the number of elements, q', in the residue field of k' is equal to q. Let n be any integer and let n = 4)(m), n +1= 4)(m') for the increasing function ifp associated with k'lk. By the remark after (7.2), m and m' are integers. Since 40(m)<O(m'), we have m < m'—that is, 0 m <m + 1 m'. Let r = cp(m + 1). Then n < r n + 1 so that G r = G n±1 by (7.5). Since G 4)(s ) = Gs,


s __ —1, it follows that

GG Gn ±1 = Gr = Gm±i , [Gn :Gn±1]=[Gm :Gm+1].

If n = 0, then m = 0, and [G° : G l] = [Go : Gil If n __ 1, then m > 0—that is, m __ 1. Since q = q', G = G° , the lemma now follows from the remark at the end of Section 2.5. •

Now, assume Theorems 7.7 and 7.13 and Proposition 7.10. As shown in part (i) of the proof of Theorem 6.8, there exists a field F and a prime element .7r of k such that

Fkur = kab, Fnkar =k, k=k„=F=kab.

Let k' be any finite extension over k in F: k ck' c F, [k' :k]< +00. Then Gal(k tion+1 = 1 for some n __ 0. Let

k" = k'14.

Since Gal(k7r/k)' 1 = 1 by Proposition 7.10 for m = 1, it follows from the Corollary of Theorem 7.7 that

Gal(k"/k)' = 1.

As k ck"c F, k"lk is a totally ramified, finite, abelian extension. Hence, by Theorem 7.13 and Lemma 7.16, we obtain

[k" :1] = [Gal(k"/k) : 1] I (q — 1)qn.

However, we know by Proposition 5.2 for m = 1 that [142,:k]= (q —1)qn. As k c Kir c k", it follows that

k c ki c Ki r c k,.

Thus every finite extension over k in F is contained in k,. Hence

F = k„, kab = kark, = Lk.

The above is the idea of the proof of Theorem 6.8 in Gold [9] and Lubin [18]. -r One sees that if Theorem 7.7 and various properties of the abelian extensions kn,lk are taken for granted, then Theorem 6.8—that is, kab = Lk—and Theorem 7.13 (Hasse-Arf Theorem) are essentially equivalent.

t For another proof of Theorem 6.8 for a local field of characteristic 0, see Rosen [201.

Chapter VIII

Explicit Formulas

In this chapter, we shall prove formulas of Wiles [25] that generalize the classical explicit formulas of Artin-Hasse [2] for the norm residue symbols of local cyclotomic fields over Q.

8.1. z-Sequences

Let (k, v) be a local field with residue field f = o/p and let .7r be a prime element of (k, v). In the following, we consider a pair (f, co) where f is a power series in the family 37,1 of Section 5.1 and where

w = {(0nIno

is a sequence of elements in the abelian group W f = Uri 1472 of Section 4.3, satisfying, for f = (Jr)f ,

a)0 E WY, a)0 0, (On = A(On4-1) = Z:f (On4-1, for all n O.

Such a pair will be called a Jr-sequence for k. Given f E „1 , there always exists a 7r-sequence (f, co) for k because W 1 = 14/7 by Lemma 4.8(iii). It also follows from the same lemma that con has the properties

(on 1 14(7-1 , Tr) = U wn ,

so that con is a prime element of kn,T = k(1417) by Proposition 5.4(ii). Let (f', co') be another .7r-sequence for k and let co' = {(0 1„}„.0 . Suppose

h(X) is a power series in o[[X]], invertible in M = Xo[[X]] (cf. Section 3.4), such that

h of 0 h -1 = f', h(con)=w, for all n O.

We shall call such an h an isomorphism from (f, (o) to (f', co') and write

h :(f, co) (f', co').

It then follows that h defines isomorphisms over o:

Ff Fr 1171 W7 , Wf 14 7 .

LEMMA 8.1. Let (f, co) be a Jr-sequence for k with co ={co,„ }n ih oe. n g Let be a power series in o[[X]] such that g(coi )= 0 for 0

i n

divisible by [.7rn']f in o[[X]]. If g(w i)= 0 for all i 0, then g(X)= O. Proof. By the Corollary of Proposition 5.4, the elements p

Wif-1 , are the conjugates of cri over k (= kul r). Since g(X) E g(coi)= 0 implies g(P) = 0 for all p E W , fi W7 1 . As this holds for 0 i n, we see that g(œ) = 0 for all cr e 1472. By the remark after Lemma 5.5, g(X) is then divisible by [elf in o[[X]]. Now, since [.7t]f =f for

Explicit Formulas 117

f E gr',1 , [.7df is contained in the maximal ideal (.7r, X) of o[[X]], generated by [7rn +If = [7.tif 0 . . 0 r _1 .7r and X. Hence 1.7rjf belongs to the (n + 1)th power

(7r, xy+1 of (.7r, X) in o[[X]]. Therefore, if g(coi)= 0 for all i __ 0, then g(X) E (.7r, X)"± 1 for all n __ 0 so that g(X)= O. •

PROPOSITION 8.2. Let (f, co) and (f', co') be .7r-sequences for k. Then there exists a unique isomorphism

h : (f, co) 2.-4 (f' , w').

Furthermore, if = co', then h(X)= X so that f = f'. Proof. Since f, f' E gr',1 , it follows from the proof of Lemma 5.1(i) that

there is a power series 0(X) in o[[X]] such that 0(X)=---- X mod deg 2, 0 of =f' o 0 (cf. (4.8)). Let co" = { w:}, where co: = O(w), n O. Then fi : (f, co) '4 (f', co"). Hence, to prove the existence of h :(f, co) (f', co'), we may assume that f =f'. In this case, since co W7 to O r_ 4 W'_! it

-- I1, n - —n, --n

follows from Lemma 4.8(ii) (iii) that there exists Lin E U = U(k) such that u • con n = ' and that such un is uniquely determined in U mod Un+1 nf

(=l+V±1). Let O. Then

u • co = u • Z ni—n • CO = .7rm—n • W I = COI urn f n u rn m m n,

whereas u• con = co n' . Hence it follows from the above remark that n f

Um =- Un mod Un±1 for 0 n m. Consequently, there exists u U such that u= un mod Un±1 for all 0, and such an element u satisfies ui.con = co n' for all n O. Let h = [u]f so that h(con )= co'n , n O. Since h of h -1 =f because f = [4, we obtain h :(f, co) (f, co'). To prove the uniqueness of h :(f, co) (f', co'), it is sufficient to show that co = co' implies h(X)= X. Let g(X)= h(X)— X. Then coi = co; = h(coi) implies g(coi)= 0 for all i O. Hence g(X)= 0 by Lemma 8.1. •

For 0 n m, let Nmm denote the norm map of the extension k,m114. A .7r-sequence (f, co) for k will be called normed if it satisfies

Nm , n(com ) = con , for all 0 n m.

We shall next see that this property of (f, co) actually depends only on f.

LEMMA 8.3. A Jr-sequence (f, co) is normed if and only if

[.7r]1 (X)= fl (X -i- y), y E WY, Y f

namely, N1(X)=X

for the norm operator N1 in Section 5.2. Proof. Assume the above equality. Let n O. By the Corollary of

Proposition 5.4, con +1 WY is the complete set of conjugates of con±1 over kna. Hence '

Nn-E1,n(Wn-E1) = (w„±i y) = Pri1(a)1) = zi.con±i = (on, Y

y E WY.


Therefore (f, co) is normed. Conversely, suppose that (f, co) is normed and let

g(X)=[z]f(X)— ( X y), y E WY. Y

Then g(wo) = — (wo -if- y ) = o

Y

because wc, 141 = 141 and WY contains O. For n 0,

g(a) n+1) = w,,1 — (a)„,1 7) = con — Nn+1,((on+i)= 0, Y

y E W.

Hence g(X)= 0 by Lemma 8.1. •

For convenience, a power series f in 3;1 will be called normed if it satisfies Nf (X) = X. Thus Lemma 8.3 simply states that a Jr-sequence (f, co) is normed if and only if f is normed.

EXAMPLE. Let (k, y) be a p-field. Let p > 2, f(X)= .7rX + Xq E „1- , and let (f, co) be a Jr-sequence for k. Then f (a) n -F1) = [Ir]f(a) n -F1) = con . Since

[ k 1 :knj = q k nn+1 = 4(0 n-F1) = k nn(a n -F1) = deg(f) by Propositions 5.2 and 5.4(ii), we see that f(X) — con is the minimal polynomial of co n+i over k njr so that N„±1,,,(— ( 0 11+1) = —con . Since q is a power of p and is odd, it follows that N n±i, n(a) n-Fi) = w,, for n O. Hence (f, co) is a normed Jr-sequence for k. In the case p =2, we can see similarly that (f, co) is normed for f(X)= 7rX — X" E

For n 0, let

Bn = (k7r ) X = the multiplicative group of the field 14,

and let B = lim B

where the inverse limit is taken with respect to the norm maps

Nm , n :(141)x (k nn) x , for 0 n m.

B is a multiplicative abelian group and an element p in B is a sequence:

13 = (Po, p,, • • • fin, • • •), Pn E Bn = (kW ,

such that Nni , n(Pm ) = pn for 0 n m. Let Nn denote the norm map of 14/k. Then Nn (fi n ) is an element of k, independent of n O. Hence we define an integer y([3) by

v(P)=

for any n 0,

y being the valuation of the ground field k. For example, if (f, (0) is a normed Jr-sequence for k with co = {coo , w 1 ,.. .}, then co belongs to B and y(w) = 1 because each co n is a prime element of kn,. Clearly

y(160') = y(0) + y([3'), for p, E B,

and the group B is generated by the fi's with y(P) = 1.


THEOREM 8.4.t (i) Let (f, co) be a Jr-sequence for k and let f3 E B with v(P)= O. Then there exists a unique power series t(X) in o[[X]]< such that

t(con ) = fi n , for all n O.

(ii) If (f, co) is normed, then for any fi E B with arbitrary e = v(0), there again exists a unique power series t(X) in Xe oRX]] such that

t(con )= fi n , for all n O.

Proof. (i) Since v(fi) = v(Nn (fi n )) = 0, fin is a unit of 14. Hence it follows from Proposition 5.11 that for each n 0, there exists tn (X) E 0[[X}]

such that

tn (w i)= pi , for0_i_n. It then follows that

t„i (coi ) — t n (w i ) = pi — 0 i = 0, for 0:5_ i n,

so that by Lemma 8.1, tn-Fl t„ mod kn+11f.

Since [e ]f E (Jr, iv) i, we see that the limit

t = lîm tn

exists in o[[X]] in the compact topology on o[[X]] defined in Section 3.4. As t clearly satisfies t t„ mod [Jr-Eli-If, it follows from Jr' co n = 0 that

t(con ) = tn (con ) = 16„, for n O.

t(X) belongs to o[[X]r (i.e., t(X)= u mod deg 1 with u E U) because t(coo) = Po is a unit of k. The uniqueness of t follows from Lemma 8.1.

(ii) Since (f, co) is normed, w = (co o , co b . .) is an element of B with v(co) = 1. Let fi E B, v(0)= e, and let

/6 ' = ar efi = (16;), 16;, .), fin' = Wnefin) n

Then fi' E B, v( 16')= O. Hence, by (i), there is a power series t'(X) in o[[X]r such that ts(co in) = f3 for all n O. It then follows that t = re is a power series in Xeo[[X]] with the property t(con ) = Pr, for all n O. The uniqueness of t follows from that of t'. •

Now, fix a normed Jr-sequence (f, w). By the above proposition, for each E B, v( 16)= e, there exists a unique power series to (X) in Xeo[[X]r such

that to (con ) = fin , for all n O.

For example, t(X) = X for co = (coo , co l , .) E B. By the uniqueness, we have

143 , = tot 0 , , for 0, E B.

Suppose v( 16) = 1 for 0 E B. Then to E X0[[Xli x so that to is invertible in M

t Compare Coleman [5].


and fp = tp of otp— i E

Since tp (con ) = pn for n 0, we see that the pair (fp , p) is a normed sr-sequence for k and

to :(f (0) (fp

By Proposition 8.2, fp is the unique power series in o][X]] such that (fp , /3) is a 7r-sequence for k. Hence fp depends only on p E B (with v(16) = 1) and it is independent of (f, (0). Furthermore, if (f', co') is any normed sr-sequence for k, then co' E B, v(co')=1. Hence (f,,,,, (0') is a normed sr-sequence for k so that f' = fw , by Proposition 8.2. Thus we see that the family of pairs:

{(j, f3)}, for all p E B with v(P) = 1,

is nothing but the set of all normed z-sequences for k.

8.2. The Pairing (a', /3 )

Let Jr be a prime element of (k, y) and let f E gri. As before, we denote the residue field of kn,, n 0, by On /P n .

LEMMA 8.5. Let n 0. (i) For each a E pn , there exists an element in mf(= pc2 ) such that zn + = cr. Let k' = kn,(4 Then k' I kn, is an abelian extension with [k' :14] qn-F1, and k' depends only on cr and it is independent of the choice of such that :0 + = cr.

(ii) In particular, if cr is a prime element of kn,, then k' I kn, is a totally ramified extension of degree qfl±l is a prime element of k', W'fi is the complete set of conjugates of over kn,, and Gal(k' I kn,):41417.

Proof. (i) An argument similar to that in the proof of Lemma 5.1(i) shows that it is sufficient to prove the results for the special case: f(X)= .7rX + Xq. In this case, kn-Fl if =fop)... is a polynomial of degree q n-Fi in 01.X1 r 1 [ 7rn ± [xl and the equation j = 0 has a solution in p o (cf. the proof of Lemma 4.6). Thus srn +l i. = cr, and 4 W7 is then the set of

all roots of jf _ = 0 in Q. As cr and W7 are contained in kn,, k' I kn, is a finite Galois extension, depending only on cr. Let a E Gal(le/kn,). Since

(= a implies zn± 1 a = , it follows from the above that

E 14/7 ç 14. Hence, for a, r E Gal(le/k7,),

(")() (a() a(r() )= (r()7.7

so that the map a ci() defines a homomorphism: Gal(V/kn,)--0 14/7. As k' = kn,(), the homomorphism is injective. Therefore, Gal(le/kn,) is an abelian group with order 5..qn± 1 = [ W7: 0].

(ii) prn-Flf A +1 mod p and p ç p„ = cron imply

a, = jrn+1 5-qn +1 mod 1:), hence, mod


Therefore, 1.1(a)= qn+1 1.4() for the valuation p, on Q in Section 3.1. As a is a prime element of 14, it follows that

e(k1114) ›. qn+1 = [Wrfi :0] e(le 14) so that

[k' : kr;E]= [W7: 0] = qn±1 = e(k 1 114),

and all statements in (ii) follow. • For n 0, let pn denote the norm residue map PE for E =

Pn:Bn = (4) x --4Gal(14,„b/Cir).

Fix f E g'n1 and let P Bn .

Let zn ±1- 3. = cry k' = k() as stated above. Since le 114 is abelian—that is, c k' = 14,aby Pn(0)1k' is an element of Gal(k114). Let

0),,,f=Pn(i3)()

Then (a, p)„,f is an element of W7 and it is independent of the choice of such that .7rn± 1 • = because if Jrn±1 • = E 4- W2, k, f f — that

P n(13)W = P n(16)()

We shall next study the properties of the symbol (cr, P)„, f. When f is fixed, the suffix f in (a, P)„,f., a P, and so on will often be omitted.

LEMMA 8.6. (i) (al 4 a2, P)„ = P)„ (a'2, fi)n,

P1132)n= (a'y /3 k),, 4- (a'y 02)n,

(a. 16)n = a • (a, 0)„, for a E O.

(ii) (a, p)„ = 0•#>,6 E N(kn,()Ik), jrn+i =

(iii) Let f be normed and let a' be a prime element of k. Then

a')n,f= o.

Proof. (i) If i, are elements of mf such that .70 ±1 • = z ni-1

follows. The third equality can be proved similarly. Since pn (OP')= pn (p)pn (P'), the second equality follows from the fact that a 1–* a() defines a homomorphism: Gal(14()/14)-- Wrfl.

(ii) It follows from Theorem 7.1 that

= '(=>Pn(i-3 ) I Cit()= 1 <=>0 EN( 14()/ 14). (iii) Let zn +l i. = cr. Since f is normed, Nf (X)= X. Hence N(X)=

NANA. • .))(X)= X so that by Lemma 5.9(i),

[jrn±l ]f = y), y E W7. Y f

= a'2, then zn ±1 •( 4- 2)= ai a2 . Hence the first equality

122


Therefore, y E

However, by Lemma 8.5(ii), ±. 14/7 is the complete set of conjugates of

over kn,. Hence cr is the norm of for the extension kn()Ikn„, and it follows from (ii) that (a, a)n,f = O. •

Let 0 n m. Then the maximal ideal pn of kn, is contained in the maximal ideal pm of k. Hence, if a E pn , then am n E pn pm .

LEMMA 8.7. (i) Let 0 s n 5m. Let CV E Pn , a' = Jet ' ci E Pn Pm , 16' E

Bm = (1c7r1)x , and p = Nm , n(13') E Bn = (k,T)x. Then

(ci', fi i )m (ci, fi)n.

(ii) Let f, f' E 3;1 and h : Fr over O. Then

h((cY, fi),)'( 11 (a), P)n,f' for a E pn , f3 E (14)x .

Proof. (i) Let gn+ 1 . •. = a. Then 7rm± l i. = a = ci'. By Theorem 6.9, pm (0') I k n,,ab p(0). Since E kryri,ab, we see

(ci', 0 1 )m= Pm(16 ')() - =P1.(f3)() =-.- (1fr, 0)n.

(ii)e±1 ).=crimpliese±1 .W)= h(a) with h(a) E p n . Hence

h((Cr, fi)n,f)= h(Pn(i3)() )= h(Pn(0)()) t.• 11 ()

= Pn(I6)(h()) (h(œ), •

For each 0, let

A n = p n = the maximal ideal of le!, and let

Af = limA n ,

where the direct limit is taken with respect to the maps

A n --4 A m , œ -_>m_" a,

for 0 -5 m. Since each A n (= PO is an o-module, the limit Af is also an o-module in the obvious manner. By Lemma 8.6(i), the map {a, p) i-* (a, f3), a pairing of abelian groups:

( , )n ,f :A n W7.

By Lemma 8.7(i), we then see these pairings ( , )n ,f, for all 0, define a pairing

( )f :Af . X B—>Wf.


More precisely, let a E A» fi = (oo, pi, .) E B, and let a be represented by an E A n—that is, an i-> a in A„ -> Af for some n O. Then ( i Nan , v n inf S

defined and is an element of W'fi Wf. Let m n O. Then a is represented in Am by am = 71-m-n ci,,, while Nm,(16,n ) = fin . Hence, by Lemma 8.7(i),

(an, fin)n,f = (am, Pm)m,fr

Therefore, for any a E Af and p E B, we can define a well-determined element (a, fi) f in Wf by

(ci, = (ci,,, On)n,»

if a is represented by an in An . It is clear that the map ( , )f :Af x defined in this manner is a pairing of abelian groups Af and B into Wf. Furthermore, Lemmas 8.6 and 8.7 show that the pairing has the properties:

(a a, ,6)f = a p)f , for a E 0,

h((a, ,6)f). (h(a), ,6)„, if h :Ff F1..

8.3. The Pairing [ci, From now on, we assume that (k, y) is a p-field of characteristic O. Let 7V be a prime element of (k, y), and f a power series in 37,1 . By Lemma 4.2, there is a unique isomorphism over k:

Xo:Ff4Ga

such that A 0(X) = X mod deg 2. For U E U = U(k), let X(X) = 0,0(X). Then A is the unique isomorphism over k:

X:Ff 4Ga

such that X(X) uX mod deg 2. All such isomorphisms, for u E U, will be called the logarithms of the formal group Ff . Since FAX, Y) E D[[X, Y]], the power series Ip(X) in the proof of Lemma 4.2 belongs to o][X]]. Hence X(X) above is a power series of the form

cn c i =u, Cn E 0, for all n 1. (8.1)

n=1 n

Let f' be another power series in 37,1 and let h:Ff., Ff be an isomorphism over U. Then h (X) = u' X mod deg 2 with U t E U. Hence

A,0h : Ff , -4 Ga , X0 h(X) uu' X mod deg 2

with LW' E U so that Ao h is a logarithm for F f'. It is customary to define the logarithm of Ff to be the unique isomorphism A0 : Ff Ga over k such that A 0(X) X mod deg 2. However, we generalized the definition slightly so that if h : F1 over U and if A is a logarithm of Ff, then A 0h is a logarithm of Fr.


LEMMA 8.8. Let A be a logarithm of Ff. Then:

(i) A(a) converges in 5-2 for any a E 11lf. If, in particular, a E pm , m __._ 0, then A(a) E k.

(ii) A(a -1- 13) = Ma) + A(13), A(a i a) = aA(a), for a, fi E Illf, a E o.

Proof. (i) For simplicity, the valuation ft on 0 (cf. Section 3.1) will be denoted by i.z. Let e = p(p)> 0 and let 13E, E --?: 0, be the exact power of p dividing n. Then pE ._ n, ti(n)= eE _.._ e logp n. Since cn E 0, It(c„) .- 0 in (8.1),

ti(Cn —n— a

n ) __._ np(a) - e logp n,

where ti(a)> 0 for a E trtf = po- . Hence (cn /n)an -*0 as n-* + oo, and

A(a) = E:=1 (cn I n)an converges in C2. The rest is clear. (ii) A :Ff 2; Ga means A(Ff (X, Y)) = A(X)+ A(Y). Hence the first equality

is clear. We also have

A 0 [a]f 0 A -1 E Endk (Ga ), A 0 [a]f 0 A -1 -=----- aX mod deg 2.

By the remark after Proposition 4.2, we then see that

Â 0 [a ]f 0 A, -1 = aX, that is, A 0 [a]f = aA.

This yields the second equality. • We now fix a normed Jr-sequence (f, co), w = {con } no, for k. Let A(X)

be a logarithm of Ff . For each p= (pe, oi, • . .) in B, we define

1 t io(con) for n __._ 0, (8.2) (506)n = AVon) to(con) '

where to (X) is the power series in Theorem 8.4 such that tp (o)„,) =13m for all m __._ 0 and where A' = dAldX, t; f3 = dtp ldX. Since t(a),,) = fin * 0 and since

OC

A'(X)= u + E cnxn-i, c,, E 0, n=1

so that pt(A 1 (con )) = 0, A 1 (co„)*0, we see that (5(f3),, is a well-defined element in kq. Note that (5(f3),, depends not only on /3 and n __._ 0 but also on (f, co) and the choice of the logarithm A for Ff.

LEMMA 8.9. (i) 6(0%, = (5(f3),, + (5(13 1 ) n , for fi, o' E B. (ii) (5(0)n E 1),T 1 for all 0 E B.

(iii) Let 0 ._n ._m and let T„,,„ denote the trace map of k114. Then

T m,n (6(fl) m ) = .7r m—n (5(/3) n , for fi E B.

Proof. (i) This follows immediately from top-, to to , and from the definition (8.2) of (5(f3),,.

(ii) Let v(p) = 1 (cf. Section 8.1). Then fi n = to (con ) is a prime element of Kir . Since /t(A 1 (con ))= 0 as stated above, it follows from (8.2) that


(5 (P)n E Ç 1 . BY (i), the same result then holds for any P E B because B is generated by the fi's with v(f3) = 1.

(iii) Again, we may assume that v(P) = 1 so that (f ' , f3), where

f' =f13 = to of 0 q l , is a normed Jr-sequence for k (cf. Section 8.1). Then, by Lemma 8.3,

kb, = fl (x -*F y'), 7' f

Since to : Ff 2', F. f' , Pri f == t13 ° [Z]f ° to-1 , and W. = t13 (14/), it follows that

to 0 [Jr] = n ( to (X) -i-, y') = n to (X 4- y), Y EW. f f

7' Y

Logarithmically differentiating with respect to X, we obtain

(

'3 ° klf) ad kif = E (.3 (x -.F 7)) -d (x -. 7). tp y tp f dX f

However, A([4(X)) = zA,(X), A,(X -j- y) = Â(X) + A(y) implies

d (A' ° [z]f) a [z],. = .7rA.' (X), A' (X 4- y)—

d (X 4- y) = A,'(X).

f dX f

Hence Jr ' t r i 1

(tttlfi ° (X

y E WY.

co n+i in the above. Since co n+1 -if- WY is the complete set of Put X=

conjugates of co n±1 over kn, (cf. the Corollary of Proposition 5.4), it follows that

Jr(5(0)n = Tn-1-1,n( 6 (f)n+1), n _._ O.

Therefore, T„,,n(b(P)„,)= :en-n(5(0) n for 0 .- n .- m. • Let n _._ 0 and let an E A n (= Pn ), /3 E B. We define

xn(a'n , fi) = 1 Tn(A(an) 6 (fi)n) zn+ 1

where Tn denotes the trace map of kn,lk and A is the logarithm of Ff mentioned above. If A is replaced by another logarithm uA, u E U, then A(ot n ) is replaced by u4an ), and (5(f3) n by u -1 •5(0) n (cf. (8.2)). Hence x n (an , /3) is unchanged. Thus xn (an , f3) is an element of k, depending on n,

an, 0, and the fixed normed Jr-sequence (f, co), but not on the choice of A.

LEMMA 8.10. Let an , ans € Al„, P, o' E B.

(i) x n (crn t cr'n, f3 ) = xn(a'n, f3 ) + xn(cr'n, f3 ) ,

xn (a Ian, 0) = axn(œn, f3 ), a E 13,

Xn (an , /3/3' ) = Xn(Crn, 13) + xn(an, P').


(ii) x„,(7rn-n f3)= Jen nxn(an, 0), for 0 n m.

Proof. (i) follows Lemma 8.8(ii) and Lemma 8.9(i). (ii) By Lemma 8.8(ii) and Lemma 8.9(iii),

1 xm(grm -ni an , 0) = m±1

1 jr .4-1

jen-n

Tm (.1,(7rm -ni.an )6(f3) m )

Tn( 71'm-n A(an)Tm,,,(1 5 (0),n))

n

1 4-1Tn(11(an)(5 ( 6)n) .7r

= 7rm — nxn( an, 0). • Now, let

a'c Af=lim A,,, E B = liM Bn ,

and suppose that a is represented by an E A n . Then, for 0 m, a is also represented by am = Jrt n n ar, in Am and xm (am , p)= z—nx n (an , p) by the above lemma. Hence, if m is sufficiently large, then xm (am , 13) belongs to o so that xm (am , p) w m is defined, com being the mth element in to =

(WO, Wi, • • .) for the fixed normed Jr-sequence (f, (0). Let m' m. By Lemma 8.10(ii),

x m ,(am ,, f3)= Irm'mx,n (04,n , (3) E 0,

xm,(cym,, f3)wm= x,n(arn, P)Jrm ' m torn' = xm(am, 13 ) tOrn•

Thus, whenever m is large enough, xm (am , (3)}. w m is defined and it is independent of m. Clearly x,(04,n , p) com E Wr W. Therefore, for any

E Af and (3 E B, we define an element [a, p]„, of Wf by

[a', P].= xm(am, P)icom

=[Jr

ni±iTm(A(am)(5(P)m)] . 'f

with any sufficiently large m. It is then clear from Lemma 8.10 that

[

:Af x wf

is a pairing of the abelian groups Af and B into Wf, satisfying

[a l a, 13],0 = a 1 [œ, (3]„ for a E O.

Note that the pairing depends only on the normed 7r-sequence (f, (0). Now, let (f', (0') be another normed Jr-sequence for k and let

h:(f, (0) (0').

By Proposition 8.2, such an isomorphism exists. Then h induces an isomorphism h : F- Ff over o and a 1-> h(a) defines o-isomorphisms

Af . Ar•


LEMMA 8.11. For a E Af , f3 E B,

h(kr , 01(0)=[h(a), 0 ] , where [ , ],,,, is the pairing defined by (f', co').

Proof. Let A :Ff Ga be a logarithm of Ff. As mentioned earlier,

= A oh 1 Ff Ga is a logarithm of Fr. For f3 E B,

ip = to h -1

is clearly the unique power series in Theorem 8.4 for (f', co') such that

(Po o h)h' = t'f3 with derivatives t'13 = dto ldX, and so on. Since h(con )= co n' , it t13 (co) = p„ for all n O. Differentiating both sides of to = t o h, we obtain

follows

co'n )h'(con )= ep(con ).

Similarly, X,0 h = A yields (c 0 'Oh ((on) = (a) n).

As h(X) vX mod deg 2 with y E U, we see that h'(X)=-----v mod deg 1,

h'(con )* O. Hence it follows from the above that

6(16)n = AVon) fin XV0 ) fin

1 t'o (con ) _

V', w (f3) the expression (8.2) for the normed Jr-sequence Therefore,

1 1 X n = Jr n+1 1n(A(an)(5(P)n Jr n +1 = 4_ 1 Tn (Â(h(a n ))6(fi) n )

for an E A n , representing a. Consequently, for large n,

h([a, 01,0 )= h(x n con ) —x h(con )= xn win

=[h(a), •

8.4. The Main Theorem

Still fixing a normed Jr-sequence (f, co) for the p-field (k, y) of characteristic 0, we shall prove the formula

(a, P)f = Pl., for CY E Af, f3 E B,

where ( , )f and [ , 1,,, are the pairings Af X B---* Wf, defined in Sections 8.2 and 8.3, respectively. We first prove some elementary lemmas.

LEMMA 8.12. Let e = v(p) = p(p)> 0, where 1.1 is the extension of v on Q and O. Let y E S-2, ft(y) . e. Then

1.1(y' 2/./(y) — e, for all integers 2.


Proof. Let j = par , a (j' , p)= 1. Then

ti(Yq.i) (Y) ti(i) /PM ae.

Hence the lemma is trivial if a = O. For a 1,

p(y//j)— 21t(y)+ e = (j — 2)12(y)— (a — 1)e (j —1— a)p(y),

where j — 1 — a — 1 — a 0 for a 1. •

LEMMA 8.13. Let o n denote, as before, the valuation ring of 14. Let y E .7Ta O n , where e a n. Then

Nn (1+ y)- 1 — 1 — L(y) MOd .7/ 3a

Tn and Nn being the trace and the norm map of 1(7,1k, respectively. Proof.

mod Z3a ,

where a, r E Gal(knjr/k) and the second sum on the right is taken over all pairs (a, T) such that a* T. Hence

\ 2

E Y OE Tn(Y), E yuyt = yu) —E y2 u) a

= i(Tn(Y) 2 Tn(Y2))-

By Proposition 7.11, .7r - no n c (k1k)' for the different 2'(1e1k). Hence y E Jra O n and Tn (2(141k) -1) o imply

T(y2) E p2a+n, Tn ( y)2 E —2a+2n Ty) E Pa+n , p . (8.3)

Since p(2)= 0 or e according as p >2 or p = 2, we obtain

E 0 mod .7r2a ±n' 0 ", "r

so that Nn (1 + y) 1 + Tn (y) mod 7r3a -e.

Using Tn (y) 2 E p2a +2n p3a , we then see that

Nn (1 + 1 — Tn (y) mod .7E 3a—e. •

LEMMA 8.14. For each a E A» there exists an integer c 0 such that for any n c, a is represented in A n by an , which satisfies

p(an )

—C.

Proof. In general, let y E Inf. Since

krif JO mod p, [Jdf E XOPC11, we see that

II( y) ___min(ti(yq), tz(zy))= min(qp(y), 11(y) + 1).

Hence, if ti(y)< 1/(q — 1) so that the minimum is qt(y), then 1.4.7ri. y) qt(y). Therefore, in any case there always exists an integer 0 such that

a,


/1(7e1 y) 1/(q — 1), and it then follows from the above inequality that

y) ti(7e Y) for all i O.

Now, let a in Af be represented by air E Ar for some j'. Putting y = c = j + j', n = i + j + j' in the above, we obtain the lemma. •

We now prove the following key lemma:

LEMMA 8.15. Let (f, co) be a normed Jr-sequence for k. Then

(a, (Of = [at, a ] ,,, for every at E Af.

Proof. Note first that co = {con },7 ,0 belongs to B. Let

n + e + 5

for e = ti(p) and for the integer c 0 in Lemma 8.14. Then a is represented by ffn E A n and

-i- = FAan, Wn) = + con+ E cgatincoln, C1 EO, i,

by (4.1). Hence, if we put

crn = w,,(1 + y„), f n

(rn , yn =--- — -t- E 1.4(4, n u' n

then p(an ko n )<I1(cgani coln-1 ) so that

ti(yn )=1/(crn )—p(oi n n —c— 1>0,

t(œ,, 4- (on) =ti(con)-

This shows that an con , as well as con , is a prime element of kn,. It then

follows from Lemma 8.6(iii) that

Wn)n,f = (an -if- con, an -if- Oin)n,f = 0.

However, omitting the suffix f, we have

(an 4- (On> atn (On)n = (an 4- (On, (On( 1 + yn))n)

— (an, (1)On (atn, 1 + yr,)„

(a)n, wn)n (a)n, 1 + yn)n•

Hence

(o)f --= (an , (On)n = (awl+ yn)n(a)n, 1 + yn)n. (8.5)

We shall next compute (otn , 1 + y„)„ and (con , 1 + yn )n . Since zn+1 • (02n+1 = wn+1, w2n+1 E kab, we see that

(wri, 1 yri)ri == Pn( 1 Yn)(W2n+1) • (02n+1, by the definition of ( , 1 ,n,

= P k(N n( 1 Y n))(( 0 2n +1) w 2,,1, by Theorem 6.9,

= (N,, (1 + y,,) 1 — 1)1 co –2n+1, by (6.4).

(8.4)


Since yn E jrn –c –1 n 0 by (8.4), it follows from Lemma 8.13 for a = n —c —1

that N„(1 + yn )' — 1 — Tn (7n ) mod arm,

where m = 3(n — c —1)— e + 2. It also follows from (8.3) that

T,,(7,,) O mod zn+ (n –c –1) , hence mod grn +1 .

Therefore, we obtain from the above that

(8.6)

(con, 1+ Yn)n = ( — Tn(7n)) • W2n +1 T(7))

= • C

1

rn Tn(7n)) • W

As before, let A denote a logarithm for Ff . Then, by (8.1)

Â(x) = E C i -=- U, Ci 0, for i 1. i=1

Hence it follows from an 4- con = con (1 + yn ) that

(8.7)

jrn +1 2n +1

00 ci fl(a'n) + A(con) = A(co n (1 + y))=> 7 w`n(1 + 7n)i

i=i where

(1+ MI 1

j2

(i — 1) 1 . Y in.

= j I j

By Lemma 8.12,

—c — 1) — e, for j 2. Therefore,

A(a'n ) + A(con ) —= A(co n ) + yn conA'(co n ) mod .7r 2(n –c e ,

where A' = dAldX. Since

6((tOn = A'((on)(0„

E

we obtain from (8.2) and from the above that

A(a'n) 6 (w)n mod z 2(n– c –1)– e –1

Consequently, by (8.3),

Tn (A(ce„)6(w) n )----- Tn (yn ) mod ,en", where

m' =2(n —c +1)—e —1+n +2

because of n_-_3c+e+ 5. Since (112'rn±l )Tn (yn ) E o by (8.6), (1/ jrn-Fi )1nwa,n)(5 ‘ .. kw n ) ) also belongs to o, and it follows from m' 2n + 2 that

1 1 jrn +1 Tn(Â(œn) (5 (w)n) .7E17+1 Tn(Yn) mod .7rn±1

1


so that by (8.7), 1

( (fin, 1+ y = "'ICY

Crni+i Tn(A(a'n) 6 ( 0)n)) ' (on (8.8)

= [a, 0 ] ,0 .

To compute (oen , 1 + yn)n, let .7rc+ l i = cro , k' = Icc,() for the integer c 0 above. Since n c, it follows from Lemma 8.7(i), that

(an, 1 + 7n)n = (ac, N„,,(1 + V,i))c = Pc(N,(1 + 'Yn))()'

However, (8.4)—that is, y,, E jrriClon-implies

N„, ( 1 + yn ) = 1 mod zn - c -1 .

By Lemma 3.5 and Proposition 7.2, N(1c7k) is a closed subgroup of finite index in (kc,r)x. Therefore, the above congruence shows that N„, ( 1 + yn ) is contained in N(k' Ikc,) whenever n is large enough. For such an n, we then have pc (N n , c(1 + yn)) I k' = 1 by Theorem 7.1 so that

(œn , 1 + yn )„= pc(N,(1+ yn ))() 0. (8.9)

Finally, we see from (8.5), (8.8), and (8.9) that for sufficiently large n,

(a, co )f = (an , 1 + yn )„ (con , 1 + yn )n

= [a, 40 . •

We are now ready to prove the following theorem:

THEOREM 8.16. Let (f, co) be any normed Jr-sequence for k. Then

(a', 0» = Pl., for a E A f , fl e B.

Furthermore, if a is represented by an in A n , n then the element

1 xn ni-l Tn(Â ( an) 6 (0)n)

belongs to 0, and

(cr, 0» = xn i on =[ TnWan)6(0)„,)]1 to,

Proof. The group B is generated by the P'S with v(0) = 1. Hence, in order to prove (cr, 0» = [ot, 010 , we may assume that v(0) = 1. By the remark at the end of Section 8.1, we have

h(= to): (f, to) -4(f', )6),

where f'=fo =to ofot o-1 . Therefore, by Lemma 8.7(ii) and Lemma 8.11, it is sufficient to prove

(a", 0», = [a", 0 ]0 , for all a" E Aft.


However, this is exactly Lemma 8.15 for the normed Jr-sequence (f', 0). Thus the first half of the theorem is proved. Let ot be represented by an E A n . For large m __._ n, we know that .,r,n (= xm (ani , 0)) belongs to o and

(cr, 0) = [cr, P ].= xml (0m.

However, since a is represented by a n E A n , it follows from the definition of (a, 0

) that (a', mf = (an, On)n,f E W. Hence, by Lemma 4.8(ii),

Jr" . y . ,•,, = n

“, rem f *-4-'rn s", .7E n±1X,n E pm± 1 .

As xni = el -nxn by Lemma 8.10(ii), we obtain

.7e1+1Xn E pm± 1 , that is, Xn E O. •

Let (f, co) now be any z-sequence for k, not necessarily normed. By Theorem 8.4, for each f3 E B with v(f3) = 0, there exists a unique power series tp (X) in o[[X]] x such that t(con ) = 0,, for all n __-_ 0. With this tp , one can define (5(0) n by (8.3) and, hence, xn (a,, 0) and [a, 0

],,, for 0 E B with

v(f3) = 0. Take a normed Jr-sequence (f', co') for k and let

h:(f, (o)(f' , co').

Since Theorem 8.16 holds for (f', co'), it follows from Lemma 8.11 that Theorem 8.16 holds also for (f, w), provided that v(0) = 0.

We shall next formulate the above result in a form more convenient for applications. For this, we need the following lemma.

LEMMA 8.17. Let 0,, E Bn = (kw. Then 0,, is the nth component of an element 0 = (p., 0,, . . .) in B if and only if Nn(fin) is a power of 21": Nn (Pn ) E (71").

Proof. Let fin be the nth component of 0 = (0° , 0,, . . .) in B. Then Nni (13,n ) = Nn (fi n ) for all m __-_ 0. Hence Nn(fin) E N(k r Ik)= (sr) by Proposi-tion 5.17. Suppose, conversely, that Nn (Pn ) E (Jr). Apply Theorem 7.6 for

k' = 14, E =14±1 , E' = Ek' = kV'.

Since Nn (f3 n ) E (.70 = N(k,lk) ç N(1(Ik), fin belongs to N(k7, 4-1 1k7r) by that theorem. Therefore,

O n — Nn+1,n(Pn+1)

for some P n+ i E Bn±i= k( njr-f -1 , x . ) Since Nn+i(On+i) = Nn(fin) E (.70, we can similarly find )6,7 4-2 E B n ±2 such that N n+2,n (8 1 8 4-1,, n+2, = • n+1. In this manner, we obtain a sequence fin, Pn+1, Pn+2, • • . , satisfying N,n+i,m(Pm+1) = Pm for all m __._ n. Putting An = N,,n (f3 n ) for 0 -_ m -_ n, we see that O n is the nth component of 0 = (fi , f3,. • .) in B. •

It is now clear that Theorem 8.16 yields the following theorem of Wiles [25 ] :

THEOREM 8.18. Let n __._ 0 and let (f, (o) be any normed Jr-sequence for k. Let an be an element of Pn (= A n ), fin an element of (k njr ) x (= Bn ) such that


Nan ) is a power of 7r, and let

1 7, 1 t io(a) „)

X n z n+1 1 n(xr ( con ) pn

where A(X) is a logarithm of Ff and 0 is any element of B such that Pn is the nth component of p. Then x„ belongs to 0, x„ is independent of the particular p chosen, and

1 e(on) (an, fin)n,f = Xn °in =L 1 in-f-iTn(A, ( c)n ) fin n f - n•

Furthermore, if /3„ is a unit of kn,, then the same formula holds for any 7r-sequence (f, co), not necessarily normed.

8.5. The Special Case for k = Qp In this section, we shall see what Theorem 8.16 states in the special case:

(k, v)= (Q, IT = p.

Suppose first that p >2 and let

f(X)=(1+ X -1 e 3 .rpi

Then (cf. the example in Section 4.2)

FAX, Y) = (1 + X)(1 + Y) — 1, [a]f = (1 + X)a — 1, for a E 0 = Zp ,

and vv7. — 1 I E Pn÷1

icq = Qp W7) =

where Wpn+i denotes the group of all pn± lth roots of unit in Q. Since

p = (1+ cr)(1 + p) — 1, the map — 1 defines an isomorphism

K :W W rf . ,

Let = {a)n}

°in — 1 E W7,

where n -1, for n

As Prlf =f,

7 r i con =f(„-1)= —1= co n _ i , for n

so that (f, co) is a 2r-sequence for k = Qp and 21- =p. Furthermore,

X -i- (' — 1) = (1+ X) —1, for — 1 E WY, that is, for E Wp .

Since p is odd, it follows that

n (X -i- ( — 1)) = (1+ — 1= f(X)= f


Therefore, by Lemma 8.3, (f, (0) is a normed Jr-sequence. (This can also be seen directly from N .,n(. — 1) = — 1 for 0 n m.)

Now, let an E A,, = pn, Pn E Bn = (kw. Since kni-lif = r o . . f • ° f = (1+ X)Pn±1 — 1, ir n±1 i = an means that

= Pn V1 + an -1. Let

p1 = Pn+V1 + an , a = pn (ign ).

Then k7r() = 14(0 , and as Wpn±i ç 14, le,zr ()1 Kir is a Kummer extension and

a(n)= ?.• I, that is, (191h + an)' =

for some in W+1. We then see that

(an , Pn)n,f = pn(13n)() = a(n — 1 ) (T1 — 1 )

=(n -1) .t. (ij - 1) = -1

because —1) -if- (ij - 1) = oi - 1. Thus, if a E Af is represented by an E An

and if 0 = (00 , 0 1 , ... , fin, . . .) E B, then

(a, 0 )f = -1= K ( )

where lc: Wpn+i :.--, 14/7 and = c n+-/1 + coa-i , a = Pn(fin).

In general, let m 1 and let k' be a local field containing primitive mth roots of unity. For any x, y E k'x , let

{x, y} = (Ç)0_l, where a = pk ,(x).

Then {x, y} is an mth root of unity, and such a symbol {x, y) is called the norm residue symbol for k' with exponent m. The above result shows that

(a, 0).f. = K({0,,, 1 + an}) (8.10)

for the norm residue symbol { , } for V, = Q p (W p n+i) with exponent p'• Thus we see that in this special case, the pairing ( , )f :Af x B —t Wf is given, up to the isomorphism lc, by the norm residue symbols of Qp (Wpn+i) for n- 0.

We shall next describe [a, 0]. For this, let

X2 X 3 A(X)= log(1 + X) = X — —

2 + —

3 — • •

Then A(X) E Qp [ [X] ] and

AM.+ X)(1 + Y) — 1) = log(1 + X)(1 + Y)

= log(1 + X) + log(1 + Y) = Â(X) + A(Y).


Hence A :Ff Ga is the unique logarithm of Ff with Â(X)=-- X mod deg 2. Since

we have

1 (x) 1 + x '

yl,(an )= log(1 + an ),

Von ) =

(5 (0 16

= t '0( (on ),

„

xn= 1Pn+ i

Tn (jr-1 tio(con ) log(1 + Pn

Therefore, for sufficiently large n,

[a, = xniwn=[xn]f(',7 — 1) = K(Vnn). (8.11)

Now, by Theorem 8.16, (a, p ) = [a, in, and (8.11) holds whenever a is represented by an E An . Hence it follows from (8.10) that if an is any element of p n (= An) and if 13,, is the nth component of an element p in B, then

{0,2, 1 + an} = xnn, x n = n Tn t io(COn 10g(1 + an)) Pn

for the norm residue symbol { , } of Qp (Wpn+i) with exponent pn ±1 • As a special case, let

Then to (X) = 1 + X, tip(X) = 1 so that for any an E pn,

» 1

1 + an} trIzy where a = pn±i Tn (log(1 + an .

Next, let P = = 1, . , 1, . .) E B.

Then to (X) = X, tifi (X) = 1 so that for a',, E p,2,

- 1, 1 + an} = where b = pnl±i Tn ( n tz 1 log(1 + crn))•

Let us now quickly discuss the case: p = 2, k = Q2, = p = 2. In this case, let

f (X) = 1 — (1— X) 2 . Then f E ,97;r and

FAX, Y) = 1 — (1 — X)(1 — Y), [alf =1 — (1— X), for a E 0,

= {1 — I E W2n+11, 14= Q(W) = Qp(W2-1),

W2+1 being the group of all 2'th roots of unity in Q. The unique logarithm

136


A : Ff ---> Ga with A(X) ---=:- X mod deg 2 is given by

X2 X 3 il(X) = —log(1 — X) = X + ---i- + —i- + • • - .

Let n2 1-1 -= nr con = 1 — fl , for n O.

Then (f, w), to = { COn } n>.0, is a normed Jr-sequence for k = Q2. Since il: (X) = 1/(1 — X), A/ (w) = W , we obtain, as for p >2,

x = 1 r

n r+1 T( _ —pn to(con ) log(1 —

On the other hand, in this case Jrn± l i. = an means

= 1 — 21/1 — ai,.

Hence, similarly as for p >2, we have the formula:

for an E Pn , 0 E B ,

with x„ as above. Here { , }, of course, denotes the norm residue symbol for Q(W2 -) with exponent 2n ± 1 . In particular, let

(Note that ( (), • • • , c,,, • . .) 0 B in this case.) Then

{— fl , 1— an } = M where a = 2n1+ 1 7'n (log(1 — an )),

and for 0 = co, we also obtain

{1 — L 1 — an } = M where b = ,,-2 1±1 Tn ( n_n_ 1 log(1 — an )).

Those formulas for { , }, for p > 2 and p = 2, are essentially the same as the classical explicit formulas of Artin—Hasse [2] for the norm residue symbols of Qp (Wpn+i) with exponent p'. Thus Theorem 8.16 may be regarded as a generalization of those classical results for arbitrary local fields of characteristic O. For the relation between Artin—Hasse formulas and Theorem 8.16, see also Iwasawa [12] and Kudo [15], and for applications of Theorem 8.16 and more recent results on explicit formulas, see Coates—Wiles [4] and de Shalit [6], [7].

Appendix

In this Appendix, we shall first briefly discuss Brauer groups of local fields which play a central role in the cohomological method in local class field theory, but which have not come up in the text in our formal-group theoretical approach. We shall then sketch, again briefly, the main ideas of Hazewinkel [11] for building up local class field theory.

Al. Galois Cohomology Groups

In this section, we shall sketch some basic facts on Galois cohomology groups. For the most part, proofs are omitted. For details, we refer the reader to Cassels-Friilich [2], Chapters IV and V, or Serre [21], Chapter VII.

Let G be a profinite group—that is, a totally disconnected compact group—and let A be a discrete G-module. We assume that the action of G on A is continuous—namely, that the map G x A -> A, defined by (a, a)1-> aa, is continuous. For n 0, let Gn denote the direct product of n copies of G: G n= GX . • - x G, and let Cn(=Cn(G, A)) be the set of all continuous maps f:Gn —> A. For n = 0, this means that G° = {1), C° = A. Note that the Cn, n- 0, are abelian groups in an obvious manner. For f E Cn, we define an element orf in Cn±' by

(&f)(ai, • • , an+i) = crif (cr2, • • • , a) n

+E (--i-)'f(cri, • • • , aicri+i, • • • , i=1

+ ( i) n-f-if(cyj , . . . , an).

Then f 1-> bnf defines a homomorphism

n 0,

such that

(5 n ° (5 n-1 = 0, 11*(5 n-1 ) ç Ker(e), for n 1.

Hence, let

Hn(G, A) = Ker(e)/Im(bn -1), for n- 0,

where we put Im (s ') = 0 if n = O. The abelian groups Hn(G, A), thus defined, are called the cohomology groups of the G-module A. For n = 0, 1, 2, bn is given as follows:

45 °(a)(a)= aa - a, for a E C° = A , a E G,

(5 1 (f)(a, r) = af(r)- f(ar)+ f(a), for f = f(a) E 0, a, T E G,

T, P) = of(T, P) - f(aT, P) + f(a, TP) - f(cr, T), for f = f(a, r) E C2, a, r, p E G.


Thus, for example,

le(G, A) = A G = {a E A I aa = a, for all a E G}.

Let G' be another profinite group, acting continuously on a discrete modulo A', and let

be continuous homomorphisms such that

a(y 1 (a1 )a)= a' (a(a)), for a' E G', a E A . (A.1)

In such a case, we call the pair A = (y, a) a morphism from (G, A) to (G', A') and write

Let f E Cl(G, A)—that is, a continuous map from G into A. Then the product

f':G'-2/-*G 1-->A 1-> A'

belongs to Cl(G' , A'), and f 1—> f' defines a homomorphism

Cl(G, A)—> C l (G', A').

Similarly, Cn(G, A)—* Cn(G', A') are defined for all n 0, and because of (A.1), they induce homomorphisms of cohomology groups:

Hn (G , A)----> Hn (G 1 , A'), for n O.

There are two important special cases of such homomorphisms. Let H be a closed subgroup of G. Then H is a profinite group and it acts continuously on A. Let i : H ----> G denote the natural injection and let id :A —> A be the identity map. Then

(i, id): (G, A)--> (H , A)

is a morphism in the sense defined above. The associated homomorphisms of cohomology groups:

res:Hn(G, A) ---> Hn (H , A), n 0,

are called the restriction maps. Next, let H be a closed normal subgroup of G and let

A" = {a € A 1 ta = a for all r E H}.

Then the factor group G/H is again a profinite group and it acts continuously on A H. Let y: G—> G I H be the canonical homomorphism and let i :A" —* A denote the natural injection. Then

(y, i): (G/H, A") —* (G, A)

is a morphism so that we obtain homomorphisms:

inf:Hn(G1H, A ll)-÷ Hn (G , A), n 0,

Appendix 139

called the inflation maps. We shall now state two important results involving res and inf; for the proofs, see the references mentioned earlier.

PROPOSITION A.1. Let (G, A) be as above and let H be a closed normal subgroup of G. Then the sequence

0 —> Hi (G/H, AH)ii-2 Hi(G, A) =5.> Hi( A)

is exact. If H l(H, A) = 0, then

0 —> H2(G/ H, A H ) ii--2L H 2(G, A) -E-> H2(H, A)

is also exact.

Next, let {U} denote the family of all open normal subgroups of G. Let U' c U for two such subgroups U and U'. Let y:GIU'—>GIU be the canonical homomorphism, and a:A u —>A u the natural injection. Then (y, a): (G / U, A u)—> (Glu', A u ) is a morphism so that we have

infuttp n O.

It is clear that if U" c U' c U, then intuit?, = infuvu,, 0 infuiu ,. Hence

lim Hn(G/U, A u) -->

is defined with respect to the maps intuit? for all U and U' with U' c U.

PROPOSITION A.2. The inflation maps Hn(GIU, in—> Hn(G, A) for open normal subgroups U of G induce an isomorphism

lim FP (G I U, A u) -4- Hn(G, A), n O. -->

We also state below two elementary lemmas which are easy conse-quences of the definition of Hi (G, A) and H2(G, A). Namely, let G now be a finite cyclic group of order n- 1. Fix a generator p of G: G = (p), pn = 1, and let

N(A)= (1+ p + • - • + pn-1)A.

Clearly, N(A) c A G c A. For each a E AG, we define g(a, r) in C2 by

g(p i, p')= 0, for 0 i, j <n, i + j <n,

=a, for 0 . i,j<n, i+j - n.

We check immediately that g belongs to Ker(6 2). Hence, let ca denote the residue class of g in H2(G, A) = Ker(6 2)/Im(6 1 ). Then:

LEMMA A.3. The map al—> Ca induces an isomorphism

A G IN(A) -4 H2(G, A). (A.2)

Note that the isomorphism depends on the choice of the generator p of G and, hence, is not canonical. It is also easy to see that there is a similar


isomorphism

A N I (1 — p)AH 1 (G, A)

where AN= {a E A I (1 + p + • • + pn-1)a = 01. Next, let H be a subgroup of the above G = (p) and let

[G:H]= m, [H :1] = d, n = md.

Then H = (pm) and GIH= ( 15) with p = pH. By Lemma A.3 for (GI H, A H ), p defines an isomorphism

(A H )G/H/N(A H ) H2(G I H, A H),

where (A H) G/H = A G,

N(A H)= (1+ p + • • • + pm - 1)AH = (1+ p + • • • + pm-1)A H.

Since

(1+ pm + • • • + pm(d-1) )A H = dAH,

(1+ p + • • + pm -1)(1+ prn + • • + p(d-l)m)= 1+ p + • -

we have

dN(A H) N(A)

so that the endomorphism of A G, defined by al—) da induces a homomorphism

d:(A H )G1H IN(A H )—> A G I N(A).

LEMMA A.4. The diagram

(A H) GIHIN(A H) H2 (GIH, A H )

A G IN(A) H2(G, A)

is commutative.

Now, let K be a Galois extension of a field F and let G = Gal(K/F). Then G is a profinite group. Therefore, if G acts continuously on a discrete module A, the cohomology groups Hn(G, A), 0, are defined. Such groups are called Galois cohomology groups. For example, G acts con-tinuously on the multiplicative group IC of K, viewed as a discrete group. Hence Hn(G, IC), n 0, are defined. For simplicity, Hn(G, IC) will also be denoted by Hn(KIF):

Hn(KIF)= Hn(Gal(KIF), O.

Let {E} be the family of all finite Galois extensions of F, contained in K:

FcEc K. FE :F1 < +00.

Appendix 141

Let U= Gal(K/E) for such a field E. Then U is an open normal subgroup of G and

Gal(E/F) = GIU, (Kx) u = Ex.

Hence it follows from Proposition A.2 that

Hn(KIF)=1irn Hn(EIF), (A.3)

where the direct limit is taken with respect to the family {E}—namely, with respect to the family { U}, as stated in that proposition.

PROPOSITION A.S. For any Galois extension KIF,

Hl(KIF)= O.

Proof. It is one of the fundamental theorems in Galois theory that 11 1 (EIF)= 0 for any finite Galois extension EIF. Hence the proposition follows from (A.3). •

PROPOSITION A.6. Let KIF be a Galois extension and let E be any Galois extension over F, contained in K: Fc Ec K. Then the sequence

0 ---* H2(EIF) iffL-* H2(KIF) 11-H2(KIE)

is exact. Proof. This follows immediately from Propositions A.1 and A.5. •

As a consequence, we can imbed H2(EIF) in H2(KIF) by means of the inflation map:

H2(EIF) c H2(KIF).

(A.3) then shows that H2 (KIF) is the union of the subgroups H2(EIF) when E ranges over all finite Galois extensions over F, contained in K:

H2(KIF)= y H2(EIF). (A.4)

Let k be any field and let Qs denote the maximal Galois extension over k—that is, the maximal separable extension over k, contained in an algebraic closure of k. Let

Br(k) = H2(Qs 1k).

Since Qs is unique for k up to k-isomorphism, Br(k) is indeed canonically associated with the given field k. It is called the Brauer group of k. In the next section, we shall discuss the structure of the Brauer group Br(k) in the case where k is a local field.

A.2. The Brauer Group of a Local Field

Let (k, y) be a local field. We shall consider the Brauer group Br(k) = H2 (Qs /k). As in Section 3.2, let K denote the maximal unramified extension

142


kur over k in a fixed algebraic closure S2 of k:

K = k„,..

Since K I k is abelian, we have

kcKc Qs .

Hence, by Proposition A.6, we obtain an exact sequence:

0 ---* H2(K1k) If1-* Br(k) 11-*Br(K). (A.5)

We shall first study H2(K1k). By Sections 2.3 and 3.2, for each integer n 1, there exists a unique

unramified extension kZ r over k in S2 with [k r : k]= n and

K= kur = U Kr. n?.-. 1

Clearly fle,'„I n , i is the family of all finite Galois extensions over k in K. Hence, by (A.3),

H2(K1k)= lim H 2(k1.1k).

Let cp be the Frobenius automorphism of k: cp E Gal(K/k). Then Gal(K,./k) is a cyclic group of order n, generated by (pn = cp I k, and by Lemma A.3, (1)„ defines an isomorphism

le I N(k1.1k) -4 H2(k,i rlk). (A.6)

Let z be a prime element of k. Then

NU(k1.1k)= U(k), N(kZ,Ik) = (2rn) x U(k)

by Lemma 3.6. Hence the isomorphism

y:kx/U(k)L. --,Z,

defined by the normalized valuation y of k, induces

le IN(K rIk) -4 ZInZ

so that (1/n)v defines an isomorphism

le I N(k1.1 k) -4-1

Z/Z. n

The product of the inverse of (A.6) and the above isomorphism then yields

1 H2(k1.1k) 4 1-71 Z/Z. (A.7)

Since this isomorphism is quite important, let us explicitly describe the map according to the above definition. For each x E le , let c, denote the residue class of g(a, r) in H2(kni,r1k)= Ker(6 2)/Im(6 1 ) (cf. Section A.1), where g is

Appendix 143

defined by

g(q9,„ Vn) = 1, for 0 j < n, i + j <n,

= x, for0i,j<n,

Then H2(k72,./k) consists of such cx for all x E V and the isomorphism (A.7) is given by

v(x) cx 1-.> mod Z, x E k x

It follows in particular that H2(klk) is a cyclic group of order n generated by c„, a- being any prime element of k.

Now, let k kin k n , n = dm. Since q)„., = (pn I k, we see from Lemma A.4 that the following diagram is commutative:

H2(kalk) kx1N(lennrIk) ZImZ - 1> ZIZ m

Id id

1 H2(k1.1k) kx1N(kZ i.lk) —=->" ZInZ, -

nZ/Z.

Here the vertical map on the extreme right is the natural injection of (1/m)Z/Z in (1/n)Z/Z. Therefore, denoting the additive group of the rational field Q again by Q, we obtain

112(K1k)=1im H 2(k1.1k) lhui ZIZ= Q/Z. ----> n

Thus:

THEOREM A.7. Let k in. (= K) be the maximal unramified extension of a local field k. Then the Forbenius automorphism q) of k defines a natural isomorphism

H2(k in.1 Q/Z.

Next, we consider Br(K) = H2 (Qs /K).

LEMMA A.8. Let L be a finite Galois extension over K. Then there exists a finite sequence of fields:

K=Ko cK i c-•-cKs =L,

such that

(i) each Ki is the maximal unramified extension of a local field k i : Ki = (ki )nr, and

(ii) Ki /K1 _ 1 is a cyclic extension for i =1, . . . , s.

Proof. It is easy to see that there exists a field F such that FK = L and that F is a finite Galois extension over k' = knir for some large n O.


Replacing k' by F n K is necessary, we may assume that F n K = k' so that Gal(L/K) 2, Gal(F/k'). By the Corollary of Proposition 2.19, there exists a sequence of fields:

k' = k o c k i c • • • c lc, = F

such that each kilki-i, 1 .i.s, is a cyclic extension. Then the fields Ki = ki K = (ki)ur, 0.i.s, have properties (i) and (ii) stated in the lemma. •

THEOREM 1.9. H2(S2s /K) = O.

Proof. By (A.3), it is sufficient to prove H2(LIK)= 0 for any finite Galois extension LI K. By Proposition A.6 and Lemma A.8, the proof can be further reduced to the case where LI K is a cyclic extension. As in the proof of Lemma A.8, let

FK = L, F n K = k' = kril,r , Gal(L/K) -4 Gal(F/k').

Let m n and let

d = [L : K] = [F :le], E = Feunr , E' = Ekr d = Fkr d .

Then KnE=ka so that both E/ k, k dlk u';. are cyclic extensions of degree d. Hence for any x E (k'unr)x , x d is the norm of x for the extensions Elk'unr and k'unr+ dIk. Therefore, x E N(E' I k'unr+ d) by Theorem 7.6. Since

E'K = FK = L, E' n K = k'unr+d ,

we then see that x E N(LIK). Hence (k'unr)x c N(LIK) for all m n and, consequently N(LI K)= Kx. By Lemma A.3, we then have H2(LI K)= o. •

By (A.5) and Theorems A.7 and A.9, we now obtain the following result:

THEOREM A.10. Let k, be the maximal unramtfied extension of a local field k. Then

inf : H2(k,1k) 4 H 2(S2,1k) = Br(k).

Hence there exists a canonical isomorphism:

Br(k) Ls; Q/Z.

The image of c in Br(k) under the above isomorphism is denoted by

inv(c).

Next, let k' be any finite separable extension over k. Since

k c k' c Qs , the restriction map

res:Br(k) = H2(52,/k)--> H2(S25 /k') = Br(k')

is defined.

Appendix 145

THEOREM A.11. Let d =[k' : k]. Then the diagram

Br(k)="—÷ Q/Z

ires id Br(k') ="--+ Q/Z

is commutative.

Proof. Let e = e(k' lk), f = f(k' lk). Let 92, q;, ' be the Frobenius automorphisms of k and k', respectively, and let y, y' denote the normalized valuations of k and k', respectively. Then, by Chapter II,

ef = d, vi I k = ev, (pi 1 K = 449f .

By (A.4) and Theorem A.10,

Br(k) =1-12(k,1k)= Ul-P(kalk), i

and similarly for Br(k'). Hence, if c E Br(k), then c E 1-12(knurlk) for some integer n 1 with f In. By an earlier remark, c = cx for some x E k> namely, c is represented by g(a, r) such that

g(T in, cp'n)= 1, for 0 i, j < n, i + j <n,

= x, for 0 i, j < n, i + j n.

Let

n' =nil; cp'„,=cp'lleurf.

Since cp' 1 K = qi, it follows from the definition of the restriction map that res(c) in 1-12 (kL7.7k) is represented by h(a' , r'), where

h(C,i, cp1) = 1, for 0 i', j' <n', i' +j' <n',

=x, for 0 i', j' <n', i' + j' -n'.

But, then

' inv(c) = y(x)

, 11' mod Z (x) inv(res(c)) = mod Z, n n'

where

v'(x) = ef

v(x) =

n' n n

Hence inv(res(c)) = d inv(c), and the theorem is proved. •

In the above theorem, let k' I k be in particular a finite Galois extension. By Proposition A.5,

I-12(k' I k)= Ker(res:Br(k)—> Br(k')),


Hence, it follows from the theorem that inv :Br(k) -4 Q/Z induces a natural isomorphism

inv:H2(elk) -4-1

Z/Z, d =[k' :k], d

which generalizes (A.7). The element c of H2(k' I k) such that

1 inv(c) = -

d mod Z

is called the fundamental class in H2(k 1 I k). By means of Theorems A.10 and A.11 and the above result, we see that separable extensions of a local field provide us a so-called class formation. However, we shall not discuss this cohomological approach here any further. Compare Serre [21].

A.3. The Method of Hazewinkel

Let (k, y) be a local field and let k' be a finite Galois extension of k. Let

K = ka„ L = le= k' K

and let k and L be their completions (in 0). Clearly LI K is a finite Galois extension and by Lemma 3.1,

KL = L, k n L = K.

Hence L/k is also a finite Galois extension and a --> a I L 1-> a I k' defines

Gal(L/k) -4 Gal(L/K) 4 Gal(k1k 0),

where 1( 0 = k' n K is the inertia field of the extension k' lk. It then follows from the Corollary of Proposition 2.19 that L/k and LI K are solvable extensions.

PROPOSITION A.12. Let N(LI1Z) denote the norm group of the finite extension LI k. Then

N(LI K)= K.

Proof. Since L/k is solvable, it is sufficient to prove the proposition for the case where L/k is a cyclic extension of prime degree. For the proof of this, see Serre [21], Chap. V, §3, where norm maps for finite extensions of complete fields (cf. Section 1.2) are discussed in detail. A slightly different proof can be found in Iwasawa [13]. •

Now, let

V(L/k) = the subgroup of U(K), generated by all elements of the form

r-1 = o-()/ for a E Gal(L/K) and E U(L).

By Lemma 3.2, y i< (= y I K) and, hence, yk (--= P I k) are normalized valuations. Therefore (K, ,u k) is a complete field in the sense of Section 1.2

Appendix 147

and it follows from Section 1.3 that L is also a complete field with respect to a normalized valuation, equivalent to pi, (= i I L). Let z' be any prime element of L. For any a in Gal(L/10, ada -1 (= o-(e)1,7e) then belongs to U(L) so that a map

Gal(L/k) U(L)/ V (L/k),

mod V(LIIZ)

is defined. It is easy to check (cf. the proof of Proposition 2.13) that the above map is a homomorphism, independent of the choice of the prime element Jr'. Since U(L)/V(L/k) is an abelian group, it then induces a homomorphism

i:Gal(L/k)a b — U(L)/V(L/k)

where Gal(L/k)ab denotes the factor commutator group of Gal(L/k). Let N (= NLIk) denote the norm map of the finite extension L/k. Then we have the following

PROPOSITION A.13. The sequence

1—> Gal(L/IZ)ab U(L)/V(L/k) U(IZ).— 1

is exact. Proof. See Hazewinkel [11]. He proved this by elementary arguments,

without using cohomology groups. However, if one is allowed to make use of the cohomology groups of Tate for the finite group Gal(L/k), it can also be derived from the long exact sequence of such cohomology groups, associated with the exact sequence

1—> U(L)—> ->Z-> 1,

defined by the normalized valuation of L. Compare Serre [22]. In any case,

the essential source of the proof is Proposition A.12 above. •

We now assume that kilk is a finite abelian extension. Then the extensions Llk, LIK, and Lk are also abelian. Hence, identifying Gal(L/k) with Gal(L/K), we obtain from Proposition A.13 an exact sequence as follows:

1—> Gal(L/K) , U(L)/V(L/10 U(k)—> 1. (A.8)

Let cp, zp, and cpc, denote the Frobenius automorphisms over k, k' and /co = k' n K, respectively. Extend these to automorphisms of L and, then, to automorphisms of L, and denote them again by the same letters cp, ip, and To. Then cp, IP, and To commute with every a in Gal(L/k) = Gal(L/K). In particular,

= for a E Gal(L/k), E U(L ).

Hence

V(L/K) -1 V(LI K)


so that the homomorphism tp - 1: U(L)--> U(L) induces an endomorphism

a = tp - 1: U(L)/V(r/k)-> U(L)/V(L/k).

Let

13 = To -1: U(k)- U (k)

and let

y:Gal(L/K)-> Gal(L/K)

be the trivial endomorphism of Gal(L/K), mapping every element of Gal(L/K) to the identity element. With these maps a, fl, and y, consider the diagram

1—> Gal(L/K) -L> U(L)IV (LI k) 2-1 > U(k) —p 1

1 --> Gal(L/K) U(L)/V(L/k) 2-1 -> U(k)—> 1,

where the rows are the exact sequence (A.8). Let

A = Ker(a), B = Ker(3), C = Coker(y), D = Coker(a).

Then we see (cf. Lemma 3.11) that the above diagram is commutative and

A = U(k)17(L1k)1V(LIk), B = U(k o), C = Gal(L/K), D =1.

Hence the maps N and i induce homomorphisms A -> B and C->D respectively, and by the Snake Lemma, there exists a homomorphism

(5:B->C

such that

AL1->B 1->CL->D

is exact. Since

Ker(6) = Im(N) = N(U(k')V(LIK))= Nkvk o(U(k'))= NU(k'l ko),

(5 induces an isomorphism

U(k 0)/NU(ki/k 0) L.; Gal(L/K). (A.9)

Checking the definition of the isomorphism, we find that if u mod NU(k' I k o)--> a for U E U(ko), then

ug' -1 mod NU(k' 1 k o)-> Tag,' a' =1.

Hence,

U(k 0) 1 c NU(k7k 0).‘

However, as k o/k is an unramified extension, one can see (cf. Lemma 3.6)

Appendix 149

that the norm map N k cjk:U(ko) —> U(k) is surjective and its kernel is U(k0)P -1 . Therefore, it induces

U(k0)/NU(le1k 0)4 U(k)INU(k'lk).

Thus we obtain from (A.9) a fundamental isomorphism

U(k)INU(k' Ik)L-_-,Gal(L/K) = Gal(k1k0), (A.10)

where k' /k is any finite abelian extension of local fields and k o is the inertia field of k' /k.

Let e=e(k' /k), f =f(k1k) for the above k' /k. Then (A.10) and Proposition 2.12 yield

[U(k):NU(Ielk)]=[k':k0] =e.

Since ef [k' k], v(N(k'lk))= fZ by Proposition 1.5, it follows that

[k:N(k'lk)]=[k:N(k'lk)U(k)][N(k'lk)U(k):N(k'lk)]

=[Z:fZ][U(k):NU(le lk)1= f • e

=[k':k].

This is the proof of the fundamental equality (Corollary of Theorem 7.1) in Hazewinkel [11]. Starting from (A.10), we can also define the norm residue map p k : kx --> Gal(kab i k) by this method. For the details, we refer the reader to Iwasawa [13].

Bibliography

[1] E. Artin, Algebraic Numbers and Algebraic Functions, Gordon and Breach, New York—London—Paris, 1967.

[2] E. Artin und H. Hasse, Die beiden Ergdnzungssâtze zum ReziprozitAtsgesetze der r -ten Potenzreste im KOrper der ln-ten Einheiten Wurzeln, Abh. Math. Sem.

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Table of Notations

General Notations

the ring of rational integers, the field of rational numbers, the field of real numbers, the field of complex numbers,

Fq a finite field with q elements, a commutative ring with identity 1*0, the additive group of R, often denoted also by R, the multiplicative group of all invertible elements in R.

Chapter I

1.1. V, vo , vp, a, p, (= alp), U, e(vi/v) (= e), f(v1v) (=f) 1.2. (k, Qp , F((T)), Jr, a, Un 1.3. e(k' lk), f(k' lk), N ielk(a)

Chapter II

2.1. (k, v), q, (Q p , vp ), (F((T)), v r), co(x), V 2.2. W 2.3. cp, /cc, 2.4. l, 2(ki lk), D(k'lk), o[w] 2.5. G„

Chapter III

tt, fly (F, IF), (Ey = OFIPF, fF = OFIPF kur, Ky a , tic = oKIPK, Pk, (Pn, Z , Vn , Vco

U(F), N(Flk), NU(Flk) R[[Xi , ,X,]], mod deg d, f 0 (g i , ,g,), M, f og, f1 ,

4-2= 0K-2/Pc2,f °

P'— Pk, fic= oic/PK, 4P

Chapter IV

4.1. F(X, Y), iF(X), MF, Ff (= f ° F ° ri ), Ga (X, Y), A(X)

4.2. R (= ok), .3;„, .5;y FAX, Y), [a]f, 0(X)

4.3. m (= a -4- py a1 a, tuf, 14/7, Wf, fn , gn ,

4.4. fin, ôn, fi, o f

Hom(F, G), End(F),

h, End(W7), Aut(W7)

3.1. 3.2. 3.3. 3.4.

3.5.

154

Table of Notations

Chapter V

5.1. om, SiT, , k m ' n L n O m ' n , o n , (5;', 5.2. 14, S y Sx, fn = on /p n , X -i- a, N(h) (= Nf (h)), N" 5.3. km, œ, L y (5, Ô ' , (5„, k, (= kliœ)

Chapter VI

6.1. kab, Lk, IN, ip, Pk 6.3. Pk, Qs, tGIH, tk'lky (k' Ik)

Chapter VII

7.1. Pelk 7.2. Gry g„ 00, iG (a) (=i(a)), ip(r), G r 7.4. c(k' lk), f(k' lk), H x , kx , f(x)

Chapter VIII

8.1. (f, (0), Nm,n , Bny B, N„, v(f3) 8.2. pn , 13)n,f, A n , A1, (cv, /3)f 8.3. 4, A., (5(0),„ Tm n , zn (ain , 13), [ce, p ] , 8.4. 8.5. K, {x, y}

Appendix

A.1. c (= Cn(G, A)), ô " , Hn(G, A), A= (cy, y), res, inf, Hn(KIF), Qs , Br(k)

A.2. inv(c) A.3. V(L/IZ)

Index

additive (formal) group, 52 Artin map, 88

Brauer group, 141

cohomology group, 137 complete field, 7 complete valuation, 5 completion (of a valuation), 7 conductor, 112 conductor-discriminant theorem, 114

different, 29 discrete valuation, 5 discriminant, 29

equivalent valuations, 4 existence theorem, 101 extension of a complete field, 12 extension of a valuation, 5

finite extension of a local field, 26 formal group, 50 Frobenius automorphism, 28, 38 fundamental class, 146 fundamental equality, 98

Galois cohomology group, 140

Hasse-Arf (Theorem of), 111

morphism of formal groups, 51

v-topology, 4 norm group, 41 norm of an ideal, 16 norm operator, 72 norm residue map, 88 norm residue symbol, 134 normalized (normal) valuation, 5 normed (power series) f, 118 normed Jr-sequence, 117

7r-sequence, 116 p-adic number field, 8 p-adic valuation, 3 p-field, 18 prime element, 8

ramification groups in the lower numbering, 33

ramification groups in the upper numbering, 104

ramification index e(v' Iv), 6 ramification index e(k' lk), 12 res (= restriction map), 138 residue degree Ay' Iv), 6 residue degree f(k' I k), 12 residue field of a valuation, 4 restriction of a valuation, 5

ideal of (k, v), 9 inertia field, 29 inf (= inflation map), 139 isomorphism of formal groups, 51

local field, 18 logarithm, 123

maximal ideal of a valuation, 4 maximal unramified extension, 37

totally ramified extension, 26, 39 transfer map, 93 trivial valuation, 3

uniqueness theorem, 101 unit group, 4 unit norm group, 41 unramified extension, 26, 36

valuation, 3 valuation ring, 4

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