Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann
441 Nathan Jacobson
PI-Algebras An Introduction
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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
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 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
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''''' 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 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
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 {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
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{ 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 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 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 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)_ 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
=