Pi-Algebras

Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann

441 Nathan Jacobson

PI-Algebras An Introduction

Springer-Verlag Berlin. Heidelberg • New York 1975

Prof. Nathan Jacobson Dept. of Mathematics Yale University Box 2155, Yale Station New Haven, CT 06520/USA

L i b r a r y of C o n g r e s s Cataloging in P u b l i c a t i o n Data

Jacobson~ Nathan, 1910PI-algebras. (Lecture notes in mathematics ; v. 441) Bibliography: p. Includes index. i. Rings (Algebra) I. Title. II. Series: Lecture notes in mathematics (Berlin) ; v. 441. QA3. L28 no.441 [Q]~°47 ] 510' .Ss [512.4] 75-6644

AMS Subject Classifications (1970):

ISBN 3-540-07143-1 ISBN 0-387-07143-1

16-01, 16-02, 16A12, 16A20, 16A38, 1 6 A 4 0

Springer-Verlag Berlin- Heidelberg. New York Springer-Verlag New York • Heidelberg • Berlin

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1975. Printed in Germany. Offsetdruck: Julius Beltz, Hemsbach/Bergstr.

Foreword

These

are lecture

notes for a course on ring theory given by the author at Yale,

September - December,

1973. The lectures had two main goals:

improved version of the theory of algebras with po.lynomial tative coefficient

first,

to present an

ident~y (over a commu-

ring) based on recent results by Formanek and Rowen and second,

to present a detailed and complete account of Amitsur's product division algebras.

construction

of non-crossed

Table of Contents

INTRODUCTION I. A s s u m e d b a c k g r o u n d

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Two results on the radical

|

. . . . . . . . . . . . . . . . . . . . . . . . .

7

3. Prime and semi-prime ideals . . . . . . . . . . . . . . . . . . . . . . . . .

9

I.

PI

-

ALGEBRAS

I. D e f i n i t i o n s and examples 2. Formal results

. . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. K a p l a n s k y - A m i t s u r

theorem

11 15

. . . . . . . . . . . . . . . . . . . . . . . . .

17

4. T h e o r e m of A m i t s u r and L e v i t z k i . . . . . . . . . . . . . . . . . . . . . . .

21

5. Central simple algebras.

27

Converse of K a p l a n s k y - A m i t s u r

6. Nil ideals in algebras w i t h o u t units

theorem .......

. . . . . . . . . . . .

7. Polynomial identities for algebras w i t h o u t units

........

. . . . . . . . . . . . . .

30 33

8. Central p o l y n o m i a l s for m a t r i x algebras . . . . . . . . . . . . . . . . . . .

36

9. Generic m i n i m u m polynomials and central p o l y n o m i a l s for finite d i m e n s i o n a l central simple algebras . . . . . . . . . . . . . . . . . . . . . . . . . .

44

10. Commutative localization

. . . . . . . . . . . . . . . . . . . . . . . . . .

48

11. Prime algebras satisfying proper identities . . . . . . . . . . . . . . . . .

54

12. PI - Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

58

13. Identities of an algebra. Universal PI - algebras . . . . . . . . . . . . . .

60

II. APPLICATIONS TO FINITE D I M E N S I O N A L A L G E B R A S I. E x t e n s i o n of isomorphisms.

Splitting fields . . . . . . . . . . . . . . . . .

67

2. The Brauer group of a field . . . . . . . . . . . . . . . . . . . . . . . . .

73

3. Cyclic algebras. Some constructions . . . . . . . . . . . . . . . . . . . . .

82

4. Generic m a t r i x algebras . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

5. D i v i s i o n algebras over iterated Laurent series fields . . . . . . . . . . . .

94

6. N o n - c r o s s e d product d i v i s i o n algebras . . . . . . . . . . . . . . . . . . . .

]04

7. A n o t h e r result on UD(K,n), K an infinite field . . . . . . . . . . . . . . .

I12

References

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

I15

INTRODUCTIO~

1.

Assumed background Our primary concern in these lectures will be with the

category

(~(K)

commutative ring

of (associative} K

with unit.

algebras with unit over a Accordingly,

will be K-ring homomorphisms mapping are understood to contain

1.

1

into

M

l,

subalgebras

Modules, unless otherwise indi-

cated, are understood to be left modules. one of these is:

homomorphisms

The definition of

is a (unital) K-module and a unital

A

module for the K-algebra A, and we have (1) for

~(ax)

= (~a)x = a(~x)

m -~K, a -CA, x ~ M . If

A

and

B

are K-algebras by an

mean a left A-module,

A - B - b i m o d u l e we

right B-module such that the following

associativity property holds: (2)

(ax)b = a(xb),

a -~A, b -~B, x ~ M . If

A

is an algebra, the opposite algebra

algebra such that we denote, as

a~b,

A° = A is

as K-module

A°

is the

and the product, which

(~)

a ~ b = ba Tensor products of modules and algebras will - unless other-

wise indicated - be taken relative to M~K

N.

If

An

A- B-

A

and

B

K°

We write

are K-algebras then

M~N

A~B

bimodule can be identified with an

for

is a K-algebra.

A~B

° - module in

which (&)

(a~b)x

Conversely, given an

A~B

= axb.

° - module this can be regarded as an

A - B - bimodule by putting (5)

ax = ( a ~ l ) x ,

xb = ( l ~ b ) x .

This gives an isomorphism of the category of A - B - bimodules with the category of

A~B

We write

°

modules.

A e = A~A

° .

Since

A

right A-module and associativity holds, Hence as

A

is an

A e-module

is naturally a left and A

is an A - A - bimodule.

A e - module defined by (A). are the ideals of

The submodules of

A

A.

We shall use the convention of writing endomorphisms of a module on the side opposite that of the algebra. an A-module then

EndAM

(or

on the right:

If ~ ~ E n d A M

applying D

x

to

x(~ ~

) = (x ~) <

Since

(ax) ~

A~(End

and ~ .

= a(x~),

M) °- module.

(~ ~

~ x

)x =

End M A ~ A

is an

End M A

° - module.

Then

M

is

M

denotes the result of

~EndAM

is defined by

is a right

EndAM-module.

A - EndAM-bimodule ,

On the other hand, if

M

hence, an

is a right

for its set of endomorphisms and

for the action of ~

D(¢x).

x~

for D , Q

M

M

is an algebra and this acts

then

In this way

A-module then we write we write

End M)

Thus if

on

is a left

x.

Also one defines

End M A - m o d u l e

and an

A module M

M

is called irreducible

are the only submodules.

are equivalent: x ~ O

in

(1)

M

(3)

The isomorphism

M

The following is irreducible

M =~JA/I

where

a m>

since

A module

M M

irreducible

irreducible Ms

is completely

(3)

of

A

Hence

in a module

M = Ax

for any

M = Ax. onto

Then one

M.

M ~ A/I.

submodules.

The kernel

Moreover

(2)

and

N

of

if it satisfies

conditions:

M = M =~M

We recall that

M = ZM s

every submodule

reducible

equivalent

submodule,

means that

and

I

is irreducible.

hence, all of the following Ms

(2)

O

is a maximal left ideal.

ax

I = ann x = {b -~A~ bx = O}.

is maximal

I

and

conditions

in (3) is obtained by taking

has the module epimorphism is

if M ~ O

(i) M = ~ M a,

a direct su~n of

M =6~)M a

for submodules

= O

s.

Ms ~ ~ m M ~ M

s

for all

is a direct sun,and,

t

there exists a submodule

one,

that is,

t

N

such that

M = N~)N

•

We shall now recall the basic results on the (Jacobson) radical and primitivity orimitive

of algebras.

if it has an irreducible module

in the sense that the map injective. ~ )

An algebra

a~>

{Ael

algebra structure.

B s = ker

suppose such that

A

set of the

As

if

the canonical map

is the projection

homomorphism

~ s ~ A.

Then

= O. m

Then

is called

which is faithful aM

is

x~>

ax, is

(formerly semi-

A

The direct product Am

-~Am

A

is a subalgebra

~s~ A

is called a subdirect of

is surjective.

{a~l - - > ae

IBm~

-~A m

endowed with the obvious

of

A/B s = A s

is any algebra and ~ B

A

completely reducible module.

A subalgebra

product of the algebras m

where

be a set of algebras.

is defined as the product

for each

aM

M

is called semi-primitive

if it has a faithful Let

Let

An algebra

and

of

~A

m

~ B~ = O.

~

As

Here into

~a As •

Conversely,

is a set of ideals in

is isomorphic

and

A

to a subdirec% product

of the algebras

As = A/B

ing characterization:

.

Using this concept one has the follow-

An algebra

is semi-simple

if and only if it

is a subdirect product of primitive algebras. If

M

is a module we write

algebra homomorphism

a-->

~.

rad A = If

M = A/I

of

A

where

contained

I

in

annAM

This is an ideal.

~h Mirred

and ~

Define

annAM .

is a left ideal,

M

for the kernel of the

denoted as

annAM I:A.

is the largest

ideal

It follows immediately

that rad A =

and

I:A

is a primitive

a primitive algebra.

~ (I :A) I max

ideal in

A

in the sense that

A/I:A

is

Hence we have also

rad A = ~ P primitive

ideal

One also has rad A =

An algebra A/tad A

~ I = ChI' I M a x left ideal It max right ideal

is semi-primitive

if and only if

rad A = O.

Moreover,

is semi-primitive.

There is an important

element

characterization

of the radical.

Consider the map ~: in

A.

a ........>...l - a

This is of period two and can be used to define the circle

composit ion aob =

= ~ (qaqb) 1

=a+b ~#e have

~l=

O.

Hence

-

(l-a)(l-b)

- ab

(A,o,O)

is a monoid isomorphic

to the

multiplicative

monoid

(A,.,l)

ciative and the elements

a

of

of

A

in the sense that there exists a form a subgroup of

A.

In particular,

o

which are quasi- regular (q.r.) b

such that

aob = 0 = boa

(A,o,O).

One has the following element

characterization

rad A = [zl az

is

q.r.

for all

a -CA 1

-- Izl za

is

q.r.

for all

a -CAl.

of the radical

Call an ideal or a one sided ideal quasi-regular elements are q.r. elements

is asso-

Then this is a subgroup

(under o).

if all its

of the group of q.r.

One has the following further characterization

of tad A: tad A

is a q.r. ideal which contains every q.r. left and

every q.r. right ideal. A one-sided or two sided ideal is called nil if every element is nilpotent.

If

with quasi-inverse

z

is nilpotent

with

zn = 0

w = z + z 2 + ... + z n-l.

then

z

is q.r.

Hence rad A contains

all nil one sided ideals. Let A' -- EndAM makes

M

M

be an irreducible A-module. is a division algebra.

Put

By Schur's lemma, ~

= A'°

Then S x ~ x$

a (left) vector space over the division algebra

A

. We

have the

DENSITY THEOREM. linear transformations -independent YI'''"Yn (6)

vectors

A M = {aM l a-CA] i_nn M

over

~

:

Xl,..., x n -CM

there exists an

a -CA

is a dense al~ebra o f Given any finite set of

and correspondin~ vector~

such that

axi = Yi' 1 _< i _< n. A set of linear transformations

of a vector space is called

dense if it has the foregoing property: of linearly independent

vectors

Given any finite sequence

xi, i < i _< n

and

Yi'l _< i < n,

thene exists an

a

in the set such that

axi=Yi'

1 < i < n.

We

have the THEOREM.

A primitive algebra is isomorphic t_~oa dense al~ebra

of linear transformation in a vector space over a division algebra. The notion of density can be formulated in topological terms and this

is sometimes useful.

set of maps of

X

into

Y.

Let

X

and

Y

be sets

yX

the

This can be endowed with the finite

topology which has as base for open sets the sets defined by a finite subset

Ixil

of

X

and map

f

by

Igl gx i -- fxil.

a vector space over a division algebra space of

VV

(~,

>

m)

n , EndAV

and this is a topological algebra: ~ -m,

(~,m)

....... >. ~ m, ~

> ~4

If x = Y =

v

is a closed subthe maps are continuous.

set of linear transformations is dense as defined above if and only if it is dense in the topological sense, that is, its closure is

End AV.

We recall also that an algebra is called (left) Artinian if it satisfies the minimum condition on left ideals. such an algebra

A, rad A

is a nilpotent ideal.

For

Also one has

the Wedderburn-Artin structure theorems that a semi-primitive Artinian algebra is a direct sum of a finite number of simple Artinian algebra and conversely.

An algebra is simple Artinian

if and only if it is isomorphic to

End A V

where

dimensional vector space over a division algebra the condition is to

Mn(~ )

A ~Mn(~°)

since

and hence is isomorphic to

EndAV

V

is a finite

~ . Equivalently,

is anti-isomorphic

Mn(~°).

A

2.

Two results on the radical If

S

is a set then we denote the set of

with entries in

S

by

an algebra.

B

is an ideal in

in

If

Mn(A ).

Mn(S ) .

If

A A

then

Mn(B)

Proof.

B m>

Mn(B )

into the set of

Mn(A ).

rad Mn(A ) = Mn(rad A).

Let

rad A

lj

be the set of

and all other rows

We claim it is

n x n

O.

q.r.

matrices with

Then

Let

Ij

j-th

is a right ideal

Zj ~ I j

have (j,j)-

t

entry

is

Mn(A ).

THEOREM i.

in

A

Mn(A)

is an ideal

Moreover it is easily seen that the map

ideals of

matrices

is an algebra,

is a bijective map of the set of ideals of

row in

n ~ n

zj

and let

be a matrix in

t

zj

Ij

be the quasi-inverse

having

of

(j,j)-entry zj .

zj.

Let

Zj r

Then

Zj o Z j

and

t

Zj OZj O = (Z

Zj Now B

has

(j,j)-entry

j o zj) 2.

is

q.r.

Then

Then

rad Mn(A ) is an ideal

ideal in

and

is an ideal in

Ao

and so

so this

Let

b -~B

Hence

is

Hence are

and

q.r.

(Zjo Z~) 2 =

q.r.

Mn(rad

has the form

This is in b

Ij.

ZjO Zj

Ij C rad Mn(A).

It follows that A

and is in

Zj o Zj

bi n = diaglb,b,...,b}. is q.r.

O

consider

Consequently

A) CradMn(A). Mn(B)

where

the matrices

Mn(B ) = rad Mn(A ) in

A.

Hence

tad Mn(A ) = Mn(B ) C M n ( r a d

A).

so it B

is a q.r.

Hence

rad Mn(A ) = Mn(rad A). Let

A[k]

with coefficients THEOREM 2.

be the polynomial algebra in an indeterminate

k

in

A. The following result is due to Amitsur.

If

A

has no nil ideals

~ 0

then

A[k ]

is

semi-primitiv @. We recall that in a commutative ring the intersection of the prime ideals is the ideal of nilpotent elements.

Also

prime ideal in a commutative ring if and only if

A/P

We use these results to prove

P

is a

is a domain.

LEMMA i.

If

+ ... + ankn

i~s ~ unit i_nn A[k]

and the other

ai

Proof.

That

A

a unit in ai~P

a

is a unit is trivial.

o

ai = 0

and A/P[k]

ao

is a unit

i > O.

in

A/P

Also if

Hence if then

and hence

is a

is a prime

~o + Z I k + ' ' ' + a n k n

which implies that every P

P

A

is

Ki = O, i > O. Thus

ai~D

P

and so is nil-

The proof of the converse is direct and will be omitted.

LEMMA 2.

has

for

~ = a + P

for every prime

potent.

i f and only i f

a° +alk

are nilpotent.

domain then every ideal in

i~s commutative, ~ p o l ~ o m i a l

A

O

If

A

i~s a_~n@rbitrary alsebra and

constant term and is ~uasi-invertible in

f(k) - ~ A [ k ] A[k],

then its

quasi inverse has coefficients i._nnthe subalgebra $ e n e r a t e d b y coefficients of Proof.

f(k). Let

g(k)

be the quasi-inverse of

( l - f ( k ) ( l - g(k)) = 1 = ( l - g ( k ) ) ( l - f(k)). n = 1,2,...

the

f(k)

so

We have for any

,

1 -

f(k) n+l

= (1-f(k))(l+

f(k)+

...+

f(k)n).

Hence I -

g(x)

:

(!-gCx))f(x)

n+l

+ 1 + f(x)+

...

+ f(X) n

and g(X) = -

(7)

If

deg g(k) = n

of

g(k)

f(k) . . . . .

f(k)n + (g(k)-l)f(k) n+l

then this relation shows that the coefficients

come from

- f(x) . . . .

- f(x) n

algebra generated by the coefficients of Proof of Theorem 2. zero element

Assume

[ai,f(k)] G r a d thah f(k).

A[k]

Hence it is

and if O.

a i.

in

~'~e may assume

~ O Then

f(k).

tad A[k] ~ O

f(k) = ao + alk + ... + amkm

minimum number of non-zero

and so are in the sub-

and choose a nontad A[k]

with the

am ~ O.

Now

it has fewer non-zero coefficients [ai,a j] = O.

This shows that the

subalgebra

B

generated by the

kf(k) G r a d

A[k]

quasi-inverse potent.

and has

0

the stated property,

since

together with

a non-zero nil ideal in

A.

3.

ideals

An ideal ideals

B

P

and

algebra if

of an algebra C

0

Proof. B

BM = M

i.

Let

and

implies

C

M

If

since

CM

and

coincide with

M.

Thus

LE~94A. (ii)

i__~sprimitive,

b =0 and let Hence

or

so either (ii)

bI = O. b = 0.

(iii).

(iii)

C ~ O.

Then

A.

Let

b = 0

Then

A

and

(BC)M = B(CM)

=

and hence

(i) A is a r ~ _ ~ a l ~ e b r a ,

o_~r c = O, is

Suppose

AbA = 0

-~-> ( i i i ) . c ~ O

-~-> ( i ) . (ii i)

is a ~

i_~S prime.

submodules

O,

or

(iv)

the right

O.

AcA = O.

Then either

I

I.

Then of

I~Ac I

is

BC = 0

for ideals

implies

B = O.

Hence

(i).

AbAcA =

be a left ideal

Let

(ii) ----->(iv) = >

the left

Then

Let

be in

(iii)

bAc = O.

Thus the left annihilator

symmetry we have equivalent.

A

righ t ideal is

(i) -----> ( i i ) .

c =O

for

BC ~ O.

of any non-zero

(AbA) (AcA) = 0

A

irreducible module for

are non-zero

implies either

Proof.

Hence we have

A.

annihilator o f any non-zero left ideal annihilator

having

BC m 0 (mod P)

The following are equiv@lent:

bAc = 0

A(k)

f(k)

C m 0 (mod P).

ideals in

BM

is nil-

Grad

form an ideal.

or

be a faithful

be non-zero

f(k)a

is prime if

B m 0

A

and

am

from polynomials 0

is a prime ideal in

PROPOSITION

let

af(k)

determined

Prim e and scmi-primc

Since

Then Lemma 1 implies that

On the other hand, am

is commutative.

constant term Lemma 2 shows that the

g(k) -~B[k].

it is clear that the

ai

A

Hence

and O B, C

~O bAc = O.

and we have and assume

is prime. (i) -~v)

By are

10

An algebra is called semi-prime if it has no niipotent ideals

~ O.

An ideal

is semi-prime.

B

Clearly,

THEOREM.

if

A A

is called semi-prime if is prime then

A

A/B

is semi-prime.

is semi-prime if and only if it is a subdirect

A

product of ~

in

algebras.

Proof.

Let

A

be a subdirect product of prime algebras

A

H

and let

N

nilpotent a, N = 0.

be a nilpotent ideal of

A a.

Hence

Conversely,

non-zero ideal in

ideal in

suppose

A.

Choose

non-zero ideal contained in Hence and

blab I ~ 0 b2 -~B.

zero

A.

Then

B~ = O. A

in

Hence

so we have an

aI

B.

Let

Then

B

A blA

then

such that

b 2 = blalbl / 0

Continuing this process we obtain a sequence of non-

B.

= bi_lai_ibi_l,

b k = biaijb j

where

aij -CA.

Since

(O) ~ [b i}= @

Zorn's lemma that there exists an ideal

such that

is maximal in the set of ideals satisfying

P

[bi} = @. be ideals of

C 1 = C + P Dp

A

we have A

Hence P~B.

P

is a prime ideal of

satisfying

If

Since P

k > i, j CIDI C

is prime.

b i C C I. then

CD + P Since

P n

P

P

Let

it of

C

and

(rood P). Then

b k = biaijbj -~CIDI it follows that Ibi} = @ B

A

and

•

CD ~ 0 [b i} C

B

is any ideal ~ 0

not containing

gh P = 0 which means that P prime product of the prime algebras A/P.

A

Similarly there exists

Hence we have shown that if

there exists a prime ideal

implies that

A.

C ~ O (mod P), D ~ 0

so there exists a

CIDI~P.

(rood P).

in

We claim

bj C D 1 = D + P.

Hence

...

The form of these elements shows that if k > i,j

follows from

a

is a

elements

contained in

D

be a

(AblA) 2 = AblAblA # O.

bl,b 2 = blalbl, b 3 = b2a2b2,...,bi

P ~

is a

Since this holds for all

is semi-prime.

bI ~ 0 B.

Na = ~a(N)

B.

This

is a subdirect

I. PI-ALGEBRAS i.

Definition and examples To define the concepts of polynomial identity for an algebra and

of a

PI-algebra we need to consider first the free algebra in a

countable set of generators over the given commutative ring X

K.

Let

be the free monoi4 generated by a countable set of elements

Xl,X2, ...

Then

X

(1)

is the set

1

.

.

.

' XllXl2

.

• Xir !

of distinct monomials in the and only if

il = Jl' "'"

x s

).

(Xil . . . .Xlr .

Xjl ... x Js

if

Multiplication is defined so that

1

is unit and (Xil "" . Xir ) ( Xjl ... Xls ) ~ Xil "'" XirX Jl ... xj s .

(2)

Let K[X} be the monoid algebra of

X

over

K.

This is called the

free algebra with ~ c ountab%e set o~f (free) generators basic property of map of K{X}

X into

into A

A

K[X}

is that if

A

x i.

is any algebra and

The ~

then there exists a unique homomorphism ~

which extends

~

x . . . .

is a of

: i

~ K{x}

is commutative, If

f~K{X],

finite subset

[Xl,..°,xm}

f = f(xl,...,Xm). homomorphism of is denoted as

f -~K[Xl,...xm} for some

the subalgebra generated by the mo

Accordingly, we write

The image of this polynomial under the K[X}

into

f(al,...,am).

A

sending

x i --> ai, 1 ~ i < ~ ,

12

DEFINITION. for all

f

is an identity for

A

if

f(al,...,am )= 0

ai~A.

EXAMPLES (i)

Any commutative algebra satisfies the identity

f : [ x l,x2] ~ x Ix2-x2x 1. (2) form

An algebra will be called almost nil if it has the

K1 + N

where

N

is a nil ideal and it is almost nil of

bounded index if every element of is, there exists an integer If

A

is almost nil

n

N

is of bounded index, that

such that

zn = 0

for all

x, y - ~ A .

Ix, y] -~N

for all

z~N.

Hence if

A

is almost nil of bounded index it satisfies an identity [Xl,X2] n for some

n.

(3)

The earliest interesting identity for algebras is

Wagner's identity for

M2(K ).

Note that if

r that is,

tra = O,

-p

then

2 ( p2+qr a -0 commutes with every matrix. we have (3)

[[ab] 2, c] = 0

Since

for all

0 p2 + qr

tr[a,b] = 0

for all

a,b,c ~ M 2 ( K ) .

Hence

(XlX 2 - X2Xl )2 x 3 - x 3 ( x l x 2 -

is an identity for

M2(K).

a,b

X2Xl )2

This is Wasner's identity published

in a paper appearing in 1936. Mn(K ) .

)

Wagner also gave identities for

Simpler ones were discovered later.

We shall indicate

these in a moment. A monomial

Xil "'" Xir

is said to occur in

has non-zero coefficient in the expression of

f

f

if it

in terms of

13

the base

X.

The

x i occurring in

f

is said to be linear in

f

is of the first degree in

linear in every f

and the

x s

xi

xi

Xil. "" Xir are

if every monomial occurring in x i.

f

is multilinear

occurring in it.

occurring in

x]•I ,...,Xir •

f

are

Suppose

f

Xl,...,x m.

if

f

is

is multilinear

Then

f

has the

form (&)

f = ~ a~l,...,~m x ~ i x ~ 2 " ' '

where ~ l . . . w m ~ K 1,2,...,m.

If

f

and

ranges over all permutations of

is mu!tilinear then

f(xl,''',Xj_l, (5)

n

Xmn

xj + Xm+l, x.j+l ,. ""%n )

= f(xl,...,x m) + f(xl,...,xj_ l, xm , Xj+l,...,x m) f(xl,...,Xj_l,

~xj, Xj+l,...,x m)

= ~f (Xl,...,Xm). These conditions imply that if A

as K-module then

[u t I

is a set of generators of

f

is an identity for

A

f(uil'Ui2'''''Uim ) = 0

for all choices of

uij

A polynomial

f(xl,...,Xm)

f

i < j.

If

is an identity if and only if

choices of distinct

ui

in

{ui}.

is alternating if

f(xl,...,Xi_l,Xi,Xi+l,...,xj_ixi, for all choices of

if and only if

f

X j+l,

... ) = 0

is multilinear and alternating f(uil,...,Uim ) = 0

in a set of K-generators

for all

Jut }" Hence

3 if

A

has a finite set of module generators

alternating multilinear polynomial of degree identity for (4) (6)

[Ul,...,Unl m

> n

then any

is an

A. The polynomial

Sk(Xl,...,x k) = E ( s g ~ )

x i x

2 ... x k

where the summation is taken over the symmetric group and the sign of the permutation

w,

sg~

is

is called the standard polynomial

14

o f degree k.

The foregoing remarks show that

for any algebra which is generated as Since eij

Mn(K )

later that (5) K-integral) that

S2n

Sn2+l

is an identity

K-module by

is generated as K-module by the

we see that

Sk

is an identity for

< k

n 2 - matrix units Mn(K).

We shall see

is such an identity.

An element of an algebra is integral (more precisely if there exists a monic polynomial

g(a) = O.

g(k) -~K[k]

1

such

It is easily seen that this is the case if and

only if there exists a finitely generated submodule containing

elements.

such that

aMCM.

If

A

A.

A

of

A

is commutative the set

of K-integral elements forms a subalgebra. for non-commutative

M

This need not be so

is called K-integral if all of its

elements are K-integral and it is K-integral of bounded de~ree if there exists an

n

of a monic polynomial

such that every element of -~K[k]

of degree

~ n .

A

Then every

element is a root of a monic polynomial of degree (7)

an+~lan-i

where the

ai

+ "'" + ~n I = O,

depend on

with respect to

b

a.

Let

is a root

n

so we have

~i~K

b 6 K.

Then taking commutators

we obtain

[an,b] + al[an-lb] + ... + an_l[ab ] = O. Next take commutators with respect to

[a,b]

to obtain

[[anb] [ab]] + Cl[[an-lb][ab]] + ... + ~n_2[[a2b][ab]] Next take commutators with respect to this process.

[[a2b],[ab]]

This leads to an identity for

A.

define

Pkl = Pkl(Xl'X2 ) = [x~x2]' k ~ 1

and continue

More precisely

15

(8) Pkj = [Pkj-IPj-I,J-I ]' j > i, j <_ k. Then the argument shows that that if we define monomial

mnn

if

f

occurs with coefficient f(xl,...,x m)

is not an identity for

identity for

is an identity for

A.

We observe

mkl = x~x2, mkj = mk,j.lmj_l,j_l, j > i,

A polynomial A

Pnn

A.

1

in

Pnn"

M2(K).

Pnn ~ O.

is called a central polynomial for A

but

[f(xl,...,Xm) , Xm+l]

Wagner's identity shows that

central polynomial for

Hence

then the

(XlX 2 - X2Xl)2

is an is a

For a long time it was an open problem

to construct such non-constant polynomials for solved quite recently by Formanek.

Mn(K) , k > 2.

This was

We shall consider this in ~ 8.

2. Formal results The notions of (total) degree and degree in a particular of a polynomial as

deg f

and

f-~K{XI degxif

all monomials of

f

homogeneous if

f

blended in

if

and

f

xi

are the obvious ones.

respectively

f

is homogeneous in every xi

We denote these

is homogeneous in

have the same degree in

xi .

f

xi . f

xi

occurring in the monomial.

The height of

of the monomials occurring in

f.

f

is called

such that:

(i)

occurring in

It is clear that

A

xi

in which

and subgroup

G-valued in the sense that f

is G-valued.

G

the hei6ht f

f

is multi-

O.

ht fj <_ ht f.

is linear,

(3)

of the additive group of fj(al,...,am) -~G

f.

xi

is a sum of blended polynomials

deg fj <_ deg f,

is linear in every algebra

f - ~ K[X}

f,

:is the maximum height

linear if and only if it is blended and has height AnY

if

is completel[

The heisht of a monomial is its degree minus the number of

L~MMA i.

xi

occurs in every monomial occurring in

is blended if it is blended in every

xi

for all

fj

(2) fj

for any, A,

fj

a i -CA

is if

16

Proof. in

f

We use induction

in which

we assume

f

t > 0

on the number

is not blended. and

f

T

f

= f(O,x2,...,Xm),

in

f

in which

which

f!

f

xI

and

f"

for

ft

in

!

= f - f

.

Then

are not blended

fv,. Then

occurring

if

t = 0

so

Put

is the sum of the terms

Hence the number < t.

is

of

fj.

xi fT

Moreover,

in and

We may assume

it holds for

We define the difference

xi

holds

x 1.

f

(1) - (3) stated for

and

of

!

does not occur.

has the properties result

The result

is not blended

~!

t

f ,,

the

f.

operator

.

in

KIX}

by

X.

i f ( Xl' " " " 'Xm) xj

= f(xl ' "''' xi-i , xi+xj, Xi+l' ... ' Xm)

(9)

- f(xl,.. .,xm) - f(xl,...,Xi_l, if

1 < i < m.

Here we are assuming,

as we may,

is regarded as contained in a subalg~bra xi. 3

is a K-endomorphism.

If

f

xj, Xi+l,...,X m) that a given

K{Xl,...,x m}

is linear

in

xi

with

f m~i.

then

X.

~jf = O. and x~

If

g

is a monomial

does not occur in

g

which

then

J

monomials

gk

in which b o t h

property that g°

if

To see t h i s

xj

xi

and

i s r e p l a c e d by

we w r i t e

g

is of degree

xi

~ x g 3

xj

xi

> i

in

x i,

is a sum of distinct

o c c u r and h a v i n g t h e

in

gk

then

gk becomes

as

g(xl,...,x m) = ax ih(x I, ...,x m) where

a

and

h

are monomials,

~ x i x g = (ax i + a x j ) [ h + h ( x l , . . . , x i _ J +~Xix.h]3

axih-

l,

xj, Xi+l,...,x

axjh(Xl, o..,xi_ l,

-- a x j h + a x i h ( X l , . . . , X i _ l , xi + axi( PX x j h ) +

d e g x i a = O, degx'lh 2> O.

axj

xj,

xj, Xi+l,...,xm) ~Xijh

Then

m) Xi+l,...,x

m)

17

X.

If

degxi h : i, ~

i h: xj

O.

Otherwise, induction on

degxi h

allows us to assume that this is a sum of distinct monomials all of which become

h

on replacing

asserted holds for deg

x~

j gk

X~g. 3

g = deg g

becomes

g

and

by

x i.

Then the result

The conditions imply that ht

g s

~x i g

<

xj

xi

xj

on replacing

obtained from distinct

xj

by

ht g.

Also the result that

implies that the

are all different.

gk

Our results

imply LEMMA 2. Then

Let

f

is a

Kaplansky-Amitsur theorem

identity for If

K

f ~ O.

A

f

is called a pro2er identity if

and some coefficient of

f

is a field this is equivalent to If

f

has a coefficient

i

are of this type.

f

is an

does not annihilat% f

is an identity

it is a proper identity

for every algebra for which it is an identity. §i

and degxj f = O.

f.

A polynomial

and

degxif > 1

and the set of coefficients of

subset of those of

A.

be blended with

xi xi ~ x3f . .is .blended, . . . deg . .~ xjf < deg f, degx i Axij f = degx i f - 1 f < ht f

3.

f

All the examples of

In this section we prove a fundamental theorem

on polynomial identities: THEOREM

(Kaplansky-Amitsur).

which has a ~

identity of degree

Let d,

A

b._~e~~rimitive al~ebra then the cente___rr C

-

A

is ~ field,

A

is simple and

[A:C] ~ [ ]2 .

o~f

18

We recall that O, A.

A is simple if

This is equivalent

We recall also that the center of

A.

to

A

EndAA = A Hence if

A ~ 0

A

is irreducible

and

A

and

as

End A A = A,

is simple

C

L~Vh~!.

If

f

is a field and

A has ~ proper i d e n t i t y

ht f > 0

If

f

is of height

terms whose

may assume that no (non-zero) Form

/~ xi f.

< ht f

and

xj

degxi

LEMMA 2. Proof.

Mn(K )

0

x jf~ = degx fm -i,

of

d

K.

Then

~l...d A = 0

we obtain

Hence assume

such that

of

deg

degx

annihilate f

f > I. i A

we

annihilates A

A.

and has height

xi ~ xjf

has no proper identities

By re-ordering

el2 , ...

we may assume

!

deg f.

The

on the height. of der ~

xi

elk O.

x

6 1 . . . d A # O.

ell,e12,e22,e23, for some

k

for every

contrary to hypothesis.

i

and

j.

....

and the pro-

Hence substituting

f(el!,el2' "'" ) = ~l...d

~i...d e i3 = 0

one:

d

we may assume

matrix units

< 2n.

it has a multilinear

.~ d. X.w l . x. 2

the

duct taken in any other order is

some

xi

coefficient

The product taken in this order is

x2-->

G = 0

coefficients

f(xl,...,x d) = ~ aw i .

Consider the sequence

regarded

A has ~ proper

we are done.

If it has such an identity

d < 2n.

A

f,

This is a proper identity for

result follows by induction

with

with

and consequently we have an

Also on dropping~the

in

Ae

~ deg f .

Applying Lemma i of §2

is blended.

E n d A e A = C,

C.

m u t t i l i n e a r i d e n t i t y o f degree Proof.

A e- module.

Hence

acts as a dense algebra of linear transformations as a vector space over

has no ideals

x!--> ell,

elk = 0 Then

for

19

LEMMA 3.

Let

F

be a field over

dimensional

vector spaqe over

formations

En~V

Proof.

x

f

Let

such that

and

V

an infinite

Then the a l~ebra of linear trans-

satisfies no proper identity.

Let

its degree.

F.

K

be an identity for

x -~V , M

2[M:F] > d.

a finite dimensional Let

the linear transformations

A = EndFM

B

in

and let

be

subspace containing

be the subalgebra

V

d

which stabilize

without unit of

M

and send a fixed

!

complement

M

n = [M:F].

Hence Lemma 2 implies that if

then

of

~B = O.

LEMMA &.

Let

F

Then

F

(over K) {e - C A

F.

c -~F.

Since

al,...,ar -~A

B ~ Mn(F)

~

where

is any coefficient

(~l)x = O.

Hence

aA = 0

~ cf = re, f -CF] of a maximal

F

of

A A

and

c

of

and

f

f

E,

contains

(F)

F

over

B:

the

then the division so this is a field

by the maximality

it follows that subalgebra

C = B ~ A

F

subfield follows by Zorn's

e -Cc~

are linearly independent

are l inearl% independent

~

=F.

is cor~nutative

be a simple in

Then

and for any such subfield

C A(F) ~ F

Let

the centralizer

of an algebra

C.

E bia i = 0

F.

E,

B

C A (F) = F.

the center of over

of

A.

Suppose

Then these elements

for

bi - ~ B

implies

b i = O. proof.

such that

The action of

m(bx)

m = Z gi<~)hi

= b(mx)

then

the density theorem an

Then

Hence it coincides with

LEMMA 5.

everz

O.

is such a subfield and

generated by

containing

into

be a division alsebra.

The existence

If

subalgebra

~

CA(F)=

proof. lemma.

V

A.

subfields

centralizer

in

This implies that

is not proper for

maximal

M

mj -CA e

mx=

on

A

extends to an action on

m -CA e, b -C B, x - C E .

Z gixh i .

Since the

implies that for any

such that

to the relation

for

Ae

we obtain

Here if are C-independent

j, 1 ! j ~ r,

mjaj = i, mja k = 0

~ bia i = 0

ai

if

bj = O.

E

x # j.

there exists Applying this

20

Proof of the theorem. linear transformations A.

Moreover,

morphisms

of

6-C

~ •

Let

bra

AT

in a vector space

the centralizer of V

of

We may assume

is the set F

A

~L

K-endomorphisms FL

of

F

(or over

over a division algebra

of scalar multiplications

V

~

xw>$x,

and consider the alge-

generated by

commute with those of

is in the center of

is a dense algebra of

in the K-algebra of K-endo-

t

bra over

V

be a maximal subfield of

the elements of FL

A

FL

A, A

and

= FL A

A.

Since

=AF L

and

V

A . FL).

Hence

A

can be regarded as an alge-

If we consider

V

as vector space over

T

F

then

i.

is an alg~br~ of linear transformations

in

V

We claim it is dense as algebra of l±near transformations

over in

F.

V/F.

!

Since

V = Ax

for any

x ~ O

in

V, V = A x.

Hence

V

is an

7

irreducible

A

- module.

Now let

c

be a K-endomorphism of

V

v

which

commutes with every element of

every

a -~A;

element of

hence,

it is in

F L . Since

F

that it is contained in

A •

~ L"

Then it commutes with

Also it commutes with every

is a maximal subfield of F L . Thus EndA,V = F

~

it follows

and we have density

t

of

A

in

EndFV.

Now suppose

A

has a proper identity

f

of degree

d.

We

T

may assume over

F

f

is multilinear.

it is clear that

f

Since

A

: FLA

is an algebra

is an identity for

is any polynomial

(in K{X}),

the map

i - ~ L = End F v

is continuous

A

.

Now if

f

( ~i''''' ~ m ) - > f ( £ 1 '

in the finite topology.

.... £ m )'

For our

f,

T

this is

0

on the dense subset

the identity 2,3 that

f

is an identity for

[V:F] ~ [~].

Since

dimensional over

~ .

formations

over ~

in

al,...,ar -~A

A

V

Since

of L,

L.

Hence it is

= FA.

Hence

we have

A = End a V

Thus

is finite

is simple.

these are linearly independent over r ~ [~]2 z

V

is a dense algebra of linear trans-

!

A

,

be linearly independent over the center

Then, by Lemma 5,

and so

It now follows from Lemma

IV:F] = [ V : ~ ] [ ~ : F ] A

0

[A:C] < [~]2

Let

C

of

F

in

A.

21

4.

Theorem of Amitsur and Levitzki We have seen that

Mn(K )

has no proper identity of degree

We shall now show that it has one of degree OF AMITSUR-LEVITZKI.

THEOR~

an identity for

2n

< 2n.

by proving the

The standard polynomia ~

S2n

is

Mn(K ).

We note first some properties of standard polynomials.

The

definition shows that Sk+l(Xl, ...,Xk+ I) = XlSk(X2, •. • ,Xk+ I) - X2Sk(Xl,X 3, ... ,Xk+ I) (i0) + ... + (-l) kxk+iSk(Xl,.-.,Xk)This implies that if Sk+ !.

Sk

is an identity for an algebra then so is

We note also that

(ii)

Sk(~l,

and if

il,...,i r

..., x k ) = sg(~) Sk( Xl,...,x k)

are distinct, 1 _< ij <_ k, and

denotes the sum of the terms of

Sk(Xl,...,x k

0 < r < k

having

mn@

S

Xll'''''Xir

as left factor, then t

(12)

S

-- +. . .Xml ..

XirSk-r(Xir+l'''''Xik) !

LEMMA i.

Let

of the terms in

r

b_~e odd, 0 < r < k,

Sk(Xl,...,Xk)

y -- xi+ I ... xi+ r

and

a

and

and let

of the form b

9 ayb

are monomials.

S

be the sum

where Then

!

(i])

S = S k _ r + l(x l,...,x i,y,xi+r+ l,...,xk). Proof.

Consider a monomial

ab -- x i... x i x (i+r+l)...x k l,..~i,i+r+l,...,k.

ay b and

~

where is a permutation of

The sign attached to this monomial in

Sk_r+l(Xl,..., xi, y, Xi+r+l,..., Xk)

is

sg(~) (- I) s

is the number of places from the position of the

(i+l) st

place.

Replacing

y

by

y

in

where ayb

Xi+l ... xi+ r

s to the

T

sign attached to

ayb = axi+ I • • Xi+rb

in

S

is

sg(w) (-i) rs.

22

Since ayb

r

is

odd,

(_~rs

are the same

n ~ 1

the theorem

and we assume

linear and alternating (ll)

it for

n - 1.

2n

(15)

S2n(eiiJl,e

For any

occurs as a subscript in (ll).

f(k)

is multi-

if

then O.

denote the number of times

in the set (ll).

n ~ f(k) k=l there

is a

occur in (li).

Call

f(k)

k

the degre e of

for

k

= in.

such that

f(k)

= O,

Then the m a t r i x units

m a t r i x units for identity

Sk

It is clear

n J2n }

• ,...,e ) = 1232 12nJ2n let

n.

Then we have

~16) Suppose

Since

distinct m a t r i x units,

k, 1 ~ k ~ n,

on

to prove that

{ e i l J l , e 1232 ,...,ei2

to

holds.

by induction

it suffices

is a set of

k

Hence the s i g n attached

in both cases and(13)

We shall prove for

= (_l)S.

Mn_I(K ).

Mn_i(K )

Since

so is

is,

k

does not

in (I$) may be regarded as

S2(n_l)

S2n.

that

Hence

is assumed to be an (15) holds

in this case.

From now on we assume (17)

f(k) > i, 1 < k < n. w

LEMMA Then

(15)

2.

in

euu

occur

Suppose

first

= 3.

eu v

or

mials

in (14) and assume

that

Then

u

is

occurs

evu, v ~ u.

in

0 e uu

In the first

in (15) are those beginning

they are those the terms

f(u)

in (5) only in the element

S2n(eilJl,...,e12nJ2n)

f(u)

m

f(u) ~ i.

is valid.

Proof. subscript

Let

~

ending

in

evueuu.

in (15) beginning

with

= 2.

euu.

and

Then Hence

(15) holds.

u

occurs as

every monomial Next let

and in one other element case the only non-zero with

e and in the second UU UV In the first case the sum of

euueuv

e

mono-

is

23

+ euueuvS2n_2 (eilJ i ,...,euu ~ A ,...,euv , ... )

where

--

deletion of the term below. the induction hypothesis = O.

u

f(u) = 3.

are in

monomials if

u

that

u

and

in (15) are

O

in the second case. f(u) = 4.

euw , v ~ u, w ~ u

so (15) holds°

$2n_2

euu , euv

and

in this case the non-zero monomials euueuv a e w u , euv a ewueuu

Hence (15)

If the occurrences then all the

The same argument applies

euu , evu , ewu , v ~ u, w ~ u.

occurs only in

aewueuueuvb.

does not occur in

Finally suppose

euu , euv

occurs in

u

indicates

implies that S2n_2(eilJl,... '~u,u' "'" ' euv' "'" )

A similar argument applies

holds if of

Since

A

ewu

Hence we may assume where

v ~ u, w ~ u.

in (15) are of the form

and the monomials

of the form

Since the number of factors

eij in

is

odd the signs attached to

e u u e u v a ewu

opposite so these

By Lemma I, and ~i), the sum of terms

of the form

cancel.

aewueuueuvb

is

and

a ewu

euvaewueuu

+ S2n_2(ewv,

...

the induction hypothesis,

since the subscript

in the

Hence (15) holds.

e~s

in

$2n_2.

Proof o f the theorem.

For, by

(16),

j = i.

we have a

r < n

idempotents (18)

eij

If this is

n

euu.

Using the multilinear

idempotent9 (15) holds.

f(u) < 4.

Then we

We now suppose (l@) con-

idempotents and we assume the result for in (14).

we use a

then the result

such that

O by

does not appear

in (li) which are n

u, 1 < u _< n

can apply Lemma 2 for the idempotmnt tains

This is

Besides the induction on

backward induction on the number of that is~ for which

u

).

are

> r + 1

property of

S2n

we have

S2n(eulUl, ... ,eur+lUr+l , ar+2, . .. ,a2n) = 0

for any

r + 1

unequal

ui

and any

at+ l,...,a2n -~Mn(K)-

We now assume for the moment that [el,e2,,..,er+l}

are any set of

r + 1

rank 1 in

Mn(K ).

such that

~] euiui = ei, 1 _< i <_ r + i.

K

is a field and we suppose

orthogonal

Then there exists an automorphism

idempotents ~

of

of

Mn(K)

It follows on applying

~] to

24

(18) that (19)

S2n(el' "'''er+l ' at+2' "'''a2n)

for all (15)

ar+l,...,a2n

for

il = Jl = l'''''ir

Case I.

i~j).

- ~ M n ( K ).

(l$)

contains

We may assume

in (14).

eij

= 0

Similarly, = Jr = r

an

eij

it suffices

where with

ik ~ Jk

if

k > r.

i > r, j > r

is listed as the ( r + l ) -

Then we have two sets of

to prove

r + 1

(and

st element

orthogonal

idempotents

of

rank i: t

e I = ell,...,e r = e r r , er+ I -- e i e I = ell,...,e r = err , er+ 1 = e i +

eij

f

Then (19).

Subtracting

S2n (el'''''er'

the first

case there

(14)

exists an

reduction

no

eij

eij

with

in (l&) with

in (14) are

to the situation

~ r < n in which

f(k) ~ 1 for l ~

also that

r = 0

of m a t r i x units

so we may assume u ~ 0

unless

every

assume

this also.

for every

i < r.

such that

f(j) < 3.

Otherwise,

either two

> ~ m

f(k)

for the other

Then,

of the next

to the case

Since a product

j

j > r

one,

(15)

of each

is trivially

occur as first and last

Also,

by (16) there

exists a

with

j > r

or for one

ones we have

j

with

is such a

satisfy" j

with

f(j) > L. g

j f(j)

j > r

More-

index.

by Lermma 2, we m a y assume

We claim that there

and for the remaining

r ~ 1.

k ~n.

is even or exactly two are odd.

the two odd indices must

therefore

trivially

if and only if ~he second subscript

is the same as the first

Other-

contrary to our

in which

over,

by

In

i < r, j > r.

case applies

satisfied

)

i > r, j > r.

We remark also that the preqeding

factor

...

from the second gives the relation

contains

wise all the subscripts earlier

) = 0 = S2n(el,...,er+l,

3.iJ ,... ) = O.

Case II. this

T~

S2n(el,...,e r,er+ !, ...

We f(i) > &

j > r with = 3

f(j) ~ 2. and

f(j)

we have f ( j ) = 3

In the first

case

25

f(i) > 6 that

for all

f(i) = 5

i < r

and in the second we have one

and hence for the remaining ones

case (16) implies that

r = i.

ell

with

and

elj

or

ejl

Then the j > i.

contrary to the fact that we have

eij

(14).

and for some

If it is

eij

e i ~ eii - eij

i ~ r

In either

This gives at most 2n-i eTs

2n.

either

we replace

and

f(i) > 6.

such

appearing in (14) are

We have therefore shown that we have a f(j) ~ 2

i < r

eli

j > r

eij

or

and

eij

er+ 1 = eij + ejj.

such that

eji

is in the set

in our set by

This gives

r + 1 orthogonal t

idempotents of rank i: el = ell ,...,el_ I = ei_l,i_l,e i, el+l= ei+l,i+l,..., T

T

e r = err, er+l o Then Since

e~ = eli-

S2n(el,...,ei_l,ei,

eij, er+ 1 = eij + ejj

. . S2n( . ..,eii,

The second term in this is 0

since it contains

The last term is

this gives

., e.13 ,... ) - S2n ( "'',eij,'" .,e.. 13,... )

+ S2n(...,eii,...,ejj,...

is

T

ei+l,...,er, er+l,... ) = O.

0

0

) - S2n(...,eij,...,ejj,

since

r + i

O

is repeated and the third

orthogonal

since we have

Hence the first is

eij

idempotents of rank i.

f(j) ~ 4

for the idempotent ejj.

and this is the required relation. t

is in the set (14) we take

... ) = O.

If

e.. 31

t

e i = eli - eji, er+ I = eji + ejj

and

argue in the same way. This completes the proof in the field case. and the result holds for

Mn(Q)

it holds for

multilinear the result holds for any REMARKS.

Since

S2n

With a set of

is

Mn(K) : M n ( ~ ) @ ~ K

Subsequently a graph theoretic proof was given

We proceed to indicate the graph theoretic formulation of

the Amitsur-Levitzki

F

Mn(~).

Mn(~ )CMn(Q )

The foregoing proof is the one originally given by

Amitsur and Levitzki. by Swan.

Since

2n

identity (first noted by Schutzenberger).

matrix units (14) we associate a directed graph

whose vertices are the subscripts of the

e

13

in (l&) and whose

26

oriented edges are the pairs set (ll). k

The number

f(k)

(i,j)

k.

Since one may assume

1,2,...,n.

eij

is in the given

we defined is the number of edges having

for initial or terminal vertex.

vertex

such that

This is called the order of the f(k) ~ l, the vertices are

To any non-zero product of the

eij

in (ll) taken in

some order one associates an Eulerian path on the graph an ordering of the edges of

~

This correspondence

between Eulerian paths and non-zero products of the

eij

vertex of the first edge of the path is

terminal vertex of the last edge of the path is ponding product of the

eij

that is

so that the end point of each edge is

the same as the beginning vertex of the next.

If the initial

~

is

ek Z

~

is k

I -1. and the

then the corres-

Pick an Eulerian path on

r

and call any other one with the same initial point and terminal point even or odd according as the permutation of the edges is an even or odd one of the one selected.

Then the Amitsur-Levitzki

identity is

equivalent to the statement that there are as many even as odd Eulerian paths on

~

with the same initial and terminal vertex.

While the graph theoretic proofs of the result do not appear to be simpler than the Amitsur-Levitzki

proof, the graph formulation

provides an improved insight into some of the steps of the foregoing argument.

(Still another partly graph theoretic proof has been given

by Amitsur.) In 1958 Kostant showed that the Amitsur-Levitzki identity is equivalent to two other results which have arisen in quite different contexts: group, and

l) a theorem of Frobenius v on characters of the alternating 2) a theorem of Dynkin's on cohomology of Lie groups.

Using cohomology theory Kostant showed also that if S2n_2(Sl,...,S2n_2)

= 0

for all skew symmetric

elementary proof of this result for all

n

n

is even then

si -~Mn(K).

An

has been given quite

recently by Rowen who obtained a number of other results on what happens if the arguments in a standard polynomial

Sk

are taken

27

to be skew symmetric not wholly

5.

or symmetric. Rowen's proofs are largely,

graph theoretic.

Central

simple al~ebras. Converse of Kaplansky- Amitsur theorem

We consider first some results that if

though

A

is simple,

its center

C

garded as an algebra over

C.

that the center of

C = KI.

A

is

on simple algebras. is a field and

We assume now that

central simple over the field

We have seen

A

K

can be re-

is a field and

In this case we say that

A

K.

The theory of tensor products over fields is especially We recall that if the maps

x-->

M

and

x~l

are injective.

If

A

N

and

y-->

and

B

subalgebras

y

Then

l~y).

l~y

of

are algebras

maps are algebra monomorphisms

with

simple.

are vector spaces over a field

the corresponding

is

M

and

over

K

so we can identify of

ab = ba

A~B

N

into

a -~A,

then

M~N

the foregoing A

and

B

x

with

(identifying

for all

K

b ~B.

with x~l,

Also we have

the independence conditions that if al,...,ar - ~ A are K independr ent then ~ aib i = 0 for b i -~B implies every b i = O. Also the 1 condition obtained by reversing the roles of A and B is valid and A~B

= AB

is generated by

A

The foregoing properties algebra over the field that

ab = ha, a -~A,

the foregoing of

A~B

K

into

E

B.

characterize

containing

b -~B,

independence

and

E

hold.

a~b-->

A

and and

Then the

be an B

B

such and on__~eof

K-linear

are finite dimensional

independence

can be replaced by the dimensionality

algebra.

then the

[E : K] = [A : K][B : K].

We now consider tensor products

involving a central simple

Lemma 5 of ~3 can be re-stated

as

map

is an algebra isomorphism.

A

relation

B

ab

A

E

We remark also that if conditions

and

Let

subalgebras

is generated by

properties

such that

A~B:

28 THEOREM i . an al~ebra of

A

E

in

Let

A

over a field

E.

Then

AB

be a central sim~le subalsebra o f K

and let

B = CE(A)

the centralizer

i~s the tensor product o f

We obtain next the ideal structure of

A

A~B

and for

B. A

central

simple. THEOREM 2.

Let

second algebra over A (~ B

t

,> I

(ii)

I

> AI

[I'}

in

B

an__dd B

B

~ B (= IA)

A~)B

= AB.

A~B

and

is

[If

is the set

I

Let

The centralizer

B. is an ideal in

{i, u~ }

B

then

be a base for

AI A.

is an ideal in Then

the set of elements ib o + u{ib I + .o. + u rbr,

AI = IA

bj ~ I .

then all

bj = O j ~ O.

Then the element is in

I.

is

The

independence property implies that if such an element is in

B

Hence

Y

AI ~ B = I.

Now let

is an ideal in Form

B

AB/I '

I

If

morphism for

A

By Lemma 5 b j

AI

u -CAB

u --> ~

the base

be an ideal in

and

homomorphism

every

write

of

AB.

is an ideal in

AB

q = u + I

into

AB/I'

and a homomorphism for

B.

Then

AB

I=I

f~ B

contained in

I .

so we have the This is a monoThe image

A

has

{~,ut ]. The elements of AB/I are % o + ~ij-bj, b~ j -~ B. of

~3,

= O.

such an element is

This implies that

T

bo,.b j -~I = I

~ B.

has the form

only if every

O

only if b o = O

b ° + ~ uojbj

-~I'

and

only if

!

of the theorem let A~B

a

a__sssubaigebras o__f

are inverses s_2oboth are bi~gctive.

Proof. If A ~B

A

is the set of ideals of

of ideals of A

Regarding

K,

T

I

of

K.

b~e central simple over a field

th___emaps

(i)

where

A

aj

Hence {vx}

I

= AI.

be a base for

~ aj vxj , aj = O.

To prove the last statement B.

-~A, and

Then every element of ~ ajvx

= O holds J If this element is in the centralizer of

29

A

then we have

2 [a]] v k

This implies every COROLLARY

If

i~s simple then A ~ B

aj A

for all

a ~A.

so the element is in

F/K

AF = F ~ K A

i~s simple.

~

If

If

B

O.

B. K

and

B

is central simple, then

field.

A

is finite dimensional central simple over

is called a splitting field for

The al~ebraic closure

Also if

A~J= M r ( ~ )

where A

is a maximal subfield of ~ Proof.

then

For any field

F

K

if

central s~nple with where

K

is ! splittin~

F/K, A F

is an algebra over

is a (central)

AF

F.

If

A~

Mn(K ).

Hence

K

A

is finite dimensional

and by the Wedderburn theorem division algebra over

F. If

then the only finite dimensional division algebra over

Then

F

is a splitting field.

[AF: F] = [A :K] ~

of

is a division alsebra and

is finite dimensional central simple then

AF = Mn(~)

A

Mn(F ).

THEOREM 3.

K.

[ a] ] =

is central sJ~nple.

a field

F = K

Then

is central simple over a field

A~B

DEFINITION. K

-~K

= 0 J

is a splitting field.

F

is

To see the

second statement we refer to the proof of the Kaplansky-Amitsur theorem.

There we showed that if

transformations in subfield of A

V/~, A

then

transformations in

F

is a maximal

is a dense algebra of linear

This applies in particular to

dimensional central simple. finite dimensional over

is a dense algebra of linear

a division algebra and

A' = FLA V/F.

A

~

In this case we may take and~

A V

finite dimensional over

independence property in Lemma 5 of §3 shows that

finite to be K. The

A'= FLA~'-- F ( ~

,

Also we saw that formations in

A

V/F.

COROLLARY.

A. K

is the complete algebra of linear transHence if

[V: F] = n

then

F~KA

~Mn(F).

The dimensionality of a finite dimensional central

sim~le algebra is ~ square.

30

We have seen that if proper identity of degree center, then

A d

[A: C] ~ [~]2.

the standard identity

is a primitive algebra satisfying a then

A

is simple and if

We have seen also that

S2n.

C

is its

Mn(K ) satisfies

We can now combine our results to

obtain the following fundamental structure theorem on primitive algebras satisfying proper identities. THEOREM OF KAPLANSKY-AMITSUR-LEVITZKI. A

~ primitive~

satisfies ~ proper identity i_~fand only if

finite dimensional over its center. for prope r identity for the center.

Moreover,

Proof. degree

d

If

A

A A

then

d

d = 2n

i_~s simple and

is the minimum degree is even and [A: C] = n 2, C

satisfies the standard identity

S d.

is primitive and satisfies a proper identity of

then

A

is simple and

let

A

be simple with

Conversely,

If

A

[A : C] ~ [~]2, [A: C]

C

finite.

Then [A: C] = n 2

and we have a splitting field

F

and hence

All the results follow from this.

A

satisfies

We shall call 6.

S2n.

n = d/2

so that

the center.

A F ~ Mn(F ) .

the degree of

Then

AF

A.

Nil ideals in al~ebras without units. In this section we consider the category of algebras without

unit over a commutative ring

K

algebras with unit.

even for these homomorphisms

However,

map units into units.

(with unit).

Its objects include need not

An important advantage of working with alge-

bras without unit is that a one sided ideal in such an algebra is a subalgebra. The concepts of prime and semi-prime algebras are the same as for algebras with unit. is nilpotent and A m = O. A

is

A

An algebra is nil if every element of

is nilpotent if there exists an

This is equivalent to: O.

A

the product of any

is called locally n$1potent

generates a nilpotent subalgebra.

m m

A

such that elements of

if every finite subset

This is equivalent to:

If

31

[bl,...,b k}

is a finite subset of

the product of any

m

of the

called locally nilpotent

A

b.

there exists an

is

O.

m

such that

A one sided ideal is

(nil, nilpotent)

if it has this property as

an algebra. LEMMA i.

Subalgebras and homomorphic

images of nil, nilpotent

o rr locally nilpotent algebras have the same property. ideal in

A

such that

nilpotent then Proof.

A

property.

A/B

is an

are nil, nilpotent or locally

has the same property.

l_~f N I

ideals in

Proof.

and

B

Clear.

LEMMA 2. potent)

B

If

A

and

N2

are nil (nilpotent,

then so is

N 1 + N 2.

(NI+ N2)/N 2 ~ NI/(N I ~

Hence

LEMMA 3.

N1 + N 2

If

[N }

local!z nil-

N2)

and

N2

have the stated

has, by Lemma i.

is a set of nil (locally nilpotent) ideals

b

in

A

then Proof.

~ N~

is nil (locally nilpotent).

Clear.

An immediate consequence of Lemma 3 is the first statement in THEOREM 1. potent)

There exists a unique maximal nil (locally nil-

ideal in any algebra,

containing every such ideal.

The

maximal locally nilpotent ideal contains every locally nilpotent one sided ideal. Proof of second statement. nilpotent ideal of

A

and let

Then

I + IA

[b i}

be a finite subset of

where

is an ideal.

ci, cij C I .

This is a subset of of

m

Let I

L

be the maximal locally

be a locally nilpotent left ideal.

We claim this is locally nilpotent. I + IA

and write

b i = ci +

Let

~ cijaij

. , aijck'aij c k ~} " Consider the finite set S = [c i, c .13 I

so there exists an

of these elements is

O.

m

Any product of

such that any product r

of the

bi

is

32

a sum of terms each a product of duct multiplied duct of

m

r

elements of

on the right by an element of

of these is

O

and

I + IA

The maximal nil (locally nil) radical

(Levitzki n i l radical)

the upper nil radical

of

S

A.

or such a proHence the pro-

is locally nilpotent.

ideal is called the upper n i l A.

It is not known whether or not

contains every nil one sided ideal.

This

question was first posed by Kothe more than thirty year ago. It is not true that the sum of all nilpotent potent ideal.

Denote this sum as

N(O).

finite sequence of ideals as follows. If

~

ideals is a nil-

We now define a trans-

Let

N(O)

be as indicated.

is an ordinal which is not a limit ordinal,

define

N(~)

to be the ideal in

sum of all nilpotent

ideals of

A

such that

A/N(8).

If

~

so

N(~)/~(~)

N(~) = U N~)o We have N(~) C N ( ~ ' ) if ~ < ~ ~<~ exists a first ordinal ~ such that N ( U ) = N ( ~ + I ) . the lower nil radical of

A.

Since

N(O)

Lemmas 1 and 2 imply that the lower nil radical

and

L(A)

ideals ~ O, A / Z n A = O.

Call this

is a nil ideal. Let

unA,

denote the upper nil radical lower nil radical

and Levitzki nil radical of

L(A/L(A))

and there

is a nil ideal

Hence this is contained in the upper nil radical, Zn A

is the

is a limit ordinal

put

N(~)

~ = ~ + i,

A.

Then

A/un

A

contains no nil

contains no nilpotent ideals ~ 0

and

Also we have ~nACL(A)

and there are examples in which THEOREM 2.

C C

un A can be replaced by C +

The lower nil radical

section of the prime ideals of

coincides with the inter-

A~ !

Proof. only if

A

also that if

The proof on is semi-prime B

p. 12

that

N

m

~ P = P priz~e carries over without change.

is an ideal the correspondence

0

if and

We recall

I - - > I/B

of the

33

ideals of

A

containing

B

arbitrary intersections: ideals of A

A/B

with the ideals of

~(I~/B)

are the ideals

containing

B.

• ( P / N ' i = O. P/N ' prime

= (~.I~)/B.

P/B

where

Applying this to

B = N

Hence

A/N'

is a niipotent ideal of

A/B

respects

Also the prime

P

is a prime ideal of we see that

contains no nilpotent ideal ~ O.

A

then (N+N')/N' ~ N/(N ~ N')

If N

is a

t

nilpotent ideal of

A/N'

so

N ~N

t

This is

0

.

Hence

N DN(O)

~N(~)

one proves in the same way that

t

defined above.

Assuming

N

!

N

contains every ideal

N~N(~)

such that

N/N(8)

is nilpotent.

f

It follows that

N

~N(q-)

the lower nil radical.

contains no nilpotent ideals ~ O, C~(P/N(~)) ideals containing

N(Z').

= 0

Since

A/N(~)

for all prime

Hence the intersection of these prime

!

ideals is If ideal.

N(~) I

and

f

N

= ~P ~ N(~). P prime is a nilpotent left ideal then

It follows that if

A

Thus

N(~)

I + IA

is semi-prime then

= S

•

is a nilpotent I

contains no

nilpotent one sided ideal ~ O.

7.

Polynomia ! identities for algebras without units Let

Xl,X2, x..

KIX}

...

be the free algebra with unit (freely) generated by

and let

KIXI T

be the ideal in

This is the set of elements of

K[X}

KIX} with

generated by the 0

constant term.

1

KIX } '

is a free algebra in the category of algebras without unit:

If

is any such algebra and

A

elements of

a i , 1 ~ i < ~, is a sequence of

A, then there exists a unique homomorphism sending t

xi m >

ai

for all

identit~ fo__r A

if

i.

If

f = f(xl,.o.,Xm) -~KIX}

f(al,...,a n) = 0

for all

then

ai -~A.

see that out unit.

f ~K{X}'

• Hence

A f

with unit then putting is an identity for

A

is an

The notion

of proper identity is the same as for algebras with unit. is an identity for an algebra

f

If f-~KIXI ai = O

we

as algebra with-

34

f -~K[X}'

will be called stron$1y resular if

non-zero coefficients of If

f

f

f # 0

and the

are units (invertible elements)

of

K.

satisfies a strongly regular identity then so does every sub-

algebra and every homomorphic image.

Strongly regular identities are

proper in the sense defined before. regular identity then

A

If

A

satisfies a strongly

satisfies a multilinear one of no higher

de gre e. THEOREM[ 1.

A nil algebra satisfyin$ ~ stron$1~ regular identity

i_.sslocally nilpotent. Proof.

Let

A

be nil with strongly regular identity

we may assume is multilinear. have to show that is

O.

A = L.

Let

L

f

which

be the Levitzki nil radical. We

Passing to

A/L

we have to show that this

It suffices to prove that the hypothesis implies that

A

contains a locally nil left ideal I ~ O. Choose a ~ 0 in A with 2 a = O. If Aa = 0 the right annihilator of A is / 0 and this is nilpotent ideal so the result is clear in this case. Aa ~ 0

Thus we may assume

and we proceed to prove that this left ideal is locally nil-

potent.

Write

f(xl,...,Xm)

the monomials in

= Xlfl(x2,...,Xm)

f2 ' do not begin with

x 1.

+ f2(xl,..o,Xm) We may assume

fl(x2,...,Xm) / 0

so its degree is

m- 1

strongly regular.

If we substitute

x I --> a, x i - - >

we obtain

0 = afl(a2,...,am)

contains a factor fl(a2,...,am) if

Z

(ba)a = O.

and this polynomial is

since every term of

annihilates every element of

the strongly regular identity

fl

Aa

in

of degree

Aa Aa

then m - 1.

Aa/Z

Z2 = 0

is locally nilpotent.

the proof.

Aa

in

f

f2(al,...,am) ai -~Aa

on the right,

on the degree we may assume it follows that

ai -~Aa

This shows that for any

is the right annihilator of

where

Aa/Z

Thus

satisfies

Using induction

is locally nilpotent.

Since

This completes

35

THEOREM 2.

Bet

identity of degree B [d/2] ~ N ( O )

A

d.

be an algebra satisfying a strongly re~ul9 ~ Then any nil subalgebr a

the sum of the nilpotent

Proof. Suppose first that integer

n

(20)

U2i_l -_ Bn-iABi-i

For

A

satisfies

A.

is nilpotent.

For any positive

,

U2i

Bn-iAB i , i _< i _< n.

=

we have

UIU 2 .. U h = (Bn-lA)hB[h/2]

and for any

j > k

(22) Let

ideals of

of

define

h < 2n

(21)

B

B

UjU k C A B n

nA.

be chosen to be the smallest positive integer such that

is nilpotent,

Since

B

is nilpotent such an

n

satisfies a strongly regular identity of degree is multilinear and on dividing by a unit in identity

f

(23)

n > [d/2]

above formulas, (24)

d

Since

A

we may assume this

we may assume the

has the form XlX 2 ... x d -

Now suppose

K

exists.

ABnK

so

Substitute

Z 2n > d

and we may take

xi - u i - ~ U i

UlU 2 .., u d -- ~ iZ

in (23).

h m d

in the

This gives

aN i... ~ dU~l -,o U~d

and by (21) and (22) we obtain (25)

(Bn-lA) d B[d/2] C A B n A

This implies that

(ABn-IA) d+l C ABnA

is nilpotent contrary to the choice of shows

that

AB[d/2]A

is nilpotent,

B[d/2]A + AB [d/2] + AB[d/2]A this is contained in Now suppose

B

N(O).

.

which implies that n.

This contradiction

It follows that

is a nilpotent ideal of Hence

is simply nil.

ABn-IA

B [d/2] + A.

Hence

B [d/2] C N ( O ) . Then

B

is locally nilpotent.

36

Let Bo

bl,...,b[d/2 ] -~B. and we have

Then these generate a nilpotent subalgebra

Bo[d/2] C N ( O ) .

Hence

bl...b[d/2 ] -~N(O).

Thus

~[d/2] C~(O). THEORY4 3. d.

Then the upper

nil radical

L.

Proof. and

Let

A

have ~ strongly regular identity o~f degree

and lower nil radicals coincide with the Levitzki

We have

Let

N

N[d/2] C N ( O ).

L [d/2] C N(O)

and

L = N(1).

denote the upper nil radical. Hence

N CN(1).

with the lower nil radical.

Then

This proves that

N~N(O) N

coincides

Since the Levitzki nil radical is con-

tained in the upper and contains the lower nil radical, we have L ~ N, L [d/2] C N ( 1 )

8.

and

L = N(1).

Central polynomia]~ for matri x algebras In this section we shall give Formanekts construction of

central polynomials for

Mn(K)

and a consequence of it due to

Am it sur. A tool we shall use in deriving the results is the Zariski topology of a finite dimensional vector space field

K.

One defines a polynomial function

is given by a polynomial, that is, if f

has the form

over an infinite

f

on

(el,...,e n)

~ aiei __> f(al,...,an)

f(~l''''' ~ n ) -~K[~l''''' ~n]' ~i

V

V

is a base then

where

indeterminates.

It is clear that

this condition is independent of the choice of the base. polynomial functions

P(V)

as one which

is an algebra over

K

The set of

under the usual

addition, multiplication, and scalar multiplication of functions. With respect to the base f( f

of

(e i)

we have the isomorphism

K[~I ,..., ~n ]

onto

P(V).

The fact that

this is an isomorphism is equivalent to the well-known theorem that if

K

is an infinite field and

f[~l'''''
is a polynomial then

37

f(~l ,. °' ,~n ) = 0

for all

~i - ~ K

defines the Zariski topology of

implies V

f([l"

", { n ) = O.

One

by specifying that the closed

sets are the sets of common zeros of sets of polynomials.

It

follows from the definition that polynomial functions are continuous if we topologize topology.

K

as one dimensional space using the Zariski

The closed sets of

K

are

K

and the finite subsets.

Any open subset is dense in the Zariski topology.

Hence a polynomial

function which vanishes on an open subset vanishes everywhere. Let Mn(K). of

If

a

again be an arbitrary commutative ring and consider

K

a = (aij) -~Mn(K ) we write the characteristic polynomial

as

(26)

~ a ( ~ ) = det(Xl- a) = k n - (tra)kn - l + .o. +(-i) n det a.

Consider the polynomial ring Let g(~l' "'''~n ) ~[~i'

"''~n]"

be a symmetric polynomial contained in

=DI"'Dn

symmetric polynomials g

and

Pl'''''Pn

h

where

is unique since the elementary are algebraically independent. We

to define a map

(27) of

indeterminates and

Then g(~i,...,~n)= h(Pl,...,pn)-~ZL[Pl,...,pn]

Pl = ~ i ' ' ' ' ' P n

use

ZL [71''''' ~ n ] ' ~ i

G: a - - > h(tra,...,deta) Mn(K )

into

K.

In particular, we can take

d = ~ [ (7i - ~ j )

2

i<j and obtain the discriminant ma~

D.

maps

are polynomial functions on

a --> tra,..., a - - > deta

Hence

G

a,

K

is an infinite field the

defined by (27) is a polynomial function.

algebraically closed and of

If

then

PI'''''Pn

If

K

Mn(K)~ is

are the characteristic roots

tra = Z pi,...,det a = pl.,°~n

G(a) = h(~ Pi'''''PI'''Pn ) = g(pl''''' ~n )"

so Thus

G

coincides with

the map (28)

a - - > g(Dl,... , pn )

defined by the symmetric polynomial

g.

In particular,

if

g-- d

38

then the corresponding map (29)

D

a -->

It is clear that

D(a) # O

characteristic roots.

is [~ ( Pi - pj) 2" i<j

if and only if

a

has distinct

This is the case if and only if

to a diagonal matrix with distinct diagonal entries.

a

is similar

The set of

matrices having this property is an open subset in the Zariski topology. We shall now proceed to construct central polynomials for ~n(K)

by Fcrmanek's method.

polynomial algebra over and let by (30)

f =

pf=E

be the free algebra over

~ (r) ~(W) ~ l

~n+l "''Dn+l ' (~)

pf

Z& generated

= (Wl' " ' ' ' ~ n + l )

we define

~i ~2 ~n ~9 n+l ~(~)x Yl x Y2 "'" x Yn x

-CK Ix, y} -- K{X, Yl,...,yn}. and

be the

If

~ F Z [ ~ I , . . . , ~n+l ] (31)

ZL [ ~ i ' ' ' " Dn+l]

ZL in indeterminates vl l' '" "' ~] n+l

~ [x, Yl,...,yn}

x'YI''°"Yn"

Let

We note that

f-->pf

is a polynomial which is linear in every

is additive x i . We now

soecialize ~32)

n x --> i~ '~i'ii' ~ e Yl --> eilJ l'''''yn --> eln3n " "

Then "qi ~J2 "Wn "~n+l x yl x Y2 "'" x Yn x --> "~i "~2 "~n Pil Pi2 "" pin

(33)

= ~ 3112 . . ~ 3213 ..... and hence n pf(~ Pieii . e . . . . . 1131'

~n+l Jn

%jlei2J2

"91 D)n ~ n+l eilJn 6Jn_linPil " "" Pin PJn • .

c. . ) = ' in3 n

~Jli2 ~J2i3 • .. 6 3n_lln f(Pil ' •.. ,Pin, pjn) ell3 . .n . This is

0

for all sequences

elnJn

(eilJl ,...,e ) in3 n

except

39

e

lll 2

,e

1213

,...

.

~eln_lln'einJ n)

in which case we get n (35) pf( Z1 Pieii,eili2,...,e In-lln . , e InJn ) = f ( P i l , " " " 'Pin' Pin) eilJn We remark that since there are only numbers in the sequence Now suppose (i) f(~l,...,~n+l)

f

n

choices of the subscripts two

(il,i2,...,in, Jn )

are equal.

satisfies is divisible by every

i - ~j' i ~ j, except

1 -~ n+l" (ii)

g(Dl''''' ~In) = f(~]l''''' hn' ~ I ) Then

is symmetric

,...,einJn) pf(X Pieii ,e i131

(eilJl,...,einJn)

except

= 0

in ]]l'''''~n"

for all choices of

(eili2, ei2i3,...,einil)

with distinct

ij, in which case pf(E Pieii,eili2 9 ...,e Inll ) = g(pl,...,Pn)eilil Now define (36)

qf(x'Yl' "'" 'Yn) =

n-1 O~ pf(x, Yi+l,...,Yi+ n)

where the subscripts are reduced ...,e.lnJn) = 0

for all choices of

(e.lll2.,el213. ~... ,einil )

(37)

mod n.

lj,

qf(E Pieii,eili2,ei2i],...,einil) f

qf(E Pieii,eilJi ,

(e i~l,...,einjn)

with distinct

The simplest choice of an

Then

except

in which case

= g(p!,...,pn ) i.

satisfying

(i) and (ii) is

Formanek t s (38)

c =

n n [[ (DI - ~i ) (~n+l - ~]i) ]~ (~]i - ~]j) 2 . i~2 i, j=2 i<j

This has integer coefficients

some of which are

+ 1

and in this case

we have n (39) d(~l,...,~n)-- C(~l,...,~n, ~]l) =

~

i <j =l

(~i - ~ j )

2,~o,

40

the discriminant polynomial we considered before. g = c

we have the following additional property:

(iii)

For any field

where

G

K

there exists an

For this particular

a-~Mn(K)

such that G ( a ) ~ O

is the map defined by (27).

If

K

is infinite this is clear since we can take

diagonal matrix with distinct diagonal entries.

If

K

a

to be

is finite we

use the fact that there exists an irredmcible monic polynomial of degree

n

in

K[X]

its characteristic

and there exist a matrix

polynomial.

Since

h(X)

distinct roots in the algebraic closure (27) is unchanged on extension of

K

K

a

having

is separable of

K.

a

h(X)

h(X) as a

has

Since the map

it is clear that

a

satisfies

O(a) J 0. We shall now prove the following result which gives a sharpening of Formanek's theorem. THEOREM i. (iii).

Then

Let

f -~[~i'''''

qf(x, Yl, ....,yn)

pol~nomial for 2.

i.

Mn(K)

There exists

and for such an

a

~n+l ]

satisfy (i), (ii) and

defined by (31) a n d (34) is a central

for any commutative ring a-~Mn(K )

there exist

such that

K.

G(a)

is not nilpotent

bl,..o,b n -~Mn(K )

such that

qf(a,b I .... ,bn) ~ O.

3.

If

f

is as in (i) and

(~l,...,~/n)~ZZ[~l,...,~n]

and

e.

is symmetric then (40)

q~f(a,b I .... ,bn) = L(a) qf(a;bl,...,b n)

for all

a,b i ~ M n ( K )

where

L

is defined as in (27) b_y e (~l''''~n j

Proof. We assume first thst field and we shall show that indeterminate

K

is an algebraically

[qf(x,y!,...,yn) , z]

is an identity for

Mn(K)

for

and (~0) holds.

equivalent to showing that the maps (&l)

(a, bl,...,bn, C) --> [qf(a,hl,...,bn),

c]

closed z

another This is

41

and (42) are

(a, bl,...,bn) --> q£f(a,bl,...,bn)-n(a)qf(a,bl,...,b O.

Since

linear in the choices

bi

bi

c

and

qf(x, Yl,...,yn) Yi

and

and in c

z

and

[qf(x, Yl,...,yn), z]

n)

are

it suffices to prove this for all

in some base for

Mn(K)

over

K.

For fixed

the maps (&l) and (42) are polynomial maps in a, 2 that is, there exist polynomials Pij' Qij in n indeterminates

~ij

such that the maps (41) and (42) are respectively

(a=

Z aijeij , bl,...,bn, c)--->

Z Pij(all,al2 i,j

(a,bl,...,b n ) - >

Z Qij(all,al2,

.. ) eij .. ) eij

Thus we have to show that the polynomial functions --> Pij(a), bl,...,bn,

(a, bl,...,bn) --> Qij(a) c

in our base).

Mn(K )

g(~l''''' ~ n ) = f(~l''''' ~ n ' ~ l ) the condition

characteristic

roots.

{eij}.

defined by

a

G(a) ~ O. Since

implies that

Hence

is similar to a diagonal matrix

a

Mn(K)

a

- ~j'

G(a) ~ O

a = ~ Pieii

matrix units

(for fixed

is divisible by every ~ i

and so applying an automorphism of diagonal:

O

It suffices to show this for all

in the Zariski open subset of

i ~ j,

are

(a,bl,...,bn, c)

has distinct

we may assume

a

is

and we may take the base to be the set of Noting that

G(a) = g(pl,...,pn )

we see

that our result follows from (37) and the considerations preceding it. K

Hence we have the result that (41) and (42) is an algebraically Next let

closure of

K.

K

be any field and let

K

such that

this is equivalent to

G(a)

this property.

a

Since

maps if

be the algebraic

Then we have an imbedding of

a -~Mn(K )

O

closed field.

Hence the foregoing results hold in we have an

are

Mn(K).

G(a) ~ O.

Mn(K)

in

Mn(~).

Also, by hypothesis, Since

is not nilpotent.

K

is a field

Suppose

a

is similar to a diagonal matrix in

(37) shows that we can choose

bl,..o,bn -~Mn(K )

so that

has Mn(K).

42

qf(a,bl,...,b n) ~ O.

By the linearity of

fact that any base for the

bi

also in

everything for

Mn(K )

Mn(K ) Mn(K)

Now let

K

K

~ (~--> of

~[q

etc.

lj!l))__>

YI = ( ~

~ O.

This proves

Consider the

in

into

K

m=(n+2)n

2

Then if a = (aij)

we have a ring homo-

sending

%ij --> aij '

This can be extended to a ring homomorphism

~(i) ij'

M n ( ~ [ ~ ij'

and the

any field.

c = (cij) -~Mn(K)

ij'''' ]

Yi

we can choose

(I) ~ ij' 1 <-- ilj <-- n. ~ij''''''


b(~

qf(a,bl,...,bn)

~ (i), ij "''' 7~(n) i j ' ~ ij ]

~[~ij'

bl = ~(~))'''13 "'bn" (b(~)' morphism of

Mn(K)

be an arbitrary commutative ring.

polynomial ring indeterminates

in the

is a base for

so that for

qf

]

.

.

into

Mn(K)

.

.

X=

sending

bl,. " ., Z = (C. ij) --> c.

Since

G

(~

and

ij)--> ao L

are

defined by polynomials with integer coefficients in the coefficents of the characteristic polynomials it follows that our homomorphism sends

L(X)

into

a field we have

L(a).

Mn(K )

Now let

a [ % ij'''" ]

[qf(X, Y1,..o,Yn) , Z] = 0

L(X)q{ (X, YI,...,Yn). into

Since

can be imbedded in

and

q{f( X, YI,...,Y n) =

Applying the homomorphism of

we obtain these same relations for

P

be a maximal ideal in

K

and let

Mn(~ [ ~ ij,-..])

a,bl,...,bn, c • F = K/P.

hypothesis there exists an

~ -~Mn(F)

a-~Mn(K )

in the canonical homomorphism of

onto

mapping onto

Mn(F ).

Then

~

G(a) = G(~)

contrary to the choise of let in

a

be any matrix in

K.

Hence

Mn(K)

Since the nil radical of

prime ideals of

A,

G(a) m O(a) + P ~ O. D = K/P exist

~.

so

and let

F

Then

K

G(a)

we have

Mn(K)

is not nilpotent. G(a)

Now

is not nilpotent

is the intersection of the

for

P

in

K

~ = a + Mn(P).

be the field of fractions of such that

Choose

niipotent implies G(a) = O

such that

G(~) ~ O

of fractions we may assume Mn(D)

G(a)

O(~) ~ O.

there exists a prime ideal

~l,...,~n -~Mn(F)

to map onto

such that

By

D.

qf(a, bl,...,bn)

bl,...,bn -~Mn(D). qf(a,bl,...,bn)

Let

Then there ~ O. Clearing

Choosing

~ O.

such that

bi -~Mn(K)

43

We now apply 3. of Theorem i c

and

~

nomials

with

f = Formanek's polynomial

taken to be successively the elementary symmetric polyPl = ~]i'

P2 =

~ ~i~j'~''Pn =~i''" ~n" Then 3. i<j and the Hamilton-Cayley Theorem give the following result which is

due to Amitsur. THEOREM 2.

Let

qo(X, Yl,...,yn) = qc(X, Yl,...,yn).

there exist central polynomials for

Mn(K )

such that for any

Then

ql(x, Yl,..o,Yn),.o.,qn(X, Yl,...,yn) a,b i -~Mn(K)

qo(a,bl,...,bn)% n - ql(a,bl,...,bn)% n-I (43) +...+(-l)nqn(a,bl,o..,bn ) = qo(a,bl,...,bn)~a(k) where H a ( k )

is the characteristic polynomial of

a.

Hence

qO(x,Yl,...,Yn)xn _ ql(x,Yl,...,yn) x n - 1

(44) +...+ (-l)nqn(X,Yl,...,yn } is an identity for

M~(K).

A useful remark on central polynomials which is due to Procesi is the PROPOSITION. Mn(K)

Any central polynomial with

i~s a__nnidentit~ for Proof.

Let

stant term for

regarded as the set of i,j = 1,...,n-1.

be a central polynomial with

and evaluate this for K

constant term for

Mn_l(K ).

q(xl,...,Xm)

Mn(K)

O

xi = u i

linear combinations of the

Then one obtains an element of

Mn(K)

of the form

and

identity for

kl, k 6 K. Mn_l(K).

It follows that

k = O

in

Mn_l(K)

eij

with

Mn_I(K).

other hand, this element is a center element of

0

On the

and so is q

is an

con-

44

9.

Generic minimum polynomials and central polynomials for finite

dimensional central simple al~ebras. We consider first the extension of the Hamilton-Cayley theorem to finite dimensional central simple algebras over a field

K.

If

K

is a finite field the only finite dimensional central division algebra over

K

is

K

itself.

This follows from the theorem of Wedderburnts

that any finite division ring is commutative.

Since any finite

dimensional central simple algebra has the form central division algebra the finite d i n e n c i o ~ l

Mn(D )

where

D

is a

central si~plc algebra~

over K are ~In(K) , n=l, . . . .

In this case we have the characteristic

polynomial of an element of

Mn(K), the Hamilton-Cayley theorem, and

the results of the last section.

We now assume the base field is

infinite and we shall prove the following THEOREM i.

Let

K

b_~e an infinite field, A

alsebra of dimensionality 1. ($5)

over

K.

T,...,N

(46)

Then:

The polynomial

are pol~nomial functions on

A

7a(X)

= Xn-T(a)xn-l+ a, T(a)

generic trace and generic

respectively of

norm

Th e generic trace

T

vanishes on all commutators homogeneous of degree T(1) = n 3.

and

n:

such that a-~A.

...+ (-l)nN(a)

generic minimum ~olynomial of

Also

n

= kn - T ~ n- 1 + ... + (-l)nN

a n - T ( a ) an-l+ ...+ (-l)nN(a) = 0 ,

2.

a central simple

There exists a unique monic p olyngmial of de~ree ~A(~)

where

n

2

and

N(a)

are called the

a.

is a linear function on

[ab] = ab - ba. N(~a) = ~nN(a)

is called the

A

which

The generic norm is

and is multiplicative.

N(1) = i.

Call an element

a ~A

roots in the algebraic closure

separable if of

K.

~a(X)

has distinct

Then the set of these

elements is a non-vacuous open set in the Zariski topology of

A.

4S

4.

Let

qo

is a central polynomial for

Then

qo(X, Yl,...,yn)

some of which are

± i

and

be Formanek's polynomial (for Mn(K)).

qo

A

with intege r coefficients

is linear in every

there exist central polynomials

Yi"

Moreover,

ql(x, Yl,...,yn),...,qn(X~Yl,...,Yn)

with the same formal properties as

qo

such that

qo(a,bl,...,bn)kn-ql(a,bl,...,bn)Xn-1

+ ...

(47) + (-l)nqn(a,bl,...,b n) = qo(a,bl,...,b n ) ~ a (k)" Hence (48)

qo(X,y!,...,Yn)xn- ql(x, Yl,...,yn) xn-l+ ... +(-1)nqn(X, Yl,...,yn )

is a__nniden_ttity fo___[r A. bi -CA

such that

F~ally,

if

qo(a,bl,...,bn)

a

i__~sseparable then there exist

~ O.

We shall make use of the following LEZ94A°

Let

infinite field

V

K,

be a finite dimensionaI vector space over an F

an extension field of

K

~' =

and let

VF

. 7

IJentify

V

with the subset of elements !

be a non-vacuous open subset of vacuous open subset of

V.

V .

1 ~Dv, v - C v T

Then

(ii) Let

f

O = O {h V

(i)

Let O

is a non-

be a polynomial function

!

on

V

such that

Then

f IV

f(v) -~K

for all

v

in an open subset of

is a polynomial function on

Proof.

Let (el,...,en)

V.

V.

be a base for

V

over

K,

K/F.

Let

hence,

&

for

V

over

F

and let

polynomial function on defines

f

relative

(~t.)

V'.

to the

be a b a s e f o r

Suppose base

f

be a

f ( ~ l ' ' ' " ~ n ) -CF [E 1,...,~n ]

(el,...,en)

~

T

f

"

iei

>

"

We can write f(~l'''''[n ) = E f~((l''''' ~ n )s t where the of the

f~ (~l,...,~n) -CK[%l,...,{n]

f~((l''''' I n )

are

# O.

If

and only a finite number

v -mY

then v--E ~iei , ~i-~K~

46

and

f(v) = ~ f%(~l,..O,~n) e u .

Since the

f (a!,...,~n) -~K,

f(v) = O

if and only if every fL(al,...,~n) = O. Thus if I) t . Of = {v'l f(v ~ O} then Of ~ V = U O f % A non-vacuous open subset is a union of sets V

is a union of sets

we assume one of the f(v) -~K

is that

Of, f # O.

Of

o (i) is now clear. To prove (ii) u eu ' say, e o = i. Then the condition that

ft (~l'''''~n) = 0

holds on an open subset of f lV

V

for every

then every

f5

is the polynomial function defined by Proof of Theorem i.

A F ~ Mn(F ). such that Let

Hence its intersection with

Let

F

- O

for

If this u ~ 0

and

fo(~l,...,~n)~{!...~n ]

be a splitting field for

We have the polynomial functions a 'n - tr(a')a'n-l+ ...+ (-i) n d e t a

T = tr ~ A,...,N = deh~A

5 ~ O,

A

tr,...,det on t

= O

for

so we have (~6) for any

a !

so AF

-CA F,

a -CA.

We

have seen that the matrices with distinct characteristic roots constitute an open subset of the elements

a -CA

such that

Mn(F ) = A F. ~a(k)

Hence, by the Lemma ( i),

= k n-T(a)kn-l+...+(-l)nN(a)

has distinct roots form an open subset

O

7a(k)

in

is the minimum polynomial of

on matrices), a

in

A.

Since

Hence

A t = A F, ~ a ( X )

~a(k)-~K[k]

and

a

of

A.

For such an

a

A' (by a standard theorem

is the minimum polynomial of T(a),...,N(a) ~ n

if

a -~0.

It follows from the Lemma (ii) that T,...,N are polynomial T functionson A. If T ,...,N are any polynomial functions such that

a n - T'(a)a n-I + ... + (-l)nN'(a) = O

the polynomial

fcr all

k n _ T'(a)k n-I + ... + (-l)nN'(a)

the minimum polynomial of

a.

For

a-CO

this is

a -CA

then

is divisible by kn-T(a)).n-l+ ....

Thus T (a) = T(a),...,N a) N(a) for ~ii a ~ O . It follows from T T (a) = T(a),... for all a. This proves the uniqueness of ~ A ( k ) . The properties of

T

and

N

stated in 2. are clear from the

properties of the trace and determinant functions of matrices.

We

have also proved 3 and statement $ follows from Theorem 2 of §8. To encompass the case in which

K

is finite and hence

A =

47

Mn(K )

we shall define the generic minimum polynomial,

trace and

norm in this case to be the characteristic polynomial,

trace and

determinant respectively in this case. Now let

A

be arbitrary finite dimensional central simple

and put (49)

T(a,b) = T(ab).

This is a bilinear form on = 0

it is symmetric.

and since

T(a,b)

trace bilinear form on THEOREM 2.

A

A.

T(a,b) - T(b,a) = T([ab])

is called the ~eneric (or reduced)

We have

The generic trace bilinear form of a finite

dimensional central simple algebra is non-degenerate. Proof.

We recall that if

vector space then

f

V

and

f(x,y)

(el,...,en)

is a symmetric bilinear form on

V

is non-degenerate if and only if the discriminant

det(f(ei, ej)) ~ O.

Hence

f

is non-degenerate on

if its extension is non-degenerate on field of the base field. bilinear form Mn(K).

is a base for a

If

tr(a,b)

a = ~ aijeij

tr(a'ekg ) = 0

for all

VF

for

F

V

if and only

an extension

Hence it suffices to prove that the trace is non-degenerate on a matrix algebra one has k,

6

tr(a,e k £ )

implies

= a 6k"

a = O.

Hence

Thus

tr(, )

is

non-degenerate. The foregoing results are applicable to primitive algebras satisfying proper identities.

If

n

is the degree of such an

algebra then the Formanek polynomial for polynomial for

A.

Mn(K )

is a central

Also any central polynomial for

A

is an

identity for every primitive algebra satisfying a proper identity of degree

< 2n.

48

lO.

Commutative localization Let

S

be a submonoid of the multiplicative monoid of

the commutative ring

K

set

(s,x), s -C S, x - C M .

S ~ M

of pairs

if there exists an (50)

and let

s ~S

M

be a K-module.

Consider the

Define (Sl,Xl)~(s2,x2)

such that

s(s2x I - S l X 2) = O.

This is an equivalence relation. and the equivalence class of

We denote the quotient set as

(s,x)

by

s-lx.

MS

We can define

s~ix I + S~Ix2 = (SlS2)-l(s 2 Xl + SlX2) (51) k(s-lx) = s-i(kx). These are well-defined and make localization o f

M

(52) of

a~% ~S

M

into

MS

S. :

MS

a K-module which we call the

We have the map X-->

i- I x

which one checks is a K-homomorphism.

The kernel

!

of

~S

that

is the set of

x s

for which there exists

sx = O, that is, ann k x ~

an

s -~S

such

S / @.

In particular we can form

KS .

Moreover,

this is a commutative

K-algebra if we define (52)

(s[ikl)(s~lk2)

The map in

KS

h) S

= (SlS2)-iklk2.

is an algebra homomorphism and

for every

s -~S

(Ks, ~ S )

~(s)

homomorphism

= i-ii = 1

D

of

K

is invertible for every ~

of

in

KS .

form a universal object for this property since

if we have any homomorphism then

is invertible

since

(l-ls)(s-ll) The pair

h)s(S)

KS

into

A

into a K-algebra s -~S

such that

A

such

then there is a unique

49

K

~

A

-0s

/1

KS

is commutative.

In fact,

(53)

s-lk If

S

~ >

is ~(s) - l ~ ( k ) .

is a submonoid of

S

t h e n we h a v e t h e homomorphism T

US

of

K

in

KS •

into

KS

and every

T

~ s ( S ), s

T

-~S

H e n c e we h a v e a u n i q u e homomorphism

,

~ SS'

is invertible : KS'

> KS

s -~S

~ S ' (s)

such that K

"~S

~ KS

KS t

Now suppose we have a situation in which for every is invertible in

KSt .

Then we have a commutative diagram K

~)S

KS

KS in which

%S'S

= IST

and

onto

KST .

is uniquely determined.

~ S S ' ~ S T S = 1S

In terms of

KS

so

~ SS'

It follows that ~ S ' S ¢ S S ' is an isomorphism of

KS

one can give an alternative definition of

50

MS, namely, LEMMA i.

We have an isomorphis ~ of

MS

onto

KS~ KM

such that (5~)

s-ix

s-ix

>

s'll~x.

Proof.

We note first that we have

in

and in particular this holds in

suppose

Ms

s[ix I = ~ i x 2.

s(s2x I - S l X 2) = O. = s [ l l ~ x I.

t((st)-ix) = (st)-l(tx) =

Then there exists

Now

Ks(x~K).

s -~S

(SSlS2)-ll~sS2Xl

= (ss2)((SlS2S)-ll)~x I

and (54) is well defined.

fication shows that the map is a homomorphism. silk I = s~lk2 s ~S

such that

Similarly, (SSlS2)-l(z~SSlX2 = s ~ l l ~ x 2 .

s[ll~x I = s~ll~x2

some

for

sI -~S, ki -~K.

Then

Hence

Direct veriNext suppose

s(s2k I - Slk 2) = 0

and this implies that s~-~klX ) = s~l(k2 x)

MS

for any

of

KS x

M

Now

x~M. into

Hence we have the map MS .

for

holds in

(s-lk,x) --~ s-l(kx)

This satisfies the conditions for a balanced T

K-product:

1 ' (s- k,k x)

and of

it is additive in both factors and for

KS@KM

into

have the same value. MS

sending

k

-CK (kts-lk, x)

Hence we have the homomorphism

s-ll~kx

= s-lk@x

> s-l(kx).

The products of the two homomorphisms are identity maps.

Hence both

are isomorphisms. It is clear that the isomorphism Also

MS

can be regarded as a

(55)

(5~)

is a K-module isomorphism.

Ks-module via

(sl-lkl)(s-lx) = (SSl)-l(klX).

When this is done

(S&)

Ks~KM

This permits us to apply properties of tensor

and

M S.

becomes also a Ks-module isomorphism of

products to obtain properties of quotient modules.

We can also go

in the other direction and apply the quotient module to obtain a basic property of

KS

as K-moah/le, namely,

KS

is flat.

We recall

51

that the functor of the category of K-modules into the category of abelian groups mapping a module

M

> N~KM

module and mapping a module homomorphism right exact.

,

where

M'

M q >

N

is a fixed

into

1N~

~

is

This means that if

t

O-->M

........>...M....

Mtt

>

>0

is exact then

N@M' is exact.

N

> NOM

''

>

> 0

is called flat if

o

N®M'

> N®M

> 0 t

is exact. then

This simply amounts to:

N~M'

> N(~M

THEOREM i. Proof. KS ( ~ M

with

monomorphism.

KS

if

~

:M

is a flat K-module.

we have to show that

If

is a monomorphism

is a monomorphism.

Using the isomorphisms of MS

> M

s-I ~](x') = 0

Ks@M'

with

~

and

s-lx t --> s-I ~] (x')

we have a

t ~S

is a

such that

!

t ~ (x') = O.

Then

q (tx') = 0

and

tx

= 0

which implies

s-ix ' = O. An important instance of localization is obtained by taking S = K - P the complementary set in this case one writes algebra:

Mp

for

MK_ P.

of a prime ideal The algebra

s-ip

condition that

in

l-lx = 0

such that

sx = O,

x ~ 0

M

that is,

and we have

ann x C P . localization

Thus

where Mp

s ~P

and

ann x C K . P.

If

Kp,

P -CP.

M ~ 0

we see that if

1-ix ~

in

M ~ 0

which

The s ~P

we can choose

Hence the ideal Then

In

is a local

is that there exists an

ann x ~ P .

be imbedded in a maximal ideal

ann x Mp

can

since

there exists a

Mp ~ O.

Of particular interest is the case in which Then

Kp

P.

it has a unique maximal ideal its radical, rad

consists of the elements

in

K

A S = AKs( = K S @

K A)

is an algebra over

A

KS .

is a K-algebra. Theorem 1

52

implies that if

B

is a subalgebra of

identified with

B KS.

implies also that if

A

then

KsB

The right exactness of the tensor functor I

is an ideal in

A

then

(A/i) Ks

An element

k -~K a = O.

is called regular for

if

implies

if

is neither a left nor a right zero divisor in bah = 0

Similarly, an element

A

a -~A

equivalent to

can be

AKs/I KS

identified with

b

can be

implies

ka = 0

b -~A

for

is regular A.

This is

a = O.

We now consider identities for

A

and

AS .

The result we

wish to prove on these is

THEOREM 2. (Rowen). for

AS

AnY identity

regarded a~s a_.nnalgebra over

element o f Proof.

S

f(xl,...,x m)

K).

is an identit X

The converse holds i f ever~

i_~s regular fo__~r A. Since any finite set of elements of

AS

can be written

with the same "denominator",

it suffices to show that if f(xl,...,x m)

is an identity for

f(s-lal,...,s-lam ) = 0

and

ai~A.

A

then

for all

s -~S

We write d

f(xl,...,x m) = ~ fj(xl,...,x m) 0 where the (total) degree of

fj

is

J.

Then

f(sal,...,sa m) = 0

gives d

sJfj(al,...,a m) = O.

Replace

s

0 successively by

l,s, .. ., Sd.

This gives the system of

equations x o + x I + ... + x d = 0

~56~

x o + SXl+ ... + sdxd = 0

d2 Xo+Sdxl÷ where

xj = fj(al,...,am).

... + s

For

t

xd = 0

an indeterminate we have

53

1

1

...

1

1

t

...

td d

(57)

,

e

t

t

1 t d

,

.

..°

h(t) -~2A[t].

cofactors of the t

by

s

( t i - t j)

i, j =0 i<j

t d2

= te(1-th(t)), where

U

i

e = (d 3 - d ) / 6

If we multiply the equation (56) by the

(j+l)-st column of the matrix obtained by replacing

in (57) and add the resulting equations we obtain the

relations se(1- sh(s))xj = O,

O < j < d

which implies that (58)

s-lxj = i-I h(s) xj .

Thus

(s-ll) ( 1 - 1 x j ) =

= (l-lh(s)) J(l-lxj)

(l-lh(s)) (l-lxj) .

Hence

(s-li) J (l-lxj)

and so

s-Jxj = l-lh(s) Jxj .

(59)

Hence f(

s_la I ' " " ,s_la m ) = od s-J fl (a!,...,am) d = 2 s-Jx. = Z l-lh(s) Jx. 0 3 J = Z 1-1~(h(S)al,...,h(s)am ) = l-lf(h(S)al,...,h(s)am ) = O.

Hence any identity for is regular for identity of

AS

A

then

in the center of S

is one for A

i~s re@ular.

Let

C

AS

and

A S . If every element of

is a subalgebra of

as algebra over

THEOREM 3.

of

A

K

is one for

be the center of CS

A.

is the center of

AS;

S

hence every

A. Then AS

Cs

is contained

if every element

54

Proof. and

If

s, t -~S, c -~C, a -~A

(t-la)(s-lc)

=~t)-l(ac).

Now suppose the elements of center of

AS .

(st)-l(ca). l-l(ca). and

c

S

Hence

then

s-lc

(s-lc)(t-ia)=(st)-l(c~

is in the center of

are regular and let

Then the calculation shows that

Then multiplication by

Since

~S

1-1(st)

is injective we have

is in the center of

A.

s-lc

AS .

be in the

(st)-l(ac) =

gives

1-1(ac) =

ac = ca

for all

Hence the center of

a

A S = CS .

66 ii.

Prime algebras satisfying prope r identities Let

A

be an algebra over

K,

C

can be regarded also as an algebra over of the multiplicative monoid of AS .

If

k E K

C

the center of C

and if

S

A.

Then

A

is a submonoid

then we can form the localization

one can define

(60)

k(s-la) = s-l(ka).

In this way

AS

is also a K-algebra.

Suppose

A

is prime and

c ~ C, c # O.

Then

cA = Ac

is a

non-zero ideal and its left annihilator is the same as the annihilator of

c.

of

C

Since

A

is prime this is

is regular.

In particular,

O.

Thus every non-zero element

C

is a domain.

monoid of the multiplicative monoid of canonical imbedding K-algebras.

of

A

into

and

AS .

0 ~ S

S

is a sub-

then we have the

This is a monomorphism of

We have

LEMMA 1. containins

~ S

C

If

0

Proof.

If then

A

i s prime and

AS

i~s prime.

Suppose

This implies

xay = 0

Then either

s-lx = 0

S

i_ss a submonoid o f

(s-lx)(t-la)(u-ly) = 0 for all or

a E A

u-ly = O.

so either Hence

AS

for all x = 0

C,

not

t-la E A S . or

is prime.

y = 0.

55

We shall be interested especially in the case in which S

=

C

IOl.

-

P = O)

The corresponding algebra will be denoted as

and called the algebra of the central quotients of

is the quotient field of the domain as

LEMMA 2.

(Amitsur).

proper identity identity

f

A0

A.

F

If

is the same thing

A

be a prime algebra satisfying

n.

Then

A

satisfies the standard

$2[n/2].

kl -~C kl

Let

of degree

Proof. Regard

by

then

with

(----F~KA).

AF

then

C

A 0 (Ap

and

A

as an algebra over its center

ka = (kl)a.

C.

identity for

Ao = A F

If k ~ K

Replacing the coefficients

we obtain a non-zero ( = proper ) identity

algebra over

C.

By Theorem 2 of §i0 (p. over the field

F.

fo

k

for

of A

f

as

63)

this is a non-zero

Thus

Ao

satisfies a

strongly regular identity so its various nil radicals coincide. Since

A°

is prime by Lemma i,

Ao[k]

is semi-primitive.

multilinear we see that degree

n.

Let

P

A°

has no nil ideal

Since we may assume Ao[X]

f

~ O.

and

Hence

fo

are

satisfies a non-zero identity of

be a primitive ideal of

Ao[X ].

Then

Ao[~]/P

is a primitive algebra over a field satisfying a non-zero identity of degree

n.

Then

$2[n/2]

is an identity for

theorem of Kaplansky-Amitsur-Levitzki. -~Ao[X] have A for

then

S2[n/2](al,...,a2~n/2] ) -~P.

S2[n/2](al,...,a2[n/2]) = 0

is imbedded in

Ao[X ]

for all

it follows that

by the

al,...,a2[n/2]

Since

N p = 0

a i ~ A o [ k ]. $2[n/2]

we

Since

is an identity

A. L~MMA 3.

satisfying r o ~ Proof. A

Hence if

Ao[k]/p

Let

A

b_~e~

identities.

s_~ubdirect p rgduet of prime algebras ~hen

A

contains n_~onil ideals ~

It suffices to prove the result for

A

prime.

Then

satisfies a standard polynomial and this is strongly regular.

Then if

A

contains a nil ideal

~ 0

it contains a nilpotent

O.

56

ideal

~ O

contrary to the primeness.

We can now prove the following important roa~it due to Rowen. THEOREM i.

Let

satisfying ~

A

be a subdirect product of prime al~ebras

identities such that the degrees o_~fthese identities

are bounded.

Then any non-zero ideal

intersection with the center Proof.

of

S2d.

of

A

Since

is an ideal in I ~ 0

and

A/P,

~ P = 0

Among these choose one

If

P

is a primitive ideal then

maximal. Mno(K)

Let

q

<_ d 2

so either

over its center.

(I+P)/P = A

there exists

PO'

P

be a central polynomial with Then

q

It follows that for any

(q(al,...,am) + P)/P Since

q

is central for A/P 0

al,...,a m E I and

is in the center of

such that

q(al,...,am) ~ O. Now let

nil ideals 3.

Then

A

~ O A[k]

and

A

We have

is an ideal in

A

is

constant term for

n ~ nO

and

q

is

A/P.

I + PO = A

Hence

q(al,...,a m) E C.

we can choose Then

q(al,...,a m) E I

Then

A

contains no

If

A

A[k]

C[~] ~ I[k] ~ 0

is and

C[k]

and

C[~] ~ I[%]

be as in the theorem and assume the center

Then

I ~ O

this is an ideal in

The center of

C ~ I ~ O.

Let

is a field.

Proof.

A[k].

Hence

COROLLARY. of

A/P 0

is semi-primitive and satisfies a standard identity.

I[k]

C

of

satisfies a standard identity by Lemmas 2 and

A[k].

I)[k].

I + P = A.

C n I ~ O.

Hence the theorem holds for

= (C~

I C P.

is either central or an

be as stated in the theorem. and

Then

al,...,a m E A,

q(al'''''am) ~ PO" Hence

nO

0

identity for every primitive algebra of degree A/P O.

or

such that

such that the degree

(e.g. Formanek~s polynomial).

central for

has a non-zero

is semi-primitive and satis-

is central simple of dimensionality

(I+P)/P

A

A.

We assume first that

fies the standard identity A/P

C

I

C,

A

i__sssimple.

is an ideal in I~

C = C.

A

then

Hence

I ~I

I ~ C ~ O. and

I = A.

Since

57

Theorem 1 and its corollary apply in particular to prime algebras.

Before stating the main theorem we give the

DEFINITION.

A subalgebra

left (right) order in inverse in

Q

Q

A

of an algebra

Q

is called a

if every regular element of

and every element of

Q

has the form

A

has an

a-lb, a , b - ~ A

(ba -I, a,b -CA). We can now state the structure theorem for prime algebras satisfying identities. THEOREM 2.

(Posner-Rowen et al).

satisfying ~ proper identitY. i.

The algebra

dimensional ~ field

F

of central quotients of

C

of

i~s ~ left and a right order in

3.

A

an~ A°

Ao

has a proper identity and

a field.

Hence

This proves i.

A°

Ao

is prime with

is simple by the Corollary~

The center is

Since

its center it is Artinian;

c-la = ac -I, c C C , is regular in

a -~A. Ao

Ao

Ao

F,

Then

is finite

as we showed

is finite dimensional over

hence any element of

a zero divisor is invertible.

that 2 holds.

is finite

satisfy the same identities.

dimensional over its center.

A

A

A o.

is primitive and so by Kaplansky's theorem

before.

prime a l~ebra

A.

A

F

b e~

Then:

2.

center

A

over its center and the center is the quotient

of the center

Proof.

Ao

Ao

Let

Any element of

Ao

Hence any element of

and so is invertible in

Ao

A o.

which is not

has the form A

regular in It is now clear

3 was proved before.

The result that a prime algebra over a field can be imbedded as a left and right order in an algebra which is finite dimensional simple over its center was first proved by Posner, in his doctoral dissertation. Amitsur.

This was generalized to algebras over rings by

The proofs of these results were based on Goldiets theorem

58

which gives an internal characterization of rings which have (classical}quotient rings that are simple Artinian (that is, can be imbedded as orders in such rings).

The present proof, due to Rowen,

side steps this and gives a clearer picture of the quotient ring. This shows that the center of the quotient ring is the quotient field of the center of the given prime ring, a fact, which have been noted by a number of other people: others.

Smal L

Martindale and perhaps

All made use of Formanekts polynomials.

The theory of semi-prime algebras satisfying polynomial identities is by no means as simple as the foregoing.

For example,

Bergman and Rowen has independently constructed such algebras which are not a left or right orders in any algebra.

12.

PI-al~ebras A number of definitions of

rings have been proposed.

PI-algebras over commutative

In this aection we consider one which

has been given by Rowen: DEFINITION 1.

An algebra

A

over a commutative ring

called a PI-algebra (or algebra with polynomial identity) exists a polynomial

f(xl,...,x m) -~K[XI

for every non-zero homomorphic image of If

S

is a subset of

K

ai~Al.

This is an ideal in

A.

f

f

Then

efficients of

f.

SfA ~ A

f

hand, if

then

Now let

SfA = A

SfA ~ 0

It is clear that

SA = IX ~iai ~ ai -~S , f

be an identity for

for the set Sf(A/SfA) = O.

is not a proper identity for then

SfA = ~

if there

A°

is an identity for every homomorphic image of

is proper if and only if

is

which is a proper identity

we write A.

K

Sf

A. Also

of coMence if

A/SfA.

On the other

for every homomorphic image

59

of

A.

Then

S~

~ O

for every

for every non-zero homomorphic

~ ~ O

image.

and

f

is a proper identity

Hence we see that Definition

1 is equivalent to: !

DEFINITION 1 • . An algebra over if there exists an identity Sf

f

for

the set of coefficients of The condition

SfA = A

K

A

is called a PI-algebra.

such that

SfA = A

for

A. is equivalent to:

there exist

ai-~A

such that (61)

ala I + . . . + ~rar = i

where

Sf = [ ~ l ' " ' ' m r I"

copies aj

A

of

A

Let

and let

in every place.

~

Ej

Then

A~

be the direct product of

be the element of

~ mjaj = 1.

Hence

~

I-[ A ~ A~

that has

is a PI-algebra.

It is clear from the definitions that a homomorphic PI-algebra PI N

is

PI.

However,

image of a

it is not clear that subalgebras are

or that tensor products with commutative algebras are is the lower nil radical of the

PI-algebra

A,

then

PI. A/N

If is a

subdirect product of prime algebras satisfying proper identities. Then, by Lemma 3 of §il (p. follows that

N

Let

d

T

Let

f

... X A , 2 n

radical of

At

satisfying

f

algebras.

I_~f A n

and

has no nil ideals

f

and put

then

~ O.

It

A.

is a PI-alsebra,

A

satisfies

m.

be an identity for

A

such that

n = [d/2].

SfA=

A.

We form

times, and use this as index set to form

, A~ = A, ~ _~ A(2n ) .

= ]~ A

At/N v .

for some

be the degree of

A (2n) = A X A

(Amitsur).

S2nm

proof.

A/N

is also the upper nil radical of

THEORem4 1. ann identit ~

67 ) ,

At/N t

If

N,

is the (upper or lower) nil

is a subdirect product of prime algebras

and this is a proper identity for all of these prime

Hence by Lemma 2 of ~ll (p. 66 ), S2n Now let

~j

be the element of

At

is an identity for

such that its value at

60

the index

c = (al,...,a2n)

S2n(~l,...,~2n)

-~N'

is

aj.

Then the element

and so there exists an

such that

S2n(~l,...,~2n )m = O.

Taking the

S2n(al,...,a2n )n = O.

Since ~ runs over all 2n-tuples of elements

of $2~

A

we have

b th

m

S2n(al,...,a2n )m = 0

is an identity for

component this gives

for all such 2n-tuples. Hence

A.

It is clear that if

S2nm

is an identity for

proper for every non-zero homomorphic

image of

A

A.

then this is

Accordingly we

have another definition equivalent to Definition i° DEFINITION 1 . satisfies an identity Definition l" is

PI.

An algebra S 2nm

A

is called a PI-al~ebra

for suitable

n

and

if it

mo

shows that every subalgebra of a PI-algebra

We can use Amitsur's result and the linearization method

to obtain a multilinear identity whose non-zero coefficients are ± i.

It follows from this that if

AC = C~KA

13.

A

I = I(A)

in the sense of the preceding

f(xl,...,x m) -~KIX}

under every homomorphism of Moreover,

K

be the set of identities of

is the subset of elements

KIX}.

K.

Universal PI-algebras

be a PI-algebra over

section and let

ideal in

is a PI-algebra then so is

for any commutative algebra

Identities of an al~ebra. Let

A

KIXI I

into

A.

such that

Evidently

is a T-ideal of

KIXI,

ideal which is stabilized by every endomorphism of clear since if

~

phism of

then ~ ~

KIXI

is such an endomorphism and is a homomorphism of

A.

~

I

Thus f-->

0

is an

that is,

KIX}.

I

an

This is

is any homomor-

K{XI

into

A,

so

61

if

f~l

then ~ ~

phism of

KIXl

f = O.

into

A

Thus

and so

~ f ~> ~ f -~I.

infinite sequence of elements of

KIX]

endomorphism of

x i n>

KIXl

0

such that

If

gl,...,gmCKIX},

gl,g2,

..

is an

then there is a (unique) gi' i = 1,2,...,

every endomorphism is obtained in this way. a T-ideal if and only if for any

under every homomor-

and

Hence an ideal

f = f(xl,...,x m) -CI

I

is

and every

f(gl,...,~n ) -CI.

We have seen that if of identities of

A

A

is a PI-algebra, then the ideal

includes a polynomial

S2nm

for some

I

n and m.

This suggests introducing the following DEFINITION. U = U I = KIX]/I for some

m

A universal PI-al6ebra is an algebra of the form

where and

Write

I

is a T-ideal in

containing

S2nm

n.

U = u + I , u-CKIXI,

a PI-algebra and PI-algebra.

KIX}

U = KIX}/I,

so

U=IGQuCKIXlI.

I = I(A),

then

It is clear also that if

is a unique homomorphism of

U

U

If

A

is

is a universal

al,a 2, .. C A

into

A

such that

then there ~i ~ >

a i.

In this section we obtain some important properties of universal PI-algebras.

We observe first that such an algebra is itself a

PI-algebra and that its ideal of identities in ideal into

I.

Let

f(xl,...,x m) -CI.

U = KIX}/I,

~

maps

Writing ~ x i = gi + I, f(~ Xl,... , ~Xm)

If

~

f(xl,...,Xm)

we know that

= f(gl,...,gm)

+ I = I. D

identity for

S2nm

In particular,

this identity has

1

is a homomorphism of KIXl into

of

Thus KIX}

of

KIX 1

into

U.

Then U

f ( ~ x l , . . . , ~Xm).

f ~ into

> O.

is mapped into

such that

Since

U,

as a coefficient it is clear that

f

Hence

for an

is an identity for

PI-algebra in the sense of the last section. identity for

is the given

f(gl,...,gm) -CI-

this holds for all homomorphisms U.

KIX}

0

Now let

U. Since U

f

is a be any

under the homomorphism

xi~->~ i = x i + I, i = 1,2, . . . .

Then

62

f(xl,..., ~ )

= 0

which implies that

is the ideal of identities of

f(xl,...,x n) -CI.

Hence

I

U.

The main result we shall obtain in this section is that the radical of a universal

PI-algebra

is a nil ideal.

This result is due to

Amitsur for algebras over infinite fields. general case is due to Rowen.

The extension to the

Since any universal PI-algebra

U

is

a PI-algebra in the sense of the last section, the upper and lower nil radicals of then

U If

U

coincide.

This implies that if

contains no nil ideals I

and

is an ideal in an algebra

(.upper nil radical.) of such that R/I (N/I) LEMMA 1.

I

A

U[k]

is semi-prime

is semi-primitive.

we define the radical

(as ideal) to be the ideal R (N)

is the radical (upper nil radical) of

If

I

is a T-ideal in

and the upper nil radical of is

~ 0

U

I.

If

K[XI t

I ~ I

of

A

A/I.

then so is the radical is a T-ideal then so

annAI T/I o Proof.

show

Let

g(xl,...,xm) -CR

that for all

the radical of

I.

We have to

fl,...,fm, f-~KIX~, g(fl,-,.,fm )f + I

is quasi-

regular in KIXI/I We may assume

f = f(xl,...,Xm).

g(xl,...,xm)f(xm+l,...,X2m ) + I

is quasi-regular in K[XI/I Hence there

exists 6 -~K[XI

Since g(xl,...,Xm)~-R,

such that

g(x l,-..,xm) f(xm+l,..-,x2m ) + ~ - g(xl,...,Xm)f(Xm+ l,...,x2m)

=- 0 (rood I) and g(xl,...,xm)f(xm+l,°..,X2m) + ~

- ~ g(xl,...,Xm)f(xm+l,...,X2m )

0 (rood I). Apply an endomorphism of Xm+ i ~ >

x i.

Since

I

K[X}

xi ---~ fi' i ~ i _~ m,

is a T-ideal this gives

g(fl,...,fm)f(xl,...,xm) +

m 0 (rood I)

such that

Z'

- g(fl,...,fn)f(xl,...,Xm) ~'

63

g(fl,...,fm)f(xl,...,xm) + Z t -

2' g(fl,...,fm)f(xl ,...,xm)

m 0 (mod I). Hence

g(fl,...,fm)f + I

g(fl,...,fm) -~R. that

g

is quasi-regular for every

f

and hence

A similar argument based on the element condition

is in the upper nil radical of

is nilpotent rood I

for any

hi,k i ~

radical of a T-ideal is a T-ideal.

I

K~}

if and only if

2 higk i

proves that the upper nil

The same type of argument applies

also to prove the last assertion. L~4A

2.

Let

le_~t B = annAI ann (I+B)/B

i.._n A/B Let

Then

A/B

PI C B

A/B

A

and

is semi-~rime and

O.

so

NI = O

Now let

A

containing

(NI) 2 = NINI C N2I = O. so

such that

and

Then

be an ideal of

N2I = O

is semi-prime. so

is N

semi-prime we have an ideal in

be an ideal in a semi-prime algebra

(lef~ annihilator).

Proof. N2 CB.

I

N C B.

(N/B) 2 = O

Then

PI = O

such that

Since

This shows that if then

annA/B(I+B)/B = ~/B.

PI 2 = O.

B

so

N/B = O. Then

A

is

N/B

Hence

is A/B

P = {p~p I ~ B }

P CB.

Hence

annA/B(I+B)/B = O. L~4A such that

3.

Let

A

be a semi-prime PI-al~ebra with center

ann A rad C = O.

completely homogeneous fi

such that

ht fi

Then any identity

f

( = homogeneous in every <-- ht f

and the

fi

of

A

C

is a sum of

xj ) identities

are blended if

f

i_~s

blended. Proof. degxlf i =.i. that

Write Let

f = f(xl,...,Xm) = c -~rad C

and let

f(c3al,a2,...,a m) = O, O < j _< d,

of Theorem 2 of ~ l D (p. 63 ), that u i = fi(al,...,am) 1 - ch(e)

and

C.

we obtain, as in the proof

c e ( l - c h ( c ) ) u i = O,

h(t) ~ Z L [ t ] .

is invertible in

d 2 fi(xl,...,Xm) whare i=O al,...,a m -CA. Using the fact

Hence

Then

ch(c) -~rad C

ceui = O.

so

This implies that

64

(cuiA)e = O. Then

Since

A

ui -~ann rad C

for

A.

Now

fi

is semi-prime this implies that so

fi(al,...,am)

is homogeneous

blended or homogeneous in any ing the process with L[Z@4A ~. Property K.

in

xj

x2,x3,..,

Let

A

Xl, if

fi

is of

f

is an identity

ht ~ ht f,

is

has this property.

Repeat-

be an al~ebra whose identities have the

Then any identity of

A

H

be a commutative algebra over

is an identity for

H~KA.

Suppose the result is false and let

of minimum height such that H~KA.

and

we obtain the result.

stated in Lemma ] and let

Proof.

= O

cu i = O.

We may assume

f

f

is an identity for

f

be a polynomial

A

but not for

is blended and using the hypothesis on

the identities of

A

we may assume

Let

f

and

f

is completely homogeneous. X.

xi

occur in

of lower height than

f

is an identity for ~t

~l p ' ' ' ' ~ m '

t _~_AH t

b t

i +

t

f.

A.

1 /~ xj f

Then

A.

is

Hence it

This inplies that if

then

a t

f(al'''''ai+l'

not occur in

and this is an identity for

A H = H ~K

b i --

T

xj

t

i' ai+l''"'am)

t

t

=

t

t

t

f(al,...,ai,...,a m) + f(al,..o,bi,...,a m) and this holds for every T

T

a 1,...,a m -~

AH '

i.

,

Using this result we see that for any

T

f(a 1,...,am)

is a sum of terms of the form

f(h I ~ a l, h 2 ( ~ a 2,...,h m ~ a where the H

h i -~H, a i -CA.

Since

f

m)

is completely homogeneous and

is commutative rI r2 f(hlQal,...,hmOam)

where

r i = degxif.

contradiction that

= hI

Since f

f

h2

rm ... h m Qf(al,a2,...,a. m)

is an identity for

is an identity for

A

this gives the

AH .

An immediate consequence of the last two results is the COROLLARY.

Let

A

be a semi-prime Pl-algebra with center

C

65

such that K.

ann A rad C - 0

and let

Then every identity for

A

H

is an identity for

This applies in particular to A[X] = A K[k]

b_~e~ commutative algebra ove_._~r

H = K[k]

and yields that

satisfies all the identities of

a PI-algebra.

Also

A[k]

AH •

A.

is semi-primitive.

Then

A[k]

is

We are now ready to

prove the THEOREM. rad U

Let

U = K[X}/I

Then

is a nil ideal. Proof. We have

and

be a universal PI-algebra.

R

R~N~I

where

N

is the upper nil radical

is the radical of the T-ideal I.

T-ideals and so is prime and since

B=annKIxIR/N.

We have that

B/N= annKIXI/NR/N , K I X } / B ~

prime by Lemma 2. show that

R=N.

N

and

KIXI/N

R

are

is semi-

(KIXI/N)/(B/N)

is semi-

Also annKIXI/B(R+B)/B = 0 (Len~na 2). We want to This will follow if we can show that

then we shall have R 2 C N

and

R ~ N

ideals ~ O. It is clear also that radical of

We know that

KIXI/B.

since

(R+B)/B

Thus if we replace

of the theorem to the case in which

U

I

U/N

R C B.

For,

has no nilpotent

is contained in the by

B

we have a reduction

is semi-prime so

N--I

and

annKIxIR/I = I.

We assume this is the case and let C

be the center

of

Let

~rad

U = KIX}/I.

b~annu(rSdU~C). c-~C~(rad c~

C

Then

b -~annu(rad (C~(rad

C).

U)b(rad U)) 2 = 0

U)b(rad U) then c ~ ( C q r a d

also

Since

U

Hence

CN(rad

Since rad U C C

c(rad U)b(rad U ) = O .

since if

U) and cb--O=bc.

Hence ( C n ( r a d

Then since

U)b(rad U ) ) 2 =

is semi-prime its center contains no nilpotent ideal ~ O. U)b(rad U) =0.

By Rowents theorem (p. 68 ) this im-

plies that (rad U)b(rad U) -- 0 and (b(rad U)) 2 = O. Since prime we have

b(rad U) = 0

our hypothesis.

Thus

so

~R

U[k] ~ U and let

~

b -~ann U tad U.

annu(rad C) = 0

is applicable to conclude that Since

C,

U[k]

Then

U

b = 0

of

by

and the foregoing corollary satisfies the identities of

these two algebras have the same identities. be a homomorphism

is semi-

KIX}

into

Let

U[k]. Since U[k]

U. f

66

has a countable set of generators there exists an epimorphism of

KIX}

onto

U[X]

and it is easily seen that we can find one

such that

~ f = ~ f.

hence for

U[X].

morphism of Thus of

~I K[XI/I

into

0

morphism U[X]

g

into

g

is an identity£or

is mapped into

U[X]

U[X].

U[~].

and so ~

g -~I

and

under any homo-

and so, in particular~

of

This maps the radical

Since ~

R = O.

K[X}

and hence for

U[X]

Hence

into U.

Then

R/I

is semi-primitive ~ f = ~ f = 0

U[X]. f -CI

Thus and

Of particular interest is the ideal Mn(K ).

0

U

~ g = O.

and we have an induced epimorphism ~ : y + I - - ~

onto

radical of

Hence

K[XI

= 0

If

f

~

y

into the maps

R/I

for every homo-

is an identity for R In

= I. of identities of

We shall study this in the next chapter.

II. APPLICATIONS TO FINITE DIMENSIONAL ALGEBRAS

In this part we shall obtain applications of PI-algebras to finite dimensional central simple algebras over a field.

We

assume some familiarity with this theory but will develop most of the results which will be needed.

i.

Extensio n of isomorphisms.

Splitting fields.

We recall that all modules for a semi-simple Artinian algebra are completely reducible and the number of isomorphism classes of irreducible modules is the number of simple components in the (unique)

decomposition of the algebra into simple ideals.

particular,

if

A

is simple we have only one.

In

For the sake of

completeness we give the proof of the foregoing statements in this case, assuming the Wedderburn-Artin theorem that we may take A = Mn(A eij

), ~

a division algebra.

the usual matrix units.

seen to be minimal and onto

Aejj

with inverse

x-->

x -~M

homomorphism

a e i i - - > aeiix

either the image is Thus the

x

0

x e IJ

are easily

gives an isomorphism of

xeji.

Now let

is a submodule. of

Aeii

...+ Aenn,

Aeii

onto

M

Aeii

be an A-module.

We have the module (Aeii)x.

or this is an isomorphism.

Now

Hence x-~Z

(Aeii)x.

is contained in a sum of irreducible modules isomorphic to

Aeii. Let

(Aeii)x

A=Aell+

The left ideals

x-->

Then for any

We have

Then A

M

is a sum of such modules.

be (left)Artinian and let

B

be a K-algebra which

m

is finitely generated as K-module:

B = Z Kbj. Then A ~ K B 1 generated as left A-module by the elements l ~ b i . Since

left Artinian it follows that tion for A-submodules.

A~B

is A is

satisfies the minimum condi-

It is clear also that any left ideal of

68

A~K

B

is an A-submodule.

We recall that if B

Hence

A

A ~K

B

is central simple

over a field

A(~KB

(p.

is simple over

K

then

also Artinian and

B

is finite dimensional

is simple Artinian. classical

and let

B

Let

A

over

Proof.

.

dimensional

o~ of

).

If

then

A

is

A ~K

B

B/K

simple subalgebra of

into

A.

A/K

can be extended to an

A

with

End ~ V

the algebra of linear

in a finite dimensional

Consider

~

(~)K B.

vector space over a division

This is simple Artinian and

be regarded as module for this in two ways: V,

over a field

A.

We identify

transformations

denote as

K

be central simple Artinian

be a finite

i_nner automorPhism o f

~

3h

and

We shall now use this to prove the following

Then any isomorphism

algebra

is simple

K

result.

THEOREM 1. K

is left Artinian.

V

In the first,which

can we

we have

(S~b) x--~bx~b6x and in the second, which we denote as ($(~b) x = $~(b)x= where

o- is the given isomorphism

identical as

~-module

- modules where

V~,

we have

c~ (b)$x

of

B

into

and are direct sums of

n = [V : /~ ].

A. n

V

and

V~

are

irreducible

Hence both are direct sums of a

finite number of irreducible modules all isomorphic to a particular one be

M. h

= kr. ~ of

Let the number of irreducible and Hence

k

respectively and let h = k

B-modules. K V

and so

V

and

components

r = [ V :~ V~

].

of

V Then

and

hr = n

are isomorphic as

Thus we have an additive group au~omorphism

such that for all

~ -~

, b -~B,

u(6b)x -- (6~ (b))ux

V~

v ~V

we have

u

69

Taking ing

b = i

6 = 1

ubu -1

we see that

we obtain

since

V

ubx = ~(b)ux

is a faithful

COROLLARY. dimensional

u6x = 6(ux)

THEORY4 algebra

L/K

2.

If

i~s ~ splitting

of an al~ebra

field for

dition holds we have Proof. A = Mr( A

)

where

V

is an

is an

L~

A

such that

CA(L)

CA(L)

where

= L.

dimensional

module

~ (b) =

of a finite

fields of a

on these

dimensional

if and onl~ i~f

the condition

such that r

A

[L : K] = rn

Suppose

aplitting

Identify

vector

L

field

is a subfield When the con-

n2 = [ ~ L

division

extension

= L.

holds:

is

central

then a finite dimensional

)

• Tak-

is inner.

is a finite

K,

A = Mr( A

Then

The first basic result

A

over a field

x.

An[ automorphism

simple algebra

algebra.

= End A V

A-module.

We consider next finite dimensional central division

u~A

for all

(Noether-Skolem).

central

so

: K].

is a subfield A

with

space over

as in the last proof.

of

End A V

~ .

Then

V

Since the

K centralizer and

of

CA(L)

Since

V

= L

L ~

as

~-module

is the set of maps

the centralizer

~

is simple

V

of

V

as

x

> ax, a-~A,

L(~) K A - module

is a faithful

completely

is

L.

reducible

K module

for

L~

irreducible. [V : L][L

~ . Since its centralizer is a field it is K 2 Also we have [V : K] = [V : ~ ] [ ~ : K] = rn =

: K]

so

from the density

V

is finite

theorem applied

dimensional to

V

over

L.

as irreducible !

module

that

L~K~

~ Mn,(L )

where

n

= [V : L].

It follows L Q

K Comparison

T

of dimensionalities aplitting

field and

over

suppose

L~K

module

L(~) K ~

.

sional vector

space over

EndLV.

V

Also

show that

rn 2 = niL

Conversely, for

K

~

: K] ~

Since L

is a vector

and

n

= n.

Thus

which gives

Mn(L ). L (i~>K ~

Let

is a

[L : K] = rn. be an irreducible

~ Mn(L),

L~) K A

space over

V

L

V

is an

n - dimen-

can be identified ~

and if

[V :~]

with

= r

70

then,

since

L

centralizer

of

L(~) K ~

in

centralizes L

in

EndLV.

/~,

L~End

End & V Since

~ V ~ Mr(~).

is contained

L~

~

The

in the centralizer

= EndLV ,

this is

L.

of

Hence

K

CEnd ~ v ( L )

= L.

The s e c o n d i m p o r t a n t

result

on s p l i t t i n g

fields

we s h a l l

need

is THEOREM 3.

Any finite dimensional

central simple algebra ha_~s

a separable splitting field. We prove first LEMMA i. algebraic i qn K

over

which Proof.

element of A ~ qK

Let

~

be a central division al~ebra which is

K.

Then if

~

is separable over

~

contains an element not

K.

~

the characteristic is p ~ 0 e satisfies an equation a p = ~, ~ -~K.

there exists a

as is easily seen,

d -~ ~

, ~K

D : x m~ Dp

D2a = ~.

dc - cd = d

Put and

such that

[dx]

in

a

c = ab-ld.

in Then

ded -I = c + i.

~

~.

purely inseparable follows that

Then

in

K(c)

Since

D ~ O.

Dc = (Da)b-ld = d

Then the subfield x ~

so

K(c)

is

dxd -I

and

is not the identity.

is not purely inseparable

Also,

b = Da ~ 0

field extension has no automorphism

K(c)

Consider

x---~ [dPx] = O.

such that

into itself by the inner automorphism

the induced automorphism

and every

dP -~K.

is the inner derivation

Hence there exists an element

mapped

K,

Otherwise,

the inner derivation

but

~

Since ~ i

a

it

contrary to our

hypothesis. COROLLARY. a maximal C &

(C ~

Assume

separable (E)) = E.

Proof.

~

is as in Lemma I and suppose

subfield hayin~ the double The n

The double

E

is a maximal

centralizer

centralizer

subfield

property

E

of

i_~s

property:

~ •

implies that ~ ' =

71

CA(E )

is central over

an element

c

not in

E. E

Hence if

K,

E

is a maximal subfield of L~MA

let

E

2.

Let

~

then

~' contains

E.

Then E(c ) ~ E

contrary to the hypothesis that

is a maximal separable subfield of

so

E

which is separable over

and is a separable field over E

~'D

~ .

Then

CA(E)

= E

and

~ .

be a central division al~ebra over

be a finite dimensional subal~ebra of

~i.

Then

K

and

C • ( C A(E))

---- E ,

Proof.

Regard

/~

We have the subalgebra

as

~o_

~)K

E(~) /~°

module for this subalgebra. irreducible as

A

A o_ module in the usual way.

of

/h (~) A °

Since A

and so

~

is a

is a division algebra

module, hence, as

E@

~ o -module.

/~ is

We claim

the endomorphism algebra of this module taken on the left ( = (EndE~AO in

fk) °)

2% is given by

~o

module is

d(ex) = e(dx) holds also.

is

CA(E).

dx = xd

~

Since the action of

e -~E

implies that

density theorem to conclude that of linear transformations of 2%

observe first that

left vector space over

E ~

be the elements of /%

that

di~ i = 0

is dense in the algebra

so this is finite.

We

can be regarded in the natural way as

f~o.

By the properties of tensor products [E : K].

Now let

Ul,U2,...,u m

which are left linearly independent over

Then if

~ o.

C A(E).

~o

implies every

independent over <_ [E : K].

~o

The converse

We now apply the

By the density theorem, there esists

Ziu j = 6ij •

as

as (left) vector space over

the dimensionality of this space is

G ~ (E).

d -~Cf~(E).

[ A : C • ( E ) ] L <_ [E : K], E~

~

The further condition that

Hence we have the result stated.

We show next that

A°

the endomorphism algebra of

acting on the left.

for

d E

di ~ d i = O.

Hence

~o

~

~i ~ E

(~ di Zi)u j

Thus the

m < [E : K]

~ i

and so

----

~ d3.

A O

such

SO

are linearly [ ~ : C Z~ (E) ]L

It now follows from the density theorem that

E ~ ~ o

72

can be identified with the algebra of linear transformations as left vector space over of

C& (C~(E))

Hence ever

where

~/~o e

over

-~E.

Then

CA(C&(E)) =

CE~

u

o

;~o

= i.

~ .

Now let

Let

c -~C(C(E)).

and since

.

Hence

of

{Ual

Then the elements

c = cu o = Z e u

CE

It is clear that the elements

are linear transformations

CA(C~(E)) K

C A (E).

of

~

over

C~

(E).

be a base for

~ o

constitute a base for

Then

c = Z e u

c = eo ~ E .

ECC~(CA(E))

we have

where Thus

Cz~(C~(E))

Eo

R~RK. By symmetry

The proof shows easily that [E : K] = [ A :

[ A : C A (E)]L = [E : K].

C A(E)]R"

We can now give the Proof of Theorem 3. sional central

It suffices to show that any finite

division algebra

Hence it suffices to show that separable

over

C A (C A (E)) = E

K.

Let

E

Thus

~

has a separable has a maximal

be a maximal

by Lemma 2.

Corollary to Lemma i.

~

Hence E

E

dimen-

splitting field.

subfield which is

separable

subfield.

is a maximal

Then

subfield by the

is a splitting field for

/k •

73

2.

The Brauer ~roup of A field Let

K

be a field and let

central simple algebras over (AI-~A2)

if

A1

K.

Mm(AI)~Mn(A2)

and

Call

Mq(A2) = Mr(A3) A2-~ A 3

then

imply

A I~

Thus

note the similarity class of of all matrix algebras and

BI-~ B 2

then

and

A2

similar

m

and

n.

This relation

Also if

Mmq(Al) ~ Mnr(A3). A 3.

~ A

Mm(AI) ~ Mn(A2)

Hence

A I-~ A 2

and

and

is an equivalence relation. by

{A}

A 2 ~ B 2.

De-

and the similarity class

Mn(K), n = 1,2, ..

A 1 ~BI~

be finite dimensional

A1

for some

is clearly reflexive and symmetric.

A2

by

Hence

i.

If

AI~A

2

{A}[B} = {A ~ B }

is a well defined binary composition on the set of similarity classes.

We denote the set of these classes as

Br(K).

It is

clear that the product we have defined is associative and commutative and

1

acts as unit.

Hence

Br(K)

is a commutative monoid.

We

shall show that it is in fact an abelian group by proving THEOREM i. the field

If

K, then

A

is finite dimensional central simple over

A ~A O ~

1.

,

Proof. Since

A~A

We consider O

as

A~A

O ~ M 2(K) n

The centralizer is

where

in the usual way.

K

and

A

is simple

[A : K] < ~o . Hence

n 2 = [A : K].

The foregoing result is that group.

° -module

is simple this is faithful and since

it is irreducible. A~A

A

{A}{A °} = i.

This result is called the Brauer group

Hence we have a Br(K)

of the field

K.

Let

L

be an extension field of

K.

If

finite dimensional central simple algebras over A1 ~ A2

then it is clear that

we have

(A~K

homomorphism of

AIL~

B) L = A L ~ L B L. Br(K)

into

A2 L,

K

and

A2

arQ

such that

Also for K-algebras

It follows that

Br(L).

A1

The kernel

[A} --> {AL} BrL(K)

A, B is a

is the set

74

of

{A}

such that

analyze this for

L L

is a splitting field for

A.

a finite dimensional Galois ( = s e p a r a b l e normal)

extension field of

K.

Let ~ be a division algebra in

{AI

and let

[L : K] = m.

Then, by theorem 2 of §i (p. 83 ),

n = mr

L

then

Of course,

We shall now

is a subfield of

A -C {/~I.

Let

A = Mr(~)

G = Gal L/K.

[/~: K] = n 2, nJm

and if

such that

If

CA(L) = L.

s -CG, Theorem 1 of

~l (p. 82 ) shows that there exists an invertible element

us ~

A

i

such that

u s ~ Us I = s(~ ), ~ -CL.

= Ust ~ Ust 1.

It follows that

Then

ustlusu$

Hence this is a non-zero element of Ps,t - C L

UsUt~utlus I = (st)(~) commutes with every

L.

~ ~_ L.

It follows that we have

such that

(i)

UsUt = Ps,tUst "

These relations and

(2)

u s~-- s(~)u s

imply that the set

A

!

(3) where

of elements of the form

Z

ZsU s

s-CO Z s -~L

is a subalgebra of

A,

The usual argument used to

prove DedekindTs lemma on the linear independence of distinct automorphisms of a field shows that

X ~ sUs = 0

implies every

~ s = O.

!

It follows that Hence

A

is an

m = IGI- dimensional vector space over

[A' : K] = [A t : L][L : K] = m 2

= r2n 2 = m 2,

we have

A t = A.

.

Accordingly

Since A

[A : K] = [Mr(~): K] coincides with the

set of elements (3). We have the associativity

(UsUt)U v = Us(UtUv).

Using (1)

this gives (4) We r e m a r k also that

Ps,tPst, v = s(Pt,v)Ps,t v . us

L.

is not uniquely determined.

This can be

75

replaced by

v s = ~sUs

where

~s

is any non-zero element of

L.

This is the only change which can be made since the centralizer of L

in

A

is

L.

Taking

v s = ~sUs , s -~G, we obtain

VsV t =

-1 v ~sUs~tUt = ~sS(~t )~sut = ~sS(~t)Ps, tUst = ~sS(~t )~st Ps,t s~t " Hence

Ps,t

is replaced by

(5)

P ' s~t = ~sS(%)%~ 1 P s~t " We recall the definition of the second cohomology group

H2(G,L *) group G

of the Galois group

L*

into

G

of

of non-zero elements of L*

as a map

that (4) holds.

If

(s,t) n > p

and

L/K L.

Ps,t

~

into the multiplicative

One defines a 2-cocycle of of

G ~ G

into

L*

such

are two such maps then so is

p

defined to be (s,t) m > Ps,t Cs,t . Also 1 : (s,t) ~ > 1 and -1 p -1 p : (s,t) ~ > s,t are 2-cocycle. In this way one obtains a group

Z2(G, L*).

a map of

G

into

It is easily seen also that if L*

then

~ : s m>

~ s S (~t) ~ s t-1

(s,t) m >

is a

These are called coboundaries and constitute a subgroup The factor group

H2(G, L*) = Z2(G, L~:~)/B2(G, L*)

second cohomology group of

G

into

~s

is

2-cocycle.

B2(G, L*).

is called the

L~ .

The main result we shall now prove is that there exists a canonical isomorphism of

H2(G, L*)

onto the subgroup

the Brauer group consisting of the classes is, such that Let

p

[AI

BrL(K) of

split by

L,

that

AL~l. be a 2-cocycle of

G

into

L* •

Then we can use

p

to construct an algebra (6)

A = (L, G, p )

called the crossed product of se_~t p.

A

the elements bijective). space.

L

with its Galois group with factor

is the vector space of elements of the form (3) us

are in

l- 1

The addition

in

correspondence with A

L

G (s ---> u s

is

is the addition of this vector

The K-structure is the restriction to

tion by elements of

where

and we define

K

of the multiplica-

76

(7)

(E ~ sUs)(~ m t u t) s

=

E Z sS(mt ) p s , t u s t s~t

It is easy to verify that this is associative. (4) 1 =

imply that

Pl,-}Ul

Also the conditions

Ps,1 = Pl,1 = Pl,s = Sl(Pll)

is a unit for

A.

•

and these imply

It follows that

A

that

is an associative

algebra. THEOREM 2.

A = (L, G, p)

[A : K] = n 2, n = [L : K].

is central simple over

Moreover,

L

K

with

is a splitting field for

A. Proof. Suppose

Let

B ~ A.

= A/B.

B

Then ~ >

Since the

are invertible in ~s~.

be an ideal in

us A.

~

and write

a = a +B.

is a monomorphism of

are invertible in Clearly

~

L

into

A, their images

consists of the sums

~

[~ : ~][~ : K] = n 2

Taking

Since

[A : K] = [A : K]

Z -~L

and let

= O.

Then

s(Z ) =~

X ~ sUn

-~L. and

E ZsU s

Thus

K

that

K

P. 83)

0

we have

CA(L) = L.

Hence

B = 0

for every

s

If

is the center and

Z • with

~-~L

c -C

This gives

A

[A : K]

Then

~(Zgs-

~ s ~ O. then

n2

is simple.

Let

s(£) ~s)Us

It follows that

~ Z s U s = ~ l U l = giPl, i

the center of

s(c) = c A

so

u--s

[~ : K] =

this gives

commute with

commutes with every

cu s = UsC.

Hence

B

L•

~s sis ,

The Dedekind independence argument shows that the

are left linearly independent over

if

A

for every

A

then c-~CA(L ) = L s -Cc.

is central simple over

is a splitting field follows from the criterion

Then

c~K.

K. The fact (Theorem 2,

for aplitting fields.

We call the factor sets

p

and

p

I

cohomol0~ous and write

!

p ~,p

if they differ by a coboundary,

same element ~s ~ L ~

[pl

of

H2(G, L~).

such that (5) holds.

which led to (5) that

that is, they determine the

If this is the case we have

It is clear from the considerations

A = (L, G, p) ~ A' = (L, Gjp').

It follows

77

that we have a well-defined map of the cohomology group into

BrL(K)

sending an element

{p }

into

{(L, G, p) I.

result we obtained at the beginning of our discussion l) is that if split by

~

is a finite dimensional

L( =

having

for a suitable

p .

L

Proof. where if

Ai

BrL(K).

~K(L, G, m) -~(L, G, pff).

The tensor product on the left has the form ala 2 = a2a I

xlY I + --. + XrY r = 0 have an isomorphism

for $-->

and we have elements

Yk -CAj $i

of

we have K.

Then we have

Let

that is, Yk = 0.

s -Cc

Li

Also we of

Ai

such that

gCL

v2vt = O-s,t 2 V s t .

is separable over L = K(e).

for

Also

then these

into a subfield

2~2v s = s ( 1 ) 2 v s,

UsU t = Ps, tlUst ,

K

only if every L

AIA 2

a i -~A i.

Aj, j ~ i,

u s ~ A l, v s -CA 2

£1u s = s ( ~ ) l U s,

L

for

are linearly independent over

elements are linearly independent over

Since

A "~ (L, G, p)

We shall show this is an isomorphism.

(L, G, p)

Xl,...,x r -CA i

(i©)

then

, >[(L, G, p)}

is a subalgebra and

(9)

(after Theorem

central division algebra

as splitting field)

{p}

THEORY4 3.

The

Hence

(~) is surjective on

H2(G, L ~)

K

f(k)

it has a primitive element, that is, be the minimum polynomial of

f(k) = ~[(X- s(@))

in

L[X]

8

over

and consequently we

S

have

f(k) =]-[(k-s(@)l)

and

f(k) = ~ ( k -

S

L2[X]

s(8) 2)

in

Ll[k]

and

S

respectively.

fs(x )

Consider the Lagrange =

interpolation polynomial

f(x)

(x- s (e) l) f' (s(e) 1 )

(ii) = t~s (X - t (e)l)/ H

t,4s

in

LI[X ].

Since

i - E fs(X) S

(S(@) l-

is of degree

<

t(e)I)

n-I

and vanishes

78

for every

k = s(8) I,

this polynomial

(12)

z fs(~) s

Also if

s / t

is

Hence we have

= i .

then

(13)

fs(~)ft(~)

~ o

(rood f(~)).

Now consider the connnutative subalgebra (i$)

LIL 2

of

AIA 2.

Put

e s = fs(@2) -~LIL 2.

Then L I.

O.

es 6 0

since

1,82,...,8~ -I

By (12) we have

~ es = i

are linearly independent over

and by (13) we have

ese t = 0

if

s

s ~ t.

Then multiplying

~ e~ = 1

by

et

gives

e~ = et.

Hence

we have (15)

e~ = es, ese t = 0

The definition of the

es

s ~ t,

E e s = i.

gives (82 - s(8)l)e s = 0

(82- s(@)l) t~s (@2-t(8)i) p2 -~L2

if

= f(82) = O.

since

It follows that for

we have

(16)

P2es = s(D)le s.

Then

LIL 2 =

~ LlS s and Lles ~ L. As is well known this des composition of LIL 2 as direct sum of simple algebras is unique. We now consider the inner automorphisms

x--> v x u~ 1

of

AIA 2

x-->

defined by the elements

uZ1

u sx

Us,V s

of

, A1

S

and

A2

LIL 2

respectively,

satisfying (9) and (i0).

and hence permute the simple components

the idempotents

es.

These stabilize

of

LiL 2

Applying these to the relations

and hence (16) and

taking into account the fact that the first ones fix the elements of

A2

and the seccnd fix the elements of

p2(UtesU~ I) = (tS)(~) 1 (Utesu~l) _I

t(p2 )(vtesvt ~) =

S(@)I (Vtesv~l)

Comparision with (16) then gives ute s = etsU t

A1

we obtain

79

(l?) vte s = e

i v~.

s t "~

5

Now put (18)

es, t : esUsU[ 1 .

Then, by (17), one verifies

n 2 es, t

that the

constitute

a set of

matrix units: (19)

es, tes,,t T = ~s,tes, t , ,

It follows that of the

es, t.

calculate hence,

AIA 2 ~ Mn(B )

where

Equally well,

this algebra.

Also,

B

is the centralizer

B ~ ell(AiA 2) ell

We know a prlorl that

[B : K] = n 2 = [ellAiA2ell

Lle I = Llell.

Z es, s = 1.

Now

: ~.

in AIA 2

and we proceed to [AIA 2 : K] = n & ,

ellAiA2ell

contains

if we put

(20)

w s = UsVsell

we have

w s = UsVse I = elUsVs -~ellAiA2ell.

Ws(Plell)

Also

= UsVsPle I = (s(P)lel)UsV s = (s(P)le I) w s

and WsW t = UsVsUtVte I = UsUtVsVte I = Ps,t!Ust~s,t2Vstel = Ps, tl~s,t2 elWst = Ps,tlO-s,tl eiWst = (Ps, t~s,t )lelwst" It follows that

ellAiA2ell

the crossed product shows that

(L, G, p ~ ) .

ellAiA2ell

(L, G, p ) ( ~

contains a subalgebra Comparison

is isomorphic

to

isomorphic

to

of dimensionalities

(L, G, p ~ ) .

Hence

(L, G, ~ ) ~ . (L, G, p ~ ) . K

We now see that the map homomorphism

of

H2(G,

L ~)

[pl ~ >

onto

[(L, G, p )}

BrL(K).

isomorphism we have to show that the kernel to

is a group

To see that it is an is

1.

This is equivalent

80

THEOREM ~. Proof.

If

i

(L, G, p)

then

We note first that,

homomorphism

of groups,

(L, G, p) = Mn(K ).

p ~ i

p ,~ i.

since {p } -->

implies

(L, G, p) ~ l .

This implies that if

(L, G, p) ~ (L, G, i).

Now let

a ~>

{L, G, p)}

a

Hence

(L, G, p) ~ T

is a

i

then

be an isomorphism

of

!

(L, G, l)

onto

field of

(L, G, p).

(L, G, p)

Then the image

and for every

L

of

s -~G = Gal L/K

!

invertible element for

~T -~L'

L

onto

L

is a sub-

we have an

Y~T

v s -~(L,

and

L

G, p)

v's v ~ = Vat' .

such that

t

vs

=

~' ~>

The isomorphism

can be extended to an automorphism

(s(Z))

~

of

~

f

vs of

(L, G, p).

!

This maps

vs

Also we have

into

vs

u s~

and we have

= s(~ ) u s

u s = ~sVs , ~s -~L,

and

and

Vs~

= s(~ )Vs, V s V t = V s t .

UsU t = Ps, tUst .

Ps,t = ~sS(~t)~s[ ~

so

p~

Then i.

We have seen that every central division algebra separable splitting field.

that is finite dimensional

~(L,

G, @)

prove that Br(K)

p.

is a torsion group,

~ K.

has a splitting field Then we have

We shall now use this to that is, every

[A}

in

has finite order. A more useful result is the more precise

Liet

THEOREM 5. signality

m2

~ •

L

Mr(~)

We can identify

A

vector space over

a base for

be a central division algebra of dimen-

= A

such that

with

End& V

~.

over

Then

[L : K] = n

{~}m

= i.

then

n = mr

CA(L) = L. Also where

V is an

r

We have [V : K] = [V :A][/h:

: K] = IV : L] mr. V

K.

be a finite dimensional Galois splitting field

We know that if

subfield of

[V : L][L

~

over the field

Proof. Let for

Galois over

for some factor set

Br(K)

has a

Since any extension field of a splitting

field is a splitting field it follows that L

~

L

Hence

and let

and

L

is a

A = (L, G, p). dimensional K] = rm 2 =

[V : L] = m. Let (Xl,...,Xm)

{u s }

be a set of elements of

be A =

81

(L, G, p)

such that

u s ~ = s(~ )Us, UsU t = Ps,tUst • n

Write

UsXi = j =Z l

mij(s) xj"

Then

n

UsUtX i = U s ~ j:l Z mij(t) xj)

= JZ,k s(mij(t))mjk(S)X k

Ps~tUstXi = Ps, t Z mik(St)Xk " This gives the matrix relation (21)

Ps, tM(st) = s(M(t))M(s)

where

M(t) = (mij(t))

and

s(M(t))

= (smij(t)).

Taking determin-

ants we obtain

Ps,tm ~st = s(~t)~ s "

(22) Since

UsUs_ 1 is a non-zero element of

is invertible and so pm.~l.

Then

~s 6 0 .

[zi}m = 1

L

it follows that every M($)

Then the foregoing equations show that

by Theorem 3.

This result shows that every element of the Brauer group is of finite order.

If the order of

exPonent of

A.

Ae~l.

A = Mr(~)

If

sionality class and

m

~t

is

e

then

e

is called the

This is the smallest positive integer such that

then

A.

[A}

m

where

A

is a division algebra of dimen-

is called the index of

(It is classical that if are division algebras then

A

or of the similarity

M r ( A ) ~ Mr' (/~ T) A__ ~

~'

.

where

This is an easy

consequence of the fact that any two irreducible modules for isomorphic. by

[A}.)

A

are

The result shows that the index is uniquely determined The foregoing theorem shows that the exponent is a divisor

of the index. prime factors.

It can be shown that these two numbers have the same It has been shown by Brauer that given any

having the same prime factors such that division algebra of exponent

e

e ~ m

and degree

m.

e

and

then there exists a

m

82

3.

C~clic algebras.

Some constructions

We consider a crossed product cyclic extension field of s

be a generator of

0 ~ k ~ nand

x-->

K,

G

(L, G, @)

that is,

and let

O

L

is a

is a cyclic group. Let

u = us.

I, (n = ~G~ = In : K]') by

for which

Then we can replace

uk

since

Usk,

x --> u ~ X ~ x ~ ) - i

ukxu -k produce the same automorphisms

sk

in

L.

Then u n

commutes with every element of L and hence it commutes with every n-i i element ~ ~i u ' ~i E L . Since this set of elements is all of %2

A = (L, G, p)

we s~e that

un = ~

-~K.

Thus

A

is the set of

elements n-i 7 ~ i ui , Z i -~L 0 for which we have the multiplication whose defining relations are (23)

(25)

u~

= s(~)u,

un = ~

We now write

A = (L~ s, ~ )

in

K.

in

K.

to indicate this and call

al~ebra defined by the generator ~ 0

0

s

of Gal L/K

A

a cyclic

and the element

Such algebras were first defined by Dickson in 1906.

The multiplication theorem for crossed products specializes to the following theorem on cyclic algebras THEOREM i.

(L, s, ~ )

~

(L, s, S ) ~

(L, s, %z$).

K

One can deduce from the criterion that

~f p ~ i

(L, G, p ) ~ i if and only

the following result on cyclic algebras.

THEOREM 2.

(L, s, "6") "~ 1

tha_.__~%is, there exists

a

c ~L

if and only if such that

~

Then we have Hence if ~

v2 =

N(c) ~

.

Vsk = v k, O < k _< n - i,

if

(L, s, ~ ) -~ 1

Vsk

= N(c -I)

is

L,

u

of

(L~ s~ ~

) by

cs(e)u 2, v 3 = cs(c)s2(c)u2,..., v n =

Taking

determined by the

is a norm in

= NL/K (c).

We note that if we replace the generator v = cu.

~

1

then v n = 1

and

(L,s, ~ ) =

we see that the factor set

and hence

then we can choose

(L~ s, ~ ) ~- I. Vsk

(L,s,l). p

Conversely,

so that the cortes-

83

ponding factor set Vs)p...,1

p = !.

= Vsn = v ~ .

v n = i.

Since

sufficient

c -~L

v = vs

it follows that

consequence

condition,

v I = i, Vs2 = v s Vs, Vs3 = Vs~¥ s =

Hence the g e n e r a t o r

v = cu,

An immediate

Then

~

satisfies

= N(c-i).

of T h e o r e m 1 and 2 is the following

due to Wedderburn,

for a cyclic algebra

to

be a d i v i s i o n algebra. THEORem4 3. norm in

L,

then

Proof. u

s,~

~ n

is the smallest

(L, s, ~')

The condition

is ~he exponent [(L,

If

of

(L, s, ~ )

~

which

is a

is a d i v i s i o n algebra.

implies,

(L, S, ~ ) .

) : K] = n 2

power o f

via Theorems

Then the index

1 and 2 that u > n.

it follows that the index is

n

Since

~ and so

is a d i v i s i o n algebra.

We shall now use this

condition to construct

some

cyclic

d i v i s i o n algebras. First, that we

can take

Ko

Ko

L = Lo(t)

be a cyclic extension of a field

with

of degree Consider algebra

t

n n

G = Gal Lo/K o = < s >

elements and

Then

s

elements,

s

the map over

of exponent

L~ .

n.

.

a

> aq.

K = Ko(t).

(L, s, t).

Then

b

To see this

amtm

~ O.

is cyclic

Then

L).

is a @ i v i s i o n

to the a s s e r t i o n

b o ÷ b I t + . . .+ bctq and

s

s (as extended to

--

at.l ~ O

L/K

We claim this

This is e~uivalent L.

Extend

Let

of

This has the form

ai,b j -r~L o,

the exten-

generates a group of a u t o m o r p h i s m s

a o + alt + . . . +

w h e r e the

For example, L°

L o.

such

L

is not a norm in

C

K°

to

with generating automorphism

the cyclic algebra

of

q

is t r a n s c e n d e n t a l

with fixed field

1 < k < n-i

element

qn

s(t) = t.

of order

Let

to be a field w i t h

where

so that

for

L°

[L o : K o] = n.

sion of

L

let

that t k

consi4er any

84

(~ ait~)(F s(ai) tl) .. (~ sn-l(ai)ti ) NL/K(c) = ( ~ b j t J ) ( T s(bj)tJ)

.. (~ S~-l(bj)tJ

N(a O) + ... + N(an)tnm N(b o) + ... + N(ba)trq Hence if

t k = NL/K(C)

then we have

N(a o) + . . . + N(am)t n m = Since

N(a m) # 0

1 < k < n-l. R

of

K~.

Then

Let

Lo

this is impossible for

b_£e~ q~qlic field extensign of 4e6ree

L = Lo(t)

and the cyclic algebra exponent

N(bq) ~ 0

Hence we have

m

THEOREM i. n

and

tk[N(bo) + ... + N ( b c ) t n q ]

i_~s cyclic of ~e~ree

(L, s, t)

n

over

is a division algebra of

n.

A special case of this construction is the following: F

be an arbitrary field and let

F(tlt2,...,tn)

rational expressions in indeterminates F.

Let

s

i~entity on

be the automorphism of F

the subfield of Then

Lo

Ko(t)

and permutes the L o = F(tl,...,tn)

is cyclic of degree

n

ti

be the field of

with coefficients in

F(tl,...,tn) ti

which is the

cyclically.

Let

Ko

of fixed elements under over

Ko

Let

be s.

and can be used in the

foregoing construction. We shall give next a conBtruction of division algebras which goes back to Hilbert.

After Hamiltonts

definition

of quaternions

this may have been the earliest construction of division algebras which are not fields. For this construction we assume we are given a division algebra D

and an automorphism

s

in

D.

In particular,

D

can be a field

We consider first the s-twisted power series algebra

E~[t, s]]. The

elements of this are the power series

85

(25) where

a o + alt + a2t 2 + ... equality

and addition ones.

is defined by equality

and multiplication

of corresponding

by elements

One defines multiplication

of

K

coefficients

are the obvious

of power series by

(a o + alt + a2 t2 + .. )(b o + b l t + b 2

t2 + ... )

(26) = Po + PI t + P2 t2 + "'" where (27)

Pi =

E aj sJ(bk). j+k=i

It is easy to verify that 1 (=i+ Ot ÷ O t 2 + . )

D[[t,

algebra We define

is a K-algebra

D

as the subalgebra

containing

a o = a o + Ot + Ot 2 + . . . . series

s]]

If

s = 1

we obtain

with unit of elements

the usual power

D[[t]]. ~ (a(t)) cr (a(t))

(28)

for

= ~ =

a(t)

if k

= a o + alt + ...

a(t)

if

by

= O

a ° .....

ak_ 1 =

O,

a k ~ O.

Th en o- (a(t)b(t)) (29)

~ ( a ( t ) +b(t)) ~(ma(t))

Hence

= ~ (a(t))

+ c~ (b(t))

~ min(~[a(t)),

~ (b(t)))

= ~ (a(t))

~ ~ 0

if

is in

K.

if we define

(30)

~a(t) l = 2 - ~ (a(t))

then we obtain a (non-archinedian) ~a(t)~ > O

and

= O

valuation: if and only if

i

~a(t)b(t)[

: ~a(t)[[b(t)~

(31) ~a(t) + b(t)l < max([a(t)~,

[b(t)[)

a(t) = O

86

~ma(t)~ = ~a(t) i

if

a ~ 0

in

Relative to the topology defined by topological algebra,

~ ~, D[[t, s]]

K

function of the variable in

are continuous (the last

D[[t, s]]

Also we have completeness relative to

I ~.

For suppose {a(~)}

D[[t, s]].

n = O, 1,2, ..

such that

ai

i ~ n

for

a(~)

there exists an > N.

for any

Define a(t)

N by

Then given any o- (a(~) - a(~)) > n

a(t) = a o + alt + ...

is the same as the corresponding # ~ N.

as a

with fixed multiplier).

is a Cauchy sequence of elements of

~,~

is a

that is, addition and multiplication and

multiplication by elements of

for all

K.

where

coefficient in

This is well defined and we have

iim a(~)

= a(t).

It is clear from the multiplication D[[t, s]] units of Since

Then

is a domain, that is, it has no zero divisors. D[[t, s]]

~ (i) = 0

Now let

are the elements

~ (a(t)) = O.

1 + z + z 2 + .. Hence

Then

= l- z

where

~ (z) ~ 1.

D[[t, s]]

The

~ (a(t)) = O.

and

a o ~ O.

Then the series

and this element is

a(t) = b(t)-la~ 1. ~[t,

s]]

non units is the set of elements We shall now localize division aIgebra.

= b(t)t k.

is a local ring whose ideal of b

such that

D[[t, s]]

We consider pairs

a(t) -~D [[t, s]],

class of

with

a(t) = a o + alt + ..

converges in

We now see that

a(t)t ~

a(t)

that

it is clear that this condition is necessary.

b(t) = a~la(t)

b(t) -1.

property of ~ ~

and we define

at

t

~(b) > O. to obtain a

(a(t), tk), k = O, 1,2,...,

(a(t), t k ) ~ ( b ( t ) t

This is an equivalence relation.

(a(t), t k)

as

a(t)t -k

~)

We denote the

and the set of these as D((t,s)).

We make this set into an algebra by defining a(t)t-k + b(t)t - ~

if

= (a(t)t ~ + b(t)tk)t - ( k + ~ )

87

(32)

a(a(t)t -k) = (ma(t))t -k,

m -~K

(a(t) t-k) (b(t) t- 4) = a(t) (s-kb(t)) t - ( k + ~ ) (where

s-k

acts on the power series by acting on the coefficients).

These are well defined. D[[t, s]] image. over

into

One has the imbedding

D((t, s))

We shall call

and we identify

D[[t, s]]

a(t)t °

of

with its

D((t, s)) a twisted Laurent series algebra

D. We can extend the valuation on

D((t, s)). b(t)t - ~ so

a(t) ~ >

We define then

b(t)t k.

is well defined.

tions (29) hold; hence, obtain a valuation.

to one on

~ (a(t)t -k) = ~ (a) - k.

a(t)t ~ =

~ ( a ( t ) ) t -k

D[[t, s]]

Hence

a(t)t "k =

~ (a(t)) - k = ~ b ( t ) ) - ~

It is immediate that the condi-

if we define ~a(t)t-kl

D((t, s))

If

= 2-~(a(t) t-k)

we

is a topological algebra relative

to this valuation and we have completeness. If we identify D((t, s))

then

a(t) -~D[[t,

D[[t, s]]

a(t)t -k

s]]

with

with the corresponding subset of

can be regarded as the product of t -k

and the latter is the inverse of

It follows that any non-zero element of in the form

b(t)t m

D[[t, s]]

and

invertible

in

tm

where

~ (b(t)) = O.

has inverse

D((t, s)).

We now assume We assume also that

D

theorem.

sn

t -m

Thus

It follows that

D((t, s))

sl C

is of order

for every

C

fixed by

d-~D.

s(u) = cu

c = s(b)-lb.

b(t)t m

where

Since

Then

Then if

Then

Co

s, [C : C o ] = n.

is

C. is

Since

by the Skolem-Noether u ~ 0

in

D

such that

sn+l(d)= s(u) s(d)s(u) -I.

c -~C.

Iteration of this gives

sn(u) = uuu -1 = u

we have NC~Co(C) =l.

Then, by Hilbertts Satz 90, there exists an element that

is a unit in

is a division algebra.

n < ~.

is an inner automorphism,

sn(u) = NCiCo(C)U •

b(t)

it is slear that

Hence there exists an element

sn(d) = udu -I

Since

can be written

is finite dimensional over its center

the subfield of elements of snlc =lc,

D((t, s))

t k.

s(bu) = s(b) cu = bu.

b

in

C

such

Hence if we replace

88

u

by

that

bu

we m a y assume at the outset that the element

sn(u)

= udu -1

satisfies

s(u) = u.

u

such

With this normalization

we have THEORem4 5.

Let

automorphism

of

fiel d o f

of s-fixed

2.

s~C

d ~D

C

D

D

over

has finite and

2 2 = m n .

K.

n.

D((t,

s))

(1,t,...,t n-l)

over

E

s

is an automorphism

s

D

and satisfies

Proof.

a(t)

ai ~ 0

centralizes

then

any element

)v -c

Co

which

E

over

D

b i -~C. where

coincides

we have then

da(t)

E

t o a Laurent by

t

so

D((t,

s))

Then

It follows v = u-lt n

of the proof is straight-forward

D

in

i - k = nin

that and

D.

a(t)t -k c i -~C.

D((t, s)). d-~D, Hence

and has the form

Conversely,

The condition

from this that the cente~ of Laurent

of

da i = aisi-k( d), d -~D.

c i - ~ C o.

to the algebra

s)) : L]

with the given

= a(t)s-k(d),

= (Co+ClV + ... ) v - £ t

rest

[D((t,

isomorphic

t(Co+Cl v + ... )v - S

isomorphic

and

-- v.

of this form centralizes

It follows

s)) i s isomorphic

t n = uv, v -~L.

si-k(ai ) = a~lda i.

where

(Co+Cl v + ...

over

the sub-

sn(d) = u d u -I ,

D((t,

first the centralizer

= a o + alt + ...

a i -- biu -n i

so that of

an

relations

o_~n E

s(v)

We consider

a(t)t -k

u

CO

s

: C] = m 2 < ~ ,

and i s generated

a -~E,

where

if

v

K,

i_~s both a left and a risht base of

ta = s(a)t,

so if

Choose

and we have the definin5

(33)

If

Assume : I.[D

Co((V))

over

over

be th e center,

contains a subal~ebra

D(v))

that

C

Then the center L

series algebr a

series alsebra

on

Let

elements.

order

s(u) = u.

to a iaurent

be a division a!~ebra

series

gives

over

L

s(ci) of

-- c i

D((t,

C o (in v).

and is omitted.

so

s)) The

is

89

4.

Generic matrix algebras Let

K

be an infinite field.

We shall be interested in

identities for finite dimensional algebras over L~MMA I. al~ebra over

Let f -~K[X] K

and let

is an identity for

L

K.

be an identity for a finite dimensional be any extension field of

(al,...,an) --> f(al,...,an)

nomial map. Hence, if it is

0

on

A,

it is

0

on

We now introduce the polynomial algebra ~ countable set of indeterminates and the matrix algebra K

de~

n

let of

Mn(~).

We call

over

K.

Let

L

sending

homomorphism of

Mn(r-4)

eij

Let

K{~ } L

A (k) = a i ~k) _~Mn(L ) ~

~(k~

into

into itself.

Then

f

K[X}

u

~

K{ ~]

•

"

be the subalgebra ~ (k) = ( ~ (k~),

be a commutative algebra over

~i k) __> aij(k) . into

Mn(L )

K

and

This extends to a

sending every matrix unit

K[~}

into

Mn(L )

is an identity for

Mn(A )

K{~ }

such that

p.i.algebra

is the T-ideal of identities of

f(~ (i), f

K[~(k)] inn

=

We have the K-algebra homomorphism

field of fractions of ~-' .

Thus

A L.

the ~eneric matrix al~ebra of

We now consider the universal In

f

is a poly-

The restriction of this homomorphism to

is a homomorphism of

where

Then

ij 1 < i, j < n, k = 1,2,3,

'

generated by the generic matrices

k = 1,2, . . . .

K.

AL .

This is clear since

over

We note first

Mn(K ).

where A

K{X} = K[X}/I n Let

f ~ I n.

-- K ( ~ i ~ k)) the

Hence we have

~(2),...,

~ (m)) = O.

is contained in the kernel of the homomorphism ~2 of

onto

K[ ~}

g -~ker ~) . A (1),...,A m

sending

Then are any

xk

into

~(k).

On the other hand, let

g(~ (I) , ~ (2) ,..., ~ (m)) = 0 m

matrices in

Mn(K )

and hence if

then g(A (1),...,A (m))

90

= 0

by (25).

Thus

g

is an identity for

We have therefore shown that THEOREM L

Let

an infinite field Mn(K ).

In = ker ~

K[X} = K{X}/I n

K

g~I

n.

and we have

defined b_~ the T-ideal

In

K{X}

of identities of

onto the 5eneric

(k) : (¢

K{<}

We shall prove next that let A

and so

the universal H I - al~ebra ove__rr

Then we have an isomorphism of

at__rix algebr__

Mn(K )

K[X] =~K{ 6 }

is a domain.

be the field of fractions of the domain

As above,

r ~ = K[ ~(k)].ij J

We

need the following L~V~L% 2. by

K[~}

is

Proof. el,...,en2

A K{< ]

the

subspace in

Mn( A )

2 { (n)

We order

Consider

< (i)

e

and re-label the indeterTninates to write

Consider

so we now have

det( ~(k~ ) _~/\.

and we can write every

e%

as a

{ (i) ..., ~(n2).

n4

as

~ Ik) =

indeterminates

Specializing

obtain the value 1 for the determinant.

matrices

spanned

Ms(A).

n2 Z ~ (k~el, 1 < k < n 2 , (~.

A

Hence

A-linear

~(k~ = 6~

det(~(~)

we

~ 0

combination of the

Clearly this implies that

/\ K{( ]

= M n ( A ). THEOREM 2. = K[X]/I n

(Amitsur).

The universal

, ~ 6 K{~]

We show first that satisfy Mn(A )

~

K{~}?

= O.

Since

KIWI

is prime and since it is a

D

K[~}

is a domain.

Proof.

quotients

PI-algebra

D

KiX] ~ I{{ ~ ] = O.

is prime.

Then by Lemma 2

is prime this implies

9 = 0

PI-algebra,

or

Let ~Mn(A) ~

~ = O.

its algebra of central

is finite dimensional simple over its center.

is a division algebra.

potent element.

Then

K{X}

means we have a polynomial

Otherwise,

Then

We claim

it contains a non-zero nil-

has a non-zero nilpotent element. f(xl,...,x m) 6 K{X}

which is not an

This

91

identity for

Mn(K)

but

fr

is an identity for

Then for any extension field f

is not.

L/K, fr

for some

is an identity for

(L,s,~)

of dimensionality

which is an extension field of and

L

K.

Since

is an extension field of

is an identity for

(L,s,~)

but

n2

f

but

K

a

over its center

C

(L,s,~) L = ( L , s , ~ ) ~ c L K,

it follows that

is not.

division algebra this is impossible.

r.

Mn(A)

In the last section we constructed for any field

cyclic algebra

= Mn(L)

Mn(K)

Hence

Since KIX}

(L,s,~)

fr is a

is a domain.

We now consider the algebra of central quotients of

KIX}.

By the preceding theorem (or its proof) this is a division algebra. We shall call this the universal division algebra of degree K

and we shall now denote it as

UD(K,n).

is the field of fractions of the center of be a central polynomial for identity for

Mn(K)

but

M~(K).

Then

[f,Xm+l]

The center of KIXI.

Let

n

over

UD(K,n)

f(xl,...,x m)

f(xl,...,x m)

is an identity for

is not an Mn(K).

Thus

f(xl,.o.,~m) ~ O, [f(xl,...,Xm),Xm+l] = O.

Since we have an endomor-

phism of

and sending Xm+l

K~}

any element of K[X}.

fixing the K~l

xi' 1 ~ i ~ m,

we see that

Conversely, suppose

f(xl,...,Xm)

into

is in the center of

f(xl,...,x m) E K{X}

has the property

that

f(~l,...,~m)

is a non-zero element of the center of

Then

f(xl,...,x m)

is a central polynomial for

Mn(K).

KIXI.

Hence we

have LEMMA 3. elements

f(E1,...,~m)

polynomial for elements

The center of

Mn(K).

K[Xl

such that

consists of

f(xl,...,x m)

The center of

g(xl,..-,Xq)f(xl,...,xm) -1

central polynomial for

Mn(K)

and

UD(K, n) where

g

0

and the

is a central is the set of

f(xl,...,x m)

i_~s~

is another on__~eo_~r O.

We prove next an important homomorphism property of T H E O R ~ 3.

Let

K[X}

be the universal

p.i. algebra

KIX}.

92

K[X}/I n

and let

A

b~e ~ central simple algebra of dimension

over an extension field

L

elements

al,a 2, ...

-CA

phism of

KIX}

A

into

of

K.

of

KIXI

there exists a unique K-algebra homomorsuch that

i~s mapped into

of non-zero elements of -~

of

K{X}

~k m >

xi m > for

Mn(E )

If where

K{X}

E A

A

and

K[X}

A.

ai

f--> 0

xi ~ >

f(xl,...,Xm)

We consider

Then

O

f(al,...,a m) -~L

and

fi ~ O,

C

of

F K[X}.

g~C

g~n _ g~i)~n-i + ... i g~i) gf~-

sending

Then

f

is K{X~

In a similar Mn(K )

for any

then it

ai~A

and

Since the center of

Mn(K )

where

the second state-

be non-zero elements of K[X}. which is the algebra

Let mi(x ) = xn _ ~i)~n-l+, . +(_l)n/li)

~ n (i) = N( fi ) ~ O.

a fixed non-zero element

Then

i ~ j ~ r.

and hence

A/L.

UD = UD(K, n)

K[X}.

are in the center

tions of the center

A

f(~l,...,~m)

fl,f2,...,fr

be the generic minimum polynomial of ~ts

into Mn(K)

f(al,.o.,am) ~ O.

imbedded in

of central quotients of

the

f9F

is central for

and the elements

Now let

K{X}

A

ai' i = 1,2,...

is a central polynomial for

ment is clear.

the center

under the homomorphism of

f -~K{X}

such that

consists of

A

Hence we have the induced homomorphism of

sending

is central for

KIXI

is an identity for

fashion one sees that if

there exist

is invertible in

is a splitting field for

xi--> a i .

into

into

More-

there exists ~ K-al~ebra homomorphis m

~fj

f -~In, f

an identity for sending

K{X}

We have the homomorphism of

ai •

ak, k = 1,2, . . . .

L, and give n any finite set Ifl,...,fr }

K{X}

such that

Proof.

2

The__~n~iyen any sequence of

over, under any K-algebra homomorphism of C

n

of

f

as element of

1

UD

UD.

Here

which is the field of frac-

Since

UD

is a division algebra

We can multiply all the

mi(x )

to obtain polynomials

where the

g~i)fn-i I + ... ± g ~i) = 0

there exists a K-algebra homomorphism

g(~) ~ C g ~

and ~

of

and

gli) ~ O.

g ~i) ~ O.

K[X}

into

Hence A

such

by

93

that

/ O.

1 ~(g)~(fi)n Here

~ (g)

We have t h e r e l a t i o n

_ ~ ( g ~ i ) ) ~ ( f i ) n - 1 + ... + ~ (gn(i)) = O.

and ~ ( g ~ i ) ) -~L

follows from this that

and

~ (fi)

~ (g)

~ O, ~1 (g~i))~ O.

is invertible in

A.

In a recent paper Amitsur showed that for certain is not a crossed product.

It

n UD(~, n)

This settled a question which had been

outstanding for forty years:

the existence of finite dimensional

central division algebras which are not crossed products.

The

starting point of Amitsurts proof is the following key result. THEOREM 4.

If

UD(K, n)

contains a maximal subfield

which is Galois over its center central division algebra extension field Galois over

L

Proof. over the

F

of

s(8)

L

of

A K

F

and

G = Gal M/F

of dimensionality

n2

then every over an

possesses a maximal subfield which

and has Galois group isomorphic to Let

8.

M = F(e)

and let

Then every

are distinct.

M

f(k)

s(e), s -~G,

Also we have a

is

G.

be the minimum polynomial is a root of us ~ 0

in

f(k)

UD

and

such that

Us8 = s(8)u s.

By multiplying by a suitable non-zero element of the

center

K[X}

Since

C

of

we may assume that the elements

s(e) -~F(@) = F [ e ]

s(@), Us -~K[XI

this element is a polynomial in

coefficients which are fractions formed from elements of we have relations

UsU t = Ps,tUst

and the

8

with

C. Also

Ps,t -~F(e)

so these

can be written as polynomials in

@

with fraction coefficients

Formed from elements of

c

be the product of all the

C.

Let

denominators of all the fractions just mentioned and the

Ps,t)

s(8)

as well as the denominators of the coefficients of

the polynomial : y ---> y'

(for the

f(k). of

K[X}

By Theorem 3, there exists a homomorphism into ~

such that

c

~ O, s(8)

t

~ O,

!

S(8) T ~ t(O)'

for

monic polynomial

s ~ t, u s ~ O. g(x) of degree

Then we obtain from n

with coefficients in

f(k) L

a such

94

that

g(e')

- o.

are distinct.

Moreover,

every

s(O)' E L I @ ' ]

We have the relations

: L(e')

uslev : s(e), u s,

x ' ---> u s' x ' (u 's)-I

that the inner automorphism

T

Since

UsU t = Ps,tUst

it follows

also that

and these

which

imply

L(O' ).

stabilizes T

!

UsU ~

and

Ust

differ

!

by non-zero

elements

restriction

to

Since

of

L( e')

s'(e)' = s(e)', s '

dimensionality

L(e')

Let Laurent

K

r

over

s ~ t.

of

~/L n

K.

is

G

< n

series

field and let

= {s'} is a

G.

with Galois

Laurent

is the

Since the it follows group

that

G T "~= G.

fields

K((t))

be the field of

We can iterate

this

construction

to form

K((tl))((t2)),

K((tl))((t2))((t3) )

etc.

steps we obtain a field which we shall denote as

extensions closed and

We wish to determine

of degree n

some standard

n

the simplest

results

in Jacobsonts

the structure

of such a field assuming

is not divisible

To handle

by the

case,

of valuation Lectures,

(non-~rchimedian)

valuation

the set of non-negative (i) (ii) (iii)

Jal = 0

vol.

algebraically

we shall need to use

III,

Chapter V.

Let

F la~

such that

if and only if

labJ = lal(b) ~a+bJ~ max(lal,

K

theory such as can be found,

I ~ : a --> reals

of algebraic

characteristic.

K((t)),

brief summary of what we shall need.

IR+

to

s

(st) ' -- s ' t ' °

Thus

isomorphic

over iterated

the fields

K((tl,...,tr)).

example~

L(e v)

be an arbitrary

series

if

of dimensionality

al~ebras

successively After

of

that if

' ' (Us) '--i then UsX

~ t'

of any subfield

is Galois

Division

It follows

x ' m>

of

group of automorphisms

5.

L(Ot).

Ibm).

a = 0

for

We give a

be a field with a is a map of

F

into

95

The range of

~I

on the set

F ~:~ of non-zero elements of

F

is a

subgroup of the multiplicative group of positive reals called the value group of The subset

~

~ ~.

We exclude the trivial case in which this is 1.

of elements such that

~a~ < 1

is a subring of

F

m

called the valuation rin~ of that

~a~ < 1

I ~.

The set

is a maximal ideal in

~

p

and

called the residue field of the valuation. field of fractions of group

~

domain)

~ .

is cyclic.

and

where

maximal ~or the values

~/p

is a field

The field

F

is the

The valuation is discrete if the value

In this case

p = (~)

of elements such

~

< i.

~

is a

p.i.d.

(principal ideal

is an element such that

~

is

A fundamental theorem of valuation

theory is Henselts len~na on fields which are complete relative to a discrete valuation.

We refer to the above reference p. 230 for

this and we shall state a consequence of this lemma which we shall need.

Suppose

g(k)

is a monic polynomial with coefficients

such that the corresponding polynomial has a root a root

p

r

in

in

If

F

dimensional

~

~

and

(g(k),

such that

~(k)') =

extension field of

F

subgroup of that of ramification index i, i

ring of

J

i

E,

P

(~ + P)/P --~/P

E e

E

and

g(k)

~/p has

E

is a finite

can be extended in one E

is complete relative

The value group of

F

is a

and the corresponding index is called the of the extension.

its maximal ideal then so the residue field

If

0

is the valuation

~ = 0 ~ F, p = P ~ F ~/P O/P

is called the residue degree

has the relation

Then

and ~ ~

cit. p. 258).

with a subfield of the residue field [0/P: ~/p]

~ ~

then

and only one w a y to a valuation on (loc.

i.

IX], ~ =

T ~ r + p = p.

is complete relative to

to this valuation

g(k) - ~

in

of of

f

F

can be identified

E. The dimensionality

of the extension.

n = [E: F] > ef. If the valuation on

then so is the valuation on

E

F

and in this case we have

More exactly one can show that if

rl,...,r f

and

One

is discrete n = el.

are elements of

O such

96

that the cosets over

~/p

rl = rl + P,...,rf = rf + P

and

O ~ j ~ e-i

P = (w)

then the

form a base for

E

ef

form a base for

elements

over

F (loc.

riwJ , i < i ~

and we have completeness. maximal ideal is

in this field is discrete The valuation ring

p = (t).

Since

is 8 set of representatives may identify

K

LEMMA i. series over

Let

K

K

[L : K((t))] = n < ~ and of K

f

L

over

K((t)).

and its

K[[t]]/p

K

and we

We can now prove the field of Laurent

be an extension field such that

and

n

be the ramification

L

(see p. lO&)

the subfield

of the residue field

be a field, K((t))

and let

F = K((t))°

~ = K[[t]]

K[[t]] = K ~ p ,

with the residue field.

f,

cit. pp. 266-267).

All of this is applicable to a Laurent series field The valuation we introduced

O/P

i~s not divisible by char K.

Let

e

index and residue degree respectively

Then there exists a subfield

L

t

of

L

over

such that i.

T

L

C~

2resentatives

the valuation ring o f

of the residue field

L,

~/P, p

L

T

is a set of re-

the maximal

ideal of

C~

!

and

[L

: K] ~ f. 2.

L

T

and

L'K~t)) ~= L' ~ K of the series convergent).

K((t))

K((t)))

are linearly disjoint over and

L'K((t))

(bo+blt+b2 t2 + ... LtK((t))

K

(that is,

i~s~ subfield consistin~

)t-m, m _> O, bi -CL' , (all

i_~s isomorphic to the Laurent series field

L'((t)). 3. Writin~

L'((t))

f°r

L'K((t))

we have

n : L'((t))(~)

!

where form~

~C

is algebraic

ke - ~ t

where

of the set of series s_~o L

over ~ ~ 0

L ((t)) is in

(bo+bl.h~ + b 2 ~ 2

i_~s isomorphic to a Laurent

~lgebraically Proof. field of

closed, i.

or/p.

LT = K

with minimum polynomial L ~ . Moreover, +

)~-m,

s~ries field over

and we ma_9_~assume

Consider the residue field Since the dimensionality

~/P

L

of the

consists

m > O, b i-C L f T

L .

If

K

is

~/ = 1. as an extension

[Cr/P : (r/p] = f

is not

97

divisible by of

~.

~(X)

char ~/p (= char K),

Hence of

p

derivative

~/P = ((~/p)(p)

over

~(X).

~/p

(Z/P

is a separable extension

where the minimum polynomial

has distinct roots and so is prime to its

Nbw K C q ,

K~F

is zero

and

K

presentatives of the elements of the subfield we have a monie polynomial morphism of ible in K

O[X]

onto

g(k)-CK[k]

(U/P)[k]

is a set of re-

~/p.

whose image under the homo-

is ~ (k).

Now

g(k)

and by Henselts lemma g(x) has a root

= r + P = p.

Then

L t = K[r]

t

is a subfield of

~

L F] P = 0

and

Consequently

r

is irreduc-

in

J

such that

L

contained in

and

~CP+KCP+L

(~,

t

P+ L

=

since

0~

t

p -CP+L

•

t

Thus

L

is a set of representatives of the residue field

of course, [Lt:K] = f. 2.

Let

This proves all the assertions in i.

(rl,. • . ,rf)

( ~ l = r l + P'''''~f= r f + P ) result quoted above if i < i<_f, O _ < j < e - l , • l,...,rf L

!

and

be a base for is a base for

P = (w)

K((t))[r]

L'

then the

form a base for

L

since

L

over

L.

r

with

LT((t)).

L'K((t)).

= f

a base for Lt((t))

and L

and

chosen so that

ri~

K((t)).

,

Hence

K.

t

Now

L K((t)) =

is algebraic over K((t))

K((t)).

it is closed

Since it contains

This set is a subfield isoK((t))

and

LT

it coincides

This proves 2.

3. Writing

[L': K]

o-/ p. Hence,by a

Hence it contains all the elements of the form

(bo+blt+b2t2+ ... )t -m, m_> O, b i -~L ~. morphic to

Then

K((t)). This implies that

Since it is a finite dimensional extension of in the topology of

K.

e f elements

are linearly disjoint over

is a subfield of

over

CZ/P over

are linearly independent over K((t))

O/P and,

L'K((t)) = LT((t)) [L : Lt((t))] = e.

ever

K((t)),

and

Also e

[LT((t)) : K((t))]

Since the elements

(i,~,...~ e-l)

L = L'((t))(~). P = (~)

we have

ri~J

is a base for

~ ej = jtJ

since

~

is the ramification index.

e

form

L

over

was Then

t

= ut

where

~u~ = i.

Since

L

is a set of representatives of

98

the residue field Then

~/I

= 1

and if

k e - v = ~e _ [ e

Cf/P there exists a ~ ~ L '

has

v =

~

[-lu,

~v~ = 1

as a root in

by Henselts lemma, there exists a u = z e ~"

so if we put

Clearly we can replace ~ge = ~ t

where

~

and

ke - ~ t

u=

/(mod

(~ = v + P etc.).

(}e - v, e ~-I) such that

Since

= i. Hence,

z e = v.

Then

then ~g e = z-e e = z-e ut =

~

= i.

P).

v --- 1 (mod P). Then

by ~C and we shall have

~ -~L'

it is clear that

z-CO

1~ = z-l~

and

(~/P

is not divisible by the characteristic

such that

Since

L=

~t.

Lt((t))

[L : L'((t))]

is the minimum polynomial of "C

(U), = e

over

t

L ((t)).

The fact that

(bo+bl~C +b 2 ~ 2 + • o. this set contains

L

is the set of series

) ? -m , m > O, b i -~L',

Lt((t))

since

t = ]~-l~e

is clear

since

and it contains

qf.

!

If

K

is algebraically

of

K,

coincides with

that

6e

=

~

closed K.

L ,

Also in this case we have a

and we may replace

We now consider a field ally closed.

Let

which is an algebraic extension

K((tl,...,

"C by

6-i~

K((tl,...,tr) ) r ))

powers where

e

to obtain where

K

be the multiplieative

zero elements of this field, K((tl,... , t r) ) # e

6-~K

such (6-11~) e =t.

is algebra~rgroup of non-

the subgroup of

is a positive integer not divisible by

eth

char K.

We

have the following

or

e omeo

constitute a set of representatives in

kr ... t r , 0 <_ ki < e = 1 of the cosets of K((tl,...,tr)) ~ e

K((tl,...,tr)) ~.

Proof. Let a ~ K ( ( t l , . . . , t r ) ) ;:~. We can write k 2 a = bot r r(l+bltr+b2tr + ... ) where k T -~Z~, b i ~K((tl,...tr_l)) b o ~ 0. Hence

We have seen that l+blt + b 2 5 2 + . . , is an eth power. kr );:~e) a --- bot r (mod K((tl,...,tr) . Using irgluction on r

99

(the case

r = I

was handled

in the last part of the proof of Lemma i)

(mod K((tl,. .. ,tr_l))$ e), O <-- kj <-- e - i, b ° m t kl I .. tr_ kr-i I

we have

t

t

i ~ j ~ r - i. Also we can write k r = ~ r e + k r where 0 ~ kr (t~r)e t;~ kV Then t r = = trr (mod K((tl,...,tr))~ e) and kl r t r _~ k -1 t kt a ~ b I ... r (mod K((tl,...,tr) ) ~ e ) . that the cosets determined 0 <_ k i _< e - i,

by distinct

kr ~ e-l~

It remains to show kI k2 kr t I t 2 ... t r ,

monomials

This will follow if we show that

are distinct.

~i gr ~ e t I ... t r -~K((tl,...,t r)

implies

every

g

= 0 (mode).

1

Hence

suppose tll ... t r

=

(b^t (l+b~t + ... ))e %)

kC~,

2.

bi -~K((tl,..,,tr_l)),

gr = 0 (mode).

Induction

L}~4MA 3. extension

of

Let

K

b_~ealgebraically

K((tl,...,tr) )

K((Wl,... , ?r ))

[K((~l,...,~i)

This relation

shows that every

i_~s finit_____~eand not divisible field

b o % O.

= 0 (mode).

closed and let

such that

L

be an

[L : K((tl,...,tr))]

by char Ko

Then L

i_~s~ Laurent

= n series

K((~l,...,~i)

~K((~l,...,~i_l,

) : K((~l,...,~i_l,ti))]=ei,

n = e l e 2 ... e r

we have relations

where

gi

implies

ti)~ and_

of the form

eil ~ ei, i-lt ~C~ ii = ~Cl "'" i-i i

(26)

where

eli = e i, Proof.

0 _< eij <_ e i - I

We use induction

on

for r.

i <_ j <_ i - i. The case

r = 1

has been

L

t

proved in Lemma i.

Now put

K

= K((tl,...,tr_l) )

t

= K ((tr)).

[L'

so K((tl,...,tr ~

t

:

K '] =

By Lemma i we have an extension f

=

L = L V ( ( ~ r ))

[L'((tr)): where'~re

K '((tr))], = ~tr,~-~L

[L: I.

L

L '((tr))]

t

of

K

such that

= e, n = ef

and

By induction we have

t

L

-- K(( t-l,..., -Cr_l))

on the subfields and the replaced by

f.

where we have the indicated relations ~j

and tj, 1 < j _< r - 1

By Lemma 2 we can write

~ = $eT1

and kI

n is kr_ 1 ''Zr-1 where

100

0 < kj < e - 1 .

If we replace

Then if we put

err -- e, eir = k r

also for

be a sequence

ible by char K. and let

sI

x1 - - > o ~ i x I

general,

we may assume

we obtain the required

8 = 1.

relation

Laurent Di

series

x i -->~ixi

nI -th

, cO i

i

which

K

sending Let

Dl((X 2,sl)) .

i < 2r.

is the identity

If

i

si on

Di+ 1 ~ Di((Xi+l,Si)).

sequence

of division algebru~' ending ~rith

D2

In

is even be the auto-

Di_ 1

and

n(i+l)/2 - root of i.

put

K((Xl,...,X2r;

series over

root of i.

is odd we let

a primitive

Let

is not divis-

of this field over

division algebra

and if

division

as follows.

the field of Laurent

has been defined for

D i - Di_l((Xi))

K

series

n -- n I ... n r

is a primitive

Di+l x Di((Xi+l)) in

such that

D 1 ~ K((Xl) )

~l

Laurent

closed field

be the automorphism

assume

morphism sends

Let

where

be the twisted

we let

~-l~ r

an iterated twisted

over an algebraically

nl,n2,...,n r

K

by

r.

We now construct algebra

~r

Then we

We obtain in this way an increasing

Sl, S3,...,S2r_l)).

D = D2r =

Repeated

application

of Theorem

of §3

shows that the center of D is F = K(tl,t2,...,t2r)) nl nl n2 n2 2 t I = x I , t 2 = x2 , t 3 = ~ , ti = x 4 ,... and [D : F] ~ n

5

where where

n = nln 2 ... n r . We now define a map non-zero

elements

of integers,

of

D

as follows.

v

of the multiplicative

into the additive Let

a I -~ D~

2 m2 r al ~ a 2 ( l + b l X 2 r + b2X2r + o.. )X2r and

a 2 # O.

where

Then we let the last entry of

: v(a l) -- ( ..... , m2r). Next write 2 m2r_ 1 a 3 ( l + ClX2r-1 + c2X2r_ 1 + ... )X2r_l

induction

m2r_l

a2

of

from the definition

of 2r-tuples

and the b i -~ D2r_l

v(al)

be

a2 = where

a3,c i -~D2r_2,

be the next to the last entry of

we obtain a definition

It is immediate

XL (2r)

D ''~ of

and write

m2r

Then we let

group

group

v(a l) •

v(al)--~ml,m2,.o.,m2r) that

a3 ~0 By

-~2z (2r).

101

v(ab) = v(a) + v(b)

(27) and

(28)

i 1,O,...,O).

v(xi) = (0,...,0, Now let

Since

L

be a maximal subfield of

F - K((tl,...,t2r)) ,

D.

Then

[L : F] = n.

Lemma 3 shows that L = K ( ( ~ , . . . , ~ 2 r ) )

where we have the relations

(26).

Since

t i = x~ [i+I/2]

we have the

equations - eilv ( r I) - ... - ei, i_l v (q~i_l) + e i i v ( ~ i ) = (29)

(0, • • .,O,n(i+i/2],

These relations

(30)

0,...,0) •

imply by induction on

v(~i)

i

that

= (fil,fi2,...,fii,O,...,O)

where (3i)

eiifii = n[i+l/2 ] .

Now let ell -e21 (32)

0

•

•

e22

0

.

I fll i

E =

-el.I -e]2. e33. O. /

Then the foregoing relations (33) Put

. . . .

)

f21

f22

0

•

•

•

~31

f]2

f]]

0

.

•

•

•

•

.

•

o/

°

•

•

•

•

•

k

(29) give

EF = diag[nl,nl,n2,n2 ,'°',nr'nr] " m =g.c.m.

(nl,n2,...,nr)

mn~l,mnl I, ...

I.

Then

FR = M = (mij)

so

M

and

E(FR)=

R : diag{mnll,mn! I ,

ml°

Hence

is lower triangular:

and we have the following relations for (3&)

0

mijejj - mij+lej+l,j

....

(FR)E = m l• mij = 0

if

Put j > i,

i ~ j ~:

- miieij = 6iJ m"

These imply that mil~mi2 tmii (%" ~ll)mil(< e21 7y2e22) mi2 ... ~i = tl ~2 "'" l = (~i ell %~ el2 ?- _el, i_ 1 _ e i i ] m i i m • •.. . . . . --i " =q~i i-i

102

d.

Then if we put

d i = m/nri+i/2 7

and ~ i = >~i :~

we have

~i n[i+I/2] = I
Let ~,

D

be the iterated twisted Laurent series

division algebra

K((Xl,...,X2r, Sl,...,S2r_l))

ali~ closed field

K

maximal subfield F

L

where the of

D

ni = qi

over a__nnalgebraic--

are primes.

The 9 agy

is an abelian extension of the center

with Galois ~roup a direct product of cyglic ~roups of orders

ql 'q2' " " "'qr" Proof.

We have

by (31), every Let

L = K((~I,..., Y2r ))

eii = q[i+i/2]

iI < i2 < ... < ir

or

be the

eii = I.

its

as before. Also

such that

~

Then,

eii = n.

eii=q[i+I/2]"

We claim that (30)

K((Z'I,..., ~'ij)) = K((tl,...,tij))( D il,..., ~ij)

where the j = I. the Also

~ i

Then the

eij

are defined as above and eii, i < il,

show that

K((tl,...,til_l)) = K((?l,...,-Cil_l)),

eilil ( = one of the

= K((~l,-.., Z~il_itil)). t

(31)

First, let

and so the relations on

K((~l,''', ~il))=K((tl,...,til))(~il)

prime degree

where

are 1

1 < j < r.

over

K((tl,..0,til))

We have

eili I iI -~ K((tl,. °-, til ) )

(dil,eilil) = 1.

q's)

and this has

and

~ il = Z~i~il

It follows that

K((qYl,..., Z'il)) = K((tl,...,til))(~]il). di Otherwise, ~ i I ~K((tl,...,til)) so ~g'il 1 ~ K ( ( t l,..,,tir)) as eili~ well as, q~ II . ~ -~K( (tl,...,til)). Since (dil, ellll • • ) = 1 this implies that ~il ~ K ((tl,...,ril)) contrary to

103

[K((~l,...,~il))

: K((tl"'"til))]

Hence (31) holds and this is the case (30) holds for

j - 1

and consider

= eili I ~ I.

j = 1 of (30). K((~l,...t~'ij))

K ( ( ~ l , ' ' ' , ~ i 3 1 )) ((q~ij_l+l,''''~ij))

Now suppose =

=

• K((tl,..., tij-l) ) (~il,...,~lj_ 1 ) (( q~ij_l+ ! ,..,, ~ij)). j = 1 shows that this is

(32)

The case

K((tl,...,ti~_l)) (~il,... , ~ij_l)((tij_l+l,...,tij))( ~ ij). !

We remark now that if of a field if

K

then

(rl,...,rf)

Lt((t))

L

is a finlte dimensional extension field

Lt((t)) ~ K((t)) ~)K L'

is a base for

L

t

over

L

This is clear since then every element of

can be written in one and only one way as a linear

combination of the

r. with coefficients in K((t)). Since 3 • . ..., ~ ) is a finite dimensional K((tl,. ",tij_l))(~11 ' ij_ 1 extension of

K' = K((tl,...,tij_l)), it follows that (32) is

identical with (33)

K((tl,''',

tij)) (~]i I'

This proves the inductive step.

..

"' ~]ij).

Thus we have (30) and using

the commutativity just indicated we have also (3~)

L = K((~I,..., ?2r)) = K((tl,''',t2r)) (D il''''' Nit)"

Also we have that

~ ij e ~ i j _~ K((tl,..., t2r))

[L : K((tl,...,t2r))] = ] ~ L/K((tl,...,t2r))

e~ij.

It follows that

is a tensor product of the extensions

K((tl,...,t2r)) (~il).

Since the base field contains all roots

of 1 and the characteristic is not a divisor of K((tl,...,t2r)) (~il) Hence

and

eijij

is a cyclic extension of degree

L/K((tl,...,t2r))

the field eilil •

is abelian with Galois group a direct

104

product of the cyclic groups of orders

eijij.

This completes the

proof. A consequence of this theorem and Theorem & of ~& THEOREM 2.

Let

where

n

UD(K, n)

is

the universal division algebra

of index

n

is not divisible by char K.

UD(K, n)

contains a maximal Galois subfield over

Then if K,

then its

Galois group i~s ~ direct product o~f c~91ic ~roups o f prince order s. Proof. Let

K

be the algebraic

closure of

K

and let

D

be the twisted Laurent series division algebra K((Xl,...,X2r, Si,...,S2r_l)) UD(K, n)

over

K.

has a maximal Galois subfield with Galois group

K((Xl' " " ' 'X2r' Sl '" .. 's 2r-i ))

G

G

then

has a maximal subfield whose Galois

group over the center is isomorphic to that

By Theorem i of §& if

G.

Then Theorem 1 implies

is a direct product of cyclic groups of prime orders.

Non-crossed product division algebra__~s In this section we shall show that certain ones ofthe universal division algebras

UD(K, n)

first the case of

K = ~

are not crossed products.

We consider

for which we shall need some results on

p-adic division algebras. Let

p

be a prime and let

Qp

be the field of p-adic numbers.

We denote the subring of p-adic integers (that is, the that ~p

l ~ p ~ i) are If

F

by

2p.

This is a p.i.d.

pk~p~ k = 0,i,2,..., n ~ i

~p~Fp__

we have a field

Fpn

such

In fact, the ideals of = X/(p).

of

pn

elements.

of non-zero elements from a cyclic group of

and these are the roots of

~ -~Qp

X p n ' l - 1 -~Fp[k].

If

pn ~

i

The set elements

is a generator

105

of

Fpn

of

co

then over

Fpn = Fp(O3). Fp.

( ~ ( ~ ) , 6(k))

Then

~ 1

Let

deg ~ ( k )

since

is monic and

Fp[k] in

be the minimum polynomial

= n, k pn - 1 = ~ ( % )

(kP n-I -i)' = -

Hensells iemma that we have g(k)

h" (k)

g(X)

Since

~

and

It follows from

k pn-I - 1 = g(k)h(X)

g(k) = ~ (k).

it is clear that

i.

0 (~)

(X)

is irreducible

in

Zgp[k]

where

is irreducible

in

2p[k]

and hence

~p[%]. We now adjoin to

W = Qp(w)

with

Qp

a root

[W : Qp] = n

and

w

of

g(k).

W

is complete with a discrete

I I

on

~p.

We have

w - ~ CY the valuation ring of

W.

Let

P

valuation which extends

~, a--> a

Cr/P. ~/p

Then we obtain

w P n-I = 1

be the maximal

the canonical homomorphism of

class field

0

so ideal of

onto the residue

is an extension field of

Fp

and

[~/P:Fp] = f < n. On the other hand, ~

is a root of

and this is irreducible

of degree

Hence this is the minimum

polynomial

Fp

Fp • of

of

Thus %

w

over

f = n

and

~p.

Let

a~>

~

polynomial so

let

W

so we can identify

g(k)

~

l)-st

of

in

over extension

w

over

_> n.

the maximal 0~ onto

ideal of

(~/P.

with the field ~"~ • Fpn

n (~

We have

Fpn

of

pn

We have

F ~'~ n[k] Hence, by He nsel's lemma, there P ~ = co and w pn-I = I. The minimum Qp[k]

is a multiple

is a factor of

we have

Since

deg g(k) = n

root of I.

O/P

to be a generator of

If ~ (k) = g(k)

deg~(k)

d

n

is an unranified

of

Hence we see that

(pn

~?

has degree

be an unramified extension of degree

w - ~ Cr such that

and

and

the canonical homomorphism

g(X) -~O~k].

oA d = 1

e = !

P

Choose

Fpn = Fp(O3),

~/P = Fp(W)

g(X) = ~(X)

e = i).

k pn-I - I = ( k - u O ) 6 ( k ) exists a

n.

(Y be the valuation ring,

[C~/P : Fp) = n elements.

and

and so

(that is, Conversely,

of

in

of

~ (CO) = O.

[W : Qp] = n

and so

we have

W = ~p(W).

pn _ I.

Then the roots of

kP n-I - 1

Thus g(k)

If w

and

Since deg g(k)_
then

is c primitive

are powers of

w

106

and so

W

is a splitting fielc~ over

is Galois over

complete field L

of

g(x)

and hence

[,'I

~p.

We recall that if

to one on

~p

K

L

is a finite dimensional extension of a

then the valuation of

K

has a unique extension

and this is given by the formula

i. (35)

lal = INLIK(a) I n

where

n = [L : K].

If

(loc. cit p. 258)

~-CGal

L/K

(35) implies that

~

(a) i = ~a~.

valuation ring of

L

and

and

(7 (p) = p.

residue field

Then

P

is its maximal

W, let

pn . I.

and if

Then

Gal

Fpn/F p

=O

~- in the

o-(w)

over

Then any Fp.

since

W = ~(w).

then corn = 1 o-= i.

so

Thus

cr such that ~(co) = u) m

are the powers

W/Qp or

m

deter-

m (w) = w m

We have is divisible by

~---> ~

is a monomor-

~

is

a - - > a p, m --> a p .

so we may take w --> w p.

(r, n) = i.

G

is

Then if

m = p.

Thus

The generators We have

~r(w)

•

We can now construct the cyclic algebra

(36)

(w, ~,

p).

We claim that this is a division algebra. applying the Wedderburn norm criterion: ~NW/Qn(a)~ unramified

- ~ a~(a) ~a~ = Ip~ k

.. ~ n ' l ( a ) ~ for some

(~/P

our map is an isomorphism.

sending where

(~ ~ G

We have

is cyclic with generator

is the automorphism of

W

Fpn

~G~ = n = ~Gal Fpn/Fp~

(w) = w m, as above,

G pr

of

and

cyclic with generator

of

~ (4

K.

G = Gal W/Qp.

~ = 1

wm = 1

phism and since Since

~

is determined by = tom

Hence

([ is the

ideal then

o- induces an automorphism

mines an automorphism

(OJ)

= N(a).

([/P. This is an element of the Galois group of

Returning to

~

N(~a))

It follows that if

~*elative to the residue field of

and

then

This will follow by Let

= ~a~ n • k.

a~W. Since

Then ~Nw/~n(a)~

Then W

is = ~p~kn .

107

It follows that the smallest power of nth

power.

Hence, by Wedderburnts

THEORem4 1. sion~ ~.

n2

over

p

which is a norm is the

criterion, we have

There exist central division algebras of dimen~

f~or a ~

n = 1,2,3,

..

We shall need some information on abelian extensions of We consider first any extension field = n p

of

is finite and its ramification index (a so-called

of

L

has

pf

elements.

coincides with

that

e

e

over

L = W(~)

such that [L : ~ p ]

is not divisible by

W

t W.

O/P

The argument used before shows that we have of degree

f

over

Qp

U/P, that is, any element of

to an element of

of degree

~p

tamely ramified extension~ The residue field

an unramified subfield

P

L

~p.

= 0 ~W.

Moreover,

~

whose residue field is congruent modulo

L

is totally ramified

The argument in the power series case shows

where the minimum polynomial of

~

over

W

is

X e - ~, ~ - C W . We now assume

L

(with

G = Gal L/gp, H = Gal L/W. Hence

H

Gal W/Qp

p ~ e)

is Galois over

We have seen that

is a normal subgroup of

G

and

and so this is cyclic of order

W

G/H f.

Qp

and put

is cyclic over

~.

is isomorphic to In other words we have

the exact sequence (37)

i --> Gal L / W - - ~ Gal n / Q p - - >

Gal W/Qp --> i

the next to the last term of which is cyclic of order ~-CG

and let

~/P. of

~

C w

~

it is clear that if

is congruent mod P ~-CH

then

contained in the kernel of the homomorphism

to an element

~ = I. ~-->

~.

Hence

Gal W/Qp

Since any automorphism of phism of

L/Qp

H

W

onto the group of automorphisms of W

over

it follows that

Qp

~-->~

is

On the other

hand, we saw that this map defined on the Galois group of isomorphism of

Let

be the induced automorphism in the residue field

Since every element of t

f.

is an ~/~.

is induced by an automoris surjective on the

108

group of automorphisms

of the residue field.

o-~>

and so the kernel has order

~

has order

kernel is ke - ~

H.

f

We note next that since

is the minimum polynomial

product of distinct roots of

ke - ~

distinct

linear factors is an

roots of unity

contained

in

W.

of

eth

L

1.

Vl,V2,...,v e.

To see this let

v

over

L[k].

root of

e.

Hence the

is Galois over

~7

in

Hence the image of

W

W, k e - ~

and

is a

The ratio of any two

Hence

L

contains

e

We claim that these are be one of the

v

and let l

g(k)

be its minimum polynomial

duct of factors where

k - vj

CYt = @ ~ W.

pt = P ~ C ~ t that

has a root

g(k)

v -~W.

L[k]

W

distinct

these constitute a cyclic subgroup

then of

~

L = W(~)

e th

Gal L/W

shows that

onto

E.

roots of

L ~;'.

We now assume

Gal L/Qp

E

of

let

~

be the element of L/@p.

lu~ = i.

roots of

Then

Gal L/W

1

are in

such W

and

and

It is well known of the Kummer

In fact

~] ~ >

v

if ~]-~Gal L/W is an isomorphism e

and (37)

is abelian and we shall show that the is contained

in

~.

Gal L/~'~ such that

~C(I])I

= IHl

so

Let

v ~E

~ (I~) = v ~

q (U)

= u I-i

and .

Let

where

Hence

C~(N) ~

1

is cyclic of order

~ < ( H ) = ~(u) n ( ~ ) =

and

T

is a cyclic extension of a cyclic group.

set

~ ~Gal

eth

E.

and the map

Thus

Gal L / ~

to

w _~

and this implies that

E of

is isomorphic

(]I) = v]-~, v - ~ E

~(k) -~(Ct/~')[~,

has a root in

and easy to see that the Galois group Gal L/W extension

g(k)-~(Y [k]

since we have a g(k)

is a proI

vj ~ C ,

it is linear, e

g(k)

polynomial

It follows that

contains

Since

and the

~ ~C~'/P ~

is irreducible,

Thus

W.

The corresponding

w ~ v (mod P).

since

in

over

=

<~

implies that

(Y and since

((v) - v (rood P)

-- ~(v)~(11)

and

D ~Gal

~(u)v

: C(v) uT~ = C(v)u.

L/W = H,

<(v) -~E.

If

~(u)vTI

All of these elements

~(u) -~ u (rood P). v

m

((v)

# v

then

Hence

109

k e - i = ( k - v ' ) ( k - v)h(k), we obtain (k e - ~

and taking images in the residue field

k e - [ = (k-~)2~(k).

~ k e-l) = [

so

k e - [ has no multiple roots in

contradiction shows that

~ - C G a l L/Qp, v -CQp. We now have

~ (v) = v.

E CQp

and

C/P.

This

Since this holds for all

E CC

ke - 1 =~

so

E C~p

(k- v)

v~E Fp = ~ p / p ~ p = ~ / ( p )

where

p ~ e,

E CQp.

Then

have the f a c t o r i z a t i o n Fp[k]

On the other hand, since

= ~

Qp.

in ~ k ] . ~e

we have

We

Then in

-~=

~

(k-V).

v-CE Since e

(ke_~,

different

~ ~-l) e th

= 1

the

roots of

of the multiplicative

~

~.

group

are distinct and so

Fp

contains

These form a subgroup of order

F ~p and

~F~I = p - 1 .

e

Hence, e ~ p - 1 .

The results we have obtained are of independent interest.

We

summarize these in the following L~A.

Let

L

be an

whose ramification index

n

e

is not divisible by

tains an unramified subfield residue de~ree o f

L

polynomial of

over

over

Qp,

order

e

~]

then

over

e

We now suppose

a n d Gal W/Qp Qp

2kq~ i ... qrkr

(d, n) = i

e~n

L = W(~)

con-

f=n/e,

the

where the minimum

ke - ~.

If

L

is Galois

f ((3?) is exact and Gal L/W f).

If

L

el p - l .

implies

e] (p-l, n).

(n, p-l) = i or 2.

k --> 0, k i > 0

and the

Write

qi

are distinct

d ~- ] (rood $) and -- 2 (rood qi ) for i< i <

r.

and so Dirichlet's theorem on primes in arithmetic

progressions there exists a prime b = 1,2,3, . . . .

of d ~ e n s i o n

L

is given and we proceed to show that the

where

Choose

Then

is c~clic of order

then

can be chosen so that

odd primes. Then

n

p.

~p

is an extension of a cyclic irou~ o f

The last condition and

n =

and

Qp

has the form

is an abelian extension of

p

over

b y a cyclic group of order

is c%clic of order

prime

W ~p,

W

Gal L/Qp

dimensional extension field of

Then

p-i

p

of the form

d +nb

= (d-l) + nb m i (rood qi )

for some

and m 2 (mod $)

110

Then no 2.

qi

divides

Also since

Under these an abelian extension

and

p = d +nb

extension

Qp

Hence

(d, n) = 1

the foregoing

of

of a cyclic

~I P -i.

and

circumstances

Such an extension groups

p - i

n

= 1

it is clear that

lemma

of degree

(n, p - ~

shows that if

then

Gal L/~p

cyclic

p#n. L

is

is an

group of order one or two by a cyclic

is either

or

group.

or it is a direct product

of two

one of which is of order two. We can now prove AMITSURIS

prime

THEOREM.

If n

is divisible

o~r b_y ~ then the universal

division

by the square algebra

UD(~,

the

are

of an odd n) is not

a crossed product. Proof.

Let

k k kr n = 2 ql "" qr

primes and suppose whose Galois

group

UD(~, is

n)

G.

where

contains

By Theorem

a Galois

... ~ Z 2 ~

Zql X

denotes a cyclic

and Theorem a

Qp

the form

Zn

extensions

(39)

or 2 and the

k i = i.

Q This

...

or

j.

By Theorem

3 of §~

just proved on the existence

of degree

Z2 ~ Zn/2

G = Zn that if

... X Zql ~

group of order

or of the f o m

It is immediate

n

r

i above and the result

whose abelian

of degre

~ ... ~ Zq r

Zj

subfield

odd

kI

~ Zq

where

distinct

2 of §5,

k G = Z2 ~

qi

G = Z2x

satisfies

n

have Galois

(n even)

group of

we conclude

that

Zn/2.

b o t h (38) and (99) then

contradicts

of

the hypothesis

k = 0, i

and proves

the

theorem. It seems likely that the foregoing K

any infinite

prove

field,

the following

proved by Schacher

or at any rate

weaker

result,

and Small.

result

if the

a special

holds

for any UD(K,n),

~h&r K ~n ~. case

~

shall mow

of which has been

111

THEOR~

2,

Let

K

be an infinite field,

n a

positive

inteser not divisible by char K but divisible by the cube of a prime. Then

UD(K, n) Proof.

is not a crossed product. Let

K

be the algebraic closure of

that there exists a division algebra of dimension K((tl~t2)).

We have seen also that if

L

K. 2 n

We have seen over

is a field of degree

n

over

K((tl,t2) ) then L = K ( ( ' ~ I , ~ C 2 ) where [ K ( ( ~ I )) : K((tl))] el = el, Z-I = t I, L =K(('~l, q72)) ~ K ( ( ~ I , t2)), [L:~((T~I , t2))] e2 lt2 ' = e2, 772 = ~C1 e2' n = ele 2 K ( ( ? l )) = K((tl))(q[1).

Then

(Lemma 3 of §5).

of

K((tl,t2) ) ~i

tains

over eI

is

(see p. 122).

K((tl,t2) )

distinct

is cyclic of degree [L : ~ ( ( t l , t 2 ) ) ]

elth eI

= e2

is cyclic of degree over

eI

K((tl,t2) )

e2

is

~I_

tl"

G

and the degree of the field Then the minimum polynomial Since the base field con-

over

K((tl,t2) ).

and q~2e2 = ~Cle21t2 over

~((~l,t2)).

then the Galois group of

it is clear that Hence if

L

over

G

L

L

is Galois

K((tl,t2) )

is

It follows as before that if

is a cyclic extension of a cyclic group.

orders.

Also since

contains a Galois subfield of degree

previously that

=tl,

roots of l, it follows that K((~l,t2))

a cyclic extension of a cyclic group. UD(K, n)

el

K((tl,t2) ) = K((tl))(~yl)((t2) )

= ~((tl))((t2))(~Cl ) = K((tl,t2))(~l) over

Since ?i

n

then its Galois group

But we had shown

is a direct product of cyclic groups of prime

To have both of these structures requires that

not divisible by the cube of a prime.

~G~

is

The theorem follows from this •

112

7.

Another result on

UD(K, n), K

an infinite field

We shall now prove a result of Rowents that exponent

n

in the Brauer group over

We ~ l l t h a t K

and

L

if

N

and

N

F,

F

= ML~LNL

has

the center of

UD(K, n).

are modules over a commutative ring

is a commutative algebra over

= L~KN))

UD(K, n)

(Jacobson,

K

then

(M~KN)

L

Structure of Rin~s, p. 103).

Iteration of this gives L

. . .

for K-modules

M i.

If the

M i = Ai

are algebras then the isomor-

phism giving (4) is an algebra isomorphism. to commutative localization. plicative monoid of M

at

(41)

Let

QM?

. . .

S

This can be applied

be a submonoid of the multi-

Then we have seen that the localization of KS is the same thing as M (P. 60). Hence we have

S

(MI(~KM2C~K

K.

"'" ~ M r )

s

~

MIS ~-]KS

"'" ~)Ks MrS

and this holds also for algebras. Now consider the universal In

PI-algebra

is T-ideal of identities of

KIX I.

We recall that

fractions

(F

C

Mn(K ) .

KIX } = K{X}/I n

where

Let

C

be the center of

is a domain and if

F

is the field of

the localization

CC# , C ~'~ = C - [0})

then

UD(K, n)

= KIX} ;~ and the center of UD(K, n) is F. Hence (41) gives C the following formula for the t-fold tensor power UD(K, n) t = UD(K, n ) O F (42)

... ~ F U D ( K ,

n), t

UD(K, n) t ~K{X}tc~ ~

:

factor:

(K{X} ~)C "'" ~ C

The dimensionality of this algebra over Now suppose eij , 1 < i, j _< n t,

U~K,

n) t ~

I.

F

is

Then we have

K {XI)c~ n

2t

.

n 2t

matrix units

Identifying the two sides of (42) we can write

eij = fij s-l' fiJ -CK[X}t'

s ~ C ~.

The relations on the matrix units

give the following relations for the elements

fij

l-1 :

113

( fij I-I) (fk~ I-I) = 8jkS fi~ I-I 2 fii I-I = sl -I. These imply that there exists an element c( fijfk~

-

c -~C ~

such that

~jkS fi~ ) ~ 0

c(X fii ) = (cs)l. We can now prove THEOP~M i. Proof. n2

UD(K, n)

Let

A

has ~

n

L

of

we have a K-algebra homomorphism

define

This maps

da ~ ~ (d)a.

C

into

Now we have the K[X} ~

L

°.. @

~

of

K{X} into

L.

If

d~C

A

KIXI into

A~C...

= ...

.

~(cs) ~ i ~

of

K{X} t

Its image under

we

~t& ~ ® ~ .. t

~cA

®~

:

... (~> A.

A) ~>

C

and we have the (A~c

"" ° ~ c A ) L

is

~it

A.

cs x~l ... ~ i is

= i~

~(cs) ~ C I ~ c

cs ~)i ...

o.. ~ C 1 and this

Hence the same is true of its image

L... ® L 1 under C .

non-zero element of

L

~ (x), x-~KIX} t

It follows that C ((cs)l)

which is the center of

same is true of the elements image

a -CA

Combining we obtain a homomorphism

(cs)l

is invertible.

such that

Then ~ (cs)

C

: K{X} t --> A x~9L .. The element

and

A.

K[X} --> A ~

is a C-algebra (subalgebra of

.., ~ L A.

A

becomes a C-algebra and

C

canonical homomorphism of ~A

D

C-algebra homomorphism

C

Also

K. By ~heorem 3 of §& (p. ii0~

In this way

is a C-algebra homomorphism of (cs)l -~L.

Br(F).

be a central simple algebra of dimenion

over an extension field

(cs) ~ O.

in

by

~(cl) ~.

and ~

A ~ L °'" ~ L A.

(sl).

is a The

Now denote the

Then we have the relations

114

cl (fijfk£ -

~jkSl fizz) = 0

c-1 (~fii) = (cs) l in

At = A ~

that the

gij

a suitable exponent

... ~)LA °

L n°

t m 0 (rood n).

Then if we put

are matrix units we may assume Then

in

that

t ~ 0 (rood n)

Hence the exponent

gij = ( ~ ) - i f i j

we see

A t . We have seen that choosing A

is a division algebra

and so of

UD(K,

UD(K,

n)

n) t ~ i is

n.

of

implies

References

i.

S

A. Amitsur,

On central division algebras , Israel Jour. of Math. vol. 12 (1972), pp. ~0~-~20.

2.

Polynomial identities and Azumaya algebras , forthcoming in Jour. of Algebra.

3.

E. Formanek,

Central polynomials for matrix rings , Jour. of Algebra, vol. 23 (1972), pp. 129-133.

&.

I. N. Herstein, Non-commutative Rinss, Carus Monograph, 1968.

5.

N. Jacobson,

Structure of Rings, AMS Colloquium series, vol. 37, 1956, 2nd ed. 196&.

6.

C. Procesi,

Rings with Polynomia ! Identity, Marcel Dekker Monograph, 1973.

7.

L. Rowen,

O_~nA15ebras with Polynomial Identity,

Yale

Dissertation, 1973. A more extensive bibliography on PI-algebras can be found in Procesl's monograph.