Provides summary of field theory that emphasizes refinements and extensions achieved in recent studies. It describes can
416 20 27MB
English Pages 567 Year 1988
Table of contents :
Preface
Contents
Chapter 1. Preliminaries
1. Notation and terminology
2. Polynomial algebras
3. Integral extensions
4. Tensor products
5. Module-theoretic prerequisites
6. Topological prerequisites
Chapter 2. Classical topics in field theory
1. Algebraic extensions
2. Normal extensions
3. Separable, purely inseparable and simple extensions
4. Galois extensions
5. Finite fields, roots of unity and cyclotomic extensions
6. Norms, traces and their applications
7. Discriminants and integral bases
8. Units in quadratic fields
9. Units in pure cubic fields
10. Finite Galois theory
11. Profinite groups
12. Infinite Galois theory
13. Witt vectors
14. Cyclic extensions
15. Kummer theory
16. Radical extensions and related results
17. Degrees of sums in a separable field extension
18. Galois cohomology
19. The Brauer group of a field
20. An interpretation of H_0^3(G, E*)
21. A cogalois theory for radical extensions
22. Abelian p-extensions over fields of characteristic p
23. Formally real fields
24. Transcendental extensions
Chapter 3. Valuation theory
1. Valuations
2. Valuation rings and places
3. Dedekind domains
4. Completion of a field
5. Extensions of valuations
6. Valuations of algebraic number fields
7. Ramification index and residue degree
8. Structure of complete discrete valued fields
A. Notation and terminology
B. The equal characteristic case
C. The unequal characteristic case
D. The inertia field
E. Cyclotomic extensions of p-adic fields
Chapter 4. Multiplicative groups of fields
1. Some general observations
2. Infinite abelian groups
3. The Dirichlet-Chevalley-Hasse Unit Theorem
4. The torsion subgroup
5. Global fields
6. Algebraically closed, real closed and the rational p-adic fields
7. Local fields
A. Preparatory results
B. The equal characteristic case
C. The unequal characteristic case
8. Extensions of algebraic number fields
9. Braridis's theorem
10. Fields with free multiplicative groups modulo torsion
11. A nonsplitting example
12. Embedding groups
13. Multiplicative groups under field extensions
14. Notes
Bibliography
Notation
Index
Field Theory
PURE AND APPLIED MATHEMATICS A Program of Monographs, Textbooks, and Lecture Notes
EXECUTIVE EDITORS
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K. Yano, Integral Formulas in Riemannian Geometry (1970) (out of print)
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J. K. Beem and P. E. Ehrlich, Global Lorentzian Geometry ( 1981) D. L. Armacost, The Structure of Locally Compact Abelian Groups (1981) J. W. Brewer and M. K. Smith, eds., Emmy Noether: A Tribute to Her Life and Work (1981) K. H. Kim, Boolean Matrix Theory and Applications (1982) T. W. Wieting, The Mathematical Theory of Chromatic Plane Ornaments (1982) D. B. Gauld, Differential Topology: An Introduction (1982) R. L. Faber, Foundations of Euclidean and Non-Euclidean Geometry (1983) M. Carmeli, Statistical Theory and Random Matrices (1983) J. H. Carruth, J. A. Hildebrant, and R. J. Koch, The Theory of Topological Semigroups (1983) R. L. Faber, Differential Geometry and Relativity Theory: An Introduction (1983) S. Barnett, Polynomials and Linear Control Systems (1983) G. Karpilovsky, Commutative Group Algebras (1983) F. Van Oystaeyen and A. Verschoren, Relative Invariants of Rings: The Commutative Theory (1983) I. Vaisman, A First Course in Differential Geometry (1984) G. W. Swan, Applications of Optimal Control Theory in Biomedicine (1984) T. Petrie and J. D. Randall, Transformation Groups on Manifolds (1984) K. Goebel and S. Reich, Uniform Convexity, Hyperbolic Geometry, and Nonexpansive Mappings (1984) T. Albu and C. Nastasescu, Relative Finiteness in Module Theory (1984) K. Hrbacek and T. Jech, Introduction to Set Theory, Second Edition, Revised and Expanded (1984) F. Van Oystaeyen and A. Verschoren, Relative Invariants of Rings: The Noncommutative Theory (1984) B. R. McDonald, Linear Algebra Over Commutative Rings (1984) M. Namba, Geometry of Projective Algebraic Curves (1984) G. F. Webb, Theory of Nonlinear Age-Dependent Population Dynamics (1985) M. R. Bremner, R. V. Moody, and J. Patera, Tables of Dominant Weight Multiplicities for Representations of Simple Lie Algebras (1985) A. E. Fekete, Real Linear Algebra (1985) S. B. Chae, Holomorphy and Calculus in Normed Spaces (1985) A. J. Jerri, Introduction to Integral Equations with Applications (1985) G. Karpilovsky, Projective Representations of Finite Groups (1985) L. Narici and E. Beckenstein, Topological Vector Spaces (1985) J. Weeks, The Shape of Space: How to Visualize Surfaces and ThreeDimensional Manifolds (1985) P. R. Gribik and K. 0. K ortanek, Extremal Methods of Operations Research (1985) J.-A. Chao and W. A. Woyczynski, eds., Probability Theory and Harmonic Analysis (1986) G. D. Crown, M. H. Fenrick, and R. J. Valenza, Abstract Algebra (1986) J. H. Carruth, J. A. Hildebrant, and R. J. Koch, The Theory of Topological Semigroups, Volume 2 (1986)
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Other Volumes in Preparation
Field Theory Classical Foundations and Multiplicative Groups Gregory Karpilovsky University of the Witwatersrand Johannesburg, South Africa
Marcel Dekker, Inc.
New York • Basel
Library of Congress Cataloging-in-Publication Data Karpilovsky, Gregory Field theory: classical foundations and multiplicative groups/ Gregory Karpilovsky. p. cm. -- (Monographs and textbooks in pure and applied mathematics ; vol. 120) Bibliography: p. Includes index. ISBN 0-8247-8029-9 1. Class field theory. 2. Groups, Theory of. I. Title I I. Series. QA247.K324 1988 512'.32--dcl9 88-18914
Copyright© 1988 by MARCEL DEKKER, INC.
All Rights Reserved
Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER, INC. 270 Madison Avenue, New York, New York 10016 Current printing (last digit): 10 9 8 7 6 5 4 3
2
1
PRINTED IN THE UNITED STATES OF AMERICA
For Helen, Suzann~, Elliott
Preface
The twin aims of this book are to provide a young algebraist with a reasonably comprehensive summary of the material on which research in field theory is based and to show how this material is used in the current research pertaining to multiplicative groups of fields.
The treatment is by no means exhaustive - this would
have been a Sisyphean task, since the subject has become so extensive and is growing almost from day to day. The book can be roughly divided into two parts, which will not preclude, however, some strong interrelations between these.
The first part (Chapters 1-3)
introduces classical foundations of the theory of fields, with an emphasis on refinements and extensions achieved in scope of relatively recent developments. Among these, we exhibit canonical fundamental units of certain classes of pure cubic fields, prove Kneser's theorem on torsion groups of separable field extensions and establish a theorem of Schinzel which provides necessary and sufficient conditions for the Galois group of a binomial
to be abelian.
xn- a
A separate
section is devoted to a result of Isaacs concerning the degrees of sums in a separable field extension. E/F,
If G is the Galois group of a finite Galois extension
an interpretation of a distinguished subgroup of
using the methods of Eilenberg and Maclane. pertaining to radical extensions.
H 3 (G,E*)
is provided by
We also include some recent results
Among them, we examine a cogalois correspon-
dence discovered by Greither and Harrison in 1986. The second part of the book (Chapter 4) is devoted exclusively to the study of multiplicative groups of fields.
After proving some preliminary results, we in-
vestigate the isomorphism class of F*, local or global.
where
F
is a distinguished field such as
Concentrating on field extensions V
E/F,
we then prove Brandi.s's
Preface
vi theorem which asserts that if F is infinite and Et F, finitely generated.
then
E*/F*
is not
The rest of the chapter is based on works of May, who made
fundamental contributions to the subject.
Numerous examples are given to illus-
trate that the results obtained are the best possible.
Special attention is
drawn to the study of fields whose multiplicative groups are free modulo torsion. These results have a number of important applications in the study of unit groups of group algebras and the isomorphism problem.
For this reason a complete ac-
count of May's contributions to the topic is given. This 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 famil-
iarity with basic ring-theoretic and group-theoretic concepts and an understanding of elementary properties of modules, tensor products and fields.
Apart from a
few specific results and the general knowledge we have presupposed, the book is entirely self-contained.
We have included, for the convenience of the reader, a
chapter on algebraic preliminaries. topics needed later in the book. too technical.
This chapter provides a brief survey of We have tried to avoid making the discussion
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. A word about notation.
As is customary, Theorem 2.1.2 denotes the second
result of Section l of Chapter 2; however, for simplicity, all references to this result within Chapter 2 are designated as Theorem 1.2.
A systematic description
of the material is supplied by the introductions to individual chapters and therefore will not be repeated here. I would like to express my gratitude to my wife for the encouragement she has given me in the preparation of this book.
For answering specific queries on
topics contained in the text I am indebted to
W. May.
Finally, my thanks go to
Lucy Rich for her excellent typing. Gregory Karpilovsky
Co~e~s
PREFACE CHAPTER l.
V
PRELIMINARIES l.
2. 3. 4. 5. 6. CHAPTER 2.
CLASSICAL TOPICS IN FIELD THEORY l. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.
CHAPTER 3.
Notation and terminology Polynomial algebras Integral extensions Tensor products Module-theoretic prerequisites Topological prerequisites
Algebraic extensions Normal extensions Separable, purely inseparable and simple extensions Galois extensions Finite fields, roots of unity and cyclotomic extensions Norms, traces and their applications Discriminants and integral bases Units in quadratic fields Units in pure cubic fields Finite Galois theory Profinite groups Infinite Galois theory Witt vectors Cyclic extensions Kummer theory Radical extensions and related results Degrees of sums in a separable field extension Galois cohomology The Brauer group of a field An interpretation of H~(G,E*) A cogalois theory for radical extensions Abelian p-extensions over fields of characteristic p Formally real fields Transcendental extensions
VALUATION THEORY l. 2. 3. 4. 5. 6.
l
7 29 32 37 41 47 47 57 63 79 85 103 117 135 148 167 172 184 192 205 214 221 242 247 263 282 303 323 333 348 353
Valuations Valuation rings and places Dedekind domains Completion of a field Extensions of valuations Valuations of algebraic number fields vii
353 368 376 389 400 411
Contents
viii 7. · Ramification index and residue degree 8. Structure of complete discrete valued fields A. Notation and terminology B. The equal characteristic case C. The unequal characteristic case D. The inertia field E. Cyclotomic ex~ensions of p-adic fields CHAPTER 4.
MULTIPLICATIVE GROUPS OF FIELDS 1. 2. 3. 4. 5. 6. 7.
8. 9. 10. 11. 12. 13. 14.
Some general observations Infinite abelian groups The Dirichlet-Chevalley-Hasse Unit Theorem The torsion subgroup Global fields Algebraically closed, real closed and the rational p-adic fields Local fields A. Preparatory results B. The equal characteristic case C. The unequal characteristic case Extensions of algebraic number fields Braridi s's theorem Fields with free multiplicative groups modulo torsion A nonsplitting example Embedding groups Multiplicative groups under field extensions Notes
414 421 421 422 427 431 436 439 439 443 449 461 463 468 474 475 481 482 487 496 501 517 519 525 531
BIBLIOGRAPHY
535
NOTATION
541
INDEX
547
Field Theory
1 Preliminaries The aim of this chapter is twofold.
First, to establish various notational
conventions for the rest of the book.
Second, to provide some preliminary re-
sults pertaining to polynomial rings.
Because we presuppose a familiarity with
various elementary group-theoretic and ring-theoretic terms, only a brief description of them is presented.
Many readers may wish to glance briefly at the con-
tents of this chapter, referring back to the relevant sections when they are needed later. 1. NOTATION AND TERMINOLOGY The aim of this section is to establish various notational conventions that we shall use throughout the book. Let A and B be arbitrary sets.
A map
f:A-.B
is a function that associates with each element a e A a unique element be B, f : a
i--+
b.
This b,
the image of a under
f: A be maps.
B and
The composite map A-. B -
f,
g: B -
is denoted by b
Let
= f(a).
C
c is denoted by gof·
Recall that
gof acts according to the rule (gof)(a) = g[f(a)]
If f: A-. B is any map, then set A'
of A.
flA'
forall
denotes the restriction of f
aEA
to a sub-
The inclusion map i : A'-->- A is defined by i(a) = a
for all
a e A'
If A,B are sets, we write Ac B if A is a su~set of B and Ac B if A is a proper subset of B.
If A~ B, B-A denotes as usual the set of elements
2
of
CHAPTER 1
B
not contatned i'n
A,
The cardinality of the set
is denoted by
A
IAI
The di'agram
f A----!•~B
th C signifies that h : C __.. B
case
are maps,
f = hog,
c
A,B and
are sets and that
f; A -
B,g:A -
c
and
The diagram commutes, or is said to be commutative, in
An arbitrary diagram is commutative if we get the same composite
maps whenever we fo 11 ow directed arrows a 1ong different paths from one set to another set in the diagram, Unless explicitely stated otherwise, all groups are assumed to be multiplicative.
We use the same symbol 1 for both the identity element and the identity
subgroup of a group
by
x
If
F*. X.
G,
is a subset of
G,
A cyclic group of order
n
wi 11 denote the subgroup of is denoted by
G.
while Let
If every element of
If
'1l. •
n
least common multiple of the orders of the elements of of
F is denoted by
The multiplicative group of a field
G is of finite order,
G
G
G
generated
ts finite, the
is called the exponent
G is called a torsion group,
G is torsion-free tf all Hs elements, except for 1, are of infinite order,
H be a subgroup of
from each 1eft co set
G.
A subset of
G containing just one element
is ca 11 ed a left transversal for
xH
H
in
G,
and right
transversals are defi'ned correspondingly, Let
{Gi}iEI. be a family of groups and let
The subgroup (£)1-'EIGi gi
of
n1-'EIGi
niEIGi
consisting of all
be their direct product
(g)
with finitely many
distinct from 1, is called the direct sum (or restricted direct product) of
the groups
Gi,
A sequence of groups and homomorphisms fi-1 fi __.. Gi-1 ~ Gi Gi+1 '''
is said to be exact at
G.
1-
if
NOTATION AND TERMINOLOGY
the sequence is called exact if it is exact at every group.
; Kerf.'2, = Imf. '!,- 1
particular, G
L
1 .- G
L
His exact if and only if
is exact if and only if
H .- 1
3
is injective, while
f
is surjective,
f
In
If we are given a 3-
terms exact sequence 1---+G
g
f
1
--a 2 --a 3
.-1
also called a short exact sequence, we see that G ~ f(G·) 1
N be a normal subgroup of
Let
morphism.
and
1
G
ea!-
3
G and let
G /f(G ) 2
l
n: G--+ G/N
be the natural homo-
Then the short exact sequence 7f
1--+N-->-G---->-G/N-->-l
is also called the natural exact sequence,
A is an abe 1 ian group.
Suppose now that finite order in
i's a subgroup of
A
A,
an
n > 0,
the map
A--->- A
We shall say that
n-divisible if
An = A.
p-divisible.
(i}
sending every element
is divisible if
An
An= A
of A
a
and
A [nl ,
for all
n
E
IN
to
and
Note that a p-group is divisible if and only if it is
Typical examples of divisible groups are:
The multiplicative group
(ii)
A
A.
if no confusion can arise.
A,
is a homomorphism whose image and kernel are denoted by
respectively.
of e 1ements of
called the maximal torsion subgroup of
It will also be called the torsi'on subgroup of For any integer
t(A)
Then the set
The group
z
00
of the
F*
pnth
of an algebraically closed field complex roots of 1 with
n
F
running over all
p
natural numbers. (iii}
The additive group ([)/'11. which is isomorphic to the group of all complex
roots of 1. Given a prime elements of
7f
we denote by
A
p
the p-component of A
A whose order is a non-negative power of p.
torsion, i.e. if A=EBpE
p,
·t(A) = A,
then there is a set
n
consisting of all
Note that if
A is
of prime numbers such that
A. p
A11 rings in this book are associative with 1 f O and subri ngs of a ring
R
CHAPTER l
4
are assumed to have the same identtty element as
Each ring homomorphism will
R.
be assumed to respect identity elements. Let
be a ring.
R
The mapping 'lL
defined by n
-->- R
homomorphism whose image is ca 11 ed the prime subring of R; ideal
m'lL for a unique m;;;, 0,
1-+
is a ring
n•l
its kernel is an
called the characteristic of R and denoted by
charR Let
be a ring.
R
positive integer n. potent, while
r
=
0,
An element x ER An ideal
is nilpotent if xn
=
in R is nil if every element of
J
is nilpotent if there is a pas iti ve integer n
J
where Jn
is the product of J with itself n
of R is idempotent if e 2
=
O for some
Two idempotents
e.
J
is nil-
such that
times.
An element e
are orthogonal if
u,v
A nonzero idempotent is primitive if it cannot be written as a sum
uv =vu= 0.
of two nonzero orthogonal idempotents. An element u of a ring R is said to be a unit if uv VER.
The set of all units of R,
denoted by U(R),
=
vu
= 1
for some
constitutes a group
called the unit group of R. Given an arbitrary ring R,
we denote by J(R)
the Jacobson radical of R,
i.e. the intersection of all maximal left ideals of R. simple if J(R)
Let
niE?i·
{Ri}iEI
=
0. be a family of rings and let R be the direct product set
We can define addition and multiplication on (x.) + (y.) ~· ~
It is straightforward to verify that direct product of the family
onto R.~
We say that R is semi-
R
R
(x.+y.)
=
~
by the rules (x.)(y.) ~ ~
~
is a ring; we shall refer to For all
(R.LiEI. ~
i EI,
is a ring homomorphism, the inj ecti ans R~. -
=
(x.y.) ~ ~ R
as the
the projection of
R
R preserve addition and
multiplication but not land so are not ring homomorphisms. Let R be a commutative ring.
Then an R-algebra is a ring A which is at
the same time an R-module such that r(xy)
=
(rx)y
=
for all
x(ry)
It follows directly from the definition that rx
=
(r•l)x
rER and x,yEA
for r ER, x E A,
and
5
NOTATION AND TERMINOLOGY that the map sending r of
to r•l
is a homomorphism of R into the centre
Conversely, if f: R -
A.
A is a homomorphism of R into
Z(A)
then
Z(A),
can be regarded as an R-algebra by setting
A
for all
rx = f(r)x
r ER, XE
A
We close this section by presenting some information pertaining to commutative In what follows
rings.
We say that
denotes a commutative ring.
R
R
is
noetherian if every ascending chain of ideals breaks off, or equivalently, if
every ideal is finitely generated.
A ring R is called artinian if every descenThe set N(R)
ding chain of ideals breaks off.
of all nilpotent elements of R
constitutes an ideal called the nilradical of R.
N(R)
= 0.
Note that N(R) '.:: J(R)
is a nil ideal.
We say that R is reduced if
N(R)
and thus
J(R)
=
if and only if J(R)
We say that R is local (respectively, semilocal) if it has
precisely one maximal ideal (respectively, if it has only finitely many maximal ideals).
An element x
y ER; in case
of R is a zero divisor if
x -f- 0 and
xy =
0 for some nonzero
is a zero divisor, we say that x
x
is a proper zero
divisor.
An integral domain is a commutative ring without proper zero divisors. ideal
P of a commutative ring
R is prime if R/P
P is primary if every zero divisor in
R/P
R5
r/s
and if, for any x,y
associated with a multiplicative set with r ER and
s
Es.
is an integral domain, while
is nilpotent.
Let S be a subset of a commutative ring R. cative if it contains
An
in
We say that s s, xy
Es.
is muUipZi-
The quotient ring
s consists of all elements of the form
By definition,
r/s = r'/s'
if there exists
Es such that
s l
s (s'r-sr')
=
0
l
Observe that if OE s,
then 0/1
we shall assume that Off. s. (r/s)(r'/s')
The mapping 1/Js : R -
Rs
=
is the only element of Rs;
for this reason
Multiplication and addition in Rs are defined by rr'/ss' ,
defined by
r/s+r'/s' = (s'r+sr')/ss' ¢ 5 (r) = r/1
is a homomorphism, called
CHAPTER 1
6
canonical;
1/Js
is injective if and only if S contains no zero divisors.
Suppose now that
A
cative subset of A.
is a commutative R-algebra and that s is a multipliThen the ring As may be endowed with an R-algebra struc-
ture by setting r(a/s)
=
r ER, a EA, s ES
ra/s
in which case 1/Js becomes a homomorphism of R-algebras.
The R-algebra As en-
joys the following easily verified universal property. 1.1. PROPOSITION.
Let
homomorphism f: A -
B
be a commutative R-algebra.
Then, for any R-algebra
B such that every element of f(S)
exists one and only one R-algebra homomorphism h : As -
is a unit, there B which renders com-
mutative the following diagram:
As
h f
A
.,B
Especially important is the case where s is the complement of a prime ideal Here one often writes RP instead of Rs. associate the expanded ideal
Ie
With every ideal
I
P.
in R we
of Rs given by
Ie = {i/sli EI, s Es}
The following property is easily verified. 1.2. PROPOSITION.
For any prime ideal
P of
R the correspondence
Ir->- Ie
is a bijection between the prime ideals of R contained in P and the prime ideals of
RP"
It follows from the above that Pe is a local ring.
is a unique maximal ideal of RP; hence RP
This ring RP is also called the local ring of R at P,
the process of forming
RP
is called localization.
and
POLYNOMIAL ALGEBRAS Finally, assume that R is an integral domain.
7
Then S
{O}
= R -
is a
multiplicative subset of R and Rs is obviously a field containing an isomorphic copy of R.
We shall refer to R8 as the quotient field of R.
We close by recording the following elementary fact. 1.3. PROPOSITION. Proof.
Let R be an artinian integral domain.
Fix O ,J:
and observe that the descending chain
x E R
Rx .::> Rx 2
xn {1-rx)
= 0.
Si nee
R
::J •••
for some n ~ 1,
so x n
= rx
has no proper zero dtvi sors,
1 - rx
= O and thus r
Hence Rxn = Rxn+i
must terminate.
Then R is a field.
n+1
or
ts a unit. • 2. POLYNOMIAL ALGEBRAS A monoid is a set G,
with an assoctatiVe binary operation, and having an iden-
tity element 1. Let RG of
R be a commutative ring and let G be a monoid. G over
The monoid algebra
R is the free R-module on the elements of
cation induced by that in
with multipli-
G,
More explicitly, RG consists of all formal
G.
linear combinations (x
Ex •g
g
g
with finitely many
x
g
,J:
Ex •g = Ey •g if and only if g g (ii) Ex •g + Z:1,1 •g = E(.x +y ) •g g ·g ' g g
(tv}
(Exg•g)(Eyh•h) = Ezt•t r(Exg •g).
=
I:(rxg ) •g
R, g
E
G)
O subject to x
(i)
(iii)
E
where for all
=
g
z
t
r
y
= E
for all
g
E
X
gh=t g
g
E G
yh
R
It is straightforward to verify that these operations define RG as an associati've R-algebra with R
and
G,
1 = 1R-1G,
respectively.
where
1R
and 1G are identity elements of
With the aid of the injective homomorphisms
CHAPTER l
8
we sha 11 in the future identify
and
R
with their images in
G
RG:
With
these identifications, the formal sums and products become ordinary sums and products. Let x
For this reason, from now on we drop the dot in xg·g. Then the support of x,
= i::x g ERG. g
Suppx
= {g E
G!xg
written Suppx,
is defined by
f O}
It is plain that Suppx is a finite subset of G that is empty if and only if X
= 0.
Let A be an R-algebra, let G be a monoid and let
2. l. PROPOSITION.
be any map satisfying iJJ(l) Then the map ijJ*
RG ->- A
=
and ijJ(xy)
1
Proof.
In particular, if
R-modules.
as a basts, then
Because
RG
for all
iJJ(x)ijJ(y)
RG ~
1jJ
is injective and A is
A.
is R-free with
G as a basis,
ijJ*
ts a homomorphism of
Let and
x = i:x g g
be two elements of
RG.
ijJ*(xy}
x,y E G
defined by
is a homomorphism of R-algebras. R-free with iJJ(G)
=
y = i:y g g
Then ¢*( Z
x ybab) =
a,bEG a
): X
ijJ(a)
aEG a
i::x y01)!(a)ijJ(b) a
[ y01jJ(b)
bEG
ijJ*(x)ijJ*(y)' as asserted. • Assume that G is a free abelian group freely generated by the set {X
.Ji E
1,
I}
Then each element of G can be uniquely written in the form
9
POLYNOMIAL ALGEBRAS n.
(n.'2,
nx.'2,
iEI
with only finitely many n.'2, ! 0.
> 0 for any
(1) in which
n.'2,
generated by
{X .. '2,
li
,z,
::l)
E
( 1)
The submonoid of G consisting of all elements is called a free commutative monoid freely
i EI
The corresponding monoid algebra over a commutative
EI}.
ring R is denoted by R[(Xi)iEil: nomial ring in the indeterminates
We shall refer to R[(Xi\Eil with coefficients in R.
X.'2,
as the poly-
For In partic-
ular, the polynomial ring in the indeterminate x with coefficients in R will n. be denoted by R[X]. > 0 and The elements of R[(Xi)iEI] (n. . '2,
niE~/
n. = 0 '2,
for all but finitely many i
definition,
R[(Xi)iEI]
are called monomials.
EI)
Thus by
is R-free freely generated by all monomials.
The degree
of the monomial (1) is defined by n.
deg ( II X •,z,} iEI
If f
is a nonzero polynomial in R[(Xi}iEil,
written
degf,
of f.
Thus
f = 0,
Let
~ n. iEI ,z,
=
,z,
degf = 0
if and only if f -
is a nonzero element of R.
We shall say that f
If
00
be a nonnegative integer and let
R[{Xi)iEI].
f,
to be the maximum of the degrees of the monomials in the support
then we say that its degree is d
then we define the degree of
be a nonzero polynomial in
f
is homogeneous of degree d
(or is a form of
degree d) if a11 monomi a1s in Sup pf a re of degree d. Let f(X)
be a nonzero polynomial tn R[Xl f(X) = a
with a.ER and with an f 0. ,z,
o
of degree n.
+ a X + ... +a 1 n
We call
a
n
r
the leading coefficient of
f
is manic if a n = 1. Let A be an R-algebra and let B be any subset of A. Then the elements its constant term.
We also say that
Then
f
(rV ER, v(b) > O) with finitely many r
V
and v(b)
This subalgebra, denoted by R[B],
distinct from zero form a subalgebra of (or simply R[a] if B
the smallest subalgebra containing B.
= {a})
A.
is obviously
For this reason, we refer to R[B]
as
CHAPTER 1
10
the subalgebra of A generated by B.
If A= R[BJ,
then we say that A is
gener>ated by B.
Let A be an R-algebra and let B be any subset of A.
We say that B is
algebr>aicaUy independent over R (or that the elements of B are algebraically
independent over R} tf the elements
(v(b) > O) are R-linearly independent,
In the special case where
B =
we see that a
{a},
i's algebraically independent over R if and only if {1,a,a 2 , . . . ,an, ... } is an R-linearly independent set.
If a
is algebraically independent over
EA
then we also say that a ts tmnscendental over R.
R,
In the case where a is
not transcendental over R, we say that a ts algebr>aic over R.
Thus a is
algebraic over
R
f{X)
such that f(a}
=
if and only if there exists a nonzero polynomial
E
R[XJ
0.
We next provide some elementary observations pertaining to polynomial rings. 2.2. PROPOSITION.
Let
Any map {x.'Z, Ji
( i)
of R-algebras.
A
be any commutative R-algebra.
E I} -+ A,
x.'Z,
In particular, if
can be extended to a homomorphism
a.'Z,
I---'---+
is generated by fo.li 'Z,
A
EI},
then
is a
A
h.omomorphic image of R[(Xi)iEIJ.
1
(ii}
The above homomorphism is tnjecti've if and only if the elements a.'Z, are
algebraically independent over (iti) A
A ~ R[(X)iEil
R
if and only if there is a generating set {ai Ii
which is algebraically independent over
only if
A
Proof.
In particular,
R.
A.~
E I}
R[XJ if and
ts generated by a transcendental element. (i)
i Let G be the monoid generated by the x., 'Z, n.
,jJ{nx/) iEI
it follows that w(1)
=
1 and w(xy)
,z,
=
=
n.'Z,
na.
iEI
Setting
(n.'Z, > O)
,z,
w(x)w(y)
EI.
for a11
x,y E
G,
(2) Now apply
Proposition 2.1 (ii}
for
Direct consequence of (2} and the definition of algebraic independence.
POLYNOMIAL (iii)
ALGEBRAS
Direct consequence of (i} and (ii).
2.3. PROPOSITION.
Let I=JUK
11
•
be a disjoint union and let S
R[(X.)
=
'EJ].
J J
Then as R-algebras. and, in particular, R [X
Proof.
1
, ...
,X ]
n
2e
and each
xk, k EK,
elements
xi,i EI
n
, ...
n-
X.
n
x1,., i
>--->-
1,
for all
l ][X ]
of R-algebras.
E I
n;;,
2
extends to a homo-
Since its image contains
the given homomorphism is surjective.
Now apply Proposition 2,2(ti).
s
Moreover, the
Let
f(X)
•
and g(X)
be nonzero polynomials in R[XJ
and bm be the leading coefficients of
at least one of a ,b n
m
f(X)
and g(X),
is not a zero divisor in R, deg (fg)
and the leading coefficient of fg Proof.
,X
of s[(Xk)kEKl are obviously algebraically independent over
2.4. PROPOSITION. a
1
By Proposition 2.2(t), the map
morphism R[(Xi)iEI]--->- S[(Xk)kEK]
R.
R [X
=
and let
respectively.
If
then
degf + degg
is ab . nm
Write +b
x1
m
Then
and the result follows. 2.5. COROLLARY. R[(Xi)iEI]
Proof.
•
Let R be an integral domain.
Then, for any set I,
is also an integral domain. We may harmlessly assume that I
Proposition 2.3, it suffices to show that R[XJ
is finite.
Furthermore, by
is an integral domain.
latter being a consequence of Proposition 2.4, the result follows. 2.6. PROPOSITION.
~~t
f(X}
and g(X)
The
•
be two nonzero polynomials in R[XJ
of
CHAPTER l
12
degrees m and n,
respectively.
leading coefficient of g(X).
Put k
max(m-n+l,O)
=
and denote by a
the
g(X) and r(X)
Then there exist polynomials
such that
ak f(X) where either r(X) in
R,
Proof.
0 or degr(X) < n.
=
then q(X)
q(X)g(X)
=
and r(X)
If m < n,
Moreover, if a is not a zero divisor
then
and we may take q(X)
k = 0
in which case k
first part, we argue by induction on m.
the leading coefficient of f(X). and r I (X).
=
I
m-n + 1.
=
has degree at most m- 1,
=
f(X).
We
To prove the
where b is
Applying induction hypothesis, there exist
such that
a(m-I)-n+i(_af(.X) -b";m-ng(X)) x where either r (x)
r(X)
= 0,
The case m = n - 1 being trivial,
Then af(X) - bI'-n g(X)
polynomials q I (X) .
r(X)
are uniquely determined.
may therefore assume that m;;. n-1
assume m ;;. n.
+
O or degr (X) < n. I
q 1 ( X)g () X + r
=
1
(X)
This proves the first assertion, by
taking
Assume that a is not a zero divisor and that
ak f(X) where either r'(X)
=
q 1 (X)g(X) + r'(X)
0 or degr'(X) < n. (q(X} - q '(X) )g(X)
If q(X} - q'(X) f 0,
and so r'(X)
=
2.7. COROLLARY.
r'(X) - r(X)
then the left side has degree at least n,
leading coefficient of g(X) since r'(X)-r(X)
Then
is not a zero divisor.
However, this is impossible
is either zero or has degree less than n.
r(X).
Hence q(X)
=
q'(X)
•
Let f(X)
and g(X)
let the leading coefficient of g(X) polynomials q(X), r(X)
since the
E
be two nonzero polynomials in R[Xl and be a unit of R.
R[Xl such that f(X)
=
g(X)q(X)
+
r(X)
Then there exist unique
13
POLYNOMIAL ALGEBRAS
where either r(Xl Proof.
If f(X)
0.
=
Let f(X)
be a nonzero polynomial in R[X]. x-r
R, f(r) = 0 if and only if
E
Proof. f(r)
< degg(X).
Direct consequence of Proposition 2.6. •
2.8. COROLLARY. given r
0 or degr(X)
=
=
is a divisior of f(X).
for some g(X)
(X-r)g(X)
Then, for any
E
R[X],
then obviously
Conversely, assume that f(r) = 0 and put g(X)
=
x- r.
Since g(X)
is a monic polynomial of degree l, it follows from Proposition 2.6 that f(X) = q(X)g(X) + r
for some r
1
as required.
0 = f{r) = r , 1
E R 1
and some q(X)
But then
E R[XJ.
•
If X is an indeterminate, an element r ER such that f(r)
0 will be
called a root of f(X) . 2.9. COROLLARY.
Let
If a , ... ,a
(i}
be an integral domain and let f(X)
are distinct roots of f(X)
m
1
R
(x-a )(x-a ) ... (X-a ) 1
(ii)
If f(X)
t
0,
m
2 ·
in R, divides
then the number of roots of f(X)
E
R[Xl.
then f(X)
in R does not exceed
degf(X).
Proof.
We first note that (ii} is a consequence of (i) from considerations of To prove (i), we argue by induction on m.
degree.
The case m =
1
consequence of Corollary 2.8, assume that the statement is true for m -
being a 1
roots.
Then
and therefore domain,
f(am) = (a -a ) ... (a -a 1 )q(a ) . m 1 m ~ m q(a) = 0 so that X-a divides q(X), m m
completes the proof.
Since R is an integral by Corollary 2.8.
This
•
Recall that a principal ideal domain
(PID)
is an integral domain in which all
ideals are principal. 2.10. COROLLARY. Proof. I
Let
F
be a field.
Then
F[X]
We know, from Corollary 2.5, that F[X]
be a nonzero ideal of F[X] and let g
is a principal ideal domain. is an integral domain.
be a nonzero element of I
Let
of smallest
CHAPTER 1
14
If f f O is an element of I,
possible degree. r
f = qg +
r· = f
-
for some q,r E F[X],
qg EI,
Let
R
where degr
degh(X) >
1,
1.
f(X)
=
g(X)h(X)
with
By hypothesis
Since R/P is an integral domain, by looking at the image of f(X)
in
(RIP)
[XI,
we deduce that g(X) for some g(x)
and
1,;;; k
h(X)
= J(mod
< n
and some
belong P,
P[XJ) a,S
and h(X) E
R-P.
hence r n
E p2 ,
2.23. COROLLARY. (Eisenstein's Criterion).
= Si1-k(mod
P[XJ)
This implies that the constant terms of a contradiction. • Let
R
be a
UFD,
let
F
be the
25
POLYNOMIAL ALGEBRAS quotient field of R and let
f(X)
=
r
r
o
+r
1
Assume that there is a prime p of
r-
1
R
such that
+ ... + r
E
n
R[X]
(n;;,,
and r.1, = O(mod p), Then f(X} Proof.
Apply Theorems 2.22 and 2.2l(it}. • Let
be a nonzero square-free integer I±
k
r-k
the polynomial
n;;,, 1,
prime divisor
p
2.25. EXAMPLE.
of Let
k p
with p 2
is irreducible over a).
Jk
is irreductble over
=
XP
-1
+
•.v-2 A"
(I).
divisible by p.
Indeed, take any •
Then the polynomial + .•. + X + 1
Indeed, it suffices to prove that f{X+l)
To this end, note that the binomial coefficients
Q).
Then, for any
1.
and apply Corollary 2.23.
be a prime number.
f(X}
over
i..; n
is irreducible in F[XJ.
2.24. EXAMPLE. integer
1-.;·
1)
(~),
is irreducible
1 ..;
t ..;
p-1
are
Since
(P) f( ·X+l)=(X+l)p-!.=xp·l+(P}xp-2 {x+ 1) - l 1. + • • . + p-1
the required assertion follows from Corollary 2.23. 2.26. EXAMPLE.
Let
F
taining F such that a fi'e 1d of
F [a I.
•
be a field and let a be an element of some field conis transcendental over F.
Then, for any integer n
r- a
Denote by K the quotient
the polynomi a1
~ 1,
is irreducible in K[Xl
This is a direct consequence of Coro 11 ary 2. 23 and the fact that the ring ts a
UFD
such that a is a prime element of
2.27. PROPOSITION. (Reduction Criterion}. domain
R,
let
F
Let
R. P
R[X]-+ R[Xl
[al
•
be a prime ideal of an integral
and F be the quotient fields of
tively, and let
R = F
R
and R =
R/P,
respec-
CHAPTER l
26
If O I f
be the natural map. irreducible in F[X l,
E
then f
is such that degf = deg]'
RIX]
and ]' is
cannot be decomposed in R [X l as a product of
two nonconstant factors. Proof.
S-uppose f(X} = g(X}h(X)
wi'th g(X} ,h(X) e R[X].
Since degg·..; degg and degh..; degh,
our hypothesis implies that we must have
equality in these degree relations.
F[Xl we conclude that Let
R::, s
Hence from the irreducibility of ]' in
or h is an element of R,
g
Then ]' = gh .
as required.
•
be commutative rings and let a 1 , ... ,an be algebraically inde-
pendent elements of
s over
Let
R.
be a variable over R[a 1 , ... ,anl.
X
Then the polynomial f(X} = (X-a} ... (X-a 1
n
l over R[a 1 , ..• ,an I
can be written in the form •.n
f(X)
X
. .n-1
- 8 X 1
+ .. , + (·1}
is a polynomial in
where each s.1, = s.{a , ... ,al 1,·1 n
Ci
1
n
S
n
, •••
,a. n
In fact
s=a+a+ ..• +a 1
9
s
The polynomials s ,, .. ,s 1
a, ... ,a. 1
n
2
n
n
1
n
2
= a 1 a 2 + a 1 a s + , .• + an-1an
= a a
1 2
a
n
are called the elementary symmetrie polynomials
It is clear thats. is homogeneous of degree i 1,
Let sn be the symmetric group of degree n.
If
cr
esn
in a, ... ,a. n
1
and
f(a , ••. ,a } e R[a , ... ,a ] , 1 n 1 n
define /5 by /5(a 1 , ... ,al= f(a cr ( 1 ), •.. ,acr (n )) n
We say that the polynomial f
is symmetrie if
/5 = f
for all
O E S •
n
It is
27
POLYNOMIAL ALGEBRAS obvious that the set of symmetric polynomials is a subring of Ria ,.,.,a J n
1
containing
and the elementary symmetric polynomials
R
The result
s , ... ,s. 1 n
below shows that it contains nothing else. Let x , ... ,x
n
1
be variables.
We define the -weight of a monomial t
ti
x to be
t
+ 2t
1
n
I
The -weight of an arbitrary polynomial
+ .•. + nt. n
2
xn g(x, ... ,x) I n
is defined as the maximum of the weights of the monomials occuring in 2.28. THEOREM.
Let
Rs s
be commutative rings and let a, ... ,a 1
cally independent elements of s over R. symmetric of degree d,
E
n
1
be algebrai-
n R[a , ... ,a] n
I
is
then there exists a polynomial g(X , ... ,X}
E
n
1
of weight .a;;
If f(a , ... ,a)
g,
R[X , ... ,X] n
1
such that
d
f(a , •.. ,al= g(s , ... ,s) 1 n I n
Furthermore, the elementary symmetric polynomials s , ... ,s l
of a , ... ,a
n
n
I
are
algebraically independent over R. Proof.
We argue by induction on n,
that the result is true for n -
The case n
variables.
1
being trivial, assume
= 1
Substituting
a
n
in
= 0
we find
I' -
(X-a) ... (X-a · n- 1 Jx =, 1
s
I'- 1 +
1,0
··· +
where s.. is the expression obtained by substituting ,z,,O
(-lt-\n- 1 , 0 X a
n
= 0
in
s .• 'Z,
that s , ... ,s , are all elementary symmetric polynomials in 1 ,O n- •,O The case To prove the first assertion, we now proceed by induction on d. being trivial, assume d degree
Note
1,
.a;; d
0)
of degree d,
such that
=g I
(s
l
,o
, ... ,s _ 10 } n '
we may find a poly-
28
CHAPTER 1
Observe that
g (s , •.. ,s 1
1
n- 1
has degree
l
and some
s '. ~. ,s 1
in
n
We have equations ss. = Z'.\. 1,,
(:\ .. E
,S.
1,,J J
1,,J
Let us bring a11 the terms in (2 l to the 1eft-hand side.
R, 1
< i,j < n)
(2)
The theory of 1i near
equations applies, and if L is the determinant s-:\
-:\ 12
11
-:\ 21
s-:\
-'.\ 22
-:\ n2
we find that
L =
integral over R. 3.2. COROLLARY.
0.
-:\ 1n I1
s-:\
2n
nn
Expanding L gives us an equation that shows that
s
is
•
If R ~ s,
then the elements of s integral over R form a
subri ng of s. Proof.
Assume that u,v ES are integral over R.
We must show that both
30
Cl:JAPTER l and
u + v
are also integral over
uv
for some finitely generated R-submodules consisting of all finite sums R-submodule of
s.
Since
virtue of Lemma 3.1.
Zmini
By Lemma 3,1,
R.
s.
of
M,N
, mi EM, ni EN
and
(u+v)MN::: MN
and
vN
Then the product
MN
uM:::: M
is a finitely generated
(uv)MN::: MN,
the result follows by
s.
Then the subring of S consisting of elements
integral over R is ca 11 ed the integral closure of R ; n S.
We say that R
is integraZly closed in S
if R is the integral closure of R in S.
3.3. PROPOSITION.
be a subring of
Let
R
is integral over
u- 1
Proof.
If u
N
•
Let R be a subrtng of
Then
c
-1
s and let
if and only if
R
ts integral over R,
u- 1
u
be a unit of
s.
E R[ul.
then
u-n + r u-(n-1) + •.. + r 1
0
n
(r. ER)
(3)
1,,
Multiply (3) through by un and rearrange to get n-1
u(r +r u+ ... +r u 1 2 n
whence
u
-1
The argument is reversible.
E R[u].
3.4. COROLLARY.
=
Proof.
U(R}
uE R,
For u
Proposition 3.3, 3.6. PROPOSITION.
In particular,
we have
R[u]
Now apply Proposition 3.7.
= R.
Let R be an integral domain contained in a field
is integral over R, Proof.
is integral over R.
if S ts integral over R.
Since
3.5. COROLLARY.
•
and let u ER be a unit of s.
Let R be a subring of S
Then u- 1 ER if and only if u- 1 Rn u(s}
) = -1
u
F.
(ii)
s s
IO -1
tn R we have that
E R[u]
Let
R
= R,
u
as required.
be a subring of
s.
- I
is integral over R.
Then the following conditions
is a finitely generated ring over R and is integral over R (i)"" (ii):
By
•
is a finitely generated R-module
Proof.
lf
then R ts a field.
are equivalent: (i)
•
This is a direct consequence of Lemma 3.1
F
INTEGRAL EXTENSIONS (ii)• (i]:
31
Assume that S ts generated as a rtng over R by u 1 , ••• ,uk and
suppose that the equation showi'ng ui to be integral has degree ni.
Then the
elements (O,.; 1'.,.; n.-1) 1,
1,
span s over R. • 3. 7. PROPOSITION.
s.
Let R c s be rings and u an element of a ring containing
Suppose that u is integral over s and that s is integral over R.
Then u is integral over R. Proof.
By hypothesis, un + a un~i + ... +a = 0 1 n
for some n R
1
Let R
~ 1,
1
=
R[a , ... ,a ,u]. 1 n
is a finitely generated R•module.
(a. E S) 1,
Then one immediately verifies that
Now apply Lemma 3,1. •
An integral domain is said to be integraUy closed tf its integrally closed in tts quottent field. 3.8. PROPOSITION,
Let
s
be a multiplicative subset of an integral domain
R.
If R is integrally closed, then so ts R8 . Proof.
Suppose that the element u in the quotient field of
gral over Rs; we have to show that u
Rs·
E
is inte-
By hypothesis,
. un + (a /s 1 )un-.1 +, .. +(an I s] = O · 1 n
Put s = s 1 s 2 ... s n and t.1, = s/s1,,.
R8
(a.1, E R,s.1, ES)
Then
n n-1 su + t 1 a 1 u + ... + tnan = O
Multi'plying the above equality by
gives an equation asserting that su
integral over R.
Hence su ER and u E Rs,
as required.
3.9. PROPOSITION.
Let R be a GCD domain.
Then R is integrally closed.
Proof.
Let F be the quotient field of R. An +
Write
\=alb
with a,b ER.
'I'
1
An-i + ... + r
n
Suppose \ = 0
is
•
E
F is such that (4)
Then there exist s,t ER such that a= (a,b)s,
32 b
CHAPTER l (a,b}t,
whence Hence
t
and
Substl'tuting
(.,,t)=l.
divides
Since
sn
(sn,t) =
:\ER and the result follows.
1,
A=s/t
in(4),weobtain
it follows that t
is a unit of R.
•
4. TENSOR PRODUCTS Let
R
be a commutative ring and u,v,w any R-modules.
A map
f:uxv-w is said to be bilinear if f f(r
u
11
+r
(l )
is linear in each argument, i.e.
u , v)
22
f(u,r 1v1 +r 2v) 2
=
r
= r
l
f(u 1 ,v)
+ r
l
f(u,v) 1
+ r
2
f( u2 ,v)
2
f(u,v) 2
Our aim is to construct an R-module M and a bilinear map :\: u xv--+ M which is universal for all bilinear mapping (1), in the sense that to any bilinear map f
as in (l) there exists a unique homomorphism
r : M->- w
which renders
commutative the following diagram:
U xV--A_ _,..._M f*
w An R-module
M
with these properties is called a tensor product of u and
and is denoted by
u ® v or simply u ® v.
v
If it exists it is clearly unique
R
up to isomorphism, and we shall speak of the tensor product. The following standard procedure illustrates the existence of M. a free R-module with
u xv as a basis and let
all elements of the form
N
Let F be
be the submodule generated by
33
TENSOR PRODUCTS (r u +r u ,v} - r (u ,v} - r (u ,v) 1122
11
22
(u,r 1v1 +r 2v) - r 1 (u,v) -r (u,v) 2 l 2 2 for u,u ,u Eu, v,v ,v Ev and r ,r ER. 2
1
factor module
2
1
Then u 0 v is defined as the R
under the natural homomorphism
Thus the R-module u 0
is denoted by u 0 v.
F--+ F/N
(u,v}
The image of
F/N.
2
consists of all finite
V
R
sums l:u.'1,
0
v.'1,
(u . '1,
V)
E U, V • E '1,
and the elements u 0 v satisfy the relations
The map
=
r 1(u1 0 v)- + r 2(u2 0 v)
u 0 (r 11 v +r 2v) 2
=
r (u 0 v) + r (u 0 v) 1
l
\ : u x V->- u ® V defined by A(u,v)
2
f : u x v-
---+
(2)
2
0 v is therefore bilinear.
= u
R
Furthermore if f' : F
(r u +r u) 0 v '1122
w is any map, then f
determines a homomorphism
w given by
(u,v)) f '(Zr . U,V If we assume that
l:rU,V f'(u,v)
=
is bilinear, then NS Kerf'
f
f*
and so the map
U®V->-W R
defined by
f'(u
0
vl
f(u,v)
is a required homomorphism. Let v and
4.1. PROPOSITION. respectively.
Then
every element
x
V® R
w
be free R-modules with bases
{v.}
and
{w.}, J
'1,
w ts a free R-module with basis
{v.
,z,
0 w.}. J
Moreover,
E v 0 w can be written uniquely as the finite sum R ( V • E V, X. E
x=i:v.®x.
i
'1,
'1,
'1,
'1,
W)
and as X =
l: j
Given
Proof.
suitable r.,r\ ER, '1,
J
v
Ev and w E w,
X 1•
J
0
(x \ E V, W. E W)
W,
J
J
we have
v
=
i:r.v. '1,
'1,
J
and w = l:r\w. for
where both sums are, of course, finite.
J J
Therefore, by (2),
CHAPTER l
34
we have v © w =
L
i ,j
r.r'.(v. 0 w.) 'I,
J
v © w is generated by
Thus the R-module
J
'I,
Suppose now that
{v. © w.}.
R
J
'I,
(r.,
Lr. , ( v . © w •) = 0 1,J .
'I,
J
and fix some subscripts, say i = k, j
= s.
there exists an R- linear map
Similarly, there exists
µ
and µ8 (w)
1.
Then the map
x
is bilinear.
Since
{v.} 'I,
Ak : V->- R such that
Ak(vk) = 1.
f
: V
S
:
Hence there exists an R-linear map
is a basis for
µ
8
(w.) J
= 0 for
defined by f(v,w)
f*:
V© w->- R
R)
v,
i f: k
0 for
:,\(vi)
with
W ->-,R
W-->- R
E
1,J
and
j ,f- s
\(v)µ 8 (w) with
Finally,
and
{v. ©w.}
is a basis for
J
'I,
V©
w.
The remaining assertions follow easily by grouping terms. then
x
isuniquelygivenasthefinitesum x=
Thus if x
Zx .. (v.©w.) 'I, J
i ,j 1,J
EV©
w,
and hence by
(2)' X
= LV. ©
i
'I,
(fa ..w.) j 1,J J .
= L(fa ..v.) ©w,
j i 1,J
as required.
•
Let B and for all
b
E
B, c
C be corrmuting subalgebras 0f an algebra E
with b.'I, EB, c.'I, E A= BC
R
w
Then the set
C.
c,
A is generated by
v and
w
i.e.
A,
let be= cb
BC consisting of all finite sums
is the smallest subalgebra containing
then we say that
Now assume that V©
J
'I,.
B and
are R-algebras.
B
and
c.
Lb .c. 'I,
If
c.
Because they are R-modules,
exists, and we assert that this R-module is also an R-algebra.
'I,
TENSOR PRODUCTS
w be R-algebras.
Let v and
4.2. PROPOSITION.
35
Then V® w is also an RR
algebra with multiplication given by (v
®
1
w ) (v
Furthermore, the maps v of V and w,
w )
i---->-
v®
2
1
2
and w
1
1 and
V®
(w
1l)
1
2
l
(v
®w
--+ 1
V E 1' 2
V w '
1l)
1'
2
E
W)
are R-a l gebra homomorphisms
which are injective if V and W are
w are commuting s uba l gebras that generate
1 ®
w. Let F be a free R-module with V x
Proof. plication in
F
,w )(v 2,w) 2
11
Define multi-
V ®
R
,w w)
= (v v
1212
Furthermore, the submodule
is obviously an R-algebra.
definition of
w as a basis.
distributively using
F
(v
Then
= (v V ) ®
2
respectively, into V ® W,
R-free, and the images v@
®
1
w is an ideal of
Therefore
F.
(v,w)
In particular, since
algebra structure of F.
N
V ® W = F/N
F
used in the
inherits the
maps to v ®w this says
that (v
®
1
w ) (v ® w 1
2
2
l
= (v v ) ® (w 12
w )
12
Finally, by the foregoing and (2) it is clear that the maps v w ,-;- 1 ®
1l)
are R-algebra homomorphisms of
Moreover, if V and are injective.
V
and w,
i-;-
v ® 1
and
respectively, into
V ®
w.
w are R-free then by Proposition 4.1, these homomorphisms
The remaining assertion being obvious, the result follows.
We shall refer to the homomorphisms in Proposition 4. 2 as canonicaZ.
• The next
result provides a universal characterization of tensor product of algebras. 4.3. PROPOSITION.
Let A ,A 1
R-algebra homomorphism, subalgebras of B.
i
=
2
and B be R-algebras and let ~-:A._,. B be an ~
1,2,
such that ~(A} 1
l
~
and ~(A}, are commuting 2
2
Then there exists one and only one R-algebra homomorphism h: A ®A ---,.3 1 R 2
which renders commutative the following diagram:
CHAPTER l
36
B i
= 1, 2
(f.1, is the canontcal homomorphism) Proof. the map h
The map
A 1
x
A
1
R
ism of R-modules.
2
iJJ
12
defined by h(a
A ® A -- B
~
(a ,a )
----->- B, 2
Furthermore, if
x = a
1
1
(a )iJJ (a )
® a ) = iJJ 2
® a
and
2
is bilinear and so
1122 1
(a )iJJ (a ) 1
2
is a homomorph-
2
y =a'® a', I
2
then
h(xy) = h(a a' @a a')= ijJ (a }iJJ (a'liJJ (a )1/J (a') 11
= [ 1/J =
22
11112222
(a )iJJ (a ) l [iJJ (a 1 ) iJJ ( a 1 )]
1122
1122
h(x)h(y)
Since h(f.(a.)) = 1/J.(a.), a.EA., 1, - 1, 1, 1, 1, 1,
so that h is a homomorphism of R-algebrgs. the above diagram ts commutattve,
Finally, the uniqueness of
quence of the fact any R-algebra homomorphism mined by its restriction to
{a
1
A
® A --. B
1 R
2
®a Ja. EA., i = 1,2}. 2
1,
h
ts a conse-
is uniquely deter-
•
1,
There ts a simple criterion for an algebra over a field to be a tensor product which is often useful.
Let A be an algebra over a field
subspaces of A,
u and
Then
V are said to be linearly disjoint over
for any linearly independent elements u. in 1,
u.v. 1,
J
in A are linearly independent over F.
natural mapping ~U®V--+A ( U
is an injective F-homomorphism.
F and let
0
V
1-----4 UV
u,v be F
if
u and v. in v, the elements J
Clearly this just means that the
MODULE-THEORETIC PREREQUISITES 4.4. PROPOSITION. B,C of A,
Let
tf (i)
be an algebra over a field
A
B and
37
F.
Given subalgebras
c are linearly disjoint over F,
(ii)
A= BC
and (iii} B and c commute elementwise, then A~ B ® c · and B n c F
By Proposition 4.3, the map B
Proof.
morphism of F-algebras. A~ B ® F
c.
Ae
c
A , b ® c ..- be
-->-
F.
is a homo-
Thts is i"njective by (i) and surjective by (ii}, hence
Assume that A e B n c.
independent, hence by (i'), Thus
®
=
1,A,A,A 2
If Ai F,
then 1,A are F-linearly
are F-linearly independent, which is absurd.
and the result follows. •
F
4.5. PROPOSITION.
Let A be a finite-dimensional algebra over a field F and
let B,C be subalgebras which commute elementwise. ly disjoint over
Then B and C are linear-
if and only if
F
dtmBC = (dimB)(dimC} F
Proof.
F
F
The image BC of the natural map B
® C-->- A, b ® c ~ be
has the
F
same dimension as
B ® C
if and only if the map is injective, as required.
•
F
5. MODULE-THEORETIC PREREQUISITES Let R be an arbi'trary ring and let M be a (left} R-module.
We say that M
is noetherian (arttnian} tf every ascendtng (descending) chain of submodules of M stops. ~
module
The ring R is said to be (left)_ noetherian (artinian} if the regular is noethertan (arttnian}; in other words,
R
is a noethertan
(arti'ntan} ri'ng if and only if every ascending (descending} chain of left ideals of
stops.
R
5. l . PROPOSITION.
Let
M
be an R-modul e.
Then the fo 11 owing con di ti ons are
equivalents (i} (ii) (iii"}
M
ts noetherian
Every submodule of
M
is finitely generated
Every nonempty collection of submodules of
M
contains a maximal element,
that is, a submodule which is not properly contained in any other submodule in the collection.
CHAPTER l
38
M
Assume that (it] is fa1se, and let
(i] • (ii):
Proof.
which ts not finitely generated.
Then, by induction, we can construct an
inftnite sequence of elements a 1 ,a: 2 , . . . then for every
so (i] •
in
such that if
N
is properly contained in
k, Mk
be a submodule of
N
Mk= Ra/ ••• +Rak'
This contradicts (i) and
Mk+l.
(it] ts true.
Assume that (tttl does not hold.
(ii} • (Hi}:
Then there exists a nonempty
collection X of submodules which has no maximal element. find submodules tained tn
tn x such that for every
M ,M , • . • 1
2
The union
Mk+i·
not necessarily a member of
w=
is properly con-
k, Mk.
00
u Mk is clearly a submodule of M although k=l
We show that w
X.
By induction, we can
is not finitely generated.
Assume the contrary, say
w for some s. the modules
Then each m. ~
= Rm
E Mk
i
we deduce that
Mk.'
l
+ ... +Rm s
for some
w = Mk.
~
~
and, chaos i ng the largest of
for some index
and hence that
ki
~
But this contradicts the way the submodules
Mk .+1 =Mk.· ~
k •,
were constructed,
Mk
~
so the implication is true. (tit]• (il:
Deny the statement.
Then there exists an infinite properly ascen-
ding chain of submodules, and ft ts clear that this collection of submodules can have no maximal element.
The proof i's therefore complete. •
Let u be a submodule of an R-module v.
5.2. PROPOSITION.
noetherian ff and only if both u and
V/U
Then v is
are noetherian.
Furthermore, if
R
is noetherian and if v is finitely generated, then v is a noetherian R-module. Proof.
If
V
i·s noetherian, then, clearly, so are u and
Convers-
V/U.
ely, assume that u and V/V are noetherian, and let
be an ascending chain .of R-submodules of v. {(An+U)/u}
Then both sequences
eventually stabilize, and hence for some integer A
n
u=
A
An
=3.
At
we infer from the modular law that
n
Accordingly, since
t
n
u and
A
n
+
u=
t
At
+
u
{A
n
nu}
and
we have for all n
~
t
MODULE-THEORETIC PREREQUISITES
39
An = An n (An +U} = An n (At+ U} = At =
Hence the sequence
At
stabilizes, and v is noetherian.
{A}
n
Finally, assume that a homomorphic image of
+ (An n U} = At + (At n U)
R
is noetl'lerian.
Then every one-generator R-module is
and hence is noethertan.
~
It therefore follows from
the preceding, by induction on the number of generators, that every finitely generated R-module ts also noetherian. • A submodule W of V ts said to be maximai, if W 1' v and there is no proper submodule of
V
strictly containing
w.
A ftnitely generated module over
any ring has an important maximaltty property, which is a consequence of Zorn's lemma. 5.3. PROPOSITION.
Let
be any ring and v a finitely generated R-module.
R
Then every proper submodule of Proof.
Write
Jf = Rv
1
ts contai'ned in a maximal submodule.
V
+ .•. + Rv . n
Given a proper submodule w of v,
w.
consider the set X of proper submodules of V which contain tnductive: module.
it contains w and if Assume uwA
=
v.
This is
cw,) ts a chain in x~ then uwA is a sub-
Then each vi EWA, for some Ai.
wµ
Let
be the
'1,
largest of the modules
WA , ••• 1
n
which contradicts the definition of X to be inductive.
Then w11 contains each vi and so v
,P.'). .
x.
=
Therefore uwA is proper and this shows
We can therefore apply Zorn's lemma and deduce that x has
a maximal element; thi's ts a proper submodule of v containing
w,
as we had to
construct. • 5.4. COROLLARY.
Let
R
be any ring.
contained in a maximal left ideal of
Then any proper left ideal of R.
In particular,
R
R
is
has maximal left
ideals. Proof.
The module
~
is generated by the single element 1.
Now apply
Proposition 5.3. • The most important source of noetherian rings is the following classical result:
wif
CHAPTER 1
40
5.5. THEOREM. (Htlbert basts theorem). then so is R[X , ... ,x J Proof.
By induction on n,
Let I
R[X].
for any integer
n
I
If
is a commutattve noethertan rtng,
R
n;;. 1.
it suffices to consider the polynomial ring
be an ideal in R[X] and denote by An
coefficients of polynomials of degree
->- X
X
of G x G into G is continuous
>->- xy - l
of G into
G
A group structure and a topology on a set
is continuous
G are said to be compatible if they
satisfy (i) and (ii) above. Let
R
be a ring.
We say that
is a topological ring if
R
R
is also a
topological space such that: The mapping
(i)
(ii}
(x,y) ....-
The mapping
(Hi}
Let
x ,--,. -x
The mapping
of Ax
y
of A
are homeomorphisms. of sets gX, x
g
E
into
A
is continuous
A
into
A
then the mappings
G,
is
continuous
n be the neighbourhood filter of the iden-
G be a topological group and let
If
A
into A is continuous of A x
(x,y} ,--,. xy
tity element l of G.
gn
x +
x
.-+
gx
and x
It follows that the neighbourhood filter of g
En and also the family
ng
of sets
Xg,
x En.
-+-
xg
is the family We close
this section by quoting the following standard properties (see Bourbaki, 1966}). 6.1. PROPOSITION.
Let G be a topological group and let n be the neighbour-
hood filter of 1.
Then
(i)
Given any x
En, there exists
(_ii l Given any x (iii).
For all
E
n, we have x- 1
g E G
and all
x
y
En such that
E
n
En, we have
gXg
-1
Y•Y
c x
En
Conversely, if G is a group and n a filter on G satisfying (i) - (iii), then there is a unique topology on
G compatible with the group structure of
G,
for which n is the neighbourhood filter of 1. 6.2. PROPOSITION. are equivalent: (i)
G is Hausdorff
Let G be a topological group.
Then the following conditions
46
CHAPTER l
(ii}
The set
(iii}
is closed
{1}
The intersection of the neighbourhoods of 1 consists only of the point 1.
6.3. PROPOSITION.
A subgroup of a topological group is open if and only if it
has an interior point. Let
X ., 1,
Every open subgroup is closed.
be an arbitrary collection of topological spaces.
i EI,
definition, the product topology on
n
the form
U.
with
U.
~I1,
1[ X.
iEI
has a basis consisting of the sets of
1,
an open subset of
1,
for all but finitely many i 6.4. PROPOSITION.
X.
for every i
and u.1,
EI,
1,
E
Then, by
=
x.1,
I.
A subspace of a Hausdorff space is Hausdorff.
of Hausdorff spaces is Hausdorff.
Every product
Conversely, if a product of nonempty spaces is
Hausdorff, then each factor ts a Hausdorff space. A collection
of subsets of a topological space x is said to have the
Y
finite intersection property if every finite subcollection
n s
~
0.
Y
1
of
Y
satisfies
For the proof of the following results we refer to Pontryagin (1977).
SEY 1
A topological space x ts compact if and only if every col-
6.5. PROPOSITION.
lection of closed subsets of
X
possessing the finite intersect ton property has
nonempty intersection. 6.6. PROPOSITION. (ii) (iii) (iv)
(i}
A closed subset of a compact topological space is compact.
A compact subset of a Hausdorff topological space is closed. A continuous image of a compact topological space is compact. Any continuous btjection x _,. Y of a compact topological space x onto
a Hausdorff space 6.7. PROPOSITION.
Y
is a homeomorphism, Let G be a compact totally disconnected topological group
and 1et u be an arbitrary neighbourhood of the identity.
Then there is an open
normal subgroup N of G such that N ~ u. 6.8. PROPOSITION. (Tychonoff's Theorem). of compact topological spaces is compact.
The product of an arbitrary collection
2 Classical topics in field theory
Our aim here is to introduce classical foundations of the theory of fields, with an emphasis on refinements and extensions achieved in scope of relatively recent developments.
Among these, we exhibit canonical fundamental units of certain
classes of pure cubic fields, prove Kneser 1 s theorem on torsion groups of separable field extensions and establish a theorem of Schinzel which provides necessary and sufficient conditions for the Galois group of a binomial abelian.
to be
A separate section is devoted to a result of Isaacs concerning the
degrees of sums in a separable field extension. finite Galois extension E/F, H3 (G,E*l
r -a
If G is the Galois group of a
an interpretation of a distinguished subgroup of
is provided by using the methods of Eilenberg and MacLane.
We also
include some recent results pertaining to radical extensions, especially a cogalois correspondence discovered by Greither and Harrison(l986}. 1. ALGEBRAIC EXTENSIONS Let E be a field and F a subfield; we write E/F for the field E considered as an extension of F. over F;
We can view E as an F-algebra and hence a vector space
its dimension is either a positive integer or an infinite cardinal and
is cal led the degree of
E
degree of E over F.
We say that E/F is a finite or an infinite extens1,"on
accordingly as
(E:F}
over
F.
In what follows we write
for the
is finite or infinite.
If S is a subset of E, we write F(S}
(respectively,
smallest subfield (respectively, subring) containing s and S =
(E:F}
F[Sl}
for the
In case
F.
{s ,s , ••• ,s }, we also write F(s , ••• ,s ) and [F s , ••• ,s 1 instead of 1.2
F(Sl and F[SJ,
n
1
respectively.
n
1
It is clear that F(s 1 , . . .,sn} 47
n
is the quotient
CHAPTER 2
48
field of F[e 1 , ... ,en I and that F[e 1 , ... ,sn J consists of all polynomials in s , ... ,s 1
F by 8
over F.
n
s , ... ,s 1
, , , , ,B
1
n
We shall refer to
F(s , ... ,s)
as the field generated over
n
1
or the field obtained by adjoining to F the elements
n
,
The field extension E/F is said to be finitely generated if E
=
F(s , ... ,s) n
1
for suitable ftnitely many elements a , ... ,s
of E. The extension E/F is 1 n called sirrrple if there is an element a in E, called a primitive element over
F,
such that
F(a).
E =
When we have a chain, also called a tower of fields: FCECK
we can regard K as an extension of E or F. E/F, K/F and K/E;
their degrees are related by the important product formula:
1.1. PROPOSITION.
Let F c E c K be a chain of fields.
bases for E/F and K/E, for K/F.
Thus we have three extensions
In particular,
respectively, then AB= {abla
(K:F}
If A and B are E
A,b
is a basis
EB}
(E:F}(K:E}.
=
m
Proof. some
is an element of
If " J
~
1.
dent over F. j=l J J = l:l,
have e. J
µ.= Z:a ..a, J i=l ~J ~
Hence A= E
i=l
n
Let
m
E e.b.
for some m ~ 1 and
(a .. EF,a.EA) ~
~J
m
E a ..a.b. which shows that
j=l ~J ~ J
K
is spanned by
It remains to show that the elements of AB are linearly indepen-
AB over F.
Then
E 'J.i .b. j=1 J J
n n
for some n
=
Furthermore, we have
µ . E E ,b . E B. J
then "
K,
E
i=l =
O, e. EE, J
1..: j ..:m.
E A..a.b.
j=l ~J ~ J
=
0 with
A •• E F. ~J
We set e . J
=
E A. .a ..
i= 1 ~J
~
and since the b. are E-1 i nearly independent, we must J
The latter in turn implies that the A •• are zero, since
the a.~ are F·linearly independent. 1.2. COROLLARY.
n
m
~J
This completes the proof.
Let F c E c K be a chain of fields.
•
Then K/F is finite if
and only if both E/F and K/E are finite. Proof.
Direct consequence of Proposition 1.1.
Let E/F be a field extension.
•
An element a of E is said to be algebraic
over F if a is a root of a nonzero polynomial over F.
If f(X)
is a nonzero
49
ALGEBRAIC EXTENSIONS
polynomial over then
F w1'th
g(X) = A-lf(X}
braic over
:,._
is the leading coefficient of
is a monic polynomial with
if and only if
F
and tf
f(ci,1 = 0
ci,
g(ci,}
=
is integral over F.
0.
Thus
ci,
f(X),
is alge-
We say that the extension
E/F is algebraic if all elements of E are algebraic over F. 1.3. PROPOSITION. F
E/F be a field extensi'on, let a
Let
and let
f(X} e F[Xl
f(X)
is irreducible
(il ( i il (iii)
F(a) = F[al
F(a).
g(X)
Proof.
(i)
F [X] with
E
and if n = degf(X)
In particular,
f(a} = 0.
be of least degree such that
di Vides each
f(X)
be algebraic over
EE
Then
g(a) = 0
then
l ,a, ... ,an-I
is an F-basis for
(F(a} :F} = n.
f(X) = f (X)f (X), f. (X) e F [X ].
Suppose that
1
2
Then
1,,
f (.a}f (.a)= 0, 1
and since
2
f (a)= 0 or f (a}= 0.
is a field, we have
F
f (al= 0.
iteness that
Since
1
1
Assume for defin-
2
degf (X}..; degf(X),
we must have
1
degf (X) = degf(X) 1
degf 2 (X} .
which forces (ii)
Let
= 0.
f(X)
This shows that
g(X) e F[Xl be such that g(a) = 0.
is irreducible.
Owing to Corollary 1.2.7, we
have g(X} = q(X)f(X) + r(X} where either r(X) = 0 r(X} = 0,
degr(X) < degf(X).
or
artinian.
Since
F[a]
of Proposition 1.1.3. linear span of with
A.1,, e F
over
F
r(a) = 0,
F
F[X]}
we must have
and not all
A.1,,
}.
Furthermore, if equal to
0,
which is impossible.
then
and hence
F(a) = F[al
by virtue
belongs to the F-
It ts clear that each element of F[a] n-·1
- E be
f(X)
has a root a
l
in F,
f(X)
so there exists g(X) e F[Xl = (X-a
1
)g(X)
If degg > l,
we can repeat this argument inductively, and express
desired form.
Note that A is the leading coefficient of f(X).
coefficients of f(X)
such that
lie in a subfield K of F,
then
f(X)
in the
Hence if the
\EK.
The following result, whose proof is due to Artin, guarantees the existence of algebraically closed fields. 1.13. THEOREM. field
E containing
Proof. F[X]
Let
F
be a field.
Then there exists an algebraically closed
F as a subfield.
We first construct an extension
of degree
~
1 has a root.
We form the polynomial ring the polynomials
f(Xf)
in
F[S] F[S].
in which every polynomial in
E /F 1
To this end, put s = {Xflf and denote by
I
We claim that I
there is a finite combination of elements in I
E
F[XJ,degf
~
l}.
the ideal generated by all ,j F[S].
Indeed, otherwise
which is equal to
l:
54
CHAPTER 2
x1 . and observe that the po lynomta 1s g.~ involve ~ actually only a finite number of variables, say X ,x , ... ,x , with m ~ n.
with
g. E F (SJ. ~
Put
X• = ~
1
m
2
Hence we may rewrite our relation as: n E g.(X , ••• ,X )f.(X.) = 1.
i=l
~
m
1
~
~
Owing to Corollary 1.11, we may choose a finite extension K/F in which each
f., 1 ..; i ~
polynomial
~
has a root, say a.. ~
n,
Substituting a. for x., 1..; i..; m, ~
Put a.= 0 for n < i..; m. ~
in our relation, we get O = 1, a contra-
~
diction. Let
M
be a maximal ideal of
F[Sl
r.
containing
field containing an isomorphic copy (F+M}/M of F, a field extension of F. ;;,,, l , xf + M E
E1
Then E
l
Hence we may regard
Furthermore, for every polynomial
is a
= F(S]/M
f E F[Xl
E
l
as
of degree
is a root of f.
Inductively, we can form a chain of fields ECEC . .,CEC
-n-
1-2-
such that each polynomial in
E [Xl
n
E be the union of all fields
and every polynomial f
1.14. COROLLARY. E/F such that
~
Let
E,
~
E
is obviously a field But then
Then there exists an algebraic extension
Let E be the algebraic closure of F in K.
1,
then f(X)
Moreover, if f(X}
has a root, say a,
in
E E [X l
F
be a field.
E containing
Hence a is algebraic
K.
F.
Thus
•
By an algebraic closure of
F such that
Then
is a po lynomi a1 of
and therefore, by Proposition l .5(ii}, a is algebraic over
a EE and the result follows. Let
E
E is algebraically closed.
is an a1gebra i c extension.
degree
Then
Let
as asserted. • be a field.
F
is algebraically closed.
over
E, n = 1,2, ... n
1 has a root in En+i·
By Theorem 1.13, there exists a field extension K/F such that K
Proof.
E/F
~
E[Xl has its coefficients in some field E. n
f E
has a root tn En+J.
of degree
E/F
F
we understand any field
is algebraic and E is algebraically closed.
Owing to Corollary 1.14, such E always exists.
It will be shown below that E
ALGEBRAIC EXTENSIONS is determi'ned uniquely, up to an F-isomorphism.
55 The following preliminary result
wtll clear our path. 1.15. LEMMA.
Let
be a field and let cr :
F
tnto an algebraically closed field L. field extension of F,
and let f(X) e F[X]
then there exist exactly t Proof.
coefficients of f(X}.
form w(a)
F
be the minimal polynomial for a.
extensions of cr
(in the algebraic closure of F), to a homomorphism F(a)
-->-
L.
t 0(x} e L[Xl be obtained from f(X) by applying cr to the
Let
t 0(x} in
be a homomorphism of
Let a be an algebraic element in a
is the number of distinct roots of f(X)
If t
of
F-->- L
Since
is algebraically closed, there exists a root
L
Given an element of F(a)
L.
with some polynomial
=
Fial,
w(x} e F[X].
B
we can write it in the
We define an extension of cr by
mapping
This is well defined, i.e. independent of the choice of the polynomial 1
w(a) (w-w )(al = 0,
w0 (x)
-
l
w01 (x), .
morphism F(a) f(Xl
whence
and thus -->- L
wO'(B). 1
=
inducing cr on
in the algebraic closure of
t 0(x) e
L[X].
a root of
w1 (a},
=
f(X) divides
l'CB) .
w(X) - w (x).
1.16. PROPOSITION.
F.
Now the number of distinct roots of
(ii)
--+ L
There exists an extension cr'
of cr is
•
be an algebraic extension and let cr
E/F
divides
is equal to the number of distinct roots of
F
the result follows. Let
f 0(x)
It is now clear that our map is a homo-
a homomorphism into an algebraically closed field (i)
Hence
l
Since the image of a under any extension F(a)
t 0(X),
used
Indeed, if w (X) e F(X] is such that
to express our element in F[al.
then
~(X)
F---+ L
be
L.
of O' to a homomorphism
E--+ L
If E is algebraically closed and L is algebraic over cr(F),
then cr'
is an isomorphism. Proof.
Let us show first that (ii}
is algebraically closed, and
L
is a consequence of (i).
is algebraic over cr(F),
Indeed, if
then cr'(E)
is alge-
E
56
CHAPTER 2
braically closed and L is algebratc over a'(E),
hence L
To prove (i), denote by s the set of all pairs field of E containing F, If
K -->-L.
and
Kc K'
and
(K,A)
Note that
A'JK = A,
ely ordered:
If
to a homomorphism
0
we write
and s
so SI 0,
(F,0} Es ,
if
(K,A),,,;; (K',A')
is a totally ordered subset, we let
{(K.,A.)} 1,,
9efine
are in s,
(K 1 ,A')
where K is a sub-
(K,A)
and A is an extension of
a'(El,
=
is inductivu
K =
K~
1,,
and
"
Then
A on K to be equal to A.1,, on each K1,,..
(K ,A)
is an upper bound
for the totally ordered subset.
Owi'ng to Zorn's 1emma, we may therefore choose a
maximal element, say
s.
claim that T
=
in
(T,µ),
Then
Otherwise, there exists
E.
has an extension to
Let
a
E
be a field.
F
0
and, by Lemma 1.15,
E-T
to E,
and we
0,
thereby contradicting the maximality of
T(a),
proves that there exists an extension of 1.17. COROLLARY.
is an extension of
µ
as required.
µ
This
(T,µ).
•
Then any two algebraic closures of
F
are F-isomorphic. Proof.
Let E and E 1 be algebraic closures of F and let
be the inclusion map. phism
E -
Then, by Propositton l.16(ii},
as required.
E',
0
0
:
F -
E'
extends to an isomor-
•
Let E/F be an algebraic field extension and let F be the
1 . 18. COROLLARY.
a1gebra i c closure of
F
con ta i ni ng
E.
Then any F-homomorph ism
E -
F
is
extendible to an automorphism of ffe. Proof.
Since F/E is an algebraic extension and
F
is algebraically close~
the result follows by virtue of Propositions 1. 16(i) and 1.12. • We close this section by proving the following observation. 1.19
PROPOSITION.
finite.
Then
Proof.
JEJ
Let E/F be an algebraic field extension and let F be in=
JFJ.
Let s be the subset of F[Xl
positive degree and, for any n
~
0,
of all polynomials of degree n +
1.
r+l + A r
so
1
+ ... + A ' A' E n
1,,
F,
consisting of monic polynomials of
let S(n)
be the subset of s consisting
The elements of S(n)
have the form
NORMAL EXTENSIONS
ls(n) I = IF x ... x Fl n factors
+-
57
IFI
=
+
Also
= I u S(n) I = IFI
Isl We now map each in
f(X)
Es
wO
into the finite set
Because every element of
E.
(possibly empty) of its roots
Kf
is algebraic,
E
UK
E =
fES f
Since each IEI.
is finite the cardinality of the collection
Kf
Therefore
IEI
l{Rf}I ,;;;; Isl
=
=
IFI
{Kf}
is the same as
which clearly implies that
IEI
=
IFI .•
2. NORMAL EXTENSIONS Let
degree ;;,, 1. of
F
{f.li 1,
be a field and let
F
By a
EI}
splitting field
such that every f.1,
be a family of polynomials in
splits into linear factors in
F[Xl
of
for this family we understand an extension
generated by all the roots of all the polynomials family of polynomials in
p[xl
f.,i 1,
and
E[X],
EI.
E
E
is
We note that any
always has a splitting field, namely the field
generated by the roots of the given family in any algebraic closure of F. turns out that the splitting field of
{f.li 1,
EI}
It
is determined uniquely up to an
F-isomorphism, as we shall see below. 2.1. LEMMA. (i)
E
(ii)
If
Let
and F
K
c
E
f(X)
E
and let
F[X]
E,K
be two splitting fields of
are F-isomorphic c F where
F is an algebraic closure of
morphism K-+ F is an isomorphism of K onto Proof.
(i}
f(X).
Let
E
exists an F-homomorphism
0
:
K--+
E.
F.
Then
E
is algebraic
Owing to Proposition 1.16, there
We are therefore left to verify that
o(K) = E
We have a factorization
then any F-homo-
E.
be an algebraic closure of E.
over F, hence is an algebraic closure of
F,
(l )
CHAPTER 2
58
f(X) with
Bi EK, 1
-
F is extendible to an
provides, by virtue of Proposition 2.5,
E
•
Kc E
c
be a chain of fields, where
L
If the number of K-homomorphisms
then every F-homomorphtsm K __.. L
has exactly
E ~ L
is a
is equal to
n extensions to homomorph-
E__,_ L
Proof.
Let
(respectively,
G be the group of all F-automorphisms of L and let G(K) G(E))
be the subgroup of G consisting of those automorphisms
of L which leave fixed every element of K (respectively, of E). that
L/F
G(E)
is a subgroup of G(K).
Let
n
G(K) = U ~.G(E) and i=l ~
m G =
u ~.G(K)
j=l J
It is clear
63
SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS be cosets decompositions of G(K]
and G. G =
U
i,j
Then
ijJ ,c/J .G(E) J 1,
is a coset decomposition of G with respect to G(E}. automorphisms any element
have distinct restrictions to
ip. J
of
ip
G
to
K
K
It is clear that the m
and that the restriction of
cotnci'des with the restriction of one of the
iJJ .. J
Since by Propositi'on 2.9, every F-homomorphism of K into L is the restriction of some automorphism of L,
it follows that K has exactly m F-homomorphisms
into L and that these are given by the restrictions of
to K.
ip , ••• ,ip 1
m
A similar argument shows that E has exactly mn F-homomorphisms into L and that these are given by the restrictions of the each c/Ji restricts to the identity on trictions to
K
if j t-
products
Thus each F-homomorphism of
k.
which are F-homomorphisms of
i'nto
E ijJ
to E.
i)!.¢. J 1,
Now
and ipj and ipk have distinct res-
K
one represented by the restriction of ipj'
to E.
nm
K
into
L,
say the
has exactly n extensions to
E
namely the restrictions of
L,
.c/l ,ip .¢ , ... ,ijJ .c/J
J 1
J n
J 2
•
3. SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS Let
F
be a field and f
a polynomial in
If
F[X].
:X.
is a root of f
in
F,
then
for some g(X) and say that
E :X.
F[Xl with
g(::\} t- 0.
is a multipZe root if m >
aim is to be able to test whether F,
We call
f
without having to go outside F.
m
the nruUiplicity of
:X.
and a simpZe root if m = 1.
in F, Our
has multiple roots in some field containing To this end, we first define derivatives
formally. Given any polynomial we define its derivative
f(X) = a +ax+ ... + a x1 o 1 n Df or f' by the rule f' (X) = a
1
+ 2a X + 2
over a commutative ring R,
CHAPTER 2
64
The mappi'ng ~R[XJ->- R[X]
~
ff-+['
enjoys the fo 11 owing properties (f +g J I = f
(i )
(ii}
+ gI
(f,g E R[Xl)
(r
(rf) ' = rf'
(ii i ) (fg) (tv)
I
I
=
E
R)
f Ig + fg I
X' = l.
Conversely, f'
is completely determined by (i) - (iv) and this provides an alter-
native definition. 3.1. PROPOSITION. f
Let F be a field, let f
in some field extension of F.
Proof.
Dividing
Then \
by (X-A) 2
f
f(X)
E
is a multiple root if and only if
we obtain
(X-A) 2 g(X) + h(X)
=
where h(X) = 0 or degh(X) < 1.
Putting
we see that f(\) = h(\) = 0.
X = A,
Differentiating and putting X
= \,
which shows that
(X-\) 2 \J(X}
H and only if f'(A)
3.2, COROLLARY.
Let f
splitting field of
f'.
E
we find that h'(\)
degf > 1.
F[X] with
are simple if and only if
f)
In particular, if
and let A be any root of
F[Xl
=
=
f 1 (A).
Hence
O. •
Then all the roots of f
(in
is prime to its derivative
f
f is irreducible, then all the roots of f are simple
if and only if f' 1 0. Proof.
By Proposition 3.1,
have a common factor. 3.3. PROPOSITION. (i) (ii)
If If
charF
=
f
has a multiple root if and only if f
•
Let F be a field and let f(X) 0,
and f'
E
F[X]
be of degree > 1.
then f' 1 0
charF=p>O,
then f'=O
if and only if f
has the form g(xP).
65
SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS
(t l
Proof.
f' = 0,
then
(ii)
If
that
f'
f
f
f( ·X) = a0 + a 1 X + . • • + a n x" 0, which is impossible.
Write
=
na
n has the form ~
f'
then obvtously
1 ~
ia. = 0,
Then
= O.
g(xP),
i
~
n,
with
a:
0.
=
n
f. 0, n ~ 1 .
Conversely, assume
t
a. = 0 if p
and hence
If
~
i.
Thus
has the form
f(X) = a in other words,
0
+ a
p
xP
+ a
2p
is now a polynomial in
f
+ arp rt;
x2 P +
xP. •
In what follows by the roots of a polynomial we mean those lying in a splitting field of that polynomial. 3.4. COROLLARY. (i}
If
O;
charF =
(ii).
If
form
g(xt'}.
charF = p
Proof. (ii}
Let
(il
then r>
F.
be an irreducible polynomial over a field
f
has only simple roots
f
0,
f
then
has multiple roots if and only if
Apply Corollary 3.2
f
has .the
and Proposition 3.3(i}.
Apply Corollary 3.2 and Proposition 3.3(ii}.
•
The next observation is a useful companion to Corollary 3.4(ii}. 3.5. PROPOSITION.
f(Xl
F[Xl
E
F
Let
be a field of characteristic p > 0
be an i'rreducible polynomial.
and let
Choose the nonnegative integer
e
such that e
f(X}
E
F[xP 1
e+l but f(X} ff F[xP . 1
e
(i) (ti}
If
f(X} =
w(xP ) ,
All roots of
then
f(X)
1/J
is 1'rreducible and has only simple roots.
have the same multiplicity equal to
f(X).
equal to the number of distinct roots of
pe
and
degi/J
is
In particular,
degf = ( degi/!)pe Propf. otherwise
( i)
1/J(y)
the assumption. (ii}
It is clear that 1/J is of the form
and hence
Hence, by Corollary 3.2,
By (i), we may split
extension of F):
g(yP}
1/J(y}
. Moreover, 1/J' f. 0, since e+l f(X) E F[xP ] , contrary to
is irreducible.
1/J
has only simple roots.
into distinct linear factors (in a suitable
66
CHAPTER 2 m
rl (y-S .)
1/J(y}
i=l
1,
Then m
i=l
e
be a root of i? - Si.
Let ai
p
e
1,
Then e
s.
(J;i
;f -
e
n (Jf -S .)
f(X} =
s. 1,
(a , ... ,a are distinct ) 1 m
1,
e
- c:l1,.
= ;f
e
=
Hence
f(X} as required.
•
The degree of the polynomial
(in the notation of Proposition 3.5) is
1)J
called the separable degree of f(X), rability of f(X}.
ial
f(Xl,
If
o:,
while e is called the exponent of insepa·
is an algebraic element over
then the separable degree of
a over F)
of inseparability of
a
over
F
=
0,
with minimal polynom-
(respectively, the exponent
is deftned to be the separable degree of f(X)
(respectively, the exponent of inseparability of f(X)). charF
F
the separable degree of a
In the case where
is defined to be the degree of a.
Thus,
in all cases, the separable degree of a ts equal to the number of distinct roots of the minimal polynomial of a. Let E/F be a finite extensi'on and let K be an algebraic closure of E. The separable degree
morphisms of E tnto K. 3.6. LEMMA.
of E/F is defined to be the number of F-homo-
(E:F) 8
This ts obviously independent of the choice of K.
Let F(a)/F be a finite extension.
the separable degree of a over morphtsms Proof.
F(a)
--+
E
F
Then
(F(a):F} s is equal to
and ts also equal to the number of F-homo-
where E is any normal extension of F containing a.
The first assertion is a particular case of Lemma 1.15 in which cr
is equal to the inclusion map. algebraic closure of E, Since E/F is normal,
To prove the second assertion, let
K
be an
Then K is obviously an algebraic closure of F(a). the image of any F-homomorphism F(a}--+ K is contained
SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS in
as required.
E,
67
•
We now introduce a property of field extensions which will play an important role in our subsequent investigations. Let F be a field and let f
E
if all irreducible factors of f field of f).
over
is separable over F
have only simple roots (in a splitting
F
In particular, an irreducible polynomial is separable if and only
if its roots are simple.
An algebraic element a of a field extension
F is called separable over
If
We say that f
F[X].
over F, we call
a separable extension.
E/F
the definition that a is separable over separable degree of a over 3.7, PROPOSITION.
Let
of
F if its minimal polynomial over F is separable.
is an algebraic extension such that all elements of
E/F
E
F
F
E
It is an immediate consequence of
if and only if the degree and the
cotncide.
be a field of characteristic 0.
F
ials over F are separable.
are separable
Then all polynom-
In particular, any algebraic extension of F is
separable. Proof.
Apply Corollary 3.4(;}. •
3,8. PROPOSITION.
Let
be a field of characteristic p,
F
braic element in a field extenston of rability of a, (F(a) :F)
(i}
= pe(F(a) :F)
(iii)
a is separable over
Proof.
(il Let
f(X)
the separable degree of a,
F
F
E
if and only if F(a)
(F(a}:F)
F(ar).
Then =
degf(X)
by Proposition 3.5(ii).
= pem,
=
F[Xl be the minimal polynomial of a and let m be
(F(a):F} and so
be the exponent of insepa-
e
s
is separable over
aP
and let
Then
e
Ci'il
F
let a be an alge-
Now apply Lemma 3.6. e
(ii}
In the notation of the proof of Proposition 3.5, aP
irreducible polynomial
iJ;
which has only simple roots.
is a root of the Hence
aP
e
is separable
over F. (iii} Assume that a is not separable.
Then, by Proposition 3.5(ii},
68
CHAPTER 2
aF is a root of g(X).
and
f(Xl = g(xPl
Hence
(F(cl) :F) ,;;;;· degg(X) < degf(X)
=
(F(a) :F)
which shows that F(aP}; F(a). Now suppose a is separable so that f(X) be the minimal polynomial of a over F(aP). distinct roots.
Then g(X)if(X),
Also a is a root of the polynomial
divides xP-aP
g(X)
has distinct roots.
that g(_X) = x-a.
=
(X-a)P.
Hence a
E
Since g(X) F(aP)
xP-aP
Let g(X)
so g(X)
has
over F(aP),
so
has distinct roots, this implies
and therefore F(a} = F(aP).
•
The next observation will enable us to take full advantage of the results so far obtained. 3.9. LEMMA.
Let F .'.:. K .'.:. E be a chain of fields, where E is an algebraic
closure of F and let A: K--+ E be an F-homomorphism. there exist exactly Proof.
(K(a}:K} . s F-homomorphisms K(a1-+
Let f(X)
(K(a):K} s
Then, for any a EE, E
extending A,
be the minimal polynomial of a over K.
is equal to the separable degree of a over K,
of distinct roots of f(X)
in E.
By Lemma 3.6,
hence to the number
Now apply Lemma 1. 15. •
As an application of the above result, we now prove 3.10. PROPOSITION.
Let K/F be a finite field extenston, say K
r
(i}
(K:F) =
(ii)
(K:F) 8
S
(iii}
n i=
(F(a , ... ,a.) : F(a ,, .. ,a, 1 )) 1 1 1,, 1 1,,8 = (K:F) if and only if each ai is separable over
If charF =
F(a , ... ,a.1,- -1 ) ,
=
F(a , ... ,a ). 1
r
F(a 1 , . . . ,ai_ 1 )
and e.1,, is the exponent of inseparability of a.1,, over
p > 0
then
1
e +_., .+e
(K:F). = (K:F) 8 p
Proof.
(i}
Put n. 1,, m. = 1,
Then, by Lemma . 3. 6, a.1,,-1 }.
=
1
r
(F(a 1 , ... ,a1,,..) : F(a l , .. , ,a.1,,- 1 )} 8
(F(a 1 , ... ,a.} : F(a , ... ,a.1,,- 1 ) l, .· 1,, . 1 n.1,,
1,;;;;
and
i ,;;;; r.
is equa 1 to the separsib 1e d_ egree of a.1,, over .. F(a , ... , . 1
This, together with Proposition l_. l, ensures that r (K:F) = 1]m. . i=l 1,,
and n.1,, ,;;;; m.1,,
(1 ,;;;; i ,;;;; r)
(1)
SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS The case r
betng obvious, we argue by i'nductton on r.
= :i.
If
ar-1 ) and E is an algebraic closure of K(hence of K1 ), there exist exactly n 1n 2 .. . n r-1 F-homomorphisms
K
it follows from Lemma 3.9 that each F-homomorphism extensions to F-homomorphisms K-+ E. morphism (K:Fl -
(_ii)
S
=
to
K->- E
1
-
Since
F(a , ... , l
K = K
(ar ) ,
1
has exactly n
E
1
1
then by induction
E.
K -
=
K
r
Because the restriction of an F-homo-
ts an F-homomorphism
K
1
69
K
-
it follows that
E,
l
n l .:.n, as required. !'
Di'rect consequence of Lemma 3. 6, ( i) and ( 1).
(ii i ) By Proposition 3.8(t}, m.'1, (_K:F}
as required.
=
e. '1,
e 1 +, . • +er
'_1:'_
1'__
r Jm. = Cl
bl'I,
hence by (1) and (i),
= p 1,n.,
= (K:F)
Jn.)p
bl'I,
e 1 + •.• +e p
r
S
•
The result above has a number of important applicattons.
Before putting it
to use, we make the followtng observation: If F c E c K ts a chain of fields and a EK ts separable over F, separ~ble over
then a is (2)
E
Indeed, if f(Xl
E
with irreducible
F[Xl
ts the mtnimal polynomial of a and f(X) = f 1 (X) .. • fn(X) then a is
fiX) E E[Xl,
must have only simple roots stnce so does 3. 11 . COROLLARY.
Let
a root
of some Thus
f(X)..
be a fie 1d extension.
E/F
fi(x)
and
f .(X)
a is separable over E.
Then the set of a 11 e1ements
of E whtch are separable over F is a subfield of E containing F. parttcular, i"f E/F
is generated over
E
'1,
In
by a family of separable elements, then
F
is separable. Proof.
Assume that a , a 1
are separab 1e over
2
{a
\ E
1
Then F(a ,a} 1
2
=
F(A,a ,a) l
2
by Proposition 3. lO(ti),
2
2
2
l
1
ts separable over F(a) l
E
by (2).
Since all elements of
F.
the result follows.
Let E/F be a field extension. understand the subfield of
1
is separable over
obviously separable over F,
and 1et
a , a +a , a-1 } ,a f O.
and a
;>,.
F
Hence, F
are
•
By the separable closure of F in E we
consisting of all elements which are separable over
CHAPTER 2
70
F.
If
J
F is called the closure of F Let
F,
ts an algebraic c1osure of separable aZosu'I'e of
F.
then the separable closure of F tn Due to Corollary 1.17, the separable
ts uniquely determined up to an F·isomorphism.
KIF be a finite extenston.
We defi'ne the inseparable degree
(K:F).
'Z,
of K over F by (K:F).'Z, = (K:F}/(K:F} 8 If
charF = 0,
charF = p > 0,
then by Propositions 3.lO(it) and 3.7,
(K:F}. . 'Z,
then by Proposi'tion 3. lO(iii),
ts a power of
3. 12. COROLLARY.
If F c E c ]{
Proof.
(K:F) 8
while if p.
= (K:F}
if
is algebraic over
F.
is a finite extension, then
KIF is separable
and only if (ii}
If KIF
(i}
(K:F}.'Z,
= 1,
(i).
is a chain of fi'elds and
Write
K = F(a. l , ••• ,a.} 'I'
KIF
is finite, then
where each
ll•. 'Z,
Then, by Propos i'tion 3, 1O( ii}, the condition that each F(a. 1 , ••• ,a.'!,• 1 }
a.'Z,
is separab 1e over
is equivalent to the condi'tton that each ai
(stmply rewrite
is separable over F
F(a. , ..• ,a.'I' } as F(a'Z-.,a. 1 , •.. ,a.'!,•1 ,0\'Z,·+1,, .•. ,a'I' } . . 1
condition, in view of Corollary 3.11, is equtvalent to
The latter
KIF being separable.
Now apply Proposi'tton 3. lO(tt). (itl
Apply Proposition 3.10(1),(iii}.
3.13. COROLLARY.
If
field extension of Proof. forces
F ts a field of characteristic p
F with p) (E:F),
(E:F}
8
=
(E:F).
If EIF
(i}
lf F ~E
is separable and
tained in some field, then
separable.
and
(i)
If KIF
Assume that
KE!/K
~
E a finite
EIF is separable.
Now apply Corollary 3.12.
rable if and only if both EIF
Proof.
then
> 0
Owing to Proposition 3. lO(iii}, the assumption that
3.14. COROLLARY.
(ii}
•
p ) (E:F}
•
K is a chain of fields, then KIF is sepa-
and KIE
are separable.
KIF is any extension with both E and K conis separable.
is seoarable, then obviously both
EIF and KIE are separable.
EIF and KIE are
If KIF
is finite, then
SEPARABLE, PURELY INSEPARABLE ANO SIMPLE EXTENSIONS
71
the result is true by vty,tue of the second equaltty in Corollary 3.12(ii). K/F is infinite, let a.EK.
Then a. is a root of a separable polynomial
Let a 0 ,a 1 , •.• ,an be the coefficients of f(X}
f(X) E E[Xl,
If
and let
E9 "' F(a 0 , ... ,an).
Then we have F -c E0 -c E0 (a.) where E0 (a.)/F ts finite and both E0/F and E0 (a.}/E0 are separable.
rable over F,
Hence E0 (a.)/F is separable and therefore a. is sepa-
proving that KIF is separable.
(ii} Every element of E ts separable over F, hence separable over K. KE is generated over K by the elements of E,
Since
it follows that KE/K is sepa-
rable by Corollary 3. 11. • If
F
c Kc E is a chain of fields with
but KIF need not be normal.
EIF normal, then EIK is normal
The following result provides a condition under
which KIF is normal. Let EIF be a normal fteld extension and let F8
3.15. PROPOSITION.
separable closure of F in E. Proof.
Let
F-homomorphism. E -
E.
F
8
IF is norma 1.
be an algebraic closure of E and let cr; F8
E.
--+
E
be an
By Proposttton 1,16, we may extend cr tq an F-homomorphism
Since EIF ts normal, we see that cr(E}
morphi sm of in F8 •
E
Then
be the
= E,
Furthermore, cr(F8 ) is separable over
Thus cr(F8 } = F8
so cr ts an f-autohence is contained
F,
and, by Proposition 2.5, F8 IF is normal. •
We next provide a class of fields over which all extensions are separable. Let
F
be a field of characteristic
p >
0 and let iP
:e
{a.Pix e F},
Then
the mapping
is obviously an injective homomorphism whose image is the subfield iP of Ifdt is also surjective (i.e .. if F is said to be perfeat,
p-th power.
F =
F,
#), and hence an automorphism, the field
Thus F is perfect if and only if every element is a
In addition, every field of characteristic O is perfect by defin-
CHAPTER 2
72
ition. Let
F
be a field of characteristic
closure of F.
For each
Then pP
-n
be an algebraic·
F
define
n;,,, 1,
pl?
0 and let
p >
-n
FlxP
= {x E
n E
F}
is obviously a subfield of F and we have a chain F
of subfields of F.
-1
-2
pP . c ·Fp
C
C
Fp
"' C
-n C
Put -oo
00
up?
-n
n=1
3. 16. LEMMA.
With the notation above,
ts the smallest perfect subfield of
pi? -
containing
F
F.
Furthermore, if
Fi'. pi?
..-00
(lC)
,
then #
is an infinite
/F
extension. Proof.
Let K be a perfect subfield of F containing F.
n
then xP E F ~ K,
so xP
n
=
E
#
-n
n =
for some A EK since K is perfect.
AP
-yt
x
If x
Hence
..-00
A EK and therefore pl?
~K,
proving that
F
-00
The fact that pl?
c K.
is perfect is a consequence of the equality (#
-n p
1
-oo·
n;,,, 1,
hence
.
(n ;,,, 1)
..-co
-oo
If pl? /F is finite, then #
Finally, assume that some
-(n-1)
= pi?
F =
(#
-n
n )P
=
We sha 11 refer to the fte l d pP
#
for
-n
a contradiction.
•
as the perfect c7,osure of
F.
By Coro 11 ary
-cp
1. 17,
pl?
is determined uniquely, up to an F-tsomorphi sm.
3.17. PROPOSITION. Proof.
Every finite field is perfect,
Direct consequence of the fact that an injection of a finite set into
itself is a bijection. • The simplest example of an imperfect field is the field functions in x over a field
F
cannot be a p-th power in F(X).
of charatteristic
p >
0.
of rational
This is so sihce x
We now provide some criteria for a field to be
perfect. 3.18. PROPOSITION.
F(X)
The following conditions are equivalent:
SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS The field
(i)
73
is perfect
F
Any finite extension of F is separable
(ii)
(ii i ) Any algebraic extension of F is separable Any po 1ynomi a1 in F[X]
(iv)
is separable.
For characteristic O there is nothing to prove:
Proof.
in that case every
field is perfect and, by Proposition 3.7, (ii), (iii) and (iv) always hold. Assume that charF
0.
= p >
that (iii} implies (ii).
It is obvious that (iii) and (iv) are equivalent and Furthermore, (ii) implies (iii) by Corollary 3.11.
We are therefore left to veri'fy that (i} is equivalent to (ii}. Let F be a perfect field and let f If
f
is not separable, then by Coro 11 ary 3. 2,
3.3(ii},
f(X)
=
g(iP),
is perfect,
F
f' = O.
Hence, by Propos i'ti on
say f(X)
Since
be any irreducible polynomial over F.
a iPr + a jP(r-l) + o 1
=
for some b.
a.= b~ ~
~
=
~
(b
r
·o
+b
1
E
+ a
r
F and hence
.f'-L + ... + br )P
But this contradicts the irreducibility of f.
Thus every finite extension of
is separable.
AE
Conversely, assume that every finite extension of
F
and consider the splitting field of iP - :\.
If
F
is separable. µ
and so the minimal polynomial of µ over F is of degree 1. A = µP,
as required.
is a root, then
Hence µ
E
F and
•
Let F be a field of characteristic extension.
Fix
p >
0 and let E/F be an algebraic
We say that E/F is purely inseparable if the only elements of E
that are separable over
F
are the elements of F.
Expressed otherwise,
E/F
is purely inseparable if and only if the separable closure of F in E is F. It is clear that E/F can be inseparable without being purely inseparable. An algebraic element a of a field containing F is said to be purely
F
74
CHAPTER 2 n
JJ e
insepa.PabZe over F tf
P for some n
;i,,-
If E/F is an algebraic
0.
fi'eld extension, then the set of purely inseparable elements of E over F is obviously a subfield of
containing
F.
We shall refer to this subfield as
the pm>e inseparabZe aZosm>e of F in E.
The next result shows that it is a
E
purely tnseparable fteld extension of F containing any other such extension K/F With
KC E.
3.19. PROPOSITION.
Let
an algebraic extension.
be a field of characteristic p » 0
F
and let E/F be
Then the following conditions are equivalent:
E/F is purely inseparable
(i)
(.ii} All elements of (.iii)
E
are purely tnseparable over
is generated over
E
F
by a family of purely inseparab 1e elements
P
(iv} There exists exactly one F-homomorphism of
E
into an algebraic closure
of E. Proof.
Since the elements of E which are purely inseparable over F form
a subfield of E containing F, (i) * (ii}:
If a.EE,
then
ti
(ii) and (iii} are equivalent. e
is separable over
F
for some e;.. 0 (Propo-
e
sitton 3.8(ii)), hence a.P E (ii} * (ivl:
Let
F.
Thus a. is purely inseparable over
be a set of purely inseparable elements of
$
and let E be an algeoraic closure of E.
E = F(S}
n
root of the polynomial
x? -
"E F[Xl
E
E -
E is a root of
(iv} * (i): rable.
for some n ~ O and some
"E F.
hence a. -
Then (F(a.) :F) = (F(a.) :F) 8
E -
E.
f(X)
Since e F[Xl
Now the image of a. under any F-homomorphism a.
proving (iv).
Assume by way of contradiction that a. E
morphisms F(a.) phism
f(X),
such that
Each element a.Es is a
all the roots of this polynomial are the same, the minimal polynomial of a. has a. as the only root.
F.
>
E,
a. If:. F
and a. is sepa-
1 and hence there are more than one F-homo-
But each such homomorphism is extendible to a homomor-
E by Proposition 2.9.
This provides the desired contradiction and
completes the proof. • 3.20. COROLLARY.
Let F be a field of characteristic p
be a chain of fields.
>
0 and let F ~ E ~ K
Then K/F is purely inseparable if and only if both E/F
SEPARABLE, PURELY INSEPARABLE AND SIMP4E EXTENSIONS
and
are purely inseparable.
K/E
Proof.
Direct consequence of Proposition 3.19(ii). •
3.21. PROPOSITION.
Let
be a field of characteristic
F
an algebraic field extension. then
If Fs
is separable and
F /F
s
Proof. Let
That Fs /F
E/F
Corollary 3.14(i).
braic extension
Let
E/F
be
is the separable closure of F in E,
is separable is a consequence of the definition of F. s
Hence
3.22. COROLLARY.
O and let
p >
is purely inseparable.
s
be the separable closure of
K
75
K = F
s
F
in
s
Then
E.
and therefore
E/F
K/F
'
is purely inseparable. •
s
be a field of characteristic
F
is separable by
p >
0.
Then every alge-
may be obtained by taking a separable extension followed by
E/F
a purely inseparable extension. Proof.
Direct consequence of Proposition 3.21.
•
The following result provides a criterion under which the order of the extensions in Corollary 3.22 can be reversed. 3.23. PROPOSITION.
Let
tic
K
p >
0 and let
rable closure of Proof.
F
be an algebraic extension of fields of characteris-
E/F
and
in
F
Then
E.
Assume that
be respectively the pure inseparable and sepa-
s
is separable if and only if
E/K
is separable.
E/K
and, by Corollary 3.20 and Proposition 3.21, Hence
Conversley, if
E = KF •
s
E =
KF
s
E = KF . s
Then E is separable over KFs E
then
is purely inseparable over
KF. s
is separable by Corollary
E/K
3.14(ii). • 3.24. PROPOSITION.
Let
be a field of characteristic
F
p >
0 and let
E/F
be
a finite extension. (i)
E/F
is purely inseparable if and only if
Proposition 3. lO(iii}, if
(E:F)
s
=
is purely inseparable, then
E/F
In particular, by
1.
(E:F)
is a power of
P· (ii}
(E:F) Proof.
(ii)
s
= (F
(i)
s
:F), where
F
s
is the separable closure of
F
in
E.
Direct consequence of Proposition 3.19(iv).
By Proposition 3.21,
E/F8
is purely inseparable, hence by (i), (E:F8 \= 1.
CHAPTER 2
76 Now (E:F)
by Corollary 3.12(ii), and
=
s
(F :F) (E:F } s
Hence
(F :F)
=
S
Let F be a field of characteristic
finite field extension of F. Proof.
Now E/F mi
by Proposition 3.19,
a~1,
we have a~1, E
F
E F
for all
s
EPF.
=
Then E/EPF is both separable and =
Let
EPF.
F
s
Since E/F is finite, we may write
is purely inseparable (Proposition 3.21) and hence,
s
forsome
s
m =
rn
0 and let E be a
Conversely, assume that E
EPF.
=
be the separable closure of F in E. F(a , ... ,a ) . 1 r
p >
Then E/F is separable if and only if E
Assume that E/F is separable.
purely i'nseparable, hence E
=
by Corollary 3.12(i).
•
3.25. PROPOSITION.
E
s s
(F :F) = (F :F) s s s
(E:F) 3
as required.
s
m.;;,,0,1.;;;i.;;;r. 1,
Setting
max{m.}, 1, m
Hence EP c
i E {1,2, ... ,r}.
-
F
But E
s
=
EPF
obviously implies
m E = EP F,
3.26. COROLLARY.
Let E/F be a finite separable extension of fields of charac-
teristic
If u , ... ,u
p
!>
0.
hence E
is separable over F.
F
s
is a basis for E over F,
n
1
=
m
•
then so is
m
up , ... ,uP n
J
for any m ~
By Proposition 3.25,
Proof. over F,
1.
and so is a basis.
E
=
EPF.
Therefore uP, ... ,uP again spans E 1
n
Now the result follows by induction on m. •
As a preliminary to the next result, we record the following observation: 3.27. LEMMA.
Let A be a finite-dimensional commutative algebra over a field F.
If A has no nonzero nilpotent elements and no nontrivial idempotents, then A is a field. Proof. J(A)
Since
of A is
0.
A
has no nonzero nilpotent elements, the Jacobson radical Hence
A
is a finite direct product of fields.
has no nontrivial idempotents, hence
A
is a field.
•
But
A
SEPARABLE, PURELY INSEPARABLE AND SIMPLE EXTENSIONS 3.28. PROPOSITION.
Let F be a fteld of charactertstic
77
O and let EIF
p >
and KIF be respectively a separable and purely inseparable field extensions. Then E 0 K is a field. F
Proof.
O ix EE@ K,
Given
we may, by Proposition 1.4.1, write
F
n
Icc0S.
x=
i=l
for some F-linearly independent a.1,, E
E
1,,
1,,
S,EK,l 1. 1, over F.
Stnce E/F ts separable, each a.1,
Now apply Proposition 3.30.
is separable
•
The result above can also be derived as a consequence of the following general assertion due to Artin. 3.33. PROPOSITION.
Let E/F be a finite-dimensional field extension, where F
is an infinite field.
Then E/F is a simple extension if and only if there are
only a finite number of intermediate fields between Proof.
Assume that E/F is simple, say E
mediate field.
Let
f(X)
= F(0)
(E:K') =
K'
coefficients
K and let
Then
K'
and
K' c K
Therefore
= degf(X) = (E:K)
is generated by the coefficients of f(X).
divides the minimal polynomial Because g(X)
e over
and the coefficients of f(X).
F
is also the minimal polynomial of e over K'.
and thus K
F.
and let K be an inter-
be the minimal polynomial of
f(X)
be the field generated by
and
E
Note also that f(X)
of e over F and both
g(X)
f(X) ,g(X) E E[Xl.
has only a finite number of distinct factors in E[Xl with leading 1,
the number of intermediate fields
is finite.
K
Conversely, assume that there are only a finite number of intermediate fields between
E
s imp 1e.
and
F.
Given a,S
To this end, let
y
have an infinite number of y exist y,o
in F,y 'I
E
F
E F
and consider the subfield Ey = F(a+yS). and a finite number of EY.
o, such that
s=
we rieed only verify that F(a,S)/F is
E E,
(y-o)
and therefore a= a+ yS-yS EE. y
EY -1
=
E0 .
We
Therefore, there
Then
(a+yS-a-cSS) E Ey
Thus F(a,S) = F(a+yS), as required.
11
4. GALOIS EXTENSIONS Let F be a field and let G be a group of automorphisms of F. FG
the subfield of F consisting of all elements x
F
such that for al 1 a
a(x) = x
We shall refer to FG as the fixed field of
E
We denote by
G.
E G
CHAPTER 2
80
Let
ca 11 ed the is a
E/F
be a field extension.
E/F
of the extension and is written Ga 1(E/F).
Galois group
is a1gebra i c over
if E
Galois extension
of the group
The group of all F-automorphfsms of
and
F
is
E
We say that
is the fixed field
F
Gal(E/F).
4.1. PROPOSITION.
Let
be an algebraic extension.
E/F
Then the following
conditions are equivalent: (i}
E/F
(ii)
E/F
(iii)
E
is Galois is normal and separable is the splitting field of a family of separable polynomials over F
Proof.
(i) • (ii):
over f.
Let a e
in E.
f
fixes g
g = (X-a
E F[Xl.
Since
In this way each rr 1
}(x-a) ... (X-a ).
Hence
2
Y'
Gal(E/F)
E
Since
E/F
Let
be all distince roots
is extendible to an auto-
permutes the
and hence
a.1,
is Galois, we deduce that gJf
proving that f =
g.
permutes the a1,. transitively,
as required.
Since
is separable,
E/F
{aJ
of separable elements.
over
F.
norma 1 ,
Y'
2
and, by the definition of g,
fig
a.EE, 1,
(ii)• (iii):
a,a , ... ,a
= 1
is irreducible, the group Gal(E/F)
g
hence each
1,
and let a
are distinct and lie in E.
By Corollary 1.18, each a e Gal(E/F)
E.
morphism of
E
be the minimal polynomial of a
f
We must show that all roots of f
E be an algebraic closure of of
and let
E
Then each
E
Ne denote by
is generated over
hence
E,
E
by a family
f.1, the minimal polynomial of a.1,
fi is a separable polynomial over
f.1, splits over
F
F.
Since
E/F
is
is the splitting field of the family
{f1,.}
of separable polynomials over F. (tit)• (i):
Assume that
separable polynomials over a EE -
F.
Since
a
homomorphism 0 : F(a) sition 2.9, Galois.
is the splitting field of a family
E
F.
Then
E/F
{f.Ji 1,
is normal and separable..
EI} of Fix
is separable, it follows from Lemma 3.6 that there is an --+ E
for which 0(a) f. a.
F-
On the other hand, by Propo-
0 is extendible to an automorphism of E,
proving that E/F is
•
Owing to Corollary 3.22, any algebraic extension may be obtained by taking a
GALOIS EXTENSIONS
81
separable extenston followed by a purely l'nseparable extension. cannot reverse the order of the tower.
Usually, one
However, there ts an important case when
it can be done. 4.2. PROPOSITION. (i)
Let
be a normal extension and let
E/F
is purely inseparable and
EG/F
(i i l lf
F
are ltnearly dtsjotnt over F. E
~
i1
®
F 8
p
n
Proof.
If charF
=
0,
Then
in
F
then
E,
= F s EG and
E
F
s'
EG
In particular,
then AP e
(iii)
= Gal (E/F).
ts separable
E/EG
ts the separable closure of
B
G
B
n
Er; = F
and some n
p
~
1.
= F B and hence, by Proposition 4.1,
E
which clearly yields the result.
F
for some prime
F
then
and
Assume that charF
= p ~
0.
EG
=F
Owing to Propo-
sitions 3.29 and 3.19, we need only verify (i). Since Gal (E/F) = Gal (E/EG), for all
a
E:
Gal (E/EG},
then 1i
it follows that if A E E is such that cr(A) = A E
i 1•
a : Ea -- E be an F-homomorphism.
i.
is normal.
ts Galois and therefore, by
E/EG
Let E be an algebratc closure of
Proposition 4.1, is separable.
automorphism of
Hence
E
and let
By Corollary 1.18, cr is extendible to an
Its restriction to E is an F-automorphism of E since
Hence cr(x)
= x
for al1
r'.
x E EJ
E/F
and the result follows by virtue of
Proposition 3.19. • We next provide a useful characterization of finite Galois extensions. 4.3. PROPOSITION.
Let E/F be a finite field extension. IGal(E/F)I
~
Then
(E:F}
and IGal(E/F)I if and only if Proof. Assume that separable. whi'le since
E/F
(E:F)
is Galois.
The first inequality follows from E/F
is Galois.
Then, by Proposition 4.1,
Since E/F is separable, we have E/F
Gal(E/F) ~· (E:F)s
is normal we have
(E:F)
s
=
E/F
(E:F)
~
(E:F}.
is both normal and (Corollary 3.12),
(E:F} 5 = IGal (E/F) I - Hence IGal (E/F} I= (E:F).
CHAPTER 2
82
JGal(E/F} I = (E:F).
Conversely, assume that Since
(E:F) 8
= (E:F),
IGa 1 (E/F} I = (E:F) B = (E:F).
is separable by Corollary 3.12.
E/F
must also be normal,
E/F
Then
Since
JGal(E/F)J
=
Thus
is Galois, by Proposition 4,1, •
E/F
(E:F} 8 ,
As a preliminary to the next result, we prove 4.4. LEMMA.
Let
ger
n;;;. 1
E/F
is finite and Proof.
be a separable extension.
E/F
Assume that there is an inte~
such that every element a of E is of degree .;;; (E:F}
over F.
Then
.;;;n.
Let a EE be such that the degree
We claim that E = F(a). such that Bi F(~),
n
Deny the claim.
(.F(a} :F)
is maximal, say
m.;;; n.
Then there exists an element BEE
and by Corollary 3.32, F(a,B)
=
F(y)
for some y E F(a,B).
!>
m over
But from the chain F ~
we see that tion.
(F(a,B) :F}
> m
F(a)
~
F(a,B)
whence y has degree
a contradic-
F,
•
We are now ready to prove the followtng classical result. 4.5. PROPOSITION. (Artin). automorphisms of F.
Let F be a field and let G be a finite group of
Then F/~ ts a finite Galois extension,
G = Gal(F/FG}
and
Proof.
In view of Proposition 4.3, it suffices to verify that
finite Galois extension and maximal set of elements of crE
G
then
(F:FG}o;;;JGJ. G
is a
F/FG
Let ~EF and let cr, ... ,cr 1
such that cr 1 (a}, ... ,cr. !' (a)
(crcr 1 (a), ... ,crcrr(a)) differs from
are distinct.
(cr 1 (a), ... ,cr:r>(a)) 1
If
by a permu-
tation, s i nee cr is injective and each crcr .(a} E fo (a), ... ,cr (a)}. ~
bea
!'
!'
Therefore
· a is a root of the polynomial
n (x-cr .(a)) i=l !'
f(X) =
and for any cr E G,
f
=
f·
~
Thus the coefficients of f
G lie in F.
Moreover,
GALOIS EXTENSIONS
f
ts separable.
It follows that every element
rable polynomtal over Fa of degree into linear factors in F.
~Jal.
83
a.
of
is a root of a sepa·
F
Furthermore, this polynomial splits
Thus F/Fa is both separable and normal, hence F/Fa
is Galois, by Proposition 4.1.
By Lemma 4.4, we have
(F:fi)..;;
Jal
and the
result follows.• Let E/F be a finite Galois extension and let a be its Galois
4.6. COROLLARY. group.
Then, for any subgroup H of
a, there exists an intermediate field
K
such that E/K is Galois and H = Gal(E/K). Proof.
Put
~ and apply Proposition 4.5. •
K =
We close by exami'ntng some properties of automorphisms of field extensions.
s be a set and
Let
a field.
F
to be linearly independent over
The maps F
:\f 1
then all
O.
:>... = 'Z,
1
f.'Z, : s----+-
F, 1..;; i
~
are said
n,
if whenever we have a relation +, •• +:X.f =O n n
(LE F) 'Z,
Recall that a monoid ts a set G, with an associative binary
operation and having an identity element.
x1 ,x 2 , ... ,Y''n be dtsttnct homomorphisms from a to the multiplicative group F* of a field F. Then x1 , ••• ,x,_. are
4.7. PROPOSITION. (.Artin}.
monoid
G
Let
1inearly independent over F.
x1 ,x 2 , ... ,y''n x = nr A ·X. for If
Proof. assume that
i=2
1
independent.
are linearly dependent, then we may harmlessly some '>..i e
F
and that
i, -i,
x2 , ••• ,x,_.
are
1i nearly
Then e a
for al 1
g
for all
g,x
( 1)
Replace g by gx in (1):
Multiply (1) by x (x} 1
Ea
and subtract the result from (2}: for all
Hence
n
r
i=2
:>..
.(x .(xl -i,
since x2 , •••
-i,
(2)
x1 (x) )xi
=
O and so \Cxi(x} •
,x,_. are linearly independent.
x1 (x))
Since xi f: \
0, 2
~
for i f:
i
g,x ~
1,
n,
it
E
G
84
CHAPTER 2
follows that ;..,
-z.,
O for if:
=
and (1) reduces to the form x (g}
1
1
0. which is
impossible. • Let f ,f , ... ,f
4. 8. COROLLARY.
1
into another field F. Proof.
Put G =
Then
f
n
Since
E J...f.
i=l
=
, •••
I
,f
G
E
are linearly independent over F.
n
and denote by X·-z., the restriction of f.-z.,
E*
x.-z., is a homomorphism of
each
be di sti net homomorphisms of a field
n
2
to G.
Then
into F* and the X·-z., are obviously distinct.
0 obviously implies
-z., -z.,
n
E ;.. .x.
i=l
=
0,
the result follows by virtue
-z., -z.,
of Proposition 4.7. • 4.9. COROLLARY. (E:F)
= n
1,
and if bf- O,(b-l)q=
(a-b)q = aq - bq = a - b, (ab}q = clbq = ab,
Let
be a field of q
F
=
pm
elements.
Then, for any
there exists a field extension K of F of dimension
namely the
n,
splitting field of n
I' Moreover, any two field extensions of Proof.
of dimension n are F-isomorphic.
F
Assume that E is a field extension of F of dimension n.
has qn elements and, by Proposition 5.1 (ii), the elements of
E
ly the roots of f1
of
F
n
-
r
splitting field of
r
over F
X
of dimension n
n - X
X
i'n an algebraic closure of
n - X
over
F.
r
e1ements and therefore has degree n The field of
q
are preciseis a
E
This shows that any two field extensions
are F-isomorphic.
are roots of
F
Therefore
IF • p
Conversely, the splitting field of
over F is the same as the splitting field of I'
all elements of
E
Then
n
n
- x. over
elements is denoted by
-
X
over
IF,
p
since
Hence it consists of precisely q F.
n
•
Jfj' •
q
For
q = p
this agrees with
FINITE FIELDS
the notatton IF
.87
i'ntroduced earlter.
p
5.3. PROPOSITION.
If q = pn,
where p
is a prime and n;;;,
a Galois extensi'on with cyclic Galois group of order n, beni us automorphism x Proof.
:-+
xP
The extension
is
generated by the Fro-
of IF • q
IF/ffi'p
being a finite extension of the perfect field
must be separable (Proposition 3.18}.
JFP
then IF/IFP
1,
On the other hand, IF/IFP
by Propositton 5.l(ii} and hence is Galois by Proposition 4.1.
is normal
Therefore, by
Gal (1E' /IF ) is of order n. Since xP = x for all X EIF , p q p the Frobenius automorphism a. of IF is an IF -automorphism of JF • Finally, q p q since Proposition 4.3,
!'
a.!'(x) = and thi's is the identity for that
a.
has order n.
r = n
xP
and not the identity for
Gal (1E' /IF ) = q p
Thus
for all 1 ..;
as required.
x e IF
q
r < n,
we see
•
All subfields of IF m are of the form IF n where nlm and, p p n there is exactly one subfie1 d of IF of order p • Furthermore,
5,4. PROPOSITION.
n Im,
for any if nlm,
min,
pm
then IF
,IIF' n
generated b~ Proof.
a.~,
is a Galois extension with cyclic Galois group of order where
Assume that nlm.
m
xJ? -
ts the Frobenius automorphism of IF m' n p Then any root of xP - X is also a root of
a.
n
Hence, by Proposition 5.l(ii}, an the roots of xJ? - X form a n Conversely, if F is a subfield of IF m unique subfi e 1d of IF m of order p . p n . p and IFI = p, then X.
m=
(JF m:F) (F:IFP) = p
provi'ng that nlm. (1E' m:IF n } = min and, by Proposition 5.3, IF rrf/Tri' n is Galois p p . p p n hence its Galois group is of order min. Since a.n is of order min and a.
By the above
fixes IF
elementwise, the result follows. • n We now turn our attention to roots of unity. p
ment
Let
F
be a field.
A E F is said to be an n-th root of 1 (or of unity) if
charF = p > 0
and
n = pms
with
(s,p) = 1,
then
"An = 1
An= 1. implies
An eleIf m (A 8 )P =
88
CHAPTER 2
or As
1.
=
Hence, if '.\
is an n-th root of unity then '.\
is an s-th root of
We shall therefore assume in what follows that the characteristic of the
unity.
field does not divide n. 5.5. PROPOSITION. (i)
Let
be a field and let
F
Then-th roots of unity in
(ii)
r -1
If
unity in
F
Proof.
F
n
> 1 be such that charF l
form a cyclic group whose order divides n.
sp 1its into li'near factors over
F,
then the n-th roots of
form a cyclic group of order n. It is clear that (i} is a consequence of (ii).
To prove (ii), note
that then-th roots of unity in F certainly form a group, say G. (r-1)
1
=
Hence
contains exactly n n-th roots of unity, i.e.
F
a.
Let m be the exponent of
r - 1,
of 1i C,
Since
nx"- 1 r 0, it follows from Proposition 3.1, that all roots of
are simple.
n.
hence
JaJ < m.
r
- 1
Jal
= n.
Then m < n and all elements of G are roots This shows that m = n and therefore
G
is cyc-
•
Let F be a field.
We say that :\
if the order of A in F*
F is a primitive n-th root of unity
E
is precisely n.
Given n
> 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 i7' - 1 over
By Proposition 5.5, if charF
F.
n-th root of unity over
always exists.
F
n-th root of unity over F,
i!7'
over F.
then F(s} n
The field F(s) n
l
n,
then a primitive
Furthermore, if sn i's a primitive is obviously the splitting field of
is ca1led then-th cycZotomic extension of E
By the maximal cycZotomic extension of F we understand the field obtained from F
by adjoining n-th roots of unity for all
5.6. COROLLARY.
Let
F
n
> 1.
be a finite field of q elements, let m > 1 be an
integer coprime to q and 1et
s be a primitive m-th root of unity over
F.
Then (i) (ii)
(F(s):F)
is equal to the order of q modulo m
Gal (F(s)/F) Proof.
is cyclic generated by s '--'" sq
It is clear that (i) is a consequence of (ii).
Let a be the n
Frobenius automorphism of F.
If q = pn, p prime, then an(s) = sP = sq.
FINITE FIELDS
89
Therefore (it} is a consequence of Proposition 5.4. 5.7. PROPOSITION.
E and
let
sions of
Then both Since
noti'ce that
n ~ 1
be a field, let
F
be such that charF
l
and
n
K be respectively then-th cyclotomic and maximal cyclotomic exten-
F.
Proof.
Let
•
charF
E/F
and
K/F
l
i1 -
1
n,
are Ga 1 ois exte.ns ions. has no multiple roots.
I' -
is the splitting field of
E
1
x" -
ting field of the family of all such polynomials Proposition 4.1).
over
and
F
1
with
It remains to K
is the split-
charF
l
(see
n
•
The following observation is often useful. 5.8. LEMMA.
Let
be a Galois extension and let
E/F
µ ,µ , ... ,µ 1
of
E
such that each
CT
E
Gal(E/F)
permutes the
2
µ , ... ,µ. n
l
n Then
be elements
(X-µ } (X-µ ) ... (X-µn) 1
is a monic polynomial over
2
If, furthermore,
F.
an integrally closed integral domain (X-µ ) ... (X-µ ) 1
and each
R
is a monic polynomial over
is the quotient field of
F
µ.
is integral over
1,,
then
R,
R.
8
Proof.
f(X) = (X-µ 1 )
Put
(X-µn} .
•••
Then
f ( X,1 = x.n - s 1 x.n-1 + s 2 x.n-2 + •.• + (-l)ks· kx.n~k +
, .. + ( - l)nsn
where s = µ + µ + .•. +µ 1
s
s
Hence
CT(s .) .
1,,
=
s.
1,,
for all .
2
n
CT
is Galois, we deduce that each al hypotheses, each integrally closed,
s.
1,,
•
1
=µµ 1
2
2
+µµ 1
=µµ ..• µ 1
E
2
Gal(E/F) s.
1,,
E
F,
is integral over
3
n
+ ... +µ
n-1
,µ
n
n
i
E
{1,2, ... ,n} .
f(X)
E
F[XJ.
and all hence R,
hence belongs to
Since
E/F
Under the addition-
R since
R is
90
CHAPTER 2
('lL/n'll,)*
In what follows we write
order of this group is denoted by
for the unit group of the ring
¢(n),
where
¢
('ll,/n'll,}* = {µ+n'll,Jl,;;;; µ < n, (µ,n) = l}
Note that
and that
n.
number of primitive n-th roots of unity is equal to
¢(n).
n ~ 1
be a field, let
F
r = ¢(n},
¢(n)
be all primitive n-th roots of unity over
coincides with
In particular, the
charF X n
be such that
The
is known as the Eiler function.
the number of generators of a cyclic group of order
Let
'll,/n'll..
F.
and lets , ... ,s, 1 r Then, by Proposition
5.7 and Lemma 5.8, r
f -J (X-s ,)
i=l
1,,
F and hence over F.
is a manic polynomial over the prime subfield of charF = 0,
more, if
Further-
then by Lemma 5.8, 1> (X} E 'll,[X]
n
We shall refer to
cJi/Xl
5.9. PROPOSITION.
Let
let
n
to a subgroup of Ga 1 (Fn/Fl cJi n (X)
(ii} (iii}
2e
(F :F) n
=
Let
a E Gal(F /F), n
a
be such that
n;,,, 1
Then
F.
F. charF X n
Gal (F /F) n
and
is isomorphic
Furthermore, the foll owing conditions are equ iva 1ent:
('11./n'll.)*. ('11./n'll,)*
is irreducible over
Proof.
more,
be a field, let
F
be the n-th cyclotomic extension of
F
( i}
as the n-th cycfotomic polynomial over
F
¢(n) s then
be a primitive n-th root of unity so that
0(2)
=
sµ
for some
is uniquely determined by
µ
1,;;;; µ < n
modulo
n
with
F
n
If
= F(s)-.
(µ,n) = 1.
and may be denoted by
FurtherCT • ).l
Then the map
l
Gal (Fn/F)->- ('11./n'll.)* CT
).l
t-->- ).l
is obviously an injective homomorphism. above and Propositions 4.3 and 5.7. Let
E/F
cyclicl if
be a field extension.
E/F
is Galois and
+ n'li. The second assertion follows from the
• We say that
Gal(E/F)
E/F
is abelain (respectively,
is abelian (respectively, cyclic).
91
FINITE FIELDS
5.10. PROPOSITION. extension of
Let Then
F.
Proof.
a ,a
Let
1
be a fteld and let
F
E
Ga 1 (E/F).
But if
E.
µ. ai(s) = s ~
required. Let
n
(µi,n)
with
Si nee
is generated over
E
(a a }(s) = (a a )Cs)
unity, it suffices to show that unity in
be the maximal cyclotomic
ts abelian.
E/F
2
E
1
2
2
is the order of Therefore
= 1.
where
1
s, ai(s)
s
by roots of
F
is a root of
is also of order
a 1 a 2 (s)
a a (s) 2
hence
n,
as
1
• n1 n2
n = p
p
1
2
nk ••• pk be the canonical decomposition of
n.
·Then
and therefore
For this reason, to determine the isomorphism class of ly assume that
n
(Z:'./nZ:'.)*,
we may harmless-
ts a power of a prime.
5.11. PROPOSITION.
(i)
If
ts an odd prime, then
p
(Z:'./pnZ:'.)*
is cyclic of
order pn- 1 (p-1} (ii)
Both
(Z:'./2Z:'.)*
Proof. (i} Since pn-l(p-1).
are cyclic, and if
(Z:'./4Z:'.}*
n;;. 3,
then
cp(pn) = pn-l(p-1) (for any prime p), (Z:'./pnZ:'.)* is of order
ts a direct product of its subgroup H of order n-1 pn-l consisting of the elements which satisfy d? = 1 and the subgroup K order
K
Hence
and
p - l
(Z:'./pnZ:'.)*
of the elements satisfying
d?- 1
are coprime, it suffices to verify that both If
n = 1,
then
(Z:'./pZ:'.}* = K
= 1. H
Since the orders of and
K
and this is cyclic by Proposition 5.l(i). 2
m + pn'll, E (Z:'./pnZ:'.)*. kp-l
which implies
k + pnZ:'. EK.
Also
(ef
and
are cyclic.
v-1
Hence we can choose an integer m such that m + pa'., m + pa'., •.. ,M n-1 distinct in Z:'./pZ:'.. Set k = rrf Because and
H
of
n-1 p-i
}
Since
( n)
= mcp P
- 1 (mod pn)
+ pa'. are
CHAPTER 2
92
n-1
= m (mod
k = rrl
p)
the elements k + p'll,, k 2 + p'll,, •.• ,kp-l + p'll, are distinct.
are distinct.
This implies that the order of
But the order of
K
is
hence
p - 1,
n
Hence also
is precisely p - 1.
k + p '1l,
is cyclic generated by
K
+ pn'll,.
k
To prove that H is cyclic, we may assume that n;;.;. 2 since otherwise H = 1 n.
Then H is a direct product of t ~ l
cyclic groups of order
Hence the number of solutions of the equation :J? fore suffices to show that the number of integers
sP = 1 (mod
pn}
then, since
sP
does not exceed p.
=s
(mod p},
p
is
= 1, x EH
ni;;.;. l.
1,,,
rt there-
pt.
satisfying
s, O < s < pn,
Now if s satisfies these conditions,
we have
s
=1
(mod p}.
Then if
s
f
1,
we may
write s
= 1
+ ypf + zpf+l
(1 ~·
f ~n-1,0 < y < p, z;;.;. 0)
in which case + (il(y+zp}pf + {1i) (y+zp) 2pd
sP = 1
+ ... + (y+zp)pppf - 1 + ypf+\mod Pft2)
If
sP
= 1 (mod
pn}
and
yp
so y = 0 (mod p}
this gives
f < n-l., f +l
= 0 (mod
contrary to O < y < s = 1 + yp
n-1
(ii)
The order of
('ll/2n'71,}*
is
)
Thus, if
p.
H
satisfies ,
1 < s < pn
This gives altogether at most
, 0 < y < p.
solutions including 1 and proves that
p
,~2
p
is cyclic. If n
¢(2n) = 2n-l.
or
=
2,
orders are 1. and 2, respectively, and there ts nothing to prove. n;;.;.
3.
We first claim that there are four distinct elements
satisfying x 2
l; tf sustained, it will follow that
=
product of at least two distinct cyclic groups n-1
a = l + 2 3
n-1
, a = -1 + 2 4
, x .. 1,,
=
n
a • + 2 '71,, 1,,
t
1.
('ll/2n'71,)*
Put a l
Then the
then these So assume that
x E
('ll/2n'71,)*
is a direct = 1,
a = -l, 2
.
xi are distinct and
FINITE FIELDS
xi
satisfy
93
which substantiates our claim.
= 1,
r
('ll./2nZ)* is a direct product of at least two cyclic groups
Since
the order of
is
('1l,/2nZ)*
2n-i, we see that, if x
E
(Z/2nZ)*,
or, what is the same thing, if m is an odd integer, then
m2
n-2
then
1 and
x2
n-2
=
= 1 (mod 2n).
2n-3 To complete the proof, we are therefore left to exhibit an x such that X f 1. n7/. 52n-3 = 5 =/- 1(mod 2n) Put x = 5 + 2 £ Observe first that, if n = 3, then n-3
but 52 k
- 1 (mod 2n- 1 ).
Now let
f;;.
be the largest integer
such that ?f-3
5Then we have k(3} y
3 and let k(f)
ts odd.
2.
=
k
= 1 (mod 2 )
Also for any
2f-3 3 we have 5
f ~
=
k(f)
1 + y2 ·
where
This implies that 52(f+l}-3 = (52!- 3/ = l + y2k(f)+l + y22k(f)
whence k(f+1)
~
so k(f)
k(f),
~
2
if f
52(!+1)-3 where
z = y
+ 2kCfl-\ 2 ts odd.
we deduce that k(f)
3.
Then the relation shows that
1 + z2k(f)+l
=
Thus
1 for al 1
= f -
~
f;;.
k(f+l) 3.
= k(f)
+ 1.
Since
k(3) = 2,
Hence
n-3
52
11 (mod 2n}
for n ;;. 3
and the result follows. • It will be shown below (Proposition 5,131 that Gal(a)/(f)}
~
('lL/nZ)*.
With
this tn mind, we now derive 5.12. COROLLARY.
Let n
k = p , p
a pri'me, let
F
be a field with
and let Fn be then-th cyclotomtc extension of F. unless p
=
2 and
k
~
3,
charF
rp
Then F/F is cyclic n
in which case Fn /F is either cyclic or Gal(Fn /F)
is a direct product of a cyclic group of order 2 and one of order 2t- 2 where t
< k.
Furthermore if p is odd if p =
2
94
CHAPTER 2 Proof.
Direct consequence of Propositions 5.9 and 5.11.
We now turn our attention to the case where 5.13. PROPOSITION.
((l)n :(Q) =
(ii }
¢
n
F = (Q.
be then-th cyclotomic extension of
(Q
n
be then-th cyclotomic polynomial over
n(X)
( i)
Let
•
and let
aJ'
variables, has a non-
r.
5.22. PROPOSITION. (Chevalley). Proof.
xF
Er
f(- End(S), sf--->-£
be the regular representation of
s
R
By the trace (norm,characteristic polynomial) of s ES we understand the
trace (determinant, characteri sttc po 1ynomi a1 of
The trace and norm of
£ )• s·
s
will be denoted by
respectively. Now assume that
E/F
is a finite field extension.
dimensional F-algebra and so for any ;\EE,
Then
E
is a finite-
we may define the trace (norm, char-
acteristic polynomial) as above. 6.1. LEMMA.
Let
E/F
be a finite field extenston, let :\EE and let f
be
the characteristic polynomial of :\. (i)
is a root of f
;\
:\ over (ii)
and, if
E =
F(:\),
then f
is the minimal polynomial of
F.
If K/F is any field extension in which
f
splits into linear factors,
n
f =
il (x-:\.), then
i=l
1,
n Z :\. i=l
1,
NORMS, TRACES AND THEIR APPLICATIONS
Proof.
Let
(i)
be the matrtx for multi'plicatton by ;\ tn
M
respect to a given F-basis of E. This implies that If E
=
F(>..),
f(">.)
and
0
when
f(X) = g(X)m,
(ii) NE/F(">.)
=
(iii} TrE/F( ">.) Proof.
Let g(X)
over F,
">.
hence the assertion.
when
and let f(X)
Then
[Nx;/1.) ]m = m [TrK/p{:\)
l
Properties (ii) and (iii) follow from (i) and Lemma 6.l(ii).
is a basis of E over F.
Let M = (aih)
tn K with respect to the basis 'A(yizj)
To
be a basis of K over F and let {zjl1 < j
Ea.,n-1 )p -1 1 n-
(b. E Z)
o
Let j
1,,
be the minimal index for
Then the number
J
s = (bl
= {b
are divisible by p.
b~s
r
i=o r
and not all the
is not divisible by
(-l)na 0
exists r ER such that
which
dividing DF;,)
p
n-1 -1
+ ... + bn-i'>--
)p
=
r- ((b/p) + (b/p);, + ... + (bj_/p):\
j-1
)
ts an algebraic integer and so ts also _.
s1 -
n-1-1 _
bi
p
- SA
n-j-1
.n A (b j+ l + b j+ zil +
-
+b
n-1
"n-j- 2) P
1
rt follows that n
(
p NF/(!') s 1
Since
p2
,!' NF/,),
)
= NF/cr:)(p1\ )
(.
n-1 ) = bl n n-1 Fft;(;,)
= NF/IJ) b}'-
the latter impli'es that plbj'
a contradicti'on. •
We close this section by providing a number of examples of the calculation of a discriminant. 7.18. PROPOSITION. n
> 2,
Let E/F be a finite separable field extension of degree
let A EE be such that E = F(X)
of :\ over
.F
and assume that the minimal polynomial
is
f(X)
=
i1
+ax+
b
rhen DE;p(X) = (-1 t(n-1)/2(nnbn-1 + (-1 t-l(n- t-1an)
Proof. µ = f' (:\) ,
By Proposition 7.8, we have
DE/F(A) =
(-lt(n-l)/ 2NE/F(f'(J,,)).
Putting
DISCRIMINANTS AND INTEGRAL BASES
0 implies n11.n-1
since 11.n +a\+ b
11. = -nb(µ
The minimal polynomial
of µ
g(X)
-1
b
1
= -na - nb\- .
129
It follows from this that
+ {n-l}a) -l
over F is therefore the numerator of .
-1
f(-nb(X+(n-l)a))
Hence we must have
The norm of
is
µ
ttmes the constant term of this po lynomi a 1 , i . e.
(-1) n
NE/F ( µ )
= n
which clearly yields the result.
nbn-1 + (-l)n-l(n-l)n-l~n ~
•
We now turn our attentton to quadratic fieZds, i.e. algebraic number fields of degree two.
An integer d
is said to be square-free if d
E 'll.
not divisible by any square except 1. d = -1
or
Id I
Thus
and d is
is square-free tf and only if
d
is a product of di sti net primes.
a square can be uniquely written in the form n We refer to s
t
Every integer n which is not
= m2 s
where s
is square-free.
as the square-free part of n.
For convenience, we assume that all quadrati'c fields are embedded in ..) = 2A-l = la,
it follows from Proposition 7.8 that
-:"'-.-.··,.·
DISCRIMlNANTS AND INTEGRAL BASES
Since
131
is square-free, i't follows from Proposttton 7.12 that
d
and
d(F} = d
l,A is an integral basis of F.
that
Assume that
d
= 2(mod
g(X) = X2 + 2X + l - d. f'(./a} = 2tl'a,
4}
or
Since
= 3(mod 4).
d
Let
f(X) = X2
-
Id and
i's the minimal polynomial of
f(X)
and
d
we have DF;tf)la} = -NF;fI}21J} = 4d
Let p
d.
be any prime dividing
Proposition 7. 17 wtth respect to and
are the same.
d(F)
by Proposi tton 7; 12.
If d
Proposition 7 .17 for
p,
satisfies the hypotheses of
hence the highest powers of p
= 2(mod
d
= 3(mod
p = 2.
Since
If
f(X)
Then
4},
4),
then
4d
d(F) = 4d
is even, hence
d
g(X}
then
dividing
sattsftes the hypotheses of
Id - 1 is a root of
and
g(X}
DF/a 1
I
2
and
x2
-
2
+bla=a
2
2·
+ b Id and
1
1
dy 2 = ± m,
then
a
a
1
2
1
>a ""'b 2
l + 15
are positi've solutions with
>b
l
2
+ b ./J
3 + /5
1
x2 !>
-
a + b la need not imply
!>
2
d
=2
or
3(mod 4)
2
5y 2 = ± 4,
l + /5
Our next result provides a useful charactertza tton of possible to treat the cases
(_2)
2
+ b la are, positive solutions of
Indeed, consider the double equation
> b •
are
la and a 2 + b 2 /if are two solutions of (1}, then
We warn the reader that tf a
b
b
and
The positive solutions will be ordered by the size of + b
1
will be called positive if a
and
but b Ed'
Then l
= b
= 1.
A1though it is
=l(mod 4)
d
2
3 + /5
simultan-
eously by applying Remark 8.2, we consider them separably for practical purposes. 8.5. PROPOSITION,
where (ti)
(i}
If
d
=2
or
3(mod 4), then
a + bid ts the minimal positive solution of x 2 If
d
= 1 (mod
dy 2 = ± 1.
-
4}, then Ed = (1/2}-(a+b./J)
where
a + bid is the minimal positive soluti'on of x 2
Proof. satisfying
(i)
Owing to Lemma 8.l(i), we may write
-
dy 2 = ± 4.
Ed= a+ bid with
a,b E 'll
UNITS IN QUADRATIC FIELDS a: 2 ~ dii2 =
Since Ed > 1,
solution of x 2
± 1
Lemma 8.4(ii) implies that a > 0
is a positive solution of x 2 -
-
139
dy 2 = ± 1.
Hence a + bid
and b > 0.
Let a:
l
+ b Id be any positive l
Then a + b Id> 1 and, by Lemma 8.l(i),
dy 2 = ± 1.
l
1
+ b/d E U(R}.
a:
Therefore, by Lemma 8.4(i), a
Because (ii}
Ed > 1,
+
l
this shows that
a
b
1
Id=
+ b Id ;;,,
1
for some n;;,, 1
End
l
+ bid,
a
as required.
Apply arguments of (i} with Lemma 8.l(t} replaced by Lemma 8.l(ii). a Observe that if N(Ed) = 1, then
if d = 2 or 3(mod 4}
and
= 1 for all units
N(u)
u
of
Hence,
R.
then by Proposition 8.5(i), the follow-
Ed= a+ bid,
ing holds: (a)
a+ bid is the minimal positive solutl'on of
according to whether N(Ed)
1 or N(Ed)
=
=
x2
-1.
-
dy 2
= 1 or x 2
In particular,
-
N(Ed}
dy 2 = 1
= -1 if
and only if the equation x2
-
-1
dy 2
has no solution in Z. Similarly, if d = 1 (mod 4)
and
Ed= (1/2) (a+b/d),
then by Propositton
8.5(til, we have (b)
a+ bid is the mini'ma1 positive solution of x 2
-
dy 2 = 4
or x 2
-
dy 2 = -4
according to whether N(Ed} = 1 or N(Ed) = -1. In particular,
N(Ed) = 1
ff and only tf the equation x2
has no solution in
Assume that d = 2 or
(i}
with a ,b EN.
Then the sequence
l
dy 2 = -4
Z.
8.6. PROPOSITION. l
-
3(mod 4}
a + b Id = (a +b Id) n
n
gives all positive solutions of x 2
1
-
1
dy 2 = ± 1.
n
and let
Ed= a 1 + b/d
(n ;;:> 1)
CHAPTER 2
140
Assume that d= l(mod 4)
(it)
and let Ea= (1/21Ca/b 1 v'd)
with a ,b EN. 1
1
Then the sequence (n;.. 1)
gives all positive solutions of x 2 Proof.
of x 2
n
;;i,
and
U(R}
E
a
n
+ bn IJ, n;.. l,
Moreover, by Lemma 8.l(i) and 8.4(ii), they are positive solutions If z = a + bid ts any positive solution of x 2
dy 2 = ± 1.
-
then z
Because Ed is of infinite order, all the
(i}
are distinct.
dy 2 = ± 4.
-
z
1 so that by Lemma 8.4(i},
l>
z.
= E~ = an +
-
= ± 1,
dy 2
bnld for some
1.
(til Apply arguments of(.,'). with Lemma 8.l(i'). replaced by Lemma 8.l(ii). • We now apply the result above to provide a rather crude method for calculating the canonical fundamental unit Ed of R. cases d
=2
or 3(mod 4)
8.7. PROPOSITION.
and d
= l(mod
It will be convenient to treat the 4)
separately.
Assume that d = 2 or 3(.mod 4}.
In order to calculate
Ed = a 1 + b 1 Id it suffices to write down the sequence
first number db 21 Proof.
1
db 2 , b = 1,2, •.. ,
of thi's sequence which differs from
Let an+bnld
(a ,b
±
1
E
lN)
and to stop at the l
by a square a 21 •
be the sequence defined in Proposition 8.6(t}.
It
suffices to verify that b
n+l
l>
b
for all
n
n
;;i,
Indeed, in this case Ed can be characterized as the unique positive solution a + bid
of x2
-
dy 2 =
±
1 with smallest possible b.
an+l + bn+lid = (a 1 +b 1 la)(an +bn Id)·
we have bn+l-=ba +ab >bn 1 n 1 n
as required.
•
Because
l
UNITS IN QUADRATIC FIELDS
To tllustrate the case d = 2 or 3(mod 41, the sequence db 2 , b a
8,
= 1
1,2, ...
=
we see that
8 +
3./T
stmilar argument shows that
Ed=
1 +
12,
7,28, 7.3 2
=
let d 82
1,
-
{7,6,2}.
E
If d
so taking b
1
=
7,
=
3 and
is the canonical fundamental unit of (Q(/7).
5 + 2/6
8 + 317
Note that both units and hence
ts
141
ts the canonical fundamental unit of (Q(/6).
5 + 2/6
and
A
1.
have norm
By taking d
2,
=
we obtain an example of a canonical fundamental unit of
norm -1. Turning to the case d = l(mod 4], observe that tt is not true in general that b
n+l
> b
for all
n
n > 1.
Indeed, it will be shown below that
the canonical fundamental unit of (Q(/51. + bn 15
a
n
then b
= b
l
= 2
21 -n(l+/S'n l
=
d
= l(mod
the
= l(mod 4}.
method of Proposition 8.7 tn order to treat the case d Assume that
In order to calculate
4}.
(a ,b 1
it suffices to write down the sequence dcl ,b term db 2
is
Hence, if
This remark explains why we have to modtfy slightly
1.
8.8. PROPOSITION.
(l/2}(1+/51
1 , 2, ... ,
=
of this sequence which dtffers from
JN)
and to stop at the first
by a square a 2
± 4
E 1
1
(if there
1
are two such squares, then as Proof.
bn +l = (l/21(a · · 1bn +b 1 an }
and so b
=ab
2
1 l
8,9. REMARK.
~
as required.
b , l
u.
a
1
posstble
+b
1
d f 5.
and that
If
(1/2}(a +b lcf}
E~ =
then
•
d = l(mod 41
Assume that
1
1
Id is the unique postttve solution of
= ± 4 with smallest
x 2 -dy 2
:xl -
Indeed, if a + bid is a postttve solution of
b . 1
with sma 11 est poss tb 1e b,
2
+b
Accordingly, 1 - db 2 = -4
2
l
+b
1
Id
or a + bid= a
Id, then b l
= b
2
and so db 2 = 5.
l
dy 2 = ± 4
then by the allove
a + bid = a
Hence, i'f a+ bid= a
11
=ab
11
+ b Id
2
2
tn which case
Because d;.,, 2,
1
a
l
=
1.
we infer that
d= . 5,
as claimed. • The remark above together with Proposition 8.7 shows that for any square-free integer d
;i;,,
2 distinct from
positive solution of x 2
-
d=l(mod4),
first term of the sequence 1
= 1 and
a = min{l,3} 1
Ed= (l/2}(a+b/J}
where
a+ bid is a unique
dy 2 = ± 4 with smallest possible b.
To illustrate the case
b
5,
db 2 ,b =
1.
let
dE{5,13,17,69}.
= 1,2, ... , is 5 = l2 + 4 Therefore
(1/2}(1+/5)
= 32
-
If
d=5,
4.
Hence
the
is the canonical funda-
UNITS IN QUADRATIC FIELDS
143
mental unit of (J?(/5), If d 13
=
then the first term of the sequence dh 2 , b
13,
=
32 + 4.
Thus
b
=
l
1 and a
fundamental unit of
1 be coprime integers.
m if x 2 = a(mod m)
for some x
Let a
We say that a is a quadratia residue moduZo E
Z.
Otherwise, we say that a is a quadratia
nonres~aue moduZo m.
For each n e group of Z/mZ.
1et ii be its image in 'll,/mZ and let U(m}
'll,,
be the unit
a
Then a is a quadratic residue modulo m if and only if
a square in U(m}. Let p be an odd prime and let a be an integer coprime to p. the Legendre symbol
(alp}
1, (alrl
of a,
relative to p,
We define
as follows:
if a is a quadratic residue modulo
p
==
-1,
ff a ts a quadratic nonresidue modulo
p
A straightforward verification shows that
(i] (ii}
If a= b(mod p),
then
(ablp} = (alrl(blp)
8.10. LEMMA,
(alrl = (blp)
(in particular,
(a 2 lp)
=
1)
Let p be an odd prime and let a be an integer coprime to p.
is
144
CHAPTER 2
Then
and, in particular, (-lip)= (-a2Jp) = (-l)(p-1)/2
Proof. 0
•
It is now easy to derive the following properties: 8.16. PROPOSITION.
If d = n 2 + 4 for some n
(t)
Ed= (n +
(ii}
If d
=
2 f m E lN,
m2 + 1 with
=m +
Ed
Proof.
( i}
/n 2 +4)/2
Suppose that
and n 2 + 4 are the same.
mE
/m
2
E
and N(Ed}
lN, =
then -1
then +l
and N(Ed} = -1
lN is such that the square-free parts of m2 + 4
Because d ts square-free, if m In,
then
n 2 + 4 < m2 + 4
and so n (ii)
2,
determine
Ed
exclusively in
terms of d. In this vein we quote the fo 11 owi·ng result due to Ri chaud (1866} and Degert (1958)_. 8.18. PROPOSITION,
Let d > 2 be a square .. free integer and write d
-n < r..; n.
divides
If r
n+IJ with Ed =
(n + /al/2
4n,
n 2 + r,
then
N(Ed) = .. sgnr
wtth N(Ed1
=
for
= .. sgnr
[(2n 2 +r) + 2n/a]/r wtth
lrl
=
for
lz,I = 4 for lrl
N(Ed} = 1
1
(except for
d=5,n=2,r=l}
t- 1,4
9. UNITS IN PURE CUBIC FIELDS A rational integer d > 1 ts said to be aube-free i'f d cube of a prtme.
The field
F =
integer and ¥1, is rea 1, ts ca 11 ed a pure aubia field. the ring of integers of the pure cubic field
F =
5,
then
divisible by
p
p,
3 % (3k+1) (3k+2)
x = ± 1
and stnce tt is clear from (4) that
x = -y.
we have
Now assume that p p
(3k+l )(3k+2}
n = 2 (mod3),
and when
xy < 0,
(x,y) = 1.
1.
y = ±
When
153
ak
d with paid but pa+l} d.
is a prime di'visor of
k > 5 :;;. 3k + 2
for
k ;;, 1 ,
which is impossible since
If
so each term in the sum ( 5) is (x,d) = 1.
we reach the same contrad1'ction.
If
Therefore
p = 3,
r°'
=
then since 2 or
5, and
d E {2,5,10}.
The informatton obtained so far shows that (n-2) 2d 2 ( n-2) 2d 3 45 + · 6 7-8 If
0
(6)
d = 10, this becomes
This equation is true for mentioned in the lemma. and obtain
n = 5,
and leads to the first of the exceptions
For other values of
n,
we may divide through by (n-5)/6
154
CHAPTER 2
n2
4n
-
+
= _
6
k+l z: (-l)k(n-6) (n-2)(n-3l(n-4)•12•10 3k-l 3k(3k+ll(3k+2)(3k+3)(3k+4)(3k+5) 71
The highest power of 5 which divides the denominator of a term in the sum is obviously at most 5(3k+5), n2
-
and because 5k+l
>
5(3k+5)
for k;;;, 2, we have
4n + 6 = (.n-2} 2 + 2
= (n 26 l (n-2 l(n-3 l(n-4). • 12 • l 0 3•4•5•6•7•8
= O(mod5) which is impossible since -2 When d
=
2 or 5,
is a quadratic nonresidue of 5.
equation (6) implies the congruence
which is again impossible, by Lemma 9,3, We are therefore left to examine the possibility Y
=
0.
The proof that this
happens only in the case of the second exception mentioned in the lemma is completely similar to what has just been done for the case z
=
0 (the only variation
lies in the fact that d may now have the sole pri'me divisor 2, so that d or 4l.
=
2
The 1emma i's therefore es tab li'shed. •
Let
d >
1 be a cube-free integer and
the rl'ng of integers of 1.
In the second case we have
But then
x,y,z
are not d1'visible by
This i's impossible, since i'f The case in which
p
=
ab[6
3,
then
so that . is fixed by all elements If
K
C
1 -
then every
K , 2
K ' l
then H
l
=E
and
is also the identity map on
K
H c H ,
K
H
cp
If
This proves (a).
K •
>, proving (b}.
H
as asserted.
2
which is the identtty map on
H c H •
elementwise if and only if
K K
then A. is fixed by all elements of
hence by
2
2-
1
fixes
E
Conversely, if :>..EE
2,
a E Gal(EIF}
i.e.
K
of
H
H
nE
cp
E
2
= K2
•
Let EIF be a Galois extension.
We shall sometimes call the group Gal(EIK)
of an intermediate fteld K the group associated with K.
We also say that a
subgroup H of Gal (EIF) · beZongEJ to an intermediate field K if H = Gal (EIK). If K ,K 1
2
are two intermediate fields, then we say that K
l
gate under an automorphism aE Gal(EIF) if
10.2. LEMMA.
cr(K) l
=
and K
2
are conju-
K, 2
Let EIF oe a normal extension and let K E>e an intermediate
field such that K/F ts normal, ~
Then the restriction map
Gal (EIF) -
l
a
Gal (KIF)
1-+
alK
induces an isomorphts~ Gal (EIF)/Gal (E/K) Proof.
Gal (KIF)
Si nee K/F i's norma 1 , the restriction of any cr E Gal (EIF)
is· an automorphism of
K,
homomorphism wtth kernel
hence alK Gal(EIK).
tion 2,9, the result follows. 10.3. LEMMA.
~
E
Gal (KIF).
Since cr
1-+
to
K
The given map is obviously a crlx is surjective by Proposi-
•
Let EIF be a Galois extension, let K ,K l
2
be two intermediate
.""':. , .
-
FINITE GALOIS THEORY
fields and 1et
H. = Gal(E/K,l,i = 1,2. 1,
1,
(i1 Then, for any given (ii)
a E Gal(E/Fl, cr(K l = K 1
is a normal subgroup of
(.i)
Proof. = 1,2.
Bydeftnttton,
Assume that
w- 1 (.y)=o- 1 (y}
K
1, 1
= aH
1
a- 1 .
and when this is so,
Then, for any
y
e K , a-l (y} e K , 2
CITcr- 1 (.y}=y,
i'.e,
,EH,
2
tfandonlyif ,(:x:)=:x: for all
1,
= a{K
H
Gal(E/F)/Gal(E/K)
,Elf.,
2
forall
:!!
if and only if
ts normal if and only if
K/F
Gal (E/F}
Gal(K/Fl
i
2
If K ts any intermediate fi'eld, then
Gal (E/Kl
169
:x:EK., 1,
hence
1
whichmeansthat
l
OTO
-1
E lf , 2
aH a-l c H
Thus
1
-
H = aH a- 1 • 2
K = a- 1 (K l
2
1,
tt follows that
o· 1H o c H, 2
-
1
i.e.
Retracing our steps, we obtain the converse,
1
(i'l'l
and since
2
Apply Ci 1 and Lemma 10.2.
•
We are now ready to record our matn result. 10.4, THEOREM, (Main theorem of fi'ntte Galots theory}.
The map
Kf---+
subgroups of H 1--r
E8.
(iil
If
G
Gal(E/Kl
be a finite
from the set of intermediate fields into the set of
is an i'ncluston.,.reversing bijection with the inverse map given by
B is a subgroup of G,
IHI= (ii il
E/F
G.
Galois extenston wtth Galois group (i}
Let
The extension
K/F
then
(E:rl
and
(G:Bl
= cE8:F)
ts norma 1 if and only if
Gal (E/K) - gi (i ,e, the restri'ction of the i .. th coordinate projection
In fact of
1[ Gi
to
satisfles this condttion,
G}
These
fi
are called canonical
homomorphisms.
(bl
If every
fij
f,. 1,,J
Assume that each Given
j EI,
is injective, then so are all the
let
is injective, and
k EI satisfy
fikh(Yl,
whence by our hypothesi's,
fjkfk(y) =
f/Yl
(cl
If
j EJ
J
for every
i: ,;;; j],
such that
and so
if we define
Now
fikfk(.x)
x
=
x,y E G.
fi(x)
Therefore
= fi(_y)
fj(x)=fjkh(x)
=
= y,
(_i .e, if for every
I
i EI there exists
then
Hm
G, 1,
lim
2cc
G.
+-jEJ J
na.,
G. C there exists a unique g Elim G. such +-j6J J - jEJ' J +-iEI 1, j E J', the j~th coordinates of g 1 and g are equal, In fact,
g1
that, for every
= fAYl for some v
= fk(y)_,
fk(xl
--i'EI
Indeed, given
v
i,j,;;; k,
j EI,
i's a cofinal subset of
fAxl
fi,
E Hm
g = (gi)
with
gi = fij(gJ) (i ,;;; j),
then
g' .-- g is a required
i'somorphism. The following result provides a universal characterization of projective 1imi ts.
11,4, PROPOSITION.
The projective limit
G = 1 im G. of the projective sys tern .,,_1-
CHAPTER 2
176
{G.,f,.\i,j EI} ~
~
of groups (rtngsl sattsftes: .
there are homomorphisms
if lI ts a group (rtngl and if
o. : H-t- G •. with commutative diagrams: ~
~
H ( 1)
(i ,;;;; j)
then there exists a unique homomorphtsm
cr : H __.. G for which all the datgrams
H - -0 ----i,..~G (2) (i
are commutati've (where characteri'zes Proof,
ft ts the canoni'ca 1 homomorphtsml.
Gi'ven
cri(_h} = fi(.a(.h}l,
h EH,
o(h1
E G,
set
cr(hl = (o,tLhl}
Thus
Th i's property
E
nai,
Because of the commu-
r:r : H--+ G i's a homomorphism satisfying
whence the commutativity of (.21 results.
also makes (21 commutative, then er=
I)
G up to isomorphism,
tat1'vity of (11,
Thus
E
fi [o(_h)l = fi[cr'(h}l
If
for all
o' : H
-+ G
h EH, i EI,
0 1•
In order to establish the second asserti'on, assume
and maps ' i : H0 - + Gi
H0
have the property formulated in the first statement of the theorem, extst untque homomorphisms
O' :
H0 -
a and o-0
: G-
H0
Then there
satisfying
fia = 'i
PROFINITE GROUPS
and T/r0 = f,c 00 0 CJ'
177
We tnfer fi = f/rcro and Ti= Tp0o for all
and o0 o are the identity maps of
G
and H0 ,
i EI,
Thus
respectively, provtng that
fs an isomorphism. • We next intrdduce the notion of homomorphism for projective systems of topolo-
gical groups (rings).
Let
be two projective systems of topological groups (rings) indexed by the same directed set I.
A homomoPphiem
{¢. : G.1,· ---+ H1,., i E.I} 1,
¢: X -
Y is a set of continuous homomorphisms
subject to the condition that the diagrams
G·l - - - - - - -fiji ) l l l a . - Gi (i..;;
be commutative. 11.5. PROPOSITION, i; ..;;·j}
... ff.
H·J
l
i ~j} and Y. = {H.,o.,l{,j EI, w 'Z, ~ be two projective systems of topo1ogtca1 groups (rings} and 1et ¢:X --
be a homomorphi'sm.
Let X
j)
= {G.,f .. li:,j EI, 1,
Then, there is a untque homomorphism (which is necessarily
conttnuousl
¢*: lima.- limn. + - 1,
such that, for every
i; E I,
¾-·-1,
the diagram
(3)
Y
178
CHAPTER 2 (f.1, ,P.1,
is commutative
denote the canoni ca 1 homomorphisms).
furthermore, if every ¢.
is injective, then
1,
Proof.
The homomorphisms ¢1,.,i EI,
¢*
In fact
is injective.
induce a homomorphism
~: na.- na. 1,
given by ¢({gi)}
1,
The commutativity of (3} implies that if
(¢i(gi)).
g = (g .}
E
1,
i:hen ¢(g l
E
1imH. ; hence define -
¢*
Wi'th thi's
qi*,
then
If \)J : limG.
p .¢* = p .\)J 1,
1,
¢*(x}=¢*(y),
injective, then
Let
=
{K.,?-. .. \ i·,j 1,
E
1,J
= (¢i)
ltmG. ,
makes (3) commutative for
+-- 1,
thus
¢* = \)J.
=qi.f.(y}]. 1, 1,
i
for all
E."
Therefore,ifevery
and thus
I
x = y.
¢.1,
•
¢
=
(i)
Y
to
be a homomorphism from x to z,
Assume that for all
i
E
Y
J,
and the
sequence
Then the induced sequence 1-+ limG. -
L
lima.
1,
L
limx. -
-1,
1
.,__1,
is also exact, Proof.
By Proposition 11.5,
cation shows that surjecttve. J
To this end, fix
Th.en all the
1,
A. J
¢*
Ker¢*= Imcp*.
A.= {(h.)\(rz.) 1,
E
is
be three projective systems of compact topolog-
I,i..; j}
a homomorphism from
is exact.
¢*
X = {G.,f .• \i,j EJ,i,;;; J'}, y = {H.,a .. \i,j EJ,i,;;; j}, 1, ~ ' ' 1, ~
ical groups (rings} and let \)!
cp.[f.(xl 1, 1,
f/Y1
limH.
-->-
1,
for every i,
then
f/x) =
11.6. PROPOSITION,
z
to
the diagram (3) becomes commutative and
-
and
-qi
to be the restriction of
-1,
is obviously continuous,
If
1,
1,
i·.e. by formula (4).
every i',
limG. , -
is injective.
A straightforward verifi-
We are therefore left to verify that \)J* (k.} 1,
in
-
limK. 1,
and consider the sets
f-\H., \)J,(h.) = k., a .. (h.) = h., i EI 1, J J J 1,J J . 1,
are closed and nonempty.
for all
i,;;; j}
Furthermore, the collection
is
179
PROFINITE GROUPS
{A .},j EI, J
possesses the finite intersection property,
Since
n
H.
iEI 1, Tychonoff' s theorem, it fol lows from Proposition 1.6.5, that each element of
n A.
result follows.
jEI J •
belongs to
is compact, by
and is mapped by
~Hi
nA. f 0.
jf::;I J l/;* to
For the rest of th.e section, we are interested in the case
Since the
(ki),
{Gi,fi)i,j EI}
is a projective system of finite groups, each endowed with the discrete topology. We call the projective li'mit
-" compact group to
gives a criterion for a Let
11.7. PROPOSITION.
G = limG.
G
a profinite group.
The following result
be profinite.
be a compact topo 1ogi ca 1 group and 1et
{N,
1,
Ii
E I}
be a family of closed normal subgroups of finite index such that Cil
for every finite subset
of
J'
there exists
I,
i
such that
EI
c: n N. jEJ J
1,-
ilN.=1
(ii}
iEI
1,
G ~ HmG/N.
Then
-
Proof. means
{13/Ni,fij li,j E I}
Conditi'on (_i} guarantees that and
N. ::i N.
" -
topologically and algebraically.
1,·
J
f •• : G/N, 1,,J J
@/N. 1,'
tive system of topological groups.
l
(where
is the natural projection) is a projec.
.
Condition (iil ensures that the map
G
.JL. ~ l. i·m
g
i-r
GIN. 1,
(gNi}
is an injective homomorphism (which is obviously continuous).
(g1,·.N1,,• } then the closed sets compact, there exists
e
N.
g .N. 1,,
1,,
g
E
E
1i m G/N1,. ,
+--
have the finite intersection property.
n g,N,.
iEI is a continuous bijection.
1,
This element satisfies
Since
6(g) = (giNi)'
G
is so
1,
Since
G
is compact and
lim G/N.
+--
(Proposition 1.6.4), it follows from Proposition 1.6,6 that as required.
If
e
is Hausdorff
1,,
is a homeomorphism,
•
We now provide some useful characterizations of profinite groups.
11.8. PROPOSITION. tions are equivalent:
Let
G be a topological group.
Then the following condi-
180
CHAPTER 2
(il
G is a proftnite group
(ii}
G
is a Hausdorff, compact group which has a bas is of open neighbourhoods of
l consisting of normal subgroups (iii)
is a Hausdorff, compact, totally disconnected group.
G
Assume that G is a profinite group.
Proof.
is compact and Hausdorff.
The normal subgroups U8 =
S
a finite subset of
I, N, 1,,
neighbourhoods of l. (ii)~ (i)_
N
iES
1,,
where
1,,
a normal subgroup of G., 1,,
form a basis of open
This proves that (i) implies (ii).
If
be the connected component of 1.
The implication Assume that (ii)
{u} Cl,
let Na= ua n N for all as
are open normal subgroups of N; a.
G
nc. X nN.,
iES
of all open normal subgroups of G,
for all
u8 n G,
ts a consequence of Propositions 11.7 and 11.3.
holds and let
NCl,
By Proposition ll.2(ii),
denotes the family a.
The groups
is connected, we must have
N
N
Cl,
= N
But then
N = nN = n(N nu)= N n(n u l = N n 1 act
a
a
a
the last equality bei'ng true by hypothesis (Hl,
a
Thus
N
=
and so (iii} holds.
The implication (iii)~ (iil being a consequence of Proposition 1.6.7, the result fol lows.
11
11.9. PROPOSITION.
Let G be a proftntte group.
(closed normal subgroup} of
Then any closed subgroup
i's the i'ntersection of some open subgroups (open
G
normal subgroups}. Proof. pose that tains H.
Let H be a closed subgroup (closed normal subgroup) of G and supg E
G belongs to every open subgroup (open normal subgroup) that con-
Then, for any open normal subgroup N of G,
1
S
of open normal subgroups of G,
s
s
n (gN. n H) = g( n N.) n H
i=l
Stnce the sets
gN
i=l
1,,
n H are closed and
and so
0.
gN n Hf
Hence for any finite family N , ••• ,N
g EHN
G
we have
t- 0
1,,
is compact, it follows from Proposition
PROFINITE GROUPS 1.6.5 that there exists h
such that h
EH
181 for all open normal subgroups
E gN
But then g-lh belongs to all open normal subgroups of G.
N.
Since G is
Hausdorff and some open normal subgroups of G form a basis of open neighbourhoods of 1 (Proposition 11.81, it follows from Proposition 1.6.2 that g-lh
1.
=
Thus g EH and the result follows. • 11.10. PROPOSITION. then
(il
If
is a closed subgroup of a profinite group G,
H
is proftntte.
H
(iil If N is a closed normal subgroup of a profinite group G,
then G/N is
profinite
(nn
The direct product
G
=
n
G.
iEI
of an arbitrary family
{G.
i
Ii
of pro-
E r}
i
fini'te groups ts profi'ntte (i'vl Every projective Hmit
(il If
Proof.
N. i
open normal subgroup of
limG. of profinite groups is profinite.
G =
_i
ts an open norma 1 subgroup of The family of all groups
ll.
l\ypotheses of Proposi'tton 11,7 and thus A closed normal subgroup
(ii')
by Propostti on 11 • 9,
G,
N
normal subgroup of
G., J
of G of finite index. ly, G ~ ltm
~-
H~
lim
then
H .
is an
n N.~
satisfies the
ll n Ni
n Ni} is a profinite group.
H/(H
ts an intersection of open normal subgroups of
Now apply Proposition 11 • 7.
Let J be a fi'nite subset of I.
(iii}
G,
Then
NJ=
Given j
r\N. x iEJ,,-J n Gi jEJ' J
E J,
let N, denote an open J
is an open normal subgroup
But the intersection of all such
ts 1,
NJ
According-
i's a proftntte group, by Proposition 11,7,
G/NJ
(ivl We know, from Proposi'tton 11.2(;}, that G ts a closed subgroup of i!Gi. By (iii-1,
1\G.
iEI
ts profi'nite,
Now apply (il,
•
i'EI i
An important result for the cohomology theory of profinite groups ts: 11,11, PROPOSITION, group of o{Hl
=
Let G be a profinite group and let H be a closed sub-
Then there exists a continuous cross-section a :
G.
1, l',e.
there exists a continuous map a:
spaces such that the composite map Proof.
G/H-+ G
G/H ~ G --. G/H
G/H--+
with
G,
of topological
is the identity map,
Let K ~ s be subgroups of G such that S/K is finite.
show that there ts a continuous cross..-sectton o : G/H __. G/K with o(H}
We first = K,
182
CHAPTER 2
Indeed, let uG be the family of all normal subgroups of G of finite index. {UKIK\U E u0} is a basis of open neighbourhoods of
Then
is fi'nite, there exists u E uG such that
(UKIK) n (SIK}
l
E
=
1.
Because SIK
GIK.
Then the natura 1
map : UK/K __.. UHIH
is a homeomorphism.
Let
be a transversal for u in
I\ = l,. . .,gk
G
and put
(n = 1 , 2,. , • , k, u
a/H to GIK with
The map cr is obviously a continuous cross-secti'on from cr(H}
U}
E
= K.
Let x be the set of all pairs {S,cr}, where s is a subgroup of is a continuous cross-section from GIB to GIS. {S,cr} < {S',cr'}
if S'
cs and
We may order x by setting
cr is induced by cr',
Indeed, let {S.,cr.},i EI be a chain in 1, 1,
X
the set {S.,cr.},i EI. 1, 1, In this case s index.
=
1.
The set X is inductive
and let s
The cross-
n Si,
=
1,"EI
sections cr.1, : GIB-->- GISi induce a continuous map a continuous cross-section cr : GIB -
and cr
B
Q" I
: GIB -
lim GIS ., -
i.e.
1,
Then {S, cr} is an upper bound for
G/S.
By Zorn's lemma, there exists a maximal element
{s,o}.
Indeed, if sf 1, then s has a proper subgroup of finite
Then, by the foregoing,
{S,cr}
is not maximal.
This implies the re-
quired assertion. • We close this section by providi-ng a number of examples, (al
If p
is a pri'me number, then the rings '11/pn'll.,n
E IN
form a projective
system with respect to the canonical projections 7/l
n
4p 'll, -
m
(n;;.. m)
'11./p '11.
The corresponding projective limit '11. = lim '11.lpn'll. p
i's the :nng of p--adic 1,ntegers.
-
Every element of '11.
p
is a sequence
x = (x.+pi'll.),i EN, 1,
where x.1,
E '11.
and x.J = x 1,.(mod pi'll.l for .
J' ;,i,
i.
A basic neighbourhood of
x
PROFINITE GROUPS
i's gi'ven by an tnteger m ~ O and
183
tt conststs of all elements
y·
=
(y .+piZ} ,i E lN 'l,
with
The map
i's an embeddt-ng and we identify Z wl'th its image in
(xil,i i'n
E
x = (xi+pill,i
lN converges to
Z. p
lN in the p-adic topology,
E
-
Z is dense
Z/piz
?(x~+piZ}
1----+
'l,
has kernel
(b}
Because the series
The canonical projection
iZ
index
zp •
i pl.
i. The ri'ngs
tions
p
Thus
Z/nZ, n
piZP
~
Z/nZ-. Z/mZ for
ibtlity
min.
lN,
x .. + 'l,
iz
i's an open subgroup of (the additive group)
ZP
of
form a projective system with respect to the projec-
min,
where the order in lN is now given by the divis-
The projecti've limit
Z=
ltm Z/nZ A
is called the Prufer ring,
The groups
nZ,n E lN,
are precisely the open sub-
~
groups of the profl'nite group
Z,
and it is easily verified that ~
(;'I
Z/nZ
£;
Z/nZ
np p of p \) np Z/p Pz \)
Given a canonical decomposftton
n =
Z/nZ
£;
by virtue of the chinese remainder theorem.
n E lN,
we have the decomposition
Passing to projective limits we
obtain a canonical decomposition A
Z=
(_c}
Let JFP
be the field of
rable closure of JFP,
p
For each
nz p
p
elements, n E JN,
p
prime,
and let iff'P
be the sepa-
I/le have a canonical isomorphism
184
CHAPTER 2 Ga 1(lF' /JF 1 9! '11./n'll. pn p
wlii'ch maps the Frobenius automorphism
of IF
¢
n
p
to
n
l
mod n'll.,
Passing to
projective limits we obta tn a canoi1i ca 1 isomorphism
which sends the Frobentus automorphism the group
of ffi
¢
to
p
,,., 1 E '71.
and therefore maps
A
to the dense subgroup '71. of 'll..
12. INFINITE GALOIS THEORY The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and tntermedi'ate fields is not valid for infinite Galois extensions (_see Example 10,5}.
The Krull topology introduced below restores this
correspondence for closed subgroups (Theorem 12.2}, Throughout, group
G can be endowed with the Krull topology defined as follows.
{E./Fli EI} 'l,
rr
G denotes the Galois group of the Galois extension E/F. Let
be the set of all finite Galoi,; subextensions of E/F,
For each
we take the cosets
E G,
crGa 1(E/E , )
(i
'l,
as a basts of neighbourhoods of rr.
t---+ CJ'
1
2
o1 o 2 Gal(E/E.) 'l,
contains the open neighbourhood 0
of
I)
o2
is cont1'nuous, since the pre-tmage of the basic open neighbourhood 1
E
The multiplicatton map
G, (rr l ,cr} 2
GX G-
of rr o
The
(o ,rr2 ),. . 1
1
Gal (E/E j . 'l,
S1'milarly, the map
becomes a topologi'cal group.
X CJ'
2
G-->- G,
Gal (E/E'l,_} o
1-+
o-l
is continuous, so
Of course, if E/F is a finite extension, then the Note also that, by Lemma 10,2,
Krull topology is discrete. toe map
i
G
G/Gal (E/Ei) - - Gal (E/F) Ga 1(. E/E·z.. ) t---- o IE'(.,
Gal(E/E.) . 'l,
- W(R}
W(R), set . - -
, r
n
: R ->- W
n
(R}
E
Wn(R)
WITT VECTORS
199
00
r
a • = (.a 0 ,a0 +a , • o • ,a0 +a + •• , + a 1 , •• , · 1 1 n-
i=o i. 13,4, THEOREM (.i).
The maps
and vn
s,sn
For x = (x 0 ,,,,,xn-·l) E Wn(R)., n-1 . .
(Ji}
preserve addition
y = (.x 0 ,x , ... ,x
n-·
l
f
~ si.(r (x.)),y = si.(r(x.}} i=o n n i. i~o i.
(a).
x =
(b).
1'n (alx = (ax 0 ,aPx1
(.c)_
n-1 r(a).y = (ax 0 ,aP\,.··•aP xn-l''"
1 , ... ) E W(R), a ER,
n-1
(tii)
(iv).
x,y
E
(vl
xn_ 1 )
for all
are such that xi = 0 or yi
W(R)_
+y
X
,aP
and rn(.ab). = rn(a)rn(b).
r(ab) = r(a)r(b} lf
, , ..
If x,yEWn(R)
= (x +y ,x +y , , , , ,x
o o
1
n-
1
are such that xi=O
l
= 0
a,b ER, n E lN
for all
i ;;;., 1,
then
+ y n- l'.,, )
or yi=O
for all
1.;;;·i,;;;n-l,
then
Ci"l
Pro0f.
We first note that s (x ,x , ••• ,x 1 ). = a s(x0 ,x , • , • ,x 1 ,.,, n o 1 nn n1
and
Stnce both
crn
vertfy that
s
and
an+l
preserve additton (Theorem l3,3(t)), i't suffi'ces to
preserves addi'tion,
To prove the latter, we may harmlessly
assume that R = A in which case s(x)_ -_ [0 ,px Col , • , • ,px (n•2) , • • • ] , s
and hence
preserves addition.
(.i·i'}
Since crn(y). = x
for x.
and
crn(r(_a).y) = rn(.a).x,
Again, we may assume that R = A, rn(a)_ =
it suffices to prove the formulas
in which case
[a,aP,. .. ,aP
n-1
and hence . . n-1-i l [0, ... ,0,p '1,a, ••. ,p '1,a'P
Therefore
(O ,;;; i ,;;; n-1)
CHAPTER 2
200
[x(o), •.• ,x(n-l)l = x
and the latter sum is
by (2).
Also
p pn-l (o) (l) (n-1) [a,a , •.• ,a ] [x ,x · , ••• ,x ]
(by (2)
and
(3))
as required. (iii}
x = rn(b)
Apply (_b) and (c) for
and
y = r(b1,
respecttvely,
It suffi'ces to prove (vl, since (iv) follows from (v) by applying
(i'vl and (v)_.
By (i)_,
rrn'
n-1 •
x +y
= =
stnce
~- = i
z s"(r
n-1 .
n-1 .
=
(x.Jl + Z s~(rn(yi)j
i=o n n "
i=o
.z
i=o
s~(rn(xi) + rn(yi))
n-1 . Z s"(r (x .+y ,1)
i~o n n ·
0 or y.i
i
i
0 for O < i < n-1. •
=
Next we introduce the followi'ng two maps 1T
~· W(R)
- 'IT_,. W(R)
I '(xi)
f-----;.
(~)
13,5, LEMMA.
(i')
If
W (R) __!!_.. W (R) n n (xo,'' • ,xn-1)
x,y e W(R),
then for all
t---T ( { , • • ' ' ~ - l)
i;;;, 0,
(1r(x+t1) )i = (1r(x) + 1r(y) )i (mod pR) (1r(xy))i = (1r(x)1r(y))i (mod pR) (i'i')_
If
x,yEWn(R),
then for all
(1rn(x+y) )i
=
iE{O,l,. •• ,n-1},
(1r{x} + 1r(y) )i (mod pR)
(1Tn(xy))i= (1r}x)1rn(y))i (mod pR) Proof,
This ts a dtrect consequence of (8),
•
From now on, we concentrate on the very important special case where charR = p.
The prime subring of R is the field IF
13.6. THEOREM.
.
Let
R
p
of
p
elements.
be an arbitrary commutative ring of characteristic
p > 0
WITT VECTORS
201
and let n;.,, 1, (tl
The maps
(ttl
IT
and IT
p"°'x=s~(IT!(x)}
(ttt)
wn (R1
and iy=si(ITi(y))
(_v)
If
R
Proof, (i'i'}
xEWn(R),yEW(R),i;;;.l
forall
W (r;? 1 = '11./pn'll, is the prime subring of Wn(R), n p is pn and the characteristic of W(R) is O, For.any xE Wn(R)
(i'v}
are rtng homomorphisms
n
Let
and yE Wn+/Rl, vn(x)y = vn(xfn,n+lrrn+l(y})
i's a field of characteristic p,
then
Direct consequence of Lemma 13, 5,
( i} X =
the characteristic of
(x0 ,x 1 , •• ,,xn~i'"' }
W(A).
E
We have
(pX). (i) = pxCi} = p(TI(x) (i-l} t- ,ix,r)
= p11{x}(i-l}
( by ( 1) )
= (_sIT(x}l(i) (mod pi+l'll,[X]}
Thus
(pXli = (1rn(x}li (mod p'll.[X] by Lemma 13.1, whtch clearly impHes have
p\
\i(y})
= s1
for all
now obtatned by applyi'ng
i ~ 1.
The pri'me subri'ng of
Wn(R)
Stnce
pnl = s~(l l = 0 and
pml
Thts proves that
Iterating the latter, we
The correspondtng assertion for
x
is
rr. n
(i'i'i}
pn.
py = srr(y),
1
is addittvely generated by
=
(1,0,.,.,0).
= s~(l l 1 0 for m 0
and
Since
µm E F.
n = m/,
Setting
= 1.
by Theorem 14,4(1).
m (hence n},
Proposition 3.8, hence the result.
= charp,
and let
(i)
E/F
E/F
t:/F
Our point of departure is the case where
14.7. THEOREM. (Artin-Schreir).
F(µ}/F
F(A)
But
=
is cyclic t F(AP ), by
•
We now turn to the case of cyclic extensions p
we have
µ=AP,
contains a primitive m-th root of unity,
F
of degree dividing
(m,p)
where
Let
F
of degree n
pn
where
= 1,
be a field of characteristic
p ~
0
be a field extension.
and
H and only if
p
is cyclic of degree
that
AP -· A E F
(ii}
The minimal polynomial over
for some
E = F(?J
A E E such
A ¢= F.
F
of any
\
E
E
with
AP - \ E F
and
/\ ff. F
is
(iii)_ over
For any given
F or splits into Proof,
then
a
a+ i
E F,
p
the polynomial
xP - x -
a
distinct linear factors over
a
We first note that if is also a root for
i
=
E F
and
a
0,1, •.• ,p-1.
is either irreducible
F.
is a root of
xP - x ·- a,
This is so since
CYCLIC EXTENSIONS
Thus
xP ..
fore
F(.al/F
X - a
splits tnto
p
dtst1'nct linear factors over
F(al
and there-
is Galois.
Now assume that E = F(A) Si'nce
209
xP -
A i's a root of
for some
A EE
X - ()l-A), E/F
i (J E {0,1, •.• ,p-1},
there exi'sts a unique
such that
AP - A E F
is Galois.
and
Ai F,
a E Gal (EJF),
Given
a{AP,..A). = AP - A + i (J .
such that
Consider the mapping
!
Gal (.E/F} (J
'll,/p'll,
i (J +
I--->-
This is obviously an injecti've homomorphism, and
we conclude that
EI F,
cyclic of degree
F,
Stnce
'll,/p'll,
is cyclic of order
ts cyclic of order
pl-A}
Thus
p.
is
E/F
is the minimal polynomial of
p
A over
This proves (ii) and the ~if" part of (t}. Assume that
Gal(E/Fl, some
xP - x -
and
p
Gal(.E/Fl
p'll,
so that
a has order p.
so that
A E F,
is cyclic of degree
F,/F
Then
"JP-\ E F,
a(tl} But
= a(.11t
cr be a generator of crp,)
Owing to Lemma 14,3(it),
= ,._P + 1
and
=" + 1
a(AP-A) = (\P+1) - {Ml)
cr(\) "' A + 1 I
since
:>,, '/; F
Let
p.
Hence
:>,,,
E =
for
= AP-A,
F(A)
and
(i) is established, To prove (itt), we need only verify that if no root of xP - X - a
F,
then
f(X} =
xP - x - a
ts trreductble over
F.
lies in
Assume by way of contra-
di'cti'on that f(X) = g(X}.h(X}. with
g,h E F[X]
1..;; degg ~ p.
and
Since
~ f(X}. = I I (X-a-i)
i=o
(_a is a root of integers
i
E
sum of terms to
-da+j
f(X}l,
we see that
{0,1,. •. ,p-1}. -(a+i)
If
g(X)
for some integer
cause the coefficients of
j.
g(Xl
then the coefficient of
d = degg,
taken over precisely But lie in
d
d
integers
I O tn F,
(X-a-i}
is a product of
F,
i.
is a
~-l
Hence it is equal
hence
a
a contradiction.
Our next aim is to characterize. the cyclic extensions
over certain
E/F
lies t n F,
be-
• of degree
p
n
210
CHAPTER 2
where p = charF,
This will be achieved by analyztng conditions for the exis~
tence of an injective homomorphism Gal (E/F)
--+ 'll./pn,lL
Precisely this was done in Theorem 14,7 in the case n =
Since 'll./pn'll.
1,
isomorphi'c to the prime subring of any ring R of characteristic pn, morphisms from
Gal(E/F}
Gal(E/Fl---+- Pn' able ring
where Pn
is the additive group of the prime subring of a suitOur ring
shall use a simplified notation pertaining to (instead of
1r
n
will be the n-th Wi'tt ring
R
of E defined tn the previous section.
wn(El
all homo-
to 'll./pn'll. are identifiable with the homomorphisms
of characteristic pn.
R
is
Since our n will be fixed, we wn(E}.,
Namely, we shall write
1r
l for the homomorphism
Wn (E}---+- Wn (El, (x 0- ,x 1 ,,,,,x 1 ~ (xP ,J?,, .. ,fn-1 } · n-1 · 0 1 and
s
(instead of sn}
wn (E)
for the shift homomorphism
Wn (El, (x ,x , , • , ,xn-1 ) -0 1
---+-
For any homomorphism a:
E---+- E,
.--+
(O ,x0 ,,, , ,xn-- 2 }
we let
cY(x) = ( cY(x 0 } ,a(x } ,. .. ,cY(x 1 ) 1 n-
1 e wn (E},
for x = (x 0 ,x ,,,.,x 1
n-1
by polynomials wi'th coeffici'ents in
er on
IF
p
and si'nce a(al .
wn (E)- to wn-(E).
is a homomorphism from duced
Since additton and multiplication are defined
Ftnally, if
for all
=
a
O'
e Gal (E/F},
a EJFp , er
the in-
wn.(E)- ts an automorphism leaving fixed the elements of the subring
wn (F).14,8, LEMMA.
Let
E
fi'ni'te order d,
be a field of characteristic p and let ere Aut(.E)
Then for some
= ~1
Let A e
Proof,
0
n
Setting y
=
have
E
be such that y = 0
x e Wn(E)
d i r a (A) f O and let
i=l
0
.
1,
Ea (A}, we see that the zeroth coordinate of y is y0 ,
Hence,
i=l
.
- -~~;:.·:.:,
CYCLIC EXTENSIONS by Corollary 13,10,
is a unit of W~(El,
y
211
Because
o(y) = y,
we therefore
have d
'
E o-i (\y- 1
1
)
i=l
and we may take
x = -\y
Given m,;;; n,
-1
•
•
note that the addi'tive order of sn-m(1)
co, ... ,o,1,0, •.• ,o)
is pm and that m n-m
p s
14.9, LEMMA.
Let
fi'nite order
Proof,
Applying
pm
be a field of characteristic
E
dividing
pn.
xE
wn (E)
Aut(E)
have
for all
(x} = 0
p
and let
CJ E
Then
Owing to Lemma 14.8, we may choose
\
E
wn (E) such that
we have
sn-m,
m
CJ(x)
p ·+1 E 1:CJ-i (sn-m(\))
i=l Pm
•
i j~ (sn-m(\11 j=1 = x + sn-m(l)
since pm sn-m(\l
=
0. •
We are now ready to describe cyclic extensions of degree dividing 14.10. THEOREM.
Let
F
be a field of characteristic
integer and let E/F be an algebraic field extension, degree
pm
dividing
l
if and only if there exist
p >
0,
let n
pn.
be an
~ 1
Then E/F is cyclic of \ 0 , \ 1 , ••• , \n-1
in
E
such
CHAPTER 2
212
E
that
F(\ o ,\ 1 ,.,.,An-1 ] and the element \=(A,\ , ... ,\n~1 ] - o 1
=
wn (El
of
sati'sfies
Furthermore, for each di·vtdi-ng
and an element
pn
Proof,
Step 1,
there exi'sts a cyclic extension
a E Wn(F),
A = (1.0 ,'.\
1
,•, •
·
,A
)
n-1
of degree
K/F
wn (K) such that
in
For the sake of clartty, we divide the proof into three steps.
Here we prove the first assertion under the assumption that
separable.
Assume that
is such that
n(\) - '.\ e -
E = F(A 0 ,A 1 , • , , ,),n~1 l,
wn (F1,
Denote by
E
where
E/F is
A= (\ 0 ,A· 1 , , •• ,An~1 ) E Wn (E)
E,
an algebraic closure of
Note
(E1, (n(x) • x) - (TI(\} - \) = 0 if and only if TI(x•\} = x - ,\ n that is, if and only i'f x - A E w (IF } , Hence {\ + i Ii e 'll./pn'll.} is the set of n p all roots of n(x) - x - (TI(A) - \) in w (E}, si nee w (JF ) = 'll.Jpn'll. by Theorem n · n p that, for
x E
13,6(iii}. under
w-_
Because
Gal (1;/p),
generated over
TI(\) - ;\ E
Hence F
E
wn (F),
i's mapped i'nto i-tself under
E/F
i's Galois. ~ Ga 1(EJF)
? cr(\) = \ + ,r;cr'
fore cyclic of order
Since
Gal(E/F},
rJ
'll./pn'll.
pm dividing
Conversely, assume that be a generator for
A + i,if!, 'll./pn'll.,
by the coordinates of
rable, we conclude that
where
these roots are mapped into themselves Gal (E/F), Si_nce
since
E/F
E
is
i-s sepa-
We have an injective homomorphism
---+ 'll.Jpn'll. I->-
i rJ
ts cyclic of order
pn, Gal(E/F)
is there-
pn,
E/F is cyclic of degree so that
o
has order
pm dtviding pm,
pn.
Let
o
Owing to Lemma 14,9,
we have for some Then
and therefore
\
E
wn (E)
CYCLIC EXTENSIONS
213
TI(A} - A E w (F). Thus F(A 0 ,A •···•" n 1 n-1 )/F is cyclic of degree pk dividing pn, by the first part of the proof, Since sn-m(l) has additive
whence
pm,
order
pk= pm
E = F(\ ,A ,, •• ,A
and
o
1
K/F of degree dividing TI(A) -
A= a.
pn
and
cm
(TI-1)\ = a
= 1,
n
and
we take
(Theorem 14,7(i)}.
A=
there exists a cyclic extension (1,. 0,1,.
,.,.,A J in 1 n-1
'.A. E
o
"a
E
~
such the
that there exists
O ..;. i ..; n-1.
K =
The requtred assertion
F(\,"i···•\_ 1 },
s to be a root of
Next, let n
n
,A 1 , ••• ,An-1 }
for
S
1,
W CK)
(TI-llA = TI(\) - A and denote by S
We prove by i'nductfon on n
wi'll then follow by Step 1 bv taking If n
E W (F),
element
~ = (1,.
such that
a
We use the notation
separab 1e closure of F.
Thus
}.
n-1 Here ~e show that, for each
Step 2,
that
a to F(A 0 ,\ 1 ,, •• ,An-1 } has order pm.
the restrtctton of
xP -
and again choose
1
and
X - a0 :>,
0
Es
K
= F(\)
such that
Then
for suitable sure of
bk Es.
If
K'
=
F(b . 1 , ..• ,bn-1 ),
then
K' and we can find, by i.nduction 1\,,,, ,µn-i
s E
ts the separable cloS
such that
(b1'"''bn_) = (TI-lH11i,,..,µn-1) = (P;,.,,,µ~-1) - (µ1'"''µn-1) But then (0,b , .. ,,b } · 1 n-1
(by Theorem 13,4(i))
(Q,µP, .. ,,µP } - (O,µ ,. .. ,µ 1 } 1 n-1 1 n-
(rr-1)(0,µ ,,,,,µ 1
n- 1
)
He now have
(a0· ,a ,.,.,a ) 1 n-1
for suitable
( TI-1} (("a, 0, •. , , 0) + ( 0 , µ 1 ,
"i E S, 1 ..; i..; n-1 ,
Thus
(TI-1)'.A =
a
.. , ,
for
µ n-i) )
A = (\,\,..
""n-1'
CHAPTER 2
214
as requi'red, Step 3,
In view of Steps l and 2, we are left to
Completion of the proof.
verifythatif E=F(A 0 ,A, .. ,,A
n- 1
1
n(A) . . - AE w n.(F),
such that
Step 2, to find
1
n
of F,
Thus
p
and therefore
n-1
)
)\=()\ 0 ,A, .. ,,A 1
is separable. such that
is separable over
that each µ., O < i < n-1, 1., µ - A E w /;IF ) •
E/F
then
µ = (µ 0 ,µ , ••. ,µ
where
),
E/F is separable,
)EW(E)
is
n
To this end, we apply
n(µ) - µ = n(A) - A and such But then
F.
A = µ - (µ-A) E wn(s),
n- 1
where
n(µ-A)
µ - A and
=
s is the separable closure
•
15, KUMMER THEORY
F be a field and n
Let
to be of exponent n for a 11
0
E Gal (E/F),
abel ian extension
E/F
n-th root of unity. F
a positive integer.
if the exponent of
A Galois extension
Gal (E/F)
divi'des n,
E/F
i.e.
if
By a Kwnmer extension of exponent n one understands an of finite exponent n,
where
F
contains a primitive
Throughout the discussi'on, the ground field
n
then
Moreover,
con ta i'ns a primitive n~th root of unity, which wi 11 be denoted
Our aim is to survey all Kummer extensions
E/F.
G endowed with the Krull topology is a compact group.
Hom c (G, < sn >) = {f : Then
will be
F
fixed, and all extensions 1ie i'n a 9iven algebraic closure P of F.
by s.
an = 1
It follows that for any such extension the characteristic of
does not divide n,
we assume that F
is said
Hom/G,)
G-
< sn > I f
G = Gal(E/F),
If
We write
; s a continuous homomorphism}
is a group under multiplication of values: Cas)(g} = a(g)B(g)
We denote by morphisms from
Hom(G, < sn >) G to
the group of a 11 (not necessarily continuous) homo-
. n
Then
Hom (G, n
C
l is a subgroup of
Hom(G, < s >) n
which can be characterized by Homc(G,) ={XE Hom(G,)I Kerx
is an open subgroup of
G}
(l)
The latter l's true, since a homomorphism from a topologi'cal group into a discrete
KUMMER THEORY
215
group ts conttnuous tf and only tf tts kernel ts an open subgroup, is finite (hence discrete), we have
Homc(G,)
In case
Hom(G,).
=
G
Some further
i'nformation is provided below, G be a finite abelian group.
Let
homomorphisms from
to ..) 2 •
>.. =
•
We are now ready to tackle the prime power case (the prime case will be excluded because of Lemma 16,3), 16.5. LEMMA.
Let
and let
p
be a prime.
(i)
p
i's odd, or
F
If
only i'f
and
charF f 2,
a (/:. F 2
and
a (/:. -4F 4 ,
(i}
Proof. n-1
xP
by
and
p = 2
If
an arbitrary element in F. n charF = 2, then xP -a is irreducible over
a
Denote by p = 2
If
a= 1l
2n
then
for some
X
µ E
- a
F,
i's irreducible over
Conversely assume that a ef:. pP. n-1 µ = ;\P Then µ ts a root of xP
xP
then
- µ.
and let
be a positive integer
2
a(/:. FP.
if and only if
(i'i)
n:;;,,
be an arbi'trary field, let
F
Let
- a,
n
~a= /\
xP
F
if and
n -µP
is dividible n
be a root of
xP
-
a
hence by Lemma 16,3,
(F(µ) :F) = p We clatm that
'.\
has degree
p
n-1
F(µ); if sustained, it wil 1 fol low that
over n
>..
has degree
degree
p n-1
over
pn
over
F(µ)
F
and
xP - a
is irreducible over
will be true by induction on
n,
F.
That
provided
A has
µ '!, F ( µ )p .
The desired conclusion is therefore a consequence of Lemma l6.4(i). (ii')
If
a E -4F 4
a
E F2
,
then obviously
and write
a= -4a 4 , a E
x
2n
- a is reducible.
F, Y
=
x2
n-2
•
Then
Suppose next that
225
RADICAL EXTENSIONS AND RELATED RESULTS Conversely, assume that a ff. that
and a ff. -4F 4 •
F2
-4a is not a fourth power in F. Since a ff.
by Lemma 16.3.
We must show that
X2 = 2
and for n
In the latter case, -µ
over F.
-+
1-
(F(µ) :F)
•
For n
2
and
= 2
this will be
this will be true by induc-
and -4µ
is not a fourth So it suffices
is a square in
-µ
-a
we have
is a square tn F(µ).
these two statements are equivalent since µ
2n
a;
> 2
to rule out the possibility that either µ or -µ
F(µ}
-
n-1
(F(:\):F(µ))
provided 11 is not a square in F(µ}
power in F(µ).
it follows
Again, let :\ be a root of x
and µ is a root of
F2
true if µ is not a square in F(µ}, tion on n,
Because a ff. -4F 4 ,
F(µ).
Now
induces an automorphism of
Hence, by applying Lemma 16.4(ii), the result follows.
•
Applying the foregoing, we now deduce the following result. Let F be an arbitrary field, let n P
16.6. THEOREM.
is irreducible over
;(I - a if and only if
Proof. a
E pP,
a If:.
If p8 -a
xP
If
Conversely, assume that a ff; Let p8
4\n
and
F.
di vi ding n
p
a=
-4:\ 4 ,
p
Hence
and if
r-a
s xP - a
is
then
:\ E F,
dividing n and a ff. -4F 4
be the highest power of a prime p dividing n.
Lemma 16.2, it suffices to show that
Then
dividing n and a ff. -4F 4 whenever 4\n.
for all primes
FP
E
F
is reduci'ble by Lemmas 16.3 and 16,5.
reducible, by Lemma 16.2.
whenever 4\n,
p
is the hi gbest power of a prime
s
then
for all primes
pP
and let a
1
is irreducible.
By
The desired con-
clusion is therefore a consequence of Lemmas 16.5 and 16.3. • Our next aim is to prove that if intermediate field
K,
Gal(K/F)
E/F
is a radical extension, then for any
is solvable.
The following preliminary obser-
vations will clear our path. Let
16.7. LEMMA. subextensions. Proof. E2 =
E/F
Then
be a field extension and let (E E ••• E 1
2
n
n
1
be radical
)/F is a radical extension.
It suffices to treat the case n
F(µ 1 , ••• ,µk)
E /F, ••• ,E /F
exhibit the assumption that
= 2. E 1 /F
Let and
E
1
=
E 2 /F
F(:\ 1 , .•• ,:\) m are radical
and
226
CHAPTER 2
extenstons.
Then
EE 1
EE /F
shows that
1
(_i)
Let
If E/F
(ii}
If
F c Kc E
1
•
be a chain of ftelds.
is radical, then so is is radical and
KIF
m
1
is a radical extenston.
2
16.8. LEMMA.
= F(\ , •• ,,\ ,µ , ••• ,µk)
2
E/K,
K over
ts the normal closure of
E
F,
then
E/F
ts radical.
(_i)
Proof.
n. E = F(A •• ,.,A) with A.1., E F(A _., ..• ,L 1l, 1,;;; i,;;; m, 1 n. m 1., 1 1.,l with A! EK{\, .. ,,\. 1 ), as required.
If
then
E = K(\- 1 , ... ,\m
Cti'l
Si'nce
is fintte, we may write
K/F
fi(X)
of an irreducible polynomial
f.(X), 1.,
then
radical.
F(A , ••• ,A) - l s
Stnce
Then
Gal(8/F)
(til
If
Let
be a prime and
p
i1' ~ 1
and
F
g
If
s.
\.1.,
is a root
is any root of
F(\ 1 , . , . , \8 )/F
a splttttng field of
is
•
x'P-1
over
a
an arbi-
splits into linear factors,
the splitting fteld of
8
charF = p,
If
i1'-a
over
F,
F.
then
then
E =
F(s),
E = F and there is nothing to prove.
then
where
s
is a prtmttive p-th root of untty.
Gal {_E/Fl
is cyclic, by Corollary 5. 12.
(iil
u
r-1!1,,
is one root of
and so
morpntsm of
a(u)
y'""x+y>x'+y'
(b)
x > O,y > 0 ""xy > 0
In discussing the ordering on R, we shall write x > and use
x = y,
for the opposite ordering.
to mean x
y
or
> y
An element x ER is said to 0.
Indeed, otherwise by Lemma 16.3,
If charF
is a power of
charF
If
Thus we may harm-
Furthermore, by passing to a normal clo-
we may assume that E/F
is Galois.
Let P be a Sylow p-
subgroup of G = Gal(E/F},
and let K be the corresponding subfield.
(K:F}
p.
(E:F}
=
is prime to
(G:Pl
is a power of
16.21. THEOREM. a square root. a root in Proof.
be an ordered field in which every positive element has
F
Assume further that every polynomtal of odd degree over Then F(i}, i 2
F.
Thus
K = F.
as required. •
p,
Let
By our hypothesis, this implies
Then
By hypothesis,
=
F
-1,
has
F
is algebraically closedo
has no finite nontrivial extensions of odd degree.
Invoki'ng Lemma 16.20, we infer that the degree of any finite extension of
F
a power of 2.
is
F(il,
We claim that the only nontrivial finite extension of
which will clearly yield the result.
let G = Gal(E/F}.
By Lemma 16.17,
passi'ng to the normal closure of
E,
charF
Let E/F be a finite extension and =
0 and so E/F is separable.
we may assume that
know that the order of G is a power of 2,
If
(E:F}
is Galois.
E/F ~
2,
F
c
F 1
c
F 2
of fields with
is the only quadratic extension of F, easily tmplies that every element in
(F :F} l
=
(F :F ) 2
we must have F
1
F
1
2.
=
1
=
We
F
2
This Since F(i}
The latter
F(i).
is a square, hence
By
then G has a sub-
group of index 4 which in turn is contained in a subgroup of index 2. yields a chain
F
is
cannot exist, a
contradiction. • Let E/F be a field extension and let M be a subgroup of E* F*M/F*
is finite.
When is it the case that
IF*M/F* I
= (F(M)
:F)?
such that An answer
to this question is given by the following result. 16.22. THEOREM. (Kneser(_l975)}.
Let E/F be a separable field extension and let
CHAPTER 2
234
M be a subgroup of
primes p,
such that FM/F*
F7
ts finite,
each p-th root of 1 which lies in F*M also lies in
i = ,J.:i_ E F
if
i E
1 ±
Proof.
(F(M} :F}
We first observe that F(Ml
consists of all F-linear combinations of
Therefore we may choose finitely many
basis for
F(M}.
Then
and that
F
Then
F* •
IF'
N
the norm of
C-11P-1aP, we have
((-l}P-1J'lq = N(alpd-p
For odd p, (61,
aP
In case
is then a p-th power of an element in F(N8 _ 11, which contradicts p = 2,
-a 2 = ;.. 2 with ;.. E F(N8 _ 11 which shows that
we have
= ad = ±i1,d.
i E F(N8 }, i ¢ F(N8 _ 11 and a 2 a== g +
Then a 2 == (g 2 -h 2 } + 2ghi
Let us write
ih
= ±i'Ad
which impli'es g 2
-=
h 2 and hence a == (l±i}g.
We conclude that
which ensures, by applying twice the induction hypothesis, that then 1
± i
E F*M and so, by hypothesis,
i E F ::_ F(N8 _ 1 ).
g E Ns- 1 •
But
This contradiction
completes the proof of the theorem. • Returning to radical extensions, we next investigate conditions under which the Galois group of a binomial
r -a
16.23. THEOREM. (Schinzel(l977}1.
Let
is abel ian. F
be a field,
n
a positive integer not
divisible by charF and let m be the number of n-th roots of unity contained in
236
CHAPTER 2 Then, for any gi'ven a
F.
only if am= \n for some \ Proof.(Wojcik(l982}}.
E
F.
For any integer t
not divisible by charF,
denote a primitive t-th root of unity over F.
Since
Let rb
If am= An for some A E F,
extension of F.
ts afielian if and
the Galots group of Jf-a
E J!,
be the maximal abelian
then Z7'a
O ~ j ~· n-1.
= sjI:!}'5::,
n
contains a primitive m-th root of unity, ~Erb by virtue of Lemma
F
15.9(.iiil.
Thus '{;fa E
r»
and F(Z7'a,s n }/F is abelian, Conversely, assume that the Galol's group of i"- a is abelian,
the m.aximal divisor of n for which am verify that k = n, that k
let st
and m\k.
of p dividing m,
be
We need only
F.
Assume by way of contradiction
Fix a prime p dividing n/k and denote by p6
n.
0.
with root A and define the
F
E
F(e )} m
I' ~.11!
(E:F} Proof.
(m,p}
f>y s
lf
Let F be an arbitrary
=
over F, n(F(e }:F} sm
then
Our first step is to reduce the genera 1 case to the case where m = n,
charF X n,
So assume that charF = p E
= F(\,e} m = F(Ap
>
O.
Then
k
,A~E) m k
wh.tch in turn is the composite of the separaf>le extension F(\P ,e )/F and a m k purely inseparable extension F(Am}/F. By Lemma 16,2, both Jfl _ a: and xP - a: k
are irreducible over
F,
Stnce Am
is a root of
iP - a, we have
CHAPTER 2
238
(F(\ml:Fl
=
l.
Hence, by Propo$ttton 3.28, k
k
(E:F} = p (F(\P ,sm}:F} k
Thus, if the result holds for i" - a (whose splitttng fteld is
We may therefore assume that m = n and hence that charF Put L
= F(E
n
1n
and s,
F(\}
Since
= (L:FL
r- a
F(\P
,sm1, then
%n, is irreducible,
(F('A} :F} = n
On the other hand, by Proposition 12,5,
It therefore follows that
and we must show that s
= s',
Now by the definition of
in Lemma 16,24,
1'
F(s1' 1 -c F(sn 1 n F(Al and hence, oy Lemma 16,24,
L
= F(\ql for
q
= L
c F(\1
-
= (F(\l:L}.
(F(\l:Fl = n,
Since
we have (L:F} = n/q =
Thts snows that \q e F(snl and q
=
n/s',
On the other hand, since \n/s e F(s 1 n F(\l n
s' Hence, by the definition of s,
= L,
we have
(F(\}:L} = q..;; n/s
whence
s
1
~
s, as requtred, •
Let E/F be a simple radical extension, say E = F(\) with power of \
in I!*.
Then we obviously have
( < ;>., > t(E*ll!*l/1!*
~ t(.E* /J!*l
;>.,
e E* and a
239
RADICAL EXTENSIONS AND RELATED RESULTS It is therefore natural to tnvesttgate ctrcumstances under whtch the equality holds, 16.26. THEOREM. (May(l979}}. sion with charFt-p. further assume that
Let p oe a prime and let E/F be a fteld exten-
Assume that
t- F(i'} {i 2 = -1}.
E
t(E* /F*}
Proof. (E:F} = p.
Let
µ
= (
xP -
Owtng to Lemma 16. 3,
where )..PEF-iP.
E=F(\l
If p=2,
Then
t(E*}F*}/F* 1..P ts trreducib 1e over
F and hence
e E* oe an element of prtme-power order modulo F*.
First
1:'
assume that the order is qr for some prtme q :j p, and 1et µq l'
is the norm map from
N
y e F*. l
E
It follows that
then N(µlq
y e F*.
=
If
1'
·,l, hence
to
F,
µ =
y E for some root of unity Ee F* which shows
=
for some
y = yq l
1
that µF* E t(E*}F*. 1'
Now assume that µ has order p:i> modulo F*, claim that y ¢ if?.
and let
Assume the contrary and write y
Let E be a prfmi ti ve p-th root of unity.
=
i/
=ye
r.
We
yP1 for some y l e F*.
Then for some m
We must have p ) m, than pr.
for otherwtse the order of
modulo
µ
For the same reason, we must have e fJ F*,
wi'th (F(e:} :F}
But
> 1.
(E:F}
=
r
would be 1ess
It follows that
p and (F(E} :Fl
divides p - 1.
This contra-
and i rt
we have (F(µ) :F}
dtction substantiates our claim. Except possibly when p = 2 ,r
!>
1
virtue of Lemmas 16.5 and 16.3, and thus r i.e. assume that p
=
2 ,r > 1,i
t
and
F
F,
= 1.
(F(µ}:F}
=
pr by
Consider the exceptional case, < 2r.
stances we know, from Lemma 16.5, that y =-4d 4 for some
Under these circum8
e
F.
From
1'
µ2
=
we see that i e E and hence E = F(i).
-4d4,
Since this case is excluded in the
hypothesis, we may therefore return to the situation where
CHAPTER 2
240
µP = y
and
(F(µ} :F} = p
Then we have
= E = F(:\)
F(µ} Hence, by Corollary 15.7, µF*
F*
E
µ = )...ka
for some
and the result follows.
E
Z and some
a
E
Thus
F.
•
We return to the study of the field ci'ble polynomial
k
F(al
where
a
is a root of an irredu-
As an easy application of Corollary 15.7, we
xP-a E F[X].
first show that the adjunction of two ~genuinely different~ p-th roots result in an extension of degree
16.27. PROPOSITION. If
a,S
p2 • Let
be a prime and let
p
arerootsoftrreduciblepolynomials
charF t- p.
be a field with
F
over
xP-a,xP-b
then
F,
(F(a,S} :F} = p 2 unless
b = c?ak
Proof.
for some
k E Z
If the polynomial
(F(a,Sl:F} = p 2 , (Lemma 16. 3 l. for some
k e
b = cP ak.
•
Assume that
i.P-b
yP =
c
e P,
a,S
b E F,
c E F,
remains irreducible over
If it is reducible over Since
z,
and some
F(a},
F(a},
then
then it has a root
in
y
it follows from Corollary 15.7 that
y = ca
are roots of irreducible polynomials a+ S,
provided
over
xP-a, xP-b
(F(a,S}:F1
=
p2?
To
answer this question, we first maRe the following preliminary observation.
elements of
Let E
be a fintte Galois extension and let
E/F
of degree
m and
n,
respectively, over
F
a
and
S be
such that
(F(a,S} :F) = mn. (_t l
For any conjugate
a,
~
of
a
and
e,.J
of
S,
there exists
0 E
Gal(E/F)
such that o(a)=a. ~
(ii)
k
Raising this equation to the p-th power, we obtain
What can be said about the degree of
16.28. LEMMA.
F(a~
and
If no difference of two conjugates of
o(s)=s. J
a
equals a difference of two
F,
241
RADICAL EXTENSIONS AND RELATED RESULTS
conjugates of
B,
then F(a,S) = F(a+S)
(i}
Proof.
Since
f E Gal (E/F).
Corollary 6.5(i), choose for some conjugate
f 1 (a)
that taktng
0
=
a
is a finite Galois extension, we may, by virtue of
E/F
Bk
of
and
f
1
B.
f(ai) = a.
such that
f(B)
Write
f 1 E Gal(E/F)
We claim that there exists
=
Bk
such
if sustained, the assertion will follow by
(Bk} = B;
(J f}- 1 •
=
I
Let
B ,.,,,B I n that the degree of polynomial over
be all conjugates of B over
F(o;l
F(al
are still
F.
B over
is still
n,
B1 , .. .,Bn •
Our hypothesis implies
so that the roots of its minimal Since
E/F(o;}
is Galois, the
required automorphtsm exists,
(_ift Let a 1 , ..• ,am be all conjugates of o; over F.
Then, by (il,
o;. + B • are all conjugates of two conjugates of
o;
by hypothesis, all Hence
mn.
a
over p2
and
over
If a coincidence occurs, then the difference of
1,
+ B.
are distinct, provtng that the degree of
J
Let
F
as asserted,
then
Thus,
a + B is
•
be a field and let
(F(a,,S}:F} = p 2 ,
Since
p
F(a+Bl
# charF, a, and
a fi'nite Galois extension
p
be a prime di.sttnct from
charF.
xP-a, Jf.J-b
= F(ct,Bl. . i,e. a,+ B has degree
with
E/F
B are separable over a,S EE.
to verify that no difference of two conjugates of conjugates of
(si - sj)a,,
B,
where
s
a,/B E F(s}.
But
a
a
has the form
Therefore, if a differ-
equals a difference of two conjugates of
(F(s}:F} m
Then deg(a+\S} Proof. Step 1.
= mn
for all
Of
AE F
For the sake of clarity, we divide the proof into a number of steps, Here we fix notation and make some preliminary observations.
243
DEGREES OF SUMS IN A SEPARABLE FIELD EXTENSION Let the fi'e l d [3 , ••• 1
be obta tned from
E
,Bn of a and
B,
Then
ts a fi'nite Galois extension
E/F
The group G acts transitively on the sets {a , •.. ,a}
and
m
1
B = {[3 ,, •• 1
Let H and K be the stabilizers in G of a and (G:Hl
and stnce
(m,nl = l,
=
m,
,B} n
B,
respectivelyo
Then
(G:K} = n
(l )
a standard argument yields (G: (.H n Kl} = mn
Hence the size of the orbit of A
x
is mn.
B
and
m
1
respectively.
and we put G = Gal(E/F). A=
by adjointng a11 conjugates a , ••• ,a
P
(a,B1 EA x B under the induced action of G on
It follows that
(.1.;; i ,;;;;· m, 1.;; j,;;;; n}.
x
A
B
and hence
ts a conjugate of a+ B
J
Since
ts transitive on
G
each a.+ B, 1.,
(.2)
[HK[ = [H[ [Kl/[H n
K\,
(3)
it follows from (l} and
(.21 that
[HK\ Hence
(.K: (H n K))
=
fPr
\~I • I~\
\G\
=
and therefore
= m
K acts transitively on A
Let W and W 1
be the F-ltnear spans of A and B,
2
V=W
and let
u = W1 n
and w ,w ,u .1
2
w. 2·
1
+W
are G-tnvariant subspaces,
2
deg(a+Bl
=
mn.
Replacing
there is nothing to prove. m0
B by
We use induction on
F-dtmenston of the normal closure of F(a,B).
are of coprtme degrees
respectively, let
Then G acts as a group of linear transformations of V
it suffices to show that
Li F,
(4)
If
[G[
=
[3 0
over L.
derive a contradiction,
if necessary,
[a\, i.e. on the then
a,B
E F
and
By induction, we may therefore assume that if a,B and n0 ,
respectively, over an intermediate field
and (a), (b} or (cl hold with respect to m0 ,n, 0
degree of a 0 +
1,
\[3,
From now on, we assume that
then m0 n 0
is the
deg(a+BJ < mn
and
CHAPTER 2
244
Owtng to (3}, not a11
s.J
a.+ 1,,
are dtsttnct and so for some k e {1,.,.,m},
s E {1, •.. ,n}
(5)
and so u t- O. Let M be the kernel of the action of G on v and let L be the fixed Then F.:: F(a,Sl::, L and L/F is Galois, so E
field of M.
L and hence
=
Thus
M = 1.
G acts fatthfully on Step 2,
(6)
Here -we prove that ther>e is no pr>oper subgr>oup
that the orbits
where m0
=
A0
IA 0
and B0 of a
and n 0
J
=
Assume the contrary.
L
@0
Stnce
L.
S under G0
(m 0 ,n0 1 = 1
Then
is the ftxed field of
respectively, over
and
G0
of
G which ac-ts so
satisfy
IB0 l,
the number of elements in the where
v
and since, by (5}, c1, + f3
-orbit of a+ f3
Now G0
+ f3s'
Gal(E/L}
=
and m0 ,n 0 are the degrees of a and 13,
@0 ,
1@0 1 < Jal,
(c} holds with respect to m0 ,n0 ,
is< m0 n 0 ,
= ak
it follows that none of (a)., (b} or
But then none of (a}, (bl or (c) holds with
respect to m ,n, a contradiction. Step 3,
Let
N neZ of the action of G on nf3 = u
fixes
N
(u
+y
2
tatn equal number of elements of A.
H ~ G0
A0 ,
Let
and hence
G0
Indeed, if a.1,, e
w1
and ts transitive on the
All of these, therefore, con-
Now a,ak ea+ u and if
A
0
be the stabilizer of the coset a+ u in
G0
is transitive on A, 0
(7)
(8)
nS
set of those cosets which contatn elements of A.
JA 0 J Im.
We now show that
ts fixed by G}
e U, ye W
Note that G permutes the sets of cosets of u in
then
A.
then
ag =
B.
We claim that
a.1,, for some
g
e G,
G0
=An (a+U), G.
Clearly,
is transitive on
Hence
245
DEGREES OF SUMS IN A SEPARABLE FIELD EXTENSION
and so g
E G0 •
G0 = G.
Thus
B
c
This establtshes transitivity and, by Step 2, we must have stabilizes
G
B + u and so B. J
=
a +
u
u.
and so Ac a+
B + u. for some
uJ,
J
By a similar argument,
Summing over
Eu,
s.J
we
EB,
derive
rs.J y,
Hence nB = u +
where
=
Eu. + nB J
J
proves (71 and (81, since
fixed by G,
2
-
tf x EL~ K,
r
1 = -
a.
a~ = 1,
rrm
Because y and
sm,
o
so
B and hence JcJ Jm!
N = 1,
of
y
It!,
charF)
(9)
we have
!>
1
are ,,;;; min(m,n).
B is
B # B. s·
G on A.
Thus
If
G
Therefore all prime divisors of
Since n
we conclude that
Be!'e oJe dePiVe a final contradiction.
charF
8
f
nm,
charF
%n,
(in particular, those of
this certainly excludes (c).
by hypothesis.
Jal Jn!
then by (8),
ts isomorphic to a subgroup of Jal
B# B ),
(recall that
and B in the above argument we obtain
IcJ
n
; A-z,.o + .1tl 'Y n
by (61.
Since (c} is excluded, we have A
1
-z,
i=l
This is a contradiction since
Completion of the pPoof.
nl are ,,;;; m.
i,
10i m
are fixed fiy G and
N be the kernel of the action of
fixes
X
;\.a,+ -
it therefore follows that
L,
IL IB = ill
N
r
L ts transitive on A,
and note that since
Summing (91 over
Let
"
1,
r
Step 5,
m
n i= 1
xEL
fixed by G.
0•
1, 1,
we have
m
i=1
IL I,
i=l
1
B = BX =
This
By (71, we have nB = u + y for some u Eu and some m As u cw, we have u = r ;\.a. for some A. E F. Now
Assume the contrary.
yE w
G.
2
then there ir, no subgPoup L c K
chaX'P ,I'
r.,ihi'Ca is t!'ansitive on A and tJuch that
w ts fixed by
w1 -~ u.
fixes all elements of
N
E
J
BePe we shaw that if charP ,I' rrm,
Step 4,
Put er
y = rs.
u = .. zu. Eu and
By interchanging
and hence all prime divisors
Hence, if charF = p > 0 and (a) holds or if
246
CHAPTER 2
charF
=
th.en cflarF) \GI.
0,
must have charF = p
>
But thts contradtcts (41 and Step 4.
We may assume that m = qe, q prtme, and let
0 and (b}.
Q be a Sylow q-subgroup of K.
it fol lows from (4) that
(K:(K n H}} = qe so K =(Kn H}Q and
Then
ts transittve on
Q
Hence we
Since charF ) q,
A.
by hypo-
thesis, we derive a ftnal contradiction by applying Step 4. • The following simple observation illustrates that to establish the best possible improvement of the above theorem, with conditions given in terms of m,n and charF, 17.2. LEMMA,
it suffices to consider only group representations Let v be a finite-dimensional vector space over a field
let a finite group Assume that u,v
G
act faithfully on
Ev are permuted oy
V
as linear transformations of
G into orbits of sizes m and n
tively and that u + v lies in an orbit of size k.
K
and
v. respec-
Then there exists a finite
Galois extension E/F (with K~ Fl and elements o.,S EE such that dego. = m, degS = n and deg(o.+S) = k Let t
Proof.
R = K[X1 , , •• ,xtJ
=
dtm v and let x ,x ,,,,,xt be indeterminates. K
1
and let E be the quoHent fteld of R. · We may identify v
wi'th the K-1 inear span of
X
l
,x , , . , ,xt in 2
R,
group of K-automorpftisms of R and hence of E. G in
E
and v.
Put
2
(so K::, F}
and let a and
Then
G
acts faithfully as a
Let F be the fixed field of
S be elements of E corresponding to u
By Proposition 4,5, E/F is a finite Galois extension and G ts the
Galois group of E/F. y under G,
Since for any ye E, degy is the size of the orbit of
the result follows. •
We close by giving some limitations on possiole improvements of Theorem 17.1. The fo 11 owing two ex amp 1es in conjunction with Lemma 17. 2 i 11 ustra te that the conclusion of Theorem 17.1 fails in each of the cases below: (il
(ti}
charF = 2, m = 3 and n charF
= 3,
m=3
and n
17.3. EXAMPLE.(Isaacs(l970)}. Let V*
= 4
=4 Let G be the alternating group of degree 4.
be a four dimensional vector space over Z/2Z and let G permute a
247
GALOIS COHOMOLOGY basis,
tn the natural manner.
{w,x,y,z},
v0 = {O, w +
x +
Put and V = V*/v0
y + z}
Then G acts faithfully on v and we put u
=
w+
v0 and
x +
v = w+
v0 •
Then u and v are permuted by a into orbits of sizes 3 and 4, respectively, However, the size of the orbit of u 17,4. EXAMPLE.(Isaacs(l970)).
with basis
'll./3'1!.,
a ,a ,a l
2
GL(V)
E 3
0 a
{w ,x ,y ,z}.
is
+ v
4 f 12.
Let V be a four dimensional vector space over Let G oe the group generated by the elements
whose matrices are
0
0
0
0
0
0
0
0
-1
0
0
0 0
a
(J'
l
0
0
0
0
2
0
0
0
0
{w,w
Put u = w and v = y.
+ x,w - x},
3
0
0
Then G is the direct product of the subgroups of order 2.
•
l
2
0
0
0
0
0
0
0
0
0
0
of order 6 and
3
Then the orbit of u under a is
and the orbit of v under a is
However,
{y,y + x,z,z + x}.
the oroit of u + v is {w + y, w + y + x, w + y - x, w + z, w + z + x, w + z -
x}
which has six elements. • 18, GALOIS COHOMOLOGY
Let
G
be a group.
A a-module is an aoelian group
A
on which
G
operates
such that (i}
la = a
for all
a EA
(it)
g(a+b) = ga + gb
for all
g E G, a,b EA
(iii)
(gg )a= g(g a)
for all
g,g
l
l
E G, a EA 1
Note that if A is a a-module, then A is a 'll.G-module via (fa g)a g
= fa
g
(ga)
(x g
E 'll., g E G,a EA)
Conversely, any given 'll.G-module A can be regarded in the obvious way as a a-
248
CHAPTER 2
The fixed module of a G-modul e A
module.
AG= {a E Alga= a
It is clear that AG If
g E G}
for all
is the largest submodule of A on which
G
acts trivially.
are a-modules, the group of all abelian group homomorphisms
A,B
is denoted by as follows:
i's defined to be the subgroup
Hom(A,B), if
The group
E Hom(A,B), g¢
¢
Hom(A,B)
A->- B
has a a-module structure defined
is the mapping
ai----g¢(g- 1a), a EA, g E G.
ln particular, if x is any abeHan group, we can form the G-module
Hom(JZ'.G,X).
A a-module of this type is said to be co-induced. Let A copies of
be any a-module. G to
A
G,
f
from the direct product of n
is cal led an n-cocha1,,"n of G in A.
said to be normalized if ti ty e 1ement of
A function
f(g , •• , ,g ) = 0 1 n
whenever any of the g.1,,
The set of a 11 n-cochatns, written
abeltan group under the multiplication of values, d1(G,AL
to the case
n = 0,
An n-cochain
c12 (G,A),
f
is
is the i denbecomes an
Extending the definition of
we put
c0 (G,A)_
= A
The formula
( 1)
determines a homomorphism
It is a standard fact that d d
n n-1
Zn(G,A)
and refer to the elements of aries, respectively.
A
for all
= 0
= Kerdn and zn(G,A)
and
n ~ 1.
Bn(G,A)
= Imdn-1
(n;;;,, 1)
as n-cocycles and n-cobound-
Bn(G,A)
Then-th cohomo'logy group
We set
~(G,A)
with coefficients in
is defined by ~(G,A)
= Zn(G,A}/Bn(G,A)
H°(G,A) = AG
(n ;;;,, 1)
GALOIS COHOMOLOGY
249
rt can be easi'ly shown that the conomology group Hn (G,A} restrict ourselves to normalized n-cochains.
is unaffected if we
Let us recall the standard fact
which says that gn(G,A) = 0
n ~ 1
for all
if
A
is co-induced
(2)
Formula (1) shows that a 1-cocycle is a crossed homomorphism, i.e. a map G'--+A
f
satisfyi'ng f(gg
I
l = gf (g 1 + f
f (g l
Similarly a 1-coboundary is a principaZ crossed homomorphism,
i.e. a map
f:G-A
for which there exists a EA such that f(gl = ga - a
In particular, if G acts trivially on A, H1 (G,A} =
then
Hom(G,A)
From (11 we see also that a 2-cocycle is a function f: G x G -
A such that (3)
while a 2-coboundary is a function map t :
G --'--->- A
f
: G
x G' -
A
for which there exists a
such that
Let E/F be a finite Galois extension with G = Gal(E/F). and multi'plicattve groups
E+
Then the additive
and EJi' of E are a-modules and we may consider
the groups for all Our first aim is to show that the group gn(G,E+l
is always trivial.
n
~ 1
This will
be derived as a consequence of a general result of independent interest, namely the normal basis theorem.
CHAPTER 2
250
Gl'ven a fi'ni'te Galois extension
E/F,
fo(a1 Ier E a EE,
for some
any F-basts gEG g
where a e
E
F)
is such that
fo(a) la
E
G}
l:: ;\ g(a} gEG g
is a normal basis.
Invoking (2)_, we
CHAPTER 2
252 are therefore left to verify tnat
Hom(U,F) For each A E Hom(a'.G,F)
and each
put Ag
g E G,
l
as a-modules
~ FG
Hom(U,F) -
Then the map
= A(g). .
FG
A i-- EA/
is a required isomorphism. • We now turn our attention to the study of ff"(G,E*), more complicated:
the ftrst cohomology is always trivial, but in contrast to
Corollary 18.2, the groups /l(G,E*} for n 18,3, THEOREM.
Then H1 (G,E*}
are not generally trivial.
> 1
Let E/F be a finite Galots extension with Galois group Go = 1
Let f: a-
Proof.
Here the situation is
E*
be a crossed homomorphism. a =
For \EE* we put
I: f(cr}crA CJEG
Because of the linear independence of the automorphisms o(Corollary 4.81, we may choose A E E* such that a ; 0, w
=
We then obtafo for ,
i ,f(crl(wAl = i f(,1- 1f(wl(,CJ"),} crEG
E (}
= f(,)- 1 a
crEG
whi'ch implies that f(, l
-1
,a_l
=
0\
Therefore f
E B 1 (.G,E*)
and so 8
1 (@,E*l
= 1,
•
In what follows, we shall f1'x the following notation: E/F G =
of
a finite Galois extension Ga 1 (EJF)., N . a norma 1 subgroup of G E B2 (G,E*l
ts a coboundary corresponding to f :
G
-+
(elf}( cr ,cr ) = er (f(cr )lf(cr cr )- 1f(cr ) 1
2
1
2
12
·
1
E*,
i.e. (cr ,cr E G) 1
2
K the fixed field of N.
The above assumptions guarantee that K/F ts a Galois extension whose Galois group is identifiable with G/N.
Each a
E
Z2 (G/N,K*)
determines a unique
253
GALOIS COHOMOLOGY
a*
z 2 (G,E*)
E
given by a* ( 0 , 0 } = a ( CT , CT ) 1
where
The map
ai = aiN' i = 1,2,
ca 11 ed inffotion map.
2
1
2
induces a homomorphism
at-->- a*
By restricting the co chains to
N x N,
we al so have a
homomorphism
ca 11 ed the restriction map.
These two homomorphisms a re connected by the fo 11 ow-
ing theorem. 18,4, THEOREM. (Hochschild(l950), Eilenberg and MacLane(l948}). l
-->-
The sequence
~ 8 2 (N,E*)
H2 (G/N,K*) J..!!f.,_ H2 (G,E*)_
is exact, Proof. Step 1.
For the sake of clarity, we divide the proof into three steps. Here we show that rnf is injective,
Assume that a
is such that a*
E 2 2 (GIN ,K*)
a*(0 ,0 1
Then
a(1,cr}
= t(1}
so that
2
1=
=
ot where t
0 (t(0 ))t(0 0 )- 1t(cr 1
2
12
l
(0 ,CT 1
t (0} = 0(t(1)),
Now set
t(d E K*.
1
i.e.
G-->- E*,
E 2
G)
Then
l
t
(0) E 1
x*
for all
implies that t
a
= 1
aos- 1 , We have
1
0
= 13*
E
c,
and t (an}
t (0) · for all
=
1
for some
1
S: GIN- K*,
it suffices to verify that a
where
0
G, n
E
E
This
N.
Setting
S*(cr) = S(CT).
is a coboundary. 1
a*= ds,
where
s(cr} = t(0}cr(t(1))- 1 •
If n ,n EN then
1
1
2
(ds)(n ,n ) = 1 1
By Theorem 18,3, we can find Let us set µ(nl = 1
µ(0)
'.\EE*
= s (0) '.\cr{'.\)- 1 •
for all
n EN,
Then
2
such that
oµ
=
os,
s(n} = n(>.)r 1
so that
a* 1
for all
= oµ.
Furthermore,
This gives (er
t.e.
µ(0n) = µ(0)
for all
n E N.
0 E G, n EN,
Also
E G,
n
E
N)
CHAPTER 2
254
Therefore n(µ(cr}} Thus we have µ
= µ(cr}
for all
for some u: G/N -
= u*
which means that µ(cr)
cr E G, n E N
E
K*,
which shows that
K*,
ct = ct oS = . er }t.. q-1 •
l
2
l
l
Owing to Theorem 18.3, we can find Because sa depends only on
2
1 . l
,cr}
Acr EE*
& and s 1
= 1,
such that we may take
Acr
GALOIS COHOMOLOGY
-
to depend only on
and ii. = 1.
(J
= a (a ,a la (ii.. )- 1\
a(a ,o} then a
If we deftne
1
1
2
1
1
2
a2
1
a a ii.~l l
2
ts obviously tn the same cohomology class as a
2
the same element of B2 (G,E*) only on a
and the coset cr
1
a (a n,a 2
1
2
l
as do a of a
2
=
mod N.
2
=
2
2
1
Cl early,
a (a ,no ]a (ii.
=
a 2 (a 1 ,al 2
1 · 1
l-1 ii.
2
1 · 0
2
1
2
cr
1-l Aa
l
a
ii.-
-
2
l
a
2
and o .
1
-1
1
2
1 2
l
l
Finally, by our construe-
lf
a2 (a l ,a 2 1
=
for some t E Z2 (G/N,K*).
= t*
2
means that the cohomology class of a belongs to
H2 (G,E*),
2
ii.-l
a2 (no 1 ,a2 la 2 (n,o 1a2 )_ a2 (.n,a 1 l
2
2
18.5. COROLLARY.
a (a ,a) depends
Moreover,
1
=
Thus a (a ,a l EK* and therefore a 2
and hence represents
1 which implies
n(a (a ,a J) = 2
l l
a ao a 2 1 2_11 l a (a n,a }a (ii. a (n,a )l A ii.: 1·1 2 1 CT 1· 2 O'CT u 1
provi'ng that a 2 (a- 1 ,a). depends only on 2 •
a (n,a}
and a.
1
a (a n,a ]a n(ii. 1
=
tion,
255
This
Im Inf, as required. •
ts of order m and if c is an arbitrary element of
N
then cm e lm(rnf)_ where
f:nf : Ii2(G/N,K*)--+ H2 (G,E*)
is the in-
flation map. Proof.
Let a E Z2 (G,E*}.
Define t : G -
t(a)
=
E*
by
naca,nl
nEN
Then we have (dt) (a
,0 ) = 1
2
= =
11 (a1 (a(a 2 ,n) la(a 1 er2 ,nl- 1 a(cr 1 ,n))
nEN
11 Ca(a1,a2n1- 1 a(cr1,a2 )a(cr1 ,n))
nEN
a(o
)m
,CJ 1
2
n (a(cr
nEN
,a n)- 1 a(a ,n)) 1
2
1
In particular, if a EN this yields 2
(8t)(a ,er}= a(a ,a )m l
2
l
2
whence cm is in Ker Res , where Res : H2 (G,E*l->- H2 (N,E*) tion map,
is the restric-
The desired conclusion now follows by virtue of Theorem 18,4. •
CHAPTER 2
256 Two Galots extensions
are sai'd to be isomorpMc H
E /F and E /F l
2
E
1
is
F-isomorphtc to E. 2
18.6. THEOREM.
let
Let
and
E /F 1
2
= Gal(E./F), i = 1,2. 1, .
G. 1,
be isomorphic finite Galois extensions and
E /F
Then there is an isomorphism
f
such that for any F-isomorphism g: E _,. E 1
p
2
a (CT ,CT g
l
Proof.
2
1=
a
then
Ag
1
l
,E*) 1
onto
g
a
1-r
a
g
is a homomorphism which maps
H 2 (G
,E*).
2
A is another F-isomorphism
If
2
E---+ E 1
t(CT}
=
-1
2
SE z (G ,E*1, SS,., 2
2
,-.
is a coboundary.
a, g
In fact,
S(CT,µ)_S(11,µ- 1CTµ).- 1 , then we have
Since
as can easily be verified. as required.
The unique isomorphism
a A = (ag ) µ ,
-1
we conclude that a,/\
is cohomo-
• of Theorem 18.6 i's called the natural isomorphism.
f
Next we introduce the following important homomorphism. 18.7. LEMMA.
for any a E cEH 2 (G,E*),
Let E/F be a finite Galois extension with Galois group Z2
(G,E*},
let
say c=a,
a
denote the cohomology class of a,
defi'ne
~
c' G -
( CT Then
c'
2
= µ, say, is an automorphism of E/F, and we have aA = (o;g)µ,
(ml (CT1,CT2 1 = S(CT ,CT }S (CT ,CT ) . ··12)112 logous to
f.
Because it has an inverse, it induces an isomor-
We now claim that, for any if we put
a
,e:)
induces the isomorphism
2
It is clear that the mapping
H 2 (G -1
I--->-
g( a(g -lCT g ,g-lCT g) 1,
coboundaries to coboundartes. phism of
the mapping
Cc 1 ,E7l- z2 (G 2
? where
2
i---+
F*/NE/F(E*) (1la(T,CT))NE/F(E*) ,EG
is a we 11-defi ned homomorphism and the map ~H 2 (G,E*)
?
--,. Hom(G,F*/NE/p{E*))
c1---rc 1
G and,
Given
,
GALOIS COHOMOLOGY
257
ts a homomorphtsm. Proof.
Given
z 2{G, E"'},
a. E
put '&(cr}
It is easy to verify that, for all
na.('r, a) TEG cr , cr E G, we have =
1
2
a (a( a } )_ = a( cr } ·1
which means that say
-1
as = olj'
a,
maps
into
G
2
2
If
F*.
SE Z 2 (G,E*)
is such that
B=
then
=
a(cr)S(cr)- 1
Thi's shows that the map
= NE;ig(a))
n,(g(cr)}g(wl- 1g(,)_
,EG
is well-deftned,
c'
By the multiplicative version of (3}, we have rn( a , a )_ = a(Tcr , a ) a( T, a a ). - 1 a( ,, cr } l
2
l
2
12
l
whence NE/F( a(cr 1 , a)) =
n
,EG
T( a( 0 , 0 l
}1
= a,( 0
2
a ) - la,( (J'
) a( - En~ (V)
?r
fr
,---+
We say that R acts dense Zy on V if for each family v ,v , .•. ,v 1
E End (V)
s
and each finite
v there exists r ER such that
in
n
2
8
8(1\1 = fl'(vi)
Let V be an R-module, let s
19.1. LEMMA.
generated as s-modu 1e.
Then
R
(1.;;; i.;;; n)
End (V)_
=
R V
acts densely on
and let V be finitely
if and only if the homomor-
phism ~ R-,,
?r
En~ (V)_
.- f r
is surjective. Proof. densely on
If the given homomorphi'sm i's surjective, then obvi'ously
Conversely, assume that
V.
v.
be a generating set for the s-module
acts densely on
R
If
8
E
End (V),
V
R
acts
and let V 1 ,.o.,Vn
then by hypothesis
s
there exists r ER with
proving that f r
=
8(v.) = f r-(v.). Thus, for any 1,· 'i n n 8( I: \ .v .) = f ( L \.v .) , i=l 1, 1, r i=l 1, 1,
in
s,
8. •
Assume that an R-module v is a finite direct sum of irreducible
19.2. LEMMA. submodules.
Then R acts densely on Put s
Proof.
=
End (v)
v.
and fix v E v,
R
E;
R.
Because
f
E End (V).
We claim that
S r1J =
for some r
:\ 1 ,\ 2 , ••. ,\n
f( V)
v is completely reducible, v =
Rv EB
w for some
R-
265
THE BRAUER GROUP OF A FIELD
w,
submodule
Let
V -
1r
!'.le tfte projecti'on.
R1!J
Then
71' E
and so
S
f( V) = f (TrV) = 'ITf ( V)
This shows that f(v}
E
as claimed.
Rv,
e E End (V} and let
Let
v , ••• ,v
and put s 1
=
End (fl}. R
Then
v,
and the assumption on
n
1
s
Define en
Ev.
fl -
i's a full matrix ring over s.
S'
one immedl'ately verifies that e
n r ER
Hence, by the first paragraph, there exists an element
by
fl
Using this
lies in End (~}. S
such that
which is what we wanted to prove. • For any F-algebra A we can form the so-called enveloping algebra A ®A 0 , F
where Ao denotes the F-algebra opposite to A. left multiplication by a on A,
and ar
i
A -,. At ::_
la
1-->-
Given a EA,
right multiplicationi En~
let ai denote Then the maps
(A)
a1
and iAo _.,Ar~ En~ (A)
?a
I-·--+ a
r
where A ={ala EA} 1
t
and Ar ={ala EA} r
are injective homomorphisms of p.-algebras.
By the associativity of for all
ab = b a Q; r r 1
Hence
A
A,
we have a,b EA
becomes a left (A® A0 )-modu1e upon defining F
(a,b,x EA}
(a ® b )x = (airl (x) = axb
19.3. LEMMA.
Let A be an F-algebra and let R
= A ®A 0 F
(t}
End (A)_ R
=
Z(Ali
:!!!
Z(Al
266
CHAPTER 2
(i't}
ts sim13le if and only if
A
(iii)
If A is simple, then
Proof.
i's an i'rreduci'ble R,-module is a field
Z(A)
The inclusion one way is clear.
(i)
f E End (A).
A
f
Then, by an easy computation,
Conversely, assume that and one then checks that
= f{l) ,Q,,
R
since
f(l) E Z(A)
End (A).
f E
R
The two-sided ideals in A are precisely the R-submodules of A
(ii)
(iii)
If
is simple, then for any Of a
A
a
as required.
A 0 A0
Z(A), Aa
= Aa = A.
Hence
U(A) n Z(A) = U(Z(A)),
•
19.4. LEMMA. (i)
E
E
Let A
be a central simple F-algebra
acts densely on A
F
A 0 A0 ~ Mn(F)
(ii)
where n = dim A.
F
Proof.
F
Put R
( i}
= A
0 A
and note that A is an irreduci b1e R-modul e by
0
F
Lemma 19.3(ii). (ii)
By Lemma 19.2, it follows that
By Lemma l9.3(i), Ind (A}~
F
acts densely on
R
A.
and so, by (i} and Lemma 19.l, the map
R
R -+ End (A)., r
f-->-
is a surjecttve homomorphism of F-algebras.
fr•
Since
F
End A
~
Mn (F),
we have
F
di'm R
dtm (End A)
= n2 =
F
and the result follows. 19.5. LEMMA. of u and
Let V,
F
•
u,v
be vector spaces over F and let
respectively,
V',V'
be subspaces
Then
(vi 0 V) n (V 0 V') F
Proof.
F
U1 0 V 1
F
F
We can write
u = u' EB for some subspaces
U",V" of
u
and
U"
v,
and
v = v' EB
V"
respectively.
Accordingly,
U © V = (U' 0 V') EB (U 1 0 V") EB (U" © V') EB (U" © V") F
F
F
Now the left-hand side of the required equality is
F
F
267
THE BRAUER GROUP OF A FIELD [(V' ® V'} EB (U" ® F
and ts clearly
r'll
n [(U' ®
F
V"l
EB (U 1' ®
F
V'll
F
U' ® V', • F
The next observati'on descrtbes a useful lattice isomorphism. 19.6. LEMMA.
Let
be a central stmple F-algebra and let
A
Then the map I -
B
be il,ny F-algebra.
is a lattice tsomorphism between the ideals of B and
A® I F
those of A~ B.
where
B
The inverse of this isomorphism is given by
ts 1'denttfied with 1 ®
Proof.
B
We claim that the given maps are inverse to each other.
This will
prove the result, for they establish a bijection which is clearly order-preserving and hence a lattice isomorphism. Let I
be an ideal in B,
Then A® I
F (A ®I) n B
ts obviously an ideal in A® B and F
= I
F
by virtue of Lemma 19,5, Conversely, let
so we need only verify that F nonzero a E J' can be written as a = a
F Jc A®
A® I c J',
and wtth
{a ,.,.,a}
acts densely on A,
A® A
0
-
1
F
® b
1
-:1-
n B,
= J
Then clearly
To this end, note that each
I, , ..
a ® b
-:1-
r
being a subset of an F~basts of A.
r
1
r
be an tdeal fo A® B and put
J
r
with O ,; b • E i
By Lemma l9,4(t},
Hence there extst xj,YJ EA for j E {1,2, •• ,,r}
F
such that Exja ,.y. = j
J
'Z,
& • , 1 .;;; i .;;;
Ex.ay.=
Therefore b 1 e JC A® I, -
I
r,
But then
'Z,
J
J
ExJa,.y.®b~=l®b EJ i; ,j 'Z, J 'Z, l
and simtlarly b '2,.. e
I
for i e
{2, ... ,p},
proving that
•
F
19,7. LEMMA. A,,i=l,2. '2,
Let
A
and
l
be F~algebras and let
2
B. '2,
® B ) = CA (B ) ® CA (B 2 }
® A (B 1
Proof.
A
Then CA
F
B
2
It is obvious that
1
F
2
1
1
2
be a subalgebra of
CHAPTER 2
268
To prove the opposite inclusion, fix an F-basis element of A 0 A F
1
of
{v.} 1,
can be uniquely written in the form
Then every
A 2
0
Ea.
1,
2
v.
1,
with
a.EA 1,
Given any b EB, we have 1
Therefore, if Hence a.1,
0
Eai
(B 1 0 B),
A
l
(B" 1 1 and so
E CA
0
vi E CA
then
for all
aib = bai
b
E B , 1
2
1
Ea • 0 v • E CA (B ) 0 A 1, 1, 1 1 F z
Applying Lemma 19.5, we i'nfer that CA
0
® A (B 1 F
F
1
2
B ) 2
C
(A
-
1
0
F
(B ) ) n
CA
(CA
2
2
1
(B ) 0 1 F
A
) 2
CA (B ) 0 CA (B ) , 2 1 1 F 2
as asserted, • 19,8, COROLLARY.
Let A
and A
l
Z(A
Proof,
be F-algebras,
2
0 A ) = Z(A
1 F
2
1
l
0 Z(A ) F
Apply Lemma 19,7 for the case B
1
19.9, COROLLARY.
Then
=
2
A
1
and B =A. • 2
2
Let A be a central simple F-algebra and let B be any F-
algebra.
(i'l A 0 B i's simple i'f and only i'f B i's simple F
( i i )_
Z(A 0 B) == Z(B) F
In particular, the tensor product of central simple F-algebras is again central simple. Proof.
Apply Lemma 19.6 and Corollary 19,8, •
We are now ready to prove 19.10. PROPOSITION.
The Brauer classes of central simple F-algebras form an
abelian group with respect to the multiplication induced by the tensor product, Proof.
Let
A
and
B
be central simple F-a 1gebras.
Then
A
0 F
is again a central simple F-algebra by virtue of Corollary 19.9,
B
== B 0 A F
Suppose that
1
THE BRAUER GROUP OF A FIELD
A "" D ® M (Fl
B ~ D
and
n
1
2
269
® M (F) F
m
Then A ® B ~ (D ® M (_F)) ® (D ® M (F)) ~ (D ® D ) ® M (F) F 1Fn 2Fm 1F2 nm
and so
® D.
A® B - D
F
1
F
Hence, if
A - A', B - B 1 ,
then
2
A'®B'-D
F
®D 1
provtng that the multiplication
F
-A®B, 2
[Al [Bl = [A® Bl
F is well defined.
F
The multiplication of Brauer classes is associative by the corresponding law for tensor products.
Taking i'nto account that
A® F"" A
and that, by Lemma 19,4
F
(iil,
A® A0 ="' M (Fl,
F
n = dim A,
F
n
we also have
[Al [F] = [A] ,
This shows that
[Fl
[A] [A 0
[F]
]
is the identity element and that
[A 0 ]
is the inverse of
[Al, •
The group of Brauer classes of central simple F-algebras is called the Brauer group of
Let
F and is denoted by
be an F-algebra.
A
field for
A
Br(FL A field extension
E)F
is said to be a splitting
if A® E =" M (El F r·
for some posi'tive integer
r.
Our next aim ts to show that
E which is a ftntte extension of
splitting field
F,
A
always has a
The following basic
result ts known as the Skolem-Noether theorem
19.11. THEOREM. suba 1gebra of
Let A.
A
be a central simple F-algebra and let
B
be a simple
Then every isomorphism of F-a 1gebras ¢ : B-,. B' ~. A
can be extended to an inner automorphism of in
A
A,
that is, there exists a unit
a
such that ¢(b ) = aba
-1
In particular, isomorphic simple subalgebras of
for all A
b E B
are conjugate and hence have
CHAPTER 2
270
tsomorpntc centraltzers, Let V be an irreductble (left} A-module, and let D = End (V).
Proof.
A
Then D ts a division algebra with centre F~lv and B 0 D is a simple F-algebra F
by Corollary l9.9(i).
We may view
V
as a
D-module via
B0 F
(b E B, d E D, v E V)
(b 0 d)v = b(dv)
Using the automorphism ¢,
we can deftne a second left B 0 D•module V 1 ,
whose
F
underlying vector space over F i's v, and witn tne acti'on of B
0 D
given by
F
(b
Then v and V'
are of tne same F·dimension.
Because B
B, d E D,
E
ts simple artin-
® D
F
e v-->- V'.
tan, we infer that there exists an tsomorphtsm of B ® D-modules
e E End
Since
F
(_v),
plication at for some a EA,
Furthermore, a E U{.Al
pnism.
is a
B ® F
D•tsomorphism, we have
a(o(dvU = ¢(o)d(avl
that ab
= 1,
e is a left multisince e is an isomor-
it follows from Lemmas 19,l and 19.2 tnat
D
Taking d
for all
b EB, d ED, VE V
and applytng the fact that A acts fattnfully on so ¢(bl
= ¢(_bla,
19, 12. COROLLARY,
VI E V )
Let
A
subalgebra with centre E.
for all
= aba- 1
b EB. •
be a central si'mp1e F..-algebra and let If dim B = k, p
v, tt follows
B
be a simple
tnen
A ~ B0 ~ CA(Bl ~ Mk(Pl ~ Mk(CiBlJ.
In particular, tf E ts a subfteld of A containing F, A® E ~ CA(El 0 Mk(F)_ , F F
Proof.
We may identify Mk(F)
ations of B, and
B
r
= (E;F)
witn the algebra of all F-ltnear transfrom-
As such it contains the subalgebras Bi of left multiplications
of right multiplications.
Clearly Bi~ B and Br~ B 0
simple,
where k
tnen
and, by Corollary 19.9, A
j
Mk(F)
is central
Now B ® F and F 0 B1 are isomorphic simple subalgebras of A® Mk(F). F
F
Hence, by Theorem 19,11 and LenJTia 19.7, CA(Bl 0 M,-.(F) ~ A ® B ~ A 0 8° F K. F r F
F
THE BRAUER GROUP OF A FIELD
19,13. COROLLARY,
271
Let A be a central simple F-algebta and let B be a stmple
subalgebra wtth centre E, (t}
CA(Bl
(ti'l
is simple with centre E
CA(C/BD = B
(.tiil dim A = {.dtm Bl(dim CA(Bl} F
Proof.
F
F
Comparing dimensions tn Corollary 19.12, we have
Bl= (dtm
(dim A}(dtm F
proving (tiil,
F
Since A® Tf
B)2 dim CA(.B},
F
F
is stmple, so ts
by Corollaries 9.12 and
CA(B}
F
9,9,
Replacing B by CiB1
tn (tit},
we therefore find
dim A = {.dim CA(BU(dim CA(C/Bll F
F
F
·
whence dim CA(C/Bll = dim B F
Since we have clearly cA(C/BU
F ~ B,
tt follows that CA(cA(B11 = B,
proving
(i i 1.
Finally, i'f K ts the centre of cA(Bl,
we also have
E ~ K,
19, 14, PROPOSITION.
t.e.
E = K.
Proof. CD(El = E,
and stnce
•
Let A be a central simple F-algebra, say A = M (D} n
D ts a divtsion rtng wtth centre
Then A.® E ~ Mnk(El F perfect square.
then K ~ E,
where k ·
=
F,
(p;:F},
and let E be a maximal subfield of D. In particular,
Applytng Lemma 19,12 for the case
A = D
dtm A= n 2 k 2 F
Hence
F
2!!
M (D ® E} ~ M (Mk(E}l n F n
£!!
is a
and taking into account that
we deduce that
A© E
where
M k(E) n
272
CHAPTER 2
as requ i'red. • Let
A
be a central simple F-algebra.
It is a consequence of Proposition
19.13 that
dim
A=
r2
F
for some integer r
and that there is a finite extension E of F such that A 0 E = Mr(E) F
Thus A always has a splitting field E which is a finite extension of F. say that
[A]
is sp tit by E,
We
or that E is a split ting field for the class
[Al,
if E is a splitting field for A.
for
[Al
That the notion of a splitting field
is well defined is a consequence of the following simple observation.
19.15. LEMMA.
Let E be a field extension of F.
Then the map
} Br(F)_--,. Br(.E) ( [Al ,..__,.
[A ©E] F
is a homomorphism whose kernel consists of those Proof,
[Al
which are split by E.
Let A be a central simple F~algebra and write A~ D ©F Mn (F). ..
Then A 0 E F
and hence
D 0 M (F) 0 E F n F
2e
D 0 M (F) F n
Thus the given map is well defined.
[A 0 El = [D 0 El,
F
2e
It is also
F
a homomorphism, since (A 0 B)_ 0 E F
2e
(.A 0 E) 0 (B 0 E)
F
F
F
The final assertion is a consequence of the definition of the splitting field for [Al•
• We denote by Br(E/F)
the kernel of the homomorphism } Br(F) (
Hence, by Lemma 19.15, [Al
which are split by
Br(E/F) E.
Br(E)
[A] f->- [A 0 E] F
is the subgroup of Br(F)
consisting of those
THE BRAUER GROUP OF A FIELD
[Al E Br(F)
We next present a criteri'on for 19.16. LEMMA.
Let
field extension.
Proof.
Let
Of c
be the division algebra in the class
D
E::. M/D).
invertible. 19.13(i),
such
[Al
n, v
C/E),
n
be
is an irreducible E-module.
is a dtvision algebra.
is the centre of
and let
[Al ,
determines a nonzero endomorphism of
c3 (E)
Thus E
BE
B = Mn(D}, v = Dn and observe that
Put
By the minimality of
c3 (E}
E
be a finite
E/F
E as a self-centralizing subfield.
is an E-module.
Now any
Br(E/F).
if and only if there exists
[Al E Br(E/F)
the least integer such that V
to be an e1ement of
be a central simple F-algebra and let
A
Then
B contains
that
273
v,
hence
c
is
Furthermore, by Corollary
l'nvoking Corollary 19.12, we also have
c3 (E) "'B 0 E - D 0 E F F E
By Lemma 19.15, this shows that
splits
D
E ts a self-centralizing subfield of
and on1y if
c8 (E)
if and only if
= E,
i.e. if
B. •
Our final observation on central simple F-algebras ts given by 19.17, LEMMA. idempotent of
Let
be a central simple F-algebra and let
A
A.
=
vector space over
D
D on which
an inner automorphism of
where
I
r
eAe
=a Mr(D).
A
A
=a Mn (Dl
Br(Fl for some
and
By changing the basis of an n-dtmensional
1.
acts irreductbly and faithfully, i.e. applying
A, we may assume that
is the identity
eAe
and so
n:;,,,
and some
in
[eAe]
By Wedderburn 1 s theorem,
divisfon algebra
be a nonzero
Then [A]
Proof.
e
=a[:r Thus
r x r-ma trix.
:J [eAel
Then
Mn(D)
[:r
[DJ
[A]
:J
[:r(D)
as required,
:J JI
CHAPTER 2
274
Let group
be a finite Ga 1ois extension and let
E/F
with respect to the natural action of
Z2 (G,E*)
remarked earlier, 2-cocycles.
G = Ga 1 (E/Fl. G on
Consider the As has been
E*.
is unaffected if we restrict ourselves to normalized
H2 (G,E*)
For this reason, we shall assume that all 2-cocycles are normalized.
For any given
ct E Z 2 (G,E*},
we shall construct a central simple F-algebra
called the crossed product of G over
This crossed product
E,
FfiG
lxG,
was intro-
duced by Noether and played a si'gni'ftcant role i'n the classical theory of central simple algebras.
r@
That the crossed products
occur in an unavoidable way in
the study of centra 1 simp 1e a 1gebras will be tll ustra ted by the fact that any central simple F-algebra is isomorphic to
such that E
A
for some
lJ,G
the cohomology class of
a
with basis
Given
G.
{g Ig
e: G}.
E Z 2 (G,E*L
In what follows, we write
& for
E Z2 (G,E*l,
ct
For the rest of this section, Galois group
is a self-centralizing subfteld of A
E/F denotes a finite Galois extension 1,Jith
ct E z 2 (G,E* l,
we denote by EctG
Thus each e 1ement of
EctG
a free 1eft E-modul e
can be uniquely written in
the form
Zxj We define the multiplication on
EaG
(xg E E, g E G}.
distributtve1y by using
(r ,r EE, x,y E G) (l} 1
2
where for all Recall that for all
A EE, g E G
x,y,z E G (2)
ct(x,ylct(xy,z). = xa(y,z)a(x,yz)
(3)
ct(x,11 = ct(l,x) = 1
Note also that, by (1),
gttg 19, 18. LEMMA.
l.
For any
ct
E
-1
=git'
xy=
Z2 (G,E*}, EaG
ct(x,y)xy
(g,x,y E G, A EE)
(4)
is an F-algebra with identity element
275
THE BRAUER GROUP OF A FIELD
It suffices to veri'fy that the multiplication given by (1} is asso-
Proof. ciative.
Indeed, in this case
is obviously a ring with identity 1 (the
EaG
latter can be seen by applying (l} and (3)_) and, by (4), morphic copy of
F
r ,r ,r
Fix
1
2
contained in
EE
and
Thus
Z(EaG).
{\•ll\
is an iso-
E F}
is an F-algebra.
EaG
We first show that
x,y,z E G.
3
for a 11
\ E E
( 5)
Indeed, applying (41 we have
x(y\jj
x(y\) =
proving (51.
-1
-1
)x
= a(x.,y)xy
\xy
-1
a(x,y)- 1
= a(x,y)xy \ a(x,y)- 1
We now have
(by (5}} = rxr a(x,y)xyr a(x,y)- 1a(x,y)a(xy,z)xyz 1
2
(by (2))
3
= [(r x}Cr 1
2
ifll Cr 3 zL
as required.• In what follows, we identify E wtth its copy 19.19. LEMMA.
For any a E Z2 (G,E*1,
Lemma 19.16,
(i}
Proof.
In particular, by
CEaaCE} = E.
[EaG] E Br(E/F).
EaG ~ Mn(F)
(ii'}
ectc.
in
the following properties hold:
is a central simple F-algebra and
(i)__ EaG
{\•ll\ EE}
if and only tf
for some n;;;. 1 Given
x = ix
g
gE
and
EaG
a is a coboundary.
\EE,
it follows from (l} and (4)
that \x
Hence
\x = x\
and
xg
t-
0
i.e.
g
=
1.
if and only tf
for some
This shows that
the other hand, if for all
g E G,
g E G,
µ E Z(EaG)
hence
µ E F,
= t\x - g g\x
g
and
= g\x
then
Thus
= Ixgg\g-
for all
x E CEa0 (E)
CEa 0 (E)
then
g
x\
= E.
g E G.
implies
Therefore, if \ = g\
It is clear that
µEE= CEa 0 (E).
Z(eGtGl = F.
for all F
:=
Z(EaG).
Furthermore, by (4)
x
t-
0
\ E E, On gµ=µ
276
CHAPTER 2 Gtven
that
x =
w:i El\;,
let !1,(xl denote the number of nonzero
Ffa.
ts a nonzero ideal of
I
choose such that has the smallest
Among the nonzero elements x Multiplying
i(x).
if necessary, we may assume that x 1 # 0.
Si'nce
y
g
Thus, by hypothesis,
1
i, e,
Ax - xA = 0
x
r.xj
of
I
Now for any A EE
f' 0 implies x . 1' 0 and since y = Of, x ,
g
=
by an appropriate g:... 1 ,
x
= Ey_q rr EI, where yg = Axg
Ax - xA
Assume
xg.
1
- x gA g
we have
E cg1iE} = E.
!L(Ax-xA) < i(x)
Hence I = Ea,G and so
ts simple,
EaG
ts a coboundary, say a = at for some t :
Assume that a
(it)
t(g}
=
1.
For each
g
E
t(g}- 1g
g=
put
G,
.
Then
{gig
E G}
G -
with
E*
is an E-basi's
of Ea,G and, for x,y E G,
xy
t(x)- 1 xt(y)- 1 y = t(xl-lxt(y)- 1 a(_x,y)xy = t(xyl- 1 Xb' =
Given
AE
E,
let
xy
k.
such that g (\) f g (\), 1
(A}s- ,,, SA = 0 1
since each
one easily verifies that
2
g
1
gE
u(A).
278
CHAPTER 2
gtves a shorter nontrtvtal relatton connecttng
Thts ts a contradtc.
tion and thus (91 ts establtshed. -1
Suppose now that mutes with each
Then the equation (81 shows that
x,y E G.
:>..EE.
Since
is its own centralizer in
E
xy
cocycle and thus we have shown that
with
implies that
A
A~ Ea.G
for some
com-
we obtain
A,
= a.(x,y}.iy
The associativity of multtplicatton in
xyxy
a.(x,y)
a. : G x G - - E* a. E Z2 (G,E*).
E*
E
ts a
•
The next property ts crucial for the proof of the main result. 19.21. LEMMA.
a.,S e z 2 (G,E*),
For all
there is a stmtlarity of F•algebras
Ea.G ® ESG - Ea.SG F
Proof.
Let
c = A ®B, where F
EB Eg and B = ia = EB Eg g€G gE-G is a central simple F-algebra by Lemma 19.20 and Corollary 19,9. A = Ea.G =
c
Then
a,'m ts to find an idempotent
e e E ®Es F
c
such that
eCe
~ & 6a,
Our
This will
yield the destred result by appealing to Lemma 19.17, We ftrst observe that the subfields
E®1
and
1 ®
E of
E ®E
commute
F
elementwtse, E = F(),1.
where
E ts separable over F,
Since Let
n = (_E: Fl.
f(X} E F[X]
there exists
be the minimal polynomial of
g
e
G - {l},
inator ts not zero stnce. A f. g(A.}
for each
also distinct from zero in
since the elements
1 inearly independent over
Then
degf(X) = n,
E ® E, F 1 ® E,
Hence
e
Observe that the denom-
g E G - {l}.
f. 0 in
E ® E,
F
f(X} = {.X-:>.)
ranges over
G - {l}.
The numerator is
{;\i ® 110
Next we note that
g
A•
such that
Now deftne
where these products are taken over all
where
A EE
n (X-g(:>..)}
Therefore, we must have
< i,;;;
n-1}
are
THE BRAUER GROUP OF A FIELD which shows that
(1 0 >.)e
=
(>. 0 l)e
i'n
E
0
279
Thus, by induction,
E,
F
>.he
(>.i 0 lle = (1 0 n-1
Because E
=
for all
i;;. 0
.
EB F>.1, i=O
and multiplication in E 0 E is commutative, we derive F
= e(l 0
(µ 0 lle = e(µ 0 1)
µ}
= (1 0 µ)e for allµ e
E
(10)
Hence e2
=
e1\(>. 01 - 10 g(;\l)/n(>.-g(>.)) 01
=
en ((>.-g(>.}) 0 1)/n (>.-g(>.)) 0 1
= e,
proving that e
is an idempotent in E 0 E. F
We are left to veri'fy that (11)
To this end, note that e(E 0 E}(; 0 y)e
?
eCe =
x,yE.G
t e(E 0 1)e•e(10 E)e(; 0 y)e
=
x,yE@
But e(l 0 E)e
=
e(E 0 l)e
by (10), and
We now compute e(; 0 y)e,
to E.
e(E 0 l)e = E'
is a field F-isomorphic
by using the formulas (µ
E E, x,y E G)
We have
where
If
x
f
From (10), we obtain
ranges over all elements of c - {1}.
z
y,
then
x(>.) - yz(>,) =
0
for
e(; On the other hand, when
x = y
z = y
0
y)e
=
x
0
we obtain e(; 0 y)e =
A stmilar argument proves that
-1
c; 0
y)e
E G-
{l},
and so
CHAPTER 2
280
e(x © yle = e(x ©
y)
The foregoing shows that
=
eCe
EBE'
g
gE-G
where and g = e(g
E'= e(E © l}e
©
g)e
(g
G}
E
and also shows that
g Clearly,
Gal (.E'/F).
e,,
g e(11 Hence, conjugation by
~fl
Thus the
=
(g © g).e
G and, for all © l).e =
e(g ©
g acts a~
µ
e(g © g).
E;
E, g E G,
g) (J.1
© l).e =
on E 1 •
g
:i:y }e
=
=
e(xy ©
=
/\ e(a(x,y}B(x,y). © l}e xy
result follows.
e(g(J.1) © l).e
E
G)
g
Moreover, for all
x,y
E
G,
= e(a(x,y)xy © B(x,y}xy)e
multiply according to the cocycle
g's
(g
aB,
This proves (.11) and the
•
We have now come to the demonstratton for which this section has been devel~ oped, 19.22. THEOREM.
Let E/F be a finite Galois extension, let G = Gal (E/F)
for any a E Z2 (G,E*).,
let
l"a
and
be the crossed product of G over E with
respect to the natural action of G on E,
Then the map
~H 2 (G,E*).->- B:r{E/F)
?
a
I-+ [~Gl
is an isomorphism, Proof.
Owing to Lemma 19. 19 (;), [EaG]
Then, by Lemma 19,21,
f
~ z2 (G,E*)
L
?
~
a
Br(E/F).
E
Consider the map
Br(E/F) [EaG]
is a homomorphism.
Furthermore, by Lemma 19, 19{tt},
THE BRAUER GROUP OF A FIELD
Kerf and thus
f(al
by
(G,E*)_
induces an injective homomorphism
f =
= B2
Now assume that
[EaG] •
exists BE [Al
f
Then, by Lemma 19. 16, there
E as a self-centralizing subfield.
B ~
voking Lemma 19.20, we deduce that
given
H2 (G,E*)-->- Br(E/F)
[Al E Br(E/F}.
such that B contains
281
EaG
f(a} = [EaG]
=
for some a
z2 (G,E*).
E
In-
Thus
[B] = [A]
and the result follows. • Our next aim is to show that every Brauer class contains a crossed product. This will be achieved with the aid of the following result. 19.23, PROPOSITION.
Every Brauer class of a field
has a splitting field which
F
is a finite Galois extension of F, Proof.
Let D be a division algebra in a given Brauer class of F,
We
claim that D contains a maximal subfield E such that E/F is a separable extension; if sustained it will follow, by Lemma 19.16, that E
is contained in some finite Galois extension
K/F, K
splits
E
Since
D,
will also split
D,
as
required, To substantiate our claim, we may harmlessly assume that charF
O.
= p >
Our first task is to show that
field extension of F, over F.
Hence, if there exists
are no separable extensions over
£
arable, i.e. that
AP
and
;\ if'. F
Z(D)
Choose
x ED,
but /(bl
=
f(c)
i.e.
and
A~ F such that
=
0,
O.
F,
A ts separable over F,
Assume by way of contradiction, that there Then, each element x e
D
is purely insep-
r
for some r = r(x).
E F
Let
E F,
contains a proper separable
D
To this end, note that each element of D is algebraic
F(Al is the required extenston.
then
DI F and that
f
be the mapping
We may therefore choose
x >--+ x\ - ;\x •
On the other band, we have fP(x) = xAP
b ED
such that
Then c
cA
=
Ac,
=/-
1
(bl has the form
Setting v
=
A- 1 c,
i
F
such
Then f f O since
= F,
f(bl f O and let
A !/'.
- Apx =
be such that
c = uA - AU
0 for all / -1 (b) # 0
and satisfies
we see that v commutes with
;\
CriAPTl:.R 2
282
u'\ .,.
C = /\ V =
Hence, putting w
=
uv
-1
we derive
,
\ -1 /\ = ( u'\-'\u,v = UV -1 /\
Thus w
+ '\w'\- 1 ,
= 1
but wq wq
=1 +
l
0,
=
-
/\UV
for some q
E F
(:\w).-l)q
a contradiction.
=
-1
= WA
pn.
-
/\W
But then
= 1 + :\r,F'\-l
(since wq
= 1 + wq
and so
/\U
E F)
Consequently, D contains a proper separable
field extension of F. Let E/F be a separable field extension in D of maximal degree. is a maximal separable extension of simple with centre E. Therefore, if cvtE} t- E, first part.
in
F
D,
By Corollary l9.13(i},
But D is a dtvis1'on algebra, hence so is
This would imply that
El'
is separable over
Thus
19.24, COROLLARY.
Let A be a central simple F-algebra.
CD(E)_ = E,
i'.e.
F,
is similar to the crossed product E°'G.
is
CD(E).
by the
contrary to the
i's a maximal subfield of D. •
E
Then there exists a
finite Galois extension E/F and a E Z2 (G',E*), where G = Gal(E/F),
Br(F}
CnfE)
then E has a proper separable ex tens ion E ',
maximality of E.
A
Then E
such that
In particular, every Brauer class in
contains a crossed product.
Proof. such that
By Proposition 19.23, there exists a finite Galois extension E/F [Al E
Br(E/F}.
Now apply Theorem 19.22, •
20. AN INTERPRETATION OF H~(G',E*} Throughout this section,
E/F
is a fi'nite Galois extension with Galois group G.
As usual, we define then-th cohomology group ~(G,E*) n-cochains.
This does not affect the isomorphism class of ~(G',E*)
useful in computations.
K/F
and is very
To motivate our discussion, we first introduce a dis-
tinguished subgroup ~(G,E*) Let
with respect tonormalized
of ~(G,E*).
be a finite Galois extension with H = Ga l(KJF) '
s=
K ::i E
Gal( K/E)_
and put
AN INTERPRETATION OF Hi(G,E*) Given
crE
H,
we put a'= aiE
283
so tn.at the map ( 1)
s.
is a surjective homomorphism with kernel aK
c!1' (H ,K*)
E
Each a
cfl(G,E*)
E
determines
given by ( 2)
aK( er , ••• ,a ) = a (er', ••• ,a') 1 n 1 n It is clear tnat if
a E
is split by K if aK
zn(G,E*), then
aK
is a coboundary.
E
We say that
zn(H,K*).
a E
zn(G,E*)
The map
izn(G,E*)--+ Zn(H,K*) (
a
aK
1--+
is obvi'ously a homomorphism (which carries coboundaries to coboundaries); hence it induces a homomorpnism
whose kernel consists of all
a
E
zn(G,E*l
which are split by K,
We put
zn(G E*) = u Ker¢ o ' - K K
where
K ranges over all finite Galois extensions of F containing E.
ed otherwise,
z~(G,E*)
consists of those
is a subgroup of
It is clear that
E.
I:f (G,E*).
containi'ng
zn(G,E*)
which are split by a
a E zn(G,E*)
suitable finite Galois extension of F containing
Express-
z~(G,E*)
We put
~(_G,E*) = Z~(G,E*)/ffl(G,E*)
Our aim is to provi'de an interpretation of of E (the group H~ (G,E*)
H~(G,E*)
in terms of the Brauer group
turns out to be trivia 1).
A precise description of
the main result is given below. For convenience, we first recall that the elements cisely those maps
G x G x G--+ E*
1
1
234
,a
of
Z3 (G,E*)
are pre-
satisfying
a(a ,a ,er )
a a(_a ,a ,o )a(a ,a
a
2
if any
= 1
3
,a ,a )
CT' )a(a 1234 123
=
a.1, =
a(a a ,a ,er )a(a ,a ,a a ) 1234
1234
1
(3)
(4)
CHAPTER 2
284
Also, for any normalized 2-cochain S: G x G ary oS: G x G x G---+ E*
cosH01,02,03 i
=
the corresponding cobound-
E*,
is defined by
01 sc02,03 1sc0102,03 r 1 sc0 1,CT2 CT3 is(0 1,CT2 i-1
(5)
is defined to be any (fi'ni'te-di'mensional) central simple E-al-
AG-normal algebra
gebra A for which every F-automorphism of E can be extended to a (rtng} automorphism of A.
It wi 11 be shown that the Brauer cl asses of G-norma 1 a1gebras
form a subgroup BrG(E)
of Br(El containing the image BrF{E)
of the homo-
morphism ~ Br(Fl. --.- Br(El
?
[A] t--->- [A ® E] F
The main result asserts that
20.1. LEMMA.
Let A and B be E-algebras such that the given automorphism CT
of E can be extended to rtng automorphisms tively,
and
f
Then there i·s a unique ring automorphi'sm
g
f ® g
of A and B, of
A® B
respec-
such that
E
Cf® glCa ® bl = f(i!I)_ ® g(bl
Proof.
Si'nce any ring automorphi'sm of
A® B
i's uni'quely determined by its
E
values on a® b, a EA, b EB, morphi'sm of
A ® B.
i't suffices to show that
Consider the map
A
x
B
--L. A
E
f ®g
® B, (a ,b)
i----+
is a ring autof(a) ® g(b).
E
Then iJ; is obviously biaddittve and, for any >.EE, iJ;(a,)J,1
= f(a)
® g(:\bl
= f(al
® 0(:\)g(b}
= CT(;\)f(a)
® g(b)
= f(>.a) ® g(b) = iJ;(:\a,b}
Hence
f ®g
A® B --+A® B
E
Since
f ®g
E
is a z-module homomorphtsm with (·.f ® g 1-l = f-1 ® g -1
obviously preserves multiplication, the result follows.•
20.2. COROLLARY.
If A and B are G-normal algebras, then so is A ®B. E
Proof,
Let CT be any P-automorphism of
E.
Then there are ring automor-
AN INTERPRETATION OF H~(G,E*) phisms
and g of
f
20,1, we see that
A
f ® g
and
285
respecttvely, whtch extend cr,
B,
is a ring automorphi'sm of
A® B
Applying Lenuna
such that
E
(f
as required.
for al 1 A E
1} = cr(J,.) ® 1 = cr(J,.)(1 ® 1)
Let A and B be E-algebras and let
f,,g. 1., 1.,
be ring automorphisms
respectively, extending the automorphism Ai of E for i = 1,2.
Proof,
E
•
20.3. LEMMA. of A,B
® g}(\ ®
By Lemma 20. l, both sides are rl'ng automorphisms of
A® B. E
(f1 ® g 1 ) (! ® g 1(a ® b 1 . 2 2
Since
(a EA, b EB) = (f
the result follows.
Then
!
1 2
® g g
1 2
1(a
® b) ,
•
As a preltmtnary to the next result, let us recall the following standard fact, Assume that {e .. } and
(i}
r =
~
1'
2
and D
D
1
2
n
1
are dtvtsion rings and let
1 and
respectively.
M (D }, r 2
and there extsts a untt u of R such that u
= ~-lD
D
1
be the matrl'x units of M (D
1.,
2
Ci i l
n
{v.J}
1.,J
Then
= M (D) = M (D 1 where
R
1
vij -- u -l eiju
for a 11
i,j E {1,2,.,.,n}
(see Jacobson (1968,p.26}1. Now assume that R = M (D) n
morphism of R.
cp(Dl
= u- 1Du
Then R
where
D
is a dt-vtsion ri'ng and let
where D' = 'f'A-(D).
= Mn.(D'l
for some unit u of R,
be an auto-
cp
Hence, by the above,
It follows that 1/J
= i'
u
cp,
where
i (x) = uxu- 1 , u
is an automorphism of
D
such that
1jJ(D} 20.4. LEMMA. (ii)
(il
= D
and 1/JIZ(D)
=
cp!Z(D)
If an algebra A ts a-normal, then so are all similar algebras
The Brauer classes of a-normal algebras form a subgroup BrG(E)
the i'mage Br>iE)
(6)
of the homomorphism
containing
CHAPTER 2
286
~ Br(F) ..- Br{E) (
[A]
[A ® E]
f---->
F
Proof.
(i)
Let D be a division algebra over E with centre E.
suffices to show that Mn(D)
is a-normal if and only D is a-normal.
G-normal, then obviously so is M)D). ma 1 and 1et
;\
of
Replacing
M
n
(iil
(D).
By hypothesis,
E G.
by
¢
It is clear that if
20.2,
;\
is a-normal, then so is A
0
¢,
Hence, by Corollary
•
Let A be a central simple F-algebra Then, the map
i)! : A ® E _,. A ® E F F
is a ring automorphism which restricts to '.\
= a® '.\(e)
as required.
is G-nor-
as i'n (52, the assertton follows.
i)!
A
If D is
extends to a ring automorphism, say
be an F-automorpntsm of E.
given by Ha® el 1 ® E,
;\
is a subgroup of Br(El,
BraCEl
and 1et
Conversely, assume that Mn(D)
It
on
•
The following observation will allow us to tie together G-normal algebras and certain E*-valued 3-cocycles, 20.5. LEMMA.
Let A be a G-normal algebra and, for each
a (rtngl automorphism of A,
define
i(ul f
E
AutA
A
extending
with
x,
by i(u)Ca) = ucm- 1 , a and
: G X G _,. U(A).
t : G
X
f(1,y)
= f(x,1) = 1
let
cr(x)
Tnen there exist maps
EA,
such that
E*
for all
x,y
E G
(7)
for all
x,y
E
G
(8} (9)
0(xl [f(y,z),Jf(x,yzl = t(x,y,zlf(x,y).f(xy,z)
ff
t(x,y,z) = 1
Proof.
The product
on the centre
0(xl0(y)0(xy)- 1
E the identity
0(1). = 1,
we may and do choose
x,y,
or
( 10)
z = 1
is an automorphism of A which induces
1 = xy(xy)- 1 ;
phism (by Theorem 19. 11 ) , say i [f(x,y }]
be
For any unit u bf
0(1} = 1.
GX G -
0(x)0{y1 = i[f(x,y)J0(xyl
x E G,
it is therefore an inner automor-
for some
f(1,y) = f(x,1),
f(x,y)
E
U(A).
Si nee
which proves (7) and (8).
Using the equation (7). and the identities i(uv). = i(u)i(v), ;\i(u) = i( ;\(u)) ;\
we may calculate the triple product
cr(xlcr(y).0(21
(;\ E
in two ways as
G,
u,v E
U(A))
287
AN INTERPRETATION OF H~(G,E*) [cr(x)cr(y)Icr(zl = i[f(x,y}f(xy,zl] cr(xyz} cr(x) [0(y)0(z)J = ifo(x) [f(y,z)J f(x,yz)}a(xyz)
These results must be tdentical; hence if we write b ,b 1
of i
on the right, we have i(b ) 1
b b- 1 = t 2
1
= i(b ) 2
2
or i'(.b b- 1 ) 1
is an element in the centre E of A,
2
for the two arguments which means that
= 1,
This proves (9).
Finally,
the special choices made above also prove (10]. • The following result will be derived at the end of the section as a consequence of some general facts of cohomology theory. 20.6. LEMMA.
Let t : G x c; x G
-->-
be as in Lemma 20.5.
E*
Then
t E Z3 (G,E*)
(i).
(ii)
For fi'xed choices cr{xl a different choice of f(x,y)
U(A)
E
replaces t
by a cohomologous cocycle, and a suitable different choice of f(x,y) t
replaces
by any specifted cohomologous cocycle
(iii)_
If the choice of the extensions cr(x)
of x
altered, and if suitable new values of f(x,y)
to automorphisms of A is
are chosen, the cocycle t
is
unaltered. In what follows we call
t
a TeichmuUer cocycle of the a-normal algebra A.
Our next aim is to provide a complete descriptton of the normality of a crossed product and the form of its Tel'chmuller cocycle,
This will be achieved with the
aid of the followtng three prelimtnary results, Let K/F Be a finite Galois extension wi'th k
s=
Gal (K/E),
For any a
E
E and put
li =
Gal(K/F),
z2 (s,K*l, we denote by Ka,S the corresponding
crossed product of s over K.
Recall, from Lemma 19, 19, that Jf's i's a
central simple E-algebra with c
rs
20.7. LEMMA.
~
(_K) = K,
With the notation above, assume that K /E is a field extension l
and that there is an isomorphtsm f: K morphism of E. Proof.
Then K and K
Put A= flE
1
such that f\E
is an F-auto-
are E-isomorphic.
and note that, since K/F is normal,
extended to an p-automorphi sm, say ly an E-isomorphism, •
K
1jJ,
of
K.
Then
fi}! : K
--->-
-1 A
K
can be is obvious-
288
CHAPTER 2
a.. norma1 tf and only tf every F-automorphi·sm can be extended to a ring automorphtsm of Ifs.
20,8, LEMMA. of
K
Proof. K,
The algebra ifs ts
Since each g E G = Gal(E/F}
can be extended to an F-automorphism of Suppose conversely that Ifs
the given conditi'on ts obviously suffi'cient.
ts a-normal. fi'eld
For each :>.EH,
K of ifs
the extension
cr(:>.'}
onto an tsomcirphtc subfield Kl
by Lemma 20,7, there exists an E-tsomorphism K
containing :>.'(E) = E, . .
The product ¢cr(A '1
morphism of Ifs whose restrtctton to K is an element, say :>. that :>. µ
l
IE
= A',
Thus
?>.A
-1
=µ
l
ts an E-automorphism of
can be extended to an E-automorphi'sm
of Ka,S.
1/J
Hence
which by Theorem 19, 11,
K,
->l
¢ of Ifs,
can be extended to an automorphism
of :>.'=:>.IE maps the sub-
ts an auto-
of H,
l
such
By Theorem 19,11,
K.
Finally,
iJ;qia(?,, 1 )
is a
ring automorphism of ifs carrying K tnto K by the automorphism ?>.?>.- 1?>.
~" = I
l
= :>.;
l
it is the required extenston of A to ifs.
•
In vtew c,f (11, we have an exact sequence 1 ;~here 11 -
--+
S
Ii
--+
G i's the restriction map,
--+
l
--+
Hence 11 may be described as a group
extension of s by G by chciostng for each v(.1)_ = 1,
G
an extension v(g} E H,
g E (}
and defi'ni'ng the automorphism 8 i---- g*s of the normal subgroup
with S
by g*3 = v (g )_s v(g)_
Then there ts a "factor set''
-1
of elements of
e(x,y)_
s
such that (x,y
v(x)v(y} = e(x,y)v(.xy)
(g
v(g)s = (g*s)v(g)
Any elements
E
G)
( 11)
G, s E S)
(12)
E
h ,h EH can be expressed uniquely as l
2
h
l
= s 1 v(g) 1
,
h
2
= s 2 v(.g) 2
G)
(13)
1e(g l ,g 2 ) )
( 14)
(s ,s ES, g ,g E l
2
l
2
Their product, in view of (11} and (12), is given by [s
1
v(g }J l
[s
2
v(.g
2
11
= s 'v(g g l
12
l
(8 , = s (g *$ l
1
l
2
AN INTERPRETATION OF H3 (G,E*)
289
0
The factor set 0 satisfies the associ'ativity condition (x ,y ,z
[x*G (y ,z )J 0 (x ,yz) = 0 (x ,y )~ (xy ,z)
G)
{15)
E S)
(16)
E
and is related to the * operation by (x ,y
20.9. LEMMA. s Es
The crossed product
Ko;S
s
E G,
is G-normal if and only if for any g
G,
E
there exist m(_g,s) EK* such that m(l,s)
=
m(g,1)
1
and ( 17)
m(g,fJS ) [v(g)o;(s,s )_] = m(g,s) [(g*s)m(g,s 1 )]a(g*S,g*s ) 1
Proof.
1
Suppose f1'rst that
1
i's G-normal,
Ko;S
Then, by Lemma 20.8, v(g)
can be extended to a ring automorphism, say a(_g),
of KaS,
with 0(1)
=
We
1,
know that multipli'cation in Ko;S l's determined by
; s a(s,s e/\ = s(';\)_$
)ss
=
l
In (191 set ;\
=
1
ES)
( 18)
(;\EK)
(19)
(s,s
l
1
v(_g)- 1µ for any µEK and apply the automorphism a(g)
to
obtain [a(g)_s] µ = [(g*s)µ] [a(g)e)
On the other hand, (19) implies that g*B-
µ = [(g*s1ril
g*s
-1
These two equations assert that l tes in
K,
[o-(g)s] g*s·
commutes with every µ
E
K,
hence
so we may write (20)
a(g)s = m(g,s)g*S
for some m(g,s) E
K*
with m(g,1)
Applying
= m(1,s) = 1.
to both sides
a(g)
of (18), we obtain m(g,s)
gn
m(g,s } g*'s 1
= 1
[v(g)a(s,s )]m(g,ss ) g*sfJ
The left side, by applying (18} and (19}, is
1
1
1
CHAPTER 2
290
Compartson with the rtght side gtves the required equation (17}. Conversely, given such a functton
s,
on
a(g)
and to defi'ne a(g} :
we may use (20) to define the effect of
m,
additively by using
Ka,S--.- KaS
a(g}(),s) = [v(g)\] [o(g)s]
(\
By reversing the above computation, one immedi'ately verifies that automorphism of
extending the given F-automorphism
Ka,S
of
g
K)
is a ring
a(g) E.
E
•
We are now ready to provtde a complete description of the normality of a crossed product and the form of its Teichmuller cocycle. 20.10.THEOREM. (Eilenberg and MacLane(19481)_, Galois extensions with and for a given
s
over K.
a
E
Gi'ven
is G-normal with
KaS
exists a cochain
let
K:: E,
(:;
z3 (H,K*)
tK E
SE C2 (H,K*)
1
of
KaS,
Assume that
Ka,S
2
(for all
2
t
i's G-normal with
f(x,y}
of
KaS
s ,s ES) 1
2
(22)
as one of its Teichmuller to an automorphism o(g}
v(g), g E G,
t
Owing to Lemmas 20.5 and 20.6,
0(1) = 1.
from a suitable unit
Then
(21)
K 1
We may extend, by Lemma 20.8, with
be defined by (2).
such that
S(s ,s )_ = a,(s ,s )
cocycles.
Gal(K/E),
=
as one of its Teichmuller cocycles if and only if there
t
8S = t
Proof.
be finite
KIF
be the corresponding crossed product Of
let
t E Z3 (G,E*},
and
Gal(E/Fl, H = Gal(K/F}, S
=
let ifs
Z2 (S,K*),
Let E/F
satisfying (71,
can be obtained
Consider the map
given by (s
n(sv(g)J = i'(s)a(g)
i(sl
where
S.
If
rs
denotes the inner automorphism of
h ,h E H, 12
say h
n(h
1
1
= s v(g ) , h 1
)n(h)
1
=
2
=
i(; i[i
1
= s 2
)a(g
2
1
then
2
2
)a(g) 2
(a(g ); Da(g )a(g) 1
l
2
S, g E G)
obtained by conjugation with
v(g ) ,
)i(;
E
l
2
AN INTERPRETATION OF H3 (G,E*)
291
0
Define S(h
1
{by (7})
i[; (cr(g }; 1f(g ,g }] a(g g )
=
1
1
2
1 '2
1
2
using the notation of (14) for the product h h ' I 2
,h ) E v(Kas), 2
=;
by
(a(g ); )f(g ,g)
{23)
~(h )~(h ) = '[S(Tz ,h )J ~(h h )
(24)
B{h ,h )s•
121
1
12
12
Then the previous equation becomes I
2
This shows that conjugation by S = S(.h ~(h )_~(h )~(h h I
SE
2
K0,S
SC1,h
2
12
1-1 which is on
I
1,,
,h 1
2
2
12
1 induces on
the identity h h (_h h )-1; in other words,
K
12
commutes with all elements of
hence
K,
S(h
g 1 = g2
= 1,
ted to
s
x
and
s
S(s ,s 1
2
12
,h ) l
l = S(h 1 ,11 = 1, so that B is a 2-cochain of
2
E K*.
}ss l 2
I
ss .
=
coincides with a,
2
then s'
2
s 1s 2 ,
=
1
It follows from (18) that
2
l
1
Clearly
over K*.
H
B choose h = s, h = s;
In the definition (23} of
an automorphism
KaS
B,
restric-
as asserted in (22},
as,
Now we calculate the coboundary
for arguments
h ,h • l
2
and
h
= 3
s v(g ), 3
3
and with the notation 03), (141, and h h 2
= 3
s'v(g g l, 2
Because multiplication in
2
2
2
* s 3 ) e(g 2 , g 3 1
= Tz (h h) = EV(g g g)
123
EE
2
ts associative,
H
{h h )Tz
where
s ' = s (g
3
123
123
S is determined from both products as E
= s'[(g g )*s ]0(g g ,g) = s (_g *s'}e(g ,g g). 1 12 3 123 112 123
Using the definition (23) repeatedly, we calculate that S(h ,h )B(h h ,h )s 12
= B(h ,h
123
12
=
B(h ,h
)S(h h ,h )s' [{g 1231
ls'"
121
(cr(g g
g
12
)*S ] 0(g 3
)s )f(g g ,g}
123
123
=
s1 (cr(g 1)s2 )f(g 1,g2 )[a(_g 1g 2)s3l f{g 1g 2,g3 )
=
xf(g 1 ,g 2 lf(g 1 g2 ,g 3 l
(25)
where X
=
S (a(g ) S )f(_g ,g ){a(g g ) S /f(g ,g 1
=
1
g ,g )
123
2
l
2
12
sI (a(g 1 )s 2 )(_a(g l )a(g 2 )s 3 )
3
I
2
rl
CHAPTER 2
292
s,
On the otfier fiand, using tfie second express i'on for
we have
[h S(h ,h )l S(h ,h h )s = [h S(h ,h )ls [o(g )s'l f(g ,g g ) 1
2
3
1
23
1
2
3
1
1
= ;; [s- 1 h S(h ,h 111
23
2
1
23
)l [o(g )s'l f(g ,g g ) 12
123
=
s o(g
=
sl o(g l } [s 2 (o(g 2 )s 3 )f(g 2 ,g 3 )l f(g l ,g 23 g )
=
X
1
1
) [S(h ,h )s'l f(g ,g g ) 2
3
2
1
23
[O(g lf(g ,g }] f(g ,g g ) 1
2
3
1
2
3
By the relation (9), this becomes xt(g1
,g2
Applying (25} and the fact that
,g )f(_g1 ,g 2 lf(g 1g 2 ,g)
lies i'n the centre
t(g ,g ,g ) 1
2
3
E
of
KCY,S,
we
deduce that t(g ,g ,g }B(h ,h )B(h h ,h ) = [h B(h ,h )l B(h ,h h ) 123
h'
But
=
1
oB
g, h' 1
= tK'
=
2
12
g, h' 2
3
=
g,
123
1
23
123
so by (51 and (21 this equation proves that
3
which is (21) in the theorem,
Conversely, assume that there exists a cochain and(22),foragiven
tEZ 3 (G,E*).
Define
BE C2 (H,K*) for
m(g,s)EK*
satisfying (21) gEG,sES
(26)
m(g,s)S(g*s,v(gl)_ = S(v(g) ,s)
rs
To show that
by definition of
by
is c~normal, we must demonstrate that
satisfies (17), which
m
m reduces to the identi'ty
B(v(g),ss ) [v(g)S(s,s }] S(g*s,v(gl} [(g*s)B(g*s ,v(g)}] 1
1
=
Since
&S = tK'
1
B(v (g l,s )[ {g*s 1B(v (g ),s
H fol lows that
lies in the subgroup
8S(h ,h ,h 1
s.
2
3
1
1
ll
B(g*s ,g*s ) S(g* (ss ) , v (g)} 1
1
vanishes when any one argument
Hence
1 = ..) = 0'(\ 1 } ® '.\
Lemma 20.l, there is an automorphism on the respective factors with
295
cr('.\ 1 )
is extendible to an automorphism of
and
'.\
of
and which has proving that
A® K,
A® K F
which agrees
cr*(l) = 1.
A® K
E
Thus
>..
is H-normal.
E
Applying Lemma 20.3 twice, we have O'*('.\)cr*(µ)
= [cr('.\ 1 } =
® '.\]
[O'(µ '} ®
= cr(\ ')cr(ii '}
µ]
®
'.\µ (by (7))
i[f('.\',µ'}l cr(\'µ')@ \µ
= (i [f(\, ,µI)] ® l )CO'((\µ) I l ® AP)
(i[f(Y,11 1 )1 ® lHcr*(>..µ).}
The automorphism i[f(\',µ')l ® l f*('.\,µ)
=
f(). ',µ ')
is conjugation in
A® K
by the element
E ®
1
and thus cr* (\)O'* (µ) = i [f* (\ ,µ)J cr* (\µ}
f*
i.e.
satisfi'es condition (7) with respect to
muller cocycle
t*
of
A ® K
H.
The corresponding Teich-
can therefore be derived from
f*
by formula (9)
E
as
t*(\,µ,v) Thus
t* = tK
t(>..',µ',v') ® 1
=
and the result follows.
•
We have now come to the demonstration for which this section has been developed. 20.12. THEOREM. (Eilenberg and Maclane(l948). TeichmUller(l940)}. fi'nite Galois extension with Gale>1's group Br(El
let
G,
BrG(E)
t(A)
iBrG(B 1___!_,. H (G,E*) 3
?
[A]
I---->-
t (A)
is a homomorphism such that KerT In particular
= BriE)
and
ImT
t(A)
of a a-normal algebra
the map
= H~(G,E*)
be a
E/F
be the subgroup of
consisting of Brauer classes of a-normal algebras, and let
cohomology class of a Teichmuller cocycle
Let
be the A.
Then
CHAPTER 2
296
Here
BrF(E)
form
[B ® E],
Br(E)
is the subgroup of where
[Bl
consisti'ng of all Brauer classes of the
Br(F).
E
F
Proof.
Let
respectively.
1
2
We claim that
A ® A I
Indeed, choose extensions satisfy (7) and (9) for
t ,t,
be two G-normal algebras with Teichmuller cocycles
A ,A
CT .(x) ~
.
E
2
l
in
~
A ®A.
to
1 E
2
2
f .(x,y)
Then, by Lemma 20, l, x
1
as its TeichmUller cocycle.
t t
and elements
i = 1,2.
an extension of the automorphism
has
to
A . , x ,y E G, ~
G(x)
G (x) ® CT (x)
=
1
is
2
Moreover,
2
f(x ,y l = f (x ,y) ® f (x ,y) l
A ®A
ts a unit of
I E
2
and
2
i[f(x,yll = i"[f (x,ylJ ® i[f (x,y)l 1
Applying Lemma 20.3, we deduce that lating (91 one deduces that parti~ular, for any
n > 1,
CT
t = t t 1
2
2
and
f
are related by (7), while calcu-
is a Tei chrnu'll er cocycl e of
I
Mn (E) - - has t
E
2
as a Teichmuller cocycle.
= 1
In
® A •
A
There-
fore, any two similar c-normal algebras have the same Teichmuller cocycles,
Thus
is indeed a homomorphism.
T
To prove that
KerT = BrF(El,
we must show that
(a)_
For any centra 1 simple F-a 1gebra
(b)
If
for
k = (E:F)
A
B, B ® E F
has a Tei chmull er cocyc 1e l B ® E ~ Mk(A)
is a c-normal algebra with Teichmuller cocycle 1, then and for some central simple F-algebra
F
.
In particular,
B.
[Al = [B ® E] F
Property (a) is the special case consider
A with
t(A) = 1,
K = E, E = F
E
To prove (b),
Using (9), we can construct an algebra
elements uniquely represented as sums g
of Lemma 20.10,
Z:a
a
~
with coefficients
a
g
B with
EA
for each
G with multiplication table determined by the distributive law and the rules
ga= xy
[CT(g )al
g
= f(x,y)xy
(g
E
(x,y
G, a E
E
A)
(29)
G)
This construction is exactly analogous to the construction of a crossed product
AN INTERPRETATION OF H~(G,E*)
algebra over a fl'eld. (_9). that
Si'nce
t(x,y,z) =
for all
1
i
is an associaUve algebra with
B
particular,
of the argument in Lemma 19.19 shows that
C/El
forward computation shows that is a central simple F-algebra.
E
{a•i\a
is simple.
B
= A;
x,y,z
G,
tt follows from
as the identity element.
is identifiable with the subalgebra
A
297
EA}.
In
A repetition
. Furthermore, a straight-
hence by (29),
= F.
Z(B)
Applying Corollary 19.12, we have
Thus
B ® E
,:!!
B
Mk(A}
F
k = (E:F),
where
which clearly implies (bl.
ItnT = H~(G,E*)
We now claim that the assertion
(K
following two statements (c)
ts a finite Galois extension of
A is split by
If the a-normal algebra
cocycle
t
(dl
t
If
A, tK
of
K ~ E,
tK
is such that
Indeed, if (c) is true, then choosing as si'tion 19.23}, we see that
A ® K
!:!!
t
ts the Teich-
K,
K a splitting field for A
sJ(G,E*). 2 I'mT,
(see Propo-
On the other hand, (dl ensures that
as claimed.
Property (c) ts a consequence of Lemma 20.11. then
then for any Teichmuller
ts a coboundary, then
muller cocycle of some G-normal algebra split by
ImT 2 HJ(G,E*).,
F}
ts a coboundary.
Z3 (G,E*)
E
is a consequence of the
for some
M (E)
E n (with arguments i'n
n ;,i, 1
Indeed, if
A
ts split by
and every Tei chmull er cocycl e for
g) is a coooundary, and Lemma 20.11 shows that
tK
K,
A ® K
E ts a co-
boundary.
To prove (.d}., suppose that Then
tK =
oS
t E z 3 (G,E*l
for some 2-cochatn
s··
of
is such that
H ,~n
K*,
is a coboundary.
tK
In parttcul ar,
dS(x,y ,z 1 = 1 for all
x,y,z
Z2 (.s,K*)..
E
so that the definition
Si
We may therefore construct the crossed product
Applying Theorem 20.9, we conclude that Teichmuller cocycles. 20.13. COROLLARY. 20.12.
a(x,yl = s(x,y)
If
K/F
Let
~S of S over K.
~s is a-normal with t
as one of its
This proves (d) and hence the result. • T: BrG(E) -
H3 (G,E*)
be the homomorphism of Theorem
is a finite Galois extension of
carries the elements of
gives an element of
Br/81
~b.ich are split by
F
with
K
2
E,
then
T
K onto those cohomology
298
CHAPTER 2
classes
t
for whi'ch t e
e H3 (G,E*).
Proof.
Z 3 {6,E*l
ts spltt by K,
Thts ts a direct consequence of properties (c) and (d} established fn
the course of the proof of Theorem 20.12. • 20.14. COROLLARY.
Further to the assumptions and notations of Theorem 20.1_2,
assume that G ts cyclic. central stmple F-algebra
Then, for any G-normal algebra A, B
there exists a
such that where k
Proof.
The general reduction theory for cohomology groups (see Etlenberg
and MacLane(l 947a,§16} 1 proves that H3 (G,E*l =
= (E:F}
8 3 (G,E*l ="' H1 (G,E*),
Hence, by Theorem 18.3,
Thus the Tetchmull er cocycl e of A ts 1 ,
1.
The destred con cl u-
s ion now follows by virtue of property (bl established tn the course of the proof of Theorem 20.12. • Turning our attention te o~(.G,E*l, we now prove 20.15. THEOREM. (Eilenberg and Maclane(1948).l.
~et E/F be a fintte Galois
extensf'on wtth Galots group a.
=
Let
Proof. K/F
aE
Then s:(G,E*l
z 02 (G,E*l, .
t.e,
aK
= &(3
1,
for some f3
E
i's a suitable ftntte Galots extenston with Galets group
s.
c1 (H,K*l,
and with
ll
Define f3 0 (sl = f3(s)
for all
and
Hence, by Theorem 18.3, there ts a constant
(3 0
e Z1 (S,K*l,
that f3 0 (slb h EH.
subgroup
=
Then
s.
s(bl (3 1
for all
s
s-
e
where K.:: E.
Then, stnce a ts normalized, of30 = l
f\
Deftne
es,
is cohomologous to
(3,
and
as
b EK*
B/h) = B(h)b/h(b)
~(3 1 = CJ.K'
such for all
whtle f3 1 ts 1 on the
In parttcular, because a ts normalized, for all
s E S
and so f3 1 (hs} = [hf3 (slJ f3 (h)., B (sh) = [sf3 (h).J S (s} . 1 1 1 1 l
Because B1 (s)
= 1, B (hs) = B (h) = B (sh} 1
Hence each value 13 1 (h} . tension all
v(.g) E H
x,y E G,
and
1
1
lies tn E*. v(.x)v(y)
ts an element invariant under s.
Now each automorphism
= nv(xy} for some
n
= n(x,y)
g E G E
s.
has an exThen, for
AN INTERPRETATION OF
H 3 (G,E*)
299
0
= aK(v(x) ,v(y)1 = &(\ (v(xl ,v(y)}
a(x,y}
= v(x)B 1
(v(yl)
ll - 1 B1 (v(x))
[S (nv(xy) 1
= xB (_v(yl} [S (v(xy) 1] - 1 B (v(x)) 1
and therefore
1
1
= oB 2 , where B2 (x) = B1 (_v(x}l is a cochain of
a
G
over E*.
So the theorem is true. • We close by showing how Lemma 20.6 can be derived as a consequence of a general cohomology theory. 20.16. LEMMA.
The following simple observation will clear our path.
Let A be a central simple algebra over a field E. Z(U(A})
E*
Proof.
It suffices to show that z(u(All c E*.
represented as the algebra of all a division algebra of
A
n x n
with centre
D
Then
The algebra A can be
matrices, with matrix units e 1,,J .. , over . Then e
E.
is the identity element
= 'Ze..
1,,1,,
and, for any r # s, (e+e
) (e-e ) rs rs -
=e
¼
e
rs
- e
rs
=e
(e-e rs l(e+e - · rs l = e - e rs + e rs = e
Proving that
Z(U(A)l,
e + e
rs
If a= 'Zd .je ,., d •.
E U(Al.
·
then a(e-te rs }
1,,
= (e+e
rs-
la or
ae
1,,J
rs
= e
ED,
1,,J
rs
a.
is any element of
Multiplication gives
E d • e • = ae = e a = 'Z d ,e , 1,,r 1,,s rs rs j SJ rJ
i
Comparing the coefficients, it follows that drr for i-! r. have O-! d
Hence, i'f d EE,
=
is defined to be a group 8 :
where
In(N)
e0 (x)
G -->-
N
E,
we
together with a homomorphism
Out(Nl
=
Aut(N}/In(N)
Each inner automorphism of
fixed, so that each coset e(x}
of Z(N).
=
0
A G-kernel
is the normal subgroup of the automorphism group Aut(N)
elements of Z(N)
=
H'
Because Z(D)
G denotes an arbitrary group.
consisting of inner automorphisms.
phism
and d.
•
For the rest of th i's section, (N,8}
d8$ for r-! s
then a = 'Z.de ii = de.
drr'
as required.
=
N
of
N
leaves the
determines a unique automor-
Thus we have a homomorphism
CHAPTER 2
300
eo : a -
Z(N)
and therefore The pair
g
Let
E/F
for all
U(A)
U(A).
Setting
e(x)
N
= U(A).
is the coset of
(U(A) .e) (N.e)
0(xl
of
N
A.
0(x).0(y) =0(xy)
g Ea.
0(g)
Then
0(xl
let
(z(N), with
0(g)
be an
is also an automorph-
Z(N)
= E*.
Thus.
U(A).
modulo _the inner automorphisms of
6(x)
G.
modulo the inner automorphsims of
then
(E* .e 0 )
is a a-kernel with centre
in
a.
(N.e).
it follows from Lemma 20.16 that
be a a-kernel with centre
automorphism
n E Z(N). g E
be a finite Galois extension with Galois group
to a rtng automorphism of
and. by (7} •
ism of
Let
e0 (g)n
be a a-normal algebra and. for each automorphism
A
extension of
if
g•n =
is called the centre of the a-kernel
(Z(N).e 0 }
20.17. EXAMPLE. let
is a G-module via
Aut(Z(N11
e 0 ).
For each
0(1) = 1.
Since
e(x).
e:
choose an
G ___,. Out(N),
is a homomorphism. 0(x)0(y)0(xy)-l E ln(N) Hence, we may choose elements
f(l,y) = f(x,1) = 1
f(x,yl E N with
0Cxl0(yl = i'[f(x,y)] a(xy)
x,y E G
for all
for all
such that
x,y E a
(30)
where
i[f(x,y)] (n) = f(x,y)n f(x,yf 1 Since
0(x) [cr(y)0(2)] = [0{x}0{y)] 0{2),
i[f(x,y)f(xy,2)] Thus there exist elements
=
t(x,y,2)
n
for a 11
E N
an application of (30) yields
i'[(0(x)f(y,2}}f(x,y2)] in
Z(N)
such that (31)
[ a(x)f(y,2)] f(x,yz) = t(x,y,2)f(x,y)f(xy,2) Thus
t
is normalized 3-cochain of
G with values in
Z(N).
Hence, in the
situation of Example 20.17, we have 20.18, LEMMA.
Each Teichmuller 3-cochain
cochain of the corresponding G-kernel Proof.
t
of a G-normal algebra
(U(Al,e)
Compare (30, (31) with (71, (9).
with centre •
(E*,8 0 ).
A
is a 3-
301
AN INTERPRETATION OF H~(G,E*)
.we can now deri've Lemma 20,6 as a consequence of Lemma 20.18 and the following general result. 20.19, THEOREM. (Eflenberg and MacLane(l947b}). wfth values tn Z(N)
correspondfng to the G-kernel
(il The cochain t (ti)
Let t
be a 3-cochain of
(N,9).
ts a cocycle
If the chotce of
f
tn (30} is changed, then t
is changed to a cohomo-
By suftable changi'ng the choice of f,
logous cocycle.
G
the cocycle t
may be
changed to any cohomologous cocycle. (tiil
If the chotce of the automorphisms a is changed, then a suitable new
selection of Proof.
leaves the cocycle t
f
unaltered,
Let us calculate the expression
(i}
A= a(x).[fo(y}f(z,u}}f(y,zu)Jf(x,yzu}
in two ways.
First using (_31} three times, and recalling that t
li'es tn Z(N),
we have A = a(_x} [t(y,z,ulf(_y,z}f(yz,ulif(x,yzu}
= a(x l [t(y ,z ,u}f(y,z 11 t(x,yz ,u}f(x,yz )f{xyz ,u} =
where
B
(t:t(y,z,u}It(x,yz,u)_t(x,y,z}B
i's given by B = f(x,y}f(_xy,z}f(::cyz,u)_
Alternatively, by first applying (30)_, we have A = f(x,y) [a(xy }f(z ,u}If(x,y 1-1 [cr{_x}f{_y,zu}J f(x,yzu} = f(x,y 1[a{xy )f(z ,u}J t(x,y ,zu}f(xy ,zu) = t(x,y,zu}t(xy,z,u)B
Comparison gives [xt(_y,z,u)J t(x,yz;u}t(x,y,z) = t(x,y,zu}t(xy,z,u)
which, by (4), shows that t
E
z3 (G,Z(N1).
(ii} Any other choice of f(x,y l in (301 has the form f 1 (x,y} = g(x,y)f(x,y}
(g(x,y}
E
Z(N})
CHAPTER 2
302
where
t'
g
from
ts any normali'zed 2-cochai"n of
f',
G tn
Z(N}.
Using (311 to calculate
we have t 1 (x,y ,z) = t(x,y ,z 1{ [xg(_y ,z 11 g(xy,z }- 1 g(x,yz )g(x,y )-1 }-l = t(x,y,z}
Cl and, by Corollary 3.13,
Lemma 21.l(i}, (E:F) = !Cog(_E/F}I, p,
i,e.
Then E/F is coseparable by
A¢ F*
and
APE F,
p!(E:F}.
there is an element 'AF
E
Since, by
Cog(E/F)
Then for allµ E F. µ + /\ ¢ F by
Then
of order
CHAPTER 2
308
(l1+11)F* f (d+"-}F*
One immediately verifies that Cog(E/F)
contains infinitely many elements
for
µ ,j, d
(µ+:\}F*, µ E F,
E
F.
Therefore
which is a contra-
diction. To pr0ve purity, let We must show that then
s = l
and
s
t-
s
sp = 1
with
s E E,
Assume first that
E F.
and there ts nothing to prove.
p
where p
p = 4.
is a prime or
ts a prime.
If
charF = p, charF t- p
Hence we may assume that
Since
1.
+ s + ••• + s p-1 -- 0 1•F* , s F* ,. • .,sp-lF*
it is not the case that (seeLemma21.l(il), Now assume that s2
t-
1.
si/sjEF(-Ct-j),
Hence some p = 4.
F*.
E
1 + s - (l+s}
distinct (see Lemma 21.l(i)). s
E
F,
choose a finite subgroup G = Cog(E/F}L
Since
sEF. charF f 2,
But
O, are
Since any of the three possible equality relati'ons purity is established.
E/F
Conversely, assume that
Cog(E/F)
l•F;., sF*, (1+s}F* of Cog(E/F}
so it is not the case that the elements
between them implies
so
Again we may harmlessly assume that (1+s )_ 4 = ~4
One checks that
are d.1s t.inc t elements in
G of
is coseparable, separable and pure. Cog(E/F}
such that
We may
EG = E (e.g, take
G is abelian, there i's a chain l = H C H c ... cH=G 0-1-· -n
of subgroups of Hence
EH /EH. 1, 1,-1
coseparable. EH. i--1
G
such that
HJ H, 1, 1,-1
is pure (since
Because
H ./H ., 1, i--1
E/F
1,
E/F.
and thus conormal.
p
0 .,,
v
1 .;;· i .;; n
is pure} and, by construction, it is also
is cyclic of order
by adjoining a p ..-th root.
since so is
is cyclic of prime order
p.,, EH 1,
• 1,
is obtained from
is separable, EH.JEE. 1, i,-1 is cogalois, conclude that EH_JEH. 1, i,-1
Note also that
Invoking Lemma 21.3, we
Hence, by inductive application of Lemma 21.4, we deduce that
E/F is also conormal.
Thus
E/F
is cogalois and the result follows.
•
With the aid of the above result, we now provide the following examples of
309
A COGALOIS THEORY FOR RADICAL EXTENSIONS cogalots extenstons, 21 . 6. EXAMPLE.
Let
be positive raUonal numbers, let n.1,
lN and
E
let n1
ns
E=(f;)(va, •• ,,f;;.}clR 1
Since E
5: JR,
E;(Q
21 .?.EXAMPLE. p
-
It is also clear that E;(Q is coseparable
E;tf) is cogalois, by virtue of Theorem 21.5,
Thus
E
is obviously pure.
8
Let p
be an odd prime, let n
be a positive integer and let
be a primitive pn ~th root of uni"ty over