Topics in Field Theory (North-Holland Mathematics Studies, 155)

TOPICS IN FIELD THEORY

NORTH-HOLLAND MATHEMATICS STUDIES Notas de Matematica (124)

Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro and University of Rochester

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD * T O K M

155

TOPICS IN FIELD THEORY G regory KARPILOVSKY Department of Mathematics University of the Witwatersrand Johannesburg, SouthAfrica

1989 NORTH-HOLLAND-AMSTERDAM

NEW YORK

OXFORD *TOKYO

' Elsevier Science Publishers B.V.,

1989

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, e!ectronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher, Elsevier Science Publishers B.V. (Physical Sciences and Engineering Division), PO. Box 103, 1000 AC Amsterdam, The Netherlands. Special regulations for readers in the USA - This publication has been registered with the Copyright Clearance Center lnc. (CCC), Salem, Massachusetts. Information can be ohtained from the CCC about conditions under which photocopies of parts of this publication may be made in the USA. All other copyright questions, including photocopying outside of the USA, shauld be referred to the publisher. No responsibility is assumed by the Publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein.

ISBN: 0 444 87297 3 Publishers:

ELSEVIER SCIENCE PUBLISHERS B.V. P.O. BOX 103 1000 AC AMSTERDAM THE NETHERLANDS Sole distributors for the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC 655 AVENUE OF THE AMERICAS NEW YORK, N.Y. 10010 U.S.A.

Library o f Congress Cataloging-in-Publication

Data

K a r p i l o v s k y . G r e g o r y , 1940T o p i c s in f i e l d t h e o r y / G r e g o r y K a r p i l o v s k y . p. c m . - - ( N o r t h - H o l l a n d m a t h e m a t i c s s t u d l e s ; 155) ( N o t a s o e maternitica ; 124) B i b l i o g r a p h y : p. I n c l u d e s index. ISBN 0-444-87297-3 1 . Fields, Aljsbraic. I. T i t l e . 11. S e r l e s . 1 1 1 . S e r i e s : N o t a s no. 124. de m a t e m b t i c a ( R i o de J a ? e i r o . B r a z i l ) QAl.il86 no. 124 O A 2 17 1 5 1 0 s--dc512/.74 88-38469

.

CIP

PRINTED IN THE NETHERLANDS

TO

THE MY

MEMORY

TEACHER

S-D-BERMAN

OF

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vi i

Preface

The present book is intended to give a systematic account of certain important topics pertaining to field theory. The author has tried to be fairly complete in what he considers as the main body of the theory and the reader should get a considerable amount of

knowledge of central ideas, the basic results, and the fundamental methods. We have tried to avoid making the discussion too technical. With this view in mind, maximum generality has not been achieved in those places where this would entail a loss of clarity or a lot of technicalities. The present monograph is written on the assumption that the reader has had the equivalent of a standard first-year graduate algebra course. Thus we assume a familiarity with basic ring-theoretic and group-theoretic concepts. For the convenience of the reader, a chapter on algebraic preliminaries is included. There is a fairly large bibliography of works which are either directly relevant to the text or offer supplementary material of interest. The following is a brief description of the content of the book. After establishing algebraic preliminaries (Chapter l), we concentrate on separable algebraic extensions (Chapter 2). Among other results, we provide a characterization of finite separable field extensions E / F in terms of the number of F-homomorphisms of E into its algebraic closure. We then characterize separability via linear disjointness and tensor products. Turning to the trace map and discriminants, we characterize separability via these notions. Chapter 3 is confined to a systematic study of transcendental extensions. We begin by introducing abstract dependence relations, which will allow us to treat in a unified manner algebraic independence and p-independence. Special attention is drawn to the investigation of the existence of separating transcendency basis. Extensions E / F with separating transcendency bases are special instances of extensions E / F which preserve pindependence. The latter extensions are characterized in a large number of different ways. The chapter culminates in the study of relatively separated and reliable extensions.

PREFACE

viii

In Chapter 4 we introduce derivations of fields and present a number of their important properties. Among these, we provide various criteria for extending the derivations and characterize separability via derivations. Given a field E of characteristic p

> 0, we also

exhibit a bijection between the set of subfields of E containing EP and the set of closed restricted subspaces of DerE. Chapter 5 is devoted to a detailed study of purely inseparable extensions. Our presentation of the theory of modular purely inseparable extensions is based on a n important work of Waterhouse (1975). The basic discovery of Waterhouse is t h a t the theory is closely related to the well developed study of primary abelian groups. After establishing some preliminary results, we develop the theory of pure independence, basic subfields, and tensor products of simple extensions. We then compute the Ulm invariants and display some complications in the field extensions not occurring in abelian groups. The final section is devoted t o modular closure and modularly perfect fields. In Chapter 6 we piesent the Galois theory which may be described as the analysis of field extensions by means of automorphism groups Special attention is drawn to the

problem of realizing finite groups as Galois groups. In particular, we show that certain types of split extensions of elementary abelian 2-group by the realizable group G occur as Galois groups of normal real extensions. The chapter ends with a brief discussion of

Galois cohomology. Chapter 7 is devoted to the study of abelian extensions, i.e. Galois extensions with abelian Galois groups. Among other results, we provide a n explicit description of all cyclic extensions of degree pn, p prime, of a given field F containing all pn-th roots of unity. We then study abelian pextensions by means of Witt vectors. After presenting Kummer theory, we finally exhibit a bijective correspondence between the subgroups of the character group of G a l ( E / F ) ( E / F is a Galois extension) and all abelian subextensions of E/F. This is achieved by applying infinite Galois theory and certain properties of character groups

of p r o h i t e groups.

Chapter 8, the final chapter is devoted to a detailed investigation of radical extensions. 1 would like t o express my gratitude to my wife for the encoiiragement she has given

me in the preparation of this book. Finally, my thanks go to Lucy Rich for her excellent typing.

ix

Contents

vii

PREFACE CHAPTER 1.

ALGEBRAIC PRELIMINARIES

1

1. Notation and terminology

1

2. Localization

6

3. Integral extensions

9

4. Polynomial rings

14

5. Unique factorization domains

24

6. Dedekind domains

32

CHAPTER 2.

SEPARABLE ALGEBRAIC EXTENSIONS

45

1. Algebraic closure, splitting fields and normal extensions

45

2. Separable algebraic extensions: definitions and elementary properties

58

3. Separability, linear disjointness and tensor products

73

4 . Norms, traces and discriminants of separable field extensions

82

CHAPTER 3.

TRANSCENDENTAL EXTENSIONS

1. Abstract dependence relations

97 97

2. Transcendency bases

100

3. Simple transcendental extensions

106

4 . Separable extensions

109

5. Weil’s order of inseparability

125

6. Separability and preservation of pindependence

131

7. Perfect ground fields

142

8. Criteria for separating transcendency bases

147

9. Separable generation of intermediate field extensions

153

10. The Steinitz field tower

156

11. Nonseparably generated fields over maximal perfect subfields

167

12. Relatively separated extensions

172

13. Reliability and relative separability

181

CONTENTS

x

DERIVATIONS

193

1. Definitions and elementary properties

193

2. Extensions of derivations

203

3. Derivations, separability and pindependence

210

4 . Restricted subspace of DerE

213

CHAPTER 4.

CHAPTER 5.

PURELY INSEPARABLE EXTENSIONS

219

1. Preparatory results for splitting theory

219

2. Splitting theory

234

3. Chains of splitting fields and complexity

244

4. Modular extensions

252

A. Introduction and preliminary results

252

B. Pure independence, basic subfields and tensor products of simple extensions 262 C. Ulm invariants of modular extensions

269

D. Ulm invariants and group algebras

276

E. Modular closure and modularly perfect fields

290

CHAPTER 6.

GALOIS THEORY

299

1. Topological prerequisites

299

2. Profinite groups

304

3. Galois extensions

315

4. Finite fields, roots of unity and cyclotomic extensions

322

5. Finite Galois theory

336

6 . Infinite Galois theory

340

7. Realizing finite groups as Galois groups

348

8. Degrees of sums in a separable field extension

363

9. Galois cohomology

368

CHAPTER 7.

ABELIAN EXTENSIONS

375

1. Witt vectors

375

2. Cyclic extensions

389

3. Abelian pextensions over fields of characteristic p

399

4. Kummer theory

408

5. Character groups of infinite abelian extensions

416

CONTENTS

CHAPTER 8.

RADICAL EXTENSIONS

xi

421

1. Irreducibility of binomials and applications

421

2. Solvability of Galois groups of radical extensions

430

3. Abelian binomials

434

4. Normal binomials

440

5. Some additional results

450

6. Cogalois extensions

454

7. A Galois correspondence for radical extensions

461

8. Duality of lattices for Gal(E/F) and C o g ( E / F )

463

9. The lattice of intermediate fields of radical extensions

467

Bibliography

477

Notation

534

Index

539

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1

1 Algebraic preliminaries

In this chapter we consider several primarily unrelated basic topics t h a t we shall need in varying degree throughout the book. These include localization, integral extensions, polynomial rings and Dedekind domains. We also establish various notational conventions that we shall use in the sequel. Later chapters will treat various aspects of t h e presented topics in greater detail and depth. Many readers may wish t o glance briefly at t h e contents of this chapter, referring back t o the relevant sections when they are needed later. Occasionally, results are stated somewhat more generally t h a n necessary for later use. Finally, because we presuppose a familiarity with various elementary ring-theoretic terms, only a brief description of them is presented.

1. N o t a t i o n and t e r m i n o l o g y .

Our aim here is to establish various notational conventions t h a t we shall use throughout the book.

All rings and algebras in this book are commutative with 1 # 0, and subrings of a ring

R are assumed t o have the same identity element as R. Each ring homomorphism will be assumed t o preserve identity elements. All modules are unital. We shall write A

-

B

for t h e complement of the subset B in the set A , while A C B will mean t h a t A is a proper subset of B . The symbol for a map will be written before the element affected and consequently, when the composition f o y of maps is indicated, g is the first t o be carried out. If f : X

+

Y is any map, then flX’ denotes the restriction o f f t o a subset X’ of X .

Let R be a ring. The mapping i 2

--t

R defined by n

++

n.1 = 1

+ . . .+ 1 ( n summands)

is a ring homomorphism whose image is called the prime subriny of R; its kernel is an ideal 7nZ for a unique m 2 0, called the Characteristic of R and denoted by char R . We shall

in future identify Z/mZ with its image in R and write n instead of n.1. An element u of

R is said t o be a unit if uv = 1 for some v E R. The set U ( R ) of all units of R is a group,

CHAPTER 1

2

called the unit group of R.

A ring R is local (respectively, semilocal ) if it has precisely one maximal ideal (respectively, if it has only finitely many maximal ideals). An element z of a ring R is a zero

divisor if z y = 0 for some nonzero y E R ; in case z

# 0 and z is a zero divisor, we say that

z is a proper zero divisor.

An integral d o m i a n is a ring without proper zero divisors. An ideal P of a ring R is p r i m e if R I P is a n integral domain, while P is primary if every zero divisor in R I P is nilpotent. Let R be a ring. For any r E R , the ideal Rr (also denoted by ( r ) ) generated by r is called a principul ideal. We say t h a t an ideal J of R is finitely generated if J =

Ral

+ . . . + R a n for suitable a l , a 2 , . . . , a ,

E R.

A ring R is noetherian if every ascending chain of ideals breaks off, or equivalently, if

every ideal is finitely generated. A ring R is artinian if every descending chain of ideals breaks off. Let M be a n R-module. If I is a n ideal of R , then IM is defined t o be the submodule of M consisting of all finite sums z r ; m , with r; E I , m; E M. The annihilator of M , written a n n ( M ) is defined to be t h e ideal of R consisting of all r E R such t h a t r M = 0. We say that M is jaithjul if a n n ( M ) = 0. Thus if I = a n n ( M ) then M is faithful as an RlI-module. 1.1. Proposition. Let M be a finitely generated R- module, let I be a n ideal of R

and let f : M

+

M be a n R- homomorphism such that f (M)

IM. T h e n f satisfies a n

equation of the f o r m

Proof. Write M = R x l we have f (xi)= c,"=,

+ . . . + Rx, for some x, E M , 1 5 i 5 n. Since f (z,)

r , J x 3 for some r t j E I , 1 6 i

E

IM,

5 n. Hence

where 6;, is the Kronecker delta. By multiplying on the left by the adjoint of the matrix (Sijf

-

rij), we conclude that det(6ijf

-

rij) annihilates each zi, hence is the zero map.

Expanding out the determinant, we obtain a n equation of the desired form.

NOTATION AND TERMINOLOGY

3

Another useful observation is given by 1.2. Proposition. Let R be a n artinian integral domain. Then R is a field.

Proof. Given a nonzero x E R , we have a descending chain R x 2 Rx2 2 of ideals of R. Hence Rz" = RY"" for some n

.. . 2

1, so Z" = rz"+' or s"(1 - r z ) = 0.

Because R has no proper zero divisors, 1 - rx = 0 and therefore r is a unit. rn Let R be a ring. An element z E R is nilpotent if xn = 0 for some positive integer

n. We say that R is reduced if 0 is the only nilpotent element of R. An ideal J in R is nil if every element of J is nilpotent, while J is nilpotent if there is z positive integer n such that J" = 0, where J n is the product of J with itself n times. An element e of R is idempotent if e2 = e. An idempotent is trivial if it is 0 or 1. Two idempotents

orthogonal if

u'u =

u,u

are

0. A nonzero idempotent is primitive if it cannot be written as a sum

of two nonzero orthogonal idempotents. The Jacobson radical J ( R ) of a ring R is the intersection of the maximal ideals of R; equivalently, J ( R ) consists of all x E R such that for all y E R , 1 - sy is a unit. A ring

R is semisimple if J ( R ) = 0. The set N ( R ) of all nilpotent elements in R constitutes an ideal called the nilradical of R. Note that N ( R ) C J ( R ) and that R is reduced if and only

if N ( R ) = 0. A prime ideal P of R is minimal if there is no prime ideal P' of R such that P'

c P.

Note that N ( R ) = intersection of all prime ideals of R = intersection of all

minimal prime ideals of R. Let R be a subring of S . For any subset A = {aili E I } of S , denote by RIA] the smallest subring of S containing R and A (for A = { a l , . . .,a,},we write R [ a l , .. . ,a,] instead of R [ A ] ) The . ring R [ A ]consists of all polynomials in a i ( i E I ) with coefficients in R . We say that S is finitely generated over R (or simply finitely generated if R is the prime subring of S ) if there is a finite subset A of S such that S = RIA). Let ( R i } i , ~ be a family of rings and let R be the direct product set

ncEIRi. We can

define addition and multiplication on R by the rules:

It is straightforward to verify that R is a ring; we shall refer to R as the direct product of

the f a m i l y { R ; } , i , ~ . For all i E I , the projection of R onto Ri is a ring homomorphism;

CHAPTER 1

4

the injections Ri

-+

R preserve addition and multiplication but not 1 and so are not ring

homomorphisms. Let R be a ring. Two ideals I , J of R are said to be coprime (or comaximal ) if

I+J=R. 1.3. Proposition. Let I I , 12,. . . , I n be ideals of R and let the homomorphism given by f(z)= (z

(i)

If I,,

Ij

are coprime for i

f :R

+

n y = , ( R / I ; ) be

+ 1 1 , . . ,z + I n ) .

n:=,I;

# j , then

= nF=lIl;

(ii) f is surjective if and only if I;, I3 are coprime for i

#j

(iii) K e r f = nFrlI,.

Proof. (i) If I , J are ideals of R , then obviously ( I + J ) ( I n J ) = I J . Hence if I , J are coprime, then

I nJ

=

I J . This proves the case n = 2. Suppose n > 2 and the result is

:::n

true for 11,.. . , I n -l, and let I = we have z;

+ y;

I , = n”,=;’.Ii

n- 1

JJz;

= n(1-y;)

=

= l(mod In)

i= 1

i=l

+I

+ In = R , 1 5 i 5 n - 1,

= 1 for some zi E I;, y; E In and therefore n- 1

Thus I ,

Because 1i

R and so n

(ii) Assume f is surjective. Let us show for example that I I , I ~ are coprime. There exists z E

R such that f(z) = (1,O.. . , O ) ; hence z 1 = (1 - 2)

l(mod

Z1)

and z G O(mod Z i ) , so that

+ z E + I,, I1

as required. Conversely, assume that I i , I3 are coprime for i example that f ( z ) = ( l , O , u,

...,0) for some

z E

R . Because I1 + I ,

+ v , = 1 for some ui E 11,v , E I;. Setting z = v 2 . . .v , ,

n(l

# j . It

we have

n

z=

-

u ; ) E l(mod 1 1 )

i=2

and z

3

O(mod I ; ) , i > 1. Thus f (z) = ( I , O , . . . ,0) as desired.

(iii) This is a direct consequence of the definition of f .

suffices to show, for

=

R

(z

> 1) we have

5

NOTATION AND TERMINOLOGY

A relation 5 on a set S is said t o be a partial order if it is reflexive, antisymmetric, and transitive, i.e. if

(i) x

5x

for all x E S

(ii) If z , y E S with z 5 y and y

5 z, then x = y

(iii) If z , y , z E S with z 5 y and y 5 z , then z 5 z

A partially ordered set is a set S together with a specified partial order 5 on S . Let C be a collection of subsets of a set S . Then we can define a partial order in C by the rule

A 5 B if and only if A

CB

This is called (partial) order b y inclusion. A directed set is a partially ordered set S such that for any i , j E S there exists k E S with i 5 k and j 5 k. In a partially ordered set S , we write x

2y

t o mean y

5 z, and

x

< y (or y > z) t o mean x 5 y but x # y .

A subset L of a partially ordered set S is linearly ordered if for every pair of elements

x,y in L , either x 5 y or z 2 y . A linearly ordered subset of a partially ordered set is also calied a chain. Let S be a set. Then a chain in S is a collection C of subsets of S such t h a t for each pair A , B E C either A 2 B or A

B.

A mazimal (minimal) element in a partially ordered set S is an element m that for all x E S , m

5 x implies x

S such

= m (z 5 m implies x = m ) . Let L be a subset of

a partially ordered set S . An upper (lower) bound for L in S is any element z E S such that

t 5 z (t 2 z) for ail t

E L. By a well-ordered set we understand a linearly ordered set

in which every nonempty subset has a minimal element.

A partially ordered set is said t o be inductive if it is nonempty and every chain in it has an upper bound. For example, a collection C of subsets of a set S is inductive in case it has t h e property: if D is any chain in S consisting of elements of C , then the union of the sets of D belongs t o C.

1.4. Proposition. (Zorn’s lemma). Every partially ordered set which is inductive has a mazimal element. Every set S has a cardinality, denoted by ISI, such t h a t there is a bijection between two sets

X and Y if and only if

1x1 5 IYI.

1x1 = IYI.

If there is an injection X

+

Y , then we write

This defines a linear order on any given set of cardinals. T h e cardinality of

W is denoted by N o (aleph-null). A set is countable if it is either finite or of cardinality No.

CHAPTER 1

6

Let R be a ring and let V be an R-module. We say t h a t V is noetherian if every ascending chain of submodules of V breaks off. This is equivalent t o the requirement t h a t every submodule of V is finitely generated.

1.5. Proposition. Let R be a ring, let V be a n R-module and let W be a submodule

of V. T h e n V is noetherian i f and only i f both W a n d VIW are noetherian. I n particular,

if R is a noetherian ring and V a finitely generated R -module, t h e n V is noetherian (and, therefore, all submodules of V are finitely generated).

Proof. Assume that V is noetherian. The lattice of submodules of W (respectively, V/W) is isomorphic t o the lattice of submodules of V contained in W (respectively, containing W ) . Thus W and VIW are noetherian. Conversely, assume t h a t W and VIW are noetherian. Let VQ C_ V1

C ... C_

V,

C . . . be

a n ascending chain of submodules of V.

Because W is noetherian, there is a n integer m such that V, n W = Vn+l n W for all n 2 m. Since V/W is noetherian, there is a n integer k such that (V, for all n

=

(Vn+*+ W ) / W

2 k. Therefore

v, + w = Vn+1 + w Take n

+ W)/W

2 max{k,rn}.

for all n

2k

We shall show t h a t V, = Vn+l. It suffices t o show that Vn+l

To this end, fix z E Vn+l. Because Vn+l

+W = V, +W , there exists y E V,

and

such t h a t z + z l = y - t z ~ .Thus x-y = z2-z1 E Vn+, n W . Note t h a t V,+,nW Hence, since z

-

y and y belong t o V,,

z E

z1, z2

V,. EW

= V,nW.

V, too, proving t h a t V, = Vn+l. Thus V is

noetherian.

Now assume t h a t R is noetherian and V a finitely generated R-module. Then there exists a free module F of finite rank n such that V

F I X for some submodule X of F .

Now apply the preceding paragraph.

2. L o c a l i z a t i o n .

Let R be a ring. A subset S of R is said to be multiplicative if 0 @ S , 1 E S , and if, for any z, y in S, zy E S .

T h e quotient ring Rs associated with a multiplicative set S consists of all elements of the form r / s with r E R and s E S. By definition r/s = r’/s‘ if there exists s1 E S such that s l ( s f r - sr’j = 0. Multiplication and addition in Rs are defined by (r/s)(r’/s’) = rr’/s.sf

r/s

+ r‘/s’ = (s‘r + sr‘)/ss’

LOCALIZATION

7

An easy verification shows that these operations are well defined and that Rs is indeed a ring. If R is an integral domain and S the set of all nonzero elements in R, then R s is a field, called the quotient field of R. Let S be a multiplicative subset of R. Then the map

f :R

-+

Rs given by

is a homomorphism, called ccnonical. The ring R s and the canonical homomorphism

f :R

--t

Rs have the following properties:

+ f ( s ) is a unit of R s f ( 7 ) = 0 + r s = 0 for some s E S

(i) s E S (ii)

(iii) Every element of Rs is of the form f ( r )f (s)-’ for some r E R and some s E S . Conversely, these three conditions determine the ring Rs up to isomorphism. This fact will be derived as a consequence of the following universal property of Rs. 2.1. Proposition.

Let g : R

-+

L be a ring homomorphism such that g(s) is a

unit of L for all s E S . Then the map h : Rs homomorphism Rs

+L

-+

L, r / s

H

g(r)g(s)-’ is a unique ring

for which g = h o f.

Proof. The map h is clearly a ring homomorphism provided that it is well defined. Assume that r / s = r’/s’ and choose t E S such that (rs‘ - r’s)t = 0. Then ( g ( r ) g ( s ’ )-

g(r‘)g(s))g(t)= 0 and, since g(t) is a unit of L , g(r)g(s)-l = g(r’)g(s’)-l. Thus h is well defined. Conversely, assume that $ : Rs Then $(r/1) = lCtf ( r ) = 9 ( r ) for all

7

-+

L is a ring homomorphism for which g = 4

of .

E R. Therefore, for all s E S ,

which implies that

as required. 2.2. Corollary. Let g : R

-+

L be a ring homomorphism which satisfies properties

(i)-(iii) with f replaced b y g and Rs replaced b y L. Then the map h : Rs + L , r / s g(r)g(s)-’ is a unique isomorphism Rs

-+

L f or which g = h o f.

++

CHAPTER 1

8

Proof. By Proposition 2.1, it suffices to verify that h is a n isomorphism. By (iii), h is surjective. Assume that h(r/s) = 0. Then g ( r ) = 0, hence by (ii) we have r t = 0 for

some t E 5'. It follows that r / s = 0/1 = 0 in R s , as required. We proceed to discuss the connection between ideals in R and R s . With every ideal

I in R we associate the ezpanded ideal I R s of Rs given by

IRs = {r/slr E I ,

s

E S}

Suppose J is a n ideal of Rs. We define the corresponding contracted ideal J c of R by

Jc = f-'(J) where f : R

--f

Rs i s t h e canonical h o m o m o r p h i s m . It i s clear that I R s is the ideal of Rs

generated by all 1-11 with r E I . 2.3. Proposition. (i) For a n y ideal J of R s , J c R s = J (ii) For a n y ideal I of R ,

Z C ( I R S ) w~i t h equality if a n d o n l y if

for a n y s

t S,

z

F

R , sx E I i m p l i e s x E I . (iii) If I i s a n ideal of R w i t h I n S = 0 a n d S + I = { s

(iv) T h e m a p I

t--t

+ Ils E S } , t h e n

ZRs is a bijection between t h e p r i m e ideals

of R which d o not meet S

and all p r i m e ideals of Rs.

Proof. s

E S, r / s

(i) By definition, J" = { r E R / r / 1 E J } . Hence, if r E J c , then for all = (r/l)(l/s) E

J and thus J " R s

J . Conversely, if r / s E J , then r j l E .7

and so r E J " . IIence r / s E J " R s and therefore J ( i i ) If r E I , then r / 1 E I R s and hence

R , sz E I implies x E I . If r r E I and therefore ( I n s ) '

C: I .

= tr' E

I for some r' t R , t , s t S. Hencc

Conversely, assume that ( I R S )i~ I and let s E S , z t R =

r / s E I R s so x E ( I R s ) '

(iii) The assumption on I ensures that --f

E ( I R s ) ' . Assume that for any s c S, z c:

(ZRs)',then t s r

be such that sx = r E I . Then x / l

homomorphism X : R

T

J'Rs.

S

+I

( R / I ) s + [ given by

i I , as required.

is a multiplicative subset of R / I . The

X(r) = ( r

+ 1 ) / 1 is such that

of (R,'I)s+r for all s E S. Hence, by Proposition 2.1, the map h : Rs by h ( r / s ) = X ( r ) X ( s ) - '

-+

X(s) is a unit

( R / I ) s - , given

is a ring homomorphism. It is clear that !t is surjective and that

9

INTEGRAL EXTENSIONS

Kerh = I R s , proving (iii). (iv) Let P be a prime ideal of R with P n S = 0. Then, by (iii), P R s is a prime ideal of

Rs. Furthermore, by (ii), (PRs)' = P. Hence, if P1 and PZ are prime ideals of R which do not meet S , then PlRs = PzRs implies PI = Pz. Finally let Q be a prime ideal of

Rs. By (i), Q = P R s for P

= Q" and

P is obviously a prime ideal of R. If s E S n P ,

then s/l E Q which is impossible since s / l is a unit of Rs. Thus S n P = 0 and the result follows. w 2.4. Corollary. If R is noetherian, then so is R s .

Proof. Let J be any ideal of R s . By Proposition 2.3(i), J = J C R sis generated by all r / 1 with r E 2". Since R is noetherian, J c is finitely generated, hence so is J . m Especially important is the case where S is the complement of a prime ideal P. We then, by abuse of notation, write Rp instead of R s , 2.5. Corollary. Let P be a prime ideal of R. Then the map I

H

I R p is a bijection

between the prime ideals of R contained in P and all prime ideals of Rp. Thus PRp is a unique rnazimal ideal of R and R p l P R p is isomorphic t o the quotient field of RIP. I n particular, if P is a maximal ideal, then

RIP Proof. Put S = R

-

Rp/PRp

P . Then a prime ideal I of R does not meet S if and only if

1 5 P . Now apply Proposition 2.3 (iv),(iii). Let P be a prime ideal of R. By Corollary 2.5, Rp is a local ring. This ring Rp is also called the local rang of R at P , and the process of forming Rp is called localization.

3. Integral extensions.

Let R be a subring of S. An element s E S is said to he integral over R if s satisfies an equation of the form

Clearly every element of R is integral over R . 3.1. Proposition. Let R be a subring of S and let s E S. Then the following are

equivalent:

CHAPTER 1

10

(i) s is integral over R (ii) R[s] is a finitely generated R -module (iii) R[s] is contained in a subring

S1

of S such that S1 is a finitely generated R -module.

(iv) There ezists a faithful R [ s ] -module M which is finitely generated as an R -module.

Proof. (i)+ (ii): Applying ( l ) ,we have

It follows, by induction, that all positive powers of s lie in ~~~~

cyzi Rs'.

Hence R[s] =

Rs' is a finitely generated R-module.

(ii)+ (iii): Take S1 = R[s] (iii)+ (iv): Note that (iv)* (i): Let f : M

S1

--f

is a faithful R[s]-module. Now take M = S1.

M be defined by f ( m ) = s m and put I

M and hence, by Proposition 1.1, f"

+ rlfn-' + ...+ r ,

= R. Then

f(M)

IM =

= 0 for some 7; E R . Since M

is faithful, s satisfies (I), as required. m Let R be a subring of

S. We say that

S is integral over R if each element of S is

integral over R. 3.2. Corollary. I f R is a subring of S , then the following conditions are equivalent.

(i) S is a finitely generated R-module

(ii) S is finitely generated over R and S is integral over R (iii) There ezist s1, s2,. . . ,s, in S such that S = R [ s l , . . . , s,] and each s, is integral over

R. Proof.

(i)

+ (ii):

Write

S = R [ s ~..., , s k ] and, for each

s s

=

Rsl

+ .. . + RSk for some s1, ...,s k

in

s. Then

E S , R [ s ] C S . Now apply Proposition 3.l(iii) for

s1= s. (ii) + (iii): Take any finite generating set of S over R (iii) + (i): We argue by induction on n. The case n = 1 is part

of Proposition 3.1.

Assume n > 1 and let Rt = R [ s l , . . . , s t ] , 1 5 t 5 n. Then, by induction, R,-1

is a

finitely generated R-module. Now R, = Rn-1[sn] is a finitely generated R,-I-module

(by

the case n = 1, since s, is integral over

&-I).

Hence R , = S is finitely generated as an

R-module. Let R be a subring of S . Then the set C of all elements of S which are integral over

INTEGRAL EXTENSIONS

11

R is called the integral closure of R in S. If C = R, then R is said to be integrally closed in S . 3.3. Corollary. Let R be a subring of S and let C be the integral closure of R in S .

Then C is a subring of S containing R.

Proof. It is clear that C 2 R. If R-module by Corollary 3.2. Iience

s1

~ 1 , E s C ~

then Rlsl,sz] is a finitely generated

f s~ and s l s z are integral over R by Proposition

3.1(iii). rn

3.4. Corollary. Let R

CS &

T be rings such that S is integral over R and T is

integral over S. Then T is integral over R. P r o o f . Given t E T , let s1,.

tn

. . ,s,

E S be such that

+ sltn--l + . . . + s,

=0

The ring S1 = R [ s l , . . . ,s,,] is a finitely generated R-module by Corollary 3.2. Furthermore, S l [ t ]is a finitely generated S1-module (since t is integral over

s,). Hence & [ t ] is a

finitely generated R-module. Therefore t is integral over R by Proposition 3.l(iii). rn 3.5.

Corollary. Let R 2 S be rings and let C be the integral closure of R in S .

Then C is integrally closed in S. P r o o f . Let s E S be integral over C. Then, by Corollary 3.4, s is integral over R. Hence s E C as required. rn The folIowing simple observation is often useful. 3.6. P r o p o s i t i o n . Let A

B be rings and let B be integral over A.

( i ) For any ideal I of B , B / I is integral over A / ( A n I ) (here

we

identify A / ( An I ) with

the subring ( A + I ) / I of B / I ) . (ii) If S is a multiplicative subset of A, then Bs is integral over A s . P r o o f . (i) If b E B then bn+a1bn-'

+. . .+a,

=0

for some a l , . . . ,a, in A . Reducing

this equation mod I , the required assertion follows. (ii) Let b / s E B s , b E B , s E S. Then the equation for b in (i) yields

(bls)"

+ ( a l / s ) ( b / s ) " - ' + . .. + a , / s n

and hence b / s is integral over A s . rn

=0

12

CHAPTER 1

As an easy application of the above result, we record 3.7. P r o p o s i t i o n . Let A

If S is

CB

be rings

and

let C be the integral closure of A in B .

a multiplicative subset of A , then C S is the integral closure of As i n B s .

Proof. By Proposition 3.6(ii), Cs is integral over A s . Conversely, if b / s E B s is integral over A s , then we have a n equation of the form (a; E

A,

S;

E S)

Pur t = s 1 . . . s, and multiply this equation by ( s t ) " . Then we see t h a t bt is integral over

A , hence bt E C . But then b / s = b t / s t E C s , as required. rn An integral domain is said to be integrally closed if it is integrally closed in its quotient field. We close by providing a class of integrally closed domains. First, however, we must introduce the notion of divisibility in R. Let R be a n integral domain. As in IL we define alb (for any a, b E R ) to mean

b = ar

for some r E R

This is equivalent to the requiremect ( b ) C ( a ) . Observe t h a t an element u E R is a unit if and only if ul l . The units are trivial divisors since they are divisors of every element of R. If alb and bla (or, equivalently, if ( a ) = ( b ) ) , then we shall say t h a t a and b are associated. It is easy to see that a and b are associated if and only if a = bu for some unit u of R. 4. nonzero element p of R is said to be a prime if ( p ) is a prime ideal. Expressed

otherwise,

a

nonzero p E R is a prime if p is a nonunit such t h a t

plab

impiies

p l a or pjb

By an irreducible element we understand a nonunit which is not a product of two nonunits.

A prime is always irreducible : if a prime p satisfies p = ab, then by definition, pja or p l b , say pla. It follows that a = p r = abr for some r E R , and a

#

0 (as divisor of p ) .

Hence br = 1, by cancellation, so b is a unit. A greatest com m on divisor ( G C D ) o f elements

a l , ... ,a,

E R is an element

such t h a t r divides each a , and r is divisible by each common divisor of

a l , . . . ,a,.

7

ER The

G C D is not unique : if r is a GCD, then so is any r' associated t o r . Conversely, if r and r' are both GCDs of a l , . . . ,a,, then r' is associated t o

7.

We shall ignore the distinction

between associates and denote any one of the GCDs of a l , . . . ,a, by

( a l , .. . ,a,).

INTEGRAL EXTENSIONS

13

Let R be an integral domain. Following Kaplansky, we say t h a t R is a G C D - d o m a i n if each pair of nonzero elements of R has a greatest common divisor. If R is a GCD-domain, then one immediately verifies that any finite number of elements of R have a G C D . In fact

(a,b,c) = ((a,b),c) = (a,(b,c)) (a; E R )

(ai,...,an) = ((al,...,an-l),an)

Let R be a n integral domain and let a , b, c E R. The following properties are immediate consequences of the definitions:

(2) If ( c a , cb) exists, then ( a , b) exists and ( c a , cb) = c ( a , b) (3) If ( a , b) = ( a , c ) = 1, then ( a , bc) = 1

3.8. P r o p o s i t i o n . A n y G C D - d o m a i n is integrally closed,

Proof. Let R be a GCD-domain and let F be its quotient field. Suppose X E F is such t h a t

A"

+ q X n - - l + ... + r , = o

(Ti

E

R)

(4)

Write X = a / b with a , b E R. Then, by (2), there exist s, t E R such t h a t a = (a,b ) s , b =

( a , b ) t , and ( s , t ) = 1. Substituting X = s / t in (4), we obtain

sn

+ r 1 s n - l t + . . . + r,tn

= 0,

whence t divides sn. Since, by (3), ( s n , t ) = 1, it follows t h a t t is a unit of R. Hence X E R and the result follows.

In order t o apply Proposition 3.8, we finally record 3.9. P r o p o s i t i o n . Let R be a n integral domain. T h e n R i s a G C D - d o m a i n i f and

only if the intersection of a n y t w o principal ideals of R i s a principal ideal.

Proof. Assume t h a t the intersection of any two principal ideals of R is a principal ideal. Given a , b E R , we then have ( a ) n ( b ) = ( c ) for some c E R; hence ab = cr for some r E R. We claim that r = ( a , b ) . Indeed, since

c

= a r l = br2 for some

r1,r2 E

R , we have

ab = a r l r = E72r and so r i a , rib by cancellation. If d is another common divisor of a and b, say a = d s l and b = ds2, put m = d s l s z . Then m E ( a ) n ( b ) , hence m = cs say, and so cr = ab = d m = eds. It follows, by cancellation, that r = d s and sufficiency is therefore

established.

CHAPTER 1

14

Conversely, assume t h a t R is a GCD-domain and fix a , b E R. Let d = (a,b) and write a = dx, 6 = dy. By (2), (x,y) = 1 and (xz,yz) = z for any z E

I?. T h e latter implies that

if xjyz, then zIz. If xzl = yz2 E (x) n (y), then x(yz2 and so xiz2. This implies

z2

E (x)

and yz2 E (xy) so that (x) n (y) = (xy). Accordingly,

as required.

4. P o l y n o m i a l r i n g s .

A monoid is a set G with a n associative binary operation and having a n identity element 1. For the rest of this section, R denotes a commutative ring and G a monoid. T h e monoid

algebra RG of G over R is the free R-module on the elements of G, with multiplication

induced by t h a t in G. More explicitly, RG consists of all formal linear combinations

with finitely many x g # 0 subject to (i) c z g . g= C y g . g if and only if xg = yg for all g E G. (4

C%.9+ C Y g . 9 = C ( X , + Yg)??

(iii) ( C . c g . g ) ( C y h . h ) = C2t.t where zt = C g h = t x g y h (iv) r ( C xg . g) = C ( r x g ) . g

for all r E R.

An easy verification shows t h a t these operations define R G as a n associative R-algebra with 1 = 1 ~ . 1where ~ , 1~ and 1~ are identity elements of R and G, respectively. With the aid of the injective homomorphisms

we shall in the future identify R and G with their images in RG. With these identifications, the formal sums and products become ordinary sums and products. For this reason, from now on we drop the dot in xg.g. Given x =

xgg E RG, the support of x, written Suppx, is defined by

POLYNOMIAL RINGS

15

It is clear that Suppx is a finite subset of G that is empty if and only if z = 0. 4.1. Proposition. Let A be a n R-algebra, let G be a monoid and let

11,:G-tA be a n y m a p satisfying $(l) = 1 and 11,(zy) = 11,(x)11,(y)f o r all z,y E G. T h e n the map

+*

: RG

--+

A given by

is a homomorphism o f R-algebras. I n particular, if 11, is injective and A is R-free with +(G) as a basis, t h e n RG

A as R-algebras.

Proof. Bearing in mind that RG is R-free freely generated by G , we see that +* is a homomorphism of R-modules. Let

be two elements of RG. Then

a,bEG

as required.

a,bEG

rn

Assume that G is a free abelian group freely generated by the set

Then each element of G can be uniquely written in the form

with only finitely many n, # 0. The submonoid of G consisting of all elements (1) in which

n,

2 0 for

all i E I is called a free commutative monoid freely generated by

The corresponding monoid algebra Over R is denoted by R[(X,),,I].

{XJi I } .

We shall refer to

R [ ( X , ) , E I ]as the polynomial r i n g in the indeterminates X, with coefficients in R . If I

CHAPTER 1

16

is a finite set, say I = {1,'2>.. . , n } , we write R [ X I , X z , . . . ,X,] instead of R[(X,),Eri In particular, the polynomial ring in the indeterminate X with coefficients in R will h r denoted by

R[X].

The elements z

nttlX,"'

of R[(X,),,l] (n,

2

0 and n, = 0 for all but finitely

mail\

E I ) are called monomials. Hence, by definition, R[(X,),,f] is a free R-module freelj

generated by all monomials. The degree of the monomial

ntElXtnLis defined by

If f is a nonzero polynomial in R [ ( X a ) i E then ~ ] , we define the degree of f , written degf, to be the maximum of the degrees of the monomials in the support of f. Thus deg f = 0

if and only if f is a nonzero element of R . I f f = 0 then we say that its degree is -a. Let d denote a nonnegative integer and let f be a nonzero polynomial in R [ ( X t ) ; E ~ ] . We refer to f as homogeneous of degree d (or as a f o r m of degree d ) if all monomials in Suppf are of degree d . Let 0

# f ( X ) E R [ X ]be of degree n. Then f ( X ) = ro

with ra € R and with r , constant t e r m

.

+ r l X + . . . + rnXn

# 0. We refer to rn as the leading coeficzent of f and

ro as its

We also say that f is m o n i e if r n = 1.

For convenience, let us recall the following information. Let B be a subset of an R-algebra A . Then the elements ( r , E R , v(b) 2 0 )

with finitely many r y and ~ ( 6 )distinct from zero form a subalgebra of A . This subalgebra,

denoted by R [ B ](or simply R [ a ]if 13 = { a } ) is obviously the smallest subalgebra containing

B. For this reason, we refer t o R [ B ]as the subalgebra of -4generated by B . If A

=

R[BI,

then we say that A is generated by 3. Let A be an R-algebra and let B be any subset of A . We say that B is algebrazcally Independent over R (or that the elements of B are algebraically independent over R ) if the

elcrnents

n

bv(b)

( 4 b ) 2 0)

bEB

are R-linearly independent. In the special case where B = { a } , we see that a is algebraically independent over R if and only if (1, a , a', , . . , a n , . . .} is an R-linearly independent set. If

POLYNOMIAL RINGS

17

a E A is algebraically independent over R , then we also say t h a t a is transcendental over

R. In the case where a is not transcendental over R , we say t h a t a is algebraic over R. Thus a is algebraic over R if and only if there exists a nonzero polynomial f ( X ) E R [ X ] such t h a t f ( a ) = 0.

We next record some basic properties of polynomial rings. 4.2. P r o p o s i t i o n . Let A be any commutative R-algebra.

(i) Any map { X ; l i E I }

4

A , X ; ++ a; can be eztended t o a homomorphism

of R-algebras. I n particular, if A is generated b y {a;li E I } , then A i s a homomorphic

image of R [ ( X i ) i € 1 1 . (ii) The above homomorphism is injective if and only i f the elements a; are algebraically

independent over R (iii) A s R[(X;);,r] i f and only i f there is a generating set {a;li E I } for A which is

algebraically independent over R. I n particular, A

E R [ X ]i f

and only if A is generated

b y a transcendental element. P r o o f . (i) Let G be the monoid generated by the X ; , i E I . Setting

it follows t h a t +(1) = 1 and +(zy) = +(z)+(y) for all z,y E G. Now apply Proposition 4.1.

(ii) This follows from (2) and the definition of algebraic independence. (iii) Apply (i) and (ii). 4.3. P r o p o s i t i o n . Let

I = J U K be a disjoint union and let S

=

R [ ( X j ) j E ~Then ].

and, i n particular,

P r o o f . Owing t o Proposition 4.2(i), the map X ;

H

X ; , i E I extends to a homomor-

I ] S [ ( X k ) k E of ~ ]R-algebras. Because its image contains S and each phism R [ ( X , ) ; ~ -+

CHAPTER 7

18

xk,

k E K , the given homomorphism is surjective. Furthermore, the elements x;,i E I of

S (( X k ) k E K ]are obviously algebraically independent over R. The desired conclusion now

follows by virtue of Proposition 4.2(ii). w

in R [ X ]and let a , and

4.4 P r o p o s i t i o n . Let f ( X ) and g ( X ) be nonzero polynomials

b, be the leading coefficients of f ( X ) and g ( X ) , respectively. If at least one of a,, b, is not a zero divisor i n R , then deg(fg) = degf

+ degg

and the leading coefficient of f g is anb,. P r o o f . If f ( X ) = a0

+ a l X + . . . + a,Xn f ( X ) g ( X )= aobo

and g ( X ) = bo

+ b l X + . . . + b,X",

then

+ . . . + anb,Xm+n

and the result follows. 4.5.

C o r o l l a r y . Let R be an integral domain. Then, for any set I , R [ ( X , ) ; E ris]

a!so an integral domain. Proof.

There is no loss of generality in assuming t h a t I is finite. Moreover, by

Proposition 4.3, it suffices to show t h a t R ( X ] is a n integral domain. T h e latter being a

consequence of Proposition 4.4, the result follows. rn 4.6. C o r o l l a r y . Let R be an integral domain. Then, for any set I , the unit groups of R and R [ ( X , ) i c l ]are the same.

P r o o f . It suffices to show that if u E R [ ( X i ) i , l ] is a unit, then u is a unit of R. But if uv = 1, then u , v E R by comparing degrees, hence the result. a 4.7. P r o p o s i t i o n , Let f ( X ) and g ( X ) be two nonzero polynomials i n R [ X ]of degrees

m and n , respectively. Put k = max{m

-

n

+ l,O}

and denote b y a the leading coefficient

of g ( X ) . Then there ezist polynomials q ( X ) and r ( X ) such that U k f ( X )= q ( X ) s ( X )+.(XI where either r ( X ) = 0 or degr(X) < n . Furthermore, i f a is not a zero divisor in R , then q ( X ) and r ( X ) are uniquely determined. P r o o f . If m < n, then k = 0 and we may take q ( X ) = 0, r ( X ) = f ( X ) . Thus we may assume t h a t m

2

n - 1 in which case k = m

-

n

+ 1.

To prove the first part,

POLYNOMIAL RINGS

19

we argue by induction on m. The case m = n - 1 being trivial, assume m 2 n. Then

a f ( X )- b X m P n g ( X )has degree a t most m

-

1, where b is the leading coefficient of f ( X ) .

Invoking induction hypothesis, we may find polynomials q , ( X ) and r l ( X ) such that

where either n ( X ) = 0 or degrl(X) < n. This proves the first assertion, by taking

q ( X ) = bam-nXm-n

+ q l ( X ) , -7(X)= r l ( X )

Assume that a is not a zero divisor and that

where either r l ( X ) = 0 or degrI(X) < n. Then

If q ( X ) - q l ( X ) # 0, then the left side has degree at least n, since the leading coefficient of g ( X ) is not a zero divisor (see Proposition 4.4). But this is impossible, because r I ( X ) - r ( X ) is either zero or has degree less than n. Thus q ( X ) = q l ( X ) and so r l ( X j = r ( X ) , as required.

4.8.

Corollary.

Let f ( X ) and g ( X ) be two nonzero polynomials i n R [ X ] and

let the leading coefficient of g ( X ) be a unit of R. Then there ezist unique polynomials q ( X ) , r ( X ) E R [ X ]such that

where either r ( X ) = 0 or degr(X) < deg g ( X ) . Proof. Apply Proposition 4.7. w 4.9. Corollary. Let f ( X ) be a nonzero polynomial in R [ X ] . Then, for any given r E

R, f ( r ) = 0 i f and only if X Proof. If f ( X ) = ( X

-

-

r is a divisor of f ( X ) .

r ) g ( X ) for some g ( X ) E R [ X ] ,then obviously f ( r ) = 0.

Conversely, assume that f ( r ) = 0 and put g ( X ) = X--7. Since g ( X ) is a monic polynomial

CHAPTER 1

20

+

of degree 1, it follows from Corollary 4.8 that f ( X ) = q ( X ) g ( X ) r l for some r l E R and some q ( X ) E R [ X ] .But then 0 = f ( r ) = rl as required. If X is a n indeterminate, then an element r in R such t h a t f ( r ) = 0 is called a root of

f(XI 4.10. Corollary. Let R be a n integral domain and let f ( X ) E R [ X ] .

(i)

If a1 . . . , a , are distinct roots of f ( X ) in R , then (X- a l ) ( X - a 2 ) .. . ( X - a,) divides f ( X )

(ii) Zf f ( X ) # 0 , then the number of roots o f f ( X ) an R does not ezceed deg f ( X ) .

Proof. We first observe that (ii) follows from (i). To prove (i), we argue by induction on m. The case m = 1 being a consequence of Corollary 4.9, assume t h a t the statement is true for m - 1 roots. Then

and so f (a,)

9(a,)

= (a,

= 0 so t h a t

X

-

-

a l ) ...( a ,

-

~,-~)9(a,).

Because R is a n integral domain,

a , divides q ( X ) by Corollary 4.9. Thus the corollary is proved. w

Recall t h a t a principal ideal domain (PID) is an integral domain in which all ideals are principal. 4.11. Corollary. Let F be a f i e l d . Then F [ X ] is a principal ideal domain.

Proof. Owing t o Corollary 4.5, F I X ] is a n integral domain. Let I be

a

nonzero ideal

of F [ X ]and let g be a nonzero element of Z of smallest possible degree. If f

#

0 is an

element of I , then by Corollary 4.8, f = qg'+ r for some q , r E F [ X ] where , degr < degg if T

# 0. But r

=f

-

qg E

I , hence r

= 0 as required.

Let R be a GCD-domain. Then a nonzero polynomial f in R [ X ]is said t o be primitive if the G C D of its coefficients is 1. 4.12. Lemma. (Gauss's lemma). Let R be a GCD-domain.

polynomials of R [ X ] .then so is f g .

Proof. Write f and g in the form

Zf f a z d g are primitive

POLYNOMIAL RINGS

21

< s1 < . . . < ,s t o < t l < . . . < t , and each a ; , b j is nonzero. We must show that if d # 0 is a nonunit of R , then d fails t o divide some coefficient of fg.

where so

If ( a 0 , d ) = ( b o , d ) = 1, then ( a o b o , d ) = 1 and so d does not divide aobo, a coefficient of fg. Thus we may assume t h a t either ( a 0 , d ) # 1 or ( b 0 , d )

#

1; say for definiteness

( a o ,d ) # 1. Now consider the sequence do, d l , . . . , d, defined by

Then do .f. 1 and, since f is primitive, there exists a smallest integer i d; = 1. Because d,-1

2

1 such that

is a nonunit divisor of d , it is enough to show t h a t d;-l

does

not divide some coefficient of fg. Hence we may assume t h a t d divides ao, a l , . .. and t h a t ( a i , d ) = 1. Again, we choose k t o be minimal with respect t o the property

( b o , b l , . . . ,b k , d ) = 1. Replacing d by d' = ( b o , b l , . . . ,bk-1, d ) (if k = 0, then d = d'), we may also assume that d divides bo, b l , . . . ,bk-1 and that ( b k , d ) = 1. Let -: R [ X ]--t [ R / ( d ) ] [ Xbe] the natural homomorphism. Since d divides ao, al,.. . , a;-l, bo, b l , . . . , bk-

1,

we have

m

n

Taking into account t h a t ( a i , d ) = ( b k , d ) = 1, we must also have ( a i b k , d ) = 1. Thus

d

a,bk so that the coefficient c ; b k of f g is nonzero. It follows that

fs = f i j # 0 and

hence d fails to divide some coefficients of fg. So the lemma is true. Let R

S be commutative rings and let a l , . . . a , be algebraically independent

elements of S over R. Let X be an indeterminate over R [ a l , .. . ,an]. Then t h e polynomial

f ( X ) = (X- a l ) . . . (X- a,) over R [ a l , .. . ,a,,]

can be written in the form

f ( X ) = X" - six"-'

+ . . . + (-l)ns,

22

CHAPTER 1

where each si = si(a1,.. . ,a,,) is a polynomial in s1 = f f 1 + a2

s, =

&I,.

. . ,a,. In fact,

+ ...+a,

f f 1 f f z . ..a,

The polynomials s l , . . . , s, are called the elementary s y m m e t r i c polynomials of It is clear t h a t si is homogeneous of degree i in

a1,. . .

, a,.

all..

. ,a,.

Let S , be the symmetric group

of degree n. If o E S , and f ( a l r ...,a,) E R [ a l , .. . , a,], define f" by

We say that the polynomial f is s y m m e t r i c if f" = f for all [T E S,. It is clear that the set of symmetric polynomials is a subring of R [ a l , .. . , a,] containing R and the elementary

symmetric polynomials sl,. . . ,s,.

It will next be shown t h a t it contains nothing else.

Let X 1 , .. . , X, be indeterminates. We define the weight of a monomial

t o be

tl

+ 2tz + . . . + nt,.

The weight of an arbitrary polynomial g ( X 1 , .. . , X,) is defined

as the maximum of the weights of the monomials occuring in g . 4.13. Theorem. Let R

be commutative rings a n d let

01,.

R. If f(al,. . . ,a,) E R [ a l , .. . ,a,] polynomial g ( X ; , . . . , X,) E RIX1,. . . , X,] of

independent elements of

ci, t h e n there ezists a

CS

S

over

. . , a , be algebraically is s y m m e t r i c of degree weight

5

d such that

Moreover, t h e e l e m e n t a r y s y m m e t r i c polynomials sl,.. . , s, of air.. . , a , are algebraically independent over R.

Proof. The case n = 1 being trivial, we argue by induction on n. SO assume that the result is true for n

-

1 indeterminates. Substituting a , = 0 in

POLYNOMIAL R I N G S

23

we find t h a t

(X - a1).. . (X - a,-l)X

= X"

- s1,0xn--1

+ . . . + (-l)-lsn-l,ox

where s;,~ is the expression obtained by substituting a , = 0 in s;. Observe t h a t s l , ~ ,... , s,-1,0

are all elementary symmetric polynomials in

a1,. .

. ,an-l. To prove the first asser> 0 and f(a1l . . . ,a,)

tion, we now proceed by induction on d. The case d = 0 being trivial, assume d

< d. Given a polynomial of degree d , there exists a polynomial gl(X1,. . . Xn-l) of weight 5 d such t h a t t h a t our assertion is true for polynomials of degree

Note t h a t g l ( s 1 , . . . , + - I )

has degree

5 d (in

has degree

al,.. . ,a,)

contains a , as a factor. But

5 d in

a1,. ..a,.

Thus the polynomial

and is symmetric. Because fl

.., a n - l , O )

fl(a1,.

is symmetric, so it contains

01..

.a, as a factor. We

deduce therefore that f l = s,fZ(al,. ..,an) for some symmetric polynomial degree is

5d-

= 0, fl

fi

whose

< d. Invoking the induction hypothesis, we may find a polynomial

g2

in

n indeterminates and weight 5 d - n such t h a t

and each term on the right has weight 5 d , proving the first assertion.

To prove the second assertion, we argue by contradiction and choose a nonzero polynomial X(X1,.. . ,X,) E R[X1,. . . ,Xn]of least degree such t h a t qs1, ...,s),

=o

We may write X as a polynomial in Xn with coefficients in R [ X l , . . . ,X,-l], X(X1,. . . , X , ) = Xo(X1,. . . ,Xn-l) and claim t h a t Xo

+ . . . + X k ( X 1 , . .. ,xn-l)x;

# 0. Indeed, otherwise we may write

(3)

CHAPTER 1

24

for some polynomial p , and thus s,p(s1,. . . ,s,)

= 0. But then p(s1,.

. . ,sn)

= 0 and fi

has degree smaller than the degree of A, contrary t o our choice of A. We now substitute s; for

X;in (3) and obtain

0 = AO(S1,.

. . ,sn-

1)

+ . . . + Xk(Sl,.

. . , Sn-1)Sk,

This is a relation in R [ a l , .. . ,a,] and we substitute 0 for a , in this relation. In this way we derive 0 = Ao(s~,o,.

. .>sn-I,O)

which is a nontrivial relation between the elementary symmetric polynomials in a n - 1 . This

al,

. .. ,

provides the desired contradiction, thus completing the proof.

5 . Unique factorization domains.

A unique factorization d o m a i n ( U F D ) is an integral domain in which every element not zero or a unit can be written as a product of irreducible elements, and given two complete factorizations of the same element z = y,y2.. . y , = 2 1 2 2 . . .Z*

then r = t and, after suitably renumbering the z:s,y;

(y,, z, irreducible)

is associated t o z,.

5.1. Lemma. Let R be a GCD d o m a i n . T h e n e v e r y irreducible elemezt of R i s a

prime.

Proof.

Suppose p is irreducible and that plab for some a , b E R. Because p is

irreducible and ( p , a ) is a divisor of p , either ( p , a ) = p or ( p , a ) = 1. Similarly, ( p , b ) = p or ( p , b) = 1. Now ( p , a ) = ( p , b ) = 1 would imply ( p , a b ) = 1, which is impossible. Thus

either pla or p l b , as required.

H

We now observe t h a t if R is a U F D , then R is a G C D domain (in particular, by

Lemma 5.1, every irreducible element of R is a prime). Indeed, giver, two nonzero elements x,y E R , we can write

where u and v are units and where

PI,.

. . , p t are pairwise nonassociated irreducible ele-

ments. Hence any common divisor of z and y is of the form wp;'& . . . p f ' , where w is a

UNIQUE FACTORIZATION DOMAINS

unit and 6;

5 min{a;,pi}. Setting

25

7, = min{a,,p,}, we conclude that (z, y ) = p?’

. . .p:’

and thus R is a GCD domain. It will next be shown that the UFD’s are precisely those G C D domains, in which the ascending chain condition for principal ideals ( A C C P ) is satisfied. 5.2. Proposition. Let R be a n integral domain. T h e n the following conditions are

equivalent: (i) Every nonzero nonunit of R is a product of primes

(ii) R is a U F D

(iii) R is a G C D domain in which A C C P is satisfied. Proof. (i) +- (ii): Because any prime is irreducible, it suffices to establish uniqueness of factorization. To this end, let e = a l ... a , = b l

...b,

be two factorizations into primes. We must show that r = s and that, after suitable renumbering the b:s, b, is associated t o a,. We argue by induction on r ; for r = 1 there is nothing t o prove (since s must then also be l ) , so let r > 1. A simple inductive argument shows that, if a product of more than two factors is divisible by a prime p , then so is one of the factors. Thus allbk for some k, say k = 1 (by renumbering the his) and so a l = b l u

for some unit u of R. Dividing a1 ...a , = bl . . . b, by It therefore follows, by induction, t h a t t h a t a , is associated t o b,, 2

i- -

bl,

we obtain u a 2 . . .a, =

62..

.b,.

1 = s - 1 and t h a t we can renumber the b:s so

5 i 5 r. Since this also holds for i

= 1, the uniqueness of

factorization is established. (ii) 3 (iii): As has been observed earlier, R is a GCD domain. Taking into account that there are only finitely many of pairwise nonassociated divisors of z E R , we conclude that

(z) is contained in only finitely many principal ideals of R. Thus A C C P is satisfied in R. (iii)

+ (i): Owing to Lemma 5.1, every irreducible element of R is a prime.

Thus, given a

# 0, it suffices t o show that z is a product of irreducible elements.

We first prove

nonunit z

that z has a n irreducible divisor. If z is irreducible, there is nothing to prove. Otherwise, let z = z l y l where z l , y l are nonunits. Then either 52,y2

z1

is irreducible or z1 = z2y2 where

are nonunits. Continuing this process, we obtain the ascending chain (z)

c ( 2 1 ) c (zz) c ...

CHAPTER 1

26

By assumption, this chain must terminate; if x, is the last term, x, is irreducible, and

x*lx. We now let x, = p l and write x = P I C l . If c 1 is irreducible, then the proof is complete. Otherwise, we have c 1 = p z c z where

p2

is irreducible. Continuing in this way, we obtain

the ascending chain

(x)L ( C l )

c (Cz) c ... Thus there exist irreducible

This terminates with a n irreducible element c , = P , + ~ . elements p l , p z , . . . ,pn+l such that

and the result follows. m 5 . 3 . Corollary. E v e r y PID is a U F D .

Proof. Suppose t h a t R is a PID. By Propositions 3.9 and 5.2, it suffices to show that R satisfies ACCP. To this end, let

be an ascending chain of principal ideals of R. Then the union is a n ideal generated by b, say, and if b E

( a k ) , then

UCan) = (b)=

(ak)

n

Hence

(Uk)

= ( a k + l ) = . . ., as required. rn

Let F be a field. By Corollary 4.6, a nonzero f ( X ) E if degf(X)

2

1. Thus f ( X ) is irreducible if and only if degf(X)

product of two polynomials of degree 5.4.

F[X] is a nonunit if and only

Corollary.

f(X)E F [ X ]of

2

1 and f ( X ) is not a

2 1.

L e t F be a f i e l d .

Then

F [ X ]is a U F D a n d e v e r y p o l y n o m i a l

positive degree c a n be w r i t t e n as

f(X) = Ufl(X)"'

. . . f,(X)".

( n , 2 1, a E F ' )

where f l ( X ) ,. . . ,j r ( X ) are d i s t i n c t irreducible m o n i c p o l y n o m i a l s u n i q u e l y d e t e r m i n e d by

f (XI. Proof. This is a direct consequence of Corollaries 5.3 and 4.11.

UNIQUE

FACTORIZATION DOMAINS

21

Let F be a field and let f ( X ) E F [ X ]be a monk polynomial of positive degree. Then, by Corollary 5.4,

where f l ( X ) ,. . . , f,.(X) are distinct irreducible monic polynomials uniquely determined by f ( X ) . We shall refer to (1) as the canonical decomposition of f ( X ) . 5.5 Proposition. Let F be a field, let f ( X ) E F I X ] be of degreee

2 1 and let

m

i=

1

be the decomposition o f f ( X ) into the product of powers of distinct irreducible polynomials. T h e n for R = F [ X ] / ( f ( X )t )h e following properties hold: 6) R

=

(4 J ( R )

F[Xl/(fz(X)'2) n:I(ft(x))/(fi(X)")

(iii) R / J ( R ) Proof.

nzlF [ X ] / ( f , ( X )and ) each F [ X ] / ( f , ( X )i)s a field. Since f [ X ] / ( f , ( X ) )is a n artinian integral domain, it must be a field by

Proposition 1.2. Hence (iii) is a consequence of (i) and (ii). Since R is a finite dimensional F-algebra, N ( R ) = J ( R ) . Therefore (ii) follows from (i). To prove (i), note that both sides are of the same F-dimension. On the other hand, the natural homomorphism F [ X ] -+

nzlF [ X ] / ( f , ( X ) e .has ) kernel (f(X)).So the proposition is true.

w

T h e next result will enable us to provide a large class of unique fact,orization domains. 5.6. Proposition. Let R be a G C D d o m a i n . T h e n t h e ring R I X I , . . . , X,] is also a

G C D domain Proof. By Proposition 4.3, we need only verify t h a t R [ X ]is a GCD domain. Let F

and T be the quotient field of R and the set of primitive polynomials in R [ X ] respectively. , U'e claim that for s E S

=

R

-

(0) and for any t E T ,

R [ X ] sn R [ X ] t= R [ X ] s t The containment R [ X ] s tC R [ X ] sn R [ X ] tis always true, so let h = f s = gt E R [ X ] sn R [ X ] t

(2)

28

CHAPTER 1

We write g = s l t l where s1 is the GCD of the coefficients of g and tl E

T. By Lemma

4.12, ttl E T and so s1 is the GCD of the coefficients of h and sIs1. This shows that

h = s l t l t E R [ X ] s t thus , proving ( 2 ) . Regarding R [ X ]as a subring of F I X ] ,we next show that

F [ x ]n ~ R [ X ]= R [ X ] t

for all t E T

(3)

Indeed, let g E R [ X ]be such that g = ft for some f E F [ X ] .Then there exists s E S and fl

E

R [ X ]such t h a t sg = t f l . Observe t h a t if

( a ) n ( b ) = ( a b ) , then a(bc implies alc

(a,b,c E R )

It follows from (2) and (4) that slfl, so g E R [ X ] tand therefore F [ X ] tn R [ X ]

(4)

R[X]t.

The opposite inclusion being trivial, (3) is established. Let f and g be nonzero polynomials in R [ X ] .By Proposition 3.9, the result will follow provided we show t h a t R [ X ] fn R [ X ] gis a principal ideal. To this end, we write f = s l t l and g = s2tz for some slsz E S and t l , t z E T . Because R is a GCD domain, it follows from Proposition 3.9 that RsI n Rsz = Rs for some s E S . Now the elements of S are units of F [ X ] ,so F [ X ] f= F [ X ] t land F [ X ] g= F [ X ] t z .Because F [ X ]is a GCD domain (Proposition 3.9 and Corollary 4.11), we deduce t h a t there exists t E T such that

F I X ] t l n F [ X ) t z= F [ X ] t

(5)

It will now be shown that R [ X ] fn R [ X ] g= R [ X ] s tand this will finish the proof.

; Since t E F [ X ] t i ,i = 1 , 2 , it follows from (3) t h a t t E R [ X ] t l hence

and thus R [ X ] s tC R [ X ] f n R [ X ] gTo . prove the reverse inclusion, write h E R [ X ] f n R [ X ] g in the form h = xy where x E S and y E T . Then silsy for i = 1 , 2 and, by (2)

R[X]sn , R [ X ] y= R [ X ] s i y .It follows from (4)that silx and thus six. On the other hand, invoking (3) and (5), we have

F [ X ] hn R [ X != F [ X ] yn R [ X ]

29

UNIQUE FACTORIZATION DOMAINS

SO

t h a t t l y . Thus s t l z y and therefore

as we wished to show.

It is now a n easy matter to provide a large class of unique factorization domains. 5.7. Theorem. Let R be a U F D . Then, for any s e t I , R[(X;);,I]

is also a L'FD.

Proof. We claim t h a t it suffices t o treat the case where I is finite; if sustained, it will follow, from Proposition 4.3, that we need only show that R [ X ] is a U F D . Assume t h a t 0 # f E R[(X;);,g]. with g1,gZ E

Then f E R [ X l , ..., X,] for some n

2

f = gig2 R[(X;),cr], then glgz E RIX1,. . . , X n , X n + l r . . ,X,] for some m 2 n. We

may regard RIX1,. . . ,X,, . . . ,X,]

1 and if

as the polynomial ring RIX1,. . . , X n ] [ X n + 1 , . . , X,],

in which case f has degree zero and hence both g1 and gz are of degree zero. This shows t h a t g1,gZ E R[XX,. . . ,X,]and hence

f is a n irreducible element of R [ ( X ; ) ; E r if] and only

if f is a n irreducible element of R [ X I , . . . , X,]. This substantiates our claim. Owing t o Propositions 5.6 and 5.2, we are left t o verify that any ascending chain

of principal ideals of R [ X ] breaks off. We may clearly assume that a!l case degfl

2

degf, = degf, case there is a

fn

# 0, in which

2 ... 2 degf, 2 .... Thus there is a n integer m 2 1 such that for all n 2 m. Hence we may assume that degfl = degfz = . . . in which nonzero r,+l in R such that r,+lfn+l = fn for all n 2 1. In particular,

degfi

if b , is the leading coefficient of f,,, then rn+lb,+l

= b , for all n

2

1, so we obtain an

ascending chain (bi)

C (bz) C . . . C ( h )C . . .

This terminates, by Proposition 5.2, so there is a n integer m for all n

2 rn.

Thus (rm+, b,,

)

= (b,+,)

for all j

2

2

1 such t h a t (b,) = (b,)

1, and hence r,+,

is a unit for all

j 2 1. Accordingly, (f,,,) = ( f m + l ) = . . ., as required. We now turn our attention t o polynomial rings over unique factorization domains. Let R be a U F D ,let F be the quotient field of R and let P be a complete set of pairwise nonassociated primes of R . Fix a nonzero f ( X ) E F [ X ] , whose nonzero coefficients are a l , . . . ,a,. Then there exist p l , . . . , p , in P such that

30

CHAPTER 1

where u; E U ( R ) , 1 5 z

5

n. We define the content of f, written c o n t ( f ) , t o be the

product cont(f) = p;'p?

.. .p$

(sk

= min{Ak,(l <_

i 5 n})

or any multiple of this product by a unit of R. Thus cont(f) is well-defined up to multiplication by a unit of R. The following properties are immediate consequences of the definition of cont(f): (a) If f ( X ) E R [ X ] then , cont(f) is the greatest common divisor of the coefficients o f f (b) cont(f) = 1 if and only i f f is a primitive polynomial of R [ X ] (c) f ( X ) = c f l ( X ) where c = cont(f) and f l ( X ) is a primitive polynomial of R [ X ] . 5.8. Theorem. Let R be a U F D and let F be the quotient f i e l d of R .

(i) Zf f , g are nonzero polynomials i n F [ X ] ,then cont(fg)= cont(f)cont(g) (ii) If 0 # f ( X ) E R [ X ]has a factorization f ( X ) = g ( X ) h ( X )an F [ X ] ,then

where cg = cont(g), C h = cont(h), g = c,gl, h = C h h l and

CgCh

E R. In particular, if

# const cannot be decomposed in R [ X ]as a product of factors of degree 2 1, then f

f

is irreducible in F [ X ] . (iii) If f ( X ) is a monic potynomial in R [ X ]and f ( X ) = g ( X ) h ( X ) ,where g ( X ) and h ( X )

are monic polynomials in F [ X ] ,then g ( X ) , h ( X )E R [ X ] .

Proof. cont(g), and

(i) By (c), we may write f = c f l and g = d g l where c = cont(f), d = fl,g1

are primitive polynomials in R [ X ] .Because cont(fg) = cdcont(flgl),

the required assertion follows by Lemma 4.12. (ii) It suffices t o notice that, by (i), C y C h is the content of

f(x)E R [ X ] ,hence C g C h

is in

R,and that, by (c), g l ( X ) , h ( X ) E R [ X l . (iii) In the notation of (ii), let

A1

and p1 be the leading coefficients of

spectively. Because g and h are monic, we have cg = A;' 1-1

-1

pl

6 R. Hence

and thus both g ( X ) and h ( X ) are in R [ X ] .

and c h = p;'

g1

and h l , reand, by (ii),

UNIQUE FACTORIZATION DOMAINS

31

Let I be a n ideal of a commutative ring R . Then the natural homomorphism R

+

R/I

induces a surjective ring homomorphism R [ X ]+ ( R / I ) [ X whose ] kernel I [ X ]consists of all polynomials with coefficients in I . We refer t o this homomorphism as the natural map. 5.9. Proposition. Let P be a prime ideal of an integral domain

+ r l X n p l + . .. + r ,

f ( X ) = roX"

E

R(X]

R and let

( n 2 I)

be such that

ro$P, rn$P2andr,gP, l < i < n Then f ( X ) cannot be written as a product of two polynomials i n R [ X ]of degree

21

Proof. Assume by way of contradiction that f ( X ) = g ( X ) h ( X )with g ( X ) , h ( X )E

R!X]and d e g g ( X ) 2 1, degh(X) 2 1. By hypothesis, f ( X ) = rOXn (mod P [ X ] ) Because

RIP is a n integral domain, by looking a t the image of f ( X ) in ( R / P ) [ X ]we ,

deduce that

g ( X ) = a X k ( m o d P [ X ] )and h ( X ) E P X n - k ( m o d P [ X ] ) for some 1 5 k

5 n and some a ,

ER

-

P . This implies that the constant terms of g ( X )

and h ( X ) belong t o P , hence rn E P 2 , a contradiction. 5 . 1 0 . Corollary. (Eisenstein's Criterion). Let R be a U F D , let F be the quotienf

f i e l d of R and let

+ r l X n p l + . .. + r ,

f ( X ) = rOXn

E R[X]

( n 2 1)

Assume that there is a prime p of R such that ro$O(modp), rn$O(modp2)andr,EO(modp), l < i < n Then f ( X ) is irreducible in F [ X ] . Proof. Apply Proposition 5.9 and Theorem 5.8(iii).

To illustrate some applications of the preceding result, we now provide three examples. 5 . 1 1 . Example. Let k be a nonzero square-free integer n

2

1 , the polynomial X n

~

#

rl. Then, for any integer

k is irreducible over Q . Indeed, take any prime divisor p of k

with p 2 1; k and apply Corollary 5.10.

32

CHAPTER 1

5.12. Example. Let p be a prime number. Then the polynomial

+ x p - 2 + . .. + x+ 1

f ( X )= xp-' is irreducible over Q

. Indeed,

it sufices t o prove that f ( X

+ 1) is irreducible

over Q

.

To

this end, note that the binomial coeficients ( f ) , 1 5 t 5 p - 1, are divisible b y p . Because

( )

f ( X + l ) = ( x + l ) p -=xP-'+ l (;)xp-2+...+ (X+1)-1

p P-1

the required assertion follows from Corollary 5.10. rn 5.13. Example. Let F be a field and let

ct

be a n element of some f i e l d containing F

is transcendental over F . Let K be the quotient field of F [ c t ] . Then, for any

such that

ct

integer n

2 1, the

polynomial X n - n:

Indeed, put R = F [ a ] and note that

is irreducible in K [ X ] CY

is a prime element of R. Since R is a U F D , the

required assertion follows b y virtue of Corollary 5.10. rn We close by recording the following useful fact. 5.14. Proposition. (Reduction criterion). Let P be a prime ideal of a n integral

domain R , let

F

be the quotient fields of fz = R I P and let -:R [ X ]+ fz[X] be the natural

map. I f 0 # f E R [ X ]is such that d e g f = degT and

7 is irreducible in F [ X ] ,then f

be written as a product of two polynomials i n R [ X ]of degree

cannot

2 1.

Proof. Suppose f ( X ) = g ( X ) h ( X )with g ( X ) , h ( X ) E R [ X ] . Then

7= gz. Because

degg 5 degg and degh 5 degh, our hypothesis implies t h a t we must have equality in these degree relations. Hence from the irreducibility of

7in F [ X ] ,we conclude that g or h is an

element of R , as required. rn

6. Dedekind domains.

Let R be an integral domain and let F be the quotient field of R. Buy a fractional ideal of R , we understand any nonzero R-submodule I of F for which there exists a nonzero d

in R such that d . I

R. Expressed otherwise, this means that the elements of I have

DEDEKIND DOMAINS

33

a “common denominator” d E R. The ordinary nonzero ideals of R are fractional ideals

(with d = 1). However, F itself is not a fractional ideal of R (unless R = F ) , since otherwise there exists 0

# d E R such that F

= Rjd-’]. Then d p 2 = rd-’

with r E R,

and so d-’ = r E A , showing t h a t F = R[d-’] = R. Any nonzero finitely generated R-submodule I of F is a fractional ideal. This follows from the fact that, if I = R x l

+ ... + Rx,

for some xi = r , / d r ( r , , d i E R ) , then d =

d l d 2 . . . d , is a common denominator for the elements of I . Conversely, if R is noetherian, then every fractional ideal I of R is a finitely generated R-submodule of F . This is so since I 2 d-’R for some 0

#d

We define the product I

E R and since the R-modules d-’R and R are isomorphic.

.J

of two fractional ideals 1 and J as the set of all finite

sums C x , y , , where x; E I and y, E J . If d l and dz are the common denominators for

I and J , then dld2 is a common denominator for the R-submodule 1. J of F . Therefore I

.J

is a fractional ideal, and so the fractional ideals of R constitute a multiplicative

monoid whose identity element is R . This monoid will be denoted by I ( R ) . The invertible elements of I ( R ) , also called invertible fractional ideals of R , constitute a group. Every

principal fractional ideal of R , i.e. an ideal of the form R x with 0 # x E F is invertible. Furthermore, the principal fractional ideals of R form a group which we denote by P ( R ) . We now introduce a class of integral domains which will be characterized by the property that I ( R j is a group. An integral domain R is called a Dedekind d o m a i n if it is noetherian, integrally closed and evsry nonzero prime ideal of R is maximal. 6.1. Proposition. A n y principal ideal d o m a i n is a Dedekind d o m a i n .

Proof.

Let R be a P I D . Then R is noetherian, by the proof of Corollary 5.3.

Furthermore, since R is a G C D domain (Proposition 3.9), it follows from Proposition 3.8 that R is integrally closed. Let ( p ) primes and p

= qt

C

( q ) be nonzero prime ideals of R. Then p , q are

for some t E R. But a prime is always irreducible, hence t E U ( R ) and

( p ) = ( q ) , as required.

To provide some characterizations of Dedekind domains we first record preliminary observations. 6 . 2 . Lemma. Let R be a noetherian ring and let I

# R be a n ideal of R . T h e n there

exist p r i m e ideals P I , .. . ,P, of R such that P, 2 I for all i = 1 , .. . ,r , and I 2 PlP2.. . P,.

34

CHAPTER 1

Proof. The assertion holds when I is a prime ideal. If I is not prime, we may find a,a' $ I such that a d E I . Put J = I

+ Ra

and J' = I

+ Ra',

so J J '

C I,

but

I c J , I c J'; hence J # R and J' # R. If the statement is already true for the ideals J , J' then it is true for I ; namely, there exist prime ideals P I , . . . ,P, containing J 3 I and &I,.

. . , Q t containing J' 3 I such that P I . . . P,

J,

&I

. . .Qt 5 J ' , so

(Pi . . . P a )(91. . . Q t ) C J J'

I

If however the statement fails for I , for example, then I is not a prime ideal and we may repeat the argument. This process must terminate, otherwise it would give rise to a strictly ascending infinite chain of ideals of R , which is impossible since R is noetherian.

1

6.3. Lemma. Let R be a Dedekind domain with quotient field F , let I be a nonzero

ideal of R and let I-' = { a E FlaI (i) I-'

5 R} = R . Furthermore, i f I

# R

prime ideals of R , then m

5n

i s a fractional ideal of R containing R and I - ' I

then I-'

2 R.

Q1.. . Q , where the { P I } and

(ii) If P I . . . P,

and the

{Qj}

Proof.(i) If 0

{Qj} are

are a rearrangement of a subset of the {Pi}.

#x

E I , then ax E R implies a E Rz-'. Thus I-'

is an R-submodule

of the finitely generated R-module Rx-' and, by Proposition 1.5, I-' generated R-module. This shows that I-'

is also a finitely

is a fractional ideal of R such that I-'

2 R.

Furthermore, R-' = R , since, if a E F is such that a R E R , then a E R.

# R , then I-' PI, . . . ,P,. such that

We now show that if I can find prime ideals

3 R. Choose 0

Choose such a set of prime ideals with minimal possible noetherian, I

#

7.

a E I . By Lemma 6.2, we

Because I

#

R and R is

P for some maximal ideal P of R. Thus

P' ... P, 2 P and we claim that P coincides with some

P; contains an element a; 4 P , and

a1

Pi,say P

= P I . Indeed, if no P; = P , then each

. . . a , E P , which is impossible. We now have

PP2 ... P , C I C P

35

DEDEKIND DOMAINS

By the minimality of r , we know t h a t Pz

p @ a R and put X

=

. . . P,

aR. Choose

p

E Pz

. . . P,

such that

a-'p. Then X @ R , but

and so X E I - ' . It is now an easy matter to prove that I - ' I = R . P u t S = I - ' I is a n ideal of R. Therefore I I - ' S - '

p E S-',

C: I - ' ,

we have I - ' p

a n R-submodule of

SS-'

=

and so I-'Pm

C

C_ R , so I - ' S - '

C I-'

for all m

3

I-'.

and note that S Hence, for any

1. But then I-'[P] is

I-' and therefore, by Proposition 1.5, is finitely generated. Note also

that RIP], being a submodule of I - ' [ p ] , is also finitely generated, again by Proposition 1.5. Since R is integrally closed, it follows that

P

E R. Hence S-'

C: R

and so S-' = R.

Applying preceding paragraph, we conclude that S = R , proving (i).

(ii) We have Pl . . . P,

C Q 1 . . .Qm C: Q1, hence, as in

(i),

&1

must be equal t o some P;,

say Q1 = P I . Then

PL'P1 whence P2.. . P,

...P,

&;'&I

.. . Q m ,

C Q2.. . Q m , and so on.

We are now ready to prove the following basic theorem, which includes results of Dedekind, Noether, Krull, and Matusita. 6.4.

Theorem.

Let R be a n integral d o m a i n .

T h e n t h e following properties are

equivalent: (i) R is a Dedekind d o m a i n . (ii) E v e r y n o n z e r o proper ideal of R i s expressible in a unique way a s t h e product of p r i m e ideals (iii) E v e r y n o n z e r o proper ideal of R i s t h e product of p r i m e ideals

(iv) T h e m o n o i d I ( R ) of fractional ideals of R i s a group.

Proof. (i)

+ (ii).

Let I # 0 be a proper ideal of R . By Lemma 6.2, I contains a

product of prime ideals:

PI . . . P, Choose a maximal ideal P such that I

C P.

cI Then, as in the proof of Lemma 6.3(i), P

coincides with some P;, say P = P I . Hence, by Lemma 6.3(i)

CHAPTER I

36

Suppose we have shown that a nontrivial ideal of R which contains a product of fewer than r prime ideals is expressible as a product of prime ideals (this is trivially true for r = 2 since prime ideals of R are maximal). We then deduce t h a t

F I I = & I . . . Qn

Q j prime

whence

I

=

PF'I

=

P Q l . . .Q ,

Invoking induction on r , we conclude that I is a product of prime ideals. The uniqueness of such decomposition follows from Lemma 6.3(ii). (ii)+ (iii): This statement is trivial (iii)+ (iv): It suffices to show that every nonzero prime ideal of R is invertible. Indeed, let

I be afractional ideal of R and let 0 # d E R be such that d I

R . Then I = ( R d ) - ' ( R d . I )

and, by hypothesis, R d . I and Rd are products of nonzero prime ideals of R. Thus if each nonzero prime ideal of R is invertible, the same holds for I . Now let P

# 0 be

a prime ideal and let 0

prime ideals, so P 2 R a = invertible, then each

P1

. . . P,

#

a E P . By hypothesis, R a is a product of

for some prime ideals Pi, 1 5 i 5 r. Because Ra is

Pi is invertible. Furthermore, since P

is prime, we have P

2 Pi

for

some i E { 1 , 2 , . . . , r } . Thus we are left t o verify that if P is an invertible prime ideal of R , then P is maximal. For this pupose, we first establish the following assertion. Let S be an integral domain, let P I , . . . , P, be invertible prime ideals and let J = PI

. . . P,.

We claim

that if J = Q1 . . . Q 8 with Q j prime ideals, then s = r and after some permutation of indices, Pi = Q; for all i = 1 , 2 , . . . , r . Indeed, let PI be a minimal ideal among P 1 , . . , P,. Since P1 2 J = Q1.. . Q s , it follows that there exists j E (1,. . . , s } , say j that PI 2 Q1. Similarly Q1

2

J = P I . .. P,, hence Q1

Thus P1 2 Pi and, by the minimality, i = 1 and PI = multiplying the equality P1 . . . P, = J = Q1 . . . Qa by

Pz.. . P,

=

PC'J

2 Pi

&I.

for some

z

=

1, such

E {l,. . . , r } .

Now, since Pl is invertible,

Pc1>we have

= Qz . . . QS

Hence, proceeding inductively, the claim follows. Let P be a n invertible prime ideal of R. By the foregoing, we are left to verify that

P is maximal. To this end, choose a E R - P and consider the ideals J = P J' = P

+ Ra'.

By hypothesis, J =

P1

+ Ra

and

. . . P,,,, J' = Q1 . . . Q n , where Pi,&? are prime

37

D E D E K I N D DOMAINS

ideals which must contain J , J ' , respectively, and hence contain P. P u t S = R / P and the iimage of a in S. Then denote by ?

where

P , = P,/P,

Q , = Q , / P are invertible prime ideals of S. Hence, by the preceding

paragraph, 2m = n and, after renumbering we have

It follows that

Q2i-1

= Q2, =

P

P

Pi, so ( P + Ra)' = P

+ Ra2 = ( P + R a ) ( P+ R a )

This implies that every element y E P 2 ,z E R.

P

aEP

-

+ . . . + x,y,

y,a E

P 2 + Ra

6 P may be written in the form z = y

-

$ P , we have

y E P and, since a

and so P = P ( P

+ Ra).

z

But P is invertible, so R = P

+ za

with

E P . Thus

+ Ra for all

A , and thus P is a maximal ideal.

(iv)+ (i): Let 0 sly1

1:

Hence z a = x

C P2 + PRa C P

+ R a 2 and thus

ZplZ

#I

be an ideal of R and let I-'

for some x, E Z , y, E I - ' .

R , so the elements

=

21,.

be its inverse. Since R = I I - ' , 1 =

If a E I , we have a

. . ,x, generate Z .

=

c:=lz,(y,a)with

Thus R is noetherian.

To prove t h a t R is integrally closed, let F be the quotient field of R and let X E F be such that A" -(rlX"-'+.

Hence J 2

+ rlArn-' + . .. + r ,

= 0 for some m

2 I,

. .+rm-lA+rm) belongs t o the fractional ideal J CJ

and, since J # 0 is invertible, then J

C R ; in

r , E R. Then A"

generated by 1 , A , .

=

. . ,A"-'.

particular, X E R and thus R

is integrally closed.

Finally, let 0 # P be a prime ideal of R , let a E R

-

P and let J = P

+ Ra.

Since J

is invertible, then I = J - ' P is such that J I = P , so 1 2 P . On the other hand, if y E I ,

then a y E P . Since a $ P , we have y E P , so Z

CP

and thus I = P. Hence P = J P and

so J = R , proving t h a t P is maximal. m 6 . 5 . Corollary.

L e t R be a Dedekind d o m a i n . T h e n Z(R) is a free abelian group

w i t h t h e collection of nonzero prime ideals of R as free generators.

Proof. As we have seen in the proof of Theorem 6.4, every fractional ideal of R can be expressed as a product of prime ideals of R and their inverses. Hence the nonzero prime ideals of R generate I ( R ) ,and Theorem 6.4(ii) shows that they generate it freely.

M

CHAPTER 1

38

Let R be a Dedekind domain. The factor group

is called the class group of R . The following simple observation ties together Dedekind domains, U F D ' s and PID's. 6.6. P r o p o s i t i o n . The following conditions are equivalent:

(i) R is a Dedekind domain and C ( R ) = 1 (ii) R as a Dedekind domain and a U F D

(iii) R is a P I D

Proof. (i)+ (ii): Let r E R be a nonzero nonunit. Then Rr is a nontrivial ideal of R and so we may write Rr = ( R p l ) . . . (Rp,) for some prime elements P I , .. . , p , , of R. Hence r = p 1 p 2 . . .pnu for some unit u of R , proving t h a t r is a finite product of primes. Thus, by Proposition 5.2, R is a U F D . (ii)+ (iii): It suffices t o show t h a t every nonzero prime ideal P of R is principal. To this end, let 0

# a E P.

Since a is a nonzero nonunit, it is a product of prime elements,

a = p l p z . . . p , . Because a E P it follows that

pi

E P for some i , hence Rp,

C P.

But

R p , is a prime ideal of R , hence it must be maximal, since R is a Dedekind domain. Thus P = R p , is a principal ideal. (iii)+

(i): If R is a P I D , then R is a Dedekind domain by Proposition 6.1. Since any

nonzero prime ideal of R is principal, we also have C ( R )= 1 as required. Let R be a Dedekind domain and let A , B be nonzero ideals of R. We say that A is divisible by B or that B divides A if A = BC for some ideal C of R . The greatest common divisor of A and B is defined t o be an ideal of R which divides both A and B and

is divisible by any other ideal with this property. Similarly, t h e least common multiple of

A , B is defined as a n ideal of R divisible by A and B and dividing any other such ideal. We say that A and B are relatively prime if they have no common prime factor. As we shall see below, g.c.d.{A, B } and l.c.rn.{A, B } both exist and are unique.

6.7. Lemma. Let A , B be nonzero ideals of a Dedekind domain R. Then B divides A if and only if B contains A . Proof. If B divides A , then A = BC for some ideal C of R and thus B 2 A. Conversely, assume that B 2 A . If B = R , then A = B A and B divides A. If B

# R , then

DEDEKIND DOMAINS

39

by Theorem 6.4, B = Q 1 . . . Qm and A = P I . . . P,,, where the { P i } and

{Qj}

are nonzero

prime ideals. Hence, by Lemma 6.3(ii), B divides A. It is now a n easy matter to prove the following result. 6.8. Proposition. L e t A , B be nonzero ideals of a Dedekind d o m a i n R , and let

where t h e { P i } are distinct p r i m e ideals, a n d Pp i s t a k e n t o be R.

(i) g.c.d.{A, B } = A (ii)

~ . C . ~ . { A , B }

+ B = nz1P?'"("''*') P ? ~ ~ ( ~ ~ , ~ . )

= A nB =

(iii) A , B are relatively p r i m e if a n d o n l y if A

+B =R

(iv) T h e nonzero ideals A l , . . . , A , , of R are pairwise relatively p r i m e if a n d only if

(v) If t h e nonzero ideals A l , . . . , A , of R are pairwise relatively przme, t h e n

n,"=, A,

=

n:=, A , . Proof. We first observe that (iii) follows from (i), (iv) from (iii), and (v) from (iii) and Proposition 1.3. T h a t g.c.d.{A, B} = A

+B

is a consequence of Lemma 6.7. This

proves (i) upon noting that A divides B if and only if a , 5 b, for all z. A similar argument proves (ii). 6.9.

P,"' . . . P:',

Corollary. Let I be a nontrivial ideal of a Dedekind d o m a i n R , say, 1 n,

--

> 0 , where P I , .. . ,Pk are distinct p r i m e ideals of R . T h e n

Proof. T h e ideals P,"', 1 5 i 5 k, are pairwise relatively prime. Hence, by Proposition G.S(iii), (v), P,"'

+ P,'"

=

R for i

#

J

and I =

n,"=,P,'".

Now apply Propositlor?

1.3.

6.10. Corollary. L e t R be a Dedekind d o m a i n , let P I , . . . , P,, be distinct nonzero

p r i m e ideals of R a n d let X I , . . . ,x, E

R. T h e n , f o r a n y given nonnegative integers

CHAPTER 1

40

e l , . . . ,en, there ezists z E R such that

Proof. Given i E (1,. . . ,n } , we have

Pie'

3 Pzes+l,so there exists a,

E P,"' - P;.+l.

+

By Corollary 6.9, there exists z E R such that z - (5, a,) E Plea+'for all i E (1,. . . ,n}. Therefore x

-

+

zt = [z- (z, a,)]

+ a, 6 Plea

-

PzeS+l.

6.11. Corollary. A n y semilocal Dedekind domain is a principal ideal domain.

Proof.

Let R be a semilocal Dedekind domain. Then R has finitely many, say

P1,. . . ,P,, distinct nonzero prime ideals. It suffices to show that each Pa is a principal ideal. By Corollary 6.10 (with

XI =

. . . = x,

i E (1,. . . ,n}, there exists an element y , Thus P;lRyi, Pi"

+ Ry, and, for j

= 0 , e i = 1,

for each

Pi - P,? and y i @ P, for j

E R such t h a t y ; E

# i, Pi # Ry,.

# i),

ej = 0 for j

# i.

Hence the decomposition of Ryi into

prime ideals is Ryi = P,, as required. m Since a Dedekind domain is noetherian, its all ideals are finitely generated. This can be improved by using the following result. 6.12. Proposition. Let R be a Dedekind domain and let A , B be nonzero ideals of

R . Then there exists a nonzero ideal C of R such that C is relatively prime to B and A C is a principal ideal.

Proof.

Write A = P:'

...Pg-,

B = P,"' ... Pin with a, 2 0, b, 2 0, where

P I , . . . , P, are distinct prime ideals. Choose a, E Pf-. Corollary 6.9, there exists a E R such that a a E Pp'

-

P,"'

+l

, so

aE

n:=,

Paas" ,1

5 i 5

a,(mod Pta'+'), 1 5

n. Then, by

t

P,"' = A (Proposition 6.8(v)) and a R + A B

fore, by Lemma 6.7, we have a R c,

=

-

+ A B = P:'

... Pi"- with

c,

2

a,. But a:

5 n. Hence

C A.

4

There-

P,"*+l, so

+ AB = A. Since a R C A , we have a R = A C for some ideal C of Therefore A C + AB = A and multiplying by A-' yields C + B = R , as

= a , and thus a R

R (Lemma 6.7). asserted. rn

6.13. Corollary. If R is a Dedekind domain, then every ideal of R may be generated

b y at most two elements: one of these elements may be arbitrarily chosen. Proof. Let A be a nonzero ideal of R and let r l be any nonzero element of R. By

DEDEKIND

DOMAIMS

41

Proposition 6.12, we may choose a nonzero ideal B of R relatively prime t o r1R such that

A B = r2R for some

r2

E R . Then we have

+ r 2 R = r l R + A B C_ A On the other hand, since B + r l R = R , 1 = p + r l a for some p E B , rlR

a E R . Hence, for

each a E A , we have

+ a p E r R + AB and so A C r l R + r2R. Thus A = r l R + r2R, as asserted. a = ar 1 a

1

We next record another consequence of Proposition 6.12. 6.14. Corollary. Let A , B be nonzero ideals of a Dedekind domain R . Then the

additive groups of R I A and B I A B are isotrorphic. I n particular, for any n

2 1, R I A

and

An/An+' have isomorphic additive groups. Proof. Owing to Proposition 6.12, we may find a nonzero ideal C of R relatively prime to A and such t h a t BC = T R for some r E R. Then the map q5 : R by +(z)

=

rx

+ AB

+

B I A B given

is an additive homomorphism, which is surjective, since

rR+AB = C B + A B = (C+ A)B = RB = B Assume that x E Ker4. Then r x E A B and rxC

C ABC

=

r A , whence x C

2 A.

+ C = R , we have 3 + c = 1 for some a E A , c 6 C. Then x = x u + xc E A . if z E A , then d ( x ) = r x + A B = A B , proving that Kerq5 = A. A

Assume t h a t R is

3

Because

Conversely,

Dedekind domain such that for every nonzero prime ideal P of

R , R I P is finite. Then we put N ( P ) = I R / P / and refer to N ( P ) as the norm of P . Note that since R I P is a finite field, N ( P ) is a positive power of a rational prime p. If I is any fractional ideal of R , say

where 1'1,.. . ,p k are distinct prime ideals of R , then the norm of I is defined by

N ( 1 ) = N(P1)"' . . . L V ( P ~ ) ~ '

-

6.15. Proposition. Let R be a Dedekind domain such that f o r every nonzero p r i m e

ideal P of R , R I P is finite. Then the map I ( R ) --t Q', I Furthermore, /or any nonzero ideal I of R .

N ( I ) is a homomorphism.

CHAPTER 1

42

Proof. T h e first assertion follows from Corollary 6.5. To prove t h e second assertion, it suffices, by Corollary 6.9, t o prove t h a t ( R / P n (= N ( P ) n for all n 6.14, we have ] P n - l / P n / = N ( P ) for all n pn-1

C

... c P

C

2

2

1. By Corollary

1. Hence, by looking a t the chain Pn

c

R , we deduce that IR/Pnj = N ( P ) " as required. w

To provide another characterization of Dedekind domains, we first record the following three elementary observations.

6.16. Lemma. Let { R , } be a family of integral domains all contained in one large domain, and suppose that each R; is integrally closed. Then nR; is integrally closed.

Proof. If X is an element of the quotient field of O R , which is integral over nRi, then X is integral over each R;. Hence X E nRi, as required. w 6.17. Lemma. Let R be a n integral domain. Then R = n R p where P ranges over all maximal ideals of R .

Proof. Suppose that z E Rp for all maximal ideals P of R. It suffices to show t h a t z E R. We may write x = a / b with a , b in R. Let I be the ideal of R defined by

I = {y E R J y a E Rb} For any maximal ideal P of R , we can find

T,S

In particular sa = rb, so s is in I . Hence I

I

= R. Therefore a E

in R with s $ P such t h a t z = a / b = r / s .

P for any maximal ideal P of R and thus

Rb and a l b = x E R. rn

6.18. Lemma. Let R be a Dedekind domain and let S be a multiplicative subset of

R. Then Rs is a Dedekind domain. Proof. By Corollary 2.4, Rs is noetherian and, by Proposition 3.7, Rs is integrally

# 0 be a prime ideal of R s . Then, by Proposition 2.3(iv), J = I R s for some prime ideal I of R with P n S = 0. Since I is a maximal ideal of R , it follows closed. Finally, let J

from Proposition 2.3(iii), t h a t J is a maximal ideal of R s , as requirez, w We say that R is a discrete oaluation ring if it is a principal ideal domain with a unique nonzero prime ideal.

6.19. Theorem. Let R be a n integral domain. Then R is a Dedekind domain if and only i f R valuation ring.

is

noetherian and for every nonzero prime ideal P of R , Rp is a discrete

DEDEKIND DOMAINS

43

Proof. Assume that R is a Dedekind domain. If P is a nonzero prime ideal of R, then Rp is a Dedekind domain by Lemma 6.18. Furthermore, by Corollary 2.5, Rp has a unique nonzero prime ideal. Hence, by Corollary 6.11, Rp is a discrete valuation ring. Conversely, assume that R is noetherian and that for any nonzero prime ideal

P of

R , Rp is a discrete valuation ring. Since Rp is integrally closed (Propositions 3.9 and 3.8), it follows from Lemmas 6.16 and 6.17 that R is integrally closed. Finally, suppose

P c Q are nonzero prime ideals of R. Then, by Corollary 2.5, PRQ C QRQ are distinct prime ideals of R Q , which is impossible since RQ is a discrete valuation ring. Thus every nonzero prime ideal of R is maximal, as required. The rest of the section will be devoted to providing a large class of Dedekind domains. 6.20. Lemma. Let R

CS

be integral d o m a i n s , S integral over R. T h e n S is a field

if a n d only if R i s a field. Proof. Assume t h a t R is a field and let 0 # s E S. If

is a n equation of integral dependence for is a n integral domains). Hence s-'

s

of smallest possible degree, then r ,

= -r;'(sn-l

+ rlsn-' + . . .+ r,-1)

#

0 (since S

E S and therefore

S is a field. Conversely, assume that S is a field and let 0

#

r E R. Then r-'

E S , hence is

integral over R , so that we have an equation

It follows that r-'

=

-(TI

' + r2 ' I .+ . . . + r k r r n - ' )

E R , as required..

Let R & S be commutative rings. If Q is a prime ideal of S, then Q n R is a prime ideal of R, since R / ( Q n R) is identifiable with the subring of the integral domain S / Q . 6.21. Corollary. Let R

iS

be commutative rings, S integral over R. If Q is a

p r i m e ideal of S , t h e n Q is m a z i m a l i f and only if Q n R is m a z i m a l .

Proof. By Proposition 3.6, S / Q is integral over R / ( R n Q) and both these rings are integral domains. Now apply Lemma 6.20. 6.22. Lemma. Let R

CS

be integral d o m a i n s , S integral over R. If I i s a nonzero

ideal of S,t h e n I n R is a nonzero ideal of R.

CHAPTER 1

44

Proof. Fix 0 # s E I and let sn

+ rlsn-' + . . . + r,

=o

(ri E R )

be an equation of integral dependence for s of smallest possible degree. Then r ,

# 0 since

S is an integral domain. Accordingly, r n = -(sn

+ rlsn-' + . . . + rn-ls) E I n R

as asserted. H We next quote the following standard fact of commutative ring theory. 6.23. Proposition. L e t R be a noetherian integral d o m a i n w i t h quotient field F and

let S be a n y ring between R a n d F .

If every nonzero p r i m e ideal of R

is mazimal, then S

i s a g a i n noetherian.

Proof. See Kapiansky (1974, Theorem 9.3). We are now ready to prove 6.24. Theorem. Let R be a Dedekind d o m a i n with quotient field F a n d let E be a

field f i n i t e d i m e n s i o n a l over F . If S i s t h e integral closure of R in E, t h e n S i s

Q

Dedekind

domain.

Proof. It is clear that E is the quotient field of S . Hence, by Corollary 3.5, S is integrally closed. Furthermore, by Corollary 6.21 and Lemma 6.22, every nonzero prime ideal of S is maximal. It remains to prove that S is noetherian. We can find a vector space basis of E over F consisting of elements of S (first take any basis, then multiply by suitable elements of R ) . Let

v1,.

. . ,v , be any such basis and

put SO = R [ v l , . . . , v , ] . By Corollary 3.2, SO is a finitely generated R-module, hence SO is noetherian. Furthermore, since So is integral over R , any nonzero prime ideal of So is maximal (Corollary 6.21 and Lemma 6.22). Since S lies between SO and its quotient field

E , S is noetherian by Proposition 6.23.

45

2 Separable algebraic extensions

Separable algebraic extensions constitute a n important class of field extensions. In this chaptcr we provide some background results concerning such extensions. These include a characterization of finite separable field extensions E / F in terms of the number of F-

homomorphis~nsof E into its algebraic closure and behaviour of separability under field extensions. We t h e n characterize separability via linear disjointness and tensor products. Turning to the trace m a p and discriminants, we also provide characterizations of scparat>ilii,y via these notions. T h e chapter also contains numerous results prrtaining to various

properties of separable algebraic extensions to b e applied in subsequent investigations.

1. Algebraic closure, splitting fields and normal extensions.

Let 1; be a field and F a subfield; we write E I F for t h e field E considered as a n exterision

of F . We can regard E as a n F-algebra and hence a vector space over F; its dimension is either a positive integer or a n infinite cardinal a n d is called the degree of E over F . In \\hat folloivs we write ( E : F ) for the degree of E over F . \Ye shall rcfcr to t h e csterisior~

/:/t+’a s being f i n i t e or injinile accordingly as I,(Tt

E . ]Ye Lvrite F ( S ) (respcctivoiy, /+’iS!) for the srrialicst sul)fieItI

S t)c a subset of

(rrspcctivcly, siibring) containing

*.T“)

S and I’. In case

and F : s l , .. . ,s,]

i v r also \vritc F’(s,, . . . , s,)

t h a t f i ’ ( . ~ , %. . .

(/< : F ) is finite or infinite.

S is finite. say .’i

~~

{sl,s z , ,

. . . ,s,~).

instead of b ’ ( S ) a n d F ’ S ] ,respectively. X o t c

is t.he quotient field of I ~ ’ l s l . . . , , s,?] and t h a t F s 1 , . . . , .s,,]

consists of

all polynomials i n s l , . . . s , over F . We shall refer to I g ’ ( s , , . . . s n ) a s t h e f i e l d g e n e r n f e d ~

I+’ by

O Z J C ~

.i1

~

. . . s,

~

or t h e field obtained b y adjoining to F tlie

Let, E I.‘ and K / F b e t\vo field extensions-. If /,; t.licn it

!.tic,

\YCI

d e n o t e 11)

er,ir,lmsi/e of /;

that E

iinil

E

anti

clciri(,rirw 5 1 ,

. . . ,s 7 .

li a r c coiirained in

soiiie Iic111

E K tile srriallcst subficld o f I , coritaiiiing l i o t l ~I;‘ a n d /i: i i i i ( I CJ! iinrl

li i n I,. T h u s

enever

l i a r c ton~iiiriotii:: s o n i r fic.lc1 /,.

writ(. i;K

ISC

nutorn;tiic;ilI)

46

CHAPTER 2

A field extension E / F is said t o be f i n i t e l y generated if E = F ( s 1 , . . . , sn) for some

finitely many elements s 1 , . . . ,sn of E . The extension E / F is called simple if there is an element

Q

in E , called a p r i m i t i v e e l e m e n t over F , such that E = F ( a ) .

Let E / F be a field extension. An element

Q

of E is said t o b e algebraic over F if

Q

is a root of a nonzero polynomial over F . If f ( X ) is a nonzero polynomial over F with f(a)== 0 and if X is the leading coefficient of f ( X ) , then g ( X ) = X - ' f ( X ) is a monic polynomial with g ( a ) = 0. Hence a is algebraic over F if and only if a is integral over F . We say that E / F is algebraic if all elements of E are algebraic over J'. By Corollary 1.3.4,

if F

CE CK

is a chain of fields, then K / F is algebraic if and only if both E / F and K / E

are algebraic. 1.1. Proposition. Let E / F be a field e z t e n s i o n , let a E E be algebraic over F and

let f ( X ) E F I X ] be of least degree, s a y n, s u c h t h a t f(a)= 0. (i) f ( X ) is irreducible a n d t h e m a p

is a n F-isomorphism. (ii) f ( X ) divides each g ( X ) E F I X ] with g ( a ) = 0 , F ( a ) = F [ Q ]a n d 1 , a , . . . , a n p 1is a n F-basis for F ( a ) . I n particular, ( F ( a ): F ) = n.

Proof. (i) Consider the natural epirnorphism FIX] 4 F [ a ]which sends g ( X ) to g ( a ) .

Since F I X ] is a P I D , its kernel is a principal ideal, say ( f l ( X ) ) . Since f(a)= 0, we have

(fix))C ( f l ( X ) ) .On the other hand, write

where either r ( X ) = 0 or degr(X) < d e g f ( X ) . Since .(a) = 0, the minimality of the degree of f ( X ) implies that fl(X) = 9 ( X ) f ( X ) . Hence ( f l ( X ) ) C ( f ( X ) )and therefore

( f ( X ) )= ( j , ( X ) ) .Thus the map

is an F-isomorphism and therefore F [ X ] / ( f ( X )is) an integral domain. But this domain is obviously artinian, hence is a field by Proposition 1.1.2. T h u s f ( X ) is irreducible and

F [ a ]= F ( a ) , as required.

ALGEBRAIC CLOSURE

47

(ii) This follows from the isomorphism of (i) and from the fact t h a t 1

(f(X)

1 1

' ' '

+ ( f ( X ) ) ,X

7-

, xn-' + ( f ( X ) )is a n F-basis for F [ X ] / ( f ( X ) ) . w

Let E / F be a field extension and let a E E be algebraic over F . Owing t o Proposition l . l ( i i ) ,there exists a unique monic polynomial f ( X )E F I X ] of least degree for which a root. We shall refer t o f ( X ) as the m i n i m a l polynomial for

SL

(Y

is

over F and t o d e g f ( X ) as

the degree of a over F . Alternatively, the minimal polynomial for a over F can be defined as the unique irreducible monic polynomial in F[X] for which a is a root. T h e following

result shows that the F-isomorphism class of F(a) is uniquely determined by the minimal polynomial of a . 1 . 2 . Proposition. Let E I F and E'IF be t w o extensions of F and l e t a and

be elements of E and E', respectively, which are algebraic over F .

0'

T h e n the following

conditions are equivalent: (i) a and a' have the s a m e m i n i m a l polynomial f ( X ) in FIX]

(ii) There exzsts a n F - i s o m o r p h i s m F ( a ) + F ( a ' ) which carries a i n t o a' Proof. (i)*

(ii): By Proposition l . l ( i ) , the map

is a n F-isomorphism. On the other hand, the m a p

is also an F-isomorphism, again by Proposition l . l ( i ) . Taking the composite of these

isomorphisms, ( i i ) follows.

+- ( i ) : If f ( X ) = X " + a l X n p l + . . . t a , is the minimal polynomial for a , then a n + a l a n p l + . . . + arL= 0 and hence (a')" + U ~ ( Q ' ) ~ - '+ , . , + a , = 0. T h u s u' is a root (ii)

of f ( X ) and so f ( X ) is divisible by the minimal polynomial g ( X ) for a', by Proposition l . l ( i i ) . A similar argument shows t h a t g ( X ) is divisible by f ( X ) , 1ier:ce t h e result. rn Let E / F be a field extension. Two elements

a,P E E

are said t o be F-conjugate if

they are algebraic over F and have the same minimal polynomial over F . 1 . 3 . Corollary. Let E I F be a field extension and l e t

the m i n i m a l polynomial f o r

is at m o s t n . :Moreover,

cy

over F is

z f p E fi

OJ

cy

E E be algebraic over F .

If

degree n , t h e n the number of F-conjugates o f u

is conjugate t o a, t h e n F ( a ) and F ( P ) are F - i s o r n o r p h c .

48

CHAPTER 2

Proof. This is a direct consequence of Corollary 1.4.10(ii) and Proposition 1.2. n 1.4. Proposition.

degree

2 I.

Let F be a field and let f l , . . , , f , be polynomials in F [ X ] of

T h e n there ezists a finite field extension E / F in which each f; has a root.

Proof. We first consider the case where n = 1 and f = f l . Let I be a maximal ideal of F [ X ]containing ( f ) . Then E = F [ X ] / Iis a field containing a n isomorphic copy

( F + I ) / I of F. Hence we may regard E as a field extension of F and as such it is obviously finite. Since

(Y

=X

+I E E

is a root of f ( X ) ,the assertion follows.

Turning t o the genera! case, we may find a finite extension E 1 / F in which

fl

has a

root. We may view f z as a polynomial over E l . Let E2IEI be a finite extension in which has a root. Proceedingly inductively, the result follows. rn For future use, we next exhibit certain properties of algebraic extensions. 1.5. Proposition. Let E I F be a field eztension.

(i) I f S i s a subset o f E sucii t h a t all elements o f S are algebraic over F , t h e n F ( S ) = F [ S ] a n d F ( S ) / F is algebraic

(ii) T h e elements of E which are algebraic over F f o r m a subfield of E containing F (iii) E I F is f i n i t e iJ and only if E / F is algebraic a n d finitely generated.

Proof. (i) We know that

CY

E E is algebraic over F if and only if

(Y

is integral over

F . Hence F ( S ) / F is algebraic if and only if F ( S ) is integral over F . By Corollary 1.3.3, F [ S ]is integral over F. If 0

#

by Proposition l . l ( i i ) . Thus

(Y-'

(Y

E

F i S j , then

(Y

is algebraic over F , hence F [ a ]is a field

E F [ a ]C F [ S ]and so F ( S ) = F [ S ] ,as required.

(ii) Apply (i) for the case where

S is the set of all elements of E which are algebraic over

F. (iii) This is a direct consequence of ( i ) and Corollary 1.3.2.

Let E / F b e a field extension. By the algebraic closure of F in E we understand the subfield of E consisting of all elements t h a t are algebraic over F . 1.6. Proposition. Let F

EE CK

be a c h a i n o f f i e l d s , let

F

be the algebraic closure

of F in E a n d let X E K . (i)

(ii)

If X is algebraic over F , t h e n X is algebraic over F .

If X is algebraic over E and E I F Proof.

i s atgebraic, t h e n X is algebraic over F .

Property ( i i ) is a particular case of (i) in which

7=

E , while ( i ) is a

A L G E B R A I C CLOSURE

49

consequence of Corollary 1.3.4. We define a field E' t o be algebraically closed if every polynomial in F I X ] of degree

> 1 has

a root in F . We observe t h a t if F is a n algebraically closed field a n d f ( X ) € F [ X ]

2

has degree n

1, then

f ( X ) = X(X

al).

~

. . ( X - a,)

for some A, a l , . . , , a , in F . Indeed, f ( X ) has a root a , in F , so thcrc exists y(X) E F : S j such t h a t

f(X)= ( X If degg

2

~

a,)s(X)

1, we can repeat this argument inductively, and express f ( X ) in t h e desired

form. Note t h a t X is the leading coefficient of f ( X ) . Hence if t h e coefficients of

f(x)

lie in a subfield K of F , then X € K . T h e following result, whose proof is due to Artin, guarantees the existence of algebraically closed fields.

1 . 7 . Theoran. L e t F be a f i e l d . T h e n t h e r e e x i s t s a n a l g e b r u z c a / ~ yclosed field 5,

F a s a subfield.

eontainzng

Proof. Let u s first exhibit a n extension E l / F in which every polynomial in F : X : of degree

2

1 has a root. To this e n d , p u t S =

{XflfE F [ X ] ,degf 2 l}.

We form tile

polynomial ring F [ S ]a n d denote by I t h e ideal generated by all t h e polynomials f(X,) i n

F [ S ] .\Ye claim t h a t I

# F j S ] . Indeed,

otherwise there is a finite combination in I which

is equal !o 1:

S l f l ( X f , ) -t...

with g n E

FIR].P u t X ,

=

+ Snfn(Xf,,) = 1

X I , and observe t h a t the polynomials q, involve

rnirnber of indeterminatcs, say X I , . . . , X,;

with

7r2

2

only a finite

\Z'c may therefore rwvrite our

71.

relation as:

c i

9; (Xl 1 . . . > X m ) f*(Xi) =- 1

1-1

T h a n k s t o Proposition 1.4, we may find a finite extension K/E' in which each polynomial fL,

1

X,, 1

5 i 5 i

n , has a root, say a,. P u t

(I, =

0 for n

< z 5

7n.

Substituting a 1 for

m, in our relation, we get 0 = 1, a contradiction.

By the foregoing, we may choose a maxirriai ideal A4 o f FILS, containing 1. Thrn El

= FISj/AIfis a field containing a n isomorphic copy (E'

+ iZI)/LIf of

b'. 'Therefore,

m a y regard E l as a field extension of E'. Moreover, for every polynomial f

dcgree 3 1 , X J t M E El is a root of f.

\\'(>

F,Sj oi

50

CHAPTER 2

Inductively, we can form a chain of fields:

such that each polynomial in E,[X] of degree union of all fields

En,n

= 1 , 2 , . ..

. Then

2 1 has

a root in En+1. Denote by E the

E is obviously a field and every polynomial

f E E [ X ]has its coefficients in some field En. But then f has a root in Enfl E, as required. rn

1.8. Corollary. Let F be a field. Then there ezists a n algebraic eztension EIF such that E is algebraically closed. Proof. Owing t o Theorem 1.7, there exists a field extension K I F such that K is algebraically closed. Let E be the algebraic closure of F in K . Then E I F is a n algebraic extension. Furthermore, if f ( X ) E E [ X ]is a polynomial of degree

3

1, then f ( X ) has a

root, say a , in K . Thus a is algebraic over E and so, by Proposition 1.6(ii), a is algebraic over F . Therefore a E E and the result follows. Let F be a field. By a n algebraic closure of F we understand any field E containing

F such t h a t E/F is algebraic and E is algebraically closed. By Corollary 1.8, such E always exists. Our next aim is t o demonstrate t h a t E is determined uniquely, up t o an F-isomorphism. This will be achieved with the aid of the following two results. 1.9. Lemma. Let F be a f i e l d and let u : F

---t

L be a homomorphism of F into an

algebraically closed field L . Let a be an algebraic element an a f i e l d extension of F and let f ( X ) E F[X] be the minimal polynomial for a. If t is the number of distinct roots o f f ( X )

(in an algebraic closure of F ) , then there ezist ezactly t extensions of u t o a homomorphism F ( a ) + I,.

Proof. Given f ( X ) E F [ X ] ,let f " ( X ) E L [ X ]be obtained from f ( X ) by applying u

t o the coefficients of f ( X ) . Because L is algebraically closed, there exists a root /3 of

f ' ( X ) in L. By Proposition l.l(ii), F ( a ) = F[a]and so a typical element of F ( a ) is $ ( a ) for some G ( X ) E F [ X ] .We define a n extension of u by mapping

This is well defined, i.e. independent of the choice of the polynomial $ ( X ) used t o express our element in F[a]. Indeed, if

$l(X)E F [ X ]is such t h a t $ ( a ) = $ l ( a ) , then ($ -

A L G E B R A I C CLOSURE

4 , ) ( a )= 0,

hence f ( X ) divides

47(X), and

thus

$“(P)

=

51

$(X)- $l(X).It follows that f“(X) divides $ “ ( X )

~

$Y(P).

It is now obvious that the given map is a homomorphism F ( a ) + L inducing u o n

F . Now the number of distinct roots of f (X) in the algebraic closure of F is equal to t h e number of distinct roots of f “ ( X ) E L [ X ] .Because the image of

cy

under any extension

F ( u ) i L of u is a root of f “ ( X ) , the result is established. 1.10. Proposition. Let E / F be a n algebraic extension, a n d let u : F

->

L be a

homomorphism into a n algebraically closed field L.

(i) There exists a n extension u’ of u to a homomorphism E

L

--*

(ii) If E is algebraically closed a n d L is algebraic over u ( F ) , t h e n u’ is a n isomorphism.

Proof. We first demonstrate that (ii) is a consequence of (i). Indeed, if E is aigcbraically closed and L is algebraic over u(F),then u’(E) is algebraically closed (since u must be injective) and L is algebraic over u ‘ ( E ) . Hence L = u’(E) as required. To establish ( i ) , denote by

S the set of all pairs ( K , X ) where K is a subfield of 6

containing F, and X is an extension of u t o a homomorphism K so

-+

L. Note that (F,G ) E S ,

S # 0. If K , A) and (K’, A’) are in S,we write ( K ,A) 5 ( K ’ ,A’) if K

If { (K,, A,)} is a chain in S , we let K

=

K’and X’IK

=

A.

U K , and define A on K to be equal t o A; on each

K,. Then ( K ,A) is an upper bound for this chain and so the partially ordered set S is inductive. Invoking Zorn’s lemma, we may therefore choose a maximal element, say ( T ,p ) , in S. Then p is a n extension of u , and we claim that T = E . Indeed, otherwise there exists

cy €

E

-

T and, by Lemma 1.9, p has an extension t o T ( a ) ,thereby contradicting

the maximality of ( T , p ) . This shows that there exists an extension of u t o E , as asserted. I t is now an easy matter to prove that the algebraic closure of F is determined uniquely,

up to an F-isomorphism.

1.11. Corollary. Let F be a field. T h e n a n y two algebraac closures of I+‘ a r e 1,’isomorp hz c .

Proof. Let E and E’ be algebraic closures of F and let u

:F

i

map. Then, by Proposition l.lO(ii), u extends t o an isomorphism E

Let F

C E 5F

E’ be the inclusion

+

E‘ as asserted.

rn

-

be a chain of fields, where F is an algebraic closure of F . Our next

airn is to show that any F-homomorphism E

i

F

is extendible to an automorphisni of

52

CHAPTER 2

-

F . This will be derived as an easy consequence of Proposition l.lO(i) and the following

observation.

1.12. Proposition. Let E I F be a n algebraic extension and let u : E

+

E be a n

F - h o m o m o r p h i s m . T h e n u i s a n automorphism.

Proof. It suffices to verify that a ( E )

E . To this end, fix cr E E and denote by

f ( X ) its minimal polynomial over F . If E’ is the subfield of E generated by all roots of f ( X ) which lie in E , then E’/F is finite by Proposition 1.5(iii). Furthermore, u maps a root of f ( X ) t o a root of f ( X ) and hence maps E’ into itself. Since (E’ : F ) = ( u ( E ’ ) : F ) , we must have u ( E ’ ) = E’. Because cr E E‘, it follows that cr E a ( E ’ ) C u ( E ) ,as desired.

m

C E C F be a chain of

1.13. Corollary. Let F

of F . T h e n a n y F - h o m o m o r p h i s m

(T

:

E

---t

F

fields, where

F

is a n algebraic closure

is extendible to a n automorphism of

F.

__

Proof. By hypothesis, F is algebraically closed and F / E is algebraic. Hence, by Proposition l.lO(i), there is an extension u’ of

(T

t o an F-homomorphism u’ : F

--t

F.

Since F / F is algebraic, the result follows by virtue of Proposition 1.12. m Let F be a field and let {ftli E I} be a family of polynomials in F [ X ]of degree

2 1.

By a splzttzng field for this family we understand an extension E of F such that every

fl

splits into linear factors in E [ X ] ,and E is generated by all the roots of all the polynomials ft,z E I. We note that any family of polynomials in F [ X ] always has a splitting field,

namely the field generated by the roots of the given family in ani algebraic closure of

F . Our next aim is to show that the splitting field is determined uniquely, up to an F-isomorphism. This will be achieved with the aid of the following preliminary result.

1.14. Lemma. Let f ( X ) E F [ X ]and let E , K he two splitting fields of f ( X )

( i ) E and K are F-isomorphic [ii) If F C E

C r;

zs a chain of fields, where

F - h o m o m o r p h i s m I<

Proof. (i) Let

E’ i s a n algebraic ciosure

F , t h e n any

7 is a n isomorphism of K onto E .

4

be an algebraic closure of E . Then E is algebraic over F , hence

ib

an algebraic closure of E’. Thanks to Proposition l.lO[i), there exists a n F-homomorphism

ALGEBRAIC CLOSURE

a :K

+

53

z.We are therefore left t o verify t h a t a ( K )= E

By hypothesis, f ( X ) splits into linear factors in K [ X ] hence ,

with

8,E K : 1 5 i 5 n, X E F .

ant1 t h u s

K

a l l . ..,a,,

Then

with a , = a @ % ) ,1 5 z

5

n, are all roots of

fix) in

E . Since

F(P1.. . . ,p,) a n d E = F ( a 1 , . . . , a , ) , we conclude that .(K) = E as desired.

:

(ii) Because

F

is a n algebraic closure of E , it suffices to p u t

= F an d apply (1).

We are now ready to prove 1 . 1 5 . P r o p o s i t i o n . Let { f t l i E I } be a f a m i l y of polynomials in F [ X ] and let K , E

be tiuo splitting fields f o r this f a m i l y . T h e n K and E are F-isomorphic. P r o o f . Denote by

a n algebraic closure of K . Th en , by Proposition l . l O ( i ) , there

exists an F-homomorphism (T : E fi

t

x.Note t h a t E contains a unique splitting field E; of

and K contains a unique splitting field K , of f l . Owing t o Lemma 1.14(ii), a ( E , ) = I(,

for all z

E I . and since K is generated by all the h’, arid E is generated by all the E,, we

deduce t h at u ( E ) 2 K .

Lct E IF be a n algebraic field extension. \I’e say t h a t E / F is normal if every irred u c i b l e polynomial of

F[X] which

has a root in E splits into linear factors in E i X ] .

1.16. P r o p o s i t i o n . Let E / F be a n algebraic field extension and let

be a n algebraic

closure of F contuining E (such 7always exists, e . g . take a n algebraic closure of E ) . ‘L’hen the following conditions are equivalent: ( i ) E/F‘ zs rzormal ( i i ) Every F-homomorphism u :

E

+

F

is a n autornorphisnz of E

( i i i ) E is the splilling field of a f a m i l y of polynomials in f<’[x’] ( i v ) E v e r y corLjugate of a n element of

E is in E .

P r o o f . T h e equivalence of (i) and (iv) is a consequence of the definition of conjugacy. LTc now show t h a t (ii)+

(iii) and (ii)*

(i).

Indeed, assume t h at (ii) holds. Given a n

54

CHAPTER 2

element X E E , let

fx(X)be

its minimal polynomial over F . Let p be a root of f x ( X ) in

-

F . By Proposition 1.2, there exists an F-isomorphism F ( X ) + F ( p ) sending X t o p . By Proposition l.lO(i), we may extend this isomorphism t o an F-homomorphism E

extension is an automorphism u of E by hypothesis, hence u(X) = every root of f x ( X ) lies in E , and splitting field of the family

-+

F.

This

lies in E . Therefore

fx(X) splits into linear factors in E[X). Thus E is a

{f~(x)}, X E- E, which

implies (iii) and (i).

Assume that (iii) holds and let { f , ( X ) l i E I } be the family of polynomials of which

E is a splitting field. If a is a root of some fi in E , then for any F-homomorphism o :E

4

F , a(&) is also a root

of f,. Since E is generated by the roots of the polynomials

f,,it follows that u maps E into itself. This proves (ii), by applying Proposition 1.12. Finally, assume that (i) holds. To prove (ii), let

(J

: E + F be an F-homomorphism.

Fix a E E and denote by f ( X ) its minimal polynomial over F . Then a(&) is a root of

f(X), hence .(a) E E . Therefore

(T

maps E into itself. By Proposition 1.12, it follows

that u is an automorphism of E. B 1.17. Remark. If E I F i s a finite eztension, t h e n condition (iii) of Proposition 1.16

is equivalent t o the requirement t h a t E is a splitting field of a single polynomial in F I X ] . Indeed, if E is a splitting field of { fili E I } a n d , f o r each f i n i t e product g of the f;, F ( g ) is the splitting field for g contained in E , t h e n

is a n ascending chain of subfields of E . Hence E

=

F ( f 1 . . . f n ) for some n

2

1, as

asserted. Let E I F be an algebraic field extension.

We define a n o r m a l closure of E I F as

a normal extension of F containing E and having the property that no proper subfield containing E is normal over F. 1.18. Proposition. Let E I F be a n algebraic field eztension. A normal closure of

E I F ezists and i s unique, u p t o a n F - i s o m o r p h i s m . Furthermore, i f E I F i s finite, t h e n so are the normal closures of E I F . Proof. For each X E E , let f > ( X )E F I X ] be the minimal polynomial of X and let Ei be the splitting field of the family Then F

CE CK

{fxlX

E E } contained in a given algebraic closure of E .

and K I F is normal by Proposition 1.16. Moreover, if K1 is such that

ALGEBRAIC CLOSURE

F

C E C K1 & K

55

and K1IF is normal, then K1 = K . This shows that K is a normal

closure of E I F . Let F

i E C

S be a chain of fields, where S is a normal closure of E I F . Then

obviously S is a splitting field of the family {fx 1X E E } . Hence S and K are F-isomorphic, by Proposition 1.15. Finally, if E I F is finite, then E = F ( a 1 , .. . ,a,) for some n in E . Adjoining to E all conjugates of a % 1 , 5 i

5 n , we

2

1 and some

a1,.

..a,

obtain a normal closure of E I F

w h i c h is finite dimensional over F . The following simple observation is often useful.

CK CE

1.19. Lemma. Let F

E‘ be a n algebraic closure

be a chatn of fields, where EJE’ is normal and let

of F containing E .

T h e n every F - h o m o m o r p h i s m K

---t

is

eztendible to a n F - a u t o m o r p h i s m of E . I n particular, every F-uutomorphism of K eztends t o a n F-aulomorphzsm of E . Proof. Owing t o Corollary 1.13, every F-homomorphism K

an automorphism of

17; its

---t

F

is extendible to

restriction t o E provides, by virtue of Proposition 1.16, the

desired F-automorphism of E . Kote thdt

if

F

CE CK

is a chain of fields such that both E / F and K I E are normal,

t h e n K,’F need not be normal. For example, in the chain

both Q ( & ) / Q

and Q

(a)

(fi)/Q are normal,

complex roots of X 4 - 2 are not in Q (*).

but Q (?2)/Q

is not normal since the

Ncvertheless we do have t h e following prop-

erties. 1.20. Lemma. (i) If F zs

CE CK

zs a chain of fields and

K/F

2s

normal, t h e n K / E

also normal

(ii)

and

If h’l/F and K2IF are normal and K 1 ,Kz are subfields of a field E , then both K I K z K 1

h’Z are normal over F .

Proof.

(1)

Let

be an algebraic closure of E containing K and let u : K

an E-homomorphism. Then

r~

is an F-homomorphism and

i

be

is a n algebraic closure of F

containing K . Hence, by Proposition 1.16(ii); 0 is an autorriorphism of K . Thus K / E is normal, again by I’roposition 1.16(ii).

CHAPTER 2

56

(ii) Let, F be an algebraic closure of F containing

K1

K2, and let

be an F-homomorphism. Then, by Proposition 1.16(ii), u restricts to an automorphism of K , , i = 1 , 2 , hence t o an automorphism of K1K2 since u ( K l K 2 ) = a ( K l ) u ( K z ) .Thus,

by Proposition l.l6(ii), K 1K 2 / F is normal. The fact t h a t K1 n K z is normal over E is a direct consequence of the definition of normality. rn 1.21. Proposition. Let F

CK 5E & L

be a chain of fields, where L I F is a finite

normal eztension. If the number of K-homomorphisms E 8'-homomorphism K

+

L has ezactly

+

L is equal to

s eztensions t o homomorphisms

E

---f

s,

then every

L.

Proof. Let G be the group of all F-automorphisms of 1, and let G ( K ) (respectively,

G ( E ) )be the subgroup of G consisting of those automorphisms of L which leave fixed every element of K (respectively, of E ) . It is clear that G ( E ) C G ( K ) and that G is finite. Let

ud,G(E)

u m

n

G ( K )=

and G =

$,G(K)

j=1

,=I

be cosets decompositions of G ( K ) and G. Then

is a coset decomposition of G with respect to G(IS). Since any K-homomorphism E

+

L

extends to a n automorphism of L (Lemmas 1.19 and 1.20(i)), it is clear that s = n. Note that the m automorphisms

GJ have distinct restrictions to K and that the restriction of

any element $ of G to K coincides with the restriction of one of the 1.19, every F-homomorphism K

--f

4,.

Since, by Lemma

L is the restriction of some automoiphism. of L , it

follows that K has exactly m F-homomorphisms into L and that these are given by the

restrictions of

$1,.

. . , G m to K . A similar argument shows that E has exactly rnn F-

homomorphisms into L and that these are given by the restrictions of the mn products

$,& to E . Now each to K if 3

4. restricts to the identity on K and $, and

# k . Thus each F-homomorphism K

restrictioil of

G3, has exactly

namely the restrictions of

+

$k

have distinct restrictions

L , say the one represented by the

n extensions to E which are F-homomorphisms of E into L ,

$J$1,$J$z,...,41& to E .

ALGEBRAIC CLOSlIRE

57

We next pruvide two applications of splitting fields of polynomials. Let R be an integral domain with quotient field F . T h e procedure for determining whether or not an element in a field extension of F is integral over R might generally be a lengthy one since there is no clear way to select th e polynomial which expresses integral dependence. The following result provides a circumstance where this procedure is simplified. 1.22. Proposition. Let €2 be a subring of a field F , let E I F be a field extension

and let X E E be algebraic over F with m i n i m a l polynomial f ( X ) E F I X ] . (i)

If X is

integral over R , t h e n the coefficients of

f(X)are integral over R

( i i ) If F i s the quotient field of R and R i s integrally closed, t h e n X is integral over R if a n d only i j l ( X ) E R [ X ] . Proof. It

IS obvious

t h a t (ii) is a consequence of (i). To prove ( i ) , let, K be a splitting

field of j ( X ) Let X I = A, A z , . . . ,A,

be all roots of f ( X ) in K and let g ( X ) E R I X be a

monic polynomial with g(X) = 0. Because f ( X ) is irreducible and has a root in common

g(X), it follows from Proposition l . l ( i i ) t h a t f(X) divides g(X) in F[X]. Hence the roots A,,. .A, of f ( X ) are also roots of g(X).T hu s X I , . . .,A, are integral over R . It

with

,

follows t h a t t h e coeficients of f ( X )= ( X

~

A,). . . ( X

-

A,) are integral over R .

1.23. Proposition. Let R be a n integrally closed integral d o m a i n with quotient field

F . Let J ( X ) E R [ X ]be a m o n i c polynomial and a s s u m e that

g(X)a n d h ( X ) are R [.Y]. where

monic polynomials in F ! X ] . T h e n both g ( X ) and h ( X ) h e in

Proof. Let E be a splitting field for f ( X ) . T hen E / P is a finite extension such t h at n

for some

f ( X ) = fl(X a ; ) ~

n,

2 1: n i E E

,=I

1,et S be the integral closure of R in E . Because f ( X ) c H I X ] is monic, an d f ( a , ) = 0 for each

7,

it followrs that, each a , E S. In E [ X ,we have n

g(X’)h(X)

=

fl(X

,= 1

~- U ‘ )

58

CHAPTER 2

whence both g ( X ) and h ( X ) are expressible as partial products n ' ( X

-

a,). Hence both

g ( X ) and h ( X ) have coefficients in S. It therefore follows t h a t the coefficients of g ( X ) and h ( X ) are in Sfl F . But Sfl F = R , since R is integrally closed. Thus both g ( X ) and h ( X ) lie in R [ X ] ,as required. rn We close by recording the following standard fact concerning degrees of (not necessarily finite) field extensions. 1.24. Proposition. Let

F & E C K be a c h a i n

of fields. If A and B are bases

f o r E I F a n d K I E , respectively, t h e n A B = {abla E A , b E B } is a basis for

KIF.

In

particular, (K: F ) = ( E : F ) ( K : E ) .

Proof. If X is an element of K , then X =

C~=l,u,b, for

some m

2

1 and some

p l E E , b, E B . Moreover, we have

for some n 2 1. Therefore, X =

cyzlxi"=, a,ja;b,

which shows t h a t

K

is spanned by

A B over F. It remains to show that the elements of AB are linearly independent over F. Assume that

Setting c, =

2

FX,ja,bj =0

CTz, Xl,a,,

we then have

with X;j E

Cy=lc,b,

= 0 , c, E

F

E , and since the b, are E-

linearly independent, we must have cl = 0, l 5 3 5 m. The latter in turn implies that the A,,

are zero, because the at are F-linearly independent. So the proof is complete. rn

2. Separable algebraic extensions: definitions and elementary properties.

Let

F be a field and let f ( X ) E F [ X ] .If X is a root of f ( X ) in F , then

for some g ( X ) E F [ X ]with g(X)

#

0. We refer to m as the rnu[tiplicity of X in F , and say

that X is a multiple root if m > 1 and a simple root if rn = 1. We say that f ( X ) is separable ouer

F if all irreducible factors

of f ( X ) over F have only simple roots (in a splitting field

59

SEPARABLE ALGEBRAIC EXTENSIONS

of f ( X ) ) . In particular, a n irreducible polynomial is separable if and only if its roots are

simple. Let

E/F be a field extension. An algebraic element a E E

over F is called separable

over F if its minimal polynomial over F is separable. A separable algebraic eztension is an algebraic extension E / F such t h a t all elements of E are separable over F. In what follows, to avoid excessive verbosity, we use t h e t e r m “separable extension” instead of “separable algebraic extension”. In future we shall extend t h e notion of separabiiity to arbitrary field

, wish t o test whether f ( X ) has multiple roots in some extensions. Given f ( X ) E F [ X ] we field containing F , without having t o go outside F . To this end, we first define derivatives formally. Given any polynomial f ( X ) =

a0

+ a l X + . . . + a,Xn

over a commutative ring R , we

define its d e r i v a t i v e D f ( X ) or f ‘ ( X ) by t h e rule

f‘(X) = a ,

$- 2a2X i . . . -t12anx-1

r .

I he mapping

{;:;

+

R[XI f’(X)

enjoys the following properties: ( i ) ( f ( X ) -t s ( X ) ) ’ = f’(x)+!?’(XI ( i i ) ( r f ( X ) ) ‘= r f ’ ( X ) , r E R

( 4 ( f ( X ) g ( X ) ) =’ f ’ ( X ) g ( X )+ f ( X ) g ’ ( X ) (iv) X‘ = 1

Conversely, f ’ ( X ) is completely determined by (i)-(iv) and this provides alternative dcfinition. \Ye are now ready t o test whether f ( X ) has multiple roots. 2.1. Proposition. L e t F be a f i e l d , l e t f ( X ) E F [ X ]a n d let X be any root of f ( X )

zn s o m e field eztenszon of F . Then X 2s a rnultzple root zf a n d o n l y z/ f’(A) Proof. \I’e first divide f ( X ) by ( X

f ( X )= ( X where h ( X ) = 0 or degh(X)

5

-

~

A)’ t o obtain

X)’J(X)

+ h(X)

1. Putting X = A, we see t h a t f ( X )

1)ifferrntiating and putting X = A, we find t h a t h’(X)= f ’ ( A )

f(X) = (X

0.

X)2!?(x) + P’(X)(X

~

’Thus

h(A) = 0 .

CHAPTER 2

60

which shows that ( X - X ) * l f ( X )if and only if f‘(X) = 0. 2.2. Corollary. Let f ( X ) E F [ X ] be of degree

2

1. T h e n all roots of f ( X ) (in a

splitting f i e l d of f ( X ) ) are simple if and o n l y if f ( X ) i s p r i m e t o i t s derivative f ’ ( X ) . In particular, if f ( X ) is irreducible, t h e n f ( X ) i s separable over F if a n d only if f ’ ( X ) # 0.

Proof. By Proposition 2.1, f ( X ) has a multiple root if and only if f ( X ) and f ’ ( X ) have a common factor. m 2.3. Corollary. Let F be a field of characteristic 0 . T h e n all polynomials over F

are separable. In particular, a n y algebraic e z t e n s i o n of F i s separable.

Proof. By definition of separability, it suffices t o show t h a t any irreducible polynomial f ( X ) over F is separable. Write f ( X ) =

uo

+ a l X + . . . + a,Xn

with a , # 0, n

2

1. If

f ‘ ( X ) = 0, then nun = 0, which is impossible. Now apply Corollary 2.2. Turning to fields of characteristic p > 0, we next record 2 . 4 . Proposition. Let F be a field of characteristic p

2

(i) If f ( X ) E F [ X ] is of degree

> 0.

1, t h e n f’(X) = 0 if and only if f ( X ) is of the f o r m

g ( X P ) ,f o r s o m e g ( X ) E F [ X ] (ii) If f ( X ) E F I X ] is irreducible, t h e n f ( X ) h a s multiple roots if a n d on1y if f ( X ) i s of the f o r m g ( X P ) f o r s o m e g ( X ) 6 F I X ] .

Proof. (i) If f ( X ) has the form g ( X P ) ,then obviously f ’ ( X ) = 0. Conversely, assume that f ’ ( X ) = 0 and write f ( X ) = i a , = 0, 1 5

i _< n , and hence a,

a0

+ alX

= 0 if p

f ( X ) = ao

i

. . . + a , X n with a , # 0, n 2 1. Then

+ i. Thus f ( X ) has the form

+ a p X ” + a * p X 2 p + . . . + a,,XTP

which shows that f ( X ) = g ( X P ) for g ( X ) = ao

+ a,X + . . . + u r P X r

( i i ) This is a direct consequence of (i) and Corollary 2.2. w 2 . 5 . Proposition. Let F be a field of characteristic p

a n irreducible polynomial. Choose the nonnegative integer

e

>

0 and let f ( X ) E F [ X ] be

such that j ( X ) E F[XP‘] but

f ( X ) 6 I.’[X1’e+’]and write f ( X ) = $ ( X ” ) f o r s o m e $ ( X ) E F [ X ] ( i ) ~ J ( Xis )a n irreducible and separable polynomial o v e r F

SEPARABLE A L G E B R A I C EXTENSIONS

61

(ii) A l l roots of f ( X ) h a v e t h e s a m e m u l t i p l i c i t y equal t o pe a n d d e & ( X ) i s equal l o the

n u m b e r of d i s t i n c t routs of f (Xi. I n p a r t i c u l a r ,

Proof. (i) It is clear that $(X) is an irreducible polynomial. Furthermore, + ' ( X ) # 0 , since otherwise by Proposition 2.4(i), $(X) is of the form g ( X P ) and hence J ( X ) E

F[XPCi'],contrary t o the assumption. Thus, by Corollary 2 . 2 ,

4(X) is separable over F .

( i i ) By ( i ) , we may split $(X) into distinct linear factors (in a suitable extension of F ) :

Then rn

f(X) =

r p p e

-A)

1=1

Let a , be a root of

XP? =

ap? =

pl. Then

8,

( a l , .. . , am are distinct)

and

Thus

as asserted.

Let F be a field of characteristic p > 0 and let f ( X ) E F I X ] be an irreducible polynomial. The degree of the polynomial +(XI E F [ X ](in notation of Proposition 2.5) is called the separable degree of f ( X ) ,while e is called the e z p o n e n t of i n s e p a r a b i l i t y of f ( X ) . if a is an algebraic element over F with minimal polynomial

f(X), then the separable

degree of a over F (respectively, the e z p o n e n t of i n s e p a r a b i l i t y of a over F ) is defined to be the separable degree of f ( X ) (respectively, the exponent of inseparability of f ( X ) ) . In case where char F = 0, the separable degree of a is defined to be the degree of a . Thus, in

all cases, the separable degree of a i s equal to t h e n u m b e r of distinct roots of t h e m i n i m a l p o l y n o m i a l of a .

62

CHAPTER 2

2.6. Proposition. L e t F be a field of characteristic p

>

0, let a be a n algebraic

element in a field e z t e n s i o n of F a n d let e be the ezponent of inseparability of a. T h e n

(i)

ape

i s separable over F

(ii) a is separable over F if a n d o n l y if F(a) = F ( a P ) .

Proof. (i) Let f ( X ) E F[X] be the minimal polynomial of a over F and let $(X) E

F[X] be as in Proposition 2.5. Then, by Proposition 2.5(i), separable polynomial $(X) E F [ X ] . Thus

ape

is a root of an irreducible

ape

is separable over F.

(ii) Assume t h a t a is not separable over F . Then, by Proposition 2.4(ii), f ( X ) = g(XP) for some g ( X ) E F [ X ] . Since

aP

is a root of g ( X ) , we have

which shows that F(@)# F(a). Conversely, assume that a is separable over F. Then f ( X ) has distinct roots. Let g ( X ) be the minimal polynomial of a over

Also

(Y

is a root of the polynomial

F(aP).

XP-ap

Then g ( X ) l f ( X ) , so g ( X ) has distinct roots.

over

F(ap),

so g ( X ) divides X P - a P

Since g ( X ) has distinct roots, we conclude that g ( X ) = X

-

= (X-a)P.

a. Hence a E F ( d ) and

therefore F(a) = F ( d ) . Let E / F be a finite field extension and let K be an algebraic closure of E. The separable degree (E : F)sof E/F is defined to be the number of F-homomorphisms of E

into K . This is obviously independent of the choice of K .

2.7. Lemma. Let F(cr)/F be a finite field eztension. T h e n (F(a) : F)s is equal t o t h e separable degree of a over F a n d is also equal t o the number of F - h o m o m o r p h i s m s

F ( a ) 4 E where E is a n y normal e z t e n s i o n of F containing a.

Proof. The first assertion is a particular case of Lemma 1.9 in which

G

is the inclusion

map. To prove the second assertion, let K be a n algebraic closure of E . Then K is obviously an algebraic closure of F(a). Since E / F is normal, the image of any F-homomorphism F(a)

+

K is contained in E , as required.

We now provide a formula which ties together the degree and the separable degree of a finite extension. 2.8. Proposition. Let F be a field of characteristic p

>

0 , let a be a n algebraic

SEPARABLE ALGEBRAIC EXTENSIONS

63

element i n a f i e l d eztension of F and let e be the ezponent of inseparability of a . Then

( F ( a ): F) = p e ( F ( a ) : F)s Proof. Let j ( X ) E F [ X ]be the minimal polynomial of

cy

and let rn be t h e separable

degree of a . Then ( F ( a ) : F ) = d e g f ( X ) = p e m , by Proposition 2.5. Since, by Lemma 2.7, m = (F(a): F ) s the result follows. To obtain some further information on separable degree, we now record.

2.9. L e m m a . Let F of F and l e t X : E

+

5E

C_ K be a chain of fields, where K i s a n algebraic closure

K be an F-homomorphism. Then, for any a E K , there ezist exactly

( E ( a ): E ) s F-homomorphisms E ( a ) --t K eztendiny A. Proof. Let f ( X ) be the minimal polynomial of a over E. Owing t o Lemmd 2.7,

( E ( a ): E ) s is equal t o the separable degree of a over E , hence t o the number of distinct roots of f ( X ) in K . Now apply Lemma 1.9. m 2.10. Proposition. Let E / F be a finite field extension, say E = F ( a 1 , . . . , a , ) ( i ) ( E : F i S = n l = l ( F ( a l , . . . , a l ) : F(cu1,. . . ,

C Y , - ~ ) ) ~

( i i ) ( E : F ) s = ( E : F) if and only if each a , is separable ouer F ( a 1 , . . . , a ; - 1 ) ( i i i ) l j chnrF = p

> 0 and e; is the ezponent of inseparability

of

a; ouer F ( a 1 , . . . ,

then

( E : F ) = (E : F ) s p e l + - - . + e r Proof. ( i ) P u t ni = F’(ni,.. . , at-i)), 1

. . . , a t ) : F(ru1, . . . , (

(F(cy1,

I Z 5 r.

~ ~ - 1 ) and ) ~

m, =

(F(a1. . . . ,n,) :

Then, by Lemma 2.7, ni is equal t o the separable degree of

at over F ( a 1 , . . . ai-1). This, together with Proposition 1.24, implies t h a t

T h e case r = 1 being obvious, we now argue by induction on r . If El = F ( a 1 , . . . , ~

~ - 1 )

and 1;: is a n algebraic closure of E (hence of E l ) , then by induction there exist exactly

n l n z . .. T

L - ~ F-homomorphisms

El

2.9 t h a t each F-homomorphism El

E El

+

+

K . Because E

+K

has exactly n , extensions t o F-homomorphisms

K . But the restriction of a n F-homomorphism E

4

El ( a r ) ,it follows from Lerrinia

=

K , hence ( E : F)s = nl . . . n,, as required.

K t o El is a n F-homomorphism

4

CHAPTER 2

64

(ii) By (i), ( E : F ) s = n:=,n,

and by (I), ( E : F ) = n:=,rn,

with n,

5 m,, 1 5 z 5 r .

Thus (E : F ) s = ( E : F ) if and only if n, = rn, for all i E (1,. . . , r } . The latter, by Lemma 3.6, is equivalent t o the separability of a , over F ( a l , . . . a , - l ) for all z E (1,. . . , r } .

(iii) By Proposition 2 . 5 , m, = p e , n t r hence by (1) and (i),

as desired. rn

To exploit this last result, we next record the following simple observation. 2.11. Lemma. Let F

E

K be a c h a i n of fields. If a E K i s separable over F ,

t h e n a is separable over E .

Proof. Let f ( X ) E F [ X ]be the minimal polynomial of a. Then f ( X ) = f l ( X ). . . f n ( X ) for some irreducible polynomials f i ( X ) E E [ X ] .Hence a is a root of some f ; ( X ) and f i ( X )

must have only simple roots since so does f ( X ) . Hence a is separable over E . rn

2.12. Corollary. Let E / F be a field eztension. T h e n t h e set of all e l e m e n t s of E which are separable over F i s a subfield of E containing F . f n particular, if E i s generated Over

F b y a f a m i l y of separable elements, t h e n E I F is separable. Proof. Let

a l , a2

E E be separable over F and let X E { a l a z ,a1 f a 2 , a ; ' } , a l # 0.

Then F ( a 1 , a ~ = ) F ( X , a l , a s ) and

a2

is separable over F ( a l ) by Lemma 2.11. Hence, by

Proposition 2.10(ii), X is separable over F . Since all elements of F are obviously separable over F , the result follows. rn Let E / F be a field extension. By the separable closure of F in E we understand the subfield of E consisting of all elements which are separable over F . If closure of F , then the separable closure of F in

F

is an algebraic

is called the separable closure of F .

Thanks to Corollary 1.11, the separable closure of F is uniquely determined up to an F-isomorphism. Let E I F be a finite field extension. We define the inseparable degree ( E : F ) ; of E over F by

( E : F ) , = ( E :F ) / ( E : F ) s If c h a r F

=

0, then E / F is separable by Corollary 2.3. Hence, by Proposition 2.10(ii),

( E : F), = 1. On the other hand, if c h a r F = p, then by Proposition 2.lO(iii), ( E : F ) ; is a

S E P A R A B L E A L G E B R A I C EXTENSIONS

65

powcr of p .

2.13. Proposition. (i) If E / F is a f i n i t e eztension, t h e n E I F zs separable if and

only if ( E : F ) s = ( E : F )

If F i E C K i s a c h a i n

(ii)

of fields and K / F is f i n i t e , t h e n

( K : F)s= ( K : E'),(E : F ) s and ( K : F ) , = (K: E ) , ( E : F ) ,

Proof.

(i) Write E = F ( a 1 , . . . , a T )where each a ; is algebraic over F. Then,

by Proposition 2.10(ii), the condition t h a t each a , is separable over F ( a l , . . . , a , - 1 ) is equivalent t o the condition that each a, is separable over E' (simply rewrite F(a1,.. . , a , ) as

E'(a;,a 1 , . . . , a,-1, ( Y ; + I , . . . , a r ) ) .By Corollary 2.12, the latter condition is equivalent

to the separability of E / F . T h e desired conclusion is therefore a consequence of Proposition 2.1O( ii) .

( i i ) Write

E

=

F ( a 1 , . . . , a , ) and K = E(PI , _ _, p_t ) . Setting n, = ( F ( a 1 , . . . , a t ):

F ( a 1 , . . . , O ~ - I ) ) ~1, 5 i 5 r , and mi = ( E ( P l , .. . , p i ) :

S(P1,..

.,/3-1))*,

15j

5 t , we

have

( E : F ) s = n1n2.. . n , and ( K : E ) s = m 1 m 2 , .. m t by Proposition 2.10(i).

E(/31,. . .

O n the other hand, since K = F ( a 1 , . . . , a , , P 1 , . . . ,jj't) and

= F(a1,.. . , a T , / 7 1 ., .. , P 3 ) , it follows from Proposition 2.10(i) that

(K

:

b'),s = ( n l . . .n,)(rnl . .m,) = ( E : E').(K :

We also have (h' : F ) , = ( K : F ) / ( K : F ) s = ( K : E ) ( E : F ) / ( K : E ) , ( E : F), = ( K : E ) i ( E : F ) * ; as rcquired.

2.14. Corollary. Let EIE' be a finite field e z t e n s i o n and let charF = p > 0. I f

p

1:

( E : F ) , t h e n E / F is separable. Proof. Since p

#

( E : F ) , it follows from Proposition 3.10(iii), t h a t ( E : F )

=

(E:

F)s.IIence, by Proposition 2.13(i), E / F is separable. We next examine t h e behaviour of separability under field extensions. 2 . 1 5 . Proposition. (i) if F

E

CK

IS

and only zj both E I F and K I E are separable.

a chain of fields, t h e n KIE' zs separable if

CHAPTER 2

66

(ii) If E / F is separable and K / F is any eztension, then KEIK is separable.

Proof. (i) If K / F is separable, then obviously E/F is separable. Moreover, by Lemma 2.11, K/E is also separable. Conversely, assume that both E / F and K/E are separable. If K / F is finite, then (E : F), = (K : E)i = 1 by Proposition 2.13(i). Hence, by Proposition 2.13(ii), (K : F); = 1 which proves that K / F is separable, by Proposition 2.13(i). If K / F is infinite, let a E X. Then a is a root of a separable polynomial f (X) E E [ X ] . Let a o , a l , . . . , a , be the coefficients of f ( X ) and let E o = F(a0,. , . , a f l ) . Then we have

F

C E o C Eo(a) where

Eo(a)/F is finite and both Eo/F and Eo(a)/Eo are separable,

by Corollary 2.12. Hence Eo(a)/F is separable and therefore a is separable over F. Thus

K/F is separable. (ii) Every element of E is separable over F, hence (by Lemma 2.11) separable over K . Because KE is generated over K by the elements of E, it follows, from Corollary 2.12, that KE/K is separable. w Let F

C K C E be a chain of fields. If E / F is normal, then E / K

is normal but K / F

need not be normal. The next result provides a condition under which K / F is normal. 2.16. Proposition. Let E/F be a normal field eztension and let F, be the separable

closure of F in E . Then F,/F is normal.

Proof. Let

be an algebraic closure of E and let a : F,

be an F-homomorphism.

-+

By Proposition 1.16, it suffices to show that u(F,) = F,. To this end, note that, by Proposition l . l O ( i ) , we may extend u to a homomorphism E

+

E.

Since E/F is normal,

it follows from Proposition 1.16 that u ( E ) = E, so u is an F-automorphism of E. Moreover,

o(F,) is separable over F, hence is contained in F,. Thus a(F,) = F, as required. w 2.17. Proposition. Let E / F be a finite field eztension and let charF

=

p

> 0.

Then E I F is separable i f and only i f E = FEP where EP as the subfield of E consisting o/ all p-powers of elements of E .

Proof. Assume that E/F is separable and let a E E. Then, by Proposition 2.6(ii), a E F E P and hence E =

FEP.

Conversely, assume that E =

FEP

and write E =

F ( a 1 , . .. ,a,). By Proposition 2.6 (i), a?' E F,, the separable closure of F in E, for some e,

2 0.

Setting m = max(e1,. . . ,e,}, it follows that

EPm

implies E = FEPm, hence E = F, is separable over F .

2 F,. But E

= F E P obviously

67

SEPARABLE ALGEBRAIC EXTENSIONS

>

2.18. Corollary. Let E J F be a finite separable field extension, let charF = p

and let m

2

1 be a n integer. If u l , . . . , u , is a basis of E over F , t h e n so i s u:'*, . , . ,

Proof. By Proposition 2.17, we have

E

=

0

II ,>"a

FEP. Hence a typical element of E is an

F-linear combination of elements in EP. We conclude therefore t h a t u : , . . . ,UP, span E over F, and hence is a basis of E over F. The desired assertion now follows by induction on m. Our next aim is t o show that any finite separable extension is simple. This will be derived as a consequence of the following two results. 2.19. Proposition. Let F be a n infinite field and let E = F ( a 1 , . . . , cr,,,p) where

the a , are separable and

p

is algebraic over F . T h e n E I F is a simple eztension.

Proof. By induction on n , it suffices t o treat the case n = 1. Let h e the minimal polynomials over F of

2 E.

of / ( X ) y ( X ) , then K

Let

71

and

a1

PI,

f(X)and

respectively. If K is a splitting field

= a l , y ~ ,. . ,ym be the distinct roots of

61 = p , 62,.. . ,6,those of g ( X ) . Because

g(X)

/(X)and

is separable over F , d e g f ( X ) = m. We may

a1

therefore assume t h a t m > 1, since otherwise,

a1

F and E = F ( P ) . Consider one of the

linear equations

This has at most one solution in E'. Taking into account t h a t F is infinite, we may therefore ci~oose3: = 7 E F such that

It will nest be shown t h a t B

=

yy, -4- h1 is a primitive element of E . To this end, consider

. g ( Q - y y l ) = g(61) the polynomial g(B-yX) which obviously belongs to F ( B ) [ X ] Then and, by (2) g(Q

-

7yl)

arid f ( X ) = n z l ( X

-

#

0, 1

5

y;) is X

z

5 m.

-

y1 = X

Thus the greatest common divisor of g(0 -

cyl.

-

=

0

rX)

But then we may choose g 1 ( X ) , g 2 ( X )in

F ( O ) / X ]s u c h t h a t

x

-

a1

= S l ( X ) f ( X ) 7- s,(X)g(Q

which implies t h a t a l E F ( 0 ) . Hence 2.20.

cyel1c.

Proposition.

p

=

B

-

-

7x1

7 a l E F ( B ) and the result follows. D

A n y finite subgroup of the multiplicative group o/ a field is

68

CHAPTER 2

Proof. Let F' be the multiplicative group of a field F and let G be a finite subgroup of F' of order m. Let n be the exponent of G. Then A" = 1 for all X E G. On the other hand, the polynomial X"

-

1 has a t most n roots in F . Since nlm, we have n = m.

Moreover, if p E G is an element of order m, then the order of the cyclic group < p > generated by p is m. Hence G = < p > 2.21.

as required. m

Corollary. Let E / F be a finite separable field eztension. Then E I F is a

simple eztension.

Proof. If F is a finite field, then E is also finite. Hence E' = < A >

for some X E E ' ,

by Proposition 2.20. Thus E = F(X). If F is infinite, then the required assertion follows by virtue of Propositions 2.19 and l.S(iii). w

We next record the following general assertion due to Artin. 2.22.

Let E / F be a finite field eztension, where F is an infinite

Proposition.

field. Then E / F is a simple eztension if and only i f there are only a finite number of intermediate fields between E and F .

Proof. Assume that E / F is simple, say E = F ( 0 ) and let K be a n intermediate field. Let f ( X ) be the minimal polynomial of 0 over K and let K' be the field generated by F and the coefficients of f ( X ) . Then K'

K and f ( X ) is also the minimal polynomial of 0

over K'. Therefore ( E : K') = d e g f ( X ) = ( E : K ) and thus K = K' is generated by the coefficients of f ( X ) . Note also t h a t f ( X ) divides the minimal polynomial g ( X ) of 0 over

F and both f(X), g ( X ) E E [ X ] . Since g ( X ) has only a finite number of distinct factors in E [ X ]with leading coefficients 1, the number of intermediate fields K is finite. Conversely, assume that there are only a finite number of intermediate fields between

E and F . Given a,P E E , it suffices to show that F ( a , P ) / F is simple. To this end, let 7 E F and let E , = F ( a

+7P).

We have infinitely many 7 E F and finitely many E,.

Thus, there exist -y,6 in F, 7 # 6 , such that E, = Eg. Then

p

2.23.

= (7 -

6)-'(a+7/3 - a

-

6/3) E E ,

Corollary. Let E / F be a finite separable eztension.

finitely many intermediate fields between E and F .

Then there are only

SEPARABLE A L G E B R A I C EXTENSIONS

69

Proof. If F is finite, then so is E and the result is clear. If F is infinite, then apply Proposition 2.22 and Corollary 2.21. w

We close this section by providing a class of fields F such that every algebraic extension EIE' is separable. Let F be a field of characteristic p

F

4

F,

z

H

> 0 and let FP

=

{zPIz

E F } . Then the mapping

zP is an injective homomorphism whose image is the subfield FP of F . If

it is also surjective (i.e. if F = FP), the field F is said t o be perfect. Thus F is perfect iT and only if every element is a p-th power. In addition, every field of characteristic 0 is perfect by definition. The simplest example of a n imperfect field is the field F ( X ) of rational functions in

X over a field F of characteristic p > 0 . This is so since X cannot be

a p-th power in F ( X ) . Let F be a field of characteristic p each n

2

>

0 and let

F

be an algebraic closure of F . For

1, define

FP-" Then FP-" is obviously a subfield of

of subfields of

F.

=

{. E F j z P " E F }

7and we have

a chain

Put

n= 1

2.24. Lemma. W i t h the n o t a t i o n above, I;"--

containing F . Moreover, ij F

is the smallest perfect subfield of

# FP--, t h e n FP-- I F is a n infinite eztension. -n

Proof. Let K be a perfect subfield of F containing F . If z E F P so

z p " = XP"

, then z P n

F

CK,

for some X E K since K is perfect. Hence z = X E K and therefore FP-"

K,

proving that FP--

C K.

Finally, assume t h a t F

E

The fact that FP-= is perfect is a consequence of the equality

# FP--. If F P - " / F is finite, then Fp-=

hence F = (FP-%)P"= FP-", a contradiction.

=

FP

-I.

for some n

2 1,

70

CHAPTER 2

We shall refer to the field Fp-"

as the perfect closure of F . By Corollary 1.11,Fp--

is determined uniquely, up to F-isomorphism. 2 . 2 5 . Lemma. Every finite field is perfect.

Proof. This is a direct consequence of the fact t h a t a n injection of a finite set into itself is a bijection. We now characterize perfect fields in several different ways.

2.26. Proposition. The following conditions are equivalent:

(i) The field F is perfect (ii) A n y f i n i t e eztension of F is separable (iii) Any algebraic eztension of F is separable

(iv) Any polynomial in F [ X ]is separable

Proof.

For characteristic 0 there is nothing t o prove: in t h a t case every field is

perfect by definition and, by Corollary 2.3, (ii), (iii) and (iv) always hold. Assume that c h a r F = p :0. It is obvious that (iii) and (iv) are equivalent and that (iii) implies (ii). Moreover, (ii) implies (iii) by Corollary 2.12. We are therefore left to verify that (i) is equivalent t o (ii). Let F be a perfect field and let f ( X ) be an irreducible polynomial over F . If f ( X ) is not separable, then by Corollary 2.2, f ' ( X ) = 0. Hence, by Proposition 2.4(i), f ( X ) =

g ( X P ) for some g ( X ) E F [ X ] say ,

Because F is perfect, a , = b: for some b, E F and thus

f ( X ) = bP,XPr+ b:XP('-') = (box'

+ . . . + b:

+ b 1 X - I + . . . + b,)',

contrary to the irreducibility of f ( X ) . Hence every finite extension of F is separable. Conversely, assume that every finite extension of consider the splitting field of XP

-

A. If

/I is

F is separable. Fix X E F and

a root, then

71

SEPARABLE A L G E B R A I C E X T E N S I O N S

and so the minimal polynomial of p over F is of degree 1. Hence p E F and X = pP, as required. m Let E / F be an algebraic extension. If the only elements of E which are separable over

F are the elements of F , then we say that EIF is purely inseparable. Expressed otherwise,

E / F is purely inseparable if and only if the separable closure of F in E is F . If charF

=

0,

then E / F is purely inseparable if and only if E =: F . To avoid trivialities, for the rest of the section we assume that c h a r F = p

> 0.

An algebraic element a of a field containing F is said to be purely inseparable o v a

F if

a p nE

F for some n 2 0 . If E / F is a n algebraic field extension, then the set of

purely inseparable elements of E over F is obviously a subfield of E containing F . \ire shall refer to this subfield as the pure inseparable closure of F in E . The following result demonstrates that it is a purely inseparable field extension of F containing any other such extension K I F with K

C E.

2.27, Proposition. Let E I F Ire a n algebraic field extension. T h e n the following

conditions are equivalent: ( i ) E/E' is purely inseparable

(ii) All elements of E are purely inseparable over F (iii) E

IS

generated over F b y a f a m i l y o f purely inseparable elements

(iv) There exists ezactly one F - h o m o m o r p h i s m of E i n t o a n algebraic closure of E

Proof. Because the elements of E which are purely inseparable over F form a subfield of E containing F, (ii) and (iii) are equivalent. (i)*(ii): If 011'-

CY

E E, then aPp is separable over F for some

e

20

(Proposition 2.6(i)), hence

E F . Thus a is purely inseparable over I;.

(ii)+(iv):

S bc a set

of purely inseparable elements of E such that E = F ( S ) and

be an algebraic closure of E . Each element a E

let Xp"

Lct

-

X E F [ X ]for some n

2 0 and some

S is a root of the polynomial

X E F. Because all the roots of this polynomial

are the same, the minimal polynomial j ( X ) E F [ X ]of a has a as the only root. Kow the image of a under any F-homomorphism E

+

E

is a root of f ( X ) , hence a

-

a proving

(iv) . ( i v ) + ( i ) : Assume by way of contradiction that a E E , a @ F and

(F(n) : F ) = ( F ( a ) : F ) s

>

CY

is separable. Then

1 and hence there are more than one F-homomorphisms

72

CHAPTER 2

-

F ( a ) --t E . But each such homomorphism is extendible to a homomorphism E

+

E

by

Lemma 1.19. This provides the desired contradiction and completes the proof. 2.28. Corollary. Let E I F be a field extension and let (Y

(Y

E E be algebraic over F . If

is both separable and purely inseparable over F , then a E F .

Proof. Since

LY

is purely inseparable over F, F ( a ) / F is purely inseparable by Propo-

sition 2.27(iii). Since

(Y

is separable over F , we must have a E F by the definition of pure

inseparability of extension.

2.29.

Corollary. Let F

C

E

C

K be a chain of fields.

T h e n K I F is purely

inseparable if and only if both E I F and K I E are purely inseparable.

Proof. This is a direct consequence of Proposition 2.27(ii). 2.30. Proposition. Let E I F be an algebraic field extension. If F, is the separable

closure of F in E , then F,/F is separable and E / F s is purely inseparable.

Proof. That F,IF is separable is a consequence of the definition of F,. Let K be the separable closure of F, in E. Then K / F is separable by Proposition 2.15(i). Hence

K = F, and therefore E / F s is purely inseparable. 2.31. Corollary. Every algebraic eztension E I F may be obtained b y taking a sepa-

rable extension followed b y a purely inseparable extension.

Proof. Apply Proposition 2.30. T h e next result provides a criterion under which the order of the extensions in Corollary 2.31 can be reversed. 2.32. Proposition. Let E J F be an algebraic extension and let K and F, be respec-

tively the pure inseparable and separable closure of F i n E . Then E I K is separable i f and only if E = KF,.

Proof. Assume that E J K is separable. Then E / K F , is separable and, by Corollary 2.29 and Proposition 2.30, E is purely inseparable over KF,. Hence E = KF,. Conversely,

if E = K F , then E I K is separable by Proposition 2.15(ii). 2.33. Proposition. Let E I F be a finite eztension.

SEPARABILITY, LINEAR DISJOINTNESS

73

(i) E I F is purely inseparable if a n d only zf ( E : F ) s = 1. I n particular, b y Proposition

2.1O(iii), if E/1' is purely inseparable, t h e n ( E : F ) is a power of p . ( i i ) ( E : F ) s = ( E I , : F ) , where F, is the separable closure of F in E .

Proof. (i) This is a direct consequence of Proposition 2.27(iv). (ii) By Proposition 2.30, E / F s is purely inseparable, hence by (i), ( E : F 6 ) s = 1. Now

( E : F ) s = (F, : F),(E :Fs)s by Proposition 2,13(ii), and ( F , : F ) s = ( F , : F ) by Proposition 2.13(i). Hence

( E : F)s= ( F , : F ) as required.

H

3. Separability, linear disjointness and tensor products. LVP begin by introducing the notion of a tensor of modules. Throughout, R denotes a

cornrnutatiLe ring and all R-algebras are assumed to be eonrnutatzve. Let I!, V .IY be any R-modules. A map

is said to be bilinear i f f is linear in each argument, i.e.

O u r first aim is to construct an R-module M and a bilinear map X : U x V

universal for all bilinear maps ( l ) ,in the sense t h a t to any bilinear map exists a unique R-homomorphism f' : M

4

+

f as in

@ I/.

(1) there

W such t h a t f = f' o A. hri R-module M

with these properties is called a tensor product of U and V and is denoted by simply I/

M which is

M

V or

If it exists, it is clearly unique u p t o isomorphism, and we shall speak of

the tensor product. T h e following standard procedure illustrates the existence of Ad. Let

F be a free R-

module freely generated by U x V and let Ar be the submodule generated by all elements of thc form (r1.1

+rzuz,~) ~

+ r2vz) -

(u,r1~1

~ I ( U I , V )

rz(u2,v)

~ 1 ( u , , u l)

rz(u,uz)

~

74

CHAPTER 2

for u , u ~ , Eu U ~ , v V , u 1 , u 2E V and

7

1

,

E~ R . Then U @ R V is defined to be the factor

module F I N . The image of ( u ,u ) under the natural homomorphism F by u '8 v. Thus the R-module U

@R

-+ F / N

is denoted

V consists of all finite sums ( u ; E U, u* E V )

and the elements u 8 v satisfy the relations

The map X : U x V if f : U x V

-+

4

U @ R V defined by X ( u , u ) = u @ u is therefore bilinear. Moreover,

W is any map, then f determines a homomorphism f ' : F

If we assume that f is bilinear, then N

+

W given by

Kerf' and so the map

defined by

f * ( u@ v ) = f ( u , u ) is a required homomorphism. 3.1.

Proposition.

Let V and W be free R - m o d u l e s with bases

respectively. T h e n V B R W i s a free R - m o d u l e with basis element z E V

@R

{ u , ~3w , } .

{u,}

and { w i } ,

Furthermore, every

W can be written uniquely as the f i n i t e s u m

and as

x=):z;@wj

(z; E V , wj E W )

3

Proof. Given ~ i ri,

u E

V and w E W , we have u = c r ; u ; and w = c r i w j for some

E R , where both sums are, of course, finite. Hence, by ( 2 ) , we have

SEPARABILITY. LINEAR DISJOINTNESS

which shows that the R-module V

@R

W is generated by {vi

@

75

w3}. Now assume that

and fix some subscripts, say i = k , j = s. Because { v ~ }is a basis of V , there exists an 12-linear map

Xk

:

V

there exists p s : W

f

:

V x W'

+

+ i

R such that X,(vZ)

R with

z # k and X,(V~) = 1. Similarly,

= 0 for

ws(wj)= 0 for j

#

s and pS(zu,) = 1. Then the map

R given by f ( v , w ) = Xk(v)p,(w) is bilinear. Therefore there exists an

H-linear map f' : V @ W

+

R with f * ( v @ w ) = X k ( v ) p L , ( w ) . Finally

proving that { u . @ w 3 } is a basis for V

@

W.

The remaining assertions follow easily by grouping terms. Thus if z E V

c,,, zt3(vt

is uniquely given as the finite sum z =

i

@

@

W , then z

w3) and hence, by ( 2 ) ,

3

as rcquricd rn

Let A be an R-algebra and let B,C be subalgebras of A. Then the set BC consisting of all finite sunis

C . If A

=

b,c, with A, E B , c , E C is the smallest subalgebra containing

B and

BC, then we say that A is generated by B and C .

Now assume that A and

B are R-algebras Because they are R-modules,

A @R B

exists, and we assert that this R-module is also an R-algebra. 3 . 2 . Proposition. Let A and B be R-algebras.

(i) A

@R

B is a n R-algebra with multiplication given b y

(ii) The maps A

+

A

@

B, a

H

a @ 1 and B

-+

A @ 13, b

1 @ b are Il-algebra

hoinomorphzsms whzch are vnjectzve zj A and B are R-jree Furthermore, A generated by the images A @ 1 and 1 @ B of A and B , respectzvely.

@R

R

2s

76

CHAPTER 2

Proof. (i) Let F be a free R-module freely generated by V x W . Define multiplication in F distributively using

(vl,wl)(vz,wz) = (vlvZ,wIwz) Then F is obviously an R-algebra. Moreover, the submodule N used in the definition of

V @ R W is an ideal of F . Therefore V @ R W = F / N inherits the R-algebra structure of

F . In particular, since ( v , w ) maps t o v

@ w ,this says that

as required.

(ii) By (i) and (2 ), the given maps A

+

A @ B and B

+

A 8 B are homomorphisms

of R-algebras. Furthermore, if A and B are R-free, then by Proposition 3.1 these homomorphisms are injective. The remaining assertion being abvious, the result follows. We shall refer t o the homomorphisms A

4

A @ B ,a

a @ l and B

+

A@B, b

-

l8b

as canonical. The next result provides a universal characterization of the tensor product of algebras.

3.3. Proposition.

Let A1,AZ and B be R-algebras and let

IJJ~

: A,

-+

B be an

R-algebra homomorphism, i = 1 , 2 . Then there ezists one and only one R-algebra homomorphism h : A l @ R A2

+

B such that $,

=h o

f,,i = 1 , 2 , where fz

: A,

+

A1 @

A2

is

the canonical homomorphism.

Proof. The map A l x A2 h : Al

@E

At

+

--$

B , ( a l , a z ) H $ 1 ( ~ 1 ) $ z ( a z ) is bilinear and so the map

B given by h(al @ u 2 ) = $l(ul)$2(a2) is a homomorphism of R-modules.

Moreover, if z = a l @ a2 and y = a1 ' 8 a2 ', then

proving that h is a homomorphism of R-algebras. Since h(f,(a,))=

$ J ~ ( U , ) , U ,E

A , , we

have $, = h o fz, i = 1 , 2 . Finally, the uniqueness of h is a consequence of the fact that any R-algebra homomorphism A l (a1

@ az/a, E A , ,

i = 1,2}.

A2

+

B is uniquely determined by its restriction to

77

S E P A R A B I L I T Y , L I N E A R DISJOINTNESS

Let A be a commutative algebra over a field F a n d let

B,C be subalgebras of A .

Then

the natural homomorphism

is a surjective homomorphism of F-algebras. We say t h a t B a n d C are linearly d i s j o i n t over F if t h e given m a p is a n isomorphism. Note t h a t if B and C are finite dimensional, then d i m F ( B @ C) = ( d i m F B ) ( d i m F C ) by Proposition 3.1. Hence, for finite dimensional

U and C , their linear disjointness over F is equivalent t o t h e requirement dimFBC = (dimFB)(dimFC) In t h e general case, we have t h e following characterization of linear disjointedness

3.4. Proposition. T h e f o l l o w i n g c o n d i t i o n s are equivalent: (i) B a n d C are l i n e a r l y d i s j o i n t o v e r F

( i i ) If a f i n i t e s e t

{ a l , . . . ,a,}

CB

is F - i n d e p e n d e n t , t h e n it is C - i n d e p e n d e n t

( i i i ) F o r u n y f i n i t e s e t s {b,} a n d { c , } of l i n e a r l y i n d e p e n d e n t e l e m e n t s in R a n d C , respec-

t i v e l y , t h e s e t { b , c j } is linearly i n d e p e n d e n t (iv) lj { b , } is a basis f o r B a n d { c , } is a basis f o r C , t h e n { b i c , } i s a basis f o r R C

( v ) T h e r e e x i s t s a basis { b ; } of B w h i c h i s C - i n d e p e n d e n t

Proof. ( i ) = + ( i i ) :Assume t h a t B arid C are linearly disjoint and let { a 1 , ., .a,} & LI be 1;-independent.

If

C,"=, a , c , = 0 , then C,"=, a , @ c , = 0.

Sincc t h e set { a l , .. . , a , , }

can b e extended t o a n F-basis of B , it follows from Proposition 3.1 t h a t all

c, =

0. Hence

{ cy 1 , . . . , a,} is C-independent . (ii)+(iii):

If A,,

Let { h i }

xt,,A,,b,c, =

=

iU

0, A,,

and { c , }

E F, then

iC

b e two finite sets of linearly independent elements.

C,b i ( C , A,;c,)

= 0 . Hence

1, AZlc,

= 0 and t h e r e f o r e

0.

(iii)+(iv): T h e assumptic?. guarantees t h a t { b 2 c 3 } is linearly independent. Sirice { b , c l } obviously spans U C , it must be a basis. ( i v - + ( i ) : By Proposition 3.1 and the hypothesis, the natural m a p B

@F

C

-+ H C

maps a

basis t o a basis, a n 3 hence is a n isomorphism.

( v ) > ( i ) : By Proposition 3.1, a typical element of H

C h, @ c , with c , C C . Sincc the 6 , are C-inder'entirnt,, of 1) @ F C onto l3C is an isoniorphisrri.

@p

C can be uniquely written as

the nat.iiral map

C 0, @ c ,

+

1b,?:

78

CHAPTER 2

(ii)+(v): Obvious. rn Let E / F be a field extension and let c h a r F = p of E and let

FP-'

Denote by

of elements z such that

be the subfield of

Lemma.

3.5.

> 0.

W i t h the above notation, a set

{@I,.

zp

F an algebraic closure

E F.

. .a,},

E E , is

ai

FP-I-

independent if a n d only if {a:, . . . ,a:} i s F - i n d e p e n d e n t .

Proof. p, = A:

E

Assume that

Cr='=, Ata,

F. On the other hand, if

algebraically closed, p , = A:

= 0 with

Cy=lp,af

A, E FP-'.

Then

Cy=,p l a y

= 0 with p z E F , then, since

for some A, in FP-I. Hence

CA:af = 0, so

= 0 for

E2F

(1

is

A,a,)p

=0

3.6. Theorem. Let E I F be a n algebraic e z t e n s i o n of fields of characteristic p

> 0.

A , a , = 0. Since A, = 0 if and only if p, = 0, the result follows. w

and

We are now ready to prove the following result.

T h e n EIF is separable if a n d o n l y if E and FP-'

-1

FP

are linearly disjoin1 over F (both E and

are regarded as F-subalgebras of a n algebraic closure

Proof. Assume that E and

Fp-'

E

o f E).

are linearly disjoint over F . Given a E E , put

n = (F(a) : F). Then { l , a ,. . . , an-'} is an F-basis of F ( a ) = F[a]and hence, by Proposition 3.4, these elements are FP-' -independent. I t follows, from Lemma 3.5, that the elements 1,a p , . . . , Q P ( ~ - ' ) are F-independent. Because there are n of these and they are contained in F ( a P ) , we conclude that F(a) = F ( a P ) . Thus, by Proposition 2.6(ii), a is separable. Conversely, assume that E / F is separable. Let { a l , .. . ,a,} be a finite F-independent subset of E, and let K = F ( a l , . . . ,a,). Then we can choose a basis { a l , .. . , a n , a n + l r

. . . , a , ) of K over F. If

i 5

r , so

p2P

=

have

p

p E K , then p is a n F-linear combination

of the a t , 1

5

,LP is an F-linear combination of the a:, 1 5 i 5 r . The same holds for

(pz)p, p3P

.. .

= (,03)P,.

On the other hand, because

0 is

separable over F, we

E F(pp) = F [ P p ] by Proposition 2.6(ii). Thus ,B is an F-linear combination of

the a f , 1

5 i 5 r . Since ( K : F )

= r , this implies that {a:,a;, . . . , a : } is an F-basis

of K . Thus {a:, . . . ,a:} is F-independent and, by Lemma 3.5,

. . ,a,} is

( ~ 1 , .

FP-I-

independent. Now apply Proposition 3.4. rn

Our next aim is to provide a characterization of separability via tensor products.

SEPARABILITY, LINEAR

DISJOINTNESS

3.7. Lemma. Let E / F be a field eztension and let f (X) E F[X] be of degree

3

1.

Then

E

@F

[F[xl/(f(x))l

E[Xj/(f(X))

as E-algebras

Proof. The map E ( X ]---t E 8 [ F [ X ] / ( f ( X ) ) ] ,XiXc

A, @ ( X '

i---t

(3)

+ ( f ( X ) ) )is

obviously a surjective homomorphism of E-algebras which maps f ( X ) t o zero. Since both algebras in ( 3 ) are of the same E-dimension, namely d e g f ( X ) , the result follows. w 3.8. Lemma. Let F

C K 1 C El

and F

C Kz C Ez

be chains of fields. Then the

m np K1 @ F

is

K2

E1 @ F Ez, a @ b H a @ b

an injective homomorphism of F-algebras.

Proof. The given map is obviously a homomorphism of F-algebras. Let {y,}

be F-bases of K 1 and K z , respectively. Then, by Proposition 3.1,

F-basis of K1 @ F K 2 . On the other hand, the elements

(2, @

(5,@

( z z } and

y3} is an

yI} regarded as elements

of E l 8~ Ez can be extended t o a n F-basis of E1 @ F E z , again by Proposition 3.1. This

shows t h a t the given map is injective as required.

H

3.9. Lemma. Let F ( a ) / F be a finite separable f i e l d eztension and let K / F be a n

arbitrary f i e l d eztension. Let f ( X ) = f l ( X ) . . . f , ( X ) be the decomposition of the minimal polynomial f ( X ) of a over F into irreducible factors over K . Then

where a, zs

Q

root of f , ( X ) .

Proof. Since F ( a ) E F [ X ] / ( f ( X ) it ) , follows from Lemma 3.7 t h a t

On the other hand, K ( a , ) E K [ X ] / ( f , ( X ) )The . desired conclusion now follows by- v i r t u e

of Proposition 1.5.5. w We are now ready to prove 3.10. Theorem. Let E / F be an algebraic extension. Then the following eonditzonr are equivalent:

a0

CHAPTER 2

(i) E / F is separable (ii) For a n y field extension

KIF, K

(iii) For a n y f i n i t e field extension

@F

E has no nonzero nilpotent elements

K / F ,K

@F

E has n o nonzero nilpotent elements.

Furtkermore, i j charF = p > 0 , t h e n each of the above conditions is equivalent t o

(iv) F P - '

BF

E has no nonzero nitpotent elements

Proof. ( i ) j ( i i ) : Assume that E / F is separable. By Lemma 3.8, it suffices to show that K

@F

F ( a ) has no nonzero nilpotent elements, for any a E E . The latter is a

consequence of Lemma 3.9. (ii+(iii): This is obvious (iii)+(i):

Given a E E, let f ( X ) be the minimal polynomial for a over F and let K be the

splitting field of f(X). Then K / F is a finite extension, so K @ FE has no nonzero nilpotent elements. Therefore, by Lemma 3.8, K @ FF(a)has no nonzero nilpotent elements. Hence, by ( 4 ) , f(X)has only simple roots and therefore a is separable. Thus E / F is separable. Now assume that c h a r F = p > 0. Then, in view of the implication ( i ) +

( i i ) , we must

also have (i)+(iv). Conversely, assume that E / F is not separable. Then, by Theorern 3.6,

E and Fp-' are not linearly disjoint over F . Hence, by Proposition 3.4(ii), there exists a finite F-independent set { a l , . . . ,a,}

C:=,

X,w, = 0.

Hence

CE

and nonzero elements A, E Fp

C Z , X p a p = 0 and, since Xp

n

n

i= 1

,=I

But, by Proposition 3.1,

CZl A,

F-linearly independent. Thus

@

a , # 0 in F P p '

--L

FY

@F

- 1

such that

E F , then

%F

E since A, # 0 and

al,.

. . . a , are

E has a nonzero nilpotent element and the resuit

follows. In what follows we assume t h a t all the fields involved are contained in one field. We close by giving a useful criterion for recognizing linear disjointness in a chain of fields. For convenience of reference, we first record the following two observations. 3.11. Lemma. Let R , S be integral d o m a i n s with quotient fields K , L respectively

and let F be a subfield of K n L. To test whether i and K are linearly di.sjoint over F , it s u f i c e s t o show that if elements are linearly independent over R.

s1,.

. . , s,

of S are linearly independent over F , t h e n they

81

SEPARABILITY, LINEAR DISJOINTNESS

Proof. Let x l , . . . ,z,b e elements of L which are linearly independent over F . Uy Proposition 3.4(ii), it suffices to show t h a t zl,. . . , z,

. . ,A, E K , not all zero, such t h a t

Assume by way of contradiction t h a t there exist XI,.

XIZl

+ . . . + Xnx, = 0

Choose nonzero r E R , s E S such t h a t r , s1,.

. . , s,

are linearly independent over A'.

=

r X , E R and

(5) s, =

sx, E

S, 1 5 i 5 n . Then

E S are linearly independent over F . O n t h e other h a n d , multiplying (5) by r s ,

we have r1s1

+ . .. + r n s ,

=0

where not all r , are zero. This shows t h a t sl,,. . ,s,

are linearly dependent over H , as

required. rn

3.12. Lemma. L e t K I F a n d L I F be f i e l d e z t e n s i o n s , l e t R be a sttbring of K s u c h

Ii' is t h e q u o t i e n t f i e l d of R a n d F C R , a n d let {x,} be a n F-basis of R . If t h e e l e m e n t s {xu) are l i n e a r l y i n d e p e n d e n t over L , t h e n I< a n d L are linearly dzsjoint o v e r F . Let r l r . .. , r ,

Proof.

be elements of I< linearly independent over F. To prove

t h a t , those elements are linearly independent over L , we may assume t h a t each r , lies

iri

!2. T h e elements r l r . . . , r , lie in a finite-dimensional vector space V over F generated by some of t h e z,, say r l , . . . ,~

~

. . z,. Hence

51,.

71,.

. .,r,

can be completed to a basis, say

,. . . , rrn of ~V over ~ F~. By, hypothesis, the elements z l r . . ,z,are linearly

independent over C, hence V ' = { X l z l

+ . . .+-A,z,\X,

is a vector space over L of dimension n. Therefore ovcr L, as required.

E L} TI,.

= {Xlrl

+ . . . i -X n r n ; X ,E L }

. . ,r , remain linearly independent

rn

3.13. Proposition.

Let F

C K C E

be a chain of j i e l t f s and iet l , / F be

11

fieid

e x t e n s i o n . l h e n L a n d E are linearly d i s j o i n t ouer F if and o n l y if I, and K ore l i r i u i r l ! j di.sjoinl over F a n d L K a n d E are linearly dzsjoint o u e r I<.

Proof. Assume t h a t L, K are linearly disjoint over F a n d LI<, I3 are linearly disjoint over K . Let (A;} a basis of

E

b e a basis of L over F , let { b , } be a basis of K over F , and let

over K . T h e n

{PJZk}

{zk)

be

is a basis of E over F . Assume by way of contradiction

t h a t L and E are not linearly disjoint over F . T h e n there exists a relation

82

CHAPTER 2

Changing the order of summation gives

contrary to the linear disjointness of E and L K over K . Conversely, assume that L and E are linearly disjoint over F . Then L and K are linearly disjoint over F , and the field LK is the quotient field of the ring K [ L ] .This ring is a vector space over K , and a basis of L over F is also a basis of K [ L ]over K . With

this remark and Lemmas 3.11 and 3.12, we see t h a t it suffices t o prove that the elements of such a basis remain linearly independent over E . But this follows from the assumption

that L and E are linearly disjoint over F , hence the result.

4. Norms, traces and discriminants of separable field extensions.

Our aim in this section is t o examine some properties of norms, traces and discriminants of field extensions. We then specialize to the case of separable extensions and provide some useful characterizations of such extensions. Let R be a commutative ring and let A be an R-algebra. We say that A is R-free if it is free as an R-module. Let A be an R-algebra which is R-free of finite rank. By an R-representation of A (or simply a representation of A ) we understand any homomorphism

of R-algebras, where V is a finitely generated free R-module. If n is the rank of V, then

EndR(V) is identifiable with the R-algebra M,(R) of all n x n-matrices with entries in R. Hence the given representation of A defines a homomorphism

Any such homomorphism is called a matrix representation of A . T h e particular case in which V = A and the image la of a E A in EndR(V) is given by

is of special importance. We refer t o the representation A

4

a

H

EndR(A)

e,

83

NORMS, TRACES AND D I S C R I M I N A N T S

as the regular representation of A . Let H be a commutative ring and let V be a free R-module of finite rank freely generated by u 1 , . . . , v n , If

f E EndR(V) then, with respect to the basis ul,.. , u , ~ , ,f

can be represented as an n x n-matrix ( a i j ) over R . The trace, d e t e r m i n a n t , and the

characteristic polynomial of f are, respectively, 11

a,,, d e t ( f )

T r ( f )=

det(a,,) and d e t ( X . I ,

:

-

(atJ))

*=l

where I , is the identity n x n-matrix. These quantities are independent of the choice of the basis. Observe also that

Now assume that R is a subring of a commutative ring

S

and that

S is R-free of firiite

rank. Consider the regular representation

S

+ EndR(S), s

H

l,

of S. F3y the trace ( n o r m , Characteristic polynomial) of s E S we understand the trace (determinant, characteristic polynomial) of

t , . The trace and

n o r m of s will be denoted

bY T r s / ~ ( sand ) Ns/R(s) respectively. The foregoing is clearly applicable to a finite field extension E I F , since E i s a finitedimensional F-algebra. Hence, for any X € E , ~ v emay define the trace (norm, characteristic polynomial) as above.

4.1. Lemma. Let E I F be a finite field extension, let X E E a n d let f ( X ) be the characteristic polynomial of A . ( i ) X zs a root of f ( X ) a n d , if

E

=

E'(X), t h e n j ( X ) is the m i n i m a l polynomial of X over

F. ( i i ) Zf K I F is a n y field eztension in which

Il:=l(x'

-

A t ) , then

fix') splits

i n t o linear factors, say f ( X )

=

a4

CHAPTER 2

Proof. (i) Denote by M the matrix for multiplication by X in E with respect t o a given F-basis of E . From matrix theory we know t h a t f ( M ) = 0. Since f ( M ) multiplies elements of E by !(A),

we conclude that f ( X ) = 0, i.e. X is a root of f ( X ) . Note that

f ( X ) is a monic polynomial over F of degree ( E : F ) . Hence, if E = F(X), then f ( X ) is the minimal polynomial of X over F . (ii) This is a direct consequence of the equality ( 1 ) .

4.2. Lemma. L e t E / F be a f i n i t e field extension, let K be a n intermediate field and let X E K . L e t g ( X ) be t h e characteristic polynomial of X over F, w h e n X is regarded as a n e l e m e n t of K , a n d let f ( X ) be t h e characteristic potynomiat of X over F , w h e n X i s regarded as a n e l e m e n t of E . T h e n

(i) f ( X ) = g(X)"

(ii)

where

m = (E :K )

NE/F(X) = [NK/F(X)lm

(4 T ~ E / F (= X )~ ( T ~ K / F ( A ) ) Proof.

Properties (ii) and (iii) follow from (i), by applying Lemma 4.1(ii). To

establish (i), let {y;ll 5 i

5 k}

be a basis of K over F and let {yll 5 j

of E over K . Thanks to Proposition 1.24, { y i z j l l

5 i 5 k,

1

5 m} be

5 j 5 m } is

a basis

a basis of E

over F . Let M = ( a i h ) denote the matrix for multiplication by X in K with respect to the basis {y;}. Then Xyi =

xh

UihYh

and thus

Ordering lexicographically the basis {y;z,}

of E over F , it follows that the matrix N for

multiplication by X in E with respect to this basis is given by

N Thus the matrix X . I,,

-

= diag(M, M , . . . ,M )

N , n = ( E : F ) , consists of m diagonal blocks, each of the form

x .Ik - M . It fO~~OWSthat f(x)= det(X'. I,, - N ) = ( d e t ( X . Ik - M ) ) "

= g(X)m:

as desired.

4.3. Corollary. L e t E I F be a f i n i t e field extension, let X E E a n d let f ( X j a n d

g ( X ) be the characteristic polynomial a n d m i n i m a l polynomial of X over F , respectively.

85

NORMS, TRACES A N D DISCRIMINANTS

Then

f ( X ) = g(X)n where

n = ( E : F(X))

I n particular, f ( X ) = g ( X ) if and only if E = F ( X ) .

Proof. P u t K = F ( X ) and apply Lemmas Q . l ( i ) and 4.2(i). m We are now ready t o provide explicit formulae for trace and norm of an element in a field extension. 4.4. Theorem. Let E J F be a finite field eztension, let

and let f ( X ) he the characteristic polynomial of X E E . Let he all distinct F-homomorphisms of E into

and u , (A), . ..,..=(A)

be a n algebraic closure of E 01,02,.

. . , u n ,n = ( E : F ) s ,

and let p e = ( E : F ) , . Then

are all conjugates of X in

E,

each repeated ( E : F ( X ) ) , times. In

particular, b y Lemma d.l(ii),

Proof. P u t t = ( F ( X ) : F ) s , K = ( E : F ( X ) ) , , r = ( F ( X ) : F ) , and a = ( E : F ( A ) ) . Then, by Proposition 2.13(ii), p e k = ar. By Lemma 2.7, there exist exactly t , say X =

XI,. . . , A t of conjugates of X in

E.

Let L be the normal closure of E / F . Then L

arid any F-homomorphism of F ( X ) into

E

CE

is in fact an F-homomorphism of F ( X ) into

L.

'Therefore, by Propositions 1.21 and 2.13(ii), each of the t F-homomorphisms of F ( A ) into C has exactly

( E : F ) , / ( F ( X ) : F ) s = ( E :F ( X ) ) , exterisions among the u t . Accordingly, each of the conjugates A, of X occurs ( E : F ( X ) ) , times in the set { u l ( X ) ,. . . ,cn(A)).We conclude therefore that

86

CHAPTER 2

Denote by g ( X ) the minimal polynomial of X over F . By Proposition 2.5(ii), all the roots

XI,. . . , A t of g ( X ) have the same multiplicity equal t o r . Thus

and, invoking Corollary 4.3, we obtain

as we wished t o show.

4.5. Corollary. Let E I F be a finite separable extension of degree n , let

be all distinct F-homomorphisms of E into its algebraic closure

z and

u1,.

. . ,u,,

let f ( X ) be the

characteristic polynomial of X E E . Then (i) f ( X ) = r I L ( X - .l(X))

and .](A),

. . . ,on(X) are all conjugates of X in

z, each repeated ( E : F ( X ) ) times.

(ii) N E / F ( X ) = n,"=, uI(X) (iii) T r E / F ( X ) =

cT=l uijh)

Proof. Because E I F is separable, we have ( E : F ) s = ( E

:

F ) and ( E : F ) i = 1.

The required assertions now follow by virtue of Theorem 4.4. H We next record some elementary properties of norm and trace.

Proof. Properties (i), (ii), and (iii) can be verified in a straightforward manner. Let a l , . . .ak be a basis of K over F and let b l , . . . , b, be a basis of E over K . Given x E E any y E K , write

xbt =

c

PI,

(+,, ya1

=

c

'ylj

(Yb,

NORMS. TRACES AND D I S C R I M I N A N T S

The products aibj give a basis of E over F and

proving ( a ) .

To prove ( b ) , we employ the notation of the proof of Proposition 1.21. We may assume that

is the identity automorphism of L. By Theorem 4.4, we have

We know, from the proof of Proposition 1.21, that the restrictions of the products

GJ4,

to E are distinct and give all F-homomorphisms of E into L. Moreover, by Proposition 2.13(ii), p a + $ = ( E : F ) , . The desired assertion now follows by applying Theorem 4 . 4 . rn T,et R be a commutative ring, and let M , N be R-modules. An R-bilinear form on

ibf x A’ is any map f :M x

-

N

t

R

having the following two properties: ( i ) For each z E M , the map y

f ( z , y ) is R-linear

CHAPTER 2

88

(ii) For each y E

N ,the map

x

H

f ( x , y ) is R-linear

In what follows we shall write < 5,y > instead of f(x,y). The bilinear form f : M x N

4

R

is said to be nonsingular if the following two conditions hold:

(i) < 2,b>= 0 for all x E M implies b = 0 (ii) < a , y >= 0 for all y E N implies a = 0 We shall refer t o a bilinear form f : M x M

--f

R as being symmetric if for all z , y E M

< x , y>=< y, x >

4.7. Lemma. Let U and V be two finite-dimensional vector spaces over a field F

and let U x V

+

F be a nonsingular bilinear form. Then dimU

= dimV = n,

to any basis u1,.. . ,un of U there corresponds just one family of elements

vl,

say, and

. . . ,v,

of V

such that

Moreover,

v1,.

.., v n

(6.. zj - 0 f.f z . # j, 6;;= 1)

6ij

=

is a basis of V.

Proof. The assignment

defines a mapping V y E V for which

4

F n which is easily seen t o be linear. T h e kernel consists of all

< u 1 , y >= . . . =< u n , y >=

0. Since the given form is nonsingular, it

follows that y = 0. Hence the mapping (2) is injective and therefore dimV 5 n = dimU. By symmetry, we also have the opposite inequality, and combining the two, we find that dimU = dimV. Since dim V = n and the map (2) is injective, it must be an isomorphism. Take the standard basis e l , . . . ,en in F" and let v , be the inverse image of e; under the mapping (2). Then < u ; , v i > =

6 i j as

claimed.

Finally, if v;, . . . ,vL is another family with this property, then each v, - u: lies in the kernel of the mapping (2) and so must be zero. Hence there is only one such family. Taking into account the v:s are the inverse image, under a n F-isomorphism, we also see t h a t they form a basis of V. So the lemma is true. Let U and V be two finite-dimensional vector spaces of dimension n over a field F and let U x V

+F

be a nonsingular bilinear form. Any two bases u l , . .. ,un and u l , . . . ,v, of

U and V, respectively satisfying =

6ii

89

NORMS, TRACES AND D I S C R I M I N A N T S

are called dual to one another with respect t o a given form. Let

EIF

be a finite field extension. Then the trace map may be used to define a

symmetric bilinear form E x E

+F

by the rule

The bilinear properties follow from Lemma 4.6(i) and the symmetric property is obvious.

We are now ready to provide the following characterizations of finite separable extensions: 4.8. Theorem. Let E I F be a f i n i t e field extension. Then the jollowzng conditions

are equivalent: (i) The bilinear j o r m < z , y > = T r ~ / , ~ ( zisy )nonsingular (ii) T T E / F ( Z#) 0 f o r some z E E

( i i i ) T T E I F: E

+

F is a n epirnorphisrn

(iv) The estenszon E I F is separable

Proof. (i)+(ii): Obvious

#

(ii)+(iii): Let T T E / F ( Z=) ,u

0 for some z E E .

TrEIF(Xp-12) = (Xp-l)TrE/F(x) =

Then, given X E F , we have

x

(iii)+(iv): This is a direct consequence of Theorem 4.4 (iv)*(i): Suppose that E / F is separable. By Corollary 2.21, E = F ( a ) for some a t 6. tIence 1.0,. . . ,

arid so

. an IS

F-basis for E for some n

2 1.

Therefore

< z, E >= 0 if and only if < z, a' >= 0 for all i E {0,1,. . . ,n -

l}. Our task is t o

prove that z = 0 under these conditions. Setting z

=

C pzcut,we have

Lct D denote the matrix (&,)

where d,, =<

the assumption < z , E > = 0 that

>.

O ' - ~ , ( Y ~ - ~ It

then follows from ( 3 ) and

90

CHAPTER 2

It will be shown that D is nonsingular which will imply that each pi = 0 and hence z = 0. Let i ( X ) be the minimal polynomial of

Q

over F and let

E

be an algebraic ciosure

of E. Because E I F is separable, f ( X ) has n distinct roots, say a =

QI, 0 2 , .

. . , a , in E.

Hence, by Lemma 4 . 1 ,

We wish t o find a similar formula for T r E / F ( Q k ) . The h e a r transformation f!, has the distinct characteristic roots

a1,. .

., a ,

in

E.

some F-basis of E ) can be diagonalized over

U-'l,U

Hence the matrix for

e,

(with respect t o

z. It follows t h a t there exists U such that

= diag(a1,. . . , a,)

Raising this to the k-th power, we infer that

U-'eaklJ = d i a g ( akL ,...,a,) k

and, by taking traces, we obtain

Now let V = V ( a 1 , .. . ,a,) denote the matrix r

1

1

ff,

ff2

ff:

ff;

... ... ...

I ffn ff',

and V t the transposed matrix. Invoking ( 4 ) , we see that the (z,j)-th entry of V V t is

k

k

It therefore follows that V V ' = D . Since V and V t have the same determinant, we deduce that d e t D = (detV)2

NORMS, TRACES AND D I S C R I M I N A N T S

91

O n the other hand, the matrix V is a van der Monde matrix and detV =

U(al

- cyj)

I>?

Since the a1 are distinct, we have detV .f- 0. Hence D is nonsingular and the result follows. w

. . , u , be 4.9. Corollary. Let E I F be a finite separable field extension and let ul,. a n F-basis of E . T h e n there exzsts a second basis v,, . . . , u n such that

Proof. This is a direct consequence of Theorem 4.8 and Lcrnrna 4 . 7 . We now exhibit a dual basis with respect to a dis?inguished basis of a finite separable field extension E I F . 4.10. Proposition. Let E I F be a f i n i t e separable field extension, let

-N

E E be

such

that E = F ( a ) a n d let f ( X ) be the m i n i m a l polynomial of a over F . Let

T h e n the dual basis of 1,a , . . . , c y n p l 2.s

m’”” PO

A - 1

f’(a)

Proof. Let a x ,... , LY,, be the distinct roots of f ( X j . We claim t h a t

(0 I T 5 n

-

1)

(5)

Indeed, let g ( X ) denote the difference of the left and right-hand side of this equality. ’Then degg(X) 5 n

-

1 and g ( X ) has n roots a l , . . . ,a,. Therefore g ( X ) is identically zero, as

claimed. T h e polynomials

92

CHAPTER 2

are all conjugate t o each other. Define the trace of a polynomial in E[X] to be the polynomial obtained by applying

TTE/F

t o the coefficients. Then, by ( 5 ) , we have

Comparing the coefficients of each power of X in this equation, we derive

as asserted.

H

Let R be a subring of a commutative ring S such t h a t S is a free R-module of finite rack n. Given

. . ,sn) E S " , the d i s c r i m i n a n t of

( ~ 1 , .

. . ,s,,), written D s / R ( s ~. .,. ,s,)

(~1,.

is defined by

D s / R ( s I , . . ,sn) = d e t ( T r s / R ( s ; s j ) )

4.11. Proposition. Let E I F be a f i n i t e field e z t e n s i o n of degree n a n d let XI,.

. . , A,

be a n F - b a s i s of E . (i) D E , I F ( A I ., .. , A),

= 0 if a n d only if

T T E / F ( Z=) 0 for all x E E

(ii) E J F i s s e p a r a b l e i f a n d o n l y i f D E / ~ ( X , .l. . , X n ) # O

Proof. (i) The "if" part is obvious. Assume that D E , F ( A I , .. . ,A,)

= 0. Then there

exist n elements P I , . . . ,b,, in F , not all zero, such that

We set z

=

C , p,X3. Then z #

0 and we have 2 ' r ~ / ~ ( X , = z ) 0, 1

5z 5

n. It therefore

follows that T r E / F ( y z ) = 0 for all y in E . If x E E , we take y = xz-' and we find that T r g / p ( x ) = 0, as required. (ii) This is a direct consequence of (i) and Theorem 4.8(ii).

H

We next exhibit two general facts pertaining to discriminants. In what follows, R denotes a subring of a commutative ring S such that S is a free R-module of finite rank n.

4.12. Lemma. If

(s;,

. . . ,s):

E S" a n d

s: =

c,"=, rlls3 with

rzJ E R , t h e n

NORMS, TRACES AND D I S C R I M I N A N T S

93

I n particular, the discriminants oj a n y t w o bases oj S over R difler by a square of a unit (hence generate the s a m e principal ideal of R )

Proof. . We have

which gives the matrix equation

where ( r q 3 ) t is the transpose of (r,,). Taking the determinants, the result follows. rn Owing t o Lemma 4.12, we may define the discriminant ideal of S over R , written

D s I R , as the principal ideal of R generated by the discriminant of any basis of S over R . 4.13. Lemma. A s s u m e that D S I R contains a n element which is not a zero divisor.

T h e n the elements sl,.. . ,sn of S f o r m a basis of S over R if and only if D S I R is generated by

D S / R (. 1 I

. .

1

sn)

Proof. It clearly suffices to prove the “if” part. Assume t h a t d = D s / R ( s ~ . . ,. ,sn)

generates D s / R . Fix a basis e l , . . . ,en of S over R and put d’ = D s / R ( e l , . . . , e n ) . We may write

in which case d = det(r;j)’d’, by Lemma 4.12. By assumption, we have Rd

=

Rd’.

Accordingly, d’ = r d for some r E R. It follows that d ( 1 - rdet(r;j)2) = 0

Moreover, d is not a zero divisor, since otherwise every element of Rd = D S / R would be a zero divisor, contrary to the hypothesis. Thus 1 - rdet(r;j)’ = 0

CHAPTER 2

94

which implies that det(rij) is a unit of R. But then the matrix (rij) is invertible and hence sl,.

. . , s n is a basis of 4.14.

X1,. ..,A,

S over R , as required. w

Corollary. Let E I F be a finite separable field extension of degree n and let E E . Then XI,.

. . ,A n

is a basis of E over F if and only if D E I F ( X ,.,. . , A n ) #

0. Proof. Apply Lemma 4.13 and Proposition 4.11(ii). We now provide a useful characterization of discriminants of finite separable field extensions. 4.15. Proposition. Let E I F be a finite separable field eztension of degree n and

let o l , . . . , u n be the n distinct F - h o m o m o r p h i s m s of E into a n algebraic closure of E . If X1,. . . , A,

are a n y n elements of E , t h e n

Proof. Since

the result follows. Let E / F be a finite separable field extension of degree n. If X is an arbitrary element of E , then the discriminant of this element with respect t o F , denoted by D E / F ( X ) is

defined by DE/F(X) =

DE/F(1,Xl...,Xn-')

Observe t h a t , by Corollary 4.14, D E / F ( X )# 0 if and only if E = F ( X ) 4.16. Proposition. Let E I F be a finite separable field eztension of degree n , let

X E E be such that E = F ( X ) , and let

into a n algebraic closure of E . T h e n

. . , u n be the n distinct F - h o m o m o r p h i s m s of

~ 1 , .

E

NORMS, TRACES AND DISCRIMINANTS

95

where f ( X ) is the minimal polynomial of X over F and

Proof. P u t L)E/F(X)

( - l ) n ( n p l ) )A,/ z=, oi(X) and b, = a,(f'(X)), 15i

c =

= det(Xi-I)*, 1 5 i , j 5 n , =

(by Proposition 4.15)

n(A;

(Vandermonde)

- XJ)2

J

5 n. Then

(61


=c

n f'(X;)

= c ( b 1 . . . b,)

(since b, =

f'(Xi))

1

Furthermore bl

. . .b,det(X:-')

= det(XI-'b,) n-I

X:as,j-l)

= det(

= det ( X i - ' ) d e t ( a s k )

and therefore

bl . . . b, = det(a,k), as required. m

4.17. Corollary. Let E / F be a finite separable field eztension. Then

D E / F ( X )= D E / F ( X+ a ) Proof. Let of E. If E

01,.

for all X E E , a E F

. . ,u n be all distinct F-homomorphisms of E into an algebraic clsoure

+ a ) and so D E / F ( X )= D E / F ( X+ u ) = 0. We may t h u s E = F(X) and hence that E = F(X + u). Because

# F ( X ) , then E # F(X

assume that

at(.\) - QJ

= OZ(X i a )

the desired equality follows by virtue of ( 6 ) . w

-

(A

+ a),

This Page Intentionally Left Blank

97

3 Transcendental extensions This chapter is confined to a systematic study of transcendental extensions. We begin by introducing abstract dependence relations, which will allow us t o treat in a unified

manner algebraic independence and pindependence. After examining some elementary properties of transcendency bases, we concentrate on simple transcendental extensions and provide a number of their important properties. We next generalize the notion of separability (introduced previously for algebraic extensions) t o arbitrary extensions. Special attention is drawn to the investigation of the existence of separating transcendency bases. Extensions E / F with separating transcendency bases are special instances of extensions

E / F which preserve p-independence. T h e latter extensions are characterized in a large number of different ways. In terms of pindependence we introduce for any extension E / F certain relative p b a s e s closely related t o separating transcendency bases. The chapter culminates in the study of relatively separated and reliable extensions. Among other results, we provide numerous connections between reliable and relatively separated extensions. It is also demonstrated t h a t relatively separated extensions can be completely characterized

in terms of reliable extensions.

1. A b s t r a c t d e p e n d e n c e r e l a t i o n s .

Lct A be a n arbitrary set. A relation < between elements of A and subsets S of A (written a < S and referred t o as “ a is d e p e n d e n t o n S”)is called a dependence relation if the following four conditions hold: (1) If a E S , then a

<S

(2) If a

< S , then a < SOfor some finite subset

( 3 ) If a

< S and every

s

SOof

S

in S satisfies s < T, then a < T

(4) (Exchange property) If a < S and a $ S-{s} for some s E S , then s < ( S - { s } ) U { a } .

A typical example of a dependence relation is as follows. If A is a vector space over a field,

98

CHAPTER 3

a E A and S a subset of A , then we say that a is linearly dependent on S if a lies in the subspace spanned by S. In future we shall encounter other types of dependence relations such as algebraic dependence, p-dependence, etc. This is the reason why we treat such relations axiomatically. We define a subset S of A t o be independent if no s E S is dependent on S - { s } . A subset B of A is called a basis of A if B is independent and if every a in A is dependent on B. 1.1. Lemma. A subset

S of A is independent if a n d o n l y if a n y finite subset of A is

independent. Proof.

Assume that SO is a finite subset of S and t h a t SO is not independent.

Then there exists so E SO with

s

<S

-

{so}, hence by ( 3 ) , So

SO

< SO - {SO}.

< S - {so}.

By (l), every s in So

- {SO}

satisfies

This shows that S is not independent. Thus

if S is independent, then so is any finite subset of S . Conversely, assume that every finite subset of S is independent. If S is not independent, then

SO

< S - {SO} for some

not containing

SO.

SO

E

S. By (2), SO < S1 where S1 is a finite subset of S

Setting S, = S1 U {so}, it follows t h a t so

< Sz - {so},

contrary to the

assumption that Sz is independent. rn 1.2. Lemma. A s s u m e that a subset S of

A

is independent a n d that a E A is not

dependent o n S . T h e n S U { a } i s independent. Proof. Since a is not dependent on S, a !$ S by ( 1 ) . Assume t h a t S U

{u}

is not

independent. Then there exists b E S U { a } such that

Furthermore, b E S,since b = a would imply that a also that b

< S - { b } , since S is independent.

< S,contrary t o hypothesis. Note

Now

b < ( S - { b } ) U { a } and b

# S - {b}

implies, by (4),that

u

< ( S - ( 6 ) ) U { b } = S,

contrary t o hypothesis. m We are now ready to prove the main result.

99

ABSTRACT DEPENDENCE R E L A T I O N S

1 . 3 . Theorem. Let A be a n arbitrary set and l e t

< be a dependence relation between

e l e m e n t s of A a n d subsets of A. T h e n (i) T h e set A h a s a basis (possibly

U ~ C U O U Ss e t )

a n d a n y t w o bases have t h e s a m e cardi-

nality. ( i i ) A n y i n d e p e n d e n t set c a n be extended t o a basis

(iii)

11 every a E A is dependent o n a subset S of A , t h e n S contains a basis of A . Proof. Let S be a subset of 4 and let

5'0

be an independent subset of A contained

i n S (such So always exists, e.g. the vacuous subset of A is independent). Consider the

collection L of subsets of A which are contained in S , contain So and are independent. Since

So E L , I, is nonempty. We order 1, by the inclusion. If (Ti12 E I } is a linearly ordered subset of L , then T = U;,,T,

obviously satisfies So

C 1' C

S . If T is not independent,

thcn by Lemma 1.1,there is a finite subset V of T which is also not independent. Because

T'y for some j E J . Since Ti is independent, this contradicts

{Yt} is linearly ordered, V

1,errirna 1.1. Thus T' is independent and T E

L. It follows that L is inductive and therefore,

by Zorn's Lemma, there exists a maximal element B in

L. We now show that if S satisfies

( i i i ) ( i n particular, if S = A ) then B is a basis of A . This will prove (ii), ( i i i ) and the first

part of ( i ) . Assume that S satisfies (iii). If each element of S depends on B , then by ( 3 ) each element of A depends on B (hence B is a basis of A ) . Let x be an arbitrary element of S.

If z is not dependent on B , then B 0 {z} is independent by Lemma 1.2, which contradicts tile maximality of B. Thus B is a basis of A . By the foregoing, we are left to verify that IBI = / C / for any two bases R and C of A . \T'e first assume that B is finite, say H c1

{ b l , . . . , b n } . \\'e claim that there is a

C such that c 1 is not dependent on ( b z , . . . , b n } . Indeed, otherwise by ( 3 ) , every

elcrnent of A is dependent on ( b 2 ,

~

< { c l r b z , . . , ,&},

,b n } . In particular, bl has this property, contrary

Since c 1 is not dependent on { b 2 , . . . , bn}, it follows from

to the independence of B.

Lernrna 1.2 that { c l , b z , . . . b,} bl

=

so every b,

is independent. Furthermore, it follows from (4) that

rcpcating this process, we obtain c2 E C such that (c1, i n taliis way, we obtain a basis { c l , . . . ,cn}

,b,} is a basis. By

c 2 , b z , . . . , b,}

is a basis. Continuing

which is a subset of C and has the same

cardinality as B. Since C is independent, C = Now assume that both / C /and

{cl, b z ,

, bn}. Hence

< {cl,b2,

(~1,.

. . c n } and thus jC/ = IB1.

lBI are infinite cardinals. Given c E C,

c is

dependent

100

CHAPTER 3

on a finite subset B, of B. Thus

We claim that B = U c E C B , ;if sustained, it will follow that ( B (5 (C(. By symmetry, ICI 5 JBI and hence IBI = ICI. Deny the claim. Then there is a 6 E B satisfies c

< UcECBc,we

have 6

-

UcECB,.

Since 6

< C and every

c E C

< UcECBcwhich does not contain 6. Because this

contradicts the independence of B , the result follows.

2 . Transcendency bases.

Let E/F be a field extension and let S be any subset of E. Recall that S (or the elements of S) is said to be algebraically independent over F if the elements

(finitely many v ( s ) # 0) are F-linearly independent. It is clear that S is algebraically independent over F if and only if so is every finite subset of S. In the special case where S = {s},

we see that S is algebraically independent over F if and only if {I, s, s 2 , . . . , s", . . .}

is an F-linearly independent set. If {s} C E is algebraically independent over F , then we also say that s is transcendental over F . Expressed otherwise, s is transcendental over F if and only if s is not algebraic over F . A field extension E I F is said to be transcendental if it is not algebraic, that is, if E

contains at least one element which is transcendental over F . It is clear that any extension of a transcendental extension of F is itself a transcendental extension of F . We say that the field extension EIF is purely transcendental if E = F ( S ) for some algebraically independcnt subset S of E over F . It will be often convenient to deal with the families of elements of E , say a family {a;}iE1.

We say that such a family is algebraically independent over F if its elements are

distinct (i.e. if ai

# aj

for i # j ) and if the set consisting of the elements in this family is

algebraically independent over F. We begin by characterizing algebraically independent sets.

101

TRANSCENDENCY BASES

2.1.

Lemma. Let E / F be a field e z t e n s i o n a n d let S = { s , } , ~ Ibe a f a m i l y of

e l e m e n t s o j E . T h e n t h e following conditions are equivalent: ( i ) S i s algebraically independent over F (ii) T h e m a p

++

si

is a n injective h o m o m o r p h i s m of F-algebras. (iii) For each s E S , s i s transcendental over F ( S ' ) , where S' =

s - {s}.

Proof. The equivalence of ( i ) and (ii) is a consequence of Proposition 1.4.2. If there

exists s E S such t h a t s is algebraic over F ( S ' ) , then s is a root of a polynomial i n

F [ ( X z ) z Gand ~ ] hence the map in (ii) has a nontrivial kernel. Converse!y, assume that S is not algebraically independent over F. Then S has n distinct elements sir.. . ,s, such that C a , ,,.,.,e,S;'S;?

. ..&"

=0

is a nontrivial linear combination of the monomials s;"sT

. . . s>.

Hence si (for some z ) is

algebraic over F ( s 1 , . . . , s , - l , s , + 1 , . . . ,sn), a s required. w We next determine the isomorphism class of purely transcendental extensions. 2.2. Lemma.

If E / F i s a purely transcendental eztension, t h e n there exists a srt I

s u c h that E is F - i s o m o r p h i c t o the quotient field of the polynomial rzng F [ ( X z ) t E ~ l .

Proof. If S is a subset of E such that E = F ( S ) ,then E is the quotient field of F L S ] .

If S is independent over F , then by Lemma Z.l(ii),

where S = {s,li E I}. Hence E is F-isomorphic t o the quotient field of F[(X,),EIl.

Let E / F be a field extension. We can introduce an ordering among algebraicdlly independent subsets of E , by ascending inclusion. A subset S of 6 which is algebraically independent over F and is maximal with respect to the inclusion ordering is called a transcendency basis of E over F . Thus a subset S of E is a transcendency basis of E over

F if and only if the following two conditions hold: (i) S is algebraically independent over F .

(ii) E / F ( S ) is algebraic

102

CHAPTER 3

Note that a n extension E / F is algebraic if and only if E has a vacuous transcendency basis over F . On the other hand, if E / F is a purely transcendental extension, say E = F ( S ) where S is algebraically independent over F , then S is a transcendency basis of E over F . We now show that transcendency bases exist and satisfy a number of important properties, all of which are consequence of the theory of abstract dependence relations. Let E / F be a field extension, let S be a subset of E and let z E E . We say that

2

is

algebraically dependent over F on S if z is algebraic over F ( S ) . 2.3. Lemma. Algebraic dependence in E / F is a dependence relation. Furthermore, with respect t o this relation, a subset S of E i s independent (S is a basis) if and only if S is algebraically independent over F (S is a transcendency basis of E over F ) . Proof. To prove the first assertion, we must verify properties (1)-(4) in the definition of a dependence relation given in Sec. 1. If a E

S,then a E F ( S ) and so

a

is algebraic

over F ( S ) , proving (1). If a is algebraic over F ( S ) , then a is obviously algebraic over F ( S 0 ) for some finite subset SO of S, proving ( 2 ) . Assume t h a t a is algebraic over F ( S )

and every s E

S is algebraic over F ( T ) . Consider the subset L of E of elements which are

algebraic over F ( T ) . Then we know that L is a subfield of E containing F ( T ) and that L is algebraically closed in E . Now S

CL

and a is algebraic over F ( S ) ,so over L. Hence

a E L which implies that a is dependent on T , proving ( 3 ) .

To prove (4), suppose a < S (i.e. a is algebraically dependent on S), but a $. T = S

-

{s} for some s E S. Let K = F ( T ) . Then a is transcendental over K and algebraic

over K ( s ) . Hence there exists a polynomial f ( X , Y ) E K [ X , Y ] ,such that

f(X,Y) # 0

and f ( a , s ) = 0. We write

~ ( x ,= Y a)o ( X ) y m+ a l ( X ) Y m p l + . . .

+ am(X)

where the a , ( X ) E K [ X ]and a O ( X ) # 0. Then ao(a) # 0 and m > 0, since a is transcendental over

K.The polynomial f(a,Y) is a nonzero polynomial belonging t o K ( a ) [ Y ]and

s is a root of f ( a , Y ) . Hence s is algebraic over K ( a ) , which implies t h a t s is algebraic

over F ( T

L.

{ a } ) . Thus s < T U { a } , proving (4).

By definition, S is independent if no s E S is algebraic over F ( S

- {s}).

Hence,

by Lemma 2.l(iii), S is independent if and only if S is algebraically independent over F . Finally, assume that S is independent. Then S is a basis if and only if every a E E is algebraic over F ( S ) , i.e. if and only if E / F ( S ) is algebraic. Hence S is a basis if and only if S is a transcendency basis of E over F .

103

TRANSCENDENCY BASES

2.4. Corollary. Let E / F be a n arbitrary field eztension

(i) The field E has a transcendency basis over F and any two such bases have the s a m e

cardinality. (ii) A n y algebraically independent set over F can be extended t o a transcendency basis

OJ

E over F . (iii) If S is a subset of E such that E / F ( S ) is algebraic, then S contains a transcendency

basis of E over F . P r o o f . This is a direct consequence of Lemma 2.3 and Theorem 1.3. Let E / F be a field extension. By the transcendency degree of E over F , we understand the cardinality of any transcendency basis of E over F . This concept is well defined, by virtue of Corollary 2.4. In what follows we write

tr.d.(EfF) for the transcendency degree of E over F . It is clear that E f F is algebraic if and only if

tr.d.( E / F )

=

0. Furthermore, purely transcendental extensions are completely character-

ized by their transcendency degree, as the following result shows. 2.5.

Proposition.

Let El / F and E 2 / F be two purely transcendental eztensions .

Then El is F-isomorphic t o

E2

if and only if tr.d.(El/F) = tr.d.(Ez/F)

P r o o f . By hypothesis, we may write E, = F ( S t ) , where S, is algebraically indeperident over F , i = 1,2. Then S, is a transcendency basis of E , over F, i = 1 , 2 . Hence, if

IS,/ =

/S21, then El is F-isomorphic t o E 2 by Lemma 2.2. Conversely, if

is a n F-isomorphism, then S: = f(S1) is a transcendency basis of

Corollary 2.4,

I S 1 1=

I S : I

=

E2

f

:

El

+

Ez

over F . Hence, by

/ S z l , as required.

2.6. P r o p o s i t i o n . Let F C_ E

CK

be a chain of fields and let S1,S2 be transcen-

dency bases of E / F and K f E , respectively. Then S1U Sz is a transcendency basis of K / 1 ' and tr.d.(K/F) = tr.d.(K/E) + tr.d.(E/F) P r o o f . Clearly S1 fl Sz = 0, so it suffices to verify that S1 u S, is a transcendency basis of K I F .

104

CHAPTER 3

Let { a l , . . . ,a,,,, b l , . . . ,b,} be any finite subset of S1 U 5’2, where we assume that the

a; are in

S 1

and the b j are in

SZ.Assume that there exists a polynomial f ( { X } , { Y } )=

f ( X 1 , . . . ,X,, Y1,. . . ,Y,) in m

+ n variables over F such that f ( { a } ,{b}) = 0. The poly-

nomial f ( { a } , Y ) in the n indeterminates Yj has coefficients in E and must be zero since the

bj

are algebraically independent over E . Because the ai are algebraically independent

over F , it follows that f ( { X } , { Y } ) ,considered as a polynomial in { Y } with coefficients in F [ { X } ] ,must be zero. Thus f ( { X } , { Y } ) = 0 and therefore S1 U

5 2

is algebraically

independent over F . By hypothesis, E is a n algebraic extension of F ( S 1 ) . Hence E ( S 2 ) is an algebraic extension of F ( S 1 ) ( S 2 )= F ( S 1 U S2). However, K is a n algebraic extension of E(S2). Thus K is an algebraic extension of F(S1 U S2) and the result follows. w

2.7. Proposition.

Let E1/F and EZ/F be two eztensions of F contained in a

common field K . Then t r . d . ( E l E ~ / E l5) tr.d.(Ez/F)

(1)

Proof. Let S be a transcendency basis of E z I F . Since ElE2 = E l ( E 2 ) and every element of E2 is algebraic over F ( S ) , we see that ElE2 is algebraic over E l ( S ) . Hence, by Proposition 2.6, any transcendency basis T of E I ( S ) / E I is a transcendency basis of

E 1 E 2 / E 1 . Since, by Corollary 2.4(iii), we may choose T as a subset of S , the inequality (1) follows.

By Proposition 2.6, we have

t7.d. ( E lEz / F ) = t7.d. ( E1 Ez / E l )

+ t7.d. ( E / F ) 1

The inequality ( 2 ) is therefore a consequence of (1). I Turning to finitely generated extensions, we next record 2 . 8 . Lemma. If a f i e l d E is finitely generated over its subjield F and

i f S a tran-

scendency basis of E over F , then S and E / F ( S ) are finite. Proof. By hypothesis, E = F ( T ) for some finite subset T of E . Furthermore, by Corollary 2.4(iii), T contains a transcendency basis of E over F . Hence S is finite by

TRANSCENDENCY BASES

105

Corollary 2.4(i). Since E / F ( S ) is algebraic and finitely generated, it must also be finite. D

2.9. Proposition. If a field E i s finitely generated over i t s subfield F and ZJ K is a n intermediate field, t h e n K is finitely generated over F . Proof. By Lemma 2.8, t r . d . ( E / F ) is finite. Suppose first that

K is algebraic over F .

1,et z l r . .. , zn be a transcendency basis of E over F . If the given elements y1,. . . , y T E K are linearly independent over F , then they are linearly independent over F ( z 1 , . . . , zn). Therefore, by Lemma 2.8,

Now we consider the general case. Let S be a transcendency basis of K over F . Then S is finite by Proposition 2.6. Furthermore, E is finitely generated over F ( S ) and K / F ( S )

is algebraic. By the algebraic case, K is finitely generated over F ( S ) , and hence over F , too. 2.10.

Proposition.

Let E I F be a purely transcendental eztension.

T h e n F is

algebraically closed in E . Proof. Let S be any transcendency bases of E I F such that E = E’(S). Since any element of E belongs to E = F ( S 1 ) for some finite subset S1 of S ,we may assume that S is finite, say S = ( 5 1 , . . . , s,}. We proceed by induction on n. Assume first that n = 1. Let X E F ( z 1 ) be algebraic over F . We may write X = f ( z l ) / g ( z l ) where , f and g are polynomials over F prime to each other. Then

This shows that if X were not in F , then

s1

is algebraic over F ( X ) (since the above

polynomial for s 1 over F ( X ) cannot vanish identically). But F ( X ) / F is algebraic, hence z1

is algebraic over F , a contradiction. Thus X is in F . Suppose now that the lemma holds for n

-

1. If X is an element of F ( z 1 , . . . , z n )

algebraic over F , it is algebraic over F ( z l , . . . , z n p l ) . By the first part, it should be in F ( z 1 , . . . ,zn-l), and hence in F (by induction hypothesis). 2.11. Proposition.

Let E I F be a field extension a n d let T be a set of elements

algebraically independent over E . T h e n E ( T ) / F ( T )is algebraic if and only iJ E I F is alge braie.

106

CHAPTER 3

Proof. It is clear that if E / F is algebraic, then so is E ( T ) / F ( T ) .Conversely, assume that E ( T ) / F ( T )is algebraic. Since the elements of T are algebraically independent over

F, T is a transcendency basis of E ( T ) / F . On the other hand, by looking a t the chain F

CE

E ( T ) , it follows t h a t T

U S,

where S is a transcendency basis of E / F , is also a

transcendency basis of E ( T ) / F . Thus S is empty, as required.

3. Sirnple transcendental extensions.

Our aim here is t o examine intermediate fields of the simple transcendental extension E / F . Namely, we show t h a t any such field properly containing F is also a simple transcendental extension. T h e above result is not valid for purely transcendental extensions E / F of transcendency degree r > 1. However, it is known (see Zariski(l958)) t h a t if F is algebraically closed, r = 2 and F

CK CE

is such that t r . d . ( K / F ) = 2 and E / K is separable, then

K / F is a purely transcendental extension. As an easy application of the lemma below, we also provide the isomorphism class of the group of all F-automorphisms of the simple transcendental extension F ( X ) / F . 3.1.

Lemma. Let F ( X ) / F be a simple transcendental extension, where F i s a n

arbitrary field, a n d let p be a n e l e m e n t of F ( X ) not in F . W r i t e p = f ( A ) g ( A ) - '

where

f ( X ) a n d g(X) are polynomials i n X with no c o m m o n f a c t o r of positive degree in A. If n = max(degf, degg), t h e n X is algebraic over F ( p ) a n d (F(X) : F ( p ) )= n.

Proof. We may write j ( X ) = (Yo where either a , f 0 or

Furthermore, a ,

-

+ ( Y I X -t . . . + (YX, ,

9, # 0.

g(X) = Po

+

PIX

The relation p = f(X)g(X)-'

pLpn = 0 since

a , or Pn # 0 and

@

6F .

+ . . . + pnXn

gives f ( A )

-

pg(X) = 0

and

Accordingly, X is a root of the

polynomial of degree n : C:=o(cy,- pLpz)X' with coefficients in F ( p ) . We are therefore left to verify that the given polynomial is irreducible. First, i t is obvious that p is transcendental over F , because X is algebraic over F ( p ) ; hence p algebraic over F implies X algebraic over F , contrary t o assumption. The ring

F [ p , X ]= F ( p ) [ X ]is the polynomial ring in two indeterminates p , X and hence the theorem on unique factorization into ireeducible elements holds in F l p , X ] (see Theorem 1.5.7).

107

S I M P L E TRANSCENDENTAL EXTENSIONS

Observe also t h a t a polynomial in F [ p ,XI of positive degree in X is irreducible in F ( @ ) ! X ]

if it is irreducible in F [ p ,X I . Now

is of degree 1 in p. Hence, if f ( p , X ) is reducible in F ( p ) [ X ]then , it has a factor h ( X ) of positive degree in X . But this implies that f ( X ) and g ( X ) are divisible by h ( X ) ,contrary to assumption.

a

We are now ready to prove. 3.2. Theorem. (Liiroth) Let F be a n arbitraryfield and let F ( X ) / F be a simple tran-

scendental extension. T h e n a n y field K with F

c K 2 F ( X ) is also a simple

transcendental

extension: K = F ( y ) , y transcendental. Proof. By hypothesis,

K

contains an element i j not in F , so by Lemma 3.1, E =

is algebraic over F ( P ) and hence is algebraic over

K 2 F ( P ) . Let

."(A)

the minimal polynomial

of X ovcr K h e

f ( X ) = X " -t y l x n - ' The

?I

have the form p z ( X ) u t ( X ) - l

+ . . . + yn

where p,, v, are polynomials in the transcendental

elenlent A. Multiplication of f ( X ) by a suitable polynomial in X will give a polynomial

f ( X , X) iri

=CO(X)X"

+

c1

(X)x"--'+ . . .

ic n ( X )

(3)

i.'[X,X:; which is primitive polynomial in X (in the sense that the greatest common

= c,(X)co(X)-' E K and not all of these are divisor of the c z ( X ) is 1). Observe also that 7%

in b' because X is transcendental over F . Thus one of the y's has the form y

= g(X)h(X)-'

where g(X),h(X) have no common divisor of positive degree in X and max(degg, degh) = rn

> 0. Owing to Lemma 3.1, g ( X ) - y h ( X ) is irreducible in F ( y ) [ X and ] ( E : F ( 7 ) )= n ~ .

Since K I, F ( y ) and ( E : K ) = n , clearly m this will prove t,hat K

=

2 n.

It will be demonstrated that rn = n and

F(7).

Since X is a root of g ( X ) - y h ( X ) and the coefficients of this polynomial are contained i n li, we have

9 ( X ) - yNX) = f ( X ) q ( X )in K!XI

We have y = g(X)h(X)--' and we can replace the coefficients of

f and

q by their rational

expressions in X and then multiply by a suitable polynomial in X to obtain a relation in

CHAPTER 3

108

FIX, XI of t h e form

where f ( X , X ) is the primitive polynoniial given in ( 3 ) . We may therefore deduce that k ( X ) is a factor of q ( X , X ) and so cancelling this we may assume the relation is

Now the degree in X of t h e left-hand side is a t most m. Because 7 = g ( X ) h ( X ) - ' with

( g ( X ) , h ( X ) ) = 1 and max(degg, degh) = m, the A-degree of f ( X , X ) is at least m. It therefore follows that it is exactly m and q ( X , X ) = q ( X ) E F [ X ] .Hence the right-hand side of (5) is primitive as a polynomial in X . This holds also for the left-hand side. By symmetry, the left-hand side is primitive as a polynomial in X also, and this ensures that

q ( X ) = q is a nonzero element of F . Then (5) implies t h a t the X-degree and A-degree of

f ( X , X ) are the same. Hence m = n and the result is established. We close by describing the automorphism group of a given simple transcendental extension F ( X ) / F . Let GL,(F) denote the group of all n x n-nonsingular matrices with entries in F . The centre of GL,(F) consists of all scalar matrices and the corresponding factor group, denoted by PGL,(F), is called the projective linear group. 3.3. P r o p o s i t i o n . Let F ( X ) / F be a simple transcendental eztension and let G be the group of all F-automorphisms of F ( X )

of F ( X ) .

(ii) The map G L Z ( F )

G given i n

(2)

is a surjective group homomorphism whose kernel

consists of all scalar matrices. I n particular.

P r o o f . (i) The above map certainly determines a unique F-homomorphism of F ( X ) into itself (namely, f ( X ) g ( X ) - ' polynomials in A). Since aX

H

f(p)g(p)-', where p = (ax +,B)(yX

+ b)-'

and f , g are

+ P,7X + 6 are both of degree 5 1 and a t least one of these

degrees is 1, it follows from Lemma 3.1, that ( F ( X ) : F ( p ) ) = 1. Since F ( p ) is the image

109

SEPARABLE EXTENSIONS

of F(X), it follows t h a t the given homomorphism is surjective. But any F-homomorphism

F(X)

---f

F(X) has trivial kernel and so (i) is established.

(ii) One verifies directly t h a t the mapping of the nonsingular matrix into the corresponding

automorphism is a group homomorphism. The kernel is the set of all such t h a t (ax + p)(7X + 6)-l = X or -yX2

+ (6 - a : ) X - p = 0.

I:

[;

E GL2(F)

The latter holds if and only

if y = p = 0 and a: = 6. Thus the kernel consists of all scalar matrices. To prove surjectivity, consider any F-automorphisrn of F(X) and denote by p the

image of A. We may write p = f(X)g(X)-', where f ( X ) and g(X) are polynomials in X with no common factor of positive degree in A. Now f ( X ) = F ( p ) and so (F(X): F ( p ) ) = 1. Hence, by Lemma 3.1, max(degf, degg) = 1. Then we have

where a

#

0 or 7 # 0 and aX

+ p, -yX + 6 have no common factor of positive degree.

easy to see that these conditions imply a6

-

It is

Py # 0 and so the proof is complete.

4. Separable extensions.

Let F be a field of characteristic p

2 0 and

let L be a n algebraically closed field containing

F . If p > 0 and n is a positive integer then the elements z of L such that form a field containing F . As usual we denote this field by

FI'.."

=

FP-".

If p

7

zP''

E

F

0, then we p u t

F . The union of all fields FP-" ( n = 1 , 2 , . . .) with be denoted by FP--. Given a

Geld extension E J F , we always choose as L an algebraic closure of E . Let E / F be a field extension. We say that E I F is separable if E and

disjoint over F . If p = 0 or, more generally, if F is perfect, then FP

-1

-- i

F P

=

are linearl)

F and hence

E / F is separable. It follows, from Theorem 2.3.6, that the general notion of separability is equivalent to the usual notion if E / F is an algebraic extension.

A transcendency basis S of E / F such that E is separable over F ( S ) is called a separating transcendency basis. If E / F has such a basis, then E J F is called separubly generated.

Our aim is twofold.

First, t o prove that a separably generated extension

is necessarily separable and that the converse is true for finitely generated extensions.

Second, w e exhibit a number of important properties of separable extensions.

CHAPTER 3

110

4.1. Lemma. Let F

CK

E be a chain of fields. If a subset S of E is algebraically

independent over K , t h e n K and F ( S ) are linearly disjoint over F .

Proof. It is clear that K and F ( S ) are linearly disjoint over F if and only if K and F [ S ] are linearly disjoint over F . Our assumption on S ensures that the set of all monomials s;' . . . sk,

E S, is an F-basis of F [ S ]and that these monomials are also

sk

K-independent. Now apply Proposition 2.3.4(v). m 4.2. Corollary. A n y purely transcendental extension is separable.

Proof. Let E I F be a purely transcendental extension. Then E = F ( S ) for some algebraically independent subset S of E over F . Since S is obviously algebraically independent over FP-', it follows from Lemma 4 . 1 t h a t E and Fp-' are linearly disjoint over

F . rn 4.3. Lemma. Let F

C K CE

be a c h a i n of fields.

(i) If E I F i s separable, t h e n so is K I F (ii) If both K I F a n d E I K are separable, t h e n so is l?/F (iii) If K / F is separably generated and E I K is separubly algebraic, t h e n E I F is separably

generated.

Proof. (i) This is clear, since the linear disjointness of E and FP disjointness of K and FP (ii)Given a set of

--I

-1

implies the linear

.

S of elements of FP-' which are linearly independent over F , the elements

S are also linearly independent over K since K I F is separable. Hence the elements of

S are also linearly independent over E since E / K is separable and since S C KP-'

. Thus

E and FP-' are linearly disjoint over F . (iii) Let

S be a transcendency basis of K / F . Since E , K is algebraic, S is also a transcen-

dency basis of E I F . Consider the chain of fields E 2 K 2 F ( S ) . Since X I F is separably generated, K / F ( S ) is separably algebraic. Hence E / F ( S ) is separably algebraic, as required.

=

R e m a r k . Let F

C

K

E be a chain of fields. 'rhen the separability of E / F need

not imply that of E I K . Indeed, let E = F ( X ) and K = F ( X P ) . Then E I F is separable by Corollary 4.2. However, E

# K

and E / K is purl-ly inseparable, hence E I K is not

separable. Note also that if K = K p ( F ) (which always holds whenever K / F is separable

SEPARABLE EXTENSIONS

111

algebraic), then the separability of E I F implies that of E I K (see Proposition 4.28). We are now ready to prove the following classical result. 4.4. Theorem. (MacLane’s Criterion). Let E I F be a field eztension.

( i ) If E I F is separably generated, t h e n E I F is separable. ( i i ) If E l F is separable a n d finitely generated, t h e n E / F

IS

separably generated.

Proof. (i) Assume that E I F is separably generated and choose a transcendency basis

S of E I F . By Corollary 4.2, the fields F ( S ) and FP let u l r . ,. , u, be elements of FP

--I

-1

are linearly disjoint over F . Now

are linearly independent over F . Then these elements

are linearly independent over F ( S ) . Since, by hypothesis, E / F ( S ) is a separable algebraic extension, it follows from Proposition 2.3.4 and Theorcm 2.3.6 that

u1,.

. . ,u , are also

linearly independent over E . Hence, by Proposition 2.3.4, E and FP-’ are linearly disjoint over F.

( i i ) We proceed by induction on the minimal number m of generators for E I F , the result being trivial for m = 0. Assume that m

> 0 and

that the result is known for all separable

field extensions with less than m generators. Write E = F ( A 1 , . . . ,A,,).

If

. . ,A,

Al,A2,.

are algebraically independent over F , then we are done. We may therefore assume that they are dependent. Let n be the unique integer such that all subsets of { A * , . . . A,},

of size n

~

1 are

algebraically independent over F but some subset of size n is dependcnt. We may ass i m e that the A:s are so labeled that

A x , . . . , A,

are dependent. Choose j ( X l , .. . , X,) E

F [ X 1 , .. . , X,j to he a nonzero polynomial of smallest total degree with M‘e claim that the exponents appearing in f ( X , , . . . ,X,)

f(A1,.

. .A,)

=

0.

cannot all he divisible hy p .

Indeed otherwise.

then g(A1,. . . A,)

I .

over

fi’p-‘.

=- 0, so the monomials in XI,.

. . ,A,

occuring in g are linearly dependenl

The assumption on linear disjointness of E and FP-‘ therefore implies that the

same monomials are F-dependent. This would yield an algebraic relation for XI,. . . , A, smaller total degree, a contradiction. Thus say some exponent for Write f ( X 1 , . . . , X , ) =

c,Xigj(Xz, X s , .

of

XIis not divisible by p .

. . , X , ) . Then, by assumption, there exists

some j’ say .j = j , with p +jo and y,, ( X 2 ,X B , .. . , X,)

#

0. Hence g,,, ( A z , .

..

,A,)

;L 0 sincc

112

CHAPTER 3

by definition of n , the elements Az,

A3,

. . , , An

are algebraically independent over F . This

ensures t h a t

g(x)= ~ x i g j ( A Z , A 3 , . . . ~ A n )

F(X2,A3,...,An)[X]

3

is a nonzero polynomial over the field

g ( X ) need not be irreducible over

F.

F

= F(X2,. . . , A n ) having

A1

as a root. Note that

On the other hand, by the property of j o , we see

that X 1 is not a root of the derivative polynomial

since otherwise

would yield a nontrivial dependence relation for

A1,

Az,

. .., A n

degree. We conclude therefore that X 1 is separably algebraic over Finally, if

F

=

F.

F(A2, As,. . . ,Am), then by Lemma 4.3(i) and induction Z / F is sepa-

rably generated. Moreover, that

over F of smaller total

E / E is separably

2 F; thus

A1

is separably algebraic over

E.

This implies

algebraic and the result follows by virtue of Lemma 4.3(iii). w

4.5. Corollary. Let E = F(A1,. . . A,),

be separable over F . T h e n the set { A l , . . . , A,n}

contains a separating transcendency basis of E I F .

Proof. We proceed by induction on the minimal number m of generators of E I F , the result being trivial for m = 1. By the final paragraph in the proof of Theorem 4.4, E = F(X2,. . .A,), is separably generated over F and EIE is separably algebraic. Let S

{ X Z , . . . A,},

be a separating transcendency basis of

basis of E I F since

E I E is algebraic.

EIF.

Then S is a transcendency

Finally, since E / F ( S ) is separable, the result follows.

w 4.6. Corollary. (F.K. Schmidt). Let E I F be a finitely generated field eztension and

let F be perfect. Then E I F is separably generated.

Proof. Since F is perfect, E I F is obviously separable. The desired conclusion is therefore a consequence of Theorem 4.4. rn

113

SEPARABLE EXTENSIONS

Our next aim is t o characterise separability via tensor products. We begin by recording the following preliminary results. 4.7. Lemma. Let F & E

CK

be a chain

of

fields a n d let X E K be transcendental

ouer E . I j E ( X ) contains a n element which is algebraic (separable algebraic) over F ( X ) and

not contained in F ( X ) , t h e n E contains a n element which is algebraic (separable algebraic) over F and not contained in F .

Proof. Let p be an element of E(X) which is algebraic over F ( X ) and let

xn+ a l X n - '

+ ... + a,

E

F(X)jX]

be its minimal polynomial. We can write a i = bi/ci where bi, c E F [ X ] (e.g. c can be taken to be the product of the denominators of the a i ) . Then a: = cp is algebraic over

F ( X ) with minimal polynomial

X" + b l X " - '

+ . . . + b,

If a = /3/y with p,7 € E ( X ) being relatively prime polynomials, then (1) implies

If y is of positive degree, then 7 has an irreducible factor and ( 2 ) shows that this is a factor of

p", hence

of

/3, contrary to the assumption on p and

y. Thus 7 is a unit a n d so

a E E [ X ] . We may therefore write a =

+ ( Y l X + . . . + a,X,

( a ;E E )

and we claim that the result will follow provided we show that

ao, a1,. . . ,a:,

are algebraic over F

(3)

Indeed, assume t h a t (3) holds. Choose p so that p !$ F(X). Then a: !$ F ( X ) and hence not every a, is in F . Thus there exists ai E E which is algebraic over F and which is not contained in F . Assume further that p is separable over F ( X ) . Then a: is separable algebraic over F ( X ) . Moreover, the field F(cuo,al,.. . , a m ) contains a separable algebraic element not in F . Indeed, otherwise c h a r F

=

p > 0 and we have a: E F for some

Then we have CYP' E F ( X ) , contrary t o the separability of our claim.

(Y

s

'. I .

over F ( X ) . T h i s substantiates

114

CHAPTER 3

To prove ( 3 ) , let 6 E F and consider the E-homomorphism E(X) + E which sends X to 6. Such a homomorphism exists since X is transcendental. Write a = .(A),

b; = b ; ( X ) E

FIX] so t h a t by (1). a(X)"

+ b l ( X ) a ( X ) " - ' + . . . + b,(X)

=0

+ . . . + b,(6)

=0

Then we have the relation

a(6)" + b l ( b ) a ( 6 ) " - '

Because the b ; ( 6 ) E F , this shows t h a t a ( 6 ) is algebraic over F . Suppose first that F

+ 1 distinct elements 6 1 , 6 ~ ,... ,6,+1. Then a ( & ) = xi"=,aj6; is algebraic for k = 1,2,. . . , m + 1. Because the Vandermonde determinant det(6:) # 0, these

contains m over F

equations for the a, have a unique solution which is given by the usual determinant formu-

.. a ,

are algebraic over F . If F is finite of characteristic

las. Thus the elements a,,

a1,.

p , then choose r so that p'

> m, and let E be a splitting field over E of XP'

F

by m

the subfield of

+ 1 distinct 6 k .

-

of elements which are algebraic over F. Then clearly

We now make the argument with these elements using

E

1. Denote

F

and

contains instead

of E and F , respectively. Then we conclude as before that the a j are algebraic over

F,

hence over F . This proves (3) and the result follows. w 4.8. Lemma, Let K I F be a purely transcendental e z t e n s i o n , let E I F be a n arbitrary

eztension, a n d let S be a transcendency basis of

KIF.

(i) K @ F E is a n integral d o m a i n whose quotient field isomorphic to E(S), where E(S) is purely transcendental over E with S as a transcendency basis. (ii) If F is algebraically closed in E, t h e n K is algebraically closed in E(S) (iii)

If every separable algebraic element of E over F is contained in F , t h e n every separable algebraic element of E ( S ) over K is contained in K . Proof. (i) We identify K and E with the subalgebras K

respectively.

@I

1 and 1 @I E of K

@I

E,

Because S is an algebraically independent set, the set M of all distinct

monomials s ; ' s y 2 . , . s L k , n

2

0 , sJ E S is an F-basis for F [ S ] . Since F [ S ]and E are

linearly disjoint, the set M is E-independent. Thus S is algebraicaliy independent in E [ S ] .

It follows from Lemma 2.l(ii) and Corollary 1.4.5 t h a t E [ S ]is an integral domain which is isomorphic t o the polynomial ring over E with S as the set of indeterminates. Thus E(S) is

the quotient field of E [ S ] .Furthermore, since S is an algebraically independent set over

E , E ( S ) / E is purely transcendental with S as a transcendency basis.

SEPARABLE E X T E N S I O N S

115

We now show that K @ F Ecan be identified with the subalgebra A of E ( S ) consisting of elements of the form zy-' with z E E [ S ] ,0

#

y E F [ S ] .Hence if we identify K

@F

f<

with A and observe t h a t E ( S ) is also the quotient field of A since A 2 E [ S ] the , assertion ( i ) will follow. The natural embedding E [ S ]4 K @F E can be extended, by Proposition 1.2.1, to a unique injective homomorphism of A into K for all 0

@F

# y E F [ S ] . To prove surjectivity, let

E since the elements y-l exist in k ' ( S ) , z

be any element of K @F E and write

z = C z,ylrx , E X , yI E E . We can write x , = t,y-' z =

(c

t,y;)y-'

where t , , y E F [ S ] .Then

where z E E [ S ]

= zy-'

It follows that z is in the image of the above homomorphism and hence A is isomorphic to K

@p

E , as required.

(ii) and (iii). By (i), we know that K = F ( S ) and F ( S ) / F , E ( S ) / E are purely transceridental extensions. It suffices t o show t h a t , if E ( S ) contains a n element which is algebraic (separable algebraic) over F ( S ) which is not contained in F ( S ) , then E contains an elcmcnt which is algebraic (separable algebraic) over F not contained in F . It is obvious t h a t if

an element of the type indicated exists in E ( S ) ,then it exists in E ( S ' ) for a finite subset

S' of S. Thus we may take S finite and a n induction argument permits us to assume that

I S 1 = 1. The desired conclusion is therefore a consequence of Lemma 4.7. rn Let E / F be a n algebraic field extension. Recall that E I F is said to be purely inseparable if t h e only elements of E which are separable over F are the elements of F . 4.9.

Lemma. Let K I F and E I F be two f i e l d extensions. Assume that at lecst one

of the following conditions holds:

( i ) K / F is purely inseparable

(ii) K / F is algebraic and every separable algebraic element of E over F is in F . Then t h e elements of K

@F

E are either units o r nilpotents.

Proof. (i) If c h a r F = 0, then K = F and K that c h a r F = p > 0. Fix z = 2 . 2 . 6 ( ; ) ,y;'

=

@F

I:= y., @ z, E K @,v

A, E F for some e,

2 0,

15 z

E

Z

E . We may therefore assume

E , yz E K ,z, E E . By Proposition

5 n. Then, setting

e = max{e,}, we ha\e

CHAPTER 3

116

Hence, either xPc = 0 or xP' has a n inverse in 1 @ E . In the latter case x is a unit of

K@FE (ii) Let L be the separable closure of F in K . Then, by Proposition 2.2.30, K / L is purely

inseparable. Furthermore, one immediately verifies t h a t

Hence, by (i), we need only verify t h a t L @ F Eis afield. If z E L @ F E ,then z = with xi E L , yi

c

c:=,z;@yi

E . Since L / F is algebraic, the x; generate a finite extension and

hence we may assume that L / F is finite. Then the separability of L / F implies that

L = F(X)

F [ X ] / ( f ( X )where ) f ( X ) is separable and irreducible in F [ X ] .By Lemma

2.3.7, L @ FE

E

E [ X ] / ( f ( X ) Assume ). that f ( X ) = g ( X ) h ( X )where g ( X ) , h ( X )E E [ X ] .

The roots of g ( X ) , h ( X )are separable algebraic over F , hence so are the coefficients of

g ( X ) ,h ( X ) . Thus g ( X ) ,h ( X ) E F I X ] and f ( X ) is irreducible in E [ X ] .This shows that

L

@IF

E is a field, as required. w

2

1, the

Proof. The extension FP-"/F is purely inseparable by Proposition 2.2.27.

Now

4.10. Corollary. Let E / F be a n arbitrary field eztension. T h e n , f o r all n

elements of FP-"

@F

E are either units or nilpotents.

apply Lemma 4.9 (i). w We have now accumulated all the information necessary to provide the following characterizations of separable extensions.

Theorem.

4.11.

Let E / F be a n arbitrary field eztension.

T h e n t h e following

conditions are equivalent:

(i) E / F

is

separable

( i i ) For a n y field e z t e n s i o n K / F , K

@F

E has no nonzero nilpotent elements

(iii) FP-.' g F E has no nonzero nilpotent e l e m e n t s (iv) Fp-' @ F E i s a field (v) FP-' @ F

(vi) Fp-"

@p

E is a n integral d o m a i n E is a field f o r all n

2

1

(vii) FP-" @ F E is a n integral d o m a i n f o r all n

21

Proof. (i)+(ii): Assume t h a t E / F is separable. We may clearly assume that E / F is

SEPARABLE E X T E N S I O N S

117

finitely generated, in which case E / F is separably generated by virtue of Theorem 4 , 4 ( i i ) . Choose a transcendency basis S such t h a t E is separable algebraic over F ( S ) . Note that

B y Lemma 4.8(i), F ( S ) @ F K is an integral domain which can be embedded in the field

K ( S ) . Thus K

@F

E can be embedded in K ( S ) @ F ( s ) E . Since E / F ( S ) is separable

algebraic, K ( S ) B F E has no nonzero nilpotent elements, by virtue of Theorem 2.3.10. lIence K c g p E has no nonzero nilpotent elements. (ii)-A(iii): Obvious

(iii)+(i): This was established in the course of the proof of Theorem 2.3.10. (iii)+(iv): This is a direct consequence of Corollary 4.10. (iv)=+(v):Obvious (v)+(vi): Since (v)+(iii)+(ii), FP-" Corollary 4.10, FP-" (vi)

@F

@F

E has no nonzero nilpotent elements. Hence, by

E is a field.

+(vii): Obvious

(vii)+(iii): Obvious. I 4.12. Corollary. Let K I F be a purely inseparable field e x t e n s i o n a n d let

separable field extension. T h e n K

@F

Proof. By Theorem 4.11(ii), K

E/F be a

E is a field. @F

E has no nonzero nilpotent elements. On the

other hand, by Lemma 4.9(i), the elements of K

@F

E are either units or nilpotents. Thus

li @ J E~ must be a field. rn As a preliminary t o the next result, we now record 4.13. Lemma. Let K I F and E I F be field extensions.

element of

If every separable algebraic

E over F belongs to F , t h e n every zero divisor of K

@F

E is nilpotent.

Proof. Let S be a transcendency basis of K over F . We have

where, by Lemma 4.8(i), F ( S ) mF E can be embedded in the field E ( S ) . Hence it suffices to show that every zero divisor of K

@F(s)

E ( S ) is nilpotent. By Lemma 4.8(iii), every

separable algebraic element of E ( S ) over F ( S ) is contained in F ( S ) . Since K / F ( S ) is algebraic, it follows from Lemma ?.9(ii) that K

%F(SJ

E ( S ) has the desired property. rn

CHAPTER 3

118

A field E is said t o be a regular extension of its subfield F if -

F

where

F

@JF E

is a n integral domain

is a n algebraic closure of F .

4.14. Theorem. Let E / F be a field extension. T h e n t h e following conditions are

equivalent: (i) E / F i s regular (ii) E I F is separable and F is algebraicaily closed in E (iii) K

@F

E i s a n integral d o m a i n f o r every field extension K / F

Proof. (i)+(ii): Since FP-I

C F , we have FP-' @ F E& F @ FE .

Hence, by Theorem

4.11, E / F is separable. Assume t h a t X E E is algebraic over F . Let f ( X ) be the minima!

polynomial of X over F . Then

is an integral domain. Hence f ( X ) is irreducible in E [ X ]and therefore f ( X ) = X

-

A. It

follows that X E F , as required. This is a direct consequence of Lemma 4.13 and Theorem 4.11.

(ii)+(iii):

(iii)+(i): Obvious. 4.15. Corollary.

EIF, K

@F

If F i s algebraically closed, t h e n f o r a n y field extensions K I F and

E i s a n integral d o m a i n .

Proof. Since

F

=

F,

F @ FE

2

E and so E I F is regular. Now apply Theorem

4.14(iii). I Our next aim is to provide a condition under which K implies K

@F

@F

E is a n integral domain

E is a field.

Let R be a commutative ring. Then the set S of all elements of R which are not zero divisors is a multiplicative subset of R . The corresponding quotient ring

Rs is called the

total quotient ring of R. Thus each nonzero divisor of R is a unit of Rs. 4.16. Proposition. If a commutative ring R is a finitely generated module over its

subfield F , t h e n nonzero divisors of R are u n i t s of R .

SEPARABLE E X T E N S I O N S

119

Proof. Assume t h a t r E R is not a zero divisor. Since R is a finite-dimensional vector space over F , there is a positive integer n such that l , r , . , . , r n - ' are F-linearly independent and r n is linearly dependent on them. Then

Furt.hermore, Xo

# 0 since otherwise r(rn-'

~

XI

-

. . . - Xn-lrn-')

= 0, hence

a contradiction. Now, applying ( 4 ) , it follows that in the total quotient ring of R , we have a relation

rn--l = X o r - ' + X1 Ilecause F is a field and

X0

#

+ x 2 r + . ..+ ~

~

-

~

r

~

-

~

0, we have

as required. rn

4.17. Corollary. Let K J F be a n algebraic field e z t e n s i o n and let E I F be a n arbitrary field eztension. T h e n every nonzero divisor in K is a n integral d o m a i n , t h e n K Proof. Let

1,

=

@F

E is a u n i t . I n particular, if K € 3 E~

E is a field.

X = C:=,a , @ b,, a ,

F ( a l , . . . , a n ) . Then X E L

@F

@fi-

E K , b, t E , be a nonzero divisor in K

B , X is a nonzero divisor in L

@F

@F

E. P u t

E and, since L / F is

a finite extension, 1,@F E is a finitely generated E-module. Hence, applying Proposition

4.16, it follows that

X

is a unit of L

@F

E and hence of K

@F

E.

We close by recording some useful properties of separable extensions required for subsequent investigations. 4.18. P m p o s i t i o n .

Let E / F be a field e z t e n s i o n a n d let T be a set of elements

nlyebraically independent over E . T h e n ( i ) E ( T ) / F ( T )i s separable if and o n l y if E / P i s separable ( i i ) E ( T ) j F ( T )is separable algebraic if and only if E / F is separable algebraic.

Proof. ( i ) Assume that E ( T ) / E ' ( T )is separable. Since F ( T ) / F is separable (Corollary 4 . 2 ) , then by looking a t the chain F

c F ( T ) C: E ( T ) ,it follows from Lemma 4.3(ii),

120

CHAPTER 3

that E ( T ) / F is separable. Hence, applying Lemma 4.3(i) t o the chain F

C E C E ( T ) ,we

deduce that E I F is separable. Conversely, assume that E I F is separable. Let {z,} is separable, {z,}

is linearly independent over FP

--I

be a basis of E over F . Since E I F

, hence

{z;}

is linearly independent

over F. But, by Lemma 4.1, E and F ( T ) are linearly disjoint over F , hence {z:}

is

linearly independent over F ( T ) . Therefore {z,} is linearly independent over F ( T ) P - ' . Since E =

C , Fz,,

we have E [ T ]=

c, F[T]z,. Setting R = C , F ( T ) z , it follow that

R F is a subring of E ( T ) whose quotient field is E ( T ) . Furthermore, R is a vector space over the field F ( T ) with basis {z,}.

Applying Lemma 2.3.12, we deduce that E ( T ) and

F(T)P-' are linearly disjoint over F ( T ) . This means that E ( T ) / F ( T )is separable, as required.

(ii) This is a direct consequence of (i) aiid Proposition 2.11. 4.19. Proposition. Let E / F and KIF be field extensions contained in a c o m m o n

field. If E I F is separable and K I F is purely inseparable, t h e n

(i) E and K are linearly disjoint over F (ii) E K

E @F K

Proof. Consider the natural homomorphism E @F K

--t

E [ K ] ,a @ b

++

ab. This

is surjective, since the image contains E and K . On the other hand, by Corollary 4.12,

E B F K is a field. Hence E @ F K

+ E [K ]

is a n isomorphism (which, by definition, means

(i)) and

E

@F

K

E

E [ K ]= E K ,

as required. 4.20. Corollary. Let E I F be a n algebraic field extension and let K be a n intermedi-

ate field such t h a t K I F is purely inseparable a n d E I K is separable. If F, is the separable closure of F in E , t h e n

E = E,K

F,

@F

K

Proof. Since K is obviously the pure inseparable closure of F in E , we have E = F s K by Proposition 2.2.32. Now apply Proposition 4.19(ii). 4.21. Proposition. The following conditions concerning a field e z t e n s i o n EIF are

SEPARABLE E X T E N S I O N S

121

equivalent:

(i) E I F is separable (ii) K / F i s separable f o r a n y f i n i t e l y generated subeztension K I F of E I F

( i i i ) K I F i s separably generated f o r a n y f i n i t e l y generated subeztension K I F of E / F (iv) E and FP--

are linearly disjoint over F

(v) E a n d FP-" are linearly disjoint over F for s o m e n

21

Proof. Note that E and FP-' are linearly disjoint, over F if and only if K and FF-'

are linearly disjoint over F , for any finitely generated subextension K / F of E / F . This shows that (i), (ii), and (iii) are equivalent, by applying Theorem 4.4. Since FP

-m

/ F is

purcly inseparable, it follows from Proposition 4.19(i) that (i) implies (iv). Since FP-" is a subfield of F P - - , (iv) implies (v). Since FP-' is a subfield of FF-", ( v ) implies ( i ) . Let E / F and K I F be two field extensions contained in a common field L . We say that E and K are free ouer F if the following condition is satisfied: for any finite subsets

. . , zn} and { y l , . . . ,y,}

( ~ 1 . .

of E and K , respectively, such that the elements of each set

arc algebraically independent over F , the elements

are also algebraically independent over F . It is a n immediate consequence of the definition

that if E I F or K / F is algebraic, then 4.22. Lemma.

E and K are free over F .

T h e following conditions are equivalent:

( i ) E a n d K are free over F

( i i ) For a n y transcendency bases

a n d the e l e m e n t s o f

S 1

a n d Sz of E I F a n d K J F , respectively, S1

S2

=

0

a n d S2 of E / F a n d KIE', respectively, S1 "1 S2

=

0

S1 U S2 are algebraically independent

(iii) For a n y transcendency bases

S 1

ocer

F.

a n d Si U Sz i s a transcendency basis of E K I F (iv) I f the e l e m e n t s of a n y given subset of E are algebraically i n d e p e n d e n t over F , t h e n

t h e y are algebraically independent over K . (v) There ezists a pair of transcendency bases S1 a n d

S2 of E / F a n d K I F , respectively,

s u c h that S1 n S2 = 0 and t h e e l e m e n t s of S1 u S2 are algebraically independent ouer

F. Proof. Since any algebraically independent set can be embedded in a transcendency

122

CHAPTER 3

basis, it is clear that (i) is equivalent t o (ii). It is obvious t h a t (iii) implies (ii) and (ii) implies (v). To prove that (ii)=+ (iii), note that each element of E is algebraic over F ( S l ) , hence over F ( S 1U S2). Similarly, each element of K is algebraic over F(S1 U Sz). Hence

E K / F ( S 1 U S2) is algebraic and therefore

5'1 U S2

is a transcendency basis of E K I F ,

proving (iii). Condition (iv) is obviously equivalent t o the requirement t h a t any transcendency basis

S1 of E / F is algebraically independent over K . As we have seen in the proof of Proposition 2.7, any such S1 contains a transcendency basis of E K / K . Thus (iv) is equivalent to the condition t h a t any transcendency basis S1 of E / F is a transcendency basis of E K I K . On the other hand, we know from Proposition 2.6, t h a t if S1 and

S2

are transcendency bases

of E K I K and K I F , respectively, then S1 U S2 is a transcendency basis of E K I F . This shows t h a t (iv) is equivalent to (iii). Finally assume that (v) holds. Let C and D be finite algebraically independent over F subsets of E and K , respectively. We can find a finite subset A of S1 such t h a t each c E C is algebraic over F ( A ) . By the proof of (ii) *(iii), S1 US2 is a transcendency basis of E K I F . Hence, by the previous paragraph, the elements of S1 are algebraically independent over

K . Therefore, A is algebraically independent over K . Hence A is algebraically independent over F ( D ) C K . It follows, from Proposition 2.6, applied to the chain

that the transcendency degree of F(AU C U D ) over F is a + d , where a = IAl and d = IDI. Since the transcendency degree of F(AU C) over F is a and C is algebraically independent over F , the transcendency degree of F ( A u C ) over F ( C ) is a - c, where c = ICI. It follows that the transcendency degree of F ( A U C

U

D ) over F ( C

U

D ) does not exceed a

-

c.

This and the formula for the transcendency degree of F ( A U C I J D ) over F imply that the transcendency degree of F ( C U D ) over F is a t least ( a

+ d ) - ( a - c ) = d + c.

It follows

that C n D = 0 and C U D is algebraically independent, as required. 4.23.

Lemma. Let E / F and K I F be field eztensions. l f E a n d K are linearly

disjoint over F , t h e n E and K are free over F .

Proof. Let S be a subset of E such that S is algebraically independent over F. Then all monomials of the form

S:'

. ..sik

(st E S, ni

2

0) are linearly independent

over F , hence over K (since E and K are linearly disjoint over F ) . This shows that S is algebraically independent over K . Hence, by Lemma 4.22(iv), E and K are free over F . m

SEPARABLE EXTENSIONS

123

4.24. Proposition. Let E I F and K I F be field eztensions. If E I F is separable anti

E a n d K are free over F , t h e n E K l K i s separable.

Proof. An element of E K has an expression in terms of a finite number of elements of E and K . Therefore any finitely generated subfield of E K containing K is contained

in E I K where El is a subfield of E finitely generated over F. By Proposition 4.21(ii), we may assume t h a t E is finitely generated over F , in which case E / F is separably generated, by Theorem 4.4(ii). Let T be a transcendency basis of E I F such that E / F ( T ) is separable

algebraic. Since E and K are free over F , it follows from Lemma 4.22(iv) that 1 ‘ is algebraically independent over K . On the other hand, E K / K ( T ) is algebraic, hence 7’ is a transcendency basis of E K I K . Since every element of E is separable algebraic over F ( T ) . the elements of E are separable algebraic over K ( T ) 2 F ( T ) . Hence E K I K is separably generated and therefore, by Theorem 4.4(i), is separable.

4.25. Proposition. Let E / F and K I F be t w o separable field extensions such that

F; and K are free ouer F . T h e n E K I F i s separable. Proof. By Proposition 4.24, E K I K is separable, On the other hand, by hypothesis. K / F is also separable. Hence, by Lemma 4.3(ii), E K / F is separable.

H

4.26. Proposition. Let E I F and K I F be field eztensions s u c h t h a t E and K are linearly disjoint over F . T h e n E I F is separable if a n d o n l y if E K I K is separable.

Proof. By Lemma 4.23, E and K are free over F . Hence, if E I F is separable, then by Proposition 4.24, E K I K is also separable. Conversely, assume that E I F is not separablc, i.e. t h a t E and FP

-1

are not linearly disjoint over F . Then E and KFP

disjoint over F . Hence, by Proposition 2.3.13, applied t o the chains F

F

C E,EK

-L

are not linearly

5 X C KFP-’

and

and KFP-’ are not linearly disjoint over K . But KP-‘ contains KFP-I, hencc

2SK amd KP-’are not linearly disjoint over K . This shows that E K I K is not separable. as required. w

4.27. Corollary. Let E / F and K I F be field eztensions such t h a t E J F is separable and K I F is purely inseparable. T h e n E K I K is separable.

Proof. By Proposition 4.19(i), E and K are linearly disjoint over P’. Now apply Proposition 4.26. m

124

CHAPTER 3

4.28. Proposition. Let

F CK

E be a chain of fields such that EIF is separable

and K = K p ( F ) . T h e n E / K i s separable.

Proof. By hypothesis, E and FP-l are linearly disjoint over F . Hence, applying Proposition 2.3.13 (with L = FP-'), it follows that K ( F p - I ) and E are linearly disjoint -1

over K . Since K ( F P ) = K p ( F ) p - ' = KP-', the result follows.

51,.

A n extension E / F is separable if and only if for a n y set

Proposition.

4.29.

. . , zn o f F-linearly independent elements of

E, t h e elements

zy,

. . . ,xP, are F-linearly

independent. Proof. Assume first that

XI,.

. . ,x, are F-linearly independent elements of E , and

C Aix;

let E / F be separable. Suppose that Choose y; E

so that

-1

F P

C;=, y;z;

such that yp = A;,

= 0. Since

E and

1

= O? where the

XI,.

.., A n

are elements of F.

5 i 5 n. Then

FP-'

are linearly disjoint, we deduce that each y; = 0,

hence each A; = 0. Thus xy, . . . , xP, are F-linearly independent. Conversely, assume that for any set

51,.

. . ,xn of F-linearly independent elements of E , the elements x;, . . . ,xx are

F-linearly independent. The latter implies that

. . ,x,

XI,.

are FP

--I

-linearly independent,

hence E I F is separable. w 4.30. Corollary. Let

elements x with

xp =

E/F be a field eztension and let E

@F

FP-' have n o nozero

0. T h e n E I F i s separable.

Proof. Let u l , . . . ,u, be an F-linearly independent subset of E , and suppose that X I , . . . A,

are elements of F such that

,= 1

Then, for t , = A?-' E F P - ' , we have n

n

t= 1

,= 1

125

W E I L ' S ORDER OF I N S E P A R A B I L I T Y

Hence

ui

(8 t i

= 0 and so each t ; = 0. Thus each A, = 0, which shows that uy, , . . ,UP, is

an F-linearly independent set. Now apply Proposition 4.29. H 4.31. Proposition. Let E be a field, let G be a group oj field a u t o m o r p h i s m s o j E

and let E G be t h e fized subfield of E , i.e. E G = { A E EIu(A) = X for all u E G } . T h e n E / E C is separable. Proof. P u t F = E G . By Corollary 4.30, it suffices to show that E nonzero elements x with X P = 0. Assume by way of contradiction t h a t 0 is such that

zp

--I

@F

FP

#zE E(

has no 8 F"' ~

= 0. Write

We suppose that x has been so chosen t h a t n is as small as possible. It is clear that n > 1 and we may harmlessly assume that

z1

= 1. Then

Now let G act by FP-I-algebra automorphisms on E u

C ~F F P - ' ,via

the factor E . Given

t G , we have n.

ik2

so that. n

Because (z- u ( z ) ) p = 0, it follows from the minimality of n that z = u ( z ) . Also it is clear from the minimality of n that the yE 's are linearly independent over F . Invoking the fact

t h a t z = o(z) for all u E G, it therefore follows that each

2,

belongs t o F . This rnakrs

n = 1, and we have a contradiction. rn

5 . Weil's order of inseparability.

In what follows, p denotes the characteristic of the given field F . If p = 0 , then all powers p" of p are to be replaced by 1. All the fields considered below are assumed t o be subfields

126

CHAPTER 3

of a n algebraically closed field L. If E , F are subfields of L , then as usual E F denotes the

composite of these fields (i.e. the least subfield of L containing both E and F ) . Let E / F be a finitely generated field extension. Our aim is to introduce a numerical invariant of E / F , called the (Weil’s) order of inseparability of E / F , which generalizes the concept of the inseparable degree ( E : F ) i in case E / F is a finite extension. We shall also exhibit some properties of the order of inseparability with respect t o extensions of the ground field F . Let E I F be a finitely generated field extension. If S is a transcendency basis of E over F . then E / F ( S ) is a finite extension by Proposition 1.5. Recall that the inseparable degree ( E : F ( S ) ) ,of E over F ( S ) is defined by

( E : F ( S ) ) i= ( E : F ( S ) ) / ( E: F ( S ) ) , By the (Weil’s) order of inseparability of E I F , written ( E : F ) t , we understand min(E : F ( S ) ) i S

where S ranges over all transcendency bases of E over F . In the special case in which

E / F is an algebraic extension, S is the empty set, F ( S ) = F and hence the order of inseparability of E / F is the inseparable degree of E over F . It is a consequence of the definition of ( E : F ) , that E / F is separably generated if and only if ( E : F ) , = 1. Thus, by Theorem 4.4, ( E : F ) %= 1 if and only if E / F is separable. Note also that, by Proposition 2.2.lO(iii), ( E : F), is a power of p . Our first task is t o show t h a t any given set {z,, . . . , zn} of generators of E over F contains a transcendency basis S such that

We begin by recording the following preliminary observations. 5.1. Lemma. L e t E I K be a f i n i t e field e z t e n s i o n , let T be a n arbitrary field a n d let

F be a subfield of K n T . Then (i) ( E T : K T ) 5 ( E : K )

(ii) ( E T :K T ) s 5 ( E : K ) s (iii) ( E T : K T ) ; 5 ( E : K ) i

w i t h equalities i f E a n d T are linearly disjoint over F .

WEIL’S ORDER

Proof. Let { X I , . . . ,A,}

OF

INSEPARABILITY

127

be a basis of E over K . Since the elements of E are algebraic

over K T , we have n

n

ET = K T [ E ]= f K T ) [ xKXj] = x ( K T ) A , 1=1

j=1

which proves (i). Let KO be the separable closure of K in E. Then K o / K is separable and E / K o is purely inseparable. Hence KoT/KT is separable and ET/KoT is purely inseparable. T h u s

KoT is the separable closure of KT in E T . It follows from (i) that

( E T : K T ) , = (KoT : K T ) 5 (KO : K ) = ( E : K ) s and

( E T : K T ) i = ( E T : KoT) 5 ( E : K O )= ( E : K ) ; , proving (ii) and (iii). Now assume t h a t E and T are linearly disjoint-over F .

It suffices t o prove the

equality in (i), in view of the inequalities (ii) and (iii) and Proposition 2.2.13(ii). To prove that ( E T : K T ) = (E : K ) , we must show t h a t X I , . . . , A n are KT-independent. Since X I , . . . , An are linearly independent over K , it suffices t o verify t h a t E and K T are linearly disjoint over K . Let { p l , p 2 , . . . , p i } be a finite set of elements of KT which are linearly independent, over

K. We

may write p , = a,/ao where a0 and the ai belong t o T [ K ] The . numerators

are linearly independent over K and we have to show t h a t they are E-independent. Thu-; we need only verify the linear disjointness of E and T [ K ]over K .

To this end, note that the ring T [ K ]considered , as a vector space over K , is spanned by the field T . We may therefore choose a basis B of T [ K ]over K such t h a t B

C

I’.

Since the element of B are K-independent, they must be F-independent. But T and E arc? linearly disjoint over F , hence the elements of B are E-independent. Thus, by Proposition 2.3.4(v), T [ K ]and E are linearly disjoint over K , as required. As an easy consequence of the above, we derive 5 . 2 . Corollary. Let E / F be a finite field eztension and let

in L which are algebraically independent over E . T h e n ( i ) ( E ( S ): F ( S ) ) , = ( E : F ) s

S be a n y set

of elements

CHAPTER 3

128

(ii) ( E ( S ): F ( S ) ) ;= ( E : F ) ;

Proof. By Lemma 4.1, F ( S ) and E are linearly disjoint over F . The desired conclusion now follows from Lemma 5.1 by setting K = F and T = F ( S ) . 5.3. Corollary. Let E / F be a finite field eztension, let E ( B ) be a finitely generated

eztension of E , and let S be a transcendency basis of F ( B ) over F . T h e n there ezists a nonnegative integer t such that (i) ( E : F ) ; = p t ( E ( B ) : J ’ ( B ) ) ;

(ii) ( F ( B ): F ( S ) ) ;= p f ( E ( B ) : E ( S ) ) i

Proof. (i) Since both ( E : F ) i and ( E ( B ): F ( B ) ) ;are powers of p , it suffices to show that ( E : F ) ; 2 ( E ( B ): F ( B ) ) ; . The latter inequality is a consequence of Lemma 5.l(iii) upon replacing K by F and T by F ( B ) . (ii) Since E / F is a finite extension, the elements of S are algebraically independent over

E . By Proposition 2.2.13(ii), we have

( E ( B ): F ( S ) ) i = ( E ( B ): F ( B ) ) i ( F ( B ): F ( S ) ) ; and

( E ( B ): F ( S ) ) i = ( E ( B ): E(S))t(E(S): F ( S ) ) z Hence (ii) follows by appplying (i) and Corollary 5.2(ii). rn Returning t o the properties of the order of inseparability of finitely generated extensions, we now prove

Theorem. Let K / F be a finitely generated extension. T h e n a n y given set

5.4. { X I , .. . x,}

of generators of K over F contains a transcendency basis S such that

( K : F ( S ) ) i = ( K :F ) ;

Proof. Let

B =

F

. . ,zn}.

(~1,.

be the algebraic closure of F in some algebraic closure of K and let Because

F

is perfect,

F ( B ) is

separably generated over

7, by

virtue

of Corollary 4.6. By Corollary 4.5, the set B contains a separating transcendency basis, say S =

. . ,zk},

{XI,.

of

F ( B ) / F . Then S

is a transcendency basis of X over F and the

elements of B are separable algebraic over F ( S ) . It will next be shown that S satisfies the required property.

W E I L ’ S ORDER OF I N S E P A R A B I L I T Y

We know that

~ k + . ~. . ,, x,, are

129

separable algebraic over F ( S ) . Therefore, Zk+l,.. . , zn

are separable algebraic over E ( S ) for some finite field extension E I F (indeed, adjoin to F those elements of

F

which appear in coefficients of minimal polynomials of

xk+l,.

. . ,z,

over F ( S ) ) .Then S is also a separating transcendency basis of E ( B ) / E and thus ( E ( B )

E ( S ) ) ,= 1. It follows from Corollary 5.3 (ii) that for some t

( K : F ( S ) ) , = pt

20

On the other hand, if U is any other transcendency basis of K / F , then by Corollary 5.3(ii), WP

have

( K : F ( U ) ) ,2 P‘ This completes the proof of the theorem. In order t o derive some properties of the order of inseparability in relation to field cxtensions, we next record the following preliminary observation. 5 . 5 . Lemma. Let E I F be a finitely generated eztension, s a y E = F ( B ) , B f i n i t e , let

KlE’ be a n algebraic e x t e n s i o n a n d let S be a transcendency basis of E I F . A s s u m e that for a n y f i e l d T between F and K s u c h that

TIF

is finite,

( T ( B ): T ( S ) ) ,= ( E : F ( S ) ) , .

Then

( K ( B ): K ( S ) ) ;= ( E : F ( S ) ) ; Proof. For any field T between F and K , let T’ denote the separable closure of T ( S ) iri ’ f ’ ( U ) . Fix a basis

{Al,A2,

T ( B ) == T ’ ( F ( B ) ) ,F’

C T’, so

. . . , An} of F ( B ) over F‘

(hence n = ( E l : F ( S ) ) , ) . Then

and therefore it suffices t o show that the A, are linearly independent over K’. By hypothesis, the A, must be linearly independent over T’ if T / F is finite. But every finite set of elements of K’ is contained in some T’ where T I F is finite. Thus the A, are linearly

independent over K’, as required. We are now ready to prove 5.6. Theorem. Let E I F be a finitely generated extension, say E = F ( B ) , B finite,

f e t K I F be an e z t e n s i o n s u c h that K a n d E are free over F , a n d let S be a n y transcendency basis of E I F . T h e n

CHAPTER 3

130

(i) ( K ( B ): K ( S ) ) ; / ( E: F ( S ) ) i = ( K ( B ): K ) i / ( E : F ) ;

Furthermore, if K / F i s finite, then the common value of both sides of ( i ) is

( K : F ) ; / ( K ( B ): E ) ; (ii) If

is the algebraic closure of F i n the algebraically closed field L containing E and

F , then

( E : F ( S ) ) ;= ( E : F ) ; ( F ( B ): B(S)); (iii) If either K and E are linearly disjoint over F or K / F is separably generated, then

( K ( B ):K ) ; = ( F ( B ): F ) ; Proof. (i) We first assume t h a t K / F is finite. By Corollary 5.3, t h e value of the ratio on the left-hand side of (i) is independent of the choice of S and is equal to ( K :

F ) , / ( K ( B ) : F ( B ) ) , . If we now let S range over the set of all transcendency bases of E / F which are subsets of B , then we deduce from Theorem 5.4 t h a t t h e foregoing ratio is equal to the ratio ( K ( B ): K ) , / ( E : F ) i . Now assume t h a t K / F is a n arbitrary algebraic extension. Let us fix a transcendency basis SO of E I F . Given any two fields F1 and F2 between F and K such that Fl

C F,,

we have

( F i ( B ): F i ( S o ) ) t 2 ( F z ( B ) : Fz(So))t by virtue of Lemma 5.l(iii). Thus there exists a field F’ between F and K such that F’IF is finite and such t h a t

( F ” ( B ) : F’’(So)), = ( F ’ ( B ) : F’(S0)) for any field F” between F’ and K which is a finite extension of F’. Applying the finite case considered in the previous paragraph, it follows that

( F ” ( B ): F ” ( S ) ) , = ( F ’ ( B ) : F ’ ( S ) ) i for any transcendency basis S of F ’ ( B ) / F and for any field F“ between F and K such

t h a t F”/F’ is finite. Invoking Lemma 5.5, we therefore conclude t h a t

( K ( B ): K ( S ) ) ;= ( F ’ ( B ) : F ’ ( S ) ) i for any transcendency basis S of F’(B)/F’. Thus ( K ( B ): K ) ; = ( F ’ ( B ) : F’)%.Taking S

to be any transcendency basis of E I F , we now have

SEPARABILITY AND PRESERVATION OF p-INDEPENDENCE

131

where p t = ( E : F ) ; / ( F ‘ ( B ): F’)i = ( E : F ) i / ( K ( B ): K ) i ,

proving (i) in case

K / F is an arbitrary algebraic extension.

Finally, assume t h a t

K / F is a n arbitrary extension. Let V be a transcendency basis of

K / F . Because K and E are free over F , it follows from Lemma 4.22(iv) that the elements of V are algebraically independent over E . Thus, by Corollary 5.2, we have

for a,ny transcendency basis S of E / F . Hence

(F(Vu B ) : F ( V ) ) i= ( E : F ) ; Now K / F ( V ) is algebraic, so by the preceding case

(F(V u B ) : F(V

u S)),= p t ( K ( B ) : K ( S ) ) , ,

where p‘ =

(F(Vu B ) : F ( V ) ) i / ( K ( D ) : K ) % ,

which proves ( i ) by applying ( I ) and (2). (ii) P u t K =.‘.

Since

K / F is algebraic, K and E are free over F. Furthermore, since K

is algebraically closed, it is a perfect field. Hence ( K ( B ): K ) , = 1. The desired conclusion now follows from (i). ( i i i ) Assume that K and E are linearly disjoint over

F. Then, applying (i) and Lemma

5.i(i) (with K = F ( S ) and T = K ) , the required assertion follows. Now assume that

K/F

is separably generated. If K / F is a finite separable extension, then the assertion follows from ( i ) . If K / F is purely transcendental, then K and E are linearly disjoint over F by 1,t:ninia 4.1 (since K 2nd E are free over E’) and we are then in the linear disjointness case. Finally, the general case is now obtained by following up a purely transcendental extension of F by a separable algebraic extension and by applying Lemma 5 . 5 .

6. Separability and preservation of pindependence.

All fields to be considered will have characteristic a fixed prime p . If F is such a field.

F”“ denotes the field of all pn-th

powers of elements from

F. Similarly, if S is

a n y set

132

CHAPTER 3

of elements of F , then SP” is the set of all p”-th powers of elements of S. If a field F and certain sets S , T , . . . are all contained in a larger field, then F ( S , T, . . .) denotes the field obtained by adjoining t o F all elements of S , of T , etc. In this section, among other results, we demonstrate that separable extensions are precisely those extensions which preserve p-independence. With any given field extension E / F there is related in a n invariant fashion a purely inseparable extension E / F ( E p ) . T h e latter may be analyzed by t h e concepts of pindependence and p b a s i s which we proceed t o develop. Let E / F be a field extension. A subset S of E is said to be relatively p-independent in E I F if

S’ C S

implies

E P (F, S’) c E P ( F S, )

A relative p-basis B of E I F is a relatively pindependent set such t h a t

E = EP(F,B) We say t h a t x E E is relatively p-dependent on S , written z

< S , if

x E FP(F,S) A subset S of E is (absolutely) p-independent if

S’ C S

implies

E P ( S ’ )c E P ( S )

An (abso1ute)p-basis B of E is defined t o be a pindependent subset for which

E = EP(B) We say that x E E is p-dependent on S if x E E p ( S ) . This is the special case of the above “relative” definitions obtained by assuming F is perfect, for then F = FP

EP, and the

field EP(F, S) used above becomes E p ( S ) . 6.1. Lemma. Relative p-dependence in E I F is a dependence relation. Furthermore,

with respect to this relation, a subset S of E is independent (S is a basis) if and only if S is relatively p-independent (S is a relative p-basis).

Proof. We begin by verifying properties (1)-(4) in the definition of a dependence relation given in Sec. 1. First, it is clear that if s E S s < S, proving (1). If x

C E , then

s E E p ( F , S ) , hence

< S we have x E EP(F,S) and, since Ep(F,S ) is the union of its

SEPARABILITY AND PRESERVATION OF WINDEPENDENCE

133

subfields EP(F, S’) where S’ is a finite subset of S,then z < S‘ for some finite subset S’of

S , proving (2). If z

E p ( F , S ) and every s E S satisfies s E E P ( F , T ) ,then

5

E EP(F,T),

proving ( 3 ) . It remains t o check the exchange property. This states t h a t , if z € Ep(E’,S ) and z $ E P ( F , S ’ ), where S’ = S

-

{s} for some s E S , then s 6 E P ( F , S ‘ , z ) . We may

assume s $! E P ( F , S ’). Consider the following chains of fields:

EP(F, S’, z,s)

2 EP(F, S‘,z) 2 EP(F, S’)

and

EP(F,S’,z,s) 2 E P (F , S ’,s ) 2 E P ( F , S ‘ ) Note that z P E E p ( F , S’) and sP E EP(F, S’). Hence we must have

( E P ( F , S ‘ , z ): E P ( F , S ’ ) )= p = ( E P ( F , S ’ , s ): E P ( F , S ’ ) ) Since z E E P ( F , S ’ , s ) ,we have E P (F , S ’, z , s ) = EP(F,S’,s) and thus

( E P ( F , S ’ , z , s ): E P ( F , S ‘ ) )= p It follows that

E P ( F , S ’ , z , s ) = E P (F , S ’, z ) = EP(F,S’,s ) and s E E P ( F , S ‘ , z ) . This completes the verification of the axioms for a dependence relation. Let S be any subset of E . If S is independent, then no s E S satisfies s E E P ( F ,S {s}). Hence, if S’ is a proper subset of S and s E

s

Ff

E P (F , S

-

S

-

S’, then

{s}) 2 EP(F,S’)

i.e. E p ( F , S ’ ) is a proper subset of E P ( F , S ) .Thus S is relatively p-independent.

Cori-

vcrsely, assume that S is relatively pindependent. Then, given s E S,p u t S’= S Then E P J F , S ’ ) # E p ( F , S ) and hence s

Ff

E p (S‘), i.e. s is not dependent on S

~

~

{s}.

{s).

Thus S is an independent subset of E .

Finally, assume t h a t B is an independent subset of E . Then B is a basis if and only if every z in E satisfies z E E p ( F , B ) , i.e. if and only if E = E P ( F , B ) . Thus B is a basis if and only if B is a relative pbasis of E I F . 6.2. Corollary. E v e r y e z t e n s i o n E I F h a s a relative p-basis and a n y two relative p-bases

of E I F have the s a m e cardinality. Furthermore,

CHAPTER 3

134

(i) A n y relatively p-independent set in E I F can be eztended to a relative p-basis of E I F (ii) If S is a subset of E such that E = E p ( F , S ) , then S contains a relative p-basis of

EIF Proof. This is a direct consequence of Lemma 6.1 and Theorem 1.3.

1

6.3. Corollary. Every field E has a p-basis and any two p-bases of E have the same

cardinality. Furthermore, (i) A n y p-independent set in E can be eztended to a p-basis of E

(ii) If S i s a subset of E such that

E

= E p ( S ) , then S contains a p-basis of E .

Proof. This is a special case of Corollary 6.2 in which F is a perfect field. w For future applications, we next provide the following characterizations of relative and absolute pindependence of sets. By Lemma 1.1, we may harmlessly assume that the sets considered are finite. 6.4.

Lemms. Let E J F be a n arbitrary field eztension and let S = { z l , . . . , x m }

be any finite subset of E . Then the vector space E P ( F , S ) is spanned over E p ( F ) b y the

elements of the f o r m

zy’zt’. . .zkm,0 5 n; < p

(in particular, ( E P ( F , S ) : E p ( F ) ) 5 p “ )

and the following conditions are equivalent: (i) S is relatively p-independent in E J F (ii) ( E p ( F , S ) : E p ( F ) ) = pm

(iii) The pm elements z;‘zJ;” ...z?,

0

5 n; < p , f o r m a basis of E P ( F , S ) over E p ( F )

(iv) The p m elements zT’zt2 . . . xkm,0

5 n; < p , are linearly independent over E p ( F ) .

Proof. Note t h a t zp E E P ( F )and hence E P ( F , x l , , . . , z I ) is spanned over E P ( F , z l ,

. . . ,z,-l)by l r z t , ... , zp-’.

By induction on m ,this obviously implies that EP(F, S ) is

spanned over E p ( F ) by the elements z;’

zc;

. . .x;m

OIn,
It is now also obvious that the conditions (ii), (iii), and (iv) are equivalent. If S is relatively pindependent, then z, $! EP(F, z1,. . . ~

( E P ( F , z i , .. . x i ) : E P ( F , z l , .. . ,xi-1)) = p

~ - 1and )

hence

for each i

which implies (ii). Conversely, assume that S is not relatively pindependent. Then z; E

E P ( F , z l , . . .x;-l,z;+l,. . .,z,,,)

for some i

S E P A R A B I L I T Y AND PRESERVATION OF p-INDEPENDENCE

and hence ( E p ( F , S ): E P ( F ) )5 pm-' 6.5.

135

< pm, as required.

Corollary. Let E be a n arbttrary field and let S = { z 1 , z 2 , .. . , x m } be any

f i n i t e subset of E. T h e n the vector space E p ( S ) IS spanned over EP by the elements oJ t h e form xy' x;'

(i) S

1s

. . . x>, 0 5 n, < p and the following conditions are equzvalent:

p-independent

(ii) ( E P ( S ): EP) = pm (iii) T h e pm elements

Z Y ' X ; ~ . . .z>,

(iv) T h e p m elements x;'

5 ; '

0

5 n, < p ,

f o r m a basis of E P ( S ) ouer EP

. . . x z m ,0 5 n, < p are linearly independent over EP.

Proof. This is a special case of Lemma 6.4 in which F is a perfect field. Let E / F be a field extension. We say that E / F preserves p-independence if every p independent set in F remains pindependent in E . Extensions preserving pindependence will provide us in future with a natural tool for investigating separating transcendency bases, largely because, given a n extension E I F which preserves pindependence, any intermediate extension K / F with K

CE

must automatically preserve p-independence. Ex-

tensions preserving pindependence will now be characterized in several different ways. 6.6. Theorem. (MacLane (1939a)). Let E / F be a field extension. T h e n the follow-

ing properties are equivalent: (i) E I F preserves p-independence (ii) There i s a p-basis B of F which is p-independent in E

n E p ( S ) f o r every finite subset S of K ( F p ) = F n K ( E P ) for every subfield K of F

(iii) F P ( S ) = F

(iv)

F

Proof. It is obvious that (i) implies (ii). Conversely, suppose that (ii) holds

To

prove (i), assume by way of contradiction that there is a pindependent subset S or 1.' which is not pindependent in E. Then there exists a n element s E S contained in E7 (S'), where S' is a finite subset of S not containing s. Since B of (ii) d

IS

a p b a s i s of F . there

IS

finite subset B1 of B such that s and S' are in FP(B1). This means t h a t each element

of S' u {s} is pdependent on B1. Applying the exchange property, we can exchange the elements s and s' E S' successively with suitable elements of B 1 , until dl1 elements of are pdependent on S', s and a remaining subset

B2

& B1. This implies t h a t

DL

CHAPTER 3

136

and that the set S’ U {s} U 3 two finite sets

B1

2

is pindependent in F . Furthermore, by construction, the

and S‘ U {s} U Bz have the same number of elements, say equal to rn.

But, by hypothesis, B1 remains pindependent in E , while the other subset S‘ U {s} U Bz is not p-independent because we assumed s E FP(S’). Applying (1) and Corollary 6.5 we have pm =

(EP(S1): E P )= ( E P ( S ’ , s , B , ) : E P ) < pm,

a contradiction. This proves t h a t (ii) implies (i).

To prove (iii)+(i), assume counter to (i) t h a t some pindependent set S of F is not pindependent in E , so t h a t again some s E EP(S’) where S’ is a finite subset of S not containing s. Then, by (iii), s E F n EP(S’) = FP(S’) which shows that S is not p-independent in F , a contradiction. The implication (iv)+(iii) may be obtained trivially by setting K = F ’ ( S ) where F’ is a perfect subfield of F . We are left to verify that (i)+(iv). To this end, let K be any subfield of F and let B be a relative p-basis of K ( F p ) / F p . Then K ( F P ) = F p ( B ) . If F n K ( E p ) # K ( F P ) , then there exists an element x of F in K ( E P )= E p ( K ) = E P ( B ) but not in K ( F p ) = F p ( B ) . Hence

3:

is not pdependent o n B in F , while B is by construction p-independent in F . It

follows from Lemma 1.2 t h a t B U {x} is a pindependent subset of F . Applying (i), we therefore deduce that B U {x} is p-independent in E , contrary t o the provious assertion that x E E P ( B ) . m

Our next aim is t o characterize separable extensions by showing t h a t they are precisely those extensions which preserve pindependence. We present two different proofs, one pertaining t o algebraic extensions and the other t o arbitrary extensions. We need the following preliminary results. 6.7. Lemma. Let E J F be a separable field extension.

(i) EpJFp i s a separable field extension (ii) EPIFP is algebraic if so is E I F .

Proof. (i) Let

X I , .. . , x,

E F be linearly independent over FP. Then

are linearly independent over F . Since E and FP-’ are linearly disjoint over F , it follows that xg-’ are linearly independent over E . Hence

21,.

. . ,5, are linearly independent over

137

SEPARABILITY AND PRESERVATION OF p-INDEPENDENCE

EP. This shows that EP and F = (Fp)p-’ are linearly disjoint over FP, i.e. that E P / F Pis separable.

+ a l X + . . . + a,X” be its is a root of g ( X ) = a: + a:X + . . .+ aP,Xn E F P i X ! ,

(ii) Assume that E / F is a!gebraic. Given X E E , let f ( X ) = a0

minimal polynomial over F . Then

XP

hence EP/FP is algebraic.

6.8. Lemma. L e t E I F be a separable algebraic extension. T h e n a n y p-basis of E a p-basts of

zs

E.

Proof. Let B be a pbasis of F . Then B is pindependent in F and F = F 1 ’ ( D ) . Since E / F is a separable algebraic extension, it follows from Proposition 2.2.6(ii) that

E = FEP. Hence E

=

F p ( B ) E p = E P ( B ) . Let S = {q, x2,.. . ,}z,

of B. Then S is pindependent in F , hence the pm elements

5;’

be any finite subset

. . . zkm,0 5 n, < p are

linearly independent over FP (Corollary 6.5). Hence, by Lemma 6.7(i), these elements are linearly independent over EP. Therefore, by Corollary 6.5, S is pindependent in E . Since

S is an arbitrary finite subset of B , we conclude that B is pindependent in E , as required. We are now ready t o prove 6.9. Theorem. (MacLane (1939a)). Let E I F be a n algebraic field eztension. T h e n

E / F preserves p-independence if and o n l y if E / F i s separable. Proof. Assume t h a t E / F is separable. If S is a pindependent set in F , we may extend

S to a p b a s i s B of F (Corollary 6.3). By Lemma 6.8, B is a p b a s i s of E , hence independent in

E . Thus S

is independent in E , proving that

E / F preserves pindependence.

Conversely, assume that E / F preserves pindependence but is not separable. Let F, bc the separable closure of F in E . Then E

# F, and, by Proposition 2.2.30, E ; F S

is

purely inseparable. Hence, by Proposition 2.2.27 (ii), E contains a X not in F, with X p in fi’s.

By Lemma 6.8, any pbasis B of F is also a pbasis of F,, so F, = F,P(B) and thus

A’’ 6 F f ( B ) . The exchange property of pindependence allows us to exchange

XP

with

some b E B , with the result that

b E F,P(B - { b } , X”)

C E P ( B- { b } )

This implies that the set B is not pindependent in E , which contradicts the assuniption t h a t the extension E / F preserves the p-independence of B . proved.

So the theorem is

CHAPTER 3

138

Applying a different approach, we now show t h a t Theorem 6.9 holds for a n arbitrary field extension E / F . 6.10. Theorem. (MacLane (1939a)). Let E / F be a n arbitrary field eztension. T h e n

the following conditions are equivalent:

(i) E / F preserves p-independence (ii) For a n y finite subset S of E , the linear dependence over F of the set SP implies the

linear dependence over F of the set S itself. (iii) E I F i s separable.

Proof. The equivalence of (ii) and (iii) is a consequence of Proposition 4.29. (i)+(ii): Assume t h a t the elements

. . . , s, of some set S have their p t h powers

SI,S~,

linearly dependent over F. Choose a pbasis B of F , so that F = F p ( B ) , and select a

C B with the property that ,:s . . . ,sP, are linearly dependent over F P ( T ) . If T = 0, then s y , . . . ,sP, are linearly dependent over FP and hence s1,. . . ,sm are linearly dependent over F . If s1,. . . ,,s are linearly independent over F , we can successively delete elements from T till we find a new T’ and a n element t such t h a t s,; . . . ,sP, are linearly finite subset T

dependent over F P ( T ’ , t ) but not over FP(T’). Then t has degree p over FP(T’) and the given linear dependence relation may be written as

(Xij

E FP(T‘))

with not all X,j = 0. Accordingly,

Now

E / F preserves pindependence, so B is pindependent in E . If rn

C i=

~;js;

#o

for some j E ( 0 , I,.. . , p

-

11,

1

then (2) shows that t is separable over EP(T’). Since t is also purely inseparable over Ep(T’), it follcws from Corollary 2.2.28 that t E EP(T’). Thus t is p-dependent on ’1” in

139

S E P A R A B I L I T Y AND PRESERVATION OF p-INDEPENDENCE

6 ,contrary to the fact that B is pindependent in E . Hence m

~,,sp =

o

for all

j E (0, I , . . . , p

--

1)

i=l

Since some A,,

#

0, this provides a linear dependence of

contradiction. Hence sir.. . ,s,

3;,

. . . ,sP, over FP(T’), again a

are linearly dependent over F .

(ii)=>(i):Assurrie by way of contradiction that E / F does not preserve p-independence. Sincc every p-independent set can be extended t o a p-basis, there exists a p-basis B of such that B is p-dependent in E . Hence there exist distinct elements b, b l , . . . , b, in

L‘

B such

that b t E“(b1,. . . ,bn). By Corollary 6.5, the vector space E P ( b 1 , . . . , b,) has over E P a basis consisting of elements cl, . . . ,,c Since b

where each c, is of the form b;‘ . , . b?,

0

5

ei

5 p-

1.

E”(h1,.. . , b,), we have

for some A, not all zero in E . Among the e1ement.s 1, A,:

. . . , A i l , of EP pick

a maximal

linearly independent subset (over F’) consisting of l , ~ : ,. .pE. . Then each A:

may b e

writtcii i n the form

Substituting

A: in

( 3 ) and collecting the coefficients of each p:, we find

Siiice b is not p-dependent on b l , . . . , b , in

F , the first term b -

a:ocl is not zero. It

therefore follows from ( 4 ) that l , f i y , , . . ,,u.f: are linearly dependent over F . The hypothesis ( i i ) then shows that l,pl,. . . , p k are linearly dependent over F . IIence l , p T , . . . ,pLpk are

linearly dependent over FP, contrary to the choice of 1, M Y , . . . , p i . 6.11.

Corollary.

L e t E I F be a separable field e x t e n s i o n , let

XI,.

. . , z,

E;

he

ulgebraically i n d e p e n d e n t o v e r F a n d let zo t E be algebraic ouer F(x1,. . . , z , ~ ) , s a y /(zo,

z1,.

. . .z n ) = 0 w h e r e t h e c o e f i c i e n t s of t h e p o l y n o m i a l f are in F . Iff i s irreducible

ouer F a s a p o l y n o m i a l in t h e i n d e t e r m i n a t e s

in the p-th p o w e r s zp0,. . . , zP,.

20,.

. . , z,

t h e n f c a n n o t be u p o l y n o m i a l

CHAPTER 3

140

Proof. Deny the statement. Then we have m

(i = 1 , .. . ,m)

e0.i

r=l

where all the coefficients X i are in F and distinct from zero. This linear dependence of yy,.

. . ,yg over F implies, by Theorem 6.10, a linear dependence of

y1,.

. . ,ym:

m

must actually appear in 9. The degree d of f in

Here not all pL1= 0, so some xj, say z, z,

is then a t least p times the degree of g in z.,

element z, of E over the field F ( z 0 , . . . ,q degree for

2,

But g = 0 provides a n equation of smaller

- 1 ) .

over that field, a contradiction.

By Theorem 1.5.8, d is t h e degree of the

H

We can now apply the result above t o prove 6.12. Theorem. (MacLane (1939a)). L e t E I F be a separable field e z t e n s i o n a n d let

B be a relative p-basis of E I F . T h e n t h e e l e m e n t s of B are algebraically independent over F. Proof. Deny the statement. Then there exist elements are algebraically dependent but with

ZO, 5 1 , .

. . ,I, in B which

. . ,x, algebraically independent over F . An . . ,2,) = 0 expressing t h e algebraic dependence

21,.

irreducible polynomial relation f (zo,z1,.

of z o r z l , . . .,x, must then as in Corollary 6.11 contain a n indeterminate, say znr such

that f is not a polynomial in the ppower

zg. This equation provides an irreducible and

separable equation for z, over the field

Since z: E EP(F, zo,.

. .,z,-1),

z,

is purely inseparable over E p ( F , zo,

. . . ,~

~ - 1 Hence, ) .

by Corollary 2 . 2 . 2 8 , ~6 ~EP(F,1 0 , . . . , x,-1) which shows that z, is relatively pdependent

on zo, . . . , z,-1.

Since {zo,

. . . ,z,}

B , we come t o the desired contradiction.

H

The next result provides a link between absolute and relative p-bases of an extension preserving pindependence. 6.13. Theorem. (MacLane (1939a)). L e t E f F be a separable field e z t e n s i o n , let C

be a subset of F a n d let B be a subset of E with B n C = 0,

SEPARABILITY A N D PRESERVATION OF p-INDEPENDENCE

(i)

If B is a relative p-basis

of E / F , t h e n B

uC

141

is a p-basis of E if and only if C i s a

p-basis of F . (ii) If C i s a p-basis of F and B U C is a p-basis of E , t h e n B is a relative p-basis of E I F .

Proof. (i) By Theorem 6.10, E / F preserves pindependence. Assume that B is a relative p-basis of E / F . Suppose further, that C is a p b a s i s of F . Since E = E P ( F , B ) and

F

=

FL’(C), we have E = E P ( B , C ) . Hence to prove t h a t B

suffices to show t h a t B

u C is pindependent

UC

is a p-basis of E , it

in E . Suppose false. Then, either an element

6 of B 1ic.s in

E P ( B {b},C) ~

E P ( F , B- { b } )

which contradicts the relative pindependence of B ,or an element c of C lies in EP(B,C -

{ c } ) . If the latter holds, then there are distinct elements b l , . . . ,b , in B such that

c E E P ( b l , . . . ,b,,

C

-

{c})

and such that ( 5 ) fails if any b, is omitted. Note that m

(5)

2 1, i.e.

at least one b, is present

in (5) since C is pindependent in F and therefore i n E , by the assumption that E / F preserves pindependence. Now both c and b1 have degree p over

and so (5) and the exchange property yield

h i E E P ( c , b z , . . . ,b,,

C

~

{ c } ) = E P ( b 2 , . . ,b,,

C ) C E P ( F , h z ,. . , b,)

which contradicts the relative pindependence of B. Conversely, in addition to the assumption that f? is a relative p-basis of E / F , suppose f u r t h e r that

BUC is a p b a s i s of E . Since C is p-independent in E , it is also p-independent

i n F . Assume by way of contradiction that C is not a p-basis of F , i.e. that there exists 1: 6

I<’ with z

$t

F P ( C ) . Since B ii C is a p b a s i s of h’,we have z E EP(l3,C). Thus there

exist distinct elements b l , . . . , b , in B such that

z E E P ( b l , . . . , b,,

C)

and such that ( 6 ) does not hold if any b, is omitted. If m

(6)

2

1, then one deduces as i n the

previoiis argument from (5) that

b,

c

E P ( x h, 2 , . . . ,6,,,

C)

E r ’ ( F , 6 2 , . . ,h,),

CHAPTER 3

142

contrary to the relative pindependence of B in E I F . Thus, by ( 6 ) , we must have z E

EP(C). Since

5

I$

Fp(C),it follows, from Lemma 1.2, that the p-independent set C U {z}

in F has become pdependent in E , contrary to the assumption t h a t EIF preserves pindependence. (ii) By hypotheses, F = FP(C) and E = EP(B,C),hence Ep(F,B) = EP(B,C)= E . Let

B’

C

B be such t h a t Ep(F,B’) = EP(F,B). Then E = EP(C,B’) and hence B’ = B ,

since B U C is pindependent in E and B’

n C = 0.

So the theorem is true. rn

As a final result, we record the following simple observation. 6.14. Lemma. Let F

K

E be a c h a i n of fields where E I K is a f i n i t e purely

inseparable e x t e n s i o n and EIF is a separable e z t e n s i o n . T h e n the (cardinal) number of elements in a relative p-basis of E / F i s the s a m e as t h e number of elements in a relative p-basis of K / F .

Proof. m7e first note that, by Theorem 6.10, E / F preserves p-independence.We may harmlessly assume that ( E : K ) = p, so that E = K ( X ) with X p E K . We claim that X p is relatively p-independent in K / F . Indeed, otherwise is a pbasis of F. Because

XP

I$

KP,

XP

XP

lies in K p ( F ) = K p ( B ) where B

can be exchanged with an element b of B , which

means that b E KP(XP,B- ( 6 ) )

C E P ( B- { b } ) ,

contrary to the pindependence of B in F and E . Because containing

XP

XP

is relatively pindependent in K I F , there is a relative p b a s i s S of KIF

(see Theorem 1.3(ii)). The replacement of

XP

by X in the set S yields (by

a straightforward verification from the definitions) a relative p b a s i s So of E / F . Since S and So have the same cardinality, the result follows. rn

7. Perfect ground fields.

Throughout p denotes a fixed prime and all considered fields are assumed to be of characteristic p. Let EIF be a field extension and let F be perfect. We know, from Corollary 4.6, that if E / F is finitely generated, then E / F is separably generated. In this section we

investigate conditions under which EIF is separahly generated without being necessarily finitely generated. Our point of departure is the simplest case in which the transcendency degree of EIF is 1.

143

PERFECT GROUND F I E L D S

7.1. Theorem. (MacLane (1939a)). Let E / F be a field e z t e n s i o n s u c h that F is

perfect, E is n o t perfect a n d tr.d. ( E I F ) = 1. T h e n E I F is separably generated. Proof. Let {A} be a transcendency basis of E I F . Then E / F ( X ) is algebraic, so if XP-"

E E for all n 2 0, then E l F ( X P

-1

, X P - 2 , . . . ) is also algebraic. But F ( X , X f l X P - ' , . . . )

is perfect and hence so is E , by Proposition 2.2.26, contrary t o the assumption t h a t f.;is not perfect. We may therefore choose the largest integer n

E. Then

pP

-1

2 0 for which

p = A"-"

is in

@ E and { p ) is a transcendency basis of E / F . We now assert that { p } is a

separating transcendency basis which will finish the proof. Assume by way of contradiction that there exists a E E which is not separable over -1

F ( p ) . P u t m = ( F ( p P , a ) : F ( p P - ' ) ) , n = ( F ( p , a ): F ( p ) ) and consider the following chains o f fields: -1

F b ) C F ( p P - ' )C F ( p P , a )

It follows from (1) t h a t ( F ( p P - l , a ) : F ( p ) )

5 nap. On the other hand, since a

(1)

is not

separable over F ( p ) , the minimal polynomial f ( X ) of a over F ( p ) can be written in Since a is a root of g ( X ) and the form f ( X ) = g ( X ) P for some g ( X ) E F(pP-')[X]. degg(X) = n / p , it follows that m

-,

which shows that pP

E F ( p ,a )

5 n / p . Hence, by

(Z),

E ,a contradiction .

I,ct E / F be a field extension and let the elements of a subset S of E be Agebraically independent over F . If f ( X ) € F [ S ] [ X ]then , there exist only finitely many. say s,, . . . , s,

of elements of S which actually appear in f and f (X)= g(s1,. . . ,s,, X ) c

F j s l , . . . , s n rX j . Choose the largest integer e

2

0 such t h a t f ( X ) can be written as a

polynomial g(s;', s 2 , . . . , s n , X ) . We shall refer to p e as the exponent of s 1 in f and say that f h a s the ezponent p e in

s1.

7 . 2 . Lemma. Let E J F be a field eztension and let the elements of a subset S

OJ

I;; be

algebraically independent over F . If a n element X E E is a root of a n irreducible separable polynomial f ( X )

F [ S ] [ X ]t h, e n a n element s of S h a s exponent f in f zf and oniy i j X

is inseparable over F ( S

- {s},sP)

CHAPTER 3

144

Proof. If the exponent p e of s in f satisfies p e > 1, then f is separable regarded as a polynomial over F ( S l , s P ) ,where S1 = S

-

{s}. Thus

X is separable over F ( S 1 , s ” ) .

Conversely, suppose that s appears with exponent 1 in f . Then f ( X ) = g ( X ,s, 5’1) has exponent 1 in s and X and is irreducible over F [ X ,s, S,] . Consider the polynomial

where

g(P)

denotes the function obtained from g by replacing each coefficient by its p-th

power. Then g(P)(XP,sp, Sp), which is the p t h power of a n irreducible polynomial g over

F I X , s, Sl], cannot be reducible over the smaller ring F [ X ,sp, S l ] ,which does not contain this irreducible factor g ( X ,s, S1). Thus g(P)(XP,sp, Sr) is irreducible over F [ X ,sp, SI]

and hence, by Theorem 1.5.8, is irreducible over F(sP, S 1 ) [ X ] Furthermore, . by Proposition 2.2.4(ii), this polynomial is inseparable over F ( s P , S , ) [ X ] .Hence A, being a root of

g(P)(XP,sp, Sr)E F(sP, S 1 ) [ X ]is, inseparable over F(sP, Sl),as required. w 7.3. Lemma. Let E / F be a field eztension, let F be perfect and let the elements of

a subset S of E be algebraically independent over F . Assume that the element X of E is separable over F ( S ) and that s E

is not separable over F ( S ) . Then there i s a n element

XP-’

S such that

(i) X is not separable over F ( S (ii) s is separable over F ( S

-

~

{s},sP)

{s},X)

but not over F ( S - {s},

XP).

Proof. We may always choose a polynomial

such that

f(X)

= 0,

f has exponent 1 in X and g is irreducible as a polynomial over F in

the variables si E S and X . If f has exponent p or greater in each s E S, then by taking the p-th root of each coefficient a; E F [ S ]of f ( X ) ,we obtain a separable polynomial over

F ( S ) having

-1

XP

as a root. Since the latter is excluded by hypothesis, we deduce that f

has exponent 1 in a t least one s E S. This shows that X satisfies (i), by applying Lemma 7.2. Since g(X,s, S’) = 0, the element s is algebraic over F(X,5”) where S’= S - { s } . The elements A, S’ are therefore algebraically independent over F (in particular X

$ F(XJ’,S’)).

By construction, g ( X ,s, S’) is irreducible as a polynomial in the variable s over F [ X ,S’].

145

PERFECT GROUND F I E L D S

Hence Theorem 1.5.8 shows that g ( X , s, S') is also irreducible as a polynomial in s over

F ( X , S ' ) . Since f has exponent 1 in s, the root s is separable over F ( X , S ' ) . F u thermore, s cannot be separable over the smaller field F ( X P , S ' ) . Indeed, otherwise

F ( s , X P , S ' ) / F ( X P ,S ' ) is separable and so is F ( s , S', X ) / F ( s , S', X P ) (since X is separable over F ( S ) ) . Hence X would be separable over F ( X P , S ' ) , although X clearly is a root of an inseparable irreducible polynomial of degree p over F(XP, S ' ) . This proves (ii) and the result follows. 7.4. Lemma. L e t a field e z t e n s i o n E I F h a v e a f i n i t e t r a n s c e n d e n c y basis S , l e t E'

be perfect a n d let K be t h e separable closure of F ( S ) in E . If EP' e

Cr K

f o r s o m e znteger

2 0, t h e n E / K is f i n i t e . Proof. It suffices t o show that E

obviously finite. Fix X K ( S p - C )and

Ape

E . Since

Ape

C

K ( S p - ' ) , since the extension K ( S P - ) ' I <

17

E K , it follows that X is purely inseparable oler

is separable over F ( S ) . By taking the pe-th roots of each coefficient of

the minimal polynomial of

XP'

over F ( S ) , we obtain a separable irreducible polynomial

over F ( S P - ' ) having X as a root. Hence X is separable over F ( S P - < )and therefore over

K ( S P - ' ) . It follows, from Corollary 2.2.28, that X E K ( S p - < )as required. w Turning t o field extensions with finite transcendency bases, we now prove 7.5.

Theorem.

(MacLane (1939a)). L e t E / F be a field e z t e n s z o n wzth a f i n i t e

t r a n s c e n d e n c y basas S a n d let F be perfect. T h e n E I F zf E P - ( S ) / F ( S as ) separable f o r s o m e znteger e

IS

separably generated zf a n d oiily

2 0.

Proof. Assume t h a t E has a separating transcendency basis B over F . Since F / F ( S) is algebraic, it follows from Proposition 2.2.6(i) that any given b E B has some power OF''

separable over F ( S ) . Since the set B is finite, we may choose e large enough such that 6"' is separable over F ( S ) for all b E B. Since E / F ( B ) is separable, E P e / F ( B P Fis ) separable

(Lemma 6.7) and hence E P ' ( S ) / F ( S ) is separable (Proposition 2.2.15(i)). Conversely, let e be the smallest integer

2 0 such

t h a t E P ' ( S ) / F ( S )is separable. If

e = 0 , S itself is a separating transcendency basis, and we are done. Assume then that e > 0. Liy the choice of e , we may choose p E E such that pPe but not

pPe-'

is separablr

over F ( S ) . Applying Lemma 7.3 (with X = p P e ) ,we may find s E S such that s is separable

146

CHAPTER 3

over F ( S

-

{s},&)

The set S1 = (S - {s}) U { p } obtained from

S by replacing s by

p is thus another

transcendency basis of E / F , such t h a t every element separable over F ( S ) is separable over F ( S 1 ) , by the transitivity of separability (Proposition 2.2.15(i)). Furthermore, p is not separable over F ( S ) but is separable over F ( S 1 ) . T h e separable closure

K1

of F ( S 1 ) in E

is thus larger than the separable closure K of F(S) in E. Because S is finite, the extension

E I K is finite by Lemma 7.4, so a repetition of this transition from S t o yield a basis

S 1

will finally

S, over which all of E is separable.

7.6. Corollary. Let E / F be a separably generated eztension where F is a perfect field and t r . d . ( E / F ) is finite. T h e n , f o r a n y intermediate field K,

KIF is also separably

generated. Proof. Pick a transcendency basis

S 1 of

KIF and a similar basis Sz of E I K . Then

the union S = S1 U S, is a transcendency basis of E I F (Proposition 2.6). The necessary condition of Theorem 7.6 applied t o this basis S then shows t h a t the sufficient condition of Theorem 7.6 must be satisfied by

S1.Hence

KIF is separably

generated, as required.

For our purposes it is especially important to note t h a t any extension of a perfect field preserves pindependence. This is included in the fo1:owing simple observation.

7.7. Lemma. Let E / F be a field eztension a n d let F be perfect. (i) E / F preserves p-independence

(ii) I f B is a p-basis of E , t h e n t h e elements of B are algebraically independent over F .

Proof. (i) Since F is perfect, each pbasis of F is void, hence necessarily remains p-independent in E . (ii) Since F is perfect, B is a relative pbasis of E I F . Since E I F is separable, the result

follows from Theorem 6.12. m We are now ready to prove 7.8. Theorem. (MacLane (1939b)). Let E I F be a field extension, let F be perfect

and let B be a n y p-basis of E . If K is the algebraic closure of F ( B ) in E , t h e n B is

a

separating transcendency basis of K I F and E = E p ( K ) . Proof. Since B 2 K is a pbasis of E , we have E = E p ( B )

C

E p ( K ) and thus

C R I T E R I A FOR S E P A R A T I N G TRANSCENDENCY BASES

E

=

147

E p ( K ) The fact t h a t the set B is algebraically independent over E' is a consequence

of Lemma 7.7(ii). Since K / F ( B ) is algebraic, B is a transcendency basis of K I F . We are

therefore left to verify that K / F ( B ) is separable. Assume by way of contradiction that K / F ( B ) is insepxablt. Then there exists an element p E K which is not separable over F ( B ) ,but pp is separable over E'(B). Therefore X

= pp

satisfies the hypothesis of Lemma 7.3

Thus there exists b E 13 such that b

srparable over F ( B - { b } , p P ) and hence over the larger field E"(B purely

inseparable over EP(B

-

( 6 ) ) and therefore 6

E

EP(B

~

~

17

( 6 ) ) . Rut b is aliq

{ b } ) , contrary to t h e

assumed pindependence of 3.m Let E l f . ' be a field extension. We say that F is algebrazcally closed in E if every element of E which is algebraic over F belongs to F. Let E I F be a field extension and let F be perfect.

7.9. C o r o l l a r y .

exists a n intermediate field K such t h a t E

=

T h e n there

E p ( K ) , K I F is separably generated and K

is algebraically closed in E .

P r o o f . Take B to be any pbasis of E , K to be the algebraic closure of F ( H ) i n I;: and apply Theorem 7.8.

8 . C r i t e r i a f o r s e p a r a t i n g t r a n s c e n d e n c y bases.

Throughout this section, all fields considered are assumed t o be of prime characteristic p . O u r aim is twofold: first to provide necessary and sufficient conditions for a given set t o

be a separating transcendency basis of a field extension; second, to obtain necessary arid sufficient conditions for the existence of some separating transcendency basis of a givcn extension. We begin by recording the following three preliminary results. 8.1. L e m m a .

Let E / F be a separably generated field e z t e n s i o n with separatzny

trunscendency basts S . If B is a p-basis of F , t h e n B U S is a p-basts of E.

P r o o f . Since E / F ( S ) is a separable algebraic extension, it follows from Lemma 6.8 that there is no loss of generality in assuming that E = F ( S ) . Since B is a p-basis of E'. we have

F

=

Fp(13) C F ( S ) p ( B ) .Hence

E

=

F ( S ) i F ( S ) " ( Bb S)= E P ( BU S)

and so we are left to verify that B u S is p-independent in E .

CHAPTER 3

148

To this end, let

be any finite subset of B U S. Assume that

for some A,,

,...,n,,kl ,...,k , E F P . Since s1,. . . ,s, are algebraically independent over F , we

then have An, ,...,n,,kl ,...,k,b?'

. . . bkm = 0

n l ,...,n,

for any given k l , . . . ,k n E { O , l , . . . , p

-

I}. But the pm elements b;' . . . h%m are linearly

independent over FP by Corollary 6.5. Hence the pm+n elements

are linearly independent over FP. Now E I F is purely transcendental, hence so is EPIFP and thus, by Corollary 4.2, EP and F are linearly disjoint over FP. It follows that these p m + n elements are linearly independent over EP. Therefore, by Corollary 6.5, B U S is

p-independent in E , as required. m Property (i) of the following observation can also be derived as a consequence of the fact t h a t a separably generated extension is a separable extension. 8.2. Lemma. Let E I F be a separably generated field eztension. T h e n

( i ) T h e e z t e n s i o n E I F preserves p-independence ( i i ) .4ny separating transcendency basis

S of EIF is a relative p-basis of E I F

Proof. (i) Let B be a p-basis of F . Then, by Lemma 8.1, B can he included in a p-basis of E, hence B does remain pindependent in E .

(i i ) Let S be a separating transcendency basis of E I F . Then, by Lemma 8.1, B u S is a p-basis of E. Since, by (i), E / F preserves p-independence, S is a relative p-basis of E I F , by virtue of Theorem 6.13 (ii). w 8.3. Lemma. Let E I F be a f i e l d eztension, let F be perfect a n d let t r . d . ( E I F ) = 1 .

T h e n a n e z t e n s i o n KIE: is separable i f a n d only iJ the algebraic closure of E i n K is separable ouer E .

149

C R I T E R I A FOR SEPARATING TRANSCENDENCY BASES

Proof. By Theorem 6.10, K / E is separable if and only if K / E preserves p-independence Suppose first t h a t the algebraic closure of E in K is separable over E. If

E

were perfect,

then K I E would preserve p-independence, by Lemma 7.7(i). We may therefore assume that

E is not perfect, in which case, by Theorem 7.1, E/E’ has a separating transcendency

basis of one element, say A. This element is also a p b a s i s of E , by Lemma 8.1. Hence,

if K / E were not t o preserve pindependence, then X would be in KP, by Theorem 6.6(ii). Then X p

- I

is in K , but is inseparable over E , contrary t o the hypothesis. Thus K / E

preserves p-independence. Conversely, suppose that K / E preserves pindependence, but some algebraic element p. i n K is not separable over E . Then E is not perfect and, as above, E / F has a separating

transcendency basis of one element, say A. The irreducible equation f ( z , A) = 0 for z = p

> 1 in z, but

over FIX] then has exponent pe

because of its irreducibility has exponent 1 in

A. Regarded as a n equation for X over F i x ] , it shows that X is separable over F(pP‘)

C Kp.

flencc. X is separable over KP. But X is also purely inseparable over KP, so A is in K’

by Corollary 2.2.28. Since the latter contradicts the assumption that K / E preserves the p-independence of the p-basis {A}, the result is established. \Ye have now accumulated all the information necessary to prove our main results. 8.4. Theorem. (MacLane (1939a)). Let

S be a transcendency basis of E I F . T h e n

and only

il S

E / F be a separable field e z t e n s i o n and l e t

S is a separating transcendency basis of E / F i/

is relatively p-independent in E I F .

Proof. By Theorem 6.10, E I F preserves pindependence. Assume that S is a separating transcendency basis of E / F . By Lemma 8.2 (ii), S is a relative pbasis of E I F . llcncc S is relatively pindependent in E / F . Corivcrsely, suppose that

S

is relatively pindependent in E I F . Assume by way of

contradiction that S is not a separating transcendency basis of E / F . Then some X t I.: is not scparable over F ( S ) , hence satisfies for

X = X an irreducible polynomial equation

f ( X , S) = 0 with exponent a t least p in X and with coefficients in L’. Owing to Corollary 6.1 1 , at least orie variable s of

S has exponent

1 i r i this polynomial f . We can regard

/(X,S) = 0 as an irreducible and separable equation for f

s over

F(S

since

~- { s } , X p ) ,

irivolves S = X only as X p . Thus s is separable over the larger ficld E P ( F , S - { s } ) .

FIowevcr, s is also purely inseparable over this field, so s must lie in the field EP(F,S

(Corol!ary 2.2.28). This means that s is relatively pdependent on S

-

~

{s})

{s}, contrary to

150

CHAPTER 3

the assumption that S is relatively pindependent. 8 . 5 . Corollary. Let a n extension E / F have a finite separating transcendency basis

S. T h e n a n y relative p-basis B of E / F i s a separating transcendency basis of E I F . Proof. By hypothesis, E / F is separably generated, hence E / F preserves pindependence by Lemma 8.2(i). Therefore, by Theorem 8.4, S is relatively pindependent. It follows that

S can be included in a relative pbasis of E / F and thus IS/5 IBI. On the other hand, by Theorem 6.12, the elements of B are algebraically independent over F . Thus IB[ 5 (SI and so IBI = IS/, proving that B is a transcendency basis of E I F . Since B is relatively pindependent in E I F , the desired conclusion follows by virtue of Theorem 8.4. The hypothesis that the transcendency degree of E / F is finite is necessary for the validity of this corollary, even if we restrict the ground field F to be perfect, as may be seen by the following example. 8.6. Example.

(MacLane (1939a)). Let F be a perfect field. T h e n there exists

a separably generated field extension E / F and a relative p-basis of E / F which is not a transcendency basis of E I F .

Proof. P u t K = F ( t o , ~ l , ~ z , ~.)~where , . . the elements t o , x l , x 2 , z 3 ,... are algebraically independent over F . Let E

ty

=

tl..l

K ( t l , t z , . . . ) where

:

+ z;

(2.

= 1,2,3,.

. .)

Then S = {tol t l , t z , . . . } is a separating transcendency basis of E / F . Furthermore, the set B = { x ~ , x z ,. . } can be shown to be a p b a s i s of E . Since F is perfect, B is then a relative pbasis of E I F . Nevertheless, this pbasis B is not even a transcendency basis of

EJF. H The next result indicates that for finitely generated extensions there is no distinction between separably generated extensions and those preserving pindependence. Part (i) of the following result is a consequence of Theorems 6.10 and 4.4. We shall provide a n alternative proof for expository reasons.

8.7. Theorem. (MacLane (1939a)). Let E / F be a finitely generated eztension. (i) E / F preserves p-independence if and only if E I F is separably generated. (ii) I f E / F i s separably generated, t h e n a subset

S of E i s separating transcendency bas&

of E I F i f and only if S is a relative p-basis of E I F .

151

C R I T E R I A FOR SEPARATING TRANSCENDENCY BASES

Proof. (i) If E / F is separably generated, then E/F preserves pindependence, by Lemma 8.2(i). Conversely, assume that E / F preserves pindependence. It will be shown below t h a t if S is a relative pbasis of E/F, then S is a separating transcendency basis of

E/F Let B be any transcendency basis of E / F and let L be the separable closure of F ( U ) in E . Since E / F is finitely generated, E/F(B) (and hence E / L ) is finite. Since L / F ( B ) is a separable zlgebraic extension, B is a separating transcendency basis of E/F. Hence, by Lemma 8.2, B is a relative p b a s i s of L/F. On the other hand, E / L is a purely inseparable extension, by Proposition 2.2.30. Since E / L is finite, it follows from Lemma 6.14 that E I F has a relative p b a s i s S1 consisting of exactly n elements, where n = IBI. T h e two relative pbases

5'1

and S have the same number of elements, by Corollary 6.2. Hence

(Si=

n.

Since, by Theorem 6.12, the elements of S are algebraically independent over F, it follows that S is a transcendency basis of E/F. But, by hypothesis, S is relatively pindependent in E / F , hence S is a separating transcendency basis of E/F, by virtue of Theorem 8.4.

( i i ) Assume that E / F is separably generated. Then, by (i), E / F preserves pindependence. Ilence, if S is a relative pbasis of E/F, then, again by (i), S is a separating transcendency

bdsis of E I F . Conversely, if S is a separating transcendency basis of E / F , then

S

is a

rclative p-basis of E/F, by Lemma 8.2.(ii). 8.8. Corollary. Let E / F be a field extension of transcendency degree 1 and let E' be

perfect. If K J E i s a finitely generated field e z t e n s i o n such that the algebraic closure of E in K is separable over E , t h e n K I E i s separably generated.

Proof. Owing to Lemma 8.3, K/E preserves pindependence. Now apply Theorem 8.7.

The hypothesis that t r . d . ( E / F )

=

1 of the above corollary is essential as the following

example shows. 8.9. Example. (MacLane (1939a)). Let E = F ( z , y) be a rationat f u n c t i o n field of

two independent variables x and y over a perfect field F. T h e n the e x t e n s i o n K J E , where

K = E ( z , u ) , z i s transcendental over E and u satisfies up = y

+2 2

is not separably generated. Proof. Assume by way of contradiction that K J E has a separating transccndcricy

152

CHAPTER 3

basis S . Since tr.d.(K/E) = 1, we have S = {s} for some s E E . Let f(u,s)= 0 and g ( z , s ) = 0 be respectively the separable irreducible polynomial equations for u and z

over E [ s ]of respective exponents p n and p m in s. Then u is separable over E ( s p " ) , z is separable over E ( s p m ) and ,

spm

is separable over E ( z ) . If n

E ( z ) ,which contradicts the equality

up

2 m,

u is separable over

= y+xzP. A similar contradiction arises if n

5 m,

hence the assertion. m We next provide necessary and sufficient conditions for the preservation of pindependence An alternative proof of the result below may be obtained by applying Proposit,ion 4.21 and Theorem 6.10. 8.10. Theorem. (MacLane (1939a)). Let E / F be a n arbitrary field eztension. T h e n

the following conditions are equivalent:

(i) E / F preserves p-independence (ii) F(S)/Fis separably generated f o r a n y finite subset S of E . (iii) F ( S ) / F preserves p-independence f o r a n y f i n i t e subset S of E .

Proof. The equivalence of (ii) and (iii) follows from Theorem 8.7(i). It is also clear that (i) implies (iii). We are therefore left to verify that (ii) implies (i). Assume by way of contradiction t h a t (ii) holds but some pindependent subset S of F becomes pdependent in E . This means that there exists s E S such that s E EP(S'), where S' is a finite subset

of S not containing s. Therefore s E F p ( A Y , .. . ,XP,, 5") for suitable elements XI,. . . ,A,

in E . This shows that S is p-dependent in the field

F ( A 1 , .. . , A n ) . But F(Xi,.. . , A,)/F is finitely generated, hence by hypothesis is separably generated and therefore, by Theorem 8.7(i), preserves pindependence. This contradiction completes the proof of the theorem. Turning t o field extensions with finite transcendency degree, we finally prove 8.11. Theorem. (MacLane (1939a)). Let EIF be a field e z t e n s i o n with a jinite

transcendency basis S . T h e n EIF is separably generated if and o n l y if t h e following two conditions hold:

(i) EIF is separable (ii) F ( E P " ,s ) / F ( s ) is separable f o r some integer n

2o

SEPARABLE GENERATION

153

Proof. By Theorem 6.10, E I F preserves pindependence if and only if E / F is separable. Assume t h a t E I F is separably generated. Then, by Lemma 8.2(i), E I F preserves pindependence. T h a t E I F satisfies (ii) folbws by repeating the argument of the first paragraph in the proof of Theorem 7.5. Conversely, assume t h a t E I F satisfies (i) and (ii), and choose a relative pbasis

B

of

E / F . Since E / F preserves pindependence, it follows from Theorem 6.12 t h a t the elements of B are algebraically independent over F . Hence we may embed B in a transcendency

basis I3 U T of E / F . By hypothesis, each element of EP" is separable over F ( S ) . Since

2n

S is h i t e , we may choose an integer m over F ( B ,T ) . Hence each element of EP

rn+l

such that each element of EP'n is separable

.

IS

separable over FP(BP,TP) C F ( B ,TI') a n d

therefore over F ( B , T P ) . This shows that

Epm+'( F , B ) / F ( B . T P )

is separable

However, B is a relative pbasis of E / F , hence

E = E P ( FB , ) = E " - ( F ,B)= . . . = EP

".+I

( F ,B )

It therefore follows that E / F ( B ,TP) is separable. Since B U T P is obviously a transcendency basis of E / F , the result follows. a

9. Separable generation of intermediate field extensions.

Let E / F be a separably generated field extension and let K be a n intermediate field. Our aim is to discover conditions under which one of the intermediate extensions K / F or E / K is also separably generated. All fields considered are assumed to be of prime characteriitic P.

9.1. Theorem. (MacLane (1939a)). L e t F

CK

E he a c h a i n of f i e l d s arid 1 ~ :

l ; / F be separably g e n e r a t e d . If tr.d.(E/F) zs finite, t h e n K I F zs a l s o s e p a r a b l y generated Proof. By Lemma 8.2(i), E I F preserves pindependence and hence so does W F . Clioose transcendency bases S1 and Sz respectively of E I K and K I F . Then, by Proposition 2.6, 5'1 J S2 is a transcendency basis of E / F . Owing to Theorem 8.11, there exists an irltegei n

2 0 such that F(EP",S1,S Z ) / F ( S l ,Sp)is separable.

Therefore F ( K p " ,SL,S,)iE'

( S , , S,) is also separable. The adjunction of the indeterminates from S1 cannot reduce

CHAPTER 3

154

any equations irreducible over F(S2). Hence F(KP",S * ) / F ( S z )is separable. Invoking Theorem 8.11, we therefore conclude t h a t K I F is separably generated, as required. I 9.2.

Corollary. Let F

K

CE

be a c h a i n of fields s u c h t h a t EIF is separably

generated a n d tr.d.(K/F) is finite. T h e n K I F is separably generated.

Proof. Let S be a separating transcendency basis of E I F . Since K I F has a finite transcendency basis, all elements of this basis are algebraic over F ( S o ) ,where SOis a finite subset of S . Hence K is contained in the algebraic closure L of F(So) in E . Since So is a finite separating transcendency basis of L I F and F

C K C L , the

result follows by

applying Theorem 9.1. rn The above result need not be true without the assumption t h a t the intermediate field

K has finite transcendency degree over F. This fact will be demonstrated in the next section. Turning t o separable generation of E I K , we now prove 9.3. Theorem. (MacLane (1939a)). Let F C_ K

E be a c h a i n of fields, let E / F be

separably generated a n d let t r . d . ( E / F ) be finite. T h e n E I K is separably generated if and only i f E I K i s separable.

Proof. By Theorem 6.10, E I K is separable if and only if E I K preserves pindependence.

If E I K is separably generated, then E I K preserves p-independence by Lemma 8.2(i). Conversely, suppose that E I K preserves pindependence. Let S be a separating transcendency basis of E / F and let integer n

2

T be any transcendency basis of E I K . Since S is finite, there is an

0 such t h a t each element of SP" is separable over K ( T ) . But E / F ( S ) is

a separable algebraic field extension, hence so is EP"/FP"(Sp") by Lemma 6.7. Thus

K ( E p " , T ) / K ( T )is a separable extension and the desired conclusion follows by virtue of Theorem 8.11. I Given a field F , its subfield

n FP" w

n= 1

is obviously perfect. Furthermore, any perfect subfield of F is contained in n:=lFP".

Thus

fir=p_, FP" is the m a z i m a l perfect subfield of F . 9.4.

Theorem. (MacLane (1939a)). Let EIF be a separably generated eztension.

T h e n F contains t h e m a x i m a l perfect subfield of E (equivalently, F a n d E have the s a m e

SEPARABLE G E N E R A T I O N

155

m a z i m a l perfect subfield.)

Proof. Let L be the maximal perfect subfield of E . Assume by way of contradiction that L

F and choose X

E

L , X !$ F . Since L is perfect, all elements XP-" lie in L C E ,

so they are all separable over F ( S ) ,where S is a separating transcendency basis of E / F .

But X = X is the root of some equation g ( X , S ) = 0 irreducible in the polynomial ring

F I X , S].Let the variable s E S be chosen with smallest exponent p e in g . Then no other s, E

S has exponent less than p e and we put So = S

-

{s}. Then g ( X , S) = g ( X ,s , SO)

is a n irreducible separable equation for X = X over F(sP', SOPe), with exponent 1 in the variable

sPc.

It follows, from Lemma 7.2, that X is not separable over the smaller field

F ( S P C + ' , S$).

We may therefore choose an integer m such that for T = SP"' all the roots A""

separable over F(?'), while not all the roots

XP-"

are

are separable over F ( 2 " ) . Let p = XP-l'

be one such inseparable root, so that p itself is not separable over F ( T p ) ,although

pp-'

is

separable over F ( T ) . The latter, by Lemma 6.7, implies that p is separable over F p ( T p ) , hence over the larger field F ( T p ) , a contradiction. 9.5. Corollary.

If the elements of a set S are algebraically independent over a field

F . then

Proof. Let

F

= UFFEO_,FP-" be the perfect closure of

is F ( S ) / F ( S ) .Hence S is algebraically independent over generated. Since

F

F

and so

is perfect, it follows from Theorem 9.4 that

whence 00

as required m

F . Since F / F is algebraic, so

00

P ( S ) / F is

separably

CHAPTER 3

156

10. The Steinitz field tower.

We consider exclusively fields of prime characteristic p . Although a given field extension

E / F need not be separably generated it may still be possible t o represent the whole field E as the union of the fields of a tower

in which each individual extension E ; / F is separably generated. For example, if E =

F ( t ,t P

--I

,t p - 2 , . . . ) is obtained from a perfect subfield F

by the adjunction of all p t h roots

of a single indeterminate t , and if E; = F ( t P - ' ) for each i 2 0, then we have a tower

This tower has additional properties: (i) Each E , / F is separably generated (in fact, E , / F is a purely transcendental extension) (ii) E L - ] = EP for all i

21

It will be shown t h a t a similar tower can be obtained for a perfect field E . We also demonstrate that even if E is not perfect, a modified such tower is possible if t r . d . ( E / F ) is finite. The precise situation will follow from the ensuing discussion.

In what follows, up to Corollary 10.6, we shall uye the following assumptions and notation:

K is a perfect subfield of E

B is a p b a s i s of E F is t h e algebraic closure of K ( B ) in E (hence 13 is a separating transcendency basis of F / K , by Theorem 7.8; in particular B is a pbasis of F , by Lemma 8.2(ii), and E / F preserves pindependence, by Theorem 6.6(ii)).

T is a transcendcncy basis of E / F For any given integer n

2 0, we put

S, = S , ( E , F ( T ) ) = { A E EIApmisseparablc over F ( T ) }

10.1. Lemma. ( i ) S , is a field, S,P

S,-1 and E is the union

of the tower

THE

(ii) For each

n,

2 1, S,

=

STEINITZ FIELD TOWER

{ A E EIXPis separable over

157

$-I}

# S,-1 for all n 2 1.

(iii) ff E # F , t h e n S,

Proof. (i) Let L be the separable closure of F ( T ) in E . If XI, X z E S,, X z # 0, then

,ign

A:",

E L. IIince

which means that X 1

-

X z , X1X;'

are in S,. Thus S, is indeed a field. That E is the

union of the chain (1) is a consequence of Proposition 2.2.6(i). Finally assume that X E S,. Then

(XP)p"-l

is separable over F ( T ) ,hence

(ii) Assume that X E E is such that

XP

XP

E Sn-l, as required.

is separable over S,-1.

Then A"''

=

(XP)P"-'

"-1

is separable over Sz_l (Lemma 6.7). But, by the definition of S n P l reach element of

Sz:-l'

is separable over F ( T ) . Hence

Conversely, suppose that X 6 S., (iii) Assume that E

XP"

is separable over F ( T ) and thereforc X E S,.

Then, by ( i ) ,

XP

E S,_1 and

XP

is separable over S,-I.

# F . Then E is not algebraic over K ( B ) ,hence is

not algebraic over

F (since F / K ( B ) is algebraic). Therefore T is nonempty. Assume by way of contradiction that S, = S,-1

for some n

2

1. Then there is no X E E s u c h that

XP"

is, but A'"-'

separable over F ( T ) . This implies E = SnP1and, by (i), we have Ep"-'

C S:-l

n- 1

is not,

C So,

which means that each element of E " - ' is separable over F ( T ) , while each element of

EP" is separable over F ( T p ) . Since B is a p b a s i s of E , we have

This implies that each element of E is separable over F ( T P ) ; in particular, any t E ?' is also separable over F(TP). But t is also purely inseparable over F ( T P ) , hence t E F ( T P ) . This is a contradiction since the elements of TP are algebraically independent over F .

In what follows, we shall refer t o the tower of fields in (1) as the S t e i n i t z field tower for E over F . In the special case where E is a perfect field, the structure of the Stcinitz tower is given by the following theorem of F.K. Schmidt.

10.2. Theorem. A s s u m e that E is perfect and that K is algebraically closed in E

(hence B is e m p t y and F = K ) . T h e n , for all n

2 0,

(i) TP-" is a separating transcendency basis of S , / F

158

CHAPTER 3

(ii) S,

=

F(S:+,) (in fact, S, = Sz+l)

Proof. (i) The elements of TP-" are obviously algebraically independent over F . If X E S,, then

XP"

is a root of a n irreducible separable polynomial over F[T]. Hence

taking the pn-th roots of the coefficients of this polynomial, we see t h a t X is a root of a separable polynomial over

F[TP-"]. Since TP-" C S,, we conclude t h a t TP-" is a

separating transcendency basis of S,/F. (ii) By Lemma lO.l(i), S:+,

C S,.

Conversely, assume t h a t X E S,. Since E is perfect,

X = pP for some p E E . Since pPn+' = A P " is separable over F ( T ) , it follows t h a t p E S,+1.

Thus X E S:,,

and S,

C S:+,,

as required.

The rest of this section will be devoted t o the investigation of the properties (i) and (ii) of Theorem 10.2 in the case E is not necessarily a perfect field. With respect t o property

(i), it is natural to enquire whether each field S, of the Steinitz field tower (1) is separably generated over F . The following easy consequence of Theorem 8.11 provides a positive answer for the case where t r . d . ( E / F ) is finite. 10.3. Proposition. A s s u m e that t r . d . ( E / F ) i s finite. T h e n

(i) E a c h field S, of the S t e i n i t t field tower (1) has a separating transcendency basis T,

over F and a separating transcendency basis B U T, over K . (ii) IT1 = IT,/ for all n

Proof.

2 0.

Since each S,/F(T) is a!gebraic and the elements of T are algebraically

independent over F, we conclude that T is a transcendency basis of S,/F.

This shows

that (ii) is a consequence of (i). Since E / F preserves pindependence, so does S,/F. Furthermore tr.d.(S,/F) is finite, since so is t r . d ( E / F ) . Because, by definition, each element of Sg" is separable F ( T ) , it follows, from Theorem 8.11, that S,/F has a separating transcendency basis T,.

Since

B is a separating transcendency basis of F / K , it follows that B U T, is a separating transcendency basis of S,/K. For subsequent use, we need the following lemma 10.4. Lemma. A s s u m e that T

# 0 and, for a n y m 2 1, p u t C,

= S,

- Sm-l(C,

0, by L e m m a lO.l(iii)). T h e n

(9

F ( C i ) = F(S$)I

F(Cn) = F(Sn)

for any

n

21

#

THE

STEINITZ F I E L D TOWER

(ii) If T is finite, then for any integer n

2 0,

159

there ezists an integer m > n such that

Proof. (i) Since C , is nonempty, we may choose X E C , with not separable, over F ( T ) . If p is a n arbitrary element of

XP'"-'

XP"

separable, but

S, not in C,, then p E S,-1

and Xp must be in C,. Since X is in C,, p is in F(C,) and thus F(S,)

C F(C ,).

Therefore

we must have F(S,) = F(C,). Similarly,

proving that F ( S E ) = F ( C i ) . (ii) By Theorem 7 . 8 , we have E = E p ( F ) = F ( E P ) . Let T, be a separating transcendency basis of S,/F (see Proposition 10.3(i)). Then T , is finite, since T is finite, and T,

C F ( R p ) . Because E is in some S,,so that there is a n integer m > n such t h a t R C S,.

Hence there is a finite subset R of E such that T, of R

Consider now any element z in

S,. Since T, is

F(Ep).

= US,, each element

Thus T,

G

F(S&).

a separating basis of S,/F,

separable over F(T,) and hence over F ( S & ) . But z is also in S,, hence in S, m

> n.

Thus

zP

z is since

is in F ( S & ) ,so that z is also purely inseparable over F ( S k ) . It therefore

follows that z is in F ( S & ) ,hence S, 2 F ( S & ) = F ( C & ) by virtue of (i). So the lemma is verified. B We now turn our attention t o the property (ii) of Theorem 10.2, which says that

S,

=

F(S:+l) if E is perfect and K is algebraically closed in E . Though we cannot assert

this fact for every S, in the general case, we can assert it if IT1 = 1. Furthermore. if T is finite, then we can obtain another tower with analogous property by omitting certain of the fields from the Steinitz tower. This is the content of the following result. 10.5. Theorem. (MacLane (1939b)). Assume that I' i s finite. (i) I f T has ezactly one element, then S , = F(S:+,)

(ii) For each k

properties:

2 0 , there

ezists

nk

2 0 such that

for all

n

2o

the fields Mk = s,, satisfy the following

160

CHAPTER 3

Proof. (i) By Proposition 10.3, each field S , has a separating transcendency basis nver F consisting of one element, say t,. the elements of X of E for which

XP

Thus, by Lemma lO.l(ii), each S, consists of

is separable over F ( t , - l ) , n

2

1. By reason of this

symmetry, it clearly suffices to prove our conclusion Sn-l = F ( S E ) only for the case n = 1. Since t l is in

S 1

and not So, tl; is separable over F ( t 0 ) and K ( B , t o ) ,but

tl

itself is not

separable Over F ( t 0 ) and K ( B , t o ) . By the exchange lemma (Lemma 7 . 3 ) , some elcment of { l o } u B can be exchanged with t:. If t o is not so exchangeable, this means, as in the

eachange lemma, that t: is separable over K ( B , t E ) , and that some z in B can be here exchanged with t;. This exchange makes z separable over K ( B - { z } ~t ; , t y ) and thus over

EP(B

-

{x}). Since z is purely inseparable over EP(B - {z}),

z

lies in EP(B - {z}),

contrary to the pindependence of the set B. It must then be possible t o exchange t o with

t y . Therefore t o is separable over K ( t ? , B ) C F ( t y ) . By the transitivity of separability, every element of So is then separable over F ( t : )

C F ( S r ) . Now every element of So is in S 1 ,

and hence is also purely inseparable over L(Sf).We condude therefore t h a t So & F(Sf), as required. (ii) If T = @,then E

=

F and there is nothing t o prove. We may therefore assume that

T is nonempty. It suffices to exhibit integers 0 = no < n l <

n 2

< ... such that for

Mk = S,,, (c) holds. Indeed, that such fields Mk satisfy (a) is trivial, while (b) follows from Proposition 10.3(i). To prove (c), we shall show by induction t h a t if the intcgers n o < nl < ... <

Rk

have already been chosen, then there is an integer

that

Mk = Sn,

where Cn = 5' ,

-

C F(D:+i)

Sn-l, as in Lemma 10.4. Since, by Lemma 4.1(ii),

the result follows. A s an easy consequence of the above, we derive

nk+l

>

n k

such

THE STEINITZ FIELD TOWER

161

10.6. Corollary. Let K be a perfect subfield of a f i e l d E such that t r . d . ( E / K ) is

finite. Then there ezists a tower of fields:

such that ( i ) F I K is separably generated

( i i ) h f k / F is separably generated for all k

20

( i i i ) F is algebraically closed i n E and E = E p ( F ) (iv)

Mk

C F(D:+,)

for all k

2 0 , where

Dk+l = Mk+l - i b f k

Proof. Define F t o be the algebraic closure of K ( B ) in E , where B is a p-basis of E . T h e n any given transcendency basis T of E / F is finite, since t r . d . ( E / K ) is finite. Hence the required properties follow from Theorems 7.8 and 10.5. rn Our next aim is t o prove t h a t if t r . d . ( E / F ) is not finite, then the fields S , of the Steinitz field tower need not all have separating transcendency bases over F . We need the following preliminary result. 10.7. Lemma. Let T be a set of elements which are algebraically independent over

some f i e l d F and let E = F ( T P w W )be the f i e l d obtained from F b y the adjunction of all elements t""

for n any integer and t i n 7'. Then, for any n 2 0 , F ( T p - " ) is ezactly the

set of all elements X of E such that

XP"

is separable over F ( T ) .

Proof. Let E , denote the set of all X in X

t

F ( T P - " ) , then A""

E F"''(T)

C En.We

shows t h a t F ( P - " )

E

E F ( T ) and

such t h a t A!"'

hence A""

is separable over F ( T ) . If

is separdble over F ( T ) . This

are therefore left t o verify t h a t

En C F ( T p - " ) .Since any

clement of E n depends algebraically on finitely many elements of 7', we may harmlessly rissii~ncthat T is finite. We now argue by induction on 17.1.

T = { t } . Fix rn > n and put

First assume t h a t

z =

t""'

We claim t h a t

(E,(tP-"') : E n ) = p"-" Indeed, since

tP""'

=

need oniy verify t h a t

tp-",

2

is a root of the polynomial XP"'-"

XP"'-" - t p - "

(2)

~

tP-'L E E,[X]. so we

is irreducible in E,[X]. Assume the contrary. Then

162

CHAPTER 3

the p t h root of tP-" is in E n , so tP Consequently, tP

-1

-1

.

is separable as well as purely inseparable over F ( t ) .

E F ( t ) which is impossible, and thus (2) is established.

Assume by way of contradiction t h a t En 9 F(tp-") and choose X E En - F ( t p - " ) . We

> n and X

may choose m large enough such that m

E F ( t P - m ) .Since F(tP-",A)

# F(tP-"),

we have

(F (tp --= ): F(tP.", A)) < ( F ( t P - ) : F(tP-")) = pm-n

En,it follows from (2) t h a t

On the other hand, since F(tP-", A)

= ( E n ( t p - m ) : En) = pmPn,

a contradiction. This establishes the case where IT1 = 1. Suppose next that the result is known when the transcendency basis has n-1 elements, and let T = T o U { t ) h a v e n elements, t

4 TO.Note t h a t E contains asubfield F1 = F(T;--)

and E = FI(tp-"). Any X in E with

XP"

separable over F ( T ) has

XP"

also separable over

F l ( t ) 2 F ( T ) . Hence X is contained in

by the proof of the previous case. Setting FO = F (tP-"),it follows t h a t X is in Fo(T;-~) and XP-" is separable over Fo(T0). Hence, by the induction hypothesis,

proving t h a t En 2 F ( T p - " ) , as required. w We have now accumulated all the information necessary t o provide a field extension for which the first field of the Steinitz tower is not separably generated. Let K be any perfect field and let

T = { t o , t i , t z r . . . >, Y = {YZ,~ 3 ,. .. } be two countably infinite sets of elements such that (a) The elements of T are algebraically independent over

K

(b) The elements yi of Y satisfy YI:

=

tn-2

+ tn-ltP,

(n = 2 , 3 , 4 , . . . )

(3)

THE

STEINITZ F I E L D

TOWER

163

Let the field E be defined by

10.8. Theorem. (MacLane (1939b)). With the assumptions and the notation above

the following properties hold: (i) The set D = ( t o } is a p-basis of E (ii) ?'he algebraic closure F o f K ( B ) in E is K ( B ) (hence TI = { t l , t z , . . . } is a transcen-

dency basis of E I F ) (iii) The Steinitz field

F ( T 1 ) is given by

S 1

= S l ( E , F ( T l ) ) of all elements

S 1=

X

of E with

XP

separable over

F ( T 1 , Y ) = K(to,TI,Y)

(iv) The f i e l d S i s not separably generated over F .

Proof. (i) The defining algebraic equations ( 3 ) can be rewritten as:

It follows, by induction on n , that each t , E EP(to),so that

and hence 15' = EP(t0). We are thus left t o verify that t o @ EP. Assume by way of contradiction that t o E EP. The n i 1 algebraically independent elements t o , 1 1 , . . . , t , are algebraic over K ( t o ,t l , y ~ , ... , y n ) by virtue of ( 3 ) . Hence the 72

~i 1 elements

20, t l , yz,

. . . , yn must themselves be algebraically independent over K . The

conclusion is that Y u {to,tl} is a set of elements algebraically independent over { t o , 1 I } are likewise algebraically independent over the subfield

K ,and

K ( Y P - = ) of E . Consider

the following subfields En of E :

Then we obviously have E = U,E,.

It follows from ( 3 ) that the t ' s in this field En can

be expressed rationally in terms of the y's and the last two t ' s . Thus

En = K ( Y P - m , t n - - l , t n )

164

CHAPTER 3

and { t n - l , t n } is a set algebraically independent over K ( Y P - - ) . Since t o E EP and

E

=

U,E,,

we have

for some n. Invoking (4) successively, we then conclude t h a t t l , t 2 , . . . and finally tn..l

are

also in EE. On the other hand, the elements tE-l,tP, are algebraically independent over

K(YP--), so that the extended field Ef: = K ( Y P - m , t c - l , t g )cannot contain a p-th root --I

t,-l

= (tEp1)P

. This contradiction shows t h a t

to $ EP, as required.

(ii) By Theorem 7.8, any element X in F is separable algebraic over K ( t 0 ) and hence over

K ( T ) . On the other hand, E is obtained from K ( T ) by the successive adjunction of p-th roots, which means that E / K ( T ) is purely inseparable. Therefore all elements X of F lie in K ( T ) . The remaining elements of T are algebraically independent of t o , hence F must be K(t0) as required. (iii) It follows from (3) t h a t each element of YP is separable over K ( t 0 , T I ) . Hence

Conversely, S1 consists of certain elements of Ep-separable over K ( T ) . Thus, by Lemma 10.7, S1

= K(TP-") whose p t h power is

K ( T p - ' ) which shows that

Applying (4) we see that M2 can also be generated as

It therefore follows that the field Mz of (5) has degree p or 1 over M I , which in turn -L

implies that S1 = M I or

S1

= Mz.

If

S 1=

Mz, then

tg

is in

S 1

CE

and hence t o E W',

contrary to (i). Accordingly, S1 = M 1 = F ( T l , Y )= K ( t o , T 1 , Y ) . (iv) We first show that for any n

2 0,

Indeed, assume that t , E: Sf for some n 3 0. Then the equation (4) solved for that t,-l t o are in

is in

Sf.A repetition of this argument implies that

Sr 5 EP, contrary to (i).

tn-2,

tn-3,.

t,-l

shows

. . and finally

THE S T E I N I T Z F I E L D TOWER

165

E Sr for some n 2 0. Then the equation (3) written i n

Now assume that t,/t,+l

tile

form

~:&zltn+L l tnltn-+l+tP,+z and hence tn+l are in

implies that

Sf, contrary to the previous paragraph. Thus

(6)

is established. dssume by way of contradiction that S 1 / F is separably generated. We rnay then choose a scparating transcendency basis over F = K ( t o ) . The adjunction of t o to this basis yields an enlarged separating transcendency basis, say Z =

(21, z g ,

. . . } of Si

over

K . O u r aim is to show that this leads to a contradiction by finding a single z the adjunction of which would simultaneously make yk and

tk

separable, in conflict with the form of ~ h c

inseparable defining equation ( 3 ) . Observe that both to and

Z,

= {z,,

. . . , z,)

tl

are separable over K ( Z ) , so there is a finite subset

so large that t o and t l are separable over K(2,).

independent elements of T cannot be dependent on this subset Z,, n

2

2, such that

tO,tl,

of z‘n, so that there is a set zk+l),

there exists an integer

. . . , t n - l are algebraic (and hence separable) over K(Z , ), while

the next element t , is not so algebraic over Z,.

K(Zk,

Because all of the

But t , will be algebraic over a larger set

z k = ( ~ 1 , . . . ,Z k ] ,

(k 2 m),for which t , is algebraic ovcr

but not over K ( Z k ) . It follows from (3) that

(a) y n is algebraic over K(tn-z,t,-l,t,)

(b) t , is algebraic over K(t,-z,t,..l,y,) Now both

and

tn-2

t,_l

are already algebraic over K ( Z k ) 2 K(Z,).

nor yn can be algebraic over K ( Z k ) ,and both

where z =

Hence neither t ,

t , and yn must be algebraic over K ( Z k , z ) ,

z k + 1.

From the equations for t , and y, over K ( Z k , z ) , it follows (from Lemma 7.2) that t h c r r are larger integers e and

f such that

( 7 ) 1, is separable over K ( Z k , z P c ) ; yis, separable over K ( Z k , z ‘ ” ) l3y Lemma 7.3, we then have (8) zp‘ is separable over K ( Z k , t , ) ; zp’

Assume that

e

2 f. Then the first

is separable over K ( Z k , y , )

statement of (7) and the second statement of (8) imply

that t , i s separable over K ( Z k , y , ) . Let L denote the separable closure of P ( Z k ) in S1.By construction, t,-z

and t,-l

are in L so that ( 3 ) makes t , purely inseparable over L(y,).

Accordingly, t , E L(y,) and t , is a rational function

CHAPTER 3

166

where we can assume that the coefficients f ( O ) , g ( O ) are not both 0. Substituting this value

o f t , in (3) yields Yfl[g(Yn)lP= t n - z [ s ( y n ) l P + t n - l [ f ( Y n ) l P Here the variable y n over L can be replaced by 0, which gives

One (and hence both) o f f ( O ) , g ( O ) are distinct from 0. Thus

contrary to ( 6 ) . Finally, in the remaining case where e


a similar argument shows t h a t y n E L ( t n )

and hence tn-2

E Lp

EP,

again a contradiction. This completes t h e proof of the theorem. D We close by providing a n example of a separably generated field extension E / K and an intermediate field F such that F / K is not separably generated. 10.9. Example. (MacLane (1939a)). Let K be a perlect field a n d Let T = { t o , t l , . . . }

be a set of elements algebraically independent over K . Define a n o t h e r set Y = {yz, y3.. .} successively b y the inseparable equations

y; = t n - 2

+ tn-1t;

( n = 2 , 3 , 4 , . . .)

Put F = K ( T , Y ) a n d E = F ( T p - ' ) . T h e n E I K is separably generated but F I K is not separably gene rated.

Proof. T h a t F I K is not separably generated follows from Theorem 10.9 (iv), where the present F appears as a field S1, and where it is shown that F has a separating transcendency basis neither over K nor over K ( t 0 ) . On the other hand, we have

Hence TP-I is a separating transcendency basis of E I K . rn

NONSEPARABLY GENERATED F I E L D S

167

11. Nonseparably generated fields over maximal perfect subfields.

All fields considered are assumed to be of prime characteristic p . To motivate our discussion, let us first record the following result. 11.1. Proposition. Let F be t h e m a z i m a l perfect subfield of a field E . If t r . d . ( E / F ) = 1, t h e n

E I F is separably generated.

Proof. Since E

# F, E

is not perfect. The desired conclusion is therefore a conse-

quence of Theorem 7.1. Our aim is t o give a counter-example t o the possible extension of the result above.

Namely, we show t h a t (a) There is a field E which is not separably generated over its maximal perfect subfield

F , but which does have a finite transcendency degree t over F . Furthermore, t may be any specified integer

2 2.

This example will also illustrate t h a t (b) The number of elements in a pbasis of a field K need not be equal to the transcendency degree of K over its maximal perfect subfield. From now on, we fix the following data.

K is a perfect field

S

= ( S ~ , S Z ,...

} a denumerable set of elements algebraically independent over K

F = K ( S p - - ) = K ( S , SP-' ,SP-',. . . ) (note that F is perfect) y , uo are algebraically independent elements over K ( S p - = ) u1, ~ 2 , . .. ,u,, n-

. . . , are defined recursively by

I

u, = yp

( n 2 1)

i-S n U n - 1 -2

E' = F ( y , U O , U ~ - ' , U ~, . . . , U P ,

-n

(1)

, . . .)

11.2. Theorem. (MacLane (1939a)). W i t h the n o t a t i o n above, the following prop-

erties hold: ( i ) E / F is not separably generated. (ii) T h e field E h a s a p-basis consisting of o n e lernent y a n d { y , u o } is a transcendency

basis of E / F .

( i i i ) F is the rnazirnal perfect subfield of E .

Proof. (i) and (ii) It is clear that {y,uo} is a transcendency basis of E / F . By ( l ) ,

168

CHAPTER 3

L

- I,

can be expressed in terms of u < - " , so E is the union of a tower of fields

u",-,

Eo &El

Ez

C ...

En

...

where

En = K ( S P - m , y , ~ E - n ) Applying (I), we observe that

which shows t h a t sl,.. . , s,, y, u, are algebraically independent over K . Moreover,

and hence En/Eois a purely inseparable extension of degree p n . Invoking the necessary condition of Theorem 7.5 (applied to the transcendency basis S = {y,

UO})

we deduce that

E / F is not separably generated. Since the extended field E ( y p - ' ) , by the equations ( l ) ,

u E - ~ , {y} -n

also contains each

is a pbasis of E.

(iii) For the sake of clarit,y, we divide the proof into a number of steps. S t e p 1 . Here we s h o w that a n e l e m e n t X of E is in En if a n d o n l y if Xpn is separable over

Eo .

E n have the desired property. Conversely, assume that

Owing to ( 3 ) , all elements X of XP"

is separable over Eo, and choose k

By Lemma 10.7, in such a field

E k

2

-k

n large enough so t h a t X E

any element X with

XP"

E k =

Eo(u; ).

separable over Eo must be in

E o ( u f - " ) . If k > n , then by (l),

-n

-n

hence E O ( U ; ) = Eo(u;-,). Applying induction, we therefore deduce that

as required. S t e p 2. Here we provide s o m e useful properties of the m a z i m a l perfect subfield L of E .

NONSEPARABLY GENERATED F I E L D S

169

Fix any element a in L. The expansion of a in terms of the generators of E can involve only finitely many of a's, z's and pn-th roots of z's. Therefore we can End a power

c

= a p " E I,

and an integer m such t h a t c E K ( S m ,y , a o ) ,

For each integer k

2 0, c

c = c i k for some

where

ck

S, =

(51,. . . ,s,}

E E and, by Step 1 , c k E E k . Therefore

E K ( S m , y , u o ) ,c E E i k

(k 2 0 )

Our next aim is to show that

C

For each k

2

E K(Ski

( k = m , m - + l...) ,

Ypkj Uk)

m , choose the smallest n such that c E K ( L 5 ' ~ - m , y p k , ~ kIf) .n > k, c is

not in the field

R

= K ( S i I , , y p k , u k ) , but is in

--t

R(sP, ) for some t . As in the argument,

hllowing ( Z ) , s, is transcendental over K ( S , - l , y , u k ) , hence over R. Since c is in the --t

purcly transcendental extension R ( s ~), it must itself be transcendental over

R. K o t e

also that, by (5) and (4), c E

c

K ( S m , y > a O ) K(Sk,YruO) = K ( S k , Y , u k )2 K ( S n - l , Y , u k )

and the last of these fields is algebraic over R. IIence c is algebraic over R , a contradiction, from which we deduce that -m

c E K(S:

,ypk,ak)

for ail k = m , m + I , . . .

\Ve now show t h a t in the expression of c in terms of these generators no roots of

SI;a r c

inrrolvc.d, which will prove ( 6 ) .

Assume by way of contradiction that the expression for c docs involve some roots 01 sk.

Choose t

> 0 so small that c is in N

'i'tie

= K ( s f t , y P k , a kb)u t not in

K(Skp-'+l,Y p k ,U k )

extension T / N is given by a tower

N o c N i c N z c ... G N

CHAPTER 3

170

where -t

No = N , N ; = N ( s ~, . . . ,

sY-')

(z = 0 , 1 , . .

Choose i such t h a t c E N; but c $! N i _ l . Because

xp

N,

c

. , k)

is purely inseparable over

-t

N;-l, and the two extensions N ; - l ( c ) and N; = N ; - l ( s : ), each of degree p , must be equal. Accordingly, sr-' E l V P l ( c ) .Setting

--t

sp

E M(s;-t+j,

contrary to the fact that sI is transcendental over M .

S t e p 3. Our aim here is to prove that

(note that Dkk is the field K ( S k , y P k , u kof ) ( 6 ) ) . It suffices to prove t h a t

Dkrn is algebraically closed in Dkk

for all k

2m

(8)

Indeed, by (5),we have

a field which is certainly algebraic over Dkm. On the other hand, c E Dkk = K ( S k ,yp*, u k ) ,

by ( 6 ) , and this field by (8) contains no elements algebraic over D k m rexcept the elements

of Dkm themselves. This shows t h a t (7) must hold.

We are left to verify (8). To this end, we first simplify the notation of (7) in the following way: k--n

Dkm = K ( S n , Y l , V ) ,

Y1

= Y p k , v = $.

(9)

171

NONSEPARABLY GENERATED F I E L D S

Therefore Dkn

C D k n + l , according t o

(9) and

(lo), and we can

go from Dkn to Dkk by a

tower

Dkn

C Dkn+l G Dkn+2 2 . . . C Dkk

Hence, t o prove ( 8 ) , it suffices t o show that Dkn is algebraically closed in Dkn+l for all n

< k. Owing to (9) and

(lo), Dkntl

=

Dkn(S,U) where, as in ( 2 ) , s is transcendental over

Ukn, while by (ii), u is inseparable and algebraic over D k n ( s ) .If Dkn is not algebraically dosed in D k n r l , choose and

in Dkn+l

-

Dkn and algebraic over Dkn. Then

x p is algebraic over Dknrso XP € Dkn. Accordingly,

x p E Dkn(s)

is purely inseparable over Dkn

and also over D k n ( s ) ,and we must have

Thus u is a rational function g ( s ) / h ( s ) of s with coefficients in D = Dkn(X). We may assume that g(0) and h ( 0 ) are not both 0. This value of u in the defining equation (11) for u yields an identity

+

h(s)P(y1

= g(s)P

SP"-",)

(12)

i n s over D . By setting s = 0, we find

wilh neither h ( 0 ) nor g(0) zero. This implies that y1 E D", hence yy argiirrient on the terms of highest degree in (12) proves that

up-'

E

--i

E D . A similar

D . But, in

of

(9), S,, y1 and u are algebraically independent over K , so that y 1 2)~ are p-independent i n

Dk,,. This, however, implies that the extension

-1

has degree p 2 . But we have just shown Dkn(yy

,u p - ' )

to be contained in D , of degree p

over Dkn. This contradiction establishes (8) and hence (7).

Step

4. Completion

of the proof. P u t L1 = K(S,)

and /I = k

-

rn. Then, by ( 7 ) , c is i n

each of the fields Dkm = L l ( ( y p ' " ) P " , u Here ~ ) . yp"' and u, are algebraically independent =

1 , 2 , . . . is L1 itself (see Corollary 9 5 ) .

cp-"

of the maximal perfect subfield L is

over L , so the intersection of all these fields, for p Thus c E K ( S m ) and the original element a =

172

CHAPTER 3

therefore in K ( S R " ) . This shows that F = K ( S p - " ) is the maximal perfect subfield of

E , as required. w Remark. Let E and F be as in Theorem 11.2. Then tr.d. ( E / F ) = 2 and F is the maximal perfect subfield of E . A field with analogous properties but with any desired transcendency degree t

2 2 over its maximal perfect

is a set of t

-2

subfield is, by Theorem 9.4, the field E ( T ) , where T

elements algebraically independent over E. Furthermore, since E / F is not

separably generated, E ( T ) / F is also not separably generated, by virtue of Theorem 9.1.

1 2 . Relatively separated extensions. Throughout, the letter p is reserved for the characteristic of the fields under discussion, and since our questions generally evaporate when p = 0 , we shall always assume that p

> 0.

We begin by introducing the following terminology due to Mordeson and Vinograde (1972,1973). Let E I F be a field extension. If B is a relative p b a s i s of E / F such that

E / E ' ( B ) is separable algebraic, then we call B a separating relative p-basis. We say that

E / F is relatively separated if every relative pbasis of E I F is a separating relative p-basis. For example, if E / F has a finite separating transcendency basis, then by Corollary 8.5, any relative p b a s i s of E / F is a separating transcendency basis of E/F. Thus E / F is relatively separated. On the other hand, if E / F preserves pindependence (e.g. F is perfect), then the elements of any relative pbasis of E / F are algebraically independent over E', by virtue of Theorem 6.12. Hence, in this case, any separating relative p b a s i s is a separating transcendency basis. In particular, if E / F preserves pindependence and is relatively separated, then E / F is separably generated. Note also that if K is an intermediate field of any extension E / F , then E = E p ( K ) whenever E / K is separable algebraic (see Proposition 2.2.6(ii)). It turns out that relatively separated extensions E / F are precisely those for which the converse of the above holds, for any intermediate field K of E / F . Before proceeding to the study of relatively separated extensions, it will be convenient to establish a number of auxiliary results, so as not to interrupt our future discussion a t an awkward stage. First we must develop our vocabulary. The imperfection degree of a field extension E I F is defined to be the (cardinal) number of elements in any relative p b a s i s of E / F . By Corollary 6.2, this concept is well defined. Note t h a t , by Theorem ll.Z(ii), even if F is perfect, the imperfection degree of E I F need not be equal t o the transcendency degree of E / F . We say that E I F is relatively perfect if

R E L A T I V E L Y SEPARATED EXTENSIONS

ils

imperfection degree is zero, or equivalently if E

=

173

EP(F).

12.1. Lemma. L e t E/E’ be a f i e l d e x t e n s i o n .

( i ) If B is a relative p-basis of EIE’, t h e n E / F ( B ) i s relatively perject. ( i i ) If E / P

is

separable algebraic, t h e n EIE’ is relatively perfect.

( i i i ) If E / F i s f i n i t e l y generated, t h e n E / F is relatively perfect if a n d o n l y if E / F i s separable algebraic.

Proof.

(i) If R is a relative p b a s i s of E / F , then E = E J ’ ( F , R )a n d therefore

I< =: E P ( F ( I 3 ) ) . ( i i ) This is a direct consequence of I’roposition 2.2.6(ii). ( i i i ) Let E / h ’ be finitely generated and relatively perfect. Ry (ii), it suffices to show thar. k ; / F is separable algebraic. If S is a transcendency basis of

E / F ,t h e n E / F ( S ) is a finitcl

extension which is relatively perfect since so is E I F . IIence, by Proposition 2.2.17, E / F ( S ) is sclparable. Thus

S is

a separating transcendency basis of

EIE’. Therefore, by Lemma

8.‘2(ii), S is a relative p b a s i s of E / F . R u t B / F is relatively perfect, hence

S is empty and

S j b . is separable aigebraic. m T,et,

E/Is. be a field extension and let K b e a n intermediate field. We say t h a t E , ’ K

1~reserzresrelative p - i n d e p e n d e n c e w i t h respect t o F if every relatively p i n d e p e n d e n t subset

.So f K i F remains relatively p i n d e p e n d e n t in E / F . 12.2. Lemma. L e t I< be a n z n t e r m e d i a t e field of a field e x t e n s i o n K / F . T h e n

ffie

jollowzng c o n d i t i o n s are equivalent: (i) 15‘:’h‘ p r e s e r v e s relative p - i n d e p e n d e n c e with respect t o F [ii)

T h e r e ezists a relative p-basis of

K l F w h i c h i s relatively p - i n d e p e n d e n t in E I F .

Proof. It is clear t h a t ( i ) implies (ii). Conversely, assume t h a t there exists a relative [I

basis 11 of

lilF

which is relalively p-independent i n h ’ / F . If there exists a set S i n l i

\vhich is rclatively p i n d e p e n d e n t in K / F , b u t is not relatively p-independelit i n E / F . t lien

there is a finite subset S’u { s } of S such that s $ 5‘ and s E El’(F, .5’). 1,et C’:-~ ij’ h c finite set such t h u l S‘ LJ { s }

KP( F’, U’). T h e n there e x i s k a subset 11” of 13‘ such t h a t

;I

174

CHAPTER 3

same number of elements, say n. Thus

E P ( F ,B’) = E P ( F ,S’, S,B’‘) and therefore

( E P ( F ,B’) : E P ( F ) = ) pn > pn-’

2 (EP(F,S’,s,B’’): EP(E’)):

which contradicts (1). rn

12.3. Lemma.

Let K be a n intermediate field of a n extension E / F , let E I K

preserve relative p-independence with respect t o F , and let B and B‘ be subsets of E such that B

n B‘

=

0 and B‘ C K .

(i) If B is a relative p-basis of E I K and B’ is a relative p-basis of K I F , t h e n B U B’ is

a relative p-basis of E I F (ii) If B’ is a relative p-basis of K I F and B u B’ is a relative p-basis of E I F , t h e n B is

a relative p-basis of E I K (iii) If B is a relative p-basis of E I K and B U B’ is a relative p-basis of E I F , t h e n B‘ is

a relative p-basis of K I F . Proof. (i) By hypothesis, E = EP(K,B ) and K = K P ( F , B’), whence

E = E P ( K P , F , B , B ’ )= E P ( F , B , B ’ ) We are thus left to verify that B u B’is relatively p-independent in E I F . If there exists

6 E B such that b E E P ( F , B ‘ , B

-

{ b } ) , then

b E E P ( F ,K , B

-

{b})

=

E p ( KB ,

~

{b}),

a contradiction. Assume that there exists b’ E B’ such that

b‘ E E P ( F ,B‘

-

Since B’ is relatively p-independent in E I F , b’ distinct elements b l , . . . , b,,

m

2 1, in B

{b’}, B )

6 E P ( F ,B‘

-

such that

b‘ E E P ( F ,B‘

-

{ b ’ ) , b l , . . . ,b,)

and b’ $! E P ( F B‘ ,

-

{ b ‘ } ) ; b ~ ,. .. b m - l )

{ b } ) . Therefore there exist

175

R E L A T I V E L Y SEPARATED E X T E N S I O N S

Hence, by t h e exchange property,

b m E E P ( F B’, , b l , . . ., b m - 1 )

2 E P ( K b, l , . . . ,b m - i ) ,

contrary to the fact t h a t B is relatively p in d e p e n d en t in E / K . (ii) By hypothesis, K = K P ( B ’ , F ) a n d E = E P ( B , B ’ , F ) .Hence

E P ( KB , ) = E P ( B B’, , F) =E lissunie t h a t there exists b E B such t h a t b E E P ( K , B - ( 6 ) ) . Th en

b E E P ( F , K P , B f ,B { b } ) = E P ( F , B f , B- { b } ) , contrary t o t h e assumption th a t B U B’ is a relative p b a s i s of E / F . Th u s

B is relatively

p-independent in E / K , as required. ( i i i ) Since U U I3’ is a relative p b a s i s of E / F , B’ is relatively p i n d ep en d en t in hcncc in K / F . Assume by way of contradiction t h a t K

E/F and

# K P ( B ’ , F ) . Th en there exists

an element X in K such t h a t B’U { A } is relatively p-independent in K / F whence in E / J ’ hy hypothesis. Thu s X Hence there exists b

4 EP(F,B‘)

X € EP(F,R’,B)

but

B such t h a t b E E P ( F , B ’ , X ,B

-

{b})

i E P ( K , B- { b } ) ,

contrary t o t h e assumption t h a t B is a relative p b a s i s of E / K . T h e following theorem will enable us t o take full advantage of t h e results so far oht ained. 12.4. Theorem Let K be a n intermediate f i e l d of a n eztension E I F .

If I.:’K

10

separable, then E / K preserves relative p-independence with respect to F .

Proof. Let S be a relatively p i n d e p e n d e n t subset of K / F . To prove t h a t S remains relatively p i n d ep e n d e n t in E / F , we may harmlessly assume t h at S is finite, say S

=

{sl,.. . , sm}. Since S is relatively p in d e p e n d e n t in K / F , it follows from Lemma 6.4(iv)

t h at the pm elements s;‘ . . . s?,

0

5 n, < p , are linearly independent over Kp(JF).

1-13 l x m m a 6 . 4 ( i v ) , wc are therefore left to verify t h at t,hese p“‘ clements a r e linearly

ir:tiependoiit ox-cr EP($‘).

176

CHAPTER 3

Let b l , . . . , b , be elements of K linearly independent over K p ( F ) . We must show that they are linearly independent over E P ( F ) . To this end, let { z a } be a basis of E over K . Since E J K is separable, { z a }are linearly independent over KP independent over K . Since E = a subring of E p [ F ]containing

and therefore, for any a ,

{zP,} are linearly

EP

=

and F . Thus E P [ F ]=

+ . . . + A,b,

for uniquely determined ;A,

, hence

c,K z a , EP caKPzP, and therefore caKP[F]zP,is

E P [ F ] is a free KP[F]-module with basis Assume that A l b l

-1

C , Kp[F]zz which

shows that

{zP,}.

= 0 for some X1,. . .,A,

in EP[(F].We may write

E K p [ F ] .Hence

c;"=, X,;b, = 0 (since {zi} are linearly independent over K ) .

Since b l , . . . ,b, are linearly independent over h'P(F), we have ;A,

=0

for all i E { l , . .. , n } and all

CL

Thus each A; = 0 and, since E P ( F ) is the quotient field of E p [ F ] ,the result follows. .4n alternative proof of a part of Theorem 6.10 can be easily deduced from the preceding result. Namely we have 12.5. Coroliary. Let E / K be a separable field e z t e n s i o n ( i ) E / K preserves p-independence

( i i ) T h e elements of a n y relative p-basis of E J K are algebraically independent over K .

Proof. Property (ii) is a consequence of ( i ) and Theorem 6.12, while (i) is a particular case of Theorem 12.4 in which F is perfect. 12.6. Corollary. Let K be a n intermediate field of a n e z t e n s i o n E J F and let EIK

be separable. (i)

If B is a relative p-basis of EIK and B' is a relative p-basis of K J F , t h e n B

IJ B'

is

a relative p-basis of E J F ( i i ) If S is a relative p-basis of K I F and E J K is separable algebrazc, t h e n S is a relative

p-basis of E I F .

177

R E L A T I V E L Y SEPARATED EXTENSIONS

Proof. (i) This is a direct consequence of Lemma 12.3(i) and Theorem 12.4. (ii) This is a special case of (i) in which B = 0 (by Lemma 12.l(ii)) and B' = S.

We are now ready to return t o our study of relatively separated extensions. 12.7. Proposition. T h e following properties of a field eztension E / F are equiualerif . (i) E / F is relatively separated.

(ii) If K is a n intermediate field of EIF s u c h that E / K is relatively perfect, t h e n E j l r

is separable algebruic.

Proof. Assume t h a t EIF is not relatively separated. Then there exists a relati\c p-basis B of E / F such that E / F ( B ) is not separable algebraic. Since B is a relative p-I>a\i$ of E / F >we have

E = E P ( F , B )= E P ( F ( B ) ) Setting K = F ( B ) ,it follows that (ii) does not hold. Conversely, assume that (ii) does not hold. Then there exists an intermediate field l i of E / F such that E = EP(K) but E / K is riot separable algebraic. Since E

=

EF((F,K ) ;

i t follows from Corollary 6.2 t h a t K contains a relative pbasis, say B , of E J F . Becauw

E / K is not separable algebraic, the same is true for E / F ( B ) . We have therefore found

d

relative p b a s i s B of E / F which is not a separating relative pbasis of E I F . Thus E: I ' is not relatively separated, as required. 12.8. Corollary.

If E I F is relatively separated, t h e n E I K is relatively sepilruterl

Jor

u n y intermediate field K of E I F . Proof. Assume that E l F is relatively separated and that K is an intermediate f i c , l t l of E J F such that E l K is not relatively separated. Then, by Proposition 12.7, there exist. an intermediate fie!d L of E / K such that E =: E p ( L ) and E / L is not separable algebraic.

'This contradicts Proposition 12.7, since L is also an intermediate field of E I F .

Next we present t h e following useful result. 12.0. Theorem. (Mordeson and Vinograde (1972)). A s s u m e that E J F is a sepcirnirlr:

field eztension. T h e n the following properties are equivalent: ( i ) For a n y intermediate field K of E I F , K I F is relatively separated. (ii) E / F is relatively separated.

(iii) E / F has a f i n i t e separating transcendency basis.

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CHAPTER 3

(iv) Every relative p-basis of E I F is a separating transcendency basis of E I F . (v) Every relative p-basis of E I F is a transcendency basis of E I F .

(vi) The transcendency degree of E I F equals the imperfection degree of E I F , and these

are finite.

Proof. (i)+(ii):

This is obvious

(ii)+(iii): Assume that E I F is relatively separated. If B is a relative p-basis of E I F , then

E / F ( B ) is separable algebraic. On the other hand, by Corollary 12.5(ii), the elements of B are algebraically independent over F . Hence B is a separating transcendency basis of

E/F. Assume by way of contradiction that B is infinite. Let Bo = {61,62,.. .} be a denumerable subset of B , let B’ = B

-

Bo and let F’ = F ( B ’ ) . Then Bo is a separating

transcendency basis of E I F ’ . Indeed, by Corollary 2.4(iii) and Proposition 2.6, the elements of Bo are algebraically independent over F’.

On the other hand, E / F ’ ( B o ) is

separable algebraic, since so is E I F ( B ) and F’(B0) = F ( B ) . By Lemma 8.2, we deduce that Bo is a relative pbasis of EIF’. Since for any n

it follows t h a t BA = {bib;,

. . . ,6,bKtl,. . . }

2 1, we have

is also a relative p b a s i s of EIF’. But, by

Corollary 12.8, E / F ’ is relatively separated, hence E / F ’ ( B A ) is separable algebraic. Thus 61

E F’(BA, 6:), by Proposition 2.2.6(ii). Thus, however, contradicts the algebraic inde-

pendence of Bo over F’. Therefore B is finite, as required. (iii)+(iv): This assertion follows from Corollary 8.5. (iv)+(v): This is a direct consequence of Corollary 12.5(i) and Theorem 8.4,

(v)+(vi): By hypothesis, the transcendency degree of E,’F equals the imperfection degree of E / F . Let B be any relative p-basis of E / F . Then, by (v)+(iv), B is a separating

transcendency basis of E / F . Hence B is a separating relative p b a s i s and therefore E / F is relatively separated. Applying implication (ii)+(iii), we deduce that B is finite, as required. (vi)+(i):

Since the imperfection degree of E / F is finite, any relative p-basis B of E / F

is finite. By Corollary 12.5(ii), the elements of B are algebraically independent over F . Since, by hypothesis,

ID?

=

t r . d . ( E / F ) , we conclude that B is a transcendency basis of

E / F . Hence, by Theorem 8.4, B is a separating transcendency basis of E / F . Therefore, by

RELATIVELY SEPARATED EXTENSIONS

179

Theorem 9.1, for any intermediate field K , K I F has also finite separat.ing transcendency basis. Invoking Corollary 8.5, we deduce that any relative pbasis of K / F is a separating transcendency basis of E / F . Hence K / F is relatively separated and the result follows. Assume that E / F is relatively separated and let K be a n intermediate field. We know, from Corollary 12.8, that E / K is also relatively separated. It is therefore natural to ask:

Is K / F necessarily relatively separated? In what follows we provide a positive answer for the case where E / K is finitely generated. This is one of the consequences (Corollary 12.12 beiow) of t h e following somewhat more general theorem. 12.10. Theorem. (Mordeson and Vinograde (1973)). L e t E J F be relatively separated

and let K be a n intermediate field. (i) If E I K is a f i n i t e purely inseparable eztension, t h e n K J F i s relatively separated. (ii) I ] E I K is separable algebraic, t h e n K I F is relatively separated.

(iii) If E / K i s purely transcendentaz, t h e n K I F is relatively separated and E I K h a s finite

transcendency degree. Proof. ( i ) We may assume that E

# K.

Let B be a relative p b a s i s of

K / F and

let S be a relative p b a s i s of E I K . Then S is finite and nonempty (Lemma lZ.l(iii)), so there is a positive integer n such t h a t SP"

i K . Since K

=

K P ( F , B ) and E = E P ( X , S ) ,

we have E = E P ( F , B , S ) . Hence, by Corollary 6.2(ii), B U S contains a relative p-basis of

E / F . Since E / F is relatively separated, we deduce that E / F ( B , S ) is separable algebraic. Hence, by Lemma 6.7, EP"/FP"(BP",SP") is separable algebraic. Since F ( B ,SP") 2

P" (Bp".

Sp"),

we conclude that F(EP",B ) / F ( B ,SP") is separable algebraic. Because

F ( B ,S p n )C: K = F'(K"', B)2 F(EP", B). it, follows that K / F ( B ,SP")is separable algebraic. Assume by way of contradiction that K / F ( B ) is not separable algebraic.

Then

F ( B , S P " ) / F ( B )is not separable algebraic and is finitely generated, since SP" is finite. Hence, by Lemma 12.1(iii), the extension F ( B , S p " ) / F ( B ) contains a nonempty relative pbasis. It is now a consequence of Theorem 12.4 and Corollary 6.2(i) that t h e r e is a relative p b a s i s of K / F ( B )containing any given rlative p b a s i s of F ( B , S p " ) / F ( B ) .Hence

K / F ( B ) is not relatively perfect, contrary to the fact t h a t B is a relative y b a s k of K / F (ii) Let B be a relative p b a s i s of K / F . Then, by Corollary 12.6 (ii), B is a relative p-basis

CHAPTER 3

180

of E / F . Since, by hypothesis, E I F is relatively separated, we conclude that E / F ( B ) is separable algebraic. Hence K / F ( B ) is separable algebraic, as required. (iii) Let T be a transcendency basis of E / K such that E = K ( T ) and let B be a relative

p-basis of K I F . Since T is a separating transcendency basis of E I K , T is a relative p-basis of E / K by Lemma 8.2(ii). Hence, by Corollary 12.6(i), B

U

T is a relative p-

basis of E I F . Since E / F is relatively separated, we deduce that E / F ( B ,I") is separable algebraic. But E = X ( T ) and F ( B , T ) = K l ( T ) for the subfield K1 = F ( B ) of K . Hence, by Proposition 4.18(ii), K / F ( B ) is separable algebraic. Thus K I F is relatively separated. Finally, since E I K is purely transcendental (hence separable) and, by Corollary 12.8, E / K is also relatively separated, it follows from Theorem 12.9 that T is finite. This completes the proof of the theorem. rn

12.11. Corollary. Let E I F be a relatively separated e z t e n s i o n , let K be a n intermediate field a n d let T be a transcendency basis of E I K . If L is the separable closure of

K ( T ) in E and ( E : L ) <

00,

t h e n K I F is relatively separated and T is finite.

Proof. By Proposition 2.2.30, E / L is a purely inseparable extension. Since, by hypothesis, E I L is finite, it follows from Theorem 12.10(i) that LIF is relatively separated. Hence, by Theorem 12.1!l(ii), K ( T ) / F is relatively separated. Therefore, by Theorem 12.lO(iii), K / F is relatively separated and T is finite.

12.12. Corollary. Let E I F be a relatively separated e x t e n s i o n a n d let ' h be a n intermediate field such that E I K i s finitely generated. T h e n K I F i s relatively separated.

Proof. This is a direct consequence of Corollary 12.11 and Lemrnz 2.8. m 12.13. Corollary. Let E I F be a field eztension. T h e n the following conditions are equivalent: ( i ) Fur a n y intermediate field K of E I F , K I F is relotively separated

( i i ) For a n y intermediate field K of E / F s u c h t h a t a n y separable element uf E I K lies in

K , the extension K J F i s relatively separated. Proof. It is clear that ( i ) implies (ii).

Conversely, assume that ( i i ) holds and let

iC be an intermediate field of E I F . Let L be the separable algebraic closure of K in E . Then L / K is separable algebraic and any scparable element of E / L lies i n 1,. Hence, by hypothesis,

LlF

is relatively separated. A4pplyingl'heorcm 12.10(ii), we conclude that

R E L I A B I L I T Y AND R E L A T I V E S E P A R A B I L I T Y

181

K/E’ is relatively separated, as required.

13. R e l i a b i l i t y and r e l a t i v e s e p a r a b i l i t y .

All fields considered below are assumed t o be of prime characteristic p . Let E / F be a field extension. Following Mordeson and Vinograde (1973), we say that E / F is reliable if E = F ( B ) for all relative pbases B of 5 / F . Since, for any relative pbasis B of

E / F , E = E p ( F , B ) , it follows that E / F is reliable if and only if

EP

F(B)

for any relative pbasis B of E / F . The condition on E / F of being reliable is much stronger than that of being relatively separated. Our aim is to provide some connections between reliable and relatively separated extensions. In fact, it will be shown t h a t relatively separated extensions can be completely characterized in terms of reliable extensions. We also

examine reliable extensions in detail and provide a number of their important properties. I n what follows, given a subset

T of E , we write ST for the separable algebraic closure of

E’(7’) in E. We begin by tying together reliable and relatively separated extensions. 13.1. P r o p o s i t i o n . (Mordeson and Vinograde (1973)). The following conditions

a r c equivalent. (i) E / F is relatively separated

(ii) For a n y transcendency basis T of E / F , E / F ( T ) is relatively separated

(iii) For a n y transcendency basis T of E / F , E I S T is reliable P r o o f . (i)+(ii): This is a direct consequence of Corollary 12.8. (ii)+(iii): Since T is a transcendency basis of E / F , E/ST is purely inseparable by Proposition 2.2.30. On the other hand, by Corollary 12.8, E / S T is relatively separated. Let U be any relative p-basis of E / S T . Then E / S T ( B ) is separable and purely inseparable,

hence E = S T ( B ) . This shows that E / S T is reliable. (iii)=+(i):

Let B any relative pbasis of E / F and let S be a transcendency basis of E / F ( B ) .

Then E / F ( B , S) is algebraic, hence B U S contains a transcendency basis T of E / F . Since

E = EP(B,F)= EP(B,F(T)),

E

=

E P ( B F, )

=

E P ( BF , ( T ) ) = E P ( B ,ST)

182

CHAPTER 3

B contains a relative p-basis of E / F ( T ) and a relative p b a s i s of E / S T . Since E / S T is reliable, E = S T ( B )and hence E / F ( T , B ) is separable algebraic. Since

it follows t h a t E / F ( B ) ( S )is separable algebraic. Hence S is a separating transcendency basis of E / F ( B ) . Invoking Lemma 8.2(ii), we deduce that S is a relative p b a s i s of E / F ( B ) .

But E / F ( B ) is relatively perfect (Lemma 12.1 (i)), hence S is empty. Thus E / F ( B ) is separable algebraic and therefore E / F is relatively separated. w Turning t o field extensions with finite transcendency degree, we next prove 13.2. Proposition. (Mordeson and Vinograde (1972)).

If E / F h a s f i n i t e transcen-

d e n c y degree, t h e n the following conditions are equivalent:

(i) E / F is relatively separated. (ii) E / S T is reliable for every transcendency basis T of E I F . (iii) E / S T i s reliable f o r s o m e transcendency basis T of E / F .

Proof. T h a t (i) implies (ii) is a special case of Proposition 13.1. Since (ii) obviously implies (iii), we are left t o verify t h a t (iii) implies (i). To this end, let B be a relative p b a s i s of E / F . Then B contains a relative p b a s i s of E / S T . Since E / S T is reliable, we have

E

= S T ( B )2

F ( T ,B )

Thus E / F ( T ,B ) is separable algebraic and, by hypothesis, T is finite. Assume by way of contradiction that E / F ( B ) is not separable algebraic.

Then

F ( T , B ) / F ( B )is not separable algebraic. Since T is finite, F ( T , B ) / F ( B )is finitely generated. Hence, by Lemma 12,l(iii), F ( T , B ) / F ( B )has a nonempty relative pbasis. But, since E / F ( T ,B ) is separable algebraic, it follows from Corollary 12.6(ii) t h a t E / F ( B ) has a nonempty relative pbasis. This contradicts the fact that E / F ( B ) is relatively perfect

(see Lemma K?.l(i)). So the proposition is true. Let E / F be a field extension. We say t h a t a n intermediate field K of E / F is proper if

K # E . We also say that E / F splits nontriuially if there are proper intermediate fields K and L such that K / F is separable, L / F is purely inseparable and E = K L . Our next aim is to tie together the concepts of reliability, relative separability and nontrivial splitting.

We need the following useful observations. 13.3. L e m m a . Let E I F be a field e z t e n s i o n

RELIABILITY AND RELTIVE SEPARABILITY

(i) E / F is reliable i f and only if E

183

# K ( E p ) f o r every proper intermediate field K

of

E/F. (ii) If E / F i s reliable, t h e n E / K i s reliable for every intermediate field K of E / F .

Proof. (i) Assume that K is an intermediate field of E / F such t h a t

X#

E and

E = K ( E P ) . Then E = EP(F,K ) , so K contains a relative pbasis B of E / F , by virtue of Corollary 6.2(ii). Thus E 3 K 2 F ( B ) and therefore E / F is not reliable. Conversely, assume t h a t E / F is not reliable. Then there exists a relative p-basis B of E / F such that

E # F ( B ) . Then E

= K ( E P ) for

K = F ( B ) # E , as required.

(ii) Assume that E / F is reliable and let K be a n intermediate field of E / F . If L is an intermediate field of E / K such that L that L

# E.

# E , then L

is a n intermediate field of E I F such

Since E / F is reliable, it follows from (i) t h a t E

# L(EP).

Hence, by ( i ) ,

E / K is reliable. rn We are now ready t o provide the following characterizations of reliable extensions.

13.4. Theorem. (Mordeson and Vinograde (1973)). Let E / F be a field eztension. T h e n the following conditions are equivalent: (i) E / F is reliable

(ii) For every transcendency basis T of E / F , E / F ( T ) is reliable. ( i i i ) E I F is relatively separated and there does not ezist a proper intermediate field K such

that E / K i s separable. (iv) E / F is relatively separated and there does not ezist a proper intermediate field K such

that E / K splits nontrivially. Proof. (i)*(ii):

T h a t (i) implies (ii) is a consequence of Lemma 13.3(ii). Assume

that (ii) holds. Let B be a relative p b a s i s of E / F and let S be a transcendency basis of

E / F ( B ) . Then E / F ( B ,S ) is algebraic, hence B

U

S contains a transcendency basis T of

E / F . Since B contains a relative p b a s i s of E / F ( T ) ,it follows from (ii) t h a t E = F(T)(BJ = F ( B ) ( S ) Thus S is a separating transcendency basis of E / F ( B ) and hence, by Lemma 8.2(ii), S is a relative p b a s i s of E / F ( B ) . Since E / F ( B ) is relatively perfect (Lemma 12.1(i)), it follows that

S

is empty. Thus E = F ( B ) and so E I F is reliable.

(i)=>(iii): Because E / F is reliable, E I F is relatively separated. Assume by way of contra-

diction that E / K is separable for some proper intermediate field L of E / F . T h e extension

CHAPTER 3

184

E / K is also reliable, by virtue of Lemma 13.3(ii). Let B be a relative pbasis of E / K . Since E

# K and E

= K ( B ) ,it follows t h a t B is nonempty. Since E I K is separable, the

elements of B are algebraically independent over K , by virtue of Corollary 12.5(ii). Now

BP+' = { b P + ' / b E B } is a relative p b a s i s of E I K . Indeed,

E = E P ( K , B ) = E P ( K , B P . B )EP(K,Bp+l) = and if Bf+'

C

BP+' is such t h a t E p ( K , B i f l ) = Ep(K,BP+l), then EP(K,B1) =

E P ( K , B ) and thus B1 = B . This shows t h a t BP+' is indeed is a relative pbasis of

E / K . However, E 3 K(Bp+') since B is algebraically independent over K . Thus E I K is not reliable, a contradiction. (iii)+(iv): Assume that K is a proper intermediate field of E / F such that E / K splits nontrivially. Then E = LM for some proper intermediate fields L , M of E I K such that

L / K is separable and M / K is purely inseparable. Then M is a proper intermediate field of E / F such t h a t E / M is separable, by virtue of Corollary 4.27, (iv)=+(i): Assume that E / F is not reliable. Then there exists a relative p b a s i s B of E / F such that E

# F ( B ) . Since E / F is

relatively separated, E / F ( B ) is separable algebraic,

) separable algebraic. Therefore hence so is Ep/Fp(Bp) (Lemma 6.7). Thus F ( E p ) / F ( B p is

E = F ( E p ). F ( B ) is a nontrivial splitting of E / K for K = F(Bp) # E . In order t o prove our next result on relatively separated extensions, we need to bring in certain properties of purely inseparable extensions. First, however, we must develop our vocabulary. A minimal generating set of an extension E I F is a subset S of E such that

E

=

F ( S ) and E

# F ( S ' ) for

all proper subsets S' of S . Let E / F be a purely inseparable

2 0.

extension. Then, by Proposition 2.2.27, for any given X E E ,

XP"

say that E / F is of finite ezponent if there exists an integer n

2 0 such t h a t EP" C F.

E F for some n

We The

smallest such integer is called the ezponent of E / F .

13.5. Lemma. Let E / F be a purely inseparable eztension and let S be a subset of

E. (i) S is a minimal generating set of EIF if and only if E = F ( S ) and S is a relative p-basis of E / F . (ii) If E / F is offinite ezponent n, then S is a minimal generating set of E / F i f and only

i f S is a relative p-basis of E / F .

RELIABILITY AND RELATIVE SEPARABILITY

185

Proof. (i) If S is a minimal generating set of EIF, then clearly E = F ( S ) and thus

E

= EP(S, F ) .

If S is not relatively pindependent in E/F, then s E EP(S

for some s E s E F(S

~

-

( s } : F )= F ( S P , S

-

{sj)

S. Thus s is both purely inseparable and separable over F ( S

{sj),

-

{s}),

hence

contrary t o the fact t h a t S is a minimal generating set. Conversely, let

E = F ( S ) and let S be a relative pbasis of E I F . If S'

C S is such

that F ( S ) = F ( S ' ) ,

then EP(F,S)= E p ( F , S') and hence S = S'. Thus S is a minimal generating set of JY/&'. (ii) Let S be a relative p b a s i s of EIF. Then E = EP(S,F) = E P " ( S , F ) = F ( S ) . The desired conclusion is therefore a consequence of (i). H 13.6. Lemma. Let EIF be a field eztension, let B , C be subsets of E a n d let n be a

positive integer.

(i) B is relativelyp-independent in EIF a n d C is a m i n i m a l generating set o f F ( E P ) ( BC , )

F ( E P ) ( B )if a n d o n l y if B

U C i s relatively p-independent

in E / F and B n C

(ii) If B U C i s relatively p-independent in E / F and B n C = 0, t h e n B U C P

-n

=

0.

i s relatively

p-independent in E ( C p - " ) I F (iii) If B U C i s a relative p-basis of E / F and B n C = 0, t h e n B U

CP-"

i s a relative

" ) 1F . p - basis of E (0-

Proof. (i) Assume that B is relatively pindependent in E / F and C is a minimal generating set of F ( E P ) ( B , C ) / F ( E P ) ( B ) .Then there does not exist c E F ( E P ) ( B ,C

(in particular, B

n C = 0). If

-

c EC

such that

{c})

there exists b E B such that

then there exists c E C such t h a t

By the exchange property, we deduce that c E F ( E p ) ( B , C - {c)), a contradiction. Thus

R

UC

is relatively pindependect in E / F and B n C = 0. The converse is immediate.

186

CHAPTER 3

(ii) By induction, it suffices t o consider the case where n = 1. If there exists b E B such that

b E F((E(CP-l))P)(B - {b},C) = F(EP)(C,B -{b}), then we contradict the relative pindependence of B u C in E/F.Therefore B is relatively pindependent in E(Cp-l)/F. Invoking (i), we are thus left to verify t h a t

CP-'

is aminimal

generating set of the extension

Assume by way of contradiction t h a t

cp-'

E

F(EP(C))(B,CP-' - {cp-l})

for some c E C. Then we have c E FP(EP*(CP))(BP, C - { c } ) 2 F(EP)(C - {c}),

which again contradicts the relative pindependence of B U C in E/F. (iii) Assume t h a t B U C is arelative p b a s i s of E / F and B n C = 0. Then, by (ii), BUCP-" is relatively pindependent in E(Cp-")/F. Since

the result follows. rn 13.7. Lemma. Let K be a n intermediate field of a purely inseparable field eztension

E j F and let E I K be finite. If K I F h a s a m i n i m a l generating set S , t h e n EIF has a m i n i m a l generating set U a n d , for a n y s u c h S and U , IS1 5 IU/.

Proof. By Lemma 13.5(i), any minimal generating set of E/F (respectively, K / F ) is a relative p-basis of E/F (respectively, K / F ) . Hence we need only verify the existence of a minimal generating set U of E/F with IS1 5 IUI. Since E / K is a finite purely inseparable extension, we have ( E : K ) = p" for some n by induction on n. Assume that n

2 0.

2 1 and t h a t

The case n = 0 being trivial, we argue

the result is true for all intermediate fields

L of E / K with (L: K ) = pn-'. Then, given such L , there exists a minimal generating set V of L I F with IS/ I /VI. Hence, if we can find a minimal generating set U of E / F with

RELIABILITY AND RELATIVE SEPARABILITY

lUI

2 IVI, then

187

the result will follow. Since ( E : L ) = p, we may harmlessly assume that

n = 1, in which case E = K ( X ) with

XP

We now distinguish two cases.

EK.

First assume that

!j K P [ F ) . Then

XP

{XP}

is a

relatively p-independent subset of K / F . Since S is relative p b a s i s of K / F , it follows from the exchange property that there exists s E S such that (S- { s } ) U {XP} is a relative p b a s i s of K / F . By Lemma 13.6(iii) (with E = K , C = { X P } and B = S - {s}), it follows t h a t ( S - {s}) U {A} is a relative pbasis of E / F . Choose rn

> 0 such t h a t

APm E

F . Then,

since E = F ( S , A), we have

which implies

E = F ( E P m ) ( S- {.},A)

=F(S

-

{s},X)(sPm)

Therefore s E F ( S - {s}, X)(s")

which shows that s is separable over F ( S Thus E = F ( S

-

{.},A)

-

{.},A)

and therefore, by Lemma 13.5(i), ( S

generating set of E / F . This settles the case where Now assume t h a t

and hence that s E F ( S

XP

XP

AP

-

{ s } ) U {A}

{s}, A).

is a minimal

$ KP(F).

E K P ( F ) . We first show that

E / F . Indeed, assume s E EP(F, S

~

-

S is relatively pindependent in

{s}) for some s E S. Since

E K P ( F ) = F P ( S P ) ( F )= F ( S P )

and

EP(F, s - { s } ) = F ( S P , X P , s

-

{s})= F ( S

{ s } , X p ) ( s " )c F ( S P , s

-

{s})(s")

~

{S},AP)(S~)

it follows that 5 E

F(S

-

and hence that s E K p ( F , S

-

{s}),

= KP(F,$-

{s})(s")

contrary t o the relative p-independence of S in K J F .

Thus S is relatively pindependent in E / F . Because E = F ( S , X ) , it suffices, by Lemma 13.5(i), t o show that S u {A} is relatively pindependent in E / F . If X E E p ( F , S), then X E K P ( F , XP, S)= K(XP)

188

CHAPTER 3

so X is separable over K , hence belongs to K , a contradiction. If there exists s E S such that s E E p ( F , S

-

{s},X),

then since s @ E P ( F , S

-

{s}),

it follows from the exchange

property t h a t X E EP(F, S ) , a contradiction. Thus S U { A } is relatively pindependent in

E I F and the result follows. w 13.8. Lemma. Let E / F be a purely inseparable eztension a n d let B = { b l , bz,. . . }

be a relative p-basis of E I F . A s s u m e that there ezist positive integers n1

< n2 < ... < n; < ni+l < .. .

such that

S={bp

".-nl

[ i = 1 , 2 , ...}

is a m i n i m a l generating set of F ( S ) / F . T h e n E / F is not reliable.

Proof. Assume by way of contradiction t h a t E I F is reliable. It is readily verified that ".+L-".

T={bsb:+l

l i = 1 , 2 , ...}

is a relative p b a s i s of E I F . Since E / F is reliable, we must have E = F ( T ) ,in which case there exists a positive integer m such that

Therefore

L , = {bf"'-"' jz = 1 , 2 , . . . , m + l} C K , Set K,+1

=

F(L,).

Then, by hypothesis, L , is a minimal generating set of F ( L , ) / F .

Note also that

{ b l b $ n ' - n ' , b 2 b ~ n 3 - " 2...,b,bP,;m:'-n=} , is a minimal generating set of -i.,/F.

It therefore follows that K,/F

ated by m elements while the intermediate field K,+l

is minimally gener-

is rninimal!y generated by m

+7

elements over F . Since the latter is in conflict with Lemma 13,7, the result follows. w The following theorem will enable us t o take full advantage of our previous results. 13.9. Theorem. (Mordeson and Vinograde (1971)). Let E I F be a purely inseparable

extension. T h e n the following conditions are equivalent:

R E L I A B I L I T Y AND R E L A T I V E S E P A R A B I L I T Y

189

(i) E I F is of f i n i t e ezponent. (ii) For every intermediate field K of E I F such that E I K preserves relative p-indepen

dence with respect to F , the field e z t e n s i o n K I F is reliable. Proof. That (i) implies (ii) is a consequence of Lemma 13.5(ii). Conversely, assume

that (ii) holds. We argue by contradiction and assume t h a t E I F is not of finite exponent. Let D ( E ) denote the set of all intermediate fields K of E I F such t h a t E / K preserves relative pindependence with respect to F . Let B be a relative pbasis of E I F . Since E E

D ( E ) ,we have E = F ( B ) . Because E / F is not of finite exponent, B is an infinite set and there exists an infinite subset { b l , b z , . . . } n, is the least integer with b:"' n, <

B such that n, <

n,+l(z =

1 , 2 , . . .), where

E F ( b 1 , . . . , b , - l ) and F ( b o ) = F . The strict inequality

can be achieved since E I F would have finite exponent if E / F ( b l , . . , b,) had

finite exponent for a finite subset { b l , . . . ,b,} of B. We now consider the sequence

{bp"=-"lli= 1,2,. . . } Assume that there exists a subsequence {b,> ij

=

1 , 2 , . . . } of { b L i z = 1 , 2 , . . . } with the

following property: If we set a3 = b,, , a1 = b l , c j = nl,, S = {a;''

-
ij = 1 , 2 , . . . }

( j = 1,2,. . . )

then

S is a minimal generating set of F ( S ) / F Let A = { a j i j = 1 , 2 , . . . }. Then F ( A ) E D ( E ) and so F ( A ) is reliable. However {a, lj

=

1 , 2 , . . . } is a sequence satisfying the conditions of Lemma 13.8 so that F ( A ) is not reliable.

This contradiction shows that the subsequence { a i l j = 1 , 2 , . . . } cannot exist. It therefore follows that the intermediate field

K = F(b~"'-"', z

=

1,2,. . .)

has the property that every relative pbasis of K / F is finite, since a relative p-basis of K / F

containing bl can be chosen from the generating set

and a subsequence of the above type does not exist. Hence if B1 is a relative pbasis of K I F , then there exists a positive integer t = t(B1) such that B:'

C F.

Hence

190

CHAPTER 3

F(KP') = F(Kp'*') which shows that F ( K P t ) / F is relatively perfect. Because K I F is not of finite exponent, F ( K P t ) / F is also not of finite exponent. However, because F ( K P t ) / F is relatively perfect, we have F ( K p t ) E D ( E ) . Therefore, by hypothesis, F ( K p t ) =

F ,a

contradiction. w 13.10. Corollary. Let E / F be a purely inseparable extension. T h e n E I F i s of finite

exponent if and only if f o r a n y intermediate field K of E I F , K I F is reliable. Proof. If E I F is of finite exponent, then so is K l F for any intermediate field K of

E I F . Hence, by Lemma 13.5(ii), any such K / F is reliable. The converse is a consequence of Theorem 13.9. We are now ready to prove the following result related t o Theorem 12.9. 13.11. Theorem. (Mordeson and Vinograde (1972)). Let E I F be a n arbitrary field

eztension. T h e n the following conditions are equivalent: (i) For a n y intermediate field K of E I F , K I F i s relatively separated. (ii)

E / F has finite transcendency degree and E/ST i s of finite exponent f o r every transcendency basis T

of

EIF.

(iii) E I F has finite transcendency degree and EIST i s of finite exponent for s o m e tran-

scendency basis T of E I F . Proof. (i)+(ii):

Let T be a transcendency basis of E I F . Then, by (i), for any inter-

mediate field K of the purely inseparable extension E I S T , K I F is relatively separated. Hence, by Corollary 12.8, KIST is also relatively separated, K I S T is actually reliable. Therefore, by Corollary 13.10, E/ST is of finite exponent. Since S T / F is also relatively separated and, by looking a t the chain F

C F(T)

S T , STIF is separable, it follows from

Theorem 12.9(vi) that T is finite. (ii)=+(iii): This is obvious (iii)=+(i):Let K be a n intermediate field of E I F , and let T be a transcendency basis of

E / F such that EIST is of finite exponent. We may extend a transcendency basis A of K / F to a transcendency basis B of E I F . Since T is finite and E / S B is purely inseparable, TP"'

SB

for some positive integer rn

Since E / S T is of finite exponent, there exists a positive integer n such that EPR

C

S T . Now s T / F ( T ) is separable algebraic, hence so is S ~ m / F P m ( T P " 'and ) therefore

RELIABILITY AND RELATIVE SEPARABILITY

191

S * / F ( T g m ) is separable algebraic. Thus S ; - F ( B ) / F ( T P " ) F ( B ) is separable algebraic. ) separable algebraic, since TPm C Sg. Hence S G m F ( B ) / F ( B ) But F ( T P m ) F ( B ) / F ( B is is separable algebraic and therefore Sgm

Ep"+"

5 S5-

C

S B . Since EP"

C

S T , we deduce that

S B , proving that E / S B is of finite exponent.

Now S B 2 S A , S B / F ( A ) is separable (since so are F ( B ) / F ( A ) and S B / F ( B ) )and

S A / F ( A ) is relatively perfect (since S A / F ( A ) is separable algebraic, see Lemma 12.1(ii)). Thus, by Proposition 4.28, S B / S A is separable and therefore K

n SB = SA.

Since E I S B

is of finite exponent, say n, we have

KP"

C K n Sg

= SA

Thus property (iii) is inherited by every intermediate field of E / F . Thus we are left to verify that (iii) implies E / F is relatively separated. Now T is finite, so S T / F has the property that every intermediate field K of S T / F is such that K / F is relatively separated (see Theorem 12.9). Since EP"

S T , we have

that F ( E P " ) / F is relatively separated. If B is any relative pbasis of E / F , B P " contains a relative pbasis of F ( E P n ) / F . Hence F(EJ'")/F(BP") is separable algebraic and therefore so is F ( E p " , B ) / F ( B ) . Since E = F ( E P " , B ) and B was an arbitrary relative pbasis of

E I F , we conclude that E / F is relatively separated, as required. 13.12. Corollary. Let E / F be finitely generated. T h e n , for a n y intermediate field K of E I F , K / F is relatively separated. Proof. Since E / F is finitely generated, it satisfies the hypothesis (ii) of Theorem 13.11. Now apply Theorem 13.11(i). 1

13.13. Corollary. Let E / F be finitely generated. T h e n t h e following conditions are equivalent : (i) E / F is reliable. (ii) l h e r e does not ezist a proper intermediate field K such t h a t E I K i s separable.

(iii) There does not ezist a proper intermediate field K such that E I K splits nontrivially.

Proof. This is a direct consequence of Corollary 13.12 and Theorem 13.4. s 13.14. Corollary. L e t E / F be algebraic and let L be t h e separable closure of F in

E . T h e n t h e following conditions are equivalent:

192

CHAPTER 3

(i) E / F is reliable. (ii) E / L is reliable and there does not ezist a proper intermediate field K of E J F such that E / K is separable. (iii) E I L is reliable and there does not ezist a proper intermediate field K of E I F such

that E I K splits nontrivially.

Proof. By Proposition 13.1, E I F is relativeiy separated if and only if E / L is reliable. Now apply Theorem 13.4. m

193

4 Derivations

In this chapter, we introduce derivations of fields and provide a number of their important properties. Among these, we examine various criteria for extending the derivations and characterize separability via derivations. We also provide a number of results which link derivations with pindependence. Given a field E of characteristic p

>

0, we exhibit a

bijection between the set of subfields of E containing EP and the set of closed restricted subspaces of DerE.

The inverse of this bijection is given by assigning to each closed

restricted subspace of D e r E its field of constants.

1. Definitions and elementary properties.

Let

H be a subring of a commutative ring S . A mapping

is said to be a derivation of R with values in S (or simply derivation of R if S = R ) if, for cvcry

5,y

E R , the map d satisfies the following properties:

d(W)=4 Y )+

Y4.1

(1)

We denote by Der(R, S ) the set of all derivations of R with values in S . Given d , d , E Der(R, S) and s E S , we put (d i d,)(s) = d(z)

+di(x)

for all x E R

and

( s d ) ( z )= s d ( z )

for all z E R

194

CHAPTER 4

Then one immediately verifies t h a t d + d1,sd E Der(R, S) and t h a t Der(R, S) becomes an S-module. 1.1. Lemma. If d E Der(R, S), then

(i) d(z") = nzn-'d(z) for any z E R and any integer n 2 1

(ii) Kerd is a subring of R (in particular, Kerd contains the prime subring of R) (iii) If R is a field, then Kerd is a subfield of R. (iv) For any given subring K of R , K

K e r d if and only if d is K-linear.

Proof. (i) The case n = 1 being obvious, we argue by induction on n. So assume t h a t d(zn-') = ( n - l)z"-'d(z).

Then

d(z") = d ( z . zn-') = zd(zn-') = z(n - l)z"-'d(z)

+ zn-'d(z)

+ zn-'d(z)

= nzn-'d(z)

as required.

(ii) We have d(1) = d ( l z ) = 2 d ( l ) , hence d(1) = 0 and 1 E Kerd. If z,y E Kerd, then obviously z - y E Kerd and z y E Kerd.Hence Kerd is a subring of R . (iii) By (ii), it suffices t o verify that if 0

and d(1) = 0, we have 0 = zd(z-')

#

z E Kerd, then z-'

+ z-'d(z)

=

E Kerd. Since 1 = z5-l

z d ( z - l ) , hence z-' E Kerd.

(iv) Let K be a subring of R. If K C_ Kerd and X E K, then for all z E R , d(Xz) = Xd(5) f zd(X) = Xd(z), proving that d is K-linear. Conversely, assume that d is K-linear. Then, for any X E K , we have d(X) = d(X .1) = Xd(1) = 0, as required. rn

For any given d E D e r ( R , S ) , we refer t o Kerd as the ring of constants for d, and if K is a subring of Kerd, the derivation d is said to be a K-derivation. When a derivation is defined on a field, the ring of constants is in fact a subfield, by virtue of Lemma l . l ( i i i ) . If

K is a subring of R , we denote by DerK(R, S) the set of all K-derivations of R in S . It is clear t h a t D e r K ( R , S ) is a n S-submodule of D e r ( R , S ) . Furthermore, by Lemma l.l(iv), D e r K ( R , S ) consists of all K-linear maps R

--f

S satisfying (1). We denote DerK(R, R ) by

DerKR. The elements of DerKR are called K-derivations of R and DerKR itself is called the derivation algebra of R over K. T h e name derivation algebra is justified by the fact

DEFINITIONS A N D ELEMENTARY PROPERTIES

195

that DerKR is a Lie algebra over R , which we proceed t o show. Let M be a n R-module. Then M is said t o be a Lie algebra over R if there exists a bilinear map M x M indicated by

(2, y

)

H

+

M,

[z, y ] satisfying

for all z,y,z E M . The identity (3) is called the Jacobi identity. If A is a n associative

R-algebra, we obtain a Lie algebra structure on the R-module A by setting [z, y] = xy -yz. In particular, if A = EndR(M) for some R-module M , then t h e corresponding Lie algebra is called the Lie algebra of R-endomorphisms of M . From now on, we concentrate on the R-module DerKR where K is a subring of R . Given d l , d z E DerKR, their Lie bracket [ d l , d z ] is defined by

which means t h a t

1 . 2 . Lemma. Let K be a subring of a commutative rang R . Then DerKR is a Lie

algebra over R with respect to the bilinear map

Proof. Since DerKR is a submodule of EndKR and the latter is a Lie algebra with respcct t o (z,y)

H

zy-yz, it suffices t o verify t h a t for all

(1). Given z , y E R , we have

as required. rn

d1,dz E DerKR, [dl,dz]satisfies

196

CHAPTER 4

1.3. Lemma. Let K be a subring of a commutative ring R. T h e n , for a n y d E D e r K R

and a n y positive integer n,

i=O

(here, by definition, d " ( x ) = z f o r all z E R).

Proof. T h e case n = 1 is just the equality (1). Applying induction on n , we have

=

2

+

(Y)d'(z)d"'+'(y)

i=o

=

5 (.

a - 1 )d'(z)d"-"(y)

i=l

(":

l)di(z)d"-'f'(y),

(since(n

1') (:>+ (. =

))

2-1

as requircd. w 1.4. Corollary. Let K be a subring of a commutative ring R of p r i m e characteristic p . T h e n , f o r a n y d E D e r K R , dP E D e r K R .

Proof. Applying Lemma 1.3 for n = p , and taking into account t h a t for ail i E ( 1 , . . . , p

-

(P)

= 0 in R

I}, we derive

Since dP is obviously a K-linear map, the result follows. w Let M be an R-module, where R is a commutative ring of prime characteristic p . A Lie algebra A contained in E n d R ( M ) is said t o be a restricted Lie algebra of characteristic p if u p E A for all a E A. Thus Corollary 1.4 can be reformulated by saying that DerKR

is a restricted Lie algebra of characteristic p . 1.5. Lemma. Let R be a subring of a field K and let F be t h e quotient field of

R. T h e n a n y derivation d

:

R

t

K can be extended in a unique way t o a derivation

D E F I N I T I O N S AND ELEMENTARY P R O P E R T I E S

d’ : F

--t

197

K . Moreover, the eztension d‘ i s given by the formula: d’(z/v) = ( Y 4 4

-

4Y))/Y2

(4)

Proof. Taking into account that 4 x 1 = d’(Y.

(Z/Y))

= Yd’(X/Y)

+ (4Y)d(Y)

it follows that the relation (3) holds for every derivation d‘ of F whic.. extends

I

This

establishes the uniqueness of d’. We now show t h a t d’ is well defined, i.e. if z / y = z l / y l , then

We may harmlessly assume that

51

=

z z , y l = zy for some z E: R. In this case we have

whence

~ i d ( z i) z i d ( ~ i= ) z ’ ( ~ ( d ( z )- zd(y)), proving ( 5 ) . A straightforward computation shows t h a t d’ satisfies conditions in the definition of a derivation. So the lemma is true. 1.6. Lemma. Let R be a ring, let d : R

+

R be a derivation and let S be fhe

polynomial ring R [ X 1 , . . . ,X,,]. For any f E R [ X 1 , .. . ,X,], let

obtained from f b y applying d to all coeficients o f f . Then the map f of

R [ X I , .. . , S,] be

fd E c-t

f

as

a derivation

S eztending d. Proof. If f is of degree zero or if f = 0, then clearly

is an extension of d. It is clear that ( f l

the set of all monomials X:l . . . X?, ti

Then

Ir

k

i=1

i= 1

f d = d(f). Thus

+ f 2 ) d = f f + f,”,for any f l , f z

the given map

E S. Let M h e

2 0, 1 5 i 5 n. Given f,g E S, we may write

CHAPTER 4

198

and hence

b

b

k

k

as required.

1.7. Lemma. Let R be a ring and let S be the polynomial ring R [ ( X ; ) ; e z ]i n the indeterminates X ; . Then, for each i E I , there is a unique derivation a / a X , of S , called the partial derivation with respect to X i , such that

and

Proof. We first consider the special case where S = R [ X ] . Then every derivation of S which is trivial on R is uniquely determined by its value on X. On the other hand, the ordinary formal derivative of f ( X ) E S satisfies the required properties. Since, for any given i E I,

R[(Xi)i~= z ]R [( X j ) j ~ ~ - {( iX}i ]) it follows that there is a unique derivation a / a X ; of S which is trivial o n R \ ( X j ) j s ~ - { ; ) ] and such that ( a / a X ; ) ( X ; )= 1. w Let F be a field. Then the partial derivations a / a X ; of F[(X;);,r] can be extended, and in a unique way, t o derivations of the rational function field F ( ( X ; ) i c i )(see Lemma 1.5); these derivations bear the same name and are denoted in t h e same way as their

restrictions t o the polynomial ring F [( X i ) i e r ] .

DEFINITIONS AND ELEMENTARY PROPERTIES

199

1.8. Lemma. Let F be a field and let E = F(X1,. . . ,Xn) be the rational f u n c t i o n

field in n variables over F . T h e n t h e partial derivations

alaXi, 1 5 i 5 n, f o r m

a basis

of t h e vector space DerF(E, K ) of F-derivations of E with values in a n y field eztension K of E . Proof. P u t D; = a/aX;, 1 5 i 5 n. Given d E DerF(E, K ) , set n

d’ = c d ( X i ) D ; i=

1

Then d‘ is an F-derivation of E which coincides with d a t the elements X I , . . . ,Xn. Since the kernel of the derivation d’

-

d is a ring, it follows that d’ coincides with d on the

polynomial ring F[X1,.. . , X,].Hence, by Lemma 1.5, d’ = d which shows that the D , span D e r F ( E , X ) . We now show t h a t the D, are linearly independent over X. Indeed, if

C X,D, = 0 (A,

E K ) , then

as required.

1.9. Lemma. L e t F be a field of characteristic p

extensions and let d : E

---t

>

0 , let EIF and K I E be field

K be a n F-derivation of E. T h e n d is a n E P ( F ) - d e r i v a t i o n

a n d , in particular,

K) DerF ( E ,K ) = D e r E , ( F ) (E, Proof. By Lemma l.l(i), we have d ( z P ) = pzP-‘d(z) = 0 for all z E EP. Hence

F , EP

C Kerd.

But, by Lemma l.l(iii), Kerd is subfield of E , hence E p ( F )

Kerd, as

required. We close by a brief discussion of a generalization of the notion of a derivation. Let R be a subring of a commutative ring S. Then a sequence of mappings

of H into S is called a higher derivation of rank m of R into S (or simply a higher derivation of rank m of R if S = R ) if

CHAPTER 4

200

for all z,y E R. A higher derivation of infinite rank is a n infinite sequence

{ d o = I , d I , . .. , d m , . .. }

of mappings of R into S such t h a t (1) and (2) hold for all n

2 0.

T h e following facts are

direct consequences of the definitions: (a) The mapping d l : R

--&

S is a derivation

(b) If { d o , d l , . . . , d m , . . . } is a higher derivation of infinite rank, then the section {do, d l , . . . , d,}

is a higher derivation of rank m.

For any positive integer m,put

and

ti =

X i +(Xm+'), 1 5 i 5 m

Then S(") is a n S-algebra which is S-free with l , t , . . . , t m , as a basis and with t m = 0. Note also that the map

c m

g : S(")

+

s,

Sit'

H

so

i=O is a homomorphism. A link between S ( m ) , g and higher derivations of rank m is provided

by the following simple observation. 1.10. Lemma.

I,/ d"') = (1,d l , . . . , d m } is a higher derivation of rank m of R into

S , then the map m

is a homomorphism such that ( g f ) ( r )= r for all r E R . Conversely, any homomorphism of R into S(") satisfying this condition has the form:

where d ( m ) = {I, d l , . . . , d,}

is a higher derivation of rank rn of R into S .

D E F I N I T I O N S AND ELEMENTARY PROPERTIES

201

Proof. It is obvious that f preserves addition. Furthermore, since dl(1) = 0 , it follows from (7) t h a t d , ( l ) = 0 for all n E { 1 , 2 , . . . ,m}, and hence f preserves identity elements. Given x ,y E

R,we have rn

m

i=O

k=O

m

proving that f is a homomorphism. Since d o ( r ) = r, we also have ( g f ) ( r )= r for all r E R. Reversing the above calculations, one immediately verifies t h a t the converse is also true. rn

1.11. Corollary. A s s u m e t h a t the ring S is of p r i m e characteristic p . If d ( " )

(1, ( 1 1 , . . . , d,}

is a higher derivation of rank m of R Into S , then

for all x E R .

Proof. Given x E R , we have m

rn

m

by applying the homomorphism f of Lemma 1.10.Thus

which clearly yields the result.

m

m

r=0

i=O

=

CHAPTER 4

202

Let S be a commutative ring and let S [ [ X ]be ] the set of all formal sums m

n=O

Given two elements of S [[ X I ] say , m

M

Eb,X"

z a n X n and

define their sum and product t o be respectively m

m

C(an + bn)Xn

CnXn

and n=O

n=O

where c, =

a;bj i+j=n

Then we see that S [ [ X ] becomes ] a ring, to which we refer as the ring of formal power

series in one variable over S . The map g : S [ [ X ]+ ] S given by

is obviously a homomorphism. 1.12. Lemma. If ( 1 = d o , d l , . . . ,d,,

.. . }

is a higher derivation of infinite rank of

R into S , then the map m

f :R

-+

S [ [ X ] ]7, H E d ; ( r ) X '

is a homomorphism such that ( g f ) ( r ) = r for all r E R. Conversely, any homomorphism of R into S [ [ X ] satisfying ] this condition has the f o r m m

r

H

Cd;(r)X' i=O

where (1 = do,dl, . . . ,d,,

. . . } is a

higher derivation of infinite rank of R into S .

EXTENSIONS OF

DERIVATIONS

203

Proof. The proof is entirely similar t o that of Lemma 10.1 and therefore will be omitted. 1.13. Corollary.

d o , d l , . . . , d,,

Proof.

.. . }

A s s u m e t h a t the ring S i s of p r i m e characteristic p . If (1 =

i s a higher derivation of infinite rank of R into S , t h e n

Repeat the argument in the proof of Corollary 1.11 with respect to the

homomorphism f of Lemma 1.12. w Let d = ( d ; ) be a higher derivation (of finite or infinite rank) of R into S . An element z

E R is called a d-constant if di(z)= 0 for all i 2 1. It is a consequence of Lemmas 1.10

and 1.12 that z E R is d-constant if and only if f(z)= z. Hence the set of d-constants of R is a subring of R which is closed with respect t o taking multiplicative inverses. In particular, if R is a field, then the d-constants constitute the subfield of R (to which we refer as the field of constants of the higher derivation d = (dz)).

2 . Extensions of derivations.

Let EIE’ be a finitely generated field extension, say E = F ( A 1 , . . .,A,) any field extension. Suppose we are given a derivation d : F

4

and let K / E be

K. The problem that

motivates this section is t o discover circumstances under which there exists a derivation d’ : E + K extending d. In what follows, we denote by I the ideal of F [ X l , .. . , X,]

consisting of all polynomials f ( X 1 , . . . ,Xn) such that

f(A1,.

. . , A,)

= 0.

2 . 1 . Theorem. Further t o the n o t a t i o n and a s s u m p t i o n s above, let p l , . . . , p n be

e l e m e n t s of K a n d let S be a n y set of generators of the ideal I . T h e n d c a n be extended t o a derivation d’ : E

t

K s u c h that

if a n d only if for all f E S ,

204

CHAPTER 4

If such eztension ezists, then it i s unique. Proof. We will first show t h a t (1) is equivalent t o the requirement t h a t

for all f E I . Indeed, assume t h a t (1) holds. Then, given f E I, we have f =

g j fj

where each fj is in S and g j E F[Xl,.. . , Xn]. Hence

since

fj(X1,.

.. , A n )

= 0 for all j. A similar argument shows t h a t

Multiplying each relation (1) by gj(X1,. . . A n ) and adding the resulting relations, we therefore obtain (2). We next observe that if d’ is any derivation of E extending d , then

for all g E F I X 1 , . . . ,Xn]. Indeed, formula d ( z y ) = z ( d y )

+ y d ( z ) shows t h a t

(3) is true

whenever g = XXF’ . . . X r n , X E F. Hence, by linearity, (3) is true for any polynomial g E F I X 1 , . . .,Xn]. Because d’(0) = 0, it follows from (3) that the relation (2) is necessary.

Moreover, (3) shows that there is a t most only one derivation d’ of

F[Xl,.

. .,An]

(and

hence, by Lemma 1.5, also a t most one derivation of the quotient field E ) such that d’ is an extension of d and

d’(Xi)

= pi, 1 5

i 5 n.

Conversely, if (2) holds for any f E I , we define d’ by

EXTENSIONS OF DERIVATIONS

This map is well defined, since if

g(X1,.

. . ,A,)

205

= h(X1,.. . ,A,),

we may apply (2) to the

polynomial g - h, and we then find t h a t

It is obvious that d‘ preserves addition and d’(A,) = p i , 1 5 z 5 n. Finally, condition d’fzy) = zd‘(y)

+ yd’(z) is satisfied since the mappings

g

H

gd and g

H

pi&

are

derivations of F [ X 1 , .. . ,X,] (Lemmas 1.6 and 1.7). Hence condition (2) is sufficient, since a derivation d’ of F [ A I , .. . ,A,]

can be extended t o the quotient field E , by Lemma 1.5.

We shall now record a number of useful consequences of the above result. 2.2. Corollary. L e t S be a set of algebraically independent e l e m e n t s over a field F ,

let K be a field containing F ( S ) a n d let d : F T)

:

S

+

K , there e z i s t s a unique derivation d‘

+

K be a derivation. T h e n ,

: F ( S ) -+

for

any map

K satisfying t h e following t w o

properties: ( i ) d’ e z t e n d s d.

(ii) d ’ ( s j = +(s) for all s E S .

Proof. First assume that F ( S ) = F(X) is a simple transcendental extension of F . Then 0 is the only polynomial f in F [ X ]such that f ( A ) = 0. Hence the desired conclusion follows by virtue of Theorem 2.1. ‘Turning to the general case, observe that if d’ satisfies (i) and (ii), then it is uniquely determined on F [ S ] ,hence also on F ( S ) , by Lemma 1.5. To prove the existence of such

S, of S such that d E S,. Since the empty

d’, we shall employ Zorn’s lemma. Let I be the set of all subsets

admits an extension d , t o F(S,) such t h a t d,(z) set belongs t o I , the set I is nonempty. If S,

=

G(z) for all

z

S,, then dp is an extension of d,.

The

latter immediately implies that the set I , partially ordered by set-theoretic inclusion, is inductive. Owing to Zorn’s lemma, we may therefore choose a maximal element M in I . Let d’ be the derivation of F ( M ) extending d and such that d ’ ( z ) = G(z) for all z E M . If

M f S , then there exists an element s in S which does not belong to M , and by the case

IS(=

1, there exists a derivation d” of the field F ( M ) ( s )which is an extension of d’ and is

such that d ” ( s ) = $J(s). Because this contradicts the maximality of M in I , we conclude that A4 = S, as required. w 2.3. Corollary. Let E = F(X) where X is algebraic over F and let f ( X ) E F [ X ] be the minimal polynomial of A. If K f E is a n y field e z t e n s i o n , p an e l e m e n t of K and

CHAPTER 4

206

d :F

4

K is a derivation, then d can be eztended t o a derivation d'

:E

+

K such that

d ' ( X ) = p if and only if

fd(X)

+ PLf'(4

=0

(4)

Furthermore, i f such eztension exists, then it is unique. Proof. In the notation of Theorem 2.1, the ideal I is the principal ideal generated by f. Thus the set of relations (1) reduces t o the single relation (4), as required. 2.4. Corollary. Let E / F be a separable algebraic eztension. Then every derivation

d of F may be eztended, in a unique way, to a derivation of E .

Proof. A straightforward application of Zorn's lemma allows us t o assume that

E = F ( X ) for some X E E. Let f ( X ) E F [ X ]be the minimal polynomial of A. Since X is separable over F , we have f'(X)

# 0.

Therefore the relation

is satisfied by a unique value of p , namely p = - f d ( X ) f ' ( X ) - l .

It follows from Corollary

2.3 that the extension d' of d t o E such that

d ' ( X ) = -fd(X)f'(X)-' is the only extension of d t o E . 2.5.

Corollary.

Let E I F be a separably generated field eztension.

Then every

derivation of F can be eztended t o a derivation of E . Proof. This is a direct consequence of Corollaries 2.2 and 2.4.

For fields of nonzero characteristic, the result above will be sharpened in the next section where it will be shown t h a t E / F is separable if and only if every derivation of F can be extended t o t h a t of E .

> 0, let E / F be a simple purely inseparable field extension with E # F , say E = F ( X ) and let e 2 1 be the smallest integer 2.6. Corollary. Let F be a field of characteristic p

such that

XP'

E F . Let K / E be any field eztension and let d : F + K be a derioation.

Then d has a n eztension d' t o E if and only i f d ( X p e ) = O., Furthermore, if d(Xpe) = 0, then the value of d ' ( X ) can be assigned arbitrarily in K .

EXTENSIONS OF

DERIVATIONS

Proof. The minimal polynomial of X over F is XP‘

207

- XP‘

. Hence relation

(4) reduces

to d ( X P c ) = 0. Now apply Corollary. 2.3.

2.7. Corollary. Let F be a field of characteristic p inseparable eztension s u c h that

EP

E F . I f S is a set

>

0 and let E / F be a purely

of generators of E j F a n d if d : F

+

I.’

is a derivation s u c h t h a t d ( S P ) = 0 , t h e n d c a n be eztended t o E .

Proof. To show the existence of extension, we shall use Zorn’s lemma. Let I be the set of all pairs (E’ : d’), where E’ is a n intermediate field of E / F and d‘ is a derivation of

E’ extending d . We set (E‘,d‘) < (E“,,”) t o mean that E’

C E”

and d” extends d‘. Then

I becomes a partially ordered set and I is obviously inductive. Furthermore, by Corollary 2.6, I is nonempty. Owing to Zorn’s lemma, there exists a maximal element (E’,d’) of I .

If E‘ # E = F ( S ) , then there exists a n element s E S such t h a t s !$ E’. Since s p E E’ and d’(sP) = 0, it follows from Corollary 2.6 that d’ may be extended to E’(s) # E’. But this

contradicts the maximality of (E’,d’), hence E’ = E, as required. 2.8. Corollary. Let F be a field of characteristic p > 0 a n d let E / F be a field

eztension. Then (i)

E P ( F ) is

ezactly the set of all elements z in E s u c h that d ( z ) = 0 f o r every F -

derivation d of E. (ii)

E P

i s ezactly the set of all elements z in E s u c h that d ( z ) = 0 f o r every derivation d

of E .

Proof. (i) Let z E E be such that d ( x ) = 0 for every 8’-derivation d of E . By Lemma 1.9, it suffices t o show that z E E P ( F ) . Assume that z 4 E p ( 8 ’ ) and consider the

extension E P ( F , s ) / E P ( F ) . Applying Corollary 2.6 to the zero derivation of EP(F) we see that there exists a n EP(F)-derivation d of EP(F,z) such that d ( z ) # 0. But, by Corollary 2 . 7 , d extends to a derivation of E , a contradiction.

(ii) This is the special case of (i) in which F is the prime subfield of E . rn

‘The rest of this section will be devoted t o the study of derivations of finitely generated field extensions. 2.9.

Theorem.

L e t E I F be a finitely generated field extension.

T h e n E I F is

separable algebraic if and only ii D e r F E = 0 . Proof. Assume that E / F is separable algebraic. Then, by Corollary 2.4, e\eq

f -

CHAPTER 4

208

derivation of E is zero, i.e. D e r F E = 0. Conversely, assume t h a t D e r F E = 0 and write

E = F(z1,. ..,z,)

for some zl,. .. ,zn E E. Suppose t h a t E I F is not separable algebraic.

Fix z o E F and choose the largest i E { 0 , 1 , . . . , n } such t h a t E/F(so,zl,. . . ,xi) is not separable algebraic. Then F ( z 1 , . . . , z ; + l ) / F ( x o ,2 1 , .

. . ,z;)

is either purely transcendental

or algebraic inseparable. Therefore, by Corollaries 2.2 and 2.6, there exists a nontrivial which is trivial on F ( z o , z l ,. . . , zi).Because E is separable derivation d of F ( z 1 , . . . ,z;+l) algebraic over F ( z 1 , . . . , z;+l),d can be extended t o E , by Corollary 2.4. This contradicts our assumption t h a t D e r F E = 0, hence the result. m 2.10. Theorem. Let EIF be a field eztension and let

51,.

.. ,xn

be n elements of

E . Then the following conditions are equivalent: (i) F ( z 1 , . . . , x n ) / F is separable algebraic. (ii) There m i s t polynomials

f l , .

. . ,f n

i n F [ X l , .. . ,Xn] such that

f,(zl,. .

.,zn) =

0,

1SiSnand

Proof. By Theorem 2.9, condition (i) is equivalent to the requirement that 0 is the only F-derivation of F ( z 1 , . . . ,zn). Assume t h a t (ii) holds and let d be a n F-derivation of

F(z1,.. . ,zn).Because d ( f;(zl,. . . ,zn))= 0, we have

af.

L ( z l r . .. , z n ) d ( z l )= 0 for all i E ( 1 , . . . , n }

ax, j=1

This gives us a system of n homogeneous linear equations in d ( z , ) , . . . , d ( z , ) with nonzero determinant. Thus d ( z 1 ) = . . . = d ( z , ) = 0 , proving t h a t d is t h e trivial derivation. Conversely, assume that (i) holds. Consider the system of homogeneous equations

obtained by letting f vary in the set of all polynomials f in FIXl,. . . , X,] such that

. ,zn)= 0. By Theorem 2.1 (see relations ( Z ) ) , it follows t h a t this 4ystern of

f(zl,..

equations has only the trivial solution u1 = u z = . . . = u, = 0. Therefore the system contains some set of n equations with nonzero determinant, as required. rn As a preliminary t o the next result, we record

EXTENSIONS

OF

DERIVATIONS

2.11. Lemma. Let F be a field of characteristic p

finite pureiy inseparable field eztension such that EP (i) There ezists a generating set { X I , .. . ,z,,)

209

> 0 and let E / F ( E # F ) be

u

C F.

of E / F such that z1

4 F , z;

@ F ( z 1 , . . . , z,-1)

for i = 2 , . . . , n and such that ( E : F ) = p".

(ii) The vector space DerpE of F-derivations of E has dimension n over E . Moreover,

there ezists a basis { d l , . . . , d n } of DerFE over E such that di(z;)= 1

and

d ; ( z j )= 0

for i

#j

(1

5 i , j i n)

Proof. (i) w e construct by induction a sequence z1,. . . ,zk,.. . of elements of E such that

z1

4

F, z; @ F ( z 1 , . . . , z;-1) for i

2

2. Since ( E : F ) is finite, this sequence must

have a finite number of terms, say n. By looking a t the chain t; C F ( z 1 ) C . . . C F ( z 1 , . . . , ~ , - 1 ) C F ( z 1 , . . . , ~ and bearing in mind that zf 6 F ( z 1 , . . . , x i p l ) , i

i - 1 ,xi)

2 2,

2';

E

C

. . . C F ( z 1 , . . . ,zn) =E

F , we deduce t h a t ( E : F ) =

Pn.

(ii) By Corollary 2.6, for each i E (1,.. . ,n} there exists an F ( z 1 , . . . , xi-1)-derivation d : of F ( z 1 , .. . ,z,)such t h a t d:(z,) = 1 (for i = 1, F ( z 1 , . . . , z t - l ) is to be interpretcd

a5

F ). We now extend d i t o a derivation d j of E = F ( z l , .. . ,zn) by imposing on d, t h e conditions

d;(z,+i) = . . . = di(z,) = 0 This is possible, as follows by successive applications of Corollary 2.6, since zp E F

for

j=z+1,

..., n

The above gives us n F-derivations d l , . . . , d , of E such that d,(z,)

7

1

d l ( z l )= 0

and

for

(1 5 2.3

z f j

'These derivations are linearly independent over E . Jndeed if

C,"=,X,d,

I n)

= 0 with A,

E E,

then we have

(c n

A, =

A,d,)(z,) = 0

for all j E (1,. . . , n }

,=I

Finally if d is any F-derivation of E , the mapping d' = d

-

C,"=, d ( z l ) d , is an F -

derivation of E which takes value 0 on the set { z l , . . . ,zn}. Hence d' takes value 0 on

E = F ( z 1 , . . . , zn),proving t h a t the derivations d l , . . . , d , span D e r r E . m

210

CHAPTER 4

We are now ready to prove the following important result. 2.12. Theorem Let E/F be a finitely generated field eztension.

(i) The dimension n of D e r F E over E i s equal t o the smallest number of elements p1,.

.. , p t

of E such that EIF(p1,. . . , p t ) is separable algebraic. Furthermore, n 2

tr.d.(E/F) and n = tr.d.(E/F) if and only if E/F is separably generated. (ii) If charF = p

>

0, then n is equal to the imperfection degree of E/F (equivalently,

p n = (E : E P ( F ) ) . (i) An F-derivation d of L = F ( p l r . . ,pi) is uniquely determined by

Proof.

d ( p l ) ,. ..,d(pt). Hence the dimension of DerFL is at most equal to t . If E / L is sep-

arable algebraic, every derivation of L has a unique extension t o E (Corollary 2.4), hence

n

5 t.

To prove the first assertion, we are therefore left t o verify t h a t there exist n ele-

ments P I , . . . ,/A,

of E such that E/F(pl,.. . ,/A,,) is separable algebraic. This is clear if

c h a r F = p is 0 , since in that case n = tr.d.(E/F), by Corollary 2.4 and Lemma 1.8. If

p

# 0, then

D e r F E = DerEp(F)E,by Lemma 1.9. We may assume t h a t E

# Ep(F),since

otherwise E/F is separable algebraic by Theorem 2.9. Since E/Ep(F) is purely inseparable, it follows from Lemma 2.11 that pn = ( E : E p ( F ) ) and t h a t there exist n. elements p1,.. . ,p n of

E and n linearly independent F-derivations d l , . . . ,d, of E such that

Let d be any F(p1,.. . ,/A,)-derivation of E.

c:=,Aid;,

Then d = 0, since we can write d =

A; E E , whence 0 = d ( p j ) = A j for every j .

E/F(p1,. . . ,/A,)

Thus, by Theorem 2.9,

is separable algebraic, proving the first assertion.

Since E / F ( p l , . . . ,/A,,) is algebraic, n 2 tr.d.(E/F). If n = t r . d . ( E / F ) , then { p L 1 ., .. ,/A,} is a separating transcendency basis of dency basis

XI,

EIF.

Conversely, if E / F has a separating transcen-

..., z t , then by the first part, t 2

n . Since n 2 t we must have n = t ,

proving (i). (ii) This was established in t h e proof of (i).

R

3. Derivations, separability and pindependence.

Throughout this section, all fields considered are assumed t o be of prime characteristic p . Our aim is to provide a number of results which link derivations with separability and

DERIVATIONS, SEPARABILITY A N D p-INDEPENDENCE

211

pindependence. In fact, all the results below will be obtained with the aid of the following theorem. 3.1. Theorem. Let F

K J F and let f : B

d :K

---t

+

5K

C_

E be a chain of fields, let B 6e a relative p-basis oj

E be a n arbitrary map. T h e n there ezists a unique F-derivation

E such that f o r all b E B

d(6) = f ( b )

Proof. By Lemma 1.9, we may replace F by K P ( F ) and so we may assume that

KP 5

C

F . Also, we may assume that F # K which means that B is nonempty. Given

E B , put B , = B

-

{z}. Then z $ F ( B , ) and

XP

E F

C F(B,).

Hence, by Corollary

2.6, there exists an F(B,)-derivation d , of K = F ( B , ) ( z ) such that d , ( z ) = f(z). If

S = ( 5 1 , . . . ,s,} is a finite subset of B , then d s = d,,

+ d,, + . . . + d,,,

is a n F-derivation of K with values in E such that

d s ( z l )= f

(~)

foralliE{1,2, ...,n}

If L is a finite subset of B containing S , then the restriction of d~ to F ( S ) coincides with the restriction of d s t o F ( S ) . Now if X is a n arbitrary element of K , then X E F ( S ) for some finite subset S of B. By the above, we can map X

H

d s ( X ) and it is clear that

the value d s ( X ) does not depend upon the choice of S such that X E F ( S ) . Hence the map d : K

--t

E given by d ( X ) = d s ( X ) is well defined. It is now immediate t h a t d is an

F-derivation of K with values in E and such t h a t d(6) = f ( 6 ) for all b E B . Furthermore, since K = F ( B ) , such d is unique. 3.2. Corollary.

Let F

K 2 E be a chain of fields and let D e r F ( K , E ) be the

vector space (over E ) of all F-derivations of K with values in E. T h e n the dimension n

of

D e r F ( K , E ) over E i s finite i f and only

If

the imperfection degree of K I F is finite.

Furthermore, if n i s finite, t h e n n i s equal t o the imperfection degree of K J F .

Proof. Let B be a relative pbasis of K / F and let V ( B , E ) be the vector space (over E ) of all mappings from B to E (by definition, ( a

+ P ) ( b ) = a(6) + p ( b ) , ( X a ) ( b ) =

X a ( b ) , a , P E V ( B ,E ) , b E B , X E E ) . Consider the map

DerF(K,E) d

-+ V(B,E) ~ d l B

CHAPTER 4

212

This map is obviously E-linear. By Theorem 3.1, it must be surjective. Furthermore, since

K = K P ( F , B ) , the map is injective and hence is a n isomorphism. Now if B is infinite, then V ( B , E ) is a n infinite dimensional vector space over E , hence rz is infinite. On the other hand, if B is finite, then the dimension of V ( B , E ) is equal t o IBI, hence t o n , as required. m

E be a c h a i n of fields. T h e n t h e following conditions

3.3. Theorem. L e t F C_ K

are equivalent: (i) E/K preserves relative p-independence with respect t o F .

(ii) E v e r y F - d e r i v a t i o n d : K

Proof. (i)+(ii):

4

E c a n be eztended to a derivation of E .

Let B be a relative p b a s i s of K/F. By hypothesis, B remains

relatively pindependent in E / F . Hence B can be embedded in a relative pbasis C of

E I F . Let d : K

-+

E be a n F-derivation. Then the restriction of d t o B can be extended

to a mapping f : C

-+

an F-derivation d’ : E

E . Applying Theorem 3.1 [with K = E and B = C) we may find --t

E such t h a t d’(b) = f ( b ) = d ( b ) for all b E B . Since d’ extends

d , ( i i ) follows.

(ii)+(i): Assume by way of contradiction t,hat B is a relative p b a s i s of K I F such that B is no longer relatively pindependent in E / F . This means t h a t there exists b E B such that

b E E P ( F ,B

-

{ b } ) . If d’ is any F-derivation of E such t h a t d ’ ( z ) = 0 for all z E B

then d’(b) = 0. By Theorem 3.1, there exist F-derivations d : K for z E B

-

--t

-

{b},

E such that d ( z ) = 0

{ b } but d ( 6 ) # 0. Hence any such d cannot be extended t o a derivation of E ,

a contradiction. 3.4. Corollary. Let F

CK

T h e n every F-derivation d :K

C_ E be a c h a i n of f i e l d s s u c h t h a t E I K i s separable.

+E

can be eztended to a derivation of E .

Proof. By Theorem 12.4, E/K preserves relative pindependence with respect t o F . The desired conclusion is therefore a consequence of Theorem 3.3. 3.5. Corollary. Let E / K be a n arbitrary field e z t e n s i o n . T h e n E f K i s separable if

a n d o n l y if e v e r y derivation d : K

-+

E c a n be eztended lo a d e r i v a t i o n of E .

Proof. Let F be the prime subfield of K . Then every derivation d : K

-t

E is an

F-derivation. Furthermore, since F is perfect, condition (i) of Theorem 3.3 is equivalent to the requirement that E / K preserves pindependence. Hence, by Theorem 6.10, condition

RESTRICTED SUBSPACE OF DERt

213

(i) of Theorem 3.3 is equivalent t o the requirement that E I K is separable. The desired conclusion is now a consequence of Theorem 3.3.

3.6. Theorem. Let E I F be a field eztension and let S be a subset of E . Then S is

relatively p-independent in E I F if and only if for every

s E

S there ezists a n F-derivation

d , of E such that

d,(s)

# 0 and d , ( z ) = 0

for all z # s in S

P r o o f . Assume t h a t S is relatively pindependent in E / F . Then S can be extended to a relative p b a s i s B of E I F . Hence, applying Theorem 3.1 (with K = E), we see that the required condition holds. Conversely, assume that S is not relatively pindependent in E I F . This means t h a t s E EP(F, S - { s } ) for some s E S . Hence d ( s ) = 0 for any F-derivation d of E such that d ( s ) = 0 for all z # s in S.

D

3.7. Corollary. Let S be a subset of a field E . Then S is p-independent if and only

if for every

s E

S there ezists a derivation d, of E such that d,(s)

# 0 and d,(z) = 0

for all x

# s in S

P r o o f . This is a special case of Theorem 3.6 in which F is perfect. rn

4. R e s t r i c t e d s u b s p a c e of DerE.

'Throughout this section, E denotes a field of characteristic p > 0 and D e r E the vector space over E of all derivations of E . Following Gerstenhaber (1964), we define a restricted

subspace of DerE t o be a subset which is a vector space over E and which is closed under t h e formation of p t h powers. Our aim is to establish a bijection between the set of subficlds of E containing EP and a certain set of restricted subspaces of DerE. The proofs of all

the results below are extracted from Ojanguren and Sridharan (1969). We begin by proving the following rather surprising result first discovered by Gerstenhaber (1964). 4.1. Theorem. Let V be a restricted subspace of DerE. Then V is a Lie subalgebra

of DerE.

214

CHAPTER 4

Proof. Since V is a subspace of E , the required assertion amounts t o proving that

for any d 1 , d z E V , [ d l , d z ] E V . We may harmlessly assume t h a t d l dl(z)

# 0 for some x E E .

#

0, in which case

Given X , y E E , we have

hence [ X d l , d z ] = X [ d l , d z ] - d z ( X ) d l . Therefore we may replace d l by X d l , where X is any nonzero element of E . In particular, by taking X = d I ( x ) - ' , we may harmlessly assume t h a t d l ( s ) = 1. For any t in the prime subfield

F, of E and any nonnegative integer m , we consider

the element 6' = (d1

+ tzmd2)P

in V

Expanding d formally as a polynomial in t , we may write

Since both dy and dg are in V , we have the relation

6'1

+ + . . . + tp-2t'p-l

EV

tb'2

for any 0 # t E

F,

Because t h e ( p - 1) x ( p - 1) Vandermonde matrix whose rows are (1,t,. . . ,tp-') ranging the p - 1 nonzero elements of I F p ) is nonsingular, we conclude that 8, E V

for all

i E {I,&. .. , p - 1)

In particular, P-'

d;xmd2dy-1-i

81 = O l ( m , d 2 ) =

EV

i=O

Since

6'1

(m

+ 1 ,d z ) = 01 ( m ,z d z ) , it follows by induction on m t h a t

(with t

RESTRICTED SUBSPACE OF DERt

215

where

Since Xo = B l ( O , d z ) E V , it follows using (1) for m = 1 , 2 , . . . t h a t X k E V for all

kE

( 0 , . . . ,m}. In particular,

( p- l)!dldz

which shows t h a t d l d 2

-

+(p

-

2 ) ! d z d l = Xp-z E V ,

d z d l E If, as required. m

We now head towards the proof of our main result which is Theorem 4.4 (below), The following two technical results will clear our path. 4.2. Lemma. Let d l , .

. .,d ,

E DerE a n d z f E be s u c h t h a t all possible L i e brackets

of all orders of d l , . . . , d , v a n i s h on z. T h e n f o r a n y p e r m u t a t i o n u of { 1 , 2 , . . . , r }

Proof. The assertion being trivial for r = 2, we argue by induction on r . Assume that the result is true for r all transpositions (i,i

-

1. We prove it for r, by showing t h a t the result is true for

+ I), 1 5 i 5 r

- d i . . . d,+ld,. . . d,(Z)

Since the r

~

-

1. Indeed, we have

+ d i ... d ; d i + i . . . d , ( z )

d i . . . [ d i , d , + i ] .. . d r ( z )

1

(2)

1 elements d l , . . . , [ d ; ,d i + l ] , . . . , d , clearly satisfy the condition of the lemma,

the right hand side of ( 2 ) is equal to d l . . . d , [ d , , d , + l ] ( z ) = 0. S o the lemma is true. I

For any subset S of DerE, define the field of constants of S to be

r\ Kerd d€S

4.3.

Lemma.

Let V be a restricted subspace of D e r E a n d let F be the field

of constants of V . If z l r . .., x, are relatively p-independent in E J F , t h e n there ezist

d l , d Z , , . . , d , € V s u c h t h a t d , ( z , ) = 6,j, 1 5 i , j 5 n , where 6;j i s t h e Kronecker delta.

Proof. The case where n = 1 being obvious, we argue by induction on n. So assume that the result is true for n. Let

21,.

. . ,z,

zn+l be relatively pindependent in E / F and

216

CHAPTER 4

let d l E V be such t h a t d i ( z j ) =

d(zi)= 0, 1 5 i

1

&j,

5 i , j 5 n. If there exists a d

5 n, and d(z,+l) # 0, then the

d: = d;

-

n

E V such that

+ 1 elements of V defined by

di(zn+l)d(x,+lj-ld, 15 i 5 n, and d',+, = d ( s n + l ) - ' d

satisfy the required properties. Assume by way of contradiction that for any d E V with

d ( z ; ) = 0, 1 5 i 5 n, we necessarily have d(z,+l) = 0.

To show t h a t the latter configuration cannot occur, we consider the element

Because d i ( z j ) = 6 i j , 1 on x k , 1 5 k

5 i , j 5 n , it follows that all the Lie brackets of all orders vanish

5 n , and hence by

our hypothesis on

z,+1

(all the Lie brackets of all orders

are in V , since by Theorem 4.1,V is a Lie subalgebra of DerE). Applying Lemma 4.2 t o d l , . . . , d n E V and zn+l E E , we have

because d' E V and d p ( z j ) = 0 for 1 5 i such t h a t d(zi) = 0 for 1

[d,dt]

5n

and hence dp(z,+l)

= 0. If now d E V is

5 i 5 n , we clearly have

[ d , d i ] ( z j )= 0 for 1 5 i,j 2 n and since

I i 5 n.

We may therefore apply Lemma 4.2 to

V , we have [ d , d t ] ( z n + ~=) 0, 1

d , d l , . . . ,d , t o infer that d ( c ) = dY-1

because d ( z i ) = 0 for 1

5i5n

. . . d;-'d(z,+l)

=0

and hence d(zn+l) = 0.

We now make a general observation t o be applied subsequently. If

d ; ( X ) = 0 for 15 i

I n and

d ( X ) = 0 for any d E V with d ( z ; ) = 0 , 1

x(X) = O

for any

xEV

Indeed, we can write n

x

X(5i)di

=

+d

i=l

where d E V satisfies d ( z ; ) = 0 for 1 5 i 5 n. Therefore

XEE

is such that

5 i 5 n, then (3)

RESTRICTED SUBSPACE OF D E R f

217

... d E - ' ( z n - l )

proving ( 3 ) . Applying (3) t o c, we obtain c = dy-'

E F.

Our next aim is t o show, by descending induction, that for any integer m with 0 5 m

5p

-

1, there exists pm E F ( z 1 , . . . ,zn) such that

d:'

for every n-tuple

(A1,.

. . ,An)

with

. . . dx,-(z,+l

-

pm) = 0

C A; 2 m and 0 5 A, 5 p - 1. To begin the induction,

we take m = n ( p - 1) and

Assume that p m has already been constructed. We proceed t o construct p m - l . For any n-tuple (A1,. . . ,A,)

with

A, = rn - 1 and 0 5 A, 5 p

-

1, we define

I t will next be shown that ~ x , . . . x , E F . Indeed,

=

d f ' . . . d : ' + ' . . . d i n z n + l- df'

=d:'

. . . d:*+' ... d?(s,+l

. . . d:'"..

. d i n p m(by Lemma 4.2)

-pm)

=o by induction. Similarly, for any 0 E V with B ( z ; ) = 0 for 1 5 i

'Therefore, by ( 3 ) , we have

UA,...X,

5 n,we have B ( ~ A , . . . A , ) = 0 .

E F

Now define

We then have d:' . . .d ~ ~ ~ ( z-,p+ml- l ) = 0 and this completes the induction. In particular,

we have z,+i

= dy..

contrary t o the pindependence of

.d:(z,+i)

= /lo t F ( z i , . . . ,zn),

. . ,z n + l in E / F . This completes the proof of the

~ 1 , .

lemma. rn Let DerE be topologized by taking as a base for the neighbourhoods of zero those subspaces V of the form Derr(-E where X is a finite extension of EP. It is immediate

218

CHAPTER 4

that the closure of a restricted subspace is again restricted, and t h a t a subspace of the form DerKE is both closed and restricted (we shall discuss many topological aspects of the theory in our future treatment of Galois theory). 4.4. Theorem. Let E be a field of characteristic p

containing EP. T h e n the m a p F of E containing

EP

H

> 0 and let F be a subfield of E

D e r F E induces a bijection between the set of subfields

and the set of closed restricted subspaces of DerE. T h e inverse of this

m a p i s given by assigning t o each closed restricted subspace of D e r E its field of constants.

Proof. For any subfield F of E containing

EP,

D e r F E is a closed subspace of DerE.

Furthermore, by Corollary 2.8, the field of constants of D e r F E is F . Conversely, let V be a closed restricted subspace of D e r E and let F be the field of constants of V. We are left t o verify t h a t

v = DerFE It is obvious t h a t V

C DerFE.

suffices t o show t h a t for any

d ( x ; ) = $(xi) for 1

21,.

5 i I n.

Since V is closed, t o prove the opposite containment, it

.. ,zn E E and $ E DerF E, there exists a d E V such that

We may assume without loss of generality t h a t zl,. . . ,x, are

relatively pindependent in E / F . Owing to Lemma 4.3, we may find d l , . .. ,d , E V such that (1 I i , j I n)

d ; ( z j ) = S;j Then

d= has the required property. m

$(s;)di E V

219

5 Purely inseparable extensions

This chapter is devoted to a detailed study of purely inseparable extensions. If E / F is a finite purely inseparable extension, we show that, for large enough extensions K of F , the K-algebra E @ F K because a group algebra of an abelian p g r o u p , whose isomorphism class is uniquely determined by certain invariants of E / F . A thorough investigation of this phenomenon constitutes the first part of the chapter. The second part is devoted exclusively to the study of modular extensions, which are the inseparable equivalent of Galois extensions. Our presentation of t h e theory of modular purely inseparable extensions

is based on an important work of Waterhouse (1975). The basic discovery of Waterhouse is tlial the theory is closely related to t h e well-developed study of primary abelian groups.

After establishing some preliminary results, we develop the theory of pure independence. basic subfields, and tensor products of simple extensions.

We then compute the Llnr

invariants and display some complications in the field extensions not occurring in abelian groups. The final section is devoted to modular closure and modularly perfect fields. O u r principal goal is to show that any purely inseparable field extension E of F is contained in a unique minimal field extension K of F , where K / F is modular ( K / F turns out t o bc also purely inseparable). We also show that an arbitrary field extension is modular if

and only if there exists a n intermediate field K of E / F such that E / K is separable and

I { / & ’ is purely inseparable modular. Finally, we provide a number of characterizations of iiiodularly perfect fields, that is fields over which all extensions are modular.

1. P r e p a r a t o r y results for s p l i t t i n g theory.

Throughout this section, all fields are assumed to be of characteristic p > 0. In what follows, we fix a finite purely inseparable field extension E / F with E

# F . Given z E E,

we write e ( z ) = e ( z , F ) for the ezponent of z over E’, i.e. e ( z ) is the least nonegative

integer n such that

zpn

E F . It is clear that

CHAPTER 5

220

( F ( z ): F )

= p+)

We define e(E : F ) to be the integer n > 0 such that ( E : F ) = p". For any given integer n

> 0, we define

( E I F ) , , ( E I F ) , , 6, = 6,(E/F) and a , = a , ( E / F )

We then have two chains of fields:

E = (E/F)O 2 ( E / F ) '

2 .. . 2 ( E / F ) " 2 .. . 2 F

F = ( E / F ) o C ( E / F ) 1C . . .L( E I F ) , , 2 . . .

E

(upper chain) (lower chain)

The following terminology is extracted from Rasala (1971). The height h = h ( E / F ) of

E I F is defined by h = {maxe(z)Iz E E }

If z E E , then z is said t o be normal in E I F if e ( z , F ) = h ( E / F ) . Thus z is normal in E I F if and only if

A sequence zl,. . .,z,of elements of E is called normal in E I F if for all i E { 1,2,. . . ,r } , z;

is normal in E/F;-l and z;

6 F;-l, where

If E = F ( z l , . . . ,zr)for some normal sequence

51,. . . ,I, is a normal generating sequence of E I F . 1.1. Lemma. Let h = h ( E / F ) . Then

zl,.. .>zr in

E / F , then we say t h a t

PREPARATORY R E S U L T S

(i) h = min{n

2 lI(E/F)" = F } = min{n 2 l l ( E / F ) " = E )

c

(ii) ( E / F ) ~ - " ( E / F ) , ,

Proof.

o 5 n 5 h.

(i) By definition,

zPh

E F for all

5 n , proving

zPh

E F for all

2'' E F for all

z E

5

E

E and therefore

C: F , by

z E

2 l / ( E / F ) n= F }

E , we have ( E / F ) h = E . Conversely, if ( E / F ) n = E , then

E , hence e ( x ) 5 n for all x E E . Thus h 5 n as required.

(ii) Fix n such that 0

since EPh

E , hence ( E / F ) h = Er"F = F .

that h = min{n

Since

z E

F . Hence e ( z ) 5 n for all

Conversely, if ( E I F ) " = F , then EP" h

221

5

n

5 h and let

z E

( E / F ) h - n . Then z E EP"-''F and hence

virtue of the first equality in ( i ) . Thus z E ( E / F ) nas required.

The next observation ensures the existence of normal generating sequences of E / F

S be a generating set of E I F .

1 . 2 . Lemma. Let

( i ) lf K is a n intermediate field of E / F , then there exists s E S s u c h that s i s normal iic

E/K ( i i ) There exists a n o r m a l generating sequence of E / F with elements c h o s e n from S.

Proof. (i) If for n

> 0 we have SP" C K , then EP"

cK

since E = F [ S ] .T h u s

max{e(z,K)jz E E } = max{e(z,K)/zE S} and therefore we can choose s E S such that e ( s , K ) = h ( E / K ) .Hence s E S is normal i n

K/K. ( i i ) If we have a normal sequence

.., x r

21,.

with z,E S , 1

5i5

r , then either F, = E

# F,. In the latter case we may, by ( i ) , extend the normal sequence such that zr+l E S and zr+lis normal in E / E , . Since E # F,., we have

arid we are done, or E by choosing

z,+1

xr,.] $ E; and therefore

51,.

. . z,+1

is a normal sequence with elements chosen from S.

'The desired conclusion now follows from the assumption that E / F is finite. w 1.3. Theorem. (Rasala(l971)). Let

X I , ., . ,z,

be a normal sequence in EIE', l e t

F, = F [ z l , .. . ,z,], Fo = F and let q; = ( F ; : F,+,), 1 5 i

5 r.

Then

CHAPTER 5

222

w h e r e f o r i = l , F [ z y ,..., z : l l ] i s t o be interpretedasF.

Proof. If r = 1, then i = 1 and the assertion is

2 ':

E F , which is true by the

definition of q1. We now argue by induction on r. Assume t h a t r assertion holds for all normal sequences of length sequence

. . ,z,-1,

z1,.

< r.

> 1 and t h a t the

Then the assertion is true for the

hence for all i E {1,2,.. . , r

F [ z y ,..., z:Ll]

-

1)

We are therefore left t o verify that

Since

z2,.

. . ,x , is a normal sequence in E / F 1 , it follows by induction t h a t

Because { z i l O 5 i

< qr} is a basis of F1 over F [ z y ] ,it follows from (2) t h a t qr-l

Xisf with X i E FIX?', . . . ,z ; : ~ ]

zzr =

(3)

i=O

Thus to prove ( l ) ,it suffices t o show that X i = 0 for 0 Write qi = p e ' , 1

5 i 5 r.

< i 5 qr - 1.

Then

and hence el

2 e 2 2 ... 2 e ,

Setting t = q z / q r , it follows from (4) and the assumption

(4) 7

> 1 , that t is

a nonnegative

power of p . Now put

M = F [ z y ] and Then F

M &N

C Fl

and {z:10

5 i < qr}

terms in ( 3 ) t o the power t t o obtain

N =F[zi] is a basis of N over M . We now raise the

PREPARATORY RESULTS

We claim that

zp

223

E M and A: E M for all i E {0,1,. . . , qr

-

1); if sustained, it will

follow from the uniqueness of a n expression in terms of a basis that z$2 = Ah and, for all

i E (1,. . . ,qr - l}, X i = 0, as required. Set h = e l , g = and t = p g - f . (a) z$2 E

e2

and

f

Then h = h ( E / F ) , g = h ( E / F 1 ) ,q2 = p g , qr = pf

= e,.

Furthermore, we also have

(E/F)g

(b) X,t E ( E / F ) g

Indeed, (a) is true since (E/F)g= hence A: E

EqZF. For

(b), note that A, E ( E / F ) f =

We are thus left t o verify that

EqrfF= E P g F = ( E / F ) g .

(E/F)g

EqrF

E M.

and

Now

M = ( F I / F ) ~= ( F l / F ) h p g = F1 n (E/F)h-g But, by Lemma 1.1,

(EIF)’ C (E/F)h-g since h = h ( E / F ) and

(E/F)’ C (E/Fi)’ = FI since g = h ( E / F , ) . Hence ( E / F ) g C M and the result follows. We now provide some applications of Theorem 1.3 in which 5 1 , . . . ,zrwill be a normal generating sequence of E/F (such sequence always exists, by virtue of Lemma 1.2(ii)). In what follows, we fix t h e following notation: z1,.

. . ,z, is a normal generating sequence of

F, = E ’ [ z ~ ,...,z,], Fo ei = e(z,,

= F, 1 5

i5

F;-l), pi = p e t , 1 5 i 5

t

E/F

t

(note that e l = h ( E / F ) and, by ( 4 ) , e l

2 e2 2

., ,

2

G)

u = (&I,.

XruJ,

. . , a , ) E z r ,a , 2 0, z a

(XCY1,

. . . z;,.

= z?’

. . . , Arur), A € z

( ( 0 1 , . . . , a t , 0 ,..., 0) E 22‘10

5

(Yk

< qk, 15 k 5

J=J,={((Y1, ...,ru,)E22r1050Lk
Z}

1
(note that {s‘ln E Ji} is on an F-basis of F, and, in particular, { z a l a E J } is an F-basis of E ) .

I, = { a E Z‘lq,a E J } 1.4. Corollary. Let n be a n integer s u c h that 0

the m a x i m a l value o/ i E { 1,.. . , r } such that e,

5n<

> n. Then

h

=

h(E/F)and l e t

?n

be

CHAPTER 5

224

Proof. (i) The first equality comes from the definition of ( E I F ) " and the fact that zl,. . . ,z, generates E / F . If -yn = r, then there is nothing t o prove. Assume t h a t 7n < r and let z be such that 7n < i Theorem 1.3, zf' E

5 r. Then e; 5 n so F [ z y ' ,. . . ,zfI1],it follows t h a t

that q;)p", say pn = q;t. Since, by

and hence t h a t

F [ z y " ,. . . ,zf"] = F[zy",. . . , z y l ] Descending from i = r t o i = 7" (ii) Set M = F [ z y " ,. . . ,zi:l]

+ 1 proves

(i).

and N = F[z;", . . . ,xi"]. To show t h a t t h e powers (z:")~,

with 0 5 a k 5 q k J p n span N over M , we need zf;" E M . By Theorem 1.3, we have z? E

Since k

F [ z ? * ., . .

5 rn,we have e k 2 n, so p n ( q k and hence

zi"

E M . To show that the given set is

linearly independent over M , we note that

is linearly independent over

Fk-1

and M &

Fk-1,

proving (ii),

(iii) If a = ( a l , .. . , a T ) then , p n a E J amounts to (a) For 1

5 k 5 yn, 0 5 p n a k < q k

(b) For -yn

< k 5 r,

ffk=

0 (since

ek

5 n and hence

qk

5 p")

The desired conclusion is therefore a consequence of (i) and (ii).

PREPARATORY RESULTS

225

(iv) Take n = e;. Then p" = q; and 7n < z . Thus, by (i),

and the required assertion follows by virtue of (iii). ( v ) This follows from (i) and (iii).

1.5. Corollary. Let y1,.

. . ,ys

be another normal generating sequence of E I F and

let e;, . . . ,el, be t h e corresponding ezponents. T h e n r = s a n d e , = e: for all i E (1,. . . , r } .

Proof. Note t h a t

70 =

given integer such that 0 by Corollary 1.4(v), e; e:

> n. Hence if e: < h

r = 6 , ( E / F ) (Corollary 1.4(v)), hence

5n
=

T

= s. Let n be any

h ( E / F ) . Then e ; > n if and only if 7 %2 z . Hence,

> n if and only if 6,,+1(E/F) 2 z . Therefore e, > n if and only if =

el = e;, then e: = e , as required.

Owing to Corollary 1.4(iv), the set

{zqzalaE

I,} is a n F-basis of F [ s y ' ,. . . ,z:lli, 1 5

i 5 r . Furthermore, by Theorem 1.3,

Hence we can write z:

uniquely in the form

We will refer t o these equations as the structure equations of E I F . Note that these equations generate all the relations among the elements 2 1 , . . . , x,.since the equations determine an I;'-algebra of dimension q1 . . . qr which is exactly the dimension of E I F Now assume t h a t A is a commutative F-algebra which has

T

distinguished elements

zl,. . . , x, such that A = F [ z l , .. . ,z,], where the defining relations for zl,. . . , z,, are of the form

with ql = p e t , e l

2 e2 2 . . . 2 e , > 1,

ut.a E F and

In this case, following Rasala(1971), we say that A is a special F-algebra.

Thus, by

the foregoing, E is a special F-algebra (with respect to any normal generating sequence XI,.

..,z,

of E ) .

CHAPTER 5

226

Now assume t h a t K / F is any field extension and consider the K-algebra E @ F K . As usual we identify E and K with their images E @ 1 and 1 @ K in E is a normal generating sequence of E / F and if z; = x; @ 1, then E

@p

@F

K . If xi,.. . ,x,

K = K [ z l , .. . ,zr]

and over K the z; satisfy the same structure equations as the zi do over F , that is

Thus E @ F K is a special K-algebra. The following observation gives a criteria for E @ F K to be a field. 1.6. Lemma. Let K / F be a field eztension and let K; be the K-subalgebra of E @ FK

defined b y K; = K [ z l , .. . , z i ] , KO = K, 1 5 i 5 r , where z ; = x, @ 1 and

51,.

. .,I,is a

normal generating sequence of E / F . Let q; be as in Theorem 1.9, 1 5 i 5 r . Then E @ FK is a field if and only if ZP'

for all i E {I,. . . , r }

4 Kf-1

Proof. By the definition of K;, we have

Assume t h a t Ki-1 is a field. It suffices to show t h a t K, is a field if and only if Now Xq'

- pi

is the minimal polynomial of z; over K;-

1,

2 ':

4 K,"_l,

hence K ; is a field if and only if

Xq' - p ; E K ; - , [ X ] is irreducible. Since the latter is equivalent t o pi $ KPP1,the result follows. m For future use, we next record the following results. 1.7. Lemma. Let h = h ( E / F ) and let n be such that 1 5 n I h. (i)

C:=,& + l - ; ( E / F )

= e ( ( E / F ) h - " :F ) I e ( ( E / F ) =: F ) =

xi"=,& , ( E / F )

(ii) e ( ( E / F ) n: F ) 5 e(E : ( E / F ) " ) with equality if and only if EP" and F are linearly

disjoint over EP" n F (iii)

C:=,b h + l - i ( E / F ) I

a i ( E / F )I

Proof. (i) Consider t h e chain

&(E/F)

221

PREPARATORY RESULTS

Then we have n

((E/F)"-n

:F) =

n ( ( E / F ) h - i : (E/F)h+'-') i=l

and hence

The inequlaity

e((E/F)h-" : F ) 5 e((E/F), : F ) is a consequence of Lemma l . l ( i i ) . The last equality being obvious, (i) follows.

( i i ) T h e stated inequality is equivalent to the inequality

I n vicw of

(E: F ) = ( E : ( E / l q n ) ( ( E / F ) :, F ) = ( E : ( E / F ) " ) ( ( E / F ) ": F ) , the inequality ( 5 ) is equivalent to

l;'urthermore, ( E I F ) " = EP"F so that (EP" : (Ep" n F ) ) 2 ( ( E / F ) ": F ) with equalit)if arid only if EP" and F are linearly disjoint over EP" n F . This shows t h a t (6) always holds and that the equality in (6) (hence in ( 5 ) ) holds if and only if Ep" and F are linearly disjoint over E P "

i? F.

( i i i ) This follows from (i), (ii) and the obvious equality

cy=l6 , ( E / F )

1 . 8 . Corollary. The following conditions are equivalent.

( i ) For all n

2

1, EP- and F are linearly disjoint over EP"

(ii) For a f i n

2

1, e ( ( E / F ) n: F ) = e ( E : ( E / F ) n )

( i i i ) For all n

2 1, c r n ( E / F ) = 6,(E/F)

nF

=

e ( E : ( E / F ) " ) .rn

228

CHAPTER 5

Proof. Assume t h a t n

>h

= h ( E / F ) . Then EPnnF = EP", ( E I F ) " = ( E / F ) n - ' =

F and ( E / F ) n = ( E / F ) n - l = E . Therefore properties (i), (ii) and (iii) hold for this n. Hence we need only verify the equivalence of (i), (ii) and (iii) with n 1

2

1 replaced by

5 n 5 h. T h a t (i) and (ii) are equivalent is a consequence of Lemma 1.7(ii). If (iii)

holds, then by Lemma 1.7(i),

c n

e((E/F)": F ) =

a i ( E / F )=

i= 1

c n

6;(E,W)= e ( E : ( E / F ) " )

i= 1

and hence (ii) holds. Conversely, assume that (ii) holds. Then n

n

i=l

i= I

for all n such t h a t 1 5 n 5 h. Hence, for all n such that 1 5 n

5 h , a n ( E / F )= 6 , ( E / F ) ,

as required. w 1.9. Lemma. Let

51,.

. . ,x,

be a generating set of E / F and let Fi = F [ x l , .. . ,xi],Fo

F , 1 5 i 5 r . Then the canonical map

is an F-isomorphism if and only if

e ( z , , F i p l )= e ( z ; , F ) for all i E {1,'2,. . . ,r } Proof. The given map is an F-isomorphism if and only if the F-dimension of both sides are the same, i.e. if and only if

Consider the chain of fields:

Since e ( z i ,Fi-1) = (F;-i(z;) : Fi-1) = (F; : F i P l ) , 1 5 i 5 r , we have

=

PREPARATORY RESULTS

229

Because e(zi,F,-1) 5 e ( z , , F ) , the desired conclusion follows by comparing ( 7 ) and ( 8 ) . rn Let K be a field and let A be a commutative K-algebra. Following Rasala we say that A is a simply truncated polynomial algebra over K (or, for brevity, a truncated algebra over K) if there exist zi E A , 1

5 i 5 r , such that

( i ) A = K [ z l , . . . ,x,] (ii) The relations among

The sequence

51,.

51,.

. . ,z, are generated by equations

. . , z,is called a truncated sequence for A , and n, is said to be the order

of x,, written n, = o ( z l ) . We also say t h a t A is of type

n1,.

. . , n r relative t o the truncating

sequence zl,. . . ,x,. Let A be a truncated algebra over K of type n l , . . . ,n, relative to the truncated sequence

,

21,.. . 5,.

Then the map

induces an isomorphism of K-algebras K [ X l , . . . ,Xr]/'(Xy', . . . ,X:.) where ( X y l ,. . . ,X,'r)

Z

A

denotes the ideal of K [ X 1 , . . . ,X,] generated by X y ' , . . . ,

.

Hence the elements

constitute a K-basis of A . It follows that the map

is an isomorphism of K-algebras. The following result provides circumstances under which

a truncated algebra is identifiable with a group algebra. 1.10. Lemma. Let K be a field of characteristic p

> 0 and let A be a truncated

algebra over K of type pel , p e 2 , . . . ,per f o r s o m e positive integers e,. Then A as isomorphic t o the group algebra K G of the abelian p-group G such t h a t

CHAPTER 5

230

Proof. Let G =<

g1

>

x

< g 2 > x . . . x < gr >, where < g; > is

order p e i generated by g; E G. P u t a; = g i - I, 1 5 i

a cyclic group of

5 r , and observe t h a t

and that F G = F [ a l ,. . . , a,] Let

51,.

. . , x, be a truncating sequence for A such t h a t .(xi)

= pe', 1 5

i 5 r . Then, by

(9) and ( l o ) , the map

A

4

KG, x;

H

a,

is a surjective homomorphism of K-algebras. Since

dimK A =

n

pe' = IGI = dimK K G ,

i=l

the result follows. B The following result due to Rasala (1971) shows t h a t the number of generators and the type of a truncated algebra are independent of the choice of the truncating sequence. 1.11. Theorem. Let A be a truncated algebra over a field K , let yl,.

. . ,ys be truncating sequences for

j 5

s.

21,.

.. , I ,

and

A and let n; = .(xi) and mi = o(yi), 1 5 i 5 r , 1 5

T h e n A is a local ring with maximal ideal I =

(21,.

.., x r )

= (yl,.. . ,y8), r = s

and a; = mi

Furthermore,

11.1,.

f o r a l l i E (1,. . . ,r }

. . ,n, are uniquely determined b y the dimensions

Proof. Let J be the ideal of K[Xl,.. . ,X,] generated by K [ X 1 , . . . ,X , ] / J Z

K

O n the other hand, the homomorphism K [ X 1 ,...,X,]

I = ( x l , .. . , x r ) , hence A / I

S

XI,. . . , X,.Then

--f

A, X ,

H

z, carries

J onto

K . Since I is a nil ideal, this shows that A is a local ring

with maximal ideal I . Applying the same argument t o the truncating sequence yl,. . . ,y,, we see that I = (yl,. . . , y s ) .

PREPARATORY R E S U L T S

231

Define f t = d i m K ( I t / I t + ' ) , for all t 2 1. Note t h a t the sets

and

are two K-bases of I t / I t + ' . In particular, by taking t = 1, it follows that y1,.

. . ,x, and

XI,.

. . ,ys both form bases of I / 1 2 . Thus r = s. Assume by way of contradiction that n; # m, for some i E (1,. . . , r } . Then we may

choose j such that

nj

# mj and such that if j < r , then n; = m; for all i > j. By nj > m i , and we put t = m j . To show t h a t this leads to a

symmetry, we may assume that

contradiction, we first note t h a t the following conditions, in which a ; E {0,1,.. . ,t-1},

1

5

i 5 7, are equivalent: (a)

C:=, a ; = t and 0 5 a; < n;, 1 5 i 5 r

(b)

C:=,a, = t

Indced, for i

and 0 5 a ; < mi, 15 i

5r

I j , the inequality 0 5 a; 5 t - 1 already implies

> j , we know that n; = mi. Next, if

ities in (a) or (b), while, for i condition 0 5 a,

the corresponding inequal-

5 t - 1 does not hold for all i E ( 1 , . . . ,r } , then

C:=,a; = t

one a; = t and all other

a , = 0. In this case, we get a t least j such sequences which satisfy (a) since

i 5 j , while we get a t most j

-

but the

1zi

> t for

1 such sequences which satisfy (b) since m; 5 t for i

2 j.

These remarks yield two different values for ft, which is a desired contradiction. Thus n;=m, foralliE{1,

. . . ,r } .

We are left t o verify that

121,.

. . ,n, are uniquely determined by f l , f2,. . . . To this

f l , f 2 , . . . , f t , . . . are known. Then f l = r . Next suppose > j . For all t 2 1, define A t t o be the number of sequenccs

end, assume that the numbers that n; has been found for i u1,.

(i)

. . , a , such t h a t

CT=,a; = t

(ii) O

5 a, < n, for i > j

(iii) 0 5 a, 5 t

-

1 for

i5j

Then the preceding paragraph shows that, for t Thus one can obtain

nj

as the least t

2 n,+l

2 n j + l , nj > t

such that f t

-

At

if and only if

<j .

f t - At

2j.

232

CHAPTER 5

We now retrun t o the finite purely inseparable field extension E / F . Let K be a field extension of E and let A = K ( z ]be a truncated algebra of type n over K . Following Rasala (1971), by a K-variation of E / F with values in A , we understand a homomorphism d :E

4

A

of F-algebras

Note that the set {zi105 i < n} is a K-basis of A and hence we may write

c

n-1

d(2) =

d;(z)zi

i=O

1.12. Lemma. Let d : E

+A

be a K-variation of EIF. Then

(i) Each d ; is a n F-linear map of E into K (ii) For all z1,z2 E E and all i E {0,1,. . . , n - l},

Furthermore d o : E

4

K is the inclusion map

(iii) { d o , d l , . . . ,dn-l} is a higher derivation of E into K .

Proof. (i) This follows from the fact that d is a n F-linear map. ( i i ) The formula

follows from the fact that d preserves multiplication. In particular, we have do(zizz) =

d o ( z , ) d o ( z z ) so that do : E

K is a n F-homomorphism. But E / F is a purely inseparable

extension, hence do is the inclusion map. (iii) This is a direct consequence of (ii). Let d : E + A be a K-variation of E / F . We say that e E E is d-invariant if d ( z ) = z (which is equivalent t o d ; ( z ) = 0 for i

#

0 or t o d ; ( z w ) = t d ; ( w ) for all i and all w E E ) .

We let E d denote the set of all d-invariant elements of E. Then E d is obviously a subfield of E containing F. This field E d is called the fixed f i e l d of d . We let F i x K ( E / F ) be

the intersection of all E d as d ranges over all K-variations of E / F . F i x K ( E / F ) as the fized f i e l d of all K-variations of E/F.

We shall refer t o

PREPARATORY RESULTS

233

We now introduce a special type of variation. Let M be a n intermediate field of E / F such that E = M [ X ]for some X E E and let e = e ( X , M ) . Denote by A a truncated algebra over E of type p e (hence A = E[z]and

zPc

= 0). Define an E-variation

d:E--tA by setting

+x

d(z) = z for z E M and d ( X ) = X

We shall refer t o d as a Taylor variation of E/F with values in A. The fixed f i e l d of all Taylor variations of EIF is defined as the intersection of all E d where d ranges over all Taylor variations of E / F . Let us show that d is indeed a variation, i.e. that d is a

homomorphism of F-algebras. Consider the M-homomorphism f : M I X ] which sends X t o X

+ x.

Since

XP'

-

p) to X

+ x.

M[x] C

= ,u E M , we have

Thus f determines an M-homomorphism (Xp'

---*

f'

: M [ X ] / ( X pc p)

But the map MIX] 4 M [ X ] / ( X P e p ) , -

4

H

A which sends

x

x + (x"'- /I)

is an

.

hf-isomorphism, hence d is indeed a variation. To calculate d, (0 5 3 < p") explicitcly, WE write pc-1

i=O

c

pp-1

u,X') =

d(

ai(X

+x)i

a.(

E M, 0

5 i < p")

i=O

whence

Thus d is j u s t the Taylor expansion of P ( X ) = that

Ed = M

u,X*

in terms of X

7

x . We finally notc

CHAPTER 5

234

2. Splitting theory.

Throughout this section, F denotes a field of characteristic p

> 0 and F

closure of F . All fields considered are assumed t o be subfields of a finite purely inseparable field extension with E

# F . All

F.

a fixed algebraic

As before, E J F is

the notational conventions

introduced in Sec. 1 remain in force. Let

21,.

. . ,z,

be a normal generating sequence of E I F and let

F, = F [ z l , . , . ,xi],FO = F , e ; = e(zi,F i - l ) ,

q; = pe'

2 e2 2 . . . 2 e,. 1.5, the sequence e l , . . . , e , does not

(1 5 i 5 r )

From (4) of Sec. 1, we have e l By Corollary

generating sequence

21,.

. . ,z,

of

depend upon the choice of a normal

EJF.Hence the numbers e l , e 2 , . . . ,e , constitute invari-

ants of E J F ; we shall refer to e l , . . . , e , as the ezponent sequence of E I F . The exponent sequence of E I F determines a unique, up t o isomorphism, finite abelian p-group G given by

G

E

npz,x npez x . . . x npc,

We refer t o G as the group attached to E I F , and say that a field extension K I F splits

E J F if the K-algebra E

@IF K

is isomorphic to the group algebra K G . By Lemma 1.10,

K J F splits E I F if and only if E

@F

K is a truncated K-algebra of type p e l , . . . , p e r . Our

first aim is t o establish the existence of a purely inseparable field extension KJF such that

K J F splits E I F and K is contained in any field L 2 F such that L I F splits E J F Let us first recall from Sec. 1 t h a t the zi satisfy t h e following structure equations:

Here I; is a multi-index set defined by

2.1. Theorem. (Rasala(l971)). Let K I F be a field eztension. Then the following

conditions are equivalent:

(i) K / F splits E I F

SPLITTING THEORY

235

(ii) E @ F K is a truncating algebra over K (iii) E

@F

K is a subalgebra of a truncating algebra over K

(iv) For all i E (1,. . . , r } and all (v) For all

R

CY

E Ii, q 6 EK

2 1, E P n ( F n Kpn) and F

are linearly dzsjoint over F

Proof. (i)+(ii): If K / F splits E / F , then by Lemma 1.10, E K-algebra of type

pel,

n KP“. @F

K is a truncated

.. . , p e r .

( i i ) +(iii) : 0bvious

(iii)+(iv): Let A be a truncated K-algebra such that E A = K [ u l , ..., u,]withu36) = 0 ,

and p u t up = u f L. . . u p for ,B =

K . Put

2,

= z, @ 1, 1

5 i 5 r.

@F

K

2 A.

Then

l < j < s

(01,.. . , P a ) E N. Then {ufll/3 E N}

is a basis of A over

Then

and over K the z; satisfy the same structure equations as the z; do over F , that is

[this implies that E

@F

constants a t , a ) . Given

K is a special K-algebra with generators cy

= (cyl,. . . , C

5

Y , ) , ~ ctk

<

qk, 15

21,.

. . ,zr and s t r u c t u r e

k 5 r , put

z C x= zp’ . . . z:r,

Since (u”L3 E N} is a basis of A over K , we may write z a uniquely in t h e form:

Then, for n

since if q P

2 0 and

4 N ,then

q = p n , we have

=

0.

236

CHAPTER 5

Now let i E ( 1 , . . . r } and let ci,p denote b,,p for the multi-index a such that za = 2,. Then

Invoking (1) and ( 2 ) , we therefore derive

Because the elements uqaP with qip E N are independent over K , we can equate coefficients

This is a system of linear equations with coefficients in Kq; which is satisfied by

(U~,~),CI~.

Conversely if ( E i , a ) a E ~ ,is any solution of the system then, by reversing the calculation, we have

But the elements

zq;,

The conclusion is t h a t

with a E Ii are independent over K , hence we must have

(ar,a)aEI,

is the unique solution in K of a system of linear equations

with coefficients in Kq, and this shows that a%,aE

Kq*

for all a E I,. Thus

for all a E I,, proving (iv).

(iv)+(i): For all i and all a , write a,+ = d:;a with d,,, E K . Setting

we then have

and

vat; E K

SPLITTING THEORY

237

Moreover, there are no other relations on the u, independent of the relations u:'

=

0 since

the relations u:' = 0 define a K-algebra of dimension q1 . . . qr = d i m x E @F K . Thus

I,'

@F

K is a truncated K-algebra of type

q1,.

. . ,qr. Applying Lemma 1.10, we deduce

that K / F splits E / F , proving (i). (iv)+(v): We first ovserve that for q = p " ,

T h u s the linear disjointness condition becomes

( ( E I F ) " : F ) = ( E q ( Fn K q ) : ( F n K q ) ) N o w t h e s e t { z q a / q a E J } ( J = { ( a l ,. . . , a , ) / O i f f k < q k , l s k l r } ) b e l o n g s t o E q a n d ,

by Corollary 1.4(iii), this set is a basis of ( E / F ) " over F . Since this set lies in E q ( F n K Y ) , condition (v) is equivalent to the requirement that for all

2

1.

{zqa/qcuE J } is a basis of E q ( F ri K q )over E' f l K Y

(3)

If ( 3 ) is true, then the expression for z : ' in the structure equations must have coefficients in F

n

Kq',

that is, for all a , a,,= E Kq., proving (iv).

Conversely, assume that for all a and i, at,= E Kq'. Then, for all

I I r n c c , for n

2

2,

we have

e, and q = p " , we have

Thercfore, by the argument of Corollary 1.4(i), we obtain

Repeating the arguments of Corollary 1.4(ii), (iii), we derive ( 3 ) . Since ( 3 ) is equivalent to ( v ) , the result follows.

From now on, we sct

CHAPTER 5

238

2.2. Corollary. The f i e l d S = S(E/F) satisfies the following properties:

(i) S/F is a finite purely inseparable eztension (ii) A field eztensian K / F splits E/F if and only if S (iii) E

CK.

S

(iv) h ( S / F ) = h ( E / F )

Proof. (i) This is a consequence of the fact t h a t S / F is generated by finitely many purely inseparable elements. (ii) This follows from the equivalence of (i) and (iv) in Theorem 2.1.

(iii) By (ii), S/F splits E/F, so that E@pS

Z

S G where G is the group attached to E I F .

Since E admits an injective F-homomorphism E

--$

E 8~S,x

S-homomorphism (hence F-homomorphism) onto S (namely, that there is a n injective F-homomorphism E

--$

H

-

x 8 1 and S G admits a n

xpg

x g ) ,we deduce

S. But E/F is purely inseparable, hence

C S as required. (iv) Since F C E C S,we have

E

M E I F ) i h(S/F) R u t S is built using qi-th roots of elements of F where q, = pe* and e ,

5 h ( E / F ) . Thus

h(S,'F) = h ( E / F ) as required. By Corollary 2.2(ii), S ( E / F ) is uniquely determined as the smallest field which splits

E/F. In particular, S(E/F)is independent of t h e choice of the normal generating sequence of E/F used for its construction. We shall refer t o S ( E / F ) as the splitting field of EIF.

By Corollary 2.2(iii), E is the smallest possible splitting field of E/F. We now present various criteria for E t o be the splitting field of E / F 2.3. Theorem. (Rasala(l971)). The following conditions are equivalent: (1)

E is the splitting field of E / F

(ii) E

F [ x l ]@ I F . . .

F [ x r ] for some

51,.

..,xr E E

-

F

(iii) F is the fixed field of all Taylor variations of E / F (iv) F i s the fixed field of all E-variations of E / F (v) For all n (vi) For all n (vii) For all n

2 1,

EP"

and F are linearly disjoint over

EPn

2 1, e ( E : (E/F)n)= e ( ( E / F ) , : F)

2 1, 6,(E/F)

=

(rn(E/F)

( v i i i ) E @ F E E E G , where G is the group attached t o

EIF

nF

SPLITTING THEORY (ix) G i v e n a normal sequence X I , .. . ,z, in E I F with

E

z,+~ E

-

239

F, = F [ x l , .. . ,xs]# E , there exists

F, such that z,+~i s normal in E / F g and e(x,+l, F,) = e ( x s + l rF )

Proof. For the sake of clarity, we divide the proof into a number of steps. S t e p 1 . Here we show the equivalence of (i), ( v ) , (vi), (vii) and (viii).

The equivalence of (v), (vi) and (vii) is t h e content of Corollary 1.8. The equivalence of (i) and (viii) follows from Corollary Z.Z(ii), (iii). Since

E P " ( Fn EP") = E P " , it follows from Theorem 2.1 applied to K = E t h a t (v) is equivalent to the requireinent that E / F splits E / F . The latter is equivalent t o (i), by virtue of Corollary 2.2(ii), ( i i i ) . S t e p 2. Here we establish the validity of the implications (i)+(ix)+(ii). Assume t h a t (i) holds. Extend yt. Let

z1,.

. . ,z,

to any normal generating sequence z1,. . . , x,,y 1 , . . . ,

y = y1 and set n = e(y, F,) = h ( E / F , ) , q = p". We seek z E E with n = e ( z , F , ) and n = e ( z , F )

Let I denote the usual index set such t h a t { z a l a E I } forms a basis of F, over F . Then the structure equation for yq has the form

yq =

b,zqa

with b, E F

qnEI

Put c, =

6. Because the formation of

S ( E / F ) does not depend on the normal gen-

erating sequence chosen for E / F and because

51,.

. . ,z,, y1,.

. . ,yt

is a normal generating

sequence for E I F , we deduce that c, E S ( E / F ) = E . Furthermore, since c$ = b, E F , we have e(c,,

F ) 5 n.

We now claim that n = e ( c p , F , ) for some

p; if sustained, z

= c p will give the element

we seek since

n = e ( c p , Fa) I e ( c p , F )

I n,

which implies that n = e ( c p , F , ) and n = e ( c p , F ) . To substantiate the claim, note that

CHAPTER 5

240

as required. This proves t h a t (i) implies (ix). Finally, assume t h a t (ix) holds. Then we can find a normal generating sequence

21,.

. . x , of E / F such t h a t , for all i E { 1 , 2 , . . . ,r } ,

This proves (ii), by applying Lemma 1.9.

Step 3. Here we complete the proof by demonstrating (ii) *(iii)+-(iv)+(v). Assume t h a t (ii) holds. By rearrangement, we may also assume t h a t X I , . . . ,z, is a normal generating sequence for E / F . Let e , = e ( z i , F ) and qi = p e ' , 1 E [ u l ,..., u,] with u y = 0. Define d : E

d ( x i ) = zi

+ ui.

4

A by setting d ( z ) = z for z E F and

Then d is obviously a homomorphism of F-algebras. If we let d; be the

Taylor variation defined as in Sec. 1 by taking M = F [ z l , .. . , and

5 i 5 r , and let A =

X = x i , then

. .. ,x , ]

X * - ~ , X ~ + ~ ,

=

Mi

we have

which proves (iii). The implication (iii)+(iv) being obvious, we are left to verify that (iv)=+ (v). Assume by way of contradiction that (iv) holds but (v) fails, say for q = p n . Choose a subset

c1,.

. . ,ct of F with the smallest number of elements t such t h a t

(a) c 1 , . . . ,c t is independent over E9 n F (b)

c1,.

. . ,ct

is dependent over

Then we can find

21,.

. . , zt

Eq

E E q with all zi

#

0 such that

necessary, we may assume that zt = 1. By (a), z;

6F

C zic; = 0.

Dividing by z t , if

for some i. By symmetry, we may

assume that z1 4 F . Consider any E-variation d : E 0=

d(C

+

E[u]and write d ( z ) = C , d,(z)ua. Then

2i.i) =

Cd(zJd(c1) = i

i

c

d(Zi)Ci

i

which implies t h a t 0 =

c

d,(z;)c;

for all

CY

i

Bearing in mind that

d ( E q ) = d(E)'?C

C Eq[u]

(4)

SPLITTING THEORY

241

E E q . Thus the equations (4) are linear equations for the c, \vith

we see that d,(z,) coefficients in E Q .

Now, since (iv) holds and since z1 $ F , we can choose d such that d ( z 1 )

#

we have d p ( z 1 )

0 for some

p#

#

zl.

Then

0. On the other hand, d g ( z t ) = dp(1) = 0. It therefore

follows from (4) that t-1

i= I

This shows that

c1,.

. . ,ctp1 are linearly dependent over E Q ,contrary to the minimal choice

oft.

For future use, we next record. 2.4. Lemma. If F 2 E

M is a chain of fields, t h e n

Proof. Put L = S ( M / F ) . Then E algebra over L since over L so that

L splits M / F .

@F

L

CM

@F

By Theorem 2.1, E

L and M

@p

@F

L is a truncatccl

L is also a truncated algebra

L splits E / F and S ( E / F ) E L , by Corollary 2.2(ii).

We close by giving a number of examples (due to Rasala(lQ71)) of field extensions

E / F for which a t least one (hence all) of the conditions of Theorem 2.3 fails. In all examples, P denotes a field of characteristic p

> 0. As a preliminary, let us note that 11

E = F [ z ] ,h = e(z, F ) , 0 < n 5 h and g = pn, then

As another preliminary, we record

2 . 5 . Lemma. Let a , b , c be algebraically independent over P , let F = P ( a , b , c ) und let

E = F [ z ,w ] w h e r e z p z = a , w p = b + c z p

CHAPTER 5

242

Proof. The assertion regarding ( E I F ) ’ is obvious. Assume t h a t ( E / F ) 1 # F [ z ~ ] . Then there exists y E E with y

$- F[zP]and yp E F . P u t d

= yp. Then, by ( 5 ) , y

$- F [ z ]

since y E F [ z ]and yp E F imply y E F[zP]. Thus E = F [ z , y ] and therefore we can write w as

w

=

c

f ; 93.I

kzi+Piyk

with

f;,j,k

E F, 0

5 i , j ,k < p

By comparison, we obtain b,c E FP[a,d]so t h a t e ( F P [ a , b , c :] F”) 5 2

M , we have e ( M [ a b,c] , : M ) 5 2. Take

Then, for any field M such t h a t FP

M

= P(aP,b”, c”)

Then M [ a ,6 , c] = P ( a , b, c) = F . However, since a , b, c are algebraically independent over

p,

e ( M [ a ,6 , c] : M ) = e ( P ( a ,6 , c) : P ( a P b, P , c ” ) ) = 3 This contradiction proves t h a t ( E / F ) , = F ( z p ] . The equality F [ z p ]= F [ w p ]follows from the assumption that wp = b+czp. Finally, the assertions regarding a’s and 6‘s follow from the computation of ( E / F ) 1 and ( E I F ) ’ . w 2.6. Example. Let E and F be as in Lemma 2.5.

Then E / F does not satisfy

condition (vii) of Theorem 2.3 and

Proof. By Lemma 2.4, we have

T h e calcuIation of S ( E / F ) is straightforward and is left to the reader. w

SPLITTING THEORY

243

2.7. Example. Here we construct E I F such that for S = S ( E / F ) , S ( S / F ) f S .

Let a , b, c , d be algebraically independent over P and let F = P ( u ,b, c , d ) , E = F [ u ,v ] where u p 3= d a n d vPZ = a

T h e n S = F ( u , z , w ] with

zPz =

Set L = F [ z , w ] .Then L

a,

wP2 = 6P

+ (bP + cPa)uP”

+ cpu, that is wp = b + czp

C S and, by Lemma 2.4,

But S ( L / F ) = F [ z ,V%, f i ] ,so S ( L / F )

we have

S and therefore S ( S / F ) # S. rn

Our final example pertains t o the conditions of Theorem 2.1. 2.8. Example. Here we construct a chain of fields F

E

K such that K I F does

not split E / F and F is the fized field of all K - v a r i a t i o n s of E I F . Let F = P ( a , b , c , d ) where a , b , c , d are algebraically independent over P and p = 3. Put E

=

F [ z , w and

K = E [ z ]where z a = a , w 3 = b + c z 3 + d z G and

5”

= c+2dz3

Then

S ( E / F ) = F [ G ,$%,$,

a]

so that E ( S ( E / F ) : E ) = 2. On the other hand, e ( K : E ) = 1. Hence S ( E / F )

K and

therefore K I F does not split E I F , by virtue of Corollary 2.2(ii). It will now he shown t h a t F is the fixed field of all K-variations of E I F . To this end. we will define a variation d : E

4

K [ u ]where u G = 0. Define d on the generators

z

and u’

of I.I as follows:

d(z) = z

+ u , d ( w ) = w + zu

’To show that d defines an F-algebra homomorphism, we must verify the following equalities (z

+ u)’

+ ZU)”

= a , (w

=b

+ c ( z + u)” + d ( z + u

) ~

The first equality is clear and we compute t o verify the second:

(w b

-

+

C(Z

+ z3u3 = b + cz” dzG + ( c + 2 d z 3 ) u 3 , + u)” + d ( z + u ) =~b + c z 3 + cu3 + d z G + 2 d z 3 u 3 +- duG = b + cz3 dz6 + ( c + 2dz3)rt3

2.)”

= W”

T

T

CHAPTER 5

244

since uG = 0. Next let M be the fixed field of d. We claim t h a t M = F which will prove t h a t F is the fixed field of all K-variations of E / F . Assume by way of contradiction t h a t M

# F.

We wish to show t h a t F [ z 3 ]= ( E / F ) 1 C M . To prove F [ z 3 ]= ( E / F ) 1 ,we observe that F[z3]

( E / F ) I and that ( E / F ) 1 cannot have exponent 2 over F , since this would imply

which would imply that E / F satisfies condition (vii) of Theorem 2.3, which will contradict the calculation of S ( E / F ) given above. We now derive the inclusion ( E / F ) ,

M by

observing t h a t , since e ( ( E / F ) 1 : F ) = 1, ( E / F ) l is the unique subfield of E of height 1 over F and so must be a subfield of any M with F

M

E . Finally, the inclusion

F [ z 3 ]C M is impossible since d does not fix z3: d ( z 3 ) = (Z

+u

) =~ 2 3

+ u3 # 23

This substantiates our claim and proves the required assertion.

3. Chains of splitting fields and complexity.

Throughout, E / F denotes a finite purely inseparable field extension of fields of characteristic p > 0 with E # F . As in the previous section, S ( E / F ) denotes the splitting field of E / F and all fields considered are assumed t o be subfields of a fixed algebraic closure of

F . We begin by introducing the following ascending chain of splitting fields:

E = S O ( E / F )& S i ( E / F ) G . . .

S,-i(E/F)

G S , ( E / F ) CS

where, for n 3 1,

S,(E/F)

S(Sn-1 ( E / F ) / F )

1

We also put Sm(E/F) =

u

S,(E/F)

n>O

It follows from Corollary 2.2(iv), t h a t h ( S , ( E / F ) / F ) = h ( E / F ) for all finite n. Furthermore, this equality is also true for n =

00,

since a union of fields of height h has height h.

C H A I N S OF S P L I T T I N G F I E L D S A N D COMPLEXITY

The above indicates t h a t there is a good chance that S,(E/F)

245

coincides with S , ( E / F )

for some finite n. T h a t this is indeed the case is a part of the following general result. 3.1. Theorem. (Rasala(l971)). The following properties hold:

(i) S,(E/F) (ii) S , ( E / F ) (iii) S,(E/F)

is a finite purely inseparable extension of F =

S , ( E / F ) f o r some n < 00

is the splitting field o f S , ( E / F ) / F

and thus S,(E/F)

is the unique min-

imal field extension K / E such t h a t K is t h e splitting field of K / F . Proof. We first show t h a t if F

CE CK

is a chain of fields with K / F finite purely

inseparable and such that K is the splitting field of K / F , then

We know t h a t E

S,(E/F)

@F

=

K CK

S o ( E / F ) C K . Assume by induction that S , ( E / F )

C

K . Then

K and K @ 3K~ is a truncated algebra over K since K splits KIE’.

By Theorem 2.l(iii), K / F splits S , ( E / F ) / F so that, by Corollary 2.2(ii),

Thus, for all n , S , ( E / F )

CK

and therefore (2) also holds.

We now show t h a t there exists a finite purely inseparable field extension K / F such that K

2E

and K is the splitting field of K / F . Let B be a p b a s i s of F , let h = h ( E / F )

dnd let q = ph. For each & E 3,put Wb = ~, let S be a finite subset of B consisting of n elements, and let K s = F [ w ~ b E, S ] . Then ( K s : F ) = qn and K s also that if z is purely inseparable over F and e ( x , F )

@ b E s F [ w b ]Note

5 h, then there exists a finite subset

S of l? such that z E K s . Thus, if E = F [ z l , .. . , z,], choose a finite subset S of B such

5 Ks

that x , E K s for all z. Then E

and K = K s is the desired field.

Let K / F be a finite purely inseparable field extension such t h a t K is the splitting field of K / F . By ( 2 ) , S,(E/F) of ( i ) .

C K , proving

(i) Property (ii) is an immediate consequence

Finally, let n < co be such that S,(E/F)

S,,,(E/F)= S ( S , ( E / F ) / F ) splitting field of S , ( E / F ) / F .

=

=

S,(E/F).

Then S , ( E / F )

S ( S , ( E / F ) / F ) , which shows that S,(E/F)

=

is the

The rest of (iii) follows from ( 3 ) . So the theorem is proved.

a By Theorem 3.1, the ascending chain (1) stabilizes after a finite number of steps. Following Rasala (1971), we define the complexzty c ( E / F ) of E / F as the least n 2 0 such

246

CHAPTER 5

that S,(E/F)

Thus c ( E / F ) = 0 if and only if E is the splitting field of

= S,,(E/F).

E I F . Our next aim is to show that c ( E / F ) 5 h ( E / F )- 1 and that this equality is the best possible. We begin by introducing the following terminology due to Rasala (1971). We call a pair ( M , N ) of intermediate fields of E / F a normal pair if E = M N and M is a tensor product of simple extensions of F (for M

# F this is equivalent to the requirement that

M splits M / F ) . The degree d ( M ,N) of the normal pair ( M ,N) is defined as h ( N / F ) . For example, the pair ( F , E ) is a normal pair of degree h ( E / F ) > 0. If E splits E / F , then the pair ( E ,F ) is a normal pair of degree 0. 3.2. Lemma. Assume that there ezists a normal pair for E I F of degree d

> 0. Then

there ezists a normal pair ( M ,N ) for E / F of degree d such that (i) The ezponent sequence e l , e z , . . . , e m of M / F satisfies e;

15 i

2 d,

5 m.

(ii) For all z E N , e ( z , M ) < d .

Proof. Let ( P ,Q) be a normal pair for E / F of degree d > 0. We may write P in the form

P where f; = e ( z ; , F ) satisfy fk

2d

fl

2

= F [ Z l ]@ F

fi

. ..@F F [ z s ]

2 . . . 2 fs.

(if no such k exists, put r = 0 ) . Then d

Let r be the maximum integer k such that

> f r + l 2 . . . 2 fs.

Now put

(for r = 0, R is to be interpreted as F).Then R is a tensor product of simple extensions of F , R .N =

. . , z 9 ]= P Q

&[z1,.

=

E , h ( N / F ) = h ( Q / F ) = d. Thus ( R , N ) is also a

normal pair of degree d. Next choose from N a maximal sequence

y1,.

. . ,yt such that if

R; = R [ y l , . . . , y i ] , Ro = R , then e(yI, Ri-1) = d for i E { 1 , 2 , . . . , t } . Set M = Rt. By Lemma 1.9, M is a tensor product of simple extensions of F. Thus ( M ,N ) is also a normal pair of degree d. The exponent sequence of M / F is

fl,.

. . , f,,d, . . . ,d, which proves (i).

Also, by the maximal choice of the sequence y1,. . . , yt we have e ( z ,M ) < d for all z E N ,

proving (ii). rn 3.3. Corollary. Suppose there ezists a normal pair f o :

a tensor product of simple eztensions of F .

E/F of degree 1 . Then E is

CHAINS

OF

SPLITTING FIELDS AND COMPLEXITY

247

Proof. Let ( M , N ) be chosen as in Lemma 3.2 with d = 1. Then, for all x E N , e ( z , M ) < 1, so t h a t x E M . Hence N 2 M and E = M . N = M . Thus E is a tensor product of simple extensions of F .

H

We are now ready t o prove 3.4. Theorem. (Rasala(l971)). Suppose that there ezists a normal pair for E / F of

degree d > 0 . Then (i) There ezists a normal pair for S ( E / F ) / F of degree strictly less than d

(ii) c ( E / F ) 5 d - 1.

Proof. We first show that (ii) is a consequence of (i). Indeed, assume t h a t (i) holds and put n = d

-

1. By repeated use of Lemma 3.2, S , ( E / F ) has a normal pair of degree

a t most 1. By Corollary 3.3, S , ( E / F ) is a tensor product of simple extensions of F. Thus,

by Theorems S.l(iii) and 2,3(ii), S,(E/F)

= S , ( E / F ) so that c ( E / F ) 5 n, proving (ii).

To prove ( i ) , choose ( M , N ) as in Lemma 3.2. Select

XI,.

. . , z m in M

so that

Next, using the fact t h a t N generates E over M and Lemma 1.2(ii),choose z,+1,. ,V such that

zm+1,.

. . ,z, t

. . ,x, is a normal generating sequence for E / M . We claim that

5 1 , . . . ,x , is a normal generating sequence for E I F

(3)

To prove (3), let

We must show that e, = h ( E / F , - , ) . For IS

i > m , this follows from the fact that x,+1,. . . ,z,

a normal generating sequence for E / M . Assume that i

5 m. Note t h a t , since S, gener-

ates E over F,-I, we have

\Ye distinguish two cases. First assume that i

5 j 5 m.

Then

248

CHAPTER 5

Now assume that j

> m. Then, by Lemma 3.2(i), e; 2 d . O n the other hand, since N has

height d over F , we have

e(sj,Fi-l) 5 e ( x j ,F ) 5 d 5 ei The conclusion is t h a t

rnaxe(xj,F;-l) = ei = h ( E / F ; - l ) , 31;

proving (3). We now examine the structure equations for

XI,.

. . ,x,.

For i

5

m, we have simply

x:* = ai E F . For i > m we have the usual structure equations

We conclude therefore that S ( E / F ) is generated over F by q$x =

Now set P = F\ V u i , ; , i

x; for i 5 m

> m ] .Then S ( E / F ) = M . P

so t h a t ( M ,P ) is a normal pair of

S ( E / F ) over F . Finally, h ( P / F ) < d since by Lemma 3.2(ii), for i > m

Hence ( M ,P ) is a normal pair of S ( E / F ) over F of degree strictly less than d, as required. w 3.5. Corollary. (Rasala(lQ71)). c ( E / F ) 5 h ( E / F ) - 1

Proof. We know that ( F , E ) is a normal pair of E / F of degreed d = h ( E / F ) > 0 . Now apply Theorem 3.4(ii). w 3.6. Corollary. Assume that xp E F f o r all x E E . Then E is the splitting field of

EIF. Proof.

Our assumption guarantees that h ( E / F ) = 1. Hence, by Corollary 3.5,

c ( E f F ) = 0 which means t h a t E is the splitting field of E I F . m We close by showing t h a t the inequality of Corollary 3.5 cannot be improved. As a preliminary, we first record the following result.

249

CHAINS OF SPLITTING FIELDS AND COMPLEXITY

3.7. Lemma. Let E l / F and E 2 / F be two finite purely inseparable field eztensions.

I/ E l and Ez are linearly disjoint over F , t h e n

C

Proof. Set L = S ( E l / F ). S ( & / F ) . Then, by Lemma 2.4, L

S(E&/F).

the other hand, since El and E 2 are linearly disjoint over F , we have E i E 2

On

El @ 3 E~2 .

'Thus

( E i E 2 )@ F

(El @F Ez) @ F L

(El @ F L ) @ L ( E l @ F L )

L of truncated algebras

over L since L splits

El and E2 over F . Thus ( E l E z ) @ F L is a truncated algebra over

L and L splits E I E i

The last algebra is the tensor product over

over F , that is, S ( E l E 2 I F ) C L. So the lemma is verified. w We are now ready t o establish our final result. 3.8. Theorem (Rasala(l971)). There ezists a field F of characteristic p

> u and

u

Jamily {E,/F}n>oof finite purely inspearable field eztensiuns such that c ( E n / F ) = n and h ( E , / F ) = n -t 1 for n

=O,],

Proof. Let P be a field of characteristic p and let { Y ; , ~ ~ O 5

2,.

z

5j <

co} be alge-

braically independent elements over P . P u t

We next define elements

Zk,,,

for 0

5 k 5 n < 03 which will be used

E n . We use induction on n. First, we put

Assume that

~

m d , for 1 5 k

k

is defined , ~ for m

5 n , we set

< n. Then we

set,

to construct the fields

CHAPTER 5

250

We now define E n by

To investigate the properties of the finite purely inseparable field extension E n / F , we introduce the element

Zk,n

defined by

= Yk,n

It follows from (6) that xk,n = Z k , n

+ xk-1,n-1

for 1 5 k 5 n

. xo,n

We also define

Mn = F[zo,n,Z l , n , . . . ,z n , n ] , Lk,n = En-k . . .En It will now be shown that

+1 (9), we first observe t h a t e ( z , , , , F ) = n + 1, hence h ( E , / F ) 2 n + 1. h ( E n / F )= n

To prove

n+1

equality, we must show that xi,, that

~ p k ’ ~ ~ , , E- ~ F .

E F for 1 5 k

5 n. By induction

(9)

To prove

on n, we may assume

Then, by ( 5 ) and (6), we have

proving (9). It follows from (9) that the coefficients

( z i l l , n - lin) (6) are elements of F .

We next show that

Indeed (10) follows from (8) and the inequality e ( E , : F )

5 n2 + nf 1 comes from the fact

t h a t E n is constructed by adjoining one pnfl-root and n pn-roots. On the other hand,

e(E, : F ) 2 e ( E n - l E n : C proving (ii).

- 1 )

= e(En-l : M , E n _ l ) = n2

+ n + 1,

CHAINS

OF S P L I T T I N G F I E L D S AND C O M P L E X I T Y

251

A s a next step, we prove that z ~ ,. ~ . . ,xn,n is a normal generating sequence of E n / F

( 12)

(4),(5) and (6) give the structure equations for x,,,, . . . , zn,n

(13)

Fix n and set P, = F[x,,,, . . . , x ~ ,for ~ 0] 5 i

5 n. Then (12) and

(13) will follow provided

we show that (a) e ( z 0 + ,

F) = n

+ 1= h ( E n / F )

(b) e(z,,, : P,-l) = n = h ( E n / P t - I ) for 1 5 z

5n

l'roperty (a) is a consequence of (9). To establish ( b ) , observe that for all j ,

whence

n 2 h(EnIPt-1) 2 e(xt,n,pz-i) = e(Pz : Pt-1) But, by ( i i ) , we cannot have e(P, : P t P l ) < n for any i, which proves (b) and hence both

(12) and (13). We next prove the following assertions

Using the structure equations, we find that

proving ( 1 4 ) . The equality (15) follows by induction on k, by applying (10). To prove (16), set Tk,n = Mn-k+l . . . kfn. F . Then ?"A,,,

is a tensor product of simple extensions

of F arid linearly disjoint from En-.k over F . By Lemma 3.6 as well as (14) and (15), we

obtain

S (Lk,nIF)

S (En-k IF)S (Tk, n / F ) =

En-k-1 M n - k . Tk,n '

CHAPTER 5

252

proving (16). We now complete the proof by showing t h a t c ( E n / F )= n. The chain

is strictly increasing so that, by (16), c ( E n / F ) 2 n. But, by Corollary 3.5 and ( Q ) , we

have

L ( E ~ / F5)h ( E n / F )- 1 = n, as required. m

4. Modular extensions.

A . Introduction and preliminary results. Throughout this section, all fields are assumed t o be of prime characteristic p. By saying that the two given fields are linearly disjoint we mean t h a t they are linearly disjoint over their intersection. Following Sweedler (1968),we say that a field extension E / F is modular if

EP"and F are linearly disjoint for all n

2: 1

Our aim in this section is to present the theory of modular purely inseparable extensions developed in an important work of Waterhouse (1975). T h e basic discovery of Waterhouse is t h a t the theory is closely related t o the well-developed study of primary abelian groups.

For convenience, we divide this section into a number of subsections. After establishing

some preliminary results, we develop the theory of pure independence, basic subfields, and tensor product of simple extensions. The following subsections are devoted t o U l m invariants, their computation, and display some complications in the field extensions not occurring in abelian groups. In the final subsection we examine the modular closure and prove a number of related results. Assume for the moment that E / F is a nontrivial finite purely inseparable extension. Then, by Theorem 2.3, the following conditions are equivalent: (a) E / F is modular (b) E is a tensor product of simple extensions of F (c) E @ F E E EG, where G is the finite abelian p-group attached t o E / F .

Condition (c) says that E is the splitting field of E / F . Thus the finite modular purely inseparable extensions play a role like t h a t of those finite separable extensions which are

MODULAR

EXTENSIONS

253

their own splitting fields, i.e. the Galois extensions. Note also t h a t E / F is modular if and only if it has complexity 0 and in this sense finite modular purely inseparable extensions have the simplest structure. The preceding discussion applies equally well to the trivial case E = F , where S ( F / F ) is to be interpreted as F and G = 1. The rest of this subsection will be devoted to recording a number of preliminary results 4.1. Lemma. L e t E and {K,} be subfields of s o m e c o m m o n field a n d let E Ire

linearly disjoint f r o m each K,.

T h e n E i s linearly disjoint f r o m K = n K,.

Proof. Assume by way of contradiction that there exist

21,.

. . , z, in K linearly

independent over E (1K but not over E . We may assume t h a t n is minimal with respect to this property. Then there is a relation

cA,z, = 0 with A,

6 E and

A1

= 1, and this

relation is unique. The z, are in each K , and are dependent over E , hence dependent over

Fin K,. Because the dependence relation over E is unique, the A, are in E n K,. Howeve1 this implies A, E E n K , a contradiction. w Let {S,} be a family of sets. We say the S, are directed by rncluszon if for any

so,s, E {s,},there exists s6 E {s,} such that

sp

C '96

and

s,

& Sg. In case

s = Js,

and the S, are directed by inclusion, we say that S is a dzrected unzon of the S, 4.2. Lemma. Let E I F be a purely inseparable e z t e n s i o n a n d let {K,} be a f a m i l y

of intermediate fields. (i) lf each K , i s modular over F , t h e n their intersection is modular over F (ii) If E is modular over each K,, t h e n E i s modular over their intersection (iii) 1J the K , are directed by i n c l u s i o n a n d each K , is modular over F , t h e n E K , zs

modular over F . ( i v ) IJ t h e K , are directed by inclusion a n d E i s modular over each of t h e m , t h e n E is

modular over UK,.

Proof. (i) By hypothesis, ICErn and F are linearly disjoint, for each a and each n

2

1.

Since the p t h power map is injective, we have

(nK,)p- = n(K:") Hence, by Lemma 4.1, the intersection of all K , is modular over F . ( i i ) By hypothesis, Ep* and K , are linearly disjoint. Hence, by Lemma 4.1, E p " and r K,, are linearly disjoint, as required.

254

CHAPTER 5

(iii) and (iv). This is a consequence of the fact t h a t linear disjointness is a condition of finite type. 4.3. Lemma. Let E / F be a purely inseparable eztension. T h e n t h e following condit i o n s are equivalent:

(i) E / F i s modular (ii) For a n y f i n i t e subeztension K I F of E I F , t h e splitting field S ( K / F ) is contained in

E. (iii) E is a directed u n i o n of f i n i t e modular eztensions of F

# F.

Proof. (i)+(ii): We may clearly assume t h a t K

Let L = S,(K/F).

Then, by

Theorems 3.1 and 2.3, L is the smallest modular extension of F containing K . Owing to Lemma 4.2(i), L n E is modular over F. Since L n E 2 K and L n E that L n E = L. Thus S ( K / F ) g L (ii)+(iii):

L , we conclude

E.

Consider the family of fields S,(K/F), where K ranges all intermediate fields

of E / F such that K / F is finite. Since S ( K / F ) C E and, by Theorem 3.1, S,(K/F)

S , ( K / F ) for some finite n, it follows that S,(K/F)

C E.

Now each S,(K/F)

=

is a finite

modular extension of F and their union is E . If M I F and N / F are two finite subextensions of E I F , then by Lemma 2.4, S,(M/F)

Thus the S,(K/F)

(iii)+(i):

C S,(MN/F)

C

and S,(N/F)

S,(MN/F).

are directed by inclusion, proving (iii).

This is a direct consequence of Lemma 4.2(iii)

4.4. Lemma. Let E / F be a purely inseparable e z t e n s i o n a n d let K be a n intermediate

field.

(a) T h e following are equivalent: (i) K and FEP" are linearly disjoint over F [ K n Ep"] for all n

2

1 a n d E / F is

modular

(ii) K n EP" and F are linearly disjoint for all n

2 1 and E / K

is modular

( b ) T h e following are equivalent: (i) KP" a n d EP" n F are linearly disjoint f o r all n 2 1 and E / F is modular

(ii) EP" and FKP" are linearly disjoint over KP"[EP" n F ] for all n modular. Proof. This is an easy consequence of Proposition 2.3.13. w

21

and K / F is

255

MODULAR EXTENSIONS

4.5. Lemma. Let E J F be a purely inseparable modular extension. T h e n

(i) E / ( E n F p - ' ) is modular (ii) FEP" and E n FP

--I

are linearly disjoint over F[EP" n F P - ' ] and, in particular,

Proof. (i) Fix a positive integer n. Then EP" and Fp-' are linearly disjoint, since so are EP""

and F . Clearly EP" and E are linearly disjoint, so EP" and E n FP-' are

linearly disjoint, by Lemma 4.1. This shows that E is modular over

E n FP

- 1

.

(ii) We have

EP" = EP" n ~

( E n FP-'

P - - I

and, by Lemma 4.1, EP" n FP-' is linearly disjoint from E . Setting K = E n FP-I, it follows t h a t the condition (ii) of Lemma 4.4 (a) is fulfilled. Hence FEP" and E n FP

-1

are

linearly disjoint over F[EP" n F P - ' ] , as required. rn

For future use, we now introduce the construction of the tensor product of an arbitrary family of algebras. First, we define the notion of a direct limit. Let R be a commutative ring. Suppose ( A , ) , E ~is a family of R- modules (R-algebras) indexed by a directed set I and that for i

5 j we are given a homomorphism

f.. . A . i3 . 1

--*

subject to the conditions (i)

fj;

is the identity map of A i for all i E I

(ii) If i

5 j 5 k, then

f j k 0 fij

= ftk

Call this setup (Ai, f,?) a direct s y s t e m of R-modules (R-algebras). Suppose ( A i , f z J is ) a

direct system of R-modules. We form the direct sum @ i E ~ Atogether i with the submodulc

L generated by all elements of the form

The direct limit lim A , of the direct system ( A z ,f,,) is defined as the factor module 4

256

CHAPTER 5

The natural homomorphism

fi

: A; + IimA; is said t o be canonical. The construction --+

guarantees t h a t for all i 5 j

(4

fj

(ii)

fi(Ai)

(iii)

lim Ai = U i c l f;(A;) +

0

fij

= fi

C fj(Aj)

for all i

5j

(directed union)

(iv) If every f;i is injective (surjective), then all the f; are injective (surjective) Suppose now t h a t ( A ; ,f;,) is a direct system of R-algebras. Then it can also be regarded as a direct system of R-modules, and so we may form the R-module ]$A;.

For any

s , y E IimA; there exists i E I such that x = f;(x;) and y = f;(y;) for some q , y ; 6 A ; . It +

-

can easily be shown that the multiplication rule on lim A, given by

is well defined and that IimA; becomes an R-algebra. This R-algebra is called the direct +

limit of the algebras A;. The next result points out the universal property of direct limits. 4.6. Lemma. Let ( A , , f,,) be a direct system of R-modules (R-algebras). Suppose a n R-module (R-algebra) A and homomorphisms f o r all

I

4, : A , -+

A are given such that

5 j . Then there exists one and only one homomorphism

: limA,

-+

$J,

=

4)o f t l

A for which

4

4, =

o

-

f, for all t 6 I . The R-module (R-algebra) lim A, together with the f, is uniquely

determined by this property, up to isomorphism.

Proof. This is routine and left t o the reader. A typical example of direct limits is as follows. Let A be an R-module (R-algebra) and Ict (A;)iEl be the family of all finitely generated submodules (subalgebras) of A . We may regard I to be a partially ordered set in such a way that i 5 j if and only if A; which case I becomes a directed set. Let

f;j : A;

---t

C A j , in

A j ( i 5 j ) be the inclusion map. Then

( A ; ,f i J ) is a direct system of R-modules (R-algebras). Furthermore, one easily verifies that the map A

--t

-

lim A , , a

H

a

+L

(where a E A is regarded as a n element of some A , )

is an isomorphism. Thus each R-module (R-algebra) i s a direct limit of finitely generated

submodules (subalgebras). Let us now recall the construction of tensor product of finitely many R-algebras. Let A;, 1 5 i

5 n be (commutative) R-algebras. Then one can define a n R-algebra structure

MODULAR EXTENSIONS

on the R-module @:=,A; by the formula

The El-algebra @:=,A; is said to be the tensor product of algebras A 1 , . . . , A n . It posses an identity element equal t o el @ . . . @ e n , where e; is the identity element of A;. Furthermore, the mapping fi : Ai

+ @:=,A,

defined by

f,(a,) = e l @ . . . @ e,-l @ a ,

e,+l @ . . . B e n

is a homomorphism of R-algebras; the homomorphisms

ft

are called canonical. The tensor

product @:=lA, is characterized up to isomorphism by the following universal property, the proof of which is straightforward and therefore will be omitted. 4.7. Lemma. Let B be a commutative R-algebra and, for all i E { 1 , 2 , . . . , n } , l e t

4, : A i

+

B be a n R-algebra homomorphism. Then there ezists one and only one R-algebra

homomorphism h : @rz1Ai---t B f o r which $; = h o f; for all i E ( 1 , . . . , n } . Now let ( A , ) i E rbe an arbitrary family of commutative R-algebras. For any finite siibsct 5’ of I , put A s = B T E s A ; .Denote by f,,s(i E S ) the canonical homomorphisrn

A,

-3

A s . If S,S’ are two finite subsets of I such that S

C S’, then

Lemma 4.7 tells us

that there exists a unique homomorphism

such t h a t fs,s,o

f , ,= ~

f,,st

for all i E S. It is readily verified t h a t ( A s , f s , ~ ,is)a direct

system of R-algebras whose indexing set is the directed set of finite subsets of 1. The direct limit lim A s of this direct system is called the tensor product of the family of R-algebras +

( , 4 ; ) ; E ~By . an abuse of notation, l i m A s is also denoted by @iE1Aieven if I is an infinite +

set. If I = { 1 , 2 , . . . , n } , then lim A s can be identified with @F=lAi.For a n arbitrary set I , +

the tensor product 8 ; e i A ; is characterized up t o isomorphism by the property described in Lemma 4.7, in which

ft

has to be interpreted as the canonical homomorphism. Ai

+

@i~rAi

We close this section by recording some information pertaining t o group algebras. Let G be an arbitrary group and let El be a commutative ring. Owing to Proposition 1.4.1, the map G

--t

1 induces a homomorphism

CHAPTER 5

258

We refer to this homomorphism as the augmentation map. The kernel of the augmentation map is called the augmentation ideal and is denoted by I ( R G ) or simply I ( G ) . It follows from the equality

c"ss=):zg(g--)+~z, that as a n R-module I ( G ) is a free module freely generated by all g 4.8. Lemma.

-

1 with 1 # g E G.

Let t,h : G + H be a surjective homomorphism of groups and let

N = Kert,h. Then, the mapping f : RG

4

R H which is the R-linear eztension of li, is a

surjective homomorphism of R-algebras whose kernel is RG . I ( N ) . In particular,

RGIRG' I ( N )

R(G/N)

Proof. That f is a surjective homomorphism of R-algebras is a consequence of Proposition 1.4.1. It is plain that RG . I ( N ) E Kerf. Consequently, f induces a homomorphism -

f : R G / R G . I ( N ) + R H . The restriction of X : [G

7,

+ RG . I ( N ) ] / R G I. ( N ) + H

is an isomorphism. Thanks to Proposition 1.4.1, X-' can be extended to a homomorphism RH

--$

R G I R G . I ( N ) which is the inverse to

7. Thus Kerf

=

RG . I ( N ) ,as desired.

Given a n abelian group G and an integer n 2 1, we put G [ n ]= {g E Gig" = I} and G" = {g"lg E G } Then G" is a subgroup of G and the map G

---t

G", g

H

g" is a surjective homomorphism

with kernel G [ n ] .Thus

G / G [ n ]E G"

4.9. Lemma. Let F be a field of characteristic p

> 0 and let G be a n abelian p-group.

The (i) FG is a local ring with I ( G ) as the unique mazimal ideal

(ii) FG . I ( G [ p ]= ) {z E FGlzP = 0 }

Proof. (i) Since F G / I ( G ) end, let z E I ( G ) . Then z =

F , it suffices to show that I ( G ) is a nil ideal. To this

Cy=,X i ( g ; - 1) for some n 2

1, X i E F, g; E G. Choose

2 59

MODULAR EXTENSIONS

rn 2 1 such that gym = 1 for all i. Then (gi - l ) P m = gp

m

-

1 = 0 and therefore

as required.

f I I ,

(ii) Consider the sequence of homomorphisms FG-FGP-FGP, g

H

q p , g E G , and

II, by X H

XP,

X E F.

If

4 = $J

where f is induced by

o f then Ker4 = Kerf = F G . Z ( G [ p ] ) ,

by Lemma 4.8. Since

4CC

ZSS)

=

C z;sp = c c

Zgs)pl

the result follows. 4.10. Lemma. Let R be a subring of a commutative ring S and let G be a group.

Then

RGBRSESSG Proof. Observe t h a t {g

@ l/g6

as S-algebras

G } is a n S-basis for RG

@R

S and the map

is a n injective homomorphism. Now apply Proposition 1.4.1. w

4.11. Lemma. Let H be a subgroup of an abelian group G and let R be a commutative ring. Then (i) For any transversal T for H in G , R G is a free RH-module freely generated b y T

(ii) RG

@RH

RG

G

R ( G / H )@ R RG as R-algebras.

Proof. (i) Let T be a transversal for H in G. Then, for any t E T , ( R H ) t is the R-

linear span of the coset H t . Accordingly, for any t l , t z , . . . , t n E T , ( R H ) t l is the R-linear span of

u,"=,H t , .

As is apparent from the definition of R G , if S1 and

are disjoint subsets of G , then their R-linear spans meet a t 0. Hence

proving (i).

+ . .. + ( R H ) t , 5'2

CHAPTER 5

260

(ii) Let T be a transversal for H in G. By (i), each element of R G @ R HRG can be uniquely

CZl(t;C3 A ; t : )

written in the form

for some n 2 1,t i , t: E T, X i E R H . Consider the map

This map is obviously a surjective homomorphism of R-algebras. Furthermore, since the elements t H , t E T, form an R-basis of R ( G / H ) , the map is injective, by Proposition 2.3.1. m

Let G be an abelian group. We say that G is an ordered group if the elements of G are linearly ordered with respect t o the relation

xz

5 and, if

for all x,y, z E G, x

5y

implies

5 yz. 4.12. Lemma. Every torsion-free abelian group G is a n ordered group.

Proof. For any subset T of G, put T-' = { t - ' / t E T}. Let L be the class of all subsets T of G that satisfy the conditions:

x,y E T implies xy E T , By Zorn's lemma,

L contains a

maximal element

1@ T

T. We claim that

Deny the statement. Then there exists a nonidentity g in G such that neither g nor g-' belongs to T. With

S = T U {tg"lt E T , n

2 1)

U {gnln

2 1)

it follows that S 3 T and that S is closed under multiplication. The maximality of T now yields 1 E S and since G is torsion-free, we see that 1 = tg" for some t E T and some n

2 1. Thus g-"

E T and replacing g by g-' in the above argument, we infer that gm E T

for some m _> 1. This substantiates our claim, as 1 = (g")"(g-")"

f

T,contrary to our

assumption. We now define x < y to mean that yz-'

T n T-' = 0, hence y

< z. Then

< y and

zy-', yx-' E

y

E

T . Our hypothesis on T ensures that

< x cannot occur simultaneously. Suppose x <

T, whence zx-l

Assume next that x,y E G with z

=

# y

(zy-')(yx-') and x

E T, proving x

< z.

# y. Then yx-' # 1,

hence yx-' E T-', by virtue of (1). Accordingly, zy-'

= (yx-')-'

y and

yz-'

E T and y

4T, < x,

261

MODULAR EXTENSIONS

proving that G is a linearly ordered set. Finally, assume that z z 6 G, yz-' = (yz)(zz)-'

< y. Then, for any

T whence zz < yz, as required. I

T h e next observation will take full advantage of the preceding lemma. 4.13. Lemma. Let G be a n abelian group and let R be a commutative ring. Then

RG is a n integral d o m a i n if and only if R is a n integral d o m a i n and G i s torsion-free.

Proof. Suppose RG is an integral domain. Then so is R , since R is a subring of RG.

If G has an element g of order n > 1, then 1 - g is a proper zero divisor since 1- g n =

(I - g)(l+ g

+ . . . + d-l) = 0

T h e necessity is therefore established. To prove sufficiency, let z and y be nonzero elements of RG. Owing t o Lemma 4.12. G is an ordered group. Let a and 6 be maximal elements in Suppz and Suppy, respectively. Then, for g E Suppz and h E Suppy, g h = a6 implies g = a and h = b. Accordingly-

as required. w

4.14. Lemma. Let R be a commutative ring, let (G;);,l be a f a m i l y of abelian groups

and let LI,,IG,

be the restricted direct product of the f a m i l y (G;)isI. T h e n

Proof. For any i E I, let E Z : RG; + R(LIG,) be the homomorphism induced by the canonical homomorphism

T T ~:

G,

4

LIG;, and let f; : RG,

i

A be a homomorphism of

R-algebras. The restriction of f,, -

f; : G; + U ( A )

is a homomorphism of groups. Therefore, thanks t o the universal property of LIG,, there

exists a homomorphism

4 : LIG, + U ( A ) such that for all i €

I, f l

=

*

0 iT;

CHAPTER 5

262

Let

4 : R(LtG;)

RG;

---t

4

A be the homomorphism induced by

4.

Because any homomorphism

A of R-algebras is uniquely determined by its restriction t o G;, we must have f; =

*

0

";

The desired conclusion is now a consequence of the universal characterization of the tensor product of algebras. w

B . Pure independence, basic subfields and tensor products of simple extensions. In what follows, all fields considered are assumed t o be of prime characteristic p. Let E / F be a purely inseparable modular extensions. Following Waterhouse (1975), we say that a subset { x i } of E

-

F is pure independent if for each n 0 5 e;

2 1 the monomials

< p " , e; < ( F ( x ; ): F )

(almost all e; = 0) are linearly independent over FEP". If EP 2 F , then pure independence reduces t o the usual concept of relative pindependence in E / F (see Lemma 3.6.4). Let

E / F be a purely inseparable extension. Following Waterhouse (1975), we call a n intermediate field K of E / F pure if K and FEP" are linearly disjoint over FKP" for all n

2

1.

Observe t h a t if K is pure in E / F , then

FIEP" n K ] = F K P "

for all n

21

and, in addition, if E / F is modular, then by Lemma 4.4(a), E / K is also modular. 4.15. Lemma. Let E I F be a purely inseparable modular eztension. If { x i } i s pure independent, t h e n the subeztension K generated b y t h e xi is pure and is t h e tensor product over F of t h e simple extensions F ( x , ) . Conversely, i f K is a pure subfield of the form @ F ( x ; ) , t h e n t h e x ; are pure independent.

Proof. By the definition of pure independence, the monomials F I z ~ ~ ( 5 O e , < ( F ( q ): F ) ) and almost all e; = 0) are independent over F .

Hence the canonical surjection

@ F ( z i ) --t K is also injective. We then have FKP" = @F(x:") and a basis of K over

F K P " is given by the monomials with e; < ( F ( z i ) : F ) and 0 5 e; < p". By definition, these are still independent over FEP" and therefore K is pure. The converse implication is straightforward.

MODULAR EXTENSIONS

263

4.16. Lemma. Let E/F be a purely inseparable modular eztension and let

Then E / F is relatively perfect (i.e.

FEP

=E)

Proof. We show, by induction on r , that

This will prove the result, since for any given X E E there exists r

2 1 such t h a t

X E J'p'

By hypothesis, (1) is true for r = 1. Assume that (1) is true for 7 and take x E E n F'

-r-,

'

.

Then d ' E E n F p - ' , so 9' E FEP"" by induction. Therefore zp is a n F-linear combination of elements yp

"+I

with y, E E. Then xp and (yp")P are in EP, and xp is in the F-span of

pn+ 1

. By modularity, xp is in the (EP n F)-span of

y, z is a

"+ 1 y,P

. Taking p t h roots we see that

( E n FP-')-linear combination of the yp", and thus

as required. Let E / F be a purely inseparable modular extension. Following Waterhouse (1975), by a basic subfield of E / F we understand an intermediate field K such that (i) K is a tensor product over F of simple extensions of F (ii) K is pure in E / F

(iii) E / K is relatively perfect

Note t h a t pure independence is a condition of a finite type and so maximal pure indcpendent sets exist. T h a t the basic subfields exist is a consequence of the following result d u e to Waterhouse (1975).

4.17. Theorem. Let E I F be a purely inseparable modular extension. Then the basic subfields are precisely the subeztensions generated by maximal pure independent sets.

Proof. Let K be a n intermediate field of E/F which is generated over F by a maximal pure independent set {x;}. Owing to Lemma 4.15, to prove t h a t K is a basic subfield of

E / F we need only verify t h a t E I K is relatively perfect. As has been remarked

after t h e

CHAPTER 5

264

definition of purity, E / K must be modular. Assume by way of contradiction that E / K is not relatively perfect. Then, by Lemma 4.16, there exists n

E n KP-'

KEP"-' but E n KP-'

2 1 such t h a t KEP"

Since E / K is modular, Lemma 4.5(ii) implies t h a t

and thus E n KP-' is generated over K by the elements zpn-' with therefore choose such a n z with

zpn-'

4 KEP".

in FKP", say a n F-linear combination of y:"

E E . We may

By purity, K n FEP" = FKP", so zPn is

with y j E K . Because F is linearly disjoint

from EP", we can take this combination with coefficients in F n EP". Taking p"-th roots we obtain z =

Since zpn-' E KEP", not all b;"-' bP" E F and

bp"-'

with

bjyj

bj

E E and bp" E F

are in KEP". Hence we have a n element b in E with

$ KEP". It will next be shown that

{z;}

u {b}

is pure independent,

which will contradict the assumption of maximality (and hence K would then be a basic subfield of E / F ) . Assume that there is a relation among the properly restricted monomials ITzt' be over

FEP". All of them have 0 5 e

< pm

and e

<

pn = (F(b) : F ) . Grouping the terms

with the same e , we clearly have a relztion among the be over KEP". means the degree of b over KEP" is

<

p m , which implies bp"-'

If m 5 n this

E KEP". The latter,

however, is impossible, because raising to the pnPmth power would give bP"-' E K E p n , a contradiction. For m > n we similarly deduce degree

< p" and bP"-'

E KEP'"

2 KEP", a

contradiction. The conclusion is that in either case the coefficients of each power be must be zero. But these now are relations among monomials in the z;, and by pure independence all the coefficients of them must be zero. Hence we have only the trivial relation, which shows that {z;}U{b}is pure independent. This contradiction proves that E / K is relatively perfect and hence that K is a basic subfield of E I F . Conversely, let K be any basic subfield of E / F . Then K is a tensor product over F of simple extensions of F , say K = @ F ( z ; ) .Since K is pure in E / F , the {z;} are pure

MODULAR

EXTENSIONS

26 5

independent, by virtue of Lemma 4.15. We are therefore left t o verify t h a t {zi} is maximal. Assume by way of contradiction that { z ; } U { b } is pure independent, with ( F ( b ): F ) = p n . Since E / K is relatively perfect, we have E = KEP" and therefore b is in the span of K over

E'E?". However, by purity, K ( b ) and FEP" are linearly disjoint over FK( b ) P" = F K ? " , so b is in K [ F K P " ] = K , which is impossible. This completes the proof of the theorem. rn

Recall that a pure inseparable extension E / F is said t o be of finite exponent if EP" for some n

CF

2 1.

4.18. Corollary. (Sweedler(l968)). Let E / F be a purely inseparable modular eztension. If E / F is of finite ezponent, then E is a tensor product over F of simple extensions of F .

Proof. Let K be a basic subfield of EIF. Since E / F is of finite exponent, we have

EP" I3

=

F

C

K for some n

2

1. On the other hand, since E / K is relatively perfect,

Kl??" = K . But, by definition, K is a tensor product of simple extensions of F ,

hence the result. 4.19. Theorem.

(Waterhouse(l975)). Let E / F be a purely inseparable modular

extension, let K be a n intermediate f i e l d such that K / F is modular of finite ezponent and let K be pure i n E / F . Then there ezists a n intermediate f i e l d L of E / F such that

Proof. It is a consequence of Corollary 4.18 that K / F is a tensor product (over F ) of

simple extensions, say K = @ F ( z ) ,z E X . By Lemma 4.15, the set in E . Since K I F is of finite exponent, we have KP"

CF

for some n

X is pure independent

2 1. P u t M

= FEP";

by purity M is linearly disjoint from K over FKP" = F . Therefore the elements of X are

pure independent over M . Now extend X to a maximal pure independent set X U Y for

E over M . Because EP"

F C_ M , we have E = M ( X u Y).Setting L = M ( Y ) ,we then

have K L = E . Because X

U

Y is pure independent over M , the basis of K over F given

by monomials in the elements of X is still independent over L. Hence E = K

@F

L as

required. As a preliminary t o the next result, we record the following observation which demon-

strates how E n FP-' controls the structure of modular extensions

CHAPTER 5

266

4.20. Lemma. Let E / F be a purely inseparable modular eztension and let K be an

intermediate pure subfield such that

Then K = E.

Proof. If

X

E E, then

XP'

6 F for some r

2

1, hence X E E n FP-'.

Therefore it

suffices t o show t h a t

The case r = 1 being true by hypothesis, we argue by induction on r. So assume that

K n FP-' = E n

FP--

and take an x E E n F p - ' - ' .

this end, we first note that write

zp

zp

E K, so by purity

zp

We need only verify t h a t x E K . To

FKP = K n F E P . We may therefore

as a n F-linear combination of yp with y i E K, in which case z is an FP-l-Iinear

combination of the y;. Because E/F is modular, x is a linear combination of the y; over

E n F P - I = K n FP-'.

But the y i are in I(,hence so is z, thus completing the proof. m

Our next aim is t o present necessary and sufficient conditions under which a modular purely inseparable field extension is a tensor product of simple extensions. We need t h e following definitions due t o Waterhouse (1975). Let E / F be a purely inseparable field extension. We say t h a t a n element

height < n if

X$

X

in E has

FEP". An intermediate field K of E / F is said t o be of bounded height

in E / F if its elements not in F have bounded height, i.e. if

K n FEP"

=F

for some n 2 1

4.21. Theorem. (Waterhouse(l975)). Let E / F be a purely inseparable modular

extension. Then the following conditions are equivalent: (i) E is a tensor product over F of simple eztensions of F (ii) E can be written as an increasing union of intermediate fields

K1

C K2 C ... K , C . . .

where each K , has bounded height in E / F and ElK, is modular.

267

MODULAR EXTENSIONS

( i i i ) E n FP-' c a n be w r i t t e n a s a n increasing u n i o n of i n t e r m e d i a t e fields

L1 where each

L,

G Lz

c ... c L,

..

h a s bounded height in E I F a n d E I L , is modular

Proof. (i)+(ii): If E is a tensor product of simple extensions, let X be a corresponding pure independent set, and let K , be the intermediate field generated by the elements o f X of degree

5

pn. Then (ii) is satisfied

(ii)=>(iii):If (ii) holds, put L, = K , n FP-I. Then each L , has bounded height in $:,IF and their union is E n F P - ' . Furthermore, by Lemma 4.5(i), E / ( E n F p - ' ) is modular, so since E / K , is modular, it follows from Lemma 4.2(ii) that E is modular over

iis

required.

(iii)+(i):

Our aim is t o construct inductively pure independent sets

such that

F(X,) n FP-' = L,

for all n

21

1 % Lemma ~ 4.20 we will then have E = F ( U X , ) . This will prove the result by applying Lemma 4 . 1 5 . Assume that we have already X,-I and expand it to a pure independent set

X,, maximal among those for which

Thc induction step will be complete provided we prove that

Assume by way of contradiction that F ( X , ) n Fp-' height, we can find r

2 0 such

that

c S,.

Then, because S , has bounded

268

CHAPTER 5

Bearing in mind that E is modular over S , and over F , it follows from Lemma 4.4(a) that

S, n FEP' = F [ S , n EPr] and therefore there is an element bp' in S, but not in F ( X , ) . We claim that X , U { b } is pure independent; if sustained it will follow t h a t

and the intersection of t h a t with FP-' is

F ( b P r )@ [ F ( X n )n F P - l ] G Sn, contrary to the maximality of X,.

To substantiate our claim, put L = F ( X , ) and N = L n FP-'

C S,.

If X , U { b } is not

pure independent, we obtain a nontrivial relation among properly restricted monomials

ns,e1be.As in the next-to-last

paragraph in the proof of Theorem 4.17 we find

This shows that there is an element in LEPr+' n S , not in L and so we are left to verify that this is impossible by showing t h a t

L E P ~ +n' s, For any given m, consider the tower F

C FLP"

cL L. Since L is a tensor product of simple

extensions, we have that FLP" is modular over F (by Proposition 4.22 to be proved later). Hence, applying Lemma 4.4(b), we conclude t h a t LP and FLP""

are linearly disjoint over

It follows that LP and FEPm+l are linearly disjoint over Lpm+'NP. Indeed, assume that the elements A: in LP are linearly dependent over FEP"+'. Since A: E F , it follows from the purity of L t h a t the A:

are linearly dependent over FLP"+'. By the previous step,

they are then linearly dependent over LP"+' NP.

MODULAR EXTENSIONS

Now let x be in LEPm n Fp-' Then z p = up

1Apup.

and write z =

269

C Atut with

Both zp and up are in FEP"'+', and

over L P ,hence by linear disjointness it is so over LP

is a linear combination of the

XP

".+I

A, E L and a, E FEP"' .

NP. Taking p t h roots, we see

that x E LPmN[FEPm] = NEP". We have thus shown t h a t

and, in particular

L E P ~ +n' s, = N E ~ +n' S, Finally, consider the chain N

C S, C E .

By purity, we know that E is modular over L ,

hence as we saw earlier E is modular over N = L n FP-' . We conclude therefore that

S, n EP"' is linearly disjoint from N for all rn. Since, by hypothesis, E / S , is modular, it follows from Lemma 4.4(a) that

On the other hand, by the choice of r , we know that S, n FEP'L'

C L , so

s,n NEP'+' c I, Thus S , ,? LEPr+'

c L as required.

C. TJ1m i n v a r i a n t s of modular e x t e c s i o n s . Throughout, all fields are assumed t o be of prime characteristic p . Assume that S / > ' is a purely inseparable modular extension.

In this section we will introduce numerical

invariants of E / F which resemble the Ulm invariants of primary abeliari groups. In fact it will be shown t h a t the corresponding theory is closely related to the well-developed study of primary abelian groups. All the results presented are due to Waterhouse (1975).

We begin by introducing the following terminology due to Waterhouse (1975). I,et

E / P be a purely inseparable extension. For any given ordinal a , we define the intermediate field ( E I F ) " (or FEP") as follows: ( i ) (I?:F)"

=

(ii) ( E / ) ' ) ' ' '

B =-

F[(E/F)"!P

(thus ( E / F ) " + ' = ( ( E / F ) a / F ) l )

270

CHAPTER 5

(iii) ( E I F ) " = n p < Q ( E / F ) oif

LY

is a limit ordinal

We also put

=nQ(E/F)a We then have a descending chain of intermediate fields

E 2 ( E / F ) ' 2 . . . 2 ( E / F ) " 2 (E/F)"+'

2 ..

This chain need not stabilize at

The following example is due to Waterhouse (1975). Let

where y and the x ; are indeterminates, and let

Then ( E I F ) " = F ( y ) and (EIF)""

= F so t h a t

(EIF)"# ( E / F ) " + ' , as claimed.

The first ordinal a for which ( E / F ) " = ( E / F ) " is called the length of E / F . In the above example, the length of E / F is w

+ 1.

Given a n ordinal a and a n element x E E , we say t h a t z is of height a if z E (E/F)"

and

z $! (E/F)"+'

By definition, the elements of ( E / F ) " are of height

00.

4.22. Proposition. Let E I F be a modular purely inseparable extension. Then, for

any ordinal a , the following properties hold:

(i) (E/F)"is modular over F (ii) E is modular over ( E I F ) " Proof. (i) The case a = 0 being true by hypothesis, we argue by induction on a .

If a is a limit ordinal, then ( E / F ) " =

n,,,(E/F)P

and each ( E / F ) P is modular over

F by induction. Hence, by Lemma 4.2(i), ( E I F ) " is modular over F. Suppose now that

K = ( E / F ) Qis modular over F . Owing to Lemma 4.3(iii), we can write K as the directed

271

MODULAR EXTENSIONS

union u Kx where each K x / F is a finite modular extension. Then FKP = U FKY, and the FKI are modular over F since Kx is a tensor product of simple extensions. Thus

( E / F ) " + ' = FKP is modular over F , by virtue of Lemma 4.2(iii). (ii) Owing t o Lemma 4.4(a), the field E is modular over ( E / F ) a if and only if ( E I F ) "

and FEP" are linearly disjoint over F [ ( E / F ) " n Ep"]for all n

2

1. If

cy

is a limit ordinal,

then the modularity holds by induction and Lemma 4.2(ii). Also, the second condition is trivial for n

2 a , as there ( E / F )n~EP" = EP"

Therefore, assuming the result for a , we must show that

Taking into account that

arid that the opposite containment is trivial, the result follows. m

We now introduce the Ulm invariants of a modular purely inseparable extension E/E'.

To illustrate the parallel with the theory of primary abelian groups, we first introduce the notion of an Ulm invariant (also called Ulm-Kaplansky invariant) of an abelian p-group G. For any ordinal a , the group

if

(Y

GP"

is defined a s follows:

is a limit ordinal. Note that if GP = 1 (i.e. if G = G [ p ] ) then , G is a n clementary

p-group, hence can be regarded as a vector space over the field

a - t h U r n (or Ulm-Kaplansky ) invariant f,(G), where

cy

IF, of integers mod

p . 'l'hc

is any ordinal is defined by

CHAPTER 5

212

Now let E / F be a modular purely inseparable extension and let a be any given ordinal.

We shall define t h e a-th U l m invariant f , ( E / F ) of E / F by imitating the corresponding definition of f"(G). T h e role of G[p] will be played by FP

--I

and GP" will correspond to

( E / F ) Q .Since GPY[p] = G[p] fl GpY, it is clear that ( E / F ) " n F P - ' must correspond t o

GPO

[ p ] . Finally, the role of IFp-dimension

will be played by the cardinality of a relative pbasis. We are thus led to define f , ( E / F ) as the cardinality of a n y relative p-basis of ( E I F ) " n FP 4.23.

-1

over (E/F)"+' n Fp-'.

Lemma. (i) f , ( E / F ) i s the cardinality of a n y m i n i m a l generating set of

( E / F )n~FP-' over (E/F)"+' n F P - ' . I n particular,

f , ( E / F ) = 0 if and only if ( E I F ) " n FP-' = (E/F)"+' n FP-'

(ii) If B is a n y relative p-basis of ( E / F ) " + ' n FP-I over F and C is a n y relative p-basis of

( E I F ) " n FP-' over (E/F)"+' n FP-I, t h e n B U C i s a relative p-basis of ( E I F ) " (1 FP-' over F (iii) For a n y finite n

2 0 , f n ( ( E / F ) " / F )= f , + " ( E / F )

Proof. (i) This is a consequence of the fact (see Lemma 3.13.5(ii)) t h a t in a purely inseparable extension of finite exponent relative p b a s e s are the same as minimal generating sets. -1

(ii) P u t El = ( E / F ) " + ' n FP-I and Ez = ( E / F ) " n F P

2 El C E,

. Then we have a chain of fields

E; F . Now B is relatively p-independent in E 2 / F and C is a minimal generating set of Ez/E1. Since

F

with Er

it follows from Lemma 3.13.6(i) that B U C is relatively pindependent in E 2 / F . Since E2 =

F ( B , C ) = E,( F , B , C ) , (ii) follows.

(iii) It suffices to show that

MODULAR EXTENSIONS

213

For n = 0, the above equality obviously holds. The general case follows by induction on

11

since if ( E / F ) " + " = ( ( E / F ) m / F ) "then ,

as required. B 4.24. Proposition. Let E / F be a modular purely inseparable eztension.

(i) (E/F)OO is relatively perfect over F a n d contains a n y intermediate field K of E/1' s u c h that K / F i s relatively perfect (ii)IJ f n ( E / F ) = 0 f o r all f i n i t e n, t h e n E / F i s relatively perfect (iii)

If there

ezists a s u c h that f a + n ( E / F ) = 0 f o r all f i n i t e n, t h e n the length of E / F is


a n y integer n pnt 1

2 0 , fn(E/F)equals t h e

5 w and

for

cardinal number of tensor generators of degree

over F .

Proof. (i) If K / F is relatively perfect, then ( K / F ) O = K for any a , hence K = ( K / F ) a2 ( E / F ) a and therefore K

2 (E/F)-.

for all

iz

Conversely, put K = (E/F)OO. We must show that K / F is

relatively perfect, i.e. that ( K J F ) ' = K . If a is the length of E J F , then K

=

(E/F)"

and ( E / F ) n = ( E / F ) " + ' . Hence

( K / F ) ' = ((E/F)"/F)' = (E/F)*+' = K , as required.

( i i ) The assumption says that f o ( E / F ) = f n ( E / F ) = 0 for all n

2

1, i.e. t h a t

Now apply Lemma 4.16.

(iii) By Proposition 4.22, ( E I F ) " is modular over F. Furthermore, by Lemma 4 23(iii),

f , ( ( E / F ) " / F ) = 0 for all finite n

2

0. Hence, by (ii), ( E / F ) " / F is relatively perfect.

Therefore, by (i), ( E / F ) m= ( E / F ) Mand therefore the length of E / F is

5

a.

214

CHAPTER 5

(iv) Let {z;liE I } be a subset of E such that

E =@,i~iF(zi) Then, for any finite n, we have

and therefore

(E/F)" = n % [ @ ; E ~ F ( $ j l It follows that ..+I

(E/F)"+' = n:Tl[@iEIF(zp proving t h a t E / F has length

)] = (E/F)",

5 w.

For each i E I , let pet be t h e degree of z; and let J,, = { j E Ilei

> n}. Then

( E / F ) n = @ j ~ j , , F ( ~ p SO "),

and

(E/F)"+' n F P - ~ = Hence {zyc'-' Ij E J n - J n + l } is aminimal generating set of ( E / F ) W F P - ' over ( E / F ) " + I n

Fp-' and therefore f , ( E / F ) = IJ,

-

J,+lI.

Since J,,

-

Jn+l is t h e set of all

i

E

I for

which z, has degree pnfl, the result follows. rn 4 . 2 5 . Proposition. Let E I F be a modular purely inseparable eztension and let n , r

be finite with r (i) fn(E/F)

2 2.

Then

fn((En F p - " - ' ) / F )

(ii) f n ( E / F )= f n ( E / ( E / F ) " + ' )

Proof. (i) Let s E {n,n+ l}. Since r 2 2, we have n + r > s. Note also t h a t , by Lemma 4.5(ii),

( E / F ) ' n FP-' = FEP' n FP-' = F[EP' n F P - ' ] Since n + r > s, we have EP' n FP-' = ( E n FP-"-,)P' n FP-' and hence

( ( E nF P - " - ' ) / F ) "n Fp-'

=

F ( E n F p - " - r ) p 'n FP-' = ( E / F ) ' n FP-'

275

MODULAR E X T E N S I O N S

Thus the same fields occur in computing both sides of (i), (ii) P u t K = FEP". We must show t h a t -1

( ( F p - ' n K ) : ( F P n F K P ) )= ( ( F K P ' ) P -n ' K ) : ( ( F K P ' ) P -n ' FKP))

(1)

Put El = FP-' n K and Ez = (FKPr)p-ln FKP. Then El n EZ = FP-' n FKP. We claim that

El and Ez are linearly disjoint

(2)

-1

Since (2) implies (E1Ez : Ez) = (El : (El n E z ) ) = (El : (FP

n F K P ) ) ,the replacement

of El Ez by the expression in (3) will yield ( 1 ) .

By Proposition 4.22, ( E / F ) " + ' = FKP is modular over F and hence F X " and Fp.' are linearly disjoint. Similarly, K and (FKp')p-' are linearly disjoint. Hence, by Lemma

4.1, we have the property (2).

To show t h a t El& is as large as claimed in ( 3 ) ,we write K as a directed union of finite modular extensions (since K = ( E / F ) " ,this can be done by virtue of Proposition 4.22 and Lemma 4.3). If the corresponding statement is true for each of these finite extensions, then it is also true for their union. We may therefore assume t h a t K / F is finite. By Corollary 4.18, we may write K / F = @ F ( z i )with

2;

of exponent e;, say. Then the elements w i t h

the p-th power in FKPr are generated by xi"' where d , = p:'-' if e;

2

r . The last type are in (FKp')p-'

n FKP, and

if ei <

T

and d, = p'-'

the others are in FP

- 1

n K , as

required. rn 4.26. Lemma. Let E / F be a modular purely inseparable extension a n d let K be a n

intermediate field s u c h that KIF is pure modular and E / K is relatively perfect. Then, f o r all f i n i t e n

2

1 a n d all ordinals a ,

fa(E/FEP"= ) fa(K/FKP")

Proof. n

2

Since K / F is pure, K and FEPn are linearly disjoint over FKP" for all

1. Furthermore E = E p " ( K ) = (FEP")K,since E / K is relatively perfect. Thus a

pure independent set giving tensor generators for K over FKP" also gives tensor generators

CHAPTER 5

276

for E over FEP". (We have used the fact t h a t both EIFEP" and KIFKP" are tensor products of simple extensions, by Proposition 4.22(ii) and Corollary 4.18). The desired conclusion is therefore a consequence of Proposition 4.24(iv). It is now a n easy matter to prove the following result.

4.27. Theorem. (Waterhouse(l975)). Let E I F be a modular purely inseparable

eztension and let S be a maximal pure independent set of E I F . T h e n , f o r a n y integer n

2 0,

the cardinal number of elements of S having degree pn+'

over F i s f n ( E J F )and

thus is independent of the choice of S .

Proof. P u t K = F ( S ) . Then, by Theorem 4.17, K is a basic subfield of E I F . Hence K is pure in E J F and E J K is relatively perfect. Furthermore K l F is a tensor product of simple extensions with S as a tensor generating set. By Proposition 4.24(iv), f n ( K J F )is the cardinal number of elements of S having degree pn+' over F . Thus it suffices t o verify that

f n ( E / F )= f n ( K / F )

for all integers n

20

To this end, we first apply Proposition 4.24(ii) (with E = K and r = 2) t o obtain

f,(K/F) = ~ ( K / F K ~ " + ' ) Next we apply Lemma 4.26, t o derive

Again, by Proposition 4.25(ii),

f n ( E / F )= f n ( E / F E P " + ' ) and the required assertion follows.

D. Ulm invariants and group algebras. In this section we will compute the Wlm invariants f , ( E / F ) of a purely inseparable field extension E / F by a process which explicitly connects them with the Ulm invariants of groups. Apart from providing some connections between modular extensions and group algebras of abelian pgroups, we also display some complications in the field extensions not occuring in abelian groups. In what follows all fields are assumed t o be of prime characteristic p .

MODULAR EXTENSIONS

277

Let F be a field and let A be any commutative F-algebra. For any given ordinal a , we define ( A / F ) ainductively just as for field extensions. Namely, we put (i) ( A I F ) " = A

(ii) ( A / F ) " + '

=

F[(A/F)a]p

(iii) ( A / F I U = n p , , ( A / F ) a if a is a limit ordinal

We then have a descending chain of subalgebras

A 2 ( A / F ) ' 2 . . . 2 ( A / F ) O 2 (A/F)"+' 2 If K / F is any field extension, then it is easy t o show inductively that ( ( Ag,F K ) / K ) a = ( A / F ) "

X

for all ordinals a

If C is an abelian pgroup, then a straightforward calculation which uses the identity

shows that

(FGIF)" = F G P ~

(1)

Now let A be any commutative local F-algebra such t h a t A / J ( A ) Z F and let the ideal A [ p ] of A be defined by A [ p ]= {u E A J a P= 0 ) Following Waterhouse (1975), we define the Ulm space Ulm(A) of A as the vector spacc over F given by

UWA)

A[Pl/J(A)A[Pl

Let U be a subalgebra of A . Then the inclusion map B Ulm(l?)

i

-+ A

induces a homomorphism

U l m ( A ) of F-spaces given by

If this homomorphism (to which we shall refer as the natural map) is injective, then \\c say that Ulni(U) injects into U l m ( A ) . From now on, if Ulm(B) injects into Ulm(A), thcii we identify

Ulm(B) with its image in Ulm(A). Assume t h a t 13

C are subalgebras of 1:

> u c h that Llm(H) and Ulm(C) inject into U l m ( A ) . Since B [ p ] C C i p ] ,we then h a v e

UIrn(R) i Ulm(C).

("1

zia

CHAPTER 5

We say that A is an Ulrn algebra if for any ordinal a Ulm((A/F)a)

injects into Ulm(A)

For any such algebra A, we have

Ulm ((A/F)"+')

Ulrn ( ( A / F ) O )

by virtue of (2). Following Waterhouse (1975), we define the Ulm invariants f a ( A / F ) of the Ulm algebra A by

f,(A/F)

= dimF Ulrn ((A/F)")/Ulm ((A/F)"+')

Note that, by Lemma 4.9, ( F G ) [ p ]= F G . I ( G [ p ] ) and

J(FG) =I(G)

U l m ( ( F G / F ) m ) = F G P U. I(G"[p])/I(G")I(G""jPl) It will next be shown that F G is a n Ulrn algebra such that

This will imply that the Ulrn invariants of G are determined by the group algebra F G . A different proof of this fact was given by May (1969), Berman (1967) and Berman and Mollov (1969). The approach we have adopted was suggested by Proposition 5.1 in Waterhouse (1975), which is a particular case of the following standard result. 4.28.

Lemma. Let G be

an

abelian p-group and let A be a subgroup of G. Then the

m aP

a@1

+

F G . I(A)/I(G) . I(A)

H

(a - 1)

i s a n isomorphism of vector spaces over

F

+ I(G) . I(A)

279

MOD U LA R E XT EN S I ON S

Proof. Let T denote a transversal for A in G containing 1. We claim that F G I ( A ) has an F-basis given by

M = { t ( a - 1)It E T , 1 f

a g

A)

Indeed, a s an F-module, F G . I ( A ) is generated by all elements of the form g ( a g E G and a E A . Because g = tal for some t E T ,

a1

-

1) with

E A , we have

g ( a - I) = t(a1a - 1) - t(a1 - 1) so that M is a generating set of the F-module F G . Z ( A) . Suppose

cultl(U1 -

1)

+ . . . + a,t,(a,

where t , E T , 1 # a, E A and a , E F . Then a , ( a ,

-

-

1) = 0

1) = 0 since F G is a free FA-module

with T as a basis (Lemma 4.11(i)). This shows that a, = 0 and substantiates our claim. It now follows that the mapping /I : F

G . I ( A ) --+ A @ZF

givcn by p ( t ( a - 1)) = a c 31

is

a11

F-module epimorphism. The identity

( 9 - l ) ( a - 1) = t ( a a 1 - 1)- t ( a l

- 1)- ( a -

I)

shows that I ( G ) I ( A )C Kerh. Let

be the induced homomorphism. Then the mapping

given by

O(U B 1) = ( U - 1) + I ( G ) . I ( A ) is

ail

b’-rnodule liomornorphism inverse to

4. So the

lemma is true.

CHAPTER 5

280

4.29. Corollary. (Waterhouse (1975)). L e t G be a n abelian p-group. T h e n t h e m a p

is a n i s o m o r p h i s m of vector spaces over F.

Proof. Put A = G [ p ] .Then A @z F = G[pl @pp F since p annihilates A . Now apply (3) and Lemma 4.28.

4.30. Corollary. (Waterhouse (1975)). L e t G be a n abelian p-group. T h e n FG is a n U l m algebra s u c h t h a t

Proof. The map Ulm((FG/F),)

4

G P " [ p ] @w F given as the composition of the

maps below

is an injective F-homomorphism, by virtue of Corollary 4.29. Hence U l m ( ( F G / F ) , ) injects into Ulrn(FG), proving that F G is an U r n algebra. Finally, f , ( F G / F ) = dimF Ulm( ( F G / F ) a / U l m ( ( F G / F ) " + ' )

[PI

= dimF(Gpa[ p ]@ F ~ F/GPU+'

BE,, F )

[PI

= dimF[(GP"[p]/GPU+' @F,F ] = d i m p p GPn [p]/GPU+' [p] = f,(G),

as required. Turning to purely inseparable extensions, we now prove 4.31. Theorem. (Waterhouse (1975)). Let E/F be a modular purely inseparable e x t e n s i o n , let

Fp-"

be t h e perfect closure of F , a n d let A be t h e

(i) There is a natural m a p

E nF P

-1

4

Fp-"

-algebra E @ F F p - m .

Ulrn(A) sending z t o t h e coset of z 8 1

-

1 @ 5.

U n d e r t h i s m a p a set {zi} goes to a basis (respectively, a n i n d e p e n d e n t set, a spanning

MODULAR

EXTENSIONS

281

s e t ) ij and only if { x i } is a relative p-basis (respectively, a relatively p-independent set, a generating s e t ) for (E fi F P - ‘ ) / F . ( i i ) If K i s a n intermediate field modular over F , t h e n the natural m a p

U l m ( K @F F p - - )

---t

Ulm(A)

i s injective. I n particular (by taking K = ( E / F j a ) , A is a n Ulm algebra over FP--

(iii) For a n y ordinal a,

fa(E/F) = fa(E @ F F p - ” ) Proof. (i) and (ii): Assume first t h a t E / F is finite and E

F f ’ . If {z,} is a

relative pbasis of E / F , then the x; are tensor product generators and E

@,F

F F - - is

BE’”-=-[ X i ] / ( X pwhere ) X i = z , @ l - l g x i . By Lemmas 1.10 and 4.14,the latter algcbra is identifiable with a group algebra of a n elementary abelian p-group with generators qi =

--

Xi + I. IIence, by Corollary 4.29, the images of the z1are a basis of U l r n ( / ? @ F F J ’ ).

Passing to a direct limit over finite subextensions, we see t h a t the same result holds for infinite E I F with E C_ F p - ’ . Moreover, any relatively pindependent set, can be expanded

to a relative pbasis, and so must go t o an independent set; similarly any generating set contains a relative pbasis, and hence goes to a spanning set. In addition, if K is an intermediate field, then a relative pbasis of K / F extends to one of E / F , and thus

Ulrn(K

@,F

F p - - ) injects into U l m ( E @ F FP

-m

) . Still assuming E

C

FP

- 1

, let {x,} be

elements whose images are independent. If the x, are not relatively p-independent, then some of the z,,say T,

for

XI,

is in the field K generated over F by the others. The images of‘

i # 1 then span Ulm(K @ F F P - - ) , so the image of

2 1

is dependent on them there

and hence is dependent on them in IJlm(E @ F Fp--).This is impossible and thus the z, are relatively pindependent. If on the other hand {z,}

span U l r n ( E @ F FP

-m

is a set of elements whose images

), let K be the subextension they generate. If K # E , then we can

choose a relative p b a s i s of K / F and extend it to a relative pbasis of E / F . Th u s there exists an element in Ulm(E B F FP-”) not in Ulm(K @ F F p - - ) , which contradicts the fact that Ulm(K @ F F p - = ) contains the images of the z,.This establishes the case where

E 5 F”’. Now let E / F be an arbitrary finite modular purely inseparable extension with tensor product generators y; having degree q; = p e s . Then E @ F FP

--

.

is @Fp--

[ x ] / ( Y : ’ where )

282

CHAPTER 5

Y,= yz @ 1 - 1 @ y,. Then Corollary 4.29 shows t h a t t h e elements Y:L'p give a basis for Ulm(E @ F

are a relative p b a s i s for (E n F p - '

However, the element

FP--).

and hence go to a basis for U l m ( ( E n F P - ' )

@F

)/F,

Fp--). Because a basis thus goes to a

basis, the map Ulm((E n F P - l ) @ F Fp--)

+

U l m ( E € 9 F~ p - - )

is a n isomorphism. Writing any modular purely inseparable extension as a direct limit of

finite modular purely inseparable extensions, we see t h a t the map is still a n isomorphism there. We have thus obtained a reduction t o E n F P - '

, where we have proved the required

assertions.

(iii) By Lemma 4.23(ii), f a ( E / F ) is equal to the dimension of

since and the latter is Ja(E @ F Fp-m),

The theorem is therefore established. rn 4.32. Corollary. Let E / F be a modular purely inseparable eztension such that

for some abelian p-group G . Then

for any ordinal

(x

Proof. By Theorem 4.3l(iii), we have

Since, by Corollary 4.30, Assume that

CL

-1

fa(FP

G / F p - ' ) = fa(G), t h e result follows.

is a limit ordinal. Then, by Corollary 4.29,

n B
Ulm((FG)B) = Ulm((FG)")

MODULAR EXTENSIONS

283

We next show that this nontrivial property is satisfied by E @F F p - - .

We need the

following preliminary observation. 4.33. Lemma. Let E / F be a n algebraic field extension, let { K g } be a family u]

intermedzate fields directed by inverse inclusion and let K

=

n K g . Suppose M

intermediate field of EIF linearly disjoint from K with ( M K : K ) finite.

zs a n

T h e n Af

1s

linearly disjoint f r o m all s u f i c i e n t l y small Kp.

Proof. Let B be a linear basis of M over M n K . It is then a basis for M K over A' and therefore is finite. Now choose any

p. Then there exists

basis for M K p over KO. Choose 7 such t h a t K ,

C KO. If

a subset

Bg of B which

is a

Bp is no longer a basis of M K ,

over K,, it can be expanded t o a larger subset B, which is. Because B is finite, successiLc expansions in this way must eventually stop. In this way, we obtain a K , and a subset of 17 such that B , is a basis of MK6 over K6 whenever K6

K,. Clearly it suffices to

verify that B , = B . Assume by way of contradiction that there exists b E B write b = x a , b , uniquely with b, E

B, and

B,

-

B,. We can

a, E K , . The same is true for each smaller

K s , so the a , are in f7W61K.5 G

K,}

But t h a t intersection is K , because the original intersection was directed. Thus we have a contradiction. We are now ready to prove 4.34.

Theorem. (Waterhouse(l975)). Let E / F be a modular purely inseparuble

extension and let a be a limit ordinal. Then

U l m ( ( E / F ) a@ F F p - - ) =

Ulm((E/F)p

@F

Fp--)

4
Proof.

We of course use Theorem 4.3(ii) to identify these spaces with subspaces

of Ulm(E @,v F p - - ) .

The inclusion

C

being obvious, we choose an element z in t h v

right-hand side. It can be expressed using finitely many elements from E n FP

-1

; among

such expressions select one using the smallest number of elements not in (E/F)af i F p - ' . Denote by M the intermediate field generated by these elements. Then M is clearly linearly disjoint from ( E I F ) "

n FP

- 1

over F , since otherwise we could drop one of t h c

CHAPTER 5

284

-m

generators of M and still span the same space together with U l m ( ( E / F ) a@ F F P

).

Invoking Lemma 4.33, it follows that M is linearly disjoint from all sufficiently small

( E / F ) P fl Fp-'. Because M is finite, the intersection of the two will also be F eventually. We may now choose a relative p b a s i s X I of ( E I F ) " n FP-' over F , expand it to a relative p-basis

X1 U X

, of ( E / F ) pn FP-' over F , and add a relative p b a s i s Xs of M / F while still

staying pindependent. In U l m ( F @ F Fp--) the images Y1 U Yz U Y3 will then be linearly independent, by Thecrem 4.31(i). The set Yl U Yz spans U l m ( ( E / F ) @@ F F p - - ) ,

SO

z is

in its span. By definition of M , we have z in the span Y1 U Y3. Thus z must be in the span of Y1, i.e. z is in Ulrn((E/F)O @ I F FP

-m

), as required.

Assume that E / F is a purely inseparable field extension. If E / F is finite, then it is modular if and only if E

@F

E

E G for some abelian p g r o u p G (Theorem 2.3). It is

therefore natural t o inquire t o what extent this is true for infinite extensions. First we show that E

@F

E

E G for some abelian p g r o u p G implies that E / F is modular (the

converse need not be true by the final result of this section). 4.35. Proposition. Let E / F be a purely inseparable extension s u c h t h a t E @ F

E

E G f o r s o m e abelian p-group G . T h e n E / F i s modular.

Proof. Let K / F be any finite subextension of E I F . Then K

@F

E is isomorphic to

a subalgebra of E H for some finite subgroup H of G. It follows, from Theorem 2.l(iii),

2 S ( K / F ) . Thus E / F

is modular,

4.36. Theorem. (Waterhouse(l975)). Let G be a n abelian p-group.

T h e n there

that E / F splits K / F . Hence, by Corollary 2.2(ii), E by virtue of Lemma 4.3(ii).

exists a modular purely inseparable field extension E I F s u c h t h a t

Proof. By Proposition 4.35,it suffices t o exhibit a purely inseparable field extension

E / F such that E

@F

E

Z

E G . Let L be any field of characteristic p . We may find a free

abelian group H and a subgroup S of H such that G Z H I S . By Lemma 4.13, LH and

LS are integral domains. Let E and F be the quotient fields of LH and L S , respectively.

R y Lemma 4.11(ii), we have

285

MODULAR EXTENSIONS

Now, since G is torsion, LH is integral over L S . Furthermore, since

it follows that LH @ L s F is a n integral domain which is integral over its subfield 1 @ F . Hence L H @ L s Fis a field and therefore the natural map L H B L ~ F + E is a n isomorphism. Thus

as required. 4.37. Corollary. Let a be any ordinal. The n there ezists a modular purely insepa-

rable eztension E / F with ( E / F ) a = F and length precisely a .

Proof. It is well known (see Fuchs (1973, p.57 and p.210)) that there is an abelian p-group G such that GPU = 1 and GPp # 1 for

< a . By Theorem

niodular purely inseparable field extension E / F such that E @F E

4.36, there exists a

EG. Then

which proves the result. 4.38. Corollary. Let E I F be a countable-dimensional modular purely inseparable

extension. T hen a

i--t

f a ( E / F ) is a function from countable ordinals t o cardinals

5 N,,

with the following properties: (i) i t is identically zero from some point on, and ( i i ) nowhere has it a n infinite string of successive zero values followed b y a nonzero value

Conversely, any such function arises from some countable-dimensional modular purely inseparable ezt e nsion.

Proof. Property (ii) follows from Proposition 4.24(iii). Because Ulm ( E @ p F P - is countable-dimensional, clearly all f , ( E / F ) are

5 No

1

(see Theorem 4.3l(iii)). Note also

that there cannot be an uncountable well-ordered chain of subspaces, so (i) holds. T h e

CHAPTER 5

286

converse holds by applying Theorem 4.36, Corollary 4.32 and the existence theorem for countable groups (Fuchs (1973, p.65)). w We close by displaying some complications in the field extensions not occurring in abelian groups. First we show t h a t the analogue of the classical Prufer's theorem fails for field extensions. Recall t h a t the Priifer's theorem states t h a t any countable abelian p g r o u p without elements of infinite height is a (restricted) direct product of cyclic groups (see Fuchs (1970, p.88)). 4.39. Theorem. (Waterhouse (1975)). There ezists a modular purely inseparable

field eztension E / F such that (i) E / F is countable-dimetasional with ( E / F l W= F (ii) E is not a tensor product of simple eztensions of F

Proof. P u t F = F p ( b , ao, a l , a z , . . .) where 6 and the ai are indeterminates. Define yn(n = 0 , 1 , 2 , .. .) inductively by -1

YO = b P

, Yn

= Yn-1

+ Zn-12,

-1

where

Z,

=

and put

Then

E,/F

X O ,2 1 , .

. . ,z,,,y n

are in FP

-1

and are relatively pindependent in FP-' / F . Hence

is a modular purely inseparable extension such t h a t the listed generators for En

are tensor generators. Because

E,-1

C E,,

it follows from Lemma 4.3(iii) t h a t E = U E ,

is a modular countable-dimensional purely inseparable extension of F . Our knowledge of

tensor generators implies that

and thus FEP' n FP-' = ~ ( y , , x,, x,.+~,x , . + ~ ,.. . ) Since a n extension without elements of exponent 1 is trivial, t o prove ( E I F ) " = F , it suffices t o verify that

n ( E F P rn F P - I ) = F r

MODULAR

EXTENSIONS

287

We obviously have

Rearing in mind t h a t FP-I =

u F ( y o , z o , . . . ,x,), it suffices to verify that

Let K be the intersection above. Considering dimensions over F and observing the composite of two fields F ( y n + l , z n + l ) and F ( y o , z o , . . . ,zn) is F ( y ~ , z o ,... , z n + l ) ,we see that

K

=

F or ( K : F ) = p. In the latter case the two fields are linearly disjoint. Now z,+I

does not belong t o F ( y o , z o , . . . ,x,) and therefore 1 and zn+l are linearly independent over F(yo,zO,. . . :z,).

We have the relation

and, by linear independence, the coefficients are unique. But if the two fields are linearly disjoint, there must be such relation over K , and by uniqueness it must be the same rclation. Hence z n , y n E K which is impossible since z,,y,

are pindependent over F and

( K : F ) 5 p . Thus we must have K = F and hence ( E / F ) W= F . Let L = F ( y 0 , z o ) . We now claim t h a t if L' is an intermediate field of ( E n F P - ' ) / L and E/L' is modular, then L' = E n Fp-'; if sustained (ii) will follow and hence the result. Indeed, if E / F were a tensor product of simple extensions of F , then any finite subextension

K of (EnF p - ' ) / F would be contained in a finite subextension K' of ( E n F p - ' ) / F with E / K ' modular. In particular, by taking K = L and K' = L' we would conclude that ( E n F p - ' ) / F is finite, a contradiction.

To substantiate our claim, we prove inductively that F ( y 0 , ZO, . . . , x,)

C L'

implies z,+~ E L'

By Lemma 4.4(a), we have L' and FEP' linearly disjoint for all r , and thus L' is linearly disjoint from

FEP'"+' n FP-' = F ( y n + l , z n i l r x n f 2 , .. .) = M , say Because by induction yo,zo,. . . ,z,are in L', we have yn also in L'. The relation

CHAPTER 5

288

expresses yn E L in terms of x,, 1 E L over M ; since x n @ M the expression is unique. Hence, by linear disjointness the coefficients must be in M n L' and consequently x,+1 E L'. This completes the proof of the theorem. 4.40. Corollary. There exists a modular purely inseparable field eztension E I F such

that (i) E

n Fp--

can be written as a n increasing union of intermediate fields

L1

C L2 C . . .

where each L , has bounded height in E I F (ii) E is not a tensor product of simple extensions of F

Thus Theorem 4.21 fails without the assumption that each E I L , is modular. Proof. Take E / F as in Theorem 4.39. Then the countability of E n FP-' auto-

matically makes it a union of chain of finite dimensional intermediate fields L , over F . Furthermore, each L , has bounded height in E I F since L,/F is finite and

This shows t h a t (i) holds. That (ii) also holds and is a consequence of Theorem 4.39. m 4.41. Theorem. (Waterhouse(l975)). Let X be any infinite cardinal. Then there is

a modular purely inseparable field eztension E I F such that (i) ( E : F ) = X and ( E I F ) " = F (ii) f , ( E / F ) = 1 for all finite n.

Proof. Given a field K , let KO be the direct product of No copies of K . Let u s show first that if lKI = A, then KO as a K-space has dimension a t least A.

Assume

that we have a basis of KO over K of cardinality < A. Obviously for this to happen X must be uncountable. The entries in the basis vectors then form a set of cardinality less than A, and thus generate over the prime field a subfield L of cardinality < A. Now a K-linear combination of elements in LO must have all its entries in a finite-dimensional L-subspace of K . However IKI = X

> ILl certainly implies ( K : L ) infinite,

so we can

produce an element in K Onot having all its entries in such a finite-dimensional space. This contradiction establishes the assert ion. Let K be a perfect field with IKI = A. Choose X linearly independent elements in

KO starting with the elements en which are 1 in the n-th place and 0 elsewhere. Denote by V the K-space spanned by these X elements. If V , is the subspace of V consisting of

289

MODULAR EXTENSIONS

elements with first n entries equal t o 0 , then we have a chain

V = V , > V , ~ V z >. . . >v,>... where d i m V = A, dimVn/Vn+l = 1 and n V, = 0 Let F = K ( V ) be the quotient field of the free commutative algebra on the vector space V . It is a purely transcendental extension of K with transcendency basis = a linear basis

of V = a relative p b a s i s of F / F P . P u t

El = FIVP-l]= F[{zP-'lzE V } ] Because K is perfect, this is the same as adjoining to F the p t h roots of the elements in a basis of V . We now define E n inductively by

Then obviously E n / F is modular and therefore E = UE, is modular over F . Clearly

( E : F) = A . Taking into account that

.4ccordingly, e:-'

is a relative pbasis of ( E I F ) ' n FP-I over (EIF)'"

I r ( E / F )= 1

n FP-' and so

for all finite r

We arc thus left to verify that ( E / F ) W= F . It suffices t o show that U l m ( ( E / F ) W @ F F J ' - )p z is trivial. Hence, by Theorem 4.34, we need only verify that

But clearly Ulm(E @ F Fp-=) is isomorphic to V @ K F P - = , with the subspaces isomorphic t o V , @ K F P - = . The required assertion is therefore a consequence of

nz,

V, = 0 . rn

CHAPTER 5

290

4.42. Corollary. There ezists a modular purely inseparable extension E I F such that

for any abelian p-group G .

Proof. Let EIF be the field extension defined in Theorem 4.41, with X > 2'0. If there exists an abelian p g r o u p G such t h a t E

@F

Fp-"

E

FP-"G, then by Theorem 4.31

(iii) and Corollary 4.32 we could conclude t h a t

GPY= 1 and f n ( G ) = 1 for all finite n But this implies (see Fuchs 1970, p.146) t h a t IGI 5 2 N 0 ,a contradiction. rn

E. Modular closure and modularly perfect fields. All fields under consideration are assumed to be contained in a common algebraically closed field of characteristic p

> 0. If two fields are said t o be linear disjoint, we mean that they

are linearly disjoint over their intersection. Our principal goal is t o show that any purely inseparable field extension E of F is contained in a unique minimal field extension K of F , where K I F is modular ( K I F turns out t o be also purely inseparable). It will also be shown t h a t an arbitrary field extension E / F is modular if and only if there exists a n intermediate field K of E / F such that E I K is separable and K / F is purely inseparable modular. This fact will be used to bring together a number of characterizations of modularly perfect fields, that is, fields F with the property that every field extension of F is modular. 4.43. Lemma. Given two fields E and K such that E K I E i s algebraic, there ezists

a unique minimal eztension L of E which is linearly disjoint f ro m K . Furthermore,

L = E [ S ] for a subset S of K and S may be chosen t o be finite whenever ( E : ( E n K ) ) is finite

Proof. We begin by fixing a basis { e , } of E over E n K . Let

{.a}

be a maximal

subset of {e,} which is linearly independent over K in E K and let {e,} be the set of elements of {e,} each e,, we have

which do not lie in { e p } . Then { e a } is a basis of E K over K and, for

MODULAR EXTENSIONS

for unique p7p in K . Setting S = { p 7 p } and

291

L = E [ S ] it, follows t h a t S is finite whenever

E / ( E n K ) is finite. We are therefore left t o verify that L and K are linearly disjoint and that L is contained in any extension M of E such that M and K are linearly disjoint. We claim that { e a } is a basis of

L

n K.

L over L n K . Indeed, the eg are linearly independent over K and hence over

By ( l ) ,E lies in the span of t h e ep (as does S

C L n K ) , thus

L = E [ S ]lies in

the span of the ep. Because { e p } is a basis of L K over K , we deduce t h a t L and K are linearly disjoint. Assume that M / E is a field extension such that M and K are linearly disjoint. Since the ep E M are linearly independent over K , it follows that the ep are linearly independent i n M over

MnX .

For each e 7 , we have (1) for unique pLrp in K . By the linear disjointness

of Ad and K over M

n K , these relations must occur over M n K which implies that p7p E M

nK

M

1 h u s M 2 E [ S ]= L as required. The following theorem is a combination of the results due to Sweedler (1968) and

Kinie (1973). 4.44. Theorem. L e t

E / F be a purely inseparable field eztension. T h e n there exists

a unique m i n i m a l field eztension L I E s u c h t h a t L / F i s modular. Furthermore, ( i ) L / F is purely inseparable

(ii) IJ E / F i s of f i n i t e ezponent n, t h e n L / F is also of ezponent n . (iii) I f E / F is a f i n i t e eztension, t h e n so is L I F Proof. Assume that the result is true whenever E / F is of finite exponent. Since

(P-' n E),'F

is of exponent p', we may then find a unique minimal field extension

Lt/(Zf'p-' fl E ) such that L , / F is modular purely inseparable of exponent p'. The mini-

mality of each L, guarantees that we have a chain

L1

c L2 c . . . c L , c .

Since each L , / F is modular, it follows from Lemma 4.2(iii) that UL,= L , say, is modular over F . Furthermore, L contains E , since

u m

FP-' n E 5 L; and E =

(FP-' n E )

i= 1

CHAPTER 5

292

Any field extension of E that is modular over F must contain all the L , (by the minimality of each L ; ) . Thus L must be the unique minimal such field. Since, by construction, L / F is purely inseparable, we are left t o treat the case where E J F is of finite exponent n .

We construct L by descending induction. Assume t h a t we have already constructed

L , 2 E such t h a t (i) L< and F are linearly disjoint for s

2m

(ii) L , is the unique minimal extension of E having property (i) (iii) L,IF

is a purely inseparable extension

(iv) L,/F

has exponent n

(v) If E I F is finite, then so is L,JF Our induction begins with L , = E and if m = 1 we stop; otherwise, we construct Lm-l.

Owing to Lemma 4.43, there is a unique minimal field M 2 LLm-l where M and F are linearly disjoint. Setting L,-1

m- 1

= M p l - = ,we then have LK-, and F linearly disjoint. By

Lemma 4.43, M = Lp,"-' [ S ]for a subset S of F . Hence, for all s

2 m, we have

and

LC-, = L c . [L<-, n F ] By looking at the chains

and taking into account that, by hypothesis L< and F are linearly disjoint for s it follows from Proposition 2.3.13 t h a t L<-,

2

m,

and F are linearly disjoint. Hence L,-l

satisfies property (i). Bearing in mind that

we see t h a t L,-1

satisfies properties (iii) and (iv) (here, we assumed n

possible since the induction begins with L,).

2

m which is

MODULAR

EXTENSIONS

293

by (v) and Lc-'/(Lc-' i !F ) must be finite. 13y

If E / F is finite, then so is L,/F

Lemma 4.43, we may assume that S is finite. Since

we conclude that L,-1/F

is finite, proving (v).

We are thus left t o verify that Lmplsatisfies (ii). To this end, assume that N 2 E and has property (i). Since

NPm

and F are linearly disjoint, we have L,,

i N . Then

L$'-l C_ NP"'-' which is linearly disjoint from F . Therefore, by Lemma 4.43, hi1 and hence L,-1

C N.

Thus the induction leads to

L1

C A''"'-'

the desired field. This completes

the proof of the theorem. w We shall refer t o the field L in Theorem 4.44 as the modular closure of E J F . Thc following result in which "separable" is replaced by "separable algebraic" was proved by Sweedler (1968). 4 . 4 5 . Proposition. (Sweedler(l968)). Let

F iE

2 I<

be a c h a i n of fields.

( i ) If E / F i s separable, t h e n E / F i s modular.

(ii) If E I F is separable algebraic, then K / F modular implies K I E modular. (iii) If K / E i s separable, then K I F is modular if a n d o n l y if so is E I F .

Proof. (i) If X E EP"

n F , then

X = pp" for some p E E . Since E I F is separable,

so is F ( p ) / F by virtue of Lemma 3.4.3(i). But F ( p ) / F is also purely inseparable, hence ,LL

E F and thus X E Fp". We conclude therefore that

'L'hus it remains t o verify that EP" and F are linearly disjoint over Fp" for all n

>.

1

Clearly, F/FP" is purely inseparable and, by Lemma 3.6.7(i), EP"/FP" is separable. Hence, by Proposition 3.4.19 (i), EP" and F are linearly disjoint over F P n . (ii) Assume that E I F is separable algebraic and that K I P is modular. Since E I I ' is

separable algebraic, it follows from Lemma 3.12.1(ii) that E = FEP" for all n

2

1. 'Thus

FK'" 2 E so that FKP" and E are linearly disjoint. By Lemma 4.4(a), this implies that K is modular over E . ( i i i ) Assume that K / E is separable. Then, by (i), K I E is modular and for all

TZ

2

1. The latter implies that

K p "

E n KP" and FEP" are linearly disjoint for

Ilence, if E / F is modular, then by Proposition 2.3.13, K / F is also modular.

E = A'" a!] n

2

1.

CHAPTER 5

294

Conversely, suppose that K I F is modular. Since KP"

nE

KP" n F = EP" n F

=

EP", we have

for all n

2I

Thus, by Lemma 4.4(b), E I F is modular. We are now ready to prove the following result contained in a work of Kreimer and Heerema (1975). 4.46. Theorem. Let E f F be a n arbitrary field eztension. T h e n E / F is modular ij

and only i f there ezists a n intermediate field K of E / F such that E I K is separable and

K / F is modular purely inseparable. Proof. Assume that K is a n intermediate field of E / F such t h a t E / K is separable and K / F is modular purely inseparable. Then E I F is modular by Proposition 4.45(iii). Conversely, assume that E I F is modular and put K = E n Fp--. Since E / F is modular,

EP" and F are linearly disjoint and consequently E and FP

-n

are linearly disjoint. Since

E n FP-" = K n Fp-=, it follows t h a t K and FP-" are linearly disjoint for all n KP" and F are linearly disjoint for all n

2. 1 and

2 1. Hence

thus K I F is modular. Note also that

K / F is obviously purely inseparable. By the foregoing, we are left t o verify that E I K is separable. 3.4.21(iv), it suffices t o show t h a t E and Kp-"

To this end, assume t h a t XI,.

..,A,

=

Fp--

By Proposition

are linearly disjoint over K .

are elements of E linearly dependent over Fp--.

Then these elements are linearly dependent over FP modular, they are linearly dependent over FP-"

-n

for some n

nE CK.

2

1. Since E / F is

Thus E and KP-- are linearly

disjoint over K , as required. w Following Kreimer and Heerema (1975), we say t h a t F is modularly perfect if every field extension E I F is modular. We wish to provide a number of characterizations of modularly perfect fields. By Theorem 4.46, a modular extension cannot be ezceptional where in the terminology introduced by Reid(1966) a n exceptional extension E / F is an inseparable extension such that

E

n FP-'

=F

This suggests the characterization of modularly perfect fields as those fields which have no exceptional extensions. A preliminary step to achieve this is the following result due t o

Kreimer and Heerema (1975).

MODULAR EXTENSIONS

295

4.47. Proposition. Let F be a field such that (F : FP)

exceptional field extension E I F such that E

@F

> p . T h e n there

ezists a n

E has n o nonzero nilpotent elements.

Proof. Because (Fp-' : F ) = ( F : FP) > p , we may find a , b in FP-' such that (F(a,b ) : F ) = p 2 Let z and y be elements which are algebraically independent over F and let

E=F(z,az+6) We claim that the subfields E and F ( y , a y

+ 6) of E ( a ,6, x, y) are linearly disjoint over E'.

To substantiate the claim, note that F ( x , y ) / F is separable and so F ( x , y ) and F ( a , 6 ) are linearly disjoint over F and

It follows that

In particular, F ( z , y, ax+b) and F ( y , ay+b) are linearly disjoint over F ( y ) . Also F ( z ,az+6) and F ( z , y) are linearly disjoint over F ( z ) ,while F ( x ) and F ( y ) are linearly disjoint over F because z and y are algebraically independent over F. The conclusion is that F ( z , a z and F ( y ) are linearly disjoint over F and hence F ( s , a z

T

6)

+ b ) and F ( y , a y + 6) are linearly

disjoint over F , as claimed.

To prove t h a t E @ F Ehas no nonzero nilpotent elements, consider the F-homomorphisrri

E

@F

6 + F ( a , b , z , y ) which sends x

1 @ (ax

+ 6) t o a y + b.

@

1 t o z, 1 @ z t o y, ( a z

+ b ) @ 1 to a z + b , a n d

Since the subfields E and F ( y , a y + b) of F ( a , b , x , y ) are linearly

disjoint over F , this homomorphism is injective and thus 0 is the only nilpotent element in E

@F

E.

Note that E I F is not separable, because { l , z , a x

F , but not over F ( a , b)

C FP

-1

+ 6)

is linearly independent over

. Finally, t o show that E / F is an exceptional extension, it

suffices to verify that F is algebraically closed in E . To this end, let X E E be algebraic

296

CHAPTER 5

over F . Then

E F ( z ) and

XP

purely transcendental, we have

XP

XP

is algebraic over F . Bearing in mind t h a t F ( z ) / F is E F . If X

4 F , then X @ 1- 1@ X is a nonzero nilpotent

element of E B F E , which is impossible. Thus X E F and the result follows. 4.48. Theorem (Kreimer and Heerema(l975)). For any given field F , the following

conditions are equivalent:

(i) ( F : FP)

5p

(ii) F is modularly perfect (iii) For any field extension E I F , the eztension E / ( E n Fp--) is separable (iv) There ezist no ezceptional field eztensions of F (v) A field extension E I F is separable if and only if E

@F

E has n o nonzero nilpotent

elements.

Proof. (i)+(ii): therefore FP

-1

=

Assume t h a t ( F : FP) 5 p . Then F = Fp(X) for some X E F arid

F(XP-'). It. follows by induction on n

2

1 t h a t FP-" = F(Xp-.") for

any positive integer n. Now let n be a positive integer and let E / F be an arbitrary field extension. Because E ( F p - " ) = E (X p -" ), we have

where m is the smallest nonnegative integer such t h a t

XPm-"

E E . Thus E and Fp-"

(hence Ep" and F ) are linearly disjoint, proving that E / F is modular. (ii)*(iii):

Let E / F be an arbitrary field extension. By Theorem 4.46, there exists an

intermediate field K of E / F such that E / K is separable and K / F is modular purely inseparable. Since K / F is purely inseparable, we have K separable, so is ( E n F P - " ) / K .

i E n FP

-m

. Since E / K is

But ( E n F P - " ) / K is also purely inseparable, hence

K = E n Fp-", as required. (iii)+(iv): Assume that E / F is a n exceptional field extension. Then E n F p - ' = F implies

E

n Fp-"

=

F and hence E / ( E n Fp--) is not separable, a contradiction.

(iv)+(v): We know, from Theorem 3.4.11, that if E / F and K / F are field extensions s u c h that E / F is separable, then K

@ g ~

E has no nonzero nilpotent elements. But if E / F is a

field extension which is neither separable nor exceptional, then there exists an element X of

E which is algebraic and purely inseparable of positive exponent over F and X @ 1 - 1 8 X is a nonzero nilpotent element of E

@F

E. This shows that (iv) implies (v).

MODULAR

297

EXTENSIONS

(v)=>(i):This is a direct consequence of Proposition 4.47. We know that any algebraic extension of a perfect field is perfect. It turns out that. a similar property holds for modularly perfect fields. Namely, we have 4.49. Theorem. (Kreimer and Heerema(l975)). L e t F be a modularly perfect field

T h e n a n y algebraic extension E of F is again modularly perfect.

Proof. P u t K = FP--nE. Since E is separable algebraic over K (Theorem 4.48(iii)), we have E = E p ( K ) . Therefore

( E : EP) = ( K ( E P ): K P ( E P )5) ( K : K P ) arid, by Theorem 4.48, it remains only t o prove that K is modularly perfect. Since any field extension L of K is also an extension of F , it follows that L is separable over

I@

I? L.

Fy--

#?I,

~

Hence, by Theorem 4.48, K is modularly perfect, as required.

4.50. Corollary. Let F be a perfect field and let E be a n algebraic eztension of a

simple transcendental e z t e n s i o n of F . T h e n E is modularly perfect.

Proof. By Theorem 4.49, it suffices t o consider the case where E

=

F ( X ) . But then

Ep = F ( X P ) and hence ( E : EP) = p. Thus, by Theorem 4.48, E is modularly perfect, as required. a We close by providing a characterization of fields F with ( F : FP)

{,he condition ( F : FP) in a p-basis of F is

5 p . Obviously,

5 p is equivalent t o the requirement that the number of elements

5 I . The required characterization will be obtained as a consequence

of a general result due to Becker and MacLane (1940). By the degree of zmperfectzon of F ,

we understand the cardinal number of elements in a pbasis of F . 4.51. Theorem, Let F be a field having finite degree of imperfection rn.

(i) E v e r y f i n i t e e x t e n s i o n E I F can be obtained by adjoining not more t h a n m elements to F

(ii) There ezists a f i n i t e extension E I F which cannot be obtained b y adjoining fewer t h a n ni elements t o F

Proof. Since F has degree of imperfection rn, we have ( F : FP) (FP-' : F ) = p -

=

pm and hence

CHAPTER 5

298

Since FP-'/F is purely inseparable of exponent 5 1 , m is the minimal number of generators of FP

- 1

over F . This proves (ii), by taking E = F p - ' .

Now let E / F be any finite field extension. Since ( E : F ) = (EP : FP), we have

( E : EP) = ( E : F ) ( F : F P ) / ( E P: F P )= ( F : F P ) = pm Thus E and F have the same imperfection degree. Let B = (61,. . . , b,}

E . Then E = EP"(b1,. . . , b,) for all n

be a pbasis of

2 1. Choose n so large t h a t for each

X E E,

AP-

is

separable over F . Then F ( E p " ) / F is a finite separable extension, hence F ( E p " ) = F ( X o ) for some

A0

E F(EP"). Since Xo is separable, we have F(X0, b , ) = F ( X ' ) for some A' E E

(see Proposition 3.2.19). It follows from E = EP" ( b l , . .),.I!,

E = F(Xo,bl,. . . >6,,,)

=

F(X', bz,.

that

. . ,b,)

This is a generation by m elements, as required. w 4.52. Corollary. For any field F , the following conditions are equivalent:

(i) ( F : FP)

5p

(ii) Every finite field eztension of F is simple.

Proof.

We may harmlessly assume that F is not perfect. If (i) holds, then the

imperfection degree rn of F is 1. Hence, by Theorem 4.51(i), (ii) holds. Conversely, if (ii) holds, then by Theorem 4.51(ii), m = 1. Thus ( F : FP) = p , proving (i). rn

299

Galois theory

In this chapter we present the Galois theory which may be described as the analysis of field extensions by means of automorphism groups. After recording topological prerequisites, we concentrate on profinite groups. The properties of profinite groups are applied to infinite Galois theory, for it turns out that the Galois group of a n infinite Galois extension E / F is a profinite group. Special attention is drawn to the problem of realizing finite groups as Galois groups. Let

F be a real algebraic number field and suppose that the group G

occurs as a Galois group of a normal real extension field of F . Using elementary methods, WF

show that certain types of split extensions of a n elementary abelian 2-group by G also

occur as Galois groups of normal real extensions of F . Among other groups, we show

that Sylow 2-subgroups of Sz- and Az- occur as Galois groups of real extensions of Q . A separate section is devoted t o a result of Isaacs concernipg the degrees of sums in a separable field extension. The chapter ends with a brief discussion of Galois cohomology.

1. T o p o l o g i c a l p r e r e q u i s i t e s .

It is assumed that the reader has some familiarity with the basis concepts of general topology. However, for his convenience we recall briefly those that are relevant for our subsequent investigations.

A set S becomes endowed with a topology as soon as there is given a collection Q of subsets of S having the following properties: ( i ) Both S and its empty subset 0 belong to R (ii) R is closed under arbitrary unions and finite intersections

T h e members of R are then referred t o as the open sets of the topology and the pair (S:[ I ) , or simply S,is called a topological space.

1,et S be a topological space. A subset S’ of S is called a closed subset if its complement

with respect to S is open. The union of a finite number of closed sets is necessarily a closed

CHAFTER 6

300

set and so too is the intersection of an arbitrary family of closed sets. Of course, it can happen that a set is both open and closed. A bijection f : S1 4 S2 of topological spaces

S, and Sz such t h a t f ( A ) is open in S2

if and only if A is open in S1, is called a homeomorphism. We say t h a t two topological spaces S1 and S2 are homeomorphic if there exists a homeomorphism of S1 onto Sz. The discrete topology on a set S is the topology for which every subset of S is open; the trivial topology on S is the topology whose only open sets are 0 and S . Suppose that S1 is a subset of a topological space S . The intersection of all the closed sets containing S1 is itself a closed set which will be denoted by closure of S1 in

S. If

71.

This is called the

s E S , then it is easy t o show t h a t s belongs t o

3 1

if and only if

every open set containing s contains a t least one element of S1. If it happens t h a t 31 = S, then one says t h a t S1 is dense in S or that S1 is a dense subset of S . Again, assume that S1 is a subset of a topological space S. It is then possible to define a topology on S1 whose open sets are precisely the intersections of S1 with the open sets of S . This is known as the induced topology. When S1 is endowed with the induced topology, we say t h a t

S1

is a subspace of S.

The elements of a topological space S are often referred t o as points. By a neighbourhood of a point s € s

S,we understand any set which contains a n open set t o which

belongs. More generally, a neighbourhood of a subset A of S is any subset of S which

contains an open set containing A . A point s E S is said t o be an interior point of a subset A of S if A is a neighbourhood of

5.

The set of all interior points is called the interior of

A . Thus A is open if and only if A coincides with its interior. Let S and S' be topological spaces and f : S

+

S' a mapping.

Then f is said

to be continuous at the point s E S if for each neighbourhood A of f ( s ) there exists a neighbourhood B of s such that f ( B ) 2 A . When f is continuous a t every point of S,we say simply t h a t f is continuous. It is easy t o verify t h a t f is continuous if and only if for every open set A in S', f-'(A)

is open in S .

Given two topologies R l and R z on the same set S , we say t h a t R1 is finer than (and that Rz is coarser than

nl), if every

C12

subset of S which is open in R z is open in R1

(equivalently, every subset of S which is closed in Rz is closed in R l ) . Assume that a topological space S has the following property: whenever

s

and s' are

distinct points there exists a neighbourhood B of s and a neighbourhood B' of s' such that B

n B'

= 0. We then say that

S is a Hausdorf space.

TOPOLOGICAL PREREQUISITES

Let

301

S be a topological space, R a n equivalence relation on S, S I R the quotient set S + S / R the canonical mapping. The finest

o f S with respect to the relation R,X :

topology on S I R for which X is continuous is called the q u o t i e n t of the topology of S by the relation R ; under this topology, S I R is called the q u o t i e n t space of S by R . It follows from the definition that a subset A of S I R is open (respectively closed) if and only if

> - ‘ ( A ) is open (respectively closed) in S. Note that if V is a topological space, then a mapping f : S / R

4

V is continuous if and only if f

o X :

S

---t

V is continuous. Thus

there is a one-to-one correspondence between the continuous mappings S / R continuous maps S

+

4

V and the

V which are constant on each equivalence class mod R.

Let S be a topological space. A collection C of subsets of S is called a base for t h e topology of S if (i) every set in C is a n open set of

S,and (ii) every open set of S is

a

union of a (possibly empty) family of sets belonging to C. Lct S and S’be two topo!ogical spaces and suppose t h a t C and C’ are bases for their respective topologies. If now A E C and A’ E C’, then A x A‘ is a subset of S x S’.It is cnsy to see that there is a unique topology on S

x S’which has the subsets A x A’ as a

base. Moreover, this topology is unaltered if we replace C and C’ by other bases for tlie open sets of S and S‘,respectively. The topological space S x S’ so obtained is called the p r o d u c t of

S and S’.The topology on the product space is referred to as the product

topology. Let S be a topological space. An o p e n c o v e r i n g of S is a family {Sili E I} of o p en subsets of S such t,hat S = U I E ISi. We say that S is c o m p a c t if for each open csvering of S thcre is a finite subcovering.

A topological space S is said to be c o n n e c t e d if it is not the union of two disjoint open sets.

A subset A of a topological space S is said to be c o n n e c t e d if the subspace A of S is

connected. Given a point s of a topological space S , the union of the connected subspaces of S which contain s is connected; it is therefore the largest connected subset of S containing s. ‘The c o m p o n e n t (or c o n n e c t e d c o m p o n e n t ) of a point of a topological space S is defincd

to be the largest connected subset of

S which contains this point. The space S

is said t o

hc totally d i s c o n n e c t e d if the component of each point of S consists of the point alone. A subset A of S is said t o be t o t a l l y d i s c o n n e c t e d if the subspace A of S is totally disconncctcd. 1,ct S be a set. A set R of subsets of S is called a f i l t e r on S if it satisfies the axiorns: ( i ) fl

;+ ~

0 and

c) fj!

R

CHAPTER 6

302

(ii) Y E R and Y

C Y1

S implies Y1 E R

(iii) Y l ,Yz E R implies Yl n Yz E R

A set

r of subsets of S

is called a filter base if (a) r

then Y3 2 Y1 n Y2 for some Y3 E

r.

such t h a t Y E R if and only if Y1

# 0 and 0 4

Every filter base

r

r

and (b) if Y1, Yz E

r

generates a unique filter It on S

Y for a t least one Yl E

r; r

is called a base of the

filter R . The set of all filters on a nonempty set S is inductively ordered by the relation

R1

C Rz; R l & R z is expressed by

saying t h a t R l is coarser than

R2,

or that

R2

is finer

than R1. Every filter on S which is maximal with respect t o this ordering, is called an

ultrafilter on S; by Zorn's lemms, for each filter R on S there exists an ultrafilter finer than R. Let S be a topological space. For any given s E S , the set of all neighbourhoods of s is a filter on S called the neighbourhood filter on s. Let G be a group. We say that G is a topological group if G is also a topological space such that (i) The mapping

(5, y

(ii) The mapping z

)

++

H

5-l

x y of G x G into G is continuous of G into G is continuous

A group structure and a topology on a set G are said t o be compatible if they satsify (i)

and (ii) above. Let R be a ring. We say t h a t R is a topological ring if R is also a topological space such that

(i) T h e mapping (z,y) (ii) The mapping z

H

(iii) The mapping (z,y)

H

z

+ y of R x R into R is continuous

-z of R into R is continuous H

z y of R x R into R is continuous.

Let G be a topological group and let R be the neighbourhood filter of the identity element 1 of G. If g E G, then the mappings x

++

g x and I

++

zg

are homeomorphisms. It follows

that the neighbourhood filter of g is the family gR of sets g X , X E R and also the family Rg of sets X g , X E R. We close this section by quoting the following standard properties

(see Bourbaki, 1966). 1.1. Proposition. Let G be a topological group and let R be the neighbourhood filter

of 1 . Then

(i) Given any X E R, there ezists Y E R such that Y . Y 2 X (ii) Given any X E R, we have X-' E R (iii) For all g E G and all X E R , we have gXg-'

ER

303

TOPOLOGICAL P R E R E Q U I S I T E S

Conversely, if G is a group and R a filter o n G satisfying (i)-(iii), then there is a unique topology o n G compatible with the group structure of G, for which R is the neighbourhood filter of 1. For this topology, the neighbourhood filter of g E G , coincides with each of the filters g R and Rg.

1.2. P r o p o s i t i o n . Let G be a topological group. Then the following conditions are equivalent:

(i) G is Hausdorff (ii) The set (1) is closed (iii) The intersection of the neighbourhoods of 1 consists only of the point 1. 1.3. P r o p o s i t i o n . A subgroup of a topological group is open if and only if it has a n

interior point. Every open subgroup is closed.

Let S,, i E I , be a n arbitrary collection of topological spaces. Then, by definition, the product topology on

n,tJS; has a basis consisting of the sets of the form niEIU; with

U, an open set of S, for every i E I, and U;= S; for all but finitely many i E I. 1.4. P r o p o s i t i o n . A subspace of a Hausdorff space is Hausdorff. Every product of IIausdorff spaces is Hausdorff. Conversely, i f a product of nonempty spaces is Hausdorff, then each factor is a Hausdorff space. 1.5. P r o p o s i t i o n . Let H be a subgroup of a topological group G and let

be the

closure of H i n G then ( i ) f 7 2s a subgroup of G

(ii)

If IT is a normal subgroup of G, then so is p. A collection Y of subsets of a topological space X is said to have the finite intersection

property if cvery finite subcollection Yl of Y satisfies

For the proof of the following results we refer to Pontryagin (1977) 1.6. P r o p o s i t i o n . A topological space

S i s compact if and only i f every collection oJ

closed subsets of S possessing the finite intersection property has nonempty intersection. 1.7. P r o p o s i t i o n . ( i ) A closed subset of a compact topological spacr: is compact

CHAPTER 6

304

(ii) A compact subset of a Hausdorff topological space is closed.

(iii) A continuous image of a compact topological space is compact. (iv) A n y continuous bijection X

---t

Y of a compact topological space X onto a Hausdorff

space Y is a homeomorphism. 1.8. P r o p o s i t i o n . Let G be a compact totally disconnected topological group and l e t

U be an arbitrary neighbourhood of the identity. Then there is a n open normal subgroup

N of G such that N

U.

1.9. P r o p o s i t i o n . (Tychonoff’s Theorem). The product of a n arbitrary collection

o/

compact topological spaces is compact.

2. P r o f i n i t e groups.

As we shall see later on, the Galois group of a n infinite (i.e. not necessarily finite) Galois

extension E / F is a profinite group. For this reason, we devote this section t o examining some general properties of profinite groups. We begin by introducing the notion of a projective (or inverse) limit which stems from topology and which has become very useful in algebra too. Let {Gi)i,l

be a system of

sets

(groups, rings, etc) indexed by a partially ordered

set I , which is directed in the sense t h a t , given i,j E I, there exists a k E I such that

i 5 k and j 5 k. Suppose that for every pair i , j E I with i 5 j , there is given (homomorphism)

f;j : G j + G; such that (i)

fit

is the identity map of Gi for each z E I

(ii) for all i 5 j 5 k in I, we have

f;j o f j k

= f;k.

Then the system

{ G t ~ ~ i j I I> ~~jE is called a projective system. The projective limit

G = limGi t

a map

305

P R O F I N I T E GROUPS

of this system is defined by

It i s clear t h a t if each G; is a group (ring), then G is a group (ring). 2.1. Example. Let G be a n arbitrary group ( a n arbitrary ring) and let {NiJi E I}

be a set of normal subgroups (ideals) of G of f i n i t e i n d e x which is closed under finite intersections.

W r i t e i 5 j if N, 2 N j .

T h e n I becomes a directed set. If z 5 j , iet

fi1 : G/ATJ+ GIN; be the natural projection, i.e.

T h e n {GIN,, f,]li,j E I } i s a projective s y s t e m of groups (rings) and t h e m a p

is a h o m o m o r p h i s m with kernel If the sets G; are topologicai spaces, we assume that t h e maps ftl are continuous. Not,e that IimG, may be empty and that t

where

Assume that each G, i s a topological space. Recall that the product topology on

ITtti G ,

niElU,, with U , an open subset of G , for every Gi for all but finitely many i I . Because limG, is a subset of ntt, G',,

h a s a basis consisting of the sets of form z E

I and U, =

E

t

w e may, and in future will, regard IimG; as a topological space. Note that if each G, is a c

topological group (topological ring), then lim G, is a topological group (topological ring). t

CHAPTER 6

306

2.2. Proposition. Let {Gi, f i j l i , j E I } be a projective system of topological spaces

and let G = lim Gi. t

(i) If each Gi is a Hausdorff space, then G is a closed subset of

n Gi.

(ii) If each Gi is a nonempty compace Hausdorff space, then G is a nonempty compact Hausdorff space.

Proof. (i) Let g = ( g i ) be an element of exist i and j with i

Ul of g; and

5 j such that

fij(gj),

fij(gj)

# gi.

Gi that does not belong to G. Then there Take open disjoint neighbourhoods U, and

respectively. Then

is an open neighbourhood of g that does not intersect G. (ii) Assume that each G; is a nonempty compact Hausdorff space. Because each G , is compact, we have

n G; also compact, by Tychonoff's theorem (Proposition 1.9). Thus,

by (i) and Proposition 1.7, G is compact. By Proposition 1.4, G is also Hausdorff. We are thus left to verify that G is nonempty.

To this end, note that the natural map prk x prl : G + Gk x Gj is continuous. Taking into account that

T

= {(gk,gj) E

Gk x GjIfjk(gk)= g j }

is a closed subset of Gk x Gj (use the Hausdorff property), it follows that Rkj =

pr,)-'(T)is a closed subset of

(pTk

x

niEIG;. Since n,,,Gi is compact, we have only to show

that the intersection of finitely many of the Rkj is nonempty. Let j(1) 5 k ( l ) ,. . . , j ( n ) 5 k ( n ) be n pairs in I. Choose t E I with k ( i ) 5 t , 1

5 i 5 n,

and g t E Gt. Define S j ( i ) = f j ( i ) , t ( g t ) and S k ( i ) = f k ( i ) , t ( S t )

For each r E I - {j(l), . . . , j ( n ) , k ( l ) , . . . ,k(n)} pick gr in G,. The element g thus defined belongs to

n:=, Rk(i),j(i)as required. H

It will be convenient at this stage to record the following elementary properties of topological groups. 2.3. Proposition. Let G be a topological group.

( i ) Every closed subgroup of G of finite index is open

PROFINITE GROUPS

307

(ii) If G i s compact, then a subgroup of G is open if and only if it is closed and of finite indez.

Proof. (i) Assume that H is a closed subgroup of G such that (disjoint union)

G = H U g 2 H U . . . U g ,,H Because H is closed, so is g i H , 1 5 i complement of

5 n, hence U b 2 g ; I I is closed. Thus, H being the

u,"=,g,H in G, is open.

(ii) Let H be a n open subgroup of G. Then H is closed, by Proposition 1.3. Because t h e

cosets of H provide a n open covering of G and G is compact, H can have only finitely many cosets in G. Thus H is closed and of finite index. T h e converse being true by virtue of (i), the result follows. We now return to our discussion of projective limits of groups (rings). Let

be a projective system of groups (rings) and let

G = lim G, c

The following simple facts are worthwhile mentioning: (a) There exist homomorphisms

f, : G

G;

(i E I )

4

s u c h that

ft I n fact j z : g

C+

= fi3

0

for all i

fJ

5j

g E (i.e. the restriction of the i-th coordinate projection of n G t to G)

satisfies this condition. These

fi

are called canonical homomorphisms

(b) If every J,J is injective, then so are all the

Assume that each

fij

I;.

is injective, and f;(s)= f;(y) for some x , y E G. Given j E I , let

k E I satisfy i , j 5 k . Now

whence by our hypothesis, fk(x) = f k ( y ) . Therefore f3(x) = for civery j E I , and so z = y.

fJkfk(z)

= fykfk(?I)

:.:

fj(Y)

CHAPTER 6

308

(c) If J is a cofinal subset of I (i.e. if for every i fE I there exists j E J such that i

5j),

then lim G j

lim G;

-a€I

Indeed, given g’ E

lim Gi

-3EJ

G j , there exists a unique g E

-2EJ

j€J

lim G,, such that, for

Cl€l

every j E J , the j - t h coordinates of g’ and g are equal. In fact, if we define g = ( g i ) with g; = f;i(gi)

(i 5 j ) , then

9’

H

g is a required isomorphism.

T h e following result provides a universal characterization of projective limits.

2.4. Proposition. The projective limit G = lim G; of the projective system {G;, fiili, c

j E I } of groups (rings) satisfies: if H is a group ( r i n g ) and if there are homomorphisms u; : H

4

G, such that (7;

5j

(1)

for all i E 1

(2)

for all i

= flj o u j

then there ezists a unique homomorphism

such that ui = fi

0u

where f; is the canonical homomorphism. This property characterizes G up t o isomorphism.

Proof. Given h E H , set u(h)= ( u , ( h ) )E n G , . It follows from (1) that u(h)E G. Thus u : H

+

G is a homomorphism satisfying u,(h)= f , ( u ( h ) ) ,proving (2). If u‘ : H

G also satisfies (2), then f , [ u ( h ) = ] f , [ a ’ ( h ) ]for all h E

H,iE I . Thus

(T

--t

= u’.

In order t o prove the second assertion, suppose that Ho and maps r, : Ho

4

G, have

the property formulated in the first statement of the proposition. Then there exist unique homomorphisms u : Ho

---f

G and

00 :

G

Ho satisfying

4

f; o u = r, and r; o uo We infer that f; = f; o u o

(TO

and

7; =

r,

o

=

ff

u o o u for all i E I . Thus u o uo and

(TO o

u are

the identity maps of G and H o , respectively, proving that u is a n isomorphism. We now introduce the notion of a homomorphism for projective systems of topological groups (rings). Let

309

P R O F I N I T E GROUPS

be two projective systems of topological groups (rings) indexed by the same directed set 1. A homomorphism

4 :X

4

Y is a set of continuous homomorphisms {I$; : G;

+ Hi,

iE I}

subject to Ui3 0

47

=

4 a- 0 f i j

for all i

5j

2.5. Proposition. Let

be two projective systems of topological groups (rinys) a n d let

4 :X

--t

Y be a homomor-

phism. Then there is a unique homomorphism (which is necessarily continuous)

s u c h that, f o r every

i E I, pi

0

4* = 4; 0 fi

(3)

where f,: p i are canonical homomorphisms. I n fact 4*((gi)) = (4i(gi)) Moreover, if every

4i is injective,

Proof. The homomorphisms

then so is

(4)

4*.

4;, i E I , induce a homomorphism

J: H G i + I I H l

It follows from (3) that if y = (yl) E limG,, then & ( g ) E limH,; hence define c

the restriction of

4'

to be

c

to limG,, i.e. by formula ( 4 ) . With this

4 * , (3) is satisfied and 4- is

t

obviously continuous. If qb : IimG,

+

c

also satisfies ( 3 ) for every i , then p ,

o

limfl; c

4-= pI o .d, for every 2 ,

If 4-(z) = q+"(y), then q4(ft(z)]= &[ft(y)i ,fi(s) = f l ( y ) for all z E 1 and thus z = y.

thus

4-= +.

Therefore, if every

4, is injective, then

CHAPTER 6

310

2.6. P r o p o s i t i o n . Let X = {G,, f i j l i , j E I , i

5 j}, Y

= {H;,u;jli,j E

I, i 5 j }

and Z = { K ; , A ; j l i , j E I, i 5 j } be three projective systems of compact topological groups (rings) and let

I$

=

(4;) be a homomorphism f r o m X t o Y and 4

=

(4;)a homomorphism

f r o m Y t o 2 . Assume that for all i E I , the sequence

is ezact. Then the induced sequence

4*l i m H ; t4’I i m. K ~ 1+ l i m G , + c

t

4

1

c

is also ezact. P r o o f . By Proposition 2.4,

4* is injective. A straightforward

verification shows that

Ker+* = Im4*. We are therefore left t o verify t h a t $* is surjective. To this end, fix ( k i ) in lim K ; and consider the sets c

A, = { ( h ; ) l ( h , )E

n

H;, + J ( h j )= k j , u ; j ( h 3 )= h, for all i 5 j }

iEI

Then all t h e A j are closed and nonempty. Moreover, the collection { A j l j E J } possesses the finite intersection property. Because follows from Proposition 1.6, that

n,

Hi,,H; is compact, by Tychonoff’s theorem, it

Aj

# 8. Because each element of

n,

Aj belongs

to lim H ; and is mapped by $J* t o (k,),the result follows. w t

For the rest of this section, we shall be interested in the case { G i , J i j l i , j E I} is a projective system of finite groups, each endowed with the discrete topology. We call t h e projective limit G = IimG, a profinite group . The following result gives a criterion for a +

compact group t o be profinite.

2.7. P r o p o s i t i o n .

Let G be a compact topological group and let { N J i E I } be a

family of closed normal subgroups of finite indez such that (i) For every finite subset J of I , there ezists i E I such that N ,

(4 ntErNz = 1 Then G

lim G / N z topologically and algebraically. c

n,

N3

311

PROFINITE GROUPS

Proof. Condition (i) ensures that { G / N , , fil li,j E I } (where z 5 j means N, 2 lVl and j z l : G / N ,

+

G / N , is the natural projection) is a projective system of topological

groups. Condition (ii) implies that the map

is a n injective homomorphism (which is obviously continuous). If (g,N,) E limG/N,, then t

the closed sets g,N, have the finite intersection property. Because G is compact, there exists g E n t E I g , N , . This element satisfies B ( g ) = (g,N,), so 0 is a continuous bijection. Since G is compact and lim G / N , is Hausdorff (Proposition 1.4), it follows from Proposition t

1.7 that 0 is a homeomorphism, as required. rn

We next provide some useful characterizations of profinite groups. 2.8. Proposition.

Let G be a topological group. Then the following conditions are

equivalent:

(i) G i s a profinite group (ii) G is a Hausdorff, compact group which has a basis of open neighbourhoods of 1 con-

sisting of normal subgroups (iii) G is a Hausdorff, compact, totally disconnected group.

Proof. Suppose t h a t G is a profinite group. By Proposition 2.2(ii), G is compact and Hausdorff. The normal subgroups U s n G, where

( S a finite subset of I )

( N , is a normal subgroup of G,) form a basis of open neighbourhoods of 1. This proves that ( i ) implies (ii). The implication (ii)*(i)

is a consequence of Propositions 2.6 and 2 3 .

Assume that (ii) holds and let N be the connected component of 1. If {U,} denotes the family of all open normal subgroups of G, let N , = U ,

nN

for all a. The groups

N,

are open normal subgroups of N ; as N is connected, we must have N, = N for all a . But then N = n,N, = n,(N n U,) = N

(n,U,)

=

Nn1

t h e last equality being true by hypothesis (ii). Thus N = 1 and so (iii) holds. T h e

implication (iii)+(ii) being a consequence of Proposition 1.8, the result follows.

CHAP.rER 6

312

2.9. Proposit,ion. Let G be a profinite group. Then any closed subgroup (closed

normal subgroup) of G is the intersection of some open subgroups (open normal subgroups).

Proof. Let H be a closed subgroup (closed normal subgroup) of G and suppose that g E G belongs to every open subgroup (open normal subgroup) t h a t contains H . Then,

for any open normal subgroup N of G , g E H N and so g N n H

# 0.

Hence, for any finite

family N1,. . ., N , o f open normal subgroups of G, we have

Because the sets gN n H are closed and G is compact, it follows from Proposition 1.6 that there exists h E H such that h E gN for all open normal subgroups N . But then g-'h belongs t o all open normal subgroups of G. Because G is Hausdorff and some open normal subgroups of G form a basis of open neighbourhoods of 1 (Proposition 2.8), it follows from Proposition 1.2 t h a t g - ' h = 1. Thus g E H and the result follows. rn 2.10. Proposition. (i)

If H is a dosed

subgroup of a profinite group, then H is

profinit e. (ii)

If N is a closed normal subgroup of a profinite group G , then G I N

(iii) The direct product G =

is profinite

niEIGi of an arbitrary family {Gili E I } of profinite

groups

is profinite (iv) Every projective limit G = lim G; of profinite groups is profinite. c

Proof. (i) If N , is a n open normal subgroup of G, then H n Ni is a n open normal subgroup of H . The family of all groups H n Ni satisfies the hypothesis of Proposition 2.7 and thus H

lim H / ( H n N i ) is a profinite group. t

(ii) A closed normal subgroup N is an intersection of open normal subgroups of G, by Proposition 2.9. Now apply Proposition 2.7. (iii) Let J be a finite subset of 1. Given j E J , let N j denote a n open normal subgroup of

G,. Then Nj x

NJ = j€J

fl G; itl-J

is an open normal subgroup of G of finite index. However, the intersection of all such NJ is 1. Therefore, G Z l i m G / N J is a profinite group, by Proposition 2.7. t

PROFINITE GROUPS

313

(iv) We know, from Proposition 2.2(i), that G is a closed subgroup of

niE1 G , is profinite. Now apply (i).

niEIG ; . By (iii),

An important result for the cohomology theory of profinite groups is: 2.11. Proposition. Let G be a profinite group and let H be a closed subgroup

G. Then there exists a continuous cross-section u : G / H

exists a continuous map u : G / H G/H4G

+

4

4

OJ

G with u ( H ) = 1, i.e. there

G of topological spaces such that the composite m a p

G / H as the identity map.

Proof. Let K & S be subgroups of G such t h a t S / K is finite. We wish t o show that there is a continuous cross-section u : G / H

+

G / K with u ( H ) = K . Indeed, let UG b e

the family of all normal subgroups of G of finite index. Then

is a basis of open neighbourhoods of 1 E G I K . Bearing in mind that S / K is finite, there

exists U E UG such that ( U K I K )n ( S / K )= 1. Then the natural map

4 : U K / K ---t U H I H is a homeomorphism. Let 91 = 1,. . . , g k be a transversal for U in G and put

The map u is obviously a continuous cross-section from G / H t o G / K with a ( H ) = K . Let X denote the set of all pairs { S , u } , where S is a subgroup of H and o is a continuous cross-section from G / H section G / H to G / S . We may order X by setting { S , u } 5 {S’,u’} if S’

C

S and u is induced by u’. The set

{ S i , a i } , z E I be a chain in X and let S =

induce a continuous map u’ : G / H

-+

n,

X

-

is inductive. Indeed. let

S ; . The cross-sections ui

:

G/H

G/S,

limG/S,, i.e. a continuous cross-section u : G / H

+

c

G/S.Then { S , o } is an upper bound for the set {Sirul},i E I . Owing t o Zorn’s lemma, there exists a maximal element { S , u } . In this case S = 1. Indeed, if S

#

1, then S has

a proper subgroup of finite index. Then, by the foregoing, { S , u } is not maximal. This

implies the required assertion. We close this section by providing a number of examples.

CHAPTER 6

314

(a) If p is a prime number, then the rings Z / p " Z , n E IN, form a projective system with respect t o the canonical projections:

The corresponding projective limit

is the ring of p-adic integers. Every element of Z p is a sequence

x = (Xi + p i z ) , i E IN, xi E

z

where zj

= zi(modp'Z)

for j

A basic neighbourhood of x is given by a n integer m y = (yi

20

2i and it consists of all elements

+ p i Z ) , i E N with = ym(modpmZ)

x, The map

{

n 4zp x

H

(x+p'iz)

is an embedding and we identify Z with its image in Z p . T h e canonical projection

has kernel p i Z p . Thus p i Z p is an open subgroup of (the additive group) iZp of index pi. (b) T h e rings Z l n Z , , n E N , form a projective system with respect t o the projections

Z/nZ

4

Z / m Z for mln, where the order in IN is now given by the divisibility. The

projective limit

2 = limZZ/niZ L

is called the PrCYer ring. The groups n g , n E IN, are precisely the open subgroups of the

profinite group

2, and it is easily verified

that

315

GALOIS EXTENSIONS

Given a canonical decomposition n = n

p p u p

of n E N, we have the decomposition

by virtue of the Chinese remainder theorem. Passing t o projective limits, we obtain a canonical decomposition

2 = p p P

(c) Let F, be the field of p elements, p prime, and let

For each n

F,

be the separable closure of

IF,.

IN, we have a canonical isomorphism

which maps the Frobenius automorphism & of IFPnt o 1 mod nZ (this property and other properties of Galois groups will be discussed in our subsequent investigations). Passing to projective limits we obtain a canonical isomorphism

which sends the Frobenius automorphism 4 of ]F, to 1 E

< > to the dense subgroup Z of

2 and therefore maps the group

2.

3. Galois extensions.

Let F be a field and let G be a group of automorphisms of F . We define F"

C F by

FG= { z E F l u ( z ) = z for all u E G} It is clear that F G is a subfield of F ; we shall refer t o F G as the fized field of G. Let E / F be a field extension. The group of all F-automorphisms of E is called t h e Galois group of the extension and is denoted by G a l ( E / F ) . We refer t o E / F as a Galois

extension if E is algebraic over F and F is the fixed field of the group G a l ( E / F ) . O u r point of departure is the following characterization of Galois extensions. 3.1.

Proposition.

conditions are equivalent:

Let E J F be a n algebraic field eztension.

T h e n the followiny

CHAPTER 6

316

(i) E / F is Galois

(ii) EIF is normal and separable

(iii) E is the splitting field of a family of separable polynomials over F. Proof. (i)+(ii):

Let a E E and let f be the minimal polynomial of a over F. We

must show that all roots o f f are distinct and lie in E. Let

E be an algebraic closure of E

and let a1 = a , a ~. .,. , ar be all distinct roots of f in D

E Gal(E/F) is extendible to an automorphism of

E . Owing to Corollary 2.1.13, each E. In this way each E Gal(E/F) (T

permutes the a; and hence fixes g = (X - .I)(X - az). . . (X - a,)

Because EIF is Galois, we conclude that g E F[X]. Hence f 1g and, by definition of g , g1f which shows that f = g . Because g is irreducible, the group Gal(E/F) permutes the a; transitively, hence each a; E E , as required. (ii)+(iii): Because E/F is separable, E is generated over F by a family {a;}of separable elements. Let f; be the minimal polynomial of a; over F. Then each f; is a separable polynomial over F. Because EIF is normal, f; splits over E , hence E is the splitting field of the family (iii)+(i):

{I;}of separable polynomials over F.

Suppose that E is the splitting field of a family {f;li E I } of separable polyno-

mials over F. Then E / F is normal and separable. Fix a E E

- F.

Because a is separable,

it follows from Lemma 2.2.7 that there is an F-homomorphism o : F(a) 4 E for which

#

.(a)

IY.Since,

by Lemma 2.1.19, D extends to an automorphism of E , the result follows.

Owing to Corollary 2.2.31, any algebraic extension may be obtained by taking a separable extension followed by a purely inseparable extension. Usually one cannot reverse the order of the tower. However, there is an important case when it can be done.

3.2. Proposition. Let E/F be a normal field eztension and let G = G a l ( E / F ) . Then

(i) E G / Fis purely inseparable and E/EG is separable (ii) If Fa is the separable closure of F in E, then E = FaEG and Fs,EG are linearly

disjoint over F. In particular,

E

E

F,@pEG

and

F 8 n E G= F

GALOIS EXTENSIONS

(iii) If X E EG - F, then charF = p > 0 and

XP"

E F for some n

317

21

Proof. If c h a r F = 0, then E = Fa and hence, by Proposition 3.1, EG = F, which clearly yields the result. Assume that c h a r F = p

> 0 . Then, by Proposition 2.2.27, (iii)

holds. Therefore, by Propositions 2.2.32 and 3.4.19, it suffices t o prove (i). Since G a l ( E / F ) = G a l ( E / E G ) , it follows t h a t if X E E is such that .(A)

= X for

all u E G a l ( E / E G ) , then X E EG. Hence E/EG is Galois and so, by Proposition 3.1,

E / E G is separable. Let

be an algebraic closure of E and let u : EG

homomorphism. By Corollary 2.1.13, u extends to a n automorphsim of

-+

be an F-

e. Its restriction

to E is an F-automorphism of E, since E/F is normal. Thus u ( z ) = z for all z E EG and the result follows by virtue of Proposition 2.2.27. We next provide a useful characterization of finite Galois extensions. 3.3. Proposition. Let E/F be a finite field eztension. Then

IGal(E/F)/

5 ( E : F)

and ( G a l ( E / F ) (= (E : F)

if and only i f E / F is Galois. Proof. The first inequality follows from IGal(E/F)I 5 ( E : F)#5 ( E : F) Assume that E / F is Galois. Then, by Proposition 3.1, E / F is both normal and separable. Because E / F is separable, we have ( E : F)8= ( E : F ) (Proposition 2.2.13(i)),while since

E / F is normal we have (E : F)8= IGal(E/F)I. Thus IGal(E/F)I = (E : F). Conversely, assume that IGal(E/F)I = (E : F ) . Since !Gal(E/F)I

5 (E

:

F)s,

we have (E : F) = (E : F ) a , hence E/F is separable (Proposition 2.2.13(i)). Since I G a l ( E / F ) / = (E : F)8,E/F must also be normal. Hence, by Proposition 3.1, E / F is Galois. rn

As a preliminary t o the next result, we next record the following simple observation. 3.4. Lemma. Let E/F be a separable eztension. Suppose that there is a n integer n

2

1 such that every element a of E is of degree

( E : F ) 5 n.

5 n over F. Then E/F is finite and

ChAPTER 6

3 18

Proof. Let a E E be such that the degree ( F ( a ) : F ) is maximal, say m

5 n.

We

claim that E = F ( a ) , which will prove the result. We argue by contradiction and choose an element

p

E F such that

p 4 F ( a ) . Since F ( a , P ) / F is separable, it is simple by

Corollary 2.2.21, hence F(a,,f?) = F ( 7 ) for some 7 E F ( a , / 3 ) . But from the chain

we see that ( F ( a , P ) : F ) > m. Hence 7 has degree

> m over F, a contradiction. D

We are now ready to prove the following classical result. 3.5.

Proposition.

automorphisms of F .

(Artin).

Let F be a field and let G be a finite group of

Then F / F G is a finite Galois eztension, G = G a l ( F / F G ) and

IGI = ( F : F G ) .

Proof. Invoking Proposition 3.3, it suffices to show that F / F G is a finite Galois extension with ( F : F G )

5

IG1. Let a E F and let

elements of G such that ol(a), . . . , u r ( a )are distinct. If

(TI,. (T

differs from ( a l ( a ).,. . , a,(a))by a permutation, because

. . ,a, be

a maximal set of

E G, then ( a a l ( a )., . . ,aa,(a)) (T

is injective and each aai(a) E

{al( a ) ,. . . ,a,(a)}. Thus a is a root of the polynomial.

and, for any u E G, f" = f. Hence the coefficients of

f lie in F G . Furthermore, f is

separable. It follows that every element a of F is a root of a separable polynomial over FG of degree

5 ]GI. Morover, this polynomial splits into linear factors in F . Consequently,

F / F G is both separable and normal, and so F / F G is Galois, by Proposition 3.1. Owing to Lemma 3.4, we have ( F : F G ) 5 IGI and the result follows.

3.6. Corollary. Let E I F be a finite Galois eztension and let G be its Galois group.

Then, for any subgroup H of G , there ezists a n intermediate field K such that E I K is Galois and H = G a l ( E 1 K ) . Proof. Put K = E H and apply Proposition 3.5. We next examine some properties of automorphisms of field extensions. Let S be a set and F a field. The maps

GALOIS EXTENSIONS

319

are called linearly independent over F if whenever we have a relation Xlfl

+ . . . + A,

(A, E F )

fn = 0

then all A, = 0. Recall t h a t a monoid is a set G with an associative binary operation and having an identity element.

3.7.

Proposition.

(Artin). Let

XI,.

m o n o i d G to the multiplicative group F'

..,xn

be distinct h o m o m o r p h i s m s from a

of a field F . T h e n

XI,.

. . ,xn

are linearly in-

dependent ouer F.

Proof. If that x1 =

. x n are

X I , ~ 2 , ..,

I:=X,xi , for some X,

linearly

dependent, then we may harmlessly assume

6 F and t h a t

xz,. . . ,x n are linearly independent.

Then

Replacing g by g z in (l), we have

We now multiply (1) by xl(z) and substract the result from (2):

C~=2X;(xi(z) - xl(z))xi = 0 and thus X;(xi(z) - xl(z)) = 0 , 2 5 i 5 n , because X Z , . . . , xn are linearly independent. But x1 # x1 for i # 1, hence A; = 0 for i # 1 Therefore

and thus (1) reduces to the form 3.8.

xl(g) =

0 , a contradiction.

Corollary. Let f l , f 2 , ...,fn be distinct h o m o m o r p h i s m s of a field E into

another field F . T h e n

fl,f2,.

. . ,f n

are linearly independent ouer F .

x, the restriction of f l to G. Then each xt is d xE are obviously distinct. Because cy==l X,Jl 0

Proof. P u t G = E' and denote by homomorphism of G into F' and the clearly implies

I:=A,xl , = 0 , the result follows by applying Proposition 3.7. m

Given a finite Galois extension E/F, any F-basis of E of the form

:

320

CHAPTER 6

for some a E E, is called a normal basis

.

3.9. Theorem. (Normal basis theorem). A n y finite Galois extension E/F has a

normal basis. Proof. Put G = Gal(E/F) and n = IGI. We first examine the case where F is finite. Then G is cyclic, say generated by a (this fact will be proved in the next section). We may regard o as an F-linear transformation of the F-space E. Let f ( X ) be its minimal polynomial. Then degf(X)

5 n

=

(E

:

F). But, by Corollary 3.8, 1,a,..., on-’ are

linearly independent over E, hence also over F , and thus degf(X) = n. Because on = 1, we have f ( X ) = Xn - 1. Thus the minimal and characteristic polynomials for o are the same. By linear algebra, there is a “cyclic vector”. That is there exists an element

CY

E E

whose images under powers of u span E over F, as required. Suppose that F is infinite and let

01,.

.. ,onbe all distinct elements of G. Since E / F

is separable, we have E = F(X) for some X E E. Let f ( X ) be the minimal polynomial of

X and let

Xi

= a;(A), 1

5 i 5 n. Set

and

Then gi(X) is a polynomial over E with

the left side is of degree at most n namely A’,.

. . ,A,

-

Xk

as a root for k

# z.

Hence

1. Moreover, (4) is true for n distinct values of X,

since gi(Xi) = 1 and gk(Ai) = 0 for k # i. Hence the left side of (4) is

identically zero. Multiplying (4) by g i ( X ) and applying ( 3 ) , we derive

Now consider the determinant

GALOIS EXTENSIONS

321

If we square it by multiplying column by column and compute its value (mod j ( X ) ) ,we obtain from ( 3 ) , (4) and ( 5 ) a determinant that has 1 in the diagonal and 0 elsewhere. Thus

D(X)’

= l(mod f ( X ) )

Because F is infinite, we may always choose a value p for X such t h a t D ( p ) # 0, p E 1; Now put

fy

= g(p). Then t h e determinant

Finally, consider any linear relation a;u;(fy)

+ a z a z ( a ) + . . . + a,u,(a)

(a ; E F )

=0

Applying the automorphism u, t o it would lead t o n homogeneous equations for the n unknowns a, E F . Invoking ( 6 ) ,we deduce that each a; = 0 and the result follows. w We close by providing two results pertaining t o the degrees of elements in Galois extensions. 3.10. Proposition. Let F be a field and let

a,P be two elements

of finite degree

over E’ such that F ( a ) / F i s Galois. Then

Proof. P u t E = F ( a ) and K = F ( P ) . Then E K = F ( a , P ) and, by the special case of a result to be proved later (see Proposition 6.5) we have

whence

as required. w Remark. The assumption that F ( a ) / F is Galois is indispensable. Indeed, let F = Q , let CY

=

fi be the real cube root

of 2 and let c be a primitive cube root of 1, say

CHAPTER 6

322

Put

p

=

ca and note that Q (p) # Q ( a )since ,D is complex and a is real. Because X 3 - 2

is irreducible over Q , we have (Q ( a ) : Q ) = (Q ( p ) : Q ) = 3. Now Q ( a )n Q

( p ) is a ( p ) whose degree over Q divides 3. Hence this degree is 3 or 1, and must be 1 since Q ( a )# Q (p). In other words, Q ( a )n Q (p) = Q . On t h e other hand,

subfield of Q

= 2 and therefore Hence ( Q (a,p) : Q (a))

as asserted.

H

3.11. P r o p o s i t i o n . Let F be a field a n d let

a,p be

two separable e l e m e n t s over

F

of coprime degrees. T h e n

Proof. Let E be obtained from F by adjoining all conjugates of a and p. Then E / F is a finite Galois extension. P u t rn = ( F ( a ): F ) , n = ( F ( P ) : F ) and G = G a l ( E / F )

Then G transitively permutes the elements of each of the sets A =

{ P I , . . . ,P,},

where a; and

p, are the conjugates of a and

and K = G p be the stabilizers in G of a and

p,

(01,.

. . ,an},B

,f?,respectively. Let

=

H = G,

respectively. Then ( G : H ) = m, (G :

K ) = n and since (rn,n) = 1, a standard argument yields (G : ( H n K ) ) = mn. But H n K = G a l ( E / F ( a , P ) ) , hence by Proposition 3.3, ( E : F ( a , p ) ) = IH n KI. It follows that

as required.

4. F i n i t e fields, roots of u n i t y and c y c l o t o m i c e x t e n s i o n s .

A field with a finite number of elements is called a f i n i t e field

, or

also a Galois field.

FINITE FIELDS, ROOTS OF UNITY AND CYCLOTOMIC EXTENSIONS

323

Assume that F is a finite field. Then c h a r F = p > 0 is a prime and the image of Z in F , denoted by F,, is isomorphic to t h e field Z / p Z consisting of p elements. It is clear that F, is the prime subfield of F , i.e. the intersection of all subfields of F. 4.1. Proposition. (i) If F i s a f i n i t e field of characteristic p a n d n = ( F : IF,),

IF1 = p n and F' i s cyclic of order p" (ii) For each p r i m e p and each n

2 1,

-

1.

there i s exactly o n e field of q = p" elements (up t o

i s o m o r p h i s m ) n a m e l y t h e splitting field of X q Xq

-

then

-

X over IF,, a n d i t s elements are roots of

X . Moreover, every f i n i t e field is isomorphic to exactly o n e of these fields.

Proof. (i) Because F is finite, n is also finite and thus (FI = p n . The second assertion follows by virtue of Proposition 2.2.20. (ii) Let F be a field of q = pn elements. Because F' has q

-

1 elements, we have

and tlierefore xq

for all x E F

=x

Because this equation has at most q roots, its roots are precisely the elements of F . We may therefore write

which shows that F is the splitting field of Xq

-

X

over IF,.

Conversely, let q = p" where p is a prime. Lot F be the splitting field of

lF,. To see that this splitting field consists precisely of of Xq

-

Xq

- X over

q elements, we note t h a t the roots

X are distinct since

(hcnce there are exactly q of them) and that they form a field: if ( a - b)q = uq

-

bq = a -

a4 =

a , bq = b , then

b , ( a b ) Q= aqbq = ab, and if b # 0, (6')s = (bq)-' = b-'.

4.2. Proposition. Let F be a field of q elements. T h e n , f o r a n y n

a field extension E of F of d i m e n s i o n n , n a m e l y the splitting field of

2 1,

there exists

324

CHAPTER 6

Moreover, a n y t w o field eztensions of F of dimension n are F-isomorphic.

Proof. Suppose t h a t E is a field extension of F of dimension n. Then E has q" elements and, by Proposition 4.1(ii), the elements of E are precisely the roots of Xq" in an algebraic closure of IF,. Therefore E is the splitting field of Xq"

-

-

X

X over F . This

shows t h a t any two field extensions of F of dimension n are F-isomorphic. Conversely, the splitting field of Xq"

-

X over F is the same as the splitting field of X*= - X over F,,

since all elements of F are roots of Xq"

-

X . Hence it contains precisely qn elements and

therefore has degree n over F . rn The field of q elements is denoted by IF,. For q = p this agrees with the notation

F,

introduced earlier. 4.3. Proposition. If q = p n , where p is a p r i m e and n

2 1, t h e n F,/F,

i s a Galois

extension with cyclic Galois group of order n, generated by the Frobenius automorphisrn x

H

zp

of

F,.

Proof. Since F,/F, is a finite extension of the perfect field IF,, it must be separable (Proposition 2.2.26). On the other hand, F,/F, is normal by Proposition 4.1(ii). Thus

F,/F, is Galois, by Proposition 3.1. It follows from Proposition 3.3 t h a t Gal(F,/F,) of order n. Because

x p =

z for all x E

IF,, the Frobenius automorphism

(Y

of

is

F, is an

F,-automorphism of IF,. Finally, we have for all z E IF,

d ( z ) = XP'

Hence a ' = 1 for r = n and a '

#

1 for 1 5 r

< n. This shows t h a t

(Y

is of order n and

thus Gal(IF,/Fp) = < a > ,as required. 4.4. Proposition. If p i s a prime, t h e n all subfields of IFpm are of the form FPm where

nlm and, for a n y nlm, there i s ezactly one subfield of Fpmof order p n . Furthermore, if nlm, t h e n IFpm/IFpn i s a G a b i s extension with cyclic G a b s group of order m / n , generated by a", where

(Y

i s the Frobenius automorphism of IFpvn.

Proof. Suppose that nlm. Then any root of XP" Thus, by Proposition 4.1(ii), all roots of XP" p n . Conversely, if

-

-

X is also a root of XP'"

-

X.

X form a unique subfield of F p m of order

F is a subfield of IFp= and IF1 = p n , then

325

F I N I T E F I E L D S , ROOTS OF U N I T Y AND CYCLOTOMIC EXTENSIONS

proving that nlm. By the above (IFpm : IFpn) = m / n and, by Proposition 4.3, IFpm/lFpn is Galois, hence

its Galois group is of order m / n . Since an is of order m / n and an fixes IFpnelementwise, the result follows. 4.5. Lemma. L e t F be a f i n i t e f i e l d of q e l e m e n t s a n d let n

c { ;ffn

1

=

a€F

Proof. Let

t

if

n $()(mod q

if

n

-

2 1 be an integer. Then 1)

= O(mod q - 1)

be a generator of F' and let X = CortF an. Then 0-2

which is the sum of a geometric progression. Therefore, if en

#

1, i.e. if n $ O(mod q

--

l),

then

A=

&q--l) tn

since

6-l

-

1

-1

=o

= 1. If cn = 1, i.e. if n z O(mod q - l ) ,then X = q

~

1, as required.

Let F be a field and let f E F [ X 1 , .. . , X , ] . By a zero of f in F n = F x . . . x E' ( n times) one means a n element ( a l , .. . ,a,) E F" such that

# 0.

By a n o n t r i v i a l zero we mean a zero ( a 1 , . . , a n ) such that some a, and an integer j

2 0 , one says that F

Given a field F

is a C,-field if every homogeneous polynomial over

F of degree d in n > d3 variables, has a nontrivial zero in F". 4.6. Proposition. (Chevalley). E v e r y f i n i t e field i s a C1-field.

Proof. Let F be a finite field of q elements, let f ( X 1 , . . . , X,) be a homogeneous

polynomial of degree d < n over F and let p = charF. Let V

f and write

CL

for a point

C F"

. . . ,a,) E F n . Then

(al,

f(a)q-l

=0

for all a E V

be the set of zeros of

CHAPTER 6

326

and, by Proposition 4.1(i), for all a E F"

f (a)B--1= 1

-

V

It follows that the polynomial g ( X 1 , .. . , X , ) = f (XI,. . . ,Xn)q-' is characteristic function of F" to IF"

-V

with values in F,. The number modulo p of points of F" - V is therefore equal

CaEFn g(a). We claim t h a t -

this sum is zero in F,; if sustained, it will follow that

V /is divisible by p . Since p I IFnl, we deduce t h a t IVI is divisible by p . This will

prove the result since IVI

2 1.

To substantiate our claim, observe that g is a linear combination of monomials M t ( X ) = X:'

. . .X>. Because

it follows from Lemma 4.5 that and multiples of q

-

CaEFfi ail . . .

vanishes unless all the t i ' s are nonzero

1. In this case, the degree t l

( q - 1 ) n . But, since g =

f-',g has degree ( q -

+ .. . + t ,

of the monomial is a t least

1)d and ( q - l ) d < ( q - 1)" by assumption.

The proof is therefore complete. We now turn our attention t o roots of unity. Let F be a field. An element A E F is said t o be a n n-th root of 1 (or of unity ) if A" = 1. If c h a r F = p > 0 and n = pms with ( s , p ) = 1, then A" = 1 implies (A")."

= 1 or

A" = 1. Hence, if X is a n n-th root of unity,

then X is a n s-th root of unity. 4.7. Proposition. Let F be a field and let n

21

be such that charF + n .

(i) The n - t h roots of unity in F f o r m a cyclic group whose order divides n.

(ii) If X" - 1 splits into linear factors over F, th e n the n - t h roots of unity in F form a cyclic group of order n.

Proof. It is clear that (i) is a consequence of (ii). To prove (ii), observe that the n-th roots of unity in F certainly form a group, say G. Since

(x"- 1)'

=n

x"-'# o

it follows from Proposition 2.2.1, t h a t all roots of X"

- 1 are

simple. Therefore F contains

exactly n n-th roots of unity, i.e. IGI = n. Let m be the exponent of G. Then m 5 n

F I N I T E F I E L D S , ROOTS OF

UNITY

AND

CYCLOTOMIC E X T E N S I O N S

321

and all elements of G are roots of Xm - 1, hence /GI 5 m. This shows t h a t m = n and therefore G is cyclic. Let F be a field. We say that X E: F is a primitive n-th root of unity if the order of

X in F' is precisely n. Given n

2 1, by

a primitive n-th root of unity over F , we mean a

primitive n-th root of unity in the splitting field of Xn- 1 over F . Owing t o Proposition 4.7, if c h a r F .+ n , then a primitive n-th root of unity over F always exists. Moreover, if

E,

is a primitive n-th root of unity over F , then F ( t , ) is obviously the splitting field

of X"

-

1 over F. The field F ( E , ) is called the n - t h cyclotomic extension of F . By the

mazimal cyclotomic eztension of F , we understood the field obtained from F by adjoining n-th roots of unity for all n 4.8. Corollary. Let

to q and let

t

2 1.

F be a finite field of q elements, let m 2 1 be an integer coprime

be a primitive m - t h root of unity over F . Then

(i) ( F ( E :) F ) is equal t o the order o f q modulo m (ii) G a l ( F ( c ) / F ) is cyclic generated b y

E H cq

Proof. It is clear t h a t (i) is a consequence of (ii). Let phism of F . If q = p " , p prime, then an(€) =

tpn

=

tq.

CY

be the Frobenius autornor-

Therefore (ii) is a consequence of

Proposition 4.4. 4.9. Proposition. Let F be a field, let n

2

1 be such that charF

4 n and

let E and

K be respectively the n - t h cyclotomic and mazimal cyclotomic eztensions of F. Then both E I F and K I F are Galois eztensions. Proof. Because c h a r F t n , X "

E is the splitting field of X " such polynomials X "

-

-

-

1 has no multiple roots. It remains t o notice that

1 over F and K is the splitting field of the family of all

1 with c h a r F ). n (see Proposition 3 . 1 ) . w

The following observation is often useful. 4.10. Lemma. Let E / F be a Galois eztension and let p 1 , . . . ,p,, be elements of E

such that each u E G a l ( E / F ) permutes the p l , ,. . , p n . Then

is a monic polynomial over F . If, furthermore, F is the quotient field of an integrally closed integral domain R and each p, is integral over R , then ( X monic polynomial over R.

-

PI).

. . (x- pm) is a

CHAPTER 6

328

Proof. P u t f ( X ) = ( X - pl). . . (X - p n ) . Then

f ( X ) = X" - q x n - 1

+ s2xn-Z + . . . + ( - 1 ) k s k X n - k + . . . + (-l)"sn

where

Hence o(q) = s; for all

(T

E G a l ( E / F ) and all z E {1,2, ... , n } . Because E/F is Galois,

we conclude that each s; E F , hence f ( X ) E F[X]. Under the additional hypotheses, each s; is integral over R , hence belongs to R since R is integrally closed.

In what follows, we write ( Z / n Z ) *for the unit group of the ring Z / n Z . The order of this group is denoted by d ( n ) ,where

and that

4(n) coincides

4 is known as the Eiler function. Note that

with the number of generators of a cyclic group of order n. In

particular, the number of primitive n-th roots of unity is equal t o Let F be a field, let n

2I

be such that c h a r F # n and let

4(n).

€1,.

..,err

7

=

4 ( n ) ,be all

primitive n-th roots of unity over F . Then, by Proposition 4.9 and Lemma 4.10,

is a monic polynomial over the prime subfield of F and hence over F . Furthermore, if

c h a r F = 0, then by Lemma 4.10,

We shall refer to Qn(X)as the n-th cyclotomic polynomial over F .

F I N I T E FIELDS, ROOTS OF UNITY AND CYCLOTOMIC EXTENSIONS

4.11. Proposition. Let F be a field, let n

21

be such that charF

329

n and l e t F,

be the n - t h cyclotomic eztension of F . Then Gal(F,/F) is isomorphic t o a subgroup of

( Z / n Z ) - .Moreover, the following conditions are equivalent: (i) Gal(F,/F) (ii)

(Z/nlZ)'

@,,(X) is irreducible over F

(iii) (F, : F ) = 4(n)

Proof. Let then u ( t ) =

EP

E

be a primitive n-th root of unity so t h a t F, = F(t). If u E Gal(F,/E'),

for some 1

5

p

< n with ( p , n ) = 1. Moreover, u is uniquely determined

by p modulo n and may be denoted by u p . Then the map

is obviously an injective homomorphism. The desired assertion follows from the above and

Propositions 4.9 and 3.3. Let E / F be a field extension. We say that E I F is abelian (respectively, cyclic ) if

E / F is Galois and G a l ( E / F ) is abelian (respectively, cyclic). 4.12. Proposition. Let F be a field and let E be the mazimal cyclotomic eztension

of F . Then E I F is abelian.

Proof. By Proposition 4.9, E I F is Galois. Let u1,uz E G a l ( E / F ) . Since E generated over F by roots of unity, it suffices to show that

of unity in E . But if n is the order of

E,

(U~(TZ)(E)

= ( u ~ u ~ ) where ( E ) t is a root

a,(€) is also of ordern, hence u,(t)=

( k t , n ) = 1. Therefore u I u Z ( t )= u 2 u 1 ( c )= ? " P z

+ with

as desired. w

Our next aim is t o determine the isomorphism class of Gal(Q n / Q ) by showing that

this group is isomorphic t o ( Z / n Z ) * .First, we concentrate on the isomorphism class of

( T L / n i z ) - . Let n = p y ' ~ ;. .~. p i " be the canonical decomposition of n. Then

U / n Z r Z/p;'n x ... x

n

and thus

( Z / n Z ) * ( Z / p y ' Z ) *x ... x ( Z / p ; " Z ) " We may therefore concentrate on the case where n is a prime power

CHAPTER 6

330

(i) If p is a n odd p r i m e , t h e n ( Z / p " Z ) * is cyclic of order

4.13. Proposition. Pn-'(P - 1)

(ii) B o t h (Z/2Z)* a n d ( Z / 4 Z ) * are cyclic, a n d i f n

2

3, t h e n (Z/2"Z)* i s i s o m o r p h i c

to t h e direct product of two cyclic groups of orders 2 a n d 2"-2. Proof. (i) Because 4 ( p " )

=

p n - ' ( p - 1) (for any prime p ) , ( Z / p " Z ) * is of order

p " - ' ( p - 1). Therefore ( Z / p " Z ) * is a direct product of its subgroup H of order p"-' consisting of the elements which satisfy the elements satisfying

zpn-'

= 1 and the subgroup

K of order p

-

1 of

= 1. Because the orders of H and K are coprime, it suffices

zp-'

t o show that both H and K are cyclic.

If n = 1,then ( Z / p Z ) * = K and this is cyclic by Proposition 4.1(i). We may therefore choose a n integer m such t h a t m Set k =

mP"-l.

+p Z ,

m2 + p Z , .

. . ,m p - ' + p l l , are distinct

Since ( r n , p ) = I, ( k , p n ) = 1 and k

+pnZ

and m

+p n Z

in Z I p Z .

E (Z/p"Z)*.

Also kP-' = (mP"-')P-' = m$(P") l ( m o d p " )

which implies k

+p"Z

6 K . Bearing in mind t h a t

k = mpn-' the elements k

m(mod p )

+ p i Z , k2 + p Z , . .. ,kp-' + p Z

are distinct. Hence also k

+ p n Z , kZ +

p n Z , . . . , k p - l + p " Z are distinct. This shows t h a t the order of k + p n Z is precisely p

But the order of K is p

-

1, hence K is cyclic generated by k

To prove that H is cyclic, we may assume that n

If is a direct product of t solutions of the equation of integers s , 0

2

+p"Z. 2

1. Hence the number of

1, z E H is p t . We are thus left t o verify that the number

< s < p " , satisfying s p 3

sp

= l ( m o d p")

does not exceed p . Now if s satisfies

s(mod p ) , we have s

l ( m o d p ) . Then, if s

may write s =1

1.

since otherwise H = 1. Then

1 cyclic groups of order p",, ni

z p =

these conditions, then, since

22

~

+ y p f + .p'+'

(l5fln-1, o
in which case

= 1+ ypf+'(mod

pf+')

#

1, we

F I N I T E F I E L D S , ROOTS OF U N I T Y AND CYCLOTOMIC E X T E N S I O N S

If

sp E

l ( m o d p") and f < n - 1, this gives

O(mod p ) , contrary to 0 < y

so y

33 1

then s = 1

+ ypn-',

0

< p. Hence, if

1< s

< p" satisfies

< y < p. This gives altogether a t most

sp

zz l ( m o d p " ) ,

p solutions including 1 and

proves that H is cyclic. (ii) T h e order of ( Z / 2 " Z ) '

is

4(2") = 2"-'.

If n = 1 or 2, then these orders are 1 and

2, respectively, and there is nothing t o prove.

So assume t h a t n

2

3. We first claim

t h a t there are four distinct elements z E (Z/2"Z)* satisfying z2 = 1; if sustained, it will follow that (Z/2"Z)* is a direct product of a t least two distinct cyclic groups a1 = 1,az = -1,aZ = 1

+ 2"-',u4

= -1

+ 2"-',zi

=

# 1. Put

a; + 2 " Z . Then the z; are distinct

and satisfy z: = 1, which substantiates our claim. Because (Z/2"Z)* is a direct product of a t least two cyclic groups

#

1 and the order

of (Z!/2"Z)" is 2"-l, we see that, if z E ( Z / 2 " Z ) * , then z2"-' = 1, or, what is the same

thing, if m is an odd integer, then mZn-' = l ( m o d 2"). To complete the proof, we are therefore left to exhibit a n z such t h a t Put z 5""-3

5

+ 2"Z.

= l ( m o d 2"-l).

# 1.

Note first that, if n = 3, then 52'-3 = 5 jL l ( m o d 2n) but

Now let f

2 3 and

Then we have k(3) = 2. Also for any f

let k(f) be the largest integer k such that

2

3 we have 52'-3

= 1+ ~

2 ~ (where f ) y is o d d .

This implies that

whence k(f

+ 1) 2 k(f), so k(f) 2 2 i f f 2 3.

where

+ 2 k ( f ) - 1 y 2 is odd. Thus k(f + 1) = k(f) + 1. Since k ( 3 ) = 2, we deduce that

2

=y

k(f) = f

-

1 for all f

2 3.

Hence

Then the relation shows that

CHAPTER 6

332

and the result follows. w We now turn our attention t o the case where F = Q 4.14. Proposition. Let Q

be the n-th cyclotomic eztension of Q and let (Dn(X)

be the n-th cyclotomic polynomial over Q (i)

(Q n : Q

.

.

Then

= 4(n)

(ii) & ( X ) is irreducible (iii) Gal($ JQ)

EZ

(Z/nZ)*

Proof. Owing t o Proposition 4.11, it suffices to verify (i). Let and let f ( X ) be the minimal polynomial of

root of unity in Q

E E

be a primitive n-th over Q . We assert

t h a t for any prime p

+ n,

obtained by raising

t o a succession of prime powers with primes not dividing n , this will

prove t h a t degf

E

cP

is a root of f . Since any primitive n-th root of unity can be

2 4(n)and hence t h a t (Q

:Q ) = 4 ( n ) .

To establish the required assertion, we first note t h a t f ( X ) divides X n - 1 , say X " - 1 = f ( X ) h ( X ) where f ( X ) and h ( X ) are monic polynomials in Q [ X I . By Theorem 1 . 5 . 8 , f ( X ) , h ( X ) E Z [ X ] . Assume by way of contradiction t h a t Then

EP

cP

is not a root of f ( X ) .

is a root of h ( X ) and hence c is a root of h ( X p ) . We may therefore write

h(XP) = f ( X ) g ( X )for some g ( X ) E Q [X I. Since both h ( X P ) and f ( X ) are monic, so is g ( X ) , hence g ( X ) E Z [ X ] by Theorem 1.5.8. Now

l p E

[(mod p ) for any integer !, so

h (XP) = h(X)P(m odp ) and thus

It follows that the images

7 and h of

common. On the other hand, X " p

-

f and h in IFp[X]have a nonconstant factor in

1 = T ( X ) g ( X )and X n - 1 has no multiple roots since

+ n. This provides the desired contradiction. H 4.15. Corollary. Let n = p k , p a prime, let F be a field with charF

#

p and let

Fn be the n-th cyclotomic extension of F . Then F n / F is cyclic unless p = 2 and k 2 3 , in which case F,fF is either cyclic or G a l ( F n / F ) is a direct product of a cyclic group of order 2 and one of order 2t-2 where t 5 k . Furthermore, Gal($

nIQ )

{

Zpk-'(p--l) x Z2k-2

Z2

Proof. Apply Propositions 4.14, 4.13 and 4.11.

if p is odd ifp=2

F I N I T E FIELDS, ROOTS OF UNITY AND

CYCLOTOMIC

4.16. Corollary. Let n , m be relatively prime integers

2

333

EXTENSIONS

1 and let

E,,

and cm be

primitive n - t h and m-th roots of unity over Q , respectively. Then

Proof. Observe that, since ( m , n ) = 1,

E

= enem is a primitive mn-th root of unity

and that en and em are primitive m-th and n-th roots of unity, respectively. Thus

On the other hand,

4(mn)= +(m)+(n) since

( m , n ) = 1 and so, by Proposition 4.14,

IIence Q (cn) and Q (em) are linearly disjoint over Q which implies immediately that

a2

(en) n

Q (tm) = Q .

We close by examining cyclotomic polynomials in detail. First, we must define the Mobius p-function on the positive rational integers. Let p l , . . . , p r denote primes, and set

4 1 ) = 1, P(P1

’ ’

.P r )

if the p i are distinct =

otherwise

It is easily verified t h a t if

(a,b) = 1

and that

Given a function f,we define its “Mobius transform” g by

4.17. Lemma. With the notation above,

CHAPTER 6

334

In particular (a) T h e Euler j u n c t i o n

(1'

4 is t h e M o b i u s t r a n s f o r m

of t h e i d e n t i t y f u n c t i o n

t = Cnlt 4(n)

Proof. Note that

where, in the sum in braces, n ranges over all divisors oft which are multiple of d. In that case, n / d ranges over all divisors of t / d ranges over all divisors of t / d and so, by ( I ) , the sum in braces vanishes except when d = t , in which case it is 1. This proves ( 3 ) . In particular, let

4 be the Euler function. Using the fact t h a t d(ab)

= + ( a ) d ( b ) if

( a ,b) = 1, one easily obtains the well-known formula

where the first product is taken over the distinct primes dividing n. This proves (a) and, by ( 3 ) , also (b). w Formulas (2) and (3) are called the Mobius inversion formulas. Formula (3) expresses

f in terms of its Mobius transform g. Conversely, from (3), it is easy t o deduce t h a t

g

is

given by (2). 4.18. P r o p o s i t i o n . Let n

21

and let Qn(X)be the n-th cyclotomic polynomial o v e r

Q . Then (1)

Xn - 1 = n d l n @d(X)

(ii) CP,(X)= n d l , ( x dl)p(n/d) (iii) If p is a p r i m e , t h e n Q P ( X )= x p - ' +

a n d , for a n y integer r

2

xp-2

+ .. . + 1

1,

Q p r ( X )= @ , ( x P ' - l )

335

F I N I T E F I E L D S , ROOTS OF U N I T Y AND CYCLOTOMIC EXTENSIONS

(iv) I f n is odd, then @Zn(X)

(v)

If p

= @n(-X)

is a prime not dividing n , then

(vi) If n = pi'p;'

. . .pi'

is the canonical decomposition of n , then

Proof. (i) This is derived by collecting together all terms ( X -

E)

of a n ( X ) belonging

to roots of unity of the same order.

(ii) Apply (i) and the Mobius inversion formula ( 2 ) , written multiplicatively. (iii) We have x p -

1= ( X - 1 ) ( X p - l +

xp-2

+ ... + 1)

Any primitive p t h root of unity is a root of the second factor on the right of this equation. Because there are exactly p

1 primitive p t h roots of unity, we conclude that these roots

~

are precisely the roots of XP-'

+ XP-2 + . . . + 1. Thus

@ p ( X )= xp-1

+ xp-2 + . . . + 1

T h e second assertion is a consequence of the fact that degQp(XP'-') = deg

that any primitive p'-th root of unity

E

Qpr

( X ) and

is a root of @ p ( X P r - ' ) The . latter can be seen by

observing that tpr-' is a primitive p t h root of unity, hence a root of Q p ( X ) . (iv) Because n is odd, 4(2n) = 4 ( n ) and a typical primitive an-th root of unity is of t h e form --c where

t

is a primitive n-th root of unity. But

--t

is a root of Q n ( - X ) , hence the

assertion. ( v ) P u t + ( X ) = @ n ( X P ) / Q p n ( X )If. t is a pIimitive pn-th root of unity, then

primitive n - t h root of unity, hence

E

cp

is a

is a root of Q n ( X P ) . Thus @ p n ( X ) l @ n ( X and P ) so

$(X) is a monic polynomial in Q [XI of degree +(n). If 6 is a primitive n-th root of unit\-, then so is 6" since ( p , n ) = 1. Hence 6 is a root of Q n ( X P ) and therefore of + ( X ) . This proves that + ( X ) =- Qrb(X), hence the assertion.

CHAPTER 6

336

(vi) The polynomial if

t

aPI

is obviously of degree d(n).On the other hand,

(XP;'-l...P:.-')

is a primitive n-th root of unity, then

~p;'-l...p:~-',

is a primitive

plp2

. . .pa-th root of

rl-l

unity. Thus

E

is a root of

~ p I p 2 . . , p a ( X...P:*-') P~

which shows t h a t it must be equal t o

@n(X).

5. Finite Galois theory.

Throughout this section, given a field E and a group G of automorphisms of E , we write

EC for the fixed field of G. Our principal goal is t o establish the main result of the Galois theory for finite Galois extensions. Because in the next section we shall deal with infinite extensions, some of the preparatory results presented are stated and proved for infinite extensions. 5.1. Lemma. Let F

CK

E be a chain of fields, where EIF is a Galois eztension.

( i ) E / K is a Galois eztension and so, i f H = G a l ( E / K ) ,then K = E H . (ii) The map K

H

G a l ( E / K )from the set of intermediate fields into the set of subgroups

of Gal(E/F)is injeetiue. Moreover, i f E I F is afinite Galois eztension, then this map is a bijection with the inverse map given b y H

H

EH.

(iii) If K 1 and K2 are two intermediate fields and Hi = Gal(E/Ki),i (a) Gal(EIKlK2)= H I

=

1 , 2 , then

n H2

(b) E = E H I n E H 2

(c) K1 & Kz if and only if Hz

E Hi

Proof. (i) By Proposition 3.1, E I F is both normal and separable. Hence, by Proposition 2.2.15(i) and Lemma 2.1.20 (i), E / K is normal and separable. Therefore E / K is Galois, by Proposition 3.1. (ii) Suppose t h a t Gal(EIK1) = Gal(E/K2) = H , say, for two intermediate fields K1 and

K2. Then, by (i), K1 = E H = K2 proving the first assertion. The second assertion follows from Corollary 3.6. (iii) An automorphism q5 of E fixes KlK2 elementwise if and only if

4 fixes

each element

of K1 and each element of Kz. This proves (a). If X E E, then X is fixed by all

elements of H I and

H2,

hence X E E H 1n E H Z . Conversely, if X E E H 1n E H 2 ,then X

FINITE

€12

C Hi.

Conversely, if Hz

C H I , then K1 = En’

as required.

357

GALOIS THEORY

E H ”= K2

4

Let E / F be a Galois extension and let K be a n intermediate field. We shall sometimes refer t o G a l ( E / K ) as the group associated with K .

We also say that a subgroup H

of G a l ( E / F ) belongs to a n intermediate field K if H = G a l ( E / K ) . If K1,Kz are two

intermediate fields, then we say t h a t K1 and Kz are conjugate under an automorphism CJ

E

G a l ( E / F ) if u(K1) = K z . 5.2. Lemma. Let E/F be a normal eztension and let K be a n intermediate field

such that K J F i s also normal. Then the restriction map

{

Gal(E/F) u

+ Gal(K/F) -a(K

induces an isomorphism Gal(E/F)/Gal(E/K)

Gal(K/F)

Proof. Because K / F is normal, the restriction of any u E G a l ( E / F ) t o K is an autoniorphism of K , hence u l K E G a l ( K / F ) . The given map is obviously a homomorphism

with kernel G a l ( E / K ) . Since a

t-t

alK is surjective by Lemma 2.1.19, the result follows.

w 5.3. Lemma. Let E / F be a Galois eztension, let K1, Kz be two intermediate f i e l d s

and l e t H,= G a l ( E / K i ) , i = 1 , 2 . (i) For any given u E G a l ( E / F ) , a(K1) = Kz if and only if Hz = u H l u - - ’

(ii) If K is a n intermediate field, then KIF is normal if and only if Gal(E/K) is a normal subgroup of G a l ( E / F ) and when this is so Gal ( K / F ) E Gal ( E/ F )/ Gal ( E / K )

Proof. (i) By definition, r E H , if and only if r ( 5 ) = z for all z E K,, z = 1 , 2

Assume that K Z = u(K1). Then, for any y E K2, a - ’ ( y ) E K1, hence r a - l ( y ) = o-’(y) for all

7

E H I , i.e. a ~ a - ’ ( y )= y, which means that u r u P 1 E H z . Thus o H : o - ’ + IT2

CHAPTER 6

338

and since

K1 = a - l ( K Z ) , it

follows t h a t a - ' H z a

H I , i.e. Hz = a H 1 a - l . Retracing our

steps, we obtain the converse. (ii) Apply (i) and Lemma 5.2. We are now ready to record our main result. 5.4. Theorem. (Main theorem of finite Galois theory). Let E / F be a finite Galois

eztension with Galois group G (i) The map K

H

G a l ( E / K ) from the set of intermediate fields into the set of subgroups

of G is a n inclusion-reversing bijection with the inverse map given b y H

H

EH.

(ii) If H is a subgroup of G , then

IHI = ( E : E H ) and (G : H ) = ( E H : F) (iii) The extension K / F is normal if and only

Gal(K/F)

E

if G a l ( E / K ) o G and when this is so G/Gal(E/K)

Proof. (i) This is a direct consequence of Lemma 5.1. (ii) By (i), H = G a l ( E / E H ) and hence, by Proposition 3.3, IHI = ( E : E H ) . Since

IG/ = ( E : F) by Proposition 3.3, we have (G : H ) = ( E : F)/(E : E H ) = ( E H : F), as required. (iii) This is a direct consequence of Lemma 5.3(ii). w

Every field F comes accompanied by a canonical Galois extension: the separable closure F/F. Its Galois group is called an absolute Galois group of F . The extension F I F has infinite degree in almost all cases. It will next be shown t h a t Theorem 5.4 does not hold for the extension F / F . More precisely, it is no longer true t h a t every subgroup of G a l ( F / F ) belongs t o some intermediate field K of E / F . 5.5. Example. Let F = F, be the field of p elements, p prime, let

F

be the separable

closure of F and let G = G a l ( F / F ) . Then G contains the Frobeinus automorphism 4 which is defined b y

4(x) = x p

f o r all z

E

F

339

F I N I T E GALUIS THEORY

and we put H

=<4>. W e claim that there is no intermediate field K such that H = Gal(F/K)

Because the fized fields of H and G are the same, namely F , it sufices to show that G

f < 4 > . To this end, choose a sequence {a,}, n E IN, of integers such that a , z a , (mod m )

i f mln

but for no integer a the congruences a,

= a(mod n )

hold for all n E N. The simplest example of such a sequence can be obtained as follows: write n = n'p"("), ( n ' , p ) = 1, and 1 = n'z,

+ p"(,)y,,

sequence. W e now d e f i n e $I, t o be the restriction of C a l ( F p n / F ) If . IFpm

C IFpn, then mln, so

a , E a,(mod

then a , = n ' x , is a required t o IFpn. Then, obviously,

+n

E

m ) , and therefore

if we recall that 4 1 ~ has order m. The conclusion is that the $, d e f i n e a n automorphism P m

$ of

IJI ! ,T =

4" for some

aE

Z,then

which implies that a,

= a(mod n )

f o r a ll n E

IN,

a contradiction. We close this section by providing an illustration of the main theorem of finite Galois t hcory. 5 . 6 . Proposition. The field C is algebraically closed.

CHAPTER 6

340

Proof. We shall use the standard fact t h a t every polynomial of odd degree in lR[X] has a root in R. Let i be a root of X z a

+ 1.

Every element of R(i) has a square root: if

+ bi E R(i), a, b E IR,then the square root is given by c + di, where

Indeed, since both a + d root in R.If b

m and

+d

--a

w are nonnegative, they have a square

> 0, then choose the signs of c and d t o be equal and if b 5 0, choose them

t o be opposite. Then it is clear t h a t

(c

+ d i ) 2 = a + bi.

Because R has characteristic 0, every finite extension of IR is separable. Thus every finite extension of IR(i) is contained in a field K such t h a t K / R is a finite Galois extension. We are therefore left t o verify t h a t K = R(z). Let G = G a l ( K / R ) and let H be a Sylow 2-subgroup of G. By Theorem 5.4(ii),

(G : H ) = (X" : IR) and thus (K" : W) is odd. On the other hand, since K R / R is a finite separable extension, K H = R ( a ) for some a E K".

Then a is a root of an

irreducible polynomial in R[X] of odd degree, But every polynomial of odd degree in

lR[X]has a root in R,hence K H = R and therefore G

=

H is a 2-group.

Because G1 = Gal(K/IR(i)) is a subgroup of G, we see that G1 is a 2-group. Assume by way of contradiction that K

# IR(i).

Then G1

#

1 and so G I has a subgroup Gz of

index 2. Hence, by Theorem 5.4(ii), ( K G 2: IR(i))= 2. But we saw t h a t every element of

R(i) has a square root and hence R(i)has no extensions of degree 2, a contradiction. w

6. Infinite G a l o i s theory.

The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and intermediate fields is not valid for infinite Galois extensions (see Example 5.5). The Krull topology introduced below restores this correspondence for closed subgroups (Theorem 6.2). Throughout E/F denotes a Galois extension and G the Galois group of E/F. The group G can be endowed with the Krull topology defined as follows. Let {E;/Fli E I} be the set of all finite Galois subeztensions of E/F. For each u E G, we take the cosets cTGal(E/E;)

(i E I)

I N F I N I T E GALOIS THEORY

34 I

as a basis of neighbourhoods of u . The multiplication map

is continuous, because the pre-image of the basic open neighbourhood u1uZGal(E/E;) of ulu2 contains the open neighbourhood

ulGal(E/E;) x uzGal(E/E;) of ( u l , u z ) . Similarly, the map G

---t

G, o

H u-’ is

continuous, so G becomes a topological

group. Of course, if E / F is a finite extension, then the Krull topology is discrete. Observe also that, by Lemma 5.2, Gal(E/E;) a G and the map

G/Gal(E/E;) uGal(E/E;)

Gal(E,/F) H

uIE,

is a n isomorphism.

Partially order the set I by defining i fiy :

5 j in case E; C Ey. Given i 5 j , let

Gal(Ei/F)

4

Gal(E;/F)

be the restriction homomorphism. We regard the groups Gal(E;/F) as discrete compact topological groups, in which case

becomes a projective system of topological groups. The significance of the corresponding projective limit stems from the following result. 6.1. Proposition. The m a p

{

G 0

--$

++

limGal(E;/F)

(OlEi)

is a n isomorphism of topological groups. I n particular, G i s a profinite group.

Proof. Consider the map

CHAPTER 6

342

Then

+ is an algebraic homomorphism, which is injective since DIE; = 1 for all i E

I

implies u = 1. For any given j E I , the sets

form a subbasis of open neighbourhoods of the product of 0,then +-'(U) = u G a l ( E / E j ) . Hence

n,,,Gal(E;/F).

If u E G is a lift

+ is continuous, and

$ ( o G a l ( E / E j ) ) = $(G) n U showing that $ : G

4

+(G) is open.

It will next be shown t h a t +(G) is closed in

n,EIG a l ( E i / F ) . To this end, given s 5 k

in I , consider the set

It is obvious that

On the other hand, if Gal(E,/F) = {ul,...,on} and S;

G a l ( E k / F ) is the set of all

extensions of u; t o Ek, then

showing that

Msk

and hence +(G) is closed. Hence 1c, is a homeomorphism of G onto the

closed subset +(G) of the compact space

niE1G a l ( E , / F ) , i.e. G is compact.

We now use the fact that G is compact t o show that

+ ( G ) = limGal(E;/F) c

which will finish the proof. If

(0;) E

limGal(E;/F), then the closed sets +

u,Gal(E/E;)

INFINITE

343

GALOIS THEORY

have the finite intersection property. Because G is compact, there exists

This element satisfies $ ( u ) =

( 0 ; ) and

the result follows. rn

We are now ready to prove our main result.

6.2. Theorem. (Main theorem of infinite Galois theory). Let E / F be a (finite or infinite) Galois eztension with Galois group G.

(i) G is a Hausdorff, compact, totally disconnected group. (ii) The niap K

H

G a l ( E / K ) is a n inclusion-reversing bijection between the intermediate

fields of E / F and the closed subgroups of G with the inverse map given b y H

++

E".

(iii) For any intermediate field' K of E / F , K / F is finite if and only i f G a l ( E / K ) is a n open subgroup of G . (iv) If h'/F is a normal subeztension of E / F (which, b y Lemma 5.3(ii), happens if and

only ij G a l ( E / K ) a Gal(E/F)), then the restriction map G a l ( E / F )

4

G a l ( K / F ) gives

rise t o the following isomorphism of topological groups : Gal(E/F)/Gal(E/K)

Gal(K/F)

Proof. (i) By Proposition 6.1, G is a profinite group. Now apply Proposition 2.8. (ii)

I11

view of Lemma 5.1, we need only verify that G a l ( E / K ) is closed in G and that if

f l is a closed subgroup of G, then H = G a l ( E / K ) where K = E H . Each open subgroup of G is closed, since it is the complement of the union of its open cosrts in G . If K / F is a finite subextension of E / F , then G a l ( E / K ) is open, because each

o i. G a l ( E / K ) has the open neighbourhood o G a l ( E / L )

C G a l ( E / K ) , where

L / F is the

riorrnal closure of KIF. If K J F is an arbitrary subextension of E/F, then

Gal(E/K)

=

Gal(E/K,)

whcre K , / F runs through the finite subextensions of K / F . Thus G a l ( E / K ) is closed. T,ct 11 denote a closed subgroup of G and let K = E H . Clearly H

5 Gal(E/K).

Con-

vc,rscly, suppose that o E- G a l ( E / K ) . If L / K is a finite Galois subextension of E J K , t h e n

o G a l ( E , L ) is a basis open neighbourhood of a in G a l ( E / K ) . The map H

G a l ( E 1 K ) is

~ - i

CHAPTER 6

344

surjective, since its image

p

has the fixed field K and is therefore equal t o G a l ( L / K ) by

the main theorem of finite Galois theory. Choosing a X E H such t h a t AIL = OIL,we have

X E H n u G a l ( E / L ) . Hence u belongs to the closure of H , i.e. u E H since H is closed, and therefore H = G a l ( E / K ) . (iii) The “only if” part was established in (ii). Assume that H = G a l ( E / K ) is an open

subgroup of G. Because G is compact, H is of finite index, by Proposition 2.3(ii). Since this obviously implies K / F is finite, (iii) follows. (iv) It is obvious t h a t the restriction map induces a homeomorphism of the topological space G a l ( E / F ) / G a l ( E / K ) onto the topological space G a l ( K / F ) . Since, by Lemma 5 . 2 , the induced map is also a n algebraic isomorphism, the result follows.

H

6.3. Corollary. Let E / F be a Galois eztension with Galois group G , let H be a

subgroup of G and let

p

be the (topological) closure of H . Then Gal(E/EH) =

Proof. By Theorem 6.2(ii), G a l ( E / E ” ) is a closed subgroup of G. It is clear that H

C Gal(E/EH).

Assume that GI is any closed subgroup of G containing H . Then, by

Theorem 6.2(ii),

GI = G a l ( E / E G 1 ) 2 G a l ( E / E H ) , as required.

H

As we shall see later in this section, the composite of two abelian extensions is again

abelian. This obviously implies that the composite of an arbitrary collection of abelian extensions is again abelian. Thus, for any field F , the composite F a b of abelian extensions of F contained in a given field extension E of F is a n abelian subextension of E / F con-

taining any other abelian subextension. We shall refer t o FQb/F as the mazimal abelian

subeztension of E/F. 6.4. Corollary.

Let E / F be a Galois eztension with Galois group G and let H

be the topological commutator subgroup of G (i.e. the topological closure of the algebraic commutator subgroup G’ of G). If F a bi s the mazimal abelian subeztension of E/F, then

Fab= EH and G a l ( F a b / F )

G / H as topological groups

INFINITE GALOIS THEORY

34 5

Proof. By Proposition 1.5, H is a closed normal subgroup of G. Hence, by Theorem 6.2(ii), H = G a l ( E / E H ) . Furthermore, by Theorem 6.2(iv), G a l ( E H / F ) F GI11 as topological groups. We are thus left t o verify that EH = F a * . Since G / H is abelian, we have EH C Fa’. Let K / F be any abelian subextension of

E/F. Since KIF is normal, G a l ( E / K ) is a normal subgroup of G such that G/Gal(E/K)

Gal(K/F)

Since G a l ( K / F ) is abelian and G a l ( E / K ) is a closed subgroup of G, we have G a l ( E / K ) 2

H . Thus

K

= EG”’(E/K) c -

EH

and so Fa* C E H , as required. m

A further property of Galois groups of infinite Galois extensions is given by 6 . 5 . Proposition.

Let E I F and K / F be f i e l d extensions, where E / F is Galots,

and assume that E , K are subfields of some other field. Then E K f K and E/(E n K ) are Galois extensions and the map Gal(EK/K)

--t

o

G a l ( E / ( E n K))

H U l E

is a n isomorphism of topological groups.

Proof. The extensions EK/K and E / ( E n K ) are obviously normal and separable, hence Galois. If u E G a l ( E K / K ) , then u l E is an F-homomorphism E element of G a l ( E / F ) , since E / F is normal. The map o

i--t

4

E K , hence is an

oIE is clearly a homomorphism.

If ajE is the identity map, then o must be the identity map on E K , since EK = K ( E ) .

Thus the given homomorphism is injective. Let S be its image. Then S leaves E r; K fixed, and conversely, if a n element X E E is fixed under S, we see that X is also fixed under G a l ( E K / K ) , so X E K and X E the Krull topology the given map u

E n K . Thus E n K is the fixed field of S. Now for

H

oIE is obviously continuous, whence its image S

is closed since G a l ( E K / K ) is compact. Hence, by Theorem 6.2(ii), S = G a l ( E / ( E 7 K ) )

arid the result follows. w 6.6. Corollary. Let E / F be a finite Galois eztension and l e t K f F be a n arbitrary

extension. Then ( E K : K ) divides ( E : F ) .

CHAPTER 6

346

Proof. By Proposition 6.5, (EK : K) = (E : ( E n K)) and the result follows. 6.7. Proposition. Let E1/F and

E2/F

be Galois eztensions and assume that El, Ez

are subfields of some other field. Then E1E2IF is a Galois extension and the map

is a n injective homomorphism. Furthermore, if El n E2 = F , then it is a n isomorphism.

Proof. Since E1/F and EZ/F are normal and separable, so is E1E2IF and thus E1E2/F is a Galois extension. The given map is obviously a homomorphism. Moreover, if o fixes El and E2 elementwise, then o is the identity map. Thus our homomorphism is

injective. Suppose t h a t El n Ez = F . Owing to Proposition 6.5, given exists a o € Gal(EIEz/EZ) which restricts t o on

E2,

01

01

E G a l ( E l / F ) , there

on El. Since o induces the identity map

G I x 1 is contained in the image of the given homomorphism. A similar argument

shows that 1 x Gz is contained in this image. Thus our map is surjective and the result

follows. 6.8. Corollary. Let E;/F, 1

5 i 5 n, be Galois extensions and let

T hen

G a l ( E 1 . . . E,/F)

G a l ( E I / F ) x . . . x Gal(E,/F)

Proof. The case n = 2 follows froxi Proposition 6.7. The general case is a striaghtforward induction on n. 6.9. Proposition.

Let G be the Galois group of a finite Galois extension E I F .

Assume that G = G1 x . . . x G , and denote b y E; the fixed f i e l d of

G1 x . . . x (1) x . . . x G , where 1 occurs i n the i-th place. Then E , / F is Galois, 1 5 i 5 n, and

E,,

1

n ( E I .. . E,) = F, E

=

E l . . . En

(1 5

i5n

-

1)

INFINITE G A L O I S THEORY

... x G, so

Proof. P u t Hi = G1 x . . . x (1) x

347

that E; = E H x ,1

5 i 5 n.

Since

Hi u G, it follows that each E;/F is Galois. Now

H,

= G a l ( E / E H * )=

Gal(E/Ei),

so by Lemma 5.l(iii),

n

n H, n

n

G a l ( E / E l . . . E,) =

Gal(E/E;) =

=1

1=1

r=l

and thus E = E l . . .En.Finally, if S; = G a l ( E / E l . . . E i ) , then E l . . . E; = E S . . Furthermore,

n Hi2 i

si=

Gi+l

j=1

and so G = H;+1S;. Thus, by Lemma S.l(iii),

as reqired. H 6.10. Proposition.

A s s u m e t h a t all fields below are contained in s o m e c o m m o n

field.

(i) If E1/F and EZIF are abelian eztensions, t h e n so i s E1E2JF. (ii) If E I F I s abelian and KIF is a n y eztension, t h e n EKIK is abelian. ( i i i ) If E J F i s abelian and K is a n y intermediate field, t h e n K I F and E I K are abelian.

Proof. (i) This is a direct consequence of Proposition 6.7. ( i i ) Since E / F is abelian, so is the subextension E / ( E n K). Now apply Proposition 6.5.

(iii) This follows from the fact that subgroups and factor groups of abelian groups are

abelian. m It is an easy consequence of Proposition 3.5 that any finite group is the Galois g r o u p of some field extension. We now generalize this fact to profinite groups. The following

preliminary observation will clear our path. 6.11. Lemma. L e t F be a field and let G be a profinite group of a u t o m o r p h i s m s of

I.' s u c h that for each x E F , the stabilizer G(z) = (0 E G / u ( z )= z}

CHAPTER 6

348

is an open subgroup of G . Then G = Gal(F/FG) Proof. By hypothesis, for every

51,.

. . ,z,

E F , the group H =

n:=, G ( z z )is open in

G. Since G is profinite, G is compact by Proposition 2.8. Hence H is closed of finite index, by Proposition 2.3(ii). Therefore the intersection N of all conjugates of H is closed and of finite index, hence open (Proposition 2.3(ii)). T h e finite group G / N can be regarded as an automorphism group of the field L = F G ( G z l , .. . ,Gz,), where Gx; = {o(z;)lo E G } , and it has F G as its fixed field. Thus L / F G is a finite Galois extension with G I N = G a l ( L / F G ) (Proposition 3.5). The field F is the union of the above L and 1 is the intersection of all above N . Hence G a l ( F / F G ) = lim G a l ( L / F G ) = lim G I N = G c

c

as required. 6.12. Theorem. (Waterhouse (1974)). Every profinite group is the Galois group of

some field eztension.

Proof. Let G be a profinite group and let S be the disjoint union of the sets G / H for all open subgroups H of G . Then G acts faithfully on S and each element of S has open stabilizer. Let K be any field, take elements of S as indeterminates and form the purely transcendental extension F = K ( S ) . Then G may be regarded as a n automorphism group of F . Since G satisfies the hypotheses of Lemma 6.11, the result follows. w

7 . Realizing finite groups as Galois groups.

A group G is said to be realizable over a field F if there is a Galois extension E / F such that G

G a l ( E / F ) . A major unsolved problem in Galois theory is whether every finite

group is realizable over Q . If a finite group G is a split extension of a normal abelian subgroup by a group which is realizable over Q , then G is realizable over Q . This fact is a consequence of an embedding theorem of Scholz (1929). By a celebrated theorem of Safarevich (1956), every finite solvable group is realizable over any given algebraic

number field. An alternative and more conceptual proof in the case of groups of odd order is contained in a fundamental work of Neukirch (1979). As a consequence of his main

REALIZING FINITE GROUPS AS GALOIS GROUPS

349

theorem, Neukirch proved that every profinite separable group G of finite odd exponent is realizable over any given algebraic number field. Here the separability of a profinite group means t h a t it has a countable basis of neighbourhoods of the identity. T h e proof of the above is dominated t o a large part by deep cohomological resuits and therefore is beyond the scope of this book. Note also t h a t many other results in the literature pertaining t o this topic employ class field theory or methods of algebraic geometry and complex function theory, together with the Hilbert irreducibility theorem. This section respresents a compromise between generality and the desire t o prove a result of interest with the techniques readily available. Namely, we show how an admittedly very restricted family of groups can be realized over Q using rather unsophisticated methods of field theory. An important ingredient of the proof of the main result (Theorcm 7. 10) is the fact that the Galois extension fields which we construct are real fields. Let F be a n arbitrary field, let f ( X ) be a separable polynomial over F and let E be the splitting field of f ( X ) over F . Then E / F is a finite Galois extension whose Galois group is called the Galois group of f ( X ) over F . If a l , . . . a , are the roots of f ( X ) in E , then

E on

=

F ( a 1 , . . . ,a,). Any automorphism

{al,.

(T

of E / F is completely determined by its effect

. . ,a,}. Moreover, a ( a , ) E { c Y ~ ., . ,a,} for all i E { 1 , 2 , . . . , n } , hence o defines

permutation T ( O )of

{&I,.

. .,a,}. It is now obvious that the map G

+

S,, o

H

~ ( ois)an

injective homomorphism and hence G a l ( E / F ) is identifiable with a subgroup of S,. This leads t o a natural question of whether S, is realizable over Q . By Hilbert's irreducibility theorem (Hilbert(1892)),for any nonzero integer a , there exist infinitely many integers b such t h a t the Galois group of

X"+aX+b is isomorphic t o

over Q

S,. But concrete determination of such integers b is a formidable problem.

Nevertheless, a recent breakthrough due to Osada (1987) reveals a remarkable fact that the Galois group of

X" is isomorphic to S , for any integer n

-

X - 1

2 2.

over Q

Note also that Xn- X

- 1 is

irreducible over Q ,

by a result of Selmer (1956). While on the subject, we mention that Schur (1973, collected works) proved that S , is the Galois group of the Laguerre polynomial L , of degree n defined by

CHAPTER 6

350

Since the roots of L , are real (and positive), it follows t h a t S, is t h e Galois group of a

normal real extension of Q . Using elementary methods, we first show t h a t for any prime p , S, is realizable over

Q. 7.1. Lemma.

Let p be any prime.

Then there exists an irreducible polynomial

f ( X ) E Q [XI of degree p such that f ( X ) has ezactly two non-real roots in Q: . Proof. Let n be a positive integer and let m > 3 be a n odd integer. Given m -- 2 even integers tl

< t 2 < . . . < t m - 2 , consider the polynomial

+

g ( X ) = ( X 2 n ) ( X - t l ) ( X - t 2 ) . . . ( X - tm--2) E Q [XI Then the real roots of g ( X ) are t l , t z , . . . ,tm-z.

By looking a t the graph of y = g ( z ) ,we

conclude t h a t it has (rn - 3 ) / 2 relative maxima. Moreover, because for any odd integer s, we have jg(s)l > 2 , the values of these relative maxima are greater than 2 . It follows that

has ( m- 3 ) / 2 positive relative maxima between t l and tm-z and thus f ( X ) has m - 3 real roots in the interval ( t l , t m - 2 ) . Furthermore, there is another real root

f(t,,-*)

= - 2 and f ( m ) =

00.

> t m - 2 , because

We have thus revealed m - 2 real roots of f ( X ) .

We now claim t h a t , with a suitable choice of n, f ( X ) has precisely m - 2 real roots;

if sustained, it will follow that f ( X ) = X"

+ alXm-' + . . . + am, where each a, is an

even integer. Since the constant term of g ( X ) is divisible by 4, t h a t of f ( X ) = g ( X ) - 2 is not divisible by 4. Hence, by Eisenstein's criterion applied t o the prime 2, f ( X ) must be irreducible over Q . This proves the result for every prime p = m

2 5.

An easy verification

shows that the result also holds for p = 2 , 3 and we are left to substantiate our claim. Let A l , A z , . . . ,A,

be all roots of f ( X ) in Q: . Then

and equating t h e coefficients of Xm-' and X m - ' , we have rn

m--2

i=l

k=l

i
REALIZING FINITE

351

GALOIS GROUPS

GROUPS A S

It follows that

Choosing n sufficiently large, we have then

EX: < 0, hence not

each A, is in

R.If

A1

xl # A 1 !$ R and thus f ( X ) has a t least two non-real roots. Because f ( X ) has m

real roots, we conclude that f (X) has precisely m

~

R,

$E ~

2

2 real roots, as required. m

7.2. Lemma. Let p be a prime and let G be a subgroup of the symmetric groups S,,

If G has a n element of order p and a transposition, then G = S,. Proof. Since every element of S p is a product of transpositions, it suffices t o verify that G contains all transpositions. We may harmlessly assume that (12) and cr = ( 1 2 3 . . . p ) are in G. Now if l , i , j are distinct, (ij) = ( l i ) ( l j ) ( l i ) .Thus weare left toverify that (1;) E

G for all i E ( 3 , . . . , p } . Conjugating (12) by u , u 2 , .. . ,a,-’,we see that (23), (34), . . . . ( p 1 , p ) and ( p l ) are contained in G. But (13) = (12)(23)(12), (14) = (13)(34)(13),. . . , ( l p ) = (1 p

~

l)(p

~

1p ) ( l p

-

l),

hence the result. rn We are now ready to prove

7.3. Theorem. I f p i s a prime, then S, is realizable over Q

.

Proof. Let f(X) E Q [XIbe as in Lemma 7.1. and let X I , . . . ,A,

be all roots of f ( X )

in 6:. Since C is algebraically closed,

f(X) = ( X and thus E = Q (Xl,X,,

..., A),

-

x l ) ( X - A,). . . ( X - A),

in Q: [Xj

is the splitting field of f ( X ) over Q . It will be sho\vn

that G = G a l ( E / Q ) is identifiable with S,. Because E 2 Q (XI)

and (Q (A,)

:Q ) =

d e g f ( X ) = p , it follows t h a t p dividrs

( E : Q ) . But /GI = ( E : Q ) . hence p divides 1G/. Thus, by Sylow’s theorem G has a n element of order p . Let u be the automorphism of E induced by complex conjugation. 2 real roots. Since cr interchanges the remaining roots,

Then u E G and cr fixes all p

-

is a transposition. Thus G =

S,, by virtue of Lemma 7.2.

i~

CHAPTER 6

352

We next show that, for any positive integer n, there exists a subfield F, of such t h a t S, is realizable over F,.

R

T h e following two preliminary results will clear our

path. 7.4. Lemma. T h e transcendency degree of R/Q i s equal t h e cardinality of R.

Proof. Let S be a transcendency basis of R/Q. For any integer rn

2

1, the set

of polynomials of degree a t most rn in elements of S over Q has cardinality max(N0, ISI}.

Therefore, if /SI < IRI, then IQ ( S ) \< IRI. But IR is algebraic over Q ( S ) ,hence

IRI.This contradiction shows t h a t IS1 = /RI,as asserted.

I$

(S))l =

H

To state our next preliminary result, we need t o recall the following terminology. Let

R be a commutative ring. If

(T

E S, and f(X1, ..., X,) E R[X1, ..., X,],then f" is

defined by

f"(X1,. . . ,Xn)= f(XU(l), . . . ,X,(,))

f"

=

alternating polynomial if f is not symmetric and

f"

The polynomial f is said t o be s y m m e t r i c if

elements of a ring containing R ,

f(A1,.

. . ,A,)

f for all u E S,. We say that f is an = +ffor all (T E S,. If X I , . . . , A , are

is called a s y m m e t r i c f o r m or an alternating

. . ,X,) is a symmetric or alternating polynomial, respectively. f o r m on X 1 , . . . ,A, if f(X1,. Note that the existence of a n alternating polynomial over R implies charR other hand if charR

# 2. On the

# 2, then D

= n ( X ; - Xi) i<J

is obviously an alternating polynomial. The following result expresses any other alternating polynomial in terms of D. 7.5.

Lemma.

Let R be an integral d o m a i n of characteristic # 2.

T h e n every

alternating polynomial over R is the product of D a n d a s y m m e t r i c polynomial.

Proof. Let f = !(XI,. . .,X,) E R [ X 1 , ... ,X,]be a n alternating polynomial. For any given u E S,, let s ( u ) E {1,-1} be defined by

f"

= s(a)f.Given ul,(~Z E S,

we

have

f"'"'

=

(f"*)"' = (s(a2)f)"' = S((T2)S(Ol)f

and thus s(alo2) = s(al)s(az).Since every transposition ( i , j ) is conjugate to (12), we have s ( ( i , j ) ) = s ( ( l , 2 ) ) = c , say. If c = 1, then s(u) = 1 for all

(T

E S,,, which is

impossible. Thus c = -1 and s ( a ) = 1 for every even permutation u E S,.

If i < j and

R E A L I Z I N G FINITE G R O U P S A S G A L O I S G R O U P S

X,,X,+,,...,X , ) ,

h = f ( X l , .. . , X j - l ,

then h = -h since for o = ( z , j ) , f" = -f.

Since 2 is not a zero divisor in R , we see t h a t h = 0. Thus X i

D divides f. Write f

353

-

X, divides f and therefore

Dg. Then, for each o E S,,

f"

=~(0 f ) = s(o)Dg

and, on the other hand, f" = Dugu = s (Q)DgU Therefore go = g, proving that g is a symmetric polynomial, as required. I

Theorem. For any given positive integer n, let r l r 7 2 , . . . ,r,, E R be alge-

7.6.

braically independent over Q and let r l , . . . ,r,,

1 5 k 5 n. Then

S,

Sk

be the elementary symmetric f o r m of degree k in

is realizable over Q

Proof. P u t E = Q ( r l , . . . , r n ) and F = Q a Galois extension such t h a t G a l ( E / F )

( ~ 1 , ..

. ,s,).

..

( ~ 1 , . ,s,).

It will be shown t h a t E / F is

S,. Consider the polynomial

fn(x) = (X- r l ) ( X - 7 2 ) . . . (X- 7,) Then

fn(X) = X" + c l x n - ' since

( - l ) k ~ k = sk

+ . . . + c,

E FjX]

for k E {I, ..., n } . Furthermore, E is the splitting field of f , ( X )

over F and hence E / F is a Galois extension. The symmetric group S , acts naturally on

{ r l , . . . , r,} and hence on E. We are therefore left to verify that the fixed field, say K of

S,, coincides with F . Because the c, are symmetric forms, we have F

C K.

Let X be an arbitrary element

of K . Then A = g ( r l , . . . ,r , ) / h ( r l , .

where g and h are polynomials in

rl,

. . . ,r ,

. . ,7),

with coefficients in Q , and such that g and h

have no proper common divisor. Now fix a n arbitrary element cr of S,. Bearing in mind that g"/h" = A" = X = g / h ,

we have g'h = hug. Therefore, since g and h have no common proper divisor, we conclude that g ) g U and hlh". Since cr preserves the degrees of polynomials, it follows that g" = Xg, and h"

=

X,h for some X u E Q

.

It is clear that ,A,

= XX ,,

for a,X E: S , and thus

CHAPTER 6

354

o

H

A, is a homomorphism of S, into the multiplicative group Q * of Q . Since fl are

the only roots of unity in Q *, we have A, = f l for all

(T

E S,. Therefore g and h are

either symmetric forms or alternating forms. If they are alternating forms, then they are divided by n,+(ri - r,) (see Lemma 7.5), contrary t o the assumption t h a t g and h have no proper common factor. Hence g and h are symmetric forms and

X = g / h E F , by virtue

of Theorem 1.4.13. Thus F = K and the result follows. The rest of the section will be devoted t o providing certain classes of groups which can be realized as Galois groups of normal real extensions of Q

.

We begin by recording

the following group-theoretic information. Suppose t h a t H is a subgroup of a finite group G and let n be the index of H in G. Then the map

is a group homomorphism whose image is a transitive subgroup of S,.

If H is core-jree,

i.e. if H contains no proper normal subgroups of G, then the above homomorphism is injective. In this case G is identifiable with a transitive subgroup of S , and we say that

G admits a faithful transitive permutation representation o f degree n. Let G be a finite group having a faithful transitive permutation representation of degree n. If W is any group, then G acts as group of automorphisms of the direct product

u = w x ... x w of n copies of W by permuting the components of U according to its action on n points. The above action of G on U gives rise to the semidirect product of U by G, which is called a wreath product of W by G and is denoted by WwrG By taking H = 1 in ( I ) , we obtain a faithful transitive permutation representation of degree n = IGI. Wreath products constructed using this action of G are called regular wreath products. The proof o f the main theorem (Theorem 7.10 below) will use some standard results from algebraic number theory. A reference for all relevant definitions and facts is Janusz (1973). For convenience, we briefly review the pertinent theory.

Let F be a n algebraic number field and let f ( X ) be a monic polynomial in F [ X ]of positive degree. Denote by E a splitting field of f ( X ) over F and let G = G a l ( E / F ) . If R , S are the rings of algebraic integers in F and E , respectively, then R 2 S. For any

prime ideal P in R , write R p for the local ring of R a t P. Then Rp is a principal ideal

REALIZING FINITE GROUPS

AS GALOIS

GROUPS

355

domain with quotient field F and PA!p is a unique maximal ideal of R p . Furthermore, there exists a unique prime integer p in P n Z and

R/P

ERp/PRp E

F,

where q is a power of p.

Let

&I,.

. . , Qn be

the prime ideals of S lying over P . Then the group G permutes

the Q, transitively. Now fix Q E { Q l , . . . , Q n } and denote by D = D(Q1P) the stabilizer of

Q in the permutation action

of G on { Q I , . . . , Qn}. Then the group D is called the

decomposition group. The finite field L = S / Q is a n extension of

IF, and it inherits

a

natural action of D as a group of automorphisms over F,. If P is unramified in S , then

and therefore D is cyclic, say D = < u > . The element u is said to be a Frobenius automorphism and the conjugacy class of u in G is called the Frobenius class. Finally, let B be integral closure of A = Rp in E . Because the coefficients of f ( X ) are in R p for all but

finitely many primes P of R , the roots of f ( X ) lie in B if we avoid the exceptional primes. The maximal ideals of B are of the form QtB and they are permuted transitively by G. Note also that

BIQB

S/Q

which shows that t h e decomposition group D acts naturally o n B / Q B and is isomorphic to G a l ( B / Q B / A / P A ) = Gal(L/IF,) in the unramified case.

After this digression into the terminology, we return to the study of Galois groups. 7.7. Lemma. Let F be a perfect field with charF # 2 , let f ( X ) be a polynomial zn

E’[X]of positive degree n and let G be the Galois group of f ( X ) over F . I f g ( X ) = f ( X 2 ) and II is the Galois group of g ( X ) over F , th e n there is a n exact sequence

where

N

i s a normal elementary ahelian 2-suhgroup of H of order dividing 2n

Proof. Let K be a splitting field of g ( X ) over F . If is

(Y.

Furthermore, -a

#

Q

since c h a r F

#

Q

is a root of g ( X ) in K , then so

2. We may therefore choose n distinct roots

al,. . . , a , of g ( X ) in K such t h a t L Y ~ -, & I , . . . ,a,, -a, are all roots of g(X) in K. Then

K

=

E ’ ( ( Y ~P,a l , . . . ,a,, -on)contains the field E = F ( a q , .. .,a:)

356

CHAPTER 6

It is clear that E is a splitting field of f ( X ) over F. Since F is perfect, it follows that E / F is a Galois extension. We thus have a chain

FLELK with Gal(K/F) = H and Gal(E/F) = G. Let N be the subgroup of H that fixes all elements of E. Since E/F is normal, N is a normal subgroup in H such that H / N

E

G

(see Lemma 5.2). Now fix any u E N . Then a fixes each a:, 1

5 i 5 n, hence (1 5

.(a;) = fa;

2

5 fl)

It follows that u2 fixes all a; and hence is the identity. Thus N is an elementary abelian 2-subgroup of G whose order is clearly at most 2n. B 7.8. Lemma. Let F be a field and let f ( X ) E F[X] have no multiple roots in a

splitting field E of f ( X ) over F. Then f (X) is irreducible over F if and only if the Galois group of f ( X ) over F acts transitively o n the roots of f ( X ) .

Proof. Let X I , . . . ,A, permutations on the set { A l , .

be all roots of f ( X ) in E. Then G = G a l ( E / F ) acts by

. . ,An}.

Assume first that f ( X ) is irreducible over F. Then,

for each i E (2,. . . ,n } , there is an F-isomorphism F(A1) + F ( A ; ) which sends

X1

to A;.

This F-isomorphism can be extended to an F-automorphism of E. Therefore we obtain an element of G which maps

A1

to A; and thus G is transitive.

Conversely, assume that G acts transitively on { A l , . polynomial for

A1

. . ,A,}.

Let g ( X ) be the minimal

over F and let a; E G be such that u;(A1) = A;,

g ( X i ) = g ( a i ( X 1 ) ) = g(X1) = 0, 1

5

z

1

5

z

5 n. Then

5 n. Since XI, ..., An are distinct, it follows that

g ( X ) must agree with f ( X ) except for a constant factor. Thus f ( X ) is irreducible over F , as asserted. w

7.9. Lemma. Let E / F be a finite Galois eztension with Galois group G and let H

be any core-free subgroup of G. Then there exists a n irreducible polynomial f ( X ) i n F[X] of degree n = (G : H ) such that

(i) E is a splitting field of f ( X ) over F (ii) H is the stabilizer of a suitable root of f ( X ) i n E

Proof. By the normal basis theorem (Theorem 3.9), we may choose X in E such that

R E A L I Z I N G F I N I T E GROUPS AS GHLOIS GROUPS

357

is an F-basis of E . Let (1 = U I , U Z , . . . , u,} be a left transversal for H in G and let pi = u i ( p )

where

(1 5 2

h(X)

p =

5 n)

hEH

It will be shown that

f(X) = (X- P l ) . . . (X- Pn) satisfies (i) and (ii). Note that p, =

CoEu,H .(A),

1

5 i 5 n.

Therefore

p1,.

. . ,p,, are linearly indepen-

is invariant dent over F and are transitively permuted by G. T h e latter implies t h a t f(X) Since under G, hence f ( X ) E F I X ] and, by Lemma 7.8, f ( X ) is irreducible in F[X]. p l , . .. , p n

are in E , E contains a splitting field K of f(X) over F. Let S be the subgroup

of G t h a t fixes all elements of K . Since K / F is a Galois extension, S is a normal subgroup

of G. But S fixes p and H is obviously the stabilizer of the root p of f(X). Hence S

CH

arid S = 1 , since H is core-free. It follows that K = E,proving that f(X) satisfies ( i ) and (ii).

We are now ready t o prove our main result in which Z2denotes a cyclic group of order 2.

7.10. Theorem. (Gow(1986)). Let F be a real algebraic number field, let h ( X ) be a polynomial i n F [ X ]whose roots in some splitting field E over F are all real and let G be the Galois group of E I F . If H is a core-free subgroup of indez n in G , then there exists a monic irreducible polynomial g ( X ) in F I X ] of degree 2n such that (i) All roots of g ( X ) are real

(ii) The Galois group of

g(X)over F is the wreath product

of order 2"IGI. The permutation action of G in the wreath product is that determined b y its action on the left cosets of H i n G.

Proof. For the sake of clarity, we divide the proof into four steps. Step 1 . Here we provide some preliminary observations which will enable us t o construct

g(X)satisfying

(la),

but not (i).

We first observe t h a t E is a real algebraic number field, since E is generated over F by the roots of h ( X ) . Owing to Lemma 7.9, E is the splitting field of a monic irreducible

CHAPTER 6

358

polynomial f(X) in

Fix]of degree n. Moreover, there exists a root

E l is the stabilizer of p . The roots

p1 = p , p 2 , . . . ,pn of

g of

f(X)such that

f ( X ) must all be real, since E is

a real field. We may assume t h a t the ordering of the p; is chosen so t h a t

Now choose a rational number c such that p1 -

c

< 0 and p ; - c > 0 for i > 1

T h e polynomial f(X+c) in FIX]is irreducible and has n real roots, precisely one of which is negative. Of course, the Galois group of this polynomial is G and H is the stabilzer of the root p1 - e . We may therefore harmlessly assume t h a t f ( X ) has exactly one negative root. In what follows, we put g(X)= f(X2), so that g(X)is rnonic of degree

212.

S t e p 2. W e now prove that g ( X ) i s irreducibfe in F [ X ] . Let the roots of g ( X ) be

AI,-AI,.

. .> An,-An

in some splitting field K , with A: = p ; . Since f ( X ) has exactly one negative root, namely p l , g(X)has exactly two nonreal roots, A 1 and - X I ,

be the Galois group of g(X)over

which are purely imaginary. Let T

F. Owing to Lemma 7.7, T contains a normal elementary

abelian 2-subgroup N such that G zz T I N

Let

u

be the complex conjugation automorphism of K . Then we have

.(Al) which shows that (.(A:)

= A:

=

for all

-XI, 2

= A, for

.(A,)

i>1

E (1,. . . ,n}. Since

(0

E = F ( X i , .. . ,A;)

and, by the

proof of Lemma 7.7, N is the subgroup of T t h a t fixes all elements of E , we conclude that uE

N. Since f(X) is irreducible, Lemma 7.8 tells us that T permutes the roots

transitively. Hence, for z > 1, there exists .*(A,)

= *Ai

E T such that u , ( A : ) = A:.

0%

pcL, of

j(X)

This implies that

for all i > 1

We conclude t h a t u;ooT1 fixes all roots distinct from *A;

and sends A; t o -A;.

This gives

n independent generators for N and thus N is of order 2n. It is also clear t h a t T permutes

the roots of g ( X ) transitively and therefore g(X)is irreducible.

REALIZING FINITE GROUPS AS GALOIS

GROUPS

359

Step 3. Here we show that g ( X ) satisfies (ii). Note that G acts faithfully as a permutation group of degree n. By Step 2, it suffices t o show that T splits over N . Let m = /GI and let the image of G in S, consist of the permutations rl,.. . , T , .

Now the permutation action of T on the roots Xp = p , is that

which is determined by G. It follows that for any

7r

E

{TI,.

. . ,rm}, there exists o in T

such that

42) = A:(*)

for a l l i c { 1 , 2 ,

..., n}

Because N has order 2", there exists a unique element 6 in N such that

O(Xa(,l) =

t,X,(,)

for all i E { 1 , 2 , . . . ,n}

which implies that

6a(A) = Now put

T'

= 6'a. Then

T'

k ( z )

for all i E { 1,2, . . . , n}

effects the same permutation on the roots XI,. . . ,A,

It follows t h a t if G I is the subgroup of T generated by on the roots p l , . . . ,/in. then G I must intersect N trivially and G1

as 7ri,

7r

effects

. . . ,T

: ~ ,

G. Thus GI is the required complement for

N in T , proving (ii).

Step 4 . Here we complete the proof b y showing that g ( X ) can be chosen t o satisfy both (iJ a n d (ia).

We keep the notation of Steps 2 and 3.

In particular, o is the complex conjugation

automorphism of K and u E T satisfies (1). Let C be the conjugacy class of T containing n

and let R , S be the rings of algebraic integers in F and K , respectively. By the Frobenius

density theorem (see Janusz(1973, p. 134)), there exist infinitely many prime ideals P in

R which are unramified in S and which have the property t h a t C is the corresponding Frobenius class. Select such an unramified prime P for which the coefficients of g ( X ) belong t o A = Rp. Then the roots of g ( X ) lie in the integral closure, say B of A in K . Denote by Q a prime ideal of S t h a t lies over P and is fixed by u . Then u generates the Galois group of L 1 = S / Q over L = R I P . Let

1be

the image of X E B under the natural homomorphism

CHAPTER 6

360

and let g be the image of g in L [ X ] .The roots of g in L1 are 3 ~ x 1 ,... , & and In they are

all distinct, because P is unramified. In the induced action of u on L1, we have o(I1) =

-X1,a(&) = Xi

for i > 1

Because u generates Gal(L1/L), we may thus express S in L [ X ]as

g = (X2 -

x:,

n ( x - Xi)(X+ 1;) i>

where

1

1 : is a nonsquare in L and 1;is in L for i > 1. It follows t h a t g is a product of an

irreducible polynomial of degree 2 and distinct linear factors. Let p be the unique rational prime of P n Z and let r E

f(X

Z be such t h a t the roots of

+ rp) are all positive, where f ( X ) is the polynomial of Step 1 having all real roots

and exactly one negative root. P u t

+

91 = f(X2 r p )

Then g1 is a monic polynomial in A[X] with all real roots and such that ?jl =i j in L[X]. Because g1 is a product of a n irreducible polynomial of degree 2 and distinct linear factors, it follows from a theorem of Dedekind (see Grove (1983, p. 285)) t h a t the Galois group of 91, over F contains a n element t which acts as a transposition o n the roots of 91. Now g1 is a polynomial in X 2 ,hence we may assume t h a t the roots of g1 in some splitting field are 61>-61,...,6n>-6n

and these roots can be numbered so t h a t t(61) = -61 and t(6;) = 6;for i

>1

It follows that t plays the role of the complex conjugation automorphism CJ of Steps 2 and 3. Thus repeating the previous arguments, we conclude that the Galois group of g1 over

F is also

Z52

w r G. Finally, because all roots of g1 are real, we may replace g by 91, thus

completing the proof. w We close by providing a number of applications contained in a work of Gow (1986). First, however, we must develop our vocabulary. Let G be a group, let F be a field and let V be a n FG-module. We say that V is cyclic if there exists v in V such that V = F G v

R E A L I Z I N G F I N I T E GROUPS AS G A L O I S GROUPS

36 1

(equivalently, such t h a t the vectors gv, g E G , span V ) . It is clear that V is cyclic if and only if V is isomorphic to quotient of the regular module F G . 7.11. Corollary. Let F be a real algebraic number field and let a finite group G occur

as the Galois group of a normal real eztension of F . If V is any cyclic FzG-module, then the semidirect product V G also occurs as the Galois group of some normal real eztension of F .

Proof. We apply Theorem 7.10 for the case where H = 1. Then the regular wreath product S = Zz wr G of order 21G11Gl is realized as the Galois group of a normal real extension of F . Let N be the normal elementary abelian 2-subgroup of S on which G acts. Then N , regarded as an FZG-module, is isomorphic t o the regular module F 2 G . Because V is a cyclic FzG-module, there exists a n FzG-submodule M of N such that

NIM

as FzG-modules

V

It follows t h a t M is a normal subgroup of

VG

S such that S/M

Since S is realized as the Galois group of a normal real extension of F , the same is true for the factor group S / M . Hence V G occurs as the Galois group of a normal real extension of F . In what follows we write GL,(2) for the group of all n x n invertible matrices over the field I F Z , S, for the symmetric group of degree n and A , for the alternating group of degree n.

7.12. Theorem. All of the groups listed below occur as Galois groups of normal real extensions of Q : (i)

iZ2w r S,, where S, permutes n copies of Zz,

(ii) A Sylow &-subgroup of Sz'L, n

21

(iii) A Sylow &-subgroup of A z " , n

21

n

21

(iv) A Sylow 2-subgroup of GL,(2)

Proof. (i) As has been remarked prior t o Lemma 7.1, S, occurs as a Galois group of a normal real extension. The desired conclusion now follow by virtue of Theorem 7.10.

(ii) Define PI = Z z and consider PI as a transitive permutation group of degree 2. Put

Pn

Z z w r P n - l , rz

2

2 , where the wreath product is constructed using the natural

CHAPTER 6

362

transitive permutation action of P,,-l of degree 2"-'.

Then, by computing the order of

P,, it follows t h a t P, is isomorphic t o a Sylow 2-subgroup of Sz-. We now show that P, is the Galois group of a n irreducible polynomial

fn

of degree 2, in Q [XIwhose roots

are all real, which will prove (ii). We argue by induction on n. For n = 1, we can take

fl(X)= X 2- 2, whose Galois group over Q is Z z

=

Pi. Assume t h a t

fn-l

has been

constructed t o have the required properties. Owing t o Theorem 7.10, we can construct a polynomial of degree 2.2"-'

= 2" over Q whose roots are all real and whose Galois group

is Z 2

wi-

P,-l

=

P,

which proves (ii). (iii) In (ii) we have realized a Sylow 2-subgroup P, of SZ- as the Galois group of a polynomial f, E Q [XIwhose roots

fX,, 1 5 i 5 2,-'

are all real. (That the roots of

fn

are of this form follows from the proof of Theorem 7.10). Now P, is a semidirect product NnP,-1,

where N , is elementary abelian. The action of a n element

(T

of N , on the roots

has the form O(Ai)

where

E,

=

I 5 i 5 2,-'

€,A,

= * l . Consider the subgroup M, of

N,,consisting of all those

o

E N , for which

i=l

Then M , is a normal subgroup of P, and has index 2 in N , . Furthermore, regarded as an FzP,,-l-module, M, is easily seen t o be cyclic. By computing the order of the semidirect product M n P n - l , we see that M,P,-,

is isomorphic t o a Sylow 2-subgroup of A z - . Now

apply Corollary 7.11. (iv) Let T, be a Sylow 2-subgroup of G L , ( 2 ) . If V , is a vector space of dimension n over

F z , then T, acts naturally on V, and one can readily verify that

T,+1

is isomorphic to the

semidirect product V , T,. It is also elementary to check t h a t V,,is a cyclic FzT,-module. Because

Tz is cyclic of order 2, it obviously can be realized

as a Galois group of a normal

extension of Q . The general case now follows by induction on n and using Corollary 7.11. H

363

DEGREES OF SUMS I N A SEPARABLE F I E L D EXTENSION

8. Degrees of sums in a separable field extension. Let F be a field and let y be a n algebraic element in a field extension of F . We write degy for the degree of y over F. Assume that

a , p are two separable elements in a field extension of F. P u t m

=

dega, n = degp and suppose that ( m , n )= 1. Then we know, from Proposition 3.11 that

( F ( a , j 3 ): F ) = mn On the other hand, since F ( a , p ) / F is a finite separable extension, it follows from Corollary 2.2.21, that F(aP) = F ( y )

for some -y E F ( a ,b )

In fact, the pioof of Proposition 2.2.19, produces (for infinite F ) an element y of the form for some X E F

y=a+xp

It follows t h a t , if F is infinite, then deg(a i- XP) = mn

for some X E F

F

is

# 0. This section is devoted to providing a solution

of

It is therefore natural t o enquire about circumstances under which the choice of X completely arbitrary, as long as X

E

this problem, by applying finite Galois theory. The following theorem is a reformulated version of a result due to Isaacs (1970).

8.1. Theorem. Let F be a n arbitrary field and let a,P be two separable elements

in a field e x t e n s i o n of F of coprime degrees m and n , respectively. A s s u m e that either ctiarF

-z 0

or

charF = p > 0 and at least o n e of the following conditions holds:

( i ) p d rnn und p (ii) p

(iii) p

> min(m,n)

m n a n d m or n is a p r i m e power

+ n and a n y p r i m e divisor of n is > m

Thcn deg(a

+ Xp) = mn

for all 0 f X t F

Proof. For the sake of clarity, we carry out the proof in a number of steps. S t e p 1 . O u r a i m is t o fiz n o t a t i o n and to record s o m e general observations.

CHAPTER 6

364

Let the field E be obtained from F by adjoining all conjugates a l , . . . ,a , and

of a and

p , respectively.

P I , . . . Pn

Then E / F is a finite Galois extension and we put

G = Gal(E/F) Then, by Lemma 7.8, the group G acts transitively on the sets A = { a l , ...,a,> and B = {PI,. . .,Pn}

If H and K are the stabilizers in G of a and (G:H)=m

P , respectively, then

and

(G:K) = n

Moreover, since ( m ,n) = 1 a standard group-theoretic argument yields (G : ( H fl K ) ) = mn

It therefore follows that the size of the orbit of

(a,P) E A

(2)

x B under the induced action

of G on A x B is mn. Thus G is transitive on A x B and so each ai (1 5 z 5 m, 1 5 j

+Pj

is a conjugate of a

+

5 n). Bearing in mind that

it follows from (1) and (2) that

Therefore ( K : ( H n K ) ) = m and thus

K acts transitively on A

(4)

Denote by W1 and W , the F-linear spans of A and B , respectively, and let

V = W1+ W , and U = W1 n Wz Then G acts as a group of linear transformations of V and Wl,W2, U are G-invariant subspaces. Replacing

by

Xp, if necessary, it suffices t o verify t h a t d e g ( a + p)

= mn.

365

DEGREES OF SUMS I N A SEPARABLE F I E L D E X T E N S I O N

We argue by induction on \GI, i.e. on the F-dimension of the normal closure of F ( ( x , p ) . If /GI = 1, then

a,P E F and there is nothing t o prove. By induction, we may therefore

assume that if a,P are of coprime degrees mo and no, respectively, over an intermediate field L

# F , and

(xO+po over

(i), (ii) and (iii) hold with respect t o mo, no, then mono is the degree of

L. From now on, we assume that d e g ( a + p ) < mn and derive a contradiction.

+ p,

Applying (3), we see that not all ai

a where a

#

ffk

+ p = (Yk + ps or

are distinct and thus

for some k E { I , . . . , m} and

E { I , ..., n }

(5)

p # ps. Then

which shows t,hat U

# 0.

Denote by M the kernel of the action of G on V and by

F

s

C F ( a , P ) & L and L / F

L the fixed field of M .

Then

is Galois, so E = L and thus M = 1. Therefore

G

acts faithfully on V

(6)

Step 2. The aim of this step is t o prove that there i s no proper subgroup Go of G which

acts so that the orbits A0 and Bo of a and p under Go satisfy

where mo = lAol and n o = IBol. Assume the contrary. Then (mo,no) = 1 and since, by ( 5 ) , a number of elements in the Go-orbit of a

+ fl

= ak

+ P s , the

+ /3 is < mono. Now Go = G a l ( E / L ) , where 1, is

the fixed field of G o , and mO,nO are the degrees of a and

p, respectively, over L. Because

jGol < IGi, it follows that none of (i), (ii) or (iii) holds with respect to mo,no. But then

none of ( i ) , (ii) or (iii) holds with respect t o m , n , a contradiction.

Step 3. Let N a G be the kernel of the action of G on A . Here we show that (u

np=u+'Y

E U , 7 E Wz is fixed by G)

(7)

Observe that G permutes the sets of cosets of U in W1 and is transitive on the set of those cosets which contain elements of A . All of these, therefore, contain equal number of elements of A . Now a , ( x k E

(x

+ U and if A0

= A n (cr

+U)

LtIAPTER 6

366

then lAol I m. Denote by S o the stabilizer of the coset n

+ U in G.

Clearly, H

Go and

therefore Go is transitive on B . We claim that Go is transitive on Ao. Indeed, if ni E

no,

then as = ai for some g E G. Accordingly, ( a + u)g = ai

+ u= a + u

which implies t h a t g E Go. This establishes transitivity and, by Step 2, we must have Go = G. Hence G stabilizes n

and thus

pj =

Therefore

np

P + uj

=u

+ U and so A C a + U . By a similar argument, B 5 /3 + U

for some

uj

E U . Summing over

+ y, where u = - C uJ E

U and y =

proves (7) and (8),because N fixes all elements of

Step 4 . W e now prove that if charF transitive o n A and such that charF

1 mn,

W1

C W1, we have

pj

E W , is fixed by G. This

2 U.

then there i s no subgroup L

K which is

+ ILI.

Assume the contrary. Owing to (7), we have nP = u fixed by G. As U

f13 E B,we derive

u =

+ 7 for some u E U and some y E W 2

C z l &a; for some A,

E F . Now if z E I,

C X,

then we have

lrn

1

P = P " = -CA,a;+-? n n

(9)

i= 1

Put 6 =

ai and observe t h a t , since L is transitive on A , we have

Summing (9) over L , we deduce t h a t

Since y and 6 are fixed by G and c h a r F desired contradiction since

1 ILI, we

infer t h a t

P is fixed

by G. This is a

p # p8.

Step 5. Here we complete the proof b y deriving a final contradiction Denote by N the kernel of the action of G on A . If c h a r F + n ,then by (8), N fixes

p and

thus N = I, by (6). Therefore G is isomorphic to a subgroup of Sm, so /GI divides m!.

DEGREES OF SUMS IN A SEPARABLE F I E L D EXTENSION

367

Hence all prime divisors of \GI (in particular, those of n ) are 5 m. Because n t h a t /3

> 1 [recall

# ps),this certainly excludes (iii).

Because (iii) is excluded, we have c h a r F and B in the above argument, we obtain IGI

5 m i n ( m , n ) . Therefore, if

charF = p

+ mn, by hypothesis. I

By interchanging A

n! and thus all prime divisors of /GI are

+

> 0 and (i) holds or if c h a r F = 0, then charF iGi.

However, this contradicts (4) and Step 4. Thus we must have c h a r F = p

>

0 and ( i i ) .

We may assume that m = q e , q prime, and let Q be a Sylow q-subgroup of K . Then

( K : ( K n H ) ) = q e so K = ( K n H ) Q and it follows from (4) t h a t Q is transitive on A .

+

Because c h a r F q, by hypothesis, we obtain a final contradiction by applying Step 4. m Our next observation shows that t o establish the best posssible improvement of the result above, with conditions given in terms of m, n and charF, it suffices t o consider only group representations. 8.2. Lemma. Let V be a finite-dimensional vector space over a field K and let a

f i n i t e group G act faithfully on V as linear transformations of V . Suppose t h a t u , u t V are permuted by G i n t o orbits of sizes m and n , respectively, a n d t h a t u -t v lies in a n orbit of size k . T h e n there ezists a f i n i t e Galois e z t e n s i o n E I F with K C Y , ~E

CF

and elements

E such that degcx = m, degp = n and deg(cx + 0)= k

Proof. Put R = K [ X 1 , .. . , Xt]where t = dimK V and let E be the quotient field or l l . We may identify V with the K-linear span of

XI,. . . ,X t

in R . Then G acts faithfully

as a group of K-automorphisms of R and hence of E . Denote by F fixed field of G in E (so K

C F ) and

by a and

p the elements of E corresponding t o

u and

u. By Proposition

3.5, E / F is a finite Galois extension and G is the Galois group of E / F . Because for any

7 i R , degy is the size of the orbit of y under G , the result follows. m

We cnd this section by giving some limitations on possible improvements of Theoreni 8.1. The following two examples in conjunction with Lemma 8.2 show that t h e conclusion of Theorem 8.1 fails in each of the following cases:

( i ) c h a r F = 2, m

:F

3, and n = 4

( i i ) c h a r F = 3 , m = 3, and n = 4 8.3. Example. [Isaacs(l970)) Let G denote the alternatzng group of degree \ ‘ - br a i-dimen5ronal vector i p a r ~01wr

IF2

and let G permute a b a s i , (wl.-.g 21

4 I,P’ iri i t r

CHAPTER 6

368

natural way. P u t

V, = (0,w

+z +y +z}

T h e n G acts faithfully on V and we p u t u = w

and

V = V*/Vo

+ x + VOand v = w + Vo.

T h e u and v are

permuted by G into orbits of sizes 3 and 4, respectively. But t h e size of t h e orbit of u

+v

is 4 r f 12.

8.4. Example. (Isaacs(lQ70)). Let V be a 4-dimensional vector space over F3, with

basis {w,x,y,z}. Let G be the group generated b y the elements u1,uz,u3E G L ( V ) whose matrices are

u1 =

1

1

0

0

0

0

0

1

; ; p ;].=

[; ; i] jl

[.

1

0

,03=

0

0

0

0

0

1

0

; ; p]

T h e n G is the direct product of the subgroups of order 6 and < u s > of order 2. Put u = w and v = y. T h e n the orbit of u under G is {w,w

+ x,w

-

+ x,z , z + x}. However, the orbit of u + v is {w + + y + x , w + y z,w + z,w+ + z,w+ z

x}, and the orbit of

v under G i s {y, y

-

Y,W

2

-

x}

which has six elements.

9. Galois cohomology.

Our aim in this section is t o establish some classical results pertaining to cohomology groups of Galois groups. For convenience, we first recall the following piece of information. Let G be a group. By a G-module one understands a n abelian group A on which G operates such t h a t

(i) la = a for all a E A (ii) g ( a

+ 6 ) = g a + 96 for all g E G, a,6 E A

(iii) ( g g l ) a = g(gla)for all g,g1 E G , a E A Observe that if A is a G-module, then A is a ZLG-module via

Conversely, any given ZG-module A can be considered in the obvious way as a G-module. The fitized module of a G-module A is defined t o be the subgroup

AG = { a

€ Alga = a

for all g E G )

GALOIS COHOMOLOGY

369

It is obvious t h a t AG is t h e largest submodule of A on which G acts trivially. If A , B are G-modules, the group of all abelian group homomorphisms A

-B

is

denoted by Hom(A, B ) . The group Hom(A, B ) has a G-module structure defined as follows: if 6 E H o m ( A , B ) , then

g4 is the mapping

In particular, if X is any abelian group, we can form the G-module H o m ( Z G , X ) . .A G-module of this type is said to be co-induced. Let A be any G-module. A function f : G x G x ... x G 4 A

( n copies of G)

is called an n-cochain of G in A. An n-cochain f is said t o be normalized i f f (91, . . . ,gn) = 0

whenever any of the gi is the identity element of G. The set of all n-cochains, written

Cn(G,A ) , becomes a n abelian group under the addition of values. Extending the definition of C n ( G , A ) to the case n = 0, we put

C"(G,A) = A The formula

determines a homomorphism:

It is a standard fact t h a t d n d n - l = 0 for all n

2 1. We set

Z n ( G , A ) = Kerd, and B " ( G , A ) = Imd,-l

(n

2 1)

and refer to the elements of Z"(G,A) and B n ( G , A ) as n-cocycles and n-coboundaries, respectively. The n-th cohomology group H n ( G , A) with coeficients in A is defined by H"(G,A) = Z"(G,A)/B"(G,A)(n H o ( G , A ) = AG

2 1)

CHAPTER 6

370

It can be easily seen that the cohomology group H " ( G , A ) is unaffected if we restrict ourselves t o normalized n-cochains. For future use, we now recall the standard fact which says t h a t

H n ( G , A ) = 0 for all n

2 1 if A

is co-induced

(2)

Formula (I) reveals that a 1-cocycle is a crossed homomorphism, i.e. a map f : G

---t

A

satisfying

f(s9') = gf(9')

+ f(9)

Similarly, a 1-coboundary is a principal crossed homomorphism, i.e. a map f : G -+ A for which there exists a E A such that

In particular, if G acts trivially on A , then

H ' ( G , A ) = Hom(G,A) It follows from (1) that a 2-cocycle is a function f : G x G

while a 2-coboundary is a function f : G x G

--t

4

A satisfying

A for which there exists a map t : G

--t

A

such that

f(s1,sz)= GJlt(9Z) t(g1gz) + t(91) ~

Let E / F be a finite Galois extension with G = G a l ( E / F ) . Then the additive and multiplicative groups E+ and E' of E, respectively, are G-modules and we may consider the groups

H " ( G , E + ) and H"(G, E ' )

for all n

21

9.1. Theorem. Let EIF be any finite Galois eztension with Galois group G. Then

H " ( G , E') = O

for all n

21

Proof. Note that the G-module E+ is isomorphic to the G-module F G , where FC is the group algebra of G over F . Indeed, by Theorem 3.9, the required isomorphism is

given by

{

FG CgEG

'!I9

-

--tE+ C g E G ',g(')

G A L O I S COHOMOLOGY

371

where a E E is such t h a t { a ( a ) l l a E G } is a normal basis. Applying ( 2 ) , we are thus left to verify that

Hom(ZG, F ) Z F G

as G-modules

For each X E B o m ( Z G , F ) and each g E G , put A, = X(g). Then the map

is a required isomorphism. w We now turn our attention to t,he study of H n ( G , E * ) . Here the situation is more complicated: the first cohomology is always trivial, but in contrast t o Theorem 9.1, the groups E I " ( G , E * ) for n

> 1 are not generally trivial.

9.2. Theorem. Let E I F be a finite Galois extension with Galois group G . Then

H 1 ( G , E ' )= 1

Proof. Let f

:

G

-+

E'

be a crossed homomorphism. For X E E' we put

Because of the linear independence of the automorphisms a (Corollary 3.8), we may choose X E E' such that a

# 0.

We then obtain for

7

E G,

which implies that

Accordingly, f t B ' ( G , A ) and so H ' ( G , E * ) = 1. rn As a preliminary t o the next result, we prove 9.3. Lemma. Let E I F be a finite Galois extension with Gaioas group G a n d , j o r

any a E Z Z ( G E , " ) , let

be the cohomology class of a . Given c E H 2 ( G ,E * ) , say

define

G a

dF*/NB/F(E-) ( ~ T E a G( 7 , a ) ) N E / F ( E * )

c =

a,

372

CHAPTER 6

Then c' is a well-defined homomorphism and the map

is a homomorphism.

Proof. Given a: E Z 2 ( G , E * ) put ,

It is easy to verify that, for all al,a2E G,we have

which means that Cr maps G to F * . If ,B E Z2(G,E ' ) is such that

p = E , say a:P-l

B 2 ( G , E * )(where (69) (a1,az)= al(g(az))g(alaz)-'g(al) for some g : G

This shows that the map c' is well-defined.

By the multiplicative version of ( 3 ) , we have

whence

This proves that c' is a homomorphism. Finally, fix c = a , d =

p in H 2 ( G , E * ) .Then, for any a E G,

+

= 6g E

E * ) , then

GALOIS COHOMOLOGY

proving t h a t the map c

H

c'

373

is a homorphism.

As an application of the lemma above, we prove 9.4. Theorem. Let E / F be a finite cyclic eztension with Galois group G. Then the

{

H*(G,E*) c

4

Hom(G,F*/NE,F(E'))

H C '

is an isomorphism.

Proof. Owing t o Lemma 9.3, the given map is a homomorphism. Assume that c' = 1. Then, if

c = 5 , we

can find an element g E C 1 ( G , E * )such t h a t

Because a is cohomologous t o a normalized cocycle, we may assume that a is normalized, in which case g(1) = 1. Then a6g-l is still normalized and, moreover, we have a6g-'(u) = 1

for all

(T

E G. We may therefore assume that a is normalized and t h a t

(Y(u) =

1 for all

u E G.

Let o be a fixed generator of G. Let us define a n element t E C 1 ( G , E * )by setting t ( 1 ) = 1 = t ( o ) and, for 1 < a


= ( E : F),

n

a- 1

t((TQ)=

a(a',a)

t=1

Then a direct computation shows that 6 t = a - l , proving that the map c

-

c'

is injective.

Finally, let X be any element in F * . We define

a(a=,O*)= 1

fora+b
X

fora+b>n

and a(oQ,(Tb) =

Then it is easy to verify that a is a cocycle and that Z(o) = A. Because a homomorphism of G into F ' / N E , F ( E * ) is determined by the image of is surjective, as required.

this shows that the map c

(T,

-+

c'

This Page Intentionally Left Blank

375

7 Abelian extensions

This chapter is devoted t o the study of abelian extensions, i.e. Galois extensions with abelian Galois groups. Any finite abelian extension is a composite of cyclic extensions of prime power degrees. For this reason, we concentrate on cyclic extensions E / F of degree p n , n 2 1, where p is a fixed prime. In case F contains all p"-th roots of unity of an algebraic closure of F , we provide a n explicit description of all cyclic extensions of degree pn.

We then concentrate on the case where c h a r F = p and investigate in detail cyclic

extensions of degree p". The next section is devoted t o the study of abelian pextensions by means of Witt vectors. After presenting Kummer theory, we finally exhibit a bijective correspondence between the subgroups of the character group of G a l ( E / F )

( E / F is a

Galois extension) and all abelian subextensions of E/F. This is achieved by applying infinite Galois theory and certain properties of character groups of profinite groups.

1. W i t t vectors

In this section, we introduce a n ingenious ring of vectors discovered by Witt(1936). This ring will play an important role in our subsequent investigations and it is therefore examined in detail. Throughout p denotes a fixed rational prime. To begin with, we introduce the polynomial ring

(ilj , k = 0,1,2

[xa,yj,zk]

A

and denote by Am the direct product of

. . .)

/IN/copies of A. The ring Am consists of

all

infinite "vectors"

z=

. . , zn-l, z,

(20,21,.

. . .)

(2,E

A)

with addition and multiplication performed componentwise. We wish to define another

CHAPTER 7

376

In what follows, we write

ring structure on the set AN.

7r(z) = (z;,z;,

. . . , Zp,-l,.

. .)

T h e elements zo, zl,. . . ,znP1 of A define elements z(O),z(l), . . .,

dn-l),. . .

of A by the following recursion formulas:

z(O)

where z = (zo,

. .,1,-1,.

21,.

z(O)

T h e elements

= z o and z('+') = (7r(z))(')

z(O),

..).

= zo and

z('),

It follows from (1) t h a t

x('+')

. . . ,z('+'),

+ paflz,+l

= z$+'

. . . of

+ pz;' + . . . + p ' + l z , + 1

A define elements

(2)

. . . ,q - 1 , . .. of A by

ZO,Z~,

the inverse recursion formulas: xo = z ( O ) and z,+~= ( l / p ' + ' ) ( z ( ' + ' )

-

zo p'+'

-

pz:'

-

.. . - p ' z f )

(3)

For convenience, we write

..

[z(O),z(l),. %z(n--l), . . .] = (zo,z1,.

where zO,x1,. . . , ~

..

~ - 1 , . are

defined in terms of

z(O),

. .,zn-l,. .. ) z ( l ) ,. . . , ~ ( ~ 1. ., by . (3) CYe have

now accumulated all the information necessary t o define a new ring structure on the set A N . Namely given

x = ( 5 0 ,X I , . we define z

. . ,zn-l,.

. .), y = (YO, y i , . . . ,~

. .) E AN

~ - 1 , .

+ y and z y as follows:

From now on, we write W ( A )for the set AN endowed with the addition and multiplication above.

WITT VECTORS

377

It is a n immediate consequence of the definition of W ( A ) that it is a commutative ring of characteristic 0 which is isomorphic t o A m under t h e mapping

Owing t o (2), for any a E A , we have ( a , O , . . .o,. . .) = [ a ,up,. . . , u p

.L-

1

, . . .]

which shows, by putting a = 0 and a = 1, that the zero and identity elements of W ( A )are ( O , O , . . . ,O,.. .) and ( l , O , . . . , O , . . .)

respectively. Let us now have a closer look a t the elements X

+Y

and X Y , where

X = ( X o ,X I , . . . , X n - l , . . .) and Y = ( Y o , Y I , .. . Yn-1,. . .) Invoking (2) and ( 3 ) , we derive

In general, it is obvious from the definitions t h a t ( X

+ Y ) , and

( X U ) , are polynomials

with rational coefficients in X o I Y o , X 1 , Y 1 ,. . . , X , , Y , . It is also clear that

where fi is a polynomial in the indicated variables. O u r first aim is t o prove that both ( X + Y ) , and ( X U ) , are polynomials in X o , Yo,X 1 , Y1,.

. . , X , , Y, with integer coefficients. This will be achieved with the aid of the following

preliminary result. 1.1. Lemma. Let j

2

1, 0 5 k 5 m

-

1 and !et z,y E W ( A ) be such that

CHAPTER 7

378

Then the following conditions are equivalent: (i) s;=y;(mod#Z[Xo,Yo

,...,Xm-l,Ym-l])

for a l l O < i < k

(ii) z f i ) = y ( ' ) ( m o d p ' + J ~ [ ~ o , ~.o. ,, ~. ~ - ~ 1~ -,

1 1 )for

aft

o
Proof. Abbreviate Z[Xo,Yo,. . . , X m - l , Y m - l ] by Z[Xr,Ys]. Since

z(O)

= 50 and

y(O) = yo, the result is clear for k = 0. We now argue by induction on k. So assume that (i) and (ii) hold for all i E ( 0 , . . ., k

-

I}. We must show that

if and only if

s f kz ) y(k)(modpk'jZ[X,,Ys]) It is obvious t h a t

Xk

z yk(mod p ' Z [ X , , Y,] if and only if p k s z pky(mod p k f j Z [ X r , Y s ] )

Invoking (l), it therefore suffices t o show that

holds under the induction hypothesis. We have

Because

('I)

O(mod p), 1

5 t 5 p - 1, this gives

5:

= yp(mod $+'),

Therefore the induction on k applies to ~ ( sand ) ~ ( y to ) give

So the lemma is true. We are now in a position t o prove 1.2. Theorem. For any i E {0,1,2,.

. .},

( X + Y)* and (XU); are in Z[Xo,Yo,. . .,Xi,Yi] Proof. Let X o Y denote X Y or X + Y. Then

( X 0 Y)o E Z[Xr,Us]

0

5

T,S

5i

0

5 i 5 k - 1.

WITT VECTORS

by virtue of (4). Suppose that ( X o Y

) k

379

E ZiXr,YZkT,] for all 0 5 k

5i

1. Because, by ( l ) ,

~

we have

p'(X

0

Y ) , = ( X 0 Y ) ( , )- (.(X

0

Y))(c-1)

(6)

it suffices to show t h a t the right-hand side is in p * Z [ X r , Y , ] .Again, by (l),

and

Y ( ' )= s(Y)("-')(mod p ' Z [ X , , Y , ] ) which ensures that

(

3

(.(X)

r)(t)

o x ( Y ) ) ( * - l ) ( m o dp'Z\X,,Y,])

(7)

Now, for any f E Z [ X , , Y , ] , we have

f(X0, Yo,. . . ,X , , Y,)P = f(X,P,Y f , . . .,xp,Y,") (mod P Z [ X r ,Ysl) Applying (8) to f

(.(X

0

=

(XoY)k, 0 5 k 5 i (.(XI

Y ) ) k

-

(8)

1, we obtain

a ( Y ) ) k (mod p Z [ X r , y s ] )

(0 5 k

5 i - 1)

(9)

It therefore follows from (9) and Lemma 1.1 that

( ~ o( Yx) ) ( ' - ' )= (.(X) Invoking (7) and

(lo), we

o

r ( Y ) ) ( * - '(mod ) p'Z[Xr,Ys])

conclude that the right-hand side of (6) is in p ' Z [ X , , Y , ] , as

required. rn

For any given i

2 0, we

put

( X + Y ) , = az(Xr,Ys), (XY), = m,(Xr,Ys) Then, by Theorem 1,2, a, ( X , , Y,) and m, ( X,, Ys)are in Z[ X O, YO . . . ,X,, Y,] and, for any z, y E W ( A ) ,

(10)

CHAPTER 7

380

Let R be any commutative ring and let

be the polynomials obtained from a ; ( X r , Y s ) and m;(Xr, Y s ) ,respectively, by replacing each coefficient c E Z by the corresponding

c

= c

. 1~

in the prime subring of R. We

have now accumulated all the information necessary t o define and investigate the Witt

ring W ( R )of R and the n - t h Witt ring Wn(R)of R , respectively. The elements of W ( R )are infinite sequences

with equality defined as usual. If a = ( a , ) ,b = ( b i ) are in W ( R ) ,then we define addition and multiplication in W ( R )by

The elements of W,(R) are finite sequences

with equality defined as usual, and addition and multiplication defined by (12). Let us introduce the following maps:

1.3. Theorem. With the notation above, the following holds: (i) W ( R ) and W,(R) are commutative rings, and u,,, f m n are surjective r f n g homomor-

phisms. The elements (0,. . . ,0, . . .) and ( l , O , . . . , O , . . .) are zero and identity elements of K’(R),while ( 0 , . . . , 0 ) and (1,0,. . . ,0) are zero and identity elements of W,,(R) (ii) { W , ( R ) ,

fmn

1 n , n E IN,m 5 n } is

a projective system of rings and the map

381

WITT VECTORS

i s a ring isomorphism. (iii) For a n y subring S of R , W ( S )and Wn(S) are identifiable with the subrings of W ( R )

and W , ( R ) , respectively, consisting of vectors whose coordinates are in S . (iv) Keral 2 Keroz 3 . . . 2 Kera,

I). . .

and

nz,

X e r a ; = 0.

Proof. (i) Consider the subring B = Z [ X t , Y , , Z k ] of A . If z = ( z z ) , y = ( y l ) and z = (zt)

are three elements of W ( R ) ,then we have a homomorphism

If X = ( X ; ) , p = ( p ; ) E W ( B ) ,then by (lo),

Thus the mapping W ( B )---t W ( R ) ,( a , ) and satisfies X

z,

++

Y

++

y, Z

H

H

( $ ( a , ) )preserves addition and multiplication,

z , where

X = (X,), Y = ( Y t ) and Z = (2,).

Then the associative, commutative, and distributive laws of addition and multiplication in W ( B )give the same rules for the elements z,y,z of W ( R ) . This proves that W ( R )is a commutative ring. The image of 0 = ( 0 , 0 , . . . , 0 ,

. . .) and 1 = ( 1 , 0 , . . . , O , . . .) under o u r

homomorphism are (0,0, . . . , 0 , . . .) and ( l , O , . . . , O , . . .), respectively. A similar argument shows that W , ( R ) is a commutative ring with specified zero and identity elements. The maps on,

fmn

are obviously surjective, and are homomorphisms by virtue of (12).

(ii) The first assertion is a consequence of definition of fmn. fmn

o a, = am for

Moreover, we also have

all n,m E IN,m 5 n. Hence, by Proposition 6.2.4, the given map is a

homomorphism which is obviously injective. If (y,) E lirnW,(R), then c

forces

yn = an(.) where z,

=

for all n

( Y ~ ) as ~ , required

(iii) This is obvious

(iv) By definition, Kera, = { ( a i ) E W(R)lao = a l = . . . = a,-l implies the required assertion. rn

= 0 } which obviously

382

CHAPTER 7

For future use, we need t o introduce some other important maps. If z = (z,) E W ( R ) , define

.. ,zn-l,. ..)

s(z) = (0, zo,.

We shall refer t o the map s :W(R)

--f

W(R)

as the shift. If a = (ao,. . . , a n - l ) E W,(R), then we define

and again refer t o s, as the shift. Given a E R , we put r(u) = (u,O,. . . , O , . . .), r,(a) = ( a , ~. ., . ,o)

This provides us with maps r :R

W ( R )and r, : R

--+

1.4. Theorem. (i) The maps

and

S,S,

v,

+ W,(R)

preserve addition

(ii) For z = (zo,. . . ,zn-l) E W n ( R ) ,y = ( ~ 0 ~ .~. ,1z n, -.l , . . .) E W ( R ) and a E R , the

following holds: (a) z = C : Z ~ s i ( r n ( z t ) ) j Y = CEO s z ( r ( ~ x ) )

(b) r,(a)z = (

n--l

U ~ O , U ~. .Z. ,~a,p

~

~

-

1

)

WITT VECTORS

383

(c) r ( a ) y = (axo, apz1,. . . , a P n - ' x n - l , . . .1 (iii) r ( a b ) = r(a)r(b) and rn(ab) = rn(a)rn(b)

forall a , b E R , n E IN

(iv) If z , y E W ( R ) are such that xi = 0 or yi = 0 for all i

2 1,

then

Proof. (i) Note first that s n ( x o , z l , .. . , % - I )

= o n s ( z o , x l , .. . ,

. . .)

and un(xo,51,.. . , x n - l ) = o n + l s ( x o , x 1 , .

. . j ~ n - 1 , .. .)

Bearing in mind that nn and on+l preserve addition (Theorem 1.3(i)), we need only show that s preserves addition. To this end, we may harmlessly assume that R = A in which case s ( x ) = [ O , p r ( O ) ,. . . , p z ( " - Z )

,...I

and thus s preserves addition. (ii) Taking into account that on(y) = x and a , ( r ( a ) y) = rn(a)x, it suffices to verify the

formulas for x. Again, we may assume that R = -4, in which case r n ( a ) = [a,aP,.. . , a p " - ' ] and thus s ' , ( r n ( a ) ) = [o,. . . , O , p ' a , . . . ,piaPn-l-'1

(0

525n

Accordingly,

c

n-1

i=O

i=O

n- 1

Sf,(Tn(Z,))= - y [ o , .. . , 0 , P i Z i , . . . ,pix:"-'-']

and the latter sum is [x(O),. . . , z

( ~ - ~=) x, ]

by (2). Also

~

1)

CHAPTER 7

384

as required (iii) Apply (b) and (c) for z = r n ( b ) and y = r ( b ) , respectively (iv) and (v). Because (iv) is a consequence of (v) by applying u,, we need only verify (v). BY (i), n-1

n- 1

i=O

i=O

n-1

i=O

since xi = 0 or y;

=

0 for 0

5 i 5 n - 1.

Our final two maps are defined as follows:

1.5. Lemma. (i) Zf x , y E W ( R ) ,t h e n for all i

(ii) If x , y E W n ( R ) ,t h e n for all i E { O , I,. . . , n

~

20

I}

Proof. This follows by a n application of (8). w We now examine a very important special case, namely the case where charR = p . The prime subring of R is then the field IF, of p elements. 1.6. Theorem. Let R be a n arbitrary commutative ring of characteristic p and let

n

2 1.

(i) T h e m a p s

F

and rnare ring homomorphisms.

385

WITT VECTORS

I . .

(11)

p'x = s L ( ~ f ( z ) ) and p'y = s'(s(y))

(iii) Wn(lFp)

E W ( R ) ,z'

for all z E W,(R),y

21

Z / p n Z is the prime subring of W n ( R ) ,the Characteristic of W n ( R )is p"

and the characteristic of W ( R )is 0. (iv) For any z E W,(R) and y E Wn+.l(R),

(v) If R is a field of characteristic p , then

Proof. (i) This is a direct consequence of Lemma 1.5. ( i i ) Setting X = ( X o , X 1 , . . . ,Xn-1,. . .) E " ( A ) , we have

Therefore

by Lemma 1.1, which clearly gives p y = s ~ ( y ) . Iterating the latter, we obtain p'y = s'(T'(Y))

for all i

2

1. The corresponding assertion for z is now obtained by applying o n .

(iii) The prime subring of W,(R) is additively generated by 1 = (1,0,. .. ,O). Bearing in mind that p n . 1 = s:(l)

= 0 and pm . 1 = sT(1) f 0 for rn < n

we deduce that 1 has additive order p". Thus Z/pnZ is isomorphic t o the prime subring of Wn(R) and charW,(R) = p". Note also that W,(F,)

contains the prime subring of

W,(R) and has pn elements. Hence Wn(IFp) E Z / p n i z is the prime subring of W n ( R ) . Moreover, the above shows that the additive order of (1,0,. . . , O , . . .) in W ( R ) is infinite which means that charW(R) = 0. (iv) Let X = ( X o , X 1 , . . . , Xn-l) E Wn(A) and Y = (Yo,Y,,. . . ,Yn)E W n + l ( A ) .We need

only verify that

CHAPTER 7

386

To this end, put

w;

= t ; (mod p Z [ X ,Y ] )

(0

5 i 5 n)

Owing t o Lemma 1.1,this is equivalent t o t(')(mod p ' + ' Z [ X , Y ] ) The above obviously holds for i = 0, because p ~ ( ~ - ' )1, 5 i 5 n

do)=

0 = t(').

Since (un(z)))(;) =

+ 1, by (2), it follows t h a t

Bearing in mind t h a t

Y ( i )= (7rn+l(Y))(i-1) +p'Y; we obtain the congruences

thus proving (iv). (v) Suppose t h a t R is a field of characteristic p . Then

Denote by

fz

an algebraic closure of R. Then W,(R) 2 Wn(R) and x, is bijective on

Wn(R). Therefore, letting u = xi(.)

and u = 7ra(y), we derive

387

M I T T VECTORS

thus completing the proof. 1.7. Corollary. W(F,) is isomorphic to the ring of p-adic integers.

Proof. This is a direct consequence of Theorems 1.3(ii) and 1.6(iii). 1.8. Theorem. Let R be a commutative ring of characteristic p and let n

2 1.

Then

the map 4 R ( z ~ > s l > . . . > z n -++ - l )zo

i s a surjective ring homomorphism whose kernel I is a nilpotent ideal (in f a c t , I" = 0).

Proof. Because W1(R) = R , our map is just

fln

and therefore is a surjective ring ho-

momorphism. Note t h a t the kernel 1 consists of all elements of the form (0, ZO,

21,.

. . ,~

~

-

and hence I = sn(Wn(R)).We claim that

If sustained, it will follow that I"

C s:(Wn(R))

= 0, as required.

To substantiate the claim, we first observe that f n , n + l v n = s,.

Invoking Theorem

1.6(iv), we therefore obtain

Because f,,,+,(y)

can be taken t o be an arbitrary element z in W,(R) and in view of

fn,n+lrn+l = n

f ","+I,

tnd therefore I'

C sn(I), proving

'or some t

2 2. Then

1'"

this gives the relation sn(z)z = s,(zn,,(z)). Then

(13) for k = 2 . Now assume, by induction, that

C_ lsb-i(Z).

Furthermore, if z , z E W,(R), then by (14)

2

)

CHAPTER 7

388

s k ( I ) , as required. w

which shows t h a t It+’

In order to provide an application of the result above, we record the following simple observation. 1.9. Lemma. Let J ( S ) be the Jacobson radical of a n arbitrary r i n g S . If 1 is an

ideal of S with I & J ( S ) , then u E S is a unit of S if and only i f u Proof. It is plain that if u is a unit of S , then u assume that u E S is such that u

+I

+I

+I

is a unit

01 S / I .

is a unit of S / I . Conversely,

is a unit of S / I . Then uu = 1

+ z and uu = 1 + y

for some z , y E I. Hence

s =I

+uus = I

+uus

and so S = U V S = u u S by Nakayama’s lemma. Now choose w E S such t h a t u v w = 1. Because 1 = ( u v ) w = w + z w , we have w = 1 - z w also, so w itself has a right inverse. Thus uu

E U ( S ) and by a similar argument uu E U ( S ) ,and therefore u E U ( S ) ,as asserted. 1.10. Corollary. Let R be a commutative ring of characteristic p . Then, for any

n

2 I , a n element

. . ,zn-l)

(ZO,Z~,.

is a unit of W n ( R )i f and only i f

20

is a unit of R.

Proof. Apply Theorem 1.8 and Lemma 1.9. w 1.11. Corollary. Let R be a commutative ring of characteristic p . Then a n element z = ( z o , z ~ ., .) . is a unit of W ( R ) i f and only i f zo is a unit of R.

Proof. Owing to Theorem 1.3(ii), z is a unit of W f R ) if and only if for all n

2

1,

un(z) = (zo,z1,. . . ,zn-l) is a unit of W,(R). Now apply Corollary 1-10. We close by proving the following result. 1.12. Theorem. Let R be a field of characteristic p and let

I = { ( 0 , ~ 1 , ~...) 2 , E W ( R ) l z ; ER } (i) W ( R )is an integral domain of characteristic 0 with unique mazimal ideal I and with

W(R)/I

R.

(ii) If R is perfect, then each nonzero element of W ( R ) can be uniquely written i n the

f o r m p k c with k Proof.

2 0 and c 4 I .

That I is a unique maximal ideal of W ( R ) follows from Corollary 1.11.

Because I is the kernel of the surjective ring homomorphism u1

: W ( R )+ W 1 ( R )= R

389

CYCLIC EXTENSIONS

we also have W ( R ) / I

R. Moreover, charW(R) = 0 by Theorem 1.6(iii). To prove the

remaining assertions, we may assume that R is perfect. Let z

= ( z g , x i , . . .)

integer with

Xk

# 0.

be a nonzero element of W ( R ) and let k

2

0 be the smallest

Because R is perfect, we may write

x = ( 0 , . . . ,o,yopk,gk,. . .) Therefore, by Theorem 1.6(ii), z = pkc where c = ( y ~ , y i...) ,

6I.

If y = p t d , d $ 1 is

another nonzero element of W ( R ) , then zy = p k + t ~ d# 0. Hence W ( H ) is an integral domain. Finally, if x = pkc = p f d with c,d

$ I and k 2 t, then pt(pkPtc - d )

d = p k L t c which is possible only in the case

k

= 0. Thus

= t and hence d = c.

2. Cyclic extensions

Let E / F be a field extension. Let us recall that E / F is called cyclic (abelian) if it is Galois with cyclic (abelian) Galois group. Our aim is t o investigate finite cyclic extensions of a given field F . It is a consequence of Proposition 6.6.9 t h a t any finite abelian extension is a composite of cyclic extensions of prime power degrees. For this reason, we concentrate

on cyclic extensions E/F of degree p n , n 2 1, where p is a fixed prime. The problem of explicit description of all such extensions is a very difficult one. However, in the case F contains all pn-th roots of unity of a n algebraic closure of F , a complete solution will he presented below. As one would expect, the cases where p=charF and p be treated separately. If p

# c h a r F , t h e corresponding result

# c h a r F should

is a special case of Theorem

2.4. The classical case where p = c h a r F and n = 1 is given by Theorem 2.7, while its

generalization to a n arbitrary n constitutes the content of Theorem 2.10. We begin by recording the following two classical results. 2.1. Theorem. (Hilbert's Theorem 90). Let E / F be cyclic of degree n with Galois

group G generated by

Then, for any z E E,

0.

NE/F(z)

=

f o r some nonzero y in E .

Proof. Given y E E , we have

1 if and only :fz = ycr(y-')

CHAPTER 7

390

Accordingly, if z = yu(y)-' for some y

# 0 in E , then

Conversely, suppose t h a t N E / F ( z )= 1. Define the map f : E

4

E by

f = 1 + zu + zu(z)u2+ . . . + za(5). . . u n - - 2 ( z ) d - ' (1 is the identity map). Owing t o Proposition 6.3.7,f is not identically zero. Thus there

exists X E E such that the element y=

+ zu(X) + zu(z)u2(X)+ . . . + zu(z)..

.un-2(z)an-'(X)

is not equal t o zero. Applying the assumption that N E / F ( z )= 1, we obtain y = zo(y).

Thus z = ya(y-')

as asserted.

H

2.2. Theorem. (Hilbert's Theorem 90, Additive Form). Let E / F be a c y c l i c ezten-

sion of degree n with Galois group generated b y u . Then, for a n y z E E ,

T T E / F ( Z=) 0 if and only if z = y

-

u(y)

f o r some y E E . Proof. Given y E E , we have n- 1

Therefore, if z

=

y

-

n-1

a ( y ) for some y E E , then

Conversely, suppose that T T E I F ( Z=) 0. Because E / F is Galois, it is separable, hence

T T E / F ( X#) 0 for some X E E (Theorem 2.4.8). Setting

we immediately derive z = y -- u(y). m We need another preliminary result. This is recorded in

CYCLIC EXTENSIONS

391

2.3. Lemma. Let F be a field and let u E A u t ( F ) be of order d .

(i) Zf z E F is such that u ( z ) = z and zd = 1 , t h e n a ( y ) = z y f o r s o m e nonzero y 6 F . (ii) Zf charF = d , t h e n ~ ( x=) z

+ 1 for some z E F

Proof. (i) By Proposition 6.3.7, we may choose X E F such that y =

xf=l

z-'ol(X)

#

0. Then d z-tut+1

u(y)=

(A) = ZY

,=1 as asserted. (ii) Choose X E F such that y = C:=,.*(A)

# 0.

Then

d

~ ( y=) y and 1 = c ~ ' ( X y - ~ ) t=l Setting x = -

d Ci=, ia'(Xy-'),

we derive

d

u(x)= -CiB'"(Xy-1)

=

i= 1

d

d

a= 1

i= 1

Cui+l(Xy-') - C(;+ l)a'+l(Xy-l)

as required. H We are now in a position t o prove 2.4. Theorem. Let n

21

be a n integer, let F be a field containing a primitive n - t h

root of unity and let E I F be a field eztension. (i) E I F i s cyclic of degree dividing n if and only if E = F ( X ) for s o m e X E E such that A" E F .

(ii) T h e m i n i m a l polynomial over F of a n y X E E with A" E F is

X d - Ad

with d l n

Proof. (i) Assume that E I F is cyclic of degree d dividing n. Then G a l ( E / F ) is generated by an e!ement, say

B, of

order d. Let 6 be a primitive d-th root of unity in F .

Owing to Lemma 2.3(i), there is a nonzero X in E such that u(X) = SX. Then the minimal

CHAPTER 7

392

polynomial of X over F has d distinct roots, namely A , a ( A ) = SX,.

. . ,od-'(X)

= Sd-'X

and hence E = F ( X ) . Moreover,

so t h a t Ad E F and A" E F .

Conversely, suppose that E = F(X) for some X E E such t h a t A" E F . Let

E

be a

primitive n-th root of unity. Then n

X"

-

A" = f l ( X

- XEi)

i= 1

which shows that E is the splitting field of X" - A "

E F [ X ] .Bearing in mind that X" - A "

has n distinct roots, we deduce t h a t E / F is Galois. We have an injective homomorphism G a l ( E / F ) A< E > sending each c E G a l ( E / F ) t o a, E < c >, where .(A)

= Xu,.

Hence

G a l ( E / F ) is a cyclic group of order dividing n , as required.

(ii) Assume t h a t X E E is such that A" E F and put K = F ( X ) . Then, by ( i ) , K / F is cyclic of degree d dividing n and Ad E F . Thus X d - Ad is the minimal polynomial of X over F . rn 2.5.

Corollary.

Let E/F be a field e z t e n s i o n of degree n a n d let F contain a

primitive n-th root of unity. T h e n E / F i s cyclic if a n d o n l y if E = F(X) for s o m e X E E such that Xn- An i s t h e m i n i m a l polynomial of X over F.

Proof. This is a direct consequence of Theorem 2.4. 2.6. Corollary. Let n be a positive integer a n d let F be a field containing all n - t h

roots of u n i t y of an algebraic closure

p

of F . If X E

i s separable over F and A" E F ,

t h e n F ( X ) / F is cyclic of degree dividing n.

Proof. If c h a r F + n ,then F contains a primitive n-th root of unity and therefore the result follows from Theorem 2.4(i). We may thus assume t h a t c h a r F = p

> 0 and n = mpt,

where ( m , p ) = 1. Setting p = Apt, we have pm E F . Because F contains a primitive m-th root of unity, F(p)/Fis cyclic of degree dividing m (hence n ) , by Theorem 2.4(i). But since X is separable over F, it follows from Proposition 2.2.6 t h a t F ( X ) = F(XP') = F ( f i ) . So the corollary is true.

393

C Y C L I C EXTENSIONS

From now on, we concentrate on cyclic extensions E / F of degree pn where F is a field of characteristic p

> 0. T h e case where n = 1 is the content of the following classical

result. 2.7. Theorem. (Artin-Schreir). Let F be a field of characteristic p

> 0 and let E / F

be a field extension. (i) E / F is cyclic of degree p if and only if E = F ( X ) for s o m e X E E such that

XP

-

X EF

and X $ F . (ii) T h e m i n i m a l polynomial over F of a n y X E E with x p-

x

(iii) For a n y given a E F , the polynomial XP

- (XP -

-

X

-a

XP -

X E F and X

$ F is

A)

is either irreducible over F or splits

i n t o p distinct linear factors over F.

Proof. Let us first observe that if a E F and a is a root of XP also a root for any i E {0,1,. . . , p (a Consequently XP - X

-

-

-

X - a, then a

+ i is

1). This is so since

+i)P

= (Yp

+ ip

=ap

+

2

a splits into p distinct linear factors over F ( a ) and thus F ( a ) / F

is Galois.

Now suppose that E = F ( X ) for some X E E such that X is a root of XP - X

- (XP -

find a unique i, E {0,1,. . . ,p

Consider the mapping

XP-X

E F and X $ F . Because

A), E / F is Galois. For any given cr E G a l ( E / F ) , we may

-

l } such that

{

Gal(E/F) u

--f

Z/pZ

~ i , + p Z

This is plainly an injective homomorphism. Because Z / p Z is cyclic of order p and E

# F,

we

deduce that G a l ( R / F ) is cyclic of order p . Hence E / F is cyclic of degree p and

XP

-

X

- (XP

-

A) is the minimal polynomial of X over F . This establishes (ii) and the

“if” part of (i) .

Suppose that E / F is cyclic of degree p . Let u be a generator of G a l ( E / F ) , so that cr has order p . By Lemma 2.3(ii), .(A)

=X

+ 1 for some X E F. Then

o(X”) = a ( X ) P =

XP

+1

CHAPTER 7

394

and U(XP -

so that

XP

A) = ( X P

+ 1) - (A + 1) = X P - x

- X E F. However, X $. F since .(A)

=

X + 1 # A. Thus E = F(X) and therefore

(i) is established. To prove (iii), it suffices t o show that if no root of XP

f ( X ) = XP

-

-

X

-

a lies in F, then

X - a is irreducible over F . Assume by way of contradiction that

with g , h E F[X] and 1 5 degg < p . Because P-1

f(X)= n ( X - a - 2 , i=O

( a is a root of f ( X ) ) ,we see that g ( X ) is a product of (X - a

i E { 0 , 1 , . . . , p - I}. If d

= degg, then the coefficient of

-

i) over certain integers

Xd-' is a s u m of terms -(a + i)

taken over precisely d integers i. Hence it is equal t o -dcu

+ j for some integer j.

But

d # 0 in F, hence a lies in F, since the coefficients of g(X) lie in F, a contradiction. m The rest of the section will be devoted t o characterizing t h e cyclic extensions E/F of degree pn where p = c h a r F

> 0. This will be achieved by analyzing conditions for the

existence of an injective homomorphism Gal(E/F)

+ Z/p"Z

This method was used in Theorem 2.7 with n = 1. Because Z / p " Z is isomorphic to the prime subring of any ring of characteristic p " , all homomorphism from G a l ( E / F ) to

Z/pnZ are identifiable with the homomorphisms Gal(E/F)

+

P,

where P,, is the additive group of the prime subring of a ring R of characteristic pn. Our ring R will be the n-th Witt ring W,(E) of E defined in the previous section. Because our n will be fixed, we shall use a simplified notation pertaining t o

write n (instead of n,,) for the homomorphism

Wn(E). Namely, we shall

395

CYCLIC EXTENSIONS

and s (instead of s,) for the shift homomorphism

For any homomorphism u : E

where z = (zo, T I , ,. . , ~

4

E , we put

~ - 1E )W,(E).

Because addition and multiplication are defined

by polynomials with coefficients in IF, and since .(a)

= a for all

a E IF,, it follows that o

is a homomorphism from W,(E) t o W,(E). Finally, if u E G a l ( E / F ) , the induced

ci

on

W , ( E ) is an automorphism leaving fixed the elements of the subring W , ( F ) . 2.8. Lemma. Let E be a field of characteristic p

> 0 and let

u

E A u t ( E ) have finite

order d . T h e n

Proof. Let X O E E be such that yo =

W,(E). Setting y =

Cid_lu'(X~)# 0 and

C,R=, u'(A),w e see t h a t

let X = ( X O , ~ , . . . , O ) E

the zeroth coordinate of y is yo. Hence, by

Corollary 1.10, y is a unit of W,(E). Since u(y) = y, we obtain

c d

1=

oi(Xy-1)

i= 1

and we may take x = -Xy-'. Given m

5 n , observe that the additive

order of ~ , - ~ (=l ()0 , . . .O,1 , 0 , . . . , 0 ) is pm

and that for all z E W,(E)

pmsn--m(z) = 0

2.9. Lemma. Let E be a field of characteristic p

> 0 and let o E A u t ( E ) have finite

order pm dividing pn. T h e n u(z)= x

+ sn-(l)

f o r some z E W , ( E )

Proof. By Lemma 2.8, we may choose X E W,(E) such that P-

CU'(X)= -1 i=l

396

CHAPTER 7

Applying

sn-'",

we have

i= 1

i=l

c Prn

=

.'(."-"(A))

i= 1 m

Setting z = Eel i u i ( s n - " ( X ) ) , it follows that

c Prn

u ( x )=

iui+l(sn-"(X))

a= 1

Prn

Prn

=

-y(i+

l)u"+'(sn-"(X))

-

i= 1

i= 1

Prn

Prn

= =yjui(s"-"(X)) j=1

=

x

Cu'+l(s"-"(X))

-

Cui(sn-m(X)) j=1

+s n - y

1) ,

since pmsn-m(X) = 0. rn We have now come t o the demonstration of our main result. 2.10. Theorem. Let F be a field of characteristic p > 0, let n

2 1 be an integer and

let E I F be an algebraic field eztension. Then E I F is cyclic of degree pm dividing p" if and only if there ezist

X o , X l ,..., Xn-l

inE

such that E = F ( X o , X i , . . . ,X n - l ) and such that the element X = (Xo, X1,. . . ,Anpi)

of W n(E) satisfies

Moreover, for each a E W n ( F ) ,there exists a cyclic eztension K I F of degree dividing pn and an element X = (XO, XI,.

. . ,X n - l )

i n W n ( K )such that

..(A)

-X =a

C Y C L I C EXTENSIONS

397

Proof. For the sake of clarity, we divide the proof into three steps. Step 1 . Here we assume that EIF is separable and prove t h e first assertion under this assumption.

Suppose that E = F(Xo,X1,.. . , & - I ) ,

W , ( E ) is such that .(A) that, for

5

-

X E W n ( F ) . Let

E

where X = (XO,XI

,..., an-^) E

be an algebraic closure of E. Observe

E Wn(E), (.(x)

holds if and only if ~ ( -zA) = z

-

-

z)

-

(.(A)

-

A) = 0

A, t h a t is, if and only if x

{ A -t ili E Z/pnZ} is the set of all roots of .(X) Wn(lFp) = Z / p " Z by Theorem 1.6(iii). Since .(A)

-

-

X - (.(A)

X E W n ( F p ) . Therefore -

A) in W n ( E ) , since

X E W , ( F ) , these roots are mapped

into themselves under G a l ( E / F ) . Therefore E is mapped into itself under G a l ( E / F ) , because E is generated over F by the coordinates of X

+ i,

i E Z / p " Z . Since E / F is

separable, we deduce t h a t EIF is Galois. We have an injective homomorphism

{

Gal(E/F) u

where .(A)

= X

+

+

Z/pniZ

H

a,

Because Z / p n Z is cyclic of order p n , it follows that G a l ( E / F ) is

2,.

cyclic of order p"' dividing p " . Conversely, suppose that E J F is cyclic of degree p"' dividing p n . Let u be a generator of G a l ( E / F ) , so that u has order p". Invoking Lemma 2.9, we obtain .(A)

=A

+ s"-(l)

for some X E Wn( E )

Then

.(.(A))

=TU(X) = T(X

+~

~ ( 1 = .(A) ) )

+ Sn--m(i)

and hence .(A)

which implies .(A)

-

-

A) = (.(A)

+ sn-"'(l))

-

(A

+ F " ' ( 1 ) ) = .(A)

X E W n ( F ) . Consequently, F ( X 0 , XI,.

. . ,X n - l ) / F

-

A

is cyclic of degree

p k dividing p " , by the first part of the proof. Because ~ " - ~ ( has l )additive order p m ,

the restriction of

E

=

5

to F ( A o , X I , . . . , X n - l )

has order pm. Accordingly, pk = pm and

F ( X 0 ,X I , . . . , A n - l ) .

Step 2. Our a i m is t o prove that, f o r each a E W n ( F ) ,there is a cyclic extension K/E' of degree dividing pn and a n element X = (Ao, XI,.

..,An-l)

in W n ( K )such that .(A)

-X

2

a.

CHAPTER 7

398

P u t (r - l ) X = .(A)

-

X and let S be the separable closure of F . It will be shown, by

induction on n, t h a t there exists X = (Ao, X i , . . . , A n - l ) with (r - l ) X = a and X i E S , 0

5 i 5 n - 1.

The required assertion will then follow by Step 1 by taking K = F ( X o , XI,.

If n

=

1 , we take Xo E S to be a root of XP

2.7(i)). Now let n

> 1. We again choose A.

-

X

and K = F(A0) (Theorem

- a0

E S such t h a t A:

. . ,An-,).

~

XO = a o . Then

for some bk E S . If K’ = F ( b 1 , . . . ,b n - l ) , then S is the separable closure of K’ and we can find, by induction, p 1 , . . . p n - l E S such that

But then

It now follows t h a t

for some A; E S , 1 5

Step 9.

F ( X o , XI,.

i 5 n - 1. Hence (r- l ) X

. . ,X n - l )

= a for X = (XO, XI,.

as required.

Completion of the proof. By Steps 1 and 2, it suffices t o show that if E =

. . ,X n - l ) , where X = (Ao, XI,. . . ,

E W,(E) is such that

then E / F is separable. To this end, we apply Step 2 to find p = ( p o , p l , . . . , p n - l ) such that r ( p ) - p = .(A) ~ ( -pA) = p

-

-

X and p

X and such that each -

pi, 0

X E Wn(IFp). Thus

5 i 5 n - 1 , is separable over F . Then

A B E L I A N D-EXTENSIONS

399

where S is the separable closure of F , and hence E / F is separable.

3. Abelian pextensions over fields of characteristic p .

In what follows, F denotes a field of characteristic p > 0. A finite Galois extension E / F is called an abelian p-eztension if the Galois group of E / F is an abelian p g r o u p . In this

section we examine abelian pextensions of F . Let E / F be a finite Galois extension with Galois group G. For any positive integer n, we write W,(E) for the n-th Witt ring of E . Recall t h a t , by Theorem 1.3 (iii), W , ( F )

is identifiable with the subring of W n ( E )consisting of vectors whose coordinates are in F . The natural action of G on E can be extended to the action of G on

It is obvious that a

H

W n ( E )by putting

g a is a n automorphism of W,(E) and that the set of these auto-

morphisms is a group isomorphic to G. This means t h a t G acts faithfully as a group of automorphism of W,(E). Therefore W , ( F ) is the f k e d subring of G, i.e.

W , ( F ) = {a E W,(E)lga = a

for all g E G}

(2)

Given any a E W , ( E ) , we define its trace T r ( a ) by

Since T r ( a ) is obviously fixed by all elements of G , it follows from (2) that

T 7 ( a )E W , ( F )

3.1. Lemma.

n

2 1.

for all a E W , ( E )

Let EIF be a finite Galois eztension with Galois group G and l e t

Then there exists a E W , ( E ) such that T r ( a ) is a unit of W , ( E ) .

Proof. If a = (ao,al, . . . , a n - l ) E W , ( E ) , then the first component of T r ( a ) is

CHAPTER 7

400

because the first components are added in forming a sum in W , ( E ) . Owing t o Theorem 2.4.8, T r E l F ( a 0 ) # 0 for some a0 E E . If

is chosen such t h a t T r E I F ( a 0 ) # 0, then for

a0

a = (ao,. . . , a,-l), we have

T r ( a ) = ( T r E / F ( a o ) ,. . .) Since the first component of T r ( a ) is nonzero, it follows from Corollary 1.10 that T r ( a ) is a unit of W,(E), as required. Applying Lemma 3.1, we now prove 3.2. Proposition. Let E f F be a finite Galois eztension with Galois group G , and

let the action of G o n the additive group of W,(E) be given b y (I). Then

H~(G,w,(E)= ) o Proof. Let f : G

---f

W,(E) be a crossed homomorphism, i.e.

Invoking Lemma 3.1, we may choose a in W,(E) such that T r ( a ) is a unit of W,(E). Setting b = W a ) - l ( Z f(g)(ga)) gEG

it follows t h a t for all z E G.

proving that f E B ’ ( G , W , ( E ) ) , as required. A s a preliminary to our next result, let us recall that, by Theorem 1.6, the map

is a ring homomorphism

ABELIAN

401

p-EXTENSIONS

3.3. Lemma. T h e m a p p : Wn(E) + Wn(E) defined by

is a n endomorphism of the additive group of Wn(E) with kernel Wn(lFp) ( b y Theorem 1,6(iii), @',(IFp)

i s the prime subring of W , ( E ) ) .

Proof. It is obvious that p is a n additive homomorphism. An element (zo, 5 1 , . . . , ~ W n ( E ) belongs t o Kerp if and only if zP = z; for all i or, equivalently, if and only if

z; E

E

~ - 1 )

IF,

for all i. So the lemma is verified. rn In what follows, given a l , a 2 , . . . , a ,

W n (E ), we write

for the field obtained from F by adjoining all coordinates of a i , 1 5 i

5

r . We have now

accumulated all the necessary information t o prove our first major result. 3.4. Theorem. Let F be a field of characteristic p > 0 , let E / F be a finite Galois

extension with Galois group G which is a n abelian p-group of ezponent p e and, f o r any given n

2 e , let Wn,p(E) = { z E wn(E)lp(z) Wn(F))

T h e n the following properties hold: (i) W,,,(E) is a subgroup of the additive group o f W n ( E ) containing W , ( F ) , and the map

where x a ( g ) = ga - a , is a homomorphism.

(ii) T h e sequence of additive groups 0 + W,(F)

--$

W n , p ( E+ ) Hom(G,Wn(Fp)) -+ 0

i s exact. I n particular,

Wn,p(E)IWn(F)

G

(iii) For a n y given a 1 , a 2 , . . . , a , in Wn,p(E),

E = F(al,aZ,... , a , )

CHAPTER i

402

if and only if

Proof. (i) Owing to Lemma 3.3, p preserves addition, hence W n , p ( E is ) a subgroup of W , ( E ) . It is clear t h a t W n ( F ) C W n , p ( E ) .Given a E W n , p ( E )and g E G, we have p(a) =

.(a) - a E W,,(F) and so g ( r ( a ) - a) = .(a) gr(a)

-

-

a or

.(a) = ga - a

which shows t h a t xa(g) E Wn(Fp). We also have x,(zy) = (zy).

-

a = (zy).

= z(ya - a ) = xa(z)

whence X , E Hom(G,W,(F,)).

proving that the map a

H

-

za

+za - a

+ ( z a - a ) = (ya

-

a)

+ (za- a)

+ xa(Y),

Given a,b E W n , p ( E )we , have

X, is a homomorphism.

(ii) For any given a E W n , p ( E )we , have x,(G) = 0 if and only if a is fixed by G. Therefore, by (2), Wn(F)is the kernel of the map a H x,. To show that the map a z,y E G, ~ ( z y = ) x(z)

Accordingly, x(g) = ga

-

H

X , is surjective, fix

Then, for any

+ x ( y ) and, because x ( y ) E W n ( F p ) , we also have

x E Z'(G,W,(E))

and, by Proposition 3.2, there exists a E W , ( E ) such that

a for all g E G. Because x ( g ) E Wn(FFp),r ( x ( g ) ) = x(g) and this yields

g ( r ( a ) - a ) = .(a)

Thus p ( a ) = .(a) surjective.

x E Hom(G,W,(Fp)).

-

-

a

for all g E G

a E W , ( F ) and so a E W,(E), proving t h a t the map a

H

x a is

403

A B E L I A N D-EXTENSIONS

Finally, because W,(Fp) is of exponent p"

2 p e , we have

Hom(G, W,(F,))

2G

and hence

G

Wn,p(E)/Wn(F),

proving (ii). (iii) Let us first assume that

Then, by (ii), Hom(G,W,(Fp)) = < x a l , . . . , x a , >. Now let E' = F ( a l , . . . , a r ) and let H be the subgroup of G corresponding t o E' (i.e. the Galois group of E I E ' ) . If h E have ha, = a,, 1 5 i 5 r , so

N ,we

xa,( h ) = 0 . It follows that

X(h) = 0

for all

x E Hom(G,W,(F,))

and thus h = 1. Hence H = 1 and therefore E' = F ( a l , . .. , a r ) = E . Conversely, let E = F ( a l , . . . , a r ) for some

al,.

. . , a , in W,,,,(E). We first claim that

x a , ( g ) = 1 for all i E ( 1 , . . . , r ) implies g = 1 . Indeed, if x a , ( g ) = 1, then ga, = a, and

thus ga = a for all a E E. But then g = 1, as claimed. We conclude therefore t h a t

In particular, if a E Wn,p(E)rthen

xa can be written in the form

Therefore, for any g E G, we have

which yields g ( a - k l a l - . . . - k,a,) = a

The conclusion is t h a t a - k la l

-

kzaz

-

-

k la l

-

kzaz - . . .

~

k ,a ,

. . . - k r a r E W , , ( F ) , as required.

3 . 5 . Corollary. Let F be a field of characteristic p

D

> 0, let E I F be a finite Galois

eztension with Galois group G which is an abelian p-group of ezponent p and let

E , = { X E EIXP - X E F }

CHAPTER 7

404

(i) E, is a subgroup of the additive group of E containing F and the map

defined b y x a ( g ) = ga

-

a, is a homomorphism.

(ii) The sequence of additive groups

04F

+

E,

+

Hom(G,Fp)

---t

0

i s ezact. In particular,

E,IF

E

G

(iii) For any given a l , . . . , a , in E,,

E = F ( a 1 , . . . , a r ) if and only if E p / F = < a ,

+ F , . .. , a , + F >

Proof. Apply Theorem 3.4 in the special case where n = 1. The rest of the section will be devoted t o the existence of abelian pextensions of any given exponent. In what follows, the map p is defined as in Lemma 3.3 and W n , p ( E is ) defined in Theorem 3.4. Note that, by the definition of Wn,,,

3.6. Lemma. Under the hypotheses of Theorem 3.4, the following holds:

Proof. Consider the homomorphism

An element a E W n , p ( E )is in the kernel of this homomorphism if and only if p ( a ) = p ( z ) for some z E W n ( F ) . This is equivalent t o p ( a - z ) = 0 which means t h a t a - z 6 W n ( F p ) . It therefore follows that the kernel of this homomorphism is W , ( F ) . Because the given homomorphism is obviously surjective, the result follows. rn

A B E L I A N D-EXTENSIONS

405

3.7. Lemma. Let a = ( a o , a l ,..., a n - l ) E W n ( F ) . Then there ezists a finite separable field eztension E I F such that E = F(bo,bl, ..., b n - l ) and the element b = (bo, b l , , . . ,bn-i) o f W , ( E ) satisfies p ( b ) = a.

Proof. To prove the case n = 1, it suffices to construct a separable extension E = F ( b ) generated by a root b of the polynomial XP

(XP - X

-

X

-

a)’ = -1

-

a. Since the derivative

the given polynomial has distinct roots. Thus any root b of XP

-

X

-

a will satisfy the

condition. By induction, we may assume t h a t we have already constructed a separable extension

L = F(b0,b l , . . . , b n - 2 ) such that the element b‘ = (bo, b l , . . . ,b,-z)

E Wn-z(L) satisfies

p(b’) = (ao,a l , . . . , an-2). Consider the polynomial ring L [ X ]and the Witt ring W , ( L [ X ] ) .

We take the vector

Y

Then there exists

=

(bo,bl,...,bn-z,X) E W n ( L [ X ] )

f(X)E L [ X ]such that

Accordingly,

Applying formula ( 5 ) of Sec. 1, we conclude that xp

= f(X)

+x +x

for some X E L. Therefore j ( X ) = XP - X - A. By looking a t the derivative of f ( X ) ,it

follows t h a t f ( X ) = a,-l

has distinct roots. If E = L ( b n - l ) where f ( b , - l )

=

a n - l , then

E is separable over L and thus E = F(b0, b l , . . . , b n P l ) is separable over F . Furthermore, by our construction, b = ( b o , b l , ..., b n p 1 ) E W n ( E )satisfies p ( b ) = a. The proof is therefore complete. rn

The following result demonstrates t h a t any finite subgroup of W,(F)/p(W,(F)) is a Galois group of some abelian pextension.

CHAPTER 7

406

3.8. Theorem. Let F be a field of characteristic p

>

0 and let S be a subgroup

of the additive group of W , ( F ) containing p(W,(F)) and such that S / p ( W , ( F ) ) is finite.

Then there ezists an abelian p-eztension EIF such that the ezponent of G a l ( E / F ) is p e with e 5 n and

s = Wn(F) n p(Wn(E))

(5)

Proof. We first note t h a t (6) follows from (3), (4) and ( 5 ) . Let s ( l ) ,s ( ' ) , . . . ,s(') be elements of S such t h a t the cosets s(')

+ p(W,(F))

By repeated

generate S/p(W,(F)).

application of Lemma 3.7, we can find nr elements b t ) , 1

5 i 5 r , 0 5 k 5 n - 1, together

with a finite separable extension E / F generated by them such t h a t

p ( b t ) , . . . ,b (4 n p 1 ) = (so(')

where s(') = ( s t ) ,

,. . . ,),'s:

i n W , ( E ) f o r a l l i E { l , ..., r}

(7)

. . . ,s")

n-1).

Let K I F be a finite Galois extension with K 2 E . We form W , ( K ) and let the Galois group G of K / F act in W,(K) as before. If g E G and b(') = ( b t ) , . . . ,b ('n1P l ) , then

p(b(')) =

gives

p(gb(')) = g p ( b ( ' ) ) = gs(') = ,(')

and so gb(') - b(') E Wn(Fp)

W , ( F ) . This implies that g ( E ) 2 E and therefore E / F is

Galois. Moreover, by (7), each s(') is in p ( W n ( E ) ) and thus

In what follows, we put G = G a l ( E / F ) . If

Z,y

E G , then zb(') = b(')

We conclude therefore that

+ c(')

and yb(') = b(')

+ d(') for some , ( * I ,

d(') E W , ( F ) .

407

ABELIAN p-EXTENSIONS

which proves t h a t G is abelian. Also zkb(') = b(') has characteristic pn. This implies t h a t p g r o u p of exponent p e with e

+ k c ( ' ) , so ~ p " b ( ~=) b(') because W n ( E )

= 1 for all z E G, so G is a finite abelian

zPn

5 n. Owing t o

( 8 ) , we are left t o verify t h a t

Let x, E H o m ( G , W n ( F p ) ) be defined by x;(g) = 96") - 6(i) (see Theorem 3.4(i)). Then it is obvious that x,(g) = 0 for all i E { 1 , 2 , . . , ,r} implies t h a t g = 1. Therefore

Accordingly, if a is any element of W n ( E )such that p ( a ) E W n ( F ) ,then we have

This implies t h a t a =

Cn,b(')+ X

for some X E W n ( F ) ,by virtue of Theorem 3.4(ii)

Then p(a) =

Cn;s(') + p ( X ) E s,

proving (9) and hence the result. rn The following observation will enable us t o take full advantage of the preceding result. 3.9. Lemma.

If Xo, X I , . . . , X n - l E F , then Xo E p ( F ) if and only if

Proof. Thanks t o Theorem 1.6(ii), we have

pn-'X

= (0,. . . ,o, A;"-')

On the other hand,

(O,...,O,Xo)-(O

, . . . ,0,x;

+ (0,. . . ,o, A);

"- I

~

+ ( 0 , . . . ,0,X:n--2) Therefore p"-'X

)=

( 0) . . . ,0 , X o ) - ( 0

( 0 , . . . ,o, $) -

+ . ..

( 0 , . . . ,o, A;"-')

E p ( W , ( F ) ) if and only if (0,. . . ,O, A),

, . . . ) 0,X;)

€

p(W,(F))

E p(W,(F)).

CHAPTER 7

408

Suppose that (0,.. . ,O, Ao) E p ( W , ( F ) ) , say

(0,...,0,Ao) = a ( a ) - a where a = ( a o , a l , .. . ,& " - I ) .

Let u = fn-l,", where

fn-l,,, is

defined prior t o Theorem

1.3. Then an(a) - .(a)

=0

=~(a) i.e., (a:, . . . ,a:-,)

Because aa = xu,we have .(.(a)) p = (ao,a1,. . . ,an-2,0),

= a ( 0 , . . . ,o, A,)

we therefore have a ( p ) = p , so

p

= (ao,. . . , a n - 2 ) . Setting

=a

-

p satisfies

x(p) - p =

(0,. . . O , Ao). Furthermore,

P = (07...,O,Pn-1) This implies t h a t that

A0

=

-

- an-l,

/3n-1 = Xo, so

A0

E p ( F ) . Conversely, if this condition holds so

then .(a) - a = (0,. . . ,0, A,)

.

for a = (0,.. , O , anpl).

We are now in a position t o provide the following generalization of Theorem 2.7(i). 3.10.

Theorem. Let F be a n arbitrary field of characteristic p > 0 . Then the

following conditions are equivalent: (i) For each

IZ

2 1, there

ezists a cyclic eztension E I F of degree pn

(ii) There exists a n element p E F which cannot be written i n the form p =

z p -

x for

any z E F (iii) There ezists a n element p E F such that XP - X - p is irreducible over F .

Proof. The equivalence of (ii) and (iii) is a consequence of Theorem 2.7(iii). Observe

# F . If (i) holds, then p ( F ) # F by Theorem 2.7(i). We are thus left t o verify t h a t p ( F ) # F implies (i).

also that (ii) is equivalent to p ( F )

Suppose t h a t p ( F ) # F and choose

A0

EF

-

p ( F ) . Let X = (Xo,X1,. ..,&-I)

A;, i > 0, are any elements of F . Owing t o Lemma 3.9, p"-'A

4

where

p ( W , ( F ) ) and this

implies that the subgroup S of W,(F) generated by X and p ( W , ( F ) ) has the property t h a t S/p(W,(F))is cyclic of order p". The desired conclusion is therefore a consequence of Theorem 3.8.

4. Kummer theory.

Let F be a field and n a positive integer. A Galois extension E I F is said t o be of ezponent

409

KUMMER THEORY

n if the exponent of G a l ( E / F ) divides n, i.e. if un = 1

for all u E G a l ( E / F )

By a Kummer eztension of exponent n , we understand a n abelian extension E / F of finite exponent n, where F is assumed t o contain a primitive n-th root of unity. It follows that for any such extension the characteristic of F does not divide n. Throughout the discussion, the ground field F will be fixed, and all extensions lie in a given algebraic closure F of F . Moreover, we assume t h a t F contains a primitive n-th root of unity, which will be denoted by

Our goal is t o provide a survey of all Kummer extensions E / F of exponent n.

E,.

If G = G a l ( E / F ) , then G endowed with the Krull topology is a compact group. We write Hom,(G, < e n > ) Then Hom,(G,

=

{f : G +<en

> I f is a continuous homomorphism}

< E , >) is a group under multiplication of values:

We write Hom(G, < E, >) for the group of all (not necessarily continuous) homomorphisms from G t o

< E, >. Then Hom,(G, < en >) is a subgroup of Hom(G, < E , >) which can be

characterized by Hom,(G, < e n >) = {x E Hom(G, < e n >)/KerXis an open subgroup of G}

(1)

T h e latter is true, since a homomorphism from a topological group into a discrete group is continuous if and only if its kernel is an open subgroup. In case G is finite (hence discrete), we have Hom,(G,

< en >) = Hom(C, < E , >)

Some further information is provided below. Let G be a finite abelian group. Then the set G' = Hom(G, C * ) of all homomorphisms from G t o C

forms a group under multiplication of values. The elements of the group G'

are called characters of G. If the exponent of G divides n , then the image of any is in the cyclic group of n-th roots of unity of Q: *. This shows t h a t

Hom(G, C *) = Hom(G, <6, >)

xE

G'

CHAPTER 7

410

where 6, is a primitive n-th root of unity in C

. Given a subgroup H

of G , we put

H I = {x E G*Ix(h)= 1 for all h E H } Then we clearly have a natural isomorphism

4.1. Lemma. If G i s a finite group, then

G

E G'

(noncanonically) a n d G*'

ZZ

G (canonically)

Proof. It is easily verified t h a t (GI x Gz)* E G; x G;. Therefore, t o prove that

G

G', we may assume that G is cyclic, say of order n. Let g denote a generator of G

and let

x

:G +

C * be a homomorphism. Then X ( g ) ,

= 1, i.e. x ( g ) is a n n-th root of

unity. Let 6, be a primitive n-th root of unity and let

be defined by for any k

2 1,

+(x) = x ( g ) .

It is obvious that

x ( g ) = b,k is a n element of G',

For each g E G ,

xE

G * , define

+ is an injective homomorphism. $J

is also surjective. Hence G

GS(x)= x ( g ) .

Then

is a homomorphism from G into G*'. If x ( g ) = 1 for all

H I = G'

S

+g

Because

G*.

E G" and the map g

x E G'

and H = < g > ,

then

(G/H)'

Hence, by the foregoing, G Z G / H which gives H = 1. Thus g = 1 and the map g is injective. Because \GI =

H

IG*l = IG**l,the result follows.

w

$,I

m

We now return t o our discussion of Kummer extensions E / F of exponent n,where n is any given positive integer. Throughout, c,, denotes a primitive n-th root of unity in F . We shall use the symbol

+,a t F , to denote any element X E F such t h a t A"

= a.

There are precisely n such elements, namely c;

X

O
We note t h a t the field F ( X ) is the same no matter which n-th root X of a we choose. This field will therefore be denoted by F (

m.

KUMMER THEORY

411

We denote by ( F * ) " the subgroup of F* consisting of all n-th powers of elements of

F'. If H is a subgroup of F' containing (F*)",we write FHfor the composite of all fields with a E H . It is uniquely determined by H as a subfield of

F(

F.

Observe also that

FH is the splitting field of the family of polynomials.

4.2. Lemma. (i) For any subgroup H of F' containing (F')",

F H / F is a Kummer

eztension. (ii) If E / F is a Kummer eztension of ezponent n, then

E = FH

with

H

=

( E * ) nn F'

(iii) E/F is a Kummer eztension of ezponent m if and only i f F contains a primitive m-th

root of unity and E is the splittingfield of a family of polynomials of the form Xm-a, a E F

Proof. (i) By Theorem 2.4(i), F( f a ) / F is cyclic of degree dividing n. Therefore, for any

0

E G a l ( F H / F ) , the restriction of u" t o F (

generated by all

fi,a E H .

is 1. Thus on = 1 since FH is

Moreover, FH/Fis abelian since it is the composite of cyclic

extensions. (ii) It is obvious t h a t FH

C E.

Observe, further, that E I F is the composite of its finite

subextensions E ' / F . Since the finite abelian group Gal(E'/F) is the direct product of cyclic groups, each of which can be understood as the Galois group of a cyclic subextension of E'IF, we deduce that E'/F (and hence E / F ) is the composite of its cyclic subextensions. Suppose t h a t K / F is a cyclic subextension of E / F . Then G a l ( K / F ) is of order dividing n , so by Theorem 2.4(i), K = F (

for some

a E (E')" n F' = H Accordingly, K

C FH and so E C F H , as required.

(iii) Suppose that F contains a primitive m-th root of unity. If E is a splitting field of a family of polynomials of the form Xm - a , a E F , then E is the composite of fields

F(

"m.Therefore, by the argument of (i), E / F is a Kummer extension of exponent m .

The converse holds by applying (ii). rn

CHAPTER 7

412

4.3. Lemma. Let E/F be a Kummer eztension of ezponent n and let H = (E*)"n

F'. Then, for any a E H , the map

is a continuous homomorphism, which i s independent of the choice of

Proof. Let X denote a fixed n-th root of a in F so t h a t distinct such roots. Then, for

proving t h a t

fi.

hi, 0 5 i 5 n - 1, are all

;/;;;= A€:, we have

xa is independent

of the choice of

G.Moreover, given

u1,uZ

E Gal(E/F),

for some k; 2 1, so t h a t

we have u;(X)/X = :6

Finally, Kerxa = G a l ( E / F ( ;/;;))

and F ( f i ) / F is finite. Therefore, by Theorem 6.6.2(iii),

KerXa is a n open subgroup of G a l ( E / F ) . Applying ( l ) ,we conclude t h a t

xa is continuous,

as required. m

We shall refer t o

xa in Lemma 4.3 as the Kummer

character of G a l ( E / F ) correspond-

ing to a. What one commonly understands under the name of Kummer theory is the content of the following result. 4.4. Theorem. Let F be a field containing a primitive n - t h root of unity

(i) The map H

H

E,

FH gives a bijection of the set of subgroups of F* containing (F'),

and Kummer eztensions of F of ezponent n. The inverse of this map is given b y

E

H

(E'),

n F*

(ii) I f E / F is a Kummer eztension of ezponent n and H = (E')"

n F',

then the map

{ :gn +

Horn,( Gal(E/F),< 6, >)

H X a

is a n isomorphism. I n particular, E / F is finite i f and only if H / ( F * ) " is finite and

i f that is the case, then Gal(E/F)

H/(F*)" and

( E : F) = IH/(F*)"I

KUMMER THEORY Proof. P u t G = G a l ( E / F ) . If a , b E H and

413

an =

a,pn = b, then ( a @ )= , ab and so

If a = A" with X E F', then xa(u) = u ( X ) / X = 1 for all u E G. Thus the map in (ii) is a homomorphism. Suppose that a E H is such that X , = 1, i.e. u(a)= a for all a E G , where an = a . Owing t o Lemma 4.2(ii), E = FH and so F(a) is a subfield of E . If a is not in F , there exists a n automorphism of F(a) over F which is not the identity. Extend this automorphism t o E and call this extension

00.

Then ao(a) # a , a contradiction. Hence

we have a n exact sequence 1 ---t H / ( F * ) , -+ Hom,(G,

< t, >)

(2)

Observe t h a t the map

G

4

Hom(H/(F*)", < e n

0

++

fc7

>

where f,(a(F*),) = xa(u),is a homomorphism. If xa(o)= 1 for all a E H , then for every generator a of E with an = a E H, we have a(&) = a and so u = 1. Consequently, we have an exact sequence 14G

-+

Hom(H/(F*)", < t, >)

(3)

Applying the first isomorphism of 4.1, it follows from (2) and (3) that E / F is finite if and only if H/(F')" is finite and if that is the case then G E H / ( F * ) , . To complete the proof of (ii), we are left to verify t h a t the given map is surjective. By the foregoing, this is true if E / F is a finite extension. If E / F is of infinite degree, we run through the finite subgroups of H / ( F * ) " and we put E , = FH,. Then let IIT,/(F')n H / ( F * ) " = uiHi/(F*)"and E = u;E; Accordingly, the groups Gal(E/E;) form a basis of open neighbourhoods of 1 E G. Since

subgroup Gal(E/E;). Now

x

x

> is open, it must contain a yields a homomorphism X : G a l ( E , / F ) +< E , > such that

the kernel of a continuous homomorphism

: G

-+<

x(a) = x(a1E;). By t h e surjectivity of the finite case,

But then

X

c,

is of the form

CHAPTER 7

414

which implies

x = xa, proving surjectivity.

To prove (i), it suffices, by Lemma 4.2(i), (ii), t o show t h a t F H ,

Hl

s H z , for any subgroups H1,H2 of F'

F H ~implies

containing (F')". If b E H i , then F ( i7%)

CFH~

and F ( i7%) is contained in a finitely generated subextension of F H ~Hence . we may assume that H 2 / ( F * ) " is finitely generated, hence finite. Let H3 be the subgroup of F* generated by

H2and b. Then F H ~= F H and ~ from what we saw above, the degree of this field over

F is precisely IH2/(F*)"I or Hence Hz = H 3 , so b E Hz and thus H1

IH3/(F*)nl

C H2,as required.

rn

We shall now derive some consequences of interest. By the mazimal Kummer eztension of F of exponent n, we understand the composite of all Kummer extensions of F of exponent n. It is clear that such a n extension is a unique largest Kummer extension of F of exponent n. 4.5. Corollary. Let E be the mazimal Kummer extension of F of exponent n. Then

H o m c ( G a l ( E / F ) , < E , > )= F * / ( F * ) " Proof. By Theorem 4.4, it suffices t o show t h a t (E*)"n F* = F ' . S o fix any X E F' and put a =

fi.Then

a E E since F ( a ) / F is a Kummer extension of exponent n. Thus

X = an E (E*)" n F' as required. rn 4.6. Corollary. Let E / F be a finite Kummer eztension of exponent n, let G =

G a l ( E / F ) and let

@ denote the subgroup

of E' consisting of all n - t h roots of elements

in F ' . Then G

Proof. P u t H = (E')"

E

n F'.

w / F * E ( @)"/(F*)" Then H = (

w)"and so the map

G + H / ( F * ) n ,5

H

z"(F*)"

is a surjective group homomorphism; its kernel is F' since F contains a primitive n-th

root of unity. Hence, by Theorem 4.4(ii),

@/F*

EH/(Ft)n

G

415

KUMMER THEORY

and the result follows. a

4.7. Corollary. Let p be a p r i m e and let F be a field containing a primitive p - t h root of unity. Suppose a is a n element of a field eztension of F such that

is such that

pP

E F , then p

=

aP

E F . If

p E F(a)

a k a for some k E Z and some a E F .

Proof. The assertion is trivial in case a 6 F . Suppose that a $ F . Thus aF $ FP since F contains all p t h roots of unity. This immediately implies that ( F ( a ): F ) = p (see Lemma 8.1.3). Hence F ( a ) / F is Kummer extension of exponent p and F ( a ) = FH where

H =< a p > ( F * ) P . If p E F , then /3 = that

ap(a-pp)

and a-PP E F . We may thus assume

p $ F . Hence F ( a ) = F ( p ) and F ( P ) = F H ~where H I =< p P > ( F ' ) p . By Theorem

4.4(i), H = HI and thus

Taking p t h roots, we get

p

= a k a for some a E F . a

It is useful sometimes t o know that the condition on F containing a primitive p t h root of unity can be deleted. Thus we have more generally: 4.8. Corollary. Let p be a prime and let F be a field with charF

is an element of a field eztension of F such that p p

E F , then

p

=

E F

a P

-

# p . Suppose a

FP If p E F ( a ) i s such that

a k a for some k E Z and some a E F .

Proof. Because

a P

$ FP, we have ( F ( a ) : F )

root of unity over F . The degree (F(t) : F ) 5 p

-

= p . Let

denote a primitive p t h

1 , hence the degrees ( F ( a ): F ) and

( F ( c ) : F ) being relatively prime implies

F ( a ) n F(t) = F Consider F ( t ) ( a )over F(t). Then Corollary 4.7, we deduce t h a t

a p

E F ( E ) ,p P E F ( c ) , and F ( c ) contains

= a k a for some k E

Z and some Q E F ( c ) . But

a = p a p k E F ( c ) n F ( a ) = F, hence the result. a

t.

Applying

CHAPTER 7

416

5 . C h a r a c t e r groups of infinite abelian extensions.

Let E / F be a Galois extension. In this section, among other results, we establish a bijective correspondence between the subgroups of the character group of Gal(E/F) and all abelian subextensions of E / F . We begin by providing some background information. A topological space is called locally compact if each of its points has a neighbourhood with compact closure. A topological group G is said to be locally compact if G is a locally compact topological space. Consider the additive group

R of all real numbers. Then IR can be topologized by

means of the standard Euclidean topology, i.e. by taking as a basis of a point zo E the open intervals containing

20.

For each

E

> 0,

IzI

< ~ / 2 IyI , < c/2 imply

Iz

and therefore addition is continuous. Similarly, one sees that the inversion z

IR all

+ y/< E H

-x is

continuous. Hence R is an abelian topological group. It is easy to verify that R is a locally compact (but not compact) group. The quotient group

R/Z is a compact topological

group with respect to the quotient topology of R. The group R / Z is also called the circle

group . The name is justified by the fact that the map

induces a homeomorphism of

R/Z with the unit circle of the complex plane equipped

with the topology induced by the standard topology of R2. We shall regard multiplicative group, by identifying

R/Z as a

R/Z with the unit circle of the complex plane. In

particular, Q /Zwill be identified (algebraically and topologically) with the group of all complex roots of unity. Let G be a topological group. A continuous homomorphism of G into R/Z is called a character of G. The set G* or Char(G) of all characters of G constitutes a group under

multiplication of values. We refer to G' as the character group of G. Assume that G is a locally compact group and let

?? be the topological commutator

subgroup of G, i.e. the topological closure of the algebraic commutator subgroup GI of G. Then it can be shown (see Hewitt and Ross(1963)) that -I

G = { g E GIx(g) = 1 for all

and that

G*

(G/G')*

x

E G'}

CHARACTER GROUPS by the map which sends

x

E G' t o

x 1 E (GIG')'

417

given by

Thus in considering the character group G' we may harmlessly assume t h a t G is abelian. Let G be a n abelian locally compact group. For any compact subset C of G and any real c

> 0, let U ( C , c )be the subset of

G' defined by

U ( C ,c) = { x 6 G * l l x ( g )- 11 < E

for all g E C}

Then G' becomes a locally compact topological group with a fundamental system of neighbourhoods of 1 consisting of all such U(C,6 ) . The following fundamental result is known as the Pontryagin duality. 5.1. Theorem. Let G be a locally compact abelian group and, for each g 6 G , let g**

E (G*)- = G*' be defined b y

(i) The map G 4 G * * , g

H

g** is a topological isomorphism

(ii) G is compact i f and only i f G' is discrete

Proof. See Hewitt and Ross(1963). Turning our attention to profinite groups, we record 5.2.

Proposition Let

x

:

G

+

H be an algebraic homomorphism

of topological

groups.

(i) If G is profinite, then (ii)

x

i s continuous if and only if Kerx is open

If G is profinite and x is continuous, then i m x

(iii) If G is profinite, then

x

is a finite subgroup of H

is continuous i f and only if

x

is continuous regarded as a

homomorphism of the topological group G into the discrete group H .

Proof. (i) If Kerx is open, then for any subset S of H containing 1, x-'(S) is a union of cosets of Kerx. Thus assume that

x-' ( S ) is open and therefore x

x is continuous. If S is any open set of H

is continuous. Conversely,

containing 1, then x - ' ( S ) is open in

G. Since G is profinite, G has a basis of open neighbourhoods of 1 consisting of (normal) subgroups (see Proposition 6.2.8(iii)). Thus Ker therefore must be open.

x contains some open subgroup of G and

CHAPTER 7

418

(ii) By (i), Ker ImX

E

x is open and so, by Proposition 6.2.3(ii), G/KerX is a finite group.

Because

G/KerX algebraically, we deduce that ImX is finite.

(iii) It follows from (i) that the continuity of

x is independent of the topology of H , hence

the assertion. 5.3. Corollary. Let G be a profinite group.

x : G R/Z is continuous i f and only if Kerx is open. any character x of G, Imx is a finite [hence cyclic) subgroup of Q /Z. I n partic-

(i) A homomorphism

(ii) For

---t

ular, G* is a torsion abelian group. (iii) G' = Hom,(G,Q /Z), where Hom,(G,Q /Z) is the group of all continuous homomorphisms of G into the discrete group Q / U . (iv) G' is the group of all homomorphisms

x :G

---t

Q /Zsuch that Kerx is open.

Proof. This is a direct consequence of Proposition 5.2. 5.4. Corollary. Let E / F be a Galois field eztension with Galois group G and let

x

be a character of G. Then there ezists a finite cyclic subeztension E,/F of E/F such that

Kerx = Gal(E/E,) and G/KerX

E

Gal(E,/F).

Proof. By Corollary 5.3(i), Kerx is open. Hence, by Proposition 6.2.3(ii), Kerx is a closed normal subgroup of G of finite index. Therefore, by Galois theory, we have Kerx = Gal(E/E,) for a unique intermediate field Ex of EIF. Since Kerx a G, EJF is normal (hence Galois) and since ImX Z G/KerX

Gal(E,/F),

E,/F is a finite cyclic

subextension of E/F, by virtue of Corollary 5.3(ii). 5.5. Proposition. Let E / F be a n abelian field eztension and, for any subeztension

K / F of E I F , let C ( K / F ) be the character group of G a l ( K / F ) . (i) C ( K / F ) can be naturally embedded as a subgroup of C ( E / F ) and

C(E/F)/C(K/F) (ii) The map K/F

H

C(E/K)

C(K/F) is a bijective correspondence between all subeztensions

KIF of E / F and all subgroups of C ( E / F ) . Proof. (i) Assume that K/F is a subextension of E/F. Then Gal ( K / F ) Z Gal ( E/ F)/Gal( E/ K )

4 19

CHARACTER GROUPS

as topological groups. Hence C ( K / F ) is identifiable with the subgroup of C ( E / F ) con-

sisting of all characters

x

for which G a l ( E / K )

C Kerx.

The restriction homomorphism

C ( E / F ) + C ( E / K )is surjective (see Pontryagin(l977)), hence

C(E/F)/C(K/F) C(E/K) (ii) The given map is obviously injective. Let H be any subgroup of C(E/F) and let

G = {nKerX/x E H } . Then, by Corollary 5.4,

G=

n

Gal(E/E,) = G a l ( E / K )

xEH

where K is the composite of all Ex and E , / F is a finite subextension of E / F . If

x fH,

5 Kerx and hence x E C ( K / F ) . Conversely, if x E C(K/F), then G = G a l ( E / K ) C Kerx. Since H consists precisely of those characters of G a l ( E / F ) which

then G a l ( E / K )

vanish on G (see Pontryagin (1977)) we conclude that

x 6H.

Thus H = C ( K / F )and the

result follows. 5 . 6 . Corollary. Let

E I F be a Galois extension with Galois group G . T h e n there is

a bijective correspondence between all abelian subeztensions of

E / F and all subgroups of

G'.

Proof. Let Fab/Fbe the maximal abelian subextension of E I F . By Corollary 6.6.4,

Gal(Fab/F)

G/G'

Hence, by (I), the character groups of G a l ( F a b / F ) and G are isomorphic. Since a subextension K / F of E / F is abelian if and only if K Proposition 5.5.

C

F a b , the result follows by virtue of

This Page Intentionally Left Blank

421

Radical extensions

This chapter is devoted t o a detailed investigation of radical extensions. After providing some basic properties of these extensions, we concentrate on abelian binomials. Among other results, we prove a theorem of Schinzel which gives necessary and sufficient conditions for the Galois group of a binomial X"

-

a t o be abelian. We also determine the

isomorphism class of Galois groups of such binomials. As a further topic, we describe all normal binomials of the form XP"

-

a E F I X ] , 7t

2 1, a E F , where p

is a prime distinct

from charF. The rest of the chapter is devoted t o a Galois correspondence and duality theorems for radical extensions. Our presentation is based on a fundamental paper of Greither and Harrison (1986). The chapter also includes some results pertaining t o the torsion subgroup of radical extensions.

1. I r r e d u c i b i l i t y of b i n o m i a l s and a p p l i c a t i o n s .

Let E / F be a field extension. Following Kaplansky (1972), we say t h a t E / F is a radical eztension if E is of the form

E = F(A1,. . . A,), with A:'

E F and

E F ( A 1 , . . . ,A,-l),

2

(11

5 z 5 m, abbreviated

by

for some n; E N. It is obvious t h a t any radical extension is a finite extension. The special

case where m = 1, i.e. E is of the form

E = F(X) with for some n

2 1 is of particular

A" E F

importance. We shall refer t o any such extension E / F as a

42 2

CHAPTER 8

simple radical eztension. Thus, if E I F is a radical extension satisfying (1) and ( 2 ) , then

F C F(X1)

F(X1,XZ)

and each F ( X 1 , . . . , X , )/F (X , , . . .

...

F(X1,. .. ,Arn) = E

is a simple radical extension.

By a n irreducible radical eztension, we understand a simple radical extension of the form F ( X ) / F such that An E F for n = ( F ( X ) : F ) . Thus E I F is a n irreducible radical extension if and only if E = F ( X ) where X is a root of a n irreducible over F polynomial of the form

(a E F)

Xn-a

Any polynomial of this form will be referred to as a binomial. Let us also note t h a t a finite extension E / F is a n irreducible radical extension if and only if E = F ( X ) for some X E E' such that the order of XF' in E * / F * is equal to the degree of X over F . 1.1. Lemma. Let E I F be a field eztension of degree n and let F contain a primitive

n - t h root of unity. T h e n E / F i s irreducible radical if and only i f E / F is cyclic.

Proof. This is a direct consequence of Corollary 7.2.5. w In order to provide a survey of all irreducible radical extensions of a given field F , we must investigate conditions under which the binomial X"

-a

is irreducible over F , for

any given a E F . T h e following simple observation gives reduction to t h e case where n is a prime power. 1.2. Lemma. Let F be a n y field, a a n elmenet in F , and m , n relatively prime

positive integers. T h e n X""

- a is irreducible over F if and only if X"

-

a and X"

-

a

are irreducible over F .

Proof. Because X""

-

a = (Xn)"- a = (X")"

-

a , if X"" - a is irreducible,

then so are X m - a and X" - a . Conversely, assume t h a t both X" irreducible over F . Denote by E the splitting field of X m "

X""

-

a. Then Am is a root of X"

-

-

-

a and X"

-

a are

a and let X E E be a root of

a and A" is a root of X" - a which shows that

( F ( X m ) : F ) = n a n d ( F ( X n ): F ) = m

Thus both n , m divide ( F ( X ) : F ) and therefore nm divides ( F ( X ) : F ) , since ( n , m )= 1. But X is a root of X m n - a , so ( F ( X ): F ) thus X""

-

a is irreducible over F . B

5 mn. This shows t h a t ( F ( X ) : F ) = mn and

IRREDUCIBILITY OF BINOMIALS

423

Turning our attention to the prime power case, we first record 1.3. Lemma. Let p be a prime and a an element of a field F . Then XP

-

a is

irreducible over F if and only if a $ FP.

Proof. If a = pP for some p E F , then p is a root of XP

-

a and thus XP

-

a is not

irreducible. Conversely, assume that a @ FP. We argue by contradiction and denote by f a n irreducible factor of XP The roots of XP and

t

-a

-

a of degree k, 1 5 k 5 p . Let c be the constant term of f .

(in some splitting field) all have the form

EU,

where u is one fixed root

is a p t h root of unity. Since i c is a product of k of these roots, we have !cc = 6uk

with 6 P = 1. Because ( k , p ) = 1, we may find integers r and s such that r k

+ s p = 1. We

then have = UTkU3P =

(+c/6)laa

Therefore ubr lies in F . However, a = (u6')P, a contradiction. rn 1.4. Lemma. Let p be a prime, let a be an element of a field F and let XP

irreducible over F . If X is a root of XP

-

-

a be

a, then

(i) If p is odd, or i f p = 2 and charF = 2, then X

$ F(X)p

(ii) If p = 2 and charF # 2, then X E F(X)' i f and only i f a E -4F4 Proof. (i) Assume by way of contradiction that X =

u p

for some w E F(X). The

case where c h a r F = p is straightforward: because w is a polynomial in X and p t h powers are taken termwise we have wp E F , which contradicts Lemma 1.3. Now assume that c h a r F # p and adjoin to F ( X ) a primitive p t h root of unity, say

E is a splitting field of XP

-

t.

The resulting field

a over F and therefore E / F is a normal extension. Any

automorphism of E / F sends X t o some clX, 0 .< z 5 p - 1, and for every a E ( 0 , 1, . . . ,p - l } , there is a n automorphism f, sending X t o ('A. P u t wl= fr(u). Then t'X = wp. The element w is in

F(X) but not in F , hence its minimal polynomial (say f ) over F has degree p and

has p distinct roots in E . If w' is any of these roots, then there is an automorphism $ of

E / F sending w g , wl,.

w

into u'.If $(A)

= c 3 X we must have $ ( w )

. . ,w p - l are all roots of f which implies that

multiply together the equations t * X = w f , t o obtain

= w j . Therefore the elements

z = wow1

. . .w P - l

E

F . We now

424

CHAPTER 8

where p = 1 . € . c2 . . .$'-'. If p is odd, p = 1 and we have a contradiction a = 9 ,proving

6).

+

(ii) Assume t h a t X = w 2 with w = a + PX ( a l p F ) . From X = (a equations a2+ P2a = 0 , 2 a p = 1. Eliminating

p, we find a = -4a4

we obtain the

E -4F4. Conversely,

+

if a = - 4 a 4 , a E F , we take p = ( 1 / 2 ) a and verify X = (a

We are now in a position t o attack the prime power case (the prime case will be excluded because of Lemma 1.3). 1.5. Lemma. Let F be an arbitrary field, let n 2 2 be a positive integer and let p be

a prime. Denote b y a an arbitrary element i n F. (i) If p is odd, or p = 2 and charF = 2 , then XP"

a

is irreducible over F i f and only i f

4 FP

(ii) If p = 2 and charF

a

-a

#

2 , then X 2 " - a is irreducible over F i f and only i f a $! F2 and

6 -4F4.

Proof. (i) If a = pP for some p E F , then XP" XP"-'

- p.

Conversely, assume t h a t a

Then p is a root of XP

-

4 FP.

-

a = XP"

-

pP is divisible by

Let X be a root of XP" - a and let p = X P " - l .

a , hence by Lemma 1.3, ( F ( p ) : F ) = p . We claim t h a t X has

degree pn-' over F ( p ) ;if sustained, it will follow that X has degree p" over F and XP" - a is irreducible over F . That X has degree pn-' provided p

over F ( p ) will follow by induction on n,

4 F ( p ) p . T h e desired conclusion is therefore a consequence of Lemma 1.4(i).

(ii) If a E F 2 , then obviously X 2 " - a is reducible. Assume next t h a t a E -4F4 and write

a = -4a4, a E F, Y = X2"-'. Then

Conversely,

x2"- a = y4 + 4a4 = ( y 2+ 2 a +~2 a 2 ) ( y 2- 2 a +~2 a 2 ) assume t h a t a 4 F2 and a 4 -4F4. Since a 4 - 4 F 4 , it follows t h a t

is not a fourth power in F . Again, let X be a root of X 2 " - a and p = A'"-'.

a

4

F2 and

p is a root of

-4a

Because

X 2 - a , we have ( F ( p ) : F) = 2 by Lemma 16.3. We must

show that ( F ( X ) : F ( p ) ) = 2"-'.

For n = 2 , this will be true if p is not a square in F ( p ) ,

and for n > 2 this will be true by induction on n, provided p is not a square in F ( p ) and -4p is not a fourth power in F ( p ) . In the latter case ,-p is a square in F ( p ) . Thus it

suffices t o rule out the possibility t h a t either p or - p is a square in F ( p ) . Now these two statements are equivalent since p

H

- p induces a n automorphism of

by applying Lemma 1.4(ii), the result follows. m

F(p) over F. Hence,

IRREDUCIBILITY OF BINOMIALS

425

We have now come t o the demonstration for which this section has been developed. 1.6. Theorem. L e t F be a n arbitrary field, let n X" - a

21

a n d let a E F . T h e n

i s irreducible over F

i f a n d o n l y if a $ FP for all p r i m e s p dividing n a n d a

4

-4F4 whenever 4/n

Proof. Let pa be the highest power of a prime p dividing n. If a E FP, then XP' - a is reducible by Lemmas 1.3 and 1.5. Hence X "

-a

is reducible, by Lemma 1.2. If 4 / n and

a = -4X4, X E F, then

Conversely, assume that a $ FP for all primes p dividing n and a $ -4F4 if 41n. Let p s be the highest power of a prime p dividing n. Owing t o Lemma 1.2, we need only verify that

XP' -

a is irreducible. This being true by virtue of Lemmas 1.5 and 1.3, the result

follows. m The rest of this section will be devoted t o providing some applications of the preceding results. Our first application is recorded in

1.7. Proposition. Let p be a p r i m e a n d let F be a field with c h a r F # p . If a,P are roots of irreducible polynomials XP - a , XP - b over F , t h e n

unless b = $ a k f o r s o m e k E Z a n d s o m e c E F

Proof. If the polynomial

XP - b

remains irreducible over F ( a ) , then (F(cr,P) : F ) =

p 2 . I f it is reducible over F ( a ) , then it has a root y in F ( Q ) , by virtue of Lemma 1.3.

Because

7 P

=b6

F, it follows from Corollary 7.4.8 that y = c a k for some k E Z,c E F .

Raising this equation t o the p t h power, we obtain b = $ a k . Assume that

a , P are roots

of irreducible polynomials

can be said about the degree of

Q

+ P,

X P -

a,

XP -

b over E . What

provided ( F ( a , P ) : F ) = p 2 ? To answer this

question, we record another preliminary result. 1.8. L e m m a . L e t E I F be a f i n i t e Galois e z t e n s i o n a n d let

E of degree m a n d n, respectively, over F s u c h that ( F ( a , P ) : F )

Q

a n d P be e l e m e n t s of

= mn.

426

CHAPTER 8

(i) For any conjugate a; of a and

Pj

ofp,

u(a)= a ,

there ezists u E G a l ( E / F ) such that and

u(p) = P j

(ii) If no digerenee of two conjugates of a equals a digerence of two conjugates of

f3, then

Proof. (i) Since E / F is a finite Galois extension, we may choose f E G a l ( E / F ) such t h a t f(a;)= a. Write

f(P,)

= p k for some conjugate P k of

p.

w e claim t h a t there exists

f l E G a l ( E / F ) such that fl(a) = a and fl(Pk) = P; if sustained, t h e assertion will follow

by taking u = ( f l f ) - ' . Let 81,. . . ,P, be all conjugates of of

P

PI,.

P

over F. Over hypothesis implies that the degree

over F(a) is still n, so that the roots of its minimal polynomial over F(a) are still

. . ,P,.

Because E/F(a) is Galois, the required automorphism exists.

(ii) Let a1,.. . , a , be all conjugates of a over F . Then, by (i),

are all conjugates of a

+ P.

If a coincidence occurs, then the difference of two conjugates

p.

of a will equal the difference of two conjugates of are distinct, proving that the degree of

Therefore, by hypothesis, all a ,

a + P is mn. Thus F ( a , P )

+Pj

= F ( a + P ) as asserted.

W

We are now ready to prove 1.9. Theorem. Let F be a field and let p be a prime distinct from chark'. Let a and

p

be roots of irreducible polynomials of the form X P - a , X P - b over F . I f ( F ( a , P ) : F ) = p 2 , then F ( a + p ) = F ( a , p ) , i.e. a + p has degree p 2 over F .

Proof. Because p # charF, a and /3 are separable over F. Therefore there is a finite Galois extension E/F with a,P E E. Owing t o Lemma 1.8, it suffices t o verify that no difference of two conjugates of a equals a difference of two conjugates of difference of two conjugates of a has the form

. (6' -

.

where

E J ) ~ ,

6

P.

Now the

is a primitive p t h root

of unity. Thus, if a difference of two conjugates of a equals a difference of two conjugates of

p, then a / P E F(E).But (F(E) : F) 5 p - 1, while the degree of a / P over F must divide

p z = ( F ( a , p ) : F ) . Thus

true. w

alp

E F, contrary t o ( F ( a , p ) : F ) = p 2 . So the theorem is

IRREDUCIBILITY

BINOMIALS

OF

427

Our nest aim is t o provide a n application of Theorem 1.6 which ensures the existence of extensions of large degree. As a preliminary result, we prove 1.10. Lemma. Let F be a field of characteristic p

be irreducible over F. If X i s a root of XP - X

-

> 0 , let a E F a n d let XP - X

a, t h e n XP - X

-

aXP-'

-a

is irreducible

ouer F ( X J .

Proof. Assume by way of contradiction t h a t XP

-

X

- aXp-'

Theorem 7.2.7(iii), it has a root p in F(X). We may write p = a0 with a , E F. Then, because

is reducible. Then by

+ alX + . . . + a,-lXP-'

+ a , we have

XP =

X

XP-'

in (3) and (4), we derive

On the other hand

Equating the coefficients of

dicts the irreducibility of XP

-

X

-

= a p - l + a , which contra-

a over F.

We are now ready t o prove the following result. 1.11. Theorem. Let F be a field a n d let p be either a n odd p r i m e o r p = 2 and

churl: = 2 . I j F h a s a n e z t e n s i o n whose degree i s divisible by p , t h e n F has a n e z t e n s i o n whose degree is divisibly by p n jor a n y given n

2

1.

Proof. If c h a r F = p and F is not perfect, then X Theorem 1.6, XP"

-X

is irreducible for all n

4

FP for some X E F. Hence, by

2 1. Therefore, if c h a r F = p , we may assume

that F is perfect. Let E / F be a field extension with p

F in E . If c h a r F = p , then K

=

1

( E : F ) and let X be the separable closure of

E since by assumption F is perfect. If c h a r F = q = 0,

then K = E . Finally, if c h a r F = q

> 0 and

q

# p , then by Proposition 2.2.30 and 2.2.33,

( E : K ) is a power of q, so p divides ( K : F ) . We may thus assume t h a t E / F is separable. By passing to a normal closure of E , we may further assume that E / F is Galois. Applying Galois theory and the existence of a n element of order p in G a l ( E / F ) , we may assume that E / F is a Galois extension of degree p .

CHAPTER 8

428

If charF = p , then by Theorem 7.2.7, E = F ( X ) with X a root of an irreducible polynomial over F of the form X P - X - a .

Thus, by Lemma 1.10, there exists an extension

of F of degree p 2 and the process can be iterated to get degree pn for any n.

Now suppose that charF

#

p and let K denote a splitting field of XP

-

1 over E. If

T is obtained from F by adjoining a primitive p t h root of unity, then KIT is Galois with Galois group cyclic of order p . By Lemma 1.1, K = T(X) with X a root of an irreducible polynomial X p - a , for some a E F . Thus a irreducible over F , for any n

$ FP and, by Theorem 1.6, XP" - a is

2 1. This completes the proof

of the theorem.

A useful companion to the result above is the following 1.12. Theorem. Let F be a field having an eztension of degree divisible b y

for any n

2 2, F has an eztension of degree divisible

4. Then,

b y 2".

Proof. If charF = 2, then the assertion follows by virtue of Theorem 1.11. Therefore assume that charF # 2. By the argument in Theorem 1.11,we may assume that the given extension E of F is such that E / F is Galois with ( E : F ) = 4. Consider the field E ( i ) with i2 = -1. Then E ( i ) / F ( i )is Galois of degree 2 or 4. In either case, E ( i ) contains a quadratic extension of F ( i ) and therefore F ( i ) contains an element a with no square root in F ( i ) . Moreover, -4a cannot have a fourth root in F ( i ) , for otherwise -a is a square and so is a. Applying Lemma 1.5(ii), we deduce that F ( i ) has an extension of degree 2" for all n

2 2.

This proves the theorem.

We now apply the foregoing results to the study of ordered fields. Let R be a commutative ring. We say that R is an ordered ring if there is a total ordering > on R such that (a) z

> z', y > y'

(b) z > 0,y > 0

+z+y zy

> z'

+ y'

>0

In discussing the ordering on R , we shall write z 2 y to mean z > y or z = y, and use

<, 5

for the opposite ordering. An element z E R is said to be positive or negative according as z

> 0 or

z

< 0. In any ring R , a cone is a subset P containing 1 but not 0 and closed

under addition and multiplication.

1.13. Lemma. Let R be a commutative ordered ring. (i) The set P of positive elements of R is a cone (called a positive cone of R ) and R =

P

u - P u (0) (disjoint union), where - P

=

(-212

E P}.

(ii) Every cone P of a commutative ring S such that S = P U - P U ( 0 ) defines an ordering

IRREDUCIBILITY

o n S b y putting x

> y if and only if x

-

OF

BINOMIALS

429

y E P.

(iii) R is a n integral domain of characteristic 0 , and the square of any nonzero element of

R is positive. Proof. T h e proof of (i) and (ii) is straightforward. Given z,y E R, z , y # 0, if z, y

> 0,then x y > 0. Similarly, if x, y < 0, then

-z, -y

> 0 and so xy

= (-z)(-y)

> 0.

> 0 > y , then 2,-y > 0, so -zy = z(-y) > 0 and < 0 < y , then xy < 0. In each case zy # 0, so R is a n integral

If z and y have opposite signs, say thus z y

< 0. Similarly, if

I

domain. In particular, if z = y , then the last two cases cannot occur and z2 > 0. Finally, note that 1*= 1 > 0 and that 1,l

+ 1,.. . are all > 0, hence charR = 0. w

With this information a t our disposal, we now prove 1.14. Theorem. Let F be a field which has a quadratic eztension but no extension of degree

4 . Then F is a n ordered field i n which every positive element has

a square root.

Proof. Let P denote the subset of F consisting of all nonzero squares. Then P is obviously closed under multiplication. Thus, in view of Lemma 1.13(ii), we are left t o verify t h a t P is closed under addition and t h a t for any 0

# a E F , either a

or

--a

is a

square, but not both. Invoking Theorem 1.11, we must have c h a r F

#

2. Accordingly, by Lemma 1.5(iii),

for every a in F , either a is a square or -4a is a fourth power; otherwise X4 - a would be irreducible over F and yield an extension of degree 4 which is impossible by assumption.

In particular, either a or - a is a square. If -1 is a square, then every element in F would be a square, and no quadratic extension of F could exist. Hence, for any 0

#

a E F , either

a or - a is a square, but not both. By the foregoing, we are left t o verify that the sum of two squares is a square. To this end, we form the field F ( i ) with i2 = -1. In F ( i ) every element must be a square, since otherwise there would be a quadratic extension of F ( i ) and hence a n extension of I: of degree 4. Applying the fact that a

+ bi

is a square in F ( i ) , we find t h a t a 2

+ bZ

is a

square in F , as required. 1.15. Corollary. (Artin-Schreir). Let E I F be a finite field eztension with E al-

gebraically closed but F not algebraically closed. Then F is an ordered field and E =

F ( i ) , i2 = -1. Proof. By assumption, E is the algebraic closure of F . Hence the degree of any

CHAPTER 8

430

finite extension of F is bounded by ( E : F). It follows, from Theorems 1.11 and 1.12, that ( E : F) = 2. Hence, by Theorem 1.14, F is a n ordered field in which every positive element has a square root, and E must be F(Z). As a preliminary t o our final result, we record 1.16. Lemma. Let F be a field such that for some prime p every nontrivial finite

eztension of F has degree divisible b y p . Then every finite eztension of F has degree a power o f p .

Proof. Let E / F be a finite field extension. If c h a r F = q > 0 and q # p , then F is perfect. Indeed, otherwise, by Lemma 1.3, F has a n extension of degree q. If c h a r F = p and K is the separable closure of F in E, then ( E : K ) is a power of p (Propositions 2.2.30 and 2.2.33). We may thus assume that E/F is separable. Moreover, by passing t o a normal closure of E over F , we may assume that E/F is Galois. Let P be a Sylow psubgroup of G = G a l ( E / F ) and let K be the corresponding subfield. Then ( K : F) = (G : P ) is prime t o p . By our assumption, this ensures t h a t K = F . Hence ( E : F ) is a power of p , as asserted.

1.17. Theorem. Let F be a n ordered field i n which every positive element has a square root. Suppose, further, that every polynomial of odd degree over F has a root in F. Then F(i), iz = -1, is algebraically closed,

Proof. By assumption, F has no finite nontrivial extensions of odd degree. Applying Lemma 1.16, we deduce that the degree of any finite extension of F is a power of 2. We claim t h a t t h e only nontrivial finite extension of F is F(i), which will clearly yield the result. Let E / F be any finite extension and let G = G a l ( E / F ) . By Lemma l.l3(iii), c h a r F = 0 and thus E/F is separable. By passing to the normal closure of E , we may assume that

E/F is Galois. We know t h a t G is a 2-group. If ( E : F ) > 2, then G has a subgroup of index 4 which in turn is contained in a subgroup of index 2. This gives a chain F c F1 c F2 of fields with (Fl : F) = (Fz : F1) = 2. Because F(i) is the only quadratic extension of F, we must have F1 = F ( i ) . T h e latter easily implies t h a t every element in

hence F z cannot exist, a contradiction. w

F1

is a square,

SOLVABILITY OF GALOIS GROUPS

43 1

2. Solvability of Galois groups of radical extensions.

Our aim in this section is twofold: first to prove t h a t if E / F is a radical extension, then for any intermediate field K , G a l ( K / F ) is solvable. Second, t o show that if F is a field of characteristic 0 and E / F is a finite Galois extension with solvable Galois group, then E can be embedded in a radical extension of F . The following three preliminary observations will clear our path.

Lemma.

2.1.

Let E I F be a field eztension and let E 1 / F , ..., E,IF

be radical

subeztensions. T h e n ( E l E 2 . . . E , ) / F i s a radical extension.

Proof.

It suffices t o treat the case n

=

2. Let E l = F ( X 1 , . .,A,)

and Ez =

F ( p 1 , . . . ,pk) exhibit the assumption that E I I F and EZIF are radical extensions. Then

shows that E l E 2 / F is a radical extension.

CK 2E

2.2. Lemma. Let F

be a chain of fields

(i) If E / F i s radical, t h e n so is E I K

(ii) If K J F i s radical and E i s the normal closure of K over F , t h e n E I F is radical. Proof.

(i) If E = F(A1,

E = K ( A 1 , .. . A,),

with A':

...,, ,A

with A':

E K(A1,. . . ,A,-l),

E

F(A1, ..., A,-I),

1

5 i 5 m, then

as required.

(ii) Because K / F is finite, we may write K = F ( a l , . . . , a a ) where a, is a root of an irreducible polynomial f , ( X ) over F, 1 5 i 5 s. If A, is any root of fi(X), then F(X1,.. .,A,) is 8'-isomorphic to K and hence F(A1,. . . ,A , ) / F is radical. Since E is the composite of a

finite number of fields of the form &'(XI,.

. . ,Aa), the required assertion follows by virtue

of Lemma 2.1. 2.3. Lemma. ( i ) Let p be a prime and E a splitting field of XP

~

1 over F . T h e n

G a l ( E / F ) is cyclic. (ii)

If F i s a field in which Xn - 1 splits i n t o linear factors,

a an arbitrary element

01 F

and E the splitting field of Xn - a over F, t h e n G a l ( E / F ) i s cyclic.

Proof. (i) If c h a r F = p, then E = F and there is nothing t o prove. If c h a r F # p , then E

=

F ( E ) where ,

Corollary 6.4.15.

t

is a primitive p t h root of unity. Hence G a l ( E / F ) is cyclic b.

432

CHAPTER 8

(ii) If u is one root of X" - a , then the general root has the form E

where

6"

= 1 and so

lies in F. It follows that E = F ( u ) , and that an automorphism of E/F is determined by

its value on u. Now all elements (T

EU

E Gal(E/F), then

E

with

E"

u ( u ) = 6a"u for some

= 1 form a cyclic group, say generated by 6. If

i, 2 1. The map Gal(E/F) +<6 >,

(T

++

6'0

is obviously an injective homomorphism, hence the result. We are now ready to prove the following result. 2.4.

Theorem. Let E/F be a radical eztension. Then, for any intermediate field

K, Gal(K/F) is solvabie.

Proof. Let K be an intermediate field of E/F and let KO be the fixed subfield of Gal(K/F). Then K / K o is Galois and Gal(K/F) = Gal(K/Ko). Because, by Lemma 2.2(i), E / K o is radical, we may assume that K = KO, i.e. K/F is Galois. If L denotes a normal closure of E over F, then by Lemma 2.2(ii), L / F is radical. Hence we may also assume that E/F is normal. Thus Gal(K/F) is a homomorphic image of Gal(E/F) and we are left to verify that Gal(E/F) is solvable. Let E = F ( A 1 , . . . ,A,) A's,

if necessary, we can arrange that in each case a prime power of A,

. . ,A;-l).

We shall now argue by induction on n . By assumption, AT E F for

inserting further lies in F ( A 1 , .

exhibit the assumption that E/F is a radical extension. By

some prime p . Let Eo denote a splitting field of XP

Eo generated by F and the roots of XP

- 1.

-1

over E and let El be the subfield of

Because Eo/F is obviously normal, Gal(E/F)

is a homomorphic image of Gal(Eo/F). It therefore suffices to verify that Gal(Eo/F) is solvable. The extension E1/F is obviously Galois and hence cyclic, by Lemma 2.3(i). Bearing in mind that Gal(E o / F )/ G a1(Eo/ E 1) Z Gal (E / F ) it suffices to show that Gal(Eo/El) is solvable. Now Eo = El(A1,..., An), for E o is generated over F by the A's and the roots of XP G = Gal(Eo/El) and H

- 1,

=

Because XP - 1 factors completely in El,E , ( A l )

and the latter are already in El. Put

Gal(Eo/El(X,)) is splitting field of XP

-

AT over El

and hence E1(A1)/Elis normal. Moreover, by Lemma 2.3(ii), Gal(E1(A)/El) is cyclic. However, E / F is normal and radical, hence Eo/EI must be normal. It follows, from Lemma 6.5.2, that H a G and G / H is cyclic. To prove that G is solvable, it remains to

SOLVABILITY

OF GALOIS GROUPS

433

show that H is solvable. This follows from over inductive assumption, since E Q / E ~ ( A is ~ ) a radical extension generated by n - 1 elements, namely Xz, . . . A,.

So the theorem is true.

a Let F be a field and let f ( X ) 6 F [ X ] .Recall t h a t the Galois group of f ( X ) is defined

to be the Galois group of a splitting field of f ( X ) over F. 2.5. Corollary. Let F be a n arbitrary field and let f ( X ) be an irreducible polynomial

over F . If there exists a radical extension E / F which contains a root of f ( X ) , then the Galois group of f ( X ) is solvable.

Proof. Let Eo be the normal closure of E over F. Owing t o Lemma 2.2(ii), E o / F is radical and hence we may assume t h a t E / F is normal. Therefore E contains a splitting field K of f ( X ) over F and, by Theorem 2.4, G a l ( K / F ) is solvable. rn Our next result provides a partial converse to Theorem 2.4. 2.6. Theorem. Let F be a field of characteristic 0 and let E / F be a finite normal

eztension with a solvable Galois group G . Then E can be embedded in a radical eztension of F .

Proof. We argue by induction on ( E : F ) . Because G is solvable, it contains a normal subgroup H of prime index p . Let K be a splitting field of XP

-

1 over E . Then K / F

is normal and, since G a l ( K / E ) is cyclic (Lemma 2.3(i)), the group G a l ( K / F ) is solvable.

Let I, denote the subfield of K obtained by adjoining t o F the roots of XP

-

1. Then

K / L is also normal. Because L / F is a radical extension, it suffices to show t h a t K can be embedded in a radical extension of L.

To this end, we first prove that Gal(K/L) is isomorphic t o a subgroup of G . Indeed, consider the map

{

Gal(K/L) u

--t

Gal(E/F)

++(TIE

Then this map is a homomorphism, which is injective since u ( E = identity implies that leaves both L and E elementwise fixed and hence

(T

0

= 1.

If Gal(K/L) is isomorphic t o a proper subgroup of G, then by our inductive assumption K can be embedded in a radical extension of L. We may thus assume that

Gal(K/L) 2 G. Let T denote the intermediate field corresponding t o H . Then (T : L ) = p , T is normal over L , and L contains a primitive p t h root of 1. Because T / L is cyclic of degree

CHAPTER 8

434

p , it follows, from Lemma 1.1, t h a t T / L is a radical extension. Now KIT is normal with

solvable Galois group H . By induction, K can be embedded in a radical extension of T . Thus K can be embedded in a radical extension of

L,as required.

H

Let F be a field and let f (X)be a polynomial of positive degree over F. Then the equation

f (XI= 0 is said t o be solvable b y radicals over F if each root of f ( X ) lies in a suitable radical extension of F . In view of Lemma 2.1, this is equivalent t o the condition t h a t a splitting over F can be embedded in a radical extension of F. field of f (X) 2.7. Corollary. (Galois's criterion). An equation f ( X ) = 0 i s solvable b y radicals

over a field F of characteristic 0 if and only if the Galois group of f ( X ) i s solvable.

Proof. Apply Theorems 2.6 and 2.4.

H

3. Abelian binomials.

Let n be a positive integer and let F be an arbitrary field. Our aim is t o provide necessary and sufficient conditions under which the Galois group of a binomial abelian. We also determine the Galois group of

X" - a ,

a E F is

X" - a in case it is abelian and X" - a is

irreducible. For any integer d 2 1 not divisible by charF, we write

for a primitive d-th

€,j

root of unity over F . The following simple observation will allow us t o assume t h a t n is not divisible by char F. 3.1. Lemma. Assume that charF = p

the Galois groups of X" - a over F and of

Proof. Let

F

> 0 and that n

X"

-

ap

-k

= pkm with ( m , p )= 1.

over F(aP-') are the same.

be an algebraic closure of F and let E be the splitting field of

over F contained in

F.

Then

X" - a

Since

it follows that E = F( A1 , . . . , A m ) , where X I , . . . , A m are all distinct roots of Xm - ap-' in

F.

Since

AT

=

ap-' E E , we deduce that E is the splitting field of X"'

-

ap-' over

F(ap-' ). But G a l ( E / F ) = G a l ( E / F ( a P - k )since ) F ( a p - ' ) / F is purely inseparable, hence the result.

435

ABEL I A N B I N O M I A L S

3.2. Theorem. (Schinzel (1977)). Let F be a field, let n be a positive integer not

divisible by eharF and let m be the number of n - t h roots o f unity contained an F . Then, for any given a E F , the Galois group of Xn- a is abelian if and only if am = A" f o r some X E F.

Proof. (Wojcik(lQ82)).Let F a b b e the maximal abelian extension of F (in a given algebraic closure of F ) . If

am =

A n for some X E F , then

"fiE

Since F contains a primitive rn-th root of unity, we have 2.3(ii). Thus

fi E

Fabby virtue of Lemma

F a b and F ( fi,c n ) / F is abelian.

Conversely, suppose that the Galois group of Xn- a is abelian. Let k be the maximal divisor of n for which am = Xk for some X E F . It suffices t o verify that k = n. Obviously

k 5 n and m l k . Assume by way of contradiction that k < n. Fix a prime p dividing n / k and denote by p a the highest power of p dividing m. P u t

t = p 3 + ' k / m , L = F ( c p d + , and ) E = F(

'"z)

Then

E

C F(i/lE,~t,) C F ( i%,€tm) C Fab

fi E F a b . By the definition of k , we have ( E : F ) = psi'. Hence & L C E . Furthermore, since p3+11n and p 3 is the highest power of p

since, by hypothesis E ~ . + I

E E and F

dividing m, we have cpd E F hut cp.+L 6 F . We now distinguish two cases. Suppose first that s = 0. Then ( L : F ) divides p

-

1. Hence L = F and cp E F , a contradiction. Now assume that s 2 1. We have

( E : L ) < pa+' and E = L ( '"fi). Thus X = A: for some XI 1.5. It follows t h a t

F(e/r;,

L = F(*)

which, by Corollary 7.4.7, implies that X = c',. A;

But then

am =

for some j

2 0,

Xz E F

with pkjn, contrary t o the definition of k . w

L , by virtue of Lemma

436

CHAPTER 8

T h e rest of this section will b e devoted t o determining t h e Galois group of case it is abelian and

X" - a

X" - a in

is irreducible.

3.3. Lemma. Let F be a n arbitrary field and let 0

X'

eztension of F such that A' E F and

# X be a n element of a field

- A' is irreducible over F. Then for any n

2

1, A" E F if and only if rln.

Proof. Our assumption guarantees that the order of XF' in F(X)*/F* is precisely r . Since A" E F is equivalent t o (XF')" = F', the result follows.

3.4. P r o p o s i t i o n . (Norris and VBlez (1980)). Let F be a n arbitrary field, let n be a

positive integer and let X be a root of X" positive integer dividing n for which

E,

- a for E F(X).

some 0

#a

E F. Let m be the mazimal

Zf X k 4 F for 1 5 k _< n and K is a n

intermediate field of F ( X ) / F ( c , ) , then !or t = (F(X) : K ) and r = ( F ( X ) : F(c,)),

we

have (i) K = F(Xt)and X t - At is irreducible over K (ii) F(X') = F(E,) and rln

Proof. Let / ( X ) be the minimal polynomial of X over K . Since X is a root of

X"

-

a E F[X] C K [ X ] , f ( X ) divides X"

-

a. Let o be a generator of all n-th roots of

unity over F. Then every root of f ( X ) is of the form 6'X for some i. Hence t

f ( X ) = n ( X - 6'iX)

for some ij

21

j=1

The element ni.=l(6iiX) = 6"Xt, s =

C:=,ij, is the

constant term of + f ( X ) , hence it

must belong t o K & F(X). Since At E F(X), we have 6* E F(X) and, by definition of

m , 6" E F(6,)

C K.

Thus A t E K . Now t = ( F ( X ) : K ) and (F(X) : F ( X f ) )5 t since

A' E F(Xt).Hence we must have that F(Xt)= K and X t

-

Xt is irreducible over K . In

particular, if K = F ( E , ) , then t = (F(X): F(E,)) = r . Since A" = a E F

E F(E,) = K ,

we have that r divides n by Lemma 3.3, as required. w Let F be a n arbitrary field. We say t h a t a polynomial

/(X)E F[X]of positive degree

is normal over F if each irreducible factor g ( X ) E F[X] of f ( X ) is such that F(X)/Fis normal for some (hence any) root X of g ( X ) . Assume t h a t the binomial X" Since c h a r F Cn, X "

-

-

a E FIX] is irreducible normal and that c h a r F .+n.

a is separable over F . Hence, for any root p of X" - a , F ( p ) / F is

ABELIAN BINOMIALS

437

Galois and cn E F ( p ) . 3.5. Lemma. Let Xn - a E F ( X ] be irreducible normal with charF cyclic Galois group. If 41n, then

€4

and with

6 F.

Proof. Let p be a root of X" F(p"/')/F.

tn

-

a. By hypothesis, F(p)/F is cyclic, hence so is

Furthermore, X4 - a is irreducible over F (Theorem 1.6). Since pn/4is a root

of X4 - a, it follows t h a t c4 E F ( P " / ~= ) F(u'/~).

Assume by way of contradiction t h a t n = 4, A = a'/'

b = e;'a1/'

€4

$ F. Then, by Proposition 3.4 applied to

and K = F(e4),we have F(c4) = F(a' /' ).

E F ( c 4 ) , and so a l l 4 = b'/'cs.

with ( F ( c 8 b ' l 2 ) : F ) = 4 and

( F ( c 8 ,b'/')

: F(csb'/'))

52

€4

E F(csb'/')). Thus

222

x Z z . This contradicts the

is cyclic of order 4 . Thus

and Gal(F(E8, b ' / ' ) ) / F ) is llrz x E

(since

: F ) = 4 or 8

If the degree is 8, then this field has Galois group Zz x

€4

= bE4 for

Now

( F ( 6 8 , b'/')

fact that Gal(F(b1/'e8)/F)

Hence a'/'

Z Z ,

again contrary to t h e fact that it must b e 224. Thus

F as required. w Let EJF be a finite field extension. We say that E / F has a unique subfield property

i f for every divisor t of ( E : F ) , there exists exactly one intermediate field K of E / F such that ( K : F ) = t.

let

# n,

3.6. Lemma. Let X "

-

a E F [ X ] be irreducible with charF

be roots of X"

-

a, Xn - b, respectively, such that F(u'/")

b'l"

F ( a ' / " ) / F has a unique subfield property, then

f o r some c E F and some integer t coprime to n.

let b E F and = F(b'/").

If

CHAPTER

438

8

We first assume t h a t n = p k , p prime. For k = 1 the assertion follows

Proof.

from Corollary 7.4.8. Therefore assume the lemma is true for k and let

7~

= pk+'.

Then

F ( a ' / p k + ' )= F(b'/P'+'). Because F ( u l / P k + l ) / Fhas a unique subfield property, we have that

b'/Pk = C ( a ' / P L ) t for some c E F and some t with ( t , p ) = 1. Therefore the ppower case follows provided c'/P E F . If cl/P

4

bl/pk+'

= ~ ~ / P ( a ' / p ' +and ')~

F , then F(c'/P) = F(al/P), so

c ' / ~= c l ( a l / P ) t l for some t l with ( t l , p ) = 1. Hence

which implies

bl/Pk+' - c1 ( a l / P k + l ) t f t i P k

with

(t

+tlpk,p) = 1

proving the prime power case. We now employ induction on t h e number of distinct prime factors in n. To this end, write n = r s , ( r , s ) = 1, r

> 1, s > 1. Because F ( a ' / " ) / F has a unique subfield property,

we have

F(u'/') = F ( b ' / ' ) , F ( a l / S )= F(b'/') It follows by induction that

Because ( r , s) = 1, we have z r

+ s y = 1 for some integers z, y. (z/s)

where ( y t l s

+ ( y / r ) = l/rs = 1/72

+ z t z r , n ) = 1, as required.

As a final preparatory result, we now prove

Hence

439

ABELIAN BINOMIALS

a E F [ X ] ,m

2

1 , p prime, be irreducible normal with

abelian Galois group. If p is odd or i f p = 2 and

€4

E F , then the Galois group of XPm- a

3.7. Lemma. Let

XPm -

is cyclic

Proof. If c h a r F = p , then the Galois group of assume t h a t p

XPm -

a is identity. Hence we may

# charF. By Theorem 1.6, XP - a is irreducible. It is also normal, since the

p is any root

Galois group of XPm- a is cyclic. Hence, if

( F ( P ) : F) = p , which implies e p E F. Also, if

(Y

of XP -a, we have

is any root of

XPm- a ,

tp

then

E F ( P ) and tpm

E F(a).

Let t be the number of pm-th roots of unity contained in F . Then t = p* for some s and XP*

- tp.

is irreducible over F for any k

hypothesis that if p = 2, then

€4

2

21

1. The latter is true by applying the

E F together with Theorem 1.6. Thus

F ( c p m : F ) = pm-'

and ( F ( a ) : F ( e p m ) )= p a

If m = s, then the assertion is obvious. We may thus assume t h a t s < m. Now consider the field F ( c p m + . ) . Then it has degree p m over F and has cyclic Galois group. If u is the generator of this Galois group, then u(cpm+.) = tbmcpm+rfor some z, and o(tpm) = e$cpm. Owing t o Theorem 3.2, we have a

UP'

= bPm for some b E F , so a = ep.bPm-' and

= C ~ . , . + ~ ~ ' / P ' .Set

r ( a ) = t6-a = c $ , ~ ~ , + . b ' / P ' (hence .(a) is a conjugate of a). By Proposition 3.4, we have t h a t F ( a P ' ) = F ( E ~ - ) , and F ( c p m )has a unique subfield property, hence

tpm

= COP',

c E F , by Lemma 3.6. It

therefore follows that .(tP".)

= .(cap") =

Note that

.(tpm)

..

= ,(,;,,)pa .

o

= t~,.epm

= a(tpm). Hence r is a n automorphism and it has the same order as LT,

proving that the Galois group of

XPm -

a is cyclic. rn

We have now come t o our second major result.

3.8. Theorem. (Schinzel(l977)). Let F be a field, let n be a positive integer not divisible b y charF and let a E F be such that Xn - a is irreducible over F with abelian Galois group G . If 4 / nand

€4

$ F , then G

U z x U n i 2 ,otherwtse G i s cyclic.

440

CHAPTER 8

Proof. (VelBz ( 1 9 8 0 ) ) . Let n =

nf=,p:' be the canonical decomposition of

n. Then

XP:' - a is irreducible by Lemma 1.2. Furthermore, XP:' - a is normal with abelian Galois group. If G, is the Galois group of XP:'- a , then G , is a p,-group and G

G

G1 X G2 X . . .X G k .

If p , is odd, then G, is cyclic by Lemma 3.7. Hence G is cyclic if and only if Gz is cyclic. If

E F and 41n, then Gz is cyclic by Lemma 3.7, hence G is cyclic.

€4

Assume that 41n and c4

4

F. Then, by Lemma 3.5, G is not cyclic. If X is any root

of X n - a , then X n l 4 is a root of X4 - a , and X4 - a is irreducible normal. Hence, by Proposition 3.4, we must have F(X"Iz) = F ( c 4 ) . Thus X"/' F(Xn12),

262 x

G

A"/'

E F ( X n / * ) and the Galois group H of F(X) over F(X"/')

€4

have G f H

-

E Z 2 Zn/2,

and hence G

E

is irreducible over is cyclic. We then

2 5 2 x Z n p . But G is not cyclic, hence

Z,,or G

as required. m

4. Normal binomials.

Let F be an arbitrary field. Recall that a polynomial f(X)E F [ X ]of positive degree

of f(X)is such that F ( X ) / F is is called normal if each irreducible factor g(X)E F[X] Thus f ( X ) is normal if and only if so is every normal for some (hence any) root X of g(X). irreducible factor of f ( X ) . Our aim is to describe all normal binomials of the form xp" -

a

( n 2 1, a E F)

where p is a prime distinct from charF. In accordance with our previous convention, for any d 2 1 not divisible by charF, we write

6,j

for a primitive d-th root of unity over F .

4.1. Lemma. Let F be a n arbitrary field with charF

are infinitely many integers t with p E

+: ;c

€ 2 ~

E

2 1 with €2'

+ cs'

# 2 . Define k to be

00

if there

E F, otherwise let k be the largest integer

F. If X E F ( c 4 ) is such that A" E F for some n 2 1, then there ezists

F such that at least one o f the following properties holds:

(i) A" = p" (ii)

R.

$O(mod 2 k ) , A" = -p"

(iii) k is Fnite, n = 2 k ( m o d 2 k + ' ) , A" = -(cZk

(iv) k is finite, n z O(mod 2k+1), A" =

Proof. Setting

c4 =

i, we have X

(cZk

= LY

+

6;:

+ 2)n/2pn

+; ;c + 2)"/'pn +pi

with a l pE F . If

i E F , then X E F and

NORMAL BINOMIALS

(i) holds for p = A. Assume that a

4 F.

Then ( a

441

+ pi)"

=7E

F implies ( a - pi)" = 7 ,

hence cy

+ pi = €;(a- pi)

for some v

20

(1)

and thus

It follows that the only conjugate of

6;

over F is e n " and the only possible conjugates of

€inare 6, (€2"

is chosen so t h a t

E&,

€2:

( 6 , = il,6 2 = *I)

= en). P u t m = ord2(2n/(2n,v))

where ordz(n) denotes the highest power of 2 dividing n. Then €2"

= eyi

with

s

= l(mod 2 )

(3)

If u is an automorphism of F ( & ) and

then we have U(€z".)

If m = 2 then

cz

= &€$,

# 1; by (1) a = pi-

€;

+1

€;

-

1

and, by (4) and (5), for all automorphisms o of F ( E ; , ) over F

Consequently 2i/(etln v is odd.

~

t;,U)

E F and, by (2) and (6),we obtain (i) if u is eve2 and ( i i ) if

CHAPTER 8

442

If m # 2,

E:

#

-1 and by (l),

pi

=

-1

€;

a-

+ 1'

€;

If u is an automorphism of F ( & ) and

then by (4) and ( 5 ) ,

u(6) = 6 Accordingly 6 E F and, because czrn

+ c?,;

# 0, it follows from (7) t h a t

A" = (-1)"(€p+ €22)"

We claim that rn 5 k

+ 1.

(3

Indeed, if rn = 0 this is obvious and if m > 0 it follows from

(3) that

+

Ezm-1

hence m

-

€&

=

+

€;"a

-

___

a

-

pi

(-)aa -+ppii

E F,

1 5 k.

Denoting by p a suitable element of F we deduce from (2) and (8) that: (a) If rn

5 k, v

(b) If rn 5 k, v

E

O(mod 2), then An = p". l ( m o d 2 ) , then A" = - p n , n

+ 1, Y E l ( m o d 2), then A" If m = k + 1, v = O(mod 2), then A"

(c) If m = k (d)

8 O(mod 2 k )

+ E;> + 2)"I2p" and n = 2k(mod2k+1) = ( t p + €2.1 + 2)n/2pn

=

-(€2h

Since the above corresponds t o the properties (i),(ii),(iii),(iv),respectively, the result follows. 4.2. Corollary. Let F be a n arbitrary field with charF

integer s u c h that

€2'

+ €2.'E F

IJ X E F(c4) is such that A'" following properties holds:

E

# 2 a n d let k be the greatest

if there are finitely m a n y of t h e m , otherwise p u t k = 00. F , t h e n there exists p E F s u c h that at least o n e of the

NORMAL

(i)

~

2

(ii)

Y

< k , Xzu

p2”

”

1

= -p2”

xZy =

(iii) v = k , (iv)

> k , Xzy

V

443

BINOMIALS

=

-(€Zk

(€2k

+ +2)2y~1pzy + + 2)2y-’pzy 6;:

6;:

Proof. This is a special case of Lemma 4.1 in which n

=

2”.

4.3. Lemma. Let F be a n arbitrary field with charF = 2 , let k be as in Corollary

e

4.2 and let

be the greatest integer such that

of t h e m , otherwise put

e= Proof. If t

and thus

e 2 k.

hence

E

that

€4

-1-2k-1

and

k

Then

+1

ifk

k,

otherwise

< 00 and

J

-(‘2k

+ €2.’ + 2 ) E F

2 2, then we have

If k < co and

F and

C2k+2

€zk+l

€

E F,

F(64), then

is conjugate over F t o

then

{

00.

E F ( q ) , if there are only finitely m a n y

E F ( E ~ then ) , by Corollary 4.2,

tZk+2

€2k+2

+ c;>+~

E F , contrary to t h e definition of k . This shows

e < k + 2. If

c2*+,

e=

t2t

E

€4

C~E;;+,

$i F!

and

or t o

C2k

is conjugate over F to

Therefore

But the latter possibility gives

-E~E;>+,.

F , contrary t o the definition of

6;:.

T.

€4€2k+l

E$!:,~-’

+

Hence the former possibility holds and

CHAPTER 8

444

So the lemma is true. 4.4. Lemma. Let F be a n arbitrary field, let p be a p r i m e with p

# charF and let

c p E F . A s s u m e that a,P are in a field extension of F a n d such t h a t

T h e n either p E F * < a > or p = 2 ,

€4

4F

and

€4

E F' < a > .

Proof. We argue by induction on n and m. If n = 0 or m = 0, then there is nothing t o prove. Assume the assertion is true for n = t

-

1 and all m. We first show t h a t it is

also true for n = t , m = 1. Assume t h a t a E F ( a P ) . Then, applying t h e induction hypothesis with a,

ml

= 1 we get either a E F * or p = 2 ,

possibility gives a E F , the latter Next assume t h a t a polynomial X p - a

p

4

€4

€4

4F

and

€4

E F*

a1 = a p ,

=

< a'>. The former

E F * < a > , so in this case lemma holds.

F(aP). Then acp 4 F ( a P ) and a is a root of the irreducible

E F ( a p ) . Let N denote the norm from F ( a ) t o F ( a P ) . Then we have N ( a ) = (-1)P-lap

On the other hand,

E F'

@P

< a >, so

distinguish two cases, namely q

p p

= uaPk+qfor 0

> 2 it follows t h a t

a p

5

q

< p, a

E F * . We now

q = 0.

> 0. Taking norms of both sides of

Consider the case q

For p

> 0 and

PP

= ucyPk+q, we get

E F(aP)P and a E F ( a p ) , which was excluded. For p = 2 , we

get -az = ( ~ ( p ) ) ' a - ' a - ' ~ , e 4 a E ~

hence

€4

hZ,p

=

a1

E F ( a ) , c4

4

F ( a 2 ) . Writing ,8 = g

( 2 ) , =p k ~c 4 ~ ( p )

+ tqh with glh E F ( a 2 ) , we obtain g2

( 1 k q ) g . Thus g 4 = -p4/4 E F' < a2 > and, by the induction hypothesis, with

= 0 2 , PI = g, ml = 2, we deduce that g E F' < a 2>,

Finally, consider the case q = 0. Let

€4

= +(1/2)p2g-2 E F' .

+ be a n automorphism of the normal closure of

F ( a ) over F ( d ) such that +(a)= acp. Since q = 0, we deduce that PP,

=

+(p) = E ~ PIt.

follows that

PP

E F(ap),+(Pp) =

NORMAL BINOMIALS

Because ppa-'p E F'

< ap >, we may apply

m = 1 t o obtain pa-'

pcy-',

possibility gives that s

2

p

E F*

and

€4

the induction hypothesis with

E F * < a P > or p = 2 ,

< a > and the proof for n

2, that lemma holds for n = t , m

induction hypothesis with

= pp,

445

€4

6 F , E E~ F'

a1 = ap,

< a 2 >. The former

= t , rn = 1 is complete. Now assume

< s and that

m = s - 1, we get

p p

pp"

E F' < a >. Using the

E F' < a > or p = 2,

E F' < a >. In t h e former case we apply the induction hypothesis with

and obtain 4.5.

p E F'

=

€4

ml

$F = 1

< a > ,which completes the induction.

Lemma. Let F be a n arbitrary field, let p be a p r i m e with charF

#

p and

let a be a n element of a field eztension of F . If aP" = p E F, epn E F ( a ) and either p

> 2 , c p E F or p

= 2, c., E F , t h e n

where w is the greatest integer such that epW E F if there are only finitely m a n y of t h e m and w = 00 otherwise.

Proof. Owing to Lemma 4.4, we have in any case. cpn E F * < a > ,

cpn = b a a , 6 E F, 1 5 i

5 pn

19)

Let

i =pkh

( h , p ) = 1,

hj

= l(mod p n P k )

Raising both sides of (9) t o the power j p n - ' , we derive

Thus k

5

w

and the lemma holds with 7 = p(1-hj)Pk-"6-l. rn

We have now come t o the demonstration for which this section has been developed. 4.6. Theorem. (Schinzel (1977)). Let F be a n arbitrary field, let p be a p r i m e with

charF # p and let a E F . T h e n the binomal

XP"- a

CHAPTER 8

446

is normal if and only if there ezists a n integer m and a n element A E F such that at least

one of the following conditions holds:

(i) apmin(u.n) = XP" (ii) p = 2 , w = 1 , n

5 k , a = -A2

+ 1, a = -Az, (iv) p = 2, w = 1, n = k + 1, a = - ( c 2 . + (v) p = 2, w = 1, n 2 k + 2, a = - ( e p +

(iii) p = 2, w = 1, n = k

+

d - ( ~ z k

6;:

+ 2) E F

6;:

+ 2)2"X2m+t ,

6;:

+ 2)2"-2X2"-'

l < m i k - 2

Here w is the greatest integer such that cpW E F if there are finitely many of them, w = 00 otherwise; k is the greatest integer such that

6Zk

of them, k = cm otherwise. Moreover, if XP"

-

+ 6;:

E F if there are only finitely many

a is irreducible over F , then

Proof. We naturally divide the proof into two parts. Step 1 . Here we establish the necessity. Suppose that Xpn- a is a product of irreducible normal factors. Let p be the least nonnegative integer such that

a = pP"-c

for some p E F

If p = 0 then theorem holds with X = /?min(n,w)

.

Therefore if p > 2 or p = 2,

-

64

E F then XP'

So assume that p > 0. Then

p is irreducible and by the assumption

normal. Denoting any of its roots by a , we obtain

Since ( F ( u ) : F ) = p" and ( F ( c p ): F )

5

p - 1, we must have ep E F . Owing t o Lemma

4.5, we have

p

=

and UP' = XP", proving (i).

€;.p-'

05s

5 min(p, w )

(12)

447

NORMAL B I N O M I A L S

Suppose now t h a t p = 2, c4

/3

= -4b4,

4

F . Then either X2'

-

p is irreducible or

p

2

2,

6 E F . In the former case we get again (11) for any root a of X2" - p ; in the

latter case, denote by d the least nonnegative integer such that

The binomial X Z d- q is irreducible in F ( c d ) , hence

is irreducible over F . Furthermore, f ( X ) is normal since it is a factor of

Let

cy

be a root of X Z d- 17, a' a root of X Z d- q', where 9' is conjugate t o 71 over F . Then

we have a' E F(a) a

-

and, on the other hand,

Thus

$ = c;,,

now have

€4

( j , 2 ) = 1 and from (13) we again obtain (ii). Applying Lemma 4.4, we

E F' < a > . Consequently,

2 = (F'

:

F')

=

( F * : F ' )

and, by squaring both sides of (14), we obtain

It follows from (10) t h a t p = n and

=

2/1-ordzt , 2

= 2p-lj,

(j,2)= I

448

CHAPTER

a

On the other hand, applying Lemma 4.5 t o the field F ( q ) ,we obtain

a = c2t0

2"-t

, 0 5 t 5 m i n ( n , l ) , 0 E F(e4)

If n 5 i! we have, by (15) and Lemma 4.3, (ii) or (iii). If n > t, then since

€4

4 F by

(15)

and Corollary 4.2, t = 0 is impossible. We have

and, by Corollary 4.2, either

or n- 1 = k =

Because

€4

e, a p ,= ( € 2 k + 6;: + 2) P-' X2"

$ F, (16) and (18) imply t

Finally, (17) in view of (15) implies t

again by (15) and Lemma 4.3, t

(17)

= 1 and then we obtain (i) or (v) respectively.

> 1 and

< k and the upper sign is excluded. This gives (iv).

Step 2. We now complete the proof b y establishing suficiency. First assume t h a t (i) holds. We argue by induction on n. The case n

5

w being trivial,

assume n > w . Then (i) gives

If s < w , then P- 1

XP" - a =

rpp-'

-

+p,+lpn-w-')

j=O

Each of the factors on the right hand side is by the induction hypothesis the product of normal factors, hence the same holds for XP"

-

a . If s = w = 0, then

NORMAL

BINOMIALS

449

where Qm(X)is the rn-th cyclotomic polynomial. Every root of

en

(x) generates over F

all the other roots, hence the result. Now assume that s = w

> 0 and let

a be a root of XP"

-

a. Then, by

(lo),

which implies €pn

= (apYA-l)J

If (ii) or (iii) holds, then 2"-

€4

Because, by Lemma 4.3,n 5

=fa

1

A-1

e and by definition

e2t

E F ( E ~ it ) ,follows t h a t

If (iv) holds, then ffZk-m

=

€;,+,X(€2k+'

+ €2;:,), ( j , 2 ) = 1

By Lemma 4.3, we have F ( c 2 c ) = F ( Q ) and thus

and €2k+l

E F(a)

If (v) holds, then a2 = cj2"(€2c+' + € & , ) A ,

(j,Z)= 1

which implies

az E

F(E2")

It will be shown that a2 has as many distinct conjugates over F as any automorphism of F ( c p ) over F . Then

€2".

Indeed, let

4 be

450

CHAPTER 8

The latter is impossible, because k-I

and thus €2"

E F(d)

Finally, assume t h a t X'" - a is irreducibIe. Then for p = 2, n 2 2 we have a for k

2 3,

u

# -A 4,

X E F. Therefore (iv) implies k

2 3,

J-(tp

+

: ;6

#

-4X4, hence

+ 2) 4 F, rn = 1;

(v) implies k = 2. So the theorem is true.

5 . Some a d d i t i o n a l results.

In this section we provide some additional results pertaining t o radical extensions. Let

E / F be a field extension and let M be a subgroup of E* such t h a t F'MIF'

is finite.

When is it the case t h a t IF*M/F*I = ( F ( M ): F ) ? An answer t o this question is given by the following result. 5.1. Theorem. (Kneser (1975)). Let E / F be a separable field eztension and let M

be a subgroup of E' such that F'MIF'

is finite. Assume that for all odd primes p , each

p-th root of unity which lies in F * M also lies in F and that i = J-1 E F i f 1 f i E F ' M . Then

IF'MIF'I = ( F ( M ): F ) Proof. We first note t h a t F ( M ) consists of all F-linear combinations of F * M . Thus we may choose finitely many

21,.

. . ,z, in F ' M which is a n F-basis for F ( M ) . Then

z I F * ,... , z n F * are distinct elements of F * M / F * and so ( F ( M ): F ) 5 IF*M/F*I

(1)

Suppose that the result is true for all Sylow subgroups of F * M / F * . If N / F * is a Sylow psubgroup of F * M / F ' of order p t , then p t = ( F ( N ): F ) I ( F ( M ): F )

ADDITIONAL

SOME

451

RESULTS

and thus

IF'MIF'I 5 ( F ( M ) : F ) Applying (l), we may therefore assume that F*M/F*is a pgroup. Choose a chain of subgroups

F' = N o C N1 C ... C Nt = F ' M

5s5

with ( N , : N,-1) = p , 1

t . We show, by induction o n s, that

and that an element of F ( N , ) (respectively, F(N,) n F'M if p = 2 and i E F ( N , ) ) whose p t h power is in N, is itself in N,. Assume that the statement is true for s

-

1. If a E N, is such that aN,-1 generates

N,/NSp1, then a is a root of the polynomial f(X) = X P - a p with coefficients in F ( N + 1 ) . Thus, if f ( X ) is irreducible, then (2) holds. If f ( X ) is reducible, then f ( X ) has a root, say b, in F ( N , _ , ) by Lemma 1.3. Hence b = induction, b E

NSp1and

hence

E

E N,

UE

with

EP

= 1 and b P = u p E N,-1.

F ' M , which implies, by hypothesis, that

By

E F.

E

It follows that a E NSp1, contrary to the assumption that aN,-1 is of order p > 1. This establishes (2). Now assume that

i E F ( N , ) ) . Then

cP

c

E F(N,) satisfies cP E N , (and c E F'M, when p = 2 and

= aqd with 0

5

q

< p and d E N,-1. We wish to show that

c

E N,.

This will be achieved by proving the case q = 0 and by showing that the case q > 0 cannot occur. Suppose t h a t q = 0 so that cP E N,-1. Since E / F is separable, so is F(Ns)/F(N,-l). Applying (2), we deduce t h a t there exists a n F(N8-1)-homomorphism f of F(N,) into the normal closure of F(N,)/F(N,-I) such t h a t f ( a ) have f ( a ) = U E with

t

#

a . Because ! ( a P ) = U P = f ( a ) P , we

being a primitive p t h root of 1. Similarly,

so f ( c ) = cc' for some 0

5

r

5

p

-

f ( c P ) = cP = f ( c ) P

1. The latter implies that f(u-'c)

= a-'c

and

and hence

that a-'c = b E F ( N a - l ) . Furthermore, b P = (cP)-'cP E N,-1 (and b E F'M in case c E

F'M) which shows, by induction, t h a t b E N,-1and hence t h a t

c E

N,.

Assume by way of contradiction t h a t q > 0 and denote by N the norm of F ( N , ) over

F ( N , - l ) . Because N ( a ) = (-l)Pp'aP, we have ((-1)P-lup)q =

N ( C ) P C P

CHAPTER 8

452

For odd p ,

ap

is then a p t h power of an element in F ( N e - l ) , which contradicts ( 2 ) . In case

p = 2, we have -a2 = X2 with X E F ( N , - l ) which shows that i E F ( N , ) , i

4

F(N,-1)

and c 2 = ad = &Ad. Let us write

+ ih

c =g Then cz = ( g 2 - h 2 )

+ 2ghi

=

with

g,h E F(N,-l)

f i X d which implies g 2 = h2 and hence c = (1f i ) g . We

deduce that g4 = - c 4 / 4

E N,-1

which ensures, by applying twice the induction hypothesis, that g E N8-1. But then 1 f i E F'M and hence, by hypothesis,

F ( N , - l ) . This contradiction completes

iEF

the proof of the theorem. rn Let E / F be a simple radical extension, say E = F(X) with X E E' and a power of X in F'. Then we obviously have

(

t ( E * ) F * ) / F *C t ( E * / F * )

where t ( A ) denotes the torsion subgroup of an abelian group A . It is therefore natural to investigate circumstances under which the equality holds. 5 . 2 . Theorem (May(l979)). Let p be a prime and let E I F be a field eztension with

charF # p . Assume that E = F(X) where

XP

E F - FP. If p = 2, further assume that

E # F ( i ) ( i 2= -1). Then t ( E * / F * )= ( < A > Proof. By Lemma 1.3, XP p E

- XP

t(E*)F*)/F*

is irreducible over F and thus ( E : F ) = p . Let

E' be an element of prime-power order modulo F'. First suppose that the order is

q' for some prime q

# p,

N(,u)'f = 7 P , hence 7

and let pq' = 7 E F * . I f N is the norm map from E to F , then

= 7;' for some

71

E F * . It follows that p =

716

for some root of

unity c E F* which shows that pF* E t ( E * ) F * . Now suppose that p has order pr modulo F', and let pPT = 7 E F * . We claim that 7

4

FP. Assume the contrary and write 7 = 7: for some

p t h root of unity. Then

71

E F * . Let

E

be a primitive

453

SOME A D D I T I O N A L RESULTS

# m, for otherwise the order of p modulo F* would be the same reason, we must have 6 4 F'. It follows that We must have p

less than p'. For

E 2 F(t)2 F with (F(c) : F ) > 1. But ( E : F ) = p and (F(E): F) divides p

-

1. This contradiction

substantiates our claim.

>

Except possibly when p = 2 , r

1 and z

4

F, we have (F(p) : F ) = p' by virtue

o f Lemmas 1.5 and 1.3, and thus r = 1. Consider the exceptional case, i.e. assume that p = 2,r

> 1,i 4 F and

(F(p) : F)

< 2'.

Under these circumstances we know, from

Lemma 1.5, that 7 = -4b4 for some 6 E F. From p2' = -4S4, we see that z E E and hence E = F ( i ) . Because this case is excluded in the hypothesis, we may therefore return t o the situation where pp=7

and

(F(p) : F )= p

Then we have F ( p ) = E = F(X) Hence, by Corollary 7.4.8, p = A k a for scme k E

pF* €

Z and some a E F . Accordingly

F' and the result follows. w

We next provide a formula for the degree of the splitting field of a n irreducible binomial. In what follows, for any n write

t,

2

1 not divisible by the characteristic of a field F , we

for a primitive n-th root of unity over F.

5.3. Theorem. (Darbi(l926), Gay and VClez(1978)). Let F be a n arbitrary field of characteristic p

2

0, f e t n be a positive integer and let m be defined by : m = n if

charF = 0 and n = m p k with ( m , p ) = 1 i f charF = p i s irreducible over

> 0. Assume that X" - a E F [ X ]

F with root X and define the integer s by s = max{tltlm a n d A"/'

E

F(r,)}

If E is the splitting field of X" - a over F , t h e n ( E :F) =

Proof. Our first step is t o reduce the general case t o the case where m = n , i.e. c h a r F t n . So assume that c h a r F = p > 0. Then

E

= F(X,E,) =

F(Xpk,Xm,cm)

CHAPTER 8

454

which in t u r n is the composite of t h e separable extension F ( X P k , r m ) / F and a purely inseparable extension F(Xm)/F. Owing to Lemma 1.2, both irreducible over F . Because Am is a root of XPk

-

X"

-

a and XPk

-

a are

a, we have ( F ( X m ) : F ) = pk. Hence,

by Corollary 3.4.12,

(E :F )

= pk(F(XPk,em) : F )

Thus, if t h e result holds for Xm - a (whose splitting field is F(XPk,em)), then

4

We may thus assume that m = n and hence t h a t c h a r F n. P u t L = F(t,)nF(X) and s' = ( L : F ) . Because X n - a is irreducible, ( F ( X ): F ) = n. On the other hand, by Proposition 6.6.5.

( E : F ( X ) ) = (F(€nr A) : F ( X ) ) = (F(E,) : L ) It therefore follows that

( E : F ) = n(F(c,) : F ) / s ' and we must show that s = s'. Let r be the maximal positive integer dividing n for which

F(+)

E, E

F ( X ) . Then

G F ( E , )n F ( X ) = L c F ( X )

and hence, by Proposition 3.4, L = F(X'?) for q = ( F ( X ) : L ) . Because ( F ( X ) : F ) = n, we have

( L : F ) = n / q = s' This shows t h a t

Xq

E F ( c , ) and q = n/s'. Thus, by the definition of s, s

other hand, since

A"/* E

n F ( X )= L ,

~ ( 6 , )

we have

(F(X): L) = q 5 n / s whence s'

2 s, as required. a

2

s'.

On the

COGALOIS EXTENSIONS

455

6. Cogalois extensions.

Throughout this section, E / F denotes a finite field extension . We put E F* for some m 2 l } C o g ( E / F ) = {XF' E E*/F*IXm

Thus C o g ( E / F ) is the torsion subgroup t ( E * / F * ) of E ' / F * .

Given a subgroup H of

C o g ( E / F ) , we put

EH = { a 6 Ela = b l

+ . . . + 6,

with b i F ' E H , 1

5 i 5 n, for some n 2 1}

(1)

Then EH is a subfield of E containing F and such t h a t

Indeed, EH is obviously a n F-subalgebra of E consisting of F-linear combinations of elements Xi, i E I, where H = {X;F*liE I} (in particular, the above inequality always holds). Because E / F is finite, E H is an artinian integral domain, hence a field. The terminology below is chosen t o reflect the duality. Following Greither and Harrison (1986), we call E / F cogalois with cogalois group C o g ( E / F ) if the following two conditions hold: (i) (Gonormality ) ICog(E/F)I 5 (E : F)

(ii) (Goseparability) E = EH with H = C o g ( E / F )

(2)

(3)

For example, one immediately verifies t h a t Q ( J ~ ) / Qis cogalois

On the other hand,

Q since ICogQ (-)/Q)l

(G)/Q is not cogalois

= 3; more precisely

In this and the next section we present a theory which is dual t o the usual Galois theory of fields. Namely, we show that (a) If

E/F is cogalois and K is an intermediate field, then E / K and K / F are cogalois.

CHAPTER 8

456

(b) The maps K

H

Cog(K/F) and H

H

EH are inverse bijections between the lattice

of intermediate fields of E/F and of subgroups of C o g ( E / F ) , respectively. We also study the connection between G a l ( E / F ) and C o g ( E / F ) in t h e special case where EIF is both Galois and cogalois. All the results presented are extracted from an important work of Greither and Harrison (1986). 0.1. Lemma. (i) If E/F is cogalois, then E/F is a radical eztension such that every

set of representatives for Cog(E/F) forms a basis of E over F . In particular. ICog(E/F)) = ( E : F)

K

(ii) If F

CE

is a chain of fields, then the sequence 1 4 Cog(K/F)

4

Cog(E/F)

-+

Cog(E/K)

is ezact.

Proof. (i) Choose

A1,.

. . ,A,

t o be a full set of representatives for C o g ( E / F ) . Then,

by (3) and ( I ) ,each element of E is a n F-linear combination of XI,. . . , A,.

Thus we have

(E : F) 5 n = ICog(E/F)I and therefore, by (2), (E : F) = n. Because E = F(A1,. . . ,A,) pi

2 1,

and A r i E F for some

(i) follows

(ii) Denote by t ( E * / F * ) the torsion subgroup of E*/F*.Because the sequence 1 -+ K ' / F *

-+

E*/F'

4

E'IK'

is exact, the induced sequence

1 -+ t ( K * / F * ) 4 t ( E * / F * ) is also exact.

-+

t(E*/K*)

H

As a preliminary t o the next result, let us record the following useful observation, in which G denotes a group.

0.2. Lemma. If 0 + A i B A C

-+

0 is an ezact sequence of G-modules, then we

have an ezact sequence

o -+ A

G~BGAC~~H~(G,A)

COGALOIS EXTENSIONS

457

Proof. We identify A with its image i ( A ) ,so that z becomes inclusion. Let

c

E CG

and suppose that b E B is such that j ( b ) = c. Then j ( g b - b) = gc

i.e. gb

--

-

c = 0,

b E A , for all g E G. Now

f : G - + A , g ~ g b - b is a crossed homomorphism and we associate with c E C G the class a c = f mod B ' ( G , A ) . The map c

H

a c is well defined, because if b' = b

+ a,

a E A , is another preimage of c ,

then the crossed homomorphism f ' ( g ) = gb'

-

b' = f ( 9 )

+ ga - a

differs from f ( g ) only by the crossed homomorphism f a ( g ) = ga - a which is contained in

B ' ( G , A ) . Observe that if a c

= 0, then f ( g ) = gb

g ( b - a ) = b - a , i.e. b - a = b' E BG and j ( b ' )

-

-j(b)

b = ga

-

a for some a E A. Hence

= c. This establishes the exactness

of our sequence. Following Greither and Harrison (1986), we say that E / F is pure if the following holds: every p t h root of unity in E ' , where p is a prime or p = 4 , lies in F * . For example Q

is pure, (fi)/Q

but Q ( a ) / Q is not. The following lemma exhibits a family of

radical extensions which are, a t the same time, cogalois. 6.3. Le m m a . Let p be a prime and Zet E = F(X) with A* = a E F , ( E : F ) = p , and

E I F i s pure and separable. T h e n C o g ( E / F ) =< XF' > and E / F i s cogalois

Proof. It obviously suffices t o show t h a t C o g ( E / F )=< XF" >. We first verify that

C o g ( E / F ) has no elements of prime order q Indeed, assume t h a t p E E therefore

Xq

~

-F

#

p

is such t h a t pq = b E F . Because q t p , ( F ( p ) : F )

(4)

#

q and

b is reducible over F . Hence, by Lemma 1.3, b = cq for some c E F so that

c is a root of X Q - b. Then ( p / c ) q = 1 and hence, by pureness, p / c E F , contrary to the

assumption t h a t p

#

F.

CHAPTER 8

458

It will next be shown that C o g ( E / F ) has no elements of order p 2

Assume by way of contradiction t h a t ( E :F )

pP

E E

-

F is such t h a t

(5) pPz = b E

F . Because

< p 2 , XP2 - b must be reducible over F. If p is odd, then by Lemma 1.5, b E FP,

say b = p p for

p

E F . Then ( p P / p ) P =

I, so p P / p E F, contrary t o pP $ F . If p

= 2 and b

has a square root in F, the same argument works. If p = 2 and b has no square root in F, then b = - 4 d 4 for some d E F (Lemma 1.5). Thus -46 = d j for some d l E F and hence ( d , / ~= ) ~- 4 . Setting t = d l / p , we have 0 = t4

and therefore either t

-

1 or t

+ 4 = ( t 2 - 2t + 2 ) ( t 2 + 2 t + 2 )

+ 1 is a fourth root

of 1. But E/F is pure, so t E F and

hence p E F, a contradiction. Let pF* E C o g ( E / F ) . Owing t o ( 4 ) and (5), pp E F and we may assume p Thus E = F(p) for any p E E

-

F. Moreover, because E / F is separable, c h a r F

Let K = F ( t ) be a splitting field of XP

-

1 over F , with

E

4

F.

#

p.

a primitive p t h root of unity.

Setting

L = E ( t ) = K(X) we have ( L : E )

5p

-

1 and ( K : F) 5 p - 1. Moreover,

p ( L : E ) = ( L : F) = ( L : K ) ( K : F ) which shows t h a t p l ( L : K ) = p and X $ K . Hence L / K is a Kummer extension, by virtue of Lemma 7 . 4 . 2 . P u t

e/K'={zEL*IzPEK*} Because ( L : K ) = p it follows from Corollary 7 . 4 . 6 t h a t Q F / K * is cyclic of order p . Since p and X are in

e/Kt and since X 4 K * , we deduce t h a t p =

for some rn E IN, k E K*

Xmk

With pP = d E F , it therefore follows that d = amkP

COGALOIS

Let G = G a l ( K / F ) and let

4 : K'

4

459

EXTENSIONS

( K * ) pbe defined by 4(z) = zp. Then the exact

sequence of G-modules 1 + < € > A K' + (K*)P+ 1

gives a n exact sequence

( K * ) G-+ ( ( K * ) p ) + G H'(G,) by virtue of Lemma 6.2. Because /GI and the homomorphism

---t

1 1

are coprime, H ' ( G ,

< t >) = 1, and

so

((K*)P)" is surjective. But ( K * ) G= F' and kP = d / a m is

in ( ( K x ) P ) ) so G, for some a E

k P = (yP

F'

Thus, d = am&' = pP and so p / ( X m a ) is a p t h root of 1. It follows t h a t p / ( X m a )E F' and hence p E XmF*,as required. 6.4.

Lemma. If F 2 K 2 E is a chain of fields and K I F and E / K are both

conormal, t h e n EIF is conormal. Proof. By definition, K I F is conormal if and only if ICog(K/F)l 5 ( K : F ) . The desired conclusion is therefore a consequence of Lemma 1.2(ii). We have now come to the main result of this section which characterizes cogalois extensions. 6.5. Theorem. (Greither and Harrison(1986)). Let

EIF be a F n i t e field extension.

T h e n E J F i s cogalois if and only if EIF is coseparable, separable and pure. Proof. Suppose first that E I F is cogalois. Then, by definition,

Assume by way of contradiction that

EIF is not separable.

EIF is coseparable.

Then F is finite, c h a r F = p

and, by Corollary 2.2.14, p l ( E : F ) . Because, by Lemma 6.1(i), ( E : F ) there is an element for all

E F, p

+X

XF' E C o g ( E / F ) of order !$ F by ( p

+ A).

= pP

X

p , i.e.

+ XP

6 F'

and

XP

>0

ICog(E/F)I,

:

E F'.

Then,

F. One immediately verifies that

E

+ X J F ' # ( d + X)F* for p # d E F . Thus C o g ( E / F ) contains infinitely many elements ( p + X ) F ' , p E F , which is a contradiction.

(p

To prove purity, let that

t E

t p

= 1 with

t

E E , where p is a prime or p

F . Suppose first that p is a prime. If c h a r F

to prove. Thus we may assume that c h a r F 1

#

p and

+ t + ...+ tp-1

t

=

p , then

#

=

0

t =

1. Because

= 4. We must show

1 and there is nothing

CHAPTER 8

460

it is not the case that 1 . F', EF*, . . . ,F I F * are distinct elements in Cog(E/F) (see Lemma l.l.(i)). Therefore some r * / d E F (i # j),so 6 E F. Now assume that p = 4 . Again we may harmlessly assume that charF One easily checks that (1

+c

) = ~ -4 E

#

2,

E'

# 1.

F'. However,

l+E-(l+€)

=o,

so it is not the case that the elements 1. F', c F * , (1

+ E)F' of Cog(E/F) are distinct (see

Lemma l . l ( i ) ) . Because any of the three possible equality relations between them implies E

E F , purity is established.

Conversely, suppose that E/F is coseparable, separable and pure. We may choose a finite subgroup G of Cog(E/F) such that Ec = E (e.g. take G = Cog(E/F)). Because G is abelian, there is a chain

of subgroups of G such that H;/Hi-l

is cyclic of prime order p i , 1

5 i 5

n. Thus

EHi/EHi-, is pure (since EIF is pure) and, by construction, it is also coseparable. Since

HiIHi-1 is cyclic of order p i , EH; is obtained from E H ; - ~by adjoining a pi-th root. Observe also that E H ; / E H ~is - ~separable, since so is E/F. Applying Lemma 6.3, we deduce that EH;/ E H ~ -is, cogalois, and thus conormal. Therefore, by inductive application of Lemma 6.4, we conclude that E/F is also conormal. Hence EIF is cogalois and the result follows.

H

With the aid of the above result, we now provide the following examples of cogalois extensions. 6.6. Example. Let a,, a 2 , .. .,a, be positive rational numbers, let n; E

Because E

lR,

El$

is obviously pure. It is also plain that

El$

N

and let

i s coseparable. Thus

E / Q i s eogalois, by virtue of Theorem 6.5. H 6.7. Example. Let p be a n odd prime, let n be a positive integer and let epn be a

primitive p n - t h root of unity over Q

.

Q ( c p n ) / Q (cp)

Then i s cogalois for

all n 2 1

461

A GALOIS CORRESPONDENCE

Indeed, Q ( c p " ) / Q( c p ) i s certainly eoseparable and separable. To show purity, one can use the fact that Q (cp") and Q ( c q ) are linearly disjoint i f q # p is a p r i m e or q = 4.

7. A Galois correspondence for radical extensions.

All the notation introduced in the previous section remains in force. We begin by recording the following piece of information. A lattice is a partially ordered set in which any two elements z,y have both a least upper bound z V y and a greatest lower bound z A y in the set. Hence the set of subgroups of a given group G forms a lattice, denoted by L ( G ) ,with inclusion as the partial ordering and with

for

Hi E L ( G ) . Similarly, if F is a field, then the set L ( F ) of subfields of F forms a lattice

with the inclusion as the partial ordering and with

A h o m o m o r p h i s m of lattices L and L' is a mapping

f:L+L' such that

for all z,y E L. It is obvious t h a t a lattice homomorphism preserves the ordering: a 5 b implies f ( a ) 5 f ( b ) , but not every order-preserving mapping between lattices is a lattice homomorphism. Let L and L' be two lattices and let f : L

---t

L' be a bijection. Then one

immediately verifies that the following conditions are equivalent: (a)

f is an isomorphism of lattices

(b) for all z,y E L, z 2 y if and only if f ( z ) 2 f ( y ) . We are now ready to prove the following result. 7.1. Theorem. (Greither and Harrison (1986)). Suppose that a f i n i t e field eztension

E / F i s cogalois and let K be a n intermediate field. Then (i) E / K and K / F are again cogalois

(ii) E'H = K for H = C o g ( K / F )

462

CHAPTER 8

(iii) Cog(Es/F) = S for every subgroup S of Cog(E/F)

(iu) The maps K

H

Cog(K/F) and S

H

Es are inverse isomorphisms of lattices oj

intermediate fields of E / F and subgroups of Cog(E/F). (v) Cog(E/K) is canonically isomorphic to Cog(E/F)/Cog(K/F).

Proof. (i) Since E / F is cogalois, it is coseparable, separable and pure, by virtue of Theorem 6.5. Therefore E/K is coseparable, separable and pure. Applying Theorem 6.5 again, we infer that E / K is cogalois. By Lemma 6.1(i), we have ICog(E/F)I = ( E : F) and

ICog(E/K)/ = ( E : K )

It therefore follows, from Lemma 6.l(ii), that

If C o g ( K / F ) = ( X I F * , . . . ,X,F'},

then XlF', . . . ,X,F* are distinct elementsof C o g ( E / F ) .

Therefore, by Lemma 6.1(i), the elements XI,.

.. ,A,

are F-linearly independent. Owing

t o ( l ) , the A; form a basis for K over F , which clearly implies t h a t K / F is cogalois.

(ii) It is obvious t h a t EH & K . On the other hand, we have

and, by (i), we also have

( K : F ) = ICog(K/F)I Consequently, EH = K as required.

(iii) It is plain t h a t S

Because (Es : F )

C o g ( E s / F ) . Note also that, by (i),

5 ISI, we have S = C o g ( E s / F )

(iv) This is a direct consequence of (ii) and (iii). (v) Owing t o Lemma 6.1(ii), there is a natural injection

Because E/F, K / F and E/K are cogalois, it follows from Lemma 6.l(i) t h a t the orders of the groups in (2) are equal. This completes the proof of the theorem.

463

DUALITY OF LATTICES

8. Duality of lattices for G a l ( E / F ) and C o g ( E / F ) .

We keep the notation of Sec. 6. In this section we examine extensions E / F which are cogalois and Galois a t the same time. T h e main result establishes a duality between the involved cogalois and Galois groups. 8.1. Lemma. Let E I F be a finite field eztension.

(i)

EIF is coseparable and Galois i f and only i f E I F

i s separable and E is the splitting

field of some polynomial

with n; E IN and a; E F, 1 5 z

5 s.

(ii) If E I F is cogalois and Galois and charF = 0, then E is the splitting field of the

polynomial ( 1 ) and for p dividing any

ni,

F contains a primitive p-th root of unity.

Here p is a n arbitrary prime or p = 4. Proof. (i) The "if" part is obvious. Conversely, suppose that E I F is coseparable and Galois. Because E I F is coseparable, there is a finite subgroup H of C o g ( E / F )such that E = E H . We may write

where the order of X,F* is equal to q; = p:', a prime power. We claim that E is the splitting field of the polynomial

with a, = A:', to show that

1

5 z 5 s. Because E

= F(X1,. . , ,A,)

and X i is a root of Xq' - a ; , it suffices

E contains a primitive q,-th root of unity for all i E { 1,. . . ,s}. Since E / F

is normal, E contains all conjugates X i e l , . . . ,Xi<, of A; over F . Here all of unity. Assume by way of contradiction that no

every conjugate of A4"p'

ti,

1

5j 5

~j are

q,-th roots

r , is primitive. Because

is a (q;/pi)-th power of a conjugate of A;, this implies that

is equal t o all its conjugates, so A;"'

E F , which is a contradiction.

( i i ) If E / F is cogalois and Galois, then E / F is coseparable and Galois. Invoking (i), we

deduce that E is the splitting field of the polynomial (1). Now let p denote a prime or p = 4 , and let pin,. Because

X", ~

a; splits completely over E and charE' = 0, Y , contains

CHAPTER 8

464

a primitive p t h root of unity, say cp. However, E / F is also pure by Theorem 6.5. Thus E,,

E F and the result follows. w 8.2. Lemma. Let E / F be a finite field eztension which is coseparable and Galois,

let G = G a l ( E / F ) and let C be a subgroup of Cog(E/F). Then the map

is well defined and satisfies the following properties:

6) f ( g , c 1 c z ) = f ( s , c 1 ) f ( s , c z ) (ii) For any c E C, the map

for all g E G , c1,cz E C

xc : G + t ( E * ) ,g

H

f ( g , c ) is a crossed homomorphism,

where t ( E * ) is regarded as a G-module in a natural way. (iii) If f ( g , C) = 1 and C = Cog(E/F), then g = 1

(iv) If f ( G , c ) = 1 , then c = 1 (v) The map C + Z ' ( G , t ( E * ) ) ,c ++

xc is an injective homomorphism.

Proof. If AF' E C, say A" = p E F * , then for any g E G , g(A") = g ( X ) " = p = A"

Thus g(A)X-'

E t ( E * ) .If p = Xp with

p E F * , then g ( p ) p - '

given map is well defined. (i) Write

c1 =

A1F' and c2 = X2F*. Then

as required. (ii) Given gl,g2 E G and c = XF* E C, we have

= g ( A ) A - ' , proving that the

DUALITY

OF

LATTICES

465

as asserted. (iii) Suppose t h a t C = C o g ( E / F ) and f ( g , c ) = 1 for all c E C. Then g(X) = X for all

X E E' such that XF' E C. But E I F is coseparable, hence generated over F by all such A. Therefore g(X) = X for all X E F , i.e. g = 1. (iv) Suppose t h a t f ( 9 , c) = 1 for all g E G, where c = XF' E C . Then g(X)

=

X for all

g E G. Because E / F is a Galois extension, we conclude that X E F ' , i.e. c = 1.

(v) Because x c ( g ) = f ( g , c ) , the given map is a homomorphism, by virtue of (i). The injectivity follows from (iv). We are now ready to prove. 8.3. Theorem. (Greither and Harrison (1986)). Let E / F be a finite field eztenszon

which is both Galois and cogalois, with G = G a l ( E / F ) and H = C o g ( E / F ) , let L ( G ) and

L ( H ) be the lattices of subgroups of G and H , respectively, and for any subgroup S of G, let E S be the fixed field of S and SJ- = { h E HI f ( S ,h) = 1)

where f is as i n Lemma 8.2. Then

E S = Esi and the map

is a duality (i.e. anti-isomorphism).

Proof. Owing t o Lemma 8.2.(i), S'

Let x E E,L so that for some n

21

x =XI Then, by ( 3 ) , .(A,)

is a subgroup of H and, by definition,

f

... + A, with XiF'

E S'

= A, for all s E S , z E { 1 , 2 ,...,n}. Thus x

ESL C E S . Assume by way of contradiction that E S

E E S and therefore

9 Esi. Because

E / F is cogalois

CHAPTER

466

8

and ES is an intermediate field, Theorem 7.l(ii) tells us ES = Ew for W

S'.

Choose

w F * E W - SL.Because w E E w = ES,we have wF' E S', a contradiction. Let L F ( E ) be the lattice of subfields of E containing F. Then, by Galois theory,

L ( G ) -+ L F ( E ) , S

H

E S is a duality. On the other hand, by Theorem 7.1(iv), the map

L F ( E )-+ L ( H ) , ES I+ Cog(ES/F) is a n isomorphism. Hence the m a p

is a duality. But, by ( 2 ) and Theorem 7.1(iii), we have

C o g ( E S / F ) = C o g ( E s i / F ) = S', hence the result. We next present the following example of a nonabelian Galois and cogalois extension. 8.4. Example. (Greither and Harrison (1986)). Let cn be a primitive n-th root of

.

unity over Q

Then

Q (€9,G)/Q (€3) is a nonabelian Galois and eogalois extension. Indeed, put

F=Q Then E = Q

s),( E

: F ) = 27 and

@ E and

tp

(€9,

t o verify that

€4

(€3) and

E = F(tg,

z)

E / F is coseparable. To prove purity, it is enough

@ E for odd primes p # 3 . This can be shown by using the

fact that F ( f i ) ( € 4 ) is quadratic over F ( f i ) but E is cubic over F ( f i ) together with the fact that p

-

1 does not divide 54 = ( E : Q ) for any odd prime p

# 3. Thus E / F is

cogalois. On the other hand, because E is a splitting field of X g - 5 over F, E / F is also Galois. Now Gal( E / F ) =< a, p > where

Because pap-'

= a4 # a , G a l ( E / F ) is nonabelian, as asserted.

467

THE LATTICE FOR INTERMEDIATE FIELDS

9. The l a t t i c e f o r i n t e r m e d i a t e fields of r a d i c a l e x t e n s i o n s .

In this section we examine the lattice of intermediate fields of certain radical extensions. Namely, we shall concentrate on extensions E / F where E is the splitting field of the polynomial

(X"'- a 1 ) ( X n 2- a 2 ) , . . (X".- a 9 )

(ni E IN,ai E F')

where a;, n; are such that for p dividing any n;,F contains a primitive p t h root of unity ( p is an arbitrary prime or p = 4). Our aim is to exhibit a distinguished subgroup C of

C o g ( E / F ) and a n isomorphism of lattices

where L F ( E ) is the lattice of all subfields of E containing F and L ( C ) is the lattice of all subgroups of C. Let r,m E W. Following Greither and Harrison (1986), we say t h a t m is related t o r ,

if m divides r , every prime p dividing r also divides m,and 41m in case 41r. If m is related to r , we put

Dr,m= K e r ( U ( Z / r Z ) Let 1

--t

H

+

---t

*

GiDrs,

U(Z/rnZ))

--t

1

be a n extension of groups. Following Greither and Harrison (1986), we say that this extension is allowable if H is a finite abelian group with H'

= 1 (here

H is a i z / r Z -

module in a natural way) and if the action of Dr,mon H determined by this extension is the natural one. The latter means that if

t : Dr,m+ G is any right inverse to $, then for any x = p

z h = t ( x ) h t ( z ) - l = hP For example, if G = H

O(

+ r Z E Dr,m for all h E H

(2)

Dr,m(semidirect product) with respect t o t h e action of Dr,,, on

H given by (2), then G is the split allowable extension of H by D7,,,. The group Dr,macts naturally on the additive group Z / r Z : for any p

+ r i z E DrI,,

CHAPTER 8

468

For convenience of reference, we first record the following standard fact 9.1. Proposition. Let A be a G-module, where G is a finite cyclic group. Then

Proof. See Weiss (1969). rn 9.2. Lemma. Suppose that m i s related t o r and let D = Dr3m.Then

(i) D is cyclic of order r / m (ii) ( Z / r Z ) D= ( r / r n ) ( Z / r Z ) (iii) H ~ ( DZ, / r Z ) = o

Proof. (i) Because m is related t o r , the canonical decompositions of m and r are of t h e form m = P1"I

. . .p:*,

r = pf' . . . p F w i t h 0

< a; 5 pi.

Bearing in mind t h a t 9

u(z/~~z)

U(Z/rZ) i= 1

and

it suffices t o treat the case where m = pa, r = p p with 0

< a < P , p prime. Now

the natural map U ( Z / p p Z ) 4 U ( Z / p " Z ) is obviously surjective. If p is odd, then by Proposition 6.4.13(i) both groups are cyclic of orders p@'(p-l),

pa-*(p-l),

respectively.

Hence D is cyclic of order PO-" = r/rn. Assume that rn = 2", r = 2 p . Because and so a

2

2, by hypothesis. If

(Y

P 2 2,417

> 2 and D is not cyclic, then by Proposition 6.4.13,

U ( Z / 2 a Z )is cyclic, which is impossible by Proposition 6.4.13. Thus we may assume that a = 2. Owing t o Proposition 6.4.13(ii),

and the element 5

+ 2pZ

generates a subgroup of order 20-'.

contained in D and l U ( Z / 4 Z ) I = 2, we must have D of order 2Bp2 = r / m , as required.

Because this element is

=< 5 + 2 p Z >. Therefore D is cyclic

THE LATTICE FOR INTERMEDIATE FIELDS

+

(ii) Let a = ( r / m ) s r Z for some s E

D. Then

/.i = 1

Z and let

z = /.i

469

+ r Z be a n arbitrary element of

+ tm for some t E Z and so z a = (1 + t m ) ( r / m ) s+ r Z = ( r / m ) s+ t r + r Z = a ,

proving that ( r / m ) ( Z / r Z C ) (Z/rZ)D. Conversely, let a = k

+r Z E

(Z/rrZ)D. Since m is related to r, (1

+ m,r) = 1 and

so z = (1+ m) + r Z E D. Because z a = a , we have mk = ( l + m ) k - k E rZ

which shows that a E ( r / m ) ( Z / r Z as ) , required. (iii) Because D is cyclic and

it suffices, by Proposition 9.1, t o show that

Now D = ((1

+ km) + rZ/O 5 k 5 (rim)- 1) and so

=

r -

m

+ -2r( - mr - 1) + r Z

(4)

r ra =-+-+rZ m 2 with a = ( r / m )- 1 E Z.If r / m is odd, then a is even and r m

-

Therefore, by ( 4 ) , r / m + r Z =

ra r +z -(m 2 m

od r)

(xdED d ) ( l + r Z ) , proving ( 3 ) .

Suppose t h a t r / m is even. Because m is related to r, 8 must divide r, and 4(m. We then have r

ra

(m +-)(I2

ma - -r + ra ra ra’m - - - ___ 2’-m 2 2 4

-)

CHAPTEK 8

470

Thus, by (4)

ma

L +rZ = (1- -)(m 2

r m

+ r2a + r Z ) E -

(c

d)(Z/rZ),

dED

which again proves (3). rn Suppose t h a t

is an allowable extension of H by Dr,,. Because Z/rZ is a D,,,-module,

we can regard

Z/rZ as a G-module via

s(k + 7221)

= *(s)(k

+T Z )

For any subgroup S of G, we put

X ( S ) = Z'

(s,Z/rZ)

B ( S ) = B'(s, Z/rZ) so that

9.3. L e m m a . Suppose that m is related t o r and that

is a n allowable eztension of H b y D = Dr9,. Then (i) IX(G)l 2 IGI

( i i ) For subgroups S C T of G , we have

IKer(X(T))*X(S)l

(iii) For subgroups S

CT

5 (T : S )

of G , the restriction map Res: X(T) ---t X(S) is surjective and

IX(S)I = IS1 (iv) For subgroups S C T of G, there is a

f o r any subgroup S of G

x E X(G) with x(S)= 0 , x(T)# 0 .

THE LATTICE FOR

FIELDS

INTERMEDIATE

Proof. (i) For each a E Z/rZ, let ,ya E B ( G ) be defined by x a ( g ) = ga the map a

++

471

-

a. Then

xa induces a n exact sequence

By definition, (Z/riz)G= (Z/rZ)Dand, by Lemma9.2(ii), we deduce t h a t IB(G)/= r / m . Thus IX(G)/ = l f f 1 ( G 9 w m l ( r / 4

Because IG/ = IH/IDJ= l H l ( r / m ) ,by Lemma 9.2(i), it suffices t o show t h a t

T h e following exact sequence comes from the Lyndon-Hochschild spectral sequence (see MacLane ( 1 9 6 3 , ~ .354)):

H 1 (G, iz/rZ)% H 1( H ,i z / r iz)

4

H 2(D , i z / ~ i z )

(6)

Observe that (a) II operates trivially on iz/riz, and TH = 0, so

H 1 ( H , Z / r i z )= H o m ( H , Z / r i z )

(7)

has exactly /HI elements. (b) The operation of G on H ’ ( H , Z / r i Z ) comes from conjugation and G operates on I1

and i z , / r i z through D by scalar multiplication. From (7) we therefore conclude that the operation of G on H ’ ( H , i z / r Z ) is trivial. Invoking (a),(b),(6) and Lemma 9.2, we conclude that (5) holds. (ii) By induction, one can perform a reduction to the case that there are no groups properly

between S and T . Then the following holds ( c ) There is an element

t E T - S with

tP

E S , p prime

(d) T = < S , t > (e) ( T :S) 2 P Now let

x E Ker(X(T)

i

X ( S ) ) .We have t o show that there are a t most p choices for

x.

Uy (d), the latter will follow provided we show that we have a t most p possible values for

X(t).

CHAPTER 8

472

Let $ ( t ) = l + m a + r Z . By repeated use of the definition of a crossed homomorphism we get 0 =X ( P ) =X(t)

+ (1+ m a ) x ( t )+ . . . + (1+ ma)P-'x(t) P- 1

=q

( t ) with

+

c = c(1 m a ) ; i=O

We claim that c

+rZ

and p

+ r Z are associated in Z / r Z ; if sustained, it will follow that

the equation 0 = c x ( t ) has at most p solutions in Z / r Z , and we are done.

To substantiate our cliam, we must establish two facts:

(f) If

a prime q

# p divides r , then q t c .

(g) If plr then pic, and if p21r then p2 + c . Suppose that a prime q

# p divides

r . Because

m is related to r , we must also have qlm. Then

c

=

c(1+

P-

1

0)

= p(mod q ) ,

i=O

proving (f). Suppose that plr (hence plm). Then

c(l+

P-

c

1

0)

= O(mod p)

i=O

Now assume that pz/r. Because p 2 / m Zthe , binomial theorem yields

c(+

P- 1

c

1

zma) = p

i=O

+ map ' 2 (mod pz) 2

If p # 2, then p2 divides map(p - 1)/2, so c

p(mod pz) and p z t c . If p = 2, then (since

41r) 4 already divides m,and again we obtain

c f 2(mod 4),

(iii) By (ii), applied to

T = S and

S = 1, we have

Again, by (ii) IKer(X(G) + X(S))i 5 (G : S)

as required.

THE LATTICE

OF

INTERMEDIATE

FIELDS

413

Taking ( 8 ) , (9) and (10) together, we deduce that all inequalities ( 8 ) , (9), (10) are in fact equalities and X(G) + X(S) is surjective. This implies that X ( T ) + X(S) is also surjective, which proves (iii). (iv) By (iii), we have IX(T)I

> iX(S)I. Therefore the restriction map X(T)

cannot be injective. T h e required assertion now follows by taking a n extension of any

CY

# 0 in

Ker(X(T))

--f

x

+

X(S)

t o X(G)

X ( S ) ) .I

We have now accumulated all the information necessary to prove the following result. 9.4. Theorem. (Greither and Harrison (1986)). Suppose that m E r E- IN and let G be a n allowable extension of

IN is related to

H b y Dr,,,. Put

X(G) = Z'(G, Z / r Z 4 ) where the action of G o n Z / r U is induced b y the natural action of Dr,,, o n Z / r U . Define

f : G x X(G) --t U / r Z

and, for any S E L ( G ) ,W E L(X(G)) ( L ( G ) and L(X(G)) are lattices of subgroups of G and X(G), respectively), put

are mutually inverse order inverting bijections.

4 74

CHAPTER 8

and W'

Proof. It is immediate S'

are again subgroups and that the given maps

are order inverting. Also, by definition, we have

Owing to Lemma 9.3(iv), if S S'

3 S'll.

cT

But obviously Sl'l

then S l 3 TI. In particular, if S

2 SL,so S l

c

S l l then

3 S * , a contradiction. Hence S = SLL

and we are left to verify that W = W l l . Every g E G defines in an obvious way an eIement g' in Hom(X(G),Z/rZ), and g' = 0 implies g = 1 by Lemma 9.3. (iv). Since X ( G ) is abelian of exponent dividing

7,

and IHom(X(G),Z/rZ)l= IX(G)I = IGI (Lemma 9.3(iii)), the map g

H

g' is a bijection from G onto H o m ( X ( G ) , Z / r Z ) . From

the duality theory of finite abelian groups, it follows that for any W

c W1 C X(G) there

exists X E Hom(X(G), Z/rZ) with X(W) = 0, X(W1) # 0. But X = g' for some g E G, so f ( g , W ) = 0 , f ( g , W1) # 0. This implies W"

= W , as required. H

We have now come to the demonstration for which this section has been developed. 9.5. Theorem. (Greither and Harrison (1986)). Let E / F be a finite field eztension

such that E is a splitting field of the polynomial

where a;, n ; are such that f o r p dividing any n;, F contains a primitive p-th root of unity, where p is a n arbitrary prime or p = 4. (i) If r is the least common multiple of the n,, then E contains a primitive r-th root of

unity, say c. (ii) Pick A; E E with A n ' = a; and denote b y C the subgroup of C o g ( E / F ) generated b y

cF* and all X,F*. Then there is a lattice isomorphism

where L F ( E ) is the lattice of all subfields of E containing F and L ( C ) is the lattice of all subgroups of C . (iii) A n y intermediate field K is generated over F b y certain elements of the form

THE L A T T I C E OF I N T E R M E DIAT E F I E L D S

475

Proof. (i) Our assumption on roots of unity ensures that c h a r F tn;,1 5 z 5 s. Thus each Xn’ - a, has only simple roots. This implies t h a t E contains a primitive n,-th root of unity (hence also a primitive r-th root of unity) and that E/F is Galois. (ii) P u t H = G a l ( E / F ( c ) ) , G = G a l ( E / F ) and D = G a l ( F ( c ) / F ) . Then the restriction homomorphism G

t

D induces an exact sequence

Observe t h a t D is canonically a subgroup of U ( Z / r Z ) (identify o E D with ,u $- rZ if O ( E )= E W ) .

Let us show that D = Dr,,, for some m >_ 1 related to r . To this end, define

mr to be the product of all primes dividing r if 4 t r and twice the latter product if 41r.

Then m’ is the smallest divisor of r related to r . Owing t o Lemma 9.2, D r , m ~ is cyclic, and by the hypotheses concerning roots of unity, D

Dr,,,,. Thus, by Lemma 9.2(i),the

order of D divides r/m’, so ID1 = r / m for some m such that m’lmlr. In particular, m is related to

T

and Dr,m

Dr,-,. Because Dr,m& is cyclic, it contains exactly one subgroup

of order r / m , and thus D = Dr,,,. It is now immediate t h a t (11) is a n allowable extension. Observe also that E / F is coseparable, since E = Ec. Because E / F is Galois and coseparable and C is a subgroup of C o g ( E / F ) , we may apply Lemma 8.2. Let f :G x C

+

t(E*)

be as in Lemma 8.2 and, for any subgroup S of G , put

SL

= {c E

CIf(S,c)= I}

Observe t h a t the image of f is contained in < E >E Z / r Z , so the conclusion of Lemma 8.2 is applicable to Z / r Z (instead of t(E*)). Therefore, there is a n injective homomorphism

C

4

Z’(G, Z / r Z ) = X ( G )

Owing to Lemma 9.3[iii), IX(G)I = IGI = (E : F ) . Because E

ICI

2 (E : F).

(12) =

E c , we must have

Hence (12) is an isomorphism and lCl = ( E : F ) , which implies that C

forms a n F-basis of E. Because C

2

X ( G ) , we have

L(C)

L(X(G))

But, by Theorem 9.4, L ( X ( G ) )is dual t o L ( G ) and, by Galois theory, L ( G ) is dual to

L F ( E ) .Thus L ( C )

L F ( E ) ,as required.

476

(iii) Let F

CHAPTER 8

G

K & E be a chain of fields and put s = ( E : K ) . Then K = E S for a

subgroup S of G of order s. By Lemma 9.3(iii), S'

= Ker(X(G) --f X ( S ) )

has index s in C, since [CI = IX(G)I = [GI. The field E s l is contained in

ES = K . Since

C forms an F-basis of E , we have

so

(E: &I)

= ( H : S*) = s = (E: K )

and hence K = E ~ IThis . means that K is generated over F by certain elements of the form tiX:lX?

.. . A t . ,

as required. rn

477

Bibliography

AKIZUKI, Y . (1936) Eine homomorphe Zuordnung der Elemente der galoischen Gruppe zu den

Elementen einer Untergruppe der Normklassengruppe, Math. Ann. 112, 5 6 8 5 7 1

ALBERT, A . A . (1930a) T h e integers of normal quartic fields, Ann. Math. 31, 381-418

(1930b) A determination of t h e integers of all cubic fields, Ann. Math. 31, 550-566. (1034) O n certain imprimitive fields of degree p 2 over P of characteristic p, Ann. Math. 35,

No. 2, 211-219. (1937) Normalized integral bases of algebraic number fields I, Ann. Math. 38, No. 4 , 923-957. (1938) Quadratic null forms over a function field, Ann. Math. 39, 494-505. (1910) On p a d i c fields and rational division algebras, Ann. M a t h . 41, 674-693. (1066) On some properties of biabelian fields, An. Acad. B r a d . Ci. 38, 217-221.

AMANO, S. (1Q58) A note on p a d i c fields, Keio University Cent. Mem. Publ. Tokyo, 881-892.

(1971) Eisenstein equations of degree p in a p a d i c field, J. Fac. Sci. Tokyo, 18, 1-21.

AMITSUR, S. (1955) Generic splitting fields of central simple algebras, Ann. Math. ( 2 ) , 62, 8-43.

B i b 1 iography

4 78

ANCOCHEA, G. (1964) Automorphismes des extensions algebriques d’un corps, Math. Ann. 157, 231-233.

ANKENY, N.C. and

CHOWLA, S .

(1949) The class number of the cyclotomic field, Proc. Nat. Acad. Sci. U.S.A. 35, 529-532. (1950) The relation between the class number and the distribution of primes, Proc. Amcr.

Math. SOC.1, 775-776. (1951) The class number of the cyclotomic field, Canad. J. Math. 3, 486-491. (1955) On the divisibility of the class number of quadratic fields, Pacif. J. Math. 5 , 321-324. (1960) A note on the class number of real quadratic fields, Acta Arithm. 6, 145-147. (1962) A further note on the class number of real quadratic fields, Acta Arith. 7, 271-272 (1968) Diophantine equations in cyclotomic fields, J. London Math. SOC.43, 67-70.

ARF, C . (1939) Untersuchungen uber reinverzweigte Erweiterungen diskret bewerteter perfekter Korper, J. Reine Angew. Math. 181, 1-44.

ARMITAGE. J.V (1957) Euclid’s algorithm in certain algebraic function fields, Proc. London Math. SOC.7, 498-509. (1963) Euclid’s algoriihm in algebraic function fields, J. London Math. SOC.38, 55-59. (1967) On a theorem of Hecke in number fields and function fields, Invent. Math. 2, 238-246.

479

Bib1 iography

AR T I N , E. (1923) Uber die Zetafunktionen gewisser algebraischer Zahlkorper, Math. Ann. 89, 147- 156. (1924) Kennzeichnung des Korpers der reellen algebraischen Zahlen, Hamb. Abh. 3 ? 319-323. (1927) Uber die Zerlegung definiter Funktionen in Quadrate, Hamb. Abh. 5, 100-115.

( 1932) Uber Einheiten relativegaloischer Zahlkorper, J. Reine Angew. Math. 167, 153-156. (1946) Galois theory, Notre Dame Math. Lectures, No. 2, Univ. of Notre Dame, NoLro Dame, Indiana. (1967) Algebraic numbers and algebraic functions, Gordon & Breach, New York.

ARTIN, E and SCHREIR, 0. (1927a) Algebraische Konstruktion reeller Korper, Hamb. Abh. 5, 85-99. [ 1927b) Eine Kennzeichnung der reel1 abgeschlossenen Korper, Hamb. Abh. 5, 225-231.

ARTIN, E and WHAPLES, G. ( 1945) Axiomatic characterization of fields by the product formula for valuations, Bull. Arner. Math. Soc. 51, 469-492. (1016) A note o n axiomatic characterization of fields,

Bull. Amer. Math. SOC.52, 245-217.

AR.WIN, A . (1929) On cubic fields, Ann. Math. 30, 1-11.

ASANO, K. (1950) 6 b e r Moduln und Elementarteiltheorie im Korper, in dem Arithrnetik

definiert ist, J a p a n . J. Math. 20, 55-71.

Bibliography

480

ASANO, K and NAKAYAMA, T. (1940) A remark on the arithmetic in a subfield, Proc. Imperial Acad. Tokyo, 16, 529-531.

AX, J (1965) On the units of an algebraic number field, Illin. J. Math. 9, 584-589.

BAER, R (1927) Algebraische Theorie der Differentierbaren FunktionkOrperJ,

S.

-

B. Heidelberger

Akad. Wiss., 15-32.

BASARAB, S.A. (1986) Transfer principles for pseudo real closed e-fold ordered fields, J. Symbolic Logic, 51,

NO 4. 981-991.

BASHMAKOV, M.I. (1968) On the field imbedding problem, Matem. Sb. 4, No 2, 137-140.

BAUER, M

[ 1939) Zur tlieorie der Kreiskorper, Acta Szeged 9, 110-112. ( 194Oa) Uber zusammengesitzte relative Galoissche Korper, Acta Szeged, 9, 206-21 1. (1940b) Uber die Zusammensetzung algebraischer Zahlkorper,, Acta Szeged 9, 212-217.

BECKER, E

[ 1979) Partial orders on a field and valuation rings, Comm. Algebra 7, 1933-1976 (1984) Extended Artin-Schreir theory of fields,

Rocky Mountain J. Math. 14, No. 4, 881-

897.

BECKER, E, HARMAN, J and ROSENBERG, A (1982) Signatures of fields and extension theory, J. Reine Angew. Math. 130, 53-75.

B i b 1 iography

481

BECKER, M.F. and MACLANE, S

( 1940) The minimum number of generators for inseparable algebraic extensions, Bull. Arrier. Math. SOC.46, 182-186.

BERMAN, S.D. (1967) Group algebras of countable abelian pgroups, Publ. Math. Debrecen, 14, 365-405

BERMAN, S.D. and MOLLOV, T . i . (1969) On group rings of abelian pgroups of any cardinality, Matem. Zarnetki, 6, 381-392.

BLANKSBY, P.E. (1970) A metric inequality associated with valuated fields, Acta Arith. 17, 217-225

BOREVICH, Z.I. and Safarevich, I.R. (1'366) Number theory, Academic Press

BOURBAKI, N

( 1950) Elements de mathernatique, Chapter 5 , Corps commutatifs, Hermann, Paris. [ 1'366) Elements de mathernatique, Topologie Generale, Hermann, Paris.

( 1'372) Commutative algebra, Addison-.Wesley, Reading, Mass.

BOUVIER, L ( 1971) Sur le 2-groupe des classes au sens restreint de certaines extensions biquadratiques tie

Q, C.R. Acad. Sci. Paris 272, A 193-196. BRANDIS, R (1'365) Uber die multiplikative struktur von korperweiterungen, Math. Z., 87, 71-73

BRINKHUIS, J (1084) Galois modules and embedding problems, J . Reine Angew. Math. 346, 141-165

Bib1 iography

BROWN, R, CRAVEN T.C. and PELLING M.J. (1984) Ordered fields satisfying Rolle’s theorem, Rocky Mountain

J. Math. 14, N o 4 ,

819-820.

BRUCKNER, G (1966) Charakterisierung der galoischen Zahlkorper, deren zerlegte Primzahlen durch binare

quadratische Formen gegeben sind, Math. Nachr. 32, 317-326. (1968) Eine Charakterisierung der in algebraischen Zahlkorpern voll zerlegten Primzahlen,

Math. Nachr. 36, 153-159.

BRUMER, A (1965) Ramification and class towers of number fields, Michigan Math. J. 12, 129-131. (1966) Galois groups of extensions of algebraic number fields with given ramification,

Michigan Math. J., 13, 33-40. (1967) On the units of algebraic number fields, Mathematika 14, 121-124. (1969) On the group of units of an absolutely cyclic number field of prime degree, J . Math.

SOC.Japan 21, 357-358.

BRUMER, A and ROSEN, M (1963) Class number and ramificaiton in number fields, Nagoya Math. J. 23, 97-101.

BULLIG, G (1936) Die Berechnung der Grundeinheit in den kubischen Korpern mit negative

Discriminante, Math. Ann. 112, 325-394 (1938) Ein periodisches Verfahren zur Berechnung eines systems von Grundeinheiten in den

total reellen kubischen Korpern, Abh. Math. Sem. Hans. Univ. Hamburg, 12, 369-414. (1939) Zur Zahlengeometrie in den reellen kubischen Korpern, Math. Z. 45, 511-532.

Bibliography

483

BUMBY, R.T. (1967) Irreducible integers in Galois extensions, Pacif. J. Math. 22, 221-229.

CARIN,

V.S.

(1054) On automorphism groups of nilpotent groups, Ukrain. Math. J. 6 , 295-304.

CARLITZ, L (1952) Primitive roots in a finite field, Trans. Amer. Math. SOC.73, 372-382.

CHASE, S (1971) On inseparable Galois theory, Bull. Amer. Math. SOC.77, 413-417

CHASE, S.U. and ROSENBERG, A (1966) A theorem of Harrison, Kummer theory, and Galois algebras, Nagoya Math. J

VOI.27. 663-685.

CHASE, S . U . and SWEEDLER, M (1969) IIopf algebras and Galois theory, Lecture Notes in Math. 97, Springer-Verlag.

CHILDS, L.N. ( 1984) Cyclic Stickelberger cohomo!ogy and descent of Kummer extensions,

Proc. A r n c r .

Math. SOC.90, 505-510.

CHINBURG, T (1085) Exact sequences and Galois module structure, Ann. Math. 11, Ser. 121, 351-376

COHEN, J . A . (1986) Extensions of valuation and absolute value topologies, Pacif. J. Math. 125, No. I 39-44.

484

Bibliography

COHEN, S.D. (1983) Primitive roots in the quadratic extension of a finite field, J. London Math. SOC.( 2 ) , 27, 221-228.

COHN, P.M. (1954) An invariant characterization of pseudo-valuations on a field, Proc. Camb. Phil. SOC. 50, 159-177. (1962) Eine bemerkung iiber die multiplikative gruppe eines Korpers, Arch. Math. 13, 344-348.

COHN, R.M. (1955) Finitely generated extensions of difference fields, Proc. Amer. Math. SOC.6, 3-5. (1956) An invariant of difference field extensions, Proc. Amer. Math. SOC.7, 656661.

CRAVEN, T and CSORDAS, G (1983a) Location of zeros, I: Real polynomials and entire functions, 111. J. Math. 27, 244-278. (1983b) Location of zeros, 11: Ordered fields., Ill. J. Math. 27, 279-299.

DAI, Z . Z . (1982) Some remarks on ordered fields, J. Math. Res. Expo., No 2, 7-10.

DARBI, G (1926) Sulla Riducibilita delle Equazioni Algebriche, Annali di Mat. pura e appl. Ser. 4 , 4 , 185-208.

DAVENPORT, H (1968) Bases for finite fields, J. London Math. SOC.43, 21-39.

Bibliography

485

DAVIS, R.D. and WISHART, E.F. (1971) Galois extensions and the ramification sequence of some wildly ramified n-adic fields,

Proc. Amer. Math. SOC.30. 212-216.

DAVIS, R.L. ( 1 9 6 9 ~ )A Galois theory for a class of purely inseparable field extensions, Dissertation, Florida

S t a t e University, Tallahassee, Florida U.S.A. (1069b) A Galois theory for a class of purely inseparable exponent two field extensions, 1 3 ~ 1 1 . Amer. Math. SOC.75, 1001-1004. (1973) Higher derivations and field extensions, Trans. Amer. Math. SOC.180, 47-52.

DELONE, B.N. and FADDEEV, D.K. (1944) Studies in t h e geometry of Galois groups, Matem. Sb.15 No 2, 243-284

DEMEYER. F.R. (1983) Generic polynomials,

J. Algebra 84, 441-448.

DEMUSHKIN, S.P. (1961) T h e group of a maximal pextension of a local field, Izv. Akad. Nauk SSR, Ser. M a t . 25. 329-346.

DEMUSHKIN, S.P. and SAFAREVICH, I.R. (1959) T h e imbedding problem for local rings, Izv. Akad. Nauk SSSR, Ser. M a t . 23, So (i, 823-840.

DEVENEY, J.K. (1973) Fields of constants of infinite higher derivations, Proc. Amer. Math. Soc. 41, 3 9 4 3 9 8 . [ 1974) A n intermediate theory for a purely inseparable Galois theory, Trans. Amer. l l a t l i .

SOC.198, 287-295.

486

Bibliography

(1976) A counter-example concerning inseparable field extensions, Proc. Amer. Math. SOC. 55, 33-34. (1 977) Generalized primitive elements for transcendental field extensions, Pacif. J. Math. 68,

41-45. (1982) Ruled function fields, Proc. Amer. Math. SOC.Vol. 86, No 2, 213-215.

DEVENEY, J.K. and HEEREMA, N . (1988) Higher derivations and distinguished subfields, Canad. J. Math. 40, N o 1, 131-141.

DEVENEY, J.K. and MORDESON, J.N. (1977a) Subfields and invariants of inseparable field extensions. Canad. J. Math. Vol. 29, No 6, 1304-1311. (1977b) Invariants of reliable field extensions, Arch. Math. 29, 141-147. (1'379a) The order of inseparability of fields, Canad. J. Math. 31, 655-662. (1079b) Splitting and modularly perfect fields, Pacif. J. Math. 83, 45-54. (1980) Distinguished subfields, Trans. Amer. Math. SOC.260, 185-193. (1981a) Distinguished subfields of intermediate fields, Canad. J. Math. 33, 1085-1096. (1981bJ Calculating invariants of inseparable field extensions, Proc. Amer. Math. SOC. 81, 373-376. (1982) Transcendence bases for field extensions, J. Math. SOC.Japan, Vol. 34, No 4, 703-707. (1983a) Maximal separable subfields of bounded codegree, Proc. Amer. Math. SOC.Vol. 88,

NO 1, 16-20. (1983b) Subfields containing distinguished subfields, Arch. Math. 40, 509-515. (1986) Uniqueness of subfields, Canad. Math. Bull. 29(2), 191-196.

Bibliography

DEURING, M (1931) Verzweigungs theorie bewerteter Korper, Math. Ann., 277-307.

( 1932) Galoische Theorie und Darstellungstheorie, Math. Ann. 107, 140-144. DICKER, R.M. (1968) A set of independent axioms for a field and a condition for a group to b e t h e

multiplicative group of a field, Proc. London Math. Soc. 3, 18, 114-124.

DIEUDONNE, J (1947) Sur les extensions transcendantes separables, Summa Brazil, Math. 2, No 1, 1-20.

D'MELLO, J.G. and MADAN, M.L.

( L983) Algebraic function fields with solvable automorphism group in characteristic p , Comm. Algebra, 11(11),1187-1236.

DODSON. B (1984) T h e structure of Galois groups of CM-fields, Trans. Amer. Math. SOC.283, 1-32

DRESS, A (1964) Z u einem S a t z aus der Theorie der algebraischen Zahlen, J. Reine Angew. Math. 2 1 0 ,

218-219. (1965) Kine Bemerkung iiber Teilringe globaren Korper, Abh. Math. Sem. Univ. Hainhiirg

28, 133-138.

DRIBIN, D.M. (1037) Quartic fields with t h e symmetric group, Ann. Math. 38, 739-746. (1938) Normal extensions of quartic fields with the symmetric

341-349.

group, Amer. J. Math. 39>

B i b 1 iography

488

EDWARDS, H.M. (1984) Galois theory, Graduate Texts in Mathematics, 101, Springer-Verlag,

New YGrk

- Berlin

- Heidelberg.

EKE, B.I. (1987) Special generating sets of purely inseparable fields of unbounded exponent, Pacif. J .

Math. Vol. 128, No 1, 73-79.

ENDLER, 0 (1972) Valuation theory, Springer-Verlag, New York

ENGLER, A.J. and VISWANATHAN, T.M. (1984) Real fields with small Galois groups, Rocky Mountain J. Math. Vol. 14, No 4, 817-818. (1986) Formally real fields with a simple description of the absolute Galois group, Manuscr.

Math. 56, 71-87.

ENNOLA, V (1958) Two elementary proofs concerning simple quadratic fields, Nordisk. Mat. Tidskur G, 114-117.

ERBACH, D.W., FISHER J. and MCKAY, J. (1979) Polynomials with PSL(2,7) as Galois group, J. Number Theory, 11, 69-75.

ERSHOV, Y.L. (1967) Fields with a solvable theory, Soviet Mathematics, Doklady 8 , 575-576. (1980) Regularly closed fields, Soviet Mathematics, Doklady, 31, 510-512. (1982) Multiply valued fields, Russian Math. Survey 37, No 3, 63-107. (1983) Absolute irreducibility and properties of Henzelizations, Algebra Logic 21, 353-357. (1984) Two theorems on regularly r-closed fields, J. Reine Angew. Math. 347, 154-167

8 i b l io g r a p h y

489

EVANOVICH, P (1973) Algebraic extensions of difference fields, Trans. Amer. Math. SOC.179, 1-22.

(1'384)Finitely generated extensions of partial difference fields, Trans. Amer. Math. SOC.

Vol. 281, NO 2,795-811.

EVANS, R.J. and ISAACS, I.M. (1977) Fields generated by linear combination of roots of unity, Trans. Amer. Math. SOC. 229, 249-258.

FADDEEV, D.K. (1954) On a hypothesis of Hasse, Dokl. Akad. Nauk SSSR, 94, 1013-1016.

FEIN, B , GURALNICK, R and SCHACHER, M (1983) Relative Brauer groups of global fields,

Arch. Math. 41, 309-318.

FEIT. W (1'359) On pregular extensions of local fields, Proc. Amer. Math. SOC.10, 592-595.

FENC,, K (1982) The rank group of cyclotomic units in abelian fields, J. Number Theory 14, 315-326.

FERNANDEZ-FERREIROS, P and MARTINEZ, J (1983) Separable valued fields,

Rev. Acad. Ci. Exactas Fis.-Quim. Nat. Zaragoza, 11, See.

38, 5-9.

FITZGERALD, R.W. (1983) Witt kernels of function field extensions, Pacif. J. Math. 109, No 1, 89-106.

FLANDERS, H ( 1953) The norm function of an algebraic field extension, Pacif. J. Math. 3, 103-1 13. ( 1 9 5 . ~ )The norm function of an algebraic field extension, 11, Pacif. J. Math. 519-528.

490

Bibliography

(1934) Elernentarteilertheorie in algebraischen Zahlkorpern, J. Reine Angew. Math. 171, 149-161.

FRAY, R and GILMER, R (1972) On solvability by radicals of finite fields, Math. Ann. 199, 279-291.

FREY, G (1973) Pseudo algebraically closed fields with nonarchimedean real valuations, J. Algebra 26, 202-207.

FRIED, M (1973) The field of definition of function fields and a problem in the reducibility of polynomials in two variables, Illin. J. Math. 17, 128-146. (1974) O n Hilbert's irreducibility theorem, J. Number Theory 6, 211-231. (1977) Fields of definition of function fields and Hurwitz families - Groups as Galois groups, Comm. Algebra 5(1), 17-82.

FRIED, M , HARAN D and JARDEN, M (1984) Galois stratification over Frobenius fields, Adv. in Math. 51, No 1, 1-35

FRIED, M and JARDEN,

M

(1986) Field arithmetic, Springer-Verlag, Berlin.

FROHLICH, A (1952) On the class group of relatively abelian fields, Quart. J. Math. Oxford, Ser. 3 , 98-106. (1954) A remark on the class number of abelian fields, J. London Math. SOC.29, 498. (1959) The genus field and genus group in finite number fields, Mathematika 6, 40-46. (1960a) A prime decomposition symbol for certain nonabelian number fields,

Bibliography

49 I

Acta Sci. Math. 21, 229-246. (19GOb) Discriminants of algebraic number fields, Math. Z. 74, 18-28. (1962) On non-ramified extensions with prescribed Galois group, Mathematika 9 , 133-13 I .

FUCHS, L (1970) Infinite abelian groups, Voi. I, Academic Press, New York and London.

[ 1'373) Infinite abelian groups, Vol. 11, Academic Press, New York and London.

FURUTA, Y (1959j On metabelian fields of a certain type, Nagoya Math. J. 14, 193-199.

(1961) A property of metabelian extensions, Nagoya Math. J. 19, 169-187. (1984) A norm residue map for central extensions of an algebraic number field, Nagoya Math.

J . 93, 61-69.

GAAL, L (1973) Classical Galois theory, Chelsea, New York.

GALOVICH, S and ROSEN, M (1982) Units and class groups in cyclotomic function fields, J. Number Theory, 14, 156-181.

GAMDOA. J.M. (1987) Some new results on ordered fields, J. Algebra 110, 1-12.

GAMBOA, J . M . and ItECIO, T (1983) Ordered fields with the dense orbits property, J. Pure Appl. Algebra 30, 237-246

GARDE, D (1985) O n the level of irreducible polynomials over Galois fields, J. Korean Math. SOC. 22,

11 7- 124.

B i b l iography

492

GARLING, D.J.H. (1986) A course in Galois theory, Cambridge Univ. Press, Cambridge - New York.

GASCHUTZ, W (1959) Fixkorper von p Automorphismengruppen rein transcendenter Korperwerturungen, von pcharakteristik, Math. Z. 71, 466.

GAY, D (1979) On normal radical extensions of real fields, Acta Arith. 35, 273-288. (1980) Normal binomials over algebraic number fields, J. Number Theory, 12, 311-326.

GAY, D , MCDANIEL A, and VELEZ, W.Y. (1977) Partially normal radical extensions of the rationals, Pacif. J. Math. 72(2), 403-417. (1979) On normal radical extensions of real fields, Acta Arith. 35, 273-278.

GAY, D and VELEZ, W.Y (1978) On the degreee of the splitting field of an irreducible binomial, Pacif. J. Math. Vol. 78, NO 1, 117-120. (1981) O n the degree of the splitting field of an irreducible binomial, Pacif. J. Math. Vol. 78;

NO 1, 117-120.

GERST. I (1970) On the theory of n-th power residues and a conjecture of Kronecker, Acta Arith. 17, 121-139.

GERSTENHABER, M (1964) O n the Galois theory of inseparable extensions, Bull. Amer. Math. SOC.70, 561-560. (1965) On infinite inseparable extensions of exponent one, Bull. Amer. Math. SOC. 71, 878-881.

Bibliography

493

(1968) On modular field extensions, J. Algebra, 10, 478-484.

GERSTENHABER, M and ZAROMP, A (1970) On the Galois theory of purely inseparable field extensions, Bull. Amer. Math. Soc. 76, 1011-1014.

GERTH, F (1987) Densities for 3-class ranks in certain cubic extensions, J. Reine Angew. Math. 381, 161-180.

GEYER, W.D. (1978) Galois groups of intersections of local fields, Israel J. Math. 30, 382-396

GILBARG, D (1942) The structure of the group of padic 1-units, Duke Math. J. 9, 262-271

GILMER, R (1972) On solvability by radicals of field extensions, Math. Ann. 199, 263-277

GILMER, R and HEINZER, W (1968) O n the existence of exceptional field extensions, Bull. Amer. Math. SOC.74, 545-547.

GIRSTMAIR, K [ 1079) Gber Konstruktive Methoden der Galoistheorie, Manuscr. Math. 26, 423-441 (I!J8:<) On the computation of resolvents and Galois groups, Manuscr. Math. 43, 289-307. (1987) O n invariant polynomials and their application in field theory, Math. Comp. 48,

No 178, 781-797. G O D W I N , H.J

( 1971) Computations relating to cubic fields, Computers Numb. Theory, 225-229

Bib1 iography

494

GOLD, R and MADAN, M (1985) Iwasawa invariants, Comm. Algebra 13, 1559-1578.

GOW, R (1986) Construction of some wreath products as Galois groups of normal real extensions of

the rationals, J. Number Theory 24, 360-372.

GRAS. G (1985) Prolongements Kummeriens dans les Z,-extensions,

Compos. Math. 55, 383-396.

GREEN, B.W. (1986) On the immediate extensions of a valued field, Quast. Math. 9, 227-244.

GREENBERG, R (1976) On the Iwasawa invariants of totally real number fields, Amer.

J. Math. 98, 263-281.

GREITHER, C and HARRISON D.K. (1986) A Galois correspondence for radical extensions of fields,

J. Pure Appl. Algebra, 43,

257-270.

GREITHER, C and PAREIGIS, B (1987) Hopf Galois theory for separable field extensions,

J. Algebra,

106, 239-258.

GROVE, L.C. (1983) Algebra, Academic Press, New York - London.

GUAN, CHI-WEN (1963) On the structure of multiplicative group of discretely valued complete fields,

Sci. Sinica 12, 1079-1103.

495

Bib1 iography

GYORY, K (1982) On the irreducibility of a class of polynomials, 111, J. Number Theory, 15, 164-181.

HADDIX, G.F., MORDESON J.N. and VINOGRADE, B

( 19GOa) Lattice isomorphisms of purely inseparable extensions, Math. Z . 111, 169-174. (1969b) On purely inseparable extensions of unbounded exponent, Canad. J . Math 21. 1526-1532.

HAJJA, M (1983) A note on monimial automorphisms, J. Algebra 85, 243-250. (1985) A note on a result of Kuniyoshi,

J. Algebra 92, 171-175.

HALL, M (1959) The theory of groups, Macmillan, New York

HALTER-KOCH, F (1971) Arithmetische Theorie der Normalkorper von 2-Potenzgrad mit Diedergruppe,

J. Number Theory 3, 412-443. ( 1972) k;in Satz iiber die Geschlechter relative-zyklischer Zahlkorper von Primzahlgrad u lid seine Anwendung auf biquadratisch-bizykiische Korper,

J. Number Theory

4 , 144-156.

HAMANN, E.A. and MORDESON, J.N. ( 1968) Pure inseparable field extensions, Math. Z. 103, 43-47 HARAN, D and JARDEN, M ( 1088) T h e absolute Galois group of a psecdo padically closed field, J . Reine Angcw. l l a t h 383. 147-206.

Bibliography

496

HARBATER, D (1987) Galois coverings of the arithmetic line, Number Theory, Semin. New York 1984/1985,

Lecture Notes in Math. 1240, 165-195.

HARRISON, D.K. (1963) Abelian extensions of arbitrary fields, Trans. Amer. Math. SOC. 106, 230-235.

HASSE, H (1950) Zum Existensatz von Grunwald in der Klassenkorpertheorie, J. Reine Angew.

Math. 188, 40-64. (1951) Zur Geschlechtertheorie in quadratischen Zahlkorpern, J. Math. SOC.Japan, 3, 45-51. (1966) Vandiver's congruence for the relative class number of the p t h cyclotomic field,

J. Math. Anal. Appl. 15, 87-90.

HASSE, H and SCHMIDT, F.K. (1931) Die Struktur diskret bewerteter Kbrper, J . fur Mathematik, 170, 4-63. (1936) Noch eine Begrundung der Theorie der hoheren Differential-quotienten in einem

algebraischen Funktionenkorper einer Unbestimmten, J. Reine Angew. Math. 177, 215-237. (1937) Noch eine Begrundung der Theorie der hoheren Differential quotienten in eineni

algebraischen Funktionenkorper einer Unbestimmten, J. Reine Angew. Math 177. 215-237.

HAUSNER, A (1961) Algebraic number fields and the diophantine equation m" = nm,

Amer. Math.

Monthly, 68, 856-861.

HAYAKAWA, K (1968) On a certain class of solvable extension fields over the rational number field, Sugaku

20, 97-98.

B i b l io g r a p h y

497

HAYASHI, H (1971) On Takagi’s basis in prime cyclotomic fields, Mem. Fac. Sci. Kyushu Univ. A25, 26;270.

HAYES, D.R. (1974) Explicit class field theory for rational function fields, Trans. Amer. Math. SOC., 189, 77-91.

HEEREMA, N (1960) Derivations and embeddings of a field in its power series ring, Proc. Amer. Math. SOC. 11, 188-194.

( 1061) Derivations and embeddings of a field in its power series ring, IT, Michigan Math. J. 8, 129-134. (1962) Derivations on padic fields, Trans. Amer. Math. SOC.102, 346-351. (1968) Convergent higher derivations on local rings, Trans. Amer. Math. SOC.132, 31-44

(1971) A galois theory for inseparable field extensions, Trans. Amer. Math. SOC.154, 193-200. (1076) p-th powers of distinguished subfields, Proc. Amer. Math. Soc. 55, 287-292 (1981) Maximal separable intermediate fields of large codegree, Proc. Amer. Math. S O C . 82,

351-354.

HEEREMA, N and DEVENEY, J (1974) Galois theory for fields K/k finitely generated, Trans. Amer. Math. SOC.189, 263-274.

HEEREMA, N and TUCKER, D (1975) Modular field extensions, Proc. Amer. Math. SOC.53, 1-6.

HEINEMANN, B (1985) On finite intersections of “Henselian valued” fields, Manuscr, Math. 52, 37-61

Bib1 iogr-aphy

498

HENSEL, K (1918) Eine neue Theorie der algebraischen Zahlen, Math. Z. 2, 433-452. (1921) Die Zerlengung der Primteiler eines beliebigen Zahlkorpers in einem auflosbaren

Oberkorper, 3 . Reine Angew. Math. 151, 200-209.

HEWITT, E and ROSS, K.A. (1963) Abstract harmonic analysis, Vol. I, Springer-Verlag, Berlin - Gottingen - Heidelberg.

HILBERT, D (1892) Ueber die irreduzibilitat genzen rationalen Funktionen mit ganzzahligen Koeffizienten,

J . Reine Angew. Math. 110, 104-129. (1894) Grundziige einer Theorie des Galoisschen Zahlkorpers, Nachr. Ges. Wiss. Gottingen, 224-236. (1897) Die Theorie der algebraischen Zahlkorper, Jber. Deutsch. Math. Verein. 4, 175-546. (1902) Uber die Theorie der relative-Abelschen Zahlkorper, Acta Math. 26, 99-132.

HOCHSCHILD, G (1950) Local class field theory, Ann. Math. Vol. 51, No 2, 331-347. (1955) Simple algebras with purely inseparable splitting fields of exponent one, Trans. Arner.

Math. SOC.79, 477-489.

HONDA, T (1971) Pure cubic fields whose class numbers are multiples of three, J. Number Theory 3. 7-12.

HOOLEY, C (1967) On Artin's conjecture, J. fur Mathematik, 225, 209-220.

499

B i bl iography

HURLIMAN,

w

(1984) A cyclotornic Hilbert 90 theorem, Arch. Math. Vol. 43, 25-26.

IKEDA, M

( 1960) Zur Existenz eigentlicher galoisscher Korper bein Einbettungsproblem, Hamb. Abh. 24, 126131.

ISAACS, I.M. (1070) Degrees of s u m in a separable field extension, Proc. Amer. Math. SOC.25, 638-634 I .

IWASAWA, K (1053a) A note on Kummer extensions, J. Math. SOC.Japan, Vol. 5, Nos 3-4, 253-262. (1953b) On solvable extensions of algebraic number fields, Ann. Math. 58, 548-572.

(1955) On Galois groups of local fields, Trans. Amer. Math. SOC.80, 448-469. (1958) On some invariants of cyclotomic fields, Amer. J. Math. 80, 773-783, correct. 81(395(3), 280. (1959) On the theory of cyclotomic fields, Ann. Math. 70, 530-561. (1960) On local cyclotomic fields, J. Math. SOC.Japan, 12, 1621. (1962) A class number formula for cyclotomic fields, Ann. Math. 76, 171-179. (1965) Some modules in the theory of cyclotomic fields, Proc. Symp. Pure Math. 8, 66-69.

(1972) On the pinvariants of cyclotomic fields, Acta. Arith. 21, 99-101. (1983) On cohomology groups of units for Zp-extensions, Amer. J. Math. 105, 189-200.

(1986) Local class field theory, Oxford University Press, Oxford.

IWASAWA, K and SIMS, C.S. \ 1Q66) Computation of invariants in the theory of cyclotomic fields, J. Math. SOC

Bibliography

500

Japan 18, 86-96.

IYANAGA, S (1929) Ueber den Fuhrer eines relativ-zyklischen Zahlkorpers, Proc. Imper. Acad. Japan,

Tokyo 4 , 108-110.

JACOBSON, E and VELEZ, W.Y. (1985) On the adkle rings of radical extensions of the rationals, Arch. Math. 45, 12-20.

JACOBSON, N (1944) Galois theory of purely inseparable fields of exponent one, Amer. J. Math. 66, 645-648. (1964) Lectures in abstract algebra, Vol. 111: Theory of fields and Galois theory,

Von Nostrand, Princeton, N.J.

JANNSEN, U (1982) Uber Galoisgruppen lokaler Korper, Invent. Math. 70, 53-69.

JANNSEN, U and WINGBERG, K (1982) Die Struktur der absoluten Galoisgruppe padischer Zahlkorper, Invent. Math. 70, 71-98.

JANUSZ, G.J. (1973) Algebraic number fields, Academic Press, New York - London.

JARDEN, M (1974) Algebraic extensions of finite corank of Hilbertian fields, Israel J. Math. 18, 279-307. (1975) Roots of unity over large algebraic fields, Math. Ann. 213, 109-127. (1979) An analogue of Artin-Schreir theorem, Math. Ann. 242, 193-200. (1980) An analogue of Cebotarev density theorem for fields of finite corank, J. Math. Kyoto

Univ. 20. 141-147.

Bibliography

501

( 1982) The Cebotarev density theorem for function fields; an elementary approach, hfath. Ann. 261, 467-475.

JEHNE, W (1962) Zur Theorie der Kreiskorper, J. Reine Angew. Math. 209, 100-103.

JENSEN, C.U. and YUI, N (1982) Polynomials with D , as Galois group, J. Number Theory 15, 347-375.

KANG, S . W . (1983) Remarks on finite fields, Bull. Korean Math. SOC.20, 81-85.

KANI, E (1986) Bounds on the number of non-rational subfields of a function field, Invent. Math. 85, 185-198.

KANTZ, G ( 1957) Eine fur die Theorie der relativabelischen Korper grundlegende Abelsche Operatorgruppe, Monatsh. Math. 61, 151-156.

KAPLANSKY, I (1947) Topological methods in valuation theory, Duke Math. J. 14, 527-541.

(1972) Fields and rings, Chicago Lecture Notes in Mathematics, The Univ. of Chicago Press. Chicago and London.

KARNI. E (1986) Bounds on the number of non-rational subfields of a function field, Invent. Math 85, 185-198.

KATAOKA, T (1980) On the integer ring of the composition of algebraic number fields, Nagoya Math. ,J. 77, 25-31.

502

B i b l io g r a p h y

KATSURADA, H (1987) On representations of the maximal unramified Galois extension of a field of positive

characteristic, Hokkaido Math. J. 16, 127-134.

KAWADA, Y (1951) On the derivations in number fields, Ann. Math. Vol. 54, No 2, 302-314. (1953) On the ramification theory of infinite algebraic extensions, Ann. Math. 58, 24-47. (1954) On the structure of the Galois group of some infinite extensions, J. Fac. Sci.

Tokyo, I, 1-18, 87-106.

KERSTEN, I and MICHALICEK, J (1981) Kubische Galoiserweiterungen mit Normalbasis, Comm. Algebra, 9, 1863-1871. (1984) Applications of Kummer theory without roots of unity, Methods in ring theory, Proc.

NATO Adv. Study Inst., Antwerp, 1983, NATO, ASI, Ser. C129, 201-205. (1987) A characterization of Galois field extensions of degree 3, COD.

Algebra, 15(5),

927-933.

KIDA, Y (1982) Cyclotomic Z2-extensions of J-fields, J. Number Theory 14, 340-352.

KIJIMA, D (1987) Cuts of ordered fields, Hiroshima Math. J. No 2, 337-347.

KIJIMA, D and NISHI, M (1987) Maximal ordered fields of rank n, Hiroshima Math. J. 17, 157-167.

KIKUCHI, T (1971) Some remarks on high order derivations, J. Math. Kyoto Univ. 11, 71-87. (1983a) Some remarks on high order derivations 111, J. Math. Kyoto Univ. 23-2, 265-267.

503

Bibliography

( 1983b) On Galois correspondence between intermediate fields and closed derivation subalgebras, J . Math. Kyoto Univ., 23-2, 281-287.

KIME, L.A. (1973) Purely inseparable, modular extensions of unbounded exponent, Trans. Amer. Math.

SOC.176, 335-349.

KINOHARA, A (1952) On the derivations and the relative differents in algebraic number fields, J. Sci

IIiroshima Univ. 16, 261-266. (1.953) On the derivations and the relative differents in commutative fields, J. Sci. Hiroshima

Univ. 16, 441-456.

KLEIMAN, H (1.971) Totally real subfields of padic fields having the symmetric group as Galois group,

Canad. Math. Bull. 14, 441-442. (1972) Normal integral bases and the Galois extensions of number fields, J. Reine Angew.

Math. 255, 83-84.

KNESER, M (1975) Lineare Abhangigkeit von Wurzeln, Acta Arith. 26, 307-308.

KOCH, H (1961) The Galois group of a local field, Dokl. Akad. Nauk SSSR, 137, 1291-1294

( I 962) Ubcr galoische Erweiterungen padischer Zahlkorper, J . Reine Angew. Math. 209,

8-11. 1 1 %3a)

Uber maximale L-Erweiterungen mit vorgeschriebenem Verzweizungsstellen, l l o [ l . Bcr. Deutsch. Akad. Wiss. , 341-344.

1!X3b) U b c r

Darstellengsraume und die Struktur der multiplikativen Gruppe eines padischrn

Zahlkorpers, Math. Nachr. 26, 67-100.

Bibliography

504

(1964) fJber Halbkorpern, die in algebraischen Zahlkorpern enthalten sind, Acta Math.

Hungar. 15, 439-444. (1965) Uber Galoische Gruppen von padischen Zahlkorpern, Math. Nachr. 29, 77-111. (1967) Beweis einer Vermutung von Hochsmann aus der Theorie der !-Erweiterungen,

J.

Reine Angew. Math. 225, 203-206 (1968) Uber die Dimension der Galoischen Gruppen maximeler pErweiterungen, Mon. Ber.

Deutsch. Akad. Wiss. 10, 5-8. (1969) Zum Staz von Golod-Schafarewitch, Math. Nachr. 42, 321-333. (1970) Galoissche Theorie der pErweiterungen, Berlin.

KOMATSU, K (1982) On a certain property of profinite groups, Proc. Jap. Acad. Ser. A58, 319-322.

KONDO, T (1963) On abelian extensions of the gaussian field, Sugaku 15, 110 (Japanese).

KOSAKI, K and YANAGIHARA, H (1970) On purely inseparable extensions of algebraic function fields, J. Sci. Hiroshima Univ.

Ser. A, I, 34, 69-72. KRAFT, H (1970) Inseparable Korpererweiterungen, Comment. Math. Helv. 45, 110-118.

KRASNER, M (1988) Local differents of algebraic and finite extensions of valued fields, J. Number Theory

28, 17-61.

KREIMER, H and HEEREMA, N (1975) Modularity vs. separability for field extensions, Canad. J. Math. 27, 1176-1182.

505

Bibliography

KRULL, W

( 1928) Galoissche Theorie der unendlichen algebraischen Erweiterungen, Math. Ann. 100, 687-698. (1930) Galoissche Theorie bewerteter Korper, S.B. Bayer. Akad. Wiss. 225-238.

( 1952) Uber unendliche algebraische Erweiterungen bewerteter Korper, Rend. circ. Palermo, 1, 164-169.

KRULL, W and NEUKIRCH, J (1971) Die Struktur der absoluten Galois gruppe uber dem korper IR(t), Math. Ann. 193, 197-209.

KIJBOTA, T (1957a) Galois group of the maximal abelian extension over an algebraic number field, Nagoya

Math. J. 12. 177-189. (1057b) Unit groups of cyclic extension, Nagoya Math. J. 12, 221-229. (1959) Factor sets in a number field and the norm residue symbol, J. Math. SOC.Japan 11,

129-138.

K U H L M A N , F.V. and PRESTEL, A (1984) Places of algebraic function fields, J. Reine Angew. Math. 353, 181-195.

KUNIYOSHI. H (1954) On pure transcendency of a certain field, T6hoku Math.

J. 6, 101-108.

LABUTE, J (1965) Classification des groupes de Demushkin, C.R. Acad. Sci. Paris, 260, 1043-1046.

LAMBERT, D (1986) Algebraic padic expansions, J. Number Theory 23, 279-284.

506

Bibliography

LA MACCHIA, M.E. (1980) Polynomials with Galois group P S L ( 2 , 7 ) , Comm. Algebra 8(10), 983-992.

LANG, S (1952) On quasi algebraic closure, Ann. Math. 55, 373-390. (1967) Algebra, Addison-Wesley, Reading, Mass.

LEBOW, A and SCHREIBER, M (1984) A formula for the separability idempotent in the tensor square of a field, J. Algebra

90, 372-374.

LEHMER, D.H. and LEHMER, E (1987) Cyclotomic resultants, Math. Comput. 48, 211-216.

LENSTRA, H.W., Jr. (1974) Rational functions invariant under a finite abelian group, Invent. Math. 25, 299-325. (1985) A normal basis theorem for infinite Galois extension, Indagationes Math. 47, 221-228.

LENSTRA, H.W. Jr. and SCHOOF, R.J. (1987) Primitive normal bases for finite fields, Math. Comput. 48, 217-231.

LESTER, C.A. (1938) A determination of the automorphisms of certain algebraic fields, Duke Math. J. 4 , 277-290.

LIANG, J (1972) On relation between units of normal algebraic number fields and their subfields, Acta

Arith. 20, 331-344.

LONG, R.L. (1971) Steinitz classes of cyclic extensions of prime degree, J. Reine Angew. Math. 250, 87-98.

Bib1 iography

507

LOWE, R.D. and ZELINSKY, D (1953) Which Galois fields are pure extensions?

Math. Student, 21, 37-41

LUBIN, J and TATE, J (1065) Formal complex multiplication in local fields, Ann. Math. 81, 380-387.

LUBOTZKY, A and DRIES L.V.D. (1981) Subgroups of free profinite groups and large subfields of

q, Israel J . Math. 39, 25-45.

MACLANE, S (1938) A lattice formulation for transcendence degree and pbases, Duke Math. J. 4 , 455-468. (1939a) Modular fields, I, Separating transcendency bases, Duke Math. J . 5, 372-393.

(1939b) Steinitz field towers for modular fields, Trans. Amer. Math. SOC.Vol. 46, 23-45.

( 1 9 3 9 ~ )Subfields and automorphism groups of padic fields, Ann. Math. 40, 423-442. (1940) Note on the relative structure of padic fields, Ann. Math. VoI. 41, N o 4, 751-753.

(1963) Homology, Grundlechren der Mathematik, 114, Springer, Berlin.

MAKI, S (1985) On the density of abelian number fields, Ann. Acad. Sci. Fenn. Ser. A I, Diss. 5.1, 104p.

MANIN, Y (1071) Cyclotomic fields and modular curves, Russian Math. Surveys, 26, 7-78.

MANN, H.B. (1954) On an exceptional phenomenon in certain quadratic extensions, Canad.

4 74-4 76.

J. Math. 6,

Bibliography

508

MA”,

H and VELEZ, W.Y.

(1975) On normal radical extensions of the rationals, Linear and multilinear algebra, 3, 73-80.

MARSHALL, M.A. (1971a) Ramification groups of abelian local field extensions, Canad. J. Math. 23, 271-281. (1971b) The maximal pextension of a local field, Canad. J. Math. 23, 398-402,

MARTINET, J (1968) Sur les extensions galoisiennes non abhliennes de degrh 2p des rationales, C.R. Acad. Sci. Paris, 266, A9.59-A962. (1969) Sur l’arithmktique des extensions galoisiennes B groupe de Galois dikdral d’orde 2p, Ann. Inst. Fourier 19, 1-80.

MARTINET, J and PAYAN, J.J. (1967) Sur le extensions cubiques non-Galoisiennes des rationnels et leur cloture Galoisienne,

J. Reine Angew. Math. 228, 1.5-37. MATSUDA, R (1964) On purely-transcendency of certain fields, T8hoku Math. J . 16, 189-202.

MATZAT, H (1979) Konstruktion von Zahlkorpern mit der Galoisgruppe

M11

iiber Q

(a),

Manuscripta Math. 27, 103-111.

MAUS, E (1967) Arithmetisch disjunkte Korper, J. Reine Angew. Math. 226, 184-203.

MAY, W (1969) Commutative group algebras, Trans. Amer. Math. SOC. 136, 139-144.

Bib1 iography

509

(1970) Unit groups of infinite abelian extensions, Proc. Amer. Math. SOC.Vol. 25, No 3, 680-683. (1972) Multiplicative groups of fields, Proc. London Math. SOC.(3), 24, 295-306. (1979) Multiplicative groups under field extensions, Canad. J. Math. Vol. 31, N o 2, 436-410. (1980) Fields with free multiplicative groups modulo torsion, Rocky Mountain J. Math. Vol. 10, NO 3, 599-604.

MCCARTHY, P.J. (1966) Algebraic extensions of fields, Waltham, Mass., Blaisdell.

MCCULLOH, L.R. (1963) Integral bases in Kummer extensions of Dedekind fields, Canad. J. Math. 15, 755-765. (1966) Cyclic extensions without relative integral bases, Proc. Amer. Math. SOC.17, 1191-1194. (1969) Cyclic extensions of prime power degree and corresponding residue systems, J. Number Theory, 1, 459-466.

MCKENZIE, R and SCHEUNEMANN, J (1971) A number field without a relative integral basis, Amer. Math. Monthly 78, 882-883.

MELNIKOV, O.V. and TAVGEN, O.J. (1085) The absolute Galois group of a Henselian field, Dokl. Akad. Nauk BSSR 29, 581-583.

MERKUR’EV, A.S. (1986) On the structure of the Brauer group of fields, Math. USSR, Izv. 27, 141-157

MIKI, H (1983) On the absolute Galois groups of local fields, I, Galois groups and their representations, Proc. Symp. Nagoya, Adv. Stud. Pure Math. 2, 55-61.

510

Bibliography

MINAC, J (1986) Poincare groups and ordered fields, C.R. Math. Acad. Sci. SOC.R, Can. 8 , 255-26(3.

MINES, R and RICHMAN, F (1984) Valuation theory: A constructive view, J. Number Theory 19, 40-62. (1986) Archimedean valuations, J. London Math. SOC.( 2 ) , 34, 403-410.

MIYAKE, K (1984) On central extensions of a Galois extension of algebraic number fields, Nagoya Math.

J. 93, 133-148.

MIYANISHI, M (1968) A remark on an interative infinite higher derivation, J. Math. Kyoto Univ. 8-3, 411415.

MOH, T and HEINZER, W (1979) A generalized Luroth theorem for curves, J. Math. SOC.Japan, 85-86.

MORDESON, J.N. (1965) Remark on coefficient fields in complete local rings, J. Math. Kyoto Univ. 4, 637-630. (1975a) On a Galois theory for inseparable field extensions, Trans. Amer. Math. Soc. 337-347. (1975b) Splitting of field extensions, Arch. Math. 26, 606-610.

MORDESON, J.N. and SCHOULTZ, W (1973) p-bases of inseparable field extensions, Arch. Math. 24, 44-49.

MORDESON, J.N. and VINOGRADE, B (1965) Extension of certain subfields to coefficient fields in commutative algebra, J. Math.

SOC.Japan 17, 47-51. (1968) Generators and tensor factors of purely inseparable fields, Math. Z. 107, 326-331

Bibliography

511

(1989a) Note on relative pbases of purely inseparable extensions, Proc. Amer. Math. SOC.22, 587-590. (19G9b) Tensor products of purely inseparable field extensions, Proc. Amer. Math. SOC. 20, 209-2 12. (1'370) Structure of arbitrary purely inseparable extension fields, Lecture Notes in Mathematics, Vol. 173, Springer-Verlag, Berlin and New York. (1971) Exponents and intermediate fields of purely inseparable extension, J. Algebra 17, 238-243. (1972) Separating pbasis and transcendental extension fields, Proc. Amer. Math. Soc. 3 1 , 417-422. (1973) Relatively separated transcendental field extensions, Arch. Math., 24, 521-526.

( 1975) Inseparable embeddings of separable transcendental extensions, Arch. Math. 26, 606-610.

( 1978) Inseparable embeddings of separable transcendental extensions, Arch. Math. 27. 42-47.

NACHEV, N.A. (1986) Invariants of separable binomial extensions, C.R. Acad. Bulgar. Sci. 39, N o 1 1 , 9-10.

NAGATA, M (1!);9/60) Note on coefficient fields of complete local rings, Mem. Coll. Sci. Univ. Kyoto 32,

91-92. (1967) A theorem on valuation rings and its applications, Nagoya Math. J. 29, 85-91 (1977) Field theory, Marcel Dekker, New York.

( 1082) A field extension with certain finiteness condition on multiplicative group extension, J. Math. Kyoto Univ. 22-2, 255-256.

512

Bibliography

NAGATA, M, NAKAYAMA, T and TSUZUKU, T (1957) An existence lemma in valuation theory, Nagoya Math. J. 6, 59-61.

NAKAI, Y (1970) High order derivations I, Osaka J. Math. 7, 1-27.

NAKAI, Y., KOSAKI K and ISHIBASHT, Y (1970) High order derivations 11, J. Sci. Hiroshima Univ. Ser. A-I, 34, 17-27.

NAKAMARA, T (1970) On the determination of the fundamental units of certain real quadratic fields, Mem. Fac. Sci. Kyusku A24, 300-304.

NAKAMULA, K (1986) Elliptic units and the class number of non-Galois fields, Proc. Int. Conf. Katata/Jap., 53-65.

NAKAMURA, Y (1986) On relatively formally real fields, Math. Rep. Toyama Univ. 9, 149-159.

NAKANO, S (1983) Class numbers of pure cubic fields, Proc. Japan. Acad. Ser. A, Math. Sci. 59, No 6, 263-265.

NAKATSUCHI, S (1968) A note on certain properties of algebraic number fields, Mem. Osaka Kyoiku Univ. 17, 1-10,

NAKAYAMA, T ( 1950) Construction and characterization of Galois algebras with given Galois group, Nagoya Math. J. Vol. 1, 11-17.

513

Bibliography

(1952) Idele-class factor sets and class field theory, Ann. Math. 55, 73-84.

NART, E and VILA, N (1979) Equations of the type

Xn+ a X

+ 6 with absolute Galois group S,,

Rev. Univ.

Santander, 2, 821-825.

NEGGERS, J (1965) Derivations on Fadic fields, Trans. Amer. Math. SOC.115, 496-504.

NEUKIRCH, J

( 1969) Kennzeichnung der endlich-algebraischen Zahlkorper durch die Galoisgruppe der maximal auflosbaren Erweiterungen, J. Reine Angew. Math. 238, 135-147.

( I073a) Einbettungsprobleme mit lokaler Vorgabe und freie Produkte lokaler Galoisgruppen, J. Reine Angew. Math. 259, 1-47.

( 1973b) Uber das Einbettungsproblem der algebraischen Zahlentheorie, Inv. Math. 21, 59.116. (1976) Zur theorie der nilpotenten algebraischen Zahlkorper, Math. Ann. 219, 65-83. (1979) On solvable number fields, Invent. Math. 53, 135-164.

(1986) Class field theory, Springer-Verlag, Berlin-Heidelberg- New York-Tokyo.

NEUMANN, 0 (1086) On the imbedding of quadratic extensions into Galois extensions with symmetric

group, Algebraic geometry, Proc. Conf. Berlin 1985, Teubner-Texte Math. 92. 285-295.

NEWMAN, M (1971) Units in cyclotomic fields, J. Reine Angew. Math. 250, 3-11.

NEWMAN, K and RADFORD, D.E. (1986) On when a separable field extension in characteristic p > 0 is determined by its

514

Bi b l iography

Frobenius structure, 11, J. Algebra, 102, 301-331.

NOETHER, E (1931) Normalbasis bei Korpern ohne hijhere Verzweigung, J. Reine Angew. Math. 167, 147-152.

NORRIS, M and VELEZ, W.Y. 11980) Structure theorems for radical extensions of fields, Acta Arith. 38, 111-115.

OHM, J (1982) Simple transcendental extensions of valued fields, J. Math. Kyoto Univ. 22-2,201-221. (1983) The ruled residue theorem for simple transcendental extensions of valued fields, Proc.

Amer. Math. SOC.89, 16-18. (1984) On subfields of rational function fields, Arch. Math. 42, 136-138 (1985) Simple transcendental extensions of valued fields 11: A fundamental inequality,

J. Math. Kyoto Univ. 25-3, 583-596.

OJANGUREN, M and SRIDHARAN, R (1969) A note on purely inseparable extensions, Comment. Math. Helv. 44, 457-461.

OKUTSU, K (1952) Integral basis of the field Q

( m, Proc. Jap. Acad. Ser. A58, 219-222.

O'MEARA, O.T. (1959) Infinite-dimensional quadratic forms over algebraic number fields, Proc. Amer. Math.

SOC.10, 55-58.

OPOLKA, H

[ 1982) Der Liftungsindex primitiver Galoisdarstellungen, J. Algebra, Vol. 74, N o 2, 535-512.

515

Bibliography

( 1083) Extensions of number fields defined by cohomology groups, Nagoya Math. J. 92, 179-186.

( 1984) Galois representations, points of inflexion on elliptic curves and automorphic representations, Expo Math. 2, 185-187. (1987) T h e norm exponent in Galois extensions of number fields, Proc. Amer. Math. Soc. 99,

NO 1 , 41-43.

ORBANZ, U (1973) Tensorprodukt und hohere Derivationen,

Arch. Math. 24, 513-520.

ORE, 0 ( 1 ~ 3 Zur ) Theorie der algebraischen Korper, Acta Math. 44, 219-314.

DE OROZCO, M.A. and VELEZ, W.Y. (1982) T h e lattice of subfields of a radical extension, J. Number Theory 15, 388-405. (1084) T h e torsion group of a field defined by radicals, J. Number Theory, 19, 283-294

OSADA, H (1987) T h e Galois groups of the polynomials X "

+ a x e+ b, J. Number Theory, 2 5 , 230-238.

OSTROWSKI, A

[ 1913) Uber einige Fragen der allgemeinen Korper theorie, J. Reine Angew. Math. 143, 255-284. (1017) Uber sogenannte perfekte Korper, J. Reine Angew. Math. 147, 191-204. (1918) Uber einige Losungen der Funktionalgeichung

d(z)d(y) = d(zy), Acta Math.

41, 271-284. [ 1919) Uber ganzwertigre Polynome in algebraischen Zahlkorpern, 3 . Reine Angew. l f a t l l .

149, 117-124.

Bib1 iography

516

(1935) Untersuchungen zur arithmetischer Theorie der Korper, Math. Z. 39, 269-320, 321-361, 361-404.

PAREGIS, B (1970) Bemerkung uber Tensorprodukte von Korpern, Arch. Math. 31, 479-482.

PARSHIN, A.N. (1984) Local class field theory, Tr. Mat. Inst. Steklova 165, 143-170 (Russian)

PAYAN, J.J. (1962a) Construktion des corps abeliens de degrC 5, C.R. Acad. Sci. Paris, 254, 3618-3620. (1962b) Entiers des corps abCliens de degrd 5, C.R. Acad. Sci. Paris, 255, 2345-2347. (1965) Contribution a 1’Ctude des COT^ abCliens absolus de degrC premier impair, Ann. Inst.

Fourier 15, 133-199. PEI, D.Y. and FENG, K (1980) On independence of the cyclotomic units, Acta Math. Sinica, 23, 773-778.

PELLING, M.J. (1981) Increasing polynomials in an ordered field, Amer. Math. Monthly, 88, 150-152.

PERLIS, G (1942) Normal bases of cyclic fields of prime-power degree, Duke Math. J. 9, 507-517.

PETERS, M (1972) Die Stufe von Ordnungen ganzer Zahlen in algebraischen Zahlkorpern, Math. Ann. 195, 309-314.

PICKERT, G (1949) Inseparable Korperweiterungen, Math. Z. 52, 81-135.

517

Bibliography

(1950) Eine Normalform fur endliche rein-inseparable Korperweiterungen, Math. 2 . 53, 133-135. (1952) Zwischenkorperverbande endlicher inseparabler Erweiterungen, Math. Z. 55, 355-363.

POLLACZEK, F (1924) Uber die irregularen Kreiskorper der l-ten und l-ten Einheitswurzeln, Math. 2. 21,

1-38.

PONOMARENKO, P (1965) The Galois theory of infinite purely inseparable extensions, Bull. Amer. Math. SOC.

71, 876-877.

PONTRYAGIN, L . S . (1977) Topological groups, Gordon and Breach, New York, London

P O P E S K Y , E.L. and P O P E S K U , N (1987) On a class of intermediate subfields, Stud. Cercet, Mat. 39, 156-162.

POSTNIKOV, M.M. (1961) Fundamentals of Galois theory, Delhi: Hindustan Pub. Co.

PRESTEL, A (1984) Lectures on formally real fields, Lecture Notes in Mathematics, 1093, Springer, Berlin.

QURESHI, M.A. (1982) A dimension formula for the tensor product of fields, Riazi J. Karachi Math. Assoc. 4 , 37-41.

RADFORD, D.E. (1982) On when a separable field extension in characteristic p

Frobenius structure, I, 3. Algebra, 127-150.

>

0 is determined by its

Bibliography

518

RAMACHANDRA, K (1966) On the units of cyclotomic fields, Acta Arith. 12, 165-173. (1969) On the class number of relative abelian fields, J. Reine Angew. Math. 236, 1-10,

RASALA, R (1971) Inseparable splitting theory, Trans. Amer. Math. SOC.162, 411-448.

REICHARDT, H (1933) Arithmetische Theorie der kubischen Korper als Radikalkorper, Monatsh. Math.

Phys. 40, 323-350. (1936) Uber Normalkorper mit Quaternionengruppe, Math. Z. 41, 218-221. (1937) Konstruktion von Zahlkorpern mit gegebener Galoisgruppe von

Primzahlpotenzordnung, J. Reine Angew. Math. 177, 1-5. REDEI, L (1959) Naturliche Basen des Kreisteilungskorpers, Abh. Math. Sem. Univ. Hamburg, 23, 180-200.

REID, J (1966) A note on inseparability, Michigan Math. J. 13, 219-223.

RELLA, T (1920) Uber die multiplikative Darstellung von algebraischen Zahlen eines Galoisschen

Zahlkorpers fur den Bereich eines beliebigen Primteilers, J. Reine Angew. Math. 150, 157-174.

REMAK, R (1952) Uber Grossenbeziehungen zwischen Discriminante und Regulator eines algebraischen

Zahlkorpers, Compos. Math. 10, 245-285.

Bibliography

519

(1954) Uber algebraische Zahlkorper mit schwachem Einheitsdefekt, Compos. Math. 12, 35-80.

RIBENBOIM, P ( i 9 6 2 a ) Rksultats et problkmes de la thkorie des corps de classes, Rev. Un. Mat. Argentina 20, 125-145.

(1962b) An existence theorem for fields with Krull valuations, Trans. Amer. Math. SOC.105, 278-294.

RICHARDS, I (1974) An application of Galois theory to elementary arithmetic, Advances in Math. 13, 268-273.

RIGO, T and WARNER, S (1978) Topologies extending valuations, Canad. J. Math. 30, 164-169.

ROBINSON, E (1987) The geometric theory of padic fields, J. Algebra 110, 158-172.

ROLAND, G , YUI, N and ZAGIER, D (1982) A parametric family of quintic polynomials with Galois group D 5 , J. Number Theory 15, 137-142.

ROQUETTE, P (1963) Isomorphisms of generic splitting fields of simple algebras, J . Reine Angew. Math.

211/215, 207-226.

ROSENLICHT, M (3983) The rank of a Hardy field, Trans. Amer. Math. SOC.280, No 2, 659-671.

ROTH, R.L. (1971) On extensions of Q by square roots, Amer. Math. Monthly 78, 392-393.

B i b l iography

520

ROTMAN, J (1960) A characterization of fields among integral domains, An. Acad. Brasil Ci. 32, 193-194.

ROZENBAUM. K (1966) On the multiplicative group of cyclic extensions of local fields, Vestnik Leningrad

Univ. 21, 80-92.

RUBIN, K (1987) Global units and ideal class numbers, Invent. Math. 89, 511-526.

RUCK, H.G. (1986) Hasse-Witt invariants and dihedral extensions, Math. Z. 191, 513-517.

RYGG, P (1963) On minimal sets of generators of purely inseparable field extensions, Proc. Amer.

Math. SOC.14, 742-745.

RYGG, P and LEHMAN, B (1969) A note on an equivalence relation on a purely inseparable extension, Canad. Math.

Bull. 12, 175-178.

SAFAREVICH, I.R. (L954a) On the construction of fields with a given Galois group of order P , Izv. Akad. Nauk

SSSR, Ser. Math. 18, 216-296. (1954b) On the prob1,em of imbedding fields, Izv. Akad. Nauk SSSR, Ser. Math. 18, 389-418. ( 1 9 5 4 ~ )Construction of fields of algebraic numbers with given solvable Galois group,

Izv. Akad. Nauk SSSR, Ser. Math. 18, 525-578. (1956) Construction of fields of algebraic numbers with given solvable Galois group,

Trans]. Amer. Math. SOC.(2) 4 , 185-237.

Bibliography

521

(1958) The imbedding problem for splitting extensions, Dokl. Akad. Nauk SSSR, 120, 12 17-12 19.

SALTMAN, D.J. (1982) Generic Galois extensions and problems in field theory, Adv. in Math. 43, 250-283. (1983) Exterior powers of fields and subfields, Nagoya Math. J. 89, 119-127. (1984a) Retract rational fields and cyclic Galois extensions, Israel J. Math. 47, Nos 2-3, 165-215. (1084b) Noether’s problem over an algebraically closed field, Invent. Math. 77, 71-84.

( I087a) Multiplicative field invariants, J. Algebra 106, 221-238. (1087b) Galois groups of order p 3 Comm. Algebra 15, 1365-1373.

SANDS. J . W . (1984) Galois groups of exponent two and the Brumer-Stark conjecture, J. Reine Angeiv

Math. 349, 129-135.

SATO, S (1969) Notes on derivations of higher order, J. Sci. Hiroshima Univ. Ser. A-I, 33, 41-46.

SCHACHER, M . M . (1968) Subfields of division rings, I, J. Algebra 9, 451-477

SCHANUEL, S (1964) On heights in number fields, Bull. Amer. Math. SOC.70, 262-263.

SCHENKMAN, E (1064) On the multiplicative group of a field, Arch. Math. 15, 282-285.

Bibliography

522

SCHILLING, O.F.G. (1935) Uber gewisse Beziehungen zwischen der Arithmetik hyperkomplexer Zahlsysteme und

algebraischer Zahlkorper, Math. Ann. 111, 372-398. (1940) Regular normal extensions over complete fields, Trans. Amer. Math. SOC.47, 440-454. (1943) Normal extensions of relatively complete fields, Amer. J. Math. 65, 309-334. (1961) On local class field theory, J. Math. SOC.Japan 13, 234-245.

SCHINZEL, A (1975a) On the linear dependence of roots, Acta Arith. 28, 161-175. (197513) On power residues and exponential congruences, Acta Arith. 27, 397-420. (1977) Abelian binomials, power residues, and exponential congruences, Acta Arith. 32, 24 5-274.

(1987) On the number of terms of a power of a polynomial, Acta Arith. 49, No 1, 55-70.

SCHMIDT, F.K. (1930) Zur Klassenkorpertheorie in Kleinen, J. Reine Angew. Math. 162, 155-168. (1933) Mehrfach perfekte korper, Math. Ann. 108, 1-25.

SCHMIDT, F . K . and MACLANE, S (1941) The generation of inseparable fields, Proc. N.A.S., 27, 583-587.

SCHOENBERG, I.J. (1964) A note on the cyclotomic polynomial, Mathematika, 11, 131-136.

SCHOLZ, A (1929) Uber die Bildung algebraischer Zahlkorper mit auflosbarer galoischer Gruppe,

Math. Z. 30. 332-356.

523

B i b 1 iography

(1936) Konstruktion algebraischer Zahlkorper mit beliebiger Gruppe von

Primzahlpotenzordnung, Math. Z. 42, 161-188. (1938) Minimaldiskriminanten algebraischer Zahlkorper, J. Reine Angew. Math. 179, 16-2 1 .

SCHUR, I (1930) Gleichungen ohne Affekt, S, -B. Berlin Acad. 443-449. (1931) Affektolose Gleichungen in der Theorie der Laguerreschen and Hermiteschen

Polynome, J. Reine Angew. Math. 165, 52-58. (1932) Einige Bemerkungen uber die Diskriminante eines algebraischen Zahlkorpers, J. Reine

Angew. Math. 167, 264-269. (1973) Gleichungen ohne Affekt, in “Gesammelte Abhandlungen”, Band 111, pp.191-197;

Springer-Verlag, Berlin-Heidelberg-New York.

SELMER. E.S. (1956) On the irreducibility of certain trinomials, Math. Scand. 4, 287-302.

SEN, S (19G9) On automorphisms of local fields, Ann. Math. 90, 33-46.

SERRE, J .

-

P.

( 1 961) Sur les corps locaux a corps rCsiduel algkbriquement clos, Bull. SOC.Math. France 89,

105-154. (1962) Corps locaux, Hermann, Paris

( 1964) Cohomologie Galoisienne, Lecture Notes in Mathematics 5 , Springer-Verlag, Berlin-Heidelberg-New York. S H AKHM AT OV , D .B

.

( 1987) T h e structure of topological fields and cardinal invariants, Trudy Moskov. Mat. Obsch.

249-254.

524

Bibliography

SHARP, R.Y. (1977) The dimension a- the tensor product of two field extensions, Bull. London Math.

SOC.9 , 42-48.

SHARP, R.Y. and VAMOS, P (1977) The dimension of the tensor product of a finite number of field extensions, J. P u r e

Appl. Algebra, 10, 249-252.

SHATZ, S (1969) Galois theory,

Lecture Notes in Mathematics, No 86, Springer-Verlag, Berlin,

146-158.

SHIH, K.Y. (1974) On the construction of Galois extensions of function fields and number fields, Math.

Ann. 207, 99-120.

SHIMURA, G (1966) A reciprocity law in nonsolvable extensions, J. Reine Angew. Math. 221, 209-220.

SMALL, C (1984) Diagonal equations over large finite fields, Canad. J. Math. 36, 249-262.

SMITH, J (1965) On class 2 extensions of algebraic number fields, Amer. J. Math. 87, 537-550.

SOUNDARARAJAN, T (1966) Completeness of Galois theories, Indian J. Math. 8 , 11-14. (1969) A note on classical Galois theory, Math. Ann. 182, 275-280.

SPACKMAN, K . W . (1981) When is the Galois group of X"

-

a dihedral?

Math. Stud. 49, N o 2, 178-183.

Bib1 i o g r a p h y

525

STAUFFER, R (1936) T h e construction of a normal basis in a separable extension field, Amer. J. Math. 58, 585-597.

STEINITZ, E (1930) Algebraische Theorie der Korper, Berlin, de Gruyter.

SUZUKI, S (1981) Some types of derivations and their applications t o field theory, J . Math. Kyoto Univ. 21-2, 375-382.

SVED, M (1970j Units in pure cubic number fields, Ann. Univ. Sci. Budapest 13, 141-150.

SWEEDLER, M.E. (1968) Structure of inseparable extensions, Ann. Math. (2) 87, 401-410. (1969) Correction to: “Structure of inseparable extensions”, Ann. Math. ( 2 ) , 89, 206-207.

TAKAGI, T (1927) Zur theorie des Kreiskorpers, J. Reine Angew. Math. 157, 230-238.

TAKAKU, A (1971) Units in real quadratic fields, Nagoya Math. J. 44, 51-55.

TAKASHI, 0 (1985) On some class number relations for Galois extensions, Proc. J a p a n Acad. Ser. AG1,

311-312.

TAKEUCHI, K (1984) Totally real algebraic number fields of degree 5 a n d 6 with small discriminant, Saitarna

Math. 3. 2, 21-32.

Bibliography

526

TATE, J (1952) The higher dimensional cohomology groups of local class field theory, Ann. Math. 56, 294-297. (1962) Duality theorems in Galois cohomology, Proc. ICM Stocholm, 288-295.

TEICHMULLER,

o

(1936a) p-Algebren, Deutsche Mathematik, vol. 1, 362-388. (193613) Differentialrechnung bei charakteristik p , J. Reine Angew. Math. 175, 89-99. (1937) Diskret bewertete perfekte korper mit unvollkommenen Restklassenkorper,

J. fur Mathematik, 176, 126-140.

TERADA, F (1971) A principal ideal theorem in the genus field, T6hoku Math. J. 23, 697-718.

TRAPPERS, D (1987) The Witt ring of a graded field, Comm. Algebra, 15(6), 1173-1186.

TSCHEBOTAROW, N (1926) Die Bestimmung der Dichtigkeit einer Menge von Primzahlen, welche zu einer

gagebenen substitutionklasse gehijren, Math. Ann. 95, 191-228. (1950) Grundzuge der Galoisschen Theorie, P. Noordhoff, Groningen-Djakarta.

TUCKER, D (1974) Finitely generated inseparable field extensions, Notices Amer. Math. SOC.21, A-607.

UCHIDA, K (1970a) Galois group of an equation

X n - aX

+ b = 0,

TBhoku Math. J. (2) 22, 670-678.

(1970b) Unramified extensions of quadratic number fields, T8hoku Math. J. 22, 138-141.

B i b l iography

521

UCHIYAMA, S (1970) On a conjecture of K.S. Williams, Proc. Japan Acad. 46, 755-757.

UCHIYAMA, S and HITOTUMATU, S

( 1072) On the irreducibility of certain polynomials, R.I.M.S., Kokuroku 155, 14-30 (Japanese)

VEHARA, T (1983) Construction of certain real quadratic fields, Proc. Japan Acad. Sec. A59, 390-392.

ULLOM, S (1969) Normal bases in Galois extensions of number fields, Nagoya Math. J. Vol. 34, 153-167. (1970) Integral normal bases in Galois extensions of local fields, Nagoya Math. J. 39, 141-148.

VAMOS, P (1977) The nullstellsatz and tensor products of fields, Bull. London Math. SOC.9, 273-281.

VAN DER WAERDEN, B.L. (1950) Modern algebra, Vol. I, Third edition, New York, Ungar.

VARMAN, J (1930) Uber die Klassenzahl Abelscher Korper, Arkiv Mat. 22, No 13, 1-47.

VARNAVIDES, P (1952) The euclidean real quadratic fields, Indag. Math. 1 4 , 111-122.

VAUGHAN, T.P. (1985) On computing the discriminant of an algebraic number field, Math. Comp. 45, 569-584.

Bib1 iography

528

VELEZ, W.Y. (1980) On normal binomials, Acta Arith. 36, 113-124. (1983) Correction to the paper “Structure theorems for radical extensions of fields” Acta.

Arith. 38(1980), 111-115, Acta Arith. XLII, 427-428. (1985) Several results on radical extensions, Arch. Math. 45, 342-349.

VILA, N (1985) On central extensions of A , as Galois group over Q

, Arch. Math.

44, 424-437.

(1988) On stem extensions of S, as Galois group over number fields, J. Algebra, 116, 251-260.

VOSKRESENSKII, V.E. (1973) Fields of invariants of abelian groups, Uspehi Mat. Nauk 28, 77-102.

WADA, H (1966) On the class number and the unit group of certain algebraic number fields, J. Fac.

Sci. Univ. Tokyo 13, 201-209. (1970) On cubic galois extensions of Q

(G), Proc. Japan Acad. 46, 397-400.

WANG, S (1950) On Grunwald’s theorem, Ann. Math. 51, No 2, 471-484.

WARNER, S (1984) Half Henselian valuations, C.R. Math. Acad. Sci. SOC.R. Can. 6 , 255-259.

(1085a) Residual fields in valuation theory, Math. Scand. 56, 203-221. (1985b) Nonuniqueness of immediate maximal extensions of a valuation, Math. Scand. 56, 191-202. (1986) Finite extensions of valued fields, Canad. Math. Bull. 29, 64-69.

Bibliography

529

WASHINGTON, L (1982) Introduction t o cyclotomic fields, Graduate Texts in Mathematics, 83, Springer-Verlag, New-York-Berlin.

WATABE, M (1083a) On certain cubic fields, 111, Proc. Jap. Acad. Ser. A59, 260-262. (1983b) On certain cubic fields, IV, Proc. Jap. Acad. Ser. A59, N o 8, 387-389.

(1984a) On certain cubic fields, V, Proc. Jap. Acad. Ser. A60, 302-305. (1984b) On certain cubic fields, VI, Proc. J a p . Acad. Ser. A60, 331-332.

WATERH OU SE, W (1074) Profinite groups are Galois groups, Proc. Amer. Math. Soc. Vol. 42, No 2, 639-640. (1075) T h e structure of inseparable field extensions, Trans. Amer. Math. Soc. 211, 39-56.

( 1087) A unified Kummer-Artin-Schreir sequence, Math. Ann. 277, 447-451. WEBER, H (1907) Uber zyklische Zahlkorper, J. Reine Angew. Math. 132, 167-188. (1011) Zur theorie der zyklischen Zahlkorper, Math. Ann. 67, 32-60.

WEIL, A (1946) Foundations of algebraic geometry, Amer. Math. Soc. Colloq. Publ. Vol. 29, Arncr

Math. Sec. Providence, R.I. (1951) Sur la theorie du corps de classes, J. Math. Soc. J a p a n , 3, 1-35. (1956) The field of definition of a variety, Amer. J . Math. 78, 509-524.

Bibliography

530

WEISFELD, M (1965) Purely inseparable extensions and higher derivations, Trans. Amer. Math. SOC.116, 435-449.

WEISS, E (1969) Cohomology theory, Academic Press, New York.

WHAPLES, G (1947) On a conjecture about infinite class fields, Bull. Amer. Math. SOC.53, 377-380. (1952) Generalized local class field theory, I, Duke Math. J. 19, 505-517. (1954a) Additive polynomials, Duke Math. J. 21, 55-66. (1954b) Generalized local class field theory, 11, Duke Math. J. 21, 247-256. (1957) Galois cohomology of additive polynomial and n-th power mappings of fields, Duke

Math. J. 24, 143-150.

WINTER, D (1974) Structure of fields, Graduate Texts in Mathematics 16, Springer-Verlag,

New-York-Heidelberg-Berlin. WITT, E (1935) Konstruktion von galoisschen Korpern der Charakteristik p zu vorgegebener Gruppe

der ordnung p f , J. fur Math. 174, 237-245. (1936) Zyklische Korper und Algebren der Charakteristik p von Grad p", Struktur diskret

bewerteter perfekter kijrper mit voIlkommenem Restklassenk6rper der Charakteristik p . J. fur Math. 176, 126-140.

(1937) Schiefkor uber diskret bewerten Korpern, J. fur Math. 176, 31-44.

WOJCIK, J (1982) Contributions to the theory of Kummer extensions, Acta Arith. 40, 155-174.

Bibliography

531

WURFEL, T

(1985) A remark on the structure of absolute Galois groups, Proc. Amer. Math. SOC. 95. 353-356.

WYMAN, B.F. (1969) Wildly ramified gamma extensions, Amer. J. Math. 91, 135-152.

YAGER, R.I. (1982) A Kummer's criterion for imaginary quadratic fields, Compos. Math. 47, 31-42.

YAKOVLEV, A.V. (1964) The field imbedding problem, Izv. Akad. Nauk SSSR, Ser. Mat. 28, No 3, 645-660.

(1967) The imbedding problem for number fields, Izv. Akad. Nauk SSSR, Ser. Mat. 31,

KO. 2. 211-224.

YAMAGUCHI, I (1983) Uber die Einheten des Kreiskorpers der Y-ten Einheitswurzeln, TRU Math. 19,

N o 1, 101-103.

YAMAMOTO, S (1070) A note on halfnorms in the cyclotomic fields of 2"-th roots of unity, Mem. Fac. Sci.

Kyushu Univ. A24, 291-299.

YAMAMOTO, Y (1970) On unramified Galois extensions of quadratic number fields, Osaka J . Math. 7 , 57-76. (1971) Real quadratic number fields with large fundamental units, Osaka Math. J. 8, 261-270.

YAMAMURA, K (1986) On unramified Galois extensions of real quadratic number fields, Osaka J. Math. 23,

471-478.

532

Bibliography

YAMASHITA, H (1983) On nilpotent factors of maximal abelian extensions of algebraic number fields, Sci.

Rep. Kanazawa Univ. 28, No 1, 1-5. (1984) The second cohomology groups of the group of units of a Zp- extension, TBhoku

Math. J. 11, Ser. 36, 75-80.

YOKOI, H (1960) On the ring of integers in an algebraic number field as a representation module of

Galois group, Nagoya Math. J. 16, 83-90.

YOKOYAMA, A (1965) On class numbers of finite algebraic number fields, T6hoku Math. J. (2), 17, 349-357. (1966) Uber die Relativklassenzahl eines relative-galoissen Zahlkorpers von

Primzahlpotenzgrad, TBhoku Math. J. ( 2 ) , 18, 318-324. (1967) On the relative class number of finite algebraic number fields, J. Math. SOC.Japan 19, 179-184.

YOSHIZAWA, M (1984) Galois groups of polynomials, Proc. Jap. Acad. Ser. A60, 127-129.

ZARISKI, 0 (1958) On Castelnuovo’s criterion of rationality p a = P2 = 0, Illin. J. Math. 2, 303-315.

ZARISKI, 0 and SAMUEL, P (1958) Commutative algebra, Vol I, Van Nostrand, Princeton, N.J. (1960) Commutative algebra, Vol. 11, Van Nostrand, Princeton, N.J

ZERLA, F (1968) Iterative higher derivations in fields of prime characteristic, Michigan Math. J. 15, 407-415.

Bib1 iography

ZYAPKOV, N.P. and YAKOVLEV, A.V. (1982) Universally concordant Galois extensions, J. Soviet Math. Vol. 20, 2595-2609.

ZYLINSKI, E

( 1913) Zur Theorie der ausserwesentlichen Discriminantenteiler algebraischer Korper, Math. Ann. 73. 273-274.

533

534

Notation Number systems. the natural numbers the integers the rational numbers the real numbers the field of q elements

Set Theory is a member of proper inclusion inclusion the cardinal of IN the cardinality of the set X the complement of X in Y the composite of g and f set union, intersection empty set the set of all a E A with

. ..

the set of all a, with i E I the smallest infinite ordinal

Maps H

correspondence mapping between sets or classes restriction of a: t o A

535

Notation

1A

the identity map on A

Kercx

the kernel of the map a

Imcx

the image of a

Number theory a divides b a does not divide b

the greatest common divisor of a and b the quadratic symbol

Cohomology theory the fixed module of a G-module A the group of all abelian group homomorphisms A

--t

B

the group of all n-cochains of G in A the group of all n-cocycles of G in A the group of all n-coboundaries of G in A the n-th cohomology group of G in A coboundary corresponding t o f

Groups

A"

= {anla E A }

A [TLI

= { a E Alan = l }

t(A)

the torsion subgroup of A

Z(P")

the group of the p"-th complex roots of unity with n running over N

4

the pcomponent of A

-

lim G,

projective limit

Rings, modules and fields

t h e unit group of R the ring of integers mod m

Notation

536

the annihilator of M the Jacobson radical of R the nil radical of R the subring generated by A direct product of rings R; the quotient ring associated with S contracted ideal expanded ideal the local ring of R at P the monoid algebra of G over R the support of x the polynomial ring in the indeterminates Xi the degree of f (X) a field extension the degree of E over F the smallest subfield containing S and F the smallest sxhring containing S and F the composite of E and K the tensor product, the characteristic of R the direct sum principal ideal domain unique factorization domain the principal ideal generated by r the ring of n x n matrices over R the F-dimension of A the Galois group of E over F the norm of

CY

a primitive n-th root of unity the trace of X the discriminant of S over R the discriminant of the field K the monoid of fractional ideals of R

537

Notation

the group of principal fractional ideals of R the class group of R the multiplicative group of F the torsion subgroup of F' the Witt ring of R the n-the Witt ring of R the shift map the shift map the separable degree the inseparable degree the transcendency degree of E over F

FP-" the fixed field of G the cogalois group of E / F the determinant of f the subgroup of E' consisting of all n-th roots of elements of 1" the degree of a the set of derivations of R into S the set of all K-derivations of R the Lie bracket the ring of formal power series over S the exponent of purely inseparable element

EP"F {z E

E(zP" E F }

e((E/F)n-l: (E/F)") e( ( E / F ) n : ( E / F ) n -11

the smallest field which splits E / F

S(Sn-,(E/F)/F)

Un>oSn ( E / F ) the complexity of E / F a normal pair for E / F the direct limit

Notation

538

the augmentation ideal of RG

E

F"E/F)*IP if a is a limit ordinal np
n, (

E m=

n z P , , ( E w

[PI)

d i m g p(GPO[ P I / G ~ ~ + ' is the a-th Ulm invariant of E / F

A Fl(A/F)"lP n p < , ( A / F ) @ if a is a limit ordinal the Ulm space of A a is dependent on S

the Weil's order of inseparability

539

Index

Absolute Galois group, 338

Cardinality, 5

Algebra, 1

Character, 409, 416

commutative, 1

Characteristic, 1

special, 225

Characteristic polynomial, 83

truncated, 229

Characterization of

Allowable extension, 467

basic subfields, 263

Algebraic closure, 48, 50

Dedekind domains, 35

Algebraically dependent element, 102

finitely generated modules, 10

Algebraic element, 17

Galois extensions, 316

Algebraically independent set, 16

GCD-domains, 13

Alternating form, 352

linear disjointness, 77

Alternating polynomial, 352

polynomial rings, 17

Annihilator, 1

profinite groups, 311

Associated primes, 12

projective limits, 308

Artin, 318, 319, 429

purely inseparable extensions, 71

Artin-Schreir theorem, 393

regular extensions, 118 separability via linear disjointness, 78

Basic subfield, 263 Becker, 297 Berman, 278 Bilinear form, 87 nonsingular, 87 symmetric, 88 Bourbaki, 302

separable algebraic extensions, 80 separable extensions, 116 simple roots, 59 symmetric polynomials, 22 UFD, 25 Class group, 38 Closed set, 299

Canonical decomposition, 27

Closure, 300

Canonical homomorphism, 7

Comaximal ideals, 4

Index

540

Complexity, 245

integral domain to be a field, 3

Cone, 428

integral domain to be integrally closed,

Conjugate elements, 47

13

Content, 30

irreducible element to be prime, 24

Continuous map, 300

minimal generation, 184

Contracted ideal, 8

modularity of extension, 254

Coprime ideals, 4

module to be noetherian, 6

Countable set, 5

normality of extension, 53, 66

Criteria for

polynomial ring to be UFD, 29

absence of exceptional extensions, 296

preservation of pindependence, 135,

algebra to be truncated, 235

137, 138

bilinear form to be nonsingular, 89

preservation of relative pindependence,

binomial to be irreducible, 425

175

element to be integral, 10

pure independence, 262

existence of cyclic extensions, 408

quotient ring to be noetherian, 9

existence of normal pairs, 247

reliability of field extensions, 181, 191,

extension of derivations, 212

192

extension of homomorphisms, 55

relative separation of intermediate fields,

extension to be of finite exponent, 189,

180

190

ring to be Dedekind domain, 42

extension to be purely transcendental,

ring to be GCD-domain, 27

101

ring to be integral closure, 12

extension to be relatively perfect, 173

ring to be integrally closed, 11

extension to be relatively separated,

ring to be PID, 20

177

separability of extension, 65

field extension to be separable, 111

separable generation, 145

field to be algebraically closed, 105

set to be algebraically independent, 17

field extension to be separably gener-

set to be independent, 98

ated, 111

simplicity of extension, 67, 68

field to be modularly perfect, 296

subgroup to be open, 307

field to be perfect, 70

surjectivity of trace, 89

field to be splitting, 238

trace to be nonzero, 89

field to be the tensor product of simple

transcendency basis to be relative p

extensions, 265, 266

basis, 148

ideal to divide another ideal, 38

Crossed homomorphism, 370

54 1

Index

Cyclotomic polynomial, 328

perfect, 69 splitting, 52

Ilarbi, 453 Decomposition group, 355 Dedekind, 35 Dedekind domain, 32 Degree, 16, 47 Dense subset, 300 Dependence relation, 97

Field extension, 45 abelian, 329, 389 abelian pextension, 399 algebraic, 46 cogalois, 455 composite of, 45 conormal, 455

Ilerivation, 193 coseparable, 455 Derivation algebra, 194 cyclic, 329, 389 Derivative, 59 cyclotomic, 327 Direct limit, 256 Direct system, 255

degree of, 45 exceptional, 294

Directed set, 5 finite, 45 Directed union, 253 finitely generated, 46 by inclusion, 253 Discriminant, 92 Discriminant ideal, 93

infinite, 45 inseparable degree of, 64 Kummer, 409

Eisenstein’s criterion, 31

maximal Kummer, 414

Elementary symmetric polynomial, 22

modular, 252

Exchange property, 97

nontrivial splitting of, 182

Expanded ideal, 8

normal, 53

Exponent of inseparability, 61, 219

primitive element of, 46

Exponent sequence, 234

proper, 182

Eulcr function, 328

pure, 262, 457 purely inseparable, 71

Field, 45

purely transcendental, 100

algebraically closed, 49

radical, 421

Cj, 325

regular, 118

fixed, 232, 315

relatively perfect, 172

Galois, 322

relatively separated, 172

modularly perfect, 290

reliable, 181

of constants, 203

separable, 109

Index

542

separable algebraic, 59

Hilbert’s Theorem 90, 389, 390

separably generated, 109

Homogeneous polynomial, 16

simple radical, 422

Homeomorphism, 300

Filter, 301 coarser, 301 finer, 301 Finite exponent, 184 Finitely generated ideal, 2 Form, 16 Fractional ideal, 32 Free commutative monoid, 1.5 Frobenius automorphism, 355 Frobenius class, 355 Fuchs, 286, 290

Idempotent, 3 Imperfection degree, 172 Independent subset, 98 Inductive set, 5 Integral closure, 11 Integral domain, 2 Integral element, 9 Integrally closed integral domain, 12 Interior, 300 Interior point, 300 Invertible fractional ideal, 33

Galois’s criterion, 434

Irreducible element, 12

Galois group of polynomial, 349

Isaacs, 363, 367, 368

Gauss’s lemma, 20 Gay, 453 GCD-domain, 13 Gerstenhaber , 213 Gow, 357, 360 Greatest common divisor, 12, 38 Greither, 421, 456, 457, 459, 461, 465, 466, 467, 473, 474 Group attached t o extension, 234 Harrison, 421, 456, 457, 459, 461, 465, 466, 467, 473, 474 Hausdorff space, 300 Heerema, 294, 296, 297 Height of an element, 220, 270 Hewitt, 417 Higher derivation, 199 Hilbert, 349

Jzcobi identity, 195 Jacobson radical, 3 Janusz, 354, 359 Kaplansky, 13, 44, 421 Kime, 291 Kneser, 450 Kreimer, 294, 296, 297 Krull, 35 Kummer character, 412 Laguerre polynomial, 349 Lattice, 461 Least common multiple of ideals, 38 Length of extension, 270 Lie algebra, 195 Linear disjointness, 77

543

Index

Linearly dependent element, 98

Multiplicity of a root, 58

Linearly ordered set, 5 Localization, 9 Liiroth’s theorem, 107 Lower bound, 5 Lyndon-Hochschild spectral sequence, 471

Natural map, 31 n-coboundary, 369 n-cocycle, 369 Nieghbourhood, 300 filter, 302

MacLane, 143, 145, 146, 149, 150, 151,

Neukirch, 398

152, 153, 154, 159, 163, 166, 167, 297, 471

Nil ideal, 3

MacLane’s criterion, 111

Nilpotent element, 3

Main theorem of finite Galois theory, 338

Nilpotent ideal, 3

Main theorem of infinite Galois theory, 343

Noether, 35

Matusita, 35

Noetherian module, 6

Maximal perfect subfield, 154

Norm, 83

May, 278, 452

Normal basis, 320

Minimal generating set, 184

Normal basis theorem, 320

Minimal polynomial, 47

Normal closure, 54

Minimal prime ideals, 3

Normalized cochain, 369

Mobius function, 333

Normal element, 220

Mobius transform, 333

Normal generating sequence, 220

Modular closure, 293

Normal pair, 246

Module, 1

degree of, 246

cyclic, 360

Norris, 436

faithful, 2

n-th cohomology group, 369

fixed, 368 G-module, 368 unital, 1 Mollov, 278 Monoid, 14 algebra, 14 Mordeson, 172, 177, 179, 181, 183, 188, 190 Multiple root, 58 Multiplicative set, 6

Ojanguren, 213 Open covering, 301 Open set, 299 Ordered group, 260 Order of inseparability, 126 Osada, 349 Partial order, 5 by inclusion, 5 Perfect clcsure, 70

Index

544

Pontryagin, 303, 419

commutative, 1

Primary ideal, 2

direct product of, 3

Prime element, 12

finitely generated, 3

Prime ideal, 2

integrally closed, 11

Prime subring, 1

noetherian, 2

Primitive idernpotent, 3

local, 2

Primitive polynomial, 20

local a t P , 9

Principal crossed homomorphism, 370

of constants, 194

Principal fractional ideal, 33

of formal power series, 202

Principal ideal, 2

of polynomials, 15

Principal ideal domain, 20

ordered, 428

Profinite group, 310

reduced, 3

Projective linear group, 108

semilocal, 2

Proper zero divisor, 2

semisimple, 3

Pure independent subset, 262

topological, 302

Pure inseparable closure, 71

Ross, 417

Purely inseparable element, 71 Safarevich, 348 Quotient field, 7

Schinzel, 435, 439, 445

Quotient ring, 6

Schmidt, 112

Quotient space, 301

Scholz, 348 Schreir. 429

Rasala, 220, 221, 225, 230, 232, 234, 238, 241, 245, 247, 248, 249 Reduction criterion, 32 Regular representation, 83 Reid, 294 Relative pbasis, 132 Relatively pdependent set, 132 Relatively prime ideals, 38 Representation of algebra, 82 Restricted Lie algebra, 196 Restricted subspace, 213 Ring, 1 artinian, 2

Schur, 349 Separable closure, 64 Separable degree, 61 of extension, 62 of polynomial, 61 Separable element, 59 Separable polynomial, 58 Separating relative pbasis, 172 Separating transcendency basis, 109 Shift, 382 Simple root, 58 Simply truncated polynomial algebra, 229

Index

54 5

Special algebra, 225

Trace, 83

Splitting field of extension, 238

Transcendency degree, 103

Sridharan, 213

Transcendental element, 17

Steinitz field tower, 157

Trivial idempotent, 3

Structure equations, 225

Truncated sequence, 229

Subspace, 300

Tychonoff’s theorem, 304

Support, 14 Sweedler, 252, 265, 291, 293

Ulm algebra, 278

Symmetric polynomial, 22, 352

Ulrn invariant, 272 Ulm-Kaplansky invariant, 271

Taylor variation, 233 fixed field of, 233

Ulm space, 277 Ultrafilter, 301

Tensor product, 73

Unique factorization domain, 24

Topological group, 302

Unit, 1

locally compact, 416 Topological space, 299

Unit group, 1 Universal property of

compact, 301

direct limits, 256

component of, 301

quotient rings, 7

connected, 301

tensor products, 76

locally compact, 416

Unique subfield property, 437

product of, 301

Upper bound, 5

totally disconnected, 301 Topology, 299

Variation, 232

base, 301

Vklez, 436, 453

coarser, 300

Vinograde, 172, 177, 179, 181, 183, 188,

compatible, 302

190

discrete, 300 finer, 300 induced, 300

Krull, 340 product, 301 quotient, 301 trivial, 300 Total quotient ring, 118

Waterhouse, 252, 262, 263, 265, 266, 269, 276, 277, 278, 280, 283, 284, 286, 288, 348 Weight, 22 Weiss, 468 Well-ordered set, 5 Witt, 375 Witt ring, 380 Wojcik, 435

546

Index

Wreath product, 354 regular, 354 Zariski, 106 Zero divisor, 2 Zorn’s lemma, 5