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English Pages 253 [265] Year 1977
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London Mathematical Society Lecture Note Series
Skew Field Constructions P.M. COHN Bedford College University of London
CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE LONDON NEW YORK MELBOURNE
27
Published by the Syndics of the Cambridge University Press The Pitt Building, Trumpington Street, Cambridge CB2 lRP Bentley House, 200 Euston Road, London NWl 2DB 32 East 57th Street, New York, NY 10022, USA 296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia ©Cambridge University Press 1977 First published 1977 Printed in Great Britain at the University Press, Cambridge Library of Congress Cataloguing in Publication Data
Cohn, Paul Moritz. Skew field constructions. l R5 • Given any S-inverting homomorphism f:R --> R', we define I
I
-1
f :R5 --> R by mapping aA to af (a £ R) and s' to (sf) , which exists in R', by hypothesis. Any relation in R5 must be a consequence of relations in R and relations expressing that s' is the inverse of sA.
All these relations still hold
in R', so f' is well-defined and it is clearly a homomorphism. It is unique because its values on RA are prescribed, as well 6
as on (SA)
-1
, by the uniqueness of inverses. •
The ring R.S constructed here 1s called the universal s-inverting ring for the pair R,S.
We have in fact a functor
from pairs (R,S) consisting of a ring R and a subset S of R (with morphisms f:(R,S) --~ (R' ,S') homomorphisms from R to R' which map S into S') to the category of rings and homomorphisms.
All this is easily checked, but it provides no
information about the structure of R8 .
In particular we
shall be interested in a normal form for the elements of RS and an indication of the size of the kernel of A, and here we shall need to make some simplifying assumptions. Let us look at the commutative case first.
To get a con-
venient expression for the elements of R8 we shall take S to be multiplicative, i.e. 1
S and a,b
E
E
S
~
ab
E
S.
Then
every element of RS can be written as a quotient a/s, where a E R, s E S, and a/s some t
E
S.
= a'/s'
if and only if as't
= a'st
for
This is not exactly what one understands by a
normal form, but it is sufficiently explicit to allow us to determine the kernel of A, viz.
ker A
(1)
{a E R
I
0 for some t
at
E
S}.
Ore's idea consists 1n asking under what circumstances the elements of RS have this form, when commutativity is not -1 assumed. We must be. able to express s a (for a E R, s E S) as a 1 /s 1 , where a 1 E R, s 1 E S, and multiplying up, we find as 1 = sa 1 • More precisely, we have as 1 /l = sa 1 /l, whence as 1 t • sa 1 t for some t E S. This is the well known Ore condition and it leads to the following result: Theorem 1.2.2.Let R be a ring and S a subset such that D.l
S is multiplicative,
D.2
For any a
E
R, s
E
S, sRnas f.(), 7
D.3
For any a
£
R, s
£
S, sa
=0
~
at = 0 for some t
S.
£
Then the universal S-inverting ring R 8 may be constructed as follows:
On R x S define the relation
(a,s) - (a' ,s') whenever
(2)
for some u,u'
E
a' u', su
s'u'
E
S
R.
This is an equivalence on R8 •
au
R x S and the quotient R x S/- is
In particular, the elements of R8 may be written as
fractions a/s =as-land ker
The proof
~s
A is given by (1).
a lengthy but straightforward verification,
which may be left to the reaaer.
It can be simplified a
little by observing that the assertion may be treated as a result on semigroups; once the 'universal S-inverting
se~
group' R8 has been constructed by the method of this theorem, it is easy to extend the ring structure to R8 • A subset S of a ring R satisfying D.l-3 is called a right denominator set. Wheii R is connnutative, p.Z-3 are automatic and may be omitted. S
=
If R is entire, D.3 may be omitted and if moreover,
R*, then D.2 reads aRnbR # 0 for a,b # 0. This was the
case actually treated by Ore [31] and R is then called a right Ore domain.
Apparently these results were found inde-
pendently by E. Noether, but not published.
There have been
many papers dealing with generalizations, e.g. Asano [49]; for a survey see Cohn [71']. It is important to observe that the field of fractions of a right Ore domain
~s
essentially unique.
that the construction is functorial.
Let us first note
Thus, given a map be-
tween pairs f:(R,S) --> (R' ,S'), i.e. a homomorphism f:R --> R' such that Sf
C S', then we have the diagram shown, and by
universali~y
of R8 there is a unique map f 1 :R8 --> R~, such that the resulting square connnutes. In particular, if f is an isomorphism, so is f 1 • 8
A
R------~
'1
A'
Rr--------------~
Rs lfl RS'
So far R,R' have been quite general; suppose now that R ~s a right Ore domain and K is any field of fractions of R,
thus we have an embedding
f:R --> K.
If S
= R*,
we have
a homomorphism £ :R --> K, which we claim is injective. -1 l s -1 -1 For if as E ker f 1 , then 0 (as )f 1 = (af)(sf) , hence af = 0, and so a = 0, because f is injective. It follows that f 1 is an embedding; the image ~s a field containing R and hence equal to K, because K was a field of fractions. Thus £ 1 is an isomorphism and we have proved Proposition 1.2.3. The field of fractions of a right Ore domain is unique up to isomorphism. •
The result is of particular interest because it ceases to hold for more general rings; we shall soon meet rings which have several non-isomorphic fields of fractions.
1.3
Skew polynomial rings Given a commutative field k, there are four important con-
structions involving an indeterminate that we can perform. We can form polynomials, rational functions, formal power series and formal Laurent series.
These constructions are
all well known in the commutative case, and the relations between the four rings so obtained may be summed up in the following commutative diagram k(x)
- - - - - - - - > k((x))
r
k
[x]
---------> k
C: x]
Each can be generalized by taking the field k to be skew, and taking the indeterminate x to be central, but that is not the most general (nor the most useful) choice. Starting from an entire ring A, let us ask for a ring R whose elements can all be uniquely expressed as polynomials 9
(1)
(a. £A).
f
1
As usual we write deg f
= n if a n # 0 in (1).
The additive
group of R is just a direct sum of copies of A (by the uniqueness of (1)) • To multiply two elements, .say f (1) and g
=
~x
1
. by a., g1ven 1
•
.
= ~xjb. we have, by distributivity, fg = ~x 1 (a.xJ)b. J
.
and so it will only be necessary to prescribe a.xJ. 1
J
1
To ensure
that R is again entire, let us assume that (2)
deg fg
deg f + deg g.
Then in particular, ax for any a
£
A has degree at most 1,
so (3)
ax
xa
0 + a ,
a.
a.
0 a ' a 1-> a are mappings of A into itself. This is already enough to fix the multiplication in R, for
where a
1-»
now we can work out ax axr
=
r
by induction on r:
2 2 (xa a. +a o)xr-1 = [x 2 a a. + x(a a.o +a oa )+a o Jx r-2 =
We derive some consequences from (3): (a+ b)x
= x(a
+ b)a+ (a+ b) 0 ,
ax+ bx
hence
(4)
and (ab)x
so 10
a
o
x(ab) + (ab) , a(bx)
...
aba
(ab) 0
(5)
a
Further, lx
xl, therefore
(6)
•
1,
cS a cS a b + ab .
0,
and since ax has degree 1 for a f 0, (7)
a
a
0
==>
a
= 0.
From (4)-(7) we see that a is an injective endomorphism of A and
cS
~s
an a-derivation of A, i.e. a mapping such that (ab) 0
(8)
6 a. 6 a b + ab •
We note that (8) entails 1° = 0, by putting a= b = 1 ~n the second equation.
Conversely, let A be an entire ring with an
injective endomorphism a and an a-derivation
o.
Then the set
of all expressions (1) can be made into a ring by defining addition componentwise and multiplication by the commutation rule (3).
The resulting ring R is again entire (because the
degree on it satisfies (2)).
It is called the skew polynomial
ring in x over A (associated with a,o) and is denoted by
A[x;a.,o]. When
o
0 we also write A[x;a] instead of A[x;a,O]; if more-
l, we obtain the polynomial ring in a central in-
over a
determinate over A, also written A[x]. In matrix notation the commutation rule (3) can be written
a(x
1)
(x
11
and the conditions (4)-(6) may be summed up by saying that the mapping of A into the matrix ring A2 defined by
(9)
~s
a
\->
c: :)
a ring homomorphism.
More precisely, it is a homomorphism
into the ring of lower 2
X
2 triangular matrices over A, which
in a suggestive notation may be written
(! ~)·
Suppose now
that A is a right Ore domain with field of fractions K, then any injective endomorphism a of A extends to a unique endomorphism of K, again denoted by a.
An a-derivation 8 defines
a homomorphism (9) which by functoriality extends to a homomorphism K
--~(~ ~),
say u
\--~{~~ ~).
is an a-derivation on K extending stead of u'.
~
Clearly u
1--~
u' 8 and we shall write u in-
Thus we have shown (using (9)) that any a-deri-
vation of a right Ore domain extends to a unique a-derivation of the field of fractions.
This remark will be useful later.
Let A be any ring with an endomorphism a, then for each c
£
A the mapping ~ :a 1---~ ac - caa c
is easily seen to be an a-derivation; it ~s called the inner a-derivation induced by c. is called outer.
A derivation which is not inner
The construction of the skew polynomial
ring shows that any a-derivation 8 on A may be made inner by going over to A[x;a,8] (which can of course be defined even if A has zero-divisors and a fails to be injective). In this ring
o is
inner on A, say o A[x;a,o]
=
induced by x; moreover, if 8 was already
= ~c
then on writing y
=x
- c we have
A[y;a], as is easily checked.
Let K be a field, then any endomorphism of K is injective, 12
for its kernel is a proper ideal of K.
Thus for any en-
domorphism a and any a-derivation 6, the skew polynomial ring R = K[x;a,o] is entire.
As in the commutative case we can
show that this is a principal right ideal domain.
For let
a be a right ideal; if a 1 0, pick a monic polynomial f say,
of least degree in a •
Then every g E a can be written as
g = fq + r, where deg r < deg f. follows that r = 0 and so g
E
fR.
Since r = g - fq e:: a , it Thus a
fR and we have
proved the first part of Proposition 1.3.1.
Any skew polynomial ring K[x;a,o] over
a field K is a principal right ideal domain.
It is a princi-
pal left ideal domain i f and only i f a is an automorphism.
To prove the second part, assume first that a is an automorphism. Then if aa = b, we have a= b 6 , where S = a-l and the commutation relation (3) can be rewritten as b 6x = xb + b 66 , i.e. (10)
xb
Now it follows by symmetry that R is a principal left ideal domain.
Conversely, assume that a is not an automorphism,
then it is not surjective and so there exists c e:: K, c
i
Ka.
We assert that RxnRxc = 0, for if not, then for some f,g e:: R*,
(11)
fx
gxc.
Comparing degrees we see that deg f = deg g f ~ xna + •.• , g
=
. in (11) we f~nd aa
n say.
Let
x~ + ••• , then by comparing highest terms = b a c, hence c = ( b -1 a )a , an d t h.~s contra-
dicts the choice of c. ~.e.
=
So we have shown that RxnRxc
R is not a left Ore domain.
=
O,
Now the assertion follows
from the fact (proved below) that every left Noetherian domain is a left Ore domain.
For if R were left principal,
RUNT LIBRARY CARNEGIE Yr~ L~1 -::.·;>~,:n PITTSBURGH, PEN~:~YlYANIA 1521
13
it would be left Noetherian and hence left Ore, but we have just seen that this is not so. • We still have to prove the assertion made in the course of the proof about left Noetherian domains.
Translating to the
right, we find that what we need is Proposition 1.3.2.
Proof.
Any right Noetherian domain is right Ore.
Let R be a right Noetherian domain and a,b
E
R*, then
Loo aibR is finitely generated, hence for some n ~ 1, 0
.
n-1
a~
bc 0 + abc 1 + ••• +a
F 0,
Since a~
bcn-l"
not all the ci vanish; let ck be the first
non-zero coefficient, then we can cancel ak and obtain
so bck e: aR, i.e. aRnbR
r
0, and this is just the right Ore
condition. • This proof (which goes back to Goldie) actually shows a In an entire ring R, if aRnbR = 0 for a,b :f 0,
little more:
then the elements a~ (n dependent over R.
=
0,1, ••• ) are right linearly in-
Fo~ if Laibci = 0, and ck is the first
non-zero coefficient, then as before we find that bck e: aR. Hence if aRn bR
=
0, we can find a free right ideal of
countable rank in R. Corollary.
This proves the
An integral domain is either a right Ore domain
or it contains free right ideals of countable rank. •
Prop. 1.3.1 shows that any skew polynomial ring K[x;a,o] over a field is right Noetherian, and hence right Ore by Prop. 1.3.2.
Therefore it has a unique field of fractions,
which we shall denote by K(x;a,o). There
~san
interesting application of the last Cor., due
to Jategaonkar [69'] and independently to Ko~evoi [70].
14
Proposition 1.3.3.
Let R be an entire ring with centre C.
Then R is either a left and right Ore domain or it contains a-free C-algebra of countable rank.
Proof.
Suppose that R is not right Ore; it will be enough
to find two elements x,y which are free, for then the elements 1
x Y form an infinite free generating set.
We choose x,y e: R*
such that xRrl yR = 0 and claim that the C-algebra generated by x and y is free.
If not, let f
=0
be a non-trivial re-
lation of least degree between x and y. (12)
a + xa + yb
This has the form
a, b e: R, a e: C.
0
Here a,b are not both zero, s1nce they have lower degree than Say, b f 0, then on multiplying (12) on the right by x we
f.
have ax+ xax + ybx = 0, i.e. ybx = x(-a-ax) e: xRrlyR, and ybx f 0, a contradiction, which shows x,y to be free over C. • The result can be used to embed a free algebra 1n a field (as both Jategaonkar and
Ko~evoi
have observed).
For let K
be a field with a non-surjective endomorphism a (e.g. the rational function field k(t) with endomorphism f(t) and form the skew polynomial ring R
=
K[x;a].
1--> f(t 2 ))
This 1s an
entire ring; by Prop. 1.3.1 it is not left Ore and so it contains a free C-algebra on two free generators, where C is the centre of R.
But R is right Ore and so it can be
embedded in a field; this then provides an embedding of the free algebra (of countable rank) in a field. In spite (or perhaps because) of its simplicity this construction is of limited use, because not every automorphism of C can be extended to an automorphism of the field of fractions, constructed here.
One application of this con-
struction, due to J.L. Fisher
[71]
is to show that C
has many different fields of fractions.
Let A= k[t] be the
usual (commutative) polynomial ring with endomorphism an: f(t)
1-->
f(tn), and consider the subring of R
=
A[x;an] 15
generated by x and y
=
xt over k, for n > 1.
in the image of k(t) under
Since t is not
n' it follows as in the proof of Prop. 1. 3 . 1 that Rx () Ry = 0, hence the subalgebra on x ct
and y over k is free, and for different n we clearly get distinct (i.e. non-isomorphic) embeddings, because x
-1
16
yx
tx
xt
n
= x(x-1 y) n = (yx-1 ) n x, so x -1 y = (yx-1 ) n •
2· Topological methods
2.1
Power series rings In the commutative case the familiar power series ring
k
[X]
may be regarded as the completion of the polynomial
ring with respect to the "x-adic topology", i.e. the topology It
obtained from the powers of the ideal generated by x.
~s
no problem to extend this concept to the ring k [ x; a J, but when The
there is a non-zero derivation we face a difficulty.
above topology may be described in terms of the order-function. o(f)
=r
if
n r r+l x ar + x ar+l+ • • • + x an
f
(a
r
+ 0).
Now it turns out that when 6 # 0, multiplication is not continuous in the x-adic topology, as the formula (1)
a.x
xa
a
+ a
6
shows, and any attempt to construct the completion directly will fail. y
=
One way out of this difficulty is to introduce
x -1 and rewrite (1) as a commutation formula for y.
We
then get (2)
ya
a 6 a y + ya y.
Owing to the inversion we now have to shift coefficients to the left (unless of course a happens to be an automorphism) and as (2) shows we cannot usually do this completely in the polynomial ring, but we can do it to any desired degree of
17
accuracy by applying (2) repeatedly: a 6a 2 o2 2 a y + a y + ya y
ya (3)
If 6 is locally nilpotent, i.e. for each a n such that
a
on
E
k there exists
; 0, this can be used as a commutation for-
mula in the skew polynomial ring. been studied by T .H.M. Smits
[68]
This kind of formula has (cf. also Cohn
[71"] ,Ch.O).
But in any case, in the power series ring we can pass to the limit in (3) and obtain the formula (4)
a oa 2 62a 3 a y +a y +a y + •.•
ya
The ring obtained in this way 1s clearly an integral domain and the set consisting of powers of y is a left denominator set, by (4), so we can
for~
the ring of fractions, which is
in effect the ring of (skew) formal Laurent series in y. If K is a field with a surjective derivation 6, then K(x;l,6) is a field with a surjective inner derivation, and we can also obtain a field with this property by taking skew Laurent series. -1 X
c
6
.
-1
Thus if the commutation formula 1s ex
6
c - c , then on writing [a,b] ; ab - ba, we have [x-l ,c]; and hence [ X-1 ,LX i
c.] 1
LXi [ X-1 ,c.
1
J;
LXi c 6.. 1
Since 6 is surjective, every element has this form and the assertion follows.
To get a surjective derivation we need
only take a function field in which every function 1s a derivative, or more algebraically, a differentially closed field (cf.e.g. Sacks [72], Shelah [72]).
This answers a
question first raised by Kaplansky [70]; he asked for a 18
field in which every element 1s a sum of commutators and B. Harris
[ss}·
answered this by constructing a field in which
every element is a commutator.
Such a field is the union of
the ranges of its inner derivations, and it can also be constructed very simply as follows (Bokut' [63]): field K, the rational function field L
= K(x)
1ndeterminate x admits the derivation f
\-->
=
1
=a
1, hence [ax,y]
for any a
in a central f'
derivative) and in the skew function field K have [x,y]
Given any
€
(the usual
= L(y;l,') K.
we
If we re-
peat this process we obtain a field K2 :::) K1 and every element of K1 1S a commutator in K2 . Thus ue can form an ascending chain K1 C K2 C .•• whose un10n 1S a field which 1S the un10n of the ranges of its inner derivations. In finite characteristic Lazerson [61] has constructed a field with a surjective inner derivation: Given any field k, of characteristic p I 0, adjoin commuting indeterminates k(x 1 ,x 2 , .. ) with derivation 8 such that 0
0
xi = xi-1 (i > 1), x 1 = 1. Put L = K(t;l,o) then o is induced by t, hence oq is an inner derivation induced by tq, for any q = pn , and it annihilates anything involving only x1, ••• ,xq-1' that [a, tq] surjective.
Thus, given a
€
L, there exists q
0 and so [ax ,tq] =a. q
=
pn such
This shows that 8 is
By taking ultra-products of such fields for
varying p we obtain fields of characteristic 0 with surjective derivations.
For other constructions see Cohn
[73'"]. There is an important generalization of the power ser1es method, to which we now turn.
Let G be a group and consider
the group algebra kG over a commutative field k. kG embeddable in a field?
When is
Clearly a necessary condition is
that it should be entire, and for this it is necessary for G to be torsion free.
For if u
€
G is of order n, then
(u- 1)(un-1 + un-2 + ••• + u + 1) =
o· 19
In the abelian case this condition on G is also sufficient. For if G is torsion free abelian, it can be totally ordered (regard G as Z -space, embed it in a Q-space and use a lexicographic ordering with respect to an ordered basis). G is totally ordered, kG is clearly entire:
When
Let a= a 1 s 1
+ ••• +as
with a. £ k, s. £ G and s 1 < ••• < sm' similarly mm 1. 1. b = b 1 t 1 + ••• + bhtn (bj E k, tj £ G, t 1 < ••• < tn)' then ab = a 1 b 1s 1 t 1 + ••• ,where the dots represent terms greater than s 1 t 1 , hence ab ~ 0. Thus we have Proposition 2.1.1. Let G be an abelian group, then the group algebra kG (over any commutative field k) is embeddable in a field if and only if G is torsion free. •
In the non-commutative case little is known; it is not even known whether kG is entire for any torsion free G. But Farkas and Snider [76] have recently proved that this
l.S
the case when G is polycyclic (i.e. soluble with maximum condition on subgroups); since kG is Noetherian in this case, it is then embeddable in a field.
In another direction
J. Lewin and T. Lewin [a] have shown (using methods of Magnus
and some results from Ch.4 below) that for any torsion free group G with a single defining relation the group algebra kG can be embedded in a field. It has long been known that Prop.2.1.1 can be generalized to non-abelian groups which are ordered.
In that case we can
form a kind of power series ring k((G)) which turns out to be a field.
This uas first proved by Hahn [07] for abelian
groups with an archimedean ordering and then generally by Mal'cev[48] and independently, Neumann [49'].
The result was
put in a general algebraic setting by Higman [52]; his proof (with some simplifications) is given in Cohn [65].
Let us
briefly describe the construction without entering into the details of the proof. Consider the k-space kG of all k-valued functions on G; it contains the group algebra as subspace, in fact a 20
= (a ) g
belongs to the group algebra precisely if its support D(a)
{g e G
is finite.
I
a
g
j: 0}
Now the multiplication on kG cannot in any
natural way be extended to kG; if a= (ag)' b = (bg)' then we should have
and there is no guarantee that the sum on the right is finite.
Let k((G)) be the subset of kG consisting of all
elements with well-ordered support (in the ordering on G). If a,b
k((G)), then the sum on the right of (5) is finite
E
for-each g, for the h such that ah f 0 form an ascending chain, so the h
-1
g (g fixed) form a descending chain and
hence only finitely many such bh-lg are non-zero.
More-
over, it is easily seen that ab s k((G)), so that the latter forms in fact a ring, with kG as subring.
The theorem
of Mal'cev and Neumann asserts that k((G)) is actually a Thus each element of k((G)) has the form Za u and
field. if u
0
u
s G
~s
the least element for which a
can write f =a
u
only of elements 0 1 + g + g
2
u
'f 0, then we 0
u (1- g), where g has support
• • cons~st~ng
0
>
1.
Now what has to be proved is that
+ ••• e k((G)); once that is established we
clearly have
f-1
=
(1 + g + g2 + ..• ) u
-1 -1 a 0
uo
In order to apply this result to embed free algebras in fields we use the fact that the free group can be totally ordered (cf. e.g. Fuchs [63] ).
Briefly the proof goes as
follows: Let F be. the free group on a set X, taken to be finite 21
for simplicity, and let b 1 ,b 2 , •.• be the sequence of basic commutators in X. Denote by yt(F) the tth term of the lower central series ofF and let b ,b , •.• ,b
be all the basic w commutators of weight < t, then as is well known, every ele1
ment a
£
2
F has a unique expression
(a.
l.
£
z)
(cf. e.g. M. Hall [59]), and we can therefore represent each a
£
F by an infinite product
Now write a> 1 whenever the first non-zero of a 1 ,a 2 , ••• is positive; this provides a total ordering of F. The same method works for free groups of infinite rank (taking the free generating set to be well-ordered). It is clear that k, the free k-algebra on X, can be embedded in kF, the group algebra of the free group on X, and since kF is embedded in the 'power series field' k((F)) just constructed, we have another embedding of kin a field. If instead of the free group F we take the free metabelian group G, i.e. the group defined by the law ((u,v),(w,x)) (where (x,y)
-1 -1
=x y
least when card (X)
=1
xy), we can still embed kin kG, at
= 2, the crucial case (Moufang [37]).
Moreover, G can again be ordered, so we have another embedding of k in a field, and these two embeddings of k in k((F)) and k((G)) are clearly distinct. Of course there are other simpler ways of finding nonisomorphic fields of fractions of k, e.g. that by Fisher described earlier.
We note that 1n the above construction
the free metabelian group cannot be replaced by a free 22
nilpotent group of any class, for the free semigroup on X cannot be kmbedded in any nilpotent group on X, by results of Mal' cev (cf. Lyapin [60]). As an application of the Mal'cev-Neumann construction we derive the one-sided principal ideal domains first obtained by Jategaonkar.
We have seen that a polynomial ring over a
field is a principal ideal domain, and it is clear that the condition is necessary, i.e. if a polynomial ring is principal, the coefficient ring must be a field.
This is true
even for skew polynomial rings relative to an automorphism, but for a non-surjective endomorphism it need not hold. precise conditions were determined by Jategaonkar Theorem 2.1.2. put R = A[x;S].
The
[69]:
Let A be a ring with an endomorphism Sand Then R is a principal right ideal domain i f
and only i f A is a principal right ideal domain and S maps
A* into U(A), the group of units of A. Proof.
If R is principal, so 1s A because it is a retract
of R, i.e. a subring which 1s also a homomorphic image (put x = 0).
Further, for any a
E
A* we have aR + xR = cR, where
c 1s the highest common left factor of a and x.
It follows
that c has degree 0 (as factor of a), sox= cf, where f has degree l, say f shows that ce
xd + e. 0 , c sd
= 1,
s
Now x = cxd + ce = xc d + ce, which . a un1"t. so c s 1s
Let au + xv = c,
then putting x = 0 we see that a is associated to c, hence a
s
. so a s 1s . a un1t, . as c l a1me . d 1s associated to c s , a un1t, . Conversely, if the given conditions hold, R is clearly entire.
Let a .be a right ideal in R.
When a
= O,
there is
nothing to prove; otherwise let n be the least degree of polynomials occurring 1n a .
The leading coefficients of
polynomials of degree n in a form with 0 a right ideal in A, generated by a say. Let f xna + • • • e: a , then aS e: U(A) n+l S . as highest coeff1c1ent. · · and hence fx x a + •.. has a un1t It follows that a contains a monic polynomial of degree n+l and so also of all higher degrees.
Now it is clear that
23
a
= fR,
hence R is a principal right ideal domain. •
This result shows that under favourable circumstances one may iterate the polynomial ring construction and still get a principal right ideal domain, and this suggests the following definitions.
By a J-skew polynomial ring one
understands a skew polynomial ring A[x;S] such that S is injective and satisfies Jategaonkar's condition: AS ~ U(A) U {0}. E.g. this condition holds whenever A is a field; what is of interest is that there are other cases.
It is easily seen
that any J-skew polynomial ring over A is entire if A is. Now let R be a ring and called a J-ring of type (a < T) such that
T
T
an ordinal number, then R is
if R has a chain of subrings R
a
(i)
R = U(R) U {0}
(ii)
Ra+l is a J-skew polynomial ring over Ra for all aa). - r
It ~s easily verified by induction that U(R ) a we have the Corollary.
U(R ), hence 0
Any J-ring (of any type T) is a principal right
ideal domain. •
It turns out that J-rings can be characterized as integra: 24
domains with Euclidean algorithm (generally transfinite) and unique remafnder (cf. Lenstra [74]). any '
there are J-rings of type T.
We shall see that for Such rings form a useful
source of counter-examples; they were constructed by Jategaonkar to provide examples of (i) a principal right ideal domain in which there are non-units with arbitrarily long factorizations (only rather special examples, in effect Jrings of type
2,
were known earlier, cf. Cohn
[67] ),
(ii) a
ring with left and right global dimensions differing by an arbitrary integer (the largest known difference had been 2 before, cf. Small
[65]),
(iii) a left but not right primi-
tive ring (such a ring was first constructed by Bergman
[56],
answering a question of Jacobson example is more direct.
[64],
but Jategaonkar's
See also Brungs
[69]
for some re-
markable properties of this construction). Skew polynomial rings over a field are J-rings of type l; J-rings of type 2 can be obtained by an ad hoc construction (Cohn
[67])
but beyond this the general case is no harder
than the finite case.
Moreover, one cannot use induction
directly, s1nce the coefficient ring depends essentially on the order type.
Jategaonkar uses an ingenious argument in-
volving ordinals; below is a direct proof based on the Mal'cevNeumann construction (cf. Cohn [71 11 ] ) . We observe that to achieve the form (6) we need a commutation rule of the form (S
=0
=> x
= 0.
Then the
ax is injective, and it 1s clearly right
K-linear, on a finite-dimensional K-space, hence it is sur-
= 1 for some b £ A. Now b 1s again a left non-zerodivisor: if bx = 0, then x = abx = 0. Hence there exists c c A such that be = 1, and so c = abc = a and this shows that ab = ba = 1, i.e. a is a unit. The rest is clear.• jective, and so ab
There is one important case where left and right degrees are the same. Theorem 3.1.2. centre.
Let K be a field of finite degree over its
Then the left and right degrees over any subfield
coincide.
Proof.
Let E be any subfield of K and denote the centre of
K by C.
By hypothesis K is a C-algebra of finite degree,
and it is clear that A subalgebra.
EC
= {Lx.y. 1 1
Jx.1 E E, y.1 E C} is a If we regard A as E-ring, we can choose a
basis of A as left E-space consisting of elements of C; then this will also be a right E-basis for A, hence
(2)
Now A 15 a c-algebra of finite degree, entire as subalgebra of K, hence A is a field.
By (1) 31
[K:C] .. [K:A]L [A:CJ Since [K:C] is finite, so is [A:C].
I f we divide by [A:CJ
and multiply by (2) we get (on using (1) again)
3.2
The Sweedler predual and the Jacobson-Bourbaki correspondence The aim of this section is to establish the Jacobson-
Bourbaki correspondence theorem, used later for Galois theory and also useful elsewhere.
We shall first prove the
Sweedler correspondence theorem on corings, see Sweedler [75]. We begin by explaining the notion of a coring. a ring.
Let A be
By an A-coring we understand an A-bimodule M to-
gether with A-bimodule
~ps ~:M
--> M ~AM, E:M --> A, such
that the following diagrams commute:
/I~
M ~ M l.E
Example.
Let
~:A-->
> M
im(U
~
V/V', the kernel of a V') ).
An element g
£(g)
=
1, 6(g)
~n
=
g
B is im(U'
~
V) +
a coring M is called grouplike if ~
g.
E.g. the standard B-coring B
has the standard grouplike 1 Proposition 3.2.1. grouplikes.
~
~
~AB
1.
Any coring map takes grouplikes to
Given a homomorphism ¢:A --> B, let C
=B
~AB
be the standard B-earing over A, then for any B-earing P there is a natural bijection (of sets)
where Hom denotes the set of B-earing maps and
{g E
Proof.
Pj
g grouplike and ga
E
A}.
The first part is clear, to prove the second we note
that under any coring map C --> P, 1 like centralizing A. Lai ~ b i
ag for all a
~
1 maps to a group-
Conversely, given g £ GA(P), the rule
j --> Laigb i defines a map C --> P which is easily
seen to be a coring map.• We shall often write gB/A for 1 ~ 1 in B ~AB. It is a natural question to ask if every B-coring is standard 33
Since B GA B ~s generated as B-bi-
over some subring.
module by the grouplike gB/A' we shall limit ourselves to corings generated by a single grouplike.
Then the answer
is 'yes', provided that B is a skew field: Proposition 3.2.2.
Let K be a skew field and M any K-coring.
Given a grouplike g c M, write D
= {x
K
£
I
xg
= gx},
then
Dis a subfield of K and the standard coring map (Prop. 3.2.1)
(1)
1-»
1 G 1
l';:K~K-·»M
is injective; i t is an isomorphism i f M
Proof.
Clearly D
~s
a field.
= La.
then there exists x
~
Suppose
~b.
~
~
1';
g
= KgK. is not injective,
0 such that
n
La.gb. 1 ~ ~
(2)
(a., b.
0
~
~
£
K),
and we assume that n, the number of terms in (2), is minimal. Moreover, 0
= s(La.gb.) = La. b., hence n > 1 and after multi~ ~ ~ ~ -1 plying by al we may take al = 1. Now by minimality a 1 = 1, a2, . , . , an are right D-independent and a2g # ga2, hence there
is a right K-space map G:M -» K such that G(a g - ga2) ~ 0. 2 Consider x' = Ln1e(a.g- ga.) ~b .. By the independence ~ ~ ~ of the bi we have x' ~ 0, but a 1 g- ga 1 0 and so x' has fewer than n terms.
c
Now let us chase x around the diagram
c ----»
M
j
j
~
where C
c
----»
= K ~K.
0.1 M ~ M ----~> K
~
34
L8(a.g)gb. l.
~
M
M
Going across and down we get 0, going down
and across we get EG(a.g)gb., hence (3)
~
o.
~
Now ~(x')
= EG(a.g1.
= E0(a.g)gb. 1. 1.
ga.) gb. 1.
1.
- E~(g)a.gb .. 1.1.
Here the first sum vanishes by (3) and the second by (2). Thus ~(x') diction.
= 0, but x'
~ 0
and x' has < n terms, a contra-
Therefore (1) is injective; clearly it is sur-
jective, and hence an isomorphism, whenever M = KgK.• Taking the case M = KgK, we get the Corollary.
Every K-coring over a field K, generated by a
grouplike is standard.
We can now prove the first correspondence theorem: Theorem 3.2.3 (Sweedler correspondence theorem). skew fields F ~ K, let M = K over F, write
C
~
F~D~K,
preserving bijection C ~>
cx:D
~:J where n:M
Proof.
1-> 1->
K be the standard K-coring
for the set of coideals in M and
set of fields D such that
D+
ker(M
J+
{x
E
I
V
for the
then there is an order-
V defined by
=K~
K
Given
K -> K
x. ~(gK/F)
~
=
K),
~ M/J.
As kernel of a coring map D+ is a coideal.
If L is
= ex} is clearly and it contains F. a subfield of K, so J is a subfield of K, + JD is the kernel of the natural aS = 1. Given D £ v, D ++ map K ~K -> K~ K and ~IF 1-> gK/D' Now x £ D 1 Q X < > X £ D. X Q 1 any K-bimodule and c
£
L, then {x
£
K I xc
+
~ex
1.
Given J e: C, put D = J+ and let n:M -> M/J be
the natural map, then c
=
n(~/F) centralizes D, by defini-
=
Moreover, c generates M/J, hence M/J K~ K by Prop. 3 . 2 . 2 , wh ere the homomorphisms of M correspond 1n + ++ this isomorphism, so J = D = J ·• In this correspondence F was any field, e.g. we can take tion of D.
35
it to be the prime subfield P of K, then we obtain a bijection between the coideals of K
~
K and all subfields of
K. To obtain a case of Jacobson-Bourbaki correspondence we need some facts on duality. we write HomA_(M,N),
Given A-bimodules M, N,
Hom_A(M~N), HomA,A(~~N)
for the set
of all left-A, right-A and A-bimodule homomorphisms.
Further
we put *M = HomA_(M,A), M* = Hom_A(M,A), *M*
= *Mn M* = HomA,A (M,A) .
E.g. if MA is free of finite rank, then *(M*)
=M,
as is
well known. Let M be an A-coring, then *M has a ring structure as follows: For f,g composition
E
*M, their product is defined as the
M
__g__> A.
Thus f.g: u I--~ I(uilui 2 ) if ~(u) = Iuil 9 uiZ' It is easily verified that this is indeed a ring structure. Simif
g
larly M* is defined as a ring: the composition of f,g is defined as
E
M*
f f.g:M ---> M 9A M g 9 1 > A 9A M = M - > A.
Here f.g maps u to I(ur1 ui 2 )f We note that for f,g s *M* both definitions reduce to Iu~ u~ thus both *M and M* 11 12' contain *M* as subring. Example.
Let C
Hom_B(B 9A B,B) C*
36
B 9A B
be a standard B-earing, then C*
Hom_A(B,Hom_B(B,B))
End_A(B),
=
Hom_A(B,B).
Thus
=
and similarly *C =End .
A-
(B), while *C* =End
latter is essentially the centralizer of
A,A
~(A)
(B)
The
'
in B.
It is clear that if f:M ~ N is a coring map, then f*: M* ---> N* and *f:*M ---> *N are ring homomorphisms.
E.g.,
for any B-earing M, e::M ---> B is a coring map, hence e:* :B ---> M*
1--->
is a ring homomorphism; explicitly, e:* :b corresponds to the map m
1--->
Ab£' i.e. b
e:(bm).
Now let K be any field and End(K) the ring of additive group endomorphisrns of K.
In End(K) we have the subrings
p(K), A(K) of right and left multiplications, and as is well known, these are each other's centralizers, thus p(K)=EndK_(K) A(K)=End_K(K).
K
Further, End(K) as subset of K can be regarded
as a topological space, the topology being induced by the K
product topology on K , taking K with the discrete topology. This is sometimes known as the topology of simple convergence; iff e: End(K), a typical neighbourhood off consists of all ~
e: End(K) such that for a given finite set c 1 , ••• ,cn £ K, c.f =c.~. In particular, this shows every centralizer to ~
.
~
be closed . We shall need one more auxiliary result: Proposition 3.2.4.
Let K be any field.
Given a subring F
of End (K) such that p (K) c;;;; F c;;;; End (K), define
D
{x e: K
I
{x e: K
I (xy)f
A centralizes F}
then (i) D = {x e: K
X
I
x.yf for all ye:K,fe:F}
xf = x.lf for all f
£
F}, (ii) the
centralizer of F in End(K) is A(D) and hence D is a subfield of K, (iii) p (K) c;;;; F ~ EndD- (K) •
Proof. (i) If xf
= x.lf,
then (xy)f
Xp f y
x.lp f y
=
x.yf 37
because p(K)C F.
(ii) Since p(K)C F, the centralizer of F
is contained in A(K); in fact the definition of D states that A f =fA , thus the centralizer is A(D). X
X
Now the rest
is clear. • Theorem 3.2.5 (Jacobson-Bourbaki).
Let K be a skew field
and End(K) its endomorphism ring, as topological ring.
Then
there is an order-reversing bijection between the subfields
D of K and the closed p(K)-subrings F of End(K), defined by the rules
(4)
D
1->
End0 _ (K) ,
F
1-> D
{x
g
K
I
xf = x.lf for f
E
F}.
Further, i f D and F correspond, then
whenever either side is flnite.
Proof.
Given F, defineD as in (4).
forK and for x y. 0
X
Let X be a left D-basis
define ox e End 0 _(K) by
E X
1
ify
0
if y "' x.
x,
Then the oX are right K-linearly independent, for if ~o X a X = 0 (a X E K), we can apply this toy EX to get a y = 0, so the relation was trivial. Moreover, if [K:D] 1 < ~, the
ox form a basis for End 0 _(K) as right K-space, because then f
= ~ox.xf,
(6)
as is easily checked.
[End0 _ (K) :K] R
if the right-hand side is finite.
38
Therefore we have
Now F ~ End 0 _ (K) , so
[F:K]R is finite if [K:D]L is, by (6). To show that
:
(7)
F = EndD- (K) ,
we use Jacobson's density theorem [56] to deduce that F, as p(K)-subring, is dense in EndD_(K) and being closed is therefore the whole of EndD_(K).
In the restricted but still im-
portant situation where [F:K]R
< ""•
(7) can be proved from
Prop. 3.2.2 by verifying that the map K
:K ~ K
1s injective.
F*,
-;:>
It is of interest that for any p(K)-subring F
the injectivity of K would follow from the density of F in Conversely, if
EndD- (K).
* (F*) -;:. * (K ~K)
= EndD- (K)
is surjective, then since F is dense in *(F*), it follows that F is dense in EndD_(K); thus the injectivity of of
th~
K
is a predual
density theorem.
To outline the proof, let [F :K]R < ""· n:F*
~ F*
->
(F
~)*,
~DW
Then the map
1->
(f 9 g
1->
(fwg)~ )
1s an isomorphism and F* has a coring structure given by _ 1
¢1--> (lF)¢, and 6:F* (mult)* > *(F*) = F, as rings. Now *K is an
~K F)* _n___> F*~
e::F* --> K,
(F
Moreover,
injective ring
homomorphism, hence Let g
=
K(g) e: F*.
~/D £
K
K =
~D
(*K)* is a surjective coring map. K be the standard grouplike and c =
Put D1 = {x e: K
cb = K(l 9 b) and be= K(b (cb)f
= bf,
(bc)f
I ~
xc =ex}, then D1 2D; if be: K, 1), hence for any f e: F,
= b.lf, so if be: D1 then bf
=
b.lf, i.e.
b e: D by Prop. 3.2.4 (i); this shows that D1 =D. Further, F* = KcK, hence K is an isomorphism by Prop. 3.2.2, therefore so is *K, i.e. F
= EndD-(K)
and [K:D]L= [F:K]R.
Conversely; given D, put F = EndD_(K) and define D1 {x
£
K I xf = x.l£ for all f e: F}, then
D 1~D
and by what we
(K), hence D1 =D. Finally if either Dlside of (5) is finite, then by (6) we have
have seen, F =End
39
l
as claimed. • Galois theory
3.3
Let K be a skew field and G the group of all its autocr + X,. morphisms. For any subfield E of K let E = {cr e: G for all x {x
£
K
I
X
+
£
x0
E} and for any subgroup H of G, put H
=x
for all a
£
H}.
Then it is clear that E+ is
a subgroup of G, H+ is a subfield of K and as in any Galois connexion,
H~H
++
and hence +++ H
+
E '
+
H .
Given a subfield E of K, we call E+ the Galois group of K/E, +
written Gal(K/E), and given a subgroup H of G, we call H the fixed field of H.
If E
H+ for some H, we shall say
that K/E is Galois. The object of Galois theory is to find which fields in K are of the formE the formE+.
= H+ and which subgroups of Gal(E/K) have
We recall that in the case of commutative
fields the finite Galois extensions are just the normal separable extensions, while every subgroup of Gal(K/E) has the form F+ for a suitable field F between K and E.
The
account which follows is based on Jacobson [56]. The commutative theory rests on two basic results: Dedekind's lemma.
Distinct homomorphisms of a field E into
a field Fare linearly independent over F.
40
If G is a group of automorphisms of a field
Artin's lemma.
E and F is the fixed field, then [E:F]
=
lei
whenever either
side is finite.
Our object is to find generalizations.
We begin with
Dedekind's lemma; here we have to define what we mean by the linear independence of homomorphisms over a skew field.
= Hom(K,L) for the
Given any skew fields K, L, we write H
set of all field homomorphisms from K to L.
Let HL be the
right L-space on the set H as basis and define HL as left K-space by the rule s
as Thus HL
sa , ~s
s
a e: K,
a (K,L)-bimodule, as
~s
element of HL defines a mapping K Es.A.: a 1--~ Ea
(1)
~
H
Hom(K,L).
easily checked. --~
Each
L as follows
s. 1
~
(a e:
A.
~
We observe that for a,S thus aSs
E
£
K, s s H, a.
K, s.~ e: H, A.~ e: L). Ss
a
sSs
~sas ~ ~
= ( aS )s ,
(a5)s and so the left K-module structure of HL
acts on K in the expected way.
Let N be the kernel of the
mapping from HL to Map(K,L) defined by (1), thus N consists s. 0 for all a. £ K. Write of all sums l:s.:\. such that l:a. l:\. 1
~
1
M = HL/N, then M is a (K,L)-bimodule whose elements have the form Es.A. (s. e: H,A. e: L), with Es.1 A.1 = 0 if and only if ~ 1 1 1 s.
=0
Ea 1A. 1
for all a e: K.
Each p e: L* defines an inner automorphism of L:
I : A 1---~ PAP
-1
fJ
,
and it 1s clear that for s s,t: K
-->
£
H, si
fJ
£
H.
Two homomorphisms
L are called equivalent if they differ by an inner
automorphism: t
= si].1 •
We note that for each s e: H, sL is a 41
(K,L)-submodule of HL which is simple as (K,L)-module, since it is already simple as L-module.
Two homomorphisms s,t
define isomorphic (K,L)-bimodules if and only if they are equivalent.
For if sL
then for all a E K,
= tL,
a.s~
say t corresponds to
that as~= ~at, hence s,t are equivalent. s,t are equivalent, then as~
= ~at
tracing our steps we find that sL to
s~
s~ (~
E
L*),
. s £'~nd s~nce as = sa , we
= s~.a t •
Conversely, if
for some ~ E L* and re-
= tL,
with t corresponding
in the isomorphism.
It follows that HL is a sum of simple (K,L)-bimodules, i.e. semisimple, hence so is the quotient M.
We recall that
a semisimple module is a direct sum of homogeneous components, where each homogeneous component is a direct sum of simple
[56]
modules of a given type (cf. Jacobson
or Cohn
[77]).
Now we have the following generalization of Dedekind's lemma. Theorem 3.3.l(i).
Given s,s 1 , ... ,sn
s.I
then s
(ii)
~
1
for some i and
~
= Hom(K,L), if
in M l='HL/N
s = l:s.A. 1
H
E
(A..
1
some~ E
E
L),
L*.
Givens E H, i f ~ 1 , ... ,~r E L are such that the
elements si
in M are linearly dependent over L, then the
~i
~i are linearly dependent over C (Ks), the centralizer of
Ks in L. Proof.
(i)
If s
=
l:s.A.., then the simple module sL lies ~ ~
in the same homogeneous component as somes., so sands. 1
generate isomorphic modules, i.e. s
=
s.I ~
~
(~ E
1
L*).
(ii) If the si~i are linearly dependent, take a relation of shortest length: p
I 1 si
~·1
A..
~
0
(A..
~
E
L) •
Then each A.i f 0 and by multiplying on the right by a suitable factor we may assume that 42
A. 1
= ~1 .
Apply this relation
to aS
E
(2)
0
K: s s -1 == L11. a. S ll· A.• 1
1
1
Next apply the relation to a and multiply by Ss on the right: 0
Lll.ct 1
s -1 s ll· A.S . 1
1
Taking the difference, we get "'p s -1 ('.as ~lll·Cl ll· 1\ p 1 1 1
s -1 ll·B ll·1 A.) 1 1
0.
The first coefficient 1s Alps - A1 Ss ; 0, hence by minimality -1
the others are also 0, so ll· A.S 11 E C (Ks). Now by (2), with cY. = i3
-1 l:]J •• ]J.
1
1
\.
1
s
s -1
-1
; S ll· A., i.e. ll· A. 11 11 = 1,
0,
and this is the required dependence relation over C(K 5 ) . • Corollary 1.
Let s 1 , ... ,sr be pairwise inequivalent iso-
morphisms between K and Land let A1 , ... ,At
E
L be linearly
independent over the centre of L, then the isomorphisms
s.I, are linearly independent over L. 1 /\. J
For if they were linearly dependent, then for some s
=
si
the siA. would be linearly dependent (because each si beJ
longs to a different homogeneous component).
C (K 5 ),
the A. are linearly dependent over J of L, which contradicts the hypothesis. •
So by (ii)
i.e. the centre
If sl' • •. ,sr are inequivalent isomorphisms between K and L, A.l, •.. ,At E L are linearly independent
Corollary 2.
over C, the centre of L and
43
(3)
(ex ••
s = L:s. I, ex ••
J
then
L),
E
1J
1 /\. 1J
s = siiA, for some i, where A= L:A}lj (ilj
Proof.
E
C).
By (i), s = siiA for some i and some A E L.
Thus
Is. I, ex •• 1 1\. 1J J
and by equating homogeneous components we can omit terms sk
r i.
with k
dent over 6.
J
E
c,
Now by Cor.l, A,A 1 , ... ,At are linearly depenbut A1 , ... ,At are independent, hence A n. a., J J
C••
Next we have to translate Artin's lemma.
Without using
= [G:E], where G, or
Dedekind's lemma the result is [E:F]
rather GE is regarded as right E-space.
We shall replace
G, a group of F-automorphisms of E, by F-linear transforma-
tions of FE.
(yx)s
Every such s E EndF_(E) satisfies yx s
X
E
E, y
This generalizes the rules (xy)s for s
E
G.
E
F.
s s
x y
y
s
= y satisfied
Given any skew field K, we consider the set
End(K) of additive group endomorphisms as a topological ring (3.2).
This set contains p(K), the ring of right multi-
plications as subring, and we recall the Jacobson-Bourbaki correspondence (p.38): There 1s an order-reversing bijection between the subfields D of K and the closed p(K)-subrings F of End(K) such that
whenever either side is finite. Given a group G of automorphisms of K, we have a right
44
K-space GK, and we need only show that this is a ring in order to be able to apply the preceding result. Proposition 3.3.2.
Thus we have
Let K be any skew field and G a group
of automorphisms of K, then GK is a p(K)-subring of End(K) and its closure GK is EndD_(K), where Dis the subset of K left fixed by G.
Proof.
In GK we have the rule ag
gag
(a
K, g
E
E
G).
Using this rule, we have
Since every element of GK is a sum of terms ga, it follows that GK is closed under products and contains 1, hence it is a ring, indeed a p(K)-ring, because p(K)~ GK. Jacobson-Bourbaki correspondence, GK
{x
E
K
I
a E
x
f
D
= x.l
a e: D
for all f
E
-
GK}.
EndD_(K), where D Thus if a
f f a = a.l for all f agf3 a.lgS for all g agf3 af3,
= a.
for all g
E
By the
E
GK,
E
G,
K, then
E
aE
K,
G. •
Combining this with Th.3.3.l, we get Proposition 3.3.3.
Let K be a skew field, G a group of
automorphisms of K, and D the fixed field of G.
Assume that
G contains every inner automorphism of K over D.
If E is a
subring of K such that D ~ E ~ K and
IE :D I1
K be a D-ring homomorphism.
Regarding E
and K as left D-spaces, we see that E is a subspace, so s can be extended to a D-space endomorphism of K, 1.e. an element of EndD_(K).
By Prop.3.3.2, EndD_(K) = GK and on any
finite-dimensional subspace s can be written as Ig.A. (g. 1 1 1 Ai e: K). In particular, since [E:D] 1 < oo, we have
E
45
G,
= Ig.A.
s
~
on E.
~
Now E is an entire D-ring of finite left degree, hence a field, and by Th.3.3.1 (i), s = g.I ~
v E K.
Applying this to a E D we have a
Thus I
v for some i and some
~s
v
hence g.I ~
v
s
an inner automorphism of K over D, and so I
E G,
v
EGis an automorphism which induces s.• Let K be a skew field with centre C and D a C-
Corollary.
subalgebra finite-dimensional over C, then any C-algebra homomorphism D --> K can be extended to an inner automorphism of K.
This follows by taking G to be the group of all inner automorphisms of K, then C is the fixed field and every inner automorphism belongs to G, so the proposition may be applied.• We now return to our initial task of finding which automorphism groups and subfields correspond under the Galois connexion.
We first deal with a condition which is obviously Let K be a field with centre
satisfied by all Galois groups.
C and let D be any subfield of K, then the centralizer of D in K is a subfield D' containing C.
Any non-zero element of
D' defines an inner automorphism of K which leaves D elementwise fixed and so belongs to the group D+ ; conversely, an ~nner
+
automorphism of K belongs to D only if it is induced
by an element of D'.
Thus we see that the a E K for which
+
Ia
~
~s
then a necessary condition for a group to be Galois and
D form together with 0 a subfield containing C.
we define: A group G of automorphisms of K is called an N-group (after E.Noether) if the set
A
46
{a
E
K
I
a
0 or I
a
E
G}
This
~s a C-subalgebra of K.
We shall call this the C-algebra
associated with G.
Clearly the associated C-algebra ~s necessarily a field. If G is any N-group with associated algebra A, and G is the 0 subgroup of inner automorphisrns I (a E A), then G is normal 1 a in G, for if x £ K, a £ A*, s £ S, then x s- las = 0( ax s-1 a -l)s = s ( s)-1 a x a = xi s• so a
s
-1
I
a
.s
I s· a
We define the reduced order of G as
With this notation we have Theorem 3.3.4.
Let K be any skew field, G anN-group of +
automorphisms of K and put D = G •
Then
(4)
whenever either side is finite, and when this is so, G++
Proof.
Suppose first that [K:D]L
(J.-B.)
[End 0 _(K):K]R
incongruent (mod G0
)
=m
. 1 , ... ,"At any elements of the =
associated C-algebra A that are linearly independent over
C.
Then the maps sir>.. are in End 0_(K) and by Th.3.3.1, J
Cor.l they are linearly independent over K; hence rt
. 1 , ••• ,"At a C-basis for A, then we know that the s.I, ~
over K.
are right linearly independent
1\.
J
We shall show that they form a basis for End 0 _(K). Lets£ G, then s = s 1 I"A say, where A£ A and so "A=
47
For any c e: K we have
l:A.ts. (ts. e: C) • J J
J
s
c
c
slr
>.cs 1 A-1 A= sl -1 l:A.~.c A J J
s -1 -1 n.c n .. A.fLA J
J
J J
l:c sl IA y., j J
where y. = J
A.~.A.
-1
spanG, e: K. This shows that the s.IA 1. •
J J
hence also GK.
J
Now GK has finite dimension over K and so
= EndD_(K) (Prop.3.3.2). [End 0 _ (K) : K] R = (K: D] L •
GK
=
GK
Next it is clear that D+ then s e: EndD_(K), hence s
++
G
It follows that !Gired ;2G.
= l:s.IA a ..• 1. • l.J
+
Conversely, if s e: D , By Th.3.3.1, Cor.
2, since s is an automorphism of K~ s = siiA for some i, where).= EA.ts.
(~.
JJ
s
E
J
e: C), but then I, e: G and s. e: G, hence 1\
0
1.
G ••
Here is an example, taken from Amitsur [54], to illustrate the need for introducing the reduced order. Let F be the Q-algebra generated by u, with defining relation u
2
+ u +
1
= 0. 2
F admits aQ-automorphism cr:u 1-> u. Put k = F[v;cr], then v 2 - 2 is central and irreducible, because the equation
has no rational solution. So we can form the skew field K = R/(v 2 - 2). Now the group generated by I has order 3, u but reduced order 2, for its fixed field is F and [K:F] = 2.
48
We note some consequences of Th.3.3.4. Corollary 1.
:If K/D is Galois (i.e. D
= G+
for some G), then
whenever either side is finite.
This follows from (4) by symmetry. • Corollary 2.
Let K be a skew field with centre C and A any
C-subalgebra of
K.
Then the centralizer
C-subalgebra of K and A"-:;;;_A.
(5)
A' of A is again a
Moreover,
[A:c]
whenever either side is finite, and when this is so,
A"
A.
=
Proof.
Suppose first that [K:A'] 1 is finite. Clearly A' 1s a subfield of K; let G be the group of inner automorphisms of K fixing A' and A1 the associated algebra, then A1 -:;;;_A and by Th.3.3.4, lclred
[A 1
:cJ
=
[K:A']L' hence A is then
finite-dimensional.
Thus we may assume that [A:C] is finite,
hence A is a field.
Now let G be the group of inner auto-
morphisms induced by A, then clearly G+ = A', and (5) follows from (4).
Moreover, (A')+= G, which means that
A" = A. • Corollary 3.
If K/D is Galois with group G and [K:D] 1 < oo,
then any p(K)-subring B of GK has the form HK, where H
=
GnB is an N-subgroup of G.
Proof.
Let H = GnB, then clearly HKc;::; B.
To prove equality
we note that HK and B are both K-bimodules contained in GK
= EndD_(K), by Prop.3.3.2.
Now GK is semisimple and
hence so is B; moreover every simple submodule M of B is isomorphic to a simple submodule of GK and hence 1s of the form M = uK, where ~u Replacing u by uy (y
= ua 8 £
for all a
£
K and some s
£
G.
K) if necessary, we may suppose that
49
l.u
and still a.u
1
ua.
s
(for somes E G).
Hence a..u: l.ua.s: a.s, i.e. u is an automorphism of K, viz. s, and since u
EndD-(K), s fixes D, i.e. s
E
E
G, so s
E
GnB:
H; this shows that B = HK. That H is a group follows because B is a centralizer (being finite-dimensional over K, by. Jacobson-Bourbaki). that H is an N-group, let I ~
0; we must show that I
to show that I hence Ia. E
E
B
a.
E
X
-1
-1
So if (i),
Conversely,
£
L be such that
E
K satisfies
ex,
ex is an automorphism of K,
we have a homomorphism
K[t;o]
--:>
L
given by t
1-->
x,
and the generator of the kernel has the form tna - 1, where 67
a satisfies (i), (ii). • If we drop the condition that wo
= w,
then wo
= wv
for
some v and it is not hard to write down conditions on v for an extension to exist.
One can also give conditions for an
outer cyclic extension of degree n if K merely contains a primitive dth root of 1, where d is a proper factor of n. As a consequence we have a form of Hilbert's theorem 90: If L/K is an outer cyclic extension of degree n
Corollary.
with generating automorphism a, and c
(4)
X
a
E
L, then the equation
XC
has a non-zero solution in L i f and only if a an-1 cc .•• c
(5)
1.
• f'les (5) . Cl ear 1y c ; a -l a 0 satls
Proo f .
holds, we have (erA )
n
c
Converse 1y, l'f (5)
; 1, where A denotes left multiplicac
tion by c, for the left-hand side maps x successively to a a a 02 a an 0 n-l x, ex, c x , .•. ,cc .•• c x ; x. Thus we have
1] ; 0.
x[(a;\ )n c
This has the form xp(a) [erA p(a); now xp(cr)
= 0
c
- 1] = 0 for some polynomial
can be considered as a differential
equation (with respect to D
=a -
1) of order n-1, so its
solution space has dimension~ n-1, hence there exists a such that ap(a) = b f o, and b(aA - 1) c so (4) holds for x = b-l. • In a similar way one can show that x 0 solution if and only if c
68
crn-1
a
... c c
1.
= o, i.e. cba =
E
L
b.
ex has a non-zero
We also note the following criterion for reducibility: Proposition 3.5.4.
Let L be a skew field with an automor-
phism o of order n and a primitive root of 1, w, in its Then for any a
centre.
€
L, tn - a is either irreducible
over L[t;a] or splits into factors of the same degree.
In
particular, i f n is prime, tn- a is a product of linear factors or irreducible according as
=
a
an-l
a XX
••• X
has a solution or not.
Proof.
Let p = p(t) be an irreducible left factor of tn- a,
then so is p(wvt), for v
= l, ••• ,n-1,
therefore tn- a=
p 1 p 2 ... prq' where p 1 = p(t) and pi= p(wvit). If r is chosen v as large as p0ssible, each p(w t) is a factor of p 1 p 2 ... pr; ~n fact this is their least common right multiple and so is unchanged by the substitution t
1-->
wt.
This means that it
~s a polynomial in tn, of positive degree, and a factor of tn
a.
Hence it must be tn - a, i.e. q
=
1 and we have
proved the first part. Now if p has degree d, then dJn, hence when n is prime, d = l or n, and now the last part follows from the identity
t
n
-
a
( t - b)(tn-l + t
n-2 an-1 2 n-1 b + •.. + b 0 b 0 .•• b 0 ) + n-l + bb 0
•••
b0
-a. •
We now turn to the case where n is a power of the charace char K. Observe that in this case (2) teristic, n = p p reduces to Dn
= 0.
Proposition 3.5.5. degree n
pe, p
Write LV
{x e L
Let L/K be an outer cyclic extension of
= char J
K, with automorphism a
x(v)
=D
+ 1,
0}, then each Lv is a right
K-space of dimension v and
69
CL
K
L,
n
L
By Th.3.5.1, [Lv:K]R ~ v
Proof.
equality.
v
v+ 1
and for v
=
l, ... ,n-1).
(v
=L'
n we have
We shall use induction on n-v, thus assume that
L 1 has a right K-basis a = l,a 1 , ... ,a. We claim that 0 \) v+ = a, then Ea.a 1• a1 ' ' ' ' ' ' a'\) are a K-basis for L\) • If Ea!a. 1 1 1 a E K and by the linear independence of a , .•. ,a 0
..
= •.•
a1
=a\!=
a,
thus al, .•.
,a~
are linearly independent;
they belong to Lv so this shows that Lv \)
Since L'
= Ln- 1 ,
Corollary 1. Corollary 2. by aPi (i
=
L~+l and
E
s
a (a
L) has a solution in
8
Ln_ 1 · •
The subspace L i is the subfield of L fixed
= O,l, •.. ,e).
P
This follows because DPi
(o - l)Pi
a Pi - 1..
Let us define the trace of a E L as tr a Then n-1
D
[Lv :K]R
we have
The equation x'
L if and only if a
we have
\)
(a _
(a - l)n (a - 1)
0 n-l
n
(J (J
=
n-1 (J\) E0 a .
- 1 - 1
n-1 v L:o a '
hence (6)
tr a
a
(n-1) ·
for any a E L.
This formula enables us to prove a normal basis theorem: Theorem 3.5.6.
=P
n
e
(p
The outer cyclic extension L/K of degree
= char
K) has a normal basis, and a tive if and only if tr a ~ a. For tr
a~ a
a(n-l)
~a
8
a i Ln-l'
L is primiThus for
any a i Ln_ 1 , a(v)E L , a(v)i L and hence a,a' , ... , (n-1) n-v n-v-1 a form a basis of L = L. • n
e
We shall only determine extensions of degree p, the case
P , e > 1, follows by repetition (for details see Amitsur 70
[54]).
We shall write/
xp - x; let us also recall the
Jacobson-Zassenhaus formula (Jacobson [62] ,p.l87(63)): xp + yp + .\ (x,y), where
~
is a sum of commutators in x and y.
It follows
that the expression V{x) defined by (7)
V(x)
(t + x)
=
p
- t
p
when evaluated in K[t;l,D] lS a polynomial in x,x I , ... ,x (p) ' since e.g. [x, t] = xt - tx = x'. We first prove an analogue of Prop.3.5.4. Proposition 3.5.7.
Let L be a field of characteristic p
with a derivation D such that oP
.
polynoaual
t
p
-
= 0.
For any a £ L, the
a is a product of linear factors over
L[t;l,D] or irreducible according as the equation V(x) + a where Vis as in
Proof.
0
(7), has a solution in Lor not.
Let h be a monic irreducible factor oft
p
-a, of
degree d say, then the polynomials h(t + v) (v = 0,1, .•• , p p-1) are factors of t - a, Their least common right mul-
p
.
tiple is of degree ~ p and is a factor of t - a, hence lt p must be t - a. Now all the h(t + v) are irreducible of the same degree, so djp and either d
(t +
p
or d
= 1.
If
O, then
V(b) + a
hence t
=p
b/ - t
V(b)
-a= (t +b)
p
=
-a,
(t + b)((t +b)
p-1
- 1) and so
tp - a splits into linear factors.
Conversely, if tp - a p has a linear factor t + b, then (t + b) - V(b) - a = 71
(t + b)h(t), hence V(b) +a hast+ bas factor. has degree 0 in t, so V(b) + a
=
But it
0. •
We can now prove an analogue of Th.3.5.3. Theorem 3.5.8.
A skew field K of characteristic p has an
outer cyclic extension of degree p i f and only if there is a derivation D in K such that (i) Dp is inner, induced by a
£
K with a
D
= 0,
but D is outer (ii) V(x) + a "'0 has no
solution in K.
When this holds, tp - a is right invariant irreducible
~n
R
= K[t;l,D] and L = R/(tp- a)R, with generating auto-
morphism
Proof.
Again (i),(ii) ensure that tp- a
~s central and
irreducible, so when they are satisfied we have an extension. Conversely, let L/K be an outer cyclic extension of degree (J
p, then by Prop.3.5.5, L has an element y such that y = y + 1. D Hence c 1-> c = cy - y..c induces a derivation on K and we have a homomorphism K[t;l,D] -> L with t 1-> y. Here yp . inner, induced by a, and a D = o, while = a £ K, so Dp ~s V(x) + a = 0 has no solution in K, by the irreducibility p of y - a over K. •
72
4 ·The ·general embedding
4.1
The category of epic R-fields and specializations We now come to the fourth method of embedding rings in
fields listed in the prologue.
It is quite general in that
it provides a criterion for arbitrary rings to be so embeddable, and also gives a survey over the different possible embeddings. Let R be any ring.
We shall be interested in R-rings
that are fields, R-fields for short.
If K is an R-field
which is generated (as a field) by the image of R, we call K an epic R-field.
In fact K is an epic R-field precisely
when the canonical map R --> K is an epimorphism in the category of rings.
Our object
~s
to make the epic R-fields
(for a given R) into a category and we must find the morphisms.
To take R-ring homomorphisms would be too restric-
tive, for if f:K -->Lis such a map between epic R-fields, then f is injective (because the kernel
~s
a proper ideal
of a field) and im f is a subfield of L containing the image of R, hence im f isomorphism.
L (because L was epic), so f must be an
To obtain a workable notion of morphism let
us define a local homomorphism between any rings A, B as a homomorphism f:A
0
--> B whose domain A 1s a subring of A 0
and which maps non-units to non-units. means that the non-units in A
0
hence A
0
is then a local ring.
understand a ring A
0
If B is a field, this
form an ideal, viz. ker f, Generally by a local ring we
in which the non-units form an ideal
111
the quotient ring A /m is then a field, called the residuea
73
class field of A .
Of course when we are dealing with R-
o
rings, a local homomorphism is understood to have a domain which includes the image of R. Let f be a local homomorphism between epic R-fields K,L. If its domain is K , then by what has been said, K is a 0
0
local ring with residue class field K0 /ker f; this is isomorphic to a subfield of L containing the image of R, and hence L. (1)
Thus
K /ker f 0
~
L.
Two local homomorphisms are said to be equivalent if their restrictions to the intersection of their domains agree and again define a local homomorphism.
This is easily veri-
fied to be an equivalence; an equivalence class of local homomorphisms from K to L is also called a specialization. It can be checked that the composition of specializations is again a specialization (i.e. composition of mappings, when defined, is compatible with the equivalence defined earlier), and so we obtain for each ring R, a category FR of epic Rfields and specializations. At first sight it looks as if there may be several specializations between a given pair of epic R-fields.
E.g. let R
k[x,y] be the commutative polynomial ring over a field, K
=
k(x,y) its field of fractions with the natural embedding and L
=
k with the homomorphism R --~ L given by x 1--~ 0,
Y 1--~ 0.
We obtain a specialization from K to L by de-
fining a homomorphism a:k[x,y] --> L in which xa
= ya = 0.
Let K0 be the localization of k[x,y] at the maximal ideal (x,y), then a can be extended in a natural way to K .
We
0
observe that there are local homomorphisms from K to L that are defined on larger local subrings than K
(we can
0
'specialize' rational functions ¢(x,y) so that x/y takes on a specified value ink), but all agree on K0 just one specialization from K to L. 74
,
so that there is
In fact this is a
general property: between any two epic R-fields there is at most onfi specialization.
This will become clear later.
Of course for some rings R there will be no R-fields at all, e.g. R
= 0,
or for a less trivial example, any simple
ring with zero-divisors, say a matrix ring over a field. For any map R - > K must be injective and this is impossible when K
~s
a field.
Even entire rings R without R-fields
exist, e.g. if R is any ring without invariant basis number (Leavitt [57], Cohn [66']); R may be chosen entire and any R-ring is again without invariant basis number and so cannot be a field. What can we say about R-fields
~n
the commutative case?
Let R be a commutative ring and K an epic R-field, then K is of course also commutative (being generated by a homomorphic image of R). R
--~
The kernel p of the natural mapping
K is a prime ideal and K can be constructed in two
ways from R and p .
Firstly we can form R/p, an integral
domain (because p is prime), and now K is obtained as the field of fractions of R/p.
Secondly, instead of putting
the elements in p equal to 0, we can make the elements outside p invertible, by forming the localization Rp• ~s
This
a local ring and its residue-class field is isomorphic
to K.
The situation can be illustrated by the accompanying
commutative diagram.
The two
triangles correspond to the two methods of constructing K. The route via the lower triangle is perhaps more familiar, but unfortunately it does not seem to generalize to the non-commutative case; we therefore turn to the upper triangle.
Even this cannot be used as it stands, for as we
have seen, the field of fractions need not be unique, which means that ~n general an epic R-field will not be determined by its kernel alone. 75
Thus to describe an epic R-field we need more than the elements which map to zero, we need the matrices which become singular.
Here we use the fact that for any square
matrix A over a field K (even skew) the following four conditions are equivalent:
A has no
left
inverse,
right A ~s a left zero-divisor.
right
A matrix A with these properties
~s
called singular, all
others (if square) are called non-singular.
Given any R-
field A:R --> K, by the singular kernel of K (or A) written Ker A, we understand the collection of all square matrices over R(of all orders) which map to singular matrices over
K.
Let P be the set of all such matrices, then we can de~
fine a localization case) as follows.
(analogous to Rp in the commutative
In 1.2 we met the notion of a universal
S-inverting ring; we shall need the corresponding construction when S is replaced by a set of matrices over R. Let Z be a set of matrices over R, possibly of different orders, but all square (this is to avoid pathologies, because we want to make the matrices in Z invertible). For every n x n matrix A in ~ we choose n 2 symbols a!. which we ~J
adjoin to R, with the defining relations (in matrix form) (2)
AA'
A'A
I,
where A'
(a!.). ~J
The resulting ring is denoted by RI and called the universal I-inverting ring.
Clearly the natural homomorphism A!R --> RI
is I-inverting, in the sense that all matrices in I map to invertible matrices over RI (an inverse being provided by (2)), and every I-inverting homomorphism f:R --> R' can be 76
factored uniquely by A (the universal mapping property). The proof is the same as for Prop.l.2.1. We can now describe the construction of an epic R-field in terms of its singular kernel.
Let K be an epic R-field,
P its singular kernel and I the complement of p l.n the set of all square matrices over R.
Thus I consists of all
square matrices over R which become invertible over K. Then the universal I-inverting ring RI is a local ring, with residue-class field K.
We shall soon see a proof of
this fact, but we note that it does not solve our problem yet.
For we would like to know when a collection of ma-
trices is a singular kernel, just as we can tell when a collection of elements of R is a prime ideal.
In fact we
shall be able to characterize singular kernels in much the sarue way in which kernels of R-fields in the commutative case are characterized as prime ideals. 4.2
The construction of epic R-fields A basic step in the construction of an R-field is the
description of its elements as components of the solution vector of a matrix equation. morphism f:R (Af)
-1
-->
Given a I-inverting homo-
R', the set of all entries of matrices
, where A£ I, is called the I-rational closure under
f of R in R'.
It is not hard to give conditions on I for
this I-rational closure to be a ring; in fact they correspond to the condition of being multiplicatively closed in the commutative case for sets of elements.
So we define a
set I of square matrices over a ring R to be multiplicative if it includes the 1 x 1 matrix 1 and for any A, B E E we C of the right size. In have ( A 0 B £ I for all matr1ces
c)
any homomorphism f:R
.
-->
R' the set of all matrices inis inver-
verted over R' is alwa~s mul~iplicati~e'[!o~)l tible and if A, B are 1.nvert1ble, so 1s 0 B , with inverse
77
The characterization of the rational closure, which is at the basis of all further development (Cohn [71"]) stems from the rationality criterion for formal power series due to Schutzenberger Theorem 4.2.1.
[62]
[68].
and Nivat
Let R be a ring and E a multiplicative set
of square matrices over R.
Given a E-inverting map f:R --> R',
the following conditions on x
£
R' are equivalent: R in R',
(a)
x lies in the E-rational closure under f of
(b)
x is a component of the solution of a matrix equation
Au + a
(1)
o,
where A
£
Ef, and a is a column
over Rf, x is a component of the solution of a matrix equation
(c)
Au
(2)
e,
where A
Ef and e is a column of the iden-
£
tity matrix. Moreover, the set of all these elements x is a subring of
R' containing Rf. Proof.
(a) and (2) is a special case of (1), so (c) => (b).
To prove (b) => (c) we note that if Au + a
= 0,
then
o, so when (b) holds, each component satisfies an equation of type (2) and so (c) holds. To prove that the rational closure is a ring containing Rf we use (b): For any c
78
£
Rf we obtain cas solution of
l.u- c
o.
Now let ul, vl be the first components of the solutions of Au + a = o, Bv + b = o, then ul - vl is the first component of the solution of =
-(UlJ
where A= (a 1 , ... ,an), u - u' . Further, first component of the solution of
1.s the
and the matrices of these systems lie in Zf, because Z is multiplicative.
This shows that the Z-rational closure con-
tains Rf and admits sums and products, hence it is a ring.• This theorem shows that every component of the rational closure can be obtained as some component u. of the solution 1. of a matrix equation Au
a.
Here A is called the denominator of u. and A., the matrix 1. 1. obtained by replacing the ith column of A by a, is called the numerator of u .. This usage is justified by Cramer's 1. rule, which states that when R is commutative,
u.
l.
det A. 1. det A
In the general case we no longer have this formula (because we do not have determinants), but we have the following substitute, still called Cramer's rule:
79
Proposition 4.2.2.
Let u. be the ith component of the solu~
tion of Au = a, where A is invertible, and let A. be the ~
matrix obtained by replacing the ith column of A by a, then
u. is a { l~ft ~
} { zero~divisor } i f and only i f · un~t
r~ght
(in the matrix ring).
Proof.
Take i
=
1 for simplicity and write again
(a,a 2 , ••• ,an)
then A1
A(Ul OJ u• I
= A [u1 0
~
U= (Ul) u' '
= (Au,Ae 2 , .•• ,Aen) =
0)I (1u• OJI ,
full matrix ring Rn) to
A. is one
Thus A1 is associated (in the
(~1 ~)
and now the result follows
because being a zero-divisor or unit is preserved by multiplying by a unit or bordering with I. • Anticipating a definition from 8.1 we may say that u 1 is stably associated to A1 . As an application let us show how to construct epic Rfields from their singular kernels. field, Let
P'
~:R -->
Let K be an epic R-
K the canonical map and
be the complement of
P in
P the
singular kernel.
the set of all square
matrices over R, then the universal P'-inverting ring, which should be written
~·
is usually written
we write Rp in the commutative case), of
P it
follows that
~
~
(just as
From the definition
can be factored uniquely by A:R -->
to give a map a:Rp --> K. with residue class field K.
We claim that
Rp
is a local ring
This will follow if we show
that every element not in ker a is invertible.
For then ker a
~s
the unique maximal ideal of Rp and its residue class field is a subfield of K containing the image of R, hence equal to K, because K was an epic R-field. Let u 1 ~ . AA
equat~on
80
Rp
be the first component of the solution of an
u + a
= O,
where A is a matrix over R which be-
Rp
comes invertible over K, and define A1 as in Cramer's rule. -h · 1nvert1 · 'bl e, h ence AAa ~ · · If Ua1 4r 0 , .~en u a1 1s 1 ~A 1s 1nverl A tible, by Cramer's rule, but this means that A1 i P, so A1 is invertible over a unit in
Rp.
~and
so, again by Cramer's rule, u 1 is
as claimed.
This result shows that any given epic R-field K can be reconstructed from its singular kernel.
What we need now
1s a simple way of recognizing singular kernels - just as 1n the commutative case the kernels of epic R-fields are precisely the prime ideals of R.
For this we need to de-
velop an analogue of ideal theory in which the place of ideals is taken by certain sets of matrices. In the first place we must define the operations of addition and multiplication for matrices; they will not be the usual ones of course, but more like the addition and multiplication of determinants. Multiplication:
As the product of two square matrices
A, B (over any ring R) we take their diagonal sum A
[~ ~).
Note that over a field A
+B
+B
is singular if and
only if either A or B is. Addition is more complicated, just as the addition of determinants is not straightforward, and in fact the latter provides the clue.
Let A, B be two matrices which agree
in all entries except possibly the first column: A= Ca 1 , a 2 , ... ,an)' B
= (al,a 2 , ... ,an)' then the
determinantal sum
of A and B is defined as the matrix
Similarly one defines determinantal sums with respect to another column or with respect to a row.
Of course it must
be borne in mind that the determinantal sum need not be defined.
As notation we shall always use A V B, indicating 81
in words the relevant column or row, when this is necessary to prevent confusion. We observe that over a commutative ring, where determinants are defined, one has det(A V B)
= det A
+ det B, when-
ever the determinantal sum (for any row or column) is defined.
Likewise, over a skew field, if two of A, B, A V B
are singular, so is the third.
Over a general ring there
is no direct interpretation, but, and this is the point, whether the operation is defined depends only on the matrices involved and not on the ring. The third operation we need is the analogue of zero, in our case a matrix which becomes singular under any homomorphism into a field. divisors, for if AB
Here one cannot just take zero-
= 0,
where A,B # 0, it may still happen
that under a homomorphism A becomes invertible and B becomes zero.
But there are some matrices that always map
to singular ones, e.g. the zero matrix, and more generally a matrix of the form
(~~- ~:J
n x n matrix A non-full if A r x n and r
1.
particular, Cr (I - BA)AnBn
0 and choosing r = n+l we find that I - BA
= C~n =
0 ••
Consider the following example due to Bergman [74]. R be defined by 27 elements, arranged as 3 U, V with defining relations UV P)U.
VU
x
Let
3 matrices P,
2
= I, P = P, UP = (I -
If R could be mapped into a field, then P could be
transformed to diagonal form, with O's and l's on the main diagonal; since P is similar to I - P, there must be equal numbers of both, but this is impossible, because the order is odd.
It is easy to see that R is entire and Bergman
shows that it satisfies Klein's nilpotence condition, therefore it is weakly finite, in particular I is full.
But by
Cor.l of Th.4.B, the unit matrix (of a certain size) can be written as a determinantal sum of non-full matrices.
Thus
R does not satisfy (ii). The conditions of Th.4.3.1 are not easy to apply; there ~s just one case where they can be checked without diffi-
culty, namely for semifirs, which were introduced in 1.1: Once the basic properties of semifirs have been derived, such a verification takes less than a page (Cohn [71"], 87
p.283), but since we have not developed this background here, we omit a detailed proof: Theorem 4.C.
Every semifir has a universal field of
fractions, obtained as the universal ring inverting all the full matrices.
We shall denote the universal field of fractions of R by F(R).
To prove the result one only has to verify (i),(ii)
of Th.4.3.1.
Here (i) follows from a form of Sylvester's
law of nullity for semifirs, while (ii) is a relatively direct calculation, based on the dimension formula dim(U + V) + dim (U
n V)
dim U + dim V
for finitely generated submodules of free modules over a semifir. To apply Th.4.C we derive a further consequence which tells us when a homomorphism can be extended.
A homomor-
phism f:R --> S between any rings is called honest if it keeps full matrices full.
Any homomorphism keeps non-full
matrices non-full, hence any isomorphism and in particular any automorphism is honest.
An honest homomorphism must be
injective, for an element c is non-zero if and only if it is full, as 1 x 1 matrix.
But an injective homomorphism
need not be full; here is an example. Let R
= k
be a free k-algebra on four gen-
erators and define an endomorphism a over k by xil--> x 1yi, y. 1--> x 2y. (i = 1,2). ·It is easily checked that a is ~
~
injective; but it is not honest, as the equation
shows.
The right-hand side
~s
not full, but the matrix to
which a is applied is full, since it can be specialized to 88
the unit-matrix which is full. Theorem 4.3;3. sendfirs.
Let f:R --> S be a homomorphism between
Then f extends to a homomorphism (necessarily
unique) between their universal fields of fractions i f and only i f i t is honest.
In particular, every isomorphism
between R and S extends to a unique isomorphism between their universal fields of fractions.
Proof.
Denote by
,
'±' the set of all full matrices over R,
S respectively, then if f
1S
honest, f
c
'±', and so the
mapping R - > s - > s'±' 1S -inverting. Hence there is a unique homomorphism f 1 :R --> S'±' such that the diagram shown commutes, i.e. f can be extended (in just one way).
Conversely,
R
----------~>
R
if an extension of f exists, any
flj
full matrix A over R becomes invertible over R and is mapped
s
-----------;>
s'!'
to an invertible matrix over SV. f
But this is the image of A , which must therefore be full. Hence f is honest, as claimed.
The rest follows since an
isomorphism is always honest. • The notion of an 'honest map' is chiefly of use for semifirs, because here the non-full matrices constitute the unique least prime matrix ideal. To get an idea of the usefulness of Th.4.C and Th.4.3.3 we really need to know how extensive the class of semifirs is.
In the commutative case semifirs are just Bezout do-
malus.
Somewhat more familiar are the principal ideal do-
mains; they may also be characterized as the Noetherian Bezout domains.
Analogously the semifirs contain as sub-
class the firs (=free ideal rings), which by definition are rings in which every right ideal and every left ideal is free of unique rank.
This class is far more extensive
than the class of non-commutative principal ideal domains. To give some examples, firs include (i) free algebras over 89
a field, (ii) group algebras of free groups, (iii) free products of skew fields. in more detail in Ch.S.
These examples will be examined For the moment we only note that
for any field k and any set X, the free algebra k is a fir; this is most easily proved by the weak algorithm, a form of the Euclidean algorithm adapted for use in free algebras (cf. Cohn [71"] ,Ch.2). Earlier in 1.1 we quoted the result from (Cohn [69]) showing the existence of n-firs from (n+l)-term relations without interference.
This can be used to show that for
any n > 1 there exists an n-fir not embeddable in a field. -
We take 2(n+l)
2
elements a .. , b .. (i,j = 1, ... ,n+l) arranged ~J
~J
as matrices A
(a .. ), B =(b .. ) with relations (in matrix
form) AB = I.
These relations satisfy the non-interference
~J
~J
condition, so we have an n-fir, but BA
~
I (by an easy nor-
mal form argument), so the ring constructed is not weakly finite and therefore not embeddable in a field.
= 1, ••• , n+l,
2(n+l) (n+2) variables a'iA.'bAi (i n+2) with relations AB = I
n+ 1
, BA
=
I
n+ 2
It we take A.
= 1, ••• ,
, we get an n-fir
with no R-field. These rings are of interest in that they enable us to answer the following problem raised by Mal'cev [73]. saw
~n
As we
1.1, the class of entire rings embeddable in fields
can be defined by quasi-identities, and we note incidentally that the conditions implicit in Cor.l,2 of Th.4.2.3 are easily put into this form.
Now Mal'cev asks whether this
class can already be defined by a finite set of quasiidentities.
The answer is 'no' (as for semigroups) and
this may be seen as follows:
Suppose there is a finite set
of first-order sentences which expresses the fact that a ring is embeddable in a field.
On replacing them by their
conjunction we obtain a single sentence A say, which ~s necessary and sufficient for a ring to be embeddable in a field. 90
Now let
Fn
be the class of n-firs for which A is
false, then Fn is an elementary class (cf. e.g. Cohn [65]) since n-firs can be defined by an elementary sentence.
Now
nFn ~s the class of semifirs not satisfying A, but every semifir n Fn
~s
= ~.
embeddable in a field and so satisfies A, hence Thus we have a family of elementary model classes
with empty intersection; by the compactness theorem logic (cf. e.g. Cohn [65], Mal'cev [73])
Fn
~n
=~for some n,
i.e. there exists n such that every n-fir satisfies A and ~n
so is embeddable
a field.
But this contradicts our
earlier findings and we conclude (Cohn [74"]): Theorem 4.3.4.
The condition for a ring to be embeddable
in a field cannot be expressed in a finite set of sentences.
In intuitive (though imprecise) terms we can say that embeddability in a field requires n-term conditions for arbitrarily large n.
This
embeddability in a group.
~s
in interesting contrast with
If R is entire, so that R*
~s
a
semigroup, a sufficient condition for the embeddability of R* in a group can be expressed in terms of 2-term conditions, for we have the following result (Cohn [71]): Theorem
4.D.
Let
R be a 2-fir in which every non-unit is
a finite product of irreducibles (i.e. R is 'atomic'), then
R* is embeddable in a group. This makes the difference between embeddability
~n
a
group and in a field rather clear, and it provides a simple answer to another question of Matcev's, whether an entire ring R exists such that R*
~s
is not embeddable in a field.
embeddable in a group but R To get an example we need
only take 24 generators A= (aiA), B = (bAi) (i = 1,2,3, A= 1,2,3,4), such that AB = 1 3 , BA = I 4 . This is an atomic 2-fir (Cohn [69]), hence R* is embeddable in a group, but there are no R-fields.
Other examples, using similar prin-
ciples, were found by A.J. Bowtell [67] and A.A. Klein [67]. L.A. Bokut [69] also gives an example of such a ring; his construction is more complicated, but unlike the other cases, his example is of a semigroup algebra. 91
5· Coproducts of fields
5.1
The coproduct construction for groups and rings Let A be any category; we recall the definition of the
coproduct.
Given any family (A.) of objects in ~
object S with a family of maps a natural correspondence f
1-->
~.:A. --> ~
A,
and an
S, this defines
~
~.f from maps S ~
-->
X to
families of maps A. --> X, thus a mapping ~
A(S,X)
-->
TI A(A.,X). ~
When this mapping is a bijection, the object S with the maps UA .. ~
~·
~
is called the coproduct of the A. and is written ~
From the definition it is easily seen to be unique
up to isomorphism, if it exists at all.
Thus for sets we
obtain the disjoint union, for abelian groups the direct sum, for general groups the free product, but we shall return to this case below. Often we need an elaboration of this idea.
Let K be a
fixed object in A and consider the comma category (K,A): its objects are arrows K --> A (A s Ob A) and its morphisms are commutative triangles
~A K
t
--------A'
This category has the initial object K _!_> K; it reduces
92
to
A when
A.
K is an initial object of
Now the coproduct
in (K,A) is ·called the coproduct over K.
E.g., for two
objects K --~ A, K --~ B, this is just their pushout. Consider coproducts over a fixed group K in the category of groups.
This means that we have a family of groups (G.) ~
and homomorphisms G.--~
a.: ~
~
~.
~
:K
G. and the coproduct C with maps
-->
~
Cis a sort of 'general pushout'.
Clearly any element of K mapped to 1 by any
~·
~
must be
mapped to 1 by every a., so by modifying K and the G. we J
~
may as well assume that each ~i
is injective; this means
that K is embedded in G. via ~
~-·
If in this situation
~
all the a. are injective, ~
the coproduct is called faithful.
If moreover,
G.a. n G.a. = ~
J J
~
K~.
~
for all
i I j, the coproduct is called separating.
These defini-
tions apply quite generally for concrete categories (i.e. categories where the objects have an underlying set structure).
Now for groups we have the following basic result
(Schreier [27]): Theorem 5.1.1.
The coproduct of groups (over a fixed group)
is faithful and separating.
This is proved by writing down a normal form for the elements of the coproduct C.
= K~. be the image of
Let K.
~
~
K in Gi and choose a left transversal for Ki in Gi of the formS. U {1}, thus G. ~
~
=
K. u S.K.. ~
~
Then every element of
~
C can be written in just one way as
(1)
u 1u 2 ••• unc
(n
~
0; u 'V e:
s.
~
'V
'
i
v-1
;. i c e: K). v'
clear how to write any element of c in this form; to prove the uniqueness one defines a multiplication of the It
~s
93
expressions (1) (which consists in a set of rules reducing the formal product of two expressions to normal form), and verifies that one obtains a group in this way. way (v.d. Waerden
[48])
A quicker
is to define a group action on the
set of expressions (1) for each group G.: ~ i
i with the understanding that u
p
f i and cg
n
=
n
=
u c' in G.' ~
i and u cg n
P
u c' in G., ~ p
~s omitted if cg s K.
~
the second case, if uncg sKi).
(or in
Now it can be verified that
these group actions can be combined to give a C-action on the elements (1), and the conclusion follows because the expressions (1) are distinct. • We have given this proof in outline since it is very similar to the corresponding proof for the coproduct of fields, which we shall soon meet. The coproduct of groups over a given group K is usually called the free product of groups with amalgamated subgroup K.
It is also possible to define a notion of free product
of groups with different amalgamated subgroups, where we have a family of groups (G.) and subgroups H.. of G. such that H.. ~J
= H... J~
~
~J
~
This can again be constructed as a co-
product, amalgamating H.. with H.. , but it will in general ~J
J~
be neither faithful nor separating (cf. B.H. Neumann [54] and the references given there). Our aim in this section is to describe the coproduct of rings.
Thus let K be a fixed ring and consider K-rings;
it is easy to see that coproducts always exist.
We simply
take a presentation by generators and defining relations for each K-ring in our family and write all these presentations together. separating.
But the coproduct need not be faithful or
Before finding conditions for it to be so we
look at some examples. 94
1.
Let k be a commutative field, K
a
k[t.J, where A is
a central indeterminate, R In R
k(>.), s = k[>.,l-ll >.l-l = S, the coproduct of R and S over K, we have
K
~ = 1.~
A-l.A~ = O,
o].
so S is not faithfully repre-
sented.
2.
The inclusion Z C Q
an epimorphism, and Q ZQ
H
=
Q.
Hence the coproduct ~s faithful, but not separating. More generally, if R --~ S is any ring epimorphism, one finds that S ~
S
=S
(cf. Knight [70
J) .
When the coproduct of rings ~s faithful and separating we shall often call it the free product. when does the free product exist?
Thus the question is:
We begin by giving a
necessary condition; for simplicity we limit ourselves to two factors. Let R1 , R2 be K-rings which are faithful, i.e. the mapping K --> R.~ is injective. If the free product is to exist, we must have
Clearly this also holds with R1 and R2 interchanged, and more generally, for matrix equations. If we take all implications of a suitable form we can obtain necessary and sufficient conditions for free products to exist, but they will not be in a very explicit form.
Syntactical criteria of a
rather different form have been obtained by D.A. Bryars. We now come to a simple sufficient condition for the existence of free products which will be enough for all the applications we have in mind.
This states essentially that
the free product of a family {RA} of faithful K-rings exists provided that each quotient module RA/K is free as right Kmodule.
A direct proof, following the pattern of the proof
of Schreier's theorem 5.1.1, is quite straightforward.
How95
ever, we shall want to prove a little more and therefore follow a slightly different approach, which will place us 1n a position to obtain Bergman's coproduct theorem in the next section.
I am indebted to W. Dicks for the presentation
of these results. We shall want to consider the coproduct of a quite arbitrary family {RA} (A E A) of K-rings. venient to write K to write A,A 1 ,A \1,1-1',
= R0
It will be con-
where we assume that 0
i
A, and
etc. for the typical element of A and
0
etc. for the typical element of AU{O}.
Theorem 5.1.2.
Let R0 be any ring, {RAIA
E
A} a family of
= u R their coproduct. If the o R A 0 quotient modules R~/R0 are free as right R0 -modules, then
faithful
R -rings and R
the coproduct is faithful and separating, and for each ll E
AU {0}, R/R
ll
(hence also R itself) is free as right
R -module. ll
Further, for each 1-1 E AU ~ llll~
R -module and put M =
{O}, let M be a free right ll
$M
ll
R.
Then the canonical R ll
ll
module homomorphism M --> M is injective and the cokernel ll
is free as right R -module. ll
(Equivalently: M
M $ M' ll
ll'
where M' is free as right R -module). ll ll Proof. For each A e AletTA U {1} be a right R0 -basis of
RA and for each Mll.
ll e A
U {0} let S be a right R -basis of ll
ll
We shall denote the disjoint union of the TA by T,
that of the Sll by S and call the elements of TA or SA associated with the index A; the elements of S
are not
0
associated with any index. We begin by proving that M
= @M
basis consisting of all products (2)
ll
~ R
s e S, ti e T, n
has a right R 0
~
0,
where no two successive terms are associated with the same ~.
96
To establish this fact, denote the set of all such
formal products by U and let F be the free right R -module 0
on U.
We shall define a right R-module structure on F and
show that F
= M.
An element u ~ U is said to be associated with A if its
last factor (an element of S or T) is associated with A. The set of elements of U not associated with A is denoted by UA Fix A E A, then we may write F = S,R @ (U\S )R , A
o
11.
o
and we can give the first summand a right RA-structure by identifying it with MA; we note that this still holds if we replace A by 0.
Now consider the free right RA-module . We h ave an R0 -l1near map: UIt RA -->
. on UA as b as1s: UARA
(U\SA)R A
o
1-->
given by formal multiplication (u,t)
£ U , t £TAU {1}).
ut (u
This map is easily seen to be bi-
jective, so we have an RA-module structure on (U\SA)R0 and in this way F becomes an RA-module, for each A £ A; all the R -actions agree, so F is in fact ari R-module. 0
Moreover, for each M
-->
~ E AU {0}
there is an R -linear map ~
F which is injective, with a free complement, so
~
there is an R-linear map M --> F.
To show that this is an
isomorphism we construct its inverse.
Let f
the canonical R -linear map and define f:U ~
if
:M ~
-->
S E
~
--> M
be
M by S • ~
By R -linearity this extends to a map f:F --> M which is 0
clearly the desired inverse.
This proves the second part,
= RA it shows that the coproduct is faithful. To show that it is separating we take M0 = R0 , MA = 0 (A E A), then we find that M = R has an R0 -basis consisting
and applied to MA
of all products t 1 t 2 ••• tn (ti ~ T), where no two successive terms are associated with the same A. Thus if 1,2 £ A, 1 f 2, then the R0 -submodule R1 + R2 has as basis T1 u T2U {1}, hence R1 nR 2
R0 and this shows that the coproduct is
separating. • 97
When R
is a field (the case of main importance for us),
0
all the conditions are easily satisfied and we obtain the The free product of any family {RA} of non-zero
Corollary.
rings over a skew field exists and is left and right free over each RA • •
The free product is known to exist under more general conditions; instead of requiring RA/R0 to be free it is enough for it to be locally free (i.e. every finite subset is contained in a free submodule) or more generally, flat, i.e. each RA is faithfully flat (cf. Cohn [59]).
But
we observe that over a semifir, 'flat' means the same as 'locally free' (Cohn [71"], Th.l.4.4,p.56). Th.5.1.2 provides a means of finding the homological dimension of certain R-modules, for a coproduct R: Proposition 5.1.3. and M
=@
(Bergman
[74]}.
Let R be a skew field 0
R, then
M ll
sup (hdR M,,). ll
Proof.
ll
1-'
Clearly it suffices to show that hdR(Ml.l
~
=
R)
Now R is left free, hence left flat over R , ll
therefore -
~
R converts a projective R -resolution of )l
ll
to a projective R-resolution of Ml.l ~ R and so hdRMl.l ~ R
Mll
-
hdR M
ll ll
~
but the latter equals hdR M , because M and M ~ R differ ll ll
ll
ll
only by a free summand.• 5.2
Projective modules over coproducts over skew fields In this rather technical section we show that the pro-
jective modules of a coproduct R
=
RRA 0
98
over a skew field
R,
R
0
are of the form 9P
module.
~
~ R, where P~ is a projective R~ ~
[74],
The results are due to Bergman
tion ~s again due to Dicks.
the presenta-
We fix a skew field R and reo
tain the notation of the proof of Th.5.1.2. For any A
E ~
M = MA 9 UARA.
we obtained a direct sum representation Thus for any u
UA we have a mapping p\u:
E
M --~ R\, namely the coordinate in RA corresponding to the term u,
Similarly, since M = UR , we have for any u
E
0
mapping p
ou and for any
:M --~ R •
For convenience we write U for U,
0
~ E ~
X of M to be {u
€
Ua
0
U {0} we define the
U~,xp
~u
+0
~-support
for some x
€
X}.
of a subset The 0-
support will also be called the support. The elements of the basis U will be called monomials; the degree of a monomial is its length.
Let us well-order the
sets S and T arbitrarily and then well-order U by degree and monomials of the same degree lexicographically, reading from left to right. Next well-order 1\U {0}, making 0 the least element, and then well-order (A U {0}) x U first by the degree of the second factor, then (for a given degree) lexicographically from left to right.
Finally let H be the set of almost-
everywhere zero functions
(~
U {0}) x U
--~
N, well-ordered
lexicographically reading from highest to lowest in (~ U {0}) X
U. Consider any non-zero element x in M; the monomials in
the support of x will be called its terms.
The greatest such
monomial (in the ordering of U) is called the leading term and its degree is the degree of x, written deg x.
If all
terms of x of degree deg x are associated with \, x is called \-pure,
If x is not \-pure, the greatest element
~n
the
support belonging to UA is called the \-leading term (this should perhaps be called the non-\ leading term).
If x ~s
\-pure for some \, it is called pure, otherwise it is impure or also 0-pure, and its leading term is then called the 099
leading term.
With these preparations we can state the main result of this section. (Bergman
Theorem 5.2.1
[74]).
Let R
=
R RA
be a coproduct,
0
where R is a skew field, and for any family {MA}, where MA 0
is a free RA-module, put M
=~
R.
MA ~
If L is any sub-
A
module of M, then for each ~ module L
E
A U {0} there is an R -sub~
of L such that the canonical map @ L
1l
1l
~
R-"' L 1l
is an isomorphism.
Proof.
Fot each A
A let LA be the RA-submodule of L con-
E
sisting of all elements whose A.-support does not contain
By L
the A.-leading term of any (non )..-pure) element of L.
0
we denote the R -module consisting of all elements of L 0
whose support does not contain the leading term of any pure element of L.
We claim that the family {L } has the desired 1l
properties. To prove that EL R choose y
E
L, y
t
1l
=
L, assume that this is not so and •
EL R so as to minimize h(y) ~
E
H, where
1 if u is ~n the ~-support of y and y is ]l-pure,
{0
h(y) (]l,u)
otherwise. Suppose first that y is pure, say )..-pure.
Since y
t
LA,
some monomial u in the )..-support of y is the )..-leading term of some non A.-pure element x of L. therefore x
Clearly deg x < deg y,
ELllR; further, there exists c
E
E
RA such that
the )..-support of y - xc does not contain u, and y - xc is either )..-pure with less A.-support or of lower degree than y, hence h(y- xc) < h(y). and so y
E
EL R. 1l
It follows that y- xc EEL R
Next if y is impure, then since y
t
1l
L , 0
some monomial u in the support of y is the leading term of some pure element x of L. 100
It follows that h(x) < h(y),
therefore x
E
EL R, and again for some c
LL R.
E
the support
0
of y - xc does not contain u, so h(y y
R
E
ll
xc)
L is injective
we shall isolate the following evident properties of the family {L } : ll
For each ll e:: A U {0},
A • ll
All elements of L
are )1-pure.
11
For all ll 1 ,11 2 e:: A U {0}, B
111].12
The lll -support of L
•
which is also the
].1 1 -leading
].11
contains no monomial u
term of a (non
xa, where x e:: L112 , a e:: R and i f Jl_l
=
)1 1 -pure)
].1 2 , deg xa > deg
element
x as
well.
Given ].1 e:: A U {0}, we choose for each monomial u that lS the leading term of some element of L
an element q
ll
E
L
ll
having this leading term with coefficient 1, and denote the set of all such q's by Q ,
From the well-ordering of U and
ll
property A
ll
it follows that Q is an R -basis of L • ll
Let A e:: A and for each u of an element of L
£
A
U
choose an element q
0
\l
0
that is the ).-leading term L
E
0
having u as
\-leading term with coefficient 1, and denote the set of all such q's by Q0 A.
By property A0 every element of R0 has a
\-leading term, hence by the well-ordering Q0 A lS an R0 -basis of L0 for each A e:: A.
We shall call the elements of QA
"associated with \", those of Q0 A "not associated with\", and write Q
=
U (Q 0 Au QA).
Further, V will denote the union
of Q and the set of all products. 0
q
£
Q,
t.
£
l
T,
n ~ 1,
where no two successlve terms are associated with the same index; thus if q
£
Q0 A then n
~
1, and t 1
£
TA. 101
We shall show that the elements of V have distinct leading terms and hence are right R -independent. 0
By the same
argument as in the proof of Th.5.1.2 we conclude that
= VR0
~ L~ 6l F
L and the proof will be complete.
From the lexicographic ordering of U it follows that if the choice of q
Q was determined by the monomial u, then
E
Thus we are led to con-
qt 1 ••• tn has leading term ut 1 ••• tn, sider an equality of the form
= u't'1''' t'n
u t 1''' t m
in
u;
if m > n say, this reduces to an equality of the form
u'. Let q
E
L
~2
, q'
L
E
~1
correspond to u, u' respectively;
there are two cases: Case 1, m > n. support of q'
E
If 11 1 ··
L
11 1
:f 0, then t
m-n
contains ut 1 ••• t
E
m-n- 1
T
11 1
, so the p 1-
which is also the
~1-leading term of the non p 1-pure element qt 1 ••• tm-n-l'
and this contradicts B
fll~2
•
If ~l
= 0,
then since the
support of q' E L
contains ut 1 ••• t which is also the m-n leading term of the pure element qt 1 ••• tm-n' we again have a contradiction to B • o
~1~2
Case 2; m = n, then u = u' is associated with A, say. Then q and q' belong to Q U QA, where ll ~s the index Oll associated with t 1 = ti if m > o, and ll o, QOJl Qo if m = 0. By the construction of the Q's, if q :f q' • they cannot belong to the same set, say q E Q011 , q' £ QA. Then the support of q
E
L0 contains a monomial u which ~s also
the leading term of a pure element q' .1, where q' this contradicts B0 A. 102
E
LA, and
This shows that the elements of V have distinct leading terms, henoe Vis an R -basis of L, as we wished to show.• 0
Corollary
1
(Bergman
[74];.
Let {RA} be any family of non-
R0 is a skew field, and R
trivial R0 -rings, where
RRA,
=
o
then
when the right-hand side is positive, and r,gl.dim,R < 1 when all the RA have r.gl.dim. 0.
Proof.
By Prop.5.1.3, r.gl.dim.R
~
supA{r.gl.dim.RA}.
Now
let M be a submodule of a free right R-module F, then by Th.5.2.1, M ~
~
M
~
].J
R, where each M
~s
].J
F, which is projective as R -module, ].J
an R -submodule of ].J
Hence hdR M
S induces a monoid homomorphism P(R)
-->
P(S), defined by
[P] J-->
[P ~
R-module by pullback along f. tor from rings to monoids. that
s],
where S is a left
In this way
P becomes a func-
Our objective will be to show
P preserves coproducts over skew fields.
More pre-
cisely we have Theorem 5.3.1 (Bergman [74]).
Let R0 be a skew field, {RA}
a family of non-trivial R -rings and R their coproduct, then 0
the map
(1)
Pit )
P(R,_)
-> P(R)
0
induced in the category of monoids is an isomorphism.
Proof.
We first prove
~njectivity.
distinct elements ~[L ], ~[M l..l
l..l
J
Thus assume that two
of the left-hand side of (1)
have the same image, i.e. there is an isomorphism (2)
ex:
Ql L
l..l
& R
--~
ED M
]J
Ill R.
We shall retain the notation of the proofs of Th.5.1.2 and 5.2.1 for M = @ M
11
L
l..l
ct
in M.
6l
R and identify L with its image 11
Since each L
~s
11
finitely generated, we can asso-
ciate with the map a an element h
s H defined by the rule
ct
1 if u is in the ]J-support h (]J 0 U)
a.
{ 0 otherwise.
We may assume the pair I[L ~n
l..l
J f
I[M
11
J
and the isomorphism a
(2) to be chosen so as to minimize h • a
With this choice
we claim that the family {L } satisfies the conditions A ]J ]J.
104
B of the proof of Th.S.2.1, and hence L a = M • It will lllll2 ll ll be important for later applications that this can be done without using the finite generation of the M nor the proJl
jectivity of the L , M • ll
ll
If A fails then some x E L is not ll 1-pure. Let lll lll u E ulll be the ]1 1 -leading term of x and consider the restriction p' = p jL (where p :M --~ R is as defined on lll u lll u lll lll u lll p.99). The image of xis a non-zero element of R , hence 0
p~ u is surjective and so splits; thus L
1 where L' lll
=
ker p'
lJlU
!& xR
•
lll lll and u is clearly not in the ll 1-support
Now x must be ll 2-pure for some llz I lllo
Taking a
!& xR and for ll I lll'llz new symbol x, we define L' = L llz llz llz put L' L There is an obvious isomorphism!& L' ~ R --~ ].l
!&
L
Jl
ll
].l
~ R sending ~ to x, and if we follow this by a we ob-
tain an isomorphism S: E& L' where hB differs from ha
~
R
--~
M.
The first place
either Cll 1 ,u) or (ll 2 ,u'), where I 0, then (ll 1 ,u) > (ll 2 ,u') by considering
If llz llz degrees, and for llz
u E u'R
ll ~s
= 0
this inequality still holds, by the
Thus ha(ll 1 ,u) = 1 > 0 hS (lJ 1 ,u), which contradicts the minimality of h , because
ordering by first components.
a
L[L'] = Z[L] f z[M ]. ].l ].l ].l
Hence A holds for all ].l
J1 E
A U {0}.
contain a moIf B fails, let the ll 2-support of L lll J.llll2 nomial u which is also the ll 1-leading term, with coefficient 1, of a (non ll 1-pure) element xa, where x
E
L
, a c R and
llz
there is a L lll y - xa (yp) (yp c R ) , unique element y¢ of M of the form y¢ lll whose J.l 1 -support does not contain u. Indeed, p = p].l' u llz• deg xa > deg x.
For each y
E
1
L ->R ; we observe that this is R -linear. Let ¢:M -~ M be lll lll ]11 the R-linear map which leaves every element of Lll (ll f J1 1 ) 105
L to y~. We claim that ~ is an ]ll automorphism, with inverse e leaving LJl (Jl ~ Jl 1 ) fixed and fixed and which maps y
E
L toy+ xa(yp). If Jll ~ !lz' then x E L is ll1 llz fixed by ~ and e. If Jll llz' then deg xa > deg x, so u is
sending y
E
~
not 1n the Jl 1-support of x and xis again left fixed by
e.
and
It follows that
~
and
e
are mutually inverse.
Now
consider the isomorphism y: ~ L & R ~> M !_> M. The first ]l place where h and h differ is (Jl 1 ,u) and it is clear that a
y
This contradiction shows that B holds. Y a ll1Jl2 we can construct V as.in the proof of Th.5.2.1 • h
~ (P @ Q ) ~ ]l
such that In
]l
]l
= n; if we choose the n and a so as to minimise h
Jl Jl a then as in the first part of the proof Rnll = P $ Q hence P is f. ]l ]l ll' ]l itely generated. Thus (1) is surjective, and hence an isomorphism. This result enables us to derive several useful consequences without difficulty (cf, Cohn [63,64,68], Bergman [74]): Theorem 5o3o2o
The coproduct of a family of firs over a
skew field is a fir,
In particular, the coproduct of skew
fields (over a skew field) is a fir.
Proof. 106
By Th,5.2.1, Cor.l, the ideals of the coproduct are
projective, and by Th.5.3.1 all projectives are free of unique rank, hence all ideals are free of unique rank. • In 4.3 we saw that any semifir (and in particular, any fir) has a universal field of fractions in which all full matrices become invertible.
So we conclude that the coproduct of
fields EA over a field K has a universal field of fractions. This will be called the field coproduct or simply coproduct ~
of the EA, written E
K F.
EA or
~n
the case of two factors E,F,
Let us recall from Cohn
[71"], ch.l that a ring is
Morita-equivalent to a fir if and only if it is an n x n matrix ring over a fir, and observe that Th.5.2.l. Cor.l,2 and Th.5.3.2 are statements about categories of modules and hence Morita-invariant.
We deduce
[74]).
Theorem 5.3.3 (Bergman
Let R
0
be ann x n matrix ring
over a skew field, {RA} a fanlily of faithful R0 -rings and R their coproduct over
(i)
R0 • {
r.gl.dim.R
Then
supA{r.gl.dim.RA} i f this is :f= 01 0 or 1 otherwise;
(ii) every projective R-module has the form
~
P
11
~
R, where
is R -projective 1
P
1l
1-1
(iii) P(R) = p*) P(RA).• 0
If we exam~ne the proof of Th.5,3.l, we find that it shows rather more than is asserted. ~:
(3)
$
M
1l
~
R
--~ i
N
1l
~
Consider a homomorphism
R.
If ~ arises from a family of R -linear maps ~ :M --> N , we 'Y
shall call it induced.
(4)
11
11
11
11
Next we observe that ~
R
(M
llz
i
R )
llz
~
107
R,
because R
~1
~
R
= R~2
~
R
= R.
An isomorphism (3) arising
by a transfer of terms as in (4) is called a free transfer isomorphism. A second kind of isomorphism arises as follows. Let e:M
--> R
~1
~1
be an R -linear functional, extended to M ~1
so as to annihilate MV for~ f ~ 1 •
Given x s M , we have a ~2
map a(x):R --> M defined by 11--> x s M ~ R~ M. Clearly v2 a(x)e = 0 if ~l f v 2 and this holds even for v 1 = ~ 2 if we add the condition x
ker e.
£
Then for any a s R the map
ea(a)a(x):u 1--> xa(ue) is nilpotent, and so 8
= 1-
ea(a)a(x)
is an automorphism of M; such an automorphism will be called a transvection, v-based in case
~l
=
~ 2 =~and
a s
R~.
The proof of Th.5.3.2 shows that every surjection (3) where M is finitely generated can be obtained as a composite of a ~
finite number of free transfer isomorphisms, transvections and an induced surjection. Proposition 5.3.4.
Let R0 be a skew field and {RA} a family
R0 -rings, then their coproduct R is entire and any
of entire
unit in R is a product of units in the RA.
We shall call such a unit a monomial unit. Proof. R
Each unit u s R defines an automorphism x 1--> ux of
= R0
R.
~
Here R is free on a single generator; any free
transfer isomorphism just amounts to renaming the generator, while a surjection is a unit in some R • V
The only transvec-
tion is the identity map, since it must be v-based for some
v,
but R
v
is entire.
This proves the second part.
Now the
assertion about zero-divisors is the special case n
=1
of
the next result. Proposition 5.3.5. integer.
Let R
0
be a skew field and n a positive
Then the coproduct of any family of n-firs over
R is an n-fir. 0
n
Proof.
For any map R --> R the image can by Th.5.2.1 be
written as ~ M ~ R, so there is an induced surjection
v
a: ~ Rn~ ~ R --> ~ M ~ R, where Ln ~
108
v
v
= n.
Since M is a
v
submodule of the projective R -module R, M ~s free of ]1 )1 rank at most n and so Ell M & R is free of rank at most n. ]1 ]1 If we repeat this argument with an isomorphism, we see that the rank must be unique. • Bergman [74] determines the units and zero-divisors of coproducts over skew fields in a more general situation. We can also obtain the elements of a coproduct which are algebraic over the ground field.
Let {RA} be a family of
entire rings over a skew field R ; their coproduct R is a 0
free product and is entire, by Th.5.1.2, Cor. and Prop.5.3.4. If a
E
R is right algebraic over R , it satisfies an equation 0
o,
+ ••• +
y.
~
R , not all o.
E
0
Without loss of generality a # 0, and since R is entire, we
# 0; on dividing by the constant term we
may assume that y
= 1.
may take y 0 ab + 1 R.
n
Then a(a
0, so a(-b)
=1
n-1
y
o
+ ••• + y
n- 1
) + 1
= o,
i.e.
and it follows that a is a unit in
By Prop.5.3.4, it must be a monomial unit and we can
write it as a if deg u
>
= p -1 up,
E
RA for some A, or
1, the first and last monomial factors of u are
in different rings RA. sume that deg up (5)
where either u
n
u PYO + u
= deg
n-1
u + deg p; then
pyl + ••• + upy n- 1 + p
= r.deg
But deg(urp)
In the latter case we may also as-
= 0.
u + deg p, so the first term in (5) has
greater degree than the rest and we have a contradiction. So this case can be excluded, and we obtain Let {RA} be a family of entire rings over a
Corollary 1. skew field R
0
and R their coproduct.
Then any element of
R which is right (or left) algebraic over R0 is conjugate to an
ele~nt
in one of the factors. •
It should be observed that not every element conjugate 109
to an element of some RA that is algebraic over R0 need itself be algebraic over R • over R0
,
p
-1
For if a satisfies an equation
0
ap may not do so; only when R0 is in the centre
of each RA and each RA algebraic over R0 does the converse of Cor.l hold. When different subfields are amalgamated in different factors Th.5.3.2 and Prop.5.3.4 no longer hold, in fact R need not even be entire.
This follows from an example for
groups due to B.H. Neumann [54].
Let k be a commutative
field and form the fields
Kl
k(x,y) with defining relation y
K2
k(y,z) with defining relation z
K3
k(z,x) with defining relation
-1 -1
-1
X
xy
X
yz
y
zx
z
-1
• • -1 -1
These fields can of course be constructed as fields of fractions of skew polynomial rings, e.g. K1 = L(y;a), where L = k(x) with automorphism a:f(x) J--~ f(x- 1 ). We form the coproduct P of K1 , K2 , K3 with amalgamations K12 = k(y), K23 = k(z), K31 = k(x). To see that this is a free product, we first form k(s,n,s) determinates. f(s,n,s)x
~n
three commuting in-
Next adjoin x subject to xf(s,n,s
-1
),
X
2
s·
Then adjoin y subject to f(x,n,s)Y
= yf(x-1 ,n,s),
y
2
n,
and finally adjoin z subject to f(x,y,s)z
= zf(x~,y
-1
.~)
z
2
s·
It is easily verified that the resulting coproduct is a 110
free product.
If e.g. there is a relation between x and y,
we can write it as a polynomial in y: +
~
Conjugating by
y
n
+ a 1 (x
+ a
2
z ~
2
0
n
a.
a. (x).
~
~
we find
n-1
)y
+ ••• + a (x n
~
2
)
o,
and by the uniqueness of the minimal equation for y, a.(x
2
~)
~
a.~ (x), hence a.~ is independent of x, soy is algebraic over k, clearly a contradiction.
Thus P is a free product, but in
P, xyz is an element of order 2: -1
xyz
thus (xyz)
5.4
yx
2
z = yz
-1 -1 x
Z
-1 -1 -1 y
X
1; this shows that P is not even entire.
The tensor K-ring on a bimodule Let K be any field, then the free K-ring on a set X, K,
may be defined by the following universal mapping property: K is generated by X as a K-ring, and any mapping X --> R into a K-ring R such that the image of X centralizes K (i.e. ya
ay for all a
E
K and y in the image of X) can be extended
to a unique K-ring homomorphism of K into R.
The elements
of K can be uniquely written as
(x.e:X,a . •
• • •x .
1
r
1
11 • •
.l.r
e:K) •
As is easily seen, K may also be represented as a coproduct K
i!!:
w
X
K[x] ,
111
where x runs over X, and since each K[x] is a principal ideal domain (and hence a fir), it follows from Th.5.3.2 that K is a fir.
We shall now outline another way of
establishing this fact which will be useful when we come to consider a generalization of K needed later. Let K be any field and M a K-bimodule. (n factors),
MfilMCil ••• CilM thus M1
We put
= M and by convention, M0
= K.
It is clear that
Mr filMs~ Mr+s, hence we have a multiplication on the direct sum T (M) = EB ~,
which turns it into a K-ring. K-ring on M,
This ring is called the tensor
It is shown in Cohn [71"], Ch.2 that this ring
possesses a weak algorithm and hence is a fir.
We shall not
repeat the proof here, but note that it yields another proof that K is a fir.
Let M
~
EB K and denote by x the X
element corresponding to 1 s K in the factor indexed by x s X. Thus the general element of M has the form Ea x (a X
almost all are 0), and it is easily seen that T(M)
X
€
K,
= K.
Let E be a field with K as subfield, and put (1)
E KK.
By Th.5.3.2 this is again a fir, but we can also obtain this ring as the tensor E-ring of a module. E-subbimodule spanned by x s X.
In EK consider the
Its elements are of the form
Ea.xb. (a.,b. s E) and it is clear from the definition that ~ ~ ~ ~ this module is isomorphic to E
~
E.
Hence we see that EK
can be described as a tensor E-ring as follows:
112
Since this is a fir (hence a semifir) we see by Th.4.C that it has a ~niversal field of fractions.
= K, the ring
When E
~
We shall write
is just the tensor ring K in-
troduced earlier, so no confusion need arise if we regard K as an abbreviation for
~,
and correspondingly de-
note the universal field of fractions by K{X}. To elucidate the relation between EK{X} and K{X} we need a lerruna. Lemma 5.4.1.
Let R, S be semifirs over a field K, then
the inclusion map
R u S
--~
R
--~
(i)
R u S is honest, (ii) the map
F (R) uS is honest.
Proof. (i) Consider the injections R --;:. R u S
--~
F (R) u S.
Let A be full over R, then it is invertible over F(R), hence also over F(R) uS, and so 1s full over Ru S, as we had to show. (ii) By (i) any full matrix over R is full over Ru S, hence we have a homomorphism F (R) - > F(R u S) (Th.4. 3. 3) and it follows that we have a homomorphism F (R) u S -> F(R
~
S).
Thus we have mappings
R u S --> F (R) u
S - > F (R U S).
Now any full matrix over Ru Sis invertible over F(Ru S) and hence is full over F (R) U S. • Proposition
5.4.2.
Let
K ~ E be any fields, then
E {X} ~ E ~ K{X}.
Proof.
Put R
K, then we have to show that
RUNT LIBRARY CARNEGIE-MELLON U"IVERSIH .
n,.,..,.n,.,
C,"l
D~ •,t·· ·,, V-','Jt,\
1~?1.
F u L --;:. F
F (E u L) •
U
Any full matrix over E u L is invertible over F U F (E u L), and hence full over F LJ L.
Thus F (E LJ L) is embedded in
F (F LJ L) and by Prop. 5. 4. 2, this is the result claimed. • Suppose that we have fields K ~ E
~
F, K ~ K'
~
F, then
we have a natural map
and when this map is honest, we obtain an embedding of EK{X} in FK 1 {X}, but in general (3) need not even be injective.
Thus let x
E
X and c
E
K1 n E, then ex - xc maps
to 0 under (3), but it is not itself 0 unless c
E
K.
Thus
a necessary condition for (3) to be injective is that (4)
K'
n
E
K.
Later in 6.3 we shall see that when K is contained in the centre of E, (4) is also sufficient for (3) to be honest. 5.5
Subfields of field coproducts Although Schreier had discussed free products of groups
114
~n
1927, it was not until more than 20 years later that
significant applications were made, notably in the classic paper by Higman-Neumann-Neumann [49].
Their main result
was the following
5.A.
Theorem
Let
G be any group with two subgroups A, B
which are isomorphic, say f:A --> B is an isomorphism. Then G can be embedded in a group H containing also an element t such that
t
-1
at
af
for all a
£
A.
We observe that this would be trivial if f were an automorphism of the whole of G: then H would be the split extension of G by an infinite cycle inducing f.
But for
proper subgroups A, B the result is non-trivial and (at first) surprising. co~sequences
It has many interesting and important
for groups and it is natural to try and prove
an analogue for skew fields. ~n
the category of fields.
What one needs 1s a coproduct However, we shall not adopt a
categorical point of view: the morphisms in the category of fields are all monomorphisms and this strictive.
~s
somewhat re-
Over a fixed ring, it is true, we have defined
specializations, but it would be more cumbersome to define them without a ground ring, and not really helpful. In this section we shall prove an analogue of the HigmanNeumann-Neumann theorem using the field coproduct introduced
~n
5.3.
But we shall also need some auxiliary re-
sults on subfields of coproducts.
It will be convenient to
regard all our fields as algebras over a given commutative field k; this just amounts to requiring k to be contained in the centre of each field occurring.
The proof of the
next result is based on a suggestion by A. Macintyre. Theorem 5.5.1.
Let K be a field and A, B subfields of K,
isomorphic under a mapping f:A --> B, where K,A,B are k-
115
algebras and f is k-linear.
K can be embedded in a
Then
field L, again a k-algebra, in which A and B are conjugate by an inner automorphism inducing f, i.e. L contains t
f
0
such that
af Proof.
t
-1
at
for all a
E
A.
Define K as right A-module by the usual multiplica-
tion and as left A-module by a.u
(1)
a
(af)u
A, u
£
E
K.
Let us form the K-bimodule K NA K, with the usual multiplication by elements of K; if we abbreviate 1 consists of all sums
~u.tv. ~
(u.,v.
~
~
E
~
~
1 as t, this
K) with the defining
relations at
(2)
t.af
.. (a
E
A).
By the remarks in 5.4, the tensor K-ring T(K
~A
and so has a universal field of fractions L.
K) is a fir,
Thus we have
embedded K in a field L in which (2) holds. • Let K be a field with k as subfield of the centre, then K is said to be finitely homogeneous over k, if for any elements a 1 , ••• ,a ,b 1 , ••• ;;b E K such that the map a. I-!> b. n n l l defines an isomorphism k(a 1 , ••• ,an) ~ k(b 1 , ••• ,bn)' there exists t
£
K* such that t
Corollary 1.
-1
a.t =b. ~
~
(i = l,ooo,n).
Every field K (over a subfield k of its centre)
can be embedded in a field (again over k) which is finitely homogeneous.
Proof.
Given a's and b's such that a.
~
1-!>
b. defines an ~
isomorphism, we can by Th.S.S.l extend K to include an element t f 0 such that t
-1
a.t =b., and the least such ex~
~
tension has the same cardinal as K or is countable. 116
If we
do this for all pairs of finite sets in K which define isomorphisms we get a field K1 , still of the same cardinal as K or countable, such that any two finitely generated isomorphic subfields of K are conjugate in K1 •
We now repeat
this process, obtaining K2 and if we continue thus we get a tower of fields
Their union L 1s a field with the required properties, for if a 1 , ••• ,an, b 1 , ••• ,bn £ Land ai 1--~ bi defines an isomorphism, we can find K to contain all the a's and b's, hence r
they become conjugate inK
and a fortiori in L.• r+ 1 A finitely homogeneous field has the property that any
two elements with the same (or no) minimal equation over k are conjugate. Corollary 2.
Hence we obtain Every field K (over a subfield k of its centre)
can be embedded in a field L in which any two elements with the same minimal equation over k are conjugate, as are any two
transcendenta~
elements.•
Let K be any field, then the group of fractional linear transformations PGL 2 (K), consisting of all mappings
x 1--~ (ax+ b)(cx +d)
-1
is well known to be triply transitive, i.e. for any two triples of distinct elements of Koo
=
K U { oo } there is a
transformation mapping one into the other.
With the help
of Cor.l we can (as P.J. Cameron has observed) construct a field on which PGL 2 is 4-transitive. We need only take the field of rational functions in one variable over GF(2) and embed it in a finitely homogeneous field L.
Given any two
117
elements a,b of L different from 0 and 1, there is an inner automorphism a mapping a to b.
Now by the result quoted
above, PGL 2 is triply transitive on L and the stabilizer of {0,1, oo} is still 4-transitive (by conjugation), hence PGL 2 00
is 4-transitive. Later we shall need an analogue of Cor.2 for matrices instead of elements.
Let A
E~
n
(K), then A is said to be
transcendental over k if for every non-zero polynomial
f
E
k[t] the matrix f(A) is non-singular.
Clearly if A is
a transcendental matrix over k, then the field generated over k by A is a simple transcendental extension of k.
The
next lemma and its application to the proof of.Th.5.5.3 are due to G.M. Bergman (cf. Cohn [73"]). Lemma 5.5.2.
Given a field K (over k) and n
~
1, let E be
a subfield of linn (K).
If F1 , F 2 are subfields of E which are isomorphic under a map ¢:F 1 --~ F2 , then there is an extension field L of K such that x¢
for all x e: F1 ,
for some T e: GL (L). n
Proof.
By Th.5.5.1 E has an extension field E' with an
element T inducing ¢.
Consider R =linn (K)
jt E'; by Tho 5. 3. 3
this is hereditary and every projective R-module is a direct sum of copies of P
~
R, where P is a minimal projective for
Since Pn ~ R =lin n (K) ~ R ~ R, it follows (by Th.l.4.2 of Cohn [71"]) that R is ann x n matrix ring over a fir,
linn (K).
say R =linn (G), where G ~s a fir containing Ko
Let L be
the universal field of fractions of G, then L contains K andlinn(L) contains the element T inducing
¢·•
Let K be a field and suppose that~ (K) contains ison
morphic subfields F1 ,F 2 ,F 3 with isomorphisms f:F 1 --> F2 , g:F 2 --~ F3 say, and such that F1 , F2 lie in a common subfield oflinn(K), as do F2 , F 3 o Then by Lemma 5.5.2 we can 118
enlarge K to a field L and obtain a unit X such that conjugation by_·X induces f.
Now F 2 ,F 3 still lie in a common subfield of9n (L) and we can enlarge L to a field M to obn
tain a unit Y which induces the isomorphism g and F 3 •
between F2
Now XY induces the isomorphism fg:F 1 --> F 3 ; in
this way the scope of the lemma can be extended.
As a
result we can prove Theorem 5.5.3.
Let K be a field (over k) and n ~ 1.
Given
two n x n matrices A, B over K, both transcendental over k, there exists a field extension L of K containing a non-singular matrix
Proof.
T such that T- 1AT
B.
Since A is transcendental, k(A)
lS
a purely transcen-
dental extension of k, thus if u is a central indeterminate over K, we have k(A)
~
k(u) and similarly k(B)
~
k(u).
We
shall take F 1 ~ k(A), F 2 ~ k(u), F 3 ~ k(B). Let K((u)) be the field of formal Laurent series in u over K, then (3)
in (K((u)) ) ll9n (K) ((u)). n
n
Now F1 , F 2 are contained in the subfield k(A)((u)) of (3), while F2 , F 3 are contained in k(B)((u)).
We can therefore
apply Lemma 5.5.2 and the remark following it and obtain an extension field H of K((u)) such that9n (H) contains a unit n
T inducing the k-isomorphism k(A) ~ k(B) defined by A
[-->
B.•
Clearly we can repeat the process for other pairs of transcendental matrices until we obtain a field K1 ~ K in which any two transcendental matrices of the same order over K are similar.
If we repeat the construction for K1 we get
a chain of fields (over k):
whose union
lS
a field with the property that any two tran-
scendental matrices of the same order are similar, thus we 119
have the Let K be a field (over k) then there exists an
corollary.
extension field L of K (over k) such that any two matrices of the same order over L and both transcendental over k are similar over L. •
This means, for example, that over L any transcendental matrix A can be transformed to scalar (not merely diagonal) We need only choose a transcendental element a of L;
form.
clearly a is transcendental as n x n matrix, therefore T-lAT =a for some T
E
GL (L). n
We shall return to this topic
in 8.4. Our next objective is to show that every countable field can be embedded in a 2-generator field.
This corresponds to
a theorem of B.H. Neumann [54] for groups. some lemmas on field coproducts.
We shall need
First we examine a situa-
tion in which a subfield of a given field is a field coproduct.
5.5.4.
Lemma
P
=K
aH(x),
Let
where xis an indeterminate centralizing H.
Then the subfield
-i
X
Kx
i
Proof.
(i
E
K be a field with a subfield H and let G of P generated by the fields K.~
Z) is their field coproduct over H.
Take a family of copies of K indexed by Z , say {K.}, ~
denote by R their coproduct over H and by U the universal field of fractions of R, thus U is the field coproduct of the K. over H. ~
By the universal property of U it follows
that the subfield Q described in the lemma is an R-specialiFrom the universal property of P = K 0 H(x), . H this specialization will be an isomorphism whenever there zation of U.
is some K-field L containing an element y # 0 such that the i -i specialization from U to L which maps K. to y Ky is an ~
embedding.
Such an L is easily constructed: the mapping
K.~ --> K.~+ 1 ~s an automorphism of R which extends to an automorphism a say, of U. Now form the skew function field U(y;a); it has all the properties required of L.• 120
We shall also need a result on free sets in field coproducts.
~iven
a field over k, by a free set over k we
understand a subset Y such that the subfield generated by
Y is free, i.e. isomorphic to the universal field of fractions of the free algebra k. Lemma 5.5.5.
Let E be a field, generated over k by a family
{eA} of elements, and let U be the field freely generated by a family {uA} over k, then the elements uA + eA form a free set in the field coproduct U
Proof.
0
k
The field coproduct G
E.
=U
0
k
E has the following
universal property: given any E-field F and any family {fA} of elements ofF, there is a unique specialization from G to F (over E, with domain generated by E and the uA) which maps uA to fA.
In particular, there are specializations
from G to itself which map uA to uA + eA (respectively to uA - eA).
On composing these mappings (in either order)
we obtain the identity mapping, hence they are inverse to each other, and so are automorphisms.
It follows that the
uA + eA like the uA form a free set.• We can now achieve our objective, the embedding theorem mentioned earlier; the proof runs closely parallel to the group case. Theorem 5.5.6.
Let E be a field, countably generated over
a subfield k of its centre, then E can be embedded in a 2-generator field over k.
In essence the proof runs as follows: Suppose that E is g enerated by e 0
=
0 '1 e '2 e ' · · · ·· we construct an extension
field L generated by elements x,y,z over E satisfying
y
-i
xy
i
= z
-i
xz
i
+ e.
1
(i
0,1, ••• ).
Then L is in fact generated by x,y,z alone. If we now ad. f"~e 1d join t such that y = txt -1 z = t -1 xt, the resu1t1ng is generated by x and t. 121
To prove the theorem, let F1 be the free fiel~ o~ x,y -l l over k; it has a subfield U generated by u.l = y xy (i = 0,1, ••• ) freely, by Lemma 5.5.4, and similarly, let F 2 be the free field on x,z over k, with subfield V freely genera-i i . 9 • ted by vi = z xz (l = 0,1, ••• ). Form K = E k F1 ; thls has a subfield W generated by w. = u. +e. (i 0,1, ••• ), l l l freely by Lemma 5.5.5. We note that w0 = u 0 + e 0 x0 = x ' so K is generated over k by x,y and thew.l (i -> 1). Let L be the field coproduct of K and F 2 , amalgamating We note that w
Wand V along the isomorphism w. v .• l
X= V 0
0
l
and that L lS generated by x,y,z and the w.l or also
by x,y,z and the v., or simply by x,y,z.
Now L contains the
l
isomorphic subfields generated by x,y and by z,x respectively, -1 -1 hence we can adjoin t to L such that t xt z, t yt = x (by Th.5.5.1).
It follows that we have an extension of L
generated by x,t over k and it contains K. • As usual we have the Corollary.
Every field over k can be embedded in a field L
such that every countably generated subfield of L is contained in a 2-generator subfield of L.
Proof.
Let E be the given field and EA a typical countably
generated subfield (always over k), then there is a 2generator field LA containing EA, by the theorem.
Let MA
be the field coproduct of E and LA over EA; if we do this for each countably generated subfield of E we get a family {MA} of fields, all containing E.
Form their field co-
product E' amalgamating E, then in E' every countably generated subfield of E is contained in a 2-generator subfield of E', namely EA is contained in LA.
Now repeat the process
that led from E toE': E C E' C E" C • • • C Ew C E w+ 1 C ••• CE, v where Ea.
122
U {E 13 J 13 < a} at a limit ordinal
a, and where v
is the first uncountable ordinal.
Then Ev is a field in
which every countable subfield is contained in some Ea (a< v) and hence in some 2-generator subfield of Ea+l~
E". • At this point it ~s natural to ask whether there is a countable field, or one countably generated over k, containing a copy of every countable field (of a given characteristic).
As
~n
the case of groups, the answer is 'no';
this is shown by the following argument, for which I am indebted to A. Macintyre. For any field K, let S(K) be the set of isomorphism types of finitely generated subgroups of K*. countable, then so is S(K). that there are c
2x
0
If K is
Now D.B. Smith [70] has shown
isomorphism types of finitely gen-
erated orderable groups.
Further, every ordered group can
be embedded in a field of prescribed characteristic, by the methods of 2.1; hence every countable ordered group can be embedded in a countable field.
It follows that there are
c distinct sets S(K) asK runs over all countable fields of
any given characteristic.
Therefore these fields cannot all
be embedded in a 2-generator field. 5.6
Extensions with different left and right degrees
In Ch.3 we examined a particular kind of binomial extension.
Given a prime number p and a primitive pth root
of 1, w say, in our ground field k, let us take a field E with an endomorphism S and an S-derivation D such that
(1)
DS
wSD
and construct fields K and L as in Th.3.4.4.
We then have
an extension L/K of right degree p and its left degree will
s
be > p if we can show that K ~ K. [K:K 5]L = ""• then [L:K]L
=
oo
More generally, if
But some care is needed here: 123
it is not enough to take [E:E 5] 1 shall have KS
0, (because S is then an inner automorphism of L, with inverse c 1--~ tct- 1). Likewise one can =
show that [L:K] 1 son [56],Ch.6).
K if D
= oo, for whatever S is, we
=
=
[L:K]R whenever K is commutative (Jacob-
To obtain the required example, let p be a prime number, k any commutative field containing a primitive root of 1, w say, and let A be any set.
When k has characteristic p this
We form the free algebra F 2 kon a family of indeterminates indexed by AxN • Let is taken to mean that w
=
1.
A~J
E
=
F(F) be the universal field of fractions of F.
On F we
have an endomorphism S defined by
This is an honest endomorphism because F5 ~s a retract of F: Let T be the endomorphism defined by
if j > 1, if J Then ST
= 1
1.
(read from left to right), hence if AS is non-
full, then so is AST
= A,
i.e. S is honest, as claimed.
It
follows that S extends to an endomorphism of E, again denoted by S. Next let D be the S-derivation of F defined by
D x>..ij
xA i+l j"
This again extends to an S-derivation of E, still denoted by D.
We now form L
= E(t;S,D) and K = E(tP;sP,nP), as in
Th.3.4.4, then L/K is a binomial extension of right degree p. 124
To show that the left degree is > p it is enough to
s
prove that K f K.
This can be shown quite easily whatever
A, e.g. we/could take A to consist of one element.
But we
are then left with the task of finding whether [K:K5]L is finite or infinite.
It is almost certainly infinite, but
this is not easy to show when
[A[ = 1,
whereas it becomes
easy for infinite A. For any
~
£
A denote by E the subfield of E generated ~
=1
over k by all x, . . such that j > 1, or j /\1]
thus we take all x's except x . 1 (i e: N). E
~
].J-
xAll
ES £
for all
~.
and E (t) ~
s
~]._ ~ L
-
~
s
K •
EJ.l (t) if and only if A f J.l•
and A f
~;
It follows that We claim that
Assuming this for the
moment, we see that the xAll are left linearly independent over K5 , for if LaAxAll = O(aA £ K5 ), and some aJ.l f 0, then we could express x].Jll in terms of the xAll' A f ].J, and so x~ 11 £
EJ.l(t), which contradicts our assumption.
That xAll
£
EJ.l(t) for A f J.l is clear from the definition.
To show that x].Jll i E~(t), writeR
EJ.l[t] (for a fixed ].J)
and observe that for any a
ta 5 + aD, hence
at - a
D
(mod R),
and so by induction on n, at If x].Jll
E
E, at
E
n _
=a
Dn
(mod R).
EJ.l(t), we would have x].Jll
Then x~ 11 g _ f then
=0
(mod R), and if g
-1
ig
, where f,g
= Lt Si' where Si
E
R.
e: EJ.l,
(mod R)
Here we have multiplied a congruence mod R by elements of R, which is permissible.
Thus we have 0
S.,S e: R. ]._
125
This is an equation in E[t], more precisely in the subring E [ t] u k (clearly this subring is a coproduct) and by 1-l 1-l~ equating cofactors of the x . 1 we see that S = s.~ = o, i.e. JH
g
= Oo But this is a contradiction, for g as denominator of
x'J..lll cannot vanish.
This proves that x'J..lll i El-l(t) and it
follows that [L:K] 1 ~ 111.1. Given any infinite cardinal a, take a set A of cardinal in this case.
ILl = a,
hence [L:K] 1 = a Thus we have found an extension with right
a and k a countable field, then degree p and left degree a.
If instead of p we have a com-
posite integer n, pick a prime factor p of n and combine an extension of (left and right) degree n/p with the previous case.
Similarly if
a is
an infinite cardinal < a, we can
start with an extension of degree Theorem 5.6.1.
a.
Thus we have proved
Given any two cardinals
a,a
of which at
least one is infinite, there is a field extension L/K of left degree a and right degree
S, and of prescribed charac-
teristic. •
Whether the left and right degrees can be both finite and different remains open.
On the face of it this looks un-
likely, but it does not seem an easy problem to decide.
126
6 · Gen-era I skew field extensions
6.1
Presentations of skew fields We have already discussed skew field extensions (in Ch.J),
but they were usually of a rather special sort, of finite degree (at least on one side). sions.
We now turn to general exten-
Of course it is no longer true, as in the commutative
case, that every simple extension of infinite degree is free, in fact we shall need to define what we mean by a free extensian. To reach the appropriate definition, consider a finitely generated field extension E
= K(a 1 , •• o,an).
As before we
shall take all our fields to be k-algebras, where k is a commutative field.
This represents no loss of generality
(in fact a gain): if k is not present we can take the prime subfield to play the role of the ground field.
Given E as
above, we have a homomorphism of K-rings: ~
(1)
--> E,
x.~
X
1-»
a. .• ~
Here ~ is the coproduct Kk k; it ~s called the tensor K-ring on X over k. of (1); since E
~san
epic
Let
P be
~-field
the singular kernel
(being generated by
the a. over K), it is determined up to isomorphism by ~
M be a set of matrices generating
P.
Let
P as matrix ideal, then E
is already determined by M; we write
(2)
E
= ~ {X;M} 127
and call this a presentation of E.
In particular we call
the a. free over K if the presentation can be chosen with 1
M
~;
this just means that (1) is an honest map, i.e. that
P consists of all non-full matrices and no others.
From
Cohn [71"] Ch.2 we know that ~ is a fir and so has a universal field of fractions, written
~
{X} and called the
free K-field on X.
Given any set X and any set M of matrices over
~,
we
can ask: When does there exist a field with presentation ~{X;M}?
(3)
Let (M) be the matrix ideal of
~
generated by M, then
there are two possibilities: (i)
(M) is improper.
Then there is no field (3), in
fact there is no field over which all the matrices of M become singular.
Here there is no solution
because we do not allow the 1-elernent set as a field. (ii) (M) is proper.
Now there is always a field over
which the matrices of M become singular, possibly more than one.
The different such fields corres-
pond to the prime matrix ideals containing (M), and there is a universal one among them precisely when the radical I(M) is prime.
In particular, this is
so when (M) is prime, and that will be the only case in which the notation (3) will be used. Let E be a field with presentation (3); we shall say that E is finitely related when M can be chosen finite, and E 1s finitely presented if X and M can both be chosen finite.
for groups we have Theorem 6.1.1
128
A finitely related field can be expressed
As
as the field coproduct of a fintely presented and a free field.
Let E = ~{X;M}, where M is finite,
Proof.
Then the set
X' of elements of X occurring in matrices from M is finite. Let X" be the complement of X' in X, then we clearly have 0
E = E'
E", where E' =~{X' ;M}, E" =~{X"}.
Here E' is
finitely presented and E" is free.• In the special case when E/K has finite degree, the above construction can be a little simplified. is surjective, not merely epic.
In that case (1)
Moreover, instead of taking
the free algebra, we can incorporate the commutativity relations as follows.
Let u 1 , ••• ,un be a right K-basis of E, then since E is a K-bimodule, we have the equations
(4)
a.u. = L:u.p .. (o;) J ~ ~J
{a. e: K) '
where a 1--~ (p .. (a)) is a homomorphism from K to K. n
~J
Let
M be the free right K-space on u 1 , ••• ,un as basis; by the equations (4), M becomes a K-bimodule, which contains K as submodule (as we see by choosing our basis of E so that u1
=
1).
Let ¢K(M) be the filtered ring on this bimodule,
constructed as in 2.5 of Cohn [71"]; by Th.2.5.1, l.c., ¢K(M) has weak algorithm and hence is a fir.
Now E is ob-
tained as a homomorphic image of ¢K(M); so we need to look for ideals in ¢K(M) which as right K-spaces have finite codimension, in fact the kernel of (1) in this case is a complement of M
~n
¢K(M).
But it is not at all clear how
this would help in the classification of extensions of finite degree. In practice most of the presentations we shall meet are given by equations rather than singularities, but the latter are important in theoretical considerations, e,g, when we want to prove that an extension is free we must check that there are no matrix singularities.
We shall return to this 129
question in 8.1. A special case occurs when E (and hence also K) is of finite degree over k.
In that case the singularity of a
matrix can be expressed by the vanishing of a norm and hence it will be enough to consider equations.
Only in the case
of infinite extensions are the matrix singularities really needed. 6.2
Existentially closed skew fields Let k be a commutative field.
k one usually understands a field (i)
k
By an algebraic closure of
k
with the properties:
is algebraic over k,
(ii) every equation over k has a solution in k. It is well known that every commutative field has an algebraic closure, and that the latter is unique up to isomorphism (though not necessarily a unique isomorphism, thus the algebraic closure is not a functor).
When one tries to
perform an analogous construction for skew fields one soon finds that it is impossible to combine (i) and (ii).
In
fact (i) is rather restrictive, so we give it up altogether and concentrate on (ii).
Here it is convenient to separate
two problems, namely (a) which equations are soluble (in some extension) and (b) whether every soluble equation has a solution in the closure. Of these (a) is a difficult question to which we shall return later, and for the moment concentrate on (b). ~o
The assertion that an equation f(x 1 , ... ,xn) has a solution can be expressed as follows:
Any sentence of the form 3a 1 , ••• ,an P(a 1 , ••• ,an)' where P is an expression obtained from equations by negation, con130
junction and disjunction is called an existential sentence. By an existentially closed field, EC-field for short, we understand a field K (over k) such that any existential sentence which holds in some field extension of K, already holds in K.
Clearly such a sentence can always be expressed
as a finite conjunction of disjunctions of basic formulae, a basic formula being of the form f = g or its negation, where f,g are polynomials
~n
x 1 , ••• ,xr, 1.e. elements of
~.
r
Now the negation of f = g, i.e. f # g, can again be expressed as an equation, viz. (f - g)y = 1, where y is a new variable, and any disjunction of equations can be expressed as a single equation, since (fl = gl)v
v(fn =
gn) holds if and only if (f 1 - g1 ) ••• (fn- gn) = 0.
So
we have reduced the sentence to a finite set of equations, and we see that K is existentially closed if and only if any finite system of equations which is consistent (i.e. has a solution in some extension field) has a solution in K itself.
For example, k itself is existentially closed
over k precisely when k is algebraically closed, but for K
~
k it may be possible for K to be existentially closed
even when k is not algebraically closed.
In fact we shall
see that every field K can be embedded in an EC-field, but the latter will not be unique in any way. Instead of the vanishing of elements, i.e. equations, we may equally well talk about the singularity of matrices. For if A= (a .. ) ~s any n x n matrix, let us write sing(A) lJ for the existential sentence
3U 1'"""' Un' V 1'"""' Vn ( I: a lj u j
= 0 A •••• A I: a . u. = 0/1.
nJ J
and nonsing(A) for the sentence
131
3x .. (i,j ~J
=
l, ••• ,n)(La. x. ~\1
VJ
= o~J ..
=
(i,j
l, ••• ,n)).
It is clear that sing(A) asserts that (over a field) A is singular and nonsing(A) -, sing(A), (where•P means 'not P'). Proposition 6.2.1.
From this it is easy to deduce
A field K over k is an EC-field if and
only if any finite set of matrices over
~
which all be-
come singular for a certain set of values of the x's in some extension of K, already become singular for some set of values in K.
Proof.
The condition for existential closure concerns the
vanishing of a finite set of elements, i.e. the singulaFity of 1 x 1 matrices, and so is a special case of the second condition, which is therefore sufficient,
Conversely, when
K is existentially closed, and we are given matrices A1 , ..• ,A
r
which become singular in some extension, then sing
(A 1 )"'.,. "' sing(Ar) is consistent and hence has a solution inK. • It is almost trivial to show that every field K can be embedded
~n
an EC-field, by first constructing an extension
K1 in which a given finite consistent system of equations over K has a solution, and then repeating the process infinitely often.
However there is no guarantee that the EC-
field so obtained will contain solutions of every finite consistent system over K.
For this to hold we need to be
assured that any two consistent systems over K are jointly consistent,
Of course this follows from the existence of
field coproducts: if equations over K, say
~l' ~ 2 ~i
are two consistent systems of
has a solution inK., then any ~
field L containing both K1 and K2 will contain a solution of ~l A ~ 2 • For L we can take e.g. the field coproduct 132
K1
° K2 ;
more generally a class of algebras is said to
possess the amalgamation property if any two extensions B1 , B2 of an algebra A are contained ~n some algebra c. An example of a class not possessing the amalgamation property is the class of formally real fields. To construct an EC-field extension of K we take the family {CA} of all finite consistent systems of equations over K and for each A take an extension EA in which CA Put ~
has a solution.
=
~ EA, then every finite con-
sistent set of equations over K has a solution in K1 • we repeat this process, we obtain a tower
whose
un~on
If
L is again a field, of the same cardinal as K
or countable (if K was finite).
Any finite consistent set
of equations over L has its coefficients in some K. and so ~
has a solution
~n
K. 1 • ~+
Thus L is an EC-field and we have
proved Theorem 6.2.2.
Let K be any field (over k), then there
exists an EC-field L containing K, in which every finite consistent set of equations over K has a solution.
When
K is infinite, L can be chosen to have the same cardinal as K, while for finite K, L may be taken countable. •
If a is any infinite cardinal, we can similarly construct a-EC-fields containing a given field K, in which every consistent set of fewer than a equations has a solution.
However, the EC-fields constructed here are not in
any way unique; even a minimal EC-field containing a given field K need not be unique up to isomorphism, as will become clear later on.
Further, it will no longer be possible
to find an EC-field algebraic (in any sense) over K. Sometimes a stronger version of algebraic closure
~s
needed, in which the above property holds for all sentences, 133
not merely existential ones.
We shall not need this stronger
form, and therefore merely state the results without proof. Let A be an inductive class of algebras (of some sort), i.e. a class closed under isomorphisms and unions of chains. By an infinite forcing companion one upderstands a subclass C of A such that F.l
Every A-algebra can be embedded in a C-algebra,
F.2
Ann inclusion C C C between C-algebras is elemen"' 1- 2 tary,
F.3
C is maximal subject to F.l,2.
Q is elementary if for every sentence a(x) which holds in P, a(xf) holds in Q.) It can (Recall that a map f:P
--~
be shown that every inductive class has a unique forcing companion (cf. A. Robinson [71], G. Cherlin [72], J. Hirschfeld-W.Ho Wheeler [75]). to skew fields.
All this applies in particular
Here we also have the amalgamation pro-
perty; but an important difference between commutative and non-commutative fields is that algebraically closed commutative fields are axiomatisable (we can write down a set of first order sentences asserting that all equations have solutions); the corresponding statement for EC-fields is false.
This follows from the fact that the class of EC-
fields is not closed under ultrapowers (J. HirschfeldW.H. Wheeler [75]). Although EC-fields do not share all the good properties of algebraically closed fields, they have certain new features not present in the commutative case.
For example,
the property of being transcendental over the ground field can now be expressed as an elementary sentence: (1)
transc(x):
J y' z (xy
yx
2
A
2 x z
This sentence is due to Wheeler (loco). 134
2
zx ;\ xz "' zx ;\ y "' 0). It states that there
is an element z commuting with x 2 but not with x, hence 2 . conJugate . 2 k(x ) C k(x); secondl y x lS to x, so k ( x) k(x 2 ),
=
J,
in particular, [k (x) :k J = [k (x 2 ) :k 2
and so, because
k(x) C k(x), the degree must be infinite.
Conversely,
when x is transcendental, (1) can be satisfied in some extension, and hence in the EC-field. Another sentence characterizing transcendental elements, in the case of a perfect ground field, was obtained by Boffa-v.Praag
[12];
3y(xy-yx
it is 1).
Wheeler has generalized (1) to find (for each n
~
1) an
elementary formula transc (x1 , .•• ,x ), expressing the fact n n that x 1 , ••• ,xn commute pairwise and are algebraically independent over the ground field; as a consequence he is able to show that every EC-field contains a commutative algebraically closed subfield of infinite transcendence degree, To describe EC-fields in a little more detail we need two results from Hirschfeld-Wheeler
[75], see also Macintyre
[75]. Lemma 6.2.3 (Zig-zag lemma) EC-fields over k, then K
~
If K, L are two countable
L i f and only i f they have the
same family of finitely generated subfields.
Proof.
Clearly the condition is necessary.
Conversely, let
K,L be countable EC-fields having the same finitely generated subfields.
Let K = k(a 1 ,a 2 , ••• ), L = k(b 1 ,b 2 , ••• ); we shall construct finitely generated subfields Kn' Ln of K and L =:> respectively such that (i) Kn C:::: Kn+l' Ln C:::: Ln+l' ( ii) Knk(a 1 , ••• ,an)' Ln ~ k(b 1 , •• , ,b n ) , (iii) there is an isomorphism between Kn+l and Ln+l extending a given isomorphism between K and L . - Since K =UK, L = UL, it will foln n n n low that K = L, by taking the common extension in (iii). 135
Put K = L = k· if K , L are defined, with an isomoro o ' n n Phism ~'~'n :Kn --~ Ln' let K'n = Kn (a n+l ), then K'n is finitely generated, hence isomorphic to a subfield of L containing an isomorphic copy of L • n
By Th,5.5.1, Cor.l, L can be emr
bedded in a finitely homogeneous extension, but L is an EC-field and hence is itself finitely homogeneous. can apply an
~nner
Thus we
automorphism of L so as to map K' onto a n
subfield 1 1 containing L , in such a way that the restricn
n
homomorphism~
tion to K is the n
n
•
Let
~ 1 :K' -~
n
n
L' be the n
isomorphism so obtained.
Now put L 1 = L 1 (b +l) and find n+ n n an isomorphic copy of Ln+ 1 in K; this will contain a subfield isomorphic to K' and by applying a suitable inner automorn phism of K we obtain an isomorphism of L 1 with a subfield n+ -1 Kn+ 1 say of K, which when restricted to 1 n1 is (~') • Now n Kn+l' Ln+l satisfy (i)-(iii) and the result follows by induction. • For the -second result let us write, for any subset S of K, C(S) for the centralizer of S in K: Proposition 6.2.4.
Let K be an EC-field over k and let
a 1 , ••• ,ar,b E K, then
This means that the formula 'bE k(a 1 , ••• ar)', not at first sight elementary (and in fact not so in the commutative case), can be expressed as an elementary sentence in an EC-field:
Vx
(2)
(a.x ~
xa. (i ~
1, ••• ,r)
~
bx
xb).
Write A= k(a 1 , ••• ,ar)' then if bE A, (2) clearly holds; if b i A, then (2) is false in K A(x) and hence Proof.
X
also inK, because the latter is existentially closed.• Taking r = O, we obtain a result which is well known in 136
the special case when k is the prime subfield: Corollary
1.
The centre of an EC-field over k is k. •
EC-fields are in some way analogous to algebraically closed groups, which have been studied by B.H. Neumann [73]; the next result is analogous to a property proved by Neumann for groups: Proposition 6.2.5
An EC-field cannot be finitely generated
or finitely related.
Proof.
Given a 1 , ••• ,an 3x,y(a.x
xa. (i
~
~
£
K, the sentence l, ••• ,n) A
xy
f yx)
k k(y);
~s
consistent, for it holds in K(x)
~n
K itself, and by Prop. 6.2.3 this means that K contains
an element y generated.
t
k(a 1 , ••• ,an).
hence it holds
Hence K cannot be finitely
In a finitely related field which is not finitely
generated, infinitely many generators occur
~n
no relation
and so generate a free factor. If x is one of them, then the sentence jy(x = y 2 ) is not satisfied inK, though clearly consistent, and this contradicts the fact that K is an EC-field.
Hence K cannot be finitely related.•
As we saw in 5.5, there are continuum-many non-isomorphic finitely generated fields, hence no countable EC-field can contain them all, i.e. there are no countable universal ECfields (Hirschfeld-Wheeler [75]).
However, it is possible
to construct a countable EC-field containing all finitely presented fields:
we simply enumerate all finitely presented
fields K1 ,K2 , ••• over k, form their field product over k and take a countable EC-field containing this product, which exists by Th.6.2.2.
The result is a countable EC-field con-
taining each finitely presented field over k. Any EC-field has proper EC-subfields, thus there are no minimal EC-fields. Theorem 6.2.6.
This follows from
Let K be an EC-field over k and c any element
137
of K, then the centralizer C of c in K is an EC-field over
k(c). Proof. C.
It is clear that k(c) is contained in the centre of
Now let f
(3)
r
0
be any finite set of equations in x 1 , ••• ,xn over C which has a solution in some extension of Cover k(c).
This means
that the solution also satisfies x c - ex
(4)
n
n
o.
Hence the equations (3), (4) are consistent and so have a solution in K.
By (4) this means that we have found a solu-
tion of (3) inC, soC is an EC-field over k(c), as claimed.• By taking intersections we get EC-fields over k itself. Such a construction can also be obtained in a more straightforward fashion: Theorem 6.2.7.
Let K be an EC-field over k and let a
E
K
be transcendental over k, then there exists b E K such that
ba
(5)
=
2
a b
r 0.
If C is the centralizer in K of such a pair a,b, then C is again an EC-field over k, and the inclusion C
~
K is an
elementary embedding.
Proof. f(a)
Since a is transcendental over k, the mapping 1-> f (a2) is an endomorphism a of k(a), so the system
(5) has a solution in some extension of K, and hence in K
itself.
Now given a,b satisfying (5), let C be their cen-
tralizer in K and let (6)
138
f
r
0
be a consistent system of equations ;n . . th e
· bl es x 1 , Since K is an EC-field, this system has a
••• ,xn over C.
var~a
solution inK.
Let c 1 , ••• ,cs s C be the coefficients occurring in (6) and consider the system consisting of (6) and (7)
x.z
= yc. •
c.y
1
J
J
This system is consistent: we form first K(y) and with the
1-->
endomorphism a:f(y)
f(y 2 ) form K(y)(z;a).
Hence (7)
has a solution in K itself; let us denote this solution also by xi' y, z.
1-->
Then the mapping y
a, z
1-->
b defines an
isomorphism
for both sides are obtained by first adjoining a central indeterminate y and then forming the field of fractions of the skew polynomial ring with respect to the endomorphism f(y)
2
1-->
that t
-1
f(y ).
By homogeneity there exists t
c.t =c., t J
x!, then x! ~
~
-1
J
yt
C and x!
£
~
a, t ~sa
-1
zt =b.
£
Now put t
solution of (6).
K such -1
xl..t
This shows
C to be an EC-field. To prove that the inclusion C
~
K is an elementary em-
bedding we need only show that every finitely generated subfield of K can be embedded in C.
Let c 1 , ••• ,cs s K and consider the system (7), but without the equations involving x .• ~
tion in K.
This system is consistent and so has a soluSince k(y,z) ~ k(a,b) with y
there exists t put t
-1
c.t J
£
K such that yt
cj, then cj
E
= ta + 0,
1-->
a, z
zt • tb.
C and k(c 1 , ... ,c 3 )
1-->
b,
If we
= k(cl, ... ,c~),
hence the result. • When K is countable, it follows from the zig-zag lemma (6.2.3) that C Corollary.
~
K and we obtain the
Every countable EC-field has a proper subfield
139
isomorphic to itself. •
An important and useful result due to Wheeler (l.c.) is that every countable EC-field has outer automorphisms; the proof below is taken from Cohn [75].
){
Every countable EC-field has 2
Theorem 6.2.8.
°
distinct
automorphisms, and hence has outer automorphisms.
Proof.
Let K be generated over k by a 1 ,a 2 , ••• , where the
ai are chosen so that ani k(a 1 , ••• ,an_ 1 ); this is clearly possible.
By Prop. 6.2.4 there exists bn commuting with
a 1 , ••• ,an-l but not with an. phism induced by b
n
Let ~n be the inner automor-
and consider the formal product
for a given choice of exponents Ei defines an automorphism on K.
= 0,1. We claim that a
Its effect on k(a 1 , ••• ,an)
is E
6
n n '
for when i > n, S. leaves k(a 1 , ••• a) elementwise fixed. n
1
Thus it is an endomorphism which is in fact invertible since each
s. l
is.
Since the E. are independent and each l
choice gives a different automorphism, we have indeed 2 K 0 distinct automorphisms; of course there cannot be more than this number.
Now a countable field has at most countably
many inner automorphisms, hence K has outer automorphisms. • This proof is of course highly non-constructive; since EC-fields themselves are not given in any very explicit form, there seems little hope of actually finding a particular outer automorphism. An important but difficult question is: Which fields are embeddable in finitely presented fields?
It would be in-
teresting if some analogue of Higman's theorem could be 140
established.
This asserts that a finitely generated group
1s embeddable in a finitely presented group if and only if it is recursively presented (Higman [61]).
6.3
A specialization lemma In this section we digress somewhat to prove a technical
result which is sometimes useful: Lemma 6.3.1
Let K be a field with
(Specialization lemma)
centre C and assume (i) C is infinite and (ii) K has infinite degree over C.
Then any full matrix over KC is
non-singular for some set of values of X in K.
Some preparations are necessary for the proof.
In the
first place we shall need Amitsur's theorem on generalized polynomial identities.
Let A be a k-algebra, then by a
generalized polynomial identity (g.p.i.) one understands
a non-zero element p of pings X
-->
A.
~
which vanishes under all map-
Amitsur [65] proved that a primitive k-alge-
bra A satisfies a g.p.i. if and only if it is a dense ring of linear transformations over a skew field of finite degree over its centre and A contains a transformation of finite rank.
We shall be particularly concerned with the case
where A is itself a skew field; in this case Amitsur's theorem takes the following form: A skew field satisfies a generalized polynomial identity i f and only i f it is of finite degree over its centre.
For the proof we refer to Amitsur [65].
A second result is the inertia theorem (Bergman [67], Cohn [71 "]).
Let R be any ring and
A
a subring, then
A
is
said to be n-inert in R if for any families (a;\) of rows in Rn and (b ) of columns in~ such that a,b ~
A,~,
h
there exists P s GL (R) -1
~
E
A for all
such that on writing a~ = aAP,
n
= p b , each product a~b' lies trivially in A, 1n the ~ A ~ sense that for each 1 = l, ••• ,n, either a~. = 0 orb'. =0 /\1 ].11
b'
~
or both
1
aH
and b'. lie in A. JU
If A is n-inert in R for all
141
n, it is called totally inert in R. Inertia theorem.
Now we have the
~ is totally inert in ~.
The theorem is proved in Cohn
[71"] (p.l03f.)
for a wider
class of rings; however the proof given there is not complete.
We therefore give a proof below (which it is hoped
is complete).
In the proof we shall need the weak algorithm;
for this we refer to Cohn [71"], and in fact, the reader willing to accept the inertia theorem will need only the following corollary in which the notion of inertia does not appear. The embedding ~
Corollary (to the inertia theorem). -:> ~
is honest.
For let C be a full matrix over F full over " F
~,
say C
= AB,
=~
which is non-
where A is n x r, B is 1\
r x n and r < n.
By inertia we can find P e GL (F) such
that on writing A'
=
AP, B'
-1
r
= P B, the product of any row
of A' by any column of B' lies trivially in F.
Since C is
full, A', B' cannot have. all their entries in F.
If the
(1,1)-entry of A is not in F, say, then the first row of B' is zero, and on omitting the first column of A' and the first row of B1 we can diminish r.
By induction on r we
obtain a contradiction; this proves the corollary, starting from the theorem.• We shall prove the inertia theorem in the following (slightly more general) form.
The proof follows Cohn-Dicks
[76]. Inertia theorem. and
"R
Proof.
Let R be a graded ring with weak algorithm, 1\
its completion, then R is totally inert in R.
For any a
E
"R the
order o (a) of a is defined as the
minimum of the degrees of the homogeneous components of a, or oo if a=O. Let
m = ~x
142
£
R
I
o (x) > 0},
then by the weak algorithm R/m is a field; moreover completion' of R in the m-adic topology. for the completion of m.
Ris
We shall write
the
m
Now m as a free right ideal of R
has, by the weak algorithm, a homogeneous basis X say, and
" can be uniquely written as a any a e: m
=
~xa
X
, where the
summation is over all x e: X and all but a finite number of the a
£
X
R are
zero.
o(a) > 1 +
(1)
We note that
mi~{o(ax)};
further~
if a e: m, then all the a lie in R. X "'r Let A F, hence a mapping E(t) --> F(t) and so an E(t)-ring homomorphism
(5)
E(t)k(t) --> F(t).
Now let A be a full matrix over Ek, then A is invertible over F, hence invertible over F(t) and by (5) it is full over E(t)k(t), as we had to show.• Let E be a field with centre k, then E(t) has
Lemma 6.3.5.
147
the centre k(t).
Proof.
Every element of E(t) has the form ¢ We shall use induction on d(~)
f,g s E[t].
= =
fg
-1
, where
deg f + deg g
to prove that if ~ is in the centre of E(t), then¢ For d(¢)
=
0 the result holds by hypothesis.
we may assume deg f
~
. deg g, replac~ng
By the Euclidean algorithm, f
= qg
+
~
by
k(t).
E
If d(¢) > O, -l 'f necessary. ~ ~
r, where deg r < deg g,
with uniquely determined q,r s E[t],
Let us write u
-1
c
= c uc,
for any c s E*, then fg
-1
q + rg
-1
q
c + r c( g c)-1 •
Since ¢ is in the centre of E(t), we have q + rg
-1
q
c
+
c -1 . r (g ) , ~.e. c
(6)
q - q
c
= r
= deg
Now v(¢)
c( g c)-1 - rg-1
g - deg f is a valuation on E(t), and the
left-hand side of (6) has value ~ 0, unless qc the right-hand side has positive value. are 0, q d(fg
-1
c
=q
and rg
-1
q, while
Hence both sides
is in the centre, but d(rg
-1
)
E(t)k(t) is honest, by Lemma 6.3.4, hence any full matrix over Ek is full over F(t)C(t) and hence over Fc.• 6.4
The word problem for free fields
The word problem in a variety of algebras, e.g. groups, 1s the problem of deciding, for a given presentation of a group, when two expressions represent the same group element.
In the case of skew fields we again have a presenta-
tion, as explained in 6.1, and we can ask the same question,
149
but the word problem is now a relative one.
Generally we
have a coefficient field K and we need to know how K is given.
It may be that K itself is given by a presentation
with solvable word problem, and the algorithm which achieves this is then incorporated in the algorithm to be constructed; or more generally, we merely postulate that certain questions about K can be answered in a finite number of steps and use this fact to construct a relative algorithm. Our aim here will be to show how to solve the word problem in free fields, and of the two alternatives described above we shall take the second, thus our solution will not depend on the precise algorithm in K but merely that it exists.
In fact it is not enough to assume that K has a
solvable word problem; we need to assume that K is dependable over its centre: Given a field K which is a k-algebra, we shall call K dependable over k if there is an algorithm which for each finite family of expressions for elements of K, in a finite number of steps leads either to a linear dependence relation between the given elements over k or shows them to be linearly independent over k. When K is dependable over k, K and hence k has a solvable word problem, as we see by testing 1-element sets for linear dependence.
Let K have centre C; our task will be to solve
the word problem for the free field KC{X}.
For this it will
be necessary to assume K dependable over C; this assumption ~s
indispensable for we shall see that it holds whenever
KC{X} has a solvable word problem. There is another difficulty which needs to be briefly discussed.
As observed earlier we need to deal with expres-
sions of elements in a skew field and our problem will be
to decide when such an expression represents the zero element.
But in forming these expressions we may need to in-
vert non-zero elements, therefore we need to solve the word problem already in order to form meaningful expressions.
150
This problem could be overcome by allowing formal expressions such as (a ~ a)
-1
; but we shall be able to bypass it alto-
gether: instead of building up rational functions step by step, we can obtain them in a single step by solving suitable matrix equations, as explained in 4.2.
In fact we
have the following reduction theorem. Theorem 6.4.1.
Let R be a semifir and U its universal field
of fractions, then the word problem for U can be solved i f the set of full matrices over
R is recursive.
Any element u 1 of U is obtained as the first component of the solution of a matrix equation Proof.
Au + a
o,
and u 1 = 0 if and only if~ = (a,a 2 , ••• ,an) is non-full. By hypothesis there ~s an algorithm to decide whether A1 is full or not, and this provides the answer to our question.• We note that it is enough to assume that the set of full matrices over R is recursively enumerable, because its complement, the set of all non-full matrices is always recursively enumerable (in an enumerable ring). In the same way one can show that for an epic R-field K the word problem can be solved if the set of all matrices inverted over K is recursive. We now come to the main result to be proved
~n
this sec-
tion: Theorem 6.4.2
Let K be a field, dependable over its centre
C, then the free K-field on a set X over C, U
= KC
{X} has
a solvable word problem and is again dependable over C. Conversely, i f the word problem in U is solvable (for an infinite set X), then K is dependable over C.
To prove the theorem we shall at first assume that C is infinite and the degree [K:C] is infinite.
Of course this
must be understood in a constructive sense: Given n > O, we 151
can
~n
a finite number of steps find n distinct elements of
C and n elements of K linearly independent over C.
Like-
wise, any other results we use will need to be put in a constructive form, e.g. the specialization lemma, and Amitsur's theorem on which it was based.
Thus given f
£
KC, f f 0, there is a method (an 'oracle') for obtaining a set of arguments for which f is non-zero in a finite number of steps. To prove Th,6,4.2, we must describe an algorithm which will enable us to decide when a matrix A over F full.
= KC is
First we observe that being full is unaffected by
elementary transformations and by taking the diagonal sum with a unit matrix. linear in the x.
~
£
This allows us to reduce A to a matrix X, the process of "linearization by en-
largement" (sometimes called 'Higman's trick', cf. Higman
[40]).
To describe a typical case of this process, suppose
that the (n,n)-entry of an n x n matrix has the form f + ab. On enlarging the matrix we can replace the term ab by separate terms a,b by applying elementary transformations, as follows: f + ab
+ ab
Here only the last two entries in the last two rows are shown. By repeated application we can therefore reduce A to the form (1)
A'
where A
homogeneous .of degree 0 and A1 of degree 1 ~n the x' s. Thus A has entries in K and A1 = l:B.x. where the B. 0 ~ ~ ~ have entries in K· moreover A' is full if and only if A ~s. 0
~s
Suppose that A'
'
~s
not full, then it will remain non-full
when the x. are replaced by 0, i.e. A must then be singular ~
152
0
over K.
Thus
~f
A
0
1s non-singular, A' (and with it A) is
necessarily full. We may therefore suppose that A
0
is singular, of rank
r < N say, where N is the order of A'.
By diagonal reduc-
tion over K (which leaves the fullness of A' unaffected) we can reduce A0 to the form
[~ ~)
(cf. e.g. Cohn[71"], Ch,8;
clearly this 1s an effective process because K is dependable over C). Let us partition A1 accordingly, then
A' where P,Q,R,S are homogeneous of degree l (and the sign of P is chosen for convenience 1n what follows). the completion
F = Kc;
Now pass to
by the corollary to the inertia A
theorem A' is full over F if and only if it is full over F. The matrix I - P is invertible over F and by elementary transformations we obtain (I- P)-lQ
)
S - R(I - P) -lQ •
To find whether (3)
S - R(I - P)
-1
Q
0
we have to check that for each v terms of degree v are 0.
=
0,1, ••• the homogeneous
Now S - R(I - P)
-1
n
Q = S - IRP Q
and equating terms of a given degree v we find that (3) is equivalent to (4)
s
a,
0
(v
O,l,ooo)o
These are equations of matrices over F and s1nce the latter is embeddable in a field, we may regard (4) as equations over 153
a field.
In that case the equations (4) follow from the
same equations with v < N.
Assuming this for now, we thus
have an algorithm for determining whether (3) holds.
When
this equation holds, the matrix on the right of (2) has at least one row of zeros and hence A' is then non-full.
If
(3) does not hold, then by Lemma 6.3.1 we can specialize the x's within K to values a. such that I - P remains non-singu~
lar and S - R(I - P)
-1
Q remains non-zero.
Translating back
to A' we find that specializing to a. we obtain a matrix of ~
rank > r. (1).
We now replace x. by x. + a. and start again from ~
~
~
This time we have a matrix A
0
over K of rank greater
By repeating this process a finite number of times
than r.
(at most N times, where N is the order of A'), we can thus decide whether or not A' is full and this completes the proof in the case where C and [K:C] are infinite. We still need to prove that the equations (4) all follow from a finite subset; this is related to the well known fact that a nilpotent n
x
n matrix A over a field satisfies
An = 0. Lemma 6.4.3.
Let P be an n x n matrix, Q a matrix with n
rows and R a matrix with n columns over a skew field K.
0 for v
(5)
If
O,l, ••• ,n- 1,
\)
then RP Q = 0 for all v.
Proof.
Let nK be the right K-space of columns with n comr
ponents.
The columns of Q span a subspace V of ~while 0
the columns annihilated by the rows of R form a subspace
~. and since RQ = 0 by hypothesis, we have VoC W. Regarding P as an endomorphism of ~ we may define a sub-
W of
space v\) of ~ for \) > 0 inductively by the equations
v
\)
154
vv
1+ PVv- 1"
Thus V
V0 + PV0 + ••• + P\IV 0 and it follows that
\)
(6)
••• c- vn- 1'
Moreover, by (5) Vv • • • • Consider the leading coefficients of f 1 v••••fn; if they are linearly independent over C then the f's are linearly independent over C'.
Otherwise we can find i, 1 2._ i2_ n, ai+l'''"'an
£
C and
v. positive integers v 1.+ 1 , ... ,vn such that f! =f. -E. 1 f.a.t J l. l. I.+ J J has lower degree than f .. Now the linear dependence over C' n
l.
of f 1 , ••• ,fn is equivalent to that of f 1 , ••• ,fi-l'fl,fi+l''''' f and here the sum of the degrees I.s smaller. Using inn
duction on the sum of the degrees of the f's we obtain the result.
155
Now return to Th.6.4.2.
If in that theorem C is finite
or more generally, there is no constructive process of obtaining infinitely many elements in C, we adjoin a central indeterminate t so that K, Care replaced by K'
= C(t).
=
K(t), C'
By what we have just seen, K' is dependable over
C' whenever K is so over C. ·
construct~ve
[K:C].
Moreover C1 is infinite in the sense ( e.g. we can t a k e 1 , t , t 2 , ••• ) and [K' • C'] =
It follows that the set of full matrices over K'c 1
is recursive, hence by Th.6.3.6 (or even the special case Lemma 6.3.4) the set of all full matrices over KC is also recursive.
Hence the word problem in U is soluble, so Th.
6.4.2 continues to hold even when the centre of K is finite. There remains the case where K has finite degree over its centre, or where no process for
~onstructing
early independent sets exists.
Instead of adjoining a cen-
infinite lin-
tral indeterminate we now form a skew extension. Let K be a field with centre C and let cr be an automorphism of K leaving C elementwise fixed. polynomial ring R K(y;o).
We form the skew
= K[y;cr] and its field of fractions K'
If no power of cr is an inner automorphism, then
the centre of K' is C.
To see this we embed K' in the field
of skew Laurent series K((y;cr)) (cf.2.1). in the centre then fy \)
all c
v
~
ca0
E
K, hence c
0
a \)
O, it follows that ,
i.e. f
= a0
E
C.
= yf, hence
= a \) c. a =0 \)
a0 \)
If f
=
Iy\la
\)
lies
= a \) and cf = fc for
Since cr\1 is not inner for except when v
=0
and a c 0
Clearly K' is of infinite degree
over its centre, e.g. the powers of y are linearly independent. Taking C'
is honest,
C in Th.6.3.6 we find that the embedding
Moreover if K is dependable over C (and if cr
is 'computable' in an obvious sense) then so is K'. 156
For
let u 1 , ••• un fi,g
E
K'; as before we write u.
E
1.
= f.g- 1 , where 1.
K[y;o] and it is again enough to test £1 , ••• ,fn for
linear dependence over
c.
This time we single out the f's
of maximal degree, £ 1 , ••• ,fr say. If their leading terms are linearly independent over C, then so are the f's. · , r Oh t erw~se 1 et f. = f. - E. £.a. (a. s C) have lower degree ~
~
~+1
J J
J
than f. and continue with £ 1 , ••• ,£!, ••• ,£ as before; again ~ ~ n this process ends after a finite number of steps. To complete the proof of Th.6.4.2 we shall need another Lemma which will also establish the second part of the theorem. Lemma 6.4.4.
Let K be a field (over k).
If for every finite
set Y the word problem for the free K-field on Y over k is soluble, then the free K-field on any set X over k is dependable over k.
Proof.
Let U
= ~ {X} be the free field; we may assume X
infinite by embedding U in a free K-field on an infinite set containing X (that such an embedding exists follows from Th.6.3.6 but is also easy to see directly). u 1 , ••• ,un
E
U, we have to determine whether the u's are
linearly independent over k. the case n
Given
We shall use induction on n,
= 1 being essentially the word problem for U.
We may assume u 1 1 0, and hence on dividing by u 1 we may suppose that u 1
=
1.
Only finitely many elements of X
occur in u 2 , ••• ,un, so we can find another element in X, y say.
Write u!
1.
= u.y 1.
- yu. and check 1.
are linearly dependent over k.
If so,
where a 2 , •• o,an E k and are not all zero, then u = satisfies yu = uy. Since u does not involve y, it that u represents an element a of k (which can be computed by suitably specializing the x's), and hence l.a- E~uiai
= 0 is a dependence relation over k. Conversely, if there is a dependence relation E~uiai = O, where not all the ai vanish, then not all of a 2 ,.o.,an can vanish (because u 1 = 157
+ 0),
and so Ln2-u!a. = 0 is a dependence relation between 1 1 u2, ' .•• ,un. ' The result now follows by induction on n.• We note that since K is a subfield of U, K is dependable
1
over C; thus the dependability of K is a consequence of the solubility of the word problem for U (on an infinite set). This completes the p:oof of Th.6.4.2 when [K:C] is infinite. When [K:C] is finite, but K has a (computable) automorphism over C, no power of which is inner, we can form the skew function field K'
K(y;cr).
Then K' 1s of infinite
degree over its centre C and K' is dependable over C, hence the set of all full matrices over K'c is recursive, and so is the set of all full matrices over K , because (7)
c
is honest.
This solves the word problem for U.
Finally if K has no automorphism cr of the required kind, we form K'
K(t) with a central indeterminate t; as we have
=
seen, the result is a field K' with centre C(t).
1-->
Now ref(t 2 ). This
peat the process with the endomorphism f(t) is computable (in any reasonable sense) and it induces an auto!
l
morphism of K(t,t 2 ,t 4 ,
••• ) ,
no power of which is inner, hence
we obtain a field K" of infinite degree over its centre C. From Th.6.3.6 taking K
=
Corollary.
we obtain the following special case by
C: Let k be any commutative field with soluble
word problem, then U
k {X} has soluble word problem.•
There still remains the word problem for a free field ~{X}
where k is not the exact centre of K.
This really
requires a sharper form of the specialization lemma, but we shall not pursue the matter here. 6.5
A skew field with unsolvable word problem As is to be expected, for general skew fields the word
problem is unsolvable; an example was given by Macintyre [73].
The account below is a (slightly simplified) version
of another example due to Macintyre. '
158
The idea is to take a
finitely presented group with unsolvable word problem and use these ~elations in the group algebra of the free group. We need a couple of preparatory lemmas. Lemma 6.5.1.
Let F
be the free group on x , ••• ,x and F 1 n y the free group on y y In the direct product F x F 1••••• n" X
X
y
H be the subgroup generated by the elements x.y.(i = 1, 1 1 ••• ,n) and elements ul, ••• ,um E Fx. Then H n Fx is the
let
normal subgroup of Fx generated by u1 , ••• ,um.
Proof. Let N be the normal subgroup of F generated by the · -1 x_l u (~ s l, ••• ,m). S1nce x. u x. = (x.y.) u (x.y.) E H, it v 1 v 1 1 1 ~ 1 1 is clear that N ~ H, hence N ~ H n F • To prove equality, X
consider the obvious homomorphism f:F
X
x F y
which maps u
~
--~
(F /N) x F
y
X
to 1 (p
= l, ••• ,m).
If wE H n F , then wf X
is a product of the (x.y.)f, and since the x's andy's comr 1
1
mute, we can write it as wf = [v(x)v(y)]f
v(xf)v(yf),
where v is a word 1n n symbols. so v(yf)
Since w
E
F , wf X
E
F /N and X
1, but the y.f are free, so v is the empty word 1
and wf = 1, hence w E ker f = N. • Let F , F , H, N be as above and consider G X
y
= FX
x
F • y
This group can be ordered: we order the factors as in 2.1 and then take the lexicographic order on G. form the power series field K
=
k((G)).
Hence we can
The power series
with support in H form a subfield L; we take a family of copies of K indexed by Z and form their coproduct amalgamating L.
The resulting ring is a fir, with universal
field of fractions D say.
If a is the shift automorphism,
we can form the field of fractions D(t;a) of the skew polynomial ring D[t;cr]. 159
With the above notation, let w e F , where
Lemma 6.5.2. F
xD(t;cr).
Proof.
X
C D; then
K
C G C K
o-
w e N if and only if wt = tw in
If w e N, then w e H by Lemma 6.5.1, hence w e L
and so wt
=
tw.
Conversely, if tw
= wt,
w is fixed under a and so lies
in the fixed field of cr, i.e. w e L. power series with support in H, so w
But L consists of all E
Hn F
X
= N, by Lemma
6.5.1..
Now let A be a finitely presented group with unsolvable word problem, say (1)
A
= u
1},
m
where u1 , ••• ,um are words in the x's. We shall construct a finitely presented field whose word problem incorporates that of A.
where (2)
Let
~
consists of the following equations: x.y. ~
tx.y. (i
J
~
u t ].l
tu
].l
(].l
~
1, ••• , n)
l, ••• ,m).
To see that this is meaningful, let P
PX be the free field
over k on x 1 , ••• ,xn and form
This is a fir and so has a universal field of fractions Q; moreover, Fx
x
Fy is naturally embedded in Q, in fact Q
~s
also the universal field of fractions of the group algebra 160
of F
X
over k
x F
y
over k.
by~.
In Q consider the subfield R generated
and let S be the field coproduct of copies of
Q indexed by Z amalgamating R. phism inS, we can form T
=
If cr is the shift automor-
S(t;cr); from its construction
this is essentially M (cf. Lemma 5.5.4).
By the universality
ofT we have a specialization from T to D(t;o).
We claim
that (3)
='l
w
£ N
D
defined by (f,a)
'
J->
f(a), where f(a) = f(a 1 , ••. ,am) ar1ses from f(x 1 , ••. ,xm) on replacing xi by ai £D. If we fix f £ F, we have a ~pping Dm - > D and if we fix a E Dm we get a mapping F --> D. morphism.
Clearly the latter mapping is a D-ring homo-
Let us write Hom(F,D) for the set of all D-ring
homomorphisms, then we have Lemma 7.1.1.
If Dis a k-algebra and F
m Hom(F ,D) a D •
Explicitly we have 4»
I-->
cp
cp
(x1 , ••• ,xm).
This follows immediately: we have seen that a
E
Dm defines 163
a homomorphism and conversely, each homomorphism
f,
where f:Dm - > D is the function on Dm defined by f.
Now
(2) is also a D-ring homomorphism if we regard the functions
from Dm to D as a ring under pointwise operations; this amounts to treating the right-hand side of (2) as a product of rings.
The image of F under 0 is written F; it is the
ring of polynomial functions in m variables on D.
The ker-
nel of 0 is just the set of all generalized polynomial ident~t~es
in m variables on D.
By identifying Dm with knm via a k-basis of D, we may view the ring of polynomial functions knm(= Dm) - > k as a
nm
; clearly G does not depend on Dm the choice of k-basis of D. Since the canonical map k ~ D Dm Dm . . . . b . --> D ~s ~nJect~ve, the su r~ng C of D generated by D central k-subalgebra G of D
and G is of the form G
~
D.
If moreover, k is infinite, G
is just the k-algebra of polynomials in mn commuting indeterminates, so C is the D-ring of polynomials in mn central Another description of C is given in
indeterminates. Theorem 7.1.2,
Let D be ann-dimensional central simple
k-algebra, then F
Dk may be expressed as the
free D-ring on mn D-centralizing indeterminates and C is the image of the evaluation map
Proof.
e.
We may regard F as the tensor D-ring on the D-bi-
module (D
164
=F
~
m
D) , and since D is central simple, the map
(4)
~:D ~ D
-->
Endk(D) ~ Dn, where (a G b)~:x
1s a D-bimodule isomorphism,
I->
axb
It follows that F 1s the tensor
D-ring on mn D-centralizing indeterminates, Now fix a k-basis u 1 , ... ,un of D and consider the dual k-basis ut, ..• ,u~ E Ho~(D,k) ~ Endk(D). For each~= l, ... ,n there exists v~
Ea~A
v~~ = u~.
F = D
ruinates v. , i = l, ••• ,m,
~
1~
(l:AblAu\, ... ,l:\bm\u\)
1->
D-ring generated by the algebra generated by
= l, •.. ,n.
Write~-
1~
= v. 8: 1~
bi].J' then it is clear that
~-
1W
.
the~-
1W
F
is the
Now G is by definition the k, hence C = GD = F as claimed.•
If we examine the role played by 8 we obtain Theorem 7 .1. 3.
If k is infini t:e and D is an n-dimensional
central simple k-algebra, then the evaluation map 8 can be expressed in t:he form
F
D ->
D [ E;.
1~
J
C
(i
~
nm
D
, where v.
1, ... , m; ~
1~
1->
E;.
1]1
1, ... , n)
hence the kernel of G is generated by the commutators of pairs of the
v.1~ ·•
The above account follows Procesi [68], with simplifications by Dicks; cf. also Gordon-Motzkin [65], who prove Th.7.1.2
when Dis a field.
For a more general treatment,
in the context of Azumaya algebras, see Procesi [73].
7.2
Rational identities The basic result on rational identities, again due to
Amitsur [66], states that there are no non-trivial rational 165
identities over a skew field which is infinite-dimensional over its centre and which has an infinite centre.
But it
is now more tricky to decide what constitutes a 'non-trivial' identity.
Here are some 'trivial' ones:
(x + y)
-1
= y -1 (x-1
+ y
[x -1 + (y-1 - x) -1]-1
-1 -1 -1 )
x
x- xyx
, (Hua's identity).
We shall give two proofs of Amitsur's result, one by Bergman [70] and one by the author, based on the results of 6.3 above (cf. Cohn
[72 ']).
Our first task is to find a means of expressing rational functions; here we shall follow Bergman [70,76].
Let D be
a skew field with centre k, then we can form D(t), the field of rational functions in a central indeterminate.
¢ s D(t) has the form ¢ in t, and we can set t
~
fg
=
u
then ¢(a) will be defined.
¢ = fg
-1
£
-1
Any
, where f,g are polynomials
k if u is such that g(u)
# 0;
Given ¢, we can choose f,g in
to be coprime, and then f,g will not both vanish
for any a
£
k.
Since we only had to avoid the zeros of g
~n
defining ¢(a) we see that ¢ is defined at all but finitely many points of k. We have to generalize this to the case of several noncentral variables.
Now we no longer have D(t) at our dis-
posal, and although we have seen in Ch.6 how to construct free fields, that construction will not be needed here. What we shall do is to build up formal expressions in x 1 , ··:•xm using+,-,~, '/.and elements of D.
The expressions
will be defined on a subset of Dm or more generally on Em, where E is a D-field. Let X be any set. a:X
--~
An X-ring is a ring R with a mapping
R; we write R or (R,a) to emphasize the mapping.
If R is a field we speak of an X-field; this is essentially 166
the same as a Z-field in our previous terminology. Given X.~ {x 1 , ••• ,xm} we write R(X) for the free abstract algebra on X with operations {0 1 ()-l + x } o' o' 1' 1 ' 2' 2 ' where subscripts indicate the arity of the operation. For each expression there is a unique way of building it up .
s~nce
no
.
relat~ons
are
.
~mposed,
thus e.g. (x-x)
-1
exists.
In contrast to 7.1 we now have a partial evaluation mapping (1)
R (X) x Rm
Thus any map a:X
R.
-:>
-->
R defines a map a of a subset of R(X)
into R, by the following rules: (i) if a
0 or 1, a a ... 0 or 1,
= xi'
(ii) if a (iii)
if a
=
= xia' ~ -b or b+c or be and ba, ca are defined, then x.a
aa= -ba or ba + ca or ba.ca. (iv) if a= b-land ba is defined and invertible in R, then a~
= (b~)- 1 •
Since a just extends a we can safely omit the bar.
In a
field invertible is the same as non-zero, hence we have Proposition 7.2.1.
Let X be a set, (D,a) an X-field and
as R(X), then aa is undefined i f and only i f a has a subexpression b
-1
, where ba
= o.•
With each a:X --> D we can associate a subset E(D) of R (X), the domain of a, consisting of expressions which can be evaluated for a.
Similarly with each f s R (X) we asso-
ciate its domain dom f, a subset of Dm consisting of the points at which f is defined; more generally we shall consider dom f in Em, where E is a D-field. is called non-degenerate on E.
If dom f ~
0, f
In this section we shall
mainly be dealing with the domains of functions f s R(X). Lemma 7.2.2.
Let D be a skew field which is an algebra
167
over an infinite field k.
If f,g are non-degenerate on a
D-field E then dam f ('dam g
Proof.
Let p
£
dom f, q
£
f
(/J.
dam g, write r
and consider f(r), g(r) e: E(t).
= tp
+ (1-t)q
Each is defined for all but
finitely many values of t in k, hence for some a e: k both are defined. • Given f,g e: R(X), let us put f - g if f,g are non-degenerate (on a given E) and f,g have the same value at each point of dam f n dam g. This is clearly an equivalence, the transitivity follows by Lemma 7.2.2.
If f,g are non-degen-
erate, so are f+g, f-g, fg; moreover they depend only on the classes of f,g not on f,g themselves, and if f f O, then f-l is defined. Theorem 7.2.3.
Thus we have Let D be a skew field with infinite centre
k and E a D-field which is also a k-algebra,
~hen
valence classes of rational functions from Em to
the equi-
E with co-
efficients in D form a skew field DE(X). •
If E is commutative, this reduces to D(X) and is independent of E.
In that case any element of D(X) can be
written as a quotient of two coprime polynomials, and this expression is essentially unique.
The dependence on E in
the general case will be examined below; now there is no such convenient normal form for the elements of DE(X). Even if we use Ch.6, a given element may satisfy more than one matrix equation Au
=
a and the relation between them
remains to be described (cf. Cohn [b]).
In terms of ex-
plicit rational expressions for f, it may be that different expressions have different domains and the resulting function is defined on the union of these domains.
Bergman
[70]
raises the question: "Whether there is always an expression for f having this whole set for domain of definition, a 'universal' expression for the rational function f." Generally the domains of functions form a basis for the open sets of a topology on Em, the rational topology on Em 168
(cf. also 8.5 below; the polynomial topology, a priori coarser, is the Zariski topology).
The closed sets are of
the form
V(P)
= {p
£
Em
I
f(p)
0 for all f £ P}
A subset S of Em is called irreducible if it is non-empty and not the union of two closed proper subsets.
Equiva-
lently: the intersection of non-empty open subsets of S is non-empty.
Thus Lemma 7.2.2 states that Em is irreducible
in the rational topology when the centre of E is infinite. A subset S of Em is called flat if p,q (1-a)q
£
S for infinitely many a
£
subset will then contain ap + (1
k.
£
S implies ap +
Of course a closed flat
a)q for all a
£
k.
Now
the proof of Lemma 7.2.2.gives us Lemma 7.2.4.
Any non-empty flat subset of Em is irreducible. •
An example of a flat closed subset is the space S defined by (2)
I:a.,x.b., = c ~ll
~
~ll
By Lemma 7.2.4, S is irreducible (if non-empty) and so as ln Th. 7.2.3 yields a skew field D5 (x) in x 1 , ••• ,xm satisfying (2), the function field of (2). In general it is not easy to decide whether a given set lS irreducible, e.g. xlx2 - x2xl = 0 for E :;2 D :;2 k. In the commutative case every closed set lS a finite union of irreducible closed sets, but this need not hold in general. It is clear that polynomially closed
~
rationally
closed; we want conditions for the converse to hold.
First
two remarks: (i)
Let S
C Em
be such that p
i S,
then there exists f
169
defined at p but not anywhere on S.
The degeneracy of
=g
f can only arise by inversion, so f
-1
where g is
non-degenerate on S and 0 at all points of S where defined, and g(p) (ii)
~
0.
=0
Any element of D(t) defined at t
can be expanded -1
in a power series. Let g = a - th say, then g a- 1 (1- tha-l)-l = ~a- 1 (tha-l)n. So we can build up any function in D(t) provided that it is defined at t Lemma 7.2.5.
= O.
Let D be a skew field which is a k-algebra,
where k is an infinite field, and let E be a D-field and If S ~ Dm is flat, then its closure in Em
a k-algebra.
is polynomially closed.
Thus for flat sets, rationally closed
polynomially
closed. Proof.
t
Let p
S; we have to find a polynomial over D which
is zero on S but not at p.
We know that there is a rational
s
function f, non-degenerate on Say f is defined at q For any x for t x
€
= 0,
€
€
and f
=
0 on
s
but f(p)
r o.
S.
Em consider f((l-t)q + tx); this is defined
so it is a well-defined element of E(t).
S, f is 0 by flatness, but for x
because it is non-zero for t
=
1.
=
If
p it is non-zero
In the power series ex-
pansion of f((l-t)q + tx), if we have to take the inverse of an expression h(t), the constant term h(O) is non-zero, because f(q) is defined, and h(O) does not involve the coordinates of x.
So the expansion f((l-t)q + tx) has coeffi-
cients which are polynomials in x; their coefficients are in D because q ~s
at least one
E
S ~ Dm.
These polynomials are 0 on S, but
non-zero at p, and this is the required
polynomial.• Corollary.
Let k,D,E be as before, and assume that D and E
satisfy the same generalized polynomial identities over k with coefficients in D, then DE(X) ~ DD(X) and for any
170
k-subfield C of D, CE(X) a CD(X).
Tqe rational closure of Dm ~n Em is polynomially
Proof.
closed by Lemma 7.2.5.
Now every g.p.~. · on Dm ho 1ds on Em, So the rat ~onal · Em , ~.e. · Dm ~s · d ense ~n · Em. ~ c 1 osure o f Dm ~s
The rest follows because CE(X) is the subfield of DE(X) generated by
c
and
x.•
Now the rational identities may be described as follows: Theorem
7.2.6. {Bergman [70]J.
Let
D be a field with centre
k then there exists a D-field E with infinite centre C ~ k such that [E:C]=oo, and for each m, any such E, C yield the same function field DE(X) (m =
Proof.
lxl).
The last part follows from Lemma 7.2.5; it only re-
If k is finite, adjoin t to get n1 = D(t). Now let F = D1 (u) with endomorphism a:f(u) 1---~ f(u 2 ) and formE= F(v;a). As in the proof of Th.6.3.6 we see that
ma~ns
to produce E,C.
the centre of E is k (t).• This theorem may be expressed by saying that for skew fields infinite over their centre (where the latter
~s
finite) there are no non-trivial rational identities.
inWe now
give another proof of this result using the methods of Ch.4 and Th.6.3.6. Theorem 7.2.7.
Let E be a field with centre C, let D be a
subfield of E and write k
D n C.
=
[E:C] =oo and (ii) C is infinite.
Assume further that
(i)
Then every element of
Dk{X} is non-degenerate on E. Proof.
We know that any p
£
Dk{X} can be obtained as a
component p = u 1 of the solution of an equation Au
= a,
where A is a full matrix over Dk, and p will be nondegenerate on E provided that A goes over into a non-singular matrix under some substitution X
---~
E.
By Th.6.3.6,
the mapping Dk ---~ EC is honest, hence A is full over 171
EC and by the specialization lemma 6.3.1, A can be specialized to a non-singular matrix over E, which is what was needed. • When D is finite-dimensional over its centre, there are of course non-trivial identities, but Amitsur [66] shows that they depend only on the degree (cf. also Bergman [70]). More precisely, if [ D:k ]
=
n
. d 2 and E ~s any
=
D with infinite centre C containing k, where [E:C] then DE(X) depends only on D,d,r, m
of (rd) 2 , •
extens~on
= lXI and not on E.
It
is shown that DE(X) has dimension (rd) 2 over its centre, hence these fields are different for different values of rd. Moreover for d 1 id 2 the field with d 1 is a specialization of that with d 2 (cf. Bergman [70]). 7.3
Specializations We now examine how rational identities change under speci-
alization.
Of course we must first define the appropriate
notion of specialization. ~s
A homomorphism between two rings
said to be local if it maps non-units to non-units.
Let
D, D' be fields, then a local homomorphism from a subring D1 of D to D' is also called a local homomorphism from D to D' with domain D1 • If ~:D ---~ D' is a local homomorphism with domain
n1 ,
then ker
~ ~s
the set of non-units of
n1 ,
hence D1 is a local ring with residue class field n1 /ker isomorphic to a subfield of D'.
~
Let (D,a), (D',a') be X-fields, then a local homomorphism ~:D ---~ D' whose domain contains Xa and such that a'
is called an X-specialization.
=
a~
Clearly this exists only if
the domain of a contains that of a'. To describe rational identities we shall need the notion of PI-degree. 2 d"
n -
.
~mens~onal
Let A be a commutative ring, then Wl (A) is over
.
~ts
n
centre (as free A-module) and it
satisfies the standard identity of degree 2n (Amitsur-Levitzki theorem):
172
Let R be any prime PI-ring; by Posner's theorem it has a ring of fractions Q which is simple Artinian and satisfies the same polynomial identities as R (cf. Jacobson [75] or Cohn [77]). Let Q be d 2-dimensional over its centre, then R satisfies
s2d
0 and no standard identity of lower degree.
We shall call d the PI-degree of R (and Q) and write d deg R.
=
PI-
If R is a prime ring satisfying no polynomial iden-
tity its PI-degree is said to be
~.
We shall also need the notion of generic matrix ring. k be a commutative field and m,d > 1.
Write k[T] for the
commutative polynomial ring over kin the family T commuting indeterminates, where i,j
Let
= l, ••• ,d, A
A = {x1J .. }
of
1, ••• , m.
Let k(T) be its field of fractions and consider the matrix rings
We have a canonical m-tuple of matrices X, I\
A
= (x1J .. ); the
k-algebra generated by these m matrices is written kd and is called the generic matrix ring of order d. the free k-algebra on X
It is
= {XA} in the variety of k-algebras
generated by d x d matrix rings over commutative k-algebras. Amitsur has shown that R
=
kd is entire (cf. e.g. Cohn
[77]: one has to find a field of PI-degree d and embed it in ffiRd(E), where Ed k).
As an entire PI-ring R is an Ore
domain; its field of fractions 1s written k{X}d; like kd it has PI-degree d, if m > 1.
Of course form= 1, kd
reduces to a polynomial ring in one variable; this is not of interest and we henceforth assume that m > 1. Let (D,a) be any X-field; we defined in 7.2 its domain E(D) as the subset of R(X) for which a is defined •. Let Z(D) be the subset of E(D) consisting of all functions which 173
vanish for a.
Any f
E
Z(D) is called a rational relation,
or k-rational relation if coefficients in k are allowed. Explicitly, we have fa
= 0 in D, but of course this pre-
supposes that fa is defined.
Now Amitsur's theorem on ra-
tional identities (7.2) may be expressed as follows:
Let
D be a field with infinite centre k, then there is an Xfield E over k such that the k-rational identities over D are the k-rational relations satisfied by X over E.
Thus we can
speak of E as the free X-field for this set of identities. Moreover, the structure of E depends only on k, m and the PI-degree of D: If PI-deg D
d, then E
=
k {X}d is the field of generic
matrices, If PI-deg D :oo, then E
k{X} is the free k-field on X.
In particular, two k-fields satisfy the same rational identities if and only if they have the same PI-degree.
For our
first theorem we need a result of Bergman-Small [75].
We
recall that a ring R is local if R/J(R) is a field (where
J(R) is the Jacobson radical of R); if R/J(R) is a full matrix ring over a field, R is said to be a matrix local ring. Theorem 7.A
(i)
(Bergman-Small [75];
If R is a prime PI-ring which is also local (or even matrix local) with maximal ideal m then PI-deg R/mdivides PI-deg R.
(ii) If R1 ~ R are PI-domains, then PI-deg R1 divides PI-deg R. We shall sketch the proof of (ii) only. PI-degrees of R1 , R.
174
Let d 1 ,d be the They are also· the PI-degrees of their
fields of fractions Q1 ,Q. Let k 1 , k be their centres; by enlarging we may assume that k :2 k. Now choose a maximal
C\
1
commutative subfield F 1 of Q1 and enlarge F 1 to a maximal commutative subfield F of Q, then [F 1 ·k · ·d es [F : k] , an d . 1J d lVl this means that d 1 1d. • With the help of this result we can describe the specializations between generic matrix rings, following Bergman [77]. Theorem 7.3.1.
c,d
1.
>
Let k be a commutative field, m > 1 and
Then the following conditions are equivalent:
E(k {X}c) ~ E(k {X}d), i.e. every rational identity in
(a)
PI-degree d is one for PI-degree c,
(b)
there is an X-specialization k {X}d --> k {X} ,
(c)
there is a surjective local homomorphism D --> D
c
c'
d
where D. is a division algebra over k of PI-degree i, l
I
c
(d)
Proof. let
~
(a)
D~ ~
then c
d.
(b)
~
(c) is clear.
To prove (c)
~
(d),
Dd be a local ring with residue class field Dc ,
= PI-deg
D c
(d)~
of Th. 7A.
I
PI-deg D'
(a):
d
PI-deg Dd by (i), (ii)
Let E be an infinite k-field.
Since cld, we can embed 9nc(E) in 9nd(E) by mapping a to diag(a,a, ••• ,a).
Then every rational identity in 9nd(E)
holds in 9n (E).
But these identities are just the rational
c
relations ink {X}d, k {X}c' hence E(k {X}c) ~ E(k {X}d)' i.e. (a). • 7.4
A special type of rational identity
As a consequence of Th.7.3.1 there are rational identities holding in PI-degree 3 but not in PI-degree 2.
We
shall now describe a particular example of such an identity which was found by Bergman [77].
From results in Bergman-
Small [75] (cf.7.3) it follows that there is no (x,y)175
specialization
Thus there must be a relation holding in PI-degree 3 but not 2, and we are looking for an explicit such relation. We shall need some preparatory lemmas; we put [x,Y]
=
XY-
YX. Lemma 7.4.1.
Let C be a commutative ring and X,Y Eftn 3 (C),
then
[x, [x, Y] 2]
(1)
(det [x, Y]). [x, [x, Yr 1
whenever [x,Y]-l is defined.
J,
For 2 x 2 matrices the left-
o.
hand side of (1) is
Put Z = [x,Y], then tr Z = 0, hence Z has the characteristic equation z 3 + pZ - q = O, where q = det z. Now
Proof.
multiply by z- 1 : z 2 + p - qZ-l
o; apply [x,-
=
. q [X,Z -1] = 0, ~.e. (1). For 2
X
2 matrices,
If we write Y'
=
z2
- q =
J:
hence [x,z 2]
o,
[x,z 2] =
o .•
[x,Y], the conclusion of the lemma can be
expressed as (2)
((YI) 2) I
for 2
((Y') -1) I
for 3 x 3 matrices.
x
2 matrices,
Here we have used the convention of writing ~ for a scalar a. Lemma 7.4.2,
a if u
av
Let X, Y Eftn 3 (C) and write b for the discriminant of the characteristic polynomial of X, then (3)
Proof. 176
det Y1 1 1
t;
det Y'.
First let X
o2 •
then~
(A. 2 - A. 3 )CA. 3 - A. 1 ),
Now an iterated commuta-
tor has the form n (A.l-A.2) yl2
0
(A.2-A.l)ny21 (A;-A.l)ny31 hence det Y(n)
0
(A.l-A.3) ny 13 (A.2-A.3)ny23 0
(A.3-A.2)ny32
n n n (Al-A2) (A.2-A3) (A3-Al) Y1zYz3Y31 +
(A.l-A.3)n(A.3-A.2)n(A.2-A.l)ny13Y32Y21
Now n
= 1,3 differ by a factor
6
2
=
~.
hence (3).
This
proves (3) for matrices over an algebraically closed field whenever A + 0; hence it holds identically. • Using (2) and (3) we can write down rational identities for 3
x 3
matrices, but most of them will hold for 2
matrices too.
x
3 matrices which fails when these com-
mutators are replaced by 0.
(4) Since
2
What we need is a relation between determinants
of commutators of 3 write Y'
x
For any X and Y let us again
[x,Y] and consider
det Y'det Y"(det(Y"-l)')(det(Y"'- 1 ) 1 ) . -1
is a derivation, det(Y )' -2 (det Y) det - Y' and (4) becomes I
a
det- y
-1
Y'Y
-1
(det Y')(det Y")(det Y")- 2 (det Y"')(det Y'")- 2 (det Yiv) = (det Y') (det Y")
-1
(det Y"')
-1
iv (det Y ) •
Applying Lemma 7.4.2, we get (5)
(det Y')(det Y")-lll-l(det Y')-l~(det Y")
1. 177
Thus we obtain Theorem 7 • 4 • 3 • (B ergman [76]) •
Let k be a commutative field
and n = 2 or 3; for X,Y £in (k) write Y' n
(YZ)' [(Y-l)']- 1 , so that by (2), o(Y')
[x,Y],
= det Y'
o(Y)
=
or 0 accord-
ing as n = 3 or 2, then there are rational identities
O(Y')o(Y") [Co(Y")- 1 )'] [Co(Y'")-l)
(6)
Proof.
']=~~
if n if n
3, 2.
By equating the left-hand side to 1 we get an iden-
tity in degree 3 but not in degree 2.
We know that this holds
if the left-hand side is defined, so we need only find X, Y for which the left-hand side is defined.
Let K be any ex-
tension of k with more than two elements and write S for the set of
matrices(~
~)
a,b
K* when n
£
00 0 a
0 0
0 0
when n
c
~)
= 2,
or
a,b,c
£
K*
= 3. Then S consists of invertible matrices and is
closed under inversion and commutation by diagonal matrices with distinct elements.
If we choose Y in S and X diagonal
with distinct entries, all terms lie in S and so (6) is defined. • 7.5
The rational meet of a family of X-rings We shall now make a closer study of specializations, fol-
lowing Bergman [76].
We shall find that for skew fields they
cannot be reduced to the situation involving only two fields, as in the commutative case.
We shall be concerned with two
basic notions: an essential term in a family of X-fields and the support relation. Given rings R1 ~ R2 we say that R1 is rationally closed 1n R2 if the inclusion is a local homomorphism. The intersection of a rationally closed family is again rationally 178
closed, so we can speak of the rational closure of X in R, which is the least rationally closed subring of R containing X.
If it is R, we call R a strict X-ring; e.g. Q (x,y) is a strict (x,y)-ring, SO l.S z[x,y,y- 1], but not z[x,y,xy- 1]. Generally, if l:: is the set of all matrices over z which are mapped to invertible matrices over R, then the rational closure of X in R is contained in the l::-rational closure of Z , in the sense of Ch.4, but the two may be distinct (if x,y,u,v
E:
X, the entries of(:
but not the former).
~)-llie
in the latter
We note that an epic Z -field,
briefly an epic X-field, is just a strict X-field.
A local homomorphism between X-fields ¢:D
--->
D' may be
described as a partial homomorphism from D to D' whose graph is rationally closed in D
x
D'; hence if there is any X-
specialization at all, the rational closure of X in D x D' is the unique least X-specializationo
So there is at most
one minimal X-specialization between two X-fields.
Our aim
is to study the rational closure of X in finite direct products; to do so we need to introduce the following basic concepts a Definition.
Let {Rs}S be a family of strict X-rings, then
their rational meet
rr s
R
S Rs
is the rational closure of X 1.n
o
s The rational meet can also be viewed as the product in the
category of strict X-ringso smaller is
S Rs'
in the sense that for T
~
S we have a pro-
~ Rs ---> ~ Rs"
E.g. whether n1 AD 2 is the graph of a specialization in one direction or the other jection
PsT:
We note: the bigger S is, the
depends on which projection maps are injective. Lemma 7o5olo
Let {Rs}S be any family of strict X-rings, then
179
For ~Rs is the set of all rational expressions evaluable in each R modulo the relation of having equal values in each s R: f - g D, (2)
Ker (D ) cj;, t
'
S
U .L T
t
Ker (D ) • S
For when (1) holds, take f En E(Ds)' f t E(Dt), then f contains a subexpression g-l such that gDt = 0 but gDs I 0 -1
for s f t.
Conversely, given such g, we find that g
be-
longs to the right but not the left-hand side of (1); the equivalence of (1) and (2) is clear.
Using this notion,
we can say when the rational meet reduces to a direct product: Proposition 7.5.2.
Let X be a set and {Ds}S a finite family
of epic X-fields, then the following are equivalent:
(a)
Each s is essential in S,
(b)
for each s
E
S, there exists es
E
~ E(Du) such that
e~t = 8st' (c)
AD
S
Proof. (a)
s
=lTD
S
~
s·
(b).
Choose f
s
in Ds but not in Dt for t f s.
defined in all D and vanishing Then gt
order) vanishes on all D's except D • t
satisfies the required condition.
180
11 srt
t
f s (in any_ 1 Now es = gs (l:tgt) =
(b)~ (c).
By (b),~ Ds contains a set of central
idempotent~'es' which shows that~ Ds; ~ Rs for someRs~
D • Now AD 1s rationally closed inTI D , hence R 1s s s s s rationally closed in D and it contains X, so R = D • s
s
s
(c) ~ (a). Givens e S, choose g en E(Dt) such that gDs; 0 but gDt f 0 fort I s.• if E(D 1 ) ~ E(D 2 ), D1 An 2 is a local ring, the graph of a specialization n1 ---~ n2 • Illustration.
Similarly if E(D 1 ) then
n1 AD 2
=
D1 x
n1 hD 2 ;
Consider ~
E(D 2 ), while if neither inclusion holds,
n2 •
For more than two factors we shall
see that AD s is a semilocal ring, 1.e. a ring R such that R/J (R) is semisimple Artinian. Lemma 7.5.3.
Let f:R
---~
R' be a homomorphism such that
Rf rationally generates R', then f is local if and only if f is surjective and ker f Proof.
~.
~
J (R).
Rf is rationally closed because f is local and
it rationally generates R'
'
hence Rf
R' •
=
If
af
=
o, then
1 + ax maps to 1, a unit, hence 1 + ax is a unit, for any X
e R, but that means that a e J (R). Dt is surjective Our next question is:
answered by
Let X be a set, {Ds}S a family of epic
X-fields and t £ S, then the following are equivalent:
(b)
any relation defined in each D8 and holding in Dt holds in all Ds,
Note that by (c) there is a local homomorphism Dt --> ~ Ds. Proof.
e
£
R(X) represents an element in ker
Ps
t if and
only if it is 1n the left-hand side of (a) and it represents 0 if and only if it is in the right-hand side of (a); this just expresses (b). • When these conditions hold we say that t supports S (or also: D supports {D }5 ). t s
More generally, if t i S, we say
183
that t supports S if it supports S U {t} in the above sense. To gain an understanding of the support relation we begin by proving some trivial facts. Proposition 7.5.6. X-fields.
(i)
Let X be a set ~nd {Ds}S a family of epic
Then
Let t e S, U
~
S.
If t supports U and U contains an
element distinct from t, then t is essential for
UU{t}. If t supports U, then i t supports U U{t}.
(ii)
(iii) If t supports s. (i ~
E
If t supports U and for each u e U, u e
(iv)
supports Su, then t supports
Proof.
't
~s
~ si.
I), then i t supports
u s u u·
su
and u
inessential for S' means: any relation defined
in all D and holding in D also holds in some D , s s
s
t
~
t.
't supports S' means: any relation defined in all Ds and Dt and holding
~n
Dt holds in all Ds.
(ii) also follows.
To prove (iii), let f be defined in Dt
and f and Ds (s e S.) 1 so t supports USi. v
E
=
=0
in Dt' then f
0 in all D, s e S., s 1
(iv) Let f be defined in Dt and Dv where
Su' for all u e U.
hence f
Now (i) is clear and
Iff= 0 in Dt then f
=
0 in Du (u e U),
~ Su.•
0 in Dv (v e Su)' sot supports
If the D are commutative and t supports S, then s either S =~or Dt specializes to some Ds (s E S). More
Corollary.
precisely: t supports {s} i f and only i f there is a specialization Dt --~ Ds.
For if t supports S and S f
~'
then either t e S or t is
inessential for SU{t}; in the latter case there exists s ~ t in S such that D is a specialization of D • • 8
t
To clarify the relation between support and essential set we have the following lemma.
Note that by (i) above, a sup-
porting index is a special kind of inessential index.
184
Lemma 7.5.7.
Let X be a set, {D } a finite famdly of pairs s wise non-isbmorphic epic X-fields and Dt an epic X-field. Then the following are equivalent:
(a)
SU{t} is a minimal set in which
(b)
S is a minimal non-empty set supported by t.
Proof.
is inessential,
t
Let us write (a ),(b) for (a),(b) without the 0
0
Then (b 0 ) ~ (a 0 ) by (i) of Prop.7.5.6.
minimality clause.
To prove that (a) ~ (b ) we know by hypothesis that S is 0
minimal subject to~ E(Ds)
=
S~t} E(Ds).
By Prop. 7.5.4,
S is the set of essential indices of SU{t}, hence the projection Psu{t}S
is surjective.
If t does not support s,
D and u e: S such that a 0 t = 0 but s aDu 1- 0. Let us write a for a Du etc. Since the map u 1\ D --;:> TT D is surjective, there exists b E: 1\ D SU{ t} s s s SU{t} s
1\
there exists a e:
such that b Then e
=
u
a
SU{ t}
-1
u '
b
s
=0
for all s "' u, t, where s e: s.
. 1\ ab is m SU{t} Ds and has value 1 in Du and 0
everywhere else, for bs
=0
for s f t and at
1s a central idempotent and so s6?t} D8
=R
=
0.
x Du.
Thus e Now write
(S\{u})U{t}, then R C ~* Ds and R is rationally gener-
S*
ated by X and rationally closed, hence R
=
~* Ds.
Further,
Psu{t}S is a local homomorphism, so pS* s\{u}is too (we have to factor by Du), therefore by Prop.7.5.4, S\{u} includes all essential indices in S, which contradicts the minimality of s.
So Dt supports {Ds}S and (b 0 ) follows.
Thus we have (a)
~
(b ) and (b) 0
~
(a ); in an obvious 0
terminology, if S is a minimal a-set, it is a b-set.
Now
take a minimal b-subset S' of S; this is also an a-set contained in S, hence S'
=
S, i.e. S was a minimal b-set.
(a) ==> (b) and similarly (b) Corollary 1.
E (D ) t
~
Thus
(a). •
:::> rl 5 E (D s ) if and only if D t supports -
185
some non-empty subfamily of {Ds}S.
For the left-hand side expresses the fact that t is inessential in SU( t}.
Now pick SaC S minimal with this property and apply the lemma to obtain the desired conclusion.• A relation 't supports S' will be called non-trivial if
s#
~.{t}.
Corollary 2.
Each s s S is essential if and only i f there
are no non-trivial support relations in S. •
The essential relations are determined by the minimal essential relations, but there is no corresponding statement for support relations.
However, Cor.2 shows that
essential relations are determined by the support relations. Let us call a set S essential if each member is essential in it. Proposition 7.5.8.
Let {Ds}SU{t} be a finite family of
pairwise non-isomorphic epic X-fields, then the following conditions are equivalent: supports S and S is essential,
(a)
t
(b)
sGtt} Ds is a semilocal ring contained in Dt (via the projection map), with residue class fields D , s there exists a semilocal X-ring R ~ Dt with residue
(c)
class fields D (s s S), s (d)
Z (D t)
n
~
E (D s)
~ ~
Z (D s) and no E(D ) contains the
s
intersection of all the others.
Proof.
(a)=> (b) by Prop~7.5.4,5, (b)=> (c)='> (d) is
trivial and (d)=> (a) is also clear.• Corollary.
Let {D 8 }S be a finite family of pairwise non-
isomorphic epic X-fields,
t
s S and suppose that t supports
S and U is the subset of essential indices, then the map
~ 186
Ds -"'
uat} Ds is an isomorphism and
t
supports U • •
7.6
The support relation We shalY now give a complete description of all possible
support relations, using the work of Bergman-Small still following Bergman [76]).
[75]
(and
We shall need Th.6.8 of that
paper, which for our purpose may be stated as follows. Theorem 7.B
Let R be a prime PI-ring and p 1 a prime ideal
of R, then PI-deg R - PI-deg R/p 1 can be written as a sum of integers PI-deg R/p (allowing repetitions), where p ranges over the maximal ideals of
R.
Let us say that an integer n supports a set M of positive integers if for each m
E
M, n-m lies in the additive monoid
generated by the elements of M. subset of {1,2, ... ,n}.
Clearly M must then be a
The Bergman-Small theorem shows the:
truth of the following: If R is any prime PI-ring, then PI-deg R supports the set
{PI-deg R/p
Ip
prime in R}.
In what follows, X will be fixed, with more than one element, so that k {X}
n
has PI-degree n.
We shall write E(n)
=
E(k {X} ), Z(n) = Z(k {X} ) for brevity. n n Theorem 7.6.1. (Bergman [76]). Let n be a positive integer and M a finite (non-empty) set of positive integers.
Then
the following conditions are equivalent:
(a)
k {X}n supports {k {X}m
(b)
Z(n)
(c)
p
(d)
n
QE(m)
-
1, each
Let A be a
commutative k-algebra which is a semilocal principal ideal domain with just s non-zero prime ideals infinite residue class field K.
~
~1 ,
•••• ~s each with
= A/~. (e.g. let ~
K
-~ k be
an infinite field extension and take a suitable localization of
K[t]).
. mn (K.).
II ~
~
Then A/ J (A)
=
T[ K.' hence m (A) I J (IDl (A)) ;; ~
~
n
n
Now for each i,IDl n (K.) has a block diagonal subring ~
isomorphic to IDl m('~. l)(K.) x ••• x IDl (' ~ m ~,ri )(K.) ~
L.
~
say.
Hence (1)
Q
=ry..._..._ L~ ~
nm
(K.)
~n~
;;wz n (A)/J
(IDl n (A)),
where Q as a direct product of simple Artinian rings is semi188
Let R be the inverse image of Q in9R (A), by the
simple.
isomorphis~
(1), then J(R)
n J (9R (A)), hence R/J (R) ~
=
n
Q.
Since R/ J (R) is semisimple (Artinian), it follows that R is semilocal and PI-deg{R/max} = ~
R and p
=
Let ~ be a prime ideal in
M.
n A, then p is prime in A, so p 1s 0 or some :1 ]_.•
= 0, then A C R/~; write F for the field of
Suppose that p
fractions of A, then since A +9Rn ( J (A)) C R C 9Rn (A), we have RA* J (A)
=
9R n (A) A*
F 0,
~must
=
9R n (F) because A is a domain and
Hence RA* is simple with 0 as the only prime, so
be 0.
If p = :1., then K.]_ = A/:1.] _C- R/1l and since R ]_ R/~
is a finitely generated A-module, ated K.-module, hence Artinian. ]_ simple, and so (d+)
='l
(b), i.e. e
~
(b).
was maximal.
is a finitely gener-
It is also prime, hence Thus R satisfies (d+).
Assume that e lies 1n the left-hand side of
=0
~sa
rational identity holding in PI-degree
n and not degenerate in PI-degree m for any m to 3how that e
=
0 holds in each PI-degree m
as in (d+); this means that for each
prime~
E
E
M.
We have
M.
Let R be
f 0 of R there is
given a map ~:X --;:. R/\V such that ea.~ is defined in R/'ll to show that all the ea.~ are 0.
Since R is semilocal, by
the Chinese remainder theorem there exists a.:X --> R inducing all the a~, maximal~,
Now ea can be evaluated (mod~) for all
hence it can be evaluated in R.
ea = 0 and so ea~
=
Since e
E
Z(n),
0 as claimed. •
To give an illustration, we have 5
=2
+ 3.
Let A be a
local principal ideal domain with maximal ideal :1, then9R 5 (A) contains the subring
and we have a local homomorphism (K
= A/:1).
ms (A)
-->
m2 (K)
X
.m 3 (K)
This gives rise to a specialization of fields, 189
replace~
because when we
n
(K) by the generic matrix ring, we
get a field with the sarre identities as
~n (K)
o
If we combine Tho7.6ol with Prop. 7o5.8, we get Corollary lo
Let n be an integer and M a set of integers,
then the following conditions are equivalent: (a)
p: M A{ }k {X}. -» k {X}
U n
~
n is injective, with residue
class fields k {X} (mE M), m (b)
k {X}
has a semilocal subring with residue class
n
fields k {X}
m
(c)
(m
E
M),
M is a minimal set supported by n. •
Let n, M be as before, then the following are
Corollary 2. equivalent: (a)
k {X}
supports a non-empty subfamily of
n
{k {X}m (b)
I mE
M},
every rational identity holding in PI-degree n holds in some PI-degree m (m
(c)
M): Z (n)
n
QE (m) ~
~ Z (m)
o
there exists a prime ring of PI-degree n with ~
{PI-deg R/max}
(d)
E
M,
n supports a subset of
M.•
To describe the connexion between prirre ideals and the support relation we shall need a couple of auxiliary lemmas. Lemma. 7.6.2.
Let R be a ring,
~ 1 ,.oo'~m any ideals in R
and \Jl 1 , ••• ,\Jln any prime ideals such that~. s e: T. To
D5 , then p:R ---> R' is
If t,t' are as in (b),
is a maximal ideal of R containing
~t
then~t'
and hence ker p, so
its image under p is a maximal ideal of R' with the same residue class field.
But if R' has a residue class field
isomorphic to Dt' then t' s Ess(T) (b)=> (a).
~
T, so (b) holds.
Since imp rationally generates R', it is
enough to show that imp is rationally closed in R', i.e. the inclusion im p
~
R' is a local homomorphism.
Let
a s R be such that ap is invertible in R', then a i Ill t for all t s T.
Now consider those s e: Ess(S) for which a s
by (b), since s
9 ~t
= ker p.
i
T,~s 12~t for all t e: T.
~
s
•
"l
By Lemma 7.6.2 there exists be: R such that
b e: ker p and for any s e: Ess (S), b i a e:
Hence~s
~s;
But then a + b lies
~n
~
s
if and only if
no maximal ideal of R and so
is a unit, and (a + b)p = ap is likewise a unit in im p. • Note that (a) shows that the residue class rings of R' at maximal ideals are just the residue class rings of R at the maximal ideals containing ker p.
We can now express
the inclusion of prime ideals in terms of the support re-
191
lation.
We shall write Supp 8 (t) for the maximal subset of
S supported by t, i.e. the union of all subsets supported by t. Theorem 7.6.4. fields, R =
Let {Ds}S be a finite family of epic Xand~
1\ D
s
s
= ker(R --::> D ) , then
s
s
l11u
Supp 8 (u) = { v e: S
(2)
Proof.
s;;~)·
Isomorphic X-fields determine the same kernel in
R, so we may without loss of generality take the Ds to be pairwise non-isomorphic.
Fix u and let T be the right-hand
side of (2), i.e. the set of all s
E
S for
which~
then T satisfies (b) of Lemma 7.6.3, so R ---> jective.
u
D is sur-
T s ~~t which contains ~u by defini-
The kernel is
tion ofT, in fact since us T, we have the map
1\
:J~,
s -
n ~ T
s
=~ • u
D --::> D is injective, i.e. u supports T. s u follows that T t;;; SuppS (u), but clearly also Supp 8 (u)
Hence It
1\
T
~
T,
so Supp 8 (u) = T. • Corollary 1.
Ess (s)n
eg
Corollary 2.
In Lemma 7.6.3, (b) just states that T:;::;>
~ Supp 8 (t). • If T, T' ~ s, then ker PsT ~ ker pST' i f and
only ~'f u T Supp 8 (t)
u
:;::;>
T'
(i)
PsT is injective
(ii)
ker PsT ~ J (R)
In general R = ~s
1\
s
Supp 8 (t).
~ Supp 8 (t)
In particular,
= s,
~ Supp 8 (t) ~ Ess(S). •
D will have prime ideals not of the form s
e.g. if C is a commutative local domain, X is a rational
n1 are the field of fractions and the residue class field respectively, then D0 AD 1 = c, but C
generating set and D0
,
may have other primes (if C is not discrete). We recall that Prop. 7.5.2 asserted that and only if S is essential.
192
= 1T D if s s s More generally we can now say 1\D
s
s = s 1u
1\ D
s
s
us r
if the S. are disjoint support sets ~n S (i.e. for any t
E
~
S., ~
Supp 5 (t)k. Si). We know that non-isomorphic X-fields may have the same kernels, e.g. k~k[t] [x;a..J, where a. :f(t)
.
Here y
= xt and tx = xt
embeddings k
->
~
~
~
(cf. page 15f.).
1->
f(ti).
The resulting
Di are distinct fori= 2,3, ••• and
none is a specialization of the others. By contrast, if R is a right Ore X-ring, the R-fields may be determined by their kernels, e.g. R = kn.
If moreover, Dt
is commutative, then Supp 5 (t) = {u E S I t supports {u}}, i.e. the set of u such that Dt ---> Du is a specialization. For let C C
.
~s
a
=
S /'--...( ) D ; we have an injection C -> Dt, so up~s t.
commutat~ve
s
~ntegral
domain with Dt as field of
Let u E Supp 5 (t), then C/~ is an integral dou main with fields of fractions D Hence the localization u ~s a local ring Lu ~ Dt with residue class ring D , at~ u u fractions.
0
i.e. we have a specialization Dt 7. 7
---~
Du.
Examples Before constructing examples let us summarize the proper-
ties of supports.
This is most easily done by introducing
the notion of an abstract support system. stand a set S with a relation on S
By this we under-
P(S) written t
x
~
U and
called the support relation, with the following properties: S.l If t e S, Uh S, then t
s.z
if t
S. 3 i f t
a:
a:
si (i
cc
e I), then t
U and for each u
E
U t cc
U, u
~ a:
a:
UU{t},
si, S
u
of(/J, then t "' U S U U • u
If in S.2 we take the index set I to be empty, the hypothesis is vacuous, hence t
a:
(/J and by S.l, t
a:
{t} always.
A special case of the support relation is that where 193
t
a:
U t cx:·{u} for all u e: U.
This is completely determined by all pairs t,u with t
a:
{u}
and if we write t < u instead of t o: { u} we obtain a preordering of
s.
Conversely, every preorder on S leads to a
support relation in this way.
Thus preorders may be regarded
as a special case of support relations.
A support relation on S induces a support relation on any subset of
s.
If a support relation is such that
so: {t} and t
ex:
{s} imply s
t,
the relation is said to be separated.
E.g. the separated
preorders are just the partial orders.
Note that this is
not the same ass e: Supp 8 (t), t e: Supp 8 (s), which may well
hold for distinct s,t in a separated support relation. We now construct all possible separated support relations on a 3-element set.
There are 10 in all, 5 of them orders
(if we allow non-separated ones and do not identify 1somorphic ones, we get 53 support systems, 29 of them preorders). We list the 10 below, the orders first, with rising arrows to indicate specializations.
Examples. 3).
(a)
X=~, D.
1
=
Z/p. (i = 1,2, 1
(b) X= {x}, xl-:> 1,2,3 inQ,
(c)
k {X}., i=3,4,5. 1
2.
Here
n1
(D 1 AD 2 )
specializes to X
n2 , n1 Ao 2 Ao 3 =
DJ (a) X= {x,y,z}, Dl = Q(x,y)
(z 1--:> 0), n2 = Q(x) (y,z 1-:> 0), 0 3 = Q(z)(x,y 1-:> 0) (b) D1 = k {X} 4 , D2 = k {X} 2 , n3 = k {X} 3 • 3.
R = D1
Ao 2 Ao 3
semilocal domain C
(o 1 An 2 ) n(n 1 An 3 ) 194
~
n1
(a)
n1
(b)
D1 =
= Q,
D. = Z/p. (i = 1,2) 1
1
o3
k {X} 6 , D2 = k {X} 2 ,
k{X} 3 •
4.
R
= D1 ,.n 2...n3
is local ring with two mini-
mal prime ideals, subdirect product of
n1 ...n3 and n2..n3 • (a)
(x (b)
s.
n 1 = Q(x) (y
1-> n1
\->
n3 = Q (x,y = Q (xI-> 0),
0),
= Q (y)
0), n 2
1-> n2
0).
=Q(x 1-> p
prime), n3 = Z /p (x 1-> 0). R = n1 ... n2 ... n3 = n1 .. n3 • (a)
n1 • Q (x,y), n2 • Q(x) (y 1-> 0),
n3
""Q(x,y 1-.~ 0).
D2
= Q(x 1-->
0),
n1
(b)
"'Q(x),
n3 - Z/p.
In each case there are commutative examples.
For the re-
maining support systems (non-orders) we have of course only non-commutative examples.
In each case s
a:
T
~s
indicated
by drawing an arrow from s to a balloon enclosing T. indicate the partially ordered set of -!>
primes~.
~
=
We also
ker(AD.
J
in each case the lowest prime is 0.
6.
195
8.
Here n 1 a: {D 3 } follows from the relations
© r ~2
shown.
~1
!nl
n1
=k
{X} 3 ,
D2
=k
{X} 2 ,
n3
=k
{X} 1 •
9. 0
n1
=k
{ x,y
I
-1 2 (x y)
= yx-1 },
II! 3
D2 = k {x,y
n3
I
(x-ly)3
= yx-1
= k.
10.
n 1 = k { x,y } ,
n 2 = k { x,y
I
-1 2 (x y)
= yx-1
},
We conclude by expressing the support relation in terms of
.
singular kernels.
Let
P be any prime matrix ideal, then
P is defined as the set of matrices all of whose first order minors lie in P.
.
Clearly P ~ P and under a homomor-
phism into a field, if P represents the singular matrices,
P represents 196
the matrices of nullity at least two.
},
7.7.~.
Lemma
any n x n
~matrix
an~
prime matrix ideals P1 , •.. ,Pr in k, A tuP. can be extended to ann x n+l matrix
Given
~
which has rank n mod
P.
~
for each i.
Write A= (a 1 , ••• ,an); if a ~s another column, we put A*= (a,a 1 , ••• ,an) and we write A* t Pi to indicate
Proof.
that A* has rank n mod P.; this means that the square matrix ~
obtained by omitting some column of A* is not in P..
We
~
use induction on r; when r = 1, A has nullity 1 and we can make it non-singular by modifying a single column.
When
r > 1, we can by induction hypothesis adjoin a column to A to obtain ann x (n+l) matrix A. such that A. ~
If for some i, Ai
wise A.
1
E
P .• 1
~
i Pi, this will show that A1
t P.J
(j 1 i).
~UP j;
other-
Now form (ex E k).
For a I 0 this is not in P1 or P2 and it lies in any Pi (i > 2) for just one value of a. Avoiding these values (which we can do, because k is infinite) we get a matrix A* such that A*
t
Theorem 7.7.2.
Let {Ds}S be a family of epic X- fields and
uP 1.• • Now the support relation is described by p
s
the singular kernel of Ds' then Dt supports Ds (s £ S)
i f and only i f
In words: every matrix which becomes singular in Dt is either singular in each Ds or of nullity > 1 in some Ds • •
Proof.
Suppose that t supports Sand let A£ Pt' A
By the lemma we can find a column a such that (a,A)
i uP. s
i UPs. A*
Hence the equation A*u = 0 defines u = (u 0 ,u 1 , •••
u ) up to a scalar multiple in any D • n s 0 in all D , i.e. Since A E Pt' u 0 = 0 in Dt and so u 0 s 197
A
E
P for all s, hence (1) holds. s
Conversely, assume (1) and let f be defined in all Ds and f
=
0 in Dt.
We can find a denominator for f, say A,
with numerator A1 , then A1 E Pt, so either A1 E n Ps, ~.e. f = 0 in all D , or A1 £ P for some s. But then A £ P s s s and this contradicts the fact that A was a denominator. •
198
8 · Equations and singularities
8.1
Equations over skew fields In the commutative case there
~s
a well known theorem,
going back to Kronecker, which asserts that every polynomial equation of positive degree over a commutative field k has a solution in some extension field of k. One effect of this result has been to try to reduce any search for solutions to a single equation.
E.g. to find the eigenvalues of a
= 0.
matrix A we solve the equation det(xi-A)
In the general case no such simple theorem exists (so far!) and in any case we do not have a good determinant function (the determinant introduced by Dieudonn~ [43] is not really a polynomial but a rational function), so the above reduction is not open to us.
In fact we shall find
it more profitable to go from scalar equations to matrices. Our first problem is to write down the general equation ~n
one variable x over a skew field K.
We cannot allow x
to be central if we want to be able to substitute noncentral values of K, but some elements of K are bound to commute with x, e.g. 1, -1 etc. and it elements form a subfield k. ax
=
xa, then a must lie
~n
~s
Moreover, if a
k [x].
k, so that
Thus we have a field
K which is a k-algebra, and a polynomial = ~ = K~
E
the centre of K if arbitrary
substitutions of x are to be allowed. p of F
clear that these
~n
x is an element
Explicitly p has the form
199
wh ere a, b i, ••• , E K• Thus even a polynomial of quite low degree can already have a complicated form, and the problem of finding soluti~ns seems at first sight quite hopeless. (But we note that for polynomials in two variables over k, i.e. elements of k, an -extension field containing solutions has been found, by Makar-Limanov [75,77]).
A
little light can be shed on the problem by trying to generalize it. p
Instead of finding extensions L of K where
= 0 has a root, let us look for L such that a given matrix
A over F becomes singular.
We shall need some definitions;
let us recall that a K-ring is just a ring R with a homomorphism K --> R and a homomorphism f:R --> R' between Krings is a K-ring map if the triangle shown commutes. Let A be any square matrix over a K-ring R· we shall say that
'
A
is
proper i f there is a K-ring homo-
morphism of R into a K-field L
K----If
~R'
such that A maps to a singular matrix over L; otherwise A is said to be improper.
To elu-
cidate this concept we note Proposition 8.1.1.
Let K be any field and R a K-ring, then
an invertible matrix over
R is improper.
When
R is commuta-
tive, the converse holds: every improper matrix is invertible, but this does not hold generally.
Proof.
If A is invertible over R and f:R --> L is a K-ring
map then Af is again invertible, hence A is then improper. When R is commutative and A is not invertible, then det A is also a non-unit and so is contained in a maximal ideal m of R.
The natural map R --> R/ m is clearly a K-ring map
into a field and it maps det A to 0, so A becomes singular over R/m. To find a counter-example we can limit ourselves to 1 x 1 matrices, thus we must find a non-unit of R which maps to a unit under any homomorphism into a field. 200
Take a ring with
no R-fields, e.g. K2 ; any element c of K2 which a unit is improper but not invertible. • Our original problem was this:
~s
not 0 or
Does every polynomial p
in F "' ~ which is a nonunit have a zero in some K-field L? We note that p has the zero a ~n L if and only if the K-ring map F
--1>
L defined by x 1--~ a maps p to zero.
Thus the question now becomes:
Is every non-invertible
element of F proper? and it is natural to subsume this under the more general form: Problem 1.
Is every non-invertible matrix over
~
proper?
We shall find that this is certainly not true without restriction on k and K, and find a positive answer
~n
certain special cases, but the general problem is still open.
As a first condition we have the following result
(Cohn [76 ']). Theorem 8.1.2.
Let K be a field with centre k.
If every
non-constant equation over K has a solution in some extension, then k is algebraically closed in K.
Proof (*)
(2)
Let a
ax- xa
E
K be algebraic over k but not in k, then (metro-equation)
1
has a solution.
Let f be the minimal polynomial for a over
k, then by (2), 0
= f(a)x-xf(a) = f'(a),
formal derivative of f.
where f' is the
By the minimality of f it follows
that f' = 0, so a is not separable over k.
In particular,
this shows that k must be separably closed in K. purely inseparable over k, say aq
E
If a
k, where q "'pr, then
on writing D for the mapping x [--> ax - xa, we have
= o.
a
~s
D~ "'Daq
t
k, D ~ 0, say bD ~ 0 for some b E K. Now a a the equation xDq-l = b has a solution x in some extension, Since a
a
o
(*) I am indebted to P.v.Praag for simplifying my original proof.
201
but then x Dq o a
= bD # 0, a contradiction. a
Hence k must be
algebraically closed in K. • Thus for Problem 1 to have a positive solution it is necessary for k to be algebraically closed in K.
To sim-
plify Problem 1 we introduce a relation between matrices. Let
R
be any ring, then two matrices A, B over
R
are said
to be associated if there are invertible matrices U,V such that A
= UBV.
For simplicity we assume that R has invariant
basis number, so that all invertible matrices are square. Then it is clear that associated matrices have to be the same size, not necessarily square.
The following weaker equiva-
lence relation is often useful:
Two matrices A,B are said
to be stably associated if for some unit matrices, A + I is associated to B + I.
Clearly this is again an equivalence
relation (for the interpretation in terms of modules see Cohn [76]); stably associated matrices need not be the same size, but if an m x n matrix is stably associated to an r x s matrix then m-n number).
r-s (at least when R has invariant basis
The following result is an immediate consequence
of the definitions. Proposition 8.1.3.
Let R be a K-ring; i f a matrix A is in-
vertible, or proper, then any matrix stably associated to
A has the same property.•
We can now view the 'linearization by enlargement', already encountered in 6.4, in a different light: us that every matrix over
~
It tells
is stably associated to a
matrix linear in x: Theorem 8.1.4. then
Let A
= A(x)
be a square matrix over
A is stably associated to Bx + C, where B,C
some n.
Proof.
E
~ 1
K , for
n Moreover, i f A(O) is non-singular, we can take
C = I.
The first part follows as in 6.4.
If A(O) is nonsingular, then so is C, the result of putting x = 0 in Bx +
and Bx +Cis associated to C-lBx +I •• The linear matrix Bx + C obtained here is called the 202
c,
companion matrix for A(x).
(3)
To give an example, let
+ ••• + a n ,
and write P = an + p 1 x, then the first step is
where we have interchanged the rows.
If we continue in this
way, we obtain the matrix X
-1
0
0
0
0
X
-1
0
0
0
0
a
n
-1
an-1 •••
This is of the form xi - A, where A
E
K , and this matrix n
or also A itself is usually known as the companion matrix of p.
As we have seen, the general polynomial is more com-
plicated than p and its companion matrix, which always exists, by Th.8.1.4, need not be of the form xi - A.
More-
over, the companion matrix of a given polynomial or matrix 1s not generally unique, but it can be shown that if xi - A, xi - B are companions for the same matrix then A and B are similar over k (cf. Cohn [76]). An element p of
~
or more generally a matrix P whose
companion matrix can be put in the form xi - A is said to be non-singular at infinity.
finity.
E.g. (3) is non-singular at in-
Generally a matrix P with companion matrix Bx + C
is non-singular at infinity if B is non-singular; in any case the rank of B depends only on P and not on the choice 203
of the companion matrix, as is easily seen. also called the degree of P.
This rank is
For a polynomial of the form
(2) it clearly reduces to the usual degree, Let A
E~
n
(K), by a singular eigenvalue of A we under-
stand an element
~ E
K such that A -
~I
is singular.
We
can now state two conjectures whose proof would entail a positive solution of Problem 1. Conjecture 1.
Every square matrix over a field K has a
singular eigenvalue in some extension field of K. Conjecture 2,
Let K be a field which is a k-algebra and
assume that k is algebraically closed in K.
Then every
square matrix A over K has a non-zero singular eigenvalue in some extension of K unless A is triangularizable
over k.
If the answer were known, we could settle Problem 1 as follows. ~;
Let A be a non-invertible square matrix over
we have to show that A is proper, and by Th.8.1.4
and Prop"8.1.3 we can instead of A take its companion matrix Bx + C. ting x
If C is singular, this must be proper, by put-
0; otherwise we can take it in the form I - Bx.
If B has a non-zero singular eigenvalue S say, then I - BS-l is singular; otherwise by Conjecture 2, there exists E GL (k) such that P- 1 BP = T is triangular, hence
P
-1
P
n
(I - Bx)P
=I
- Tx (here we have used the fact that the
entries of P lie ink). ment a, then I - Ta
-1
If T has a non-zero diagonal ele-
is singular; hence all diagonal ele-
ments of T are 0 and so (Tx)n
0; it follows that I - Tx
is invertible, hence so is A, a contradiction. Before we can describe the special cases in which these conjectures can be settled we need to introduce another kind of eigenvalue, which always exists and which can be used to accomplish a transformation to Jordan canonical form.
204
8.2
Left and right eigenvalues of a matrix One of the main uses of eigenvalues in the commutative
case is to effect a reduction to diagonal form (when posLet A be a square matrix over a field K and sup-
sible).
pose that A is similar to a diagonal matrix D an).
=
diag(a 1 , ••• ,
Then there is a non-singular matrix U such that
Au
un.
If we denote the columns of U by u 1 , ••• ,un' this equation can also be written as Au.
~
u.a. ~
(i
~
l, ••• ,n).
This makes it clear that we have indeed an eigenvalue problem, but the a.l. need not be singular eigenvalues of A, since they do not in general commute with the components of u .. ~
Let K be any field and A
E
K ; an element a n
E
K is called
a right eigenvalue of A if there is a non-zero column vector u, called an eigenvector for a such that (1)
Au
ua.
Similarly a left eigenvalue of A is an element 6
E
K for
which there exists a non-zero row vector v, an eigenvector for 6, such that vA
=
eigenvalues of A
called the spectrum of A, spec A.
l.S
Let c e: K*• if Au
'
Sv.
= ua,
The set of all left and right then A.uc
= uac = uc.c-1 ac.
This
shows that the right eigenvalues of A consist of complete conjugacy classes; similarly for left eigenvalues. If -1 -1 -1 P E GL (K), then P AP.P u = P Au = p- 1ua, hence a is also n a right eigenvalue of P- 1AP. In other words, right (and left) eigenvalues are similarity invariants of A.
For sin-
gular eigenvalues this is not in general the case; in fact 205
it is easy to see that the three notions of eigenvalue coincide for elements in the centre of K ( 11 central" eigenvalues), but in general there is no very close relation between them.
Thus it is possible for a matrix to have a
right but no left eigenvalue (Cohn [73 11 ] ) , but as we shall see later, over an existentially closed field the notions of left and right eigenvalue coincide. G.M. Bergman has observed (in correspondence) that left and right eigenvalues are special cases of the following more general notion:
If A
£
K then n
~ £
K is called an
inner eigenvalue, more precisely an r-eigenvalue (where
1
ax, pb:x
1-->
xb,
then (2) may be written m.
(3)
and f(pb)
= pb :>. a
and by hypothesis f(\ ) is a unit a Now define the polynomial ~(s,t) in the
We note that :>..apb
= 0.
commuting variables s,t by cp(s,t)
f(s) - f(t)
=
s - t
then ¢(\ ,pb)(\
a
a
-
p )
b
f (\ ) • a
207
Since f(Aa) is a unit, Aa - pb has a two-sided inverse and it follows that (3) has a unique solution in M. • The significance of the lemma lie~ in (t:i:): GiveRn R,S,MS as in the lemma, the set of all matr~ces 0 s , r s , s s , m s M, is a ring under the usual matrix multiplication, and (2) shows that
(~ ~)
(: :)· (: :) (~ :)
· (a m)b · ·
'1 ar to a 'd'~agona 1' matr~x. · ~s s~m~ 0 Theorem 8.2.3. Let K be any field and A s K • ~.e.
n
Then spec A
cannot contain more than n conjugacy classes, and when it consists of exactly n classes, all except at most one algebraic over the centre of K, then A is similar to a diagonal matrix.
Proofo
We have seen that spec A consists of conjugacy
classeso
Let r be the number of classes containing right
eigenvalues and s the number of the remaining classes in spec A, then the space spanned by the columns corresponding to right eigenvalues is at least r-dimensional and the space of rows orthogonal to this is at least a-dimensional, hence r+s
~
n, and r+s is just the number of conjugacy classes in
spec A. Assume now that r+s
n; let a 1 , •• o,ar be inconjugate right eigenvalues and u1 ,oo.,ur the corresponding eigenvectors, while
=
s1 ,oo•,Ss
are the left eigenvalues not con-
jugate among themselves or to the a's, with corresponding eigenvectors v 1 , ••• ,v 5 • By Prop.8o2.1 the u's are right linearly independent, the v's are left linearly independent and v.u. 0 for all i,jo Write ul for then X r matrix J
~
consisting of the columns u 1 ,.o.,ur and matrix consisting of the rows v 1 , ••• ,vs.
v2
for the s x n Since the columns
of u1 are linearly independent, we can find an r x n matrix 208
V1 over K such that v1 u1 =I and similarly there ~san n x s matrix U2 such that v 2u2 =I. Put U = (u u ), V 1 2
(~; )-
then
vu The matrix on the right is clearly invertible and since one-sided ~nverses over a field are two-sided (i.e. a field is weakly finite, cf. Cohn
[71")), we have U(VU)-l =
v- 1,
so
VA 6 "v s s
It follows that VAV-l
=
(~ ~).
where a= diag(a 1 , ••• ,ar),
6 = diag(6 1 , ••• ,6s) and T s rKs. Now all the a's and 6's are inconjugate and all but at most one are algebraic over the centre of K, hence their minimal equations are distinct (cf.3.3).
If only right or only left eigenvalues occur, we
have diagonal form; otherwise let 61, ••• ,6s be algebraic, say.
Taking f to be the product of their minimal polynomials
we have f(S)
0 while f(a) is a unit.
can find X s rKs such that aX - XS matrix by
(~ ~)
=
By Lemma 8.2.2 we
T and transforming our
we reach diagonal form. •
The restriction on the eigenvalues, that there is to be only one transcendental conjugacy class (at most)
~s
not as
severe as appears at first sight, but is to be expected, since K can be extended so that all transcendental elements are conjugate (cf. 5.5). 209
8.3
Canonical forms for a single matrix over a skew field As before let K be a field which is a k-algebra; our
task is to find a canonical form for a matrix under similarity transformation.
The results are not quite as pre-
cise as in the commutative case, but come very close, the main difficulty being the classification of polynomials over K, i.e. elements of K[x]. Let A e K ; in 5.5 we called A transcendental over k if n
for any f e k[x]*, f(A) is non-singular.
If there is a non-
zero polynomial f over k such that f(A)
O, A is said to be
algebraic over k.
Of course when K
= k (the classical case)
every matrix is algebraic, by the Cayley-Hamilton theorem. In general a matrix is neither algebraic nor transcendental, e.g. diag(a,l), where a is transcendental over k, but we have the following reduction. Proposition 8.3.1.
Every matrix A over K is similar to the
diagonal sum of an algebraic and a transcendental matrix.
Proof.
We can interpret A as as an endomorphism of Kn;
clearly being algebraic or transcendental is a similarity invariant, and so may be regarded as a property of the endomorphism.
More precisely, V
= Kn can be regarded as a
(K,k[t])-bimodule, where tis a central indeterminate, with
= Av.
the action vt
Then the restriction of A to an A-
invariant subspace W of V is algebraic if and only if W is a k[t]-torsion module.
v,
Let V0 be the torsion submodule of
then V is a K-subspace, hence we can find a complement 0 of v in V: 0 (1)
v
v
0
~
vl.
Now A restricted to V is algebraic, while the transfor0 mation induced on V ~ V/V is transcendental. Hence if 1 0 we use a basis adapted to the decomposition (1), A takes the form 210
where A0 ~s algebraic and A1 is transcendental. By applying Lemma 8.2.2 we can reduce A' to 0 and so obtain the desired decomposition. • It is not hard to see that the algebraic and transcendental parts are in fact unique up to similarity, so that by this result, we need only consider algebraic or transcendental matrices. The transcendental part with.
~s
in many ways simpler to deal
For by suitably extending K, we can always transform
a transcendental matrix to scalar form, as we saw in 5.5. Of course over K itself we cannot expect such a good normal form.
We state the result as
Proposition 8.3.2.
Let K be existentially closed (over k).
Then any transcendental matrix A is similar to al, where a is any transcendental element of
K. •
To describe the algebraic part, let V space, with an algebraic endomorphism 8.
= Kn
as right K-
Writing R
= K[t]
with a central indeterminate t, consider V as right R-module by letting
~tic.
~
correspond to
~eic .• ~
If A is the matrix
of 8 relative to a basis of V, we shall call V the R-module associated to A.
It is clear that two matrices are similar
if and only if the associated R-modules are isomorphic. Now R is a principal ideal domain and every matrix over R is associated to a diagonal matrix (c£. e. g. Cohn Ch.8), thus there exist P,Q
£
[71"].
GL (R) such that n
where A. 1 is a total divisor of A. (i = 2, ••• ,n). This ~~ means that for each i there is an invariant element c of R 211
( i.e, cR = Rc) such that A.1-1 R -~ cR -~ A.R. 1
The A.1 are just
the invariant factors of ti - A and as right R-module V is isomorphic to the direct sum
We observe that this holds for any matrix A, algebraic or. not.
In fact we now see that A is algebraic if and only
if A divides a polynomial with coefficients in k. n
Let us
take k to be the precise centre of K, then a polynomial over K is invariant if and only if it is associated to a poly-
[71
nomial over k (Cohn
p.297).
11 ] ,
It follows that A is
algebraic if and only if A divides an invariant polynomial, n
i.e. (by definition) if and
on~y
if A is bounded. n
To find when A is transcendental we recall that an element of R is said to be totally unbounded if it has no bounded factor (apart from units).
Suppose that A has a bounded n
factor p say, then the R-module V has an element annihilated by p and hence by p*, where p* is the bound of p (i.e. the least invariant element divisible by p).
Now p*
=
p*(t) is
invariant and p*(A) is singular, so A cannot be transcendental. Conversely, if A is not transcendental, V has an element annihilated by an invariant polynomial, so some invariant factor
A. has a factor which is bounded, and hence A then has a n
1
bounded factor.
This proves most of
Proposition 8,3.3.
A
Let K be a field with centre k and let
Kn have invariant factors A1 , ••• ,An' Then (i) A is algebraic over k if and only if A is bounded, (ii) A is E
n
transcendental over k if and only i f A is totally un-
= ••• = An-1 = 1.
bounded, and then Al
n
Only the last part still needs proof:
Each A. (i < n) is 1 a total divisor of A , so there is an invariant element c such that A.1
I
c
I
n
An •
ding An is 1, hence Ai 212
But the only invariant element divi-
=1
(i
= l, ••• ,n-1).•
To obtain a normal form for algebraic matrices we need a result on tbe decomposition of cyclic modules over a principal ideal domain R.
We recall (cf. Cohn [71"] p.229) that a
cyclic R-module R/aR has a direct decomposition
where each qi is a product of pairwise stably associated bounded atoms, while atoms
~n
different q's are not stably
associated, and u is totally unbounded. result to (3) and observe that A. for i
If we apply this
We shall call a free over K if the map x isomorphism~
an
~
{x}
K(a).
a defines
Then we have the
n
Let a e E , then a is free over K i f and only i f
Corollary.
A + LAiai is singular only when A + LAixi is not full in ~.
•
Th. 8.5.1 also makes it clear how specializations in projective space should be defined. Each point of the projective space P n(E) ~s described by an (n+l)-tuple I; = and 1;,11 represent the same point if and only if n 11.>- for some A e E*. At first sight it is not clear ~
(I; , ••• ,!; )
o
I; .
~
=
how specialization in projective space is to be defined; instead of polynomials we would have to consider rational functions in the x. and (in contrast to the commutative ~
case) there is no simple way of getting rid of the denominators.
However, with Th. 8.5.1 in mind we can define
I; --> 11 (over K) if and only if n
E
0
A.~. ~
singular
~
~
EnA.n. singular, 0
~
1.
for any matrices A , ••• ,A over K. o n The condition of Th. 8.5.1 can still be simplified if we are specializing to a point in K: Theorem 8.5.2.
Let E/K be any extension, a e En, A e
~'
then a--> A i f and only i f for any matrices Al, • • • ,An over
K, I Proof.
EA.(a.1.
1.
).. ) is non-singular. 1
Assume that a-->
>-·• if
I -
LA. (x. ~
~
-
A.) becomes ~
singular for X = a, it must also be singular for X = A, but then it reduces to I, a contradiction, hence I - EA.~ (a.~ -
A.) is non-singular. 1.
Conversely, when the condition is satisfied, let the singular kernel of the map ~ show that under the map x
1-->
-->
K(a).
1- every matrix of
P be
We must
P becomes
singular, and here it is enough to test matrices of the 225
Thus let A + IA.a. be singular, we have to
form A + IA.x .• 1
show that C
1
1
=A
1
is also singular; note that C has + IA.A. 1 1
If C were non-singular, we could write
entries in K.
A+ EA.a. =A+ EA.A. + EA.(a.- A.) 11 11 11 1
=C+
EA. (a. - A.) 1
1
1
-1
C(I- IB.(a.- A.)), where B.= -C 1 1 1 1
A.• 1
Here the left-hand side is singular, and by hypothesis the right-hand side is non-singular, a contradiction, which shows that A+ EA.A. is in fact· singular. • 1
1
With the help of the specialization lemma 6.3.1 we again get a criterion for a point to be free: Corollary.
Let K be infinite-dimensional over its centre
k, where k is infinite, then for any a
£
En the following
conditions are equivalent: (a)
a is free over K,
(b)
every point of Kn is a specialization of a,
(c)
I - EA.(a.- A.) is non-singular for all A. over 1 1 1 1 K and all A. £ K. 1
Proof.
By Th. 8.5.2, (b)
To prove (b)
~
D
167
Z (D)
zero-set of a:X ---> D
173
kd
generic matrix ring
173
J (R)
Jacobson radical of R
174
rational meet
179
set of all m x n matrices over R
189
~n
Ess(S) set of essential suffixes
191
Supp 5 (t)
192
maximal subset of S supported by t
support relation
193
233
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118, 206
The range of derivations on a skew field and the equation ax- xb
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25
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115, 123, 158
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250
Index
Algebraic
109, 210
Algebraic dependence
230
Amalgamation property
133
Artin's lemma
EC-field
131
Eigenvalue, central
206
41
Bezout domain
89
Binomial extension
61
inner
206
inverse
226
left, right
205
singular
204
Elementary divisor
213
Elementary mapping
134 73
Central eigenvalue
206
Central extension
61
Essential index
180
Coideal
33
Essential index set
186
203
Existentially closed field
131
Companion matrix Coproduct
92
Coring
32
Cramer's rule
79
Crossed product
54
Dedekind's lemma Degree
40 30, 99, 204
Denominator Denominator set
79 8
Epic R-field
Extension
30
Faithful coproduct
93
X- field
166
Field coproduct
107 1
Field of fractions
30
Finite extension
116
Finitely homogeneous
89
Dependable
150
Fir
Dependence relation
230
Flat subset
169
Derivation
11
Forcing companion
134
Determinantal sum
81
Free field
128
Dickson's theorem
54
Free transfer isomorphism
108
Differential equation
65
Free point
225 251
Full matrix
82
226
Level
28
Lie algebra Galois group
40
Galois group, outer
52
Generalized polynomial identity (g.p.i.) 141, 163 Generic matrix ring Group like
173
152, 202
Local homomorphism
73, 172
Local ring
73, 174 227
Locus
33
Hilbert's 'theorem 90'
68
Honest homomorphism
88
Hua's identity
166
Improper matrix
200
Inductive class
134
Inert, inertia theorem
Linearization by enlargement
14lf.
Infinite forcing companion
134
Inner derivation
12
Inner eigenvalue
Mal'cev-Neumann construction
20f. 82
Matrix ideal Matrix local ring
174
Metro equation
201 99
Monomial
108
Monomial unit Multiplicative set n-fir
7, 77 3
N-group
46
206
N-invariant
50
Invariant factor
212
Non-singular at
Inverse eigenvalue
226
Numerator
oo
203 79
S-inverting, E-inverting 6, 76 Irreducible subset
169
Order Order function
Jacobson-Bourbaki correspondence
Ore condition, domain 38
Jacobson-Zassenhaus formula 71 Jordan canonical form J-skew polynomial ring
17 7f.
Outer derivation
12
Outer Galois group
52
216 24 PI-algebra
Klein's nilpotence condition
142
PI-degree 5
Presentation Prime matrix ideal
162 172f. 128 82
Laurent series
18
Projection
179
Leading term
99
Proper matrix
200
252
Pseudolinear extension
56
Tensor K-ring
127
Pure
99
Totally unbounded
212
Pure extension
61
Trace
Quadratic extension
56
Quasi-identity, quasivariety
70
Transcendental
210
Transvection
108
Trivial, trivializable
2£.
4
6
Unit Rational closure
77, 179
Rational meet
179
Rational relation
174
Rational topology
168, 228£.
Reduced order
47
Regular ring
5
Residue class field
166
Semifir
Universal EC-field
137
Universal R-field
85
Universal S-inverting ring
7, 76
Weakly finite
87
Zigzag lemma
135
3
Semilocal ring
181
Separating coproduct
93
Singular eigenvalue
204
Singular kernel
76
Singularity support
229
Skew polynomial ring
11
74, 172, 224
Specialization lemma
141
Spectrum
205
Stably associated
202
Strict X-ring
179
Strongly regular ring
4£.
Support Support relation
108
73
X-ring
Specialization
Unit, monomial
99 184, 187
Sweedler correspondence
35 253