Local Class Field Theory (Oxford Mathematical Monographs)

Local Class Field Theory KENKICHIIWASAWA Princeton University OXFORD UNIVERSITY PRESS· New York CLARENDON PRESS . Oxfor...

Author: Kenkichi Iwasawa

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Local Class Field Theory KENKICHIIWASAWA Princeton University

OXFORD UNIVERSITY PRESS· New York CLARENDON PRESS . Oxford 1986

Oxford University Press Oxford New York Toronto Delhi Bombay Calcutta Madras Karachi Petaling Jaya Singapore Hong Kong Tokyo Nairobi Dar es Salaam Cape Town Melbourne Auckland and associated companies in Beirut Berlin Ibadan Nicosia

Copyright

© 1986 by Kenkichi Iwasawa

Published by Oxford University Press, Inc., 200 Madison Avenue, New York, New York 10016 Oxford is a registered trademark of Oxford University Press All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Oxford University Press. Library of Congress Cataloging-in-Publication Data Iwasawa, Kenkichi, 1917Local class field theory. (Oxford mathematical monographs) Bibliography: p. Includes index. I. Title. II. Series. 1. Class field theory. QA247.I95413 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

246897531 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 example, 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 [11]. 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. Degenhardt of Oxford University Press for his help in publishing this book. Princeton September 1985

K.1.

Contents Chapter I. Valuations

3

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

Chapter II. Local Fields 2.1. 2.2. 2.3. 2.4. 2.5.

12

18

General Properties 18 The Multiplicative Group k X 22 Finite Extensions 25 The Different and the Discriminant Finite Galois Extensions 32

29

Chapter III. Infinite Extensions of Local Fields 3.1. 3.2. 3.3. 3.4. 3.5.

Algebraic Extensions and Their Completions 35 Unramified Extensions and Totally Ramified Extensions The Norm Groups 40 Formal Power Series 43 Power Series over Ok 45

Chapter IV. Formal Groups Ff( X, Y) 4.1. 4.2. 4.3. 4.4.

35

Formal Groups in General Formal Groups Fj(X, Y) The o-Modules Wi 57 Extensions in (k 61

50 50

53

Chapter V. Abelian Extensions Defined by Formal Groups 5.1. Abelian Extensions L n and k,:,n 65 5.2. The Norm Operator of Coleman 69 5.3. Abelian Extensions Land kJr 75

Chapter VI. Fundamental Theorems 6.1. The Homomorphism Pk 6.2. Proof of Lk = kab 84 6.3. The Norm Residue Map

36

80

88

80

65

Contents

V111

Chapter VII. Finite Abelian Extensions 7.1. 7.2. 7.3. 7.4.

Norm Groups of Finite Abelian Extensions 98 Ramification Groups in the Upper Numbering 101 107 The Special Case k,;,njk Some Applications 110

Chapter VIII. Explicit Formulas 8.1. 8.2. 8.3. 8.4. 8.5.

.7r-Sequences 116 The Pairing (£1', (3)f The Pairing [£1', f3]w The Main Theorem The Special Case for k

Appendix

116

120 123

127 = Qp

133

137

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

Bibliography

151

Table of Notations Index

98

155

153

141

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 satisfies the following conditions: (i) (ii)

v(x) is a real number for x For any x, y in k,

=1=

E

k, is called a valuation of k if it

0; and v(O)

min(v(x), v(y)) (iii)

0, define a function Il(x) on k by 11 (x ) = av(x),

for all x E k.

Then 11 is again a valuation of k. When two valuations v and 11 on k are related in this way-namely, when one is a positive real number times the other-we write v~1l

and say that v and 11 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 lYE k, v(y - x) > a}.

This is a subset of k containing x. Taking the all a ERas a base of neighbourhoods of x in topology on k, which we call the v-topology field in that topology; the valuation ring 0 is ideallJ is open in k. A sequence of points, Xl,

family of subsets N(x, a) for k, we can define a Hausdorff of k. k is then a topological closed in k and the maximal X2, X3, ... , in k converges to

5

Valuations X E

k in the v-topology-that is,

lim Xn = X, n-+ OO

if and only if lim v(xn - x) =

+ 00.

n-+ oo

When this is so, then lim v(xn) = v(x). n-+ OO

In fact, if x =1= 0, then v(xn) = v(x) for all sufficiently large n. A sequence XlJ X2, X3, ... in k is called a Cauchy sequence in the v-topology when

v(xm

-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 i

00

2: Xn = lim 2: Xn

n=l

i-+oo n=l

converges in k if and only if v(xn)~

+00, as

n~

+00.

A valuation v of k is called discrete if v(kX) is a discrete subgroup of R+-that is, if

v(kX) = Zf3

=

{nf31 n

=

0, ±1, ±2, ... }

for some real number f3:> 0. If f3 = 0, then v is the trivial valuation Example 1. When f3 = l-that is, when

Vo

in

v(kX) = Z = {O, ±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 v' a valuation of k'. Let v' I k denote the function on k, obtained from v' by restricting its domain to the subfield k. Then v' I k is a valuation of k, and we call it the restriction of v' to the subfie1d k. On the other hand, if v is a valuation of k, any valuation v' on k' such that

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

Local Class Field Theory

6

Let v' 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 0', lJ', and f' denote the valuation ring, the maximal ideal, and the quotient field of v', respectively: 0'

= {x' E k' I v'(x'):> O},

lJ' = {x'

E

k' I v'(x') > O},

f'=o'/lJ'·

Then 0= 0'

n k,

so that f = o/lJ = o/(lJ'

n 0) = (0 + lJ')/lJ' c 0' IlJ' = f'.

Thus the residue field f of v is naturally imbedded in the residue field f' of v'. On the other hand, v' I k = v also implies v(kX) c v'(k'X) c R+.

Let

e = e(v' Iv)

= [v'(k'X): v(kX)],

f =f (v' Iv) = [f' : f],

where [v'(k'X): (kX)] is the group index and [f': f] is the degree of the extension f'/f. e and f are called the ramification index and the residue degree of v' lv, respectively. They are either natural numbers 1,2,3, ... , or

+00. The following proposition is a fundamental result on the extension of valuations. 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 v' of k': v' I k = v. If, in particular, k'ik is a finite extension, then v' is also complete, and PROPosmON

1 n

v'(x') = - v(Nk'lk(X'»,

for all x'

E

k',

where n = [k': k] is the degree and N k'ik is the norm of the extension k' / k. • Proof. We refer the reader to van der Waerden [23]. Let k'lk, v, and v' be as stated above and let a be an automorphism of k' over k. Then v' 0 a = v'-that is, COROLLARY . .

v'(a(x'» = v'(x'),

for all x'

v(o') = 0',

a(lJ') = lJ'·

E

k',

so that Hence a is a topological automorphism of k' induces an automorphism a' of f' over f:

in

the v' -topology, and it

a':f' ~f'. Proof.

The proof can easily be reduced to the special case where k' Ik is

7

Valuations

a finite extension. The first part then follows from Nk'lk(a(x'» = Nk1k(X'). 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 kisomorphism a: k' .::; kIf 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, Xl, X2, ••• , in k in the v' -topology: X,

= l'1m x n • n-+ oo

Then v'(x') = lim v'(x n) = lim v(xn). n-+oo

n-+ OO

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

f' = f, so that e(v'/v) = !(v'/v) = 1

in this case. It is also clear that if (k', v') is a completion of (k, v) and if Jl = av, a>O, then (k', Jl'), with Jl' = av', is a completion of (k, 11)' 4. Let K be an extension of k and let Jl be an extension of v on k to the extension field K: JlI k = v. Suppose that Jl is complete. Let k' denote the closure of k in K in the Jl-topology. Then k' is a subfield of K, and (k', v'), with v' = Jl I k', is a completion of (k, v).

EXAMPLE

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'(k'X) = v(kX) = Z by t Some authors call a field a complete field if it is associated with a complete valuation, not necessarily normalized.

8


Section 1.1. Since v'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 v', 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 vp(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

where - 00 «n indicates that there are only a finite number of terms an Tn with n < 0, an =1= O. Then k = F((T» is an extension field of F in the usual addition and multiplication of Laurent series. Let yeO) = + 00 and let vex) = i if 00

x

=1=

0,

x

= L.J ~

an Tn ,

with ai =1= 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]]/TF[[T]] = F, and veT) = 1. The field k contains the subfield F(T) of all rational functions of T with coefficients in F, and the restriction v I F(T) is the normalized valuation ofF(T) in Example 3, Section 1.1. Furthermore, (k, v) is the completion of (F(T), v I F(T».

Now, let (k, v) be any complete field and let 0, lJ, and f = o/lJ 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(kX) = Z, there exists an element n in k such that v(n) = 1.

Any such element n is called a prime element of (k, v). Fix n. Since lJ = {x E k I vex) ~ I} in this case, lJ

= (n) = on.

Hence, for any integer n :> 0,

lJn = (nn) = onn = {x

E

k I v(x):> n}.

In general, let a be an o-submodule of k, different from {O}, k. As a =1= k, the set {v(x) Ix E a, x =1= O} is bounded below in Z, and if n denotes the

9

Valuations

minimum of the integers in this set, then

a = {x

E

k I v(x):> n} = onn.

Such an o-submodule a of k, a =1= {O}, 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 lJ. An ideal of (k, v), contained in 0, is nothing but a non-zero ideal of the ring 0 in the usual sense. Hence the sequence {O} c ...

C

lJn

C

•••

C

lJ2 C lJ

C

lJo =

0

(1.1)

gives us all ideals of the ring o. Since lJn = (nn), 0 is a principal ideal domain. Furthermore, in this case v(x):> n ~ vex) > n -1 for any n E Z. Hence all the ideals lJ n , n :> 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 lJn = (nn) =1= {O}, the field k is totally disconnected and non-discrete as a topological space. Next, we consider the multiplicative group k X of k. Let (n) denote the cyclic subgroup of k x , generated by a prime element n. Then the isomorphism k XI V.:; v(kX) = Z in Section 1.1 implies k = (n) x V,

(n)=Z.

Let VO= V,

and let f+ and fX denote the additive group and the multiplicative group of f = 01 lJ, respectively. Then we have a sequence of subgroups of k x:

{I}

C

.•• C

Vn

C

••• C

VI

C

Vo = V

C

kx

(1.2)

such that (1.3) In fact, the canonical ring homomorphism o~ f = o/lJ induces a homomorphism of multiplicative groups, V ~ f X , which in turn induces Vol VI':; fX. On the other hand, for n:> 1, if we write an element of Vn = 1 + lJn = 1 + onn in the form 1 + xnn with x E 0, then the map 1 + xnn mod Vn+l ~x mod lJ defines an isomorphism Vnl V n+1 .:; f+. The multiplicative group k X is a topological abelian group in the topology induced by the v-topology of k. The groups Vn , n :> 0, are open and closed subgroups of k x 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/lJ in o-that is, a subset of 0 such that each residue class of 0 mod lJ contains a unique element in A. We assume that A contains 0, namely, that 0 is the representative of lJ in A. For

10


each nEZ, fix an element nn in k such that venn) = n,

and consider an infinite sum of the form

where the an's are elements of A and the sum over - 00 «n means the same as in Example 6 above. Since v is complete and since v(annn) = v(a n) + v(nn):>n so that v(annn)~+oo as n~+oo, 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=

2: -oo«n

If x

=1=

annn,

with an EA.

0 and if aj =1= 0, an = 0 for all n < i, then vex) = i.

(ii) Let

Then, for any integer i, vex - y) :> i ~ an = b n for all n < i. Proof. We first prove (ii). If an = b n for all n, then x = y, vex - y) = + 00, and statement (ii) is trivial. Hence, assume that there is an integer m such that am =1= b m, an = bn for all n < m. Then 00

x- y=

2:

n=m

(an - bn)nn·

Therefore, for n > m, v((an - bn)nn):> venn) = n > m and it follows that vex - y)

= v((a m -

= v((am -

bm)nm),

bm)nm) = m.

Statement (ii) is then obvious. The argument also proves the uniqueness and the formula vex) = 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 =1= 0, v(x) = i < + 00. Now, by the definition of A, 0=

A

+ lJ = {a + lJ I a E A}.

= {x E k I v(x) :> n} = onn for nEZ, it follows that lJn = Ann + lJ n+1 = Ann + Ann+l + ... + Anm + lJm+l, for all m

Since lJn

:>

n.

11

Valuations

As X E lJ i , we see that there exists a sequence of elements ai, ai+l, ai+2, ... , such that

10

A,

j

x=

2: anJrn mod pi+I,

for any j

n=i

:>

i.

It then follows that j

X

= lim j __ oo

oc

•

2: anJrn = n=i 2: anJrn-

n=i

oo

Let A denote the set of all sequences (ao, aI, a2, ... ), where an are taken arbitrarily from the set A defined above. Thus A is the settheoretical direct product of the sets An = A for all n :> 0: 00

Introduce a topology on A oc as the direct product of discrete spaces An, n:>O. COROLLARY OF PROPOSITION

1.2.

The map 00

(ao, aI, a2, ... ) ~

2: anJrn

n=O

defines a homeomorphism of A onto the valuation ring 0 of (k, v). Proof. (i) shows that the map is bijective and (ii) implies that it is a homeomorphism. • 00

Let Jr be a prime element of k : v( Jr) = 1. Then we may choose Jrn as Jr n in the above proposition. Hence we see that each x E k can be uniquely expressed in the form x = 2: anJr n, with an E A, (1.4) -oc«n

and that if x

=1= 0,

ai =1= 0, and an = 0 for n < i, then vex) =

V(~i anJr n ) = i.

It follows in particular that

EXAMPLE 7.

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

so that each x

E 0 =

= {O, 1, ... , p -

I},

Jr

=p

Zp; that is, each p-adic integer x can be uniquely

12


expressed in the form 00

x=

2:: anpn,

n=O

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, For x

EO,

JC=

T.

the expansion

mentioned above, is then nothing but the formal expression for x as an element of the power series ring 0 = F[[ T]]. Let k be a field with a non-trivial, discrete, complete valuation J-l on it. Then J-l is equivalent to a unique normalized valuation v of k: J-l ~ v, and v is again complete so that (k, v) is a complete field. Since the valuation ring, the maximal ideal, the residue field, and so on, are the same for J-l and v, to study a complete field is essentially the same as investigating a field k with such a valuation J-l. REMARK.

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 v' on k is equivalent to

v: kck' ,

In such a case, we shall also say that (k', v') is a complete extension of (k, v), or, in short, that k'ik is an extension of complete fields. Let (k, v) and (k', v') be as above and let J-l = v' I k,

e = e(v' IJ-l),

f = fey' IJ-l).

e and f are then denoted also by e(k' Ik) and f(k' Ik), respectively, and they are called the ramification index and the residue degree of the extension k' Ik of complete fields: f = f(k' Ik) = fey' IJ-l). e = e(k' Ik) = e(v' IJ-l), Let J-l = av, a> O. Then J-l(k X) = av(kX) = aZ so that e = [v'(k'X): J-l(k X)] = [Z: aZ] = a. Therefore

v' I k

=

ev.

This characterizes e = e(k' Ik) and it also proves that e(k'lk) re~v(xi):> r for i = 1, ... ,n.

For r = 0, this means that {aI' ... , an} is a free basis of the o-module 0': 0' = Oll'l EB ... EB o an' Let k n (resp. on) denote the direct sum of n copies of k (resp. 0) 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 k n :::;. k' , n

(Xl, ...

,Xn )

~x' =

2: Xill'i, i=l

is a topological isomorphism and it induces a topological o-isomorphism (1.8) Let (k, v) and (k', v') be the same as in Proposition 1.5, and let 0 and 0' be their valuation rings. Then the trace map and the norm map of the finite extension k' / k: PROPOSITION

1.6.

Tk'lk:k'~k,

Nk'lk:k'~k,

are continuous in the v-topology of k and the v'-topology of k', and T k'lk(O') co, Proof.

Let X'

E

k' and let n

x' a·I = L.J ~

X··ll'· I] ] '

j=l

for the basis {a}, ... , an} stated above. Then x ij depends continuously on x' E k'. Since Tk'lk(X') and Nk'lk(X') are the trace and the determinant of the n x n matrix (Xij), respectively, the first half is proved. If x' EO', then x ij EO for all i, j by (1.8). Therefore, the second half is also clear. • Now, let 0' be an ideal of (k', v')-that is, an o'-submodule of k' different from {O}, k' (cf. Section 1.2). We define the norm Nk'lk(O') of

17

Valuations

a' to be the o-submodule of k generated by Nk'lk(X') for all x' E a'. Then Nk'lk(a') is an ideal of (k, v) and it follows from the formula for v'(x') in Proposition 1.5 that if a' = ~'m, m E Z, then Nk'lk(a') = ~mf,

Hence, in particular,

with f = f(k' Ik).

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/~ is a finite field. Let q be the number of elements in {-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

c~,

that is, v(p . 1k ) > 0

where 1k 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/~ =

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, VT) is a complete field and its residue field is f = o/~ Therefore (k,

VT)

= F[[T]]/TF[[T]] = F.

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 0 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

2.1.

Let (k, v) be a local field. Then k is a non-discrete, totally disconnected, locally compact field in its v-topology. The valuation ring o( = ~O) and the ideals ~n of 0, n:> 1, are open, compact subgroups of the additive group of the field k, and they form a base of open neighborhoods of in k. Furthermore, 0 is the unique maximal compact sub ring of k. Proof. Let A be a complete set of representatives of f = o/~ in 0, PROPOSITION

°

°

containing (cf. Section 1.2). Since f is a finite field, A is a finite set. Hence the set A = n~=o An, An = A, in the Corollary of Proposition 1.2 is a compact set as a direct product of finite sets An, n:> 0. Therefore, by the same Corollary, 0 is compact in the v-topology. We already stated in Section 1.2 that each ~n, n :> 0, is at the same time open and closed in k and that the ~n's in (1.1) gives us a base of open neighbourhoods of in k so that k is a non-discrete, totally disconnected topological field. Since 0 is compact, k is locally compact, and since ~n is closed in 0, ~n is also compact. Let R be any compact subring of k. Then the compactness implies that the set {v(x) I x E R} is bounded below in the real field R. If x E R, then xn E R and v(xn) = nv(x) for all n :> 1. Hence, by the above remark, vex) :> O-that is, x E o. This proves that R coso that 0 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 n be a prime element of a local field (k, v): v(n) = 1. Since n ~n = onn, n:> 0, the map x ~ xn , x E 0, induces an isomorphism of o-modules: o/~ ~ ~n /~n+l,

Since f = Fq ,

[o:~] =

for n :> 0.

q, it follows that [o:~n]=qn,

forn:>O.

Thus o/~n is a finite ring with qn elements. Since intersection of all ~n, n :> 0, is 0, we see that

0

is compact and the

where the inverse limit is taken with respect to the canonical maps o/~m~ o/~n for all m:> n:> 0. Therefore, if (k, v) is a p-fie1d so that q is a power of p, then the additive group of 0 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/~ = Fq for a p-field (k, v). Then, for each x

E 0,

the

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

20


limit w(x) = lim xqn n-+oo

exists in

0,

and the map w: 0 ~

w(x) =x

mod~,

has the following properties:

0

w(x)q = w(x),

w(xy) = w(x)w(y).

By induction, we shall prove the congruences n n-l x q =x q mod ~n

Proof.

for all n :> 1. For n = 1, x q = x mod ~ follows from the fact that f = o/~ is a finite field with q elements. Assume the congruence for n:> 1 so that 1 xqn = x qn - + y with y E ~n. Then x qn + 1 =

q

2: (~)xiqn-lyq-i. l

i=O

For 0 < i < q, the integer

1) ( ~) = ~ (~1-1 l

I

is divisible by p so that (nyq-i is contained III ~n+l. Since the same is obviously true for i = 0, we obtain x qn + 1 xqn mod ~n+l.

=

Now, we see from these congruences that {xqn} n~l is a Cauchy sequence in o in the v-topology. As v is complete and 0 is closed in k, the sequence converges to an element w (x) in o. It is clear that the congruences yield xqn = x mod ~ for all n :> 1. Hence w (x ) =x mod ~. We also see 1 w(x)q = lim x qn + = w(x), w(xy) = lim xqnyqn = w(x)w(y). • n-+ cc

PROPOSITION

V

n-+cc

2.3.

=

Let (k, v) be as stated above and let q q {x E k I x - 1 = 1}, A = V U {O} = {x E k I x =x}.

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

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

=

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

21

Local Fields

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

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

REMARK.

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 = pI, for the residue field f = o/~. Then (k, v) is a complete extension of the p-adic number field (Qp, v p ), and PROPOSITION

[k : Qp] = ef < + 00. Furthermore, the valuation ring ef=[k:Qp]. Proof.

0

of (k, v) is a free Zp-module of rank

Let A= v I Q. Since (k, v) is a p-field and p = p . 1k

* 0,

0< A(P) = yep) < +00. However, it is known that such a valuation A on Q is equivalent to the p-adic valuation vp 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 ~ vp that k' = Qp and v I Qp ~ vp. Hence (k, v) is a complete extension of (Qp, v p) in the sense of Section 1.3. Since vp(P) = 1, v(P) = e, and v I Qp = e(kIQp)vp, we obtain e(kIQp) = e. On the other hand, f = Fq, q = pI, and ZplpZp = Fp imply f(kIQp) = f < +00. Therefore, [k: Qp] = ef by Proposition 1.4. That 0 is a free Zp-module of rank ef is a special case of (1.8) in Section 1.4. • 2.5. Let (k, v) be a p-field of characteristic p with f = o/~ = Fq and let (Fq(( T)), V T) denote the p -field of Laurent series in T stated in Example 1. Then k contains Fq as a subfield and there exists an Fqisomorphism PROPOSITION

(k, v):; (Fq((T)), v T ), namely, an Fq-isomorphism k :;Fq((T)), which transfers the valuation v of k to the valuation VT of Fq((T)). q Proof. Since k is a field of characteristic p, the set A = {x E k Ix = x} in Proposition 2.3 is a subfield of k with q elements. Then F q = A c k. Now, fix a prime element Jt of k and apply Proposition 1.2 for A = Fq and n Jrn = Jr , n E Z. We then see that the map n

n

defines an Fq-isomorphism (k, v) :;(Fq((T)), t Compare van der Waerden [23].

VT).

•

22


2.2. The Multiplicative Group k

X

Let (k, v) be again a p-field with residue field f = o/~ = Fq • We shall next study the multiplicative group k x of the field k. 2.6. The group k X is a non-discrete, totally disconnected, locally compact abelian group in the topology induced by the v-topology of k. The unit group V (= Vo) and its subgroups Vn = 1 + ~n, n:> 1, are open, compact subgroups of k X , and they form a base of open neighbourhoods of 1 in kX. Furthermore, V is the unique maximal compact subgroup of kX.

PROPOSITION

Proof. By Proposition 2.1, Vn = 1 + ~n, n:> 1, are open, compact subgroups of kX. Since VI VI = fX by (1.3), VIV1 is a finite group. Hence V is also an open, compact subgroup of kX. That V is the unique maximal compact subgroup of k X can be proved similarly as the fact that 0 is the unique maximal compact sub ring of k. The rest of the proposition is clear • from what we mentioned in Section 1.2.

Now, let Jr be a prime element of (k, v). By Section 1.2, we have (Jr)=Z, VI VI = f

X ,

Since f = F q' it follows that [ V : VIl = q - 1,

so that (2.1) We also know that the canonical ring homomorphism o~ f = o/~ induces the isomorphism V I VI::; fX. However, by Proposition 2.3, the same ring q 1 homomorphism induces V::; f X , where V = {x E k I x - = 1} is a subgroup of V = {x E k I vex) = O}. Hence we obtain V= V x VI,

k X = (Jr)

X

V X VI,

(2.2)

where (Jr) = 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 VI· In general, let G be a finite abelian p-group. Then G can be regarded as a module over Zlpnz whenever n is sufficiently large. As Zlpnz::; Zplpnzp, we see that we can canonically define a structure of Zp-module on such a group G. Let G, now, be an abelian pro-p-group-that is, G = lim G;, ~

where {G;} is a family of finite abelian p-groups. Since every G; is a Zp-module in the natural manner, the inverse limit G can also be made into a Zp-module, and one checks easily that the structure of Zp-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 Zp-module in a canonical manner. Now, just as 0 is the inverse limit of o/pn, n > 0, the compact group VI is the inverse limit of finite abelian p-groups V 11Vn with respect to the canonical maps VI/Vm~ V 1 /Vn for m >n >0:

Thus VI is a Zp-module by the above remark, and one also sees that the Vm n > 1, are Zp-submodules of VI. We shall next study the structure of the Zp -module VI. Let (k, v) 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(kIQp) Let

= v(P), and where f = f(kIQp) is the exponent of q = pl.

W = the set of all roots of unity with p-power orders in k. Since (2.2) induces k X I VI = Z EB Z/(q - l)Z, we see that W is a subgroup of VI; in fact, W is the torsion submodule of the Zp-module VI. 2.7. Let (k, v) 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 VI = 1 + p and VIIW is a free Zp-module of rank d = [k: Qp]: PROPOSITION

V 1 1W = Z:'

UI = W EB Z:,

W = ZlpaZ,

a > O.

Proof. Let n > e = v(p) and let U~ = {x P I x E Un}. (U~ is not the direct sum of p-copies of Un-) U~ is the image of the compact group Un under the continuous endomorphism x ~ x P so that it is a compact subgroup of VnNow, since n > e, U~ =

where ppn = pn+e, p2n

C

(1 + pny = 1 + ppn mod p2n,

pn+e+l. Hence we have

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

Vn+e

= U~Um'

for all m

>

n + e + 1.

This shows that Un+e is the closure of the compact subgroup U~ of Vn so that Un + e = and it follows that

V~,

for n

>

e,

24


In general, it is easy to see (the proof is left to the reader as an exercise) that if A is a compact Zp-module (in additive notation) such that A/pA is finite, then A is finitely generated over Zp. Since

[VI: VI{]

[VI: V~] = [VI: Vn][Vn : V~] < +00,

0, d = [k: Qp]. X

Hence V x W is the group of all roots of unity in k and it is a finite cyclic group of order (q - l)pa. REMARK.

The proposition can be proved also by using the p-adic logarithm

for k:t x2

x3

10g(1 +x) =x -~+3-

"',

We now consider the Zp-module VI = V1(k) for a p-field (k, v) of characteristic p. By Proposition 2.5, we may assume that

where q=pt, f=[Fq:Fp]:> 1. Let {WI,"" Wt} be a basis ofFq over the prime field Fp' For each integer n :> 1, prime to p, let An denote the direct product of f copies of Zp. Take any element a- in An: a- = (a 1, . . . , at), ai E Zp' We define an element gn(a-) in Vn = 1 + pn by t

gn(a-) =

IT (1 + WiTnt;· i=1

Let bi E Z, ai

=bi modpZp for 1 - 0 and f3n E An, f3n f pAn· It then follows from (2.4) that

*

gn( a'n)

d: 'f

Um+l,

*

for m = m(l'Vn) = nps. l.4

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

*

*

gn'( a'n')

E

Vm+l1

*

*

*

with m = m( a'n) < m( a'n')'

Therefore, and this proves that g is injective. Thus g:A ~ VI' Since A is obviously the direct product of countably infinite copies of Zp, the proposition is proved. • By (2.2) and Propositions 2.7 and 2.8, the structure of the multiplicative group k X of a local field k is completely determined.

2.3. Finite Extensions Let (k, v) be a local field with residue field f = olp = F q • Let k' be any finite extension over k. By Proposition 1.5, there is a unique normalized valuation


26

v' on k' such that v' I k ~ v, and (k', v') is then a complete extension of (k, v) with

[k': k] Let f' =

0' Ip'

e

= ef,

= e(k' Ik),

f

= f(k' Ik).

be the residue field of (k', v'). Then

[f': f] = f(k' Ik) - 1, there exists an unramified extension k'ik 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 xqn - 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'lk) induces an automorphism a' of f'/f, and the map a ~ a' defines an isomorphism PROPOSITION

Gal(k' Ik)

~

Gal(f'If).

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[ Xl 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) = X" + ~?==-01 aiXi, ai EO, then w" = - ~?==-01 aiw i, ai EO,

28


and this implies that v'(w) >- O-that is, WE 0', 0' being the valuation ring of (k', v'). Hence, let w denote the residue class of W in f' = 0' 1'r,J' for (k', v'). Then clearly g(w) = O. As g(X) is irreducible in f[X], it follows that [f( w): f] = degg(X) = n. On the other hand, f(w) = 0 implies [k': k] = [k(w): k] - 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 w ~ wq,

for w E f'.

Let qJ denote the automorphism of k' I k, corresponding to the above automorpl.ism of f'/f under the isomorphism Gal( k' I k) ~ Gal(f'If) in Proposition 2.11. Then qJ is a generator of the cyclic group Gal(k'lk) and it is uniquely characterized by the property that qJ (y )

- yq mod 'r,J' , for all YEO' .

This qJ is called the Frobenius automorphism of the unramified extension k' I k; it will play an essential role throughout the following. Next, let k'ik 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, q' =qf. 2.12. The splitting field ko of xq' - X over k in Lemma 2.10 is the unique maximal unramified extension over k, contained in k'. k' Iko is a totally ramified extension and PROPOSITION

[k':k o] = e,

[ko:k] = f.

Proof. We already know that kol k is an unramified extension with [k o:k] = f. Let k 1 be any unramified extension of k, contained in k', and let fl =f(k 1 Ik) = [k 1 :k], q"=qf By Proposition 2.11, kl is the splitting field, in k', of the polynomial xq" - X over k. As f = f(k' Ik) = f(k' Ik 1 )f(k l lk), fl is a factor of f so that q" - 1 divides q' - 1, and xq" - X divides xq' - X in k[X]. Therefore, kl is contained in the splitting field ko, in k', of xq' - X over k. This proves that ko is the unique maximal unramified extension of k, contained in k'. Now, f(kolk) = [k o :k] = f = f(k' Ik) implies f(k'lk o) = 1namely, that k' Iko is totally ramified. As ef = n = [k': k], f = [k o:k], we also have [k' : ko] = e. • l•

Local Fields

29

The field ko in the above proposition is called the inertia field of the extension k' I k.

2.4. The Different and the Discriminant Let k'ik still be a finite extension of local fields and let f = o/p, f' = 0' Ip', and so on, be defined as above. Throughout this section, we assume that k'ik is a separable extension. By Proposition 1.6, the trace map Tk'lk: k' ~ k is continuous and Tk'lk(O') co. Let T = Tk'ik for simplicity, and define m = {x'

E

k' I T(x'o') co}.

Since k'ik is separable, there exists an element x' E k' such that T(x') =f=. O. Hence T(k') = k, and it follows from the definition that m is an 0'submodule of k', different from k'. Furthermore, T(o') co implies that 0' em, m =f=. {O}. Therefore m is an ideal of (k', v') (cf. Section 1.2), containing 0' so that its inverse Tk"lk(ZO")

CO¢:> Tk'lk(Tk"lk'(ZO")O') C 0

= nlUo' ¢:> Tk"lk,(n,-azo") c ¢:> n,-az Em' ¢:> Z E nlUm' = mm'. ¢:> Tk"lk'(ZO") em

0'

•

Therefore, m" = mm'-namely, Cjj)(k"lk) = Cjj)(k"lk')Cjj)(k'lk).

We now consider the different Cjj) (k' I k) in the case where k' I k is either unramified or totally ramified. Suppose first that k'ik is unramified. Let w and f(X) be as in Proposition 2.14 and let w denote the residue class of w mod p': W E f' = 0' Ip'. Since 0' = 0 ED ow ED ... ED ow n - 1 and [f': f] = [k' : k] = n, we see that f' = f( w) and the minimal polynomial g(X) of w over f is the reduction of f(X) mod p. As f'/f is separable, g'( w) =f=. O. Hence !' (w) ::t= 0 mod p', and it follows from Proposition 2.14 that Cjj)(k'lk) = 0'.

Next, let k'ik be totally ramified and let e = n = [k': k] > 1. By the Corollary of Lemma 2.13, we have then 0' = 0 ED on' ED ... ED on ln - 1 for any prime element n' of (k', v'). Let n-l

f(X)

= xn + 2:

aiXi,

ai E

0

i=O

be the minimal polynomial of n' over k. From f(n') = 0 and v' I k = ev = nv, we then see that v(a;) > O-that is, ai E p, for 0 - O. Furthermore, if a =f=. 1, then there exists y in 0' such that a(y) =f=. y, so a(y) =1= y mod pin + 1 for sufficiently large n. Therefore, Gm = 1 for all large m, and we have a sequence of normal subgroups of G:

1 = ... = Gm

C

Gm -

1 C

••• C

G1 C Go c G.

These groups Gn are called the ramification groups (in the lower numbering) of the Galois extension k' I k. 2.18. Let ko be the inertia field of the extension k'ik (cf. Proposition 2.12). Then PROPOSITION

GIGo = Gal(kolk) ~ Gal(f' If)

Go = Gal(k' Iko),

so that G IGo is a cyclic group and [G: Go] = f.

[Go: 1] = e,

Proof. Each a in G induces an automorphism a' (= ao) of f' = 0' Ip', and a' obviously fixes elements of f. Hence a ~ a' defines a natural homomorphism G = Gal( k' Ik) ~ Gal(f' If), and by definition, Go is the kernel of this homomorphism. For the Galois extension kol k, we have a similar natural homomorphism Gal( kol k) ~ Gal(fo/f), where fo denotes the residue field of ko. Since k c ko c k', f c fo c f', there also exist canonical homomorphisms Gal(k' Ik)~ Gal(kol k) and Gal(f' If)~ Gal(fo/f) , and those maps define a commutative diagram

Gal(k' Ik)

~

Gal(f'/f)

~

Gal(fo/f).

lp Gal(kolk)

1

However, by Proposition 2.12, f(k'lk o) = 1, f' = fo and by Proposition 2.11, Gal( kol k) ~ Gal(fo/f). Hence Go = Ker( a-) = Ker(fJ) = Gal(k' Iko).

34


The rest of the proposition then follows immediately from Propositions 2.11 • and 2.12. 2.19.

PROPOSITION

For each n::> 0, there exists an injective homomorphism Gn1Gn+1~ V~/V~+b

where

V~ =

V' is the unit group of k' and V;

= 1 + lJ,i

for i::> l.

Proof. Let J'C' be any prime element of k'. Then a E Gn implies a(J'C') = J'C' mod lJln+l so that a(J'C')J'C,-1 = 1 mod lJln-that is, a(J'C')J'C,-1 E V~. If J'C" is another prime element of k', then J'C" = J'C' u with u E V', and a(u) = u mod lJln+l implies a(u)u- 1= 1 mod lJln+l so that a(J'C')J'C,-1 = a(J'C")a,,-1 mod Hence the map

V~+I.

An: Gn ~ V~I V~+I' a~ a(J'C')J'C,-1 mod V'n+l

is independent of the choice of the prime element J'C'. Let r E Gn. Then J'C" = r(J'C') is another prime element of k'. Hence, it follows from

are J'C ')J'C ,-I = ( a( J'C")J'C,,-I)( r( J'C ')J'C ,-1) that

An(ar) = An(a)An(r),

namely, that An:Gn~V~/V~+1 is a homomorphism. Let An(a) = 1 for a E Gn. This means a(J'C')J'C,-1 === 1 mod lJln+l so that a(J'C') = J'C' mod lJln+2. Let ko be as in Proposition 2.18 and let 0 0 be the valuation ring of k o. Since k' Iko is totally ramified by Proposition 2.12, it follows from the Corollary of Lemma 2.13 that o'=oo[J'C']. Hence, for aEGncGo=Gal(k'k o), a(J'C') = J'C' mod lJln+2 holds if and only if a(y) = y mod lJln+2 for every y E o'-that is, a E Gn+1. Thus Ker(An) = Gn+1, and An induces an injective • homomorphism GnIGn+l~ V~/V~+I.

A finite Galois extension k' Ik of local fields is always a solvable extension-that is, Gal(k'lk) is a solvable group. COROLLARY.

Proof. GIGo is cyclic by Proposition 2.18 and Gn1Gn+1 is abelian for n::> 1 by Proposition 2.19. Since Gm = 1 for sufficiently large m, G = Gal(k' I k) is a solvable group. • Now, suppose that k is ap-field so that both q and q' are powers ofp. By (1.3) for (k', v'),

Hence, it follows from Proposition 2.19 that

GOIG1 = a cyclic group, [Go: GIl I (q' - 1), Gn1Gn+1 = an abelian group of type (p, ... , p),

[Gn : Gn+1l I q',

In particular, [Go: GIl is prime to p and [G 1: 11is a power of p.

for n ::> l.

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 infin{te 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-Frohlich [3], Chapter V.

3.1.

Algebraic Extensions and Their Completions

Let (k, v) be a local field with residue field f = o/'p = Fq • Let Q be a fixed algebraic closure of k, and 11 the unique extension of von Q (cf. Proposition 1.1). We denote by (Q, fi) the completion of (Q,Il). Let F be any intermediate field of k and Q: kcFcQcQ.

The closure

F of F in

Q in the fi-topology is a subfield of Q. Let

Then IlF is the unique extension of v on the algebraic extension F over k, and (F, 11ft) is the completion of (F, IlF). 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, 11 F) and (F, 11 ft) as stated above. Let OF = the valuation ring of IlF' 'pF= the maximal ideal of IlF' ff= OF/'pF = the residue field of IlF'

and let Oft, 'pft, and fft be defined similarly for 11ft. Since f(llft/IlF) injection OF~ Oft identifies fF with fp:

= 1,

the

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 IIp-topology, and by continuity, it can be uniquely extended to a topological automorph-

36

ism


a of F in the Ilrtopology. Ilpoa=llp,

and a and

We then have

a(lJF) = lJp,

a(op)=op, a(Oft) = Oft,

a(lJft) = lJft,

a induce the same automorphism of fp =

n ::> 1,

fft over f.

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

c

Fc E

c

Q c Q, and

EF=E. Furthermore, if E/ F is a separable extension, then EnF=F. Proof. Clearly F c EF c E and EF / F is a finite extension. Since Ilft is complete, it follows from Proposition 1.1 that {t I EF is a complete valuation on EF so that EF is closed in Q in the {t-topology. Hence EF = E. Assume now that E/ F is separable. Since E/ F is separable, there is a finite Galois extension E' over F, containing E: F c E c E' , 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 E/ F itself is a finite Galois extension. Then E = EF 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] Since F c E

1, there exists a unique un ramified extension k~r over k in Q with degree [k~r: k] = n, namely, the splitting field of the polynomial xqn - X over kin Q (cf. Proposition 2.11). Since k~r/k is a cyclic extension, one sees immediately that

I

k~r c k'::,.¢:,; n m,

Hence the union kur of all

k~n

for n, m

>

1.

n > 1, is a sub field of Q:

For simplicity, we shall often write K for k ur : K = k ur .

It is clear that kurl k 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( a) is a finite extension of k in F so that k' I k is unramified. Hence k' = k~r for n = [k' : k] and a E k~r c k ur . Thus

k cF ckur . Therefore kur 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~n respectively, and let fK = oKI'pK be the residue field of K = k ur · PROPOSITION 3.3. fK is an algebraic closure of the residue field f (= fd of k. Each a in Gal(kurlk) induces an automorphism a' of fKIf, and the map a ~ a' defines a natural isomorphism

Gal(kurl k) .::; Gal(fKIf). Proof. If min, then k c k'::,. c k~r c kur so that f c fm C fn C fK. Since OK is clearly the union of on for all n > 1, fK is the union of fn for all n > 1. Since [fn : f] = [k~r: k] = n, and since the finite field f has a unique extension with degree n in any algebraic closure, it follows that fK is an algebraic closure of f. Now, it is clear that

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

Gal(fK/f) = lim Gal(fn/f), ~

where the inverse limits are taken with respect to the canonical maps Gal(k'::,.lk)~Gal(k~/k), Gal(fmlf)~Gal(fnlf) for n 1m, m, n > 1. As explained in general in Section 3.1, each a in Gal( kurl k) induces an automorphism a' in Gal(fKlf). However, by Proposition 2.11, the map a~ a' induces an isomorphism Gal(k~rlk).::; Gal(fn If) for each n > 1.


38

Hence it follows that f).

0 ~ 0'

defines an isomorphism Gal(kurl k) .::; Gal(fKI ..

Since f = Fq , the map (jJ ~ (jJq, (jJ E fK' defines an automorphism of fK over f. Let qJ denote the corresponding element in Gal( kurl k) under Gal(kurlk).::; Gal(fKlf)-namely, the unique element in Gal(kurlk) satisfying

As qJ is uniquely associated with the local field k (with Q fixed) in this manner, it is called the Frobenius automorphism of kurl k, or, of k, and is denoted by qJk. It is clear that qJk induces on each k~n n > 1, the Frobenius automorphism qJn of k~rlk (cf. Section 2.3). Since Gal(k~rlk) is the cyclic group of order n generated by qJn, the map a mod n ~ qJ~, a E Z, defines an isomorphism ZlnZ.::; Gal(k~rlk).

For n I m, let ZlmZ~ ZlnZ be the natural homomorphism defined by a mod m ~ a mod n, a E Z, and let

Z=

lim ZlnZ

~

with respect to those maps for n I m. Since the diagram

qJm

Ikn =

qJk

Ik n =

qJn

for n I m,

ZlmZ ~ Gal(k:"lk)

1

1

ZlnZ ~ Gal(k~rlk)

is commutative. Hence we obtain an isomorphism of profinite (totally disconnected, compact) abelian groups: (3.1) Now, the natural homomorphisms Z~ ZlnZ, n > 1, induce a monomorphism Z ~ Z 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 qJk of k so that (3.1) induces

n~qJZ

between the subgroups. Since Z is dense in Z, the cyclic group (qJk > is dense in Gal(kurlk). Hence k is the fixed field of qJk in kur and the topological isomorphism (3.1) is uniquely characterized by the fact that 1 ~ qJk. REMARK.

For each prime number p, let Z; denote the additive group of

Infinite Extensions of Local Fields

39

all p-adic integers. It is easy to see that Z is topologically isomorphic to the direct product of the compact groups Z; for all prime numbers p. Let k be a p-field so that q is a power of p, and let Voo be the multiplicative group of all roots of unity in Q with order prime to p. For n > 1, let Vn denote the subgroup of all (qn - 1)st roots of unity in Q. Then

Since k: r is the splitting field of

xqn - X

over k in

we have

Q,

(3.2) On the other hand, we see from Proposition 2.3 that the canonical ring homomorphism OK~ fK = oKI'pK induces an isomorphism

Voo:; f~.

(3.3)

Hence it follows from qJk( TJ) = TJ q mod 'p K that qJk( TJ) = TJ q ,

for TJ

E

(3.4)

Voc.

The above equality also uniquely characterizes the Frobenius automorphism qJk 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'kur = k'(Voo). Hence k'kur is the maximal unramified extension k~r over k' in Q: I

I

Let qJk' be the Frobenius automorphism of k' and let 1= I(k ' Ik). Then fl=Fq , with q'=qffor the residue field f' of k'. Hence qJk'(TJ) = TJ q '= TJ q ! for all TJ E Vx , and it follows that qJk,l kur

= qJr, with I = I(k ' Ik).

Now, let F be an algebraic extension of k in Q: k c F c Q. Similarly as for unramified extensions, we define F Ik to be a totally ramified extension if every k such that k c k C F, [k k] < + 00, is a totally ramified extensionthat is, I(k'lk) = 1. Clearly, if k c F' c F and Flk is totally ramified, then F'lk is also totally ramified. Let k' be any finite extension of k in Q and let ko denote the inertia field of the extension k Ik (cf. Proposition 2.12). Then ko = k'n kur and [k o: k] = I( k Ik) by Proposition 2.12. Therefore k Ik is totally ramified if and only if k'n kur = k. It follows that in general F Ik is totally ramified if and only if I

I

I :

I

I

I

F n kur = k. Let L be an algebraic extension of k such that

L = Fk un

k = F n k ur .

Then LI F is a Galois extension, and the restriction map a ~ a I kur defines

40


an isomorphism

Gal(L/ F).::; Gal(k ur / k). Hence there exists a unique element ljJ in Gal(L/ F) such that ljJ I kur = qJb the Frobenius automorphism of k. In other words, qJk has a unique extension ljJ in Gal(L/ F). Let F c F' c L. Then it follows from the above isomorphism that

[F'

n kur:k] =

[F' :F).

Hence F' / k is totally ramified only when F' = F, and we see that F is a maximal totally ramified extension over k contained in L. 3.4. Let E be a Galois extension over k, containing k ur . Let ljJ be an element of Gal(E/k) such that ljJ I kur = qJk and let F be the fixed field of ljJ in E. Then LEMMA

Fk ur = E,

F

n kur =

k,

Gal(E/ F).::; Gal(kur / k).

In particular, F is a maximal totally ramified extension over k in E. Proof Clearly F n kur is the fixed field of qJk = ljJ I kur in k ur . Hence F n kur = k. Let M be any field such that F c M c E,

= n < +00, F n kur = k, we have [M : F)

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

3.3. The Norm Groups For each algebraic extension F / k, k c F c Q, let U(F) denote the unit group of F: U(F) = Ker(IlF : F X~ R+). 3.5. Let k' /k be a finite extension: k c k' c Q. Then Nk'lk(U(k')) is a compact subgroup of U = U(k), and Nk'lk(k'X) is a closed subgroup of kX. Proof By Proposition 1.6, the norm map Nk'lk:k'~k is continuous, and by Proposition 2.6, U(k') is compact. Hence Nk'lk(U(k')) is a compact subgroup of k X and is closed in kX. Furthermore, the formula for v'(x') in Proposition 1.5 shows

LEMMA

Nk'lk(U(k')) = Nk'lk(k'X) n U(k) c U = U(k). Since Uis open in k X, it follows from the above that Nk'lk(k,X)/Nk'lk(U(k')) is a discrete subgroup of kX/Nk'lk(U(k')). Hence Nk'lk(k'X) is closed in kX. •


41

For any algebraic extension F I k, we now define N(F Ik) =

n N k'lk(k'X),

n N k'lk(V(k')),

NV(Flk) =

k'

k

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

[k': k] < +00.

k ck' cF,

We call N(Flk), and NV(Flk) the norm group and the unit norm group of the extension Flk, respectively. Clearly, if Flk is finite, then N(Flk) = NF1k(F X), NV(Flk) = NFlk(V(F)). In general, it follows from Lemma 3.5 that N(Flk) is a closed subgroup of k X and NV(Flk) is a compact subgroup of V = V(k). In fact, NV(Flk) = N(Flk)

n V(k).

It is clear from the definition that if k c F c F' c Q, then N(F'lk) c N(Flk),

NV(F'lk) c NV(Flk).

Furthermore, if {k i } is a family of fields such that kckicF,

[ki:k] 1. Now, as k'ik is unramified, J'C is also a prime element of k' (cf. Section 2.3). Hence it also defines similar isomorphisms for k': V'IV'1 . . . . f'X , ~

. . . . !til + , V i'IV'i + 1~

s: lor

l• ::> -

1,

where V' = Vb = V(k'). V; = 1 + lJ li, i::> 1. By Proposition 2.11, there is a natural isomorphism Gal(k' Ik) ~ Gal(f'/f). Therefore, we obtain the following commutative diagrams:

llill i i +1

-----==--.,... -----"

*+ !

42


where N = N k'/b and T', N' are the trace map and the norm map of the extension f'lf, respectively. However, since f'lf 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 N k'lk( V')Vj = V,

for all i ::> l.

Since N k'lk( V') is compact, hence, closed, in V, we obtain N k'lk( V') =

u

PROPOSITION

•

3.7.

Let Flk be an unramified algebraic extension. then

NV(Flk) = V(k). If, furthermore, F I k is an infinite extension, then

N(Flk) = V(k). In particular,

N(kurlk) = NV(kurlk) = V(k). Proof. The first part is an immediate consequence of Lemma 3.6. Let k' I k be the un ramified extension of degree n::> 1: k' = k~r. Then a prime element J'C of k is also a prime element of k' so that k'x = \J'C) x V(k'). Hence Nk'lk(k'X) = \J'C n) X Nk'lk(V(k')) = \J'C n) X V(k)

ck x = \J'C) x V(k). If F I k is an infinite extension, there exists k' such that k c k' c F with arbitrarily large n = [k' : k]. Hence the second part is proved. •

3.8. An algebraic extension F I k is totally ramified if and only if N(Flk) contains a prime element of k. Proof. Suppose first that F is not totally ramified-that is, F n kur =1= k. Then k c k~r c F for some n::> 2, and the proof of Proposition 3.6 shows that

PROPOSITION

Since n ::> 2, N(F Ik) contains no prime element of k. Suppose next that F I k is totally ramified. For each finite extension k' I k such that k c k' c F, let S(k') denote the set of all prime elements of k, contained in N(k'lk). Let J'C' be a prime element of k'. Since k'ik is totally ramified, N k'lk(J'C) is then a prime element of k (cf. Section 2.3), and we see S(k')

= N k'lk(J'C' V(k')) = N k'lk(J'C')NV(k' Ik).

Hence S(k') is a non-empty compact subset of the compact set S(k) = J'CV(k), J'C being a prime element of k. Furthermore, if both kUk and k~/k are finite extensions in F, then clearly S(k~k~) c S(k~) n S(k~). Therefore, it follows from the compactness of S(k) that the intersection of S(k') for all k'

43


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

3.4. Formal Power Series In general, for any commutative ring R with identity 1 =f=. 0, let S = R[[Xl1 ... , Xn]]

denote the commutative ring of all formal power series

I(Xt , ... , Xn) =

2: ail, ... ,inX~1 ... X~,

a·lb- .. ,I.n ER '

i

where Xt, ... ,Xn are indeterminates and i = (it, ... , in) ranges over all n-tuples of integers >0. For I, g E S and for any integer d > 0, we write

I=gmoddegd if the power series 1- g contains no terms of total degree less than d. Let Yt , . . . , Ym be another set of indeterminates that let gt, ... , gn be power series in R[[Yt , ... , Ym ]] such that gi = mod deg 1 for 1 O.

(3.5)

Since a E lJg, such sums converge in Og and oc

2: f3i Xi

g(X) =

i=O

•>

satisfies f(X) = (X - a)g(X).

In the above, let a be algebraic over k-that is, a E lJg (a E Q, f-l(a) O)-and let f(X) be a power series in the subring o[[X]] of Og[[X]], 0 being the valuation ring of k. Take a finite extension k' over k such that a E k' c Q. Then k' is a local field and the infinite sum (3.5) converges in the valuation ring 0' of k'. Hence f(X) = (X - a)g(X) holds with g(X) E 0' [[X]].

LEMMA

3.10.

Let h(X) be a monic polynomial in o[X] such that m

h (X) =

n (X -

ai)

i= 1

with distinct aI, ... , am in

lJg.

Let f(X) be a power series in o[[X]] such that


45

f( ai) = 0 for 1 < i < m. Then f(X) = h(X)g(X) with a power series g(X) in o[[X]]. Proof. Let k' = k(al' ... ,an). Then k' is a finite Galois extension over k in Q, and by the above remark, f(X) is divisible by h(X) = n~l (X - ai) in 0' [[X]]-namely , f(X) = h(X)g(X) with g(X) in 0' [[X]]. For each a in Gal(k'lk), let fO(X) denote the power series in o'[[X]] obtained from f(X) by replacing each coefficient a of f(X) by a( a). Then we have fO(X) = h O(X)gO(X), where gO(X) and h O(X) are defined similarly as fO(X). Here, fO = f, h ° = h because f, g E o[[X]] c k[[X]]. Hence it follows from f = hg = hgo that g = gO. As this holds for every a 10 Gal(k'lk), g(X) is a power series in o [[X]]. •

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 ur • Let CPk be the Frobenius automorphism of k : cP k E Gal( K I k), and let ip k be the natural extension of cP k on the closure (completion) K of K in Q (cf. Section 3.1). For simplicity, we shall denote both CPk and ipk by cp. By definition, cP induces the automorphism OJ ~ OJ q on the residue field fK = fk = Ok/'pk so that

a fP ( = cp( a» == a q mod 'pk,

for all a

E

Ok.

Now, let Ok also denote the additive group of the valuation ring Ok and let U(K) be the multiplicative group of units in the complete field K. We consider the endomorphisms cP-1:0k~Ok, a~(cp

cP -

-1)(a) = cp(a) - a,

1: U(K)~ U(K), ~~ ~fP-1 = cp(~)/~.

LEMMA

3.11.

The following sequences are exact: ~o,

1 ~ U(k) ~ U(K)

11'-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, fk (= f K ) is algebraically closed. Hence the maps fk~ fk defined by OJ ~ OJ q - OJ and q 1 OJ ~ OJ are both surjective. This implies that (3.6)

46


Let ~

V(R). We shall define, by induction, a sequence of elements {1/n}n~O in V(R) satisfying E

1/n = 1/n+1 mod lJ7/\

for all n ::> o.

(3.7)

By (3.6), there exists 1/0 E V(R) such that ~ = 1/6- 1 mod lJK. Hence, suppose that we have found 1/0, 1/1, ... ,1/n' n::> 0, which satisfy the required conditions. Let n be a prime element of k. Since Klk is unramified, n is then also a prime element of K and of R: lJk = nOk. Hence, by (3.7): ~1/~-qJ = I + an n+\ with a E Ok. By (3.6), there exists f3

Then, clearly, 1/n+1 n E k, we also have 1/~;1

such that a = (q; -1)f3 mod lJk. Let 1/n+1 = 1/n(1 + f3n n+1).

E Ok

V(R) and 1/n

E

= 1/n+l mod lJK+l.

== 1/~-1(1 + (q;(f3) - f3)nn+1) =

1/~-1(1

SInce q;(n) = n for

+ ann+l) =

~

mod lJl+2.

Thus the existence of the sequence {1/n}n~O is proved. Since R is complete, it then follows that 1/ = lim n _ oo 1/n exists in V(R) and satisfies 1/qJ-l = ~. Hence the exactness of V(R)~ V(R)~ I is verified. It is obvious that V(k) is contained in the kernel of V(R)~ V(R). Suppose that ~qJ-1 = I-that is, q;(~) = ~-for an element ~ in V(R). It follows from (3.3) that the set A = {a} U Voo is a complete set of representatives for fk (= f K ) in Ok. With n as above, it follows from Proposition 1.2 that the element ~ in V(R) C Ok can be uniquely expressed in the form 00

~

=

2: ann n,

n=O

with an

E

A

= {a} U V

x •

Applying q;, we obtain x

~

However, by (3.4), umqueness,

= q;(~) =

q;(a) = aq

E

2:

n=O

A

an = q;(a n) = a~,

q;(an)nn.

for every a EA. for all n

::>

Hence, by the

o.

Therefore, either an = 0 or an is an element of the cyclic group V of order q - I in Proposition 2.3. Consequently, ~ E k n V(R) = V(k), and the exactness of I ~ V(k)~ V(R)~ V(R) is also proved. • Now, for each power series

t(XI ,

... ,

Xn)

=

2: ail ..... inX~l ... X~, i

47


defined by

ffIJ(Xl' ... , Xn)

=

L 1. Then the above equation for a can be written as a - ya fP = f3

f-lk(Y) =

(3.11)

where f3 and yare known quantities. However, since f-lK"CY):> 1, a = f3

+ yf3 11' + Y1 + 11'f3 11'2 + . . .

(3.12)

converges in Ok and it obviously satisfies (3.11). Furthermore, if both a1 and a2 are solutions of (3.11), then a1 - a2 = year - ai), f-l k( al - (2) = f-l k( q;( al - (2» = f-l k( ar - a1'),

so that f-lk(a1 - (2) = +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 Hn+l satisfying (3.10), and so F'n+l satisfying (3.9). Now, starting from F1 = L, we can define successively a sequence of polynomials Fn, n:> 1, in Ok[X1, ... , Xm] such that deg Fn 1, let F~


49

denote the sum of the terms of degree 1. Hence F' = F. • Now, let k' be the unramified extension of degree n over kin Q: k' = k: n k c k' c K c K, and let 0' denote the valuation ring of k'. Let Jl'1 and Jl'2 be prime elements of k'. Since K/k' is unramified, Jl'1 and Jl'2 are also prime elements of K and K. In the above proposition, suppose further that 11,12 E o'[[X]] and L E 0'[X1, ... , Xm]. Since k' is complete and cp(k') = k', the element ll'in (3.11) belongs to 0', and it follows that the power series F in the proposition has coefficients in 0': F(Xl' ... , Xm)

E

This remark will be used often later.

o'[[Xl1

... ,

Xm]].

(3.13)

Chapter IV

Formal Groups Fj(X, 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 =1= 0 and let X, Y, and Z be indeterminates. A power series F(X, Y) in R[[X, Y]] is called a formal group over R if it satisfies the following conditions: (i) (ii) (iii)

F(X, Y) = X + Y mod deg 2, F(F(X, Y), Z) = F(X, F(Y, Z», F(X, Y) = F(Y, X).

Note that (i) implies F(O, 0) = 0 so that both sides of (ii) are well-defined power series in R[[X, Y, Z]]. In the general theory of formal groups, a power series such as F(X, Y) above is called, more specifically, a onedimensional, 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(O, Y) = Y. Therefore, 00

F(X, Y) = X

+Y+

2:

i,j=1

CijXiyj,

(4.1)

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 . . umque power senes 00

iF(X) = -X +

2: biXi,

i=2

bi

E

R,

such that F(X, iF (X) ) = O. t For the theory of formal groups, see Frohlich [8].

(4.2)

51

Formal Groups Ff(X, Y)

For f, gEM = XR[[X]], let

f

+g = F(f(X), g(X». F

t

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 g, with F inverse iF(f) for f We denote it by M F. 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), fey»~.

(4.3)

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

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

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

foF=Gof In general, if F(Xl' ... , Xm) is any power series in R[[Xu ... , Xm]] and if f EM = XR[[X]] is invertible in M : f- 1 EM, then we define a power series Ff (Xl' ... , Xm) in R[[Xu ... , Xm]] by

Ff(Xv ... ,Xm) = foFof-l = f(F(f-\X1 ),

•••

,f- 1 (Xm»).

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

HomR (F, G) = the set of all morphisms f : F ~ G, End R (F) = HomR (F, F). For simplicity, these will also be denoted by Hom(F, G) and End(F), respectively. 4.1. Hom(F, G) is a subgroup of the abelian group M G , and End( F) is a ring with respect to the addition f F g and the multiplication f og. Proof Let f, g E Hom(F, G) and let h = f t g. Then LEMMA

+

hoF = foF

i goF = Gof i Gog = G(Gof, Gog),

where G of = G(f(X), feY»~, Gog

= G(g(X), g(Y». Using (ii), (iii), for G,

52


we obtain

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

= G((f Hence h = f

±g)(X), (f ±g)(Y))

=

Goh.

+g belongs to Hom(F, G). As stated earlier, iG(f) = iGofis the G

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

G(G(X, Y), G(iG(X), iG(Y)) = G(G(X, iG(X)), G(Y, iG(Y)) = G(O, 0) = 0, that is, Go iG = iG 0 G. Hence

iG (f) 0 F = iG 0 f 0 F = iG 0 G of = G 0 iG 0 f = G 0 iG (f)

°

so that iG(f) E Hom(F, G). Since clearly 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, hEM, then

fo (g

+h) = f(F(g(X), h(X))) = F(f(g(X)), f(h(X))) F

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

Let f

E

•

tha t is, F ° = G.

End(F) :fo F = Fo f. Then we have

fOoFo = FOofo,

fO

E

End(G).

Since for f, g

E

End(F), we see that the map f ~ fO defines a ring isomorphism () : End(F) ~ 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. 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 LEMMA

such that A(X) = X mod deg 2. Proof. Let Pi = 8F / 8Y. Differentiating, with respect to Z, both sides of

Formal Groups Fj(X, Y)

53

(ii) in the definition of F(X, Y), we obtain F;.(F(X, Y), Z) = Fl(X, F(Y, Z»Fl(Y, Z).

Since F(X, 0) = X, it follows that F;.(F(X, Y), 0)

= F;.(X,

(4.4)

Y)F;.(Y, 0).

Now, as F(X, Y) = X + Y mod deg 2 implies Fl(X, 0) a power series 1/J(X) in R[[X]] such that

= 1 mod deg 1, there is

00

1/J(X) = 1 +

2: an X" , n=l

Let 00

A(X) = X +

2: an X n,

. han -ER,

WIt

n

n=l

n

so that dAldX = 1/J. Then (4.4) can be written as 1/J(F(X, Y»Fl(X, Y)

8

= 1/J(Y),

8

that is, 8y A(F(X, Y» = 8yA(Y).

Therefore, A(F(X, Y» = 8(X) + A(Y) with a power series 8(X) in R[[X]]. But, putting Y = 0 in this equality, we find that A(X) = 8(X). Thus A(F(X, Y» = A(X) + A(Y),

To see the uniqueness of such A, it is sufficient to consider the case F = Ga. Hence, let A: Ga .:; Ga be any automorphism of Ga , satisfying A(X) = X mod deg 2. Then ,10 Ga = Ga 0 A-that is, A(X) = X mod deg 2.

A(X + Y) = A(X) + A(Y),

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

A'(X) = 1 mod deg 1.

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

•

Thus A is unique.

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. REMARK.

4.2.


Let (k, v~ be a local field with residue field f = o/p = Fq and let Q, Q, K = k un K, fk = Oklpk, CfJ = CfJk = ipb and so on, be the same as in Section 3.5. We shall next define a special type of formal group Fj(X, Y) over

R=

Ok.


54

For each prime element Jr of K, {lk(Jr) = 1, let :¥jf denote the family of all power series f(X) in R[[X]] such that

f(X) = JrX mod deg 2,

f(X) =

xq mod l:'k·

For example, the polynomial JrX + xq belongs to :¥jf' It is also clear that if f E :¥jf, then ffP E :¥ fP(jf)' The union of the sets :¥jf, for all prime elements Jr of K, will be denoted by :¥. 4.2. For each f E :¥jf, there exists a unique formal group Fj(X, Y) over R such that f E HomR(Fj, FJ)-that is,

PROPOSITION

foFj = FJof. Proof. Apply Proposition 3.12 for Jr1 = Jr2 = Jr, f1 = 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 + Ymod deg 2,

(4.5)

Let

F1(X, Y, Z) = F(F(X, Y), Z), Fi(X, Y, Z) = F(X, F(Y, Z». It follows from (4.5) that

l\(X, Y, Z) = F(X, Y) + Z = X + Y + Z mod deg 2, fo F1 = FfP(f(F(X, Y», f(Z» = FfP(FfP(f(X), f(Y», f(Z» = Fro f. Similarly,

F2(X, Y, Z) =X + Y + Z mod deg 2, Hence the uniqueness of Proposition 3.12 with L(X, Y, Z) = X + Y + Z, m = 3, implies F1 = Fi-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, 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 FfP is also a formal group over R. Writing Fj for F, we see that Fj is the unique formal group over R such that fo Fj = FJ 0 f-that is, f E Hom(Fj, Fl)· • Now, it follows from f 0 Fj = Flo f that

ffP 0 FJ = (Fl) fP 0 f CfJ ,

Formal Groups fj(X, Y)

55

and this implies (4.6) Let f be again a power series in :¥ and let a be any element in the valuation ring 0 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 0 cp = cp rp of. We denote this power series cp(X) by [a ]f : cp = [a It· Thus [a It is the unique power series in R [[X]] satisfying [a]f

=aX mod deg2,

fo[a]f=[alJof.

(4.7)

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

o~

Proof.

EndR(Fj).

Let cp = [a ]f. Then f

0

= cp rp f cp = FJ f

cp 0 Fj

f 0 Fj 0 cp 0

0

0

Fj = cp rp 0 FJ 0 f

= (cp cp rp of = (Fj

0

Fj) rp of,

= FJ cp)rp Fj= Fjo cp = a(X + Y) mod deg 2. 0

0

cp

0

0

0

f,

Hence, by the uniqueness of Proposition 3.12, that is, cp = [a It E EndR(Fj).

cp 0 Fj = Fj 0 cp,

Let a, b

E 0,

then

[a ]f

t [b ]f = [a It + [b ]f = (a + b )X mod deg 2.

By the Definition (4.7) for [a

+ b ]f' we then see

[a]f

t [b]f

=

[a

+ b ]f·

Similarly, Hence a ~ [alt defines a ring homomorphism [a]f = aX mod deg 2, the homomorphism is injective.

o~ EndR(Fj).

Since •

Let nand n' be prime elements of K and let f E :¥ll' f' E :¥ll'. We shall next compare the formal groups Fj and Ff' over R = 0 k. Since I-l k( n) = I-l k( n') = 1,

n' = n~,

with ~ E U(K).


56

By Lemma 3.11, there exists an element 1J of U(K) such that ~=1JqJ-l.

Let L(X) = 1JX. Then

Te'L(X)

=

TeL qJ(X).

Hence, applying Proposition 3.12 for fl = f', f2 = f, Tel = Te', Te2 = Te, L(X) = 1JX, m = 1, we see that there exists a unique power series e(X) in R [[X]] such that

f'oe

e(x) = 1JX mod deg 2,

=

Since 1J E U(K), e(X) is invertible in M = XR[[X]]: PROPOSITION

eqJ°f.

(4.8)

e- I EM.

The above e(X) has the following properties:

4.5.

e: Fj ~ Fr ,

that is, F7 = Fr , for a E o.

[a17 = [alr' Proof.

f' eo Fj = e qJ f Fj = e qJ Fl f = (e Fj) qJ f, f' Fj e = Ff f' e = Ff e qJ f = (Fro e) qJ f, eo Fj= Fj e = 1J(X + Y) mod deg 2. 0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

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

eo Fj = Fr 0 e,

=

1J(X + Y),

that is, F7 = Fr.

•

The proof for [a 17 = [a lr is similar.

It follows in particular from the above proposition that the formal groups Fj over R are isomorphic to each other for all power series f in the family ::IF.

The isomorphism e(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 Te = p. Then

f(X)

= (1 + xy -

belongs to the family

~.

1 = pX + (~)X2 + ...

+ XP

Let

F(X, Y) = (1 + X)(1 + Y) - 1 = X + Y + XY. One checks immediately that F is a formal group over Zp, hence over R = Ok, satisfying f 0 F = F qJ 0 f. Therefore,

Fj=F=X+Y+XY. For each a E Zp, define

57

Formal Groups F[(X, Y)

where a) _ a(a - 1) ... (a - n + 1) , EZp (n n.

for n >0.

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

[aIr = (1 + X)" -1 =

~1 (:)X",

for a E Zp.

For n > 0, let (cf. Section 4.3 below) Wi = {a

E lJg

I [pn+lltC a) = O}.

Since [pn+lltC X ) = (1 + xyn+l - 1, Wi = {C - 1 ICE Q, Cpn+l = 1}. Hence Qp(Wi) 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 Qp(Wi) is an abelian extension over Qp and that the Galois group of Qp(Wi)/Qp is naturally isomorphic to the factor group Z;/(1 + pn+lzp). In the following sections, we shall show in general that similar results can be proved for an arbitrary local field (k, v) and for the extension k(Wf), defined by means of the formal group Fj(X, Y) introduced above.

4.3. The o-Modules

Wi

As before, let Og/lJg be the residue field of the completion Q of a fixed algebraic closure Q of k. For simplicity, let m denote the maximal ideal lJg of Og:

m = lJQ = {a: E Q I ji( a) > O}. Let [( be the completion of K = kur and let Fj(X, Y) be the formal group over R = Ok in Section 4.2, associated with a power series f in the family :¥. For a, [3 in m and for any a in the valuation ring 0 of the ground field k, let (4.9)

£1'+[3 = Fj(a, (3), f

[a ltC X) being the power series in (4.7). By the general remark in Section 3.4, both Fj(a, (3) and [altCa) are well defined in oQ. Furthermore, it follows from Fj(X, Y) = 0, [a ltCX) = mod deg 1 that those elements again belong to m c OQ:

°

a

t [3, a a j

E m,

for a, [3 Em, 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 £1'+[3, the inverse of a being f

given by iF(a) for F = Fj in (4.2). By Proposition 4.4--that is, [alt°Fj =

58


~0

[a lr, and so on- we then also see that the map ox

m~m,

(a, a)~aja

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

I

~n+l j a = {a j a a E ~n+l},

for a Emf'

Wi= {aEmf I ~n+lja=O}.

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

{O} = Wjl

C

WJ

C ...

c Wi c ... c

"'I,

where "'I denotes the union of Wi for all n :> -1. Now, let f' be another power series in the family gj. Let 8(X) denote the power series in XR[[X]], satisfying (4.8). Then 8(X) has an inverse 8- 1(X) in XR[[X]] and F! = F r , [a = [a 1r by Proposition 4.5. Hence it follows from (4.9) that 8 defines an isomorphism of o-modules:

1!

8: mf ~ mr, a~

8(a),

which induces o-isomorphisms ll. o.

LEMMA

4.6.

wnf --')"" wnr,

n:> -1.

(4.10)

For f E gj and i :> 0, let F,' -_ fCPi, Ji

F o f i-lo . • • oJ(h F gi=Ji

Then Wi={aEmlgn(a)=O},

forn:>-1.

Since Wjl = {O}, g-l(X) = X, the statement is clear for n =-1. Let n :> O. Let f' and 8 be as stated above so that f' 0 8 = 8 cp 0 f Then we have Proof

' f .00

llcpi _

and

I

llcpi+1

-0

f '' -- 0

/;

llcpi+1

O. "

I

/;

ll-cpi

0.00 "

n:>O.

Since 8(m) = m, 8(Wi) = Wi" we see that if the lemma holds for f, then it also holds for f'. Now, let :re be a prime element of (k, v). Since K / k is unramified, :re is then also a prime element of K and K. Hence f(X) = :reX +

belongs to the family gj. Since :re follows from (4.7) that

[; = f,

EO,

xq

f = :reX mod deg 2 and f 0 f = fCP 0 f, it


59

As lJ n + 1 = oJr n + 1, we obtain

W;

= {a E mf IlJ n + 1 ill' = O} = {a Em I gn(a) = O}.

By the earlier remark, the lemma then holds for an arbitrary f in :¥.

•

Fix an integer m :> 1 and let k' denote the unique un ramified extension of degree mover k in Q:

k '=k mur Let f' = 0' /lJ' be the residue field of k' and let Jr be any prime element of k'. Then

f(X) = JrX + xq is a polynomial in o'[X], belonging to the family :¥. 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' [[X,

Y]],

[a ]AX)

E

o'[[X]],

for a E

o.

(4.11)

Let Then so that

gn(X)

= hn(X)hn-l(X) ... ho(X)X.

4.7. (i) hn(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) gn(X) is a monic separable polynomial of degree qn+l in o'[X] and W; is the set of all roots of gn(X) in Q. Hence the order of is qn+l and k'(Wf) is a finite Galois extension over k'. (iii) Let hn( £1'0) = 0 for n :> 0, £1'0 E Q. Then t:f W n - 1 (q - 1)qn = [k"(a o): k'] < [k'(W;): k'] £1'0 E WI' £1'0 -

1

fJP.

ilK is an abelian extension and there exists a topological

isomorphism

b: U = U(k) ~ Gal(il K) which induces the homomorphism b n : U ~ Gal(in I K) in Proposition 4.11 for each n > o. Proof. It is clear from Proposition 4.11 that b n is the product of b n+1 and the canonical homomorphism Gal(in+l I K)~ Gal(in I K). Hence we


64 have

(): U = lim UIU n+ 1 ~Gal(LIK) = lim Gal(LnIK). ~

~

•

Since the structure of the compact group U = U(k) is explicitly described in Section 2.2, the structure of the Galois group Gal(LI 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 = kur of a local field (k, v). In the last chapter, we constructed certain abelian extensions of K by means of the formal groups Fj(X, 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 fJP. We shall then study the properties of such abelian extensions.

5.1.

Abelian Extensions L nand kr;:,n

For each integer m > 1, let om again denote the valuation ring of the unique unramified extensions k7:,. of degree mover k in Q. Let n be a prime element of k7:,. so that it is also a prime element of K = kur and of K. Let fJPn be the family of power series in R[[X]], R = Ok, as defined in Section 4.2, and let

fJPr;: = fJPn n om [[ X]]. For example, f(X) = nX + xq in Lemma 4.7 belongs to fJPr;:. We put

LEMMA

5.1.

fJPm

= the

r

= the union of

Let m

>

union of fJPr;: for all prime elements n of k7:,.,

1, n

>

fJPm for all integers m

>

l.

o.

Let f E fJPr;:. Then the field k7:,.(W;) depends only on m, n, and n, and is independent of the choice off in the family fJPr;:. (ii) Let f E fJPoo. Then the field K(W;) depends only on n and is independent of the choice off in r. Proof. (i) The proof is similar to that of Lemma 4.10. Namely, let f(X) = nX + xq and let f'eX) be any other power series in fJPr;:. Since we may put n = n', ; = 11 = 1 in the proof of (4.8), it follows from the remark (3.13) that in this case, the power series 8(X) in (4.8) belongs to om[[ X]]. By Lemma 4.7(ii), k7:,.(W;) is a finite Galois extension over k7:,.. Hence it is a local field and is closed in Q. Since 8(X) E om[[x]], we see from (4.10) that (i)

WI'

= 8(W;) c

k:reW;),

k

c

k7:,.(W;,)

c

k7:,.(W;).

It then follows that k7:,.(W;,) is again a local field so that k7:,.(W;) 1

since also 8- (X)

E

om[[x]]. Hence k7:,.(W;) = k7:,.(W;,) for any f'

c

k7:,.(W;,)

E

fJPr;:.


66 (ii) Let fEr. Then f belongs to fJP'; for some m

E

= K (W7) =

:>

1 and n

E

k7:,.. Let

K . k:'r(W7)

and let E denote the closure of E in Q. By (i) above and by Lemma 4.7, k7:,.(W7)/ k7:,. is a finite Galois extension for any f E fJP';. Hence E / K is also a finite Galois extension, and by Lemma 3.1 and the definition of in,

K=EnK. Let f' be another power series in r and let E' = K(W7.). Then E' / K is again a finite Galois extension and E' = in. Let M = EE'. Since both M / E and M / E' are finite Galois extensions, it follows from Lemma 3.1 that E= M

nE =M n in =M nE' = E',

•

namely, K(W7) = K(W7.). In view of the above lemma, we put

fJP,;, for any fEr,

for any f

L n = K(W7),

E

n:> -1, n:>-1.

Note that

L -1 = kur because Wj1

(= K)

= {OJ.

5.2. Let m :> 1, n:> 0, and let n be a prime element of k:'r The field in in Section 4.4 is the closure of L n in Q, and

PROPOSITION

(i)

Gal(in / K) ~ Gal(L n / K). (ii)

L n = Kk,;,n, k:'r = K n k,;,n so that k,;,n is a maximal totally ramified extension over k7:,. in L nand Gal(Ln/ k7:,.) = Gal(L n/ k,;,n) x Gal(L n/ K) ~

(iii)

Gal(K/k:'r) x Gal(k,;,n/k7:,.).

L n/ k, L n/ K, k,;,n/ k, and k,;,n/ k7:,. are abelian extensions and

Proof. It was already proved above that both L n/ K and k,;,n / k7:,. are finite Galois extensions and that E = EK, K = En K for E = K(W7) = Ln. Hence Gal(in / K) ~ Gal(Ln / K), [Ln: K] = [in: K] = (q - l)qn by Proposition 4.11, proving (i). It is clear that

so that

(q - l)qn = [Ln: K] = [k,;,n: K n k,;,n] < [k,;,n: k7:,.], Let f(X) = nX + xq, n

E

(5.1)

k7:,., and let hn(CKo) = 0 as in Lemma 4.7(iii). Then

W7 = 0 i CKo = {[a]tC CKo) I a EO}

67

Abelian Extensions Defined by Formal Groups

by Lemma 4.8(ii). However, since f E om[[x]], it follows from (4.7) and the remark (3.13) that [aIr belongs to om[[x]]. Hence WJ

c

k:(ao),

and, by Lemma 4.7(iii),

= [k:(ao):k:] = (q -l)qn. Comparing this with (5.1), we obtain k: = K n k:"n. This proves (ii). The [k:"n:k:]

statement on L nI K and k:"n Ik7:r in (iii) are consequences of (i), (ii), and Proposition 4.11. To see that Lnlk is abelian, take a prime element Jr of the ground field k. By (i), (ii), Gal(Ln I K) ~ Gal(L nI K) ~ Gal(k~nl k~r)' where k~r= k. Hence k~nlk is abelian by Proposition 4.11. Since L n = Kk~n and K I 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 n :> 0, there exists a homomorphism b n : U = U(k)~ Gal(Lnl K)

such that for each fEr, each u bn(u)(a)

E

U, and each a

E

WI'

= Uj a( = [u]tCa)),

and it induces an isomorphism

U I Un + 1 ~ Gal(L n I K).

(ii) Let m :> 1 and let homomorphism

Jr

= U(k)~ Gal(k:"nlk7:r) E ;g;:" each u E U, and each a E WI' b~(u)(a) = u j a( = [uIr(a)), b~:

such that for each f

be a prime element of k7:r Then b n induces a

U

and b~ in turn induces an isomorphism UIUn+1 ~Gal(k:"nlk:). COROLLARY.

Let O-1. These k':r, 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 = o[[X]] Sx = {g(X)

E

S I g(O) =1= 0 mod tJ,

that is, g(O) E V}.

Since the valuation ring 0 of the local field (k, v) is compact, S is a compact ring in the topology introduced in Section 4.1. Let n be a prime element of k and let f(X) be a power series in the family ~~ = ~7r n S. t Compare Coleman [5].


70

LEMMA 5.5.

Let g(X) be a power series in S such that g(WJ) = 0,

that is, g(y) = 0,

for all Y E WJ.

Then g(X) is divisible by [n]f in S:

heX) E S.

g(X) = [nyh(X),

First, consider the case where f(X) = nX + xq. Since n follows from (4.7) that Proof.

E 0,

it

f(X) = [n y.

By Lemma 4.7(ii), WJ is the set of all roots, in Q, of the separable polynomial f (= go) in o[X]. Hence, if g(WJ) = 0, then g is divisible by f= [nyin S by Lemma 3.10. In general, letf'(X) be any power series in @i~ and let 8(X) be the power series in (4.8). As stated in the proof of Lemma 5.1(i), 8(X) then belongs to S = o[[X]]. Now, let g'(X) be a power series in S such that g'(WJ,) = O. Since WJ, = 8(WJ) by (4.10), we then have (g' 0 8)(WJ) = 0 with g' 0 8 E S. Therefore, by the above,

g' 0 8 = [n ]fh,

g' = ([ nf] 0 8- 1 )(h 0 8- 1),

with h, h

0

8- 1 E S.

However, [n]f 0 8- 1 = 8- 1 0 [n]/" by Proposition 4.5, and 8- 1(X) is divisible by X in S. Hence g'(X) is divisible by [n]/, in S. • REMARK. By a similar argument, we can also prove that if g(Wi) g E S, then g(X) is divisible by [nn+l]f in S.

In the following, we shall fix a power series f in @i~ = for simplicity, the suffix f in [a ]f' Wi, and so on. Let

@in

=0

for

n S and omit,

n:> -1,

be the residue field of k':r = k(wn) and let £1' E wn em = ~Q. Since £1' E ~n = m n k':r, it follows from the general remark in Section 3.4 that F[(X, £1') is a well-defined power series in on[[X]]. Let X-+- £1'( = X

t £1') = F[(X, £1').

Then X-+- £1' = £1' 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 -+- £1') 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 Woo

Then there exists h (X) in S such that g=ho[n]. Proof.

Let ao = g(O)

E 0

and let hI = g - ao. Then h1(y)

= g(y) -

g(O) =


o for

all

y E Woo

71

Hence, by Lemma 5.5, hI is divisible by [n] in S: g = ao + [n]g I ,

hI = g I [ n ],

gI

E

S.

Since [n](X -+- y)

= [n](X) -+- [n](y) = [n](X),

it follows that gl(X -+- y) = gI(X) for all y E Woo Therefore, by the same argument, so that g = ao + aI[n]

+ [n]2g2

where g2(X) again satisfies g2(X -+- y) = g2(X) for all y E Woo Thus we can find a sequence, ao, aI' a2, ... in 0 and a sequence g I, g2, ... in S such that g = ao + aI[n]

Hence g = h 0 [n] REMARK.

+ ... + an[n]n + [nr+Ign+l1 for all n > O. with h(X) = ao + aIX + ... + anXn + ... in S.

•

Again the lemma can be generalized for wn and [nn+I].

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 [n] for g, h ideal of k. Then, for each n > 0,

E

S and let ~ be as before the maximal

g =Omod~n~h

=Omod~n.

Consequently, g=O~h=O.

Proof ¢: is obvious. We prove ~ by induction on n. For n = 0, this is trivial. Hence, let g = 0 mod ~n, n > 1. Then g = 0 mod ~n-l, g = nn-lgl , with gi E S. By the induction assumption, we have h = 0 mod ~n-l, h = nn-ih i with hI E S, and it follows from g = h 0 [n] that gi = hI 0 [n]. However, g 0 mod ~n implies gI 0 mod~. Since [n] = f(X) xq mod~, we see

=

=

hI(xq) = hI

so that hI(X) = 0 mod~. Hence h

0

=

=

[n] = gI = 0 mod ~ nn-Ih I = 0 mod ~n.

•

REMARK. The proof shows that the lemma holds also for power series g, h in R = 0 k[[ X]] and for the maximal ideal ~ k of k. Now, let h(X) be any power series in S = o[[X]] and let hI(X) =

Il h(X -+- y),

y

E

Woo

y

Then hI(X) is a power series in Oo[[X]] ,

00

being the valuation ring of


72

k~ = k(WO). By the Corollary 1 of Proposition 5.4 for m Y =1= 0, are conjugate to each other over k. Hence

hf =

hv

for all a

E

= 1, all

y

E

WO,

Gal(k~1 k)

so that hl(X) belongs to S = o[[X]]. Furthermore, since (X -+- y) -+- y' = X-+- (y -+- y'), hl(X) satisfies hl(X -+- y)

= hl(X) for

for y, y', y -+- y'

E

WO,

all y E Woo Therefore, by Lemmas 5.6

and 5.7, hl=h 2 o[Jr]

with a unique power series h 2(X) in S. We shall denote this h2 by Nf(h), or simply, by N(h). Thus N(h) (= Nf(h)) is the unique power series in S such that N(h) ° [Jr]

= IT heX -+- y),

(5.2)

y

The map N(=Nf):S~S

is called the norm operator on S, associated with f prove some basic properties of the operator N. LEMMA

E

~~. We shall next

5.8.

(i) N(hlh2) = N(h l )N(h 2), for hv h2 E S. (ii) N(h) == h mod p, for h E S. (iii) h E XiS Xfor i > 0 ~ N(h) E XiS X, SX being the multiplicative group of the ring S. (iv) h = 1 mod pi, i > 1 ~ N(h) = 1 mod pi+l. Proof. (i) is obvious from the uniqueness in (5.2). (ii) [Jr] = f(X) =xq mod p implies N(h) ° [Jr]

On the other hand, y X

E

WO c

+ y = X mod Po,

=N(h)(Xq) mod p.

Po and [Wo: 0] = q, f = o/p = Fq imply

IT heX -+- y) = h(X)q = h(Xq) mod Po· =

Hence, it follows from (5.2) that (Nh)(Xq) h(Xq) mod p so that (Nh )(X) heX) mod p. (iii) Let h E SX-that is, h(O)=I=O mod p. Then N(h )(0) h(O) =F- 0 mod p by (ii) so that N(h) E SX. For heX) = X, X-+-O = X implies

=

=

N(X)

= Xhl(X),

with hI

Hence, by (5.2),

[Jr](h l ° [Jr]) = X

IT (X -+- y), y*O

E

S.

(5.3)

73


Dividing the both sides by X and putting X = 0, we obtain

By the Corollary (i) of Proposition 5.4, the product on the right is the norm of y =1= 0 in WO for the extension k~/ k. By Proposition 4.2(ii) and Proposition 5.4(ii) , such a y is a prime element of the totally ramified extension k~ over k. Hence the norm of y-namely, nhl(O)-is a prime element of k. Therefore, h1(0) is a unit of k and it follows from (5.3) that N(X) EXSx. By (i), we now see that h EXiS X implies N(h) EXiSx. (iv) Write h in the form h = 1 + nihl' i > 1, hl E S. Then N(h)

0

[n] =

Il (1 + nih1(X -+- y)) =(1 + nih1(X))q mod ni~o Y

== 1 + qnih1(X) + ... + niqh1(X)q = 1 mod ni~o. Let N(h)

= 1 + h2'

h2 E S. Then it follows that h2 0 [n] = 0 mod ni~o, hence, mod ~i+l.

Therefore, h2 = Omod ~i+l-namely, N(h) = 1 mod ~i+l_by Lemma 5.7 . n

We now define the iteration N of the norm operator Non S by ~(h)

•

Nn(h) = N(Nn-l(h)) = N(N( . .. (N(h )) ... )), for n > 1. LEMMA 5.9. (i) Nn(h) 0 [nn] = TIn' h(X -+- ll'), for ll' E wn-\ n > o. (ii) If h E XiS X , i > 0, then Nn+\h)/ Nn(h) E SX and

= h,

Nn+\h) = Nn(h) mod ~n+\

n>

o.

(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 Wn-1/WO in W n- 1: W n- 1= A -+- WO = {ll' -+- y Ill' E A, Y E WO}. Then Proof.

Il

n'EW n -

h(X-+-ll')= 1

Il Il

h(X-+-ll'-+-Y)=

n'EA YEW O

Il N(h)([n](X-+-ll')), n'EA

where [ n ] (X -+- ll') = [n] (X) -+- [n] ( ll') = [n] (X) -+- n . ll',

Since W n -

2

ll' EA.

= n . W n - 1 = n . A, it follows that

Il

n'EW n -

h(X -+- ll') = 1

Il

N(h )([n ](X) -+- {3),

~Ewn-2

and by the induction assumption, the last term is equal to Nn-\N(h)) 0 [nn-l]([n](X)) = Nn(h)

0

[nn].

(ii) By Lemma 5.8(iii), h E XiS X , i > 0, implies Nn(h) E XiS X for all n > O. Hence Nn+l(h)/ Nn(h) E Sx. By Lemma 5.8(ii), we have hl = 1 mod~,

for hl = N(h)/h

E

Sx.


74

Hence, by (iv) of the same lemma, we obtain successively

N(h l ) = 1 mod lJ2,

...

,Nn(h l ) = 1 mod lJ n + l .

Since Nn(h l ) = Nn+l(h)/ Nn(h), the assertion in (ii) is proved.

•

We still keep fixed a power series f( X) in @P~ = @PlT: n S and choose an element a such that a E W n, aft Wn-t, n :> O. Let ai = lC n- i . a = [lC n- i]( a), for 0 < i < n.

Then ai E Wi C k~, ai ft W i- I so that by Proposition 5.4(ii) for m = 1, a is a prime element of k~, and for the valuation ring and the maximal ideal of k~. LEMMA 5.10. Let f3i E lCn-ilJoOi for 0 < n. Then there exists a power series heX) in S = o[[X]] such that

h(ai) = f3;, Proof.

for 0 < n.

Let

0< i <no

gi(X) = [lCn+l][lCi]/[lCi+l],

Since [lC j] == lC jX mod deg 2 for j :> 0, [lC n+l ] = [lC n- i] 0 [lC i + l] lCn-i[lC i+l ] mod deg 2 in [lC i+l ].

=

Hence gi(X) = (lC n- i + al[lC i+l ] + a2[lC i+I]2 + so that gi

E

... )[lCi],

S. Now, it follows from [lCi](ai) = ao, [lCi+I](ai) = 0 that

gi( ai) = lC n- iao· If 0 <j < i, then [lC i]( aj) = 0 implies gi( aj) = 0; and if i < j < n, then [lCi+I](aj) = aj-i-I 0 and [lCn+I](aj) = 0 imply again gi(aj) = O. Since f3i E lCn-ilJoOi, where lJOOi = aoo[ ail, we may write f3i in the form

*

f3i = lCn-iaohi(ai), Then

with hi(X)

E

o[X].

n

h

=

L gihi

i=O

belongs to S and it clearly satisfies h(ai) = f3i for 0 < n.

•

We now prove a main result of Coleman [5]. 5.11. Let lC be a prime element of k ( = k~r) and let f E @P~ = n i @PlT: n o[[X]]. Fix an element a E Wi, a ft Wi-I, n:> 0, and let ai = lC - j a for 0 < i < n. Let; be any element of U(k~) and let PROPOSITION

o< n,


75

where Nn,i denotes the norm from k~ to k~. Then there exists a power series h (X) in 0 [[ X]] such that

;i = Proof.

Since

U(k~) e

; =

h(ai ),

for 0 0, the homomorphisms b n and of Proposition 5.3. Proof. Similar to the proof of Proposition 4.12.

b~,

respectively, •

Note that by Proposition 5.3, b has the property: b(u)(a) = u

+f a,

U E

U,

for any n > 0 and for any f in g;x. Similarly for b ll · For m = 1 and n E k = k~n let k~oo be denoted by k ll :

kll

= k~x = the union of k~

( = k~n) for all n > -1.

As a special case of the above proposition, we see that

so that kll is a maximal totally ramified extension over k in Land Gal( LI k) = Gal( LI k ll ) x Gal( LI k ur ) .:; Gal( kurl k) x Gal( kll Ik),

(5.4)

bll : U.:; Gal(klllk). Now, fix a prime element n of k and an integer m

>

1. Let

k' = k'::,., cP'

= CPk' = the

Frobenius automorphism of k'.

Clearly cP' = cpm, cP = CPk being the Frobenius automorphism of k. Let n' be any prime element of k'. Since n is also a prime element of k', we have n' =

Let

n;,

with;

E

U(k').

77


The following lemma relates the action of the Frobenius automorphism on the formal groups for lC and lC' to the action of the endomorphism [u Jr. LEMMA 5.14. Then

Let f

E @P~,

f'

E @P';,

and let 8(X) be the power series in (4.8).

8 CfJ' = 8 0 [u ][. Proof. Let gm-l and g:n-l be defined for f and f' as in Lemma 4.6. The proof of the same lemma shows that ' o8-8CfJ'og g m-l m-l'

(5.5)

Let x=

Nk'lk(lC')

= lCmU.

It is clear from the definition of g:n-l that g:n-l(X) = xX mod deg 2. On the

other hand, since f' E @P';" f'CfJ' = f', we see from the definition of g:n-l that f' 0 g:n-l = g::f-l 0 f'. Hence, by the uniqueness in (4.8) and Proposition 4.5, g:n-l = [x]!, = [x]!, However, since f belongs to @P~, we have f = [lC Jr, gm-l it follows from (5.5) that 80 [u][ 0 [lC m ][ = 8 CfJ' 0 [lC m ][.

m

= [lC ][.

Therefore,

As [lC m ][ = [lC][ 0 • • • 0 [lC ][, the equality of the lemma follows from the remark after Lemma 5.7. • LEMMA 5.15. Let lC' be another prime element of k and let u E Un+ 1 • Then n >--1. k n:rr -kn :rr', Proof. a' E WI"~ 5.14,

lC'

=

lCU

with

Apply the preceding lemma for m = 1, cp = cp', lC' = lCU. Let By (4.9), WI' = 8(W;) so that a' = 8(a) with a E W;. By Lemma 8CfJ( a) = 8([ u Jr( a».

But, as u E Un + 1 , a E WI' it follows from Proposition 5.4(ii) that [u][(a) = a so that 8CfJ( a) = 8( a). On the other hand, since kur n k~ = k by Proposition 5.2(ii) for m = 1, the Frobenius automorphism cp of kurl k can be extended to the Frobenius automorphism of L n = kurk~ over k~ and, hence, to an automorphism of its closure in over k~. As a E Wi c k~, we then obtain 8(a)CfJ = 8 CfJ(a) = 8(a), Therefore, a' = 8( a) Lnlk~. Thus

E k~

lC

E

Ltl = kur(Wil

because cp is the Frobenius automorphism of k~,

Since

for a' = 8(a)

= k(W;') c

k~.

= U-1lC', u- 1 E Un+ 1 , we obtain similarly

k~ c k~,.

Hence k~ = k~,


78

for Jr' = JrU, U E Un+1. By Proposition 5.4(i) both Jr and Jr' are contained in N(k~1 k) = N(k~,1 k). Consequently, U = Jr' I Jr E N(k~1 k) for every U E Un+1-namely, Un+1 C N(k';;lk). Hence (Jr) x Un+1 C N(k~/k). •

For any prime element Jr of k,

PROPOSITION 5.16.

N(k~/k) =

Proof.

(Jr) x Un+1, for n:>-l.

Since we know N(k~/k) c k

(Jr) x Un+1 C

X

(Jr) x U,

=

it is sufficient to show that NU(k~/k)

= U n N(k~/k) c Un+1.

Let U E NU(k~/k)-namely, U = N'(;), where; E U(k~) and N' denotes the norm of the extension k~1 k. Fix an element a E Wi, a ft Wi -1. As in the proof of Proposition 5.11, we then have ; =

h1(a),

with hl(X)

E

SX.

By Lemma 5.9(ii),

Nn(h 1), Nn+1(h 1) E SX, Let Ul = Nn(h1)(O), U2 = Nn+l(hl)(O). Then it follows from the above that

Ul, U2 E U = U(k), However, by Lemma 5.9(i) Ul =

IT

!3EW n - 1

h 1({3),

so that

U21ul =

IT h 1({3),

where {3

E

Wi,

(3 ft

Wi-I.

!3

Therefore, by the Corollary (i) of Proposition 5.4 for m = 1,

U21ul = N'(h1(a»

= N'(;) = u.

Hence U2 = Ul mod ~n+l implies U = 1 mod ~n+l-namely, U E Un+1. Thus NU(k~/k) c Un+1· • PROPOSITION 5.17.

For any prime element Jr of k, N(knlk)

=

(Jr).

More generally, if F is a totally ramified extension over k in k n : k c k n c F c Q, then N(Flk) = (Jr).

Q,

containing

Proof. Since the intersection of Un+1 = 1 + ~n+ 1 for all n :> 0 is 1, the first part follows immediately from Proposition 5.16. Let F be as stated


7

above. Then N(Flk) c N(knlk) = (n).

However, since Flk is totally ramified, N(Flk) contains, by Proposition 3.8, a prime element of k. Since n is the only prime element of k in (n), it follows that n E N(Flk) so that N(Flk) = (n). • The fields k,;,n in Section 5.1 were introduced by de Shalit [6] as generalizations of the abelian extensions k~ in Lubin-Tate [19]. He also generalized the result of Sections 5.2 and 5.3 and proved, for example, that REMARK.

N(k,;,Xlk) = (Nk'lk(n»

for any prime element n 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 k~ for the sake of simplici ty.

Chapter VI

Fundamental Theorems Let kab denote the maximal abelian extension, in Q, of the local field (k, v). Using the results on the abelian extensions L n, k,;,n, and k~ 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: k X~ Gal(kab/ 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 k c L c kab'

where kab is, as stated above, the maximal abelian extension over k in Q. For each prime element lC of (k, v), we also defined in Section 5.3 an abelian extension k n of k, contained in L. Let CfJk denote, as before, the Frobenius automorphism of k: CfJk E Gal(kurlk). Since Gal( LI k n )

.:;

Gal( kurl k)

by (5.4), there exists a unique element tjJn in Gal(Llk) such that

I

tjJn kur

= CfJb

I = 1.

tjJn k n

(6.1)

As (CfJk) is dense in Gal(kurlk), (tjJn) is a dense subgroup of Gal(Llkn ) so that k n is the fixed field of tjJn in L. Now, fix a prime element lCo of (k, v) and, for simplicity, write tjJ for X tjJno' Since k X = (lC o ) X U, each x in k can be written uniquely in the form x =

lC';U,

m E Z,

U E

U.

In fact, here m = vex). For such an x, let

p(x)

= tjJmb(u- 1),

(6.2)

where b: U.:; Gal(Llkur )

is the isomorphism in Proposition 5.13. It is then clear that the map x ~ p(x) defines a homomorphism of abelian groups

p: kX~ Gal(Llk)

Fundamental Theorems

81

satisfying

p(x) I kur = tjJm I kur = cpr;:, LEMMA 6.1.

with m

= vex).

(6.3)

Let lC' be a prime element of k' = k':,., m:> 1, and let x = Nk'lk(lC').

Then p(x) is the unique element a of Gal(LI k) such that

a I km,oo :rr' - 1•

a I k ur = cP mk ,

Proof. Since cpr;: is the Frobenius automorphism CPk' of k': cpr;: = CPk' E Gal(kurlk'), it follows from Proposition 5.12 that there is a unique element a in Gal(Llk) with the properties stated above. As p(x) satisfies (6.3), it is sufficient to prove that p(x) I k';:oo = I-namely, that p(x) I k,;;n = 1,

for all n :> 1.

Now, the prime element lCo of k is also a prime element of the unramified extension k' over k. Hence with some; in U(k'), so that

x

=

Nk'lk(lC')

= lC~u,

Let f E g;~o' f' E g;,;" and let 8(X) be the power series in (4.8) for f, f'. Then, by Lemma 5.14,

8CP' = 8 ° [u ]f' where cP' denotes the extension of cP k' = cpr;: on the completion K of K = kur- Let a' E Wi, = 8(Wi) and let a' = 8( a), a E Wi. By (6.2) and Proposition 5.3,

p(x)(a) = b(u-1)(a) = [u-1]f(a). Using the fact that 8(X) from a' = 8(a) that

E

Ok[[X]] and that p(x) extends to cP' on K, we see

p(x)(a') = 8CP'(p(x)(a» = 8o[u]fo[u-1]tCa)

= 8(a) = a'.

Since k,;:n = k'(Wi,), we obtain p(x) I k,;;n = 1. REMARK.

•

The above proof shows why we define p(x) by (6.2) instead of

p(x) = tjJmb(u). PROPOSITION

6.2.

There exists a unique homomorphism Pk: k X

~

Gal(LI k)

such that for every prime element lC of k. Proof. Applying Lemma 6.1 for m = 1, lC' = lC, we see that p(lc) is the

82


unique element of Gal(Llk) such that

p(1[) I kur = f/Jk,

p(1[) I k n

= 1.

Hence p(1[) = t/Jn by (6.1). Thus the homomorphism p:kx~Gal(Llk), defined above by means of a fixed prime element 1[0 of k, has the property mentioned in the proposition. Let u E U and let 1[' be a prime element of k. Then 1[" = 1['U is again a prime element of k and u = 1["11['. 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.:; Gal(LI k ur ), (6.4) u~ b(u- 1 ) on the subgroup U of kX. Let

1[

be a prime element of k and let m=V(X)EZ,

UEU,

for an element x of k x. Then

= Pk(1[)m pk (U) = t/J';b(u- 1 ), Pk(X) I kur = t/J'; I kur = erik, m = v(x). Pk(X)

(6.5)

Thus (6.2) and (6.3) hold not only for the particular prime element 1[0 but also for any prime element 1[ of k. (This is also clear from the fact that P = Pk is independent of the choice of 1[0·) X

6.3. (i) Pk is injective and is continuous in the v-topology of k and Krull topology of the Galois group Gal(Llk). (ii) The image of Pk is a dense subgroup of Gal(LI k) and consists of all elements a in Gal(LI K) such that a I kur = f/J'!: for some integer m. In particular, if a I kur = f/Jb then there is a unique prime element 1[ of k such that a = Pk(1[). Proof. (i) Let Pk(X) = 1 for x = 1[ mU in (6.5). Then f/J'!: = Pk(X) I kur = 1, 1 and since = Z, it follows that m = 0, x = U, b(u- ) = Pk(U) = Pk(X) = 1. As b is an isomorphism, we then see that u = I-namely, x = 1. Thus Pk is injective. The continuity of Pk follows from the fact that it induces a topological isomorphism U.:; Gal( LI k ur ) on the open subgroup U of k x . (ii) By (5.4), Gal(LIK) = Gal(Llkn ) x Gal(Llk ur ). Hence the first part of (ii) is a consequence of (6.5), Pk( U) = Gal(LI k ur ), and the fact that is a dense subgroup of Gal(Llkn ). The second part is also clear from (6.5) for m = 1. • PROPOSITION

Now, let k' 1k be a finite extension of local fields and let

be the associated homomorphisms given by Proposition 6.2. We shall next prove an important preliminary result on the relation between Pk and Pk'·

83


For simplicity, let L

= Lb L' = L k·, and let E=LnL'.

Since kcEcL and Llk is abelian, Pk(X), for xEk x , induces an automorphism of Elk. Let x' E k'x. Then Pk'(X') is an automorphism of L' over k', and hence, over k. Therefore, Pk'(X') also induces an automorphism of Elk. 6.4. fields. Then LEMMA

Suppose that k' Ik is a totally ramified finite extension of local

Pk'(X') I E = Pk(Nk·/k(X'» I E,

for all x'

E

k'X.

Proof Since k' x is generated by the prime elements of k', it is sufficient to prove the above equality for a prime element Jr' of k'. Extend Pk.(Jr') in Gal(L'lk') 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 t/Jn' = Pk·(Jr') in L' is k~. by (6.1), F n L' =

k~.,

so that Flk' is totally ramified and k' c

k~. c

F. Hence, by Proposition 5.17,

N(Flk') = (Jr').

(6.6)

On the other hand, r I k~r = Pk·(Jr') I k~r = CfJk' by (6.5) and CfJk·1 kur = CfJk because k' /k is totally ramified so that Gal(k~r/k') ~ Gal(kur/k). Hence

r I kur

= (r I k~r) I kur = CfJk·1 kur = CfJk·

Let a = r I L. Then a I kur = CfJk by the above. Therefore, by Proposition 6.3(ii) , a = Pk(Jr) for some prime element Jr of k. Since k n is the fixed field of Pk(Jr) in L by (6.1), k c k n c F = the fixed field of r in Q,

and since both F Ik' and k' Ik are totally ramified, F Ik ramified. Hence, again by Proposition 5.17,

IS

also totally

N(F Ik) = (Jr). It then follows from (6.6) that

Nk·/k(Jr')

E

N(Flk)

= (Jr).

However, as k'ik is totally ramified, Nk·/k(Jr') Therefore, we see from the above that

Nk·/k(Jr') = Jr so that

IS

a prime element of k.


84

Let k' Ik be a totally ramified finite extension of local fields such that k c k' eLk' Then COROLLARY.

Pk(N(k'lk)) I k' Proof.

= L n L'.

Clearly k c k' c E

Pk(N(k'lk)) I k'

6.2.

= 1.

Hence

•

= Pk,(k'X) I k' = 1.

Proof of Lk = kab

In this section, we shall prove that the abelian extension L (= Ld over k introduced in Section 5.3 is actually the maximal abelian extension kab of k. We first prove the following key lemma: Let (k, v) be a p-field and let (k', v') be a cyclic extension of degree p over k-that is, [k': k] = p. Then LEMMA

6.5.

N(k'lk)=I=k x . Proof. Let G = Gal(k'lk) and let Gn , n > 0, be the ramification groups of k' Ik defined in Section 2.5. Since G is a cyclic group of order p, each Gn 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(Nk'lk(X')) is divisible by p for every x' E kX. Hence a prime element n of k is not contained in N(k'lk), and consequently N(k'lk)=I=k x . Next, suppose that Go = G so that k'ik is totally ramified by Proposition

2.18. By the remark after the Corollary of Proposition 2.19, the index [Go: Gd is prime to p. Hence there is an integer s > 1 such that G s + 1 = G s + 2 = ... = 1.

Fix a prime element n' of (k', v') and let f(X) denote the minimal polynomial of n' over k. By the corollary of Proposition 2.13 and Proposition 2.14, the different qj) = qj) (k' I k) of k' I k is given by qj) = f'(n')o', where f' = dfldX and 0' is the valuation ring of (k', v'). Clearly f'(n') = (n' - a(n')) with a ranging over all elements =1=1 in G. Since G = G G s + 1 = 1 implies v'(n' - a(n')) = s + 1 for all a =1= 1, we obtain

no

H

v'(f'(n'))

= (p

- l)(s + 1),

qj)

= tJ'(p-l)(s+l),

tJ' being the maximal ideal of (k', v'). Now, let x' x' = 1 + y', y' Ey,s+l. Then Nk'lk(X')

= IT (1 + a(y')),

E

aE

o

=

U;+l

= 1 + tJ,s+l and let

G,

(6.7)

1 + ~y'A+Nk'lk(y')' A

where A ranges over all elements of the form A = a 1 + ... + at, 1 < t

o.

By (6.4) and Gal(Llkur) ~ Gal(kJt1k), Pk induces an isomorphism

V

~ Gal(kJt 1k ).

Let V' be the subgroup of V such that

V' ~ Gal(kJt 1k ') under the above isomorphism. Since VIV' ~Gal(k'lk), VIV' is a cyclic group of order p'\. Her.ce there is a character X of V I V' with order pS, which we view as a continuous character of the compact group V with Ker(x) = V'. We shall next show that there is a continuous character}. of V with order ps+ I such that

X=

}.p.

Consider first the case where k contains no primitive pth root of unity. By Proposition 2.7 (for a = 0) or Proposition 2.8, VI = 1 + p is isomorphic to the direct product of finitely or infinitely many copies of Zp. Since V = V X VI by (2.2) and since X(V) = 1 for the group V of order q - 1, the existence of }. is clear because the character group of Zp is isomorphic to QplZp , a divisible group. Suppose next that k contains a primitive pth root of unity In this case, the p-field k has characteristic O. Since k' is contained in a cyclic extension E of degree pS+ lover k, it follows from Lemma 6.7 that E N(k' /k). Hence, by the Corollary of Proposition 6.4, Pk('p) I k' = 1, so that 'p E V', X('p) = 1. It then follows again from Proposition 2.7 and (2.2) that there is a character}. of V satisfying X = }.p. (iv) Let X and }. be as above. Let V" = Ker(}.) and let kIf be the subfield of kJt, corresponding to V":

'po

'p

U" '" Gal(kJt I kIf) , ~

V I V" ~ Gal(k" I k).

88


Then k c- k' c- k" c_ kJt'

Enk"=k' ,

and both Elk and k" I k are cyclic extensions of degree pS + 1 with [E: k'] = [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 kJt = L n F. Hence E c k*k" C kJt , and this contradicts k' = E n kJt =1= E. Thus the assumption F =1= kJt in (iii) leads to a contradiction and it is proved that

THEOREM 6.8.

For each local field (k, v),

namely, for any prime element

1C

of k.

EXAMPLE. Let (k, v) = (Qp, vp). In Example 2 of Section 5.1, we have seen that in this case n > -1,

with

Wpn+l

the group of all pn+lth roots of unity in Q. Hence kp

= Uk; = Qp(Wp"')' n

W poo being the group of all p-power roots of unity in Q. On the other hand,

by (3.2), kur = Qp(Voo),

where Voo is the group of all roots of unity in Q with order prime to p. Therefore, for (k, v) = (Qp, vp), 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, v) is associated a homomorphism

Pk: k X ~ Gal(kab l k). Pk is called the norm residue map, or, Artin map, of the local field (k, v). Some of the properties of Pk were already given by Proposition 6.3 and

89


(6.4). For convenience, we shall restate such properties of Pk: kX~ Gal(kab/ k) below: Pk is injective and is continuous in the v-topology of k X and Krull topology of Gal(kabl k). (ii) The image of Pk is a dense subgroup of Gal(kabl k) and consists of all elements a in Gal(kabl k) such that a I kur = ffJ'k for some integer m. In particular, if a I kur = ffJk1 then there is a unique prime element n of k such that Pk(n) = a. (iii) Pk induces a topological isomorphism (i)

U 2; Gal(kab/ k ur ), U~D(U-l)

on the subgroup U of k X , where D: U 2; Gal(kabl k ur ) is the isomorphism in Proposition 5.13. For any prime element n of k, it also induces U 2; Gal(k}flk), u~

D}f(u- 1 ),

where D}f 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: k X ~ Gal(kab l k), Pk': k'x ~ Gal(k~bl k')

be their norm residue maps. Since kabk' Ik' is an abelian extension, we have k c kab c kabk' c

so that the map a~ a I kab' a

E

k~b

Gal(k~blk'), defines a homomorphism

res: Gal(k~blk')~ Gal(kab1k). 6.9. Let Nk'lk: k'x ~ k X be the norm map of k' I k. Then the following diagram is commutative: THEOREM

k'X

~ Gal(k~bl k')

lres

IN kX

~

Gal(kab 1k ).

In other words, Pk'(X') I kab = Pk(Nk'lk(X)),

for all x'

E

k'x.

Proof. Let ko be the inertia field of the finite extension k' Ik so that kol k is unramified and k' Iko is totall ramified (cf. Section 2.3). It is clear that if the theorem holds for the extensions kol k and k' Iko, then it also holds for

90


k' I k. Hence it is sufficient to prove the theorem in the cases where k' I k 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 = kab' Hence, suppose that k'ik is unramified and let k' = k'::,., m = [k': k]. As usual, it is enough to show that for any prime element n' of k', Pk,(n') I kab = Pk(Nk'lk(n')). By Proposition 5.4(i), k,::nlk'::,. is a totally ramified finite abelian extension and n' is contained in N(k,::nlk'::,.). Therefore, applying the Corollary of Lemma 6.4 for k' = k'::,. c k,::n eLk' = k~b' we see that Pk,(n') I k,::n = 1 for all n > o. Noting that k c k,::n c kab by Proposition 5.2(iii), we find

a I k'::oo = 1,

for a = Pk,(n') I kab.

On the other hand,

a I kur = Pk,(n') I kur = qJk' = qJk by (6.5). Hence, by Lemma 6.1, Pk,(n') I kab COROLLARY.

•

= a = Pk(Nk'lk(n')).

Let k' I k be a finite extension of local fields and let x

E

k x.

Then Pk(X) I (k' Proof.

n k ab ) =

Let x = Nk'lk(X'), x' Pk(X) I (k'

E

1 ¢:>X

E

N(k' I k).

k'x. Then

n k ab ) = Pk'(X') I (k' n k ab ) = 1

because Pk'(X') E Gal(k~b/ k'). Conversely, suppose that Pk(X) I (k' n k ab ) = 1. Since Gal(kabk' / k'):::;. Gal(kab/ k' n k ab ), Pk(X) can be extended to an automorphism in Gal(kabk' I k') and, then, to an automorphism a of Gal(k~bl k'): a I kab = Pk(X). Let m>1.

Then Pk(X) I k'::,. = 1 so that Pk(X) I kur is, by (6.4), a power of the Frobenius automorphism qJk over k'::,.. Since Gal(k~J k'):::;. Gal(kurl k'::,.), a I k~r is then a power of the Frobenius automorphism qJk' over k', and it follows again from Proposition 6.3(ii) that

a

=

Pk'(X'),

for some x'

E

k'x.

Therefore, by the theorem above, Pk(X) = a I kab = Pk(Nk'lk(X')). Since Pk is injective by Proposition 6.3(i), it follows x = Nk'lk(X')

E

N(k' I k).

•


91

In Proposition 6.2, we defined the map Pk by means of l/Jn in (6.1), which in turn depends on the field k n constructed by the formal group Fj(X, Y) for f E @P~ (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: k X ~ Gal(kab 1k )

with the following two properties: For each prime element n of k, pen) I kur = ({Jb ({Jk being the Frobenius automorphism of k. (ii) For each finite abelian extension k' over k, p(N(k' /k)) Ik' = 1. Proof. By (6.5) and the Corollary of Theorem 6.9, Pk has properties (i) and (ii). Let p:kx~Gal(kablk) be any homomorphism satisfying (i) and (ii), and let n be any prime element of k. Since n E N(k~1 k) by Proposition 5.4(i) for m = 1, it follows from (ii) that pen) I k~ = 1 for all n > O. Hence pen) I k n = 1. As pen) I kur = ({Jb we see from (6.1) that pen) = l/Jn = Pk(n). Therefore, p(x) = Pk(X) for every x E kX-that is, P = Pk. •

(i)

We can see now that Pk is naturally associated with the local field (k, v) in the following sense. Namely, let a:(k, v).:; (k', v')

be an isomorphism of local fields-that is, an isomorphism a: k .:; k' such that v = v' 0 a. We can extend a to an isomorphism of fields

and define an isomorphism of Galois groups: a*: Gal(kab1k).:; Gal(k~blk'), a~

ara-I.

We then have the following theorem: THEOREM

Proof.

6.11.

The following diagram is commutative:

It is clear that a:kab.:;k~b and a*:Gal(kablk).:;Gal(k~blk')

induce a*: Gal(kurlk).:; Gal(k~rlk')

and


92

on

k~r.

Let

oa-l·k,x~Gal(k' Ik') · ab, P '=a*op k and let n' be any prime element of k'. Since v=v'oa, n=a-l(n') is a prime element of k. Hence

(Pk 0 a-l)(n') I kur

Pk(n) I kur

=

=

({Jk1

p'(n') I k~r = a*(({Jd = ({Jk'· Let E' be any finite abelian extension over k' and let E = a-I(E'). Then Elk is a finite abelian extension over k and

a:N(Elk) ::::;.N(E' Ik'), Hence

a*: Gal(Elk)::::;. Gal(E' Ik').

(Pk 0 a-I)(N(E' /k')) I E = Pk(N(E/k)) I E = 1, p'(N(E' /k')) I E' = a*(pk(N(E/k)) I E' = 1.

Therefore, by Theorem 6.10, P' = Pk'· COROLLARY.

Let a be an automorphism of a local field (k, v). Then Pk(a(x))

=

apk(x)a- l ,

for all x

E

•

kX,

where a on the right denotes any automorphism of kab' which extends the given a:k ::::;.k. Proof. Apply the theorem for k = k'. • Theorem 6.11 can also be proved directly by going back to the definition of Pk in Proposition 6.2. REMARK.

To explain the next result on Pk, we need some preparations. Let Q s denote the maximal Galois extension over the local field (k, v) and let k' be a finite separable extension over k so that n = [k' :k]

< +00.

Let G = Gal(QsI k),

G' = Gal(QsI k ab ),

H = Gal(Qslk'),

H' = Gal(QsI k~b).

Then H is an open subgroup of the compact group G with index [G :H] = [k' :k] = n, and G' and H' are the topological commutator subgroups of G and H, respectively. Let {TI1 ... , Tn} be a set of representatives for the left cosets HT, T E G: n

G=

U HT;.

;=1

Then, for each a E G and each index i, 1 -< i -< n, there exist a umque element h; E H and a unique index i', 1 -< i' -< n, such that


93

Denoting the above hi by hi( a), we define an element t G1H ( a) HIH' = Gal(k~blk') by

III

n

tG/H( a) =

IT h

i(

a)H',

for a

E

G.

(6.9)

i=l

It is known from group theory that tG/H( a) depends only upon a E G, namely, that it is independent of the choice of the representatives 1'v . . . ,1'n for the left cosets H1', and that a ~ tG/H( a) defines a homomorphism tG/H:G~HIH',

which is called the transfer map from G to H. t In the following, we shall denote tGIH also by tk'ik and call it the transfer map from k to k'. Since H / H' is an abelian group, the homomorphism tk'ik (= tG/H) can be factored as G~GIG'~HIH'.

For convenience, the induced map GIG' ~ HI H' will also be denoted by tk'ik. Thus for each finite separable extension k'lk, tk'ik denotes homomorphisms of Galois groups Gal(Q)k)~ Gal(k~blk'),

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

(2) For a

c

E

k'

c

k", then

G = Gal(Qslk), n = [k' :k] = [G :H],

tk'lk(a)G' = anG',

that is, tk'lk(a) I kab = an I kab·

(3) Let k' Ik be a Galois extension and let a

tk'lk( a) =

E

IT 1'a1'- H' = (IT 1'a1'1

r

= Gal(QsI k').

H 1

)

Then

k~b'

r

where l' ranges over a set of representatives for the factor group GIH = Gal(k'lk). (4) Let k' I k be a cyclic extension and let aH be a generator of GIH = Gal(k'lk). Then tk'lk(a) = anH'-that is, 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 [10].


94

diagram

k X ~ Gal(kab 1k )

1

(k'lk)

lt

k ik '

k'X ~ Gal(k~blk') where the vertical map on the left is the natural injection of k X into k'x. We shall prove below that the above diagram (k'lk) is commutative. We first establish some preliminary results. Suppose that (k' I k) is commutative. Then tk'lk: Gal(kab l k)~ Gal(k~blk') is injective. Proof. Let tk'lk( a) = 1 for a E Gal(kabl k). Extend a to an automorphism of Gal(QsI k) and call it again a. Then, by (2) above, a n(= an I k ab ) = tk'lk(a) I kab = 1, n = [k': k]. LEMMA

6.12.

Since Gal(kurlk)::;Z and Z is a torsion-free abelian group (cf. Section 3.2), it follows that a E Gal(kab1kur)' Hence, by (6.4), a = Pk(U) for some u E V. As (k' I k) is commutative, we then have Pk'(U) = tk'lk(Pk(U)) = tk'lk(a) = 1.

Since Pk' defines V' ::; Gal(k~blk~r), it follows that u = 1, a = Pk(U) = 1. LEMMA

6.13.

•

Let k c k' c kIf.

(i) If both (k"lk') and (k'lk) are commutative, then so is (k"lk), (ii) If both (k"lk') and (k"lk) are commutative, then so is (k'lk). Proof. Consider the diagram

k X ~ Gal(kab 1k )

1

(k'lk)

1

k'X ~ Gal(k~blk'). (k"lk')

1

1

By (1) above, the outside rectangle of the above diagram is (k"lk). Hence (i) is obvious. If (k" I k') is commutative, then tk"lk' : Gal(k~bl k') ~ Gal(k~bl kIf)

is injective by Lemma 6.12. Hence (ii) follows. LEMMA

6.14.

Let k'ik be a finite Galois extension and let x Pk'(X) = tk'lk(Pk(X)),

E

N(k'lk). Then

95


Proof.

Let x = N k'lk(X'), x'

E

k'x, namely,

X=

n rex') T

where r ranges over all elements of Gal(k'lk) = GIH. Extend each r to an automorphism in Gal(QsI k) = G and denote it again by r. Then it follows from (3) above that for o=Pk,(x')EGal(k~blk') and for tk'lk:Gal(kabl k) ~ Gal(k~bl k'), T

a I kab

=

Pk'(X') I kab = Pk(N k'lk(X')) = Pk(X)

by Theorem 6.9. On the other hand, on k~b' Pk'(X) =

n Pk,(r(x')) n rpk,(x')r- n rorI

=

T

=

T

I

T

by the Corollary of Theorem 6.11. Hence Pk'(X) = tk'lk(Pk(X)).

•

6.15. If k' Ik is unramified, then the diagram (k'lk) is commutative. Proof. Since k x is generated by the prime elements of k, it is sufficient to prove

LEMMA

Pk,(n) = tk'lk(Pk(n)) for any prime element n of k. In this case, k~bl k is a Galois extension. Hence, extend tpJr = Pk(n) to an automorphism tp of k~b over k and let F denote the fixed field of tp in k~b. Since tpl kur = tpJr I kur = CfJb it follows from Lemma 3.4 that F

n kur = k,

Furthermore, because F n kab Proposition 5.17,

IS

the fixed field of tpJr

= tpl

kab

III

kab. Hence, by

N(Flk) = (n). Let k'

= k~n n = [k' : k], and let F' = Fk nun

Then F' is the fixed field of tp' in

k~b

and

Therefore, by Proposition 6.3, there exists a prime element n' of k' such that F' =k~"


96

and by Proposition 5.17, N(F' Ik')

= N(k~';k') = (n').

Now, let E be any finite extension of k in F: k c E c F, [E: k] < + 00. Since F n k' = k, we then have [Ek' :k'] = [E:k],

N(Elk) c N(Ek' Ik').

As F' (= Fk') is the union of all such fields Ek', it follows that (n) = N(Flk) c N(F' Ik') = (n'). Hence n = n' because both nand n' are prime elements of k'. Now extend tJI to an automorphism a of Q s over k, so that a I kab = tJI I kab = Pk(n). Then it follows from (4) above that tk'lk(Pk(n))

= tk'lk( a) = an I k~b = tJI' = Pk,(n') = Pk,(n).

•

We are now ready to prove the following THEOREM 6.16 the diagram

Let k' / k be a finite separable extension of local fields. Then

k X ~ Gal(kab lk ) (k'lk)

1

k'X

tk'ik Pk'

~ Gal(k~blk')

is commutative. In other words, Pk'(X) = tk'lk(Pk(X)),

for all x

E

kX.

Proof. Since k' Ik 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' I k is a finite Galois extension. But, by the Corollary of Proposition 2.19, Gal(k'lk) is then a solvable group. Hence, by Lemma 6.13(i), we may even assume that k'ik is an abelian extension. Let n=[k':k],

E'=Ek' ,

and let n be a prime element of k. Then Pk(n) I E (= CPk I k: r ) is a generator of Gal(Elk) and PE(n) = tE1k(Pk(n)) by Lemma 6.15. Since k c k' ck ab c Eab , [k': k] = n, it follows from (4) above applied for Elk that PE(n) I k'

= tE1k(Pk(n)) I k' = Pk(nr I k' = 1.

Hence PE(n) IE' = 1 for the finite abelian extension E' over E, and it follows from the Corollary of Theorem 6.9 that n E N(E' IE). Therefore, by Lemma 6.14, PE,(n)

= tE'IE(PE(n)) = tE'IE(tElk(Pk(n))) = tE'lk(pk(n)).

97


Since k x is generated by prime elements, this shows that the diagram (E' I k) is commutative. Since (E'lk') is commutative by Lemma 6.15, it follows • from Lemma 6.13(ii) that (k'lk) is also commutative. COROLLARY.

The transfer map tk'lk:

Gal(kabl k) ~ Gal(k~bl k')

is injective. Proof.

Since (k'lk) 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 X Pk: k ~ Gal(kab l 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 kab' [k': k] < +00. We denote by Pk'ik the product of Pk and the canonical restriction homomorphism Gal(kablk)~

Gal(k' Ik) = Gal(kab1k)/Gal(kab1k').

Thus X Pk'lk: k ~ Gal(k' Ik).

7.1. Let k' Ik be a finite abelian extension of local fields. Then the above homomorphism Pk'ik induces an isomorphism k XI N(k' I k):; Gal(k' I k). THEOREM

Furthermore, N(k'lk) = p;\Gal(kablk'»

and Gal(kabl k') is the closure of Pk(N(k' I k» in Gal(kabl k). Proof. By the Corollary of Theorem 6.9, the kernel of Pk'ik is N(k'lk). By (ii) at the beginning of Section 6.3, the image of Pk is a dense subgroup of Gal(kab1k). Since Gal(kablk') is an open normal subgroup of Gal(kab1k), Pk'ik is surjective. Hence kXIN(k'lk):;Gal(k'lk) and N(k'lk)= p;1(Gal(kab1k'». The isomorphism shows that Pk maps each coset of kXmodN(k'lk) into a coset of Gal(kab1k) mod Gal(kablk'). Since the cosets are closed and since the image of Pk is dense in Gal(kabl k) and Pk(N(k'lk» c Gal(kablk'), it follows that Gal(kablk') is precisely the • closure of p(N(k'lk» in Gal(kab1k). COROLLARY.

For a finite abelian extension k' Ik of local fields, [k X:N(k' /k)] = [k': k].

The above equality is called the fundamental equality in local class field

99

Finite Abelian Extensions

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 k x modulo the norm group N(k'lk) onto the Galois group Gal(k'lk). PROPOSITION

7.2.

(i) Let k'ik be any finite extension of local fields. Then

N(k'lk) = N((k'

n kab)lk),

[kX :N(k'lk)]

1, n

>

0 and let

JC

be a prime element of a local field

mlrn E = k ur"-lr

with the abelian extensions k':,. and respectively. Then

k~

N(Elk) =

over k, defined in Sections 3.2 and 5.2,

(JC

m

) X

Un + 1 •

Proof. By Proposition 5.4(i) for m = 1, k~ is a totally ramified abelian extension over k and JC is the norm of a prime element JC' of k':r. E = k':,k~ is then an unramified extension of degree mover k':r so that N(Elk':r) =

(JC,m) X U(k~),

as shown in the proof of Proposition 3.7. However, NU(k~/k) = Un + 1 by Proposition 5.16. Hence, taking the norm from k~ to k, we obtain from the above that We note in passing that the field E = 5.1 for JC in k.

k':,.k~

above is actually k,;,n of Section • THEOREM 7.5. For each closed subgroup H of k X with finite index, there exists a unique finite abelian extension k' Ik such that

H=N(k'lk). Proof. Let m = [k x : H] < + 00. Then JCm E H for a prime element JC of k. Since H n U is a closed subgroup of U = U(k) with [U: H n U] 0 such that Un + 1 C H n U c H. Hence, by Lemma 7.4, N(Elk) = (JC m ) X U + C H c k X n 1

for E = k':,.k':r. Now, by Theorem 7.1, PElk induces

k x IN ( Elk) 2; Gal ( Elk). Therefore, there exists a field k' such that

k c k' c E,

HIN(Elk) 2; Gal(Elk')

(7.1)

101


under the above isomorphism (7.1). However, by Proposition 7.3, N(k' I k)1 N(EI k) 2; Gal(EI k')

also under (7.1). Hence H = N(k' I k). The uniqueness of k' follows from • Proposition 7.2(ii). 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 k x. Therefore, we see from Theorem 7.5 that the map k' ~ 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 x . ,

Furthermore, by Proposition 7.2(ii), this correspondence reverses inclusion, namely, if kl ~ HI, k2 ~ H2, then kl c

k2~HI

C

H 2.

Hence N(k' k"l k)

=

N(k' I k)

n N(k"l k),

N((k'

n k")1 k) =

N(k' I k)N(k"l k)

for any finite abelian extensions k' I k and k" I k. 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 kX. See Artin [1]. THEOREM 7.6. Let k'ik be a finite extension of local fields. Let E be any finite abelian extension of k and let E' = Ek'. Then N(E'lk') = {x'

E

k' I Nk'lk(X')

E

N(Elk)}.

Proof. Note first that E' I k' is a finite abelian extension and that k c E c E' C k~b' Let x' E k'x, x = Nk'lk(X'). 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' / k') ~ Pk'(X') I E'

= 1 ~ Pk'(X') I E = 1

~Pk(X) I E = 1

7.2.

~X E N(E/k).

Ramification Groups in the Upper Numbering

Let k' I k be a finite Galois extension of local fields and let G

=

Gal(k' I k).

•

102


In Section 2.5, we defined a sequence of normal subgroups Gn , n > 0, of G: 1 c ... c Gn c Gn - 1 c ... c Go c G, called the ramification groups (in the lower numbering) for the extension k' /k. Let k c- kIf -C k' , H = Gal(k' / kIf) c G and let H n , n > 0, be the ramification groups for k' / kIf. Then it follows immediately from the definition of Gn and Hn that

Hn

=

Gn nH,

for n >0.

Suppose now that kIf / k is a Galois extension so that H is a normal subgroup of G and \G/H = Gal(k"/k). Let (G / H)n, n > 0, be the ramification groups for the extension kIf / k. A natural relation one may expect between Gn and (G / H)n might be that (G / H)n is the image of Gn under the canonical (restriction) homomorphism G = Gal(k' / k)~ G/ H = Gal(k"/ k).

However, this is not true in general, and in order to obtain a simple relation between Gn and (G / H)n, we have to change the "numbering" of the ramification groups, as discussed below. Let Gn , n > 0, be as above. For each real number r > -1, we define a subgroup Gr of G as follows:

G- 1 = G, Gr = Gn ,

if n - 1 < r -< n, nEZ, n

>

0.

Let

gr = [Gr : 1],

for r > -l.

Then gr = gn for n - 1 < r -< n, n > 0, and

go = [Go: 1] = e (k' / k ) by Proposition 2.18. We consider a real-valued function q>(r) for r > -1 with the following two properties: (i) (ii)

q>(r) is a continuous, piecewise linear function of r > -1 with q>(0) = 0, The derivative q>'(r) of q> at r ft Z is equal to gr/gO'

Drawing the graph of q>(r), we see easily that there exists a unique such function q>(r) and that in fact

q>(r)=r, 1

for -l- (r) is an integer, then so is r. For each a in G = Gal(k' /k), let

°

v'(a(w) - w),

for all YEO',

and it follows that i(a)=v'(a(w)-w),

(7.3)

foraEG.

It is then clear from the definition of Gr that Gr = {a

E

G I i( a) > r + 1}

= {a E G

I v' ( a( w) -

Consider now a function A(r), r

>

w)

>

r + 1},

r >-1.

-1, defined by

1 A(r) = -1 +min(i(a), r + 1). go aeG

2:

Clearly A(r) is continuous and piecewise linear because each min(i( a), r + 1) is such a function. Furthermore A(O) = -1 + (go/go) = 0, and for r in the open interval i - 1 < r < i, i E Z, i > 0, A(r) is given by b A(r) = a +- (r go

+ 1),

where a and b are constants and b is the number of a's in G such that i(a»r+1-namely, b=[Gr :1]=gr. Hence A'(r)=gr/g(), and it follows from the definition of (j>(r) that (j>(r) = A(r)-that is, 1 (j>(r) = -1 + min(i( a), r + 1), go aeG

2:

for r

>

-1.

Since (j>(r) is, by (7.2), a strictly increasing function of r denote the inverse function of (j>(r): (j>(ljJ(r)) = ljJ({j>(r)) = r,

>

for r >-1.

We then define subgroups G r of G by GCP(S) = G S )

for real r,

S >

-1.

(7.4) -1, let ljJ(r)


104 For example,

G- 1 = G- 1 = G, These subgroups G r, r > -1, are called the ramification groups, in the upper numbering, for the finite Galois extension k' Ik.

Now, as mentioned above, let k" be a Galois extension of k, contained in k', and let GIH = Gal(k"lk). H = Gal(k'lk"), Then we have the following theorem: THEOREM 7.7. Let G r and (G IHY, r > -1, be the ramification groups in the upper numbering for k' Ik and k" Ik, respectively. Then (G IHY is the image of G r under the canonical homomorphism G ~ G IH: for r >-1.

(GIHY = GrHIH,

We shall next prove this theorem in several steps. (i) For each 0' E G I H,

where i GIH denotes the i-function for k"lk, e' = e(k' Ik"), and the sum on the right is taken over all elements a E G such that a 1-4 a' under G~ GIH-that is, a' = aH. Proof If a' = 1, then both sides are +00. Hence assume a' 1. Let 0" be the valuation ring of the normalized valuation v" on k". By Lemma 2.13, there exist elements wand z such that

*

0' =

o[w],

0" =

o[z).

Since v' I kIf = e' v", it follows from (7.3) that iG(a) = v'(a(w) - w),

Fix an element

0 E

i G1H ( a')

= v"( a'(z) -

z)

= ~ v'( a(z) e'

z).

G such that a' = aH. Since

1 ( IT (or(w)-w) ) , -;e1 L iG(or)=-;e1 L v'(ar(w)-w)=-;v' e T:EH

T:EH

T:EH

the equality to be proved is equivalent to the following: v'(a(z)-z)=v'(IT (ar(w)-w»). T:EH

Let s = [H : 1] = [k' : kIf] and let g(X) denote the minimal polynomial of w over kIf. Since k' = k"(w) and v'(r(w» = v'(w) >0, we see that g(X) =

IT (X -

r(w»

T:EH

= Xs

+ al xs-l + ... + as'

aj EO",

1 -< i -< s.

105


Let

ga(x)

= Xs + a(al)X

S

=

IT (X -

-

1

+ ... + a(as),

a(ai)

E 0"

ar(w».

T:EH

Then

ga(w) - g(W)

= ga(W) = IT (W - ar(w». T:EH

However, the coefficients a(ai) - ai of ga(x) - g(X) are divisible by a(z) - z in 0" = o[z). Hence we obtain

v'

(IT (w -

ar(w»)

:>

v'( a(z) - z).

T:EH

On the other hand, it follows from z polynomial heX) in o[X) such that

z

E 0" c 0' =

o[ w) that there

IS

a

= hew).

As heX) - z is a polynomial of o"[X) that vanishes for X

heX) - z = g(X)gl(X),

with gl(X)

E

=

w, it follows that

o"[X).

Applying a to the coefficients of both sides and using heX) we obtain

E

o[X)

c

k[X],

heX) - a(z) = ga(X)g?(X), where g?(X) is a polynomial of o"[X), defined similarly as ga(x). Put X = w in the above. Then

z - a(z) = ga(w)g?(w)

=

IT (w -

ar(w»g?(w),

T:EH

where g?( w)

EO'.

Hence

v'(z - a(z»:> v'

(IT (w -

ar(w»).

T:EH

•

This completes the proof of (i). (ii) For a'

E

GIH, let

j( a')

=

max(i e ( a) I a

E

G, aH = a').

Then i elH ( a') - 1 = cfJ k'lk,,(j( a') - 1),

where cfJk'lk" denotes the cfJ-function for the extension k'lk". Proof. We may again assume a' 1. Fix an element a aH = a', j(a') = ie(a). Let n = j(a'):> O. By (7.4) for k'lk",

*

1

E

G such that

cfJk'lk,,(j(a') -1) = cfJk'lk,,(n -1) = -1 +----;- L min(iH(r), n). e T:EH


106

Suppose iH ( r) :> n so that min(iH( r), n) = n. Then r E H n - 1 = H n Gn - 1 • Since iG(a) = n, so that a E Gn - 1 , it follows that ar E Gn - 1 , iG(ar):> n. By the choice of a, we then obtain iG( ar) = n = min(iH( r), n). Next, suppose iH(r) < n. Then r f. Hn - 1 = H n Gn - 1 • Since a E Gn - lJ it follows that iG( ar) = iH( r) = min(iH( r), n) also in this case. Therefore, by (i), 1

L

cfJk'lk,,(j(a') -1) = -1 +----;iG(ar) = -1 + iG1H(a'). e T:EH

•

(iii) (Herbrand's Theorem). Let s = cfJk'lk,,(r), r:> -1. Then

GrHIH = (GIH)s· Proof.

a'

E

GrHIH ¢:>iG(a):> r + 1 for some a ¢:>j(a'):> r + 1

E

G such that a' = aH

cfJk'lk,,(j(a') - 1):> cfJk'lk,,(r), because cfJk'lk" is an increasing function ¢:>i GIH(a')-1:>s, by(ii) ¢:>

¢:>a'

E

(G/H)s'

•

(iv) cfJk'lk = cfJk"lko cfJk'lk'" tJ1k'lk = tJ1k'lk"o tJ1k"lk' Proof. Since tJ1 is the inverse function of cfJ, it is sufficient to prove the first equality. Let A = cfJk"lko cfJk'lk'" It is clear that A is continuous, piecewise linear, and A(O) = O. Let r:> -1, r f. Z. Then cfJk'lk,,(r) f. Z by the remark after (7.2), and the derivatives satisfy

A'(r) =

cfJ~"lk(S)cfJ~'lk,,(r),

with s

cfJk'lk,,(r),

=

where

cfJ~"lk(S) =!, [(GIH)s: 1], e' with e'

=

e(k' Ik"), e" = e(k"lk). However, by (iii) above, (GIH)s

=

GrHIH.::; Grl(Gr n H)

=

GrlHr.

Since e'e" = e = e(k' Ik), it follows that

A'(r) =! [Gr : 1] = e

cfJ~'lk(r).

Therefore, by the uniqueness of the function cfJ(r), cfJk'lk(r) = A(r) for r:> -1. • We are now ready to complete the proof of Theorem 7.7. Let s = tJ1k'lk(r), r:> -1, so that Gr = Gs. By (iii) above,

GrHIH = GsHIH = (GIH)(J By (iv),

with t = cfJk'lk"(S),

107


Hence

•

7.7. Let both kl/k and k21k be finite Galois extensions of local fields (in Q) and let k' = k 1 k 2. Then

COROLLARY OF THEOREM

Gal(kl/ kY = 1, Gal(k2/ kY = 1 ¢:> Gal(k' / kY = 1. Proof.

By the theorem, the left hand side yields Gal(k' /kY

Gal(k' / k 1 ),

c

Gal(k' /kY c Gal(k' /k2).

However, since k' = k 1k2' the intersection of Gal( k' / k 1) and Gal( k' / k 2) is 1. Hence Gal(k' /kY = 1. The converse is clear. •

7.8. Let k' Ik be a finite Galois extension of local fields. Let lJ' denote the maximal ideal of k' and let LEMMA

q;(k'lk)=lJ/a,

a:>O,

for the different q; (k' I k) of the extension k' I k. Then a=

L

oc

L

i(a) =

(gn - 1),

a

E

G = Gal(k'lk),

where gn = [Gn : 1], n = 0, 1, 2, .... Proof. By Proposition 2.14, q;(k'lk)=f'(w)o',

where w is an element of 0' as in Lemma 2.13 andf'(X) is the derivative of the minimal polynomial f(X) of w over k. Since f(X) =

n (X -

a(w»,

aE

G,

a

it follows from (7.3) that a = v'{f'(w» =

L

v'(w - a(w» =

Let g~ = gn - 1. Since i( a) = n if and only if a

L

E

L

i(a).

Gn- 1 , a f. Gn, we see that

oc

i(a) =

L

n(g~-l - g~) = (gb - g~)

+ 2(g~ - g~) + ...

= go'+ gl'+ g2' + ....

7.3.

The Special Case k,;,n / k

Let m

:>

1, n

:>

°

and let

Jr

•

be a prime element of k':r Let

for the finite abelian extension k';' n over k, defined in Section 5.1. We shall next determine the ramification groups Gr and G r , r:> 1, for the extension


108

By Proposition 5.2, kur n k,:,n = k':,.-namely, k':,. is the inertia field of the extension k,:,nlk. Hence Go = H so that for all integers i :> O.

Gi = Hu

Therefore, we also have G r = H r,

for all real r:> O.

Since

G- 1 = G- 1 = G,

Gr

= Gr = H , for -1 < r

-1, denote the ramification groups in the upper numbering for k' I k. Then

Gr = Pk'lk(Ui ),

for i-I >

0, 1,

for sufficiently large m > 1 and n > O. By definition, Pk'ik is then the product of PElk:kx~ Gal(Elk) and the canonical (restriction) homomorphism Gal(Elk)~Gal(k'lk). However, by Theorem 7.7, Gr (=Gal(k'lkY) is the image of Gal(ElkY, r > -1, under the same homomorphism Gal(EI k) ~ Gal(k' I k). Therefore, it is sufficient to prove the equalities of the theorem for the extension Elk instead of k'ik. Now, by Lemma 7.4 and Proposition 7.10, N(k,:,i-l Ik) =

N(k7:~~-l Ik) =

Gal(ElkY = Gal(Elk,:,i-l), with k,:,-l = k-;/

(nm) X Ui'

for i - 1 < r < i, i > 0,

= k. On the other hand, by Proposition 7.3, PElk(N(k,:,i-1/k» = Gal(Elk,:,i-l).

Since (nm) is contained in the kernel N(Elk) = (nm) X Un+ 1 of PElk: EX~Gal(Elk), we obtain PElk(Ui ) = PElk((n m ) X Ui) = Gal(E/k,:,i-l) = Gal(E/kY. • Let k' / k now be any finite Galois extension of local fields and let G = Gal( k' I k). The definition of the subgroups Gn r > -1, of G in Section 7.2 states that Gr = Gi if i is the integer satisfying i - 1 < r < i, i > O. Hence it is natural to ask whether similar equalities: (7.5) Gr = G i, for i - 1 < r < i, i > 0, i E Z hold also for the ramification groups Gr in the upper numbering. This is not true in general. t However, for an abelian extension k' I k, we have the following theorem of Hasse-Arf, which is an immediate consequence of the above Theorem 7.12: 7.13. Let k' Ik be a finite abelian extension of local fields and let r be a real number > -1. Then Gr = G i, for i - 1 < r < i, i E Z, i > O.

THEOREM

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 Un

C

N(k' Ik) ~ Pk'lk(Un) = 1 ~Gal(k' Ikr

= 1,

n

:>

o.

Since any of these equivalent statements holds whenever the integer n is sufficiently large, let c(k' I k) denote the minimal integer n :> 0 for which the above conditions are satisfied, and define f(k'lk) = pc(k'lk),

p being, as usual, the maximal ideal of k. The ideal f(k'lk) of k is called the conductor of the abelian extension k' Ik. It follows immediately from the definition that f(k' /k)

= o~c(k' /k) = O~Gal(k' /k)O = 1 ~k' /k

is unramified,

namely, by Proposition 2.16, f(k'lk)

= o~qJJ(k'lk) = o'~D(k'lk) = 0,

o and 0' being the valuation rings of k and k', respectively. If k' I k is ramified, then f( k' I k) c p and it is the largest ideal of 0 such that Pk'lk(1

+ f(k' Ik» =

l.

We shall next discuss the relation between f(k'lk) and D(k'lk) for finite abelian extensions k' I k . Let Gn again denote the ramification groups, in the lower numbering, for the abelian extension k' I k. Let gn = [Gn : 1] and let Q> be the increasing function associated with k'ik (cf. Section 7.2). Let k LEMMA

7.14.

c-

kIf C- k' ,

H = Gal(k'lk"),

GIH = Gal(k"lk).

Suppose Gn $ H, Gn+ 1 c H for an integer n :> -1. Then c(k"lk) = 1 + Q>(n) 1

= - (go

go

+ g 1 + ... + gn),

where the last sum is meant to be zero if n = -l. Proof.

By Theorem 7.7, GrHIH = GrfJ(r)HIH = (GIH)rfJ(r).

Since Gr = Gn + 1 for n < r < n the assumption that (G I H)rfJ(n) =1= 1,

+ 1 and

since Q> is continuous, it follows from

(G I H)rfJ(n)+e = 1,

for small

E

> o.

Hence,'by Theorem 7.13, Q>(n) is an integer, and (GIH)rfJ(n) =1= 1,

(GIH)rfJ(n)+l = l.

Therefore, by (7.2) and by the definition of c(k"/k), 1 c (k" / k) = 1 + Q> (n) = - (go + g 1 + ...

go

+ gn).

•

113


COROLLARY.

Let Gn

* 1, Gn+

1

= 1 for an integer n > -1. Then

c(k'lk) = 1 + Q>(n) 1

+ g 1 + ... + gn).

= - (go

go

REMARK.

Q>(n) = c(k'lk) -1 motivated the definition of the function Q>(r).

Now, for each character X of the finite abelian group G = Gal(k'lk), let

Hx = the kernel of the homomorphism X : G ~ ex, kx = the fixed field of Hx in k' ,

f(X) = f(kxl k), the conductor of kxl k. THEOREM 7.15. Let D (k' 1k) be the discriminant of the finite abelian extension k' 1k. Then

D(k'lk) =

nx f(X),

where the product is taken over all characters X of Gal( k' 1k). Proof Let qj)(k'lk) be the different of k'ik and let qj)(k'lk) = 'Pta, a >0. Then, by Sections 1.3 and 2.4,

= N k'lk(qj)(k' 1k» = 'Pat, where f = f(k' Ik) = glgo with g = [G: 1] = [k': k], go = [Go: 1] = e(k' Ik). By D(k' 1k)

Lemma 7.8, 00

a=2: (gi- 1),

forgi=[G i :l], i>O.

i=1

Hence the equality of the theorem is equivalent to

where c(kxlk) is defined as above. Now, fix a character X, and for each i > 0, let X( Gi ) =

~ 2: X( a),

go

with a

E

Gi ·

a

Then

x(Gi ) = 1, = 0,

*

if X I Gi = 1, that is, Gi C Hx' if X I Gi 1, that is, Gi d: Hx'

*

*

Let X 1 so that Hx G and Gn d: Hx' Gn+ 1 c Hx for some n Lemma 7.14, we then see 1

00

c(kxlk) = - 2: gi(l- X(Gi». go i=O

>

-1. By


114

X(Gi ) = 1 for all i > o. Hence the above equality holds also in this case. When i is fixed and X

If X = 1, Hx

= G, then

kx

= k,

c(kxlk)

= 0, while

ranges over all characters of G,

L X(Gi ) = the number of characters X of G such that Gi c

Hx

x

=

the number of characters of G I Gi

=glgi.

Therefore, 100

(

)

00

Lc(kx/k)=-Lgi g-g =K L (gi- 1). x go i=O gi go i=O

•

This theorem is called the conductor-discriminant theorem for finite abelian extensions of local fields. EXAMPLE. Then

Let k' I k be a cyclic extension of degree I, a prime number. X=l~Hx=G,

X* l~Hx = 1,

f(x) = f(k/k) = 0, f(X) = f( k k ). I /

Hence D ( k I Ik) = f( k I Ik )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,:,nlk. 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 k'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: f = Fq. Suppose that (7.5) holds for the ramification groups Gr of k I Ik for all real r > -1. Then

so that [G: G n+ 1] I (q - l)qn,

for every n

>

o.

Since k I k is totally ramified, G = Go = GO and the number of elements, q in the residue field of k' is equal to q. Let n be any integer >0 and let n = cp (m ), n + 1 = cp (m ') for the increasing function cp associated with k'lk. By the remark after (7.2), m and m ' are integers. Since cp(m) < cp(m'), we have m < m'-that is, 0 <m < m + 1 <m'. Let r = cp(m + 1). Then n < r < n + 1 so that Gr = G n+ 1 by (7.5). Since Gt/>(s) = Gs , Proof

I

I,

115


s

>

-1, it follows that

If n = 0, then m = 0, and [Go: G I ] = [Go: Gd. If n > 1, then m > O-that is, m > 1. Since q = q', G = GO, 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 JC of k such that F

n kur = k,

Let k' be any finite extension over k in F: k Gal(k' Ikr+ 1 = 1 for some n > 0. Let

c

k'

c

F, [k': k] < +00. Then

k" = k' k':r.

r+

Since Gal( k':rl k 1 = 1 by Proposition 7.10 for m = 1, it follows from the Corollary of Theorem 7.7 that Gal(k"lkr+ 1 = 1.

As k c k" c F, k" I k is a totally ramified, finite, abelian extension. Hence, by Theorem 7.13 and Lemma 7.16, we obtain [k": 1] = [Gal(k"lk): 1] I (q - l)qn.

However, we know by Proposition 5.2 for m = 1 that [k':r: k] = (q -l)qn. As k c k':r c k", it follows that

Thus every finite extension over k in F is contained in kjf' Hence

The above is the idea of the proof of Theorem 6.8 in Gold [9] and Lubin [18]. t One sees that if Theorem 7.7 and various properties of the abelian extensions k':rl k 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 [20].

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 Qp.

8.1.

.7C -Sequences

Let (k, v) be a local field with residue field f = o/lJ and let JC be a prime element of (k, v). In the following, we consider a pair (f, w) where f is a power series in the family ;¥~ of Section 5.1 and where

W=

{wn}n~O

"f = Un W7

is a sequence of elements in the abelian group satisfying, for f = (JC)f'

Wo E

WJ,

Wo =1= 0,

Wn

=

f( Wn+l)

=

JC j Wn+l,

of Section 4.3,

for all n >

o.

Such a pair will be called a JC-sequence for k. Given f E ;¥~, there always exists a JC-sequence (f, w) for k because JC j W7+ I = W7 by Lemma 4.8(iii). It also follows from the same lemma that Wn has the properties

Wn d:.'F W f - ' so that Wn is a prime element of k':r = k(W7) by Proposition 5.4(ii). Let (f', w') be another JC-sequence for k and let w' = {w~}n~O. Suppose h(X) is a power series in o[[X]], invertible in M = Xo[[X]] (cf. Section 3.4), such that n

hofoh-I=f',

I

h( Wn) = w~,

for all n

>

o.

We shall call such an h an isomorphism from (f, w) to (f', w') and write

h:(f,

w)~(f',

w').

It then follows that h defines isomorphisms over 0:

Fj ~ Ff"

W7 ~ W7·,

"f ~ WI'.

8.1. Let (f, w) be a JC-sequence for k with W = {wn}n~O. Let g(X) be a power series in 0[[ X]] such that g( Wi) = 0 for 0 < i < n. Then g(X) is divisible by [JCn+l]f in o[[X]]. If g(w;) = 0 for all i > 0, then g(X) = O. Proof. By the Corollary of Proposition 5.4, the elements f3 E Wj, f3 ft Wj-I, are the conjugates of ai over k (= k~r). Since g(X) E o[[X]], g( Wi) = 0 implies g(f3) = 0 for all f3 E Wj, f3 ft Wj-I. As this holds for o< n, we see that g(a) = 0 for all a E W;' By the remark after Lemma 5.5, g(X) is then divisible by [JCn+l]f in o[[X]]. Now, since [JC]f = f for LEMMA

117

Explicit Formulas

f E @P;, [nIt is contained in the maximal ideal (n, X) of 0 [[ X]], generated by nand X. Hence [nn+lIt = [n]/ o • • 0 [nIt belongs to the (n + l)th power (n, Xr+l of (n, X) in o[[X]]. Therefore, if g(w i) = 0 for all i > 0, then g(X) E (n, Xr+l for all n > 0 so that g(X) = O. • 8.2. Let (j, w) and (j', w') be n-sequences for k. Then there exists a unique isomorphism PROPOSITION

h : (j, w):::; (j', w').

Furthermore, if w = w', then h(X) = X so that f = f'. Proof. Since f, f' E @P;, it follows from the proof of Lemma 5.1(i) that there is a power series 8(X) in o[[X]] such that 8(X) = X mod deg 2, 8 0f=f'08 (cf. (4.8)). Let w"={w~}, where w~=8(wn)' n>O. Then 8: (j, w):::; (j', w"). Hence, to prove the existence of h : (j, w):::; (j', w'), we may assume that f = f'· In this case, since Wn, w~ E WI' Wn, w~ ft WI-I, it follows from Lemma 4.8(ii) (iii) that there exists Un E U = U(k) such that UnjWn=w~ and that such Un is uniquely determined in UmodUn+1 (= 1 + lJn+l). Let m > n > O. Then

whereas Un j Wn = w~. Hence it follows from the above remark that Urn = Un mod U n+1 for 0 0, and such an element u satisfies u j Wn = w~ for all n>O. Let h=[u]1 so that h(wn)=w~,n>O. Since h o f o h- 1 =f because f = [n ]/' we obtain h : (j, w):::; (j, w'). To prove the uniqueness of h: (j, w):::; (j', w'), it is sufficient to show that w = w' implies h(X) = X. Let g(X) = h(X) - X. Then Wi = wI = h(wj) implies g(w i ) = 0 for all i > O. Hence g(X) = 0 by Lemma 8.1. •

=

For 0

= nj

Hence g(X) =

Wn+l -

°

n (Wn+l t y)

=

0,

Wn - Nn+1,m(Wn+l)

=

0,

YEWJ.

y

•

by Lemma 8.1.

For convenience, a power series I in f.Fl will be called normed if it satisfies NtCX) = X. Thus Lemma 8.3 simply states that an-sequence (f, w) is normed if and only if I is normed.

Let (k, v) be a p-field. Let p > 2, I(X) = nX + xq E f.F~, and let (f,w) be a n-sequence for k. Then I(Wn+l) = [n]tCwn+l) = Wn" Since k~+l = k( wn+d = k~( wn+d, [k~+l: k~] = q = deg(f) by Propositions 5.2 and 5.4(ii), we see that I(X) - Wn is the minimal polynomial of Wn+l over k~ so that Nn+1,n( -wn+d = -Wn" Since q is a power of p and is odd, it follows that N n+l,n( wn+d = Wn for n > 0. Hence (f, w) is a normed n-sequence for k. In the case p = 2, we can see similarly that (f, w) is normed for I(X) = nX - xq E f.F~. EXAMPLE.

For n

>

0, let

Bn

= (k~)X =

the multiplicative group of the field

k~,

and let where the inverse limit is taken with respect to the norm maps

Nm,n: (k';)X~ (k~)X,

for

° O.

(ii) If (j, w) is normed, then for any. f3 E B with arbitrary e = v(f3), there again exists a unique power series t(X) in xe o[[X]] such that t( Wn) = f3n, for all n > O.

Proof (i) Since v(f3) = v(Nn(f3n)) = 0, f3n is a unit of k~. Hence it follows from Proposition 5.11 that for each n > 0, there exists tn(X) E o[[X]] such that It then follows that

tn+1 ( Wi) - tn( Wi)

=

f3i - f3i

=

0,

for 0 O. Then f3' E B, v(f3') = O. Hence, by (i), there is a power series t'(X) in o[[X]]X such that t'(w~) = f3~ for all n > O. It then follows that t = Xet' is a power series in xe o[[X]] with the property t( Wn) = f3n for all n > O. The uniqueness of t follows from that of t'. • Now, fix a normed n-sequence (j, w). By the above proposition, for each f3 E B, v(f3) = e, there exists a unique power series tf3(X) in xeo[[X]]x such that tf3(W n) = f3n, for all n > O. For example, tw (X) = X for W = (wo, WI, . . . ) E B. By the uniqueness, we have tfj f3' = tfjt w, for f3, f3' E B. Suppose v(f3) = 1 for f3 t Compare Coleman [5].

E

B. Then tf3

E

Xo[[X]r so that tf3 is invertible in M

120


and ~~.

ff3 = tf3 of 0 ("Ii 1 E

Since tf3(w n)=[3n for n>O, we see that the paIr (ff3,[3) Jr-sequence for k and

IS

a normed

tf3 : (f, w) ~ (ff3' (3). By Proposition 8.2, ff3 is the unique power series in o[[X]] such that (ff3' (3) is a Jr-sequence for k. Hence ff3 depends only on [3 E B (with v([3) = 1) and it is independent of (f, w). Furthermore, if (f', w') is any normed Jr-sequence for k, then w' E B, yew') = 1. Hence (f w', w') is a normed Jr-sequence for k so that f' = f w' by Proposition 8.2. Thus we see that the family of pairs: for all [3

{(ff3' (3)},

E

B with v([3) = 1,

is nothing but the set of all normed Jr-sequences for k.

8.2. The Pairing (a', f3)f Let Jr be a prime element of (k, v) and let f residue field of k~, n > 0, by onl'Pn.

E

~~. As before, we denote the

Let n > 0. (i) For each a E 'Pn, there exists an element ; in rntC = 'Pg) such that Jr n+1 j; = a. Let k' = k~(;). Then k' / k~ is an abelian extension with [k': k~]

qn+1 = [Wi: 0]

>

[k' : k~]

>

e(k' / k~)

so that [k' : k~] = [Wi: 0] = qn+1 = e(k' / k~),

and all statements in (ii) follow. For n > 0, let Pn denote the norm residue map PE for E =

k~:

•

Pn : Bn = (k~)X ~ Gal(k~,ab/ k~).

Fix f

E

@P~ and let

Let nn+1 j; = a, k' = k~(;) as stated above. Since k' / k~ is abelian-that is, k~ c k' c k~,ab' Pn([3) I k' is an element of Gal(k' / k~). Let

Then (a, (3)n,f is an element of Wi and it is independent of the choice of ; such that nn+1.f f::,t = a , because if nn+1.f f::,t' = a ,f::,t' E f::,t f wnf' wnf -c k n so m that

+

i ;' = Pn([3)(;) i ;.

Pn([3)(;')

We shall next study the properties of the symbol (a, (3)n,!, When f is fixed, the suffix f in (a, (3)n,f, [3, and so on will often be omitted.

at

LEMMA

8.6.

(i) (a 1

+a2, (3)n = (aI, (3)n +(a2' (3)n, (a, [31(32)n = (a, (31)n +(a, (32)n, (a . a, (3)n = a . (a, (3)n,

for a E o.

(ii) (a, (3)n =0~[3 EN(k~(;)/k~), nn+1.; = a. (iii) Let f be normed and let a be a prime element of

k~.

Then

(a, a)n,f= O. Proof. (i) If ;1,;2 are elements of m f such that nn+1';1 = aI, nn+1 . ;2 = a2, then nn+1. (;1 ;2) = a1 a2' Hence the first equality follows. The third equality can be proved similarly. Since Pn([3[3') = Pn([3)Pn([3'), the second equality follows from the fact that a~a(;).!..; defines a homomorphism: Gal( k~( ;)/ k~) ~ Wi. (ii) It follows from Theorem 7.1 that

+

+

Pn([3)(;) = ; ~ Pn([3) I k~(;) = 1 ~ [3

E

N(k~(;)/k~).

(iii) Let nn+1 j; = a. Since f is normed, Nf(X) NtCNtC· . ·))(X) = X so that by Lemma 5.9(i), [nn+1]f =

n (X t y), y

= X. Hence Ni(X) =

YEWi·

122


Therefore,

a

=

nn + 1 j ;

=

n (; t y),

YEW;'

y

However, by Lemma 8.5(ii), ;

+t WI is the complete set of conjugates of ;

over k~. Hence a is the norm of ; for the extension from (ii) that (a, a)n,f = 0.

k~( ;)/ k~,

°

and it follows •

Let < n < m. Then the maximal ideal t.Jn of k~ is contained in the maximal ideal t.Jm of k';. Hence, if a E t.Jn, then nm-n j a E t.Jn c t.Jm.

LEMMA 8.7. (i) Let O-

0,

at

Proof. (i) For simplicity, the valuation {t on Q (cf. Section 3.1) will be denoted by 11. Let e = 1l(P) > 0 and let pE, E >- 0, be the exact power of p dividing n. Then pE < n, Il(n) = eE < e logp n. Since Cn E 0, Il(c n) >- 0 in (8.1), Il( ~an)

>-

nll(a) - e logp n,

where 11 (a) > 0 for a E mf = pg. Hence (c nIn) an ~ 0 as n ~ + 00, and A(a) = ~:=1 (cnln)a n converges in Q. The rest is clear. (ii) A: Fj ~ Ga means A(Fj(X, Y)) = A(X) + A(Y). Hence the first equality is clear. We also have

AO [a]foA -1 = aX mod deg 2. By the remark after Proposition 4.2, we then see that A0 [a ]f 0 A-1

=

aX,

that is, A0 [a ]f = aA.

This yields the second equality.

•

We now fix a normed n-sequence (f, w), W = {w n }n2:0, for k. Let A(X) be a logarithm of Fj. For each 13 = (13o, 131, ... ) in B, we define (8.2) where tf3(X) is the power series in Theorem 8.4 such that tf3( Wm) = 13m for all m >- 0 and where A' = dAI dX, t~ = dtf3 IdX. Since tf3 (w n) = f3n =1= 0 and since oc

A'(X) = u +

2: cnX n- 1,

n=1

so that Il(A' (wn)) = 0, A' (w n) =1= 0, we see that D(f3)n is a well-defined element in k~. Note that D(f3)n depends not only on f3 and n >- 0 but also on (f, w) and the choice of the logarithm A for Fj. 8.9. (i) D(f3f3')n = D(f3)n + D(f3')n, for f3, 13' E B. (ii) D(f3)n E p;1 for all 13 E B. (iii) Let 0 O. Therefore, T m,n( D(f3)m) = Jrm- nD(f3)n for 0 < n < m. Let n :> 0 and let an

E

An (= Pn), f3

•

E

B. We define

1

xn(an, f3) = Jr n+1 Tn(A(an)D(f3)n) where Tn denotes the trace map of k~/ k and A is the logarithm of Fj mentioned above. If A is replaced by another logarithm UA, U E U, then A(an) is replaced by uA(an), and D(f3)n by u- 1 D(f3)n (cf. (8.2)). Hence xn(an, f3) is unchanged. Thus xn(an, f3) is an element of k, depending on n, an' f3, and the fixed normed Jr-sequence (f, w), but not on the choice of A. LEMMA

8.10.

Let an' a~ E An' f3, f3'

(i) Xn(an -+- a~, f3) = xn(an, f3) f

E

B.

+ Xn(a~, f3),

Xn(a jan, f3) = axn(an, f3), a E 0, xn(an, f3f3') = xn(an, f3) + xn(an, f3').

126


(ii) xm(.rr m- njan, f3) = nm-nxn(an, f3),

for 0 - 0. Differentiating both sides of tJ3 = tJ3 oh, we obtain (t~oh)h'=t~ with derivatives t~=dtJ3/dX, and so on. Since h(wn)=w~, it follows Similarly, X0 h = ). yields i.'(w~)h '(Wn) =

)"(wn).

As h(X) == vX mod deg 2 with v E U, we see that h '(X) = v mod deg 1, h'(w n ) *0. Hence it follows from the above that t~(wn)

1

b(f3)n

=

)"(wn)

~n

1

=

t~(w~)

X'(w~) ~n

-

=

b(~)n'

l>(~)n

being the expressIon (8.2) for the normed n-sequence (f', w'). Therefore,

for an

E

An, representing a. Consequently, for large n, h([a, ~]w) = h(xnjwn) =xnj' h(w n) =xnj' w~ =

8.4.

[h(a), ~]w'.

•

The Main Theorem

Still fixing a normed n-sequence (f, w) for the p-field (k, v) of characteristic 0, we shall prove the formula

(a, ~)t = [a, ~]w,

for a

E

At, ~

E

B,

where ( , )t and [ , ]w are the pairings At x B ~ "1, defined in Sections 8.2 and 8.3, respectively. We first prove some elementary lemmas. 8.12. Let e = v(P) = f.l(P) >0, where f.l is the extension of v on Q and Q. Let y E Q, f.l(Y) >- e. Then LEMMA

f.l(yi/j)

>-

2f.l(Y) - e,

for all integers j

>-

2.

128


Proof.

Let j

= paj',

>-

0, V', p)

= 1.

Then

f.l(yj Ij)

= jf.l(Y) -

f.lV)

= jf.l(Y) -

a

Hence the lemma is trivial if a = O. For a

where j - 1 - a

>- pa

- 1- a

>-

0 for a

>-

ae.

1,

1.

>-

•

8.13. Let on denote, as before, the valuation ring of k:. Let Y E naon, where e < a - c in the above, we obtain the lemma. • We now prove the following key lemma: LEMMA

8.15.

Let if, w) be a normed n-sequence for k. Then (a, w)f = [a, w]w

for every a E A f . Proof Note first that W = {w n}n2:0 belongs to B. Let n >-3c

for e = 1l(P) and for the integer c by an EAn and

>-

+e+5

0 in Lemma 8.14. Then a is represented i, j

>-

1

by (4.1). Hence, if we put i, j

>-

1,

then Il(anlw n) < ll(cija~w~-l) so that

Il(Yn) = Il(an) -Il(w n) >- n - c - 1> 0, This shows that an

+f Wn,

Il(an +Wn) = Il(W n). f

(8.4)

as well as Wn, is a prime element of k~. It then

follows from Lemma 8.6(iii) that

(Wn' wn)n,f= (an

t Wn, an t wn)n,f= O.

However, omitting the suffix f, we have

(an

+Wn, an +wn)n = (an +Wn, wn(1 + Yn))n, = (an, wn)n +(an, 1 + Yn)n +(wn' wn)n +(wn' 1 + Yn)n'

Hence

(a, w)f = (an, wn)n =

...!..

(an, 1 + Yn)n...!.. (wn' 1 + Yn)n'

(8.5)

We shall next compute (an, 1 + Yn)n and (wn' 1 + Yn)n' · SInce n n+1 . W2n+1 -- Wn+l1 W2n+1 E k ab, we see th a t

(Wn' 1 + Yn)n = Pn(1

+ Yn)(W2n+1)...!.. W2n+11 by the definition of ( = Pk(Nn(1 + Yn))(W2n+1)...!.. W2n+11 by Theorem 6.9, = (Nn(1 + Yn)-l - 1) j W2n+11 by (6.4).

, )n,

130


Since Yn that

E

nn-c-1 0n by (8.4), it follows from Lemma 8.13 for a = n - c - 1

+ 2.

where m = 3(n - c - 1) - e >- 2n

7;, (Yn) == 0

It also follows from (8.3) that

mod nn+(n-c-l),

hence mod nn+1.

(8.6)

Therefore, we obtain from the above that 1

(Wn' 1 + Yn)n = (-Tn(Yn»' W2n+1 = (- nn +1 7;,(Yn»)' nn+l. W2n+1 (8.7)

1 =

(nn +1 7;,( Yn») . Wn'

-!..

As before, let A denote a logarithm for Fj. Then, by (8.1) oc

2: ~i Xi,

A(X) =

i=l

C1 = U,

Hence it follows from an

+Wn = wn(l + Yn) that

where (1

for i >- 1.

Ci EO,

l

l

(i -

~ 1) 1 . . 1 ~y~. j=2 ] ]

+ Yn)i 1

-'---.~=~+Yn+ L.J l

By Lemma 8.12, l1(y~/j) >- 211 (Yn) - e >- 2(n -

C -

1) - e,

for j >- 2.

Therefore,

A( an) + A( Wn) == A( Wn) + YnWnA' (wn) mod n 2(n-c-l)-e, where A' = dAI dX. Since

we obtain from (8.2) and from the above that

'( an ).st.( ) U W n == Yn mo d n 2(n-c-1)-e-1 .

I\,

Consequently, by (8.3),

7;,(A(an)b(w)n) == 7;,(Yn)

mod n m',

where

m' = 2( n -

C

+ 1) - e - 1 + n >- 2n + 2

n >-3c + e + 5. Since (l/nn+1)7;,(Yn) E 0 by (8.6), (1/ nn+1)Tn(A(an)b(w)n) also belongs to 0, and it follows from m' >-2n + 2 that because

of

1

~

n

1

7;,(A(an)b(w)n) =~ 7;,(Yn) n

mod nn+1

131

Explicit Formulas

so that by (8.7),

(wn' 1 + Yn)n

=

-=- (nn\l Tn( Yn)) . Wn

=

1 -=- (nn +1 Tn(A(an)- 0 above. Since n >- c, it follows from Lemma 8.7(i), that

(an' 1 + Yn)n = (ao Nn,c(l However, (8.4)-that is, Yn

E

Nn,c(l

+ Yn))c = Pc(Nn,c(l + Yn))(~) -=-~.

nn-c-lon-implies

+ Yn) = 1 mod nn-c-l.

By Lemma 3.5 and Proposition 7.2, N(k' / k~) is a closed subgroup of finite index in (k~)x. Therefore, the above congruence shows that N n,c(l + Yn) is contained in N(k' / k~) whenever n is large enough. For such an n, we then have Pc(Nn,c(l + Yn)) I k' = 1 by Theorem 7.1 so that

(an' 1 + Yn)n

=

Pc(Nn,c(l

+ Yn))(~) -=- ~ = o.

(8.9)

Finally, we see from (8.5), (8.8), and (8.9) that for sufficiently large n,

(a, w)f = -=- (an' 1 + Yn)n -=- (wn' 1 + Yn)n = [a, w]w-

•

We are now ready to prove the following theorem: THEOREM

8.16.

Let (f, w) be any normed n-sequence for k. Then (a, f))f = [a, f)]w,

for a

E

Furthermore, if a is represented by an in An, n

belongs to

0,

A f , f) >-

E

B.

0, then the element

and (a, f))f = Xn j Wn = [nn\l Tn(A( an)- n, we know that Xm (= Xm (am' (3)) belongs to 0 and (a, (3)t = [a, {3]w = Xm j Wm' However, since a is represented by an E An, it follows from the definition of (a, {3)t that (a, {3)t = (an' (3n)n,f E WJ. Hence, by Lemma 4.8(ii), As Xm = Jrm-n xn by Lemma 8.10(ii), we obtain Jrm+l x n

m 1 E +" h +

• not necessarily normed. By

that is'xn E' 0

Let if, w) now be any Jr-sequence for k, Theorem 8.4, for each (3 E B with v({3) = 0, there exists a unique power series tf3(X) in o[[X]r such that tf3(w n) = {3n for all n >- O. With this tf3 , one can define D({3)n by (8.3) and, hence, xn( a, (3) and [a, {3]w for (3 E B with v({3) = O. Take a normed Jr-sequence if', w') for k and let h : if, w)::::; if', w').

Since Theorem 8.16 holds for if', w'), it follows from Lemma 8.11 that Theorem 8.16 holds also for if, w), provided that v({3) = O. 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 {3n E Bn = (k~) x . Then {3n is the nth component of an element {3 = (13o, (31, ... ) in B if and only if Nn({3n) is a power of Jr: N n({3n) E (Jr). Proof. Let {3n be the nth component of {3 = (13o, (31, ... ) in B. Then Nm({3m) = N n({3n) for all m >- O. Hence N n({3n) E N(kJr/ k) = (Jr) by Proposition 5.17. Suppose, conversely, that Nn({3n) E (Jr). Apply Theorem 7.6 for k' =knJr'

E = k nJr + 1 ,

Since Nn({3n) E (Jr) = N(kJr/k) c that theorem. Therefore,

N(k~+l/k),

(3n belongs to

N(k~+l/k~)

by

(3n = N n+l,n({3n+l) for some {3n+l E Bn+l = (k~+l)X. Since N n+1({3n+l) = Nn({3n) E (Jr), we can similarly find (3n+2 E Bn+2 such that N n+2.n+1({3n+2) = {3n+l' In this manner, we obtain a sequence {3n, {3n+b (3n+2, ... , satisfying N m+l.m({3m+l) = 13m for all m >- n. Putting 13m = N n.m({3n) for 0 -< m -< n, we see that {3n is the nth • component of {3 = (13o, (31, ... ) 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 if, w) be any normed Jr-sequence for k. Let an be an element of l:In (= An), {3n an element of (k~)X (= Bn) such that

133

Explicit Formulas

Nn(f3 n) is a power of n, and let

1 (1

Xn = nn+l Tn A'(W ) n

t~( Wn) ) f3n A(an) ,

where A(X) is a logarithm of Fj and f3 is any element of B such that f3n is the nth component of f3. Then Xn belongs to 0, Xn is independent of the particular f3 chosen, and

_

_[1 (1 t~( Wn) )] nn+l Tn A'(W ) f3n A(an) jWn·

(an' f3n)n,f-X n jW n -

n

Furthermore, if f3n is a unit of k~, then the same formula holds for any n-sequence if, w), not necessarily normed.

8.S. The Special Case for k = Qp In this section, we shall see what Theorem 8.16 states in the special case:

n =p. Suppose first that p > 2 and let

f(X) = (1 + xy -1

E

:¥;.

Then (cf. the example in Section 4.2)

Fj(X, Y) = (1 + X)(1 + Y) - 1, and

w; = {~- 11 ~ E Q, ~pn+1 = I}, k~

= Qp(W;) = Qp(Wpn+I),

where W pn+1 denotes the group of all pn+lth roots of unit in

a

+t f3 = (1 + a)(1 + f3) -

Q.

1, the map ~ ~ ~ -1 defines an isomorphism K:Wpn+I~W;,

Let

Wn

= ~n -

1E

W;,

where ~o =1= 1,

~g

~~ = ~n-l'

for n

>

1.

= f,

As [n]t

n j Wn = f( ~n - 1) = so that

= 1,

if,

X

~~ -

1 = ~n-I

-

1 = Wn-I,

for n

>

1,

w) is a n-sequence for k = Qp and n = p. Furthermore,

+t (~ - 1) = (1 + X)~ -

1,

for ~ - 1 E WJ, that is, for ~

Since p is odd, it follows that

f1C (X +t (~- 1)) = (1 + xy -

1 = f(X) = [n]t.

E

Wp.

Since


134

Therefore, by Lemma 8.3, (j, w) is a normed Jr-sequence. (This can also be 1) = 'n - 1 for 0 1 and let k' be a local field containing primitive mth roots of unity. For any x, y E k'x, let

{x, y}

= (Vy)O-t, where 0 = 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 (8.10)

for the norm residue symbol { , } for k~ = Qp(W pn+l) with exponent pn+l. Thus we see that in this special case, the pairing ( , )f: Af x B ~ "f is given, up to the isomorphism K, by the norm residue symbols of Qp(Wpn+l) for n:> O. We shall next describe [a, f3]w- For this, let A(X) = 10g(1 + X) = X -

Then A(X)

E

X2

X3

2 + 3 - ....

Qp[[X]] and

A«1.+ X)(l + Y) - 1) = 10g(1 + X)(l + Y) = 10g(1 + X) + 10g(1 + Y) = A(X) + A(Y).

Explicit Formulas

135

Hence A: Fj .:; Ga is the unique logarithm of Fj with A(X) = X mod deg 2. Since A'(X) =

1 1 +X'

we have

g:

f>(fJ)n =

(g: 1;'(

1

Xn =pn +1 Tn

W n) log(1

1;'(Wn),

+ -O.

~

We also state below two elementary lemmas which are easy consequences of the definition of Hl(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-l)A.

Clearly, N(A) cA G cA. For each a EA G , we define g(a, T) in C 2 by

g(pi, pi) = 0,

for 0 - 0, are defined. Such groups are called Galois cohomology groups. For example, G acts continuously on the multiplicative group K X of K, viewed as a discrete group. Hence Hn(G, K X), n >- 0, are defined. For simplicity, Hn(G, KX) will also be denoted by Hn(KI F): Hn(KIF) = Hn(Gal(KIF), K X),

n

>-

O.

Let {E} be the family of all finite Galois extensions of F, contained in K: FcEcK.

fE :Fl

< +00.

141

Appendix

Let U = Gal( K / E) for such a field E. Then U is an open normal subgroup of G and Gal(E/F) = G/U,

Hence it follows from Proposition A.2 that Hn(K/ F) = lim Hn(E/ F), ~

(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 K/ F, Hl(K/F) = O.

Proof. It is one of the fundamental theorems in Galois theory that Hl(E/ F) = 0 for any finite Galois extension E/ F. Hence the proposition

•

follows from (A.3).

A.6. Let K / F be a Galois extension and let E be any Galois extension over F, contained in K: F c E c K. Then the sequence PROPOSITION

O~H\E/F)~H\K/F)~H\K/E) is exact. Proof.

This follows immediately from Propositions A.I and A.S.

•

As a consequence, we can imbed H\E/ F) in H\K/ F) by means of the inflation map: H\E/F) cH2(K/F).

(A.3) then shows that H\K/ F) is the union of the subgroups H2(E/ F) when E ranges over all finite Galois extensions over F, contained in K: H\K/F) =

U H2(E/F).

(A.4)

E

Let k be any field and let Q s 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) = H\Qs/k).

Since Q s 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, v) 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 Q of k: K = k ur .

Since K I k is abelian, we have kcKcQs·

Hence, by Proposition A.6, we obtain an exact sequence: o~ H2(Klk)~ Br(k)~ Br(K).

(A.5)

We shall first study H2(KI k). By Sections 2.3 and 3.2, for each integer n >- 1, there exists a unique un ramified extension k~r over k in Q with [k~r: k] = nand K

= kur = U

n~l

k~r'

Clearly {k~r}n~l is the family of all finite Galois extensions over k in K. Hence, by (A.3),

Let CfJ be the Frobenius automorphism of k: CfJ E Gal(Klk). Then Gal(k~rlk) is a cyclic group of order n, generated by CfJn = CfJ I k~n and by Lemma A.3, CfJn defines an isomorphism k X I N(k~rl k) ~ H2(k~rl k).

Let

j{

(A.6)

be a prime element of k. Then

by Lemma 3.6. Hence the isomorphism v:kxIU(k)~Z,

defined by the normalized valuation v of k, induces k X I N(k~J k) ~ ZlnZ

so that (lIn)v defines an isomorphism 1 k x IN (k~rl k) ~ - Z/Z. n

The product of the inverse of (A.6) and the above isomorphism then yields 1

H2(k~rlk) ~ - Z/Z.

(A.7) n Since this isomorphism is quite important, let us explicitly describe the map according to the above definition. For each x E k X , let ex denote the residue class of g(a, r) in H2(k~rlk) = Ker(b 2)/Im(b 1 ) (cf. Section A.l), where g is

143

Appendix

defined by g( lP~, lP~)

=

for 0 nand, consequently N(LIK) = KX. By Lemma A.3, we then have H2(LIK) =

O.

•

By (A.5) and Theorems A.7 and A.9, we now obtain the following result: THEOREM

A.10.

Let kur be the maximal un ramified extension of a local field

k. Then Hence there exists a canonical isomorphism: Br(k) ~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) is defined.

=

H2(Q sIk)~ H2(Q sIk')

=

Br(k')

145

Appendix THEOREM

A.Il.

Let d = [k': k]. Then the diagram Br(k)

~

Ires

Q/Z

I

d

Br(k')

~

Q/Z

is commutative. Proof. Let e=e(k'lk), f=f(k'lk). Let qJ, qJ' be the Frobenius automorphisms of k and k', respectively, and let v, v' denote the normalized valuations of k and k', respectively. Then, by Chapter II, v' I k = ev,

ef = d,

qJ' I K = qI.

By (A.4) and Theorem A.la,

Br(k) = H2(k ur lk)

= U H2(k':,Jk) , n~l

and similarly for Br(k'). Hence, if c E Br(k), then c E H2(k: r lk) for some integer n > I with fin. By an earlier remark, c = Cx for some x E kX_ namely, c is represented by g (a, i) such that

g( qJ~,

qJ~)

a n'.

But, then . inv(c) = vex) mod Z, n

inv(res(c)) =

v'(x) mod Z, n'

where

v'(x) = ef vex) = d vex) . n' n n

• be in particular a finite Galois extension.

Hence inv( res( c)) = d inv( c), and the theorem is proved. In the above theorem, let k' I k By Proposition A.5,

H\k' Ik)

= Ker(res: Br(k)~ Br(k')),

146


Hence, it follows from the theorem that inv: Br(k) ~ Q/Z induces a natural isomorphism inv'. H2(k' Ik) ~ --!d Z/Z '

d=[k':k],

which generalizes (A.7). The element c of H2(k' Ik) such that . mv(c)

1

= dmodZ

is called the fundamental class in H\k'lk). By means of Theorems A.I0 and A.ll 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, v) be a local field and let k' be a finite Galois extension of k. Let L=k'ur =k'K

and let K and L be their completions (in Q). Clearly L/ K is a finite Galois extension and by Lemma 3.1,

L,

KL =

KnL=K.

Hence LI K is also a finite Galois extension and a ~ a I L ~ a I k' defines Gal(LI K) ~ Gal(LI K) ~ Gal(k' Iko),

where ko = k' n K is the inertia field of the extension k'ik. It then follows from the Corollary of Proposition 2.19 that LI K and LI K are solvable extensions. PROPOSITION

A.12.

Let N(LI K) denote the norm group of the finite

extension LI K. Then N(LI K) = K.

Since LI K is solvable, it is sufficient to prove the proposition for the case where LI 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]. • Proof.

Now, let V(LI K) = the subgroup of U(K), generated by all elements of the form

;0-1 =

a( ;)1; for a

E

Gal(LI K) and; E U(L).

By Lemma 3.2, 11K (= 11 I K) and, hence, Ilk (= iii K) are normalized valuations. Therefore (K, 11 k) is a complete field in the sense of Section 1.2

147

Appendix

and it follows from Section 1.3 that L is also a complete field with respect to a normalized valuation, equivalent to J-l L (= ill L). Let lC' be any prime element of L. For any a in Gal(LI K), lC,a-l (= a(lC')1 lC') then belongs to U(L) so that a map Gal(LIK)~ U(L)IV(LIK), a~ lC,a-l

mod V(LI K)

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 lC'. Since U(L)IV(LI K) is an abelian group, it then induces a homomorphism i: Gal(LI Kt b ~ U(L)IV(LI K) where Gal(LI Kt b denotes the factor commutator group of Gal(LI K). Let N (= N Did denote the norm map of the finite extension LI K. Then we have the following PROPOSITION

A.13.

The sequence

1 ~ Gal(LI Kt b -4 U(L)IV(LI K).14 U(K)~ 1

is exact. 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(LI K), it can also be derived from the long exact sequence of such cohomology groups, associated with the exact sequence Proof.

I~U(L)~LX~Z~I,

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 k' I k is a finite abelian extension. Then the extensions Llk, LI K, and LI K are also abelian. Hence, identifying Gal( LI K) with Gal( LI K), we obtain from Proposition A.13 an exact sequence as follows: 1 ~ Gal(LI K)-4 U(L)IV(LI K).14 U(K)~ 1.

(A.8)

and qJo denote the Frobenius automorphisms over k, k' and ko = k' n K, respectively. Extend these to automorphisms of L and, then, to automorphisms of L, and denote them again by the same letters qJ, 1jJ, and qJo. Then qJ, 1jJ, and qJo commute with every a in Gal(LI K) = Gal(LI K). In particular, Let

qJ, 1jJ,

Hence


148

so that the homomorphism 1jJ - 1: U(L)~ U(L) induces an endomorphism

a

= 1jJ - 1: U(L)IV(LIK)~ U(L)IV(LIK).

Let {3 = CPo -1: U(K)~ U(K) and let y: Gal(L/ K)~ Gal(L/ K) be the trivial endomorphism of Gal( LI K), mapping every element of Gal(LI K) to the identity element. With these maps a, {3, and y, consider the diagram 1~ Gal(LIK)~ U(L)IV(LIK)~ U(K)~ 1

l~

lY

lp

1 ~ Gal(LI K) ~ U(L)IV(LI K) ~ 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')V(LIK)IV(LIK),

B = U(ko),

C = Gal(LI K),

D=l.

Hence the maps Nand i induce homomorphisms A ~ Band C ~ D respectively, and by the Snake Lemma, there exists a homomorphism b:B~C

such that

is exact. Since Ker(b) = Im(N) = N(U(k')V(LI K)) = Nk'lko(U(k')) = NU(k' Iko), b induces an isomorphism U(ko)/NU(k' /ko) ~ Gal(L/ K).

(A.9)

Checking the definition of the isomorphism, we find that if u mod NU(k' I ko)~ a for u E U(k o), then Ucp-l mod NU(k' Iko)~ cpacp-1a- 1= 1. Hence, U(kO)cp-l c NU(k' Ik o).

However, as kolk is an unramified extension, one can see (d. Lemma 3.6)

Appendix

149

that the norm map Nkolk: U(ko)~ U(k) is surjective and its kernel is U(koyp-l. Therefore, it induces U(k o)/ NU(k' / k o) ~ U(k)/ NU(k' / k).

Thus we obtain from (A.9) a fundamental isomorphism U(k)/ NU(k' / k) ~ Gal(L/ K) = Gal(k' / k o),

(A.10)

where k' / k is any finite abelian extension of local fields and ko is the inertia field of k' / k. Let e=e(k'/k), f=f(k'/k) for the above k'/k. Then (A.10) and Proposition 2.12 yield [U(k): NU(k' / k)]

Since ef = [k': k], v(N(k' / k)) [k: N(k' / k)]

= [k': k o] = e.

= fZ by Proposition 1.5, it follows that

= [k: N(k' / k)U(k)][N(k' / k)U(k): N(k' / k)] = [Z :fZ][U(k): NU(k' /k)] = 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 Pk: k X ~ Gal(kab/ 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 Erganzungssatze zum Reziprozitatsgesetze der In-ten Potenzreste im Korper der In-ten Einheiten Wurzeln, Abh. Math. Sem. Hamburg, 6 (1928), 146-162. [3] J. W. S. Cassels and A. Frohlich (eds.), Algebraic Number Theory, Thompson Book Co., Washington, D.C., 1967. [4] J. Coates and A. Wiles, On the conjecture of Birch and Swinnerton-Dyer, Inv. Math., 39 (1977), 223-235. [5] R. F. Coleman, Division values in local fields, Inv. Math., 53 (1979), 91-116. [6] E. de Shalit, Relative Lubin-Tate groups, Proc. Am. Math. Soc., 95 (1985), 1-4. [7] E. de Shalit, The explicit reciprocity law in local class field theory, to appear in Duke Math. Jour. [8] A. Frohlich, Formal Groups, Springer-Verlag, New York-Heidelberg-Berlin, 1968. [9] R. Gold, Local class field theory via Lubin-Tate groups, Indiana Univ. Math. Jour., 30 (1981), 795-798. [to] M. Hall, The Theory of Groups, Macmillan Co., New York, 1959. [11] M. Hazewinkel, Local class field theory is easy, Advances in Math., 18 (1975), 148-181. [12] K. Iwasawa, Explicit formulas for the norm residue symbol, Jour. Math. Soc. Japan, 20 (1968), 151-164. [13] K. Iwasawa, Local Class Field Theory, Iwanami-Shoten, Tokyo, 1980 (in Japanese), MHP, Mockba, 1983 (Russian Translation). [14] K. Kato, A generalization of local class field theory by using K-groups, I, Jour. Fac. Sci. Univ. Tokyo, lA, 26 (1979), 303-376; II, ibid., 27 (1980), 603-683; III, ibid., 29 (1982), 31-43. [15] A. Kudo, On Iwasawa's explicit formula for the norm residue symobl, Mem. Fac. Sci. Kyushu Univ., Ser. A., 26 (1972), 139-148. [16] S. Lang, Algebra, Addison-Wesley Publishing Co., Reading-Menlo Park-LondonOntario, 1971. [17] S. Lang, Algebraic Number Theory, Addison-Wesley Publishing Co., Reading-Menlo Park-London-Ontario, 1970. [18] J. Lubin, The local Kronecker-Weber theorem, preprint. [19] J. Lubin and J. Tate, Formal complex multiplication in local fields, Ann. Math., 81 (1965), 380-387. [20] M. Rosen, An elementary proof of the local Kronecker-Weber theorem, Trans. Amer. Math. Soc., 265 (1981), 599-605. [21] J.-P. Serre, Corps Locaux, Hermann, Paris, 1962. [22] J.-P. Serre, Sur les corps locaux a corps residue I algebriquement clos, Bull. Math. Soc. France, 89 (1961), 105-154. [23] B. L. van der Waerden, Algebra, I, II, Springer-Verlag, New York-Heidelberg-Berlin, 1964. [24] A. Weil, Basic Number Theory, Springer-Verlag, New York-Heidelberg-Berlin, 1974. [25] A. Wiles, Higher explicit reciprocity laws, Ann. Math., 107 (1978), 235-254.

Table of Notations General Notations Z Q R C Fq R R+ Rx

the ring of rational integers, the field of rational numbers, the field of real numbers, the field of complex numbers, a finite field with q elements, a commutative ring with identity 1 =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. 1.2. 1.3.

lJ, f (= o/lJ), U, e(v'lv) (= e), f(v'lv) (= f) (k, v), Qp, F«T)), JC, a, Un e(k'lk), f(k'lk), Nk'lk(a')

v, Yo, vp ,

0,

Chapter II 2.1. 2.2. 2.3. 2.4. 2.5.

(k, v), q, (Qp, vp ), (F«T)), VT), w(x), V W

C{J, ko qj), qj)(k'lk), D(k'lk), o[w] Gn

Chapter III 3.1.

3.2. 3.3. 3.4.

3.5.

J-l, Q, ii, (F, J-lF), (F, J-l p), fF = 0 FIlJF' fp = opllJp k: n k un K, on, f", fK = oKllJK, C{Jb C{Jn, Z, Vn, V"" U(F), N(Flk), NU(Flk) R[[Xl' ... ,Xn]], mod deg d, f 0 (gl, ... ,gn), M, fog, f-1, fg Og/lJg, fa C{J( = C{Jb ijJk), fk = OkllJk, frp Q,

=

Chapter IV 4.1.

F(X, Y),

iF (X) ,

Ga(X, Y), A(X)

4.2. 4.3. 4.4.

M F,

Ff(=foFof- 1 ),

Hom(F, G),

End(F),

R (= Ok), :¥jp :¥, flt(X, Y), [a 1t, 8(X) m (= lJg), a' /3, aia', mf' WI' "1, fn, gn, hn' End(W;), Aut(WI )

+

in, Dn, i, Df

Table of Notations

154

Chapter V 5• 1• 5.2. 5.3.

n km,n !ii.n , U;r !ii.n ;r' L , om,n , U n k;, S, SX, fn = 0nllJn' X a, N(h) (= NAh )), N km,oo, L, D, Dn, D;r, k;r (=k~OO) m

O

,

O];m O];m ,

.:T;r,.:T

0];00

. .:T,

+

Chapter VI 6.1. 6.3.

kab' Lb 1/J;r, 1/J, Pk Pb Qs, t G1H , tk'/b (k'lk)

Chapter VII 7.1. 7.2. 7.4.

Pk'ik

Gn gn cjJ(r), iG( a) (= i( a)), 1/J(r), G r c (k' Ik ), f( k' Ik ), Hx' kx' f(x)

Chapter VIII 8.1. 8.2. 8.3. 8.4. 8.5.

(j, w), Nm,n, Bn, B, Nn, v(f3) Pn, (a, f3)n,f, An, At, (a, f3)t Ao, A, D(f3)n, Tm ,n1 xn( an, f3), [a, f3]w Tn1 N n K, {x, y}

Appendix A.1.

en (= en(G, A)), Dn, Hn(G, A), A= (a, y), res, inf, Hn(KIF), Qs, Br(k)

A.2. A.3.

inv(c) V(LIK)

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

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 .7r-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'/v), 6 ramification index e(k' Ik), 12 res ( = restriction map), 138 residue degree f( v' I v), 6 residue degree f(k' Ik), 12 residue field of a valuation, 4 restriction of a valuation, 5

Hasse-Arf (Theorem of), 111 ideal of (k, v), 9 inertia field, 29 inf ( = inflation map), 139 isomorphism of formal groups, 51

totally ramified extension, 26, 39 transfer map, 93 trivial valuation, 3

local field, 18 logarithm, 123

uniqueness theorem, 101 unit group, 4 unit norm group, 41 unramified extension, 26, 36

maximal ideal of a valuation, 4 maximal unramified extension, 37

valuation, 3 valuation ring, 4

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