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London Mathematical Society Lecture Note Series: 92
Representation of rings
over skew fields A.H. Schofield
Fellow of Trinity College, Cambridge
CAMBRIDGE UNIVERSITY PRESS Cambridge London
New York
Melbourne
Sydney
New Rochelle
CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo
Cambridge University Press The Edinburgh Building, Cambridge C132 8RU, UK
Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521278539
© Cambridge University Press 1985
This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1985 Re-issued in this digitally printed version 2008
A catalogue record for this publication is available from the British Library
Library of Congress Catalogue Card Number: 84-22996 ISBN 978-0-521-27853-9 paperback
CONTENTS
Preface
vii
Part I: Homomorphisms to simple artinian rings 1
Hereditary rings and projective rank functions
2
The coproduct theorems
22
3
Projective rank functions on ring coproducts
41
4
Universal localisation
50
5
Universal homomorphisms from hereditary to simple artinian rings
68
Homomorphisms from hereditary to von Neumann regular rings
89
Homomorphisms from rings to simple artinian rings
94
6
7
3
Part II: Skew subfields of simple artinian coproducts 8
The centre of the simple artinian coproduct
137
9
Finite dimensional divisions subalgebras of skew field coproducts
143
10
The universal bimodule of derivations
166
11
Commutative subfields and centralisers in skew held coproducts
178
Characterising universal localisations at a rank function
185
Bimodule amalgam rings and Artin's problem
196
12
13
References
219
Index
224
To Nicky
vii
PREFACE
The finite dimensional representations of a ring over commutative fields have been studied in great detail for many types of ring, for example, group rings or the enveloping algebras of finite dimensional Lie algebras, but little is known about the finite dimensional representations of a ring over skew fields although such information might be of great use. The first part of this book is devoted to a classification of all possible finite dimensional representations of an arbitrary ring over skew fields in terms of simple linear data on the category of finitely presented modules over the ring. The second part is devoted to a fairly detailed study of those skew fields that arise in the first part and in the work of Cohn on firs and skew fields.
As has been said, the main goal at the beginning is to study finite dimensional representations of a ring over skew fields. An alternative view of this is that we should like to classify all possible homomorphisms from a ring to simple artinian rings; such a study was carried out in the case of one dimensional representations which are simply homomorphisms to skew fields by Cohn who showed that these homomorphisms are determined by which sets of matrices become zero-divisors over the skew field and gave a characterisation of the sets of matrices that could be exactly those that become singular under a homomorphism to a skew field. This theory has a particular application to firs, rings such that every left and right ideal are free of unique rank to show that they have universal homomorphisms to skew fields. This applies to the free algebra over a commutative field, and the ring coproduct of a family of skew fields amalgamating a common skew subfield, and gives the free skew field on a generating set and the skew field coproduct with amalgamation.
In order to classify homomorphisms from a ring to simple artinian rings, it is necessary to investigate what type of information a homomorphism gives on the ring. The most obvious point is that it induces certain rank
viii
functions on the modules over the ring; over a simple artinian ring, where
S = M (D), n
is a skew field, every module is a direct sum of
D
copies of the simple module, and the free module of rank one is the direct sum of
copies of the simple module, so we can assign a rank to the
n
finitely generated modules over the ring taking values in 1 2z so that the n
free module on one generator has rank 1. If there is a homomorphism from R
to
we may assign ranks to the f.g. projectives or more generally
S,
R
the finitely presented modules over MGRS.
by
p(M)
is the rank over
S
of
Considering the rank functions induced on finitely generated pro-
jectives is important for constructing universal homomorphisms from an hereditary ring to simple artinian rings, whilst rank functions on finitely presented modules are precisely what is needed to classify all homomorphisms from the ring to simple artinian rings. The main result states that a rank function
on the finitely presented modules over a k-algebra
p
R
taking
values in 1 2l arises from a homomorphism to a simple artinian ring if and n only if it satisfies the following axioms:
1. p(R1) = 1; 2. p (A®B) = p (A) + p (B) ; 3. if
is an exact sequence of finitely presented modules
A 3 B 3 C -> 0
then
p (C) 5 p (B) If
R
<_
p (A) + p (C) .
is a ring that is not a k-algebra, it is necessary to
have a fourth axiom:
4. p(R/mR) = 0 or 1 for any integer Two homomorphisms
m.
ai:R -
Si
induce the same rank function if
and only if there is a commutative diagram of rings:
R
'Z
SI
Sz
S
2
In chapter 1, we begin the study of hereditary rings and rank functions on finitely generated projectives over them. In the main, it is a study of the category of finitely generated projectives and the ranks that the rank function induces on the maps in the category. It is shown that this behaves in a very similar way to the rank functions on von Neumann regular rings, which is where the notion of a rank function came from; this analogy
ix
is developed to its logical conclusion in chapter 6, where it is shown that a rank function taking values in the real numbers defined on the f.g. projectives over an hereditary ring must arise from a homomorphism to a von Neumann regular ring. Chapter 2 sets forth the first of the ring constructions that are needed in order to construct homomorphisms, the ring coproduct amalgamating a semisimple artinian subring. On the whole, it is a summary without proofs of Bergman's coproduct theorems. Chapter 3 shows how projective rank functions behave under the coproduct construction. It is also shown that if a module
M
over
the ring coproduct still requires
modules over
n R2
R1
requires
R'
R1
of
n
and
generators then the module R2
MRRR'
amalgamating a skew subfield
over F
generators provided that there are finitely generated requiring arbitrary large numbers of generators; the
condition is clearly necessary. This may be regarded as the analogue of the Grushko Neumann theorem. The results on projective rank functions are applied to prove a recent theorem due to Linnell; a finitely generated group is accessible if there is a bound on the size of finite subgroups. Chapter 4 presents the second important construction, adjoining universal inverses to maps between finitely generated projectives over a ring; this was studied by Cohn for matrices in order to construct homomorphisms to skew fields, but it has usually been regarded as a difficult technique, although it has arisen, usually in disguised form, in a number of contexts. For example, one of the methods used for showing that some finite dimensional algebra is of wild representation type amounts to adjoining a universal inverse to a suitable map. There are a number of ways of studying this construction developed recently which make it a little easier to calculate with and to think about, and the aim of this chapter is to present them. At the end, the algebraic K-theory of a universal localisation is discussed; there is an exact sequence for the algebraic K-theory that generalises the Bass, Murthy sequence for central localisation. Chapter 5 pulls together the various pieces presented in the first four chapters in order to construct universal homomorphisms from an hereditary algebra with a rank function on its finitely generated projectives to a simple artinian ring. The idea is fairly simple; given a rank function p
on an hereditary ring
R,
we ask which maps between finitely generated
projectives have a chance of becoming invertible under a homomorphism from R
to a simple artinian ring that induces the given rank function; if
a:P - Q
is such a map, then
p(P) = p(Q)
and
a
cannot factor through a
x
projective of smaller rank. Such maps are called full maps. The universal localisation of an hereditary ring at all full maps with respect to a rank function taking values in 1 a n
is a perfect hereditary ring and it is
simple artinian in a large number of interesting cases. Chapter 6 completes this circle of ideas by showing that if
R
is an hereditary ring with a
rank function on its finitely generated projectives taking values in the real numbers, there is a homomorphism from
R
to a von Neumann regular ring
with a unique rank function that induces this rank function on
R.
This
theorem actually holds provided that all countably generated right and left ideals over
R
are projective, which means that it applies to a von Neumann
regular ring with a rank function. Chapter 7 contains a number of results on homomorphisms to simple artinian rings beyond those that were discussed above. The space of all possible rank functions on finitely presented modules over a k-algebra that satisfy the axioms we stated earlier form in a natural way a I-convex subset of an infinite dimensional vector space. Given two rank functions that satisfy the axioms given, so does the rank function ql
and
are positive rationals such that
q2
where g1p1 + g2p2 q1 + q2 = 1. It is shown in
the course of chapter 7 that every rank function in the form pi
Eq,p.
where
q.
p
has a unique expression
are positive rationals such that
q. = 1
and
are rank functions that cannot be written as the weighted sum of different
rank functions. So, the space of all possible rank functions is a sort of locally finite dimensional 0-simplex. The methods and theorems developed in the first part of this book are of great use in studying the skew fields constructed by Cohn, and the second half of this book is a fairly detailed investigation along these lines.
In chapter 8, we investigate what is known about the centre of the skew field and simple artinian coproduct. We have a complete answer when we amalgamate over a central subfield; however, the results are rather incomplete for simple artinian coproducts where none of the factors are skew fields. Chapter 9 continues with a detailed discussion of the finite dimensional division subalgebras of skew field coproducts and a number of other related skew fields. As an example of the odd results that occur, it is shown that if
and
E1
E2
over the central subfield E1
and
E2
amalgamating
field extension
L
of
K,
are skew fields containing no elements algebraic k,
k
then
El o EZ k
the skew field coproduct of
can sometimes contain a finite dimensional
but if it does,
[L:k]
must be divisible by two
xi
different primes; there is an example where example of a skew field
with centre
D
k
There is also an
EL:k7 = 6.
such that
DOkks
and
DMkkP are skew fields where ks is the separable closure of k, and ' kP for some skew is the inseparable closure of k but Dokk = MP(D') field
D',
where
is the algebraic closure of
k
this settles a
k;
question of Cohn and Dicks. Chapter 10 develops the technique of the universal bimodule of derivations in order to distinguish between various non-isomorphic skew
fields. In particular, it is shown that the free skew field on m
generators
cannot be isomorphic to the free skew field on
m x n.
n
generators for
It also gives a way for recognising when a skew field is a universal localisation of an hereditary subring. Chapter 11 continues the investigation of the skew subfields of a skew field coproduct; we are particularly interested in the commutative subfields of such skew fields and in centralisers in matrix rings over a free skew field. In the first case, it is possible to bound the transcendence degree of commutative subfields of a skew field coproduct in terms of the transcendence degree of commutative subfields of the factors and the amalgamated skew field of the coproduct. For centralisers, it is shown that a skew subfield
D
with transcendental centre of
free skew field over
M (F) n
where
has a finitely generated centre over
k
is a
F k
of trans-
cendence degree 1, its dimension over its centre is finite, and this dimension must divide
n2.
At the end of the chapter, it is shown that a
2 generator skew subfield of a free skew field must either be free on those 2 generators or else it is commutative. Chapter 12 develops the characterisation of the universal localisations of hereditary rings that are skew fields which was developed in chapter 10 into a characterisation of simple artinian universal localisations of hereditary rings; then it is shown that if
T
is a subring of a
simple artinian universal localisation of the hereditary ring contains the image of then
T
R
so that the map from
is itself a universal localisation of
R R.
to
T
R
that
is an epimorphism,
It follows from this
result that epic endomorphisms of the free algebra over a commutative field are isomorphisms; this is the non-commutative analogue of the Jacobian conjecture.
The final chapter presents among other things a solution to an old problem; it is shown that for any pair of integers exists an extension of skew fields
E 0 F
a,b > 1,
there
such that the left dimension of
xii
E
over
F
is
a,
whilst the right dimension is
b.
By extending the
construction, it is possible to construct a new class of hereditary artinian rings of finite representation type. In order to effect these constructions, we develop a new type of hereditary ring construction, the bimodule amalgam rings; these are rings generated by two simple artinian rings and
subject only to conditions on the
S'
SS' = {Esis': si a S, si a S'}.
S, S'
S
bimodule
When we are able to show that these
hereditary rings have a rank function, their properties are of particular interest. In addition to the results mentioned above, they also allow us to construct isomorphisms between skew fields that at first glance appear to be quite different. As an example, it is shown that if division subalgebras of the skew field where
k
is a central subfield, then
F
such that
F 0 E1 k
E1
and
E2
are
CE1:k] = CE2:k]
is isomorphic to
F
E2.
k
There are a number of people that I should like to thank for their encouragement and help during the proving of these results and subsequently during the time that I was writing them down. The first person I should like to thank is Warren Dicks with whom I have discussed most of the results of this book; his care and accuracy have been of great assistance to me and many of the results have arisen out of conversations between us. I should also like to thank Paul Cohn for his interest and encouragement; I owe him a particular debt for having proven the first results in this area. I should also like to thank Rufus Neal for bearing with me despite the length of time that it has taken me to get this book into his hands at Cambridge University Press, and I am very grateful to Diane Quarrie for typing this book so well from a partial typescript of poor quality.
PART I
Homomorphisms to simple artinian rings
3
1 HEREDITARY RINGS AND PROJECTIVE RANK FUNCTIONS
Definitions and preliminaries In this chapter, we introduce the two main subjects of the first part of this book; hereditary rings and the projective rank functions on the rings, which we need in order to study their homomorphisms to simple artinian rings.
A left hereditary ring is one such that all left ideals are projective modules. We shall be interested in a number of variants of this definition; a left semihereditary ring is one such that all finitely generated left ideals are projective and a left Xo-hereditary ring is one such that all countably generated left ideals are projective. We shall often need to consider the two-sided properties, whose definitions we leave to the reader; our results tend to work most often in the case of two-sided X0-hereditary rings. We shall tend to miss out the words 'two-sided', when using these conditions. There is a two-sided condition implied by all of the one-sided conditions above; a ring is weakly semihereditary if, for all pairs of maps
a : PO -> Pl,s
modules such that in
Pi
aR = 0,
and the kernel of
:
P1 -- P2
then S
between finitely generated projective
P1 = P1 ® P1,
contains
Pi.
where the image of
a
lies
This is a two-sided condition
because of the duality between the category of finitely generated left projectives and finitely generated right projectives induced by HomR(_,R). We shall abbreviate 'finitely generated' to f.g., 'finitely presented' to f.p., and, in the case of vector spaces over a skew field, we abbreviate 'finite dimensional' to f.d.
Lemma 1.1
A left semihereditary ring is weakly semihereditary.
Proof: Suppose that a
:
PO -r P1,$
P1 -s P2
are two maps such that as = 0,
is a finitely generated projective for
where
Pi
of
is a projective module, so
6
:
P1 = ims$ker$,
i = 0, 1, 2.
The image
and the image of
a
lies
4
in the kernel of
B.
A good reason for introducing the notion of a weakly semihereditary ring is the following
Theorem 1.2
theorem due to Bergman.
Every projective module over a weakly semihereditary ring is
a direct sum of finitely generated projective modules.
We refer the reader to 0.2.9 of Cohn (71) for a proof of this result.
We shall work as long as possible with weakly semihereditary rings; however, we shall eventually be forced to restrict our attention to two-sided X,-hereditary rings. This class of rings draws much of its initial interest from the fact that all von Neumann regular rings have this property. By a von Neumann regular ring, we mean a ring R,
there exists an element
y
such that
R,
such that for all
xyx = x;
x
in
we shall see that there
are interesting connections between these classes of rings. Much of the work of this chapter is just a study of the category of finitely generated left projective modules over a weakly semihereditary ring. This has been done with a great deal of success for semifirs and firs by Cohn (71); a fir is a ring such that all left ideals and right ideals are free of unique rank, and a semifir is a ring such that all finitely generated left ideals (and so, all such right ideals too) are free of unique rank. In this case, the arguments work well because we have a good notion of the size of a finitely generated projective, and so we would like to have a generalisation of this idea for other rings. The relevant idea comes from the theory of von Neumann regular rings. Given a ring,
R,
we associate to it the abelian monoid
P®(R)
of isomorphism classes of f.g. projectives under direct sum. We may also associate to it a pre-ordered abelian group, the Grothendieck group,
K0(R).
It is generated by the isomorphism classes of finitely generated left projective modules [P],
subject to the relations
pair of isomorphism classes [P], [Q].
[P®Q] _ [P] + [Q],
for every
The pre-order is given by specifying
a positive cone, by which we mean simply a distinguished additive submonoid of positive elements, and, in this case we take the isomorphism classes of finitely generated projective modules, [P]. universal group associated to to be stably isomorphic when
P(R).
It is clear that
Two projectives
[P] = [Q];
P
and
K0(R)
Q
is the
are said
this is equivalent to the existence
5
of an equation P ®Rn = Q $ Rn. A projective rank function on a ring pre-ordered groups, p By definition,
p(P)
we shall call a rank function faithful if
We shall often simplify the notation by
P.
We note that a projective rank function is a
p([P]).
for
is a homomorphism of
the real numbers, such that p([R1]) = 1.
KO(R) --1R,
p([P]) ? 0;
for all non-zero
p([P]) > 0,
writing
:
R
left, right dual notion, because of the duality
HomR(-,R).
A partial projective rank function is a homomorphism of preordered groups p [R1],
:
A
where
A ---P,
is a subgroup of
and the partial order is that induced from
K0(R),
K0(R)
containing
by restriction.
We recall theorem 18.1 of Goodearl (79):
Theorem 1.3
Every partial projective rank function extends to a projective
rank function on
R.
This result allows us to characterise those rings that have a projective rank function. We say that a ring has unbounded generating number if for every natural number requiring at least
there is a finitely generated module,
n,
M,
generators. It is an easy check that this equivalent
n
to the condition that for no m is there an equation of the form Rm = R(m+l) ®P; and this is a left, right dual condition, which justifies the two-sided nature of our definition. Cohn mentions this class of rings in (Cohn 71) under the guise of rings such that for all matrix cannot be written as an
n
by
(n-1)
n,
the
matrix times an
n by n (n-1)
identity
by
n
matrix. We leave it to the reader to check the equivalence.
Theorem 1.4
A ring has a projective rank function, if and only if it has
unbounded generating number.
Proof: Certainly, if
R
has a projective rank function, it must have unbounded
generating number.
Conversely, if of
KO(R)
R
has unbounded generating number, the subgroup
generated by [R1] is isomorphic to
Z,
and under the isomorphism,
no stably free projective module can have negative image. So this isomorphism defines a partial projective rank function on
R,
which must extend to a
projective rank function by 1.3.
We have shown that most rings have a projective rank function; in
6
fact, projective rank functions arise quite naturally on rings and one is forced to study them in order to solve certain types of problems. If
some skew field
is a simple artinian ring, it has the form
S
and so,
D,
K0(S)
M
n
for
(D)
can be identified in a natural way with
so in this case, we have a unique rank function. If we have a homo-
1 Z; n
morphism from a ring K0(S),
R
to
this induces a homomorphism from
S,
which is naturally isomorphic to
to
K0(R)
therefore, homomorphisms to
n Z;
simple artinian rings induce rank functions to
n
and we shall need to
Z,
consider such rank functions in order to study homomorphisms to simple artinian rings. More generally, many von Neumann regular rings have rank functions so that in order to study homomorphisms to von Neumann regular rings we shall need to consider quite general projective rank functions. These projective rank functions appear naturally in the representation theory of finite dimensional algebras, for if algebra over the field
k,
M
and
this defines a homomorphism from
[M:k] = m,
a rank function
R
R
to
on
R;
it is easy to see that if
projective module over
R,
P = Re,
P (P)
p
=
is a finite dimensional
is a finite dimensional module, Mm(k)
and so determines is a principal
P
then
[Me:k] [M:k]
which determines the projective rank function, since all f.g. projective modules are direct sums of principal projective modules for an artinian ring. Another class of rings with a projective rank function that occurs naturally are the group rings in characteristic function on the group ring and
G
where
FG,
is a group given by
F
0.
We have a trace
is a field of characteristic
tr(Ifigi) = f0,
where
g
i
0
0
is the identity
element of the group. We extend this to a trace function on the ring
M (FG)
in the natural way and then we define the rank of an f.g. projective
P
be the trace of an idempotent
e
in
Mn(FG)
such that
well known that this is well-defined, taking values in
FGne = P. Q,
n
to
It is
and that it is
a faithful projective rank function. This will turn out to be useful to us later on in proving results due to Linnell on accessibility of f.g. groups.
Trace ideals It is often useful to be able to work with a faithful rank function on a ring rather than one that is not; so we should like to have a way of
7
getting rid of the projectives of rank zero. There is a standard way of dealing with this problem; we define the trace ideal of a set of f.g. projective modules,
closed under direct sum to be the set of elements,
I,
T,
that lie in the image of some map from one of these projectives to the free
module of rank
It is easy to see that this set is an ideal, in fact, an
1.
idempotent ideal known as the trace ideal of the projectives in is the universal R-ring such that
R/T
R/T4aR P = 0,
for all
and that
I,
in
P
I.
We wish to study the behaviour of this construction.
Theorem 1.5
Let
sum over a ring
be a set of f.g. projective modules closed under direct
I
and let
R
For a f.g. projective,
is the quotient of
R/T
P S Q = PQ, where Suppose that
Proof:
Q
be the trace ideal of this set of projectives.
R/TO R
Q,
summand of an element of
T
= 0,
if and only if
Q
is a direct
The monoid of induced projective modules over
I.
Pe(R)
and
by the relation Q'
P- P',
if and only if
are direct summands of elements of
then every element of
R/TQ RQ = 0,
image of a map from an element of
I
to
since
Q;
there must be a surjective map from an element of
lies in the
is finitely generated,
Q I
Q
I.
to
which proves
Q,
the first assertion.
Suppose that R/T;
R/TORa:R/TORP i R/TORP'
is an isomorphism over
so there is a surjection:
a06: PQ -> P',
where Q is a direct summand of an element of I. Therefore, 0 -> ker aes - PQ - P' 3 0 is a split exact sequence, where R/TQ Rae 0
over
is an isomorphism, since it equals R/T,
So
R/T0 R a.
kera e S
and must be a direct summand of an element of
as required, we have an equation of the form
becomes
Therefore,
I.
The converse is
P $Q = P' 9Q'.
clear.
We have the following consequence:
Theorem 1.6
Let
R
be a ring with a projective rank function
be the trace ideal of the projectives of rank projective rank function on
Proof:
0;
then
p;
let
extends to a
p
R/T.
By the last theorem, there is a partial projective rank function
defined on the image of
KO(R)
in
KO(R/T),
1.3, this extends to a rank function on
R/T.
induced by
p.
By theorem
T
8
We can do rather better than this on a weakly semihereditary ring. First, we need the following result on the behaviour of
on passing
P®
to the quotient by a trace ideal over a weakly semihereditary ring.
be a weakly semihereditary ring, and let
R
Let
Theorem 1.7
trace ideal of the set of f.g. projectives, R/T
then
Q
and
by the relation
Proof:
P - P'
We denote passage to Let
then over So, over
P',S
P
a
a$ = yd,Y
R,
:
R/T
P"
P'
P -* Q, d
:
P'®Q',
P ®Q
if and only if
by bars, so
is the quotient
P®(R/T)
are direct summands of an element of
Q'
be the
closed under direct sum;
I,
is a weakly semihereditary ring, and
P®(R)
of
T
where
I.
R = R/T.
be a pair of maps such that where
Q 3 P",
as = 0;
is an element of
Q
I.
we have
R,
(a Y)
R1=0
( and we note that
(ay) = a,
R
Since
im(ay) c P1,
and
.
is weakly semihereditary,
ker
ima c Pker
and \-6/ = S
P' 9 Q
Ptherefore,
P1 ®P2,
P1 ®P2,
where
where
that is, the weakly semihereditary condition is satisfied
for maps between induced projectives. Let
e
:
Rn -* Rn
So, by the previous argument,
but the image of
be a map such that Rn = P1 ®P2,
is equal to the kernel of
e
e2 = e;
where
that is,
e(1-e) = 0.
im e c Pl c ker(1-e);
(1-e),
so
im e = P11
shows that all f.g. projectives are induced, and, in consequence,
R/T
which is
weakly semihereditary. The rest follows from theorem 1.5.
This allows us to pass from a weakly semihereditary ring with a projective rank function to a weakly semihereditary ring with a faithful projective rank function, simply by killing the projectives of rank
0.
We
summarise this special case:
Theorem 1.8 function then
R/T
Let
p;
let
R
be a weakly semihereditary ring with a projective rank T
be the trace ideal of the f.g. projectives of rank
is a weakly semihereditary ring, and
p
induces a faithful
0;
9
projective rank function on
R/T.
If
takes values in
p
Z,
is semihereditary on either side.
Proof:
then
R/T
n
All is clear except for the last remark. In this case,
R/T
is a
weakly semihereditary ring with a faithful projective rank function, taking
Let M be a finitely generated left ideal and let
values in 1 Z.
n a f.g. projective over
be
of minimal rank such that there is a surjection
R/T
lies in the kernel of this surjection, we have a sequence:
x
If
a : P + M.
P
R/T-+P-+M c R, whose composite is
so
0;
P = P1 ®P2,
the kernel of the surjection. Hence, p(P2) < p(P)
unless
where
aIP2
:
:
which is in
P1,
is a surjection. Since
P2 -+ M,
we deduce that a
x = 0,
is in
x
P -+ M must be an isomorphism.
We have already noted that trace ideals must be idempotent; curiously, the converse is true for left hereditary rings as we see next.
Theorem 1.9
Let
ideal; then
I
Proof:
R
be a left hereditary ring, and let
As a left module over
R/I ®RI = I/I2.
contain
I,
be an idempotent
I
is a trace ideal.
R,
I
I = I2
is projective and since
Hence, the trace ideal of the projective module I must
but it can be no larger, since its image in
R/I
is trivial.
The inner projective rank If we have a partial projective rank function,
the subgroup A PA
of
KO(R),
pA'
defined on
we define the generating number with respect to
of a finitely generated left module
M
over
R
by the formula:
g.pA(M) = inflP]{pA(P): [P] is in A, ] a surjection P i M}
If all stably free modules are free of unique rank, and of
K0(R)
where
pA'
generated by
ER1],
A'
is the subgroup
the generating number with respect to
is the unique rank function defined on
A
pA''
is the minimal number
of generators of a module. So we could hope and we shall show that our more general notion is a useful refinement. We intend to use a projective rank function
p
to analyse the
10
category of f.g. projectives over a ring
We have a rank associated to
R.
each object of the category, so, our next aim is to give each map a rank. Let
a
:
P - Q
be a map between two f.g. projectives; we define the inner
projective rank of the map with respect to
p(a) = inf[P,]{p(P')
to be given by the formula:
p
3 a commutative diagram
:
P\ -' Q}
P'I
it is sometimes useful to have a related notion to hand; we define the left nullity of
a
defined by
p(Q) - p(a). p(P) = p(Q), the nullity of
to be
Similarly, the right nullity of
p(P) - p(a).
is
a
is
a
p(P) - p(a).
We may relate the inner projective rank of a map and the generating number of suitable modules.
Lemma 1.10
Let
be a ring with a projective rank function
R
inner rank of a map
a
P - Q
:
infM {g.p(M): a(P) E M c Q,
p
Proof:
I
then the
is equal to the following:
where
In particular, if
p;
R
M
is a f.g. submodule of
Q}.
is a left semihereditary ring,
P
(P
Q}
If there is a commutative diagram P
then
a(P) c S(P') c Q
P'
g.p (R(P')) : p(P'),
and
Conversely, if P' -> M,
so
p(a) ?infM{g.p(M):a(P) cMcQ}. and there exists a surjection
a(P) S M c Q,
we have a commutative diagram
P -w Q,
since
P
is projective.
P'
Hence,
P (a) = infM{g.P (M) :a (P) cMcQ} , A map
p(a) = p(P), to
p
a
:
P + Q
is said to be left full with respect to
and it is right full if
p(a) = p(Q);
p
if
it is full with respect
if it is left and right full. The reason for considering full maps
with respect to a projective rank function is that the only maps to have a chance of becoming inverted under a homomorphism to a simple artinian ring are the full maps with respect to the induced projective rank function; of course, it is in general rather unlikely that they all do; however, we shall find that for a hereditary ring there are homomorphisms for each projective rank function that invert all full maps.
11
We see that entirely the same theory may be set up on the dual category of f.g. right projectives, where the rank of a f.g. right projective module is that of its dual module. It is clear that the dual of a left full map is right full and vice versa. The important fact about the inner projective rank on a weakly semihereditary ring is an analogue of Sylvester's law of nullity. We say that a ring satisfies the law of nullity with respect to ively, that the projective rank function
p,
is a Sylvester projective rank
P
function, if for every pair of maps between f.g. projectives B
:
such that as = 0,
P1 - * P2
or, alternat-
then P (a) + P (S)
< _ p (PI) .
a
If
PO -* P11
:
is a
R
ring such that all f.g. projectives are free of unique rank, and this rank is a Sylvester projective rank function,
Theorem 1.11 p;
then
p
Let
R
be a weakly semihereditary ring with a rank function
for
as = 0
weakly semihereditary ring, then
P".
and the kernel of
Hence
is a Sylvester domain.
is a Sylvester projective rank function.
Proof: Recall that if
in P',
R
:
and
PO -* P1,
P1 = P' ®P",
contains
s
p (a) S p (P') ,
a
so that
P',
and p (s) <_ p (P") ,
a
:
s
over a
P1 -* P2
where the image of
a
lies
factors through
so that p (a) + p (g) < p (P1) .
There are a few results that we can deduce from the law of nullity for a projective rank function on a ring. On the whole, they are a little technical, but since we shall need them later, it seems better to bore the reader now than to break up the flow of later proofs. Their point is to demonstrate the analogy between these rings with Sylvester projective rank functions and simple artinian rings with the standard rank function.
Lemma 1.12
Let
be a ring with a Sylvester projective rank function
R
then for any pair of maps
a
P(as)2,p(a) + p(s) - p(P1).
:
PO a P11 B
:
P1 -* P2'
In particular, this holds for weakly semi-
hereditary rings for any projective rank function. as Proof:
the maps
Suppose that
(al Y)
:
P .P is a commutative diagram. Then we have
P0 ; P ®Q, 1
-d
'
P1 ®
Q
-, P 2 ,
and
(alY) ( -) = 0
p;
12
so, by the law of nullity, p(ay) + p(6) :- p (P1) + P(Q). p(a)
p(s) < p( 6)
and
<_ p(aly)
;
therefore
But
p(a) + p(s) - P(Pl) 5 P(Q)
Taking the infimum on the right proves the lemma.
We have the following corollary:
p.
If
then
R
Let
Corollary 1.13
be a ring with a Sylvester projective rank function
is right full, then
a
p(as) = p(s); dually, if
is left full,
a
In particular, the composite of left full maps is left
p(sa) = p(s).
full, and the composite of right full maps is right full.
Let
Proof:
then
a
PO
:
be right full, and let
P1
s
:
So,
p(s) ? P(as) ? p(s) + p(a) - P(P1) = P(s).
P1 -> P2
be some map;
There-
p(as) = p($).
fore, the composite of right full maps is right full. The rest follows by duality.
Let
Lemma 1.14 Let
a
:
P ( a )
Proof:
Let
P
be a ring with a Sylvester projective rank function
R
P1 -r P2,
and
s
:
Q1 -* Q2
s) <_ p(a) + P(Q1), P(s) + P(P2).
+ P ( s ) <
be a f.g. projective through which
have an equation
(Y S Since
61e2 = 0,
r a
I
2)
(el
S)
factors; then we
/
`
=
e2)
the law of nullity shows that
p(a) + p(s) < p(gl) + p(e2) < p(P),
p(dl) + p(c2) 5 P(P);
and taking the infimum on the right
shows the first inequality.
If a = Sc, a 0
so pry
p.
be a pair of maps; then
d :
P1 -> P,c
: P-P2, then
0) (E
(a s) _ (0 <_ p(P) + p(Q1),
and taking the infimum over
s )
P(y ) <_ p(a) + P(Q1)
p(P)
s)
shows that
so
13
d'E', d'
similarly, if
:
Q1 -r Q, E'
:
Q +
then we have the equa-
2,
tion:
(Y So
a O
p
\Y 5 \O
B
s p(Q) + p(P2),
O'
and taking the infimum over
p(Q),
we see that
Y
a
P(Y B) < P W + p(P2) We have one more dull lemma to put behind us:
Let
Lemma 1.15 if
all
a
a
R
be a ring with a Sylvester projective rank function p; Ol p(a) + p(B), Also, for is left full, pa B
is right full or
and
= -
Y
P(O S) = P (a) + p(5).
B,
/
Proof: We use the notation of the last lemma. p(a) + p(B) < p (Y $) < p(a) + p(Ql); and so
P(B) = P(Q1), a
a01 p(a) + p(B) = p YBI,
if
a
is left full,
A similar argument works if
is right full. If
a = dl d2,
and
B = E1E2,
we have the equation:
(O B) - (01 62)(01'2) It follows that
p(0 B) = p(a) + p(B)
In order to get fairly decisive results, it is necessary to
restrict our attention to a two-sided
X.-hereditary ring with a faithful
rank function. Dicks pointed out that the proof of the next main theorem,
originally stated for two-sided hereditary rings actually holds in the greater generality. It is the central point in the proof that a rank function on a two-sided X 0 -hereditary algebra arises from a homomorphism to a von
Neumann regular ring, and the reader may well wish to read it only when he comes to this theorem in chapter 6.
Theorem 1.16
Let
R be a two-sided X,-hereditary ring with a faithful
projective rank function. Then every map between f.g. projectives factors
14
as a right full followed by a left full map.
First, we show that every map has a non-trivial right factor that
Proof:
is left full. If
a
P1, im (a) c P1 c Q,
1.10, there exists and
such that
p(P1) < p(P), p(Q),
p(P1) - p(a) < 1. a1 : P1 } Q
If
at the nth stage, if P
is not a left full embedding, we may find
P1 S P2 c Q, p(P2) < p(P1),
such that
P
is not left full or right full, then, by Lemma
P -> Q
:
c Q,
S P
a
P
:
p(P2) - p(a1) < /;
< p(P
), and p(P ) - p(a n n n-1 n n-1 n-1 if this process does not terminate, we obtain the chain:
n'
ima c P1 C P
c .......... c P 2
UPn
P!
is a countably generated submodule of
some
N.
We claim that the embedding PN
Since PN c Pm
for all
uPn
P1 ®P2 a ....
N
is finitely generated, it lies in
ima
PN c Q
is finitely generated,
m 2 n.
n
so it is projective,
Q,
is a f.g. projective module.
Since
<
)
c ...... c Q.
n
and must be a direct sum of f.g. projectives by 1.2. So
where each
P2
in general,
is not a left full map, we choose
-)' Q
n-1 n-1 such that p(P
and
PN = e P', i=1
for
is a left full map.
PN c Pn
for some
n,
and so,
Moreover, it is a direct summand of each such
Pm
since it is a direct summand of their union. Consider a f.g. submodule,
such that
Q',
PN c Q' c Q;
consider
the split exact sequence:
0 -> Pm n Q' Pm n
PN
,
so
P + Q' - Pm
Pm n Q' = PN a Q",
-
0
,
for some module
Q",
and so
p(Pm n Q') ? P(PN). Also,
P-1 C P
c P m + Q',
so that
p(Pm + Q')
>_ P(Pm - m + p(Q') = p(P + Q') + p(P n Q'), and we m m have just shown that the right hand side is greater than or equal to 1 so that we can deduce p(Q') >_ p(PN) but this holds P(P") + P(Pm) - m' N M; for all m > n, and so we conclude that PN S Q must be a left full map
m From the exact sequence,
that is a right factor of
m
p(P
a
:
P - Q;
by construction,
p(P") < p(Q).
15
By duality, we deduce that we can always find a right full left factor of pl
:
a
P -Q,
:
where
P -* Q1,
when
where
is not left or right full, a = ply,
a
p(Q1) < p(P).
Since a sequence of right full maps is
right full, by 1.13, we may assume that the right full map is to some sub-
module
of
Q1
yi
.
where the induced map Q1 C Q
is not left full, we Q1 C Q2
and
p(Q2) < p(Q2),
such that
Q2, Q1 C Q2 c Q
can find
ima c Q1 c Q,
So we have
is right full. If the map
: P i Q1
a'
Q.
is a
right full map. We assume that this process does not finish and obtain a contradiction. If it continues, we have an ascending chain:
ima c Q1 C Q2 c .... c Q,
where
P(Qi+l) < p(Qi)
is a countably generated submodule of Q, so it is projective and, iQi by 1.2, vQ. Z sQ!, where each Q is f.g. projective. For some m,
Q c
1- j=1 j
®Q';
summand of
Q1 c Qn
again, for some
Qn+l,
n,
j=1Q' j-c Qnn
so that we have
m
factors through
m p3(®1Qj)
Q.+,,
and
®Q' j=1 j
is a direct Since
P(Qn+l) < p(Qn)'
J
it cannot be right full, but it is the
composite of the right full maps
Qi C Qi+1'
so that it is right full by
1.13. This contradiction shows that our process must end at a finite stage;
Qm c Q
that is,
is left full for some
m.
But
P i Q
m
is the composite
of right full maps, and must be right full. So, we have a factorisation of a
:
P -+ Q
as a right full followed by a left full map.
When every map factors as a right full then a left full map with respect to a projective rank function, we shall say that the ring,
R,
has
enough right and left full maps with respect to the projective rank function. We describe a factorisation of a map as a right full by a left full map as a minimal factorisation.
Corollary 1.17
Let R be a two-sided X0-hereditary ring with a faithful
projective rank function a
:
P - Q,
Proof:
p(a) = mine,
p.
Then for any map between f.g. projectives
{p(P')
:
ima c P' C Q}.
This follows at once from 1.16. For
right full and
a2
is left full. By 1.12,
a = a1a2,
where
a
P -* P'
is
p(a) ? p(a1) + p(a2) - p(P');
so
:
p (a) = P (P') . There are occasions when this result is automatically true for a
16
weakly semihereditary ring; for example, if we have the descending chain condition on the numbers
p(P).
It fails in general for weakly semihereditary
rings.
The torsion modules for a projective rank function We pass from the study of quite general maps in the category of f.g. projectives to the study of the full maps with respect to a projective rank function. We shall assume that the projective rank function is faithful throughout this section. The sensible way to study the full maps with respect to a faithful Sylvester projective rank function is to look at the full subcategory of modules that are cokernels of full maps with respect to the projective rank function; we shall call this the category of torsion left modules with respect to the rank function. If lemma shows that
and
a
B
coker a = coker S,
Schanuel's
are stably associated, that is, we have a
matrices of maps equation:
1 (0 11)
1 0 OI )
P
where
and
y
Theorem 1.18
Let
rank function to
are invertible maps.
6
p;
R
be a semihereditary ring with a faithful projective
then,
T,
the category of torsion modules with respect
is an abelian category.
p,
Let
Proof:
M1
and
M2
be torsion modules with presentations:
a
0 i Q. # P. -+ M. -+ 0, where ai
is a full map, and let
some map. We have a commutative diagram:
0-+
1 0'
0-+Q2 -+ P2 -+ M2 -+0 We have two presentations of
0-+Q2-+ imO' +Q2+imo+0
0i Qi-+P1+ imO-+0
1
imo:
0 : Ml -> M2 be
17
is finitely generated, so it is projective, and
P' = imo' + Q2
where
Q2 ®Pl = Qi ®P2,
Schanuel's lemma shows that tive.
Q1 S Qi
p(PZ)
>_ p(Q2);
P1,
S
Q2 S PZ c P2,
and
on putting this into
so that
and so has rank at least
P1
module with respect to whilst
p.
is also f.g. projec-
p(Qj)
and
Qi
p(Q!) = p(P1),
The kernel of
so
and
lies between
P1
Q1
is a torsion
imo
and is also torsion,
Q1/Q1,
is
0
>_ p(Q1),
we see that equality
Q2 ®P1 = Q1 ®P2,
must hold; moreover, any f.g. module between and
Qi
so
which is torsion. Finally, finite direct sums of
coker 0 = P2/P'
torsion modules are torsion, which completes the proof that
is an abelian
T
category.
Clearly, this result will have consequences for the factorisations of full maps. In particular, we shall need the following lemma later on.
Lemma 1.19
Let
R
be a semihereditary ring with a faithful projective rank
fY1 function
Let
p.
O
Yn) where
as =
a,6 and
are all
i = 1 to n,
yi.,
:::-.
full maps with respect to
p;
then there exists an invertible map
0
such
that
where
a.i
and
$i
are full maps such that
It is sufficient to prove this for
Proof:
n = 2
= yi
.
since the general case
O
y
follows by induction. So, consider
aisi
a$ = dl
y
;
coker yl embeds in coker as
with quotient module isomorphic to coker y2; coker as maps onto coker a, and the image of coker y1 in coker a is a torsion module of the form coker all where
a1
is a left factor of
yl'
whilst the quotient of coker a by a
is a torsion module of the form coker a2 where a2 is a left factor
coker a1
of y2; therefore, there is an invertible map
0
such that
0 aO =(al 0 ) a2
;
inspection shows that
{b 1B
takes the form I S2
2/
In certain cases, we can show that this category of torsion
18
modules must be artinian and noetherian, which allows us to apply the JordanHolder theorem in order to deduce the unique factorisation of maps into atomic full maps. Our method of proving this is to show that the category is noetherian in certain cases and then use a duality between the left torsion and right torsion modules in order to prove that it is artinian too. This duality occurs as a special case of a duality we find in a number of contexts, in particular, in the representation theory of finite dimensional algebras, so we shall develop it in some generality.
R be some ring, and let
Let
1B
be the full subcategory of
f.p. modules of homological dimension 1, such that
we call
HomR(M,R) = 0;
these the left bound modules; similarly, we define the right bound modules, that are the objects of the category if
M
is in
such that
and
a
ax
The categories
Theorem 1.20 functors
Proof:
1B
Our conditions simply mean that
is the cokernel of some map
a
:
P -* Q
are injective; it is clear that this should
Qx -* Px
:
lead to a duality between
B1.
M
if and only if
1B
and
B1
and
1B
by sending coker a to coker ax.
B1
are dual with respect to the
Extl(_,R).
Let
a 0 -* P i Q -> M -> 0
be a presentation by f.g. projectives of an a} Px -- Ext1(M,R) - 0,
element of
1B;
Ext1(M,R),
which we shall call the transpose of
then,
0 -* Q
TrM
it is clear that
lies in
M,
is a presentation of and write as
TrM;
and that
B1,
Tr(TrM) = ExtR(TrM,R) = coker a = M.
ExtRR) M -* N
is a contravariant functor of the first variable, so
induces a map
Tr a: TrN -* TrM;
we write this out explicitly in
order to show that the diagram below is commutative:
Tr(Tr6) Tr(TrM)
Tr(TrN)
IIZ
112
M
N
,
where the vertical maps are simply
given by the isomorphisms coker (ax)x = coker a. a--
Let 0-*P1
M and
presentations of O
P
1
a
a
0-*P2 -
Q1 -M-0, and O-*P2142 Q2+N+O be then we have the commutative diagram below:
N;
Q1 - M-*O B
Q2
+N -*0
19
Dualising gives us:
alx
0-}Q2 i P2->TrN->O Ix
O-*Q1
a -*
Ix
1
Trs
P1 ->TrM-r0
which allows us to exhibit the map
Trs
:
TrN -> TrM.
Dualising again gives,
xx
O - P1
al
->
a2x
O r
2
-r
Q1 -> Tr (TrM) -+ 0 TrTrs Q2 + Tr(TrN) - 0
The natural equivalence on the category of f.g. projectives between the identity and the double dual induces a natural equivalence
between the identity and the double transpose on 11and on
Hom(M,R) = 0 R
is a semihereditary ring and
R
If
is equivalent to
M's
M
B1.
is a f.g. module,
having no projective direct summand. If
is a semihereditary ring with a faithful rank function
cokernel of a full map with respect to
p,
then M
p
lies in
and M 1B,
is the
since the
dual of a full map is a full map, and full maps must be injective. It is clear that
TrM
is also the cokernel of a full map; consequently,
Tr
restricts to a duality between the categories of left and right torsion modules. In fact, it is clear that if we have a ring
R
with a faithful
Sylvester rank function, we may define the category of left and right torsion modules and show that they are dual in the above way, since, under these circumstances, all full maps are injective.
We wish to show that these categories of torsion modules with respect to a projective rank function on a hereditary ring are both artinian and noetherian, and it is clear from the duality that we have found that we shall need to prove only one of these conditions, since the other is a direct consequence. So we shall need a certain type of ascending chain conditions on a ring.
Lemma 1.21
Let
rank function
R p.
be a left X,-hereditary ring with a faithful projective Let
PO C P1 2 P2 c ...... c Q
be an ascending chain of
20
f.g. submodules of the left projective module a full map with respect to
for all
p
Q
such that
Pi S Pi+l
is
then the chain must stop after
i;
finitely many steps.
which is countably generated and so projective. ThereUP. i 1 uPi = ® Q., fore, by 1.2, where each Q. is f.g. projective. J i J J m m S P Eventually, P S ® is a for some m and n. ® Q. 0 n j=1 j=1 Proof:
Consider
,
m
direct summand of
and so, of
u1Pi,
Pn;
hence
p( ® 9,) 5 p(Pn) j=1
with
m equality only when P
c Pn for
m
m' > n, If
a
,
n
since the rank function is faithful. However,
is a full map, and equality must hold. Since this is true for all
0
P ,
Q. = P
j=1 J
P - Q
R
the chain must stop.
is a ring with a faithful projective rank function
is a full map with respect to
p,
p
and
we define it to be an atomic
full map if, in any non-trivial factorisation,
P (P') > P (P) = P (Q) = p(a). We shall show that, over an X0-hereditary ring, any full map has a finite factorisation as a product of atomic full maps, that any two such factorisations have the same length, and the atomic full maps in one factorisation may be paired off with those in the other so that the corresponding maps are stably associated, which, as we saw earlier, is the same as their cokernels being isomorphic.
Theorem 1.22 rank function to
p
Proof:
Let p.
R
be an X0-hereditary ring with a faithful projective
The category of left or-right torsion modules with respect
is an artinian and noetherian abelian category.
By theorem 1.18, it is an abelian category and, by theorem 1.22, it
is a noetherian category; it is dual to the category of torsion modules on the other side by 1.19 and the subsequence discussion, so it must be artinian too.
Theorem 1.23
Let
R
be an X0-hereditary ring with a faithful projective
21
rank function
p.
Then the atomic full maps with respect to
p
are exactly
the maps whose cokernels are simple objects in the category of torsion modules. Consequently, every map has a factorisation into atoms; if a = a1a2 .... an, m = n,
associated to
Proof:
and
a = 6182
and there is a permutation
Sm,
where
in
ai, Sj
such that
Sn
are atoms, then ai
is stably
aa(i)
It is clear that only the cokernels of atoms can be simple in the
category of torsion modules with respect to
p,
and that the cokernels of
atoms are simple.
Every object in an artinian and noetherian abelian category has a finite composition series consisting of simple modules, any two such series have the same length and the simple modules may be paired off between the two series so that two paired together are isomorphic: remembering that if the full map
ai
has the same cokernel as
6.,
that they must be stably
associated, and that a composition series for the cokernel of a full map in the category of torsion modules corresponds to a factorisation of the map as a product of simples, we deduce the theorem.
It follows at once from this that if
a
is an atomic full map,
then the ring of endomorphisms of coker a is a skew field, since it is a simple object in an abelian category.
22
THE COPRODUCT THEOREMS
2
The purpose of this chapter is to present statements of Bergman's coproduct theorems (Bergman 74) and then to use these results as Bergman does in (Bergman 74') to study a number of interesting ring constructions. Both of these papers by Bergman are of great importance and this chapter is not designed to obviate the need to read them, but rather to interest the reader in their results and to make them plausible; however, we shall summarise all that we need from them for the majority of this book. At the end of the chapter, there is a discussion of the commutative analogues of some of the constructions considered.
The basic coproduct theorems For ease of notation, we shall consider right modules in this chapter. We begin by running through a number of definitions that we need in order to state the coproduct theorems. An RO
to
of RO-rings
{R x :
A E A},
which we shall write as of rings
is a ring
R-0 ring
R with a specified homomorphism from
it is a faithful R -ring if this is an embedding. Given a family
R;
{RA
X E A}
there is a coproduct in the category of
the general case that this ring is not the trivial ring. We call base ring of the coproduct and the rings coproduct use
u
R.
R0-rings,
R = u R., and call the ring coproduct of the family R O amalgamating R0. There is no reason to suppose in
It is technically convenient to label A u {O}
for a general element of
M.
R
0
as
M,
Most of the time, we shall take
to be a semisimple artinian ring, and each
the
are the factor rings of the
RA
RA
we shall see that under these conditions each
and to
RO
will be a faithful RA
R -ring; 0 embeds in the coproduct
R.
Ideally, we should like to be able to reduce all problems about the module theory of the coproduct to the module theory of the factor rings; in the case where
R
0
is semisimple artinian and each Rx
is a faithful
R
0
23
ring we can go a long way towards doing this; in fact, most practical problems about such coproducts may be solved although the proof is often rather technical.
An induced module over the coproduct R v
V
M
where
e M @R R
or an
MOOR R
module; an induced module of the form
R
is a right U
u
is a module of the form
U
module of the form
RA
is called a basic module.
MOORO RA
O
ea ®RU R : e MUOR R a e N OR R UU U U
We say that a map of the form
is an induced
U uV
map.
There are certain induced modules which are clearly isomorphic over
M = e M OR R be an induced module and suppose that for some
let
R;
u
N = e NVO N
u
u
we have an isomorphism
a1,
=
e
MX
MA = NX
1
=
4
a (MO OR RX
for
);
then M = N,
where
1
O
A x A1,A2; NA1 = MBA
and
;
Such an isomorphism of induced modules is called a
(MOOR RX ). O 2
2
X2
where
R,
MX
basic transfer map; if
R
is a skew field it is usually called a free
0
transfer map.
There is one further type of isomorphism that we shall need to consider, a particular sort of automorphism of an induced module. Let and assume that for some
M = e M OR R, u
e
:
u
Extend
Mul -> R u l.
e(M) = 0
for
and write
1a : R
for left multiplication by
Mu2,
and let
the composite
elay
has square
x
specifying that
y
: R - M 0,
take
and if
a.
to
1
pi = u2
should lie in the kernel of
x
e
: M -r R
by setting
and then extending by linearity. Let
u x u1,
R
to a linear functional
e
be in
let
there is a linear functional
V1
u
e.
be in
a
R,
Finally, for some x.
Then, if
u2'
ul x y2,
we ensure this by The map
IM - elay
is
invertible; we call such an automorphism a transvection.
We may now state the main theorems. We shall assume that a semisimple artinian ring, and that each
RX
RO-ring,
is a faithful
is
R
0
for
the rest of the chapter.
Theorem 2.1
be a semisimple artinian ring, and Let R {Ra 0 If M = e M OR R family of faithful RO rings. Let R = u R
A E A},
:
R
U
0
module, each
embeds in
M11
M,
and
M
11
is isomorphic as an
u
R -module to 11
a direct sum of
M
and a basic module. u
So we can write
Ru c R,
and
M,1 c M.
a
is an induced
.
24
Let
Theorem 2.2
be semisimple artinian, and {Rx a family of X EA) 0 faithful R0-rings. Let R = u RX . Then any R-submodule of an induced module Ro R
:
is isomorphic to an induced module.
Theorem 2.3
R = u RV where
Let
RG
is semisimple artinian, and each
R
RD
is a faithful RD ring. Let
f
: M -r N
be a surjection of induced modules;
then there is an isomorphism of induced modules
g
:
which is a
M' = M,
finite composition of basic transfers, and transvections such that the composite
gf
:
M' -> N
is an induced map.
These three theorems allow us to deduce all the rest of those results that we shall prove in this chapter; however, in later work, we shall need technical versions of these theorems, and in order to state these results, we shall need to set up a certain amount of the machinery to prove the co-
RD
product theorems. We shall do this under the assumption that
is a skew
field, and indicate the modifications needed to deal with the general case.
We assume that RQ RG
is a skew field; then each
module, and we can choose a basis for
T = T..
For each NX
RX
of the form
in the induced module
N = ® N OR R u
P
basis,
Su.
to
if
A;
Write
S = USA.
is in
s
S.,
If
t
is in TV
is a free right
U {1} U T .
Write
we pick an
RG
U
we say that it is associated
we say that it is associated to
A;
if
is in
s
we say that it is associated to no index. A monomial is a formal product
So,
stIt2...tn...s a S, ti e T1
or else an element of
successive terms in the series U
Let
Every element of
A.
SG.
U
that are not associated to
Theorem 2.4 (see theorem 2.1) {Ra
We denote by
if
A,
is
U
those
U,,,,,
A.
Let R0 be a skew field, and let
be a family of R rings. Let
A E A}
is associated to
U
associated to some index except for those in elements of
such that no two
S1
are associated to the same index.
s,t1,...tn,
be the set of monomials; an element of
and only if its last factor is associated to
then the induced module
R = u R.;
R0
N = $ Nu0 RJR u
R
has for a right RQ-basis the set
U,
defined in the foregoing
u
material. For each a free right
A,
N
is the direct sum as right RX module of
R,-module on the basis
Given
A
and
u E U,,,A,
Nl
and
U,,,A.
we denote by
CAu
: N -* RA,
right 'co-efficient of u' map given by the decomposition of
N
the RX-linear in the last
25
theorem; for
we denote by
u E U,
cou
the R0-linear 'right co-
: N -> R0,
efficient of u' map given by the decomposition of last theorem. For
AE A,
N
the decomposition of such that
U E U_A
it lies in element
as an RO module in the x
N
in
relative to
in the last theorem, is the set of elements is not
cAu(x)
x
0;
has empty A-support if and only if
The O-support (or support) of an element consists of those
NA.
such that
u E U
N
the A-support of an element
is not
cOu(x)
The degree of a monomial
O.
is defined to be
st1...tn
and
(n+l),
is 1; the degree of an element of
N
is the
maximum degree of a monomial in its support. We define an element
x
in
the degree of an element of
S
N
to be A-pure, if all those monomials in its support of maximal degree are associated to
It is O-pure if it is not A-pure for any
A.
Next, we well-order the sets induces a well-ordering on reading from
S
and
T
A.
in some manner; this
first by degree and then lexicographically,
U,
left to right.
N
The leading term of an element in
(with respect to the fixed
ordering) is the maximal element in its support. We call this the O-leading term. If some element
is not A-pure, then some elements in its support
x
of maximal degree will not be associated to term of
Theorem 2.5 (see theorem 2.2) Let
be the induced module
N
to be the R -submodule of u
and let
® Nu0R R,
L,
S,T
and
U
u R R be defined R =
X.
u
as in the foregoing discussion. Let u
we define the A-leading
RO be a skew field and let
Let
u
L
A;
to be the maximal element of this sort.
x
L
be some R-submodule of
N.
Define
consisting of those elements whose u-
support does not contain the u-leading term of some non-u-pure element of Then
L
® L u
L.
in the natural way.
R,
u Ru
We shall outline the adjustments needed to deal with the case is a semisimple artinian ring. The first step is to use Morita R 0 equivalence to pass from this case to the particular situation that RO is that
a direct sum of skew fields. This technique is of interest and use in its own right.
be a semisimple artinian, and let Let R 0 modules such that S. = S complete set of simple R
S1, 52.... Sn
,
0
jective module
P = e Si
be a
i = j.
The pro-
RO;
so by Morita
J
is a projective generator over
i
equivalence the category of modules over
RO
is naturally equivalent to the
26
category of modules over R0 = EndRO(P). RO
are mutually non-isomorphic of
Since each
S.
is simple and they
is the direct sum of the endomorphism rings
and so, it is a direct sum of skew fields.
S.,
P 0
be a faithful RD ring; then
RX
Let generator over
RX
is a projective
so that the category of modules over
R.,
RA
is naturally
equivalent to the category of modules over R = EndRX (P OR RX), O is a natural embedding of RO in R..
and there
We give an example to clarify what is happening here. Let in this case
RO = M2(k) x k;
R1
is simply
R0
Let R1 = M3(k),
k x k.
where
is an RD ring via
\ d, e
((:
;
b c
d
O
O
O
e
and it is an R ring via
R1 = M2(k),
Then,
(a,b) -}
(Oa
01
b/I Returning to the general case, we are given a family of faithful RQ rings,
{RA
R rings,
{R1
:
X E A)
:
X E A}
and we have associated to this a family of faithful We wish to study the ring
.
R
R =
and, as
RO
before, we form the
R = EndR(P OR R); O
-ring,
0
it is clear that this is
u RA. RO
simply the ring
Therefore, in order to study the category of as well study the category of
R modules, we may
R modules, which is a coproduct over a direct
sum of skew fields. All the statements of the next two theorems are Morita invariant and so they translate well. So for the present, we assume that fields,
RO
X Ki; i=l
{e1
Let
:
gonal central idempotents. Any sum of vector spaces over as a vector space over
for
M
n 0 e1RX; i=l
RO module as
be the complete set of ortho-
n ® Mel; i=1
we choose a basis
We call the n-tuple of bases
B1
for each
{B1}
a basis
R0. If
RX
K1.
is a direct sum of skew
R0 module has a decomposition as a direct
K1, M =
Met
over
i = 1 -> n}
RO
RX
is an R ring, it decomposes as a right RA-module,
we write
1RA = e1RA.
n 1R = ® 1Re3; i=1
In turn,
again, we write
-1 RA
1R
decomposes as a right = e1RXe3,
we note that the
27
usual formalism of matrix units works with respect to these superscripts. We as right K3 module, 1TA for i x j, and R for the remaining cases. We note that this gives us a basis for
pick a basis for 1T1u{e1} RX
R0-module in the sense defined earlier.
as
Let
{N
R -modules. We choose an
u e M} be a family of
R -
11
basis for each
u
iTAD,
and
i
S = u S1 and T = u 1 T For each t0 in a i,u u A,irj are respectively the right and the left index of t; A
j
u
is the A-index of
does not lie in
,
A member of
t.
U
has the right index
S1
i,
and if it
it has the natural A-index.
NO,
Let s e S, t
Let
N ,{S1}.
be
and the set of formal products
S
where
st1,...tn,
where adjacent terms do not have the same A-index, but the
a T,
1
right index of any term is equal to the left index of the next term. The right index or A-index of any such element is the right or A-index of its last term. So we may partition
U = uU1,
where
U1
is the set of elements
module
N1
on
i
associated to the
module
RO
we form the free
i;
K1
U1
and consider
N = e N 1. i
U'
Let
may form the
be those elements of
not associated with
U1
We
A.
NA®(i®1U1K1OR RA).
module
RA
O n
Theorem 2.6
RO be a direct sum of skew fields, iX1Ki. Let
Let
be a family of faithful RO rings, and let Ru-modules. Then
® u
R,
U1
set of
R =
u
above as RO module, and as where
where
A e A}
be a collection of
module it is isomorphic to
RX
u
to
e1.
N
NA
N
as defined
(®U'KIQRORA),
with the same sub-
u e U1,
is isomorphic to
: N -> R
c
uu i
uR
e1R
via a map
Consequently, the definition given in 2.6 allows us to
define co-ordinate functions
is the right index of
As we did when T.
u E M}
is isomorphic to
in this last representation is identified in
We note that for
and
Ro
:
:
U.
sending
where
IN,
RA,
{RA
RO
and then well-order
,
which takes values in
e1R,
u
u.
was simply a skew field, we well-order
S
length lexicographically. We define degree,
U
u-purity, the p-leading term and so on as we did previously. To recover a version of coproduct theorem 2.5, we need one more concept, that of homogeneity. Given an
RO module
M,
we define
m E M
to be i-homogeneous if
met = m. An element is homogeneous if it is i-homogeneous for some
i.
28
be a direct sum of skew fields
Let
{RA
be a family of faithful R rings, and let R = uRORA.
:
X E A}
R 0
n x Kl; i=l
Theorem 2.7
® N u
be the induced module
u Ru
be the R submodule of
L1
and let
R,
L
let
Let N
be an R-submodule of
Let
N.
consisting of those i-homogeneous elements of
L
whose u-support does not contain the p-leading term of some homogeneous n then ® L1 is an R submodule L of L, and non-p-pure element of L; u u i=1 u L L,
u
L0 R. u Rp
In the more general case, where ring, each
is a faithful R0-ring,
R
RD
is a semisimple artinian R
R = u u
RO
R-submodule
L,
u
with an
u
we pass by Morita equivalence denoted by bars to the case
described by the hypotheses of the last theorem. We then find distinguished submodules
Lu such that
by Morita equivalence, we have sub-
L = e LueR R; ii
modules
Lu
L 2 e LueR R.
such that
u
u
We note a small corollary of 2.7, which will have some consequence later.
Lemma 2.8
Let
R = u R , RO A
is a faithful R e N QR R; u
u
where
Let
0-ring.
R
is semisimple artinian, and each
0
R A
be an R-submodule of the induced module
L
then, in the decomposition of
L
given in 2.7, and the following
u
discussion, L = ® L P
u Ru
R,
we find that L n N c L
u- u
By the process of Morita equivalence, it is enough to show this, when
Proof:
RO is a direct sum of skew fields. In this case, it is clear, for, if 1E L n Nu,
lel
is i-homogeneous and has empty p-support so it satisfies the
conditions to be in
L V
There is another result of this technical type that we shall need later on.
Lemma 2.9
Let R = - RV where RO
RO module of an induced module
N = e N OR R; u
theorem 2.5, and assume that each
nx
must be in
L.
is a skew field. Let
L
0
u
write
u
is empty. Then if
be an
L
R sub-
L = e L 0R R given by u
End E L
u
for
nx e NX,
29
We may assume that at least two of the
Proof:
is 0-pure. Write the support of
are non-zero; so
nA
Ena
with respect to our well-ordered basis
Enx
in descending order. Given two finite strings of basis elements in descending order
and
{ai}
{bi}
the first place at which they differ string of
{ai}.
{bi}
we shall say that
We choose
or else
bi < ai
in
Enx
L
so that
is an initial
{b
and the associated
nx E L
string of basis elements is minimal in the above sense. Enx 4 L0;
if for
{ai}
is less than
is 0-pure but
EnA
so, it contains the leading term of some pure element which is
forced to take the form n for some n e N
;
by subtracting an appropriate
x
multiple of n from
EnA
we reduce the support and obtain a contradiction.
It is clear that the information we have built up in the preceding results allows us to answer a number of natural questions about ring coproducts, amalgamating a common semisimple artinian ring. We begin with a few such applications.
Theorem 2.10
Let
R =
UR
RD
where
is semisimple artinian and each
is a faithful Ro ring. Then the homological (or weak) dimension of
Rx
M OR R a
over
Proof;
R
is equal to the homological (or weak) dimension of
Given a resolution P- M- 0
resolution of
M OR R over
by 2.1, this resolution of
R,
over
R
is flat over
considered over
RA.
contains
RA
as a direct summand, so that the homological dimension of on passing to
over
R,, POR R i M OR R- 0
since by 2.1,
M OR R a
M
M
RA.
is a Again,
P -> M -
0
cannot decrease
M OR R.
From this, we are able to determine the global and weak dimensions of a ring coproduct.
Theorem 2.11
Let
R = uRA,
where
R0
is semisimple artinian, and each
RA is a faithful RD ring; then the right global dimension of
the supremum of the right global dimensions of the RV of them is not
0.
If each
RA
R
is equal to
provided that one
is semisimple, however, the global dimension
may be 0 or 1. A similar result holds for weak dimension.
Proof:
Consider any submodule of a free
R
module; by 2.2, it is an induced
30
module. Moreover, by 2.1, if
M = ® M QR R. u
u
M
each
u
is a submodule of a u
basic module, which are projective modules. From theorem 2.9, it follows that the global dimension of Ru,
R
is the supremum of the global dimensions of the
provided that one of them is not semisimple artinian. If each of them
is semisimple artinian, it follows that the global dimension is at most
1.
Exactly the same proof holds for the weak dimension.
R0
If
is a skew field, it is fairly clear that
R
cannot be
a semisimple artinian ring; however, this is most easily shown after we have developed some more properties of the ring coproduct, so we shall deal with this towards the end of the chapter. However, we present an example to show that when
R0
is only semisimple artinian, the ring coproduct may be simple
artinian.
Let RO = k x k x k, Rl =M 2 (k) x k,
RV + R
by
(a,b,c)
RO ; R2
by
(a,b,c)
I/a
a
It is clear that
and R2 = k x M2 (k) , ,
O
,(b
0
O
c
where
and
Rl U R2 = M3(k). RO
One can show by the same method that the coproduct of right semihereditary rings is right semihereditary and a similar result holds for X0-hereditary.
Before passing from the study of the category of modules over the coproduct to the more particular study of the category of f.g. projectives, we leave as an exercise for the reader the details of the next example, which answers a question of Bergman.
Example 2.12
We wish to find an extension of induced modules that is not
induced.
Consider the free ring
R = k<x,y> = k[x]
k[y].
The module
k
R/xyR
is an extension of the induced module
R/yR
by the induced module
R/xR, but it is not an induced module itself. We start to look at the category of f.g. projectives over the
31
ring coproduct. We already know all the objects of this category. We have the next result.
Let
Theorem 2.13
R = Li R.,
where
is semisimple artinian, and each
RC
R RX
is a faithful R
® PuaR R,
0-ring.
where each
Then, a projective module over is a projective module over
P11
isomorphism classes of f.g. projectives over
R
R
RP.
has the The monoid of
is isomorphic to the commu-
tative monoid coproduct of the monoids of isomorphism classes of f.g. projectives over
product of
amalgamating the monoid of isomorphism classes of f.g.
RA,
projectives over
Consequently,
R0.
amalgamating
KC(Rx),
is the commutative group co-
K0(R)
KO(RC).
Every projective module lies in a free module and so must be iso-
Proof:
morphic to an induced module by theorem 2.2. By theorem 2.9, we know that if P= 0 MuR3, R,
M
each
must be a projective
R
u
module. V
Theorem 2.3 tells us that any isomorphism between two f.g. induced modules is the composite of a finite sequence of basic transfers and transvections followed by an induced map which, over
R,
is an isomorphism and
so must have been an isomorphism on each summand. Transvections do not alter the form of an induced module; basic transfers simply amount to the amalgamation of the images of
P0(R0)
in the different
P49 (RA),
which proves the
result for the monoids of f.g. projectives. The result for since
K0
KO
is the universal abelian group functor applied to
follows
P®.
We should also like to have information on the maps between f.g. projectives; that is, we should like to be able to explain all isomorphisms and all zero-divisors over the coproduct in terms of the factor rings. We already know how to explain all isomorphisms since 2.3 explains all surjections, so we look at the zero-divisors next.
where R is a semisimple artinian ring and R0 is a faithful R0-ring. Let a P - P', and B P' -> P" be maps
Theorem 2.14 each
RX
Let
R = - R,
between f.g. projectives over commutative diagram:
:
R
such that
:
aB = 0.
Then there exists a
32
P.
II
ePa PRR
eaOR
0P' a u
u
u
S0R R
Ru
u
UI
0P"
RU
u
R
where each direct sum in this diagram is a finite direct sum of f.g. proausu = 0, for all
jectives and
Proof:
ims
First,
V.
is an induced module by 2.2; so
P' + im$
is a sur-
jection of induced modules, and by 2.3, there exists a commutative diagram:
a
6
P'
P
P"
UI
eker6 a R=- a P' OR R u
u
u
u
'sims a R
=
ims
P Ru
u
So the problem is to replace
kersu,
and
by suitable f.g
ims u ,
projectives.
ima'
is a f.g. submodule of
a kerb a R u
u
submodule generated by finitely many elements from an induced map from
e PIJRR to P
11
a kergua R,
and so lies in an R-
kers u.
where each
Thus, there is Pu
is a finitely
11
generated free Ru-module, such that the image of this induced map contains the image of
a'.
Since
is a projective module, we have a commutative
P
diagram:
P
P'
II
ePa u
u Ru
u
!' Ru
p"
UI u Ru
R
We miss out the bottom right hand corner of this diagram
(e imBuaRR); U
dualise the rest of the diagram and then fill in the bottom
left hand corner of this diagram as we did in order to replace a kers a u
R.
u Ru
The diagram we obtain is the dual of a diagram that suits our needs.
Universal ring constructions These results on the coproduct construction allow us to study a
33
number of interesting constructions on the category of f.g. projectives over a ring. Given a k-algebra,
the ring coproduct S . k[x] may be regarded k
S,
as the S-ring with a universal map on the free module of rank 1 that centralises the k-algebra structure. Suppose that projectives over the ring an S-ring,
with a universal map from
T,
P1
and
P2
are f.g.
we are also interested in finding and studying
S;
P 0 T
P 0 T
to
1 S
2
that centralises
S
the k-structure. In the language of functors, we are trying to find the Sring in the category of k-algebras that represents the functor We say that an object
S' i HomS1(P1aSS., P2aSS').
represents a covariant functor lent to
HomC(O, ).
F : C - Sets
if
in the category
0
F(
)
C
is naturally equiva-
We should like to investigate other universal construc-
tions; thus, we wish to find an S-ring,
with a universal isomorphism between
T1,
P 0 T 1
S 1
in the category of k-algebras and
P20ST1.
That is,
T1
represents the functor that associates to each S-ring that is a k-algebra the set of isomorphisms between to find an S-ring,
on the projective
P1aSS'
and
P2aSS'.
Again, we should like
that is a k-algebra with a universal idempotent map
T2,
POST2;
that is,
T2
represent the functor on the category
of S-rings that are k-algebras, which associates to each object of idempotent maps on
S'
the set
POSS'.
It is quite easy to see that there are rings with these universal properties, essentially by generator and relation constructions. First, we note that if summand of
A Fn
is an additive category such that every object is a direct for some object
F
in the category, then this category is
a full subcategory of the category of f.g. projectives over the endomorphism ring of
F;
a ring over
and adjoin some set of maps to the category of f.g. projectives
R,
R
becomes the free module of rank 1. So, if we start off with
F
subject to some set of relations, and construct the additive cate-
gory that they generate, the objects are still all direct summands of for suitable
n,
Rn
and therefore, the category is a full subcategory of the
category of f.g. projectives over the endomorphism ring of the object Clearly, there is a hanomorphism from
makes
E
R
to this endomorphism ring
R1. E,
which
into the universal R-ring having this additional set of maps satisfy-
ing the specified set of relations. Clearly, the constructions in the preceding paragraph all have this form, and we shall discuss in chapter 4 another construction which may also be shown to exist by this method. For our purposes in this chapter it is more useful to study them in a different way. One construction of a slightly different nature that we wish to study is the formation of the universal k-algebra
T'
with a universal homo-
34
morphism from
M (T').
to
S
Here,
n
category of k-algebras,
represents the functor on the
T'
Homk-alg(SIMn( ))'
The constructions on the category of f.g. projectives are all approached in a similar way. We shall explain the case of adjoining a universal map between two projectives in some detail, and since the other cases use similar arguments, we shall consider them more briefly. be a homomorphism between k-algebras, and let
0 : R i S
Let
and Q2 be f.g. projectives; then, if R' is the universal R-ring, and Q1 it is clear that R ' u S k-algebra, with a universal map from Q1 to Q2, R to
is the universal S-ring and k-algebra, with a universal map from Q a S 1 R Q2aRS.
For suitably large
and
such that
Mn(S),
in
e2
Mn(S),
that is a k-algebra. Let
S',
e1 + e2 + e3 = 1 in
which makes
e1Mn(S')
to
e2Mn(S')
k x k x k -+ Mn (S)
and that induced by the second summand is
seen that the
given by
into a faithful
Mn(S)
ring. The projective module induced by the first summand of e1Mn(S)
for an
be that idempotent such that
e3
we have a map
Mn (S) ;
(a,b,c) -+ e1a + e2b + e3c,
S
so it is enough to show that there
eiMn(S),
is an M(S)-ring with a universal map from S-ring,
ell
ei(nS) = Pi; under the Morita equivalence of
becomes
ei(nS)
there are orthogonal idempotents,
n,
k x k x k-
k x k x k
e2Mn(S).
is
It is easily
k x k-ring with a universal map from k x 0 to 0 x k is the 0) T2(k) =(k where k x k embeds along the
lower triangular matrix ring
diagonal, and 2
is the universal M (S)-ring n
M (S)
kxkxk
a map from
induced the universal map. Therefore, the ring coproduct,
e21
u
T (k) x k
n
to
e1Mn(S)
e2Mn(S).
S-ring with a universal map from T = Sk
that is a k-algebra with
By Morita equivalence, we have a universal P1
to
P2.
We shall use the symbol
for this construction. We notice that it is made for
studying in terms of the ring coproduct.
Theorem 2.15 over
S.
map from
Let
Let
P a T 1 S
as
T
S
S
be a k-algebra and let
T = Sk to
then
P20ST;
P1
T
1.
P2
be f.g. projectives
has the same global or weak dimension
except when the global or weak dimension of
has dimension
and
be the k-algebra and S-ring with a universal
S
is
All f.g. projectives are induced from
the ring homomorphism from
S
to
T
in this case,
0;
induces an isomorphism
S;
in fact,
P®(S) = P®(T).
35
Proof: We retain the notation of the preceding discussion. We see that
u
Mn(T) = Mn(S)
has global
T2 (k) x k; T2 (k)
kxkxk and weak dimension 1, so that except when
that of
S
sion
1
by
over
T2(k) x k
M
n
has global or weak dimension equal to
(T)
has dimension
S
in which case,
0,
has dimen-
T
The monoid of isomorphism classes of f.g. projectives
2.10.
is isomorphic to the monoid of isomorphism classes of f.g.
projectives over
simply by the map induced by the ring homomorphism
k x k x k,
described earlier; so, from 2.12, we deduce that
P9(Mn(S)
u
T2 (k) x k)
k x k x k
P0(Mn(S),
from which it follows by Morita equivalence, that
k x k-ring with a universal isomorphism between
The is
O x k,
where
M2(k),
k x k
are orthogonal idempotents in
e2
such that
Mn(S)
ring with a universal isomorphism between
M (S) n
U
is given by
M2(k) x k
have an S-ring,
where
e3
(a,b,c) -(a
0\
O
b)
c\
is and kxkxk to
is 1 - e1 - e2, ,
By Morita equivalence, we
.
which has a universal isomorphism between
T,
is just
e2Mn(S)
M (S) n
kxkxk
(a,b,c) - ae1 + be2 + ce3,
of this coproduct. We shall use the symbol
Theorem 2.16 over as
Then
S.
S
relation
0
be a k-algebra and let
S
T = Sk P2>
except when
dimension
Proof:
Let
or
S
1.
Sk
and
P2>
for this ring.
be f.g. projectives
P2
has the same global or weak dimension
has global or weak dimension P®(T)
S
in the first factor
Mn(k)
P1
and
P 0 T 1
by taking the centraliser of the copy of
P20ST,
and
el
the M(S)-
ei S = Pi, and
e1Mn(S)
where the map kxkxk to
M2 (k) x k,
and
k x O
embeds as the diagonal matrices; the elements
are the universal map and its inverse. Therefore, if
e12
and
e21
P®(T) = P®(S).
0.
Here
is isomorphic to the quotient of
T may have P®(S)
by the
[P1] = [P2].
We use the notation of the preceding discussion.
Mn(T) =
Mn(Sk
_). P2>) = M2(k) x k
u
Mn(S)
kxkxk The global or weak dimension of if this not equal to
0,
Mn(T)
by 2.10. If it equals
0,
must equal that of
Mn(S)
however, it can be at
36
most 1; we see that it may be 1 by adjoining a universal isomorphism between kl and kl over k, obtaining the Laurent polynomial ring
other hand, it may be
as we find for the ring
0,
adjoin a universal isomorphism between
and
k x O
On the
k[t,t-1 ].
where we
k x M2(k),
O x e11M2(k),
obtaining
M3(k). For the last statement of the theorem, we note that
is the quotient of since
P®(Mn(T))
we find that the result holds for
P®(k xk xk),
The result holds for
and
P®(M2(k) x k)
when we adjoin universal isomorphisms between
P6(Mn(T)), e2Mn(S).
[k x O x O] = [O x k x O];
is the commutative monoid coproduct of
amalgamating
P®(Mn(S))
by the relation
P®(k x k x k)
P®(M2(k) xk)
and
e1Mn(S)
by Morita equivalence.
P®(T)
Next, we deal with the adjunction of a universal idempotent map on a f.g. projective free module of rank
If we adjoin a universal idempotent map to the
P.
over the field
1,
fore, if the idempotent
in
e
Mn(S)
we obtain the ring k x k. Theren satisfies e S = P, the S-ring that k,
is a k-algebra with a universal idempotent map on
P.
P, e = e2>
Sk<e:P
is obtained by taking the centraliser of the matrix units in the ring co-
product Mn (S) k x kk x k x k, where k x k maps to Mn (S) by (a,b) - ae + b(l-e), and k x k embeds in k x k x k by (a,b) -> (a,a,b). Once again, we may deduce homological information: Theorem 2.17
The global or weak dimension of
equal to that of where
T = Sk<e:P ; P, e = e2>
except, possibly, when the dimension of
S
T may have dimensions
are two more generators
0
or
and
[Q1],
P®(S)
1.
[Q2]
embeds in
S
is
is
0,
and there
P®(T),
subject to the relation
[Q1] + [Q2] _ [PAST].
This may be left to the reader since it in no way differs from that
Proof:
of the last two theorems.
It is clear that these constructions may be put together in a number of interesting ways. One construction of some interest is the adjunction to
k
to a pair of matrices
call this ring, n
by
n
matrix,
We m This may be obtained by adjoining universal an idempotent
R.
E,
to
k
mAn
and
nBm
such that
AB = I
.
and then adjoining to this ring a universal
isomorphism between the image of
E
and the free module of rank
m.
We are
able to deduce that this is a hereditary ring and that it has exactly one
37
new indecomposable projective,
up to isomorphism, and
Q,
Q ® Rm
Rn
is
the only relation that it satisfies.
Our final construction is a k-algebra that represents the functor Horn
));
k-alg(S,Mn(
Theorem 2.18
following Bergman, we call this ring
Consider the ring
which is the centraliser of the first
T
factor in the ring coproduct Mn (k) uk S. Given a homomorphism
Proof:
0 : T -
which determines a homomorphism g
:
S -> Mn(A),
M (k) to n
M
n
01
we have a map
centralising
(k)
Then
T.
k<S -> M > n
we have homomorphisms
A,
u S ° Mn (T) -> Mn (A)
S -> Mn (k)
map
k<S -> Mn>.
:
g'
:
in M
A
Conversely, if we have a
> Mn(A).
S
n
Mn(k) (A)
k
that sends
S } Mn(A)
and acts as
g
ing to the centraliser of the matrix units determines a map
on g
Restrict-
S.
T - A.
:
These processes are mutually inverse, so they demonstrate the natural equi-
Homk-alg (T,
valence of
and the functor
)
Homk-alg
(S,Mn( )).
Before we describe the homological properties of this construction, we consider some of its interesting ring-theoretic properties.
Theorem 2.19
Proof: If
is a domain, and the group of units is just
k<S -> M >
n
are in
a,b
k<S - M
n
>
and
ab = 0,
endomorphisms of the induced projective module,
we may regard
where
M (k) u S,
n
k
R = M (k) -S n k
in the following.
a
k.
M W. We write
is the simple module over
P
n
By 2.14, we have a commutative diagram:
P'M (k)R-
b
n
_ PAM n (k)R
FBOR Q1% (k)R n
n
(k)R
aOR
PM (k)R n
and
b
as
P QM (k)(Mn(k) - S), over the
n ring
a
kx.
Q2®M (k)R
n
38
where
are defined over
S,a
M
n
(k)
and compose to zero.
We know that the middle term of the bottom row is
PQM
)R,
n
(k
since this is the only representation of it as an induced module.
However, the only way that a pair of maps over
and then a:P - Q2
Mn(k), B:Q1 - P,
can compose to zero for one of the maps to be zero itself,
which, in turn, implies that one of
a
and
b
are
0.
The result on units is a consequence of 2.3. If we have a unit in
it defines an automorphism of
k<S -- Mn>,
all basic transfers
P®M (k) R;
and transvections must be the identity map on P% n(k)R; PAM (k)R
group of automorphisms of morphisms of
over
P
over
so, by 2.3, the
R mustnbe the group of auto-
M (k), nwhich is simply
kx.
n
It is clear that the same argument shows that if Mn(T) = Mn(D)D R,
where
R
is a D-ring,
D
is a skew field, and the iso-
morphism sends the obvious set of matrices to the obvious set of matrices, is a domain and the group of units of
T
then
T
Dx.
is just
This argument allows us to determine the global dimension of a ring coproduct over a skew field.
Theorem 2.20
Let
RO
be a skew field and let
an R0-ring; then, the global of dimensions of
R
R
R = U R RO
where each
R. is
equals the supremum of the global
except when they are all zero; then it becomes 1.
Proof: By 2.10, we have this result except possibly when the rings are all semi-simple. Consider the ring of endomorphisms of an induced projective module
P®R R,
where
P
is a simple module for the semisimple ring R.
A
The group of units is simply the group of automorphisms of the module and by the same argument as in the last theorem we see that this is just the group of automorphisms of of
P&Ra R
P
over
RX.
However, the ring of endomorphisms
is larger than the ring of endomorphisms of
over
P
that this ring of endomorphisms is not a skew field. Therefore,
R,,
R
so
cannot
be a semisimple artinian ring since over a semisimple artinian ring, an indecomposable projective module is simple, and so its ring of endomorphisms must be a skew field.
We see that the global dimension of
k<S -
M
n
>
is equal to that
39
of
by 2.10, except when
S
is semisimple artinian in which case we may
S
apply 2.20 to show that it is equal to 1.
It is interesting to examine some of these constructions in the
category of commutative algebras. We shall not provide detailed proofs since this last part is meant to be only illustrative.
We begin with a commutative ring projectives
P1
C[a:P1 + P2]
and
over
P2
and a couple of f.g.
C
We wish to find a commutative C-algebra
that represents the functor on the category of C-algebras
A 3 HomA-mod
The functor
(P10CA,P2QCA).
equivalent to the functor P2;
C.
HomA(PlQCA,P20CA)
where
HorA((PloCP2)&CA,A),
PZ
is naturally is the dual of
but it is clear that the C-algebra representing this functor is just
the symmetric algebra on the module
PlaCP2
over
It is an immediate
C.
consequence that this algebra is geometrically regular over for each prime ideal Q(C/p)
p
of
C,C[a:Pl i P2]aCQ(C/p)
C;
that is,
is a regular ring where
is the algebraic closure of the ring of quotients of
C/p.
We shall find that this holds for each construction examined and we shall be able to obtain each construction in a sufficiently explicit form to be able to determine the dimension of each geometric fibre over example, in the above example, the fibre over the prime Q(C/p)[x13:i = 1 -- ml, j = 1 i m2],
where
mi
p
For
C.
is
is the local rank of
at
Pi
P_
We examine next the construction of adjoining a universal isomorphism from P1 to P2. This is only possible when the local rank of the same as that of
P2.
If such is the case,
P1
P1
is
becomes isomorphic to
P2
on a Zariski cover and by further refinement we see that there is a Zariski cover which is the union of finitely many open and closed subspaces such that the restriction of tion of
P2
P1
to each of these subsets is isomorphic to the restric-
and both are isomorphic to free modules on each subspace. If on n,
then the space of
all isomorphisms is a principal homogeneous space for GL n.
Therefore, we see
a particular subspace the restrictions have local rank
that the fibre at a prime of space of isomorphisms of the local rank of
P1
C
of the spectrum of the ring representing the
with
P2
is just
where
GLn(Q(C/p)),
n
is
Pi.
If we wish to adjoin an idempotent map
e:P i P
to
C,
it is
reasonable to specify the local rank of the idempotent also, which must be constant on connected components and dominated by that of
P.
We may as well
40
restrict to a situation where the local rank of that of the idempotent at idempotents of rank
on which
P
m
and
n
EndA(PMCA).
in
becomes free of rank
potents of rank m
in Mn (A)
P
is constant and so is
respectively. So we wish to represent
m
n.
Over
C,
there is a Zariski cover
In a particular algebra, any two idem-
present free modules with free complement on
a suitable Zariski cover, so that with respect to the Zariski topology all these elements are conjugate. Moreover, the centraliser of an idempotent whose image is free of rank is just
M (A) X M
(A);
m n-m potents of rank m in
m
and whose complement is free of rank
(n-m)
the consequence of this is that the set of idemEndA(PQCA)
becomes on a Zariski cover the functor
GL /GL X GL Since this last functor is geometrically regular, so is the m n one we are interested in, and it is easy to check that the dimension of a n-m.
fibre is
2mn.
41
3 PROJECTIVE RANK FUNCTIONS ON RING COPRODUCTS
The purpose of this chapter is to investigate how projective rank functions behave under the coproduct construction. On the way we shall give an application of this theory to the problem of accessibility for f.g. groups.
Also, we shall prove some interesting results on the behaviour of the number of generators of an induced module over a coproduct. The main applications of the results of this chapter will be to hereditary rings in the following chapters.
The Generating Number on Ring Coproducts When investigating partial projective rank functions on a ring coproduct,
where
R = R1 U R2, RO
RO
is a semisimple artinian ring and
R.
is a faithful RO-ring, it is sensible to assume that it is defined on the image of KO(RO)
Lemma 3.1
R = R1 - R2 RO
in
KO(R).
In this situation, we have the following lemma:
The partial projective rank functions on a ring coproduct where
is a semisimple artinian ring and
R 0
R0 -ring, that are defined on the image of
pairs of partial projective rank functions projective rank function on
R.
KO(RO) (pl,p2)
where
is a faithful
are given by
KO(R)
defined on the image of
they induce the same projective rank function on
Proof:
in
R.
pi
K0(R0)
is a partial such that
R0.
This is clear since by 2.12, we have a push-out diagram:
KO(RO) ± KO(R1) 1
!
KO(R2) } KO(R)
We shall usually describe a partial projective rank function of
42
this type by the pair p,
is defined on
of partial projective rank functions where
(p1,p2)
R..
We wish to show that the inner projective rank of a map a:P -r Q
defined over
R1
same as the inner rank of
with respect to a rank function over
R:POR R -> QQR1R
c
1
1
respect to the projective rank function
is just the
p
R = R1 U
with
R2
Ro In order to do this, we
(p1,p2).
need to investigate how the generating number with respect to a rank function behaves under the coproduct. Before we do this we present a result that is independent of rank functions.
In chapter 1, we have already encountered the notion of a ring with unbounded generating number; that is, there can be no equation of the m+1R1 mR1 If however such an equation holds, a similar equa® P. form R1-rings, and in particular for a ring coproduct,
tion holds for all
R = R1 u R2.
So the number of generators of a module induced from
may
R2
Ro
well change. We show that this is the only way that things can go wrong.
Theorem 3.2
Suppose that
generating number, where of generators over
R
is a skew field, and
0
and
R1
R2
number of generators of the module
Proof: Suppose that
are R0-rings. Then the minimal number
MR R1 R
is equal to the minimal
over
MOR1 R
R.
may be generated by m
there is a surjection of induced modules is an isomorphism
M
of the f.g. module
R1
S:P 9Rl R ® P 1
2 R1
has unbounded
R2
a:mRl&
MR,@
R ->
Rl
elements over
R -> MR
Rl
R.
By 2.3, there
which is a finite composi-
R1 R ,
tion of free transfers and transvections such that the composite P1
induced map; that is, it maps that
P1
onto
M
is an m-generator module and so is
and
then
R;
P2
to
0.
aB
is an
We shall show
M.
We recall from theorem 1.4 that if
R2
has unbounded generating
number, we have a well-defined partial rank function
p2
on stably free R2-
modules. Since any stably free module has non-negative rank, the rank of a free summand of a stably free module
P
is bounded by the rank
p2(P).
If after a series of free transfers and transvections we have
passed from
mRJOR R 1
Pl 6 p2(P2)R1 = mR1.
to
P1QR R ®P 2®R1 R,
we shall see that
1
We assume that this is true at the nth stage; if our
43
next step is a transvection, it alters nothing; if it is a free transfer from
P1
to
it is still true; if it is a free transfer from
P2
we can transfer at most
P1
p2(P2)
to
P2
factors, so it is still true. It is
true at the beginning so by induction it is true at the end. We have shown the hard direction of our theorem; the other is trivial.
We pass to the more technical versions we need for partial projective rank functions in general.
Theorem 3.3 Let each
and
R1
Let
K0(Ri).
M2
be partial projective rank
p1, p2
m1
be f.g.
and m2
Then the generating number of
p2.
is a semisimple artinian ring and
R0
respectively, defined and agreeing on
R2
and
M1
with generating numbers
is
where
is a faithful R0-ring; let
R.
functions for in
R = R1 U R2, R0
and
R1
K0(RD)
modules respectively
R2
respectively with respect to
and
p1
with respect to
M1OR1R a M2OR1R
(pl,p2)
m1 + m2.
Proof:
Certainly, it is at most
ml + m2.
Conversely, suppose that we have a surjection over (P1OR1R) $ (P20R2R) -
(MIOR1R) 0 (M2OR2R),
where
R,
is defined. By
p2(Pi)
2.3, after a series of basic transfers and transvections on
P1OR R 6 P 20R R, 1
2
we obtain an induced surjection:
PjOR R ® P
R R - MIOR R ® M 2 Q R R
1
1
Since
pi
2
is defined on the image of
K0(R0)
agree there, the basic transfers produce modules on which Since
Pi
maps onto
Mil
pi(Pi) ? mi,
so the rank of
K0(Ri)
pi
is defined.
P1aR R ® P2aR R 2
1
with respect to
(pi,p2)
must be at least
and
in
m1 + m2.
At this point, it is of some interest to show how this theory may be applied to provide a fairly simple proof of a result of (Linnell 83) on accessibility for f.g. groups.
We begin with a summary of the background to the problem, and refer the reader to (Dicks 80) for the details of the subject. Let
X
be a finite connected graph with edge set
E
and vertex
44
set
and end
we allow the beginning re
V;
to
vertex. We label the edges and vertices of
X
and
G
edge
we have embeddings
e
G
-> G
Te
of an edge to be the same with groups, and for each these
Te = te,
if
-> G
e
'
embeddings may well be different. This is known as a graph of groups; we associate to this a group, the fundamental group of a graph of groups in the following way.
Let
be a maximal subtree of
T
which we regard as a subtree
X,
of groups by the appropriate labelling of the elements of T
vertex of
with only one edge,
incident with it, so
e,
and
a
for the vertex,
e,
without
v = Te,
loss of generality. We pass to the tree with one fewer vertex, by omitting
be a
v
let
T;
obtained
T',
which we label by the same groups as before except
Te,
which we label with the group coproduct of
G
and
re
Ge' Gre * Gte. By induction, we eventually reach a single point with a group GT associated to it, with a specified homomorphism G1e amalgamating
Gv -. GT
for each vertex GT
clearly,
T;
G
-> G
e
re c -G T
and
G
The resulting group,
e
in
GX,
we have two embeddings of
X - T,
e
in
Ge
in
c G so we form the multiple ANN extension T X - T with respect to these pairs of embeddings.
e - G e
over all edges of
of GT
for each edge
GTe n G1e = Ge
is universal with respect to this property.
For each edge G T,
such that
v
;
is the fundamental group of our graph of groups;
it can be shown that it is independent of the maximal subtree chosen. Also, if
is a full connected subgraph of
X'
from of
X
X
by shrinking
X'
we may express
X,
fundamental group of a graph of groups on
where
X,
X
as the
GX
is the graph obtained
to a point. The groups associated to the elements
are the same as those in
X
for those that are not affected and that
associated to the point to which we shrunk
is the group
X'
GX'.
In the following, we shall assume that edges groups are finite,
and also that if there is an edge then
e
such that
G
e
= G
or
Te
G
e
= G
to
re = te.
Given a f.g. group, of expressing
G
G,
we are interested in the various ways
as the fundamental group of a graph of groups with finite
edge groups. It is possible to show that if a graph of groups on
Xl
and
X2,
is the fundamental group of
G
then these two representations have a
common refinement, that is, there is a representation of mental group of a graph of groups on a graph
X
G
as the funda-
such that the two preceding
representations arise by collapsing suitable subgraphs of
X
to points. So
the question arises whether there is a representation from which all others arise by collapsing suitable subgraphs. If there is, the group is said to be
45
accessible. This is equivalent to finding a representation where all the vertex groups are neither HNN extensions over some finite group nor nontrivial coproducts over some finite group. Since it is also easy to see that the number of generators of the fundamental group of a graph of groups on a graph
E(X) - V(X) + 1
is at least
X
(consider the homomorphism to the
fundamental graph of groups on the graph
where all vertex groups and
X
edge groups are trivial, which is the free group on the stated number of generators), we know that if the vertex groups keep on decomposing they must eventually decompose only as coproducts over finite groups, so that we shall only consider this possibility. It is clear that this cannot happen for f.g. torsion-free groups since the number of generators of sum of the number of generators of
H1
and
H2
is the
H1*H2
by the Grushko-Neumann
theorem. In order to prove a more general theorem we need to find a substitute for the number of generators that grows in a satisfactory way for the coproduct with amalgamation over a finite group. Linnell (83) was able to do this for those groups whose subgroups are of bounded order; we shall present his theorem, using partial rank functions instead of the analysis he used.
We noted in the first chapter, that on every group ring in characteristic
0
there is a faithful projective rank function on f.g. pro-
jectives induced by the trace function. We are interested in the subgroup of
KO(KG)
which is generated by the f.g. projectives induced up
K0(KG)
from the subgroups of finite order, where We shall denote by trace on
Kf(KG).
pf
is a field of characteristic
*
then
G2,
KG = KG1 U KG2;
GD
finite group, then that
KF
Lemma 3.4
Let
is a
is semisimple artinian and we should like to show
G = G1 F GZ
just the subgroup of
F
if
KGO
is just the partial projective rank function
pf
0.
the partial projective rank function induced by the G = G1
If
K
Kf(KG)
where
F
f
(p1 ,p2).
is a finite group; then
generated by the images of
is the partial projective rank function
Kf(KG)
KD(KGi).
is
Con-
sequently,
pf
Proof:
If
P
of
we know that
H
is a conjugate of some subgroup
or G2; consequently,
P
is isomorphic to a projective induced from H' and
G,
(pI,p2).
is a f.g. projective induced up from some finite subgroup
must lie in the image of Kf(KG1)
or
Kf(KG2).
So
Kf(KG)
H'
of either
H G1
is generated by
46
the images of
and
KO(KG1)
Since
and
pf
KO(KG2). p2
agree on the image of
KO(KF)1
lemma 3.1
shows that our last statement is true.
This result allows us to show Linnell's theorem; we shall use the generating number with respect to in
to measure the size of
KG
Theorem 3.5
Let
p
of the augmentation ideal of
G
G.
G be a f.g. group such that the number of elements in
finite subgroups of
is bounded; then
G
G
is accessible.
We recall that it is sufficient to show that we cannot keep on decom-
Proof:
posing such groups as a group coproduct amalgamating a finite subgroup in a non-trivial way. If
where
G - G1 F*, G2,
IFI
we know that
< -,
wG = (wG1)G ® (wG2/wFG2)G ,
where
is the augmentation ideal of a group
wH
KG = KG1
KG2
H
in
KH.
and it is clear from the above equation that
wG
is an induced module. By 3.3 and 3.4, we see that
g.pf(wG) = 9.pl(wG1) + g.pz(wG2/wFG2).
and it is clear that if G,
then
g.p2(wG2/wFG2)
m
is the bound on the order on finite subgroups of
>_ .
m Consequently, if
g.pf(wG) = q,
there is no decomposition of
G
as the vertex group of a tree of groups with finite edge groups having more than
(m:q)
vertices, which proves Linnell's theorem.
Sylvester projective rank functions on ring coproducts We return to the general theory. Consider a ring homomorphism 0 : R-* S;
suppose that we have a projective rank function on
inducing a projective rank function
pR
on
R.
honest with respect to the projective rank functions map a
:
P -+ Q
pR
in the category of f.g. projectives over
and
in
S, pS,
We say that the map pS
0
is
if for any
R, pR(a) = pS(anRS).
47
This clearly reduces to the usual notion of honesty for firs (Cohn 71, p.264).
Theorem 3.6 and each on
R.
Let
R = R1
R2,
R
O is a faithful
R.
where
is a semisimple artinian ring
RO
R0-ring. Let
be projective rank functions
pi
that induce the same projective rank function on
p = (pl,p2) embedding R1 -> R
Proof:
Let
a
clear that
:
is honest with respect to
p1 and
Let
RO.
be the projective rank function they define on
Then the
R.
p.
P -+ Q be a map between f.g. projectives over
R1.
It is
>_ p (° R1R).
p1 (a)
So let
M
be a f.g. R-module such that
By 1.10, we need to show that
g.p(M)
a(P)@RlR c M C QQR1R.
>_ p1(a).
By 2.8, the decomposition of
M
given by 2.7 has the form
M = (MOOROR) ® (M1QR1R) a (M2OR2R).
where
a(P) c M1.
By 3.3, the generating number of
is at least the generating number of
M1
M with respect to
with respect to
By 2.1, the embedding of R1-modules
p
p1.
Q c QQR R
splits; so consider
1
the image of
Ml, M',
of
%,
P
pl(a).
under
in
under the splitting map. It contains the image
Q
so its generating number with respect to g.p(M) 2 g.p1(M1)
We have shown
p(aQR1R) = pl(a),
is at least
p1
>- g.pI(M') >_ P1(a),
so
as we wished to show.
This result will be of great use to us later; for the present, we give a couple of interesting consequences.
R = R1
Let
Theorem 3.7
RO
R2,
where
R0
R.
is a faithful RO ring. Let
on
R.
pi
is a Sylvester projective rank function.
Proof:
Then
p
be a projective rank function
is a Sylvester projective rank function if and only if each
Suppose that
p
i
Consider a couple of maps such that
p = (pl,p2)
is semisimple artinian, and each
a$ = 0.
is a Sylvester projective rank function on a
:
P -+ P',
and
$
:
P' } P"
defined over
Then by 2.13, we have a commutative diagram:
R
i
.
R
48
a
P
i
i
P"
i
i
are maps defined over
ai,si
where
$
P'
By the law of nullity,
such that
R.
p(a) + p(8) <_ ip(ai) + pi(8.) projective rank function too.
Conversely, suppose that a
jectives over
such that
R1
and
P -> P',
B
P' -
:
P"
Since
for any maps defined over
p(a®R1R) = p1(a)
summing, we
is a Sylvester
p
is a Sylvester projective rank
p
as = O.
so that
p(P'),
be maps between f.g. prois a Sylvester projective
p
p(aQR1R) + p(SOR1R) 5 p(P'QR1R);
rank function,
function
:
<_ pi(P');
pi(ai) + pi(6i)
find that
function, and let
aisi = O.
R1.
but, by the last theorem,
So the projective rank
is Sylvester.
pl
Theorem 3.8
R = R1
If
where
R2,
RO
is a skew field, then
R
is a
RO Sylvester domain if and only if
Proof:
R1
and
R2
are Sylvester domains.
This is an immediate consequence of the last theorem when we notice
that all the f.g. projectives are free of unique rank over if this is true for each
R
if and only
R.. 1
We have seen in theorem 3.6 that if
R = R1
RZ
for semisimple
RO artinian
and faithful RO-rings
RO,
projective rank function on full maps with respect to
R,
and
R1
and
R2
p = (p1,p2)
that full maps with respect to
when they are induced up to
p
R.
p1
is a
remain
In the case
where the rank functions are all Sylvester, we are able to prove that the factorisations of a full map induced up from come from
Theorem 3.9 R.
Let
R = R1 R1 R2,
where
is a faithful RD ring. Assume that
projective rank function on on
R ,). i
R1
to
R over
R
essentially
R1.
Then, if
R
aOR1R = Sy
RO
is semisimple artinian and each
p = (p1,p2)
is a Sylvester faithful
(and, in consequence, the same holds for
p. i
is a factorisation of a full map as a product
49
of full maps over -1 C
there is an invertible map,
R,
are defined over
y
Proof:
such that
e,
and
Be
R1.
Suppose that we have the factorisation over: aOR1R
B
R1R
PAR R
where
B,Y
a P'
are full maps with respect to
p,
the rank function on
R.
So
p(P') = pl(P) = p1(Q).y must be an embedding, since it is a full map with respect to a faithful Sylvester rank function. Since by 2.8 that the decomposition of
y(P') 2 a(P),
we know
given by 2.7 takes the form
y(P')
Y(PI) = MOOR R) $ (M1OR 1R) ® (M2OR2R),
where
a(P) 2 M1.
or M2 is not zero, its rank with respect to the MO and so relevant rank function is not zero; consequently, p1(M1) < pl(P)
If one of
aOR1 R
could not be full since its image lies inside
M OR R. 1
1
So
Y(PI) = MlOR R. 1
Now all we need to do is to show that e
to be the identification of
the induced map
P'
with
M10R 1R,
M1 E Q,
for then we take
and we see that
ftR R,
where
B
is the map
yOR R,
where
y
is the inclusion of
Be
is
and ey is
P -> a(P) E M1
1
the induced map
M1
in
Q.
1
Since
we see that d:M
B
p1(P) = p1(M1)
B
is a factor of the full map
is a full map. We consider the map over
c_ QQR1 R + (QOR R)/Q, 1
and
and notice that
Bd = 0;
R1
hence, since the rank
1
function is faithful
6 = 0,
so that
M1 E Q
a,
as desired.
50
4.
UNIVERSAL LOCALISATION
One of the first constructions developed in the theory of rings was localisations for commutative rings as a way of passing from a commutative domain to its field of fractions. When the theory of non-commutative rings was developed, it was noticed that an analogue of this was possible for suitable sets of elements in a ring provided that they satisfied the ore condition. This method was shown to be of particular importance when Goldie showed that for a prime Noetherian ring the set of non-zero-divisors of the ring satisfied the Ore condition and the Ore localisation at this set was a simple artinian ring.
One construction that was considered but rejected on the grounds that at the time nothing could be proved about it was the construction of adjoining universally the inverses to a subset of elements of the ring. Of course, the ore localisation is a special case of this construction. However, in studying the homomorphisms from rings to skew fields, Cohn was forced to study a generalisation of this construction. He showed that the set of a
matrices over a ring skew field
F
R
that become invertible under an epimorphism to a
determine the epimorphism; more specifically, the ring obtained
by adjoining universally the inverses to these matrices is a local ring
L
whose residue skew field is
is
F.
Of course, the homomorphism
R i L -r F
our original epimorphism. For the first time, therefore, the construction of adjoining universally the inverses of some set of matrices required serious consideration; this was taken a step further by Bergman who considered the construction of adjoining universally the inverses to a set of maps between f.g. projectives.
By now, there is a powerful but secret body of knowledge about this construction that is of fundamental importance for the study of homomorphisms to simple artinian rings. The purpose of this chapter is to present this theory which has been substantially simplified and extended recently. First, we shall give a simple existence proof for the construction;
51
this allows us to develop standard expressions for the elements of the universal localisation together with a simple description of the equivalence relation on these expressions given by their equality in the universal localisation. This is based on and generalises work of Cohn (71) and Malcolmson (83). Then we generalise an argument of Dlab and Ringel (84) to find simple homological information on the universal localisation; this allows us to give a very simple proof of a result of Dicks and Bergman (78); a universal localisation of a right hereditary ring must be right hereditary. Also, there is a short discussion of an interesting way to study the ring coproduct in terms of the universal localisation of a suitable ring. In the last section, we prove an exact sequence in algebraic K-theory that generalises the exact sequence of Bass and Murthy (67) for ore localisation.
Normal forms for universal localisation Let
be a ring, and let
R
projective modules over
that the maps
R a a E
Theorem 4.1
Let
R
let
R;
wish to construct a ring for
RE
be some set of maps in the category; we
E
that is universal with respect to the property are invertible.
a EE
R be a ring and
RE
for
R a a E
be a set of maps between f.g. left
E
projectives. Then there is an R-ring property that every element
be the category of f.g. left
P(R)
R
universal with respect to the a EE
has an inverse.
Proof: We use the observation that if we take the category of f.g. projectives over a ring
R,
and adjoin a set of maps between various objects subject to
some set of relations and consider the additive category generated by these maps and relations and our original category, then this is a full subcategory of the category of f.g. projectives over the endomorphism ring of the object that was the free module of rank 1 over case, we adjoin a set of maps adjoin the relations that
a:Q -
as = IP,
Clearly the map from
R
P and
R
in our initial category. In our
for each map
as = IQ
a:P ; Q
for all
a
in
E
in
E.
and
to the endomorphism ring of the object
that was the free module of rank 1 over
R makes this ring into an R-ring
with the correct universal property. We shall call this ring
RE.
It is
clear that the category of modules that we have constructed consists of the f.g. projectives over
RE
that are induced from such projectives over
R.
Now that we have shown the existence of this construction, we
52
should note that there is no difference between adjoining universally the inverses of maps between left f.g. projectives or between right f.g. projectives since we adjoin an inverse to ax:Qx Px. an inverse to +
Q
a:P
if and only if we adjoin
We have shown the existence of the universal construction, but this does not allow us to show in special cases that the ring we have is nontrivial; therefore, we need a concrete representation of the elements of the ring together with a way for determining when to such representations are equal; this is our next goal.
First of all, we show that every map in the category of induced f.g. projective modules over a universal localisation
represented in the form fyg, where
R and
where each
y =
a
c maps and for the maps If
a:Q + P
f1y11g1
and
f, g
of
R may be
This is clearly true for induced
.
i
for each f2y21g2
R£
and y are maps defined over
a:P + Q,
where
a
are both maps from
is in
RE0RP
to
E.
REQRO,
then we find the equation
YYO2-1 1
f1Y1 91 - f2y2 g2 = (f1
1
f2)
1
O (-g 2
l
if
f1Y1191:RE 2 P
R
1
+ R 0 P E R 2
and
f2Y21g2:REORP2 +
REORP3
are a couple of maps, then -1
-1
.
flyl 91.f2Y2 g2 : REIRPl + REORP3 is given by
-1 fly1g1.f2y21g2 = (f1
1
-gY2f 2
0
2
Y
0)
g2
We see that every map must have this form since all induced maps are got by successive composition or taking the differences of maps previously found. The reason for considering such representations of the maps is that we are able to write down a criterion for two such expressions to be equal maps
53
in
RE.
Theorem 4.2 (Malcolmson's criterion) Let of maps between f.g. projectives over projectives over
over
R,
and
has the form
RE
y =
I
R
R. fy-lg
for suitable
be a ring and
E
a collection
Then every map between induced f,g, and
for maps
y
defined
further,
ai e E;
O n f
1
y
1g1 1
1g2
= f y
2 2
defined over
if and only if there is an equation where all maps are
R.
We refer the reader to Malcolmson (83) for a proof in the case
where all elements of of the ring
RE
E
are matrices over
R;
he only considers elements
instead of all maps between induced projective modules; the
transformation to this present form is purely mechanical. The only apparent difficulty arises from the fact that addition and multiplication are not defined everywhere; however, a moment's thought shows that this causes no problem.
There is another representation of the maps in the category of induced f.g. projectives over
R
that is often useful for performing certain
calculations; this is a generalisation of techniques of John.
Theorem 4.3 (Cramer's rule) Let R be a ring and
E
a collection of maps
between f.g. left projectives. Then in the category of induced f.g. projectives over form:
RE,
every map
a:RE0RP -> R 0 Q E R
satisfies an equation of the
54
so
=s'
ot' 1-01)
where
8,a'
I
ai a E,
over
Proof:
are both maps defined over
R,
where
and
is an identity map on a suitable projective, and
a'
is some map
RE.
We use the generator and relation construction of the category of
induced f.g. projectives over Certainly, for
a
RE
given in the preceding discussion.
in
E,
the element
satisfies
a
as=l which is of the required form; for
a
defined over
R,
the equation
Ia = a is of the required form.
Next, given
we find that for
ai:P -* Q
al - a2,
al
ai
0
o
B.
a
i = 1,2
satisfying equations:
we have the equation
Bi
0
B2
s2
Finally, suppose that we have equations for i = 1,2, a! i a.
for maps
a1:P1 + P2
and
= ails
a2:P2 - P3;
then we construct the equation
55
0'2
0
0
a2
a2
-al al
o
al
I O
a2
aia 2
0
a a
0
a2
a2
=
-al
al
1 2
Since every map in the category of induced f.g. projectives may be obtained from the generators by these
two processes of differences and
compositions, this proves the theorem.
Of course, in the last theorem there are no good reasons why the representation of
in the equation
a
should be in any way
= a'
a
unique. We shall see however that Malcolmson's criterion gives us a great deal of control over such representations. At this point, it is useful to introduce a pair of formal definitions that we need later. We say that a set of maps jectives is lower multiplicatively closed if 0$) E E
for arbitrary suitably sized
Y.
E
a, ae E
between f.g. proimplies that
Similarly, we may define
(Y
upper multiplicatively closed. A set of maps every map between f.g. projectives over
R
is associated over
E.
to an element of
R
E
is said to be saturated if
that become invertible over
RE
We may define the lower and
upper multiplicative closure of a set of maps in the obvious way; it is convenient to be more careful when defining a saturation of a set of maps E:
a saturation of
E
is a lower and upper multiplicatively closed set of
maps such that every map between f.g. projectives over invertible over
RE
R
that becomes
is associated to an element of the saturation. For
Cramer's rule, it is useful to use lower multiplicatively closed sets of maps and for Malcolmson's criterion it is better to use upper multiplicatively closed ones.
We note the following consequences of Cramer's rule.
Corollary 4.4
All maps between induced f.g. projectives over
RE
are stably
associated to induced maps.
Proof: Simply look at the form of Cramer's rule.
Corollary 4.5
All finitely presented modules over
RE
are induced from
56
finitely presented modules over
R.
m by n matrix; by 4.4, this
A f.p. module is the cokernel of an
Proof:
is stably associated to an induced map between f.g. projectives, and cokernels are isomorphic for stably associated maps. So all f.p. modules over RE
be an induced
are cokernels of induced maps. Let RE0Ra:RE0RP + RE2RQ
map; then by the right exactness of the tensor product, the cokernel of
RE®a
is isomorphic to
REOR(cokera)
proving the result.
In particular, this applies to the f.g. projectives over
however, we cannot conclude that all f.g. projectives over from
for example, consider Z + 2M2(Z) c M2(Z);
R;
RE
RE;
are induced
all f.g. projectives
are free; however, the central localisation of this ring at 2 is isomorphic to
and there are some new projectives in this case.
M2(Z2),
One use of 4.4 is to show that iterated universal localisations are in general universal localisations.
Theorem 4.6 R,
Let
and let
E'
jectives over
E
be a collection of maps between f.g. projectives over
be a collection of maps between stably induced f.g. proThen
RE.
(RE)E,
is a universal localisation of
suitable set of maps between f.g. projectives over
Proof:
over
Let a RE.
such that
:
P' + Q'
Since
and
stably associated to
at a
be a map between stably induced f.g. projectives
and
P'
P' ® Rn
R
R.
Q'
are stably induced, there is an integer
Q' ® Rn
a ® In,
are both induced modules. Clearly,
n is
a
which is a map between induced modules and as
such is stably associated to an induced map by 4.4. Consequently a
is
stably associated to an induced map. Adjoining a universal inverse to a map however has exactly the same effect as adjoining the universal inverse of a map stably associated to it, so we may replace all the elements of a set (RE)E,
E
of induced maps that are stably associated to elements of is just
(RE)I
and this is clearly
REUE
E'
by
E'.
.
Homological properties of universal localisation Let
E
be a set of maps between f.g. left projectives over
It is clear that the homomorphism from
R
to
RE
R.
is an epimorphism in the
category of rings, since there can be at most one inverse to a given map
57
a:P -+ Q
of right
in the category of modules over a ring. Consequently, the category modules may be regarded as a full subcategory of the category
RE
of right R modules. It turns out that this subcategory has some interesting properties that form the content of this section which were worked out with Warren Dicks. Our next result is a development of an argument due to Dlab and Ringel (84).
The category of right
Theorem 4.7
in the category of right for
RE
Proof:
M
modules
R
and
induced map
RE
modules amongst the
RE
a e E, a
and for each
P - Q,
:
R modules by the RE
module
M,
the
is bijective. It is now clear that an exten-
1'%Ra:MQRP -+ MORQ
R modules of a pair of
sion in the category of
Ext1R(M,N) = Ext11 (M,N) RE
N.
We may characterise the
property that for each
modules is closed under extensions
RE
modules. Therefore,
module. It follows trivially that
RE
modules must be an
Ext1R(M,N) = Ext1R(M,N)
for
RE
E
modules M
and
N.
Our next result gives a number of consequences of the conclusion of this result.
Theorem 4.8
Let
be an epimorphism in the category of rings; then
R -+ S
the following conditions are equivalent: a/ Ext1 = Exti on left S modules; b/ Tori(S,S) = 0;
c/ TorR(M,N) = Tori(M,N) for S modules M and N; d/ Extl = Extl on right S modules.
Proof: a = B:
let
R modules where
M - F + S -+ 0
0 -
F
be a short exact sequence of right
is free. Then, we have the exact sequence:
1/ 0 -+ TorR(S,S) -+ MORS -+ FORS -+ SOBS = S -+ 0 we also have the push-out diagram: 2/ 0 -+ M -+ F -+ S -+ O j
O -+MMRS
by assumption,
II
-+N-+S;O
N must be an
S
module, and so, we may factor 2/ through
1/ in the following sense; we have a canmutative diagram with exact rows: O
-+ M
0 -+ TorR(S,S) -+ 1
-+
F
-+ S+ O II
M0+
S -+ F® S -+ S -+ 0
11 R
O -+ MGRS -+
JP-
N
II
-+
S -+ 0
58
It follows that
let M be a right S module and let
c:
b
TorR(S,S) = O.
sequence of
S
modules where
0 -> A -* F -> M -+ 0
be an exact
is free; then we have the exact sequence:
F
O = Torl(F,S) -+ Torl(M,S) -+ AM RS a Fe S
MGRS -+ O.
R Since S
BORS = B
is an epimorphism in the category of rings
R -+ S
module,
B,
0 -+ C -+ G -+ N -+ 0
Let N be a left
TorR(M,S) = O.
so
be an exact sequence of
S
for any
module, and let
S
modules where
G
is free;
then
O = TorR(M,G) -+ TorR(M,N) -+ MORC -* M®RG -+ MORN -+ 0
is an exact sequence. For any left me SD
by the natural map since
exact sequence shows that M
and
and
module
S
is an epimorphism; therefore, the last
D,
TorR(M,N) = Tori(M,N)
MORD
for arbitrary S modules
N.
c 4 a: let M
R -
is isanorphic to
S
N
0 -+ M - A - N -+ 0 are
S
be an exact sequence of right R modules where
modules; then because
we have the
TorR(N,S) = 0,
commutative diagram with exact rows:
M = MO RS
O
-+ M
O
-ML
and
-+
A
-+
N;O
-+ A RS -+ NORS -+ O
N = NORS,
so by the 5-lemma
must be an S module. It follows that
A = AORS,
ExtR = Exts
which shows that
A
on right S modules.
We have shown that a,b, and c are equivalent; by symmetry, d is equivalent to them too.
In particular, a universal localisation satisfies all these conditions. It is clear that an epimorphism satisfying the conditions of theorem 4.8 need not be a universal localisation, since if of a ring
R
such that
I = 12,
seldom a universal localisation of
then
Tor1(R/I,R/I) = 0,
I
but
is an ideal R/I
is
R.
The preceding results allow us to prove a theorem due to Dicks
59
and Bergman.
The universal localisation of a right hereditary ring is right
Theorem 4.9 hereditary.
Proof:
A ring
is right hereditary if and only if
R
functor in the second variable. If on
Exti
then
RE
modules is isomorphic to Ext1
RE
is a right exact
Ext1
is a universal localisation of
R,
which is right exact in
E
the second variable. Therefore,
RE
must be a right hereditary ring.
Universal localisation and ring coproducts There is a useful way to obtain the ring coproduct of two rings amalgamating a common subring by considering a suitable universal localisation.
Theorem 4.10
the map a
Let
and
B
B) + (A
(0
:
A
be R-rings; let
T =
A
AORB
O
B
and consider
given by left multiplication by CO 0
ARRB)
101
0 then
M2(A
Ta
R
B).
The elements
Proof:
(1
(O O(
e11
0),
e22
O
O ,
O
1)
a,
and
a-l
of
Ta
form
a set of matrix units; the centraliser of these matrix units is isomorphic to which is generated by a copy of
e11Tae11,
to the relation that
R
A
and a copy of
B
subject only
is amalgamated.
This has the following consequence.
Theorem 4.11 R:
then
Proof:
S1
S1
M2(S1
Sl'RS2 S
0
2
Let S2
R
R
S2)
S1
and
S2
be semisimple artinian R-rings for some ring
is hereditary.
is a universal localisation of the hereditary ring
60
Algebraic K-theory of universal localisation We mentioned earlier that ore localisation is a special case of universal localisation at sets of maps between f.g. projectives; in particular, central localisation is a special case. Bass and Murthy (67)
R with the K-theory of
found an exact sequence connecting the K-theory of for a central subset
RS
S
of
from
R,
down to
K1
and later,
K0,
Gersten (75) was able to generalise this to a long exact sequence for central localisation. Recently, Cohn (82) and Revesz (84) were able to calculate the
of the free skew field and it was possible to rephrase this
K1
as Revesz did in terms of an exact sequence connecting
K0
and
of the
K1
free algebra with the K-theory of the free skew field. Here we shall find a common generalisation of the Bass-Murthy result and the Cohn-Revesz theorem. The reason for proving such a result lies not just in allowing us to calculate of a universal localisation, but also in its giving us a way of quantify-
K1
ing Malcolmson's criterion.
First of all, we define a useful category; the definition of this category is very natural in the context of universal localisation though
R be a ring and
rather less obvious in the Ore case. Let of maps between f.g. projectives such that
R
embeds in
E
RE.
a collection Let
be a
E
collection of maps between f.g. projectives that become invertible over such that all maps between f.g. projectives over are associated to some map in
RE
all elements of embed in
that have inverses in
is just R. Further,
are injective because of our assumption that
E
T
Let
modules over
R Rf
clearly,
E;
R
should
be the full subcategory of the category of f.p.
whose objects are the cokernels of elements of
R
RE
category is independent of our choice of in the category of modules over
since
R,
E.
This
and is closed under extensions
E
E
may be chosen to be lower
multiplicatively closed.
Theorem 4.12
Let
be a ring and
R
projectives such that
embeds in
R
a collection of maps between f.g.
E
RE;
let
and
E
be as we defined
T
in the foregoing discussion. Then there is an exact sequence,
K1(RE)
K1(R)
where
map a
r [?
and
u
K0(T) } KO(R) i KO(R
are induced by the ring homomorphism,
] + [Q] - [P],
is an element of
E,
where and
0 s
P
a
Q ± Ma -
0
t
is given by the
is an exact sequence, and
will be defined in the course of the proof.
61
Throughout this proof,
Proof:
Let over
RE;
and
S
B'
=
a
are defined over
S'
KO(T)
is defined over
a1
so is
RE,
y',
RE.
Since
so it is associated to
We attempt to define a map from such an invertible map
E.
by
and
R
all -are invertible over
\0 an element of/ to
be an isomorphism between induced f.g. projectives
a:P -)- Q
6 (0,
and
a.
by Cramer's rule, we have an equation:
al
where
will denote the cokernel of
M6
We need to show that this is well-
s(a) = [MS,] - [Ms].
defined and its restriction to automorphisms defines a homomorphism from K1(RE)
to
Assuming that we have proven this, it is not too hard to
KO(T).
complete the proof of the theorem, so we shall demonstrate this first.
Exactness at K0(R)
1
Certainly,
to = 0;
where Ma has a presentation morphism over
Ma a T,
for given
t([Ma]) = [Q] - [P],
since
0 ; P i Q } Ma -*'0;
a
becomes an iso-
u(Q) - u(P) = O.
RE,
Conversely, if
u([P2] - [P1]) = 0,
R R between induced f.g. projectives
so there is an isomorphism
n;
RE,
we have an equation over
a:RE0R(P1 ® Rn) I REgo R(P2 ® Rn),
and so,
by Cramer's rule, we have an equation:
al
(I.
= S'
s
O
a
for maps defined over 6,6'
a1
R,
8:Q' -> Q e P1 ® Rn,
are associated to elements of is some map defined over Hence,
RE,
(since
E
and
IQ
6':Q' -> Q ® P2 a Rn, a
where
is invertible over
is the identity map on
RE),
Q.
t([MS,] - [Ms])
[Q] + [P2] + n - [Q'] - [Q] - [P1] - n + [Q'],
which is
[P2] - [P1],
so that we have exactness at this point.
2
Exactness at Let
Then we defined
KO(T) a:P -* P
be an isomorphism on an induced f.g. projective.
s(a) _ [Ms,] - [Ms],
where we have an equation:
62
al IQ
6
B'
O
where
8
and
a
8':Q' + Q ® P
are both associated to maps in
E.
So we find
that
st(a) = [Q] + [P] - [Q'] - [Q] - [p] + [Q'] =
Conversely, suppose that
t([Mal ] - EM
a2
0
.
then if we have
]) = 0;
a.
presentations
0 - P. a Q. -> Mi + 0
so for suitable
for
we have an equation,
n,
construct from this, presentations for for suitable
and Q. a 81$I
P
0 + P1 $ p2 $ Rn and
M1
and
We
s. O + pi + Qi + Mi + O
M2
This is easy for
I®a2$1
+
0 + P1 ® P2 $ R
1Q1] - [P1] _ [Q2] - [P2],
,
Q1 ® P2 ® Rn = Q2 ® P1 ® Rn.
Q1 9 P2 ® Rn + M1 + 0
1 +
n
i = 1,2
is a presentation for
n
P1 ® Q2 9 R + M2 + 0
Ml
is a presentation for
M2.
Consider the map over Because
82(62 1 61) = B11
RE
defined by
62181:RE8RQ + REIRQ.
we see that s(6 281) _ [MB ] - [M6 ] = [Ma] - [Ma ]. 1
So we have shown exactness at
3
1
2
2
KO(T).
Exactness at K1(RE)
This is rather harder to show than the previous two parts. Certainly, R,
the equation
rs = 0;
l.a = a
for, if
a:P - P
is an automorphism over
satisfies the condition of Cramer's rule, so
rs(a) = EMa ] - EMa I = O. Conversely, suppose that induced f.g. projective over
RE
a:P + P
such that
is an automorphism of an s(a) = 0;
we have an equation:
al
Q
O
where
8:Q' - Q ® P,
we deduce that
= B' a
B':Q' + Q ® P
are associated to elements of
f
and
0 = s(a) = [M8,] - [M8].
We wish to show that the class of
a
in
K1(RE)
lies in the
63
image of
KI(R)
in
and our equation shows that the class of
Kl(RE),
is equal to that of
in
K1(RE)
if
[Ms] = [MS,]
in
B-ls';
so we need to examine what it means
KO(T).
We define a recursive relation category
that amounts to the relation
E,
-
on the objects of an exact
[X] _ [Y] in
KO(E).
We start our recursive relation by defining that next, if
X = Y;
a
X - X',
X - Y,
if
and we have exact sequences
Z - Z'
and
O-+X->Y-+Z->O, and 0-X' ->Y' -Z' -+
then
0,
similarly,
Y - Y';
if we have the first exact sequence, and in place of the second we have the
exact sequence 0 -+ Z' - Y' -+ X' -+ Next, if
X - X',
and
Y - Y'.
we extend the relation to
0
Y - Y',
and we have exact sequences
O; X -+Y -+Z + O O -+ X' -+ Y' } Z' - 0,
then
Z - Z'; and if we have exact sequences
O-+X-+Y-+Z;O O-
Z ' - + Y ' - + X' - * 0 ,
then Z ^- Z' .
We repeat these operations as often as they apply, and finally we make sure that the relation is transitive by defining that if X' - X",
then
X - X'
X - X".
We examine what this means in the category
that is, we
T;
attempt to find the equivalence relation defined on the maps by a, 8 in
and
Ma - M$
for
E.
Ma = M$
if and only if
a
is stably associated to
$.
Our
first operation for generating the equivalence relation on the modules induces the relation on maps that if M(O'
M
M - M,, a a
and M
O\
a
(0
y
M
then
,
Y
Our second operation for
Y
generating the equivalence on modules induces the relation on maps that if
Ma - M a
'
,
and
M M(O°
then
(O
MY ... MY , .
We have that MS - MS,
for the maps
B
and
B':Q' -+ Q 0 P,
so
we know that we can pass along some chain of simple operations of the type 0 described above to show this equivalence; by passing to and 0 Im/ I S for suitably large m and n, we may assume that any stable I
associations that occur in this chain are actually associations. We wish to
show that this equivalence on maps forces the image of to lie in the image of
K1(R)
in
KI(RE).
0'-16
in
KI(RE)
64
If
If
RE
K1(R)
al S
I
a2
a2
O
0
and since the class of
al that the image of (
and the image of
1(0
O )-1/al
a2
t0
O
a
/al
O
al
O
O
I
O
I
al 0 ,
(I
and
0)
Y
a2
Y
=
0l
O
I
O
I 0
0
I
Y
I
O
() () () K1(R£),
are trivial in
K(R£) 1
is equal to the image of
(
2
a2
we see
al
O \I
O
a
2
So we need to check that the image of
).
I
0) in
B
in
)
lies
a.la':Q. - Q.
S
Y
(al
I
I)
1I
K1(R).
then from the equations
K1(RE),
in
satisfy
i = 1,2
for
-). Q,
the class of the map
R,
clearly lies in the image of
K1(RE)
in
a.,a!:P.
in the image of
O
are associated over
a,a':P1 -r Q1
over
a:Ql -+ Q1
a2
K1(RE)
lies in the image of
K1(R),
which is clear.
a2 1
Finally, if
1
(0 K1(RE)
a ) 2
a
\
lies in the image of
does too, then so must
and all al
which shows that the
a21a2,
second operation generating the equivalence relation the property that
_1Y' Y
- MY'
M
preserves
Y
lies in
K1(R).
We are left with showing that our map
s
from isomorphisms of
induced f.g. projectives is well-defined and induces a map from KO(T).
in
K1(R)
2
K1(RE)
to
This is where Malcolmson's criterion is most useful.
Suppose that
B:R?RP -. REQRQ
is an isomorphism between induced
f.g. projectives and that we have two equations defined by Cramer's rule:
(aIa')
and
(alai),
(ala'),
then we know that
(YIYl)
and
(YIY1)( 0 (YIY')
62\ =
(YIi')
S
are associated to elements of
E;
65
sl
(a al
0
I\O
Y
Yl
s
Ca
B
so,
(0
0
0
I) a a1 0 0
I
Y1
-1
a,
Yl
Y
tO
=
2
s-s
a
O
Y1
Y
I
a'
Y
0
=
Y'
by Malcolmson's criterion, we have an equation: al
a
0
0
0
0
Y
Yl 0
0
a
j10 0
0
0
61
0
0
0
0
0
0
62
E
O
O
O
I
E
O
O
0
Y1
and
1
U1
U2
Y
0
=
(v I T)
U
3 U
2
4
O
are associated to elements of
r
E.
3 I
U4
From this, we construct two equations:
U1 1
/'2 3
I'
U4
O
0
O
a 1 Y1
0
0
0
Y
0
0
0
Yl 0
0
0
0
0
0
0
a
0 v
0
O
I
_
0 0
0 0
0
d
0
0
01
62
0 0
I
E1
0
0
0
Y'
and also
0
l Y
Y
0 0
0
0OO6
0
0
0
O
OO
0
a
a
O
Y
O
0
First of all, it is clear that and
[MRHS(1)]
[MRHS(2)]
=
=
O
I
RHS(1)
-a'
l
0 0 62
0 0 -E 2
0
el 0 and
[M61] + [M62] + CM(aal)] + CM(YY')]'
0
RHS(2)
lie in
whilst
[M61] + [M62] + [M(aa')] + EM(YYl)I. If
X,Y E E, [MXy] = [MX] + [MY] when
XY
is defined. Since
E,
66
Ov 01
lies in
ul 0 u2 Y' u3 0 0
£,
must be associated to an element in
and
£
u4
0 0 [MV -T ]_ [MLHS(2)]
[MLHS(1)] = [MU] + [MV] = [MU]
0
I
We deduce that CM(aaI + [M(YY')I = CM(aa')I + CM(YY1)], which shows that
is well-defined.
s
We need to show that it induces a homomorphism from KO(T),
s(o 0) = s(a),
and to do this, we need to show that
s(ay) = s(a) + s(y).
K1(R£)
to
and
It is convenient to show that
Ia
for arbitrary
s 061 = s(a)
al.
Suppose that we have equations:
7I g2\1 I=
(alai)
(ala')
//I a3
(Ylyl
then we find that
/
$3
O
al
a
0
1
\
0
a1 1
I
a2
O
a
O I
(YIY')
I
0 a
y
yl
0
\0 0
a
Y' a'
operating by a permutation matrix internally on the left hand side of this equation does not alter
[M
]
in any way and allow us to change the
00100
( 00 aal 0
6
0
T
R
0
0
a
second matrix on the left hand side to the form /I
s /I 61
O a
=
C
Y]
Y Y l
O Ol a a')
CMY
Y1
0 ] _ [M(aa')]
(O Ol a al) 0
so 2
l
[M(aa )]
=
d(B)
1
as we wished to show.
In order to show that under the assumption that
a e T.
s(ay) = s(a) + s(y),
we first show it
67
s(I Y) 06
s(By)
by our last step; it is clear that
an isomorphism over
R;
d
is
s(1 O)
I O O /B
(a al)
a
O
Y 4o,l'
so
if
so we find that
s(BY) = s(I BY)
If
s(SE) = s(E)
a
)I
1
O
1
0
s(B O) = [MB] + [M(aa,)] - EM(aal)] = s(B) + s(y),
O\ Y1
I(a
1
and
s(BY) = sY) = s(B) + s(Y)Finally,
we deal Assume
s(d1d+s(0 BY) so s
withh the general case.
(d1dl)
= (dld') then, by the last step,
1
\ OIBB / s (d1d1) \O
BY)
s(BY) = s(1 B0O) = s(Y) + s(6
')
=s
(d1d')
o -6YYSOW) W') + s(Y)
s(dd1) = s(Y) + s(B)
induces a homomorphism from K1(Rto KO(T),
which shows that
and so completes the
proof of our theorem.
The Bass-Murthy sequence is a special case of this result; it would be of interest to know whether this sequence may be extended to a long exact sequence of K-groups for universal localisation as Gersten does for central localisation. It appears very likely for special cases such as the
passage froma fir to its universal skew field of fractions. This leads to the following conjecture for some set
X
K2(F),
where
F
over an algebraically closed field,
is the free skew field on k.
K2(F)
should be
x
where A is the set of stable association classes of full A matrix atoms over k<X>. K2(k) x X k ,
68
5
UNIVERSAL HOMOMORPHISMS FROM HEREDITARY TO SIMPLE ARTINIAN RINGS
In this chapter, we shall prove our main theorems on universal homomorphisms from hereditary rings to simple artinian rings; we shall find that these arise as universal localisations at the set of maps between f.g. projective modules that are full with respect to a projective rank function In doing so, we shall also develop techniques
that takes values in 1 Z. n
for studying intermediate localisations, which generalise the results known to hold for firs, and which will lead in later chapters to a detailed analysis of the subring structure of these universal localisations of hereditary rings. However, the most interesting result of this chapter in the short term is the construction of the simple artinian coproduct with amalgamation. This is given by constructing a universal homomorphism from the ring
Sl S SZ so
to a
Sl U S.,, where S. is simple artinian. In order to O assist the reader in understanding the method, we give a brief outline of simple artinian ring
what we shall do in this case.
tion
p
S1 SO S2 is a hereditary ring with a unique projective rank funcby 3.1. We consider what maps between f.g. projectives have a chance
of becoming invertible under a suitable homomorphism to a simple artinian ring; firstly, the homomorphism must induce the projective rank function
p
S1 S S2 since this is the only rank function. Therefore, if a:P - Q O is a map that becomes invertible under a homomorphism to a simple artinian
on
ring,
p(P) = p(Q),
rank; that is,
a
and
a
cannot factor through a projective of smaller
is a full map with respect to the rank function. At this
point, we consider the ring obtained by adjoining the universal inverses of all full maps with respect to the rank function, since Cohn (71) has shown that in the case where
S.
are all skew fields that the result is a skew
field; we are able to show in this case that this universal localisation is simple artinian as we wanted it to be. We have been calling such constructions by such names as the 'simple artinian coproduct' or the 'universal homo-
morphism from the hereditary ring
R
to a simple .artinian ring inducing the
69
projective rank function
we shall not justify these names immediately,
p;
but in, chapter 7 when we discuss homomorphisms to simple artinian rings in
greater generality, we shall see that all other homomorphisms from hereditary rings to simple artinian rings inducing the given projective rank function are specialisations of the one referred to as universal.
In fact, the general method applies to constructing universal homomorphisms from rings with a Sylvester projective rank function to von Neumann regular rings, so we shall deal with this generality in the opening stages of this chapter; we shall complete the discussion of homomorphisms to von Neumann regular rings in the next chapter.
Universal localisation at a Sylvester projective rank function Theorem 5.1
R be a ring with a Sylvester projective rank function
Let
p
having enough right and left full maps (see 1.16, and the following definition). Let
be a collection of full maps with respect to
E
projectives: then the projective rank function projective rank function on
p
p
between f.g
extends to a Sylvester
that has the same image as
p
and has
enough right full and left full maps. The kernel of the map from
R
to
R,
pE,
lies in the trace ideal of the projectives of rank
First of all, we may assume that
Proof:
E
RE
0.
is lower and upper multiplicat-
ively closed (see the discussion before 4.6), since all elements of the lower and upper multiplicative closure of and become invertible in
E
are full with respect to
p
by 1.15
The use of this observation is that it puts
RE.
us in good shape for both Cramer's rule and Malcolmson's criterion. We prove the last statement of the theorem first, since it is a simple step. Let
r e R be in the kernel of
R -> RE;
then, by Malcolmson's
criterion, we have an equation:
100 1
0
0100 00a1 0
000 a2 1Os10 where
ai,di e E
and
r O
_
al (d21 u)
O S2
0
are maps defined over
R.
70
This shows that the nullity of LHS(l) is 1; by 1.15, and the fact that
are full with respect to
l,al,a2
we may deduce that the
p,
is 1, so that the nullity of
nullity of
is 1, or that its inner
r
(1l or)
projective rank with respect to
is
p
0,
which implies that
in the trace ideal of the projectives of rank it is clear that
RE
r
with respect to
0
must lie p.
So,
is not the zero ring.
Cramer's rule and Malcolmson's criterion give representations of maps over
RE,
and ways of characterising when two representations are the
same. We use these in order to define the rank of a map between induced f.g. projectives over
and then to show that it is well-defined; then we
RE,
show that this rank function on maps between induced projectives is the inner projective rank associated to a projective rank function that satisfies the law of nullity for these maps; it is not a hard step to show that the rank function must actually satisfy the law of nullity for all maps between f.g. projectives. Given a map,
between induced f.g. projectives,
S:RL RPl -> RE0RP2
Cramer's rule gives us an equation:
(aIal)
2
= (ama') 0
where and
(aa')
is a f.g. projective over R,1
Q
(aal) EE,
is defined over
is defined over
RE,
R.
It is clear that if
extends to
p
we should define
RE,
pE(S) = p(aa') - p(a) = p(aa') - p(Q) = p(aa') - p(aai) + p(P1)
The first definition shows that
is non-negative, if it is
p(S)
well-defined. Suppose that we have a second equation: IQ 3
(YIY1)
,a 2
= (YIY')
o where RE,
(yyl) E E,
and
(yy')
is a f.g. projective over
Q'
is defined over
R.
R,
S2
the equation:
al
4
yl (0,
sl
0 y
S
S2
a-a
is defined over
Then, from (2) and (3) we construct
a10
= 0
yl
Y
a' Y'
71
where
is an identity map on a suitable f.g. projective over
I
Over
0 = (0 0 0
RS,,
I)
I
(00100 -1 O
R.
a'
Y'
Yl Y Yl
By Malcolmson's criterion, we have an equation:
a al O O 5
0
0 Yl Y Yl 0
0 0
00
SZ
OO
O
00 0 1 where
01:
00 00 61 O
61 0
(p /(v1 T)
_
O
` O
Si,p, v e E
We see that the right nullity of LHS(5) is
p(P2). Since
61
and
SZ
are
is
p(P2).
al O O a' full, we see from 1.15, that the right nullity of
(aO
Y1 Y Y1 Y'
00 OI
O
In turn, we look at the bottom row of this matrix of maps and apply 1.15 to
deduce that the right nullity of ( Y1 Y Y
is
p(P2).
Since
P2
is the
1
codomain of
(.,:)
and
al
Y
OY
0)
is clearly right full (because
O Yl Y Y
YJ a
is full), we may write a minimal factorisation of
(a0
6
where
0 Y1 Y
p
and so
Y,)
1
/
as:
Y:) = (T2) al
p(a al O ) = p codomain CO Yl Y/I
pYlY
must be a full map. From (6), we construct two equations
O)
7 2
8
Y1
(XlIX2)
I
2
X1
(a al O O
0)
(T1 OY)
Y
1
(ol
IZ)
(aO
Yl
YI 0')
Using lemmas 1.15 and 1.13, we see that
P(act l) + P(YY') = p1 1 ,1 = P(aa') + p(YY1) 2
/
72
Therefore,
p(yy') - P(YY1) + P(P1) = p(aa') - p(aal) + p(P1)
that our definition of
pE
PE( S) = pE(a) + pE(R)
It is useful to show next that that when
is defined
as
which shows
is well-defined.
pE(c
So suppose that
PE(a),pE(R).
)
and
/
and
(YY1)(O a1) = (yy')
(661)(0 Si)
with the usual conditions obtained from Cramer's rule. Then, if
aR
= (66') is
defined we construct the equation:
= (6
61 0
0 -y' y
a1R
6'
O
I
aR
61 0 6
6
'
+ p(dom a);
- p
O -Y' Y 0 f6
0 -Y' Y Y1
a' 0 -6
61 0 6'
pI
11
1= P
0 - Y'
61 0 0
6
PE(aR) = p
so
(0,
YO
0
Y
so by lemma 1.14,
;
Y
< p(66') + p(codom (yy'))
P
pE(a6)
therefore,
p(66') + p(codom(yy')) - p(661) - p(yy') + P(doma)
p(66') - p(661) + p(codoma) since p.
Since
codom a = dom R, 0(06
Again, PE(aR)
pE(aR) 5 PE(R).
d' 0 -6 1 O
is full with respect to
(yyl)
y
< p(yy') + p(dom (66'))
and so we find
'
p(yy') + p(dom(66')) - p(661) - p(yyl) + p(dom) p(YY') - P(YY1) + p(doma) = pE(a)
For arbitrary
since
(661)
is full.
we have the equation:
a,R,
R1
d
0 0
O
d
O
Y Y1
O
O
y
0
0
0
a O l
from which it is clear that p E
0
B)=
(a) +
p
E
p
(R)
O
Y.
d' O
73
Given a f.g. projective
over
P
and the natural thing to do is to define idempotent map defined over
where
f.g. projective
on aniinduced
RE
we wish to define
RE,
pE(P) = PE(eP)
R Q Q
is isomorphic to
eP
P.
such
R
E
that the image of
pE(P)
is an
eP
We check that this is well-
defined. If
e,f
are idempotents on REQRQ1,
such that their images are isomorphic to RE
such that
and
aR = e,
Sa = f;
so
P,
and
RE6dRQ2
there are maps
a,s
pE(P)
over
pE(e)
pE(e) = pE(as) <_ PE(Sasa) = P"E(Ra) = PE(f) < PE(czaa) = PE(e)
equality must hold and
respectively
so that
is well-defined.
It is easy to show that the rank
pE
that we have defined on
maps between f.g. induced projectives is the inner projective rank function with respect to the projective rank function
pE
that we have just defined
on f.g. projectives over RE. Consider the diagram: a
RE`RPl
Q
where
R Q P E R 2
is some f.g. projective. Let
suitable maps
S:REIRP1 - RE®RP,
Q = (REORP)e;
y:RE®RP -> RERP2.
conversely, we recall that the definition of
9
(S B1)
( IP \\ O
a
1
a)
pE(a)
then So,
a = Sey
for
0,(a) 5 PE(e) = PE(Q);
gives us an equation:
= (S S')
where the usual conditions of Cramer's rule hold, and, by definition,
PE(a) = PO B') - P(P). Let tion of
(S S') = yd, y:P' + P", then
(S g');
I- O 10
P
0 consequently,
d:P" - P'" be a minimal factorisa-
)=
a
(S
S1)-ly
I p -a 1 0
I
REORP' = (REIRP) 0 Q, REORP" _ (RE0RP) ® Q"
RERP "' = (R ERR P) ® Q;
and also,
and partitioning (10) according to these decompositions
74
over
R,
so, over
we fin d the equa tion:
I- O P
I-
E
0
E2
E3
P
a
where
a = 112V2
RE,
I-
1
p
1
2
3
O
I-
ul
112
Vl
0
P
V2
but we have the equations:
112:Q
which shows that
p(P) = p(a),
- p(P) = p(S
P(V) = p(P")
IP
p
on maps
between f.g. induced projectives is the inner projective rank function associated to the projective rank function
and also that there are
p,
enough right and left full maps with respect to this rank function for maps between induced projectives. Let
6:RE&RP2 + REORP3
and
a:REORPl 3 REORP2
maps between f.g. induced projectives such that
as = 0;
be a pair of we have the usual
equations deduced from Cramer's rule:
S
(0,
(Y Y1)
(Y Y')
.
(d dl)
=
1
(d d')
s
then
/a al O
0
-Y' Y
a
/I
o
Sl
=
6i
0
6')
-Y'
Y
0
(06
als
Yl I
a6
0 dl
O
O
-Y'
Y
Yl
d
so
d'
by Malcolmson's criterion 0
we have an equation over d
12
where el
-1
0 = as = (0 0 011)
and
E2
R:
0
0
0
0
6'
0
6i -Y'
Y
Yl
0
0
0
0
0
0
0
E1
0
0
0
0
0
0
0
E2
2
0
0
0
I
0
0
Ei,u,v
lie in
O
E;
m
l
)
=
(ta ) (V
so the right nullity of LHS(12) = p(P3).
are full, we deduce from 1.15 that the right nullity of
Since
75
6l 0
d
O -Y' Y
0
0
0
0
6'
Y1
0
I
0
is
and, in turn, the right nullity of
p(P3)
-Yl
P
is p(P3);
Y
(O
so, because
0
61 0 6')=pI
6
-61
(
0')
(0 -Y' Y O
is is right full, /
Y'
61 0 )
6
O -Y' Y
We see from 1.14 that
d 6' O -61
P(6 6') + P(Y Y') < P
O0 Y
P(6 61) + (Y Yl So
61 O 6'
d
\0 -Y' YO
Y'
dl O
d
=P
= p1
0 -y' y
- P (P1)
PE(a) + PE(B) = P(6 6') - P(6 61) + P(P2) + P(Y Y') - P(Y Yl) + P(P1)
S P(6 6) + P (Y Y1) - P (P1) - P(6 61) + p(P2) - P (Y Yl) + p(P1) = P (P2) , which proves the law of nullity for maps between f.g. induced projectives. Given an arbitrary map we define
RE,
function that for
on f.g. projectives. If
pE
a6 = 0,
a:Q1 - Q2,
then
(0
O)
:
with respect to the rank are maps such
Q3
B:Q2
Q1 ® Q1 - Q2 ® Q.
Q2 ® Q2 - Q3 ® Q2 ® (Q3 ® Q2)
0) 02
a
we find induced f.g. projectives such that RP = Q. ® Q'
i = 1,2,3;
(O
between f.g. projectives over
a:Ql -+ Q2
to be the inner rank of
pE(a)
are maps between f.g. induced
:
Q'2
(cz O
projectives such that
so
O) (O OIQ1 0) -
0 P2(00
)+P2B OIQ, 01_p(Q2) 0(O
2
0 PE(O
O) =
PE(a) '
pE(01
01
pE(a) + PE(B) <_ pE(Q2)
rank function on
+ PV E(Q)
but
//lI
IOQ
2
O) = PE(B) +
so we deduce that
pE(QZ)
as we wished to show. So
pE
defines a Sylvester
RE.
Finally, we know that all maps between induced f.g. projectives factor as a right full followed by left full map. So
the notation of the last paragraph) and it follows that factors through
0
0
1 (and has the same rank as
IO
0 )
(C,
a
does (taking
must since it
O)'
It is useful to be able to prove that all f.g. projectives over
76
RE
are stably induced from
R
in the circumstances of theorem 5.1; this
is not in general true, but what does hold is sufficiently good for our purposes.
Theorem 5.2
R be a ring with a Sylvester projective rank function
Let
having enough left and right full maps; let maps with respect to
then any f.g. projective
p;
p
be a collection of full
E
over
Q
satisfies
RE
an equation of the form:
Q ® R ® QO = REaRP
where
p(QO) = 0,
Proof:
Let
e
and
P
is an f.g. projective over
be an idempotent in
Mm(RE)
R.
such that
then by
Rme = Q;
Cramer's rule, we construct the commutative diagram below: RE0RPl
R & S
R
E R
y
RE0RP2 _
R A P R
E
R REORS
eQ In
11?
m+n RE
where
a = ay
Rm+n
tion of
Q ® Rn
and
i
m+n a
over
P p
R,
p
so in the equation
as we wished to show.
on the image of
is an f.g. projective such that where
p(P) = n,
Q ® Rn a QO - REORP, of Q under any rank function extending
in theorem 5.2 for arbitrary extending
is the projec-
p.
Incidentally, this shows that the rank function extension of the rank function Q
ARE
Q ® Rn must'be surjective; moreover, pE, PE(Q ® Rn) = p(P2);
R E'RP2 = Q ® RE ® QO ' PE(Q0) = 0,
if
n
RE
is a left inverse to
The map
by the definition of
11th
i
is a minimal factorisation of onto
a
ER
n Q®RE
P
3
Q
pE(Q) = 0,
KO(R)
pE
in
is the unique KO(RE);
for
there is an equation
from which it follows that the rank p
must be
0;
the equation proved
shows that there is a unique rank function
p.
Let
R
be a ring with a Sylvester projective rank function
p
77
and let
be the collection of full maps with respect to
E
we call the
p;
P
ring
the universal localisation of
RE
R
at
and we shall in general
p,
P
write it as
In the case where
R . p
1 Z,
takes values in
p
we are able
n
to prove a great deal about such a ring.
Theorem 5.3
be a ring with a Sylvester projective rank function
R
Let
taking values in 1 Z;
R
then the universal localisation of
n
at
R
p,
is a perfect ring with a faithful Sylvester projective rank function all f.g. projectives over
are stably induced from
R
p
p
and
p
The kernel of the
R.
P
map R i R
is the trace ideal of the projectives of rank
and
0,
R -> R
P
P
is an honest map.
Proof: We shall need the last half of this theorem in order to prove the first part. Since
takes values in 1 Z
p
it is clear that there are
n
enough left and right full maps, so we may apply 5.1 and 5.2; in particular RP
has a Sylvester projective rank function
and the map
P
functions
PE
taking values in
Z,
must be honest with respect to the pair of rankn
R -> R
p, pE.
Let
P
be an f.g. projective module over
such that
R P
there is an idempotent
PE(P) = 0;
PE(e) = pE(P) = 0,
so
e
Mn(Rp)
in
PE(In - e) = n.
such that
and
Rne = P
By Cramer's rule, we have an
equation:
(aI a1) (O
where
RE;
is a full map and
(aa1)
has rank
(aa')
n,
therefore,
e = 0, P = 0,
zero over
(ala')
-
I-e)
pE
is defined over R,
since
R;
In - e
and so, it is invertible over
must be invertible and so the identity map. Hence,
In - e and
(act')
is a full map over
is a faithful rank function. All projectives of rank
must be killed by the homomorphism, so by 5.1, the kernel of
R
the map is precisely the trace ideal of the projectives of rank all f.g. projectives over
are stably induced from
R
R.
0.
By 5.2,
It is easy to see
P
from this that all maps over
that are full with respect to
R
pE
are
P
actually invertible. For, if over
RP,
a:Q1 -).Q2
there is an integer such that
a ® In:Ql ® Rn -> Q2 $ Rp P
is a full map with respect to Q. $ Rn = RP®RP.
pE
and
is still a full map with respect to
pE
but this
78
time it is a full map between induced projectives. By Cramer's rule, we have an equation:
where (O
is a full map over
(Sg1)
O
is full with respect to p
I
over
R ,
which shows that
a
R
and
is defined over
(BB')
since
R;
pEmust be full and so it is invertible itself is invertible.
P
Next, we show that the descending chain condition holds on f.g. left ideals over
of bounded generating number with respect to
RP
p
;
E
clearly, this implies that
is right perfect, and the left perfect
R P
condition will follow by symmetry. If there are any infinite strictly descending chains of f.g. left ideals of bounded generating number, we choose one
10 Z) I1 Z) ....
where all
modules have the same generating number, and this is minimal for such a descending chain to exist. We find f.g. projectives with surjections such that
P. -> I.
is the generating number of
pE(Pi)
Ii,
and construct
the diagram below using projectivity.
I0
D
11
D
IZ
a1
p2
......
1
0. a0
Pf
Since
Ii x Ii+l'
the map we construct from
Pi+l
to
P.
cannot be full since it is not even surjective so there is a projective such that in
I.
pE(Qi) < pE(Pi)
is a left ideal
less than that of
Ii,
Ii
and
ai
factors through
containing
Ii+1
Qi;
the image of
Q, i
Q,
of generating number strictly
which is contradictory since we may choose a strictly
descending chain of left ideals from the sequence
{I'}
such that all modules
1
have the same generating number less than that of the sequence
{Ii}.
As noted before, this argument shows by symmetry that
R P
also left perfect, so our theorem is complete.
is
79
Constructing simple artinian universal localisations Most of the time, we shall be able to deduce from this last result that
is actually simple artinian as we shall see in the next two
R P
theorems. First of all, however, we give an example of a ring with a faithful Sylvester projective rank function that cannot be embedded in a simple artinian ring at all. We consider the endomorphism ring of the abelian group where Cn is the cyclic group of order n. This ring has two nonp trivial idempotents corresponding to projection on C 2 and C ; we assign C 2 ® C
the ranks
2
P
to the first, and
3
to the second idempotent. It is a fairly
3
simple matter to check that this gives a Sylvester projective rank function on the ring. It is clear that it cannot be embedded in a simple artinian ring.
The next theorem was proven in a rather different way in (Dicks, Sontag 78); we present it here as a simple corollary of the last theorem.
Theorem 5.4
Let
R
taking values in
be a ring with a Sylvester projective rank function then
72;
R
is a skew field and the map
R i R
P
p
is P
honest.
Proof:
We know that R
is a perfect ring with a faithful Sylvester prop
jective rank function RP
such that
must be
0,
ab = 0;
pE
taking values in
let
72;
then by the law of nullity
which implies that
a
or
a,b pE(a)
b must be zero.
be elements of or else, So
PE(b)
is a domain
R P
and a perfect ring which forces it to be a skew field.
This is the classical case, and it is quite sensible to recall at this point some of the examples to which it applies; all our examples are firs, which have the unique rank function on the f.g. projectives since these are free of unique rank. The free algebra on a set one sees by writing it as the ring coproduct X.
X
We generalise this slightly to the ring
X, k<X>,
k[xi] E<X>
as
is a fir, as
runs through
xi
generated freely by a set
of E-centralising elements; this is a fir by the same argument. There is
a further generalisation of this example; let the skew field
E;
M
be an
E,E
bimodule over
we may form the tensor ring on the bimodule
which is a graded ring whose nth grade has the form
n Q M.
M,
E<M>,
It is shown in
(Cohn 71) that this is a fir. The unique universal localisation that is a
80
EX),
skew field for these firs are written as
and
Our next theorem gives different circumstances under which
R P
is forced to be a simple artinian ring; this will be exactly what we shall need to show that the simple artinian coproduct exists.
Theorem 5.5
Let R be a ring with a unique projective rank function
Assume that
p
precisely 1 z;
then the universal localisation of
n
simple artinian ring of the form
We know that
Proof:
p.
is a Sylvester projective rank function whose image is
where
Mn(F),
R
at
R
p,
is a
p
is a skew field.
F
is a perfect ring such that all f.g. projectives
R P
are stably induced from
so any rank function on
R;
R
is determined by P
its values on the image of
There is a unique rank function on
K0(R).
N
If
so there is a unique rank function on R.
R,
is the nil radical of
R ,
has a unique rank we know that R IN is semisimple artinian; since R p p function, RP/N must actually be simple artinian, and since the image of
the rank function is precisely 1 Z,
and all projectives in
projectives over R,
M (D) for some skew field D. The n so RP Mn(R'), where by Morita
n
equivalence,
R'
rank function R'
must be
R /N
p
p
matrix units lift from
RP/N
to
RP,
R /N
lift to
p
is a perfect ring with a faithful Sylvester projective
taking values in Z,
so, by the same argument as for 5.4,
is a skew field, which completes the proof.
This theorem applies well to the ring coproduct of simple artinian rings amalgamating a common simple artinian subring.
Theorem 5.6
Let
S1
and
be a couple of simple artinian rings with
S2
common simple artinian subring a unique rank function p.
the simple artinian coproduct of Si
Mn (Di),
i
for skew fields
Si
Sl Di,
(S1
and (S1
and
D
is an hereditary ring with
52)p
is a simple artinian ring,
S
amalgamating
S2,
M(D),
S2)
S.
If
where
is a skew field.
From 3.1, we know that
Mn (Di),
the image of
p
has a unique rank function
S1 S S2
is precisely
i
(S1
S1 S S2
P
n = l.c.m, fnl,n2},
Proof:
Then
S.
Therefore,
52)p = Mn(D)
for some skew field
D.
n Z
so, by 5.5,
p;
if
81
Following Cohn's notation and terminology for the skew field coproduct, we call this ring the simple artinian coproduct of amalgamating
S
S1 S S2.
and we denote it by the symbol
and
S1
S2
Of course, we may
construct the simple artinian coproduct of finitely many simple artinian rings,
Si,
amalgamating a common simple artinian ring in exactly the same
way; for infinitely many
Si,
we have to be more careful. In this case, the
last theorem is proved in the same way provided that
l.c.m.{n
exists,
}
i
where
S. = Mn (Di);
at
exists but it will be a von Neumann regular ring, that is not simple
when this does not exist, our universal localisation
i p
artinian.
Another generalisation of 5.6 that it is worth mentioning at this point occurs if we amalgamate a semisimple artinian ring rather than a simple artinian ring; by 3.1, we know that if embeddings of the semisimple artinian ring
a: '0 i Si, i = 1,2,
are
in simple artinian rings
RO
RO S2 has a rank function if and only if al and a2 induce the same rank function on RO. When there is a rank function, it is unique and we S1
S2 S1 at this RO rank function is a simple artinian ring, which we shall call the simple
may apply 5.5 to show that the universal localisation of
artinian coproduct of S1
S2.
S1
and
amalgamating
S2
R0,
and write as
It is important to bear in mind that this exists only if the
same rank function is induced on R
0
by the maps from
RO to
S1
and
S2.
Given a ring with a Sylvester projective rank function to 1 Z, n
we have already seen that it need not arise from a map to a simple artinian ring. However, if our ring is a k-algebra, we can show that there is an honest map to a simple artinian ring inducing the given projective rank function.
Theorem 5.7 p
to 1 Z; n
Let
R be a k-algebra with a Sylvester projective rank function
then there is an honest homomorphism from
artinian ring inducing the rank function
Proof:
We saw in 3.1 that
Mn(k) k R,
p
R
to a simple
p.
extends to a projective rank function
which still takes values in
Z;
on
p
since the rank functions on
the factor rings are Sylvester, that on the n coproduct must also be Sylvester by 3.7, and the map
R ; Mn(k) k R
is honest by 3.6.
Mn(k) K R = Mn(R')
and the rank function on the coproduct induces a rank function
which takes values in Z. R'
to a skew field
F
p'
on
R'
Consequently, by 5.4, we have a homomorphism from
that is honest and induces the rank function
p'
on
82
R'.
We have a map
which is the composite of honest maps and must
R + Mn(F)
be honest.
R
In particular, we consider when
is a k-algebra for a ring P
R with a Sylvester rank function there is an embedding of
RP
in
p
taking values in
Mn(F)
we see that
$;
for some skew field
R must be nilpotent of class at most
radical of
n 1
so the nil
F,
n.
Intermediate universal localisations Before leaving these matters for the time being, we shall develop a few results on intermediate localisations of a ring
R
full with respect to a Sylvester projective rank function
at some set of maps p.
We wish to
have useful criteria for the embedding of such rings in the complete localisation at all the full maps. This question is not immediately important, so the reader may wish to skip these results until they are referred to later on.
and
is a ring with a Sylvester projective rank function
R
If
is a collection of full maps with respect to
E
RP must be the universal localisation of if and only if
R
pE
at
RE
pE.
p,
So
p,
it is clear that embeds in
RE
is a faithful projective rank function. There is a
P
useful way of determining when this happens. We define a set of maps be factor closed, if any full left factor of an element of in
Let
taking values in
RE,
then
R
be a ring with a Sylvester projective rank function Z;
embeds n in
RE
let RP
E
the extension of
pE,
to
p
is a faithful projective rank function, which is true if and only if the
RP
Proof:
E,
E,
is factor closed. The image of
is always a universal localisation of
We have already seen the first equivalence. Next, suppose that
are full maps
S:P 3 P', y:P' + P".
REMR ' = REMRP ® coker(REQRS).
Over
RE
R.
is a faithful projective rank function and let
so,
p
be a collection of maps full with respect to
if and only if
lower multiplicative closure of in
is invertible
E
RE.
Theorem 5.8
p;
to
E
RE, $
a e E, a = By
where
pE
$,y
has a right inverse, and
pE(coker(REaR$)) = 0
so it must itself
be zero if the rank function is faithful, and, in this case,
S
has an
inverse.
Conversely, suppose that
E
is factor closed, and let
e
be an
83
idempotent such that
I -I o
(aa1)
$
I
(a al)
where
pE(e) = 0;
0
Y
(aa') = 6(YY)
:1) = (6 al) \O
\
is a full left factor of
(Sai)
invertible over
RE;
where the co-
we have an equation:
a;
Y Y')
I
I/
O
that
of the form
(aa')
has the same rank as the codomain of
B
(B al)
where
and we must have a
pE(e) = 0, p(aa') = p(a),
minimal factorisation of domain of
_ (a a')
e
since
a E;
l
we have an equation by Cramer's rule:
-
(aal)
cancelling it over
O
/
and so, by assumption, is shows that
RE
e = 0,
which shows
is a faithful projective rank function.
pE
We are left with the last sentence of the theorem; the image of RE
in
is obtained by killing the f.g. projectives of rank zero, which
RP
is the universal localisation of
RE
at idempotent matrices, whose kernels
are isomorphic to f.g. projectives of rank localisation of
O.
By 4.6, this is a universal
R.
It is clear that if
R -> S
is a ring homomorphism from
R
a simple artinian ring inducing a Sylvester projective rank function the set of maps
E
between f.g. projectives over
factor closed and saturated; therefore
RE
R
to
embeds in
R
then
p,
invertible over
is
S
by the last theorem P
This remark gives us a useful sufficient condition for a localisation of a ring at a set of maps full with respect to a Sylvester projective rank function to embed in the complete localisation.
Lemma 5.9
Let
E
be a collection of full maps with respect to the Sylvester
projective rank function
p
over the ring
R,
and let
RE - S
be an
embedding into a simple artinian ring such that the composite map from to
S
induces
Let
Proof:
over
S;
E
p.
Then
RE
embeds in
R
Rp.
be the collection of maps over
then we have the diagram of ring maps:
R
that become invertible
84
R-> R£-RRE -+ R p
S
The map from in
RE
to
S
.
is an embedding, so
embeds in
RE
which embeds
R£
R P
We can simplify 5.8 in the case that
is a weakly semi-
R
hereditary ring, since in this case, it is easily seen that a set of maps full with respect to a projective rank function are factor closed if and only if their lower multiplicative closure is factor closed.
Theorem 5.10
Let
taking values in p;
then
RE
R be a weakly semihereditary ring with rank function let
Z;
embeds n in
RP
p
be a collection of maps full with respect to
E
if and only if
Proof: It follows from 1.19 that lower multiplicative closure of
£ E
E
is factor closed.
is factor closed if and only if the is factor closed.
5.8 completes the
proof.
Finally, in the case of a faithful rank function on a two sided hereditary ring, we know by 1.23 that we have unique factorisation of full maps into finitely many atomic full maps. Consequently, a universal localisation at a factor closed set of maps is entirely determined by the full maps that are factors of elements of the factor closed set. Further, if
El,E2
are collections of atomic full maps such that any element of one is stably associated to an element of the other,
R
El
=- R E2.
So, if we wish to classify
all the intermediate localisations that embed in the complete one, we can do so by the sets of stable association classes of atomic maps that are exactly those inverted in some universal localisation. We shall show next that any collection is possible.
Theorem 5.11 function R
p
Let
R
be an hereditary ring with faithful projective rank
taking values in 1 Z; n
then the intermediate localisations of
that embed in the universal localisation
RP
are in 1 to 1 correspondence
with collections of stable association classes of atomic full maps.
Proof: We have shown everything except for the result that if
E
is a
85
collection of atoms then any atom invertible in an element of
is stably associated to
RE
In order to deal with this, we look at the maps we defined
E.
in the proof of the K-theory exact sequence of chapter 4 for the universal
localisation R
of
R.
Let
T
be the category of torsion modules with
respect to this complete localisation; we have a well-defined map from isomorphisms between induced f.g. projectives over
RP
to
and
K0(T)
K0(T)
is the free abelian group on the set of stable association classes of atomic full maps, and our well-defined map sends a full map
exists in
R
[coker S].
to
a
Suppose that we have an atomic full map
a-1
such that
a:P -. Q
then, we have an equation of the Cramer rule type,
;
E
a(a-1)
which shows that the image of
= I,
However, by Cramer's rule for
RE,
a-1
is [- coker a].
KO(T)
in
we have an equation:
a a
where
lies in the lower multiplicative closure of
B
full map. So, the image of
a-1
since our map is well-defined,
is
E,
and
[coker g'] - [coker S].
B'
is a
Therefore,
[coker a] + [coker a'] = [coker s]. Since
each side is a positive sum of generators of a free abelian group, and is a generator, it equals one of the generators involved in the
[coker a]
right hand side, which all have the form a
is stably associated to an element of
for
[coker ai]
ai
in
so,
E;
E.
In the course of the proof of the theorems in this chapter, we have often tried to prove that all the f.g. projectives over a universal localisation are stably induced; it is often useful to be able to show that all f.g. projectives are actually induced; we end this chapter with a consideration of this problem. Let
and let
E
be a collection of maps between f.g. projectives full with
respect to p; if
a
exists
be a ring with a Sylvester projective rank function p
R
and a',
S
we say that
E
is factor complete if it is factor closed and
are maps defined over
R
such that
as
a map between induced f.g. projectives over
is invertible over
lies in RE,
E,
there
such that
RE.
-/ Theorem 5.12
Let
R
be a ring with a Sylvester projective rank function
p;
86
let
be a lower multiplicatively closed factor complete set of maps between
E
f.g. projectives; then all f.g. projectives over
Proof: Let
be an idempotent in
e
RE
are induced from
R.
Mn(RE); by Cramer's rule, there exists
an equation:
IP (a a1)
1
e) = (a
O
where
(a a1)
a')
and
(a a') :PO -> P e Rn
a E,
is defined over
(a a')
R,
(a a'):PO ->P eRn. Let
y:PO -*-Q
and
be a minimal factorisation over
(a a') = y(6 6')
(6 6'):Q } P e Rn; then (y a1) \6
factor completeness, there exists Y al is invertible. E1 E2) Counting ranks shows that p(Q) = p(P) + pE(e);
(Cl
R;
0) = (a al) so, by
E2):RE&P'\ 3 REEORQ a R£
p(P') = p(Q) + n - p(P0)
such that
and also that
p(P') = pE(e) + p(P) + n - p(P0) = pE(e).
therefore,
We intend to show that there is a map from
REaRP'
onto
Re,
and hence that they must be isomorphic since there are no projectives of rank
0
by the factor closure of
E.
First we rewrite 1 as:
(y a)
6)
l
e=
I,/\0
(O
(y
I
0
6
a
6'
(6 0
O
so
(y a ) 1
kI
P
(0
6S
(O
e
v) 0
\
E1
(y al)
-O
EZ
(O
(0
(O
I
O
e
68e
e)=
=e
0
87
On the one hand, (e
l
\1
2
1
REaR(PO ® P') to
REORP'
E2)
bsl
l
al
(O
E1
O
has image
e)
Rne
since
is invertible; on the other hand, it is a map from
vanishing on
Rne,
I )(0
RE'RPO;
so it defines a surjective map from
which completes our proof.
We may refine this a little in the case where
R
is weakly
semihereditary.
Theorem 5.13 let
p;
over E, E,
Let
R
be weakly semihereditary with a projective rank function
be a factor complete collection of maps between f.g. projectives
E
full with respect to
R
p;
then the lower multiplicative closure of
is also factor complete and so, all f.g. projectives over
induced from
R.
It is sufficient to show that if
Proof:
E2 = {(°l a
are
RE
I
:
ai a E}
is factor complete then
E
is also factor complete; the rest follows by
2
induction. So, suppose that
(Y21(61 62) _ s( 1
J,
the weak semihereditary property codom(y1) = dom(62)
01 then 1162 = 0,
there exists a decomposition of
so that the product
1162
is trivially
the above equation with respect to this partition of
(Y21
Y22) (a'1211
so by
622/
\S1
By factor completeness, there exists Yll
0
Y21
Y22
Y '
0
y'
0;
we rewrite
codom(y1):
a2/
and
y"
such that
rY11) 1\
Y
and
/I
(122)
are invertible: then
0
is invertible.
Y"
Finally, there is the following special case that we shall need later on.
Theorem 5.14
Let
R be a hereditary ring with a faithful projective rank
function
let
a:PO > P1
p;
be an atomic full map with respect to
p
such
88
that if
then
ima c Q c P
ima
is a direct summand of
Q;
then all f.g.
1
projectives over
Proof:
Ra
are induced from
It is enough to show that
{a}
R.
is factor complete; it is certainly
factor closed, and the remaining part of the definition of factor completeness is trivial.
89
6. HOMOMORPHISMS FROM HEREDITARY TO VON NEUMANN REGULAR RINGS
In the last chapter, we investigated the universal localisation of a ring at a Sylvester rank function taking values in
!-Z; n
the resulting
ring is always a perfect ring. Since the degree of nilpotence of the radical may grow with the integer n, it is likely that if the rank function takes values in the real numbers, we are not going to be able to say a great deal about the universal localisation at the rank function in general. This suggests that rather than investigating epimorphisms we should investigate homomorphisms, and we shall see that every rank function on an hereditary ring arises from a homomorphism to a von Neumann regular ring with a rank function. Under suitable hypotheses we shall be able to show that the universal localisation at the rank function is von Neumann regular. In chapter 1, we showed that over a two-sided X,-hereditary ring
with rank function
p,
every map factors as a right full followed by a left
full map; we described this as having enough right and left full maps. We shall require a complementary condition; we say that maps with respect to a Sylvester rank function
p
R has enough full
if every left full map is
a left factor of a full map and every right full map is a right factor of a full map.
Theorem 6.1
Let
R
be a ring with a Sylvester rank function
such that
p
R has enough left full, right full and full maps with respect to the universal localisation of
R
at
p,
R
,
Then
p.
is a von Neumann regular ring.
P
All f.g. projectives over function extends to
RP.
R
are stably induced from
R
and the rank
The kernel of the homomorphism from
R
to
RP
is
the trace ideal of the f.g. projectives of rank zero.
We have seen all of this result in 5.1 and 5.2, except for the fact
Proof:
that
is a von Neumann regular ring. To show this, we need to show that
R P
every principal left ideal of
RP
is a direct summand of
RP.
Let
a E RP
90
then right multiplication by
a:P - Q
where and
is left full. Since
y
is stably associated to some map
a
is a map defined over
where
a = Ry,
R.
RP®Ra,
is right full
a
is right full, it is a right factor of some
$
full map which has an inverse over
has a left inverse,
Rp0RB
so
Rp,
that is, it is a split surjection. Similarly,
R
has a right inverse,
db Ry
P
and must be a split injection. The consequence of this is that the image of is a direct summand of
RPQRa
RpQRQ,
and so, the cokernel of
projective. This is the cokernel of right multiplication by is a direct summand of
R a
is f.g.
however, so
as we wished to show.
R
P
RP®Ra
a,
P
Of course, it may be hard to check whether there are enough full
maps; however, we have already seen that over an
Xo-hereditary ring there
are enough right and left full maps. We should like to be able to embed any k-algebra with a Sylvester rank function
honestly in another k-algebra
p
with a Sylvester rank function such that the second k-algebra has enough full maps. It turns out that we can do this by adjoining a large number of generic maps.
Theorem 6.2
Let
R be a k-algebra with a Sylvester rank function
values in the reals; then the embedding of
R
an infinite set is an honest map for the rank functions
on the coproduct
(p,r)
on
k<X>.
Further,
where
R v k<X>,
r
for
P
of
R
X-Y
k,
p ,
defined over
where
R t k<X>.
to define a generic map from
Q
to
for
p
(p,r).
We may use the elements
X.
over
P
R k k<X>
and the
should be a full map with respect to
a
The
Then it is actually defined
is a finite subset of
Y
composition of this with
and
a:P -+ Q be a left full map
idea of the proof is a fairly simple one; let
over
(p,r).
we take liberties with
R,
k<X>)ORP,
(R
is
and
R,
R k k<X> has enough full maps with respect to
the notation by writing
with respect to
on
X
is the unique rank function
Since all projectives are induced from
Proof:
p
taking
p
where
R V k<X>,
in
p,
since
any other answer would involve some kind of degeneracy. A dual argument deals with right full maps. There is some integer
n
such that both
generator projectives, and neither is of rank there are idempotents
eP, eQ
so that our left full map
in
a:P -> Q
Mn(R)
n
P
and
are
Q
with respect to
such that
RneP = P,
and
may be represented by a matrix
p;
re-
so
RneQ
Q
91
such that
a E Mn(R k k)
For convenience in writing the
epa = a = aeQ.
proof, we pass by Morita equivalence to the ring
k
Mn(R k k)
the map sends
k[x] = Mn(R k k
to the matrix
x
k<xij>),
k
i,j = 1
where
We call the ring
(mi,).
and
Mn(R u k)Rl,
13
and
R1 k k[x]
Mn(R of
This may be regarded as a subring of
S.
by identifying the set
k<X>)
k
is called
with some subset
{xi.: i,j = 1 -* n) j
The rank function
X - Y.
function
on Mn(R k k<X>)
pn
shall still call
on
pn
R1
induces by Morita equivalence, a rank
p
which in turn induces rank functions that we and
and all maps mentioned above are
S,
honest since all maps represent one ring as a factor in a coproduct over pn(R1ep), pn(R1eQ) < 1
of the other.
must be respectively
p(P)/n
left full map a:P i Q map with respect to R1eQ a
R1
The element
a
represents the
so that right multiplication by
a
is a left full
pn
from
p(Q)/n.
R e
defines a left full map from containing
pn(I)
a,
Over the ring
map from
and since the embedding of
R1eQ,
to
1 P
is split injective and so left full, right multiplication by
R1
in
and
k
since, by Morita equivalence, they
R1
to
R e I
R e 1 P >_ pn(R1eP).
to
so for all left ideals of
R1;
the map
S = Rl k k[x]
so we consider the map
P
which we intend to show is a full map over that it defines a full map over
Mn(R
k
is a kind of generic from
by our remark that the inclu-
sion is honest, and so the map Morita equivalent to it over full map having
a:P -
Q
to itself,
SeP
once we have this, we know
S;
k<X>)
xeP
axeP
is a
R k k<X>
as a left factor. In order to deal with this, we
shall need the details of the coproduct theorems.
What we need to show is that any f.g. left ideal inside SeP containing axeP pn.
M
Let
has generating number at least
be such a submodule of
SeP;
with respect to
pn(Sep)
then, if
we find that in the decomposition given by 2.7,
gpn(M)
<_ P(Se) < 1,
m = SokMO ® SoR M1 ® sok[x]M2' 1
where we take MO
and
RO
must be
M2
greater than
1
of 2.7 to be 0,
by 3.3. Therefore,
We choose the basis 1<x<x2<.... element
module of
;
k,
R1
to be
and
RI
M
k[x],
M would be
M = SoR M1.
l,x,x2...... for
we well-order some basis for
R1
and order it by
k[x]
over
as smallest element. Then by theorem 2.7,
ep
to be
R2
for if not the generating number of
containing the
k
M1
is the
R1-sub-
consisting of the elements whose 1-support does not contain
the 1-leading term of some non-l-pure element of
M;
if
axeP E Mi,
its
92
1-support
must be the 1-leading term of some non-l-pure element of
xeP
and the nature of the ordering of be
where
xeP + rl
lies in
r1
and we see that
in the 2-support of
xeP + rl
element of
is empty, so
M2
However,
must lie in
eP
For, some element
M.
the element must actually be
eP,
to be the smallest element of a basis of
eP
xeP + rl
must be the 2-leading term of some non-2-pure
If this 2-leading term is
M.
since we chose
eP
forces the form of this element to
k[x]
R1eP.
does not lie in M2
M,
the contrary case we reduce the support of
obtaining
r1
In
R1.
xeP + r2.
The
same argument applies but we cannot produce an infinite sequence of terms
xeP + rn
where the maximal element in the support of
rn
continues to
decrease since our ordering is a well-ordering, so eventually to be in
gpn(M) = pn(Sep).
and therefore
M,
This leaves the case that theorem 2.6 that the structure of R e
1 P
0 R1xeP 0 B,
direct summand
where
R1xeP
to
R1;
R1
pn(R1eP) = pn(SeP),
axeP
M
M1
a
M1
onto the
M1
R1xep
contains
ideal containing since
so the generating number of
so that
we recall from
M1;
as an R1-module has the form
Sep
the image of
xeP;
3.3 the generating number of pn(Sep),
lies in
is a basic module. We project
B
already noted that any left at least
axeP
according to this decomposition.
module on the generator
is forced
eP
is a free
we have
axeP;
has generating number
a
defines a left full map from is at least
is at least
pn(R1ep)
pn(R1eP),
R1ep
and so by
which is just
must be a full map, as we wished to show. R k k<X>
This shows that every left full map over
is a left
factor of a full map; the dual result must hold for right full maps by a dual argument.
An immediate corollary of 6.1 and 6.2 is the following theorem.
Theorem 6.3
Let
R
be an Xo-hereditary k-algebra with a rank function
taking values in the real numbers; then there is an honest map from von Neumann regular ring
V
with a rank function
p
R
p
to a
v.
First of all, we form the ring coproduct R k<X> which is Xak hereditary by the remarks after 2.10, so, by 1.16, it has enough right and Proof:
left full maps; by 6.2, it has enough full maps. Therefore, by 6.1, the universal localisation of Neumann regular ring
V.
R k k<X>
at the rank function
The map R + R k k<X>
p
and the map
are both honest, so their composite is also honest.
is a von
R k k<X> 3 V
93
We can say a little about the monoid of f.g. projectives of
V,
when it is constructed in the manner used in 6.3.
Theorem 6.4
Let
be a ring with a Sylvester rank function
R
image is the subgroup localisation of then
of the additive reals. Let at the rank function
be the universal
V
extended to
p
is naturally identified with the positive cone of
P®(V)
Proof:
A
R k k<X>
Certainly, the image of
on V
p
by theorem 5.1, which is the image of a map from
P®(V)
p
is the image of on
to the positive cone of
R.
whose
p
R k k<X>; A.
on R k k<X>
p
The rank function induces
which we wish to show is an
A,
isomorphism.
Suppose that pV(P) <_ pV(Q);
n
and
are f.g. projectives over
Q
then there are idempotents
is a suitable integer larger than
VneQ
of
P
Q.
eP
and
eQ must lie in
and
eP
eQ
in
Vne P
such that
pV(P)
(R k k)p
V
such that where
Mn(V)
and
P,
for some finite subset
Y
X. We look at the ring
S = Mn(R k k)
kix7
= Mn((R k k)
k
inclusion of it in X - Y.
Mn(V)
The map from
Sep
is honest by taking to
SeQ
pV(P) < PV (Q).
such that the
Mn(V)
to be a subset of
{xij}
given by right multiplication by
is seen to be a full map by the argument of 6.2, if left full if
as in the last
k<xi.>)
k theorem, which may be defined to be a subring of
pV(P) = pV(Q);
Full maps become isomorphisms over
left full maps become split injective, so we see that if projectives of the same rank over rank less than that of
Q
shows that the map from since all elements of
A
V,
and
Q
they are isomorphic, whilst if
it is a direct summand of P®(V)
P
V,
A
and are f.g. P
has
is injective;
are differences of elements of the image of
the second statement shows that the map is surjective.
it is
The first statement
Q.
to the positive cone of
ePxeQ
P®(V),
94
7. HOMOMORPHISMS FROM RINGS TO SIMPLE ARTINIAN RINGS
Introduction Now that we have studied special homomorphisms from hereditary rings to simple artinian rings, we are in a good position to study all possible homomorphisms from an arbitrary ring to simple artinian rings. In order to see what type of theory to look for, we should look at the special case of homomorphisms from an arbitrary ring to skew fields, which were classified by Cohn in chapter 7 of (Cohn 71). First of all, given a homomorphism a skew field
F,
then we regard two homomorphisms skew subfields of
F,
i
$.:R -
from a ring
4:R + F
we can talk of the skew subfield of F., i = 1,2,
R
generated by
F
to R;
as equivalent if the
are isomorphic as R-rings. We should like to be able
to characterise the equivalence classes in some way; this we do by using the notion of the singular kernel of the homomorphism of square matrices over P
1/
R
that are singular over
Such a set of matrices
F.
must satisfy the following axioms;
it includes all non-full matrices; these are the
n
arbitrary
n,
that can be written as the product of an
an
(n-1)
by
n
2/
1 j P;
3/
(A
4/
if B)A = (a..)
e P
if and only if
where
aij = bij
by
matrices for
n
by
n
(n-1)
and
matrix;
and
A
B = (b..),
or
B E P;
A,B E P,
1]
1]
(cij) e P if
this is the set
¢:R + F;
cij = aij
j x k,
then
for
and
a.. = b 1)
i x k,
(dij) e P,
and
where
i x k,
dij = aij
then
i]
ckj = akj + bk,; for
similarly,
j x k
and
cik = aik + bik.
Any set of matrices over a ring satisfying axioms 1 to 4 is called a prime matrix ideal; Cohn showed that the equivalence classes of homomorphisms
95
from a ring to skew fields are in 1 to 1 correspondence to prime matrix ideals, where an equivalence class is paired with the associated singular kernel. We shall present a proof of these results later in this chapter. There are two problems involved in a generalisation of this theory: first of all, we cannot talk of the simple artinian subring of a simple artinian ring generated by a subring since there need not be a unique minimal simple artinian ring containing a given subring; therefore, we shall have to find a new way to define an equivalence relation on homomorphisms from a ring to simple artinian rings; secondly, we must find some analogue of the prime matrix ideals that applies to simple artinian rings and not just to skew fields.
The first problem can be met in a fairly simple way; Bergman proposed that two homomorphisms
f.:R -> S., i = 1,2,
where
is simple
S.
artinian, should be regarded as equivalent if there is a commutative diagram:
is simple artinian. If there is such a commutative diagram, we
where
S
write
¢1 - 2. This relation reduces to the standard one if
Si
is a skew
field. This is easily shown to be an equivalence relation using the simple artinian coproduct.
Lemma 7.1
Proof:
- is an equivalence relation.
Certainly, it is reflexive and symmetric by definition. If we have
homomorphisms
4i:R -
Si, i = 1 to 3,
and
a commutative diagram of ring homomorphisms:
01 - 2' '2 - 031
we construct
96
R
-S
S
1
S O S, S2
from which it is clear that -
is also a transitive relation.
Next, we have to see what concept should replace the prime matrix ideal. Here, we use an idea due, in the case of homomorphisms from a ring to
skew fields, to Malcolmson (80). If we have a homomorphism :R + S = Mn(E), where
E
is a skew field, we can define a rank on f.p. modules over
the formula
p(M) = ps(MeRS),
which takes values in
R
by
Such a rank
Z.
function must satisfy the following axioms, which simply n express that
SRS
is a right exact functor:
= 1;
1/
P
2/
p(A ® B) = p(A) + p(B)
3/
if A - B - C - 0 is an exact sequence,
(R1)
p (C)
<_
p (B)
<_ p (A) + p (C) .
A rank function on f.p. modules satisfying these axioms is called a Sylvester module rank function. Malcolmson (80) showed that a Sylvester module rank
function taking values in Z
is equivalent information on the ring to a
prime matrix ideal. The main theorem of this chapter is that the equivalence classes of homomorphisms from a k-algebra to simple artinian rings are in 1
to 1 correspondence Sylvester module rank functions taking values in 1 Z n
for some
n.
In order to prove this theorem and also to show how the notion of a prime matrix ideal links with the concept of a Sylvester module rank function, we need another type of rank. A homomorphism from a ring artinian ring if
a:P - Q
S
is a map between f.g. projectives over
p(a) = p5(a&RS)
is the rank as
R
to a simple
also determines a rank on maps between f.g. projectives;
where
pS(S)
for a map
a
R,
we define
between f.g. modules over
S
S-module of the image. A rank on maps induced by a homomorphism
to a simple artinian ring must satisfy the following axioms:
97
1/
P(I1) = 1;
2/
P(G S) = p(a) + P(R);
3/
P(a R) > P(a) + P(S);
4/
P(aa) 5 p(a), P(R)
A rank function on maps satisfying these axioms is called a Sylvester map rank function.
The notion of a Sylvester map rank function is equivalent to that of a Sylvester module rank function in the following way. First, we suppose that we have a Sylvester map rank function,
a Sylvester module rank function by a:P + Q; p,
p(a)
p;
we extend this to
p(coker a) = p(IQ) - p(a)
for a map
next, we suppose that we have a Sylvester module rank function,
from which we define a Sylvester map rank function by = p(Q) - p(coker a)
for a map
a:P + Q.
It is not hard to check that
these functions are well-defined and satisfy the axioms required. As we have seen, Sylvester module and map rank functions are equivalent notions; we shall therefore usually refer to a Sylvester rank function,
p,
which is
defined on f.p. modules and on maps between f.g. projectives and restricts respectively to a Sylvester module and a Sylvester map rank function. In certain situations, we shall be able to construct a rank
function on matrices taking values in 1 Z n
for some
n
that satisfies axioms
1 to 4 for a Sylvester map rank function; we shall call such a function a Sylvester matrix rank function. There is the following useful observation.
Lemma 7.2
A Sylvester matrix rank function extends uniquely to a Sylvester
map rank function, having values in the same subset of the additive group
of R. Proof:
Given a Sylvester matrix rank function,
module rank function by
p(coker
A) = n - p(A),
p,
we define a Sylvester
where
A:mR + nR.
That
this defines a Sylvester module rank function is the same proof as was needed to show that a Sylvester map rank function determines a Sylvester module rank function. In turn, the Sylvester module rank function determines a Sylvester map rank function that extends the Sylvester matrix rank function.
This allows us to see the equivalence between prime matrix ideals
and Sylvester rank functions taking values in Z
in the following way. We
98
construct from the prime matrix ideal, by defining
p(A) = n,
tains an
by
n
where
n
minor not in
n
a Sylvester matrix rank function
P,
A
is the maximal integer such that
con-
this determines a Sylvester rank func-
P;
tion as we discussed before. Conversely, if we have a Sylvester rank function, p,
which takes values in Z
P = {A: p(A) < n,
we may define a prime matrix ideal by
A
where
is an
by
n
n
matrix}. We shall leave the
checking of the details to the reader.
Characterising the homomorphism by the rank function We take our first step by describing our previous equivalence relation on homomorphisms to simple artinian rings by the associated Sylvester map rank functions. In order to do this, we need to find out more exactly what a Sylvester map rank function induced by a homomorphism from simple artinian ring
S
tells us about the functor
R
to a
QRS.
We have already seen that a Sylvester map rank function determines a Sylvester module rank function, and it is clear that this is the Sylvester
module rank function induced by the homomorphism from
R with presentation
is an f.p. module over
R
P a Q + M -
for if
M
to
S,
0,
then denoting
the Sylvester map and module rank functions induced by the homomorphism from R
to
by
S
p,
we have the equation
further determine the rank of R
M&RS
p(a) + p(M) = p(Q) = p(IQ).
for an arbitrary f.g. module
can be represented as the direct limit of a family of f.p. modules
minimal rank of some
NiaRS,
rank of
anRS
by the formula
if we have a map n RS
over
N
a:M -
N,
M
agRS
from M
N
Finally,
as the directed union of all the
that contain the image of
to
p(Ni).
is f.g., we can determine the rank
is the minimal rank of the maps from a
{Ni}
we may determine the
a, N = VN.; 7
induced by
M
is equal to the
p(agRS) = p(coker agRS) = p(N®RS).
a:M 3 N where only
in the following way; we write
f.g. submodules of
M&RS
that is, the minimal value of
Given a map between f.g. modules
of
M
in terms of the Sylvester module rank function. For every f.g. module
where all the maps are surjective, and the rank of
of
We can
M
to
N.
then the rank
J
for varying
j
N.
We see that if two homomorphisms
+1:R i S1
and 2:R ; S2
induce the same Sylvester map rank function, then they must agree numerically in all the ways determined above; as we shall see this happens because there is a commutative diagram of ring homomorphisms:
99
S1
R
S
3
S2
where
is simple artinian.
S3
Theorem 7.3
R be a ring with two homomorphisms
Let from
2:R -)- S2
R
to artinian rings
S1
and
S2
¢1:R -)- S1
and
that induce the same
Sylvester map rank functions; then there is a commutative diagram:
where
Proof:
is simple artinian. Conversely, if there is such a commutative
S3
diagram,
and
S1
S2
induce the same Sylvester map rank function on
R.
The last sentence is clear, since they must both induce the same
Sylvester map rank function as
S3.
We wish to find a non-zero homomorphism from the ring
S1 '9 S2
to a simple artinian ring. At first sight, this appears a hopeless task until one notices that this is an hereditary ring, as will become clear in a moment. After that, it is simply a matter of showing that the ring has unbounded generating number.
In order to study the ring, we consider the upper triangular matrix ring
T = S1 O
S1aRS21 S2
If we adjoin the universal inverse to the map
to
defined by left multiplication by /0
IS1 1\O
(0
O
SlaRS2/
we obtain is isomorphic to S1 R S2
M2(S1 R S2),
from
a
laR1)
0
0
0
S2
the ring
O
as we saw in 4.10. Consequently,
is hereditary, and by 5.1, it has a rank function (which must be
unique) when this map is full with respect to the rank function on
T,
p,
100
that assigns the rank
/
to both
S1%'2`.
and
/00
0S
happens, the universal localisation of
(
1/
0
If this
functi/on
at the` rank
T a
is simple
artinian by 5.5; but this is a universal localisation of
M2(Sl R S2); has a simple artinian universal localisation. So, it
S1 R S2
therefore,
remains to show that
is a full map with respect to
a
S1%S2
First of all, we show
YR 4- S1
p.
is not zero. The rank of the map
as a map of R-modules is 1, where by rank, we mean the rank func-
tion induced by the homomorphisms to
and
S1
1'RS1.S1 -
R with
we find
S1;
composing with the multiplication map gives us the
S1aRS1;
identity map, so the rank of has rank 1 over
we show this by consider-
S2;
ing the map obtained from it by tensoring over
must be 1. Therefore, the map
in particular Sl%S2
S2;
We wish to show that
is non-zero.
is a full map with respect to
a
also
01%S2
In
p.
order to do this, we need to show that the minimal rank with respect to of an f.g. projective submodule of
O a
p
S1&S2 containing the image of
S1
0
is /. So, we need to know what the submodules of
S1
look
S1
S21 0
O
like. We leave it as an easy exercise for the reader to show that such a submodule
eS1db S
\0
0
O
S2-complement of
is the rank of
Si modules over
O
/pI(eS1) + /p2(M)
C0
containing the image of
a
0
0
M 0
q
less
1-e;
over
such that it has rank
0 S1'RS2/
than . Then consider the map S2
where
eS1aRS2) ®( 0
/eS1
0 S1
is an
Si.
Suppose that there is some submodule
of
M
in S1%S2.
R i Consequently, its rank is equal to
pi
M) where
eS1QRS2 ® (0
takes the form (eS1
M
we find that the rank of
2(q - /P1(eSI)),
(l.e).
y®RS2
M1
which sends
So, the rank of
of rank over
S1
to
is at most
M under left
yRSl:Sl -> (l-e)S1aRS1
is equal
2(q - /p1(eSI)).
under
S1
1
lies in the image of
S2
which is at most
That is, the image of in some submodule
(1-e)S1
y%S2:S2 -; (1-e)S1aRS2
since the image of
multiplication by to the rank of
y:R -)-
yQRS1
at most
in
(1-e)S1%S1
2(q - 4p1(eS1));
lies
therefore,
101
O
19RS1/
O
O
S1
lies in (eSl O
S1QRS1
O
eSIaRSl) 0(0
inside the ring
M1), O
O
This shows that the rank of the map defined by left
.
Sl
multplication by
O
19 1 R
O
O
from (0
O
O
S
to
S1
S
% S 1 1R
with
0
O
respect to the rank function that assigns these modules the rank / is at most
q,
since its image lies in
eS1
eS1aRSl
® (0 M1
O
0
O
with respect to this rank function is at most
whose rank
O
/p1(eSl) + q - /p1(eSI) = q.
However, it must be full, since it becomes invertible under the homomorphism to
o
given by
M2(S1)
S1aRS1
where the map from
SlQRSl to S1
i(Sl S1) is the multiplication map. Therefore, we have a contradiction if we assume that a full map. So,
Ta
MZ(S1 R SZ)
the universal localisation of
S1 R SZ
call the simple artinian coproduct of write as
S1
R
is not
a
has a unique rank function by 5.2, and is a simple artinian ring, which we S1
and
S2
amalgamating
and
R,
S2'
Universal localisation and Sylvester rank functions Our method for constructing homomorphisms from a ring to suitable simple artinian rings will be to adjoin the universal inverses to some set of maps between f.g, projectives; provided there are enough maps of a suitable sort the ring we shall obtain will be local with simple artinian residue ring. In order to prove these results, we shall need to investigate ways of extending Sylvester module and map rank functions from a ring to a universal localisation of the ring. This theory allows us to prove Cohn's results on homomorphisms from a ring to skew fields which we shall need in preparation for the general case.
Theorem 7.4
Let
R be a ring with a Sylvester rank function,
be a collection of maps between f.g, projectives over
p;
let
E
R whose rank is equal
to the rank of the identity map on the domain and codomain. Then, the universal localisation of
map rank function on
R
R
at
E,
R£,
does not vanish, and the Sylvester
extends to a Sylvester map rank function on
R, pE,
102
that takes values in the subgroup of the reals that
Proof:
Once we show how to extend
is well-defined, it follows that
to
p
RE
RE,
p
does.
and show that this extension
does not vanish.
First, we note that we can assume that
E
is upper multiplicat-
ively closed, since the rank of all elements of the upper multiplicative closure of
is equal to the rank of the domain and codomain.
E
As we have done in similar situations, we define the extension of
p
to
by Cramer's rule, and then, we show that it is well-defined
RE
by Malcolmson's criterion. Let
be a map between f.g. induced projective R Then, by Cramer's rule, there is an equation:
S:RE0RP -+ R a Q E
modules over
RE.
IP, Ql (a al)
o
We attempt to define
by
pE(a)
pE(a) = p(a a') - p(P').
Suppose that we
have two equations of the above form:
IP, (a al)
0
1
= (a a' )
,
Ip S2
(y 1
/
(Y Y,
)
0
Then,
a
aal0
al 0
0YlY
0Yl Y
so, by Malcolmson's criterion, there is an equation
a
al
10
0
0
0
0
Y1
Y
Yl
O
O
O
0
0
0
61
0
O
0
0
0
0
62
O
O
O
I
E1
0
From this, we construct two equations:
)
103
al
0
0
0
0
10 Y1
a
Y
Yl
0
0
0
0
0
0
61
0
0
0
0
0
0
62
p
0
0
I
el 0
a
al
0
0
0
0
l0 Y1 y 1 0
0
1.
0
and
2.
1 2
O
O
O
0
61
O
0
0
0
0
01
62
0
0
0
I
E1
0
S,
then p(x1 X2) = p(X1) + p(X2),
Note that if
Xl
lies in
O
4
for arbitrary
, since by axiom 4 for a Sylvester map rank function, it is at least that, whilst
X
G1
X2/
\ol
0)1 i \ I01 \0
0X2/
so that axiom 2 shows that it cannot be greater. A similar argument applies
for arbitrary '.
p(X1) + p(X2)
to show that
Since
I X1X2 0
X2
I ),
is associated to 1\
is defined, and
\
we deduce that
p(X1X2) = p(X2)
whenever
X 1////
X1X2
X1 C E.
So, p(LHS1) = p(6l) + p(62
+ p(P) + p(aa1) + p(yy');
whilst
p(LHS2) = P(61) + P(d2) + p(P) + p(yyl) + p(aa') But
p(RHS1) = p(RHS2),
since both
O ) and
lie in
(
IT) E;
so,
equal to
p(RHS1) = p(LHS1) = p(LHS2) = p(RHS2). p(yy') - p(YY1),
So,
p(aa') - p(aa
from which it is clear that
p
)
is
1
is well-defined.
We have to be careful what we mean at this point since this rank function is not as yet defined on association classes of maps over
RE
but only up to
104
R
multiplication by invertible maps over
on the left. We shall however
show that our rank function defined on maps between induced f.g. projectives over
RE
satisfies axioms 1 to 4 for a Sylvester rank function from which
it follows that it is well-defined on association classes of maps over
RE.
It follows that it extends to a Sylvester map rank function since it restricts to a Sylvester matrix rank function and so it extends to a Sylvester map rank function by lemma 7.2.
Axioms 1 and 3 are clear, so we are left with 2 and 4 which are rather less obvious.
First, we show that
p((I
al
for arbitrary
= pE(a) + p(Q),
E\OQ a
al.
Suppose that we have equations:
IQ
a
of (RIsl)
a 1)
= (BIB')
,
(YIYl)
(IQ2 O
then,
We note that
0
I
0
a"
is associated to
O
0
00
s
0
I
Q2
Y Yl
s'
0
a"
0
a'
Q2
IQ3 0
al
0
IQ
1
0
0
I93
0
0
0
Q1
0
0
is in
E.
0
since
(YY
a
This equals
p(Q3) + PE(a)
Next, we show that pE(a6) <_ PE(a), PE(B)
map. Also, we shall show that if
a
is in
E,
then
a
as we stated.
when
a
is an induced
pE(a6) = pE(B).
105
Let
be an induced map over
a:PO -> P1
be a map between induced projectives over P1
P1
=p
R)
pE(a
«)=
1
o
and let
R:P1 _* P2
pE(aR) + p(P1)
-IP
E(R
= p
o
RE,
Then
RE.
where the last equations arise from
,
a
o
multiplication on the left by invertible maps over
R.
IP' R1 _ (y Y') then
If (Y Y1)
(o IpI Bl O
Y Y1 0
(0
-I
O
0
IP
O
IQ'
( Y Y' -Y1
=
00
0 a
IQ
O
-1l
1
=
0
a
IP
0
a /J
O
Y Y' 0
where 0
1
is the domain of
Q'
IP 1
(YY1
E(R -I So,
P
oa
P(a) + P(Q') - PCP'
- PCP')
P
O0 Since
P(YY') + P(P1) - p(P').
a
is in
(YY1
E,
p(Q') = p(YY1) = P(P') + P(P1).
p(a) + p(Q') - p(P') = p(a) + p(P1).
Therefore, we see that So,
(Y Y. -Y1)
-
pE(aR) + P(P1) =PEIO a)
PE(a) + P(P1), PE(R) + P(P1)
as
we want.
If we assume
a)
PE (
=PI
a
is in
OO
al
E,
/-
P (P' )
Hence,
= P(a) + PE(R) = P(P1) + PE(R).
= P (YY') + P W - P (P' )
PE(aR) = PE(
In general, suppose that we have maps over f.g. projectives
a:PO + P11
and
a) - P(Pl) = PE(R)
R between induced
Suppose that we have an equa-
R:P1 -* P2.
tion:
(IP al) _ (Y Y')
(Y Yl)
O
a
I
then p (I P ) + P E (a R) = P E < PE(YY')
pE(aa) 5 P
Let
0
PE (Y Y'
1
aR
so
p(P) + pE(R) (R)
I
aR P
pE(aR)
aR P
(0
1 aR
PE (Y Y'
<_ P(y ') - P(P) = PE(a),
and
which is axiom 2. and
a1,a2
projectives such that
(S1
tions:
I\
a
be maps over
RE
between induced f.g.
is defined. Suppose that we have two equa-
a
2)
106
I
(Y Y1)
= (Y Y')
al
O
a"
I
a'
P1
,
(6
61)
= (S d')
P2 0
a2
Then
I
0
O'
a
+ p (Pl P2) = PE O O a
P
0
06 0 0
00602
(aloot2
0 Sls S' = PE
Y Y.
O y
0
O O Sl a2
00 so that the problem reduces to the
0 dls 6 d'
O
case where
y yl
PE
Y'
I a' O O0a 0
S1
- PE
are induced maps.
al,a2
So, suppose that we have an equation: P
(e e1)
S1
= (e e') IP
S P.
O
al
B 01
al O then
+ p(P) = PE
p E
a
S
2
)
(O
S
0
E 0 el = PE
a 02
IO
) 0
=P(
0 al
)
EL0 Cl =
El Oe 602
IP
- P( 1
O
o
O
al 0
O
S
a
2)
0
p(al) + p(a2) + p(IP
(e e1
OP a
2
which proves axiom 4.
We have shown that our rank function defined on maps between induced f.g. projectives satisfies the axioms of a Sylvester map rank function. It follows that it extends uniquely to a Sylvester map rank function as we have outlined before.
In the case where the Sylvester map rank function on values in R
2Z,
R
takes
we can say a great deal about the universal localisation of
at all the maps between f.g, projectives whose rank is equal to the rank
of the identity map on the domain and codomain. In fact, we obtain Cohn's classification of homomorphisms to skew fields.
Theorem 7.5 values in
Z;
Let
R
be a ring with a Sylvester rank function
then the universal localisation of
R
at
E,
p
taking
the collection
107
of those maps between f.g. projectives whose rank is equal to the rank of the domain and codomain is a local ring whose residue ring is a skew field. The kernel of the map from
from
to the residue ring consists of those elements
R
which define maps of rank
Therefore, the equivalence classes of maps
O.
to skew fields are in 1 to 1 correspondence with the Sylvester rank
R
Z, or equivalently, with the prime matrix
functions taking values in ideals.
Sylvester rank function on
taking values in
RE
a + b = (1 1)
subgroup because
Let
Z.
consisting of those elements whose rank is
RE
of
exists and
RE
By the last theorem, we know that
Proof:
p
induces a
I
be the subset
then it is an additive
0;
0) (1) and it is closed under
(Oa
b
1
multiplication so it is an ideal. If
is not in
x
I,
its rank is
1.
Let
be some equation given by Cramer's rule; then because
must be in
So,
E.
has rank 1,
(a a')
is invertible.
x
So, as we stated
R -
RE/I
morphism from
RE
is a local ring with maximal ideal
I,
and
is the one we began with. Conversely, if we start with a homoR
to a skew field,
R -
inducing a Sylvester map rank
F,
there is a map from
RE
to
and the kernel of this map is
I,
for if
function F,
x
= (aa')
is a skew field. Clearly, the Sylvester map rank function induced by the
RE/I
map
X
(a al)
p,
zero image in (a a1)
F,
extending the map from
F
a
is in
I,
and
a
R
to
has non-
we consider some equation given by Cramer's rule: Since
I x/ = (a a').
side is invertible over
a
has non-zero image in
F,
the left hand
and so must be the right hand side; however, the
F
right hand side is not invertible, since its rank given by the map rank function induced by the homomorphisn from
R
F
to
is not equal to the rank of
the identity map on its domain and codomain. We have a contradiction, so lies in the kernel of the map from F
as the skew subfield of
F
RE
to
F.
Therefore,
generated by the image of
The remark on the kernel of the map from since the map from
R
to
F
R
I
embeds in
RE/I R.
to
F
is trivial,
induces the Sylvester rank function
p.
Ring coproducts and rank functions In this section, we shall show how Sylvester module and map rank functions on a k-algebra
R
may be extended to Sylvester module and map rank
108
functions on the ring
This will allow us to show that every
Mn(k) k R.
Sylvester module rank function on a k-algebra arises from a homomorphism to a simple artinian ring by showing that the Sylvester module rank function
taking values in 1 Z n
Mn(k) k R for some
extends to a Sylvester module rank function on
also taking values in
Mn(k) k R
n Z;
is isomorphic to
Mn(R')
and by Morita equivalence there is a Sylvester module rank
R'
function on
taking values in Z; we have shown that this is induced by
R'
a homomorphism to a skew field,
and this shows that the original rank
F,
R + Mn(k) k R
function must be induced by the homomorphisms:
Mn(R') -r Mn(F).
It remains to find a way of extending the Sylvester map rank function on
to one on
R
The idea is to mimic what we know the
Mn(k) k R.
result would have to be if there were a homomorphism from artinian ring
S
have homomorphisms:
M n(k)
a:P -> Q
S -> M n (k) o S,
R -> Mn(k)
which induce a
k
k
k
Mn(k) k R
Sylvester map rank function on map
to a simple
R
inducing our Sylvester map rank function. In this case, we
extending that on
R.
Mn(k) k R,
between f.g. projective modules over
Given a the rank of
aa(M (k) o S) is the minimal rank of an f.g. projective module over n k Mn(k) k S containing the image of aQ(Mn(k) k S). It is hoped that this
discussion will help to motivate the definition we shall propose later on for the map rank function on
Mn(k) k R.
Unfortunately, before we can begin, we need to go through a certain amount of technical work on coproducts over a skew field. Rather than referring the reader to an earlier chapter, we shall reproduce some of the
definitions here; also, we shall prove one of the coproduct theorems here, since it is precisely the technical details of the proof that we need to examine.
Let
R0
Set M = A u{O}.
be a skew field and
induced module which has the form right basis over u,
R0
of the form
we choose a basis over
for
R R and consider an For each
R R.
A,
we choose a
u
u
for the ring
{1} U T
R0, Su,
R =
O
N = e N u
a family of R0-rings.
{RA:A u A}
We form the ring coproduct
Write
N . u
and for each
RA,
S = US
uu
,
and
T = UT
If
t e TA,
it is associated to
if
s E S0,
it is associated to no index. A monomial is an element of
a formal product
A;
st1t2...tn , s E S
if
S E SA,
and
terms are associated to the same index. Let element of
U
is associated to
A
ti E T U
it is associated to
as
A;
S
or
such that no two successive
be the set of monomials; an
if and only if its last factor (in
S
or
109
T)
is associated to
Every element of
A.
except for the elements of
not associated to
We denote by
S0.
those elements that are
U_X
We recall again without proof theorem 2.4.
A.
Let all terms be as above. Then a right basis for
Theorem 2.4
N
is the direct sum as
is the set
U.
For each
and a free
RA
module on the basis
Given
A
A,
and
For
efficient of
u'
the theorem.
For
we denote by
U-,
such that
empty A-support if and only if it lies in x
s
N
of
S
is
cXu(x)
R0 module in
as
N
of
is not
is the has
x
0;
The 0-support (or support)
NA.
st1...tn
is
is not
cOu(x)
0.
and the degree
(n + 1),
The degree of an element of
1.
RA
the
linear 'right co-
x
consists of those monomials such that
The degree of a monomial
of an element
R0
the
cou:N } R0
the A-support of an element
finite set of monomials in
of an element
cAu:N * RX
map given by the decomposition of A E A,
RO
N
map given by the decomposition of the
u'
we denote by
u e U,
over
N
module of
RA
U_A.
U E U_A,
linear right 'coefficient of theorem.
is associated to some index
U
N
is the
maximal degree of an element in its support. We define an element
x
in
N
to be A-pure if all the monomials in its support of maximal degree are associated to the index for any index in
It is O-pure if and only if it is not A-pure
A.
A.
We well-order the sets order
U
and
S
T
in some way, and then we well-
by degree and then lexicographically reading from left to right.
Next, we well-order M,
making
0
the least element. We well-order
M x U
first by the degree of the second factor, and afterwards lexicographically from left to right. Let from
M x U
lowest in
H
be the set of almost everywhere zero functions
to M well-ordered lexicographically reading from highest to M x U.
Given any element
x
in
N,
its leading term is the largest
element in its support; its A-leading term is the maximal element in its Asupport (if it has any).
Given a homomorphism of f.g. induced modules
a:$ M n R u
wish to find an isomorphism that the image of
a
g: ® M' R N Ru
is isomorphic to
- N,
U Ru
we
W M Q R of induced modules such u U RL ® (as)M'® R. It turns out that we Ru u u
may find such an isomorphism composed of transvections and free transfer
110
maps. In order to present the proof, we introduce following Bergman the notion of a well-positioned family of submodules of
N.
N, {L
} is said to be wellu positioned if and only if the following conditions are satisfied:
A family of
AU: VIEM,
R
submodules of
u
all elements of
BU1U2: the U1-support of
are U-pure;
L11
LU1 contains no monomial
U1-leading term of sane non-Ul-pure element
which is also the
u x
xa,
LP
in
a
,
in
and
R,
U21
deg xa > deg x.
if U1 =
2
It is not hard to show that if is naturally isomorphic to
ELUR
of the isomorphism is that if
is a well-positioned family
{L }
The idea for the construction
® LU®R R.
is not a well-positioned family,
{a(MU): p E M}
s :e M' e R - ® M ® R so that U l U URU 11R {asl(MU): p E M} is a 'better positioned' family; by using the well-ordering,
we find some free transfer or transvection
we can make sure that this process terminates and so, it gives us the isomorphism that we were looking for.
Theorem 7.6
Let
be a homomorphism of f.g. induced modules,
a: ® MU®R R -+ N u
where
is as described above. Then, there is an isomorphism of induced
N
modules
which is a finite composition of free
s: ® Mu0R R -- ® MU0R R u
u
transfers and transvections such that family of submodules of
We associate to the map
Proof:
h(p,u) =(1, if
u
is a well-positioned
{as(MU):U E M}
N.
ha:M x U a {0,1}
a function
a
is in the U-support of
by
a(MU);
O, otherwise.
This is an almost everywhere zero function since
M
is f.g. and is nonU
trivial for only finitely many Suppose that
p.
is not X1-pure for some
a(Mx )
then there
al;
1
in
is some
a(x)
the map
c:M -> R
a(M)
with A1-leading term
cXu(x) = 1;
such that
u
1
where
A1
is an
R
split surjection; so,
x1
M
x1
= ker c
e xR x
1
x
is free of rank 1. We perform the free transfer between
xR 1
MU0R R M.
to
and
® Mu0R R,
where
M' = Mx
for
A x al;
M0 = MO e xRO;
u
al
= ker
M0,
c
maps
atl' ker
It is the identity map on
cx u
to
ker
1
that if
Sl
is this free transfer,
cA u
M' a
for
and sends
A x A
x to
1 has :M x U -> {0,1} 1
1
;
x.
it maps
M'
0
It is clear
is a smaller function
111
in the well-ordering of such functions that we described earlier. If, however, in
cOu; M'X
M1 = MX, A z a1; MO = ker
effect. Again, it is clear that has
si: ® M' QRu R -r M R, u R u u U
a xRX .S1
= MX 1
a(x)
cOu(x) = 1;
such that
u
so, we perform the free transfer
MO = ker cOu e xRO;
where
is not O-pure, there is some element
a(M0)
that is X1 pure with leading term
a(M0)
has the obvious
1
is less than
under these condi-
ha
tions. Therefore, after a suitable finite sequence of free transfers, we may assume that each
a(MU)
u-pure.
is
Next, suppose that for some pair
fails for the
u1,u 2, B
ul2 u family of modules
that is, there is an element
{a(MU)};
a(x) in
a(MU 1
ul-support contains a monomial
such that its
where
a(y)a
term of a non-ul-pure element
that is the
u y
is in
M. ,
a
ul-leading is in R,
2
and if
ul = u2,
deg (ya)
cu u((y)a) = 1.
> deg (y). We may assume that
1
We have a functional
1
M = ® MUOR R
to a functional on
given by
- R11
M11
which we extend
cu
1
1
u
in the way we described when describing
u
transvections. Left multiplication of this functional by an endomorphism of
M
of
t
of square
so
0;
$ = IM - t
ya
now gives us
is an automorphism
M which is a transvection. It is an easy check to show that
smaller function than
transvections and (possibly more) free transfers,
is a
s = IIs; i
we may ensure
is a well-positioned family of submodules.
{as(MU):u E M}
that
has
therefore, after a suitable finite sequence of
ha;
Now we shall show that a well-positioned family justifies its name.
Theorem 7.7 N;
{M,,:u E M}
EM R = ® M 0
then
Proof:
Let V
Given
u
uR
u e M,
term of some element of
R
be a well-positioned family of submodules in
in the natural way.
we choose for each monomial M
an element
q
in
M11
u
that is the leading
having this leading term
with co-efficient 1; we denote the set of such q's by
Q Q.
It is clear from
112
and monomial
in M0
so QOA is an RD basis of
A-leading term
The elements of
V
We consider the set where
gtl...tn,
q
lies in
M0
Every element of
QOX.
M0
for each
are associated to
QX
are associated to all indices in
QOa
A E A,
we choose
M0,
that has this term as A-leading term with co-efficient
and we denote the set of such q's by
1,
For each
that is a A-leading term of an element of
u
q
an element
is an RO basis of M.
Qu
the well-ordering that each
has a
X.
and the elements of
A,
A - {A}.
consisting of all elements of the form
U(QAu QOA)
and no two successive terms in the
above monomial are associated to the same index, together with N
shall show that these elements in
UQ 11P
We
.
have distinct leading terms, and so,
are rightly R0 -linearly independent. Therefore, they form an R0 basis of
from which our theorem is clear.
EM R,
uu
By the well-ordering of
where
utl...tn,
where
q
lies in
q
lies in and
QOA
M
in
from
u'
,
M
in
q'
V2
tion, q'
If
If
is in
If
BOX.
the support of
QoA,
is in
comes from
u
m >- n;
A x V.
u1-support of
q'
Q0
for
or
M11 1,
in
q
in
MA,
which
and we obtain as above a
q E Q0.
cannot lie in
we see that if contains
M
QOA'
contains a mono-
since m = n = 0;
QA
Once more, we have a contradiction to
is in
q'
q'
is in
q'
q' E MO
and, by cancellation, we find
m > n,
Since
then
QOA,
q' E Q0,
so,
A,
m = n = 0,
Q0,
The same applies if
is in
QA,, So,
is in
q'
BOX.
q
If
so the
or
q,
So, we consider
q.
we obtain by cancella-
by construction. Therefore,
QX
q'
contradiction to
q
where
Assume m = n;
.
which is the leading term of a pure element
contradicts
so
is the leading term of
is associated to
u
q c QA,
cannot lie in
for A' x A. u'
u
is
ill
u = u'.
mial
and
is the A-leading term of
utl...tm = u'ti...t'
an equality of the form q
QX u
gt1...tn
the leading term of
U,
BOA,.
utl...tm-n = U'.
ul x 0, tm-n
utl...tm-n
is in
u
(or
Tµl
if
ill
l
m = n + 1).
gtl...tm-n
This also is the u1-leading term of the non-u1-pure element (or
q
if
m = n + 1),
which contradicts
B
.
ul = 0,
If
ulu2
then since the support of of the pure element
q'
gtl...tm-n
in
M0
contains ut1...tm-n
the leading term
we again have a contradiction to
B
.
ulu2
This completes the proof, since we see that the elements of
V
113
are independent and span
EM UR. P
Bergman also uses a direct way of finding well-positioned families of submodules that generate a given submodule of an induced module. Thus, let
L c N;
we define
R
to be the
L u
submodule of
u
L
consisting of the elements whose U-support does not contain the u-leading term of some non-p-pure element of
By construction, the family
L.
{L
is well-
}
u
positioned; it is not hard to show that is naturally isomorphic to
is a f.g. sub-
L
as being generated by some finite family
submodules of
RU
uu
In the case where
u
module, we may regard L f.g.
® LVOR R.
which as we have just seen
L = EL R,
{MU} = F;
L,
by
{0,1}
associate a function
hF:M x U -'
hF(u,u) _ 1 if
is in the u-support of some element of
O
u
of
To such a family, we
EMUR = L.
so,
F
M
otherwise.
We shall characterise the family of submodules
{L
where
}
is the set
L
u
u
of elements whose u-support does not contain the p-leading term of some nonu-pure element of
by the property that the associated function is minimal
L
in the well-ordering of such functions.
Theorem 7.8
Let
be an f.g. submodule of
L
have considered throughout this chapter. Let
N
N,
where
L
be the
is the module we R
submodule u
u
consisting of those elements whose u-support does not contain the p-leading Then, the function
hF
associated to
term of some non-U-pure element of
L.
the family of submodules
is minimal over all possible finite
F = {L
}
u
families of submodules that generate
Assume that we have a family with smaller associated function than
Proof: hF.
L.
Given a family of submodules
F1 = {M
that generate
}
L,
we have a
u
map a:$ MU0R R -* N
whose image is
L.
The proof of the last theorem showed that the method used there of passing to a map
such that the family
a': 9 M ® R -* N u u RU
u
is well-positioned always forces the function
hF2
So, we may begin by assuming that our family Therefore,
L = ®M 0R R u
F2 = {a'(M )}
to be less than
{a(M )}
h F1.
is well-positioned.
in the natural way.
u
u1-support of
Suppose that the
M
contains the u1-leading u1
term
u
of some non-p
1-pure element
We note that the degree of
x
x
in
L.
We assume that
cUu(x) = 1.
is less than the degree of an element of
M ul
114
whose
u.
u1-support contains
L = EM'R
We wish to show that where
M' = B(M ) ul ul
uu
S(m) = m - xc
of degree less than equal to
ulu
M' = M
where
u
So,
(m).
whilst
u x u, u
is the identity on elements
a
x.
< h so, if EM'R = L, the h F F uu well-ordering of our functions imply that after finitely many steps (which M},
F' = {M':u
Clearly, if
may include more operations that change F = {M
we reach a family
}
,
,
into a well-positioned family)
F'
M
where no element of
u term of some non-p-pure element of
and
L,
contains the p-leading
u
Clearly,
EM R = L.
uu
M
= L u
u
Since our associated functions have decreased at each step, this will prove our theorem.
We show that
x
lies in
EM R
from which it follows that this
u
must equal
L.
We use the notation of 7.7. We look at an expression of
R0
with respect to the of
basis
can occur since all elements of V have distinct leading terms. So,
x
all elements in this expression of from
x
No elements of degree greater than that
V.
Q
either lies in
and so must
must lie in
x
Mu = M',
for
u x ul,
since their term
EM R,
uu
or else it is fixed by
g;
x.
We can begin the proof that every Sylvester map rank function on a k-algebra tion on
R
taking values in 1 Z
R' = Mn(k) k R
R1 = Mn(k);
and
extends to a Sylvester map rank func-
that also takes values in
Set
Z.
R2 = R.
R0 = k;
n
The first point to notice about a Sylvester map rank function is a:P - Q
that the rank of a map for, if a'
depends only on the image of
has the same image then
a':P' -> Q
factors through
a,
a
factors through
a
in
Q;
and
a'
so they have the same rank. So, our problem is to
assign to a given f.g. submodule of an f.g. R' submodule of an f.g. projective
R'
module a rank so that the associated map rank function is Sylvester. Let
P = e PQ R
be an f.g.
u Ru
u
R'
module, where we may assume
is 0; we identify P with the module N that we have been P 0 discussing in this chapter, and so, we identify N with P P. So, we have
that
u
bases T2 u{1}
Qu of
R
over
Pu
of each over
k,
bases
k,
orderings previously defined of
U
q
or
M(k)
of
and consequently, a basis
consisting of monomials of the form
to the natural numbers.
T1 u{l}
gt1...tn.
U
of
over P
over
and
k,
k
We also have the well-
and of the functions from
{0,1,2} x U
115
L c P
Let RU
be some f.g.
submodule of elements of
R'
submodule of
whose U-support does not contain the p-
L
leading term of some non-U-pure element. For RU
module, where
pU
be the
LU
let
P;
U = 0,1,
has a rank as
L11
are the usual rank functions on
k
and
For
Mn(k).
U = 2,
the map rank function allows us to define the rank of the inclusion
of
in
L2
as a map of
P
modules (recall that we earlier how to extend
R2
a Sylvester map rank function in a canonical way to give us the rank of maps from an f.g. module to an arbitrary module). We define the pre-rank of to be
p0(L0) + p1(L1) + p2(L2 E P).
L
The rank of
L
is defined to be the
minimal possible pre-rank of an f.g. submodule of
P
that contains
the rank of a map by the formula:
a:p' - P p(a) =
between f.g. projective
min {pre-rank (L)}.
So,
L.
modules is defined
R'
We shall show that
is a
p
LD in a Sylvester map rank function. If we have a map
a: P'
R - P,
we could assign what we might
U
11
p0((Pp)) + pl((Qi)) + p2(alpi);
regard as a pre-pre-rank, which is
next lemma, which shows that the pre-rank is the minimal possible rank' as we consider the composition of
with other induced modules is the main step of the proof that
our
pre-pre-
®
with isomorphisms of
a
2
RP
R
is a
p
Sylvester map rank function.
Lemma 7.9
Let
{MU:U = 0,1,2}
L c P be
RU
be some f.g.
submodules of
submodule of
R'
L
such that
let
P;
EMUR' = L;
then the
11
L c P
pre-rank of
Proof:
is the minimal value of
Theorems 7.6 and 7.8 give us a finite sequence of operations that
pass from any given trio of modules where
{LU}
p0(M0) + p1(M1) + p2(M2 c P).
LU
{MU}
that generate
L
to the trio
is the set of elements whose U-support does not contain
the p-leading term of some non-U-pure element of
L.
So, we simply need to
show that these operations cannot increase the 'pre-pre-rank'.
So, suppose that M0 M0,
if
where x
x
is not O-pure; then there exists
is 1- or 2-pure, its leading term is
is 1 pure, we replace:
M0 by ker cOUl M0; M1
by
M1 + xRl;
M2
by
M2.
u,
and
x
in
cOu(x) = 1;
116
The pre-pre-rank does not increase. If
x
is 2-pure, we replace
MO
by
ker cOu1MO;
M1
by
Ml;
M2
by
M2 + xR2
and again the pre-pre-rank cannot increase, since the contribution of the 0-term has decreased by 1, whilst the 2-term has increased by at most 1.
an R0
R1
module to
free module so the pre-pre-rank cannot increase.
leading term xR2
is not 2 pure, there is an element
M2
if
where
is not 1-pure, we simply transfer a free
M1
If
such that
u
direct summand of
since
P,
M2;
c2u (x) = 1,
M2
with 2-
M2 = ker c2u+ xR2
in this case
c2u (x) = 1;
is a free direct summand of
in
x
it is also, however, a free
so,
p2(M2 c P) = P2(ker(c2ulM2) SP) + 1. In this case, our operation replaces:
M0
by
MO + xRO;
M1
by
M1;
M2
by
ker(c2ulM2)
and by our previous remarks, we see that the pre-pre-rank cannot increase.
Thus, we have shown that our efforts to make each
M11 u-pure do
not increase the pre-pre-rank. We consider next the transvections we need to make them well-positioned. If
element
xa
contains the u1-leading term
M11
for
in x
then for
ul = 0,1,
Mp ,a
in
R'
and if
u
for some non-p1-pure
pi = u2,
deg xa > degx,
2
we know that the transvection takes
M
to a homo-
ul
morphic image of itself whilst fixing the other more care is required. We assume that
M M.
c2u(xa) = 1.
If, however,
pi = 2,
We define a functional
C2u
-
on
M2
by
f:M2 c P
M2
to
MZ
the image of
the transvection fixes
R2;
LM
- xaf; 2
M0
and
M1
and sends
this map has as a left factor the
117
inclusion of
M2
in
consequently, by axiom 2 for a Sylvester map rank
P;
p2(MZ c P)
function it follows that
p2(M2 c P)
is at most
and the pre-
pre-rank does not increase.
consider the case where
Finally, we
positioned family of submodules of
11
contains in its
ul = 0,1,2
some
c
ulu
but M p
V
x
in
for 1
p1-support some monomial
u1-leading term of some non-p1-pure element that
is a well-
(M,,:u a M}
EM R' = L
such that
L
that is the
u
we also may assume
L;
(x) = 1.
If
we see that the new MP
]Il = 0,1,
is a homomorphic image 1
of
Mu
,
whilst the remaining
MP
do not change. If
we fix M0
2,
ul =
1
Ip-x C2u
and send
M1
and
is clearly an
M2
to the image of
linear map; again
R2
M2 S P
M2 S P
P,
-*
where
IP - xc2u
is a left factor so the pre-
pre-rank cannot increase, which completes the proof of the lemma, since there are no more operations that we have to worry about.
Theorem 7.10
taking values in rank functions
be a k-algebra with a Sylvester rank function
R2 = R
Let
RO = k,
let
a Z;
pO,pl
and let
respectively; let
function on maps between f.g. projectives over Then,
p
Rl = Mn(k)
R' = R1 R R2; O R'
with the standard let
p
Sylvester map rank function
be the rank
defined in the foregoing.
is a Sylvester map rank function taking values in
there is a homomorphism from R
p
Z.
Consequently,
to a simple artinian ring inducing the n
p.
We need to show that axioms 1 to 4 for a Sylvester map rank function
Proof:
actually hold for
p.
1 is clear.
We consider 2. Let where we have an expression of we need it. The image of
a:Pl - P2, P.
S:P2 -* P3
as an induced module up our sleeve when
lies in the image of
aR
g,
is the pre-rank of some f.g. submodule containing
p(a)
L = ® L u
R' c P
u Ru
2
where
be a pair of maps
and so,
<_ p(6).
im a
is the set of elements whose p-support does
L 11
not contain the p-leading term of some non-p-pure element of p(a) = pO(LO) + pl(Ll) + p2(L2 so, by the last lemma,
p(as)
P);
the image of
as
L.
lies in
That is, E(LU)R',
118
+ P2($(L2) S P).
P((X8) 5 PO(B(L0)) + P1(R(L1
The map L2 S P2
B
has
P3
as left factor so
L2 C P2
P(as) <_ PO(LO) + P1(L1) + P2(L2 c P2) = P(a).
So, 2 holds.
For the time being, we remark only that
p
(a
S) <_ p(a) + P(S);
the rest of 3 follows from 4.
The reason that a Sylvester map rank function works is essentially because of axiom 4; so, one would expect this to be the most difficult part of the proof. It is.
0\:(P)
Consider a map
(a Let
-
P
be an f.g. submodule of
L
(QQ) that contains the image of
Q ® Q'
a O Y B)
such that
p(Y
is the pre-rank of
B p:Q 0 Q' + Q
projection
on
We consider the action of the
L.
the idea is to find
L;
R
submodules of u
L,
that
such that {p(L')} are a well-positioned family, which implies u u p(L) _ p'(LU)ORUR; also, we wish the pre-rank of L to equal
{L '}
p0(LD) + pl(L1) + p2(LZ c Q ® Q'). 11 as being
a2 O p2(Y2
Q
P"
(aY 2 O
for a map
where
S2/
2 5 2I:
(impIL!)R' + (ima2)R'
corollary of 4 for
Let
p2
on
im a.
and also,
From this, 4 follows as a
R2.
be the set of elements of
L
are f.g. R2
Pi
PZ/ --((D
(kerpILi)R' + (im82)R' D im S,
projectives,
p2(LZ c Q ® Q')
This allows us to express
L
whose p-support does not
u
contain the p-leading term of some non-p-pure element of that
L = 0 L OR R' u
u
L:
in the natural way, and the pre-rank of
then, we know L
is equal to
u
pO(L0) + P1(L1) + p2(L2 c Q (D Q').
Let
p:Q ® Q' -+ Q
be the projection on
Q;
if
{p(LU)}
is not
a well-positioned family, we perform a series of free transfers and transvections on the family of modules
{L
}
until their image do form a well-
u
positioned family. It is clear that the free transfers do not increase the pre-pre-rank of the family of modules
{LU)
at any stage, so, we are left
119
to worry about the transvections. These arise when the image all p-pure but some element
contains the u-leading term
p(L
cuu(xa) = 1,
chosen so that
xa
u = 2.
and if
It is clear that the only case that worries us
u = u', deg(xa) > deg(x).
occurs when
p(LU)
is in
x
are
p(L )
of a non-p-pure
u
In this case, we alter
to the image of
L2
I - x'a(pc2u)
L2 C Q ® Q'
where
Q 0 Q',
Since this map factors through
is a pre-image of
x'
{L'}
pre-rank. So, eventually, we reach a family of modules {L'}
pre-pre-rank associated to began with and the images {LU}
modules
{LU}
family
generate
{L
we
form a well-positioned family. Since the
their pre-pre-rank must be at least that of the
by lemma 7.9. Since
p(L) = 0 p(L')QR'. RU
p.
such that the
is at most that of the family
P(Lu')
L,
under
x
it cannot give a greater pre-
L2 c Q 0 Q'
form a well-positioned family,
p(LL)
p restricted to
So, the kernel of
L
is
11
11
0 kerplL'® R.
0 + L' n Q' + L
We have exact sequences
u RU
u
for
-+ p(L') -+ 0 u
u
U = 0,1,2.
From this, we wish to deduce that pu(L')
= pu(L' n
p2(L2
Q
which is clear; and also, that
+ pu(p(L')), u = 0,1,
In fact, it is not obvious that we can define L2 n Q'
which needs an argument.
p2(LZ n Q' s Q') + p2(p(LZ) S. Q),
((L' n Q') c Q')
p
2
2
because
need not be f.g. Instead we approximate it by f.g. modules that are
good enough.
Let
a2:P1 + Q
be some
tive module whose image equals lifting
a2,
linear map from an f.g.
R2
then, we have a map from
P
R2
p(L2);
we write its composite with
to
Y2
L2
is a f.g.; therefore, we can find a f.g.
submodule of
Q'
L2 n Q',
it generates the whole of
(a21P `Y
such that together with the image of
R2
LZ
fa2\:P + \Q
as
L2 S. Q ® Q'
projec-
M2 L
2
we may further assume that image of
(L'O n Q')R' + (L1 n Q')R' + M2R')
contains the
$.
We find some
R2
linear map
and we construct the composite map:
p2:P2 + Q'
whose image is
a2
(Y2
\
SP2
02):(P:)-.(Q Q, a2 0
By construction, the image of
62is
Y2
L2'
so that
M2,
120
p2 (a2 S) > p2(a2) + P2(62).
is
P2(LZ c Q ® Q ' )
2/I
1\Y2
Since
P0(L' n Q') + p1(Li n Q') + p2(B2),
P(B) P(a)
im B E (L'0 n Q')R' + (Li n Q')R' + (im B2 )R',
and similarly,
<_ P0(P(L')) + P1(p(L')) + P2(a2). So,
L = p0(LD) + p1(Li) + p2(LZ c Q ® Q')
) = pre-rank of
P(OY
p0(LD) + pl(L') + P2(Y2 S
which is
1
which is greater than or equal to
2
pl(Li n Q') + p1(p(Li)) + p2(a2) + p2(B3),
P(L, n Q') +
p(a) + p($).
is at least
which
This is axiom 4, and also completes the proof of
3.
We have extended Mn(k) k R,
on
p
Mn(k) k R = Mn(A) coproduct.
p2
and it is clear that
A
where
So, by Morita equivalence,
A
p
to a Sylvester map rank function
takes values in
n Z.
is the centraliser of the first factor of the
tion on A that takes values in Z. tion of
R2 = R
on
p
induces a Sylvester map rank func-
By theorem 7.5, the universal localisa-
at the set of maps between f.g. projectives whose rank is equal
to the rank of the identity map on the domain and the codomain is a local ring with a skew field
F
for residue class ring. Moreover, the map A -* F
induces the rank function on
R - Mn(k) k R
A.
Mn(A) 3 Mn(F)
Consequently, the composite map
induces the rank function
p2
on
saw in theorem 7.5, the kernel of the map consists of the elements R
such that
Theorem 7.11
Z
p(R/rR) = 1,
or equivalently, such that
As we
R.
r
in
p(r) = 0.
A Sylvester rank function on a k-algebra R
arises from a homomorphism to a simple artinian ring,
taking values in Mn(D),
where
nis a skew field. The kernel of the map consists of the elements of rank
D 0.
We summarise the contents of this chapter so far with regard to homomorphisms from k-algebras to simple artinian rings.
Theorem 7.12
Let
be a k-algebra; then there is a 1 to 1 correspondence
R
between equivalence classes of homomorphisms from rings of the form
M (D), n
where
functions that take values in
n
D Z.
R
to simple artinian
is a skew field, and Sylvester rank
121
This result has the following interesting consequence
Theorem 7.13
Every right artinian k-algebra embeds in a simple artinian
ring.
Proof:
Call the right artinian ring
Let
A.
be the length of
s
A
as a
module over itself. We define a module rank function by p(M) _ (length of M)/s.
Clearly, this is a Sylvester module rank function.
By theorem 7.11, there is a homomorphism from S,
which induces this rank function on
morphism is the set of elements
a
in
only element for which this is true is
A
A
to a simple artinian ring
and the kernel of this homo-
A,
such that
p(A/aA) = p(A); the
0.
In order to find the modification necessary to extend theorem 7.12 to a theorem about homomorphisms from an arbitrary ring to simple artinian rings, it is useful to see what can go wrong in general. We look at possible Sylvester map rank functions on artinian rings. If ring,
then
GD(R) = Zk,
where
t
A
is an artinian
is the number of non-isomorphic simple
modules. We assign the simple modules ranks in the non-negative rationals, and it is clear that this defines a rank function on Sylvester module rank function on on
A,
G0(A)
that induces a
and so, a Sylvester map rank function
A provided that we normalise it to ensure that the rank of the free
module of rank 1 has rank 1. In the case where
A
is a k-algebra, we have
just shown that these must all arise from a homomorphism to a simple artinian ring, and it is easy to check that they arise from the representations given by the f.g. projective left A modules. However, for a general artinian ring, it is easy to see that they do not all arise from a homomorphism to a simple artinian ring. For example, consider the ring simple module
Z/p2,
which has the unique
to which we assign the rank '/. It has no embedding in
Z/p,
a simple artinian ring, but this rank function is faithful. What has gone wrong in this example is that we have the rank / associated to multiplication
by p
on the free module of rank 1, whereas in a simple artinian ring, the
rank of an integer must be or
0.
0
or
1,
depending on whether it is invertible
This turns out to be the only modification we need to our previous
theory.
Theorem 7.14
Let
R
be a ring. Then the equivalence classes of homomorphisms
122
to simple artinian rings are in 1 to 1 correspondence with the
R
from
Z
Sylvester map rank functions that take values in of an integer is
0
or
1.
such that the rank
n
All that is left to do to prove this is to show that every Sylvester
Proof:
map rank function of the right type arise from homomorphisms to simple artinian rings. Let
be such a rank function on the ring
p
Assume that
R
the set of elements in
such that
that this forms an ideal of
m
and
one of m
n,
Consider
Z.
R.
n must lie in
and
R/I,
which is a
Z/p
mn
a
in
lies in
for if
I
with Z
I
by
R/I
R
to
Conversely, assume that Then, by 7.4, the rank function function on
RZ*
p
induces
S
p(n) = 1 on
R
p
R/I - S
Consequently, it is induced by some homomorphism artinian ring. The map from
I
be
I
on
p
for integers
is a prime ideal
p(a) = p(a)
it is simple to check that
R/I;
Let
p(m) = 1,
algebra for some prime
define a Sylvester map rank function on the image of
R.
n.
We have already remarked
p(r) = O.
Moreover, if
Therefore, the intersection of
p(mn) = p(n). of
for a non-zero integer
p(n) = 0
R
p.
We
where
a
is
is well-defined.
to a simple clearly.
for all non-zero integers
n.
extends to a Sylvester map rank
since all the elements in Z*
identity map on their domain and codomain. But
have the same rank as the RZ*
is a p1-algebra, so the
rank function on it is induced by a homomorphism to a simple artinian ring, S.
The induced map from
R
to
S
induces on
R.
Maximal epic subrings and dominions in simple artinian rings One of the difficulties we encountered at the beginning of our study of a homomorphism fiom
maximal epic R-ring in
S
R
to a simple artinian ring
S
was that the
was not simple artinian. In this section, we shall
study an individual homomorphism from a ring to a simple artinian ring more closely; we shall be interested in the maximal epic R-ring and the dominion of R
in
S.
Our principal results state that they must both be semiprimary
rings.
We had better define the terms in the last paragraph. Given a
A
homomorphism from a ring
A-rings in
B;
to a ring
B,
we consider the set of all epic
since the ring generated by two epic A-rings is an epic A-
ring, and the union of epic A-rings is an epic A-ring, there is a unique
maximal epic A-ring in D
of
B
B.
The dominion of
such that homomorphisms from
B
A
in
B
is the maximal subring
that agree on the image of
A
in
123
B
must agree on
it is clear that the maximal epic A-ring in
D;
in the dominion of
A
in
R
in the next lemma. First, we define a useful construction. If
whose by
B
Proof:
Let
4:A - B
be a ring homomorphism; then the dominion of
is the centraliser of
1 0 1
B,B
in the
bimodule
the first is the identity map from They agree on
b + b a 1 - 1 M b.
d 0 1 = 1® d
shows that
for
to
B
in
D.
A, 1 and 2,
the
B,B
bimodule 1 0 1
so that the centraliser of
B '2A B,
to
which
D,
to a ring
B
41(B)02(B)
lies in
B aAB;
b e B
the second sends
B;
so, they must agree on
d
by
B
A,
Conversely, if we have two homomorphisms from that agree on
B 0AB.
to the trivial extension of
B
A
for we have the following
Certainly, the dominion centralises 1 ® 1,
two ring homomorphisms from
of
is a ring
is the ring
and whose multiplication is defined
R ® M,
bimodule structure is
R
lies
M2 = 0.
Lemma 7.15 in
R by M
is an R bimodule, the trivial extension of
M
and
B
The dominion has a simple description given
B.
C
is a quotient
D.
In order to give the reader some better idea of the dominion, we prove the following interesting lemma. The double centraliser of a subring A of
of a ring A
in
is the ring of elements of
B
B.
Lemma 7.16
Let :A } B
M(k),
the dominion of
Proof:
Let
A
A
in
in M(k)
morphisms from
B
agree on the image of consequently,
D
in
b A,
M
n
(k)
prove the converse. We use the last lemma. Mn(k)OAMn(k).
some integer
t.
As
Let
in
B;
and so, they agree on
B
Mn(k), Mn(k)
c1,...ct
in
B
is a k-subalgebra of
we define two homo-
D
It is clear that they the dominion of
lies in the double centraliser of
In the case where
in
A
b + x(bc - cb).
to
B
A
one of which is the identity map, the
BCx:x2 = 01
to
other of which sends
If
B.
is its double centraliser.
lie in the centraliser of
c
A
be a ring homomorphism. The dominion of
lies in the double centraliser of
D;
that centralise the centraliser
B
and A D
bimodule,
A
in
A
in
B.
is a k-subalgebra, we may
is the centraliser of
1 a 1
Mn(k)QAMn(k) = tMn(k) for
be the images of
1 0 1
in the direct
summands on the right of this isomorphism; then the dominion of
A
in
M
n
(k)
124
cl,...ct
is the centraliser of these elements, and
A
centraliser of
all lies in the
Mn(k).
in
A under a homomorphism
We wish to study the dominion of a ring A
from
to a ring
Lemma 7.17 in
Let
4:A - B
be a ring homomorphism; then the dominion of
is the endomorphism ring of the module
B
M
where
is given by the exact sequence:
b) = (0
a (O
our next lemma gives us a method of attack.
B;
Proof:
10AB).
(0
0 -* (0
B)
M
ar
lift to endomorphisms of
The endomorphism ring of
left multiplication by elements of such that
over the ring
B
BLAB
O
B
BOAB) -s M - Q
(B
lob).
All endomorphisms of
normalise
M
A
tM1 = let.
so we look for elements
B;
that
is defined by t
of
B
Hence, our lemma follows.
is semiprimary, so is
B
If
BMAB)
(B
BOAB)
(B
B
BOAR
O
B
;
therefore, we wish to
show that the endomorphism ring of a finitely presented module over a semiprimary ring is itself semiprimary. This has been shown by Bjork (71), who also had results in the direction of the corresponding theorem for left perfect rings; we shall present a proof of both these results next, generalising Pjork's theorem. Theorem 7.18
let M ring of
Proof:
Let
R
be a left perfect ring (or a semiprimary ring), and
be a finitely presented right module over M
over
R
R;
then the endomorphism
is left perfect (or semiprimary respectively).
First, we see that because M
is finitely presented it has the
descending chain condition on submodules with a bounded number of generators, since this is true for its projective cover by the left perfect condition. Consequently, it is the direct sum of finitely many indecomposables. Let sequence:
morphism of
P
be the projective cover of
0 -; I - P - M + 0
M
where
I
lifts to an endomorphism
intend to show that every endomorphism of of
P
so there is an exact
M;
is finitely generated. Any endoof
a
M
P
such that
a(I) S I.
We
lifts to a nilpotent endomorphism
or acts invertibly on a direct summand of
M.
125
a:M - M
Let integer
such that
n
is surjective, and endomorphism of
P,
map onto
P
M
as
lift to
anM = an+1M. Since a:anP _>. anP n is left perfect, a P = eP for some idempotent
anP = an+1P
EndR(P) e.
If
and
anP = 0,
then
a:anM i anM
presented; further,
such that cover of
Since
anM;
therefore,
is finitely
an M
is a surjective map; we shall show that
Let
anM.
be the kernel of
J/eI
S:eP + eP
so, there exists an endomorphism
eP/J = eP/eI;
RJ = eI.
I c radP, ei c rad(eP) a
cannot
(1-e)P
M.
which shows that
this forces it to be an isomorphism on this map; so,
since
anM = 0,
is the projective cover of
(anP + I)/I = eP/I n eP = eP/eI,
an M
There exists an
P, a(I) c I.
a:P -
and
is the projective
eP
must be an invertible map since it is inducing
an isomorphism module the radical of
we obtain an
$J = eI c J,
if
P;
infinite descending chain of modules of bounded number of generators, J x SJ x .... z ei ..., isomorphism on M
from
onto
n
which is impossible,
with inverse
a M
so
We see that
y.
Next, we show that if
gives a projection
y a
M
is an indecomposable module, its endo-
is indecomposable, every endomorphism on
M
R
is semiprimary).
elements hence,
that lift to nilpotent endomorphisms
are themselves nilpotent and so,
a.
al ... anP +I c al ... an-1P + I
The number of generators of a1 ... a P c I,
eventually,
n
that the radical of
EndR(M)
al
Let
a1 .., a P + I n
which implies that
EndR(P)
aN = 0 of
M
that is not invertible EndR(M) If
N.
M
aN = 0.
is nilpotent, so
hence, if
P;
R
so
This shows is left
is semiprimary.
a
N
such that
is an endomorphism
By the Nagata, Higman theorem, the EndR(M)
is semiprimary.
is an arbitrary f.g. module, we write
are indecomposable; by theorem 2.4 of Tachikawa (73)
perfect if
n,
EndR(M)
is semiprimary, so that there exists an integer
for any nilpotent endomorphism of
radical of
R
The
... an-1 M;
al ... an = 0.
is left T-nilpotent and so
be P.
al " ' an-1P 4 I.
is bounded for all
perfect by Tachikawa (73). This leaves the case where Then
al
as we
a, i
on
ai
.. anM z
provided that
M
of
P
have just shown. First, we deal with the left perfect case. EndR(M)
Since
is invertible or else it
lifts to a nilpotent endomorphism of the projective cover
elements of
is an
a
which proves our dichotomy.
anM,
morphism ring is left perfect (or semiprimary if M
and
J = eI, n n
is, and it is semiprimary if
R
M =
®=1M, ,
End(M)
is semiprimary.
where is left
126
Putting this theorem together with the preliminary discussion, we deduce what we were after.
Theorem 7.19 ring
Proof:
Let
be a homomorphism from a ring
+:R -+ S
then the dominion of
S;
R
in
to a semiprimary
R
is also semiprimary.
S
By lemma 7.17, the dominion is the endomorphism ring of a suitable
finitely presented module over a semiprimary ring; so it is semiprimary by theorem 7.18.
This theorem has a number of interesting consequences but we begin with the following lemma.
Lemma 7.20
Let
{D
be a family of subrings of a ring all of which are
.}
1
their own dominion in
Proof:
then
R;
The dominion of
nD,
1 1
in
D.
is its own dominion in
1 1
R
By considering the natural map from
is the centraliser of ROnD R
to
ROD R,
1
centraliser must lie in
D.
1
and so in
nD,; 1
R.
101
in
RO
nDi
R.
we see that this
1
therefore
nD,
1
is its own
dominion.
A semisimple artinian ring is always its own dominion, so there is the following result.
Theorem 7.21
The intersection of a collection of semisimple artinian subrings
of a ring is always semiprimary.
The kernel of a derivation d:R -+ M
is always its own dominion,
as one sees by considering ring homomorphisms from sion of
R
by
M;
morphism of a ring
R
to the trivial exten-
similarly, it is clear that the fixed points of an endoR
is a dominion.
It follows that the kernel of a family
of derivations or the fixed points of a family of ring endomorphisms on a semiprimary ring is also semiprimary. Similar results are true for perfect rings. The following corollary will be needed in the next section.
Corollary 7.22 of
R
Let
S
is semiprimary.
be semiprimary with subring
R;
then the centraliser
127
The centraliser is the kernel of the inner derivations determined by
Proof:
the elements of
R;
so we can apply the foregoing discussion.
R
We should like to study the maximal epic R-subring of a ring under a homomorphism from
to a semiprimary ring,
R
R
maximal epic R-subring must lie in the dominion of
First of all, the
S.
in
and this
S,
suggests that we attempt to find it in the following way. Let dominion of
R
in
S;
in general, let
be the dominion of
Di+l
at all stages the maximal epic R-subring of
Di;
so we hope that at some stage epimorphism and
D.
R
in
is the maximal epic R-subring of
R
R
in
must lie in
S
which implies that
Di = Di+l,
be the
DO
in
Di,
is an
R -> D.
This is
S.
what we shall show and it follows that the maximal epic R-subring of a ring in a semiprimary ring is always semiprimary, since we know that each
is
D,
i
semiprimary.
Let :R - S
Theorem 7.23 ring
be a ring homomorphism from
then the maximal epic R-subring in
S;
Proof:
As above, we let
we let
Di+l
S
DO be the dominion of
be the dominion of
R
in
R
to a semiprimary
is semiprimary.
R
in
S,
and in general
By theorem 7.19, each
Di.
D.
is
semiprimary. At each stage the number of elements in a maximal set of orthogonal idempotents is finite and decreases as the subscript increases; therefore it is eventually constant at some integer
s.
Once this stage has been
reached, the number of non-isomorphic principal indecomposable projective modules can only increase as the subscript increases, and it is bounded by s;
therefore it is eventually constant at some integer
we have reached this point for the subring Let
Nk
Nk = Nm n Dk.
that
be the radical of
D
k
D ;
t;
we assume that
.
m then for
Certainly, m n Dk C Nk.
we shall show
k > M,
For the converse, we note
that by construction a maximal set of orthogonal idempotents in
a maximal orthogonal set of idempotents in m; Dk/Nm n Dk
inside the semisimple artinian ring
Dm/Nm.
to s}
If
D IN
Dk/Nm n Dk
modules; consequently, if M m is a maximal set of orthogonal idempotents in
Dk/Nm n D D IN
mm
eiDk/Nm n Dk = e.Dk/Nm n Dk = x M
i vi (E.), i
for skew fields
if and only if E.,
i
is also
Further the number
of non-isomorphic principal indecomposable projectives over equal to the number of simple
Dk
this continues to hold for
is
{e :I = 1 i
eiDm/Nm = e.Dm/N.
it is now clear that
128
Dk/Nn n Dk
where
x v (Fi) i
is a semiprimary subring of the skew field
F.
1
and consequently a skew field itself; it follows that Nk = Dk n Nm.
Ei
Nu-1
Suppose that
m Dm+u
is the dominion of
Dm+l
c D D IN IN m+u m+u - m+l m+l
= o.
in
Dm,
Dm+lINm+l
Dm+uINm+u = Dm+lINm+l'
m m
is contained in the
Dm+uINm+u
dominion of the semisimple artinian ring
Since
D IN .
in
Dm/Nm;
so,
In general, assume that then
D /Nj n D /N3 n D = D m+u m m+j' m+u m+j m
D
/Nj+l n D
m+u m
m+u
c D
m+j
/N &+l n D
m
m+j
and they differ only in the socle of the rings which implies that as right modules over
M
where D
m+u
is a semisimple module; this implies that the dominion of
/N j+1 n D
m
A c B
$ M, = D /N j+1 n D D /N j+1 n D m+j m m+j m+u m+u
D /Nj+l n D , m+u m+u m
m+n
is itself and must equal
and B= A$ M
as right
in
B.
mOl x l8b
By induction, we find that
is the maximal epic R-subring in
/NJ+1 n D m+j+1'
m+j+l m
For, if
module, then
A and
BOAR = A $ MOAB = A®AB ® MOAB,
D
for any
m
in
M
and
b
which proves that D = D m+u-1 m+u-l m+u Therefore, the maximal epic R-subring
D S.
is semiprimary.
An odd consequence of this result is that a ring has a homomorphism to a simple artinian ring if and only if it has an epimorphism to
a simple artinian ring: let :R -> S artinian ring, and let
E
be some homomorphism to a simple art
be the maximal epic R-subring in
S;
then
semiprimary and so it maps surjectively to some simple artinian ring, the composite map from
R
to
S'
E
is
S';
is an epimorphism.
The simple artinian spectrum of a k-algebra There has been much talk about a theory of non-commutative algebraic geometry. It is not my intention here to add to this, but rather to point out that our preceding theory does give us a functor from rings to topological spaces which is a simple summary of the information on possible homomorphisms from the ring to simple artinian rings. It would be possible to equip this space with a sheaf of rings, and to represent modules over the ring as a sheaf of modules over this sheaf of rings; however, in the absence of any obvious use for this machinery, I shall leave it to future mathematicians of greater insight. Finally, the theory here is stated only for k-algebras over a field
k;
it is possible to present the theory for
arbitrary rings, but the care required in order to avoid simple pitfalls
129
makes the discussion cumbersome.
We begin by considering a particular pre-ordered abelian group functorially associated to a ring
R.
We define
to be the abelian
M(R)
group generated by isomorphism classes of f.p. modules over We define an ordering on
[A $ B] = [A] + [B].
tions
a positive cone.
[A] > 0
for all f.p. modules
is an exact sequence, we define If
4:R -
S
functor.
M(R)
Given
defines an order-
n
to
M(S),
R
S
is a right exact such that
are simply order-preserving homomorphisms
for varying integers
M
R
It is now clear that our Sylvester module
A.
m
such that
such Sylvester module rank functions,
n
[A] + [C] - [B] > 0.
[A] -s [A®RS]
2
to 1 Z
M(R)
A -. B -. C - 0
since
M(R)
rank functions on a k-algebra from
and if
has an order unit since there exists
for any module
[nR] > [A]
by specifying
and
[B] - [C] > 0,
is a homomorphism of rings,
preserving homomorphism from
A,
with rela-
R
M(R)
p
,
[R]
goes to 1.
we form a family of
i
{Egipi.Egi = 1,
This qi > 0, qi E @}. arises from the ring theory in a natural way, for if the homomorphism
Sylvester module rank functions
YR - S.
gives rise to the Sylvester rank function
morphism from
R
to
xSs.,
pi,
we have a homo-
and the various homomorphisms from this semi-
simple artinian ring to simple artinian rings give rise to the rank functions So, our space of equivalence classes of homomorphisms from
Egipi'
R
to
simple artinian rings has the structure of a (Q-convex subset of the space of
all order-preserving homomorphisms from
M(R)
the subspace topology (once the space of all
to the reals, and can be given order-preserving maps from
M(R) to the reals has been given a suitable topology). A particularly important set of points in this space are the extremal points which are those Sylvester module rank functions that do not lie in the linear span of some other set of Sylvester module rank functions. We should like to be able to show that every Sylvester rank function arises in a unique way as an element in the linear span of extremal rank functions. In order to prove this it is convenient to look at homomorphisms to simple artinian rings in a slightly different way.
If we have a homomorphism from a ring ring
M
n
(D)
where
module as an R,D
D
R
to a simple artinian
is a skew field, we may regard the simple left M
bimodule
M
such that
[M:D] = n;
a bimodule, we automatically have a homomorphism from
n
(D)
conversely, given such R
to
M
n
(D).
The
Sylvester module rank function associated to such a bimodule is given by p(A) _ [AODM:D]/[M:D].
We note that the endomorphism ring of
M
as an R,D
130
bimodule is the centraliser of R,D
in
M
n
A decomposition of
(D).
then either
as an is
M.
If this happens
allows us to write p = ([M1:D]p1 + [M2:D]p2)/[M:D].
pi
M
where the rank function associated to
M = M1 ® M2
bimodule,
R
is not extremal or else it is extremal and the image of
p
is a distinct subgroup of
p
We shall show a sort of converse to
1/[M:D]p.
this statement.
Any Sylvester rank function lies in the linear span of some
Theorem 7.24
set of extremal rank functions.
Proof:
values in
be a ring and let
R
Let
$;
some skew field a for
arises from a homomorphism from
p
we prove our theorem by induction on
be the simple left M
n
for
Mn(D)
it is clear
module considered
(D)
bimodule.
Under the assumption that artinian rings
S1,S2
and
S
p
is not extremal, there exist simple
and homomorphisms
the composite map induces the rank function different to
rank function
p
artinian ring
M (D) o S = M (E), n R np
p.
p
R -+ S.
induces the
By theorem 7.3, there exists a simple and from this we form the simple artinian
let M be the simple left
module regarded as an R,F bimodule. Since
M = (MODF)R
such that
R -> S1 X S2 -; S
whilst
i
Mnp(D) Ml(D)Mnp(E) = Mnp(F);
Mnp(F)
to n;
since all such arise from homomorphisms to skew fields which
n = 1,
ring
R
so D;
are extremal points. Let M as an R,D
be a Sylvester rank function taking
p
as
R,F
bimodule; on the other hand,
R
lies in
Mnp(D),
R
lies in
S1 X S2
and
so the centraliser of
R in M (F) contains an idempotent e such that np the rank function associated to eM is the rank function p1; consequently, eM
is not isomorphic to a direct sum of copies of MO F;
since
M
is a
D
module of finite length, the Remak, Krull, Schmidt theorem implies that M11DF
M1 ® M2
for some
R,F
bimodules
M1
and
M2.
Therefore,
p
in the linear span of the rank functions associated to the bimodules however,
[Mi:F] < n,
and so the theorems follows by induction on
lies M,; i
n.
By applying the argument used in this theorem in the light of the information it gives us we can strengthen our conclusion yet further.
Theorem 7.25
Any Sylvester rank function has a unique expression as the
weighted sum of extremal points; further, if
p = 1/mEmip
i
where
Emi = m,
131
h.c.f.{mi} = 1, then
and
maps onto
p
is an extremal rank function mapping onto
pi 1
Let
Proof:
be the ring on which we have the Sylvester rank function
R
Further, let p = 1/mEmipi extremal points p .,
S. = M
(D.) 1 Si
(D
pi
maps onto
pi
as a weighted sum of
p
so that to each rank func-
1/s
R
p.
to a simple artinian ring
1
1
si
be some expression of
where
we have a homomorphism from
tion
M
1/s,Z,
1/ Em1,s1,.
where
D, 1
is a skew field; let
module regarded as an
.)
1
R,D
bimodule. Let
,
1
is a skew field, be a homomorphism inducing R,D
regarded as
M (D) n
be the simple left
M. 1
p.
-> MN (E)
where
(D),
n
D
be the simple left
M
Let
bimodule. Let N = Es.m,.
morphism R -> xMSi(D.) -> xM sim1 (Di)
R -. M
We also have a homo-
1 1 1 for some skew field that induces
i
the rank function
on
p
in order to do this, it must assign the rank
R;
(D,), and we we have specified xM i Simi 1 We form the simple artinian ring MN(E) o Mn(D)
to a simple module over the ring
1/n
homomorphisms from
to
D.
E.
R
which is isomorphic to simple artinian ring the simple left xM
lies in k
1 Si
MN (E') p
for some skew field
(E') MNp(D) 14n?D)MN t p
MN (F) P
MN lE)MNP
module considered as an
(E) Z MN
M
Let
(F).
be
P R,F
bimodule. Since
R
M (xMm1 (k))
where
its centraliser contains a copy of
(D.),
we form the
E';
p
is the centre of
F,
and therefore,
M = ® (Mi0D F)mip
Since the rank function induced by the bimodule
M.OD F
R,F
as
bimodule.
is extremal and maps
1
onto
where
l/siZ
also lies in as
Mn(D),
divides
p,
and
p maps onto
1/t
divide
so, on setting
is indecomposable; however,
q = Np/n,
we see that
D' t,
McDF = ® (M.OD F)m1P/q,
therefore,
N
divides
n.
there is a homomorphism to a simple artinian ring
is a skew field that induces but it is clear that
p
p;
we have just shown
takes values in
1/N Z
R
M = (MeDF)q
bimodule. By the Remak, Krull, Schmidt theorem, we see that
R,F
where
MiQDF
si = EM.:D.],
q If
Mt(D')
N must
so that the last
statement of the theorem follows. The first statement of the theorem is clear by now, since for any skew field
G
containing
F,
M®FG = ® (MiOD G)miP/q 1
is the unique representation of MFG as a direct sum of indecomposable bimodules, which shows that any representation of
p
as a weighted sum of
other rank functions is refinable to this representation in terms of extremal points.
132
There is one further point we should mention in this discussion of
bimodules; the last result shows that the extremal points correspond
R,D
to bimodules that remain indecomposable under 'extension of skew field'; we should like to be able to characterise those extremal points that arise from epimorphisms to simple artinian rings in some similar way. Before we can do this, we need to link the notion of epimorphisms on rings to the associated functor on categories of modules over the rings. The next result is due to Silver (67). A functor from a category full if it is an embedding, and if functor, and
a
Theorem 7.26
Let :R - S
If :R + S
MGRS
be a homomorphism of rings; it is an epimorphism
MASS = M:
McS(SORS)
O:mod S + mod
map it induces from
MGRS
to
Conversely, if S
a(s) = sal;
S
is a full functor.
R
module
S
modules
M,
M,N
an
R
linear as we see by considering the
NORS.
we have an
is a full functor,
R linear map from
so this defines an
s®l = a(s) = a(l)s = les,
R + S
S
:mod S - mod R
HomR(S,S&RS) = Hom (S,SORS) ;
given by
so for any pair of
N must be
to
is said to be
is in the image of the functor.
a
is an epimorphism, then for any
linear map from m
D
are objects in the image of the
is a map between them,
if and only if the forgetful functor
Proof:
to a category
C
M,N
S
to
S
SORS
linear map, and SORS = S,
which implies that
which shows that
is an epimorphism.
The following consequence is noted by Ringel (79).
Theorem 7.27
Let
be a ring, and let
R
EndR(M)
is a skew field
from
R
to M
way.
M
Proof:
then if and
(D)
M
and
be an epimorphism from
is the simple
[M:D] = n where
skew field, and
[M:D] = n;
R
such that
then the corresponding map
Mn(D) module.
S
module,
S = M (D). n M is an
Conversely, if
Mn(D)
be a module over
is an epimorphism, and all epimorphisms arise in this
n is the simple
Let R + S
D,
M
[M:D] = n,
R
R module such that
we have a map from
module, its structure as an
to a simple artinian ring
EndR(M) = EndR(M) = D,
R
to
M
EndR(M) = D, n
(D)
S;
a skew field,
a
and given any
R module is a direct sum of copies of
133
M;
so it is clear that all
R module maps between
module maps, which shows that the map from
R
S
modules are
is an epimorphism
Mn(D)
to
S
by 8.4.
So this shows that epimorphisms ought to correspond to simple bimodules. Our next theorem proves that this is precisely the case.
Theorem 7.28 values in
Let
be a Sylvester rank function on the ring
p
then
!-Z; n
p
ring if and only if there is a skew field M,
D
and a simple
that induces this rank function on
[M:D] = n,
R
taking
is induced by an epimorphism to a simple artinian R,D
bimodule
R.
R + M (D) is an epimorphism where D is a skew field then the m simple left M (D) module M considered as an R,D bimodule is simple. m Conversely, if M is a simple R,D bimodule for some skew Proof:
If
field,
[M:D] = m,
the radical of N = 0,
and
let
R',
NM
R'
R'
then
R,D
if
N
is
therefore,
is a direct sum of simple artinian rings, since it is a
R'
semiprimary ring (theorem 7.19); if e x 1,
R in M (D); m sub-module of M;
be the dominion of
is a distinct
em
e
is a central idempotent in
is a distinct sub-bimodule of
M
and so
must be simple artinian, and the homomorphism from
R
R',
Hence,
e = O.
to
R'
is an
epimorphism.
The results of this section show that the space of all Sylvester rank functions on a ring form a sort of infinite dimensional
T-simplex since
every point in it may be expressed in a unique way as a weighted sum of extremal points; this suggests that some detailed study should be made of the homomorphisms from a ring to simple artinian rings that induce extremal rank functions on it; at present, little has been done in this direction. In addition to the simplex structure on the space of all Sylvester rank functions, there is also a topology reflecting the notion of specialisation; we define a rank function rank function
p2
if
p1
p2(M) < p1(M)
to be a specialisation of the Sylvester for all f.p. modules
M;
it is easily
checked that this generalises the standard notion of specialisation. The closed sets in our topology on the space of Sylvester rank functions are those that are closed under specialisation. We may also reformulate a version of the support relation discussed by Bergman (76) in terms of the simplex structure on the space of Sylvester rank functions and the specialisation
134
topology. We say that the extremal Sylvester rank function extremal Sylvester rank functions
{pi:i = 1 to n}
some point on the interior of the face spanned by we have little information on this structure.
p
supports the
if it specialises to {pi}.
Again at present
PART II
Skew Subfields of Simple Artinian Coproducts
137
THE CENTRE OF THE SIMPLE ARTINIAN COPRODUCT
8
In this chapter, we shall begin a fairly specific study of the simple artinian coproduct with amalgamation; we shall attempt to find the centre of such simple artinian coproducts in terms of the factors and the amalgamated simple artinian subring. Unfortunately, our results are at present incomplete, so we shall begin by stating the conjecture, and then we shall summarise the cases for which it is known to be true, before we prove them.
These results will be used in the next chapter to study the f.d.
division subalgebras of a skew field coproduct.
Conjecture 8.1
The centre of
S1 o S2
lies in
S,
except possibly when
S
both S,
and
S1
are of rank 2 over
S2
in this case, the centre lies in
S;
or is the function field of a curve of genus
with
0
over its intersection
S.
The function field of a curve of genus 0 over a field field K,
and
such that
F t
LOKF = L(t),
where
L
K
is a
is a suitable f.d. extension of
is a transcendental. We shall present examples to show that all
curves of genus
0
arise as centres of skew field coproducts at the end of
this chapter.
This conjecture is known to be true when subfield of
S1
and
S2.
S
is a common central
It is also true for skew field coproducts
amalgamating an arbitrary skew subfield, except possibly when there are only two factors both of which are of dimension 2 over the amalgamated skew subfield. It may also be proven if one of the factors is simple artinian.
The result that we use to prove our conjecture when we can is due to Cohn (84); it summarises the connection between the centre of a fir,
138
and the centre of its universal skew field.
Let
Theorem 8.2
T
be a non-Ore fir with universal skew field of fractions
then the centre of
U;
is the centre of
U
T.
The proof is too long to include here; the interested reader should look at Cohn (84). Our first special case of 8.1 is a direct application of this.
lies in
E1OMm(E2)
The centre of
Theorem 8.3
E,
provided that
[E1:E]
E
or
is greater than 2.
[Mm (E2):E]
Proof:
El 0 Mm (E2) E
El E Mm(E2) = Mm(R),
is the universal simple artinian ring of fractions of where
R
is a fir, since, by Bergman's coproduct
theorems (see chapter 3), and the Morita equivalence of every submodule of a free dimension imply that
Mm(R)
and
R,
module is free. Further, the assumptions on
R
cannot be Ore, and so, by 10.2, the centre of the
R
universal skew field of fractions of
R
is the centre of
So, the centre
R.
M (R) is the centre m and, in particular, consists of units of R.
of the universal simple artinian ring of fractions of M (R), m
of
which lies in
is the ring of endomorphisms of the
R SoM (E
m of
R,
where
Mm(R),
is the unique simple
S
M
M
module
(R)
Mm(E2)
module, and the units
2
R
induce the automorphisms of this module. By 2.18, the group of auto-
morphisms of this module are induced by the automorphisms of module; so the units of the centre of
R
Mm(E2) - E,
lie in the copy of
lies in the copy of
Mm(R) = E1 E Mm(E2) of
R
and
must also lie in f
E2 E2.
is an element of
cannot be central. Therefore, the centre of lies in
E.
E2
S
Mm(E2)
as
inside it; therefore,
inside it, and the centre of However, if E1 - E, MM(R)
e
of x fe,
is an element so that
which lies in
E2
e
actually
But this is the centre of its universal simple artinian ring of
fractions, as we saw at the end of the last paragraph.
Since this is all that is used in the following chapter, and the remaining proofs of results towards conjecture 8.1 are complicated, the reader may wish to bypass them for the time being; in this case, he may wish
139
to look only at the examples which show that all function fields of curves of genus
0
arise as the centre of a suitable simple artinian coproduct.
We wish to find the centre of the simple artinian coproduct of two simple artinian rings over a common central subfield. The next two theorems deal first with the non-Ore, and then with the Ore case. Of the two, the nonOre case is more complex.
Theorem 8.4
The centre of
that one of
[Mm (E1):k] 1
Proof:
is equal to
Mm (E1) o Mm (E2) k 1 2 or
k,
provided
is greater than 2.
[Mm (E2):k] 2
The idea of the proof is to reduce to the previous theorem by re-
placing m (E1)
by a skew field closely related to it; more precisely, the
1
skew field will be a twisted form of
Mm (E1).
A simple artinian ring
S
1
is said to be a twisted form of a simple artinian ring central subfield
k
SOkL = S'OkL.
such that
S',
if we have a
of each of them, and a f.d. field extension
L
of
k
For a detailed discussion of twisted forms of
algebraic structures, the reader should consult Waterhouse (79). A twisted form of
that is a skew field need not exist;
Mm (E1) 1
however, the case where one does exist is an important special case:
Special case: there exists a central division algebra [D:k1 = m1, DOkE1 E 0 L i k
such that
is a skew field, and
is simple artinian for
D
such that
has a Galois splitting field
D
L
i = 1,2.
In this case, our idea is to show that ((Do El) o M
km2 (E2
that
He L = (Mm (E1) o Mm (E2))OkL. k k 2 1
(DOkE1) o Mm (E2) k 2
Mm (E1) 0 Mm (E2 ), k 2 1
has centre
if
k;
C
From theorem 10.3, we know is the centre of
it follows from the isomorphism that
COkL = L,
so
C=k. Consider the ring centre
k.
Therefore,
DokE1 o Mm (E2); k 2
S = (Do El o Mm (E2))&kL k
k
2
then by theorem 8.3, it has is simple artinian and it
140
(DOkEl k Mm2(E2))0kL = (DOkElOkL) L Mm2(E20 L)
is a universal localisation of
Mm (E10kL)
which is isomorphic to
an hereditary ring with
L Mm (E20kL) 2
1
unique rank function, since each of the factors is simple artinian. On the (M m
other hand, this ring is visibly isomorphic to
(E1)
2
(Mm (E1) o Mm (E2))akL
has the universal localisation
k
1
which is semisimple
2
artinian and hence weakly finite, and so must be a subring of must be epic over it, and so it is actually the whole of centre
the centre of
L,
Mm (E1) o Mm (E2) 1 k 2
is
division algebra
Since
S.
until there is.
k
be the commutative field
K
Let
k(xl,...,xm),
where ¢
D =
let
C
be the centre of
DokE
and so,
Let
D.
has a skew field of fractions.
be the skew field of fractions of
Ei
be the skew field of fractions of
Mm(F') = Mm (E') o Mm (E'). 1
k
LOkC
then
Mm (E1) o Mm (E2 ),
k
that the centre of
M
(E
m
)
1
o M
k
(E
m2
)
is
If
k L
2
is central
Mm (Ei) o Mm (E'). C 2 1
Mm (Ei) o Mm (E2) 1
F'
2
C
is central in
2
once we know that the centre of
and let
Mm(F) = Mm (E1) o Mm (E2).
where
FOkC,
C
E.
1
Then we have the equation
1
is
is a Noetherian domain for any skew
DOkE
It is easy to see that
in
{xi}
of this field
given by the cyclic permutation of the indices. We form the
mi,
E,
has
S
k.
a set of commuting indeterminates. There is an automorphism
field
S
of the sort required for the application of the special
D
case, we extend the centre
skew field
But
S.
In order to deal with the situation where there is no f.d.
General case:
of order
which
k Mm (E2))OkL,
1
is
C,
So,
we shall know
2 k.
2
1
However, DOkEl,
and
KOCEi
case, the centre of
[D:C] = ml2, DOkEj
Mm (E1) o Mm (E') C
1
Mm (E1) 0 Mm (E2) 1 k 2
is the skew field of fractions of
is actually a skew field for
to be
k,
is
C,
1 = 1,2,
so, by our first
which forces the centre of
2
as we wished to show.
We wish to deal with the remaining case of the coproduct of two
141
simple artinian rings
and
S1
such that
S2
[Si:k] = 2;
clearly, both
are commutative fields.
Theorem 8.5
Let
the centre of over
E1
and
be a pair of quadratic extensions of
E2
Then
k.
is purely transcendental of transcendence degree
E1 o E2
1
k
k.
This was also shown by Cohn by a different method.
Proof:
By Bergman (74') or a simple calculation, the centre of has the form
is a free module of rank 4 over
El k E2
and
kit],
El k E2,
must be the central localisation of
El JC E2
4 over the central subfield
E1 k E2 k[t].
and must be of rank
Since it is not commutative, this must
k(t).
be the whole centre.
We finish off this chapter by shdwing that all function fields of curves of genus
0
over
k
occur as the centre of a suitable simple
artinian coproduct with amalgamation. We shall need the next lemma in the proof of this.
Lemma 8.6
M2(L) LXL M2(L) = M2(L[t,t-1])
All embeddings of
Proof:
L x L
in
for a commuting indeterminate
t.
M2(L), are conjugate to the embedding
along the diagonal; so we may assume that the first factor has matrix units ell, e12, e21, f21' and e22.
whilst the second factor has matrix units
and e22,
Since
lies in
f12
e11(M2(L) L
M2(L))e22'
ell' f12'
it takes the form
X -L
I\(O O Ot)
when it is written as a matrix over the centraliser of the first
factor, and similarly,
f21
has the form
from which it is clear
(s 00),
that
Since the centraliser of the first factor is generated by
s = t-1.
the entries of these matrices, the centraliser is isomorphic to where
t
L[t,t-1]
is an indeterminate, since the dimensions of the coproduct we are
considering is infinite.
Suppose that over
k;
that is,
Galois group
G.
FOkK Then
F
is the function field of a curve of genus K(t)
F
for some Galois extension
is a twisted form of
k(t)
K
of
k
0
with
and so by twisted form
142
H1(G,PGL2(K))
theory, it corresponds to a suitable element of PGL2(K)
is the automorphism group of
Waterhouse (79)). over
K,
PGL2(K)
Since
this element
quaternion algebra
DF
a
k;
the universal splitting field of commutative subfield of
Theorem 8.7
DF;
K
(for example, see
is also the automorphism group of
moreover, Roquette (62) shows that DF.
Let
M2(K)
corresponds to some central
H1(G,PGL2(K))
of
over
over
k(x)
since
F
is
be a maximal separable
L
[L:k] = 2.
M2(k) o DF = M2(F); L arise as the centre of a suitable
Name everything as in the last paragraph; then
so all function fields of curves of genus
0
simple artinian coproduct with amalgamation.
Proof:
u M2 (L) = M2(L[t,t 1]) by the lemma above. So, LxL since both sides of this equation are obtained
(M2 (k) a DF)QkL = M2 (L)
L (M2 (k) o DF)OkL = M2(L(t)), L
by central localisation of the corresponding sides in the first equation. Let
F'
be the centre of
M2 (k) o DF;
then
F'OkL = L(t),
so
L
that
F'
is either purely transcendental or else it is the splitting field
of a suitable quaternion algebra
DF,.
However,
M2(k) o DF
F'MkDF,
so
L
that
F'
cannot be purely transcendental since it splits
DF.
Roquette
(62) shows that the universal splitting field of a central quaternion algebra can split no other central simple algebra, so which is what we want.
F'
must be isomorphic to
F,
143
FINITE DIMENSIONAL DIVISION SUBALGEBRAS OF SKEW FIELD COPRODUCTS
9
We intend in this chapter to make a detailed study of division subalgebras of skew field coproducts; since the method developed may be applied in much greater generality, we shall develop the results in this context and then specialise the results to skew field coproducts. In later sections of this chapter, we shall briefly consider its applications to related rings.
Division subalgebras of universal localisations For the purposes of this section, we shall assume that the rings R,F,K,k
satisfy the following conditions:
Assumptions: tive field.
R p
n
R
centre
is a simple artinian ring
,
p K,
k
is a commuta-
is a rank function on the f.g. projectives over such that the universal localisation of
values in I 7 tion,
is a left semihereditary k-algebra, where
where
K
M
n
(F)
where
is a regular extension of
F
R
R
taking
at the rank func-
is a skew field with
k.
On the whole, there is nothing to this beyond naming the rings that we are interested in.
An extension of commutative fields if
KOkL
K D k
is said to be regular
is a domain for all commutative field extensions of
k.
It is useful to reformulate our problem; we wish to find whether a f.d. division algebra with centre
K,
D
over
k
embeds in a simple artinian ring
a regular extension of
and ask for what numbers does
D
k.
embed in
Mn(F)
We can broaden this a little M (F). m
This has a useful re-
formulation.
Lemma 9.1 Let
S
Under the previous assumptions,
be the simple module for
D
0
kF,
Do0kF and let
is a simple artinian ring. [S:F] = s;
then
D
embeds
144
M
in
M
if and only if
(F)
Let
Proof:
divides
s
be the centre of
C
m.
zero ideal intersects non-trivially with so any non-zero ideal of
COkF
D°OkF = D QC(COkF),
then
D;
C&kF.
COkF = (C®kK)IIKF,
In turn,
COkK which
intersects non-trivially with
is a field by our assumption on
K.
D°®kF
So,
so any non-
must be simple artinian.
We are left with the last sentence of the lemma. If D°
M (F),
in
m
embeds in
M (F°),
embeds
D
which is the endomorphism ring of the
F
M
module
Consequently, we can give
Fm.
module, and so,
Fm
Fm
the structure of a left FOkD
is a direct sum of copies of
The argument is
S.
reversible.
It is important to bear this in mind, as most of the embedding theorems will be stated in terms of the dimension over D°0kF
embeddings of
D
in matrix rings over
It is also worth remembering
F.
M
that the same information is conveyed by the rank as MM(F)®kDo
simple
FOkD°
module of a
(F)
M module, since the Morita equivalence of
MM(F)&kD°
shows that the dimension of a simple F®kD module over
times the rank of a simple
f.d. division algebra over is equal to
R
MM(F)QkD0
as
Mm (F)
F
and
is
m
module.
Suppose that our previous assumptions hold, and let
Theorem 9.2
over
of a simple
F
module, since this actually gives us information about all possible
D
be a
then the rank of a simple RP0kD module
k;
h.c.f.M{p(M)}
M
as
runs through f.g. left RekD
P
submodules of free
ROkD° modules.
Proof: That the rank of a simple
D module as an
R
P
where
M
°
R 0 D p
RPdb RM,
module must
R
P k
divide this h.c.f. is easily seen by considering the
k
modules,
is an RMkD module. We are left with the converse. R QkD°
First, we see that
is a universal localisation of
P
ROkD°
at the full maps with respect to
Cramer's rule. Let
e
p
be an idempotent in
isomorphic to the unique simple module
S
over the ring
RpQkD over
R.
such that
0kD
R
.
So we can use (Rp0kD )e
Then, by 4.3,
is e
P
is stably associated to some map between f.g. projectives induced up from some map between f.g. projectives over R®kD , a:Q -+ P;
we may choose this
map to be associated to it e The image of is isomorphic to of O o t (Rp0kD ® S, so this is also isomorphic to the image of our map R 0 a P R .
)
145
between f.g. projectives over
R 0kD°O. P
M c P
Let
this map extended to
be the image of
Rp0kD° R
centralises the action of
over ROkD ;
a
(RpOkD )M c (RPOkD
is
and
then the image of
Since D
D°P.
)®R
P t
we may actually write this equation
RP,
At this point it is useful to recall that 1.10 shows that
as
R M c R 0RP.
if
Npc Rs pis a f.g. submodule of R,
R N
the rank of N'
P
to find the rank as is isomorphic to the rank of
(RPOkD°)t ® S,
as an
S
module of
Rp
N c N' c RS.
such that M c P,
the case we are examining, we have an RMkD module
module
R
as
P
R modules
is equal to the minimal rank of
and we wish
RPM c RPa P, since we know that RPM R and we shall be able to deduce from this
R M
module. We know by 1.10 that the rank of
R P
R
p
In
as
P
module is the minimal rank of an
R
module
such that M c M' c P; M having
M'
our final step is to show that there is an ROkD module above this minimal rank; in fact, we shall see that if
so does DM'
which is an
RMkD° module since
D0M' = EdiM'. diM'
centralises
be a basis of D over
{1 = d0,d1...d}
Let
has the minimal rank,
M'
D°
is isomorphic to
M'
as
R
k;
R.
then
module because the action
i
of D commutes with that of PAkD° p
module. If
containing
M,
M1
and
and
R,
M2
d M' i
contains
M
M
since
is an
are both submodules of minimal rank
q
in
consider the exact sequence:
0+M1 f1 M2->M1®M2+M1+M2+0 which is split as a sequence of
2q = p(M1 n M2) + p(M1 + M2);
So, q,
R modules since
is left semihereditary.
but both terms on the right are at least
which implies that they are exactly
induction shows that
R
q.
So
p(D °
We see that the rank as an equal to the rank as an
Rp
module of
(Rp0kDo)t 0 S
R module of the ROkD module, D M',
a f.g. submodule of a free module. Consequently, h.c.f.M{p(M)} through the f.g. as an
R
and an easy
p(M1 + M2) = q,
is
which is as
M
runs
ROkD submodules of free modules divides the rank of
S
module. The converse has already been noted. P
Division subalgebras of simple artinian coproducts In this section, we apply 9.2 to the problem of finding the f.d.
division subalgebras of a simple artinian coproduct amalgamating a simple artinian subring under the assumption that conjecture 8.1 holds; since we
146
have verified this conjecture in chapter 8 for a number of cases including the simple artinian coproduct amalgamating a central subfield, this is not unreasonable.
There is a spurious generality in dealing with the amalgamation over a simple artinian ring, since this may be reduced to amalgamation over a skew field instead; whilst performing this reduction we set up the notation and names for this section. (E.) for i = 1,2, where S and S = M (E), and S = M n 1 n. 1 is a skew field. are simple artinian rings and S lies in Si, and E.
Let
S.
Since
.
embeds in
S
the centraliser of S1 S S2 = Mn(Mm (E1) 1
M 1
(E 1) 0Mm (E2), E 2
divides
n
Si,
Mn(k)
Mm (E2)),
E
so we need only consider the ring
S1 S S2
since
R modules where
is the
(E
by
n
localisation of
then
ring,
on
M (F), M
at
R where
p
p,
modules
.)
R = Mm (E1) E Mm
.
1
and the unique rank
will be called
(E2)
Let
p.
2
takes values in Rp,
p
1
.
and the universal
7l,
was shown in 5.6 m to be a simple artinian
is a skew field; we use
F
matrix ring over it.
n
1
1
m = l.c.m. {ml,m2};
then on taking
we see that
2
We shall call the rank function on M m function on
mi = ni/n;
so let
nil
S1 S S2
in
p'
for the rank function
modules.
Mm (F)
be the intersection of the centre of
with E; m we recall that our conjecture states that either this is the centre of our Let
C
M (F)
coproduct, or else, the centre is a function field of a curve of genus over
C.
Theorem 9.3
C.
We use the preceding notation; further, we assume that the simple
artinian coproduct considered satisfies conjecture 8.1. Let division algebra over M m(F)
0
At any rate, we are assuming that it is a regular extension of
module is
then the rank of a simple
C;
h.c.f.i
{pi(A,.)} rj
as
D°aCMM(F)
runs through f.g.
A..
7
be a f.d.
D
module as 0
D ®CM m (E.) 1
7
modules.
Proof:
We have
(El) E Mm R = M M 1
(E2)
and
2
R
= M p
m1
(E) OM (E ) 1 E m2 2
= M (F). m
Our conditions fulfil those required for theorems 9.2 to apply. Therefore,
the rank of a simple Mm(F)QCD° module is
h.c.f.M{p(M)}
as
M
runs through
147
ROkD°
f.g.
submodules of free RokD modules. In order to change this to
the form of our theorem, we have to deal with a special case.
Special case: Assume that
has centre. C.
E
Then
ROCD E
Since
(MM1(E1) E Mm2(E2))OCD = (Mml(E1)OCD) EUCDo(Mm2(E2)OCD ).
has centre
C,
is simple artinian, which allows us to use Bergman's
E11CD0
ROCD°
coproduct theorems. By 2.2, any
submodule of a free module has the
M12(Mm (E1)QCDo)(RACD ) ® M20(Mm (E2)QCD°)(ROCD°),
form
1
where
is a
M.
2
submodule of a free m (Ei)QCD module. We see that
p(M) = p1(M1) + p2(M2),
1
and so, since the rank of a simple equals
h.c.f.M{p(M)}
it is also equal to of free
M
as
Mm(F)OCD° module as
Mm (F)
module
runs through f.g. submodules of free
h.c.f.i.j{pi(Bij)}
as
Bij
R11CD0
runs over f.g. submodules
Mm (Ei)0CD modules. 1
Mm (Ei)0CD°
Given any f.g.
module,
Ai.,
it has a presentation:
J
1
-' AiJ . 0
O -+ Bij -+ s(Mm (Ei)®CD ) 1
and so, it is clear that f.g. module over
h.c.f. ..{p 1J
Mm (Ei)QCD
as
.(A 1.,J )}
1
is equal to
A 1.,J
runs through every
h.c.f.{pi(B,j)}
as
runs
Bi
1
J
through f.g. submodules of free modules.
Before passing to the general case, we note the following corollary which is of importance for a reduction to the case we have just dealt with.
Corollary 9.4 Let
S
division algebra over S8C(x)
module is
be a simple artinian
then the rank of a simple
C;
h.c.f. ,{pS(A .)), J
modules, and
A
,
algebra, and let
C
where
pS
D
be a f.d.
(SoC(x))OCD° module as C
is the rank function on
runs through the f.g.
SOCD° modules. In particular, if
J
the centre of (S
8
S
is a regular extension of
C(x))OCD module over
module as
S
module.
S
]
S o C(x) C
C,
the rank of a simple
equals the rank of a simple
SMCD°
148
We saw in theorem 8.3 that the centre of
Proof:
so, must be
lies in
S o C(x) C
C,
and
therefore, we can apply the first case to prove our corollary.
C;
SOCD
The last sentence follows since
is simple artinian in this instance.
General case
We return to the general case. We assumed that the centre of
Mm (E1) o Mm2
is either
(E2)
or else the function field of a curve of
C
E
1
genus
over
0
In either case, it is a regular extension of
C.
by the last part of 9.4, the rank of a simple
2
E
(E1) o m (E2) 0 C(x))QCD module as an
(Mm
E
1
2
2
E
module is the rank of a simple
Mm (E1) o Mm (E2) 1
and so,
(Mm (E1) o Mm (E2))QCD0 1
module as
C,
Mm (El) 0 Mm (E2) o C(x)) E
1
C
2
C
module.
Mm (E1) o M
We shall show that
the ring
(M
(E
m1
)
1
o C (x) )
0
(E
(M
E o C (x)
C
2
C
o C (x)) .
)
2
m2
is isomorphic to
(E2) o C(x)
M
E
1
C
C
We note the two factors in this coproduct and also the amalgamated skew field all have centre
bg 9.3, so that we may apply 9.4 to each of
C
the factors, and then our first case to the whole coproduct in order to find the rank of a simple
(Mm (F) o C(x))QCD°
as
Mm (F) o C(x)
C
Mm(F)OCD° module as
the rank of a simple
module, which is
C
module, the number we wish
Mm(F)
to find.
Mm (E1) 0 Mm2 (E2) o C(x) 1
hereditary ring
E
T = Mm (El) U Mm (E2) u C[x] 1
on
is the universal localisation of the
C
E
2
T.
M
Consider the subring
mi
which has the universal localisation M
m, 1
at the unique rank function
C
(E.) i C[x] 1 C
(E.) u C[x] 1 C
M
m
1
= M
mi
(E.) u (E u C[x]), C 1 E
(E.) o (E o C(x)). 1 E C
Since
has a unique universal localisation that is simple artinian,
this shows the isomorphism M
mi
(E.) o C(x) i
C
M
mi
(E,) o (E 0 C(x)). i
E
149
Let
be the collection of full maps between f.g.
i = 1,2
for
E.
i
projectives with respect to the unique rank function on Mm (Ei) u C[x], C
i
and let ring
M
and the ring V, where
TE
(E
mi
be the union of these maps induced up to
E
o C(x))
)
1
U
E o C (x)
C
(E
(M
m2
is the ring
T'
We show that
o C(x)).
)
2
We consider the
T.
is isomorphic
T E
C
C
to V. There is a homomorphism from TE to
sending
T',
Mm (Ei)
to
i
M m
(E
by the identity map and
)
i
elements of onto
E
x
to
are inverted, so that it extends to a homomorphism from M
since its image contains
T',
Conversely,
m
(E.) o C(x) 1 C
i
Mm (E1) o C(x) i C
which have as common subring
C
2
fore, there is a homomorphism from
T'
to
TE
TE
i = 1,2.
for
is generated by a copy of
TE
Mm (E2) o C(x),
a copy of
Under this homomorphism, all
x.
E o C(x); C
and
there-
which is clearly inverse to
the homomorphism in the last paragraph. T'
M
ml
(E
has the universal localisation
o C(x))
)
1
o
(M
E o C (x)
C
(E
0 C(x))
)
which must be the universal
C
2
m2
C
localisation of
that is simple artinian; this is
T
Mm (E1) o Mm (E2) 0 C(x), 1
E
2
C
which proves the isomorphism we need.
Since
E o C(x) has centre C, we may apply the first part of C the proof of this theorem to conclude that a simple (M
ml
(E
)
1
o C(x)) C
o
(M
E o C (x)
m2
(E
2
o C(x))® D° module has rank as
)
C
C
C (M
ml
(E
)
1
(M o C(x)) 0 C m2 E o C(x) C
h.c.f. {p1(A1), p2(A2)} (Mm (Ei) o C(x))OCD°
(M
mi
(E
)
i
where
)
o C(x)) modules. C
o C(x)) module equal to C
2
A.
module, and
C
i
(E
is the unique simple pi
is the rank function on
150
By 9.4,
is equal to
pi(Ai)
h.c.f.
,{pi(AiJ.)}
AiJ
as
,
J
f.g.
over
M
m
i
(E)0 D°
i
,
runs
modules.
C
Putting this together, we find that the rank of a simple module as
(Mm (F) o C(x))QCDo C
runs through f.g.
Mm(F)
Mm (EI)®CD°
module is
h.c.f.i/j{pi(Aij)}
as
Aij
modules. We have already remarked that if
i
conjecture 8.1 holds for
Mm1 (E1) ° Mm2
then this number is the same
(E2 ),
E
as the rank of a simple Mm(F)QCD module as
module by 8.4. This
Mm(F)
completes the proof.
As we saw in 9.1, this result gives us all the embeddings of in
Ms(F)
for varying
s.
D
The major part of the rest of this chapter
illustrates what exactly this theorem means. Before ending this section, we notice the following simple corollary.
Theorem 8.5
Suppose that the simple artinian coproduct Mm (E1) o Mm (E2) 1
E
2
satisfies the conditions of theorem 8.3; then the rank of a simple D0IDC(Mm (El) o Mn(E2)) E 1
simple
1
DO C(M m (El) o Mm 1
Proof:
module over m (E1) o m (E2)
C
(E2 ))
module over
2
E
is the rank of a
2
Mm (El) o Mm (E2). 1
2
C
Compare the formulas given by 8.3; they are identical. Finally, we could have stated 8.3 for the simple artinian co-
product of arbitrarily many simple artinian E-rings
Mm (Ei.)
provided that
i
l.c.m. {mi}
exists. The proof does not differ from that given in 8.3.
Division subalgebras of skew field coproducts Here, we shall discuss more explicitly the consequences of theorem 9.3; we shall organise this by discussing when a division subalgebra of a skew field coproduct must be conjugate to a division subalgebra of one of the factors. This is in analogy to what is known to hold for group coproducts. It usually fails here, but our investigation does have some interesting consequences.
By 9.5, we may as well restrict our attention to skew field coproducts over a central subfield
k,
since the division subalgebras of skew
151
field coproducts amalgamating other skew subfields may be reduced to this case whenever we can calculate them. By 8.3, the only case that at present eludes us is the skew field coproduct of two skew fields E1 and E2 both of which are of dimension 2 on either side over the amalgamated skew subfield E.
For much of this section, we shall further restrict our attention to the
case where each of the factors of the skew field coproduct over the central k
subfield
are f.d. over
Theorem 9.6 let
k.
then
divides
[D:k]
divides
[D:k]
[A. .:k] 1J
= [A..:E.][E.:k] 1
D
and
k,
is a f.d. division subalgebra of
o Ei, k
t.
Proof:
13
be f.d. division algebras over
{Ei: i = 1 to n}
Let
t = l.c.m.i {[E.:k]}. If
Aij
as
[Aij :k]
which divides
D°akEi
runs through f.g.
D
If
[Aij:E.1 ]t,
modules.
embeds in
oEi, k
1
h.c.f.i'j{[Aij:k]} = 1,
and so,
[D:k]
divides
t.
The next result has slightly more general hypotheses, and clearly applies to the cases we are considering.
Theorem 9.7
Let
i = 1 to n}
{E1.:
identity, and let
D
be skew fields satisfying a polynomial
be a f.d. division subalgebra of
then the poly-
oE.;
k 1
nomial identity degree of
D
divides
l.c.m.{p.i. degree E, 1
Let p = l.c.m.i {p.i.degree Ei}.
Proof:
and let
m.. = [A..:E,]. 1J
13
is isomorphic to
m..
(E°), 1
13
A13 .
be a D OkE.
Then the centraliser of the action of
1
M
Let
1
and
D°
p.i.degree M
m .,
(E
(p.i.degree E.). 1
that
D
If
p.i.degree D divides
A.. 1J
embeds in
oE., k 1
.)
which is
1
1J
1J
on
embeds in it. By Bergman, Small (75),
p.i.degree D = p.i.degree D° must divide
just m
module, E. 1
{M,.} = 1;
h.c.f..
1,J
so
1J
p.
Our next theorem is more by way of example to show that the bounds in 9.6 and 9.7 are in general best possible.
Theorem 9.8 [Dl:kJ
Let
D1
is co-prime to
and
D2
[D2:k];
be f.d. division algebras over then
D 0 D 1 k 2
embeds in
k
D1 o D2. k
such that
152
i x j;
for
D.
Di0k(D10kD2)° module of dimension
We have a simple
Proof:
by 9.3, there is a simple
dimension 1 over
D1 o D2;
(D1 k D2)8k(D1'kD2)° module of
D1®kD2
that is,
over
[D.:k]
embeds in
D1 o D2.
k
k
Our next result in contrast to the last one gives us a number of cases where a division subalgebra of a skew field coproduct must be conjugate to a division subalgebra of one of the factors.
Theorem 9.9 of
Let
be a f.d. (possibly non-commutative) Galois extension
E
any f.d. division subalgebra of
then
k;
E o k(x)
is conjugate to a
k
division subalgebra of
Proof: Let
C
be the centre of
with Galois group over
E.
where
G,
then
E;
is a Galois extension of
C
k
is the group of outer automorphisms of
G
E
k.
D °0 C = x.Si,
k
the group
G
where each
Si
is a simple algebra over
and
C,
i
acts transitively on this set.
D°0kE
k
G
i
i
permutes the simple algebras
and the group
x(Si0CE),
(D @ C)OCE Z-
transitively. C [Ai:E] = [A, :k] where A. S Q E i
Therefore,
module, and the dimension of a simple
the dimension of any simple D OkE same. So
embeds in
D
E o k(x) k
is a simple
module,
S 0 E i
C
module must equal
D ®k (E o k(x)) k
by 9.3, since they are all the
if and only if
D
embeds in
E.
Since the Noether, Skolem theorem states that all embeddings of a f.d. division algebra in a central simple artinian k-algebra are conjugate, and
E o k(x)
k
has centre
to its embedding in
k,
any embedding of
D
in
E o k(x)
k
is conjugate
E.
We shall restrict our attention for the time being to the coproduct of commutative f.d. extensions of fields in 1 variable over
k.
k,
and of rational function
9.7 shows that any f.d. division subalgebra
of such a coproduct must be commutative. We are able to extend 9.9 a little in this context.
Theorem 9.10 of
k
in
Let
E o k(x) k
E
be a normal extension of
k;
is conjugate to a subfield of
then any f.d. extension E.
153
Let
Proof:
be a commutative subfield of
C
where
h.c.f. {[Ai:E]} = 1,
E o k(x), [C:k] < ' ; k runs through field joins of E and
A.
But since
inside the algebraic closure of
k.
unique such field join over
and so,
k,
then C
is normal, there is a
E
[Ai:E] _ [A .:E] = 1.
That is,
C
J
embeds in
E.
Conjugacy follows from the Noether, Skolem theorem as before.
Before we use these results to get further restrictions on the f.d. division subalgebras of skew field coproduct of commutative extensions we need some embedding lemmas for general skew field coproducts. In order to explain the proof cleanly we shall need the next definition. Consider a tree whose edges are labelled by skew fields and whose vertices are labelled by rings so that if v
and
D
e
Rv
is associated to the vertex
is the skew field associated to an edge
e
on which
v
lies
there is a specified embedding of
D in R ; the tree coproduct associated e v is generated by copies of the vertex rings R v together with the relations that if ie and Te are the vertices of the
to this tree of rings
edge
e
then
it is easy to see that this tree coproduct
R1e n RTe = De;
may be got by a series of coproducts amalgamating skew fields and so if all the vertex rings are skew fields the tree coproduct must be a fir by Bergman's coproduct theorems.
Lemma 9.11
2/
of
{Fi: i = 1 to n}
{Ei:i = 1 to n},
1/ Let
fields such that
E. 2 F.; 1
1
then
oE.
k1
embeds in
over a finite indexing set embeds in
k
be families of skew
OF..
k1
E o k(x).
k
embeds in u F.; the universal localisation of U F. k 1 k 1 at the full matrices over uE. is the tree coproduct of rings associated to k 1 Proof:
The ring u E. k 1
the tree of rings: F
n
k E.
F. 1
154
which is a fir since the vertex groups are skew fields; its universal skew field must be
oF.
since this is the only universal localisation of
oF.
that is a skew field.
For the second part, of
over a finite indexing set embeds in
k of k
over the countable indexing set M;
this has the automorphism
a
of
infinite order that shifts the indexing set by 1; we form the skew field (oE)(x:(Y)
k
which it is easy to see is isomorphic to
E o k(x).
k We can put these results together to find a useful restriction
on the f.d. extensions of
k
lying in the skew field coproduct of f.d.
commutative extensions.
Theorem 9.12 of
k,
Let
and let
D
{Ei:i = 1 to n}
be a f.d. division subalgebra of
commutative, its dimension divides the normal closure
Proof:
E
then D is k 1 l.c.m.{[E1.:k]}, and it can be embedded in
of a field join of the fields
oE.;
E .. 1
The first two conditions follow from 9.5 and 9.6. We have shown that
embeds in
oE.
be a finite collection of f.d. extensions
and now we can apply 9.10.
E o k(x), k
k 1
We give an example to show that these conditions are not sufficient for a field to embed in a given skew field coproduct.
Example 9.13
Consider
o
Q (V2,,13)
is
Q itself, and
is a subfield satisfying the consequences of 9.11. But and
are both fields, so that by 9.3,
does not embed in
o
As has become clear, the coproduct of separable extensions does not behave particularly well with respect to the property that f.d. division subalgebras are conjugate to a subfield of one of the factors; it comes as a pleasant surprise that purely inseparable extensions work extremely well in this context.
Theorem 9.14
Let
oEi
be the skew field coproduct of the skew fields
E.; 1
if
C
is a purely inseparable extension of
k
lying in
oEi, k
it is
155
conjugate to a subfield of some
Proof:
Ei.
Let p be the characteristic of
sion over
E.
embeds in
oEi
of a simple
we shall show that the dimen-
k;
module is a power of
COkE.
if and only if it embeds in some
p,
so that
C
as usual, the Noether,
Ei;
Skolem theorem completes the proof.
Let CekC.,
E
then the centre of
;
extension,
of
Ci,
C'
k.
is a power of
C',
with any field
is a purely inseparable extension of
modulo its radical, which implies that the dimension of a simple module over
E.
must divide
[C':C.]
which is a power of
p.
so
Ci,
is the simple artinian image of
C'OkEi
p.
is
is a purely inseparable
C
and so, has a unique field join,
k,
COkEi
i
which is a local artinian ring, because
extension of
[C':Ci]
be the centre of
C.
COkEi
CekEi
We conclude
the proof as was indicated previously.
This fits well with earlier results.
Theorem 9.15
Let
oEi k
be a skew field coproduct of finitely many f.d.
purely inseparable extensions of
then any f.d. subfield of
k;
oEi k
is
conjugate to a subfield of one of the factors.
Proof: By 9.12, any subfield must lie in the normal closure of a field join of the fields extension of
But this is simply the unique purely inseparable field
Ei.
k
generated by the fields
E
So, any such subfield is
..
i
purely inseparable, and, by 9.14, it must be conjugate to a subfield of one of the factors.
There is a further situation, where one could hope to prove that a f.d. subfield of a skew field coproduct of commutative fields
would be
conjugate to a subfield of one of the factors; when each factor has dimension p
over
k,
where
p
is a prime. This turns out to be false in general, but
the set of primes for which it is true is an interesting collection. The next result depends on the classification of finite simple groups, since we use Cameron's classification of doubly transitive permutation groups (Cameron 81).
Theorem 9.16 q
Let
p
be a prime not equal to 11 or
is a prime power, and
t > 2.
field extensions each of dimension
Then, if p
over
{Ei} k,
(qt-1)/(q-1),
where
is a finite collection of any f.d. subfield of
oE.
k 1
156
is conjugate to one of the factors. For the proscribed primes, there are counter-examples.
Suppose that
Proof:
any
[E:k] = p;
By 9.5,
i.
oE., [E:k] < -,
embeds in
E
and
E * Ei
and that
for
cannot be purely inseparable by 9.14.
E
So, it is separable. Since
n
EOEi =
embeds in
oE., k i
Since
E * Ei,
n ? 2.
x Fi., j=l
G
lies in the normal closure,
for any
Because
j.
Let
k.
G
lies in
E
of
N,
i
1k
by 9.10.
E.
which
E. 0 E.
So,
N
be the Galois group of
E.
over
k.
acts faithfully on the roots of an irreducible polynomial for a over
Ei
k,
p divides
IGI,
but
we see that the subgroups
[Ei:k] = p = [E:k], E
so that
i.
E
generator of
and
for which
i
[F..:E.] i 1
h.c.f. {IF:E.]} = 1, n 1
is a separable extension of Since
there is some
7
E[F..:E] = p, 17 i implies that
E
p2
H'
and
respectively are inconjugate p-complements in
Hall's theorem,
does not. Since that fix
H
E.
therefore, by
G;
cannot be soluble.
G
The actions of
on the right cosets of
G
non-isomorphic faithful transitive actions of
G
H
and of
define
H'
on a p-element set. By
Burnside's theorem (11) we see that transitive actions of an insoluble group on a p-element set are doubly transitive. Cameron (81) shows that this can happen only when q,
and
p = 11
(qt-1)(q-1)
for a prime power
t > 2.
For
11 points; for PGL(q,t)
or is of the form
p = 11,
there are two different actions of
p = (qt-l)/(q-1),
PSL(2,11) on
the actions are given by the action of
on the points, and dually, on the hyperplanes of projective space. We have reached a contradiction provided that for a prime power
(q -l)/(q-1)
q
and integer
t > 2.
p x 11,
or
So we need to
examine what actually happens in these cases. Given a group
G
having two
such faithful actions on p-element sets, we have two inconjugate p-complements. We may find a Galois extension of fields
N D k,
having Galois group
G,
and the two p-complements give us two non-isomorphic field extensions of
k,
both of dimension
E
and
E';
then
occurs when
E
p = 7.
p
over
k
embeds in
with isomorphic normal closure. Call them E' o E', k
and vice versa. Our first example
It is of some interest that at present we are unable to tell
157
E o E, E o E' and E' o E' where E and E' k k k are as described in the last paragraph of the proof above are isomorphic.
whether the skew fields
The invariants that we shall develop in chapter 10 are not good enough to settle this point.
Transcendence in skew fields One question that arises naturally is whether the coproduct of skew fields having no elements algebraic over
k
also has this property.
We shall present a counter-example to this in the course of this section, as well as giving a counter-example to a conjecture of Cohn and Dicks (80), connected to suitable ideas for transcendence in skew fields. We shall also give the definitions for notions of transcendence that do behave well for the skew field coproduct. First, however, we present a positive result in the direction of our original question.
Theorem 9.17 [C:k] = pn
Let
be a commutative subfield of
C
for some prime
elements algebraic over
such that oE, k 1 then one of the skew fields Ei contains
p;
k.
We lose nothing by assuming that
Proof:
C
is a primitive extension of
k,
C = k(a).
Let
be the centre of
Li
irreducible polynomial of
a
over
k
E
if
;
i
C®kLi
is not a field, the
splits over Li
and the co-
efficients of the factors generate an algebraic extension of So, assume that
CO L
is a field; then
k i
in
Li ,.
is simple
y
artinian, and so, the dimension of a simple [C:k] = pn,
k
C"kEi = (COkLi)QL E.
and must itself be a power of
C@kE, p.
module over
Since
C
divides
E.
embeds in
oE k it
11.4 shows that the h.c.f. of these numbers is 1, and therefore, there is some
i
for which the simple
C embeds in
is,
COkEi
module has dimension 1 over
[C:k]
that
i
Of course, we cannot conclude that when
Ei;
E ..
C
embeds in some
E. i
even
is prime, as we saw in 9.16.
We can re-phrase 9.17 in the form that an algebraic extension of k
lying in the skew field coproduct of two skew fields having no algebraic
elements must have dimension divisible by at least two primes. We shall find such a subfield of dimension 6.
158
Example 9.18
be an extension of fields such that
E m k
Let
[E:k] = 6,
and there are no intermediate fields; this occurs, for example, if the Galois group of the Galois closure of E1
pair of skew fields neither
nor
E1
such that
E2
and
is
k
We shall construct a
S6.
embeds in
E
have any elements algebraic over
E2
Let
over
E
be the skew field such that
E1
although
E1 o E2 k
k.
and
M2 (E 1) = M2 (k) o E,
k
be the skew field such that
E2
let
M3(E2)
Certainly,
M3 (k) o E.
E
k
E1 o E2 since there is a simple ERkEI of dimension 2 over El, k and a simple E®kE2 module of dimension 3 over E2. Since h.c.f.{2,3} = 1,
embeds in
it is clear that E2
E1 o E2. k
embeds in
E
have no algebraic elements over Let
Since
E1
be a commutative subfield of
C
lies in
C
It remains to show that
M2(E1) = M2(k) o E,
and
El,
E1
and
k.
such that
[C:k] = n.
there is a simple
k
module of rank
M (k) 0 E
2= h.o.f. {n/2, [C.:Ell
/
as
over
must be odd, and the equation
M (k) 0 E,
so that by 11.4,
2
1 = h.c.f. {n,[C .:E]}
modules. So
COkE
runs through simple
C.
n
holds. Since
J
[COkE:E] = n,
is a sum of the numbers
n
and so,
:Ell
EC i
which is equivalent to the hypothesis that
1 = h.c.f.{[C .:E]},
embeds
C
J
in
E 0 E
and
n
is odd. By 9.6,
n
divides 6, so
n = 1,3.
Since
k
it follows in either case that for some
h.c.f.{[C.:Ell = 1, that is,
so that
C embeds in
E.
By assumption,
k
[C.:E] = 1,
has no non-trivial subfields,
C = k.
We have shown that of
E
j,
E1
has no non-trivial algebraic extensions
inside it; an entirely similar argument shows that
trivial extensions either. Thus
E1 o E2 k
of which have no algebraic elements over
E2
has no non-
is a coproduct of skew fields both k
except for those in
k,
but
the coproduct does have non-trivial algebraic elements.
We give some variants of the definition of transcendence that do behave well with respect to the coproduct construction, and discuss the connections of these notions with those in (Cohn, Dicks 81). We define a skew field to be n-transcendental over a central subfield
k
dividing
if for all f.d. division algebras n,
DokE
transcendental over
D
over
k
of p.i. degree
is a skew field. We define a skew field to be totally k
if it is n-transcendental for all
n.
It is clear
159
that n-transcendental implies that there are no algebraic elements over that are not in
The skew fields of example 9.17,
k.
E1
and
k
are not
E2,
even 1-transcendental.
Theorem 9.19
The coproduct of n-transcendental skew fields over
is n-
k
transcendental. Therefore, the coproduct of totally transcendental skew fields is totally transcendental.
Proof:
The proof is clear from 9.3.
We consider the definitions of Cohn and Dicks. A skew field is said to be regular over k, fields
K over
EOkK
if
is a domain for all commutative
This is shown in their paper to be equivalent to
k.
is a skew field where
k
E
is the algebraic closure of
k.
EQkk
So regular is
equivalent to 1-transcendental.
Cohn and Dicks define if
E
EOkL
to be totally algebraically closed in
k
is a skew field for all simple algebraic extensions
L
of
k.
_
They ask whether this together with the condition that
EOkkp
is a skew
_ field where
p
is the characteristic of
purely inseparable extension of
k
and
k,
imply that
E
kp
is the maximal
is 1-transcendental. We
shall find that there are a number of counter-examples to this that arise as a consequence of the next theorem, which is a perhaps more natural result.
Theorem 9.20
F D k
be a normal extension of commutative fields such
that
where
p
Let n [F:k] = p
M q(D) = Mq(k) o F,
where
is a prime. Let
D
be the skew field such that
Then for any algebraic extension
q = pn-l
E
k
of
k,
Proof:
M
q
(E)
DOkE
Let
has zero-divisors if and only if
[E:k] = e;
module as
M
q
(k)
E D F.
as usual, we shall use 9.3. The rank of a simple module is
correspond to field joins of join up to isomorphism over
E
k
e/q.
and
since
F,
F
The simple modules for
EOkF
and there is a unique such field is a normal extension of
k.
We
have the following diagram, where the dimension of the extensions is the label of the edges.
160
EF
p
n-m E
- F
n-m
p
We see that
rank of a simple
and so the rank of a simple
[EF:F] = e/pm
Mq(D)QkE module is less than
e/p or vice versa, this number is less than divides
and is not equal to it. But
e/q
happens exactly when contains
m = n.
Let
n [F:k] = p
where
Mq(k) o F, k
that M q(D)
q = pn-l
is totally algebraic closed in
k
transcendental over
so that
E
k
Let
be the skew field such
Dokkp
Then
D,
that is neither simple
k D
although
D
is a skew field, is not 1-
k.
EQkD
In 9.20, we showed that
Proof: of
e/p so this
q = pn-l,
[EF:k] = e,
be a normal extension of
F n k
nor purely inseparable, and let
E,
e/q divides
if and only if and
m <_ n,
In this case,
Since
e/q.
e/q
F.
Theorem 9.21
and
M (D)Q kE
has zero-divisors if and only if the
DOkE
h.c.f. {e/q, a/pm}.
module is
if and only if
E
has zero-divisors for a f.d. extension,
contains
F.
E
If
contains
F,
it cannot
be a simple extension nor can it be purely inseparable. Therefore, is a skew field, and
is totally algebraically closed in
k
not 1-transcendental over
D,
DOkk D
but
is
k.
We have used skew fields
D
of the form
M t(D)
M t(k) 0 E k
where
E
is a f.d. commutative extension of
k
as our main construction
recently. Before leaving them, we mention one further property that has some interest. Let t
divide
n;
[E:k] = n,
where
and let
be the skew field such that
D
E
is a simple extension
E = k[a].
Let
Mt(D) = Mt(k) 0 E. k
DMkE module over
D
is
from 9.3. That is, the minimal polynomial of over
k
splits over
Then the dimension of a simple
n/t factors of length
t.
t
as one checks D
into
161
Generic partial splitting skew fields Amitsur (55) and Roquette (62) developed the notion of the generic splitting field of a central simple f.d. algebra over a field
k.
This construction has a natural non-commutative analogue, which we shall develop and study in the course of this section. We shall investigate its f.d. division algebras using the techniques developed earlier for such questions.
be a central division algebra over the field
D
Let [D:k] = n2,
m divide
and let
splitting skew field of
D
n2.
A skew field
DOkE = Mm(S)
if
k,
is said to be an m-
E
for some simple artinian ring
S.
A commutative field can only be an m-splitting field when m divides
n;
however,
D
0
is an n2-splitting skew field of
There are universal
D.
or generic m-splitting fields that are commutative for
dividing
and
n,
we intend to show that there are suitable skew analogues of these constructions. We shall construct a ring that represents the functor Hom(Mm(k),DQk_)
in the category of k-algebras, and then show that it has a
universal skew field of fractions, which will be the skew field that we are interested in.
The functor
Theorem 9.22 R,
for
where D
Proof:
D k m(k) = DO R.
if and only if
Let
In particular,
E
is an m-splitting skew field
for some k-algebra
Then
A.
is a homomorphism and must take the centraliser of
-> DMkA
in the first ring to
A,
which gives us a map from
Conversely, given which induces a map
is representable by the k-algebra
is not empty.
Hom(R,E)
a e Hom(M(k)),DOkA)
l u a:D k Mm(k) D
Hom(Mm(k),D®k-)
B:R -+ A,
we have a map
R
to
A.
1 OS:D&kR -* DOmA,
S:M (k) - M (k) u D = DOkR S DOkA. m
m
k
The two processes above are mutually inverse, so
R
represents
the functor as we wished to show. The last sentence is trivial. We shall look at
R
a little more closely, but first we tighten
up the notation. The ring representing the functor
denoted by
Theorem 9.23
Hom(Mm(k),DOk-)
is an hereditary domain whose monoid of f.g. projectives
R(D,m)
is generated by the free module of rank 1, and a projective n2
the relation
is
R(D,m).
R
p
,
P
satisfying
and no other relations. Therefore, it has a unique
162
rank function on its f.g. projectives, and this rank function takes values So, it has a universal skew field of fractions.
Z.
in
DOkR = D k Mm(k),
Proof:
so
morphism
R
where in this iso-
Mn2(k) U Mm(D
Mn2(R) = D Ok(DO kR) = D Ok(D k Mm(k))
Do
is the centraliser of the first factor. By Bergman's coproduct theorems, 2.18,
is a domain and all
R
f.g. projectives are induced up from the factors in the coproduct representation
Mn2(R) = Mn2(k) Do
Morita equivalence shows that the monoid
m(D).
of f.g. projectives is as stated, and
R
is hereditary. 2
We have shown all but the last sentence. Since rank under any rank function can only be
Pm = R n
its
,
so that there is a unique
n2/m,
rank function, and the universal localisation of
R
at this rank function
is a skew field.
Let
be the universal localisation of
U(D,m)
R(D,m)
at the
unique rank function. We shall call it the universal m-splitting skew field of
D.
We wish to investigate the f.d. division algebras over
k
that embed
in it.
Theorem 9.24
Let
be the universal m-splitting skew field of
U(D,m)
then the f.d. division subalgebras of division algebras p
particular, if U(D,m)
p,
E.
of
D°
such that
divides m
has no elements algebraic over
Proof: We recall from 9.23 that fractions of Mn2(U)
over simple
h.c.f.{[Ei:k],n2/m} = 1.
implies that
where
R,
is given by
E
[S:D ] U,
divides
[S: D] = 1,
and
k.
is the universal skew field of Hence,
where
S
module
is the unique
[E:k].
we know that this number is
is equivalent to the two statements:
In
for all primes
except for those in
h.c.f.{[E:k]/n2,[S:D°]/m}
embeds in
happens if and only if
n2/m
By 9.3, the rank of a simple E kMn2(U)
E 0kD module. So, If
divides
M n2(R) = M 2(k) uM (D ). n Do M
Mn2(k) o Mm(D ). D° Mn2(U)
p k
U = U(D,m)
D;
are isomorphic to the f.d.
U(D,m)
1/n2,
h.c.f.{[E:k]/n2,1/m} = 1/n2,
which which
163
embeds in
1/
E
2/
h.c.f.{[E:k],n2/m} = 1.
and
Do,
This proves all but the last sentence, which is an easy corollary. It occurs for example, if
n = m.
Division subalgebras of the universal skew fields for rings with weak algorithm This final section is rather technical; we use 9.2 to describe the f.d. division subalgebras of the universal skew field of fractions of a ring with weak algorithm. For the record, we define what a ring with weak algorithm is; however, the reader is more likely to gain some understanding of the notion by reading chapter 2 of (Cohn 71). Rings with weak algorithm are a generalisation of tensor rings of bimodules over skew fields.
R
Let
a function
be a ring with a filtration over N;
from
v
to N
R
v(x-y) <_ max{v(x), v(y)};
Let
such that
v(xy)
for
v(x) >_ 0
be a subset of the elements of
{ai:i E I}
that are almost all
{b1.:i E I}
v(Eaibi) < maxi {v(ai) + v(bi)}
v
that
on the set
{a
of
;
R;
we say that
if there exist
v
ai
is
O.
is right dependent with respect
R
almost all 0 such 1 v(a- Ea.b.) < v(a) whilst v(a.) + v(b.) <_ v(a) for each i. i l I 1 1 We say that R satisfies the weak algorithm with respect to
the filtration
v,
.}
a
,
1
{b1.:i E I}
if given any set of elements
dependent with respect to some
a
if there exist
v(O) _ -
such that
0,
or else, some
We say that an element to
x x 0;
v(1) = O.
<_ v(x) + v(y);
it is right dependent with respect to the filtration elements
that is, we have
v
such that
{ai:i = 1 to n}
v(a.) <_ v(a.) 1
is right dependent with respect to
for
i < j,
right then
J
on the preceding set
v
{al,a2...ai-1}.
We shall assume the results of chapter 2 in (Cohn 71) throughout this section. In particular, fractions we call
together with
0
U.
of
R = R0[x;a,d],
(l,a)-derivation. When
U
is a fir whose universal skew field of
the set of elements of whose filtration is
RO,
forms a skew field, and
when it takes the form is a
R
R
where
0
is a two-sided Ore domain only a
is an automorphism and
d
is not a two-sided Ore domain, the centre
R
by 8.1, since RO is the group of units of R. When R 0 we shall need to use different techniques which we present at
lies in
R= RO[x;a,8]
the end of the section. In the case where be the centre both of
R
and
U.
R
is not two-sided Ore, let
K
164
Theorem 9.25 U
R be a ring with weak algorithm with universal skew field
Let
and Oth term in its filtration
Then the dimension over h.c.f. {[N.:RO]}
as
let
RO;
D°OKU module is equal to
of a simple
U
runs through the f.g. D aKRO modules.
N.
By 9.2, the dimension of a simple
Proof:
h.c.f. {[M.:R0]}
as
be a f.d. division K-algebra.
D
module is given by
D°OKU
runs through f.g. D QKR
M3
submodules of free
modules. For ease of notation, we shall consider left
right
D,
bi-
R
modules on K-centralising generators, since these are clearly the same thing.
We recall that
is filtered by
R
URi = R.
R0 c R1 C R2 c ...... c R,
Let
be a free left
F
right
D,
R
bimodule on the set
X,
which we filter by
DXRO C DXR1 C DXR2 C ...... C DXR = F.
For
we define
f E F,
right
R
Let M
v(f) = min {r:f E DXRr}.
sub-bimodule of
M
from which it will be clear that the dimension of sum of the numbers
for
[N.:ROI
RO bimodules
D,
We choose our basis by induction. Let the ith stage we have a elements Mi-1,
{a.}
lying in
DXR
and no element of
subject to containing
M
submodule of
R
D,
be a f.g. left
M
We construct a basis for
F.
{a,}
such that
n
M - Mi-1
lies in
D,
as an R-module
as an R-module is a Ni.
MO = 0.
Suppose that at
R module by
generated as
is a basis of this module, where
DXRn,
DXRn
is minimal
a j,
Let M
A, = {a:a E M - Mi-1, v(a)
are invariant under the action of
is a left
D,
right
weak algorithm that if R0
D,
is minimal}.
Since
it is clear that
Mi-1
and
A. + Mi-1/Mi-1
bimodule; moreover, it is a consequence of the
R0
M. = A.R + M 1
together with the set
i-1,
1
{a.}
A
a basis for
is a basis for
Mi
.
+ M 1.-1 /M
1
i-1
over
R module.
as
J
Since for all
i.
sion of some D,
U
M
is f.g.,
M = Mm
But it is clear that D,
R0
for some
m,
[Mi+1:R] - [Mi:R]
and
[Mi:R]
is finite
is equal to the dimen-
bimodule. So, we see that the dimension of a simple
bimodule is equal to the
h.c.f.{[M:R]}
bimodules, which must be divisible by
for f.g. left
h.c.f.{[Ni:R0]}
D,
for f.g.
right D,
R0
R
165
bimodules by our argument. However, given a struct the
U
D,
and so, the dimension over h.c.f.{CNi:R0]}
U
bimodule
R0
D,
It is clear that
NOR U.
bimodule
o
of a simple
O
bimodule must divide
U
D,
we con-
N,
[NOR U:U] = CN:RO],
which forces equality.
To complete this section, we shall deal with the two-sided ore case. We recall that in this case, a
is an automorphism and
Theorem 9.26 Let
Let
U
be the intersection of the centre of
K
R0
where
is a skew field,
derivation.
(l,a)
be the universal skew field of fractions of
division K-algebra lying in
Proof:
R = ROCx;a,d]
is a
6
U
U
with
R0.
R0Cx;a,61.
Then any f.d.
is isomorphic to a skew subfield of
We shall show that there is a valuation on
residue class skew field is identified with
R0.
trivial on
U If
D
R O
whose
is a f.d. division
K-algebra in
U,
the valuation must be trivial on
extension of
K,
and so, the valuation induces an embedding of
D
R0.
since it is a f.d. D
in
R0.
The construction of the valuation essentially occurs on p.18 of Cohn (77). We summarise it briefly here. and write out the commutation formula:
y = x-1
Set
a rx=xr +r , so yr= ry+yry.
a
6
6
The set of power series over
R0
in
y
with co-efficients on
the left and the stated commutation rule (which allows us to re-write any expression in
R0
and
y
as a power series of the given form) is a
principal valuation domain; the ring of fractions by is
R0,
x -> y-1.
R0Cx;a,6]
and the valuation is trivial on
is complete.
embeds in its skew field
The residue class skew field of the valuation R0
as we wished, so our proof
166
THE UNIVERSAL BIMODULE OF DERIVATIONS
10
When studying a homomorphism of commutative rings :R + S, is often useful to look at the module of relative differentials of R.
S
it
over
There is a non-commutative analogue of this construction, the universal
bimodule of derivations, which, in many situations of interest to us here is very powerful. It will enable us to find useful numerical invariants of certain skew field coproducts, that allow us to distinguish between some of them; in particular, we shall be able to distinguish between free skew fields on different numbers of generators. On the way, we shall be able to characterise those epimorphisms from an hereditary k-algebra to skew fields that arise as universal localisations by the associated map on the bimodule of derivations over
k.
The results in the first section of this chapter from 10.4 to 10.6 that calculate the universal bimodule of derivations for certain universal constructions are all from the work of Bergman and Dicks (75,78).
Calculating the universal bimodule of derivations In the commutative case, we look at derivations from the commutative ring to modules over it, that vanish on a given subring; the natural non-commutative generalisation of this is to look at derivations to bimodules over the ring; if
is an
R
vanish on the image of
derivation
6:R i C
to a bimodule
RO
R0.
(R);
R0-ring we are interested in derivations that On general principles there is a universal such that is, the functor
the set of derivations from
M
is naturally isomorphic to the functor
Der (R,M) R0
R
r
6s = 0,
for
that vanish on
R0
where the
a:CR (R) - M
is the
We can construct the bimodule in the following way: for each
6a.
element
M
Horn R-bim,(CRO(R),M)
derivation associated to a particular homomorphism composite
to
which associates
of
R, s
we have a generator
in the image of
R0,
6r,
and we impose the relations
6(r1 + r2) = 6(r1) + 6(r2),
and
167
6(r1r2) = r16(r2) + 6(r1)r2.
It is clear that the map
has
6:R -> QRO (R)
the desired universal property; however, the nature of this bimodule is opaque from this description, so we shall first give an alternative description, and then we shall use it to simplify our presentation above. Our new description gives us a useful connection between the universal bimodule of derivations and homology.
Theorem 10.1
Let
Owhere m
R
RO
be an RO-ring; then there is an exact sequence:
(R) +RoROR m R1 O,
is the multiplication map, and the universal derivation in this
representation is given by
Proof:
6(x) = x0l - l&x.
Certainly, the map from
is a derivation vanishing on
R
to
given by
ker(m)
d:x -* x0l - lox
So the composite of this derivation with
R0.
HomR-bimod(ker(m),-)
any bimodule map gives a natural transformation from
m
DerR (R,-), which must be injective since the kernel of O is generated by the elements x0l - lox.
to the functor
Given any derivation vanishing on an
R,R
bimodule, we define a bimodule homomorphism from
by the formula d'
d':R - M
R0,
ad,(Exieyi) = Ed'(xi)yi = -Ex id'(yi),
is a derivation and
ker(m)
M to
is
M
which works because
From these equations, it is clear that
Exiyi = 0.
this map is a bimodule homomorphism, and that d:R - ker(m)
where
d' = dad,;
so the derivation
has the correct universal property, and by the uniqueness of
an object representing a functor, we see that our theorem holds.
We can simplify our previous description a little using this result. From the relations, 6(r1 + r2) = 6(r1) + 6(r2),
by the set of elements
X,
6(r1r2) = rI6(r2) + 6(r1)r2 it is clear that if then
(R)
S2R
R
and
is generated over
is generated by
6(X).
RO
We wish
O
to determine the relations imposed, so first we consider the case where the set
X
is a free generating set.
Theorem 10.2 X
Let
R = RO u 7l<X>,
subject to no relations; then
the ring generated over (R)
S2R
O
RO
by the set
is the free bimodule over
R
on
168
the set
X.
Let
Proof:
be any
M
vanishing on
R,R
bimodule; then any derivation from
is determined by the image of the elements
R0
derivation; moreover, the elements of
R
M
to
under the
X
may be mapped anywhere by a
X
derivation. So, the functor functor
HomSets
bimodule over
(X,M);
R
on the set
extending the map functor,
X -+ 6X;
6X,
which has a derivation from
R
to this free bimodule is the one mentioned earlier.
Given an arbitrary R0-ring, of elements
to it
R
by the uniqueness of an object representing a
must be isomorphic to this bimodule and the universal
QR (R)
O derivation from
is naturally equivalent to the DerR (R,M) O one object that represents this functor is the free
R,
there is a surjection
X,
RO
generated over
RO
by a set
so we should like
72<X> -+ R;
to describe how the universal bimodule of derivations changes under surjective homomorphisms between R0 -rings. This is quite simple to describe as we shall see in the next theorem.
First of all, we introduce some notation, which simplifies the equations of this chapter substantially. In many situations, we shall have and also a particular
a specific homomorphism of rings
R - S
module
bimodule
we may form the
M;
S,S
general write as V; if the rings
SORMORS,
and
R
R,R
bi-
which we shall in
are constructed from other
S
rings in a manner which is reflected in the names of
R
and
S,
SORM&RS
will be quite unwieldy. For example, we shall prove later on the following formula:
QR (R1 U R2) = 0
Since Q
(RI.)
we mean
QR
0 must be an
QR (Rl Ru R2) O
QRO
RO
R1
O
is an
Ri,Ri
(R l 1
Theorem 10.3
2
0
Let
an exact sequence:
R2, R1
RO
R2
bimodule, whilst
O
bimodule, it is clear that by the symbol
00
(Ri)
)0 Ri (QRO (R.))O (R uR ). 1 Ri 1R0 2)' R
be an R -ring, and O P
I/I2 + 0
R0
(R)0
02
RO
I
an ideal of
(R/I)
by the ring homomorphism, and the map from
-
I/12
0,
where
then we have
R;
p
is induced
to the bimodule
00R (R) 0
O
is induced by
d.
169
R0
If
is a semisimple artinian ring, our sequence may be
extended to an exact sequence:
Proof:
O - I/I2 -> 0S2 (R) Q RI
We have the exact sequence:
0 + S2
be split exact as a sequence of left
RO
(R)
-> RO
$2
RO
(R/I)
-> 0.
R a R + 0, RO
which must
modules; consequently, we have the
R
exact sequence:
R/I2 RO (c R (R)) -> R/I0 ROR - R/I -> 0.
O-
0
We look at part of the long exact sequence of
TorR( ,R/I):
TorR(R/IO R,R/I) -+ Tor1(R/I,R/I) - a0R (R) 0 } R/IOR R/I -> R/I -> 0, O
all terms are
R/I,R/I
where
O
O
bimodules.
It is well-known (and we shall see this soon) that Tor1(R/I,R/I) = I/I2.
The map
R/ISCR R/I -> R/I
so its kernel is isomorphic to sequence:
R/I&R R
I/12 - R Q
R
by 7.2, which gives us the exact
0Rpp(R/I)
(R) 0 + QR (R/I) -> 0;
RO
if
is semisimple artinian,
O
O
is projective, so that
exact sequence:
is the multiplication map,
0 -), 1/1 2 +
Q
Tor1(R/I&R R,R/I) = 0,
(R) 0 -> QR (R/I) -> 0
0
R
O We need to show that the map from
and we have the
as stated.
O
I/12
is the O one we said; in order to show this, we describe the isomorphism between I/I2 and
R
Tor1(R/I,R/I).
R --R/I -0
-
0 TorR( ,R/I)
- R/I --O
R
0 --+R/IORQR (R)
O
to it gives us a commutative diagram:
0 - TorR(R/I,R/I) i I0RR/I = 1/1
II
which shows that
00R (R) 0
Consider the map of exact sequences:
0-I Applying
to
rR(R/I,R/I) 1 I/12
-- 0Q
RO
1/1
2
-> R/I
(R)o
Tor1(R/I,R/I),
demonstrates that the map from
2
to
whilst the right hand column 0SiR (R) a
O
is what we want it to be.
170
This result gives us a presentation of
in terms of
QR (R)
generators and relations whenever we have a presentation of If the set
R
generates
X
over
universal bimodule of derivations of module by the elements
6X
R
over
subject to the relations
R0,
over
R
R0
fi
the
is generated as a bi-
subject to the relations
is the formal differential of the element
R0.
{fi}
6fi = 0,
where
RO
of the ring
6fi
Z<X>.
We need a few results about the behaviour of the universal bimodule of derivations under various universal constructions; we begin with the coproduct construction.
Theorem 10.4
Q_
be a family of R0-rings; then
{Ri:i e I}
Let
OR
® (0QR(Ri 0
P ID
Proof:
This is most easily seen using our generator and relation construction
of the universal bimodule of derivations described after theorem 7.3. It is clear that
modules
SR
(R Ri)
O
is generated as a bimodule by the image of the bi-
0
and there are no further relations, so the theorem must
QR (Ri) O
hold.
We can also calculate the universal bimodule of derivations of a tensor ring.
Theorem 10.5
M
Let
be an
R,R
bimodule; then
derivation induces the identity map on
Proof:
where the
M.
We look at the generator and relation construction of the universal
bimodule of derivations again. only relations are given by R,
J1 R(R<M>) = % 0,
and m
in
is generated by
QR(R<M>)
6(rm) = r6(m),
and
6(M),
6(mr) = 6(m)r
and the for
r
in
So the result follows.
M.
We also wish to study how the universal bimodule of derivations behaves under the process of adjoining universal inverses.
Theorem 10.6 the R0-ring
Let R;
E
then
be a collection of maps between f.g. projectives over QR (RE) _ O
0
(R)®.
Q
RO
171
M
Given a bimodule
Proof:
over a ring
additive structure is isomorphic to given by
and whose multiplication is the trivial extension of
(t,m)(t',m') = (tt', tm' + mt'),
we write this ring as
M;
we can form the ring whose
T,
T $ M
by
T
(T,M).
We wish to construct a derivation from extends the universal derivation from
R
to
RE
OR (R). 0
homomorphisms:
that O So, consider the ring 0$2R (R)0
to
-=(R,OR0(R))
lR
(RE,a0
R E
RO
(R)
we wish to complete this to a commutative diagram of ring homomorphisms by a map from s
+(s,ds)
from
R
to
RE
(RE,
where
to
d
(R)
S2R
a
Q
R
because such a map must take the form
(R)0),
is a derivation extending the universal derivation by the commutativity. However, the set of maps
E
are
O
(RE, ®0R (R)since they are invertible modulo the nil-
invertible over
0
potent ideal
(O,QS2R (R)0);
(RE,0QR (R))
therefore there is a unique map from
RE
to
completing the diagram.
O
We wish to show that this must be a universal derivation. Given
a derivation d':R -> M where M vanishes on restricted to
R0,
R
is an
RE,RE
bimodule such that
we know from the universal property of factors through
(R),
S2R
S2R
(R)
d'
that
d'
O so we have a diagram:
O
where we have shown that the top triangle is commutative and we wish to show that the bottom triangle must also be commutative to complete the proof. We have two homomorphisms from and the second via
s -
RE
to
(s,da(s)).
(RE,M)
the first via
These agree on
R,
s -> (s,d'(s)),
and since
R -
RE
172
is an epimorphism they must be the same map.
Generators for the free skew field For a free group on
generators it is well known and easy to
n
see that any generating set of
elements must be a free generating set,
n
and the corresponding result for a free k-algebra was shown by Cohn and independently by Lewin (69). It is natural to ask whether the corresponding result holds for free skew fields, which arise as the universal localisation of the free k-algebra; J.Wilson asked whether the free skew field on
generators could be isomorphic to the free skew field on m
n
generators, and
a negative answer to this question would follow from the first result. We should also like to know that if subfield of
generated by
to DM, and if
M
of
is an
generated by
E(N)
E
is a skew subfield of
D
and
D
X
M
and
then the skew subfield
N,
is naturally isomorphic to
In the first of our problems we shall show that if tl,...tn ¢
on
generate
k(x1.....xn given by
k
then the skew
is isomorphic in the natural way
subbimodule of
E
E,
as a skew field over
(yi) = ti
k,
n
elements
the homomorphism
induces an isomorphism:
Qk(Mxl.....
It follows from our next theorem that the skew field
k(y1....yn
¢
must extend to an isomorphism of
with
as we wished to show.
Our next result gives us a way of recognising those epimorphisms from a right hereditary k-algebra to a skew field that are universal localisations.
Theorem 10.7
Let
E
R be a right hereditary E-
be a skew field, and let
ring. The ring homomorphism :R - F universal localisation if and only if
from
R
to a skew field
F
is a
induces an isomorphism
0 QE(R) = QE(F).
Proof:
If
F
is a universal localisation of
Conversely, if the map
R,
is an isomorphism,
5
the result is clear.
2R(F) = 0,
at least an epimorphism. Therefore, by theorem 7.5, if of square matrices over
R
that become invertible over
prime matrix ideal, and the induced map RE i F
and so
F,
they form a
is surjective, where
is a local ring, and the kernel must be the radical.
is
is the collection
E
RE
173
Since
RE
is a universal localisation of a right hereditary
ring, it is right hereditary by 4.2, and the kernel of be a free module, so that
except when
I x 12
R. + F,
must
I,
I = 0.
From 7.3, we have the exact sequence-
also we have the commutative diagram
0 + i/12 + 0S2E(RE)0 + SZE(F) + 0;
E (RE)
where the top row and the first slanting arrow are isomorphisms which implies that the second slanting arrow must also be an isomorphism, and follows from the exact sequence above. Hence,
I = 0,
and
I/12 = 0
R. = F.
Next, we show that the skew subfield of E(X generated by and
where
X
D
is a skew subfield of
E
is isomorphic to
D
the
proof is simply to apply the last result.
Theorem 10.8
Let
D
E(X generated by D
and
E.
E(X
By the remarks after 10.3,
ScE(E<X>)
generated by the elements in
then the skew subfield of
E;
is isomorphic to
X,
and
e
in
6X
and
D
subject to the relations
that is,
E;
E-centralising generators generated by
6X.
Let
homcmorphism from
K
K
is
S2E(E<X>)
e6x = 6xe
for all
is the free bimodule on the
be the skew subfield of E(X
0D(K)
K,K
bimodule generated by
6X
induced by the ring
in
This bimodule is just the free bimodule on
to
the D-centralising generators on the D-centralising set induces an isanorphism
12E(E(W
and consider the
X,
This is the image of
in
so, by
is the universal localisation of the fir
10.6,
x
X
First, we calculate the universal bimodule of derivations of E(X
Proof:
over
be a skew subfield of
6X,
2
(K)
6X.
However,
0
D
(D<X>)
is the free bimodule
and so, the natural map from 2' QS2D(D<X>)0;
since
D<X>
D<X>
to
K
is a fir, 10.7
D
shows that
K
must be isomorphic in the natural way to
This result gives us a useful handle on our first problem about the generators of free skew fields. Since we are looking at bimodules, it is
174
likely that we shall have to think a little about the enveloping algebra of a k-algebra,
R,
which is defined as R okR;
our last result shows that
the enveloping algebra of a free skew field embeds in a skew field, and so, the number of generators of a free bimodule over a free skew field is an invariant. In fact, we shall consider a more general result than simply generators for free skew fields, since the method of proof is no harder.
Theorem 10.9
Let
be a skew field with central subfield
E
E o k4xl....xn
be elements of
over
k Then, if the enveloping algebra of
E.
sending
to
yi
E o kjxl....xn
E u k
weakly finite, the natural map from
k
to
over
is
k
E o kix ....x 1
n
E o k4yl....yn)
with
n
1
extends to an isomorphism of
ti
Let
k.
that generate it as a skew field
tl.... tn
k
k
E o k'xl....xn k
The universal bimodule of derivations of
Proof:
E k k<xl....xn>
is the free k-centralising bimodule on the generators
dxi,
the same holds for the universal bimodule of derivations of
over
E
so, by 10.6,
E o k
over
E.
Since
tl....tn
E o kix .... nx as a skew field over
generate
k 1 So they are free generators because 2 (E o k k we assumed that the enveloping algebra of E o kixl....xn was weakly E,
dti
generate
finite. Hence, the map from
yi - ti
E o kjyl....yn k
to
k
E o
E o k4xl....xn k
yn>)
V
So, by 10.7, the map sending
morphism from
to
E U k
k induces an isomorphism Q (E
yi
to
® -r c
ti
given by
(E
k extends to an iso-
kjxl....xnk
No example is known of a tensor product of skew fields that is not weakly finite, so that it is yet possible that 10.9 may apply for all skew fields. We shall prove only that the enveloping algebra of E o kjxl....xn}
is weakly finite when
E
is finite-dimensional over
k.
k
We shall prove this by embedding
E o kixl....xn}
for a
in
k
suitable set
X
and natural number
m.
It is useful to have a more general
result on embeddings of simple artinian rings.
Lemma 10.10 such that
Let
{E i
for
i = 1 to n}
l.c.m,{[E;k]} = m; then
suitable set
i
Z.
be f.d. division algebras over
oE, o k4X> k 1 k
embeds in
M (k)
m
k,
for a
175
Proof:
S.
embeds in
MH(k),
so, by 9.11,
embeds in
o Ei
k
for
o Mm (k)
k
M (k), which embeds in o M (k) for countably many copies m k m M (k), which we index by Z. We have an automorphism, o, of this m defined by sending the ith copy of Mm(k) to the (i + 1)st, so we form
copies of
n
of
the skew Laurent polynomial ring
which is a prime
m (o M(k))[x,x-1;a], k
principal ideal ring, and so, has a simple artinian ring of fractions. However, we may construct this ring as a universal localisation of k Y.Y -1 ], which is clearly a universal m (k) u k[y]. Form the ring g M m(k) M k k[y], and consider the subring generated by the localisation of Mm(k)
k by the powers of
conjugates of
M
the copies of
MM(k),
m
(k)
y;
there can be no relations between
so this ring is just
for countably many
kMm(k)
M (k). We adjoin the universal inverses of all full maps between m inside Mm(k) k[y,yand the ring we f.g. projectives over kMm(k) k find must be our skew Laurent polynomial ring above. The simple artinian copies of
ring of fractions of this ring must be a universal localisation of Mm(k)
k[y],
ring
so it can only be
So far, we have embedded
k
k
m
just M (k) where m the isomorphism sends 1Jx
Therefore,
k
This is a
Z
M (k) k<X'> for X' = X u y. However, this is m k is the set {z.. i,j = 1 to m, x e X'} where :
x
in
universal localisation of
.
o E, o kW in the simple artinian k 1k
which is just M (k) o kEXuy).
M (k) o k(y) o m
z..
m(k) o k(y). k
k
1Jx
X' to the matrix
M (k) o k(X' m k
{z
}
1Jx
is isomorphic to
whose ijth entry is
M (MZ)). m
We can prove the theorem we wanted now.
Theorem 10.11
Let
of the skew field
be a f.d. division algebra over
E
E o k(xl....xn
k;
then
n
elements
that generate it as a skew field over
E
k
are free generators.
Proof: set
Z.
E o kExl.... xn embeds in Mm(k(S ) k Consequently, its enveloping algebra over k embeds in
By the last lemma,
Mm2(kEZ'°akk(Z)).
for some
So we need to show that the enveloping algebra of k("
is weakly finite from which our lemma follows. However, we know that the enveloping algebra of
morphism to a skew field correspondence with from 10.9.
Z.
embeds in a skew field, because it is a simple
k4Z'
ring (the centre of k("
is
k,
k(Z-°(Z'a
as we stated in 8.2), and we have a homowhere
Z'
is a set in bijective
So this is a weakly finite ring. Our theorem follows
176
There is another interesting result that we can prove using the techniques of this section, which gives us an invariant of skew field coproducts of f.d. division algebras over
We consider the universal bi-
k.
module of derivations of such a skew field. After we have shown that the enveloping algebra of a skew field coproduct of f.d. division algebras over k
is a weakly finite hereditary ring with a unique rank function, it follows
that the rank of the universal bimodule of derivations, which may be considered as a f.g. projective module over the enveloping algebra, is an invariant of the skew field. Since this number may be computed without any difficulty from any coproduct representation, this is quite useful.
Theorem 10.12
Let
algebras over
k;
be a finite collection of f.d. division
{Di: i = 1 to n}
then the enveloping algebra of
o D i
is a weakly finite
k
hereditary ring with a unique rank function taking values in natural number
m 1
Z
for some
m.
M
embeds in
for some set Z; by the m argument of 10.11, its enveloping algebra must be weakly finite since it Proof:
By 10.10,
o D. k 1
Mm(k4Z.).
embeds in the enveloping algebra of The enveloping algebra of
is a universal localisation of
o D. k 1
L o((o Di)o0kDi).
which is isomorphic to
(o Di)° 0k (k Di), k
(o D.)
k
k
Each ring of
o D. k 1
is
k
or
(o Di)°ID D.
k(t)
is simple artinian, because, the centre
for some transcendental
t
by 8.3; so our
enveloping algebra is a weakly finite universal localisation of a ring with a unique rank function. It must be a localisation at some set of maps full with respect to the rank function, and so, it has a unique rank function itself by the fact that all f.g. projectives are stably induced from the ring of which it is a universal localisation by 5.10.
Once we have this theorem, we know that the universal bimodule of derivations of
o D. k
over
k
is a projective bimodule, since it is a
sub-bimodule of a free bimodule, because of the exact sequence:
0-Q k(o Di) k
oD k
i Qk ok Di -* ko Di - O.
It is f.g. projective from the fact that
S2k(o Di)
k
® aS1 k(D.
177
by 10.4 and 10.6. Its rank is an invariant since there is a unique rank function.
m . -l The rank of
Theorem 10.13
S2
We have the formula
Proof:
(o
kk
1
where m
1 i mi
is
z
= [D :k].
i
i
2Qk(Di
0k (o Di)
®
k
From the exact sequence 0 -+ 2 k(Di) -+ Di0kDi + Di -. 0, we find the exact sequence 0 -+ (o D.) oD c2 (D,)-* (o D.) D. -+ o D. -+ 0 by tensoring k
on the left by
k
1
k
As a sequence of
(o Di)QD1.
bimodules, it is a
o Di, D. k
(o D.)OkD o
split exact sequence of f.g. projective bimodules since
is
simple artinian. The middle module is free of rank 1, the right module has rank
so
1/mi;
has rank
(o Di)@DS2k(Di)
1 - 1/mi
module. This implies that lthe rank of the bimodule 1 - 1/m1.,
as a
Di
o Di,
bi-
QS2k(Di)®k must be
and summing gives us the result we want.
We shall discuss in a later chapter the question of distinguishing more fully between skew field coproducts with f.d. factors, as well as providing examples of isomorphisms between certain of them that have the appearance of being quite different.
As a last result, we are able to show that certain skew subfields of
have the form they should have by the methods of this
E(N)
chapter.
Let M C N
Theorem 10.14 E;
be an extension of bimodules over the skew field
then the skew subfield of E(N generated by
E
and
M
is isomorphic
in the natural way to
Proof:
and
E<M>
are both firs and their universal skew fields of
E
fractions are
00E(E<M>)
By 10.5 and 10.6,
and
and a similar statement holds for
QE(E(Nf);
M
moreover, the universal
derivation takes the form of the identity map from
M
to M
in this
representation.
Let we wish to find bimodule of
K
be the skew subfield of
S2E(K),
2E(E(N))
generated by
M
and we have a homomorphism from it to the generated by
However, this must be simply isomorphism, and, by 10.8,
OM0;
6M,
since
so the map
K must be
EM.
M
generates
K
&S2E(E<M>)0 -+ S2E(K)
over
E;
K,K over
subE.
is an
178
11
COMMUTATIVE SUBFIELDS AND CENTRALISERS IN SKEW FIELD COPRODUCTS
In this chapter, we shall find out what we can about the commutative subfields of skew field coproducts, and centralisers in the matrix rings over the free skew field; in the first problem, the transcendence degree of commutative subfields of the skew field coproduct is shown to be essentially determined by that of the factor skew fields; in the second problem we find that skew subfields of cendental over
must divide
k
n2.
that have centres trans-
f.d. over their centre and in fact this dimension
are
In addition, we shall prove an odd result on skew sub-
fields of the free skew field; we shall see that every 2 generator skew subfield of the free skew field is either free on those 2 generators or else it is commutative; there is an analogous result to this known for the free algebra.
The basic result we need to prove our theorems is a characterisation of the transcendence degree of commutative subfields of matrix rings over a given skew field due to Resco (80) which we shall not prove.
Theorem 11.1 M1(E)
sion of
The maximal transcendence degree of a commutative subfield of
for varying
n
is the first integer
EOkk(x1...xm+l)
is
where
m,
m
such that the global dimen-
is a set of independent
{xi}
commuting variables. If there is no such integer, the maximal possible transcendence degree is infinite.
At present, it is unknown whether there can be no commutative subfield of transcendence degree in
E;
Mn(F)
m
inside
M
n
(E)
when no such field lies
however, there is a candidate for a counter-example. We consider which is isomorphic to
Mn(k) o k(xl...xm).
It is unclear whether
k
k(xl...xm)
can be embedded in
F.
We shall apply 11.1 to the study of commutative subfields of simple artinian coproducts; the author's original proof of the main theorem
179
in this direction was missing a step, which was supplied by Dicks who pointed out the next theorem.
be a weakly semihereditary k-algebra, and let
R
Let
Theorem 11.2
a simple artinian universal localisation of SOkA
of
Proof:
RQkA
is bounded by the global dimension of
-
If the global dimension of ROA is
or
be some
SMkA module, and let
resolution of
Tensoring over
with
R
A.
the result is
ROkA
is
Let
n >- 1.
be a
0 '' Pn+l + Pn + .... + Pl + M + 0
RokA module, where each
as an
M
for any k-algebra
0,
trivial. So, we assume that the global dimension of
M
be
S
then the global dimension
R;
is projective.
P.
gives us the sequence
S
0 + Pn+l RS + P 0 S + .... + P1ORS -+ MORS + 0 which we shall show is an n R R exact sequence. For Tori(M,S) = 0 for i 2 2, since R is weakly semihereditary, and so, of weak dimension 1; whilst Tor1(S,S) = 0 since S is R a universal localisation of R and so, Tor1(M,S) = 0, because M is an S-module.
P.QRS
module, and
SOkA
as
Me RS = M
which is
A(SOkA) k
SOkA
a projective
M
module. So,
has global dimension at most
as we
n,
wished to show.
This applies to the study of commutative subfields of simple artinian coproducts in the following way.
Let
Theorem 11.3
be simple artinian, and let
S
artinian S-rings. Let global dimension of
and
S1
be simple
S2
be a common central subfield. Suppose that the
k
SokE = n,
and the global dimension of
0 E
S i
then
n, a n.
n, = n
If
(S1 o S2)akE
is
n
or
for
n+l.
i = 1,2, If
ni > n
is
k
ni
then the global dimension of for
i = 1 or 2,
then the global
S
dimension of
is the maximum of
(S1 o S2)QkE
nl
and
n2.
k
Proof:
Let
A c B
right free over dimension of
provided that to
A A
SokE c S'akE,
A;
be a pair of rings such that
B
is left free and also
then by 9.39 of Rotman (79), we know that the global
is less than or equal to the global dimension of
B
has finite global dimension. In particular, this applies where
S'
is any S-ring. It also applies to show that
180
the global dimension of
is at least the maximum of
(S1 o S2)QkE k
nl
and
n2.
However, by the last theorem, the global dimension of (S1 o SZ)0kE k
is bounded by the global dimension of
(S1 8 S2(a k E _ (S1a kE)
subring
U (S20kE); Sia E SokE
(S1 S S2 )a E.
is free on either side over the
so, by a theorem of Dicks (77), the global dimension of
SOkE,
is bounded by the maximum of
(S1 S S2)OkE larger than n,
and if
and
n1
n2
if one of these is
then it is bounded by
nl = n = n2,
n+l.
The
theorem follows.
If we take
to be
E
k(x1...xm)
for varying
and apply
m,
11.1 and 11.3, we deduce:
Theorem 11.4 k
Let
and
n,n1
be the maximal transcendence degree over
n2
Mt(S), Mt(S1)
of commutative subfields of
where all rings are k-algebras, and is larger than
if
Proof: of
for varying
Mt(S2)
is an S-ring. Then, if
S.
n1
t,
or
n2
the maximal transcendence degree of commutative sub-
Mt(S1 o S2) k
fields of n2;
n,
and
n1 = n = n2
over
k
it may be
for varying n
or
t
is the maximum of
nl
and
n+l.
By 11.1, the maximal transcendence degree of commutative subfields for varying
Mt(S1 o S2)
t
is equal to the maximal global dimension of
S
(S1 o S2)okk(x1...xm)
for varying
m.
The theorem follows at once from
S
11.3.
Both possibilities may occur for S = k(x1...xm)
First, consider
n1 = n = n2.
Si = S o k(yi); then k S1 o S2 = k(y1) o k(y2) o k(x1...xm). By 11.4, the transcendence degree of and
k
k
maximal commutative subfields of On the other hand, let Then the centre of
Mt(S1 o S2) and
S = k(xl...xm),
S1 o S2
contains
S,
is at most m Si = S(ai)
in this case.
where
and also the element
ai = x1. al a2 + a2a1,
S
which is transcendental over of commutative subfields is
S.
So, here, the maximal transcendence degree
n+l.
We pass from considering the commutative subfields of skew field coproducts to the study of centralisers. There have been a number of
181
interesting results on centralisers in skew fields and rings. The most interesting of these is due to Bergman (67), who showed that the centraliser of a non-central element in the free algebra over a field is a polynomial ring in one variable. Another result of interest is due to Cohn (77') who showed that the centraliser of a non-central element in the free skew field is commutative; together with the earlier results of this chapter, we also know that its transcendence degree over
k
is
We shall prove that it
1.
must also be f.g. as a special case of results on centralisers in skew subfields of
M
We shall show that centralisers of elements trans-
n
cendental over
k
in such skew fields are f.g. of p.i. degree dividing
and having transcendence degree
1
over
number of skew fields may be embedded in
n
We have already seen that a
k.
In particular, we showed
Mn(k(X)').
in 7.12 that all skew field coproducts of f.d. division algebras over
may
k
be so embedded.
Theorem 11.5
Proof: X.
C
be an arbitrary commutative field extension of
k;
is an hereditary noetherian prime ring.
k(X
By 10.8,
is the skew subfield of
Since kW has centre
k,
COkk(X
generated by
c(X)
k
and
is a simple ring and so, must embed
which shows that it is a domain. It must be an Ore domain, because
in
C
Let
COkMn(MXf)
then
is commutative, and therefore, by representing
f.g. fields
COkk<X.
C
as a direct limit of
may be represented as a direct limit of noetherian
domains.
CokkW is a universal localisation of
C<X>,
and so, it must
be hereditary by 4.9. However, an hereditary Ore domain must be noetherian by Robson (68). Therefore,
is an hereditary
CekMn(k(X)) = Mn(Cokk(X9)
noetherian prime ring as we wished to show.
This gives us another handle on commutative subfields of
Any commutative subfield of
Theorem 11.6
degree at most
Proof:
Let
C
1
over
n
is f.g. of transcendence
k.
be a commutative subfield of Mn(kW);
is hereditary and noetherian. So,
COkC
by 11.5,
CokMn(k(X)
must be noetherian, since
is faithfully flat over it (see Resco, Small, Wadsworth 79)).
CokMn(kEX))
Therefore,
M
M (k4X}). n
C
must be f.g. That the transcendence degree is at most 1 follows
182
from 11.4.
This already shows that centralisers in
of non-central
k(Xi
elements are f.g. and so, f.d. over the field generated by the element they centralise. We extend this a little.
Theorem 11.7 an element
M
be a skew subfield of
E
Let
transcendental over
y
then
k;
Proof: Let M be a maximal commutative subfield of is f.g. of transcendence degree algebraic extension of If
finite. But
M' = M,
M 2 k(y), and so,
E.
M
Therefore,
k.
is a f.d.
and is finite;
E, [M:C] <_ [M:k(y)]
is the centraliser of
M'
M
in
[E:M'] = [M:C]
E,
and so, [E:k(y)] = [E:M][M:k(y)]
So, skew subfields of over
over
is finite.
k(y).
is the centre of
C
therefore, if
1
whose centre contains
n [E:k(y)]
M M(X)')
is
is finite too.
whose centres are transcendental
n
are f.d. over their centre. It turns out that we can get a good bound
k
on the p.i. degree of such skew fields. The result we prove yields Cohn's theorem that centralisers in the free skew field are commutative.
Theorem 11.8
be a skew subfield of
E
Let
Then the p.i. degree divides
Let
Proof:
EOkk
so,
be the algebraic closure of
k
embeds in If
is f.g. over
k If
embeds in
E
k;
M (k(X)),
and
n
Mn(k(Xf).
has finite p.i. degree,
E
of finite p.i. degree.
Mn(k(X.)
n.
E
is f.d. over its centre, which
E6kk must be artinian.
by 12.6, and so,
where each
EMkk/rad(EQkk) = XSi,
is simple artinian,
Si
i
each
is a central extension of
S.
E,
and so, has the same p.i. degree.
Further, the transcendence degree of the centre of each is isomorphic to
so, by Tsen's theorem,
S.
degree of
is the centre of
E,
and
Eokk/rad(EOkk) ideal,
Ekk
C.
Mm(xC.),
E,
S. i
where
is at most
m
1,
is the p.i.
Therefore,
and since matrix units lift modulo a nilpotent
i
M
m
(A)
for some artinian ring
M (k) in M (k(X)), m n which shows that m divides n.
preserving embedding of of
S..
Mm(Ci)
A.
Therefore, we have a unit-
where
m
is the p.i. degree
We summarise the last few results in the following form.
183
Theorem 11.9
be a skew subfield of
E
Let
transcendental over
then
k;
transcendence degree 1 over
E k,
whose centre is
Mn(kfX3)
is f.d. over its centre which is f.g. of and the p.i. degree of
divides
E
n.
We come to the last theorem of this chapter which is on 2 generator skew subfields of the free skew field. At present, it is quite unclear whether all skew subfields of the free skew field must themselves be isomorphic to free skew fields on some number of generators; we have already seen that commutative subfields must be f.g. of transcendence degree 1 but the methods are clearly too weak to show that they actually must be rational. It therefore comes as a surprise that we are able to prove anything at all about 2 generator skew subfields.
Theorem 11.10 then either
Let
be a 2 generator skew subfield of
F
over
k4X}
k;
is commutative or else it is free on the 2 generators.
F
is a weakly finite hereditary ring such that all
Proof:
projectives are stably free of unique rank, because it is a universal
localisation of F <X>
which embeds in
We have an exact sequence of
where the left action of F
Since
skew field by ds
and
dt;
F
s
St.
and
bimodules:
on kW is given by the embedding of F
is generated by elements
6s,
k(X3
FOkk4X + kW i 0
0 + Qk(F)0
by elements
F,
s
By theorem 10.7, either t
and F
t,
2k(F)
in
is generated
is freely generated as a
or else there is a non-trivial relation between
in the latter case,
2k(F)0
is a non-zero 2 generator pro-
jective bimodule with some non-trivial relation between the 2 generators;
so it has rank 1 considered as a projective considered as an
F
module, and k(X
module via the left action of
F
on
k4X3
described above is a simple module that is torsion with respect to the rank function.
We have another torsion F0gkk4X module,
Fok(s)k(X
and this
maps onto our simple module via the multiplication map. The category of torsion modules is an abelian finite length category by 1.22; therefore, as a torsion module,
has a unique
largest semi-simple quotient torsion module and the endomorphism ring of
184
must act on this semisimple module. This semisimple torsion
F'Ok(s)k(X)
module maps non-trivially onto our simple module
it has a n
unique largest direct summand of the form morphism ring of map from
k(s)
FOk(s)k(X)
which lies in the endomorphism ring of )n
endomorphism ring of centraliser of
F
transcendental over are not in commutative.
k
which is
in kW; consequently, k,
and the endo-
must also act on this module. So, we have a
Mn(C)
F"k(s)k(X
where
C
[C:k] = - because
to the
is the s
must be
and since centralisers of elements of k(X that
are commutative,
F,
which centralises
C,
must be
185
12
CHARACTERISING UNIVERSAL LOCALISATIONS AT A RANK FUNCTION
Simple artinian universal localisations In chapter 10, we saw that if
is a right hereditary k-
R
algebra, we can characterise those epic skew fields over
R
that are
universal localisations by the property that the associated map on the universal bimodule of derivations
0c
0K(F)
(R) 0
is an isomorphism. As
we shall see, this is equivalent to the property that
Tor1(F,F) = 0,
which
is a condition that we have already discussed in chapter 4; this is precisely the property that we need to generalise to a characterisation of epic simple artinian rings that are universal localisations. This condition fits nicely into the theory of f.d. hereditary algebras and allows us to characterise those epimorphisms from f.d. hereditary algebras to f.d. simple algebras that arise as universal localisations.
In order to prove these results, we have to discover a lot of information about the module structure of epic simple artinian rings that are universal localisations at a rank function on a hereditary ring,
R;
we
are able to turn this information into results about epic R-subrings of such simple artinian rings; they must all be universal localisations of
R.
Such a result was already known in the Noetherian case where, however, it was stated in terms of Silver localisation. One consequence of this result on intermediate rings is that an epic endomorphism of the free algebra on n
generators over a field
k must be an isomorphism.
It may well be true that if
the map R - R' R'
R
is a right hereditary ring, and
is an epimorphism of rings such that
is forced to be a universal localisation of
R.
Torl(R',R') = 0
then
The author has no way
of attacking this question, however, and the only results are those in this chapter apart from the result that if hereditary ring,
R,
then
a universal localisation of
i = I2
for an ideal
I
in a right
is a trace ideal which shows that
I
R
R/I
is
as we saw in chapter 4. Such a result cannot
be true for all semihereditary rings as one sees by looking at a local Bezout
186
domain,
since
whose maximal ideal,
L,
is idempotent;
M,
Tori(L/M,L/M) = 0,
M2 = M.
Lemma 12.1
Let
4:R - S
be an epimorphism between K-rings, where aQK(R)a, if and only if 0K(S) a
K
is a
semisimple artinian ring; then Tori(S,S) = 0.
Proof:
By 7.1,
is an exact sequence split as a
0 - 0K(R) - RQKR -> R - 0
sequence of left R-modules, so the sequence is also exact, and since
K
0 - SORS2K(R) -)- MR -r S -r 0
is semisimple artinian,
SOKR
is a projective
R-module. Therefore, we have an exact sequence:
O } Tor1(S,S) -)- Q0K (R) a -> SoKS -> S -> 0. We extract the exact sequence:
O -> Tori(S,S) -> a S2 K(R) a - S2K(S) -* 0,
from which our lemma follows.
Before setting about the main proofs, we isolate a useful lemma.
Lemma 12.2
Let
R be a ring of weak dimension 1 (in particular, a right
hereditary ring), let M
and
respectively , with submodules
be a right and left module over R, R M' and N'; then if Tor(M,N) = 0, N
Tori(M',N') = 0.
Proof:
Apply the long exact sequence of
O -> M' -> M - M/M' -* 0,
which shows that
TorR(
N)
to the sequence
Tori(M',N) = 0,
and then apply
the long exact sequence of TorR W, ) to the sequence 0 -r N' -r N -. N/N' -+ 0 to complete the proof.
We can begin on the first result.
Theorem 12.3
Let
R
epimorphism from localisation of
Proof:
R
be a right hereditary ring and let :R + S to a simple artinian ring if and only if
R
S;
then
S
be an
is a universal
Tori(S,S) = 0.
Universal localisations of a ring
R
f.g. projectives always satisfy the condition
at some set of maps Tori(RE,RE) = 0
E
between
by Bergman
and Dicks (78) and 4.7. The strategy for proving the converse is to find the structure of
S
as a right and left R module.
First of all, we can divide out by the trace ideal,
T,, of all
187
f.g. projectives of rank R
on
by the map to
with respect to
0
p,
the rank function induced
this is a universal localisation of
S;
maps all of which become invertible over
of a right hereditary ring is right hereditary, so by 1.7 and 1.8, all f.g. projectives over the rank function T
on
p
lies in the kernel of
R
R
at suitable
By 4.9, a universal localisation
S.
R/T
R/T
is right hereditary;
are induced from
induces a faithful rank function on Tori(S,S) = 0,
and
R -).S
and
R,
Therefore, if we may prove our theorem on the assumption that
Since
R/T.
Tori/T(S,S) = 0
too.
is a
p
faithful rank function, it follows in general.
Let M that Q
be a right R
Tori(M,S) = 0.
submodule of
S;
then, we see from 12.2
0 - P a Q i M - 0 be a presentation of
Let
is f.g. projective, and
where
M,
must be projective, and therefore, a direct
P
sum of f.g. projectives by 1.2. Then, since
Tori(M,S) = 0,
the sequence
below is exact: a0S 0 -> PORS
Since
p
QQ_RS -+ MGRS -+ 0
is a faithful rank function, and
generated, and were not,
-+
agRS
p(P) < p(Q).
Also,
a
MGRS x 0,
P
must be finitely
must be a left full map, for, if it
factors through a module of smaller rank, and cannot be
injective.
We define the presentation rank of a f.p. module by
p. p (M) = p (Q) - p (P) ,
where
0 -+ P -+ Q -* M - 0
is a presentation of
it is well-defined by Schanuel's lemma. Our aim is to show that
M;
as a
S
right module is a directed union of f.p. modules of left full presentation of presentation rank 1, having no submodules of presentation rank
O.
There
is an analogous result on the left.
In order to carry this out, we need to see how left full maps with respect to
behave under
p
QRS.
We wish to show that they become
injective; this is a little harder to show immediately than that right full maps become surjective, which is our next step. Let over
R;
a:P -+ Q be a right full map between f.g. right projectives
then all f.g. submodules of Q
rank at least that of kernel of a map from
Q
containing the image of
to
S.
S
have
Since our first step showed that all such
kernels are finitely generated, we deduce that all modules over
P
and so, by our first step, cannot occur as the
Q,
Hom(cokera,S) = 0;
are projective, this shows that
cokeragRS = 0,
since
and
188
must be surjective.
09 RS
Since all right full maps between f.g. right projectives become split surjective over
all left full maps between left f.g. projectives
S,
become split injective over
by duality. If we had assumed two-sided
S
hereditary, we would know that left full maps between f.g. right projectives become split injective, and we would proceed as we do in the latter half of the proof; as it is, we are able to deduce these properties in the right hereditary case with a little more work. So, let Tori(S,M) = O.
Let
M be a f.g. left
R
0 - F -s Q + M - 0
be a presentation of
is f.g. projective, and, since submodules of
R
submodule of
then by 12.2,
S,
are projective. Our intention is to show that
F
where
M,
Q
is left semihereditary by 1.8, all f.g. F
is a
directed union of f.g. projectives where all inclusions in the system are left full, and the rank of each module in the system is less than that of Q.
Let
in such a P1j
P2i
E I1}
{Pli:i
mal possible rank
be the set of f.g. submodules of
if there are f.g. submodules of
q1;
we consider the set of f.g. submodules
Pli,
for any
and the rank of each
i,j
P2i
is
F
F
of mini-
that do not lie
{P2i:i E I2} q2,
where
the minimal
possible. In general, at the nth stage, either all f.g. submodules are inside some {Pni qn,
F
:
Pki
i e In}
for
k < n,
such that
or else, we form the set of f.g. submodules
Pni 4
for
k < n,
and the rank of
Pni
is
pkj
the minimal possible. We see that in the limit, every f.g. submodule of
must lie in some
of numbers in 1 mi,
Pni,
since the ranks
qn
form an ascending sequence
and so, are eventually bigger than the rank of any f.g.
m is the directed union of these modules; moreover, if
module. So
F
Pkl C Pi,,
the inclusion is a left full map, since there are no intermediate
modules of smaller rank. Consequently,
SMRF
is the directed union of the system of f.g.
modules
{SORPki:k a M, i e I } and all the maps in this system are k injective, because we have shown that left full maps between f.g. left
projectives become injective. As noted above,
Torl(S,M) = 0,
so the
sequence below is exact:
0 -> SORF -+ SORQ - SORM - 0 . Therefore, Q,
since
SORF
SORM
must be finitely generated and of rank less than that of does not equal
0.
Consequently, the rank of each
Pki
must
189
be less than the rank of
since S%Pki embeds in
Q,
our directed
SORF,
system must have stopped at some finite stage, and it represents
as a
F
directed system of f.g. submodules where all maps are left full, and the modules all have rank less than that of a:P + Q
Let
as we wished.
Q,
be a right full map between left f.g. projectives;
then all f.g. submodules containing the image of of
Q,
so that the image of
to
S.
Therefore,
a
have rank at least that
a
cannot lie in the kernel of a map from
HomR(cokera,S) = 0,
and
SORcokera = 0.
Q
Right full
maps between left f.g. projectives become split surjective, and so, by duality, left full maps between left f.g. projectives become split injective, as we wished to show originally.
Let M c N
be a pair of f.p. right modules of left full
presentation such that the presentation rank of any module and
N
is at least
between M
M1
then we show next that MRS + NORS
p.p(M);
is an
embedding. For consider the commutative diagram with exact rows:
a Q1 +M+0
O -> P
IIpj
nl
O+P -; Q2 ->N -r 0, obtained by pullback from a presentation of
maps. Then, any
such that
Q
Q1 E Q E Q2
the presentation rank of the image of assumption. So,
Q1 -> Q2
Q
where
N,
satisfies in
N
a,s
are left full
p(Q) a p(Q1)
is left full. Tensoring our diagram with
us a commutative diagram with exact rows:
since
M by
is at least that of
gives
S
0 + PIDRS + Q10RS + MGRS -0 11
nl
since
a
and
B
1
O + PORS + Q 0 S + NORS + 0, 2 R are left full. It follows that the map MGRS + NORS must
be an embedding. we use this to examine the structure of
S
R
as right
module in a similar way to our method of studying a left submodule of a free left module earlier on.
First of all, we note that every right containing with
S,
R
submodule of
R has presentation rank at least 1; for on tensoring we obtain
S = ROBS + MRS + SORS = S,
S
R c M c S
where the composite map is
the identity, so that the rank of MoS is at least 1; but we have seen that R Tor1(M,S) = 0,
so that
p.p(M) = pS(M@RS)
>_
1.
We consider the set of all f.g. right such that
Mli
contains
R,
and
p.p(Mli)
R
submodules
Mli
of
S
is minimal, which implies that
190
it is 1. Our intention is to show that {M2i:i E 12}
modules. If it is not, let S
that contain
be the sets of f.g. submodules of
do not lie in any Mli
R,
and have minimal presentation
subject to these conditions. In general, our nth step is to take
q2,
rank,
is the directed union of these
S
the set of f.g. submodules of
that contain
S
S
must lie in some
since we have seen that all f.g. sub-
Mki
modules have a presentation rank; consequently,
no f.g. submodule of that of
Mli
Mkj0RS -> Mli0RS
If
Mkj
must be an embedding. This shows that
module is isomorphic to
modules
{Mki }.
so, we have shown above that the map obtained by tensoring
Mkj;
s,
S
is the directed union
S
Mli, there can be S such that its presentation rank is less than
of the directed system of submodules
as
for
Mki
and have minimal rank subject to these conditions. Every f.g. sub-
k < n,
module of
with
do not lie in
R,
in particular, each of the ranks
Mki®RS;
PS(Mki'RS) = 1.
this is just the presentation rank of Mki (Tori(Mki,S) = 0), must have stopped at the first step; that is, the f.g. right
R
submodules
Since
our process
is the directed union of
S
{Mli}.
From here, it is not too hard to see that localisation of
which
S
is the directed union of the sub-
SORS
must he a universal
S
We know that all full maps between f.g. right projectives
R.
become invertible, since they are both left and right full. So we have a map from the universal localisation of
R
at the rank function,
Rp,
to
S,
which we shall show is surjective.
Let tion rank
s E S,
of
1
and let M
be a right R
containing both
S
R
and
submodule of presenta-
we consider the commutative
s;
diagram with exact rows:
O -+P ->P $R-+R-r0
0--P
Q
-).
+
M-.). O
obtained by pullback along P 0 R and
and
M
R c M
from a presentation of
M.
We know that
have the same rank, and since all f.g. modules between
Q
R
have presentation rank at least 1, the middle column is a full map.
So, over
it is invertible, and its inverse induces a map from M
R
to
P
whose composite with the homomorphism from
P embedding
of
M
in
S
RP
to
induces the
S
we began with. So the map from R
to
S
is sur-
P
jective.
We know that R
is a perfect ring by 5.3, so the surjective P
191
maps from R
R
to simple artinian rings arise as
N
is the nil radical, and
artinian ring
R IN.
If
e,
is an idempotent of
R
it is clear that
at the map
whose image in
R IN P
P
eR IN = eR e P
R
where
P
P
is a central idempotent in the semisimple
e e
P
is
+ R IN + eR IN, P
P
so
e,
is the universal localisation of
P
is a universal localisation as we wished to show,
S
P
because all f.g. projectives over
are stably induced from R by 5.3,
RP
and so, the iterated universal localisation theorem, 4.6, applies. Of course, after the event, we know that localisation of
R
S
at the rank function, so that R
is the universal is a rather better
P
ring than we knew during the proof.
It is possible to give a description of the right structure of
module
R
in the situation of theorem 12.3 essentially by abstracting
S
the relevant information from the proof. First, we assume the rank function is faithful by passing to the quotient by the trace ideal of the f.g. projectives of rank
0
if necessary. Next, we see that
union of the f.g. submodule of
M
rank 1. If
that contain
S
S
is the directed
and have presentation
is such a submodule, all modules between
presentation rank at least 1. Conversely, if R
R
a:R + M
R
and M
have
is an embedding of
in a f.g. module of left full presentation having presentation rank 1,
and all f.g. modules between R
M
and
have presentation rank at least 1,
we form the commutative diagram with exact rows obtained by pullback from a presentation of
M;
0
P® R+ R
P
0
ni
O + P + Q + M0 1
1
It is clear that the embedding of
P ® R
in
Q
is full, and so, it is
invertible over
S,
which leads to a homomorphism from
the map from
to
S.
module of a:R + M
M
R
of presentation rank
from
R
M
to
S
extending
The kernel of this map is the unique maximal subO.
So we wish to see how all the maps
to f.g. submodules of presentation rank 1, such that all
submodules have presentation rank not
0,
and those that contain
R have
presentation rank at least 1, having left full presentations, fit together to form a map
S.
If we have two such maps
a:R + M
ai:R + Mi.
the pushout map gives us
to a module of the right form apart from the problem that
it may have a submodule of presentation rank
0,
quotient by the unique maximal such submodule of commutative diagram:
R + M
1l
M2-* M
so we pass from
M
to the
M, M; this gives us a
192
so that we may form the directed system of all such maps. The direct limit is
S.
We may combine the last theorem with 12.1.
Theorem 12.4
be a right hereditary K-ring, where
K
is semisimple
be a map from
R
to a simple artinian ring
is the universal localisation of
R
at the rank function induced by the
artinian; let S
R
Let
R -> S
map if and only if
Proof:
then
S;
cK(S) = a9K(R)a.
We simply combine the equivalences given by 12.1 and 12.3.
In chapter 1, we defined the transpose of a f.g. module of homological dimension 1 such that the dual of the module is trivial by TrM = Ext1(M,R).
This defines a duality between the categories of such
modules on the left and on the right as we showed in 1.19. A pre-projective module,
is an indecomposable module such that
M,
for some non-negative integer
n;
indecomposable module such that
(TrD)nM
is projective
(DTr)nM
a pre-injective module,
is an
M,
is injective for some
clear that these are dual notions with respect to the duality also defines a module
Ringel
D.
over a f.d. hereditary algebra to be a brick if
M
is a f.d. division algebra and
EndR(M)
It is
n.
pre-projective and
Ext1(M,M) = 0;
pre-injective modules are examples of bricks.
Theorem 12.5
R
Let
be a f.d. hereditary k-algebra and let
indecomposable module over algebra such that
R
[M:D] = m;
with endomorphism ring
D
be a f.g.
M
a f.d. division
then the associated map from
R
to
is a universal localisation at the rank function it induces on R
M
only if
M
m
(D)
if and
is a brick; in particular, for all pre-projectives or pre-
injectives, the associated homomorphism from
R
to a f.d. simple artinian
ring is always an epimorphism.
All M (K)
Proof:
copies of
M;
for
modules considered as M,
R modules are direct sums of
ExtI(M,M) = 0 = ExtM1(k)(M,M)
and so,
m ExtM1(k) = EXt1 n
for all
TorR(M (k),M (k)) = 0, 1
m
m
Mm(k)
modules; by theorem 4.8,
and so by theorem 8.3,
m (k) m
is a universal
193
localisation of
R.
We shall use the last result in the next chapter to construct some interesting isomorphisms between simple artinian coproducts, but, for the present, we shall investigate the subring structure of simple artinian universal localisations of right hereditary rings more closely.
Epic subrings of simple artinian universal localisations of hereditary rings In this section, we shall show that all epic subrings must themselves be universal localisations. The principal application of this result is to show that all epic endomorphisms of the free k-algebra on n generators must be isomorphisms; this was shown for
by Dicks and
n = 2
Lewd; (82) . R
Let
Theorem 12.6 such that
R
then
T
be a right hereditary ring with a rank function
is a simple artinian ring; let
T
be an epic
p
subring of
R
P
R
;
is the universal localisation of
R
at those maps between f.g.
P
Proof:
that become invertible over
R
projectives over
T.
The method of proof is entirely similar to that of 12.3 except for
the need in one or two places for closer attention. We may assume as we do in 12.3, that
is a faithful rank function.
p
We have already seen in the course of 12.3 that all f.g. submodules of
are f.p. of left full presentation and if
R
M
is such a
P
TorI(M,R
module,
)
R
Further,
= 0.
P
union of f.g.
Let
under tensoring with
over
Rp
M
where
R c M c T,
P
MRp = RP,
since
has presentation rank 1; then, we R p
} R ® R
p R p
is isomorphic to
MORRp
for
MO R
RP
P. c M.
Tort(
R
)
associated to
P
the short exact sequence: 0 -)- M -> R -*R /M -} 0; we find the exact p
P
0 = Tor I (R
demonstrates that
P '
R
p
)
since
and the image of MO R R P P R Consequently, Torl(Rp/M,RP) = 0, as
we see by considering the long exact sequence of
sequence,
= R,
Tor1(M,R ) = 0, 1
is
is a universal
becomes an isomorphism R;
M has presentation rank 1, and RP
T
R.
find that the inclusion M c R
in
is also a directed
T
submodules of presentation rank 1, from which it will
R
follow in a similar way to the proof of 12.3, that localisation of
R
is the directed union of f.g. P
submodules of presentation rank 1; we shall show that
-r Tor I (R
P
) - M Rp
/M,R
Tor1(RP/M,RP) = 0.
P
R
-+
RpR ORp,
which
By lemma 12.2, we see that
194
which allows us to show that MORT -r TORT + T
Tor(T/M,T) = 0,
MORT -> TART a T
morphism; certainly, Torl(T/M,T) = 0,
contains
T
and
RT = T,
is an epic
R
that contain
T
tion rank,
tion rank
{M2i
:
R
j
submodules
and have minimal presenta-
Mli,
submodules of
do not lie in any
Mki
for
{Mni
T
:
i e In}
kl < k2
then
over all
Mki
that
and have minimal presenta-
k < n,
subject to these conditions. It is clear that
qn,
c_ Mk i, 1
R
is larger than 1. In general, at the nth stage,
q2
directed union of the complete set of
Mk
i E 11}
is 1; we wish
Mli
be those f.g.
i e 12}
do not lie in any
R,
we consider the set of f.g. R,
T {Mli
submodules of
subject to these conditions; we saw in the course of the
q2,
proof of 12.3, that
contain
which
MT
is the directed union of these modules.
T
If it is not, let of
R
and the presentation rank of each
R c Mli
to show that
ring; the image is just
R c M.
since
We consider the set of f.g. such that
is an iso-
is an injective map because
k.
T
is the
We know that if
and any module between the two of them has
2
rank at least equal to
qk
consequently, as we saw in the course of the
;
1
proof of 12.3,
Mk jfARRp i Mk i0RRP 1
qk
rank (since
1 T
modules
in one of rank
qk .
Therefore,
TORR
is an epic
R
that contain
R
P
P
P
ring) must be the directed union of the system of
and so, each such
Mki0RRP.
= TOTR = R
TORTOTR P
2
implies as we wanted, that Mli
must be an embedding of a module of
2
RP
module has rank at most 1, which
T must be the directed union of the submodules
and have presentation rank 1.
By the same method as we employed in the proof of 12.3, we
deduce that if R
is the collection of maps between f.g. projectives over
E
that become invertible over
these maps in over
RP,
E
T,
the map from
must be full with respect to
RE p,
is surjective. All since they are invertible
so we may apply the result proved in 5.8, that the image of an
intermediate localisation in the complete localisation at all full maps is still a universal localisation; in particular,
T
is a universal localisa-
tion.
Together with the theorems at the end of chapter 5, we see that the epic R-subrings of the simple artinian universal localisation of a hereditary ring
R
are precisely the universal localisations of
factor closed set of full maps between f.g. projectives.
R
at a
195
It is not a great deal of effort to prove from this theorem that all epic endomorphisms of the free algebra on
generators must be iso-
n
morphisms.
Theorem 12.7
:k<x1....xn> ; k<xl....xn>
Let
be an epic endomorphism,
then it is an isomorphism.
The elements
Proof:
4(xi)
generate the free skew field
k(x1....xn
consequently, by 10.11, they must be free generators; therefore k<xl....xn
is an epic
k<@(xl)...4(xn)>
subring of its universal skew
field of fractions, and, by 12.6, it must be a universal localisation of
k
at some set of maps. In order to show that it can only be
the trivial localisation, we consider the induced map on the functor
K1,
K1($).
Since
the identity map on
K1(k<x1.... xn>) = k*, k
;
by Gersten (74),
K1(¢)
must be
on the other hand, we have an exact sequence of
K-groups associated to the universal localisation by 4.11:
K1(R) + KI(RE) -r KO(T) -> K0(R) -> K0(RE) , where
R
and
RE
are isomorphic to
category of modules of the form is injective, and
K1(4)
but this implies that that
k<xl....xn>
E
k<x1....xn>
cokera
where
a
is an isomorphism so that
and
T
is in
is the full subE;
KO(T)
KO(R) -> K0(Rp)
must be zero;
must be trivial as required because it implies
is equal to
k.
So
0
is an isomorphism.
196
13
BIMODULE AMALGAM RINGS AND ARTIN'S PROBLEM
One of the purposes of this chapter is to construct a new class of skew fields generalising the skew field coproduct with amalgamation. They are interesting to us for several reasons. Many arise naturally as skew subfields of skew field coproducts; also, the methods that apply naturally to them apply just as well to the skew field coproduct giving us results that would not have been clear without this greater generality; further, we may use the new construction in order to show a number of interesting isomorphisms between apparently different skew field coproducts, and to study the simple artinian coproducts of the form
However, the major interest
M (k) o M (k). m
k
n
in the construction is the solution it leads to for Artin's problem. Artin asked whether there are skew field extensions, and right dimension of [E:F]1 x [E:F]r.
E
over
F,
E D F,
respectively
[E:F]
such that the left [E:F]
r
< - but
We shall see that for arbitrary pairs of integers greater
than 1 occur as the left and right dimension of a skew field extension. In recent work of Dowbor, Ringel and Simson (79), it was shown that the hereditary artinian rings that have only finitely many indecomposable modules correspond to Coxeter diagrams in the same way that hereditary artinian algebras (f.d. over a central subfield) correspond to Dynkin diagrams; they were unable to show however that any Coxeter diagram that is not a Dynkin diagram actually had a corresponding hereditary artinian ring, since the existence of such a hereditary artinian ring required the existence of an extension of skew fields having different but finite left and right dimension together with further conditions; at the end of the chapter, there is an example of an hereditary artinian ring of finite representation type corresponding to the Coxeter diagram
I2(5).
Bimodule amalgam rings Given a couple of skew fields cyclic
E1, E2
bimodule to be a pair
E1
(M,x)
and
E2,
where
M
we define a pointed is a
E1, E2
bimodule,
197
x
is in
of
For example, if
M = E1xE2.
and
M,
and
E1
the pair
E2,
is a common skew subfield
F
is a pointed cyclic bimodule on
(ElQFE2,1®1)
an F-centralising generator.
The importance of this idea for us lies in the observation that is the universal ring containing a copy of E1 E1 E2 is a quotient of such that the F pointed cyclic bimodule (E1,E2,1)
the ring coproduct and
E2
(E10FE2,101). If
tions for the generator
E1
the universal ring containing a copy of is a quotient of E1
x)
and
E2
such that
then
(EIE2,1)
is clearly the ring
(M,x)
E2 = <E1,E2
(Mu
eij a E1, eij a E2,
jeijxeij = 0;
are
x
bimodule and the rela-
E1, E2
is a pointed cyclic
(M,x)
0 J
J
if
J
Eei
j
ij
xei. = O> J
The first question to arise is whether this ring is not the trivial ring. We shall show that inside this ring to
and so, it can only be trivial if
(M,x)
that
E1
(M x)
M
(E1E2,1)
is isomorphic
is. Further, we shall show
is a fir. Therefore, it has a universal skew field of
E2
fractions, which we shall denote by
E1 (M, x) E2. It is fairly clear that the isomorphism class of the ring
depends in general on the generator x that we choose; E1 (M x) E2 surprisingly, this is not true for the universal skew field of fractions; E1 (M,x) E2
is actually independent up to isomorphism of the generator
x,
E1 o E2 where M is a M cyclic bimodule, though in this notation there are no specific embeddings
so that it makes sense to talk of the skew field
of
E1
and
and into
E2
in
E2
E1
E1 o E2, M
(M,x)
E2.
whilst there are specified embeddings of
E
This result allows us to prove a number of
interesting isomorphism theorems.
The method that we develop applies just as well to simple artinian rings
S1
and
S2
in place of
E1
and
E2,
so our policy will be to work
in this generality whilst pointing out what occurs in the skew field case.
Let (M,x)
S1
and
S2
be a couple of simple artinian rings, and let
be a pointed cyclic S12 S2
bimodule. A good way to study such a
situation is to consider the upper triangular matrix ring
R =
S1
M
O
S2
This ring is a particularly pleasant sort of hereditary ring, so, if we can pull the ring
S1 (M,x) S2
out of it in some way, we shall be able to show
198
that it too has good properties.
Theorem 13.1
M
Let
be right multiplication by
at
R =
O
x\.
O
0
is isomorphic to M2(S
a
S1, S2
be a pointed
both simple artinian rings; let
cyclic bimodule, where
S1
M
O
S2
and let
a: S1 O
are
S.
O
O
M
o
o
S2
Then the universal localisation of
S2).
1 (Mx)
R' _ /Sl
If
where
M ® N\
R
N
,
\0
is some
(Si
(M;x)
S2
bimodule, then the universal localisation of
where
M2(T),
is
S2
S1,
T
is the tensor ring over
S2)as1NOS2(S1
Both of these ring constructions are
S2).
(M;x)
a
on the bimodule
S2
S1 (M,x)
at
R'
hereditary.
In the ring
Proof:
representing fore
aS2a and
where
R
bimodule arises as a quotient of aS2a-1
suitably); therefore
If
R. = (,]. M®N
is generated by
occur state that
S1,
S1
R
R
is -generated by
and
S1
.
(S1aS2a1,1)
as pointed
S1,
(once we identify
(S1xS2'x)
is isomorphic to
S1
(Mx)
S2.
S2
Of
is not the trivial ring. then
I
2/I aS2a1,
and
and those
2 by 2 /matrix units; there-
set of
O O a O O and the only relations arise because
course, we do not know that \R
R'
a`
R
is
(p 0)
and
the elements (1 0)
and its inverse form
a
R a = M2(R),
aS2a -1 ,
Ra,
Ra - M2(R')
and
Na -l,
where
O) Ra(O O);
where the only relations that
generate a-copy of
aS2a-1
I
R' = (
S1
S2
and that
(t "'x)
Na
is isomorphic as left
S1,
bimodule (again we identify
S2
tensor ring over (S1
S1 KU
S2
(M, x) S2)0 S1N&S2(S1 Mx) Since
hereditary ring
M2(Sl (M,Ix) R,
right and
aS2a S2a-1).
bimodule to
N
as
S1,
S2
This is clearly just the
on the bimodule S2).
S2)
arises as a universal localisation of the
it must itself be hereditary by 4.9. So
S1
S2
is hereditary, and similarly, our tensor ring is hereditary.
When considering an upper triangular matrix ring of the form
199
(S1 o M/ S
where
S1
and
are simple artinian rings, we shall call the
S2
projective rank function defined by setting
P(01 O)
p(O
the
S2)
standard u.t. rank function; we have singled this rank function out solely because we shall use it most often, not because it has any special significance. If the map to
(0
S
1
given by right multiplication by
a
(0
from
O)
(Si 0)
Ra
is a full map with respect to this rank function /, then
will
2
inherit many of the good properties of
as we shall now show. In most
R
applications, it will be clear that this map is full; for example, this is true if
S1
Theorem 13.2 N
or
Let
be some other
is a skew field.
S2
be a pointed cyclic
(M,x)
S1, S2
bimodule. Let
Sl,
S2
bimodule, and let
R = (S1 S®N), and let
be the
p
2
defines
standard u.t. rank function. Then, if right multiplication by (0 to (0
a full map from (S1 0)
the tensor ring,
MON),
(S1 (M,x) S2)0S1NOs (S1 (t4,r) S2)
over the ring
T,
on the0)bimodule is an heredi-
S1 (M,x) S2
tary ring with a unique projective rank function whose image is generated by the images of the projective rank functions on the subrings
In particular, this applies when N Sl (M x) S2
Proof:
is
S1
and
S2.
we find that the ring
0;
has these properties.
We know that
is the universal localisation of
M2(T)
given by right multiplication by we are assuming is full. in the map from
KO(R)
(0
from
O)
(01 0)
R
at the map 52N),
to
which
(0
KO(R) _ Z ® Z,
and there is a non-trivial kernel
KO(M2(T)),
so that there is at most one partial
to
rank function defined on the image of
projective rank function on
M2(T)
K0(R)
in
K0(M2(T))
must agree with
p
on
so that any K0(R);
by the
remarks after 5.2, the rank function must be unique, if it exists and there is an extension since
a
is a full map.
In the case described by the conditions of 13.2, we deduce that the universal localisation of the ring by 5.5. We denote this by of
S1
(S1S2,1)
and
S2
S1 (Mop) S2,
S1
(M,x)
S2
along the pointed cyclic bimodule
is isomorphic to
(M,x)
must be simple artinian
the universal simple artinian amalgam (M,x).
as pointed bimodule; for
Also, we see that (O1
)
embeds
2
in the universal localisation at the rank function, so it certainly S embeds
200
but we have an isomorphism
in (OS 1 S )a, 2
(S1aS2a-1,1) = (S1xS2,x)
(M,x),
which is what we wanted. We can improve our result in the case where and
are skew fields.
S2
Theorem 13.3
bimodule
The map
E
R = (O1 \
E
and
E2
bimodule; then
be skew fields, and let
(M,x).
given by right multiplication by (0
a
be a pointed
(M,x)
is a fir. The pointed cyclic
E2
El
is isomorphic to
(E1E2'1)
(00
to
E1
Let
El, E2
cyclic
Proof:
S1
O)
from (01 0)
is full with respect to the standard u.t. rank function on
I
2
and it is factor complete by 5.14; so all f.g. projectives over
I
2
E2) = R
M2(E1
are induced from
Since
R.
become isomorphic over the universal localisation E1 (M,x) E2
and
101 0) `at
(0
E2)
it follows that
a,
must be a fir.
Isomorphism theorems Theorem 13.4 cyclic
Let
S1, S2
and
S1
as a bimodule such that
(O1 0)
to
(0
S2)
rank function. Then
Proof:
S2
be simple artinian rings, and let
bimodule./Assume that 10 0)
and
x \and
(O1 S2)
S2 S1 1 (M0 (Mx)
is isomorphic to
S2)
M
S
1 (Mx) S2. ,
os1 S
at the given rank function is at the unique rank
M2(S1 (Mux) S2)
function and the universal localisation of (M,x)
be a
with respect to the standard u.t.
on the ring
both the universal localisation of
M2(S1
M
are both generators of
both define full maps from
(0 0)
The universal localisation of
function; so,
y
M2(S1 (MU
M2(S1 (M,y) S2);
52)
at its rank
since both sides of the
isomorphism are simple artinian rings, we can deduce that S1 (M,x) S2 = S1 (M,y) S2.
For skew fields, we can eliminate a hypothesis from this theorem.
Theorem 13.5 bimodule; then
Let
E1
and
E2
be skew fields; let M be a cyclic
E1 (MOx) E2 = E1 (Moy) E2
E1, E2
for any bimodule generators
x
201
y
and
of
M.
y = Ee.xe',
If
from
E1 (M,y) E2
El (M,y) E2 E1
and
E1
to
with
E1
it is possible to calculate that a specific map
o E (Mx)
(M,x)
2
that extends to an isomorphism of
is given by
E2
E1Eeiei.
is mapped to
F1
We see from the last theorem that if skew subfields of
by the identity map,
E2 -> E2
and
El
E2
such that
and
ElIF E2 - E111F E2 1
bimodules, then
E1 Fo E2
El Fp E2.
1
are common
F2
as
El. E2
2
Our next result simply states some
2
interesting special cases of this result.
Theorem 13.6
Let
be a f.d. simple artinian k-algebra, and let
S1
a simple artinian k-algebra such that and
T2
then
Sl
S2
Sl
T1
since the ring
O
S00kS2
is simple artinian, and
2
both bimodules are free Sl
T1
[T1:k] = [T2:k],
S2.
1
R =
is simple artinian. If
T2
Sl0T S2 = Sl0T S2,
Proof:
S1®kS2
are common f.d. simple artinian k-algebras such that
be
S2
SlOT1S2
S2
S2
modules of the same rank. So
Si T2S2
S1
S2
O
O
T 1
Right multiplication by the element
1
defines a full map with
O \
respect to the standard u.t. rank function from the localisation of
R
is isomorphic to
R(
M2(Sl
0) S2)
to
R(0 1),
since
by 13.2, which has
T1
a homanorphism to a simple artinian ring inducing our rank function on
R.
OA1 Similarly, the element
T2
O Sl
T1 S 2
defines a full map; therefore, by 13.4,
O
= S1 T2 S2.
Our last result gives us a number of isomorphic simple artinian coproducts with the same factors but different amalgamated simple artinian subrings; our next result uses this to provide examples of isomorphic simple artinian coproducts that have different factors and the amalgamation takes
202
place over the same central subfield.
Theorem 13.7
Let
be a f.d. simple artinian k-algebra, and let
S
S2
be simple artinian subalgebras of
S
Sl = S o S2.
k
S
and
S1
[S1:k] _ [S2:k];
such that
then
k
= S o (S
S o S
Proof:
k
1
S
2
S o (S o S S1 2k 1
o S
2k
1
)
by 13.6, since the centre
is k or else purely transcendental over of degree 1, as we k saw in chapter 10, so that S°Qk(S1 o S2) is simple artinian. However,
of
S1 U S 2
k
S o (S2 o S1)
S
k
1
S2.
k
We can prove a stronger theorem than 13.7 in the case where is a central simple artinian k-algebra, that is, when the centre of exactly
Let
be a central simple k-algebra such that
S
be a simple artinian k-subalgebra such that mml where IXI = n2 S o S1 = k is the universal localisation of
M2 (S o S1)
[S:k] = n2
[SI:k] = m.
S1
Proof:
is
S
k.
Theorem 13.8 let
S
S
SekSl
Then
at the
k O
S1
S,
m
me
SMkS1 =
standard u.t. rank function. As a bimodule,
copies of the
i=l
bimodule
where
S
acts on the left in the obvious way, and
S
on the right via the embedding of
S1
in
S1
acts
m
S.
By 13.1, the universal localisation of
S
at the map
IDS
i=1
O O
from
to
(00)where
O
S 1
given by right multiplication by
IDS
S1
O
s
is
M2(T),
T
is the tensor ring over
S o S1 = S
0
(1,0,0...)
0
0
on the bimodule
S1 m
m
e So S$ S, S1 i=1 S
which is
bimodules; so
T
identify with
2 m1 n
a
S$
i=1
S<X>
where
S.
This is the direct sum of
S1
X
is a set of
mn2 m-1 m
simple
n2 m1 elements which we m
S-centralising generators of this bimodule. Hence, the
m universal localisation of
T
at its unique rank function is
203
is a universal localisation of
Therefore,
that
SQkSl
S
O
S1
is simple artinian, and the rank function induced on this ring is
so
p;
S o S1. k
We can use our methods in a rather more complicated way in order Mm(k) o M n(k); k
to investigate simple artinian coproducts of the form
we
should like to show that these are all suitably sized matrix rings over a free skew field on a suitable number of generators. We are not able to deal with this degree of generality at present, but there are a number of special cases of interest where we can prove this result; we shall also show that they are all suitably sized matrix rings over stably free skew fields. A skew field,
is said to be stably free, if
F,
F o k<X = k
for a
k
finite set
m
Of course, if
X.
divides
our last theorem applies to
n,
n for suitable X; Mm(k) o M n(k) =` M(k(X3) this actually allows k us to prove the next theorem with little effort. Before we do this, we note
show that
Mm(k) o Mn(k) = Mp(F);
a few generalities. We write
since the rank function
k
on of
has as image p where
Mm(k) k Mn(k) in
and
and
n,
M (F)
p
at this rank function, it follows that m
and
Suppose that
p
k
n
where p = l.c.m.{s,t};
i
Msn(k) o Mn(k) Z k
for some set
by 13.8. So
Z1
°
(o
M
n
is isomorphic to M (k) n
Mn(Ms(k(Z1j) o Mt(k)) k inside it.
M (k(Z ?)
0
M (F).
M (k) p s
k)Mtn(k);
this
n
as we see by taking the centraliser
Ms(k(Z1.) o Mt(k) =
k
Y.
k
M (k)Mtn(k)
s
k
m
for a suitable finite set
Msn(k) k Mtn(k) - Msn(k) k Mn(k)
of
M (k) u M (k)
is the least common multiple of
Ms(k) k Mt(k) = Mp(F),
Msn(k) o Mtn(k) = M n(F o
Proof:
is the least common multiple
n.
Theorem 13.9 then
p
is the universal localisation of
Ms
Ms (k) Taking the centraliser of
o Mt(k)
k M (k),
s
we find that
2o4
this is the ring Ms(k(Z
where
0 MM.(F)),
p' = p/s.
k
o Mp(F)
k1Z
is the universal localisation of the hereditary
k
ring
U
k u Mp,(F) = k u Mp,(k)
k
k where
Z2
p' (k)
written as matrices
Z1
MP,(k) u k. k
in
MP, (k)
Mp,(F),
MP' (k)
is the set of components of elements of
over the centraliser of
u
Mp,(F) = Mp,(k)
It is clear that this
Mp,(F u k), which is a hereditary ring whose only universal k Tracing the argulocalisation that is simple artinian is Mp,(F o
ring is just
Msn(k) o
ment back shows that
= Mpn(F
MM tn(k)
k
In order to go a little further, we look at suitable universal localisations of the ring
R =
m(k)OkMn(k)
Mm (k)
M
0
n
, which is a f.d.
(k)
hereditary k-algebra; we shall use 13.7 to find good universal localisations
of related rings, which allow us under suitable restrictions on m to construct universal localisations of rank function assigning the rank that have the form
R
and
n
at full maps with respect to the
to both indecomposable projectives,
/
M2P (k4X)).
We recall from the last chapter that the epimorphism from a f.d. hereditary algebra,
associated to a pre-projective or pre-injective
R,
module right module is always a universal localisation. For our purposes, we s
take
R
(k
to be the ring
O
k k
,
where
s
function associated to a particular module p
((O
):kJ/rM:k],
)R) _ [M(0
and
p
2!
2.
We wish to find the rank This is given by
M.
)R) _ [M(
((O
):k]/[M:k].
We
summarise from Dlab, Ringel (76), what rank functions occur associated to pre-projective or pre-injective modules over O\
) :k],[M
([M
O
k])
O
R;
the pairs
that occur are the positive real roots associated
01):
C01
t to the graph,
where there are
s
arrows from the first point to
3 second. Rather than developing here the precise theory which gives us these
positive real roots, we give an ad hoc description of them. We begin with the pair
(0,1);
our inductive procedure is to pass from the pair
(a,b)
205
to the pair then
(b,sb - a);
M,
(a,b)
is a pair constructed above,
is also a pair.
(b,a)
If
module
finally, if
is a particular positive root corresponding to a
(a,b)
then
CM:k] = a + b,
the epimorphism given by M
(so that the rank function `associated to
is
O)R)
P((O
as+b
= aalb
P ((O
,
s k0
is, we have an epimorphism from
That
.
11)R)
/
k
to
Ma
which must be a
+ b(k)
above rank function.
universal localisation at the
By Morita equivalence, we must have an epimorphism from to
Mb(k)
Ss
0
Ma (k)
which is a universal localisation. Here,
M2ab (k),
is the unique simple left
Mb(k),
right
Ma(k)
S
bimodule. The rank function
associated to this universal localisation is the standard u.t. rank function.
Consequently, we consider the ring
Mb(k)
Mb(k)
0
Ma (k)
We see from the above that we have a universal localisation of this which by 13.1, has the form
M
where
( T ) ,
T = M
ab(k)<(M
2
ab
(k)
®
Sab-s
Mb
% (k)Mab (k)>.
(k)
A dimension check shows that the bimodule in the brackets is isomorphic to Mab
(k)ab(ab-s)
as
bimodule. Consequently,
Mab(k)
T = Mab(kEW,
where
IXI = ab(ab-s).
Therefore, the universal localisation of
Mb (k)
Mb(k)®kMa(k)
0 at the rank function we are considering is M2(Mb(k) o Ma(k));
therefore,
Ma(k)
0 k
M2ab
Ma (k)
,
However, so is
Mb(k) =
We summarise what we have shown in the next theorem.
Theorem 13.10
Let
where there are
Ma(k) o Mb(k) k
be a positive real root associated to the graph
(a,b)
s
arrows from the first point to the second. Then
where
IXI = ab(ab-s).
206
the pairs
s = 2,
If
exactly the pairs
or
(n,n+l)
that occur as positive roots are
(a,b)
so that
(n+l,n),
for a suitable finite set
Mn(k) o Mn+l (k) M - n(n+l) k
We use this
X.
to prove the last result of this section.
Theorem 13.11
m
Let
and
MM(k) o Mn(k) = Mp(F), n,
k and
F
be arbitrary positive integers; then
n
where
and
is a stably free skew field.
By 13.9, we may assume that m
Proof:
m
is the least common multiple of
P
reduce to this case. Let
We know that
and
s
t
and
are co-prime, since we may
n
be integers such that
sm = to + 1.
Mmst(k) o Mnst(k) - Mmnst(F o
k
k
by 13.9.
But M
mst
(k) = M (k) o M k nst mst
M
which is isomorphic to
for a finite set
Z,
mst
(k)
since
(k)
must
sm = to + 1. o
Ma (k) o Mb(D) = Ma (k) o Mb(k)
k
k
M
o
Mms
Mb(D)
M
o
(k)
M ms
(k)
(k4Z))
o
M
If
ms
(k) O M (k) k nt
M (k)
nst
nst
nt
by 13.10,
b,
which is
Mb(D),
Mb(k)
13.8; this is isomorphic to
(k)
nt
divides
a
(k)
M
o
M
by
Mb(k) 0
Mb(D o k
the proof that
Mmst(k) o Mnst(k) =
where
Z'
is a finite
k
set.
We have no examples of a skew subfield of a free skew field that is not itself free, so certainly, we have no examples of stably free skew fields that are not free. At present, it is not clear how these matters will eventually settle themselves.
Artin's problem for skew field extensions The purpose of this section is to construct extensions of skew fields
E D F
such that the left dimension of
the right dimension of
E
over
F, [E:F]
r,
E
over
F,
[E:F]1,
and
are an arbitrary pair of integers
greater than 1. In fact, we shall consider a slightly more general problem; what are the possible pairs of integers for the left and right dimension of
a simple artinian ring M
n
(E)
over a skew subfield
F?
We shall see that
(k),
207
arbitrary pairs of integers greater than 1 and divisible by the left and right dimensions are divisible by tion that
Mn(E)
as a left or right module over itself is the direct summand
isomorphic simple modules.
n
of
occur; that
n
follows from the observa-
n
We shall actually prove something rather stronger than this, and the extra strength will be important in the final section where we shall construct hereditary artinian rings of representation type
In order
12(5).
to set up the notation used throughout this section, we state precisely what we shall prove. Throughout this section, the skew fields elements
{eij:i = 1
the elements
to
h = 1 to b}
{tih : i = 1 to n
and
E
the
F,
{skj:k = 1 to a, j = 1 to n},
the elements
n},
and the integers
n, a
and
b
will be named as in the following theorem.
Theorem 13.12 Let M(E) D F,
M
n
an, bn > 1:
{tih: i = 1 to n, h = 1 to b} over
F
such that
eiitih
F n Mn(E) = F,
right basis for
= tih.
M
n
over
F
M
(E)
be elements of kjeJ..7
and let
= skj
that right independent
Mn(E)
E
E,
n
(E)
F over
Mn(E)
F
is
and a
{skj }
{tih}.
is
In order to construct
be integers
UU
a left basis for
Mn(E)
s
a,b
Then there exist skew fields
F where
Let
Mn(E).
such that
F
be elements of
and a diagram of rings:
F D F
are skew fields. Let
F
{skj:k = 1 to a, j = 1 to n}
let
that are left independent over
(E)
and
E
be a set of matrix units in
{eij: i,j = 1 to n) such that
where
E
and
in theorem 13.12, we shall need
F
an intermediate construction.
Theorem 13.13 and
F'
where
Given the data of theorem 13.12, there exist skew fields
and a diagram of rings:
F' n Mn(E) = F,
the pointed bimodule
a basis for (Mn(E)F',1)
Mn(E')
E'
Mn(E)
U
U
F'
F
F'Mn(E)
over
F'
is isomorphic to
Notice that the extension of rings
is
{skj }
whilst
(Mn(E)OFF',101).
Mn(E') D F'
still satisfies
all the conditions originally stated for the extension of rings
M (E) D F. n
There is also a version of theorem 13.13 interchanging the role of left and right, of
a
and
b,
and of
{skj }
and
{ti },
which we leave
208
to the reader to formulate; we shall refer to this as theorem 13.13' when we need to mention it.
We begin by showing how theorem 13.12 follows from 13.13 and 13.13'.
We begin with the extension of rings
purposes we set
for inductive
F;
F = F0.
and
E = E0
Mn(E)
At an odd stage in our construction we assume that we have a diagram of rings:
Mn(E2
' F2
)
m
U
U m
FO
Mn(E0)
and
F2m n Mn(E0) = F0
such that
are skew fields such that the
E2m, F2m
conditions of theorem 13.12 are satisfied on replacing and
by theorem 13.13, we construct skew fields
F2m;
E
E2m
and
F
by
E2m
+ l' F2m + 1
with a diagram of rings:
F
Mn(E2m + 1) U
+ 1 U
F2m
Mn(E2m)
such that
F2m
{ski}
is
F2m + 1
+ 1 n Mn(E2m) = F2m,
whilst
a left basis for
(Mn(E2m)F2m
Mn(E0) n F2m + 1 = F0
the conditions of theorem 13.12 are satisfied with E
and
+ 1Mn(E2m)
over
+ 1, 1) - (Mn(E2m) OF2mF2m + l'
as pointed bimodule. It follows also that
replacing
F2m
E2m
+ 1
and
101)
and that
F2m + 1
F.
At an even stage of the construction, we assume that we have an Mn(E2m
extension of rings:
-
1)
D F2m - 1
U
U
Mn (Ep) such that
F2m
- 1
n Mn(E0) = F0
D
F0
where
E2m
- 1
and F2m - 1
are skew
fields such that the conditions of theorem 13.12 are satisfied on replacing E
and
F
skew fields
by
E2m - 1
E2m
and
and F2m F2m
- 1;
using theorem 13.13', we construct
with a diagram of rings:
209
Mn(E2m)
F2m U
U
F
M n (E 2m
-1
such that Mn(E2m F2m
over
is
-
whilst
{tih}
a right basis for
n F2m = F2m - 1,
1)
(F2mMn(E2m
as pointed bimodule. It follows that
1),1) - (F2mF
-
Mn(E0) n F2m = FO
tions of theorem 13.12 are satisfied with
Elm
2m-1
- 1)F2m
1),le1)
Mn(E2m -
and that the condi-
F2m
and
Mn(E2m
replacing
and
E
F.
On setting
Mn(E) 0 F
such that
for some integer
E = UE,, F = UF,, Mn(E) n F = F.
s e Mn(E),
If
over
{tih}
F2p
dependence relation between the elements
{skj }
dependence relation actually occurs over
Fq
+ 2'
on the left over
for some
over
Mn(E)
and the elements
F,
{skj}
Therefore, the elements
F.
over
{skj }
If there is a
q;
cannot happen by construction; similarly, the elements independent over
Mn(E2p
it lies in
consequently it is left dependent on
p;
and right dependent on
F2p + 1
we have an extension of rings
the
f,
however, this remain right
{tih}
are a left basis for
form a right basis, which completes
{tih}
the proof of theorem 13.12 assuming that we can prove theorem 13.13.
It remains to be seen how to prove theorem 13.13. There are two problems to overcome; firstly, we must construct a skew field an
F',
M
n
with basis
[M:F'] = an
artinian ring
as pointed
(F'Mn(E),l) ° (M,x)
and
such that
x
secondly, we must construct a simple
{xskj};
containing
Mn(E')
F' n F
M with an F-centralising generator
bimodule
(E)
and
F'
Mn(E)
Mn(E)
F',
such that
bimodule. We have already seen
an approach to the second construction; we set
M (E') = F' n
for
M (E); n
o (M,x)
the first construction, we set units where let
and
a
B
M = g11
11Man(F')
run over pairs and
generator of the bimodule show that
elements
x = g11
{xs
fa a F';
a
}
kj
11'
Certainly,
as the elements
{skj }
If the elements
M.
is a basis for M
{xskj }
the elements
Man(F') = Man (F) o Mn(E); F
over sa
for
k = 1
to
a
and
we have yet to show that [M:F'] = an, F'.
be matrix
gas
let
j = 1 to n; x
is a
and we intend to
For the time being we rename
since the indexing sets are the same.
are not a basis, there is a relation
rewriting this equation in the ring
Ef xs = 0 a a
Man(F')
for
we find that
210
the row whose ath entry is Man (F)
f
a
kills the matrix which lies in
This implies that the element
Mn(E), Eg
lies in a
Eg
F
left ideal of rank less than 1 where the rank is the unique one on
in turn, /O /
Man (F) u Mn (E) ; F
Egallsa\ of T = 1
0 must lie in a
inside
TI 0
0
/
/Man (F)
\0
ideal of standard u.t. rank <
take the form /0
0 )I
Man (F) ®FI\
0
I
\0
/
Mn (E)
Left ideals of
T
N \ where I
is a
0
left ideal of Mn (E) and N is an Man (F) submodule of Its standard u.t. rank is /(p(I) + p(N)). Let J=
0
O
Man(F)o I
0
N
0
I
0
0
0
<
T
Man (F)OFMn (E) .
containing
.
0 We have a map from
From the inequality [g11
be a left ideal of
of standard u.t. rank
Egall
M an(F)O Mn (E)
/
11N:F]
p(I) + p(N) < 1,
< an(l - p(I));
<_ Man(F)O FI ® N
g11 11'FEFsa
so we find that
[aFSa n I:F] > anp(I).
notation rather than
kj
a.
Since
I n Mn(E)eii = 0,
p(I) = m/n < 1
for some integer
I n EFski = 0; k
and so,
[(I + Mn(E)eii) n
and also
N.
g11
we deduce that
It is convenient to return to the indexing by
to
m,
then for some
i,
p(I + Mn(E)eii) _ (m + 1)/n
then
E Fskj:F] > anp(I) + a = an(p(I + Mn(E)eii)); k,j
we set
I'
to be
I + M (E)e n ii
by induction, we eventually find
and if
I' S M (E), n
an = CMn(E) n
E Fskj:F] > anp(tin(5)) = an. k,j
This contradiction shows that the element zero-divisor in
Man(F')
a basis for M = g11 bimodule on the generator
and so, the elements 11Man(F')
over
F'.
So,
we repeat the argument;
M
Egallsa
is not a
form {g11 llskj} is a cyclic F', Mn(E)
x.
We consider the simple artinian ring
M (E') = F'
n
0
(M,x)
M (E);
n
211
we already know that bimodule; so
Mn(E)
F
lies in
s
that the normaliser of
x
Man(F) 0 Mn(E),
and in
M
gives it
g11 11sg11 = g11 lls;
F.
It remains to show that (Mn(E)OFF',101);
xs = fx
Man (F')
Another way of stating this is
s = f.
is
F',
{skj}.
such that
n
rewriting this equation in
F';
so, inside
g11 11s = fg11 11'
the set
F'
s e M (E)
for let
n
in
f
as pointed
(M,x)
has as left basis over
F'Mn(E)
F' n M (E) = F;
Also,
for some element
follows that
is isomorphic to
(F'Mn(E),l)
(M
n
(E)F',l)
is isomorphic to
in order to prove this, it is simpler to prove the follow-
ing more general result.
Theorem 13.14
Let
rank function
p
be a semihereditary ring with faithful projective
R
taking values in
let
Z;
map between f.g. left projectives; let N n regard as a subring of
E(P) = EndR(P)
be an atomic full
a:P
be the normaliser of
and
E(Q) = EndR(Q)
which we
a
via the
embedding as left and right normaliser respectively; then, in the category of f.g. projectives over as pointed bimodule to
and of
R a
Rp,
is isomorphic
(E(Q)a-lE(P),a-1)
(E(Q)ONE(P),1M1).
It is immaterial whether we consider the bimodule in the category of
Proof:
f.g. projectives over
Ra
or over
of maps and so by theorem 5.8
RP
Ra
since
embeds in
Rp.
Certainly, there is a natural map from E(Q)a-1E(P)
sending
101
to
a-l.
is a factor closed set
{a}
E(Q)%E(P)
to
It remains to check that there is no
kernel.
There are two natural maps from
l:g
Sa&1
from
E(Q)@NE(P)
a(8a) = (as)a in
to
in
a8
E(Q)%E(P);
to
However, we have the relation
E(Q)a-lE(P).
from which we see that the element of
goes to
E(Q)
HomR(Q,P)
and their images agree under the natural map
and *2:8 -r 1Ha6,
E(P);
N
that goes to
it follows that the maps
01
and
8a
*2
agree; it is an embedding since the composition with the map to E(Q)a-1E(P) c Hangp
(R
regard
as an
1/
HanR(Q,P)
n E Sla i=1
where
pR
Q,RpQ
R
P)
Consider a relation in y1 +
6 e HomR(Q,P);
is the embedding of
E(Q),E(P)
Hom (Q,P); R
we shall
sub-bimodule via this map. E(Q)a-1E(P);
0
we shall show by induction on
n
that the relation
212 n
+ 6 = 0
Ea i=l
holds in
E(Q) @N E(P);
so far we have shown this to be
ey
true if
n = O.
From the relation 1/, we find that the map has nullity
p(P) = p(Q).
We write an equation expressing this defined over
o
a u
and
v
must be full maps such that
Vv =
By 1.19, there is an O
invertible map a such that
for full maps
u,,v.
ue
= I
0
R:
a
and ev =
we may adjust our previous equation to obtain the
equation:
Since
a
is an atomic full map, we may further assume that one of u,,v,
is
213
an identity map whilst the other is N
ul = a,
If
a.
Y1 = aTi
which lies in
and this allows us to shorten our relation by using a relation holding in by induction we deduce that our original relation is a
E(Q)& E(P);
consequence of relations in
Similarly, if
E(Q)®NE(P).
un
is an identity
= a a and again our relation is forced to be a n consequence of a relation in E(Q)QNE(P). So, we may assume that u1 is
map and
v
n
= a,
then
the identity map on
P
B
n
and
Vk + 1
un = a; let
be the first
ui
that is
equation with this information:
Since the top left hand corner of the leftmost map on the right hand side is invertible, we may adjust our equation again to obtain:
0
0
C Bn
61
1
-d
al
.
ak
.
.
.
From this, we have the relations so
mi
and
wi
. an via + aWi = O:qi a E(P),Wi E(Q);
lie in the left and right normaliser of
a
respectively.
once again, this
aTk + 1; Yk + 1 = allows us to shorten our relation by one that holds in E(Q)%E(P),
Further, we have the relation
induction, the relation 1/ is a consequence of a relation in We have shown that in all circumstances
and by
E(Q)OnE(P).
EBiOYi + d = 0,
which
is what we needed to prove our theorem.
To complete the proof of theorem 13.13, we need to show that this
214
last result implies that inside
0
F'
(M,x)
M (E), n
(M (E)F',l) n
is isomorphic
(Mn(E)QFF',1®1).
to
We regard
M (F'
the matrix ring
M \ at the standard u.t. rank function. Let
T = IF,
\O O\
P = IF
as the universal localisation of
M (E)) IX) n
(M0
2
Mn(E)/ o = k
and let
M
and
a:P -r Q
be right multiplica-
(E)
tion by
between
O
x
O
O
F'
o (M,x)
F'
to
EndR(P)
;
then
EndR(Q) = Mn(E)
M (E) n
and
and
Mn(E)
End to
calculate the pointed bimodule
Since
P T
P)
aEndR(Q)a-1.
Therefore, we need to
(aEndR(Q)a-lEndR(P),l) a
the isomorphism
is induced by the map that sends
as
is an injective map, this is the
(End R(Q)a-1EndR(P),a-1)
as
EndR(Q), EndR(P)
bimodule.
is an atom, our last theorem shows that this isomorphic to
a
(EndR(Q)ANEndR(P),101) a;
(T
bimodule; since
aEndR(Q)a-1, EndR(P)
same as calculating
TP
EndR(P) = F';
and
as pointed bimodule, where
this is just the normaliser of
Therefore,
(Mn(E)F',l)
x c M
is isomorphic to
N
is the normaliser of
which we have shown is
F.
(Mn(E)OFF',l0l).
This completes the proof of theorem 13.13 and, by symmetry,
theorem 13.13' follows; we have already seen that theorem 13.12 follows from these.
An hereditary artinian ring of representation type 12(5)
In (Dowbor, Ringel, Simson 79) they showed that hereditary artinian rings of finite representation type, by which we mean that there are only finitely many f.g. indecomposable modules, correspond to Coxeter diagrams in the same way that hereditary artinian algebras f.d. over a central subfield correspond to Dynkin diagrams. The Dynkin diagrams are all realised by suitable f.d. hereditary algebras, however, there were no known examples of hereditary artinian rings whose representation type corresponded to a Coxeter diagram that was not Dynkin since such example required an extension of skew fields over
F
E D F
where the left and right dimensions of
E
are finite and different; we shall construct an hereditary artinian
215
ring of representation type corresponding to the Coxeter diagram 12(5). At present, there are no examples known corresponding to the other Coxeter diagrams apart from H3 and H4 which arise as a trivial consequence of the existence of suitable bimodules for I2(5).
skew fields is a
v
bimodule for
ie
are labelled by bimodules
e
the beginning and
M
where
e
M
e
the end of the edge
re
The tensor algebra of the species is a tensor algebra of the bimodule
e.
® M
and whose edges
D
D1e, DTe
are labelled by
v
A species is a directed graph whose vertices
e e
XD
over the ring
McDv = 0
vv
v x
if
e.
,
If
where
R
D M
ve
= 0
if
v x i
e
and similarly
is an hereditary artinian ring of finite
R must be Morita
representation type, it is shown in the paper cited that
equivalent to the tensor algebra of a species; so, the species may be associated to
R
in a natural way; next the bimodules that occur in the
species of such an hereditary artinian ring of finite representation type
may be assigned an integer in a natural way (corresponding to suitable linear data on the bimodule which we shall not explain in detail); replacing
the species by the underlying undirected graph with the edges labelled by the integer associated to the given bimodule, we always obtain a graph whose connected components are Coxeter diagram. Hereditary artinian rings corresponding to the same Coxeter diagram have very similar module categories.
Next, we shall state precisely what conditions we need on a bimodule to construct an hereditary artinian ring of type I2(5). Let F
be skew fields and let
bimodule MR = HomF(M,F); cardinals
M
E,F
be an
if
[M:F]r = n,
E
and
bimodule; we may form the F,E then
also. If the
[MR:F] = n
[M:F]r, [MR:E]r, [MRR:F]r ... are all finite, the bimodule is
said to have finite right dualisation; the sequence of cardinals is always
known as the right dimension sequence. We shall be interested in constructing a bimodule such that the sequence of cardinals begins 2,1,3,1,...; in (Dowbor,
Ringel, Simson 79) it is shown that if such a bimodule exists the rest of the dimensions are determined; further, the ring
is an hereditary E F artinian ring of representation type I2(5). First, we turn this information
on the bimodule into information on the skew fields
Let M be a [M:F]r = n,
G,H
the action of
in turn we may recover M
the simple left M its embedding in
n
(H)
M (H); n
as
G
E
bimodule for skew fields on
M
and
F.
G
and
induces an embedding of
from an embedding of
G
into
G
Mn(H)
G,H bimodule via the action of
H;
G
if
into
Mn(H);
by regarding induced by
by the duality of rows and columns in a matrix ring,
216
is isomorphic as
MR = HomH(M,H)
it is clear that
consequently as
H,G
a bimodule sequence beginning to an embedding of
into
G
E,F
into
F
the embedding of CM3(F):E]1 = 3
given by the left action of
M3(F)
on itself
E
It remains to construct such an
[M3(F):E]r = 3.
and
whilst for
[E:F]1 = 2, [E:F]r = 3,
such that
E
into
E
which has a
this corresponds to an
right bimodule sequence beginning 2,1,3,1,... embedding of
M
bimodule
corresponds
whilst
[Mb(H):G]1 = ab
such that
In turn, if we have an
[Mb(H):G]r = cb.
module. Therefore,
bimodule M
H,G
for an
a,b,c ... Mb(H)
bimodule and
H,Mn(H)
bimodule to the simple right Mn(H)
extension of skew fields.
The construction EO D FO
Let CEO:F01
and
=2,
be an extension of skew fields such that
[EO:FO]r = 3;
{l,e}
let
be a left basis whilst
is a right basis. The left action of
{1 = el,e2,e3}
with the given right basis induces an embedding of x e EO
is sent to the matrix
(fi
such that
)
M3(F0)
where
Let
3
be matrix units corresponding to the basis chosen. Then E0, F
(M 3(F 0)g ill 911)
EO
i
>
is isomorphic as pointed embedding of
into
xe. = Eeifi..
>
{g,.: i,j = 1 to 3}
on itself together
E
EO
into
M3 (F
)
to
bimodule via the given
0
(EO,l).
O
We shall construct skew fields
such that
EO n F = FO,
basis for
E
over
F
a left basis for
U
U
E0
FO
over F
E
{l,e},
is
under the embedding of
{e1, e2, e3};
is
F
E
a right E
into
M3(F) given by the choice of basis (and therefore extending the embedding of
EO
M3(F0))
into
a left basis for
M3(F)
over
E
{g ii: i = 1 to 3}
is
which is also a right basis.
At an odd stage of the construction, assume that we have an extension of skew fields
Elm
F2m U
U
EO D F such that a left basis for fell e2, e3}.
Elm
0
over
F2m
is
Under the embedding of
E2m
into
choice of basis we have the extension of rings:
{i,e}
a right basis is
M3(F2m)
given by this
M3 (F2m)
E2m
U
U
M3(FO)
EO
217
such that a left basis for
Elm
over
M3 (F2m)
F2m
theorem 13.12, there exist skew fields
{gii:i = 1 to 3}.
is
By
and a diagram of
+ l' E2m + 1
rings:
E2m + 1
M3(F2m + 1
U
U
M3(F2m)
such that a left basis for
E2m
E2m
over
M3 (F 2m + 1)
+ 1
E2m + 1, f' -+ f
E2m.
into
where
g11f' = fg11
Under this embedding,
bimodule is isomorphic to is a right basis for
(E2m
+ l'
1)
(M3(F2m + 1)g11,g11).
over
M3 (E 2m)
is
{g ii: i = 1 to 3)
into + 1 which extends the embedding of
which is also a right basis. We have an embedding of
F2m
as pointed Since
F2m
E2m
+ l'F2m + 1 {e1911,e2g11'e3g11}
and therefore a basis for
F2m
{e1,e2,e3}
M3(F2m + 1)g11 = M3 (F2m)gll IF2mF2m + 1'
is also a right basis
since it is not over F2m Also, + 1. E2m n F2m + 1 = F2m + 1 and CE2m:F2mI 3. It follows that EO n F2m FO. + 1 = r =
E2m
for
E2m
At an even stage of the construction, assume that we have an extension of skew fields:
E2m
F2m - 1
- 1
U
E
EO n F2m - 1 = FO,
such that
U
0
=
and a right basis for
So, we have an embedding of
F2m - 1
is
{el,e2'e3}
M3(F2m -
1)
extending our embedding of
there exist skew fields
.
E2m
and
F2m
E2m - 1 E2m - 1
over into
into M3(F0). By theorem 13.12, E 0 and a diagram of skew fields:
F2m
E2m
U
U
E2m
FO
F
-1
such that {l,e}
-1
a left basis for E2m over F2m is E2m - 1 n F2m = F2m - l' {el,e2'e3} is a right basis. Once more, we note that the
whilst
embedding of
E2m
embedding of
E2m - 1
into
M3 (F2m)
into
induced by this right basis extends the
M3 (F 2m - 1 )'
218
Consider the extension of skew fields,
E
F,
E = UEi,
where
and
i
Any element of
F = UFi.
be left dependent on
lies in
E
Elm
- 1
there were a dependence relation between 1 and over some for [E:F7 E
n
= 2,
into
e
which we know does not happen; so,
F
F
and similarly
and
CE:F}r = 3.
over
E
for some integer m
and right dependent on
l,e
M3 (F)
{el,e2,e3}
over
over
this would occur
F,
{l,e}
E2m;
if
is a left basis
is a right basis. So,
This choice of basis induces an embedding of
which is just the union of the embeddings of
En
into
that we have considered at each stage of the construction. Any
M3(Fn)
element of
M3(F)
lies in
M3(F2m)
both left and right dependent on
for some integer
{gii:i = 1 to 3}
m;
over
there can be no dependence relation between the elements over
and must
el,e2,e3
E.
So,
[M3(F):E]
= 3 = [M3(F):E]
as we wished.
therefore, it is E2m
+ 1'
Clearly,
{gii:i = 1 to 3}
219
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222
INDEX
base ring 22 basic module 23 basic transfer map 23 bound module 18 brick 111
n-transcendental 158
dominion 122 doubly transitive 155
pre-pre-rank 115
enveloping algebra 174 factor closed 82 factor ring 22 faithful R-ring 22 faithful projective rank function 5
fir 4
free transfer map 23 full map 10
generating number 10 generating spartial splitting skew field 159 geometrically regular 39 Grothendieck group 3 honest 81 induced map 23 induced module 23 inner projective rank function 10
law of nullity 11 left bound modules 18 left full map 10 left hereditary ring 3 left nullity 10 left semihereditary ring 3 lower multiplicatively closed 39 maximal epic subring 122 minimal factorisation 14
partial projective rank function 5 perfect ring 77 pointed cyclic bimodule 196 pre-injective 192 pre-projective 192 pre-rank 115 prime matrix ideal 95 projective rank function 5 regular field extension 143 regular skew field extension 159 ring coproduct 22 R-ring 22 saturated 55 saturation 55 semifir 4 simple artinian coproduct 81, 101
singular kernel 94 species 215 stably associated 16 stably free skew field 203 stably isomorphic 4 standard u.t. rank function 199 Sylvester domain 11 Sylvester map rank function 97 Sylvester matrix rank function 97 Sylvester module rank function 96 Sylvester projective rank function 11 Sylvester rank function 97
tensor ring 79 torsion left modules 16 totally algebraically closed 158 totally transcendental 158 trace ideal 7 transpose 18 transvection 23 tree coproduct 153 tree of rings 153
223
twisted form 139 unbounded generation number 5 universal localisation at rank function 77 upper multiplicatively closed 55 von Neumann regular ring 4 weak algorithm 163 weakly semihereditary ring 3 well-positioned 110
