Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, ZUrich
237 Bo StenstrOm University of Stockholm, Stockholm/Sweden
Rings and Modules of Quotients
$ Springer-Verlag Berlin. Heidelberg- New York 1971
AMS Subject Classifications (1970): 16A 08
ISBN 3-540-05690-4 Springer-Verlag Berlin. Heidelberg. New York ISBN 0-387-05690-4 Springer-Verlag New York- Heidelberg. Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1971, Library of Congress Catalog Card Number 70-180692. Printed in Germany. Offsetdruck: Julius Bdtz, Hemsbach
Contents
Chapter I. Torsion theory I.
Preradicals
I
2.
Torsion theories
4
3.
Topologies
12
4.
Stable torsion theories
20
5.
Topologies for a commutative noetherian ring
23
6.
~-injective modules
29
Chapter 2. Categories of modules of quotient s 7.
Construction of rings and modules of quotients
33
8.
Modules of quotients and F..-injeetive envelopes
41
9.
Coreflective subcategories of
44
I0.
Giraud subeategories and the Popescu-Gabriel
Mod-A
theorem
48
Chapter 3. General properties of rin~s of quotients II.
Lattices eli-pure submodules
58
12.
Finiteness conditions on topologies
68
13.
Flat epimorphisms of rings
72
14.
Maximal flat epimorphic extension of a ring
82
15.
1-topologies and rings of fractions
86
Chapter 4. Self-injective rings 16.
The endomorphism ring of an injective module
93
17.
Coperfect rings
97
18.
Quasi-Frobenius rings
I01
Chapter 5. Maximal and classical rin~s of quotients 19.
The maximal ring of quotients
20.
The maximal ring of quotients of a non-singular ring
IIO 113
21.
The maximal ring of quotients of a reduced ring
118
22.
The classical ring of quotients
123
References
130
Introduction
These notes are intended to give a survey of the basic, more or less well-known, results in the theory of rings of quotients. An effort has been made to make the account as self-contained and elementary as possible. Thus we assume from the reader only a knowledge of the elements of the theory of rings and of abelian categories. We will briefly describe the contents of the notes. Chapter 1 treats the necessary preliminaries on torsion theory. The main result here is the establishing of a I-I correspondence between hereditary torsion theories and topologies on a ring (Gabriel [31]
and Maranda [51] ) In Chapter 2 we construct rings and modules of quotients with respect to an additive topology, following the approach of
briel ([10],bl]). This construction is a special instance of the construction of associated sheaves of presheaves for an additive Grothendieok topology! in these notes, however, we will not pursue that course. Rings and modules of quotients are then described in terms of injective envelopes (Johnson and Wong [114] and Lambek [46]). The main result of this chapter is the theorem of Popescu and Gabriel [62] which asserts that every Grothendieck category
C
is the category of modules of quotients for a
1
suitable topology on the endomorphism ring of a generator for The proof we give for this theorem is a simplified version of Popescu's proof [12], due to J. Lambek [48] and J.E. Roos [unpublished].
C .
VI
In Chapter 3
we treat some aspects of rings of quotients
related to finiteness conditions. In particular we prove a theorem of Popescu and Spircu [104] which characterizes flat epimorphisms in the category of rings as a special class of rings of quotients. In this context we also discuss rings of fractions, i.e. rings obtained by inverting elements of some multiplicatively closed subset of a ring. Chapter 4 contains some material on self-injective rings. Various well-known characterizations of quasi-Frobenius rings are given here. These results are used in Chapter 5 where ws discuss maximal rings of quotients and classical rings of quotients. Necessary and sufficient conditions for these rings to be regular, semi-simple or quasi-Frobenius are given (results due to Gabriel [ 3 ~ , Goldie [91], Mewborn and Winton
[541,
Sandomierski [681, and others). The references at the end of each section are intended to tell the reader where he can find a further discussion of the treated topics. Their purpose is not to record credit for the results of the section. I am grateful to J.E. Roos for allowing me to use his preiimina~j manuscripts for [66 I. Stockholm, December 1970 Bo StenstrSm
Some notation All rings have identity elements. modules are right
E(M) If
MA
A-modules is denoted by
to indicate that
M
is the injective envelope of L
is a submodule of
J(A) ~ or simply
denotes a ring and all
A-modules, unless otherwise is stated.
The category of right we write
A
M
and
Mod-A , and
is in this category. MA . x ¢ M , then
(L:x) =
J , denotes the Jacobson radical of
A .
Chapter I.
Torsion theory
§ I. Preradieals. A preradioal r(M)
r
of
Mod-A
assigns to each module
in such a way that every homomorphism
r(M)--~r(N)
by restriction.
M-~N
The class of all preradicals of
Mod-A
Mod-A
for all modules
the obvious ways. If preradicals
rlr 2
and
(rl,r2)(M)Irl(M) idempotent if if
r(M/r(M)) = 0
Proposition I.I.
Proof. To obtain
and rl:r 2
as
r
(ri)
r
and
is called r:r = r , i.e.
M . r
smaller than
larger than
of pro-
one defines
rlr2(M ) = rl(r2(M))
For every preradioal A
means that
( ~ r i , defined in
r 2 are preradioals,
for every module
~
rl~r 2
and is called a radical if
idempotent preradical smallest radical
and an infimum
r2( Irl( )) . A preradical
rr = r
.
M , and any family
~ ri
rI
induces
is a complete lattice,
because there is a partial ordering in which
radicals has a supremum
a submodule
In other words, a preradical is a
subfunctor of %he identity functor of
rI(M)Cr2(M )
M
there exists a largest
r , and there exists a
r .
A
r , we define a sequence of preradicals by trans-
finite induction as follows: r ~ = rr 8-I , and if
~
if
8
is not a limit ordinal we put
is a limit ordinal we put
r8 =
N
In this way we obtain a decreasing sequence of preradicals and we put
~ =Nr
~ . It is easy to see that the preradical
ra . r~ A
r
is
~dempotent, and that i% in fact is the largest idempotent preradical smaller than
r .
is obtained in a dual fashion. set
r
= r_l:r
, i.e.
r
If
~
is given by
is not a limit ordinal we r (M)/r _I(M ) = r ( M / r _ l ( M ) )
,
and for a limit ordinal
~
we set
ascending sequence of preradicals
Lemma 1.2.
If
r
r(M)nL
= L , so
the canonical homomorphism
so r(MIL)Cr(M)IL Proposition 1.3.
=
. This ~ v e s
ra(M)
~
~r , and we set
is a radical and
Proof. The canonical homomorphism with kernel
r8
LOt(M) M-¢M/L
, then induces
r(M)/LCr(M/L)
M/L-~M/r(M)
~ -~r
r(M/L)
.
r(M)/L .
=
r(M)--~r(M/L)
. On t h e o t h e r
maps
an
r(M/L)
hand,
into zero,
.
(i)
If
r
is idempotent, then so is also
(ii)
If
r
is a radical, then so is also
~ . r .
Proox. To prove (i) we have to shows (I) If
rI
and
r2
(2) If
ri
(i~I)
are idempotent, then so is also are idempotent, then so is also
rl,r 2 . Z ri •
For (ii) we have to show: (3) If
rI
and
(4) If
r.
(i~I)
i
(i): We have this gives
r2
are radicals, then so is also are radicals, then so is also
rl(M) C (rlzr2)(M) C M , and since rl(M) C r l ( r l z r 2 ) ( M ) C rI(M ) , so
Therefore we obtain from
Consequently (2): From
rlzr 2
ri(M)C
~r.. 1
rI
is idempoten~
rl(rlsr2) = r I .
(rl:r2)(rl~r2)(M)/rl(rl:r2)(M)
r2((rl:r2)(M)/rl(rl:r2)(M)) = r2((rl:r2)(M)/rl(M))
rlr 2 .
that
(rlzr2)(rl:r2)(M)/ rl(M )
rlr2CM/rlr2(M))
=
= r2r2(M/rl(M)) = r2(M/rl(M)) = (rl:r2)(M)Irl(M).
is idempotent. ~r~(M) C M
j~
we get
ri(M ) = ri(~jrj(M)) , so
3
(3): Since
=
rlr2(M) C r2(~) , we get from Lemma 1.2 ~hat = rlCr2(M)/rlr2CM))
= 0 .
(4): Since
j~rj(M) C
Nri(~Inj rj(~))i
preradical r
(b)
If
, we get from Lemma 1.2 that
= Nri(M)l~ rS(N~i
Proposition 1.4.
(a)
ri(M)
" o.
The following assertions are equivalent for a
r :
is a left exact functor. L CM
, then
r(L) = r ( M ) n L
.
The proof is easy. A consequence of this result is that a left exact preradical always is idempotent. To each preradical
L = (MI
r
we associate two classes of modules, namely
r(~') = M }
~r = T~ I rCM) = 0 }. Clearly: Proposition I.~. =r
~r
is closed under quotient modules, while
is closed under submodules.
Corol!ar ~ 1. 6 .
If
M ¢ ~r
and
N c ~r ' then
HomA(M,N) = 0 .
Examples: I. Let
A
be an integral domain. For every module
denote the torsion submodule of maximal divisible submodule of radicals, 2. Let s(M) of
A
t
M . Beth
t
d(M) and
we let
d
are idempotent
be an arbitrary ring. For each right module
M , and let
submodules of
r(M) M .
s
t(M)
denote the
is also left exact, which is not the case for
denote the socle of
radioal.
M , and let
M
M
d . we let
M , i.e. the sum of all simple submodules
denote the intersection of all maximal proper is a left exact preradical, while
r
is a
Exercisesz I. Show that every preradical commutes with direct sums, and that a left exact preradical commutes with directed unions of submodules. 2. Suppose
r
is an idempotent preradical. Show that
the largest submodule every
L' ~ L
.
3. Show that if
r
L
of
M
such that
is a left exact preradical,
~(M)
r(L/L') $ 0
is for
then also
is left exact. 4. Show that a preradical idempotent and
T =~
r
is left exact if and only if
r
is
is closed under submodules.
References: Radicals have been studied by many authors, usually in the context of general categories or the category of rings. The case of module categories has been treated e.g. by Amitsur[~|, Kurosch [ ~
and Maranda
the russian literature,
[ S ~ . For some further references to see Math. Reviews 35, no. 234.
§ 2. Torsion t h e o r i e s Definition. A torsion
theo,r~
for
Mod-A
is a pair
(~,~)
of
classes of right A-modules such that
(i)
~om(e,~) = 0
(ii)
Tm
and
The modules in in
F
F
for all
~ 6T,
F ~F,
are maximal classes having property (i). T
are called torsion modules and the modules
are torsion-free.
Any given class
of modules generates a torsion theory In
C
w
in the following way:
=
I H
(c,F)
o
all rorall
Clearly this
C c F
is a torsion theory, and we also note that
(~, F)
is the smallest class of torsion modules containing
T
C .
S
Dually, the class such that
(T, F)
is the smallest class of torsion-free modules
F
containing
o o6enerates a torsion theory
~
~ .
We can characterize those classes of modules which may appear as the class of torsion modules for some torsion theory! such a class of modules will be called a torsion class . (Classes will henceforth be assumed to he closed under isomorphisms). Proposition 2.1. A class of modules is a torsion class if and only if it is closed under quotients, direct sums and extensions. Proof. A class
C
is said to be "closed under extensions" if
for every exact sequence in
0-bL-eM
-- N - e 0
with
L
and
N
~ , also M e ~ .
Suppose
(~,
~)
is a torsion theory.
~
under quotients and direct sums, since Let
O-~L-oM-#N-vO
be axaot with
is torsion-free and factors over
MGT
=sM-VF
N . But also
, then
Hom(QTi?F) L
~
is obviously closed
and
N
in
is zero on
Hom(N,F) = 0 , so
=~Hom(Ti,F) T . If L , so
F =
a = 0 . Hence
•
Conversely, assume
C
and extensions. Let
to be closed under quotients,
(~, ~)
be
the
direct sums
torsion theory generated by
C . We want to show that
_C = T , so suppose
for all
be the sum of all submodules of
F s F . Let
C
Horn (T, F) = 0 T
•
belonging to
~ . C
then also belongs to
~ , since
under direct sums and quotients. To show that to show that
T/C e F . Suppose we have
The image of
a
is a module in
get a submodule of C , since
C
T
is closed
C = T , it suffices
a:C'-eT/C
C , and if
~
where
C' • C .
a = O , then we would
which strictly contains
C
and belongs to
is closed under extensions. This would contradict
the maximality of
C , and thus we must have
a = 0 , and
T/C e F .
Dually one shows ( by working in an abelian category one obtains this immediately by duality): Proposition 2.2.
A class of modules is a torsion-free class if
and only if it is closed under submodules, direct products and extensions. Let
(7, ~)
every module to
be a torsion theory. It follows from 2.1 that M
has a unique largest submodule
T ~ called the torsion submodule of
a preradical
t
of
t(M)
belonging
M . In this way we obtain
Mod-A , and it is easily verified that
t
is an idempotent radical. Conversely, given any idempotent radical
t
of
Mod-A ~ the pair of classes
(~t' ~t )
defined
in § I is a torsion theory. As a result we obtain: Proposition 2.3.
There is a I-I correspondence between torsion
theories and idempotent radicals. Proposition 2.4.
Let
r
be an idempotent preradical. ~
is the
idempotent radical associated with the torsion theory generated
by Proof. We know from I.I and 1.3 that
~
is the smallest idempotent
radical containing
r . It must therefore correspond to the
smallest torsion class containing
Proposition 2.~.
Let
C
~r ' which is just
M
Proof. Let
C
T(~)
has a
of modules
M
satisfying the ~ , and if
..~, then it is easily seen that
. It therefore remains for us to show that
a torsion class. Suppose
(Mi)
T(C_)
is a family of modules in
M i . Since
does also O-eL~
azM-~M'
a(Mi)
N . T(~)
T(~)
is
is clearly closed under quotients.
is a non-zero epimorphism,
Let
M
obviously contains
is a torsion class containing
on
consists of all
.
be the c l a s s
T(~)
announced condition.
~T(~)
C
such that each non-zero quotient of
non-zero submodule in
.
be a class of modules closed under
quotients. The torsion class generated by modules
~
choose
Mi
T(~)
. If
such that
~
a:QMi-~N is non-zero
contains a non-zero submodule in
~ , so
is thus closed under direct sums.
M-eNdO
be exact with
is a non-zero epimorphism.
L , N
in
T(~) . Suppose
We get an exact commutative
diagram 0
0
0 --~ K ~ L - - ' 4
L 0 -'~
L
--~
L
K f~L # L , then
so has then also
K --~ K t ~
L
0-'~L/Kf~L-~
If
0
M --J N
l
I N' --m 0
t
J
t
0
0
M' . If
has a non-zero submodule in
---D O
M'-*
0 L/Kf~L
0
L
has a non-zero submodule in LCK
, then
C . Hence
M' ~ N'
and so
M e T(_C) .
C M'
and still
Of particular interest for us will be the torsion theories (T, F)
where
~
is closed under submodules.
Pro~sition 2 6
(~, ~)
~et
be a torsion thsory and let
t
be the corresponding radical. The following assertions are equivalent: (a)
T
is closed under submodules.
(b)
F=
is closed under injective envelopes.
(c)
t
is left exact.
(d)
If
Proof.
LCM
, then
(c)4~(d)
(a)~(d):
If
t(L) = L N t ( M )
was remarked in 1.4.
LCM
, then certainly
is a torsion module,
t(L) = L ~ t ( M )
(d) ~ ( b ) :
Suppose
is essential in
(b)~a):
Suppose
t(L)CLNt(M)
since it is a submodule of
must have
F
.
. But
L~t(M)
t(M) , so we
.
F e F . Then
t(E(F))f%F = t(F) = 0 . But
E(F) , so this implies
T c T
and
CCT
t(E(F)) = 0 .
. There is a commutative
diagram C
t
;
T
o/t(c),,,, ~(c/t(c)) But
E(C/t(C))
and hence
is torsion-free,
so
~ = 0 . This implies
a = 0
C = t(C) c T .
A torsion theory with the properties stated in 2.6 is called hereditary. Applying 2.6 to Proposition 2.3~ we obtain: Corollary 2.7.
There is a I-i correspondence between hereditary
torsion theories and left exact radicals.
Proposition 2.8.
A hereditary torsion theory is generated by
the family of those cyclic modules Proof. A module for all
M
A/I
which are torsion modules.
is torsion-free if and only if
Hom(A/I,M) = 0
A/I c T .
A hereditary torsion theory is thus uniquely dstermined by the family of right ideals In the next §
I
for which
A/I
is a torsion module.
we will characterize these families of right ideals.
Prp~ositi0n 2. 9 .
Let
~
be a class of modules closed under
quotients and submodules. The torsion theory generated by is hereditary. Proof. We will show that the class of torsion-free modules is closed under injective envelopes. Suppose and there exists a non-zero Im a ¢ C to
and
Ff%Im a
a:C-~E(F)
F
is torsion-free
with
C ~ ~ . Then
is a non-zero submodule of
F
belonging
C , which is a contradiction.
Example s: 1. Let
A
be an integral domain. The "standard" torsion theory
is hereditary. An example of a non-hereditary torsion theory is given by
~ = [divisible modules), ~ = ~reduced modules}.
The next two examples generalize the standard torsion theory to arbitrary rings. 2. Let
A
be an arbitrary ring and let
modules of the form of
M/L
where
L
~
be the class of
is an essential submodule
M . The torsion theory generated by
torsion theory. Since
~
~
is called the Goldie
is closed under quotients and submodules,
the Goldie torsion theory is hereditary and is generated by the
10
family of cyclic modules ideal of
A . M
A/I
where
I
is an essential right
is a Goldie torsion module if and only if each
non-zero quotient of
M
has a submodule
@ 0
in
~
which we
may assm, e cyclic (2.5). Hence: Proposition 2.10. M
is a Goldie torsion module if and only if
for each submodule (L:x)
L~M
there exists
0 @ x ¢ M
such that
is an essential right ideal.
3. The torsion theory cogenerated by
E(A)
will be called the
Lambek torsion theory. It is hereditary, because we have: Proposition 2.11. Hom(L,A) = 0 Proof. If a
M
is a Lambek torsion module if and only if
for every submodule
LCM
and
a:L-~A
may be extended to
L
of
M .
is a non-zero homomorphism, then
M--*E(A)
. So every Lambek torsion module
has the mentioned property. The converse follows from the fact that if so
a
-I
a:M-~E(A) (A)"~A
is a non-zero homomorphism, then
Im a ~ A
= 0
is non-zero.
Exercises: I. Show that in the definition of a torsion theory
(7, [)
one
may replace the axiom (ii) by: (ii)': Each module
M
has a submodule
M/T ~ ~ ! moreover
=T and
T c T
such that
F= are closed under
isomorphisms. 2. Let
~
be any class of modules. Let
of modules
M
T(~)
be the class of
such that each non-zero quotient module of
has a non-zero submodule in
C . Show that:
M
11
(i)
T(C_) is a torsion class!
(ii) if
~
is a torsion class containing
(note that 3. Let
T
~T(~)
~ , then
~DT(~)
in general).
be a hereditary torsion class, generated by a class
I
C
closed under quotients and submodules. Show that if
T'
--
is
Z
a hereditary torsion class contained in generated by
C~'
T , then
T'
is
. (Hint: use 2.5).
4. (i) State and prove the dual of Proposition 2.9. (ii) Show that a hereditary torsion theory by the family of injective modules
(~, ~)
E(A/I)
is cogenerated
such that
A/I e F . (iii) Let
E
be an injective module. Show that the torsion
theory cogenerated by ~ E~ 5. A module
M
there exists (i)
is hereditary.
is called torsionless if for every a:M-eA
Show that
M
such that
a(x) $ 0 .
is torsionless if and only if
submodule of a direct product of copies of (ii)
0 $ x e M
Show that the class
L
M
is a
A .
of torsionless modules is closed
under submodules and direct products. (iii) Let
T
be the class of modules
M
for which
Hom(M,A) = O
W
Show that the following assertions are equivalents (a)
E(A)
(b)
L
is torsionless.
is closed under injective envelopes.
(c)
%)
(d)
(~, ~)
is a hereditary torsion theory. is the Lambek torsion theory.
If direct products of projective modules are projective, show that these conditions are equivalent to:
12
(e)
E(A)
is projective.
(Hint: show
3o show
suppose given
O~
choose maximal
K
and
K + L
L~
M ~M/L'O0
with
K~L
is essential in
References: Alin [11, Dickson
with
= 0
L , M/L
in
and remark that
~ !
KCM/L
M ).
~21], Mishina and Skornjakov
Exercise 5 is based on a paper by Wu, MOohizuki and Jans
[~]. [~].
§ 3. Topologies
In this §
we will establish a i-i correspondence between
hereditary torsion theories and certain families of right ideals, called "topologies". Definition. A right additive topology (or Just topology) is a non-empty family T I. If
I ~ ~
T 2. If
I
~
of right ideals of
and
a s A , then
A , satisfying:
(l:a) C ~ .
is a right ideal and there exists
(l:a) 6 ~
for every
a ~ J , then
J ~ ~
I ~ ~ .
Note that T i , together with the asst~mption that implies
such that
~
is non-empty,
A C F . i
Lemma 3.1. A t o p o l o ~ T 3. If
J ¢ F
T 4. If
I
and
Proof. T 3- If T 4: If by
and
JCI
J
are in
is a filter, i.e. satisfies: , then
IftJ ~ F
I s F .
F , then
a ¢ J , then
a c J , then
T I, so
F
(l:a) = A ¢ F , so
((IHJ):a) by T 2.
l(~J e F . I s F
by T 2.
= (l:a)i~(J,a) = (l:a) s F_
IS
The axioms T I, 3 and 4 state that in which
F
A
is a topological ring
is a fundamental system of neighborhoods of zero.
A non-empty family of right ideals satisfying T I, 3 and 4 is called a pretopology. Proposition 3.2. There is a i-I correspondence between: I) Non-empty families of right ideals of 2) Non-empty classes
u
C
define
M
is in
x c M . C
M
F
m
if and
C .
for which
satisfies T I, for if
multiplication by 0 -~ which shows that
a
Ann(x) ~ F_
obviously satisfies (2). COnversely, if
is the given class of modules, we put
This
M ~ C
is a family of right ideals satisfying T I, we
_C as the class of modules
for every C
F
satisfying T I.
of modules such that
only if every cyclic submodule of Proof. If
A
I c F
F--~I~A/I
and
~ CI.
a e A , then left
induces an exact sequence
(I:a) --0 A .-e A/I A/(l:a) C A/I .
It remains to verify that these correspondences are the inverses of each other. Starting with all
x S M~,
for all
F
we get
C = ~ M~ Ann(x) s F
and from this we get (I~ A/I s C ~ = ~ I ~
a S A~ = F
_c, we f i r s t get
(I:a) ~ Z
by T I. On the other hand, if we start with
Z
and then
A
(x)
~ M | each cyclic submodule e C ~ = C .
Proposition 3.3. There is a I-I correspondence between: l) Pretopologies on
for
A .
2) Classes of modules closed under submodules, quotients and direct sums. 3) Left exact preradicals of
Mod-A .
14
Proof. (I) 4-~(2): This is a particular case of 3.2. For it is easy to see that if modules
M
F
for which
is a pretopology,
then the class of
Ann(x) ¢ ~ , all x e M , is closed under
submodules (by 3.2), quotients (by T 3) and direct sums (by T 4). Conversely, if A/I e ~
~
is the given class of modules, then
~ ={I
is a pretopology, for T I is satisfied by 3.2, T 3 is
obviously satisfied and so is also T 4 because submodule o f (2)*=#(3): If
A/I(~A/J ~
is a
.
is the given class of modules, we let
be the larges~ submodule of
M
belonging to
exact preradical. Cbnversely, if then the class
A/I~J
~r
r
C . r
r(M)
is a left
is a left exact preradical,
is closed under quotients and submodules.I%
is also closed under direct sums, for if
r ( ~ M i ) ~ M j = r(Mj) = ~j
f o r each
M i c ~r ' then
J , so
r(~Mi)
=@M i
(or use § I, Exercise I). The correspondence between 2 and 3 is clearly I-i. Note in particular that given the pretopology r
is obtained as
r(M)
ffi{ x C
IAnn(x) ¢
~ , the preradical
}.
Theorem 3.4. There is a I-I correspondence between: I) Topologies on
A .
2) Hereditary torsion theories for 3) Left exact radicals of
Mod-A .
Mod-A .
Proof. We have already established the I-I correspondence (2)4-@(3) in Corollary 2.7. ~f
F
is a topology, then the corresponding
m
left exact preradical
t , defined above as
Ann(x) s [~, is a radical. In fact, if then ¢~},
Ann(~) = { a i xa ¢ t(M)~ = { a ~ and i% follows from T 2 that
t(M) ={ x c M
~ S M/t(M)
Ann(xa) ¢ [ ~ = I Ann(x) c ~
and
Ann(E) s [ ,
a~ (Ann(x):a)
and hence
~ ffi 0 .
15
O~ t h e o t h e r hand, i f
(~, ~)
i s a hereditary t o r s i o n t h e o r y ,
then the corresponding pretopology T 2. For if a ¢ J
I
for some
A/IuJ
and also
(Isa) e ~
for all
¢ ~
---eA/l--~ A/lu J --~ 0
s i n c e i t i s a q u o t i e n t module o f
~/I~J e T
A/I e ~
satisfies
J c F , we consider the exact sequence
= (Isa) ~ F . Since that
I~A/I e T}
is a right ideal such that
0 --@~/l~J
where
Z'(
since T
a e J
implies
A/J e ~ ,
((I~J):a) =
is closed under extenaions, it follows
and hence
I c ~ . This concludes the proof of
t h e theorem. If
F__i and
F_2 are topologies on
,eakor than F_2 (and
A , we say that
is stronger
Zl)if
FI
is
Zl C
It is clear that any intersection of topologies is a topology, so if
E
is any family of right ideals of
weakest topology
3(E)
oontaining
A
,
there exists
a
~ . The c o r r e s p o n d e n c e
between p r e t o p o l o g i e s and c l a s s e s o f modules d e s c r i b e d i n 3.3 preserves inclusions. It follows in particular that if a pretopology, then
_E is
J(E_) is the topology corresponding to the
hereditary torsion theory generated by ~ A / I ~ I 6 E }. Examples, I. The family
E
of essential right ideals is a pretopology,
but does not always satisfy T 2. The corresponding left exact preradioal is usually denoted by sin~ular submodule of
Z , and one calls
Z(M)
M . The hereditary torsion theory
the
16
corresponding to
J(~)
is the Goldie torsion theory (§ 2,
Example 2). The Goldie torsion radical containing
~
is the smallest radical
Z . The transfinite process which leads from
Z
to
G
(Proposition i.i) gets stationary at an early stage! in fact we have, in the notation of I.I: Proposition 3.~.
G = Z2 .
Proof. We first remark that G(M)
for every module
then
Z(L) = L , Z(M) = 0
implies so
Z(M)
is an essential submodule of
M . For if and
L = 0 . We now have
L
L C G(M)
and
L N Z(M) = 0 ,
is thus torsion-free, which
Z2(M)/Z(M ) = Z ( M / Z ( M ) ) D G(M)/Z(~) ,
z2(M) ~ G(M) .
~ 0 ~ o e i t i c n !.6. If then
is the family of essential right ideals,
J(~) = { I ~ there exists
(I:a) e ~
for all
Proof. If
I
If
~
such that
ICJ
and
a e J}.
satisfies the condition, then
I e J(~) , then Z2(A/I) . Ws R i t e
~ves
J e ~
A/J = Z(A/J)
A/I
r ¢ J(~)
is a Goldie torsion module, so
Z(A/I) = J/I . The f o = u l a a~
by T 2.
hence
= Z2(A/I) = Z(A/I) = J/I , so
J c ~. (l:a) s ~
defi~ing
~e also have for all
A/I = Z2
Z(J/I) =
a e J .
2. T"he right ideals of the topology corresponding to the Lamhek torsion theory are called dense. We denote this toplogy by Proposition 3.7. A
~
is the strongest topology on
A
~ .
such that
is torsion-free.
Proof.
A
is trivially torsion-free for
hereditary torsion theory such that
~ . If
A 6 F , then
(~, ~) T
is any
contains
the Lambek torsion modules and the corresponding topology is weaker than
u
D .
17
A more explicit description of
D
is given by:
Proposition 3.8. A right ideal
I
is dense if and only if
(~r:a) has no left annihilators for any Proof. Suppose
(Iza)
never has left annihilators. ~f there
exists a non-zero homomorphism 0 @ x c E(A)
such that
0 ~ xa @ A , If
xI = 0 . Let
A
with
, then there exists
a @ A
be such that
xa~ - 0 , and hence
xa
is
(I:a) , a contradiction.
Suppose conversely that in
A/I-@E(A)
b ¢ (l:a) , then
a left annihilator of
a , b
a c A .
I
is dense, and suppose there exist
b(l:a) = 0 . We get a commutative diagram
A/(I:a) ~
f A
h:
where
f(~)
= bc
injective.
I
,
g{~)
= a.
dense implies
and
h = 0
h
e x i s t s because and henoe
ideal
A
is
b = 0 .
Corollary 3.~. Every dense right ideal is essential in The ring
E(A)
A .
is called right n o , s i n g u l a r if the singular right
Z(A)
is zero.
Proposition 3.10. If
A
is right non-singular, then every essential
right ideal is dense, and the Lambek torsion theory coincides with the Goldie torsion theory. Proof. If
I
is an essential right ideal, then also
essential for every (Y:a)
(l:a)
is
a c A , and so non-singularity implies that
can have no non-zero left annihilators.
18
3. Let
F
be the family of those right ideals which contain a
non-zero-divisor.
F
is a topology if and only if
the "right Ore condition", and arbitrary b C A
a ~ A
such that
4. Let
S
satisfies
i.e. for every non-zero-divisor
there exist a non-zero-divisor
t
s
and
at = sb .
be a multiplioatively closed subset of
we also understand that ideals
A
I i for every
I c S ). The family
a c A
there exists
A
(O
for some
Every hereditary torsion theory on
n .
A-Nod
is stable. Proof. Let
M
be a torsion module. We decompose
E(M) = e E ( A / , ~ )
• S i n c e each
with the torsion module
E(M)
as
has non-zero intersection
M , we must have
from 5.9 and 5.11 that each and hence
A/~.
E(M)
E(A/~i)
~i s ~ " rt follows
is a torsion module,
is a torsion module.
Example: Let
A
be a commutative noetherian ring and let
preradical assigning to each module
M
its socle
s
be the s(M)
(§ 3,
Example 5). From Proposition 4.3 we get that the corresponding torsion submodule
s(N)
in
~(M)
is the maximal essential extension of
N
Exercises:
Let
A
be a commutative noetherian ring with a topology
i. Show that a module Ass(N) C ~ 2. L e t
M
M
~.
is a torsion module if and only if
•
be a f i n i t e l y
ideal. Show that
g e n e r a t e d m o d u l e and l e t
InN = O
for some
n> 0
I
be a n y
if and only if
.
29
I
is contained in every
References: Gabriel
~ c Ass(M) .
~1~
(oh. 5, §4 and 5), Matlis
[g2], [$3].
§ 6. ~-in~eetive modules
Let
~
be a topology on
Definition. A module for every
I ¢ F
E
A .
is
F_-in~ective if
ExtI(A/I,E) = 0
.
Before giving some alternative descriptions of
F_-injectivity,
we will introduce a convenient terminology. If
M
a submodule for every of all
L
of
M
is called an
x ~ M , i.e. if
F_-submodules of
M
and T4) which we denote by
Lemma 6 . 1 .
If
K c F_(L)
M/L
and
M/L
F-submodule if
is a filter (as a consequence of T3 ~(M) . Note that T 1 implies
and
L ~ ~(M) , t h e n
E(A) = ~ .
K ¢~(M)
.
O-~L/K-eM/K-eM/L-eO
are torsion modules. Then also
Proposition 6.2.
(Lzx) e
is a torsion module. The family
Proof. We have an exact sequence L/K
is a module,
M/K
where
is torsion.
The following properties of a module
E
are
equivalent: (a)
E
is
~-injective.
(b)
ExtI(M,E) = 0
(c)
If
for every torsion module
L ¢ ~(M) , then every homomorphism
to a homomorphism Proof. (a) =@ (o), Let
M ° L--~E
may be extended
M--*E . L c F_(M) and
f:L--,E
be given. In the
usual way we may assume that there is a maximal extension f':L'--~E
of
f , where
T 3. Suppose there exists
L C L' C M . Then also x c M
such that
L' e ~(M)
x ~ L' . Let
by I =
SO
= (L':x) C ~ , and let By (a) we may extend that
F
~
to
be the homomorphism
A , i.e. there exists
a(a) = ya . We may then define
= f'(z) + ya . g (c)~
¢:I-9E
extends
y ¢ E
g:L' + x A - ~ E
as
.
such g(z + xa) =
f' , which is a contradiction.
(b): We choose an exact sequence
is free. Since
a(a) = f,(xa)
Ann(x) ¢ ~
O-~K-~F-~M-,O
for every
x c M , we have
where K e ~(M)
.
The exact sequence
Hom(F,E)---, ~om(K,E)
;ExtZ~M,E)
together with (c), shows that (b) =~(a)
An
ExtI(M,E) = 0 .
F_-in~ective envelope of
such that
E
is
[-inJective and
Propositi0n 6.3. The su~od~e is an
,
is trivial.
Definition. M C, E
, 0
~-injective
envelope of
M
is a monomorphism
M ¢ ~(E)
.
{x ~ E(M) I (,,x) c ~ } of E(M) M .
Proof. Clearly the subset in question is a suhmodule of It suffices to show that Suppose we are given
E' = { x C E(M)~ (M:x) C ~ } is
f:l-~E'
extended %o a homomorphism
with
g:A-*E(M)
I ¢ F . f
E(M)
.
[-injecti,.
may in any case be
, so we get a commutative
diagram O ---~ I
o ~
o-., If
A
E'~E(M)
L
E,/~ ~
h % 0 , then
t
E(~)/M ~ E(M)/E'
: A/I
~
0
;E(~)/E,
,
0
,
0
U
E(,)/E,
contains a non-zero torsion submodule,
and from the lowest row we see that this contradicts the fact that
3!
E'
is the maximal submodule of
F-submodule. Hence
E(M)
h = 0 , and
g
containing
M
as an
actually maps
A
into
It follows from this result that the M
is ~ i q u e
up to isomorphism!
We will give another u s e f ~
F_-injective envelope of
it A l l
he denoted as
description of the
~(M)
~oposition
6.4.
If
C
to
~oof.
mod~e
the
for all
f,~(M)~0
with
fCM) = O ~
M .
Suppose
x c ~(M)
f(M)} = 0 . There exists = f(xI) = 0
cogenerati~
F , then
~CM) = { x ~ E(H) I f(x) = o for e v e ~
is
(§ 2, Exercise 4).
is an injective m o d ~ e
torsion theory a s s o o i a t ~
.
F_-injective
envelope. Recall first that a hereditary torsion t h e o ~ cogenerated by some injective m o d ~ e
E'
and
and consider any I ~ ~
f(x) = 0
Conversely, a s s ~ e
x
such that
since
C
f:E(M)--~C
with
xI C M . Then
f(x)I =
is t o r s i o ~ f r e e .
has the p r o p e r ~
that
f(x) = 0
for
f:E(M)~C
with
f(M) = 0 . We want to show that the right ideal
I = (M:x)
is in
g:A/I~C
is given, we may extend it to give a commutative dia~Tam
f
, i.e. to show that
A/~
i
Hom(A/I,C)
= 0 . If
~ E(M)/M
C
where implies
i(~) = xa . Since g = 0 .
hf(M) = 0 , we have
hf(x) = 0
which
$2
Examples: i.
Proposition 6.2 holds also when
If e.g.
~
filter
~(M)
the
F
is only a pretopology.
is the family of essential right ideals, then consists of the essential submodules of
the
M , while
[-injective modules are precisely the injective modules.
2. Let D_(M)
D
be the family of dense right ideals. The members of
are usually called dense submodules of
M .
Exercises:
1. An e x a c t sequenoe for e~ex-J onto
x
x G N
L
with
such that
submodule of
0--~ L ~
M-, N---0
Ann(x) c F
Ann(y) = Ann(x)
is called
there exists . L
F-~ure if y c M
is then an
F_-pure
M . Show that the following properties of a module
are equivalent:
(i)
L
(ii)
Every exact sequence
(iii)
is
L
mapping
F-injective.
is an
0 -, L -~ M -~ N-, 0
F-pure submodule of
References: Walker and Walker
[8|].
E(L) .
is
F-pure.
Chapter 2.
Categories of modules of quotients
§ 7. Construction of rin~s and modules of ~uotients
Let
F
be a right additive topology on the ring
right module respect to
M F
.
A
.
For each
we will define its module of quotients with This we will do in two steps. The first step we
take is to define M(F) = ~
HomA(I,M )
,
I e F ,
where %he direct limit is taken over the downwards directed family of right ideals. We want to give ring and Lemma 7.1.
then
M([)
that of a right
If
I , J c F
and
A([)
the structure of a
A(~)-module. For this we need: =:Ir-~A
is a homomorphism,
a"l(J) c F .
Proof. For each
=(j,,(a)) c Z
a ¢ I
hyTl.
We define a pairing x c M(~) , a ¢ A(~) we then define
we have
so -l(j) c Z
are represented by
l=(ah) ¢ J } =
by T2.
M(~) " A(~)---~ M(~)
xa c M(~) -l(j)
(a-l(j):a) = { h
as follows: ~:J-~M
and
suppose a:I--~A !
to be represented by the composed map
~ J-"~M
,
using Lemma 7.1. It is easy to see that
xa
is well-defined,
i.e. is independent of the choice of the representing homomorphisms , a . One also easily verifies that the pairing is biadditive. When
M = A , this makes
in the general case it makes
M(~)
A(~)
into an
M([)- A(Z)--~M([ )
into a ring, and A(~)-module.
34
There are canonical homomorphisme
~M
' ~ ~" H°mA(A'M) ---" lira HomA(I,M )
In particular~
~A
~A
is a ring homomorphism.
we may consider each
M(F_)
A-linear. The assignment
M~-P M(F_)
which is left exact, because are exact in
as an
Hom
t
A-module.
~M
is a fm~ctor
Mod-A . We denote by
Mod-A-'~Mod-A(F_)
LsMod-A--~Mod-A
xl = 0
Lemm~ 7.3.
= 0 ~ ~hen
and
F .
x
goes to zero already in some
M
a~xa
is a torsion module if and only if
M(F ) = 0 , then
represented by
M
~:J-PM
will follow that
is zero on
t(M) = K e r ~ M
by 7.2. Suppose on
is a torsion module. Let . If we can show that
x = 0 . For each
~(a)I a = 0.
= M
Put
a c J
M(F ) . 0 .
x e M(F )
Ker~ c F , it
there exists
K = ~" aI a . Then
be
KCKer~
Ia ¢ F
, and
aeJ for each
a e J
follows that
Lemma 7 . 4 .
we have
(K:a) D I a , so
K e _F by T 2. Hence
If
x G M(F__)
then the diagram
i
c---p
~(a) = xa .
(K:a) C F
and it
Ker ~ e F .
is represented
M
commutes, where
I
x e t(M) . This arguzent may he reversed.
the other hand that
such that
this functor
Mod-A(F_)--@ Mod-A .
Hom(I,M) , I ~_F . This means that the map
Proof. If
is then
Ker(M ~M-~-~M(F)) = t(M) .
Proof. If ~M(x)
i.e.
By pullback along
denote the torsion radical associated to
Lemma 7,.2.,,
•
is left exact and direct limits
followed by the forgetful functor Let
M(F_)
A
I
by ~ : I - e M
, I e F ,
,
35
Proof. If
a e I , then
A = (I:a)-@l-@M
xa
is represented by the composed morphism
given by
b~(ab)
= l(a)b , so
xa = ~ M l ( a )
.
An immediate consequence of this isl Lemma 7.~.
Coker~M
is a torsion module.
As the second step in our construction of modules of quotients we
apply the functor
L
once more to
AF
(called the ring of quotients of
an
AF-mOdule
MF
for each
M([) . This results in a ring A
A-module
with respect to
[ ) and
M . The description of this
is considerably simplified by, Lemma 7.6.
L
isomorphism
(Kit(M))(~) ~ ~ .
carries the monomorphism
Proof. Apply the left exact funotor
M/t(M)C~ M(_F) into an
L
to the exact sequence
0 --* MIt(M) --* M(_F)--~ C o k e r ~ M --~ 0 and apply Lemmas 7.5 and 7.3.
We have thus obtained the formula M F = lim H O m A ( I , M/t(M)) One verifies that the ring structure of structure of let
x e MF a e AF
MF
I
AF
S F
.
and the module
aregiven by the following pairing
be represented by -,, -
~:J-~M/t(M)
MFXAF--~M F :
,
a:I--~A/t(A) ,
induces J/t(J)~/t(X) exactness of
,
% ! xa e ~
and we have ~/t(J)~A/t(A) is represented by
u
.-ICJltCJ)) --~ItCJ)--~MltCM)
by left
36
For each
f:M-~N
in
Mod-A
one ge%s
fF:MF-eNF w
which gives a funotor
~
in
Mod-A F ,
- -
m
q:Mod-A-~Mod-A F . There are canonical J
homomorphisms TA:A'@A F
~M:M-*MF
of
A-modules!
in particular
is a ring homomorphism. For each
f e HomA(M,N)
we
m
g e t a commutative diagTsm f
M
~ N
MF -
.~ N F
--
Note t h a t
fF
--
Ker TM = t(M)
and %hat
also is a torsion
Coker ~M
module. When the torsion theory is stable (§ 4), the formula for simplifies somewhat:
7.7.
~oposition = I~
When the torsion theory is stable, one has
HomA(I,M )
Proof. Since
I~
for every module
M .
is an exact functor, the sequence
O---*~(M) - ~
M -~
M/t(M) - ~
0
induces an exact sequence 0-~
-
lim ---@ Hom(I,t(M))--~ li~ Hom(I,MI---~lim Hom(I,M/t(M))--~
lim
tl(1,t(M)) .
The first term is zero by Lemma 7.3. If of
t(M) , then
0-
E
E
is an injective envelope
is a torsion module by hypothesis. The sequence
t(M)
,E
:o
induces the exact sequence lim Hom(I, EItCM))----~ lim Extl(I,t(M))
; 0
where the first term is zero, again by Lemma 7.3. Hence the last term of the long exact sequence is zero.
37
We w a n t t o s t u d y
the image category
of the fm~ctor
q . For this
purpose we i n t r o d u c e r Definition.
MA
is
F-olosed if the oanonioal maps
M ~ Ho~A(A,M ) ~ are isomorphisms for all Thus and M
M
is
HomA(I,M )
I c F .
F-closed if and only if
M
is both torsion-free
~-injective (as defined in § 6). For every we get an isomorphism
Propositiom7.8.
MF
that if
M
x e M(F_)
T M Z M ~---~MF . Conversely we have:
is
Proof. To show that
F-closed for every module
MF
is torsion-free,
is torsion-free, and
xJ = 0
F__-elosed module
for
then
some
M(_F)
MA .
it suffices to show is torsion-free.
J ¢ F_ . L e t
x
Suppose
be represented
:I - ~ M . BY Lemma 7.4 we have a commutative diagram I
(
r
A
= xa M so
~M~
so
Next we show that with
H/t(xl
J
~|INJ MF
= 0
is
and
If~J s F . But
~M
x = 0 .
F_-injective.
Suppose we are given
I e F . Consider the pullback diagram j
w h e r e also
N(F-)
is zero when restricted to
a monomorphism,
f-I-~M F
P
?K
c
~
'
I
",-
is a right ideal. We have
I/J ~= Coker ~ =
is
by
38
= Coker~M ~at
, which is a torsion module, so it follows from T 2
J ¢ F . Lemma 7.4 now tells us that we may extend
a homomorphism
h:A-~M F . h
is also an extension of
g
to
f , because
m
h~I
and
f
are equal on
J , and therefore their difference
factors over the torsion module then implies
I/J , and
~
torsion-free
hJI = f .
Corollar~ 7.~.
The f u l l
subcategory of
Mod-AF
consisting
of
m
modules o f t h e form of
Mod-A
Let
C
MF
is
consisting of
equivalent
to the full
subcategory
F-closed modules.
be the full subcategory of
Mod-A
consisting of
m
F-closed modules. We have a number of interesting functorsz m
q
-e.
is the forgetful functor, q
is the functor
M~M
~t
is t h e f u n c t o r
M~MQA
i
is the inclusion functor,
F , AF '
is the functor
M~
M?
considers each
F_-closed module as an
AF-mOdule and is full
a n d faithful. We h a v e
Ja
=
q
and
i
=
'T~,J
• ~
induces a natural
a i ~ - - I d C . On t h e o t h e r h a n d , i a = LL = T m q .
equivalence
39
PTo~osition 7.10. Proof.
a
is a left adJoint of
i .
We must show that the canonical map HomA(MF,N ) --# HOmA(M,N ) m
is an isomorphism when
N
is
F-closed.
~M M
•
r y~
! i
l
f~
IfF
N
~
P
N~
It clearly is an epimorphism, so it remains to show that it is a monomorphism. The exact sequence o
--,
M F --P C o k e r T M ~
0
induces
where the first term is zero since
Coker~M
is a torsion module.
The desired conclusion now follows from the observation that
Hom(
It(M),N)
Hom(M,
)
.
We will show in the following sections that the category
_C~ of
F-closed modules is very well-behaved, in fact it is an abelian category with exact direct limits, although the inclusion functor i
is not exact.
Examples AD
Let
D
be the family of dense right ideals. The ring
is called the maximal (or complete) ring of quotients of
and will he denoted by have
~
= ~
Qm " Since
Hom(I,A) , I ¢ D .
A
is
_D-torsion-free, we
A
40
Exercises, I. Show t h a t morphism
if
E CF
are topologies,
there
is a ring
homo-
A E --~ A F .
2. Show that if
A
is a commutative ring, then also every
AF
is commutative.
(Hint: one can reduce the problem to showing
that if
and
I ~ F
=, ~ : I - ~ A
, then
a~(x) = ~=(x)
for
xcI2). 3. Let
A = K[X,Y]
maximal ideal I D m_n
where
K
m_ be the
_m = (X,Y) . Consider the topology
for some
F w (I
n~. Show that,
AF = A .
(i)
M = A/(X),
(ii) If
class of
Y
epimorphism, functor
q
then
in then
fF
b e a t o p o l o g y on
(so
A
is a suhring of , put
(i) If
KIT,I/T],
where
I:A-~M
r
is the
is the canonical
is not an epimorphism, and the
is n o t e x a c t ) .
F
~
NF =
M . (So if
4. L e t
of
is a field, and let
I~=~q
I ¢ F(AF)
A
and assume
A
F-torsion-free
AF). For each right
A-submodule
I
c AF ~ qICA}.
(of. § 6), show that
I~
HomA(I,A ) .
m
(ii) Call and
I
F-invertible if there exist
ql'""" 'qn ~ I4t such that
following properties of (a) I
is
(C) I
is a finitely
I
al,..,a n e I
1 - ~ aiq i . Show that the
are equivalent:
_F-invertible.
generated
projective
module and
I e _F(~). m
(Hint, of. [131, p.
41
5. Show that for any ring
A
and module
MA
one has
E(M)IZ(M) = lim Hom(I,M) where
I
runs through the downwards direo%ed family of
essential right ideals of
A .
References, Bourbaki [lO], (p. 157 and following), Gabriel [ 31],
(p
4n
and fonowing), Gol~an [33], ~oos [66] (oh. 1).
§ 8. Modules °f quotients andS-in'co rive envelopes Let
F
be a topology on
Pro~qsition 8.!. ' If MF~
EF(M )
Proof. 7.5, so
MF
M
A .
is a torsion-free module, then
as
AF-mOdules.
is
~-injective and
MF
is an
M~M
is a torsion module by
~-injective envelope of
M . If
EF(M) =
= ~x e E(M) I (M:x) e ~ } for a fixed injective envelope of
M , then the isomorphism
M F C EF(M )
Let us desribe explicitly the
for a torsio~free module is represented by
(note that
S~se
x ~ ~(.)
. Since
y e ~(M)
EF(M )
AF-linear by 7.9.
AF-mOdule structure of
=:I-~A/t(A)
there exists a unique a e I
M
is
~(M)
such that
is a module over
E(M)
, and
is
~ (M )
q ~
F_-closed,
xa(a) = ya
for all
A/t(A) ), and then
xq = y . This description of the module structure is applicable also for the
F-closed module
Proposition 8. 2.
If
injeotive envelope of
M
E(M) .
is torsion-free, then MF
in
Mod-~ m
.
E(M)
is an
42
Proof. The inclusion map
MF= EF(M)--~E(M )
Since it is an essential monomorphism in essential also in
over
AF-linear by 7.8.
Mod-A , it is obviously
Mod-A F . It only remains to show that
is injective as an
~-module.
E(M)
This follows from:
Every torsion-free injective
Le~aa 8.3.
is
A-module is inJeotive
AF . M
Proof. Let N'--~N
E
be torsion-free injective over
is any monomorl~ism in
~-linear.
f
extends to a homomorphism
g'(q) = g(xq)
and
and coincide on Coker ~ A - - V E E
and
x C N , consider the two maps
For each
and
Mod-AF
A . Suppose f:N'--~E
g,N--~E AF-aeE
. But
g'-g"
Coker ~ A
is torsion-free, so
A-linear
factors over a homomorphism
is a torsion module (Lemma
g, = ~' , and
g
is
F
7.5)
~-linear.
For the remaining part of this § we will assume that torsion-free. Thus
Mod-A .
given by
~'(q) = g(x)q . They are both
A . Hence
in
is
is contained in the the topology
AA
is
D
of
dense right ideals, and the torsion theory is cogenerated by an inJeotive module
C = E(A) ~ F
ring of the module The ring of
C
ring of
CA , and consider
HomH(C,C )
. Since
C
C
as an
. Let
is
H C
be the endomorphism as bimodule
HCA .
is usually called the double centralizer F_-closed,
H
is also the endomorphism
AF-module. It follows that there is a commuta-
tive diagram of canonical ring homomorphisms
i ""'* Ho (C,c)
43
Theorem 8.4.
~
centralizer Proof.
is
of the
We w i l l
an isomorphism cogenerating
exhibit
e HomH(C,C ) . Let
ftE(A)--pC
Extend .
f
~(A) ~F
is
to
psC-~E(A)
~sC-~C
AF
and the
double
C .
of
~ . Suppose
be the canonical projection.
by using Proposition 6.4.
A-linear map such that by defining
f(p~(1))
H-linear.,H~ce
/*
p~(1) e ~ ( A )
be any
. Then
inJective
an inverse
We want to show that Let
between
f(1) = 0 .
T(F) - 0 , where
. ~(~(1))
~ ~(f(1))
ps(1) e ~ ( A ) = ~
~ 0
C~=
since
. We may now define
as ~C~) = p~(1) c ~ i s an a d d i t i v e and i t
~k=
o n l y r e m a i n s t o show t h a t
we m u s t show t h a t
there exists ~(x)
map. C l e a r l y
=
~(h(1))
if
h e H .
8(1)
=
0
Then
~X~-~
,
i s a monomorl0hism . Thus
e F , then
such that
h(~(1))
~
id..
8 = O . For every
h(1) = x
and
x e C
h(F) = 0 . Then
.
Exaaples. I. Let
_F be the Goldie t o p o l o g y (§ 3, Example l). For each
torsion-free (i.e. non-singular) module we g e t 2. The maximal right ring of quotients of centralizer of
A
I ~ - E(M) .
is the double
E(A) .
E x e r o i see:
Let MA
E
be an inJectiva module. The
is said to be
~n
exists an exact sequence
(notation:
E-dominant dimension of E--dora.dim M ~ n ) if there
44
0
--*
where each
E. 1
M
.......:..... E l - - - - +
.......
----*E n
is a direct product of copies of
the hereditary torsion theory cogenerated by
(i)
M
is torsion-free if and only if
(ii)
M
is closed if and only if
is
Lamhek [471,[48], [49],
[79],
Turnidge
E . Show that:
E-dom.dlm M ~ I .
E-dom.dim M $ 2 . (Hint:
~-injective if and only if
References:
E . Consider
E(M)/M
Morita
is
['57],
M
F_-torsion-free).
Tachikawa
[74],
Wong and Johnson [114].
§ 9. C0reflective Subcate6ories of
Mod-A.,
We prepare the study of the category of
F_-closed modules by
a consideration of a more general situation: Definition. A full subcategory if the inclusion functor
C
of
Mod-A
i:C-PMod-A
is called oorefleo~ive
has a left adjoint
a .
In such a case there exists a natural transformation ~ :l-~ ia such that the bijection
where
N ¢_C , is given by
a~a~M
.
Examples: I. If
F
is a toplogy on
A , then the category of
F-closed
m
modules is coreflective in 2. If
(T, =F)
Mod-A .
is a torsion theory for
Mod-A , then the category
of torsion-free modules is coreflective in
~,(M)
=
M/t(M) .
Mod-A , with
45
Let
C
be corefleotive in
we may choose Lemma 9.1.
a(M) = M
Mod-A . If
M
is a module in
with the identity map as
= IM '
Proof. From the preceding remark it follows that ~a(M)
is the
~M
=:a(M)-@M
such that
"
=~M
then
If there exists
~M
C ,
is an isomorphism.
identity map. The morphism
a
induces by naturality of ~
a
commutative diagram
l='a(M) [
['M
2(~)
--
, a(~)
H,noe ~M~ = a(a)~a(M ) . la(M) , and so a is the invers of ~ M
Proposition 9.2~
A corefleotive subcategory of
Mod-A
has
arbitrary limits and colimits. Proof. Let lim iG
~
he a small category and
exists and we denote it by
G:~-~
M , and let
denote the canonical projections. Since ~d:afM) -~ G(d)
such that
X~ I a(~) The family for if
= ~d' ~ M
~ 8d}dsD
X-d-.~,
and hence
a functor. Then =d:M-~G(d)
G(d) 6 ~ , there exist
~ d ~ M = =d "
"~
G(d)
is compatible with the morphisms in i.
_D, then
G(k)~dT~
= G(k)=d
D ,
= ~d' =
G(k )Sd = 8d' " It is therefore induced a
46
map
8:a(M)-~M
~d ~ M ~M
= ~d~M
such that = ad ' so
is an isomorphism,
limit for
G
in
~d ~ = ~d 8~M
for all
d c ~ . Then
= IM " From 9.1 it follows that
and it is then clear that
a(M)
is a
~ . Note that we have obtained the formula
i( m G) . lim __ iG which also follows from the fact that a right adJoint functor always commutes with limits (when these exist). To prove that left adjoint
a
lim __~ G
exists in
C
preserves colimits and we therefore have
a(lim iG) = lim aiG = lim G , since
In other words, limits in while colimits in into
is easier, because the
C
C
ai ~ I .
may be computed in
are taken in
Mod-A
Mod-A ,
and then coreflected
C .
Proposition 9.3. a:Mod-A-*~
If
~
is coreflective in
preserves kernels,
then
~
Mod-A
and
is an abelian category
with exact direct limits and a generator. Proof.
C
is preadditive since it is a full subcategory of
Mod-A
. We have proved that
that
C
C
has limits and colimits. To prove
is abelian, it only remains to show that if
is a homomorphism in
~ , the~ the canonical map
~:Coker(ker a ) - - - ~ K e r ( c o k e r is an isomorphism. Mod-A and
a:M--~N
~)
If we denote kernels and cokernels ~aken in
by underlining them~ we have
Coker(ker a) = a(Coker(ke F a))
Ker(coker a) = Ker(a(coker a)) = a(Ker(coker a))
preserves kernels. ~
is therefore an isomorphism.
since
a
47
Next we show that direct limits are exact. Let directed category and morphism
G-~G'
G , G':~-~
it follows that
Mod-A , and since ~
G-elim__e G'
Finally, it is easy to see that for
is
D
be a small
two functors with a mono-
. The induced morphism
a monomorphism in
u
lim i G - e l i m iG'
a a
a(A)
is
preserves monomorphisms,
monomorphism in
C
.
will be a generator
~.
Definition. A coreflective subcategory of
Mod-A
is called a
Giraud subcategory if the left adjoint of the inclusion functor preserves kernels. Thus if adjoint
i:C-*Mod-A a
of
i
is a Giraud suboategory, then the left
is an exact functor. It is important to notice
that the inclusion functor morphisms in
C
i
is in general not exact! epi-
are not necessarily surjective maps. An
abelia~: category -with exact direct limits and a generator is usually called a Grothendieck category. Proposition 9.3 thus states that every Giraud subcategory is a Grothendieck category. Conversely, the Popescu-Gabriel theorem (Theorem 10.3) states that every Grothendieck category is a Giraud subcategory of Mod-A , where
A
is the endomorphism ring of some generator of
the category.
References: Mitchell [102] (ch. V:5).
48
§ i0. Giraud subcate~ories and the Popescu-Gabriel
Proposition !0.i.
if
[
is a topology on
modules form a Giraud subcategory of Proof. The category of
i
that
is full and faithful, a
A
~-closed
Mod-A.
L
and
the relation
i
preserve kernels
ia = LL
implies
preserves kernels.
Theorem 10.2. on
A , then the
F-closed modules is coreflective by
Proposition 7.10. Since the functors and
theorem
There is a i-I correspondence between topologies
and equivalence classes of Giraud subcategories of
Mod-A .
Proof. We already know how to associate a Giraud subcategory to a topology. Conversely, Mod-A
and let
i:C--#Mo~-A
a
. Let
let
a
be a Giraud subcategory of
be the left adjoint of the inclusion functor T
be the class of modules
a(M) = 0 . We verify that Since
C
T
M
for which
is a hereditary torsion class.
is exact, it is clear that
T
is closed under
I
extensions,
submodules and quotient modules. Since
right adjoint,
a
T
is
there corresponds the
C .
We will now show that the two maps P {topologies on A} --'--=* {Giraud subcategories of
Mod-A )
are the inverses of each other. We first show that ~
T
{I ~ A/I C =T~, which is the topology we associate
with the given Giraud subcategory
Let
has a
commutes with direct sums and hence
closed also under direct sums. To topology
a
be a toplogy and let
~
be the category of
modules. We must show that if
a(M) = 0 , then
F_-torsion module. But
certainly implies
Y~ = 0
~ ~ = id..
M
F-closed is an
M = t(M) .
49
It remains to show of
Mod-A
a' . Let
~=
id..
with the inclusion
Let
~
be a Giraud suheategory
i':~-~Mod-A
and its left adjoint
[ he the toplogy of right ideals
I
for which
a'(A/l) = 0 . We wan% to show that the category modules is equivalent to
~
of
~-closed
_D.
Mod-A a'i i
ai' C
We first note that
a'ia ~ a' , for if
M ¢ Mod-A
then the
exact sequence
0
• t(M)--~
M--~M F
~ CokerTM--*O
m
gives
a'(M) ~ a ' ( ~ )
. Similarly we have
ai'a' = a , beoause
the adjointness transformation ~:id.--~i'a'
gives the exact
sequence ~M
i'd'
and
a'(~M)
are
F-torsion modules, and it follows that
isomorphism
is an iso.o~his., so Ker ~ M
and a(~M)
is an
a(M) ~ ai'a'(M) .
From these two natural equivalenoes we obtain a'i' ~ id.
Coker ~ M
and
ai'.a'i ~ ai ~ id. , and
~
thus equivalent. We may now state the Popescu-Gabriel theorem.
a'i -ai' and ~
are
50
Theorem 10.3. U . Put
T(C)
Let
be a Grothendieck category with a generator
A = HOmc(U,U ) HOmc(U,C )
=
~
an~ let
Then:
.
(i)
T
is full and faithful.
(ii)
T
has a left adjoint
(iii) T
Proof.
of
which is exact. ~
and a Giraud
Mod-A .
Let us first see how (i) and (ii)
be the full suboategory of form
S:Mod-A-@~
induces an equivalence between
subcategory
be the functor
T:~-IpMod-A
Mod-A
(iii). Let
imply
Im T
consisting of modules of the
T(C) , C e C . We have a commutative diagram T
C
.~ Mod-A
Im T
and (i) states that Mod-A-@Im T . a Im T
T'
is an equivalence. Define
is exact and is a left adjoint of
is thus a Giraud subc~tegory of
Proof of (i):
T = Hom(U,.)
is
is faithful since
A-linear, then
by (ii).
is a generator.
C , D 8 C ~
and
is of the form
for some ~ C * D .
morphisms
U - ~ C . There is a corresponding exact sequence
where
~
UI
f -------~ C
is the direct sum of
~(fi)'u ~ of
~_a~
D
induce a mo~hism
we set
(fi)I
U
@(f) = ~ f
o_~K
Let
i
Mod-A .
To see that it is full, we must show that if ~:HOmc(U,C)--~HOmc(U,D )
a = T'S:
be the set of all
~0 I
copies of
U . The morphisms
h, ~--~ D . ~or each s ~ n d
K i = Ker fi = K f%U i .
Ui
51
0--.-,K.
gi
,
1
t
0
)K L
f"I
) U
,0
JI
) ~
g
~ C
~ c - - o
1,
D For every
s e Hom(U, Ki)
o.
s) _ - ~ ( f i ) g i
#(figi
generator. for some
we get by the
~ , ~o ~ ( f i ) g i
It follows that ~:C-~D
= hui = ~ f i
A-linearity of
= o
hg = 0 , and
. For each
fi:U~
beca~, h
C
~
u
i~ a
factors as
we then have
that
h =~f ~(fi)
"
Proof of the easy part of (ii), namely that a left adjoint
max
exists: For this we
S
either use general existence theorems
for adjoints
([102], oh. V,§3), or we max proceed as follows.
Consider
as a preadditive category
A
A
and define in the obvious way a functor object
with only one object) A u:A-*C
with image
U . By a standard result in elementary category theory
([I02], p.106) functor
=
we can extend
S:Mod-A--~C
of a functor
i~A
u
to a colimit preserving
. Since every module
i
(i ¢ I), where each
HomA(M,T(C)) = HOmA(I~Ai,Hom~(U,C)) " 4--lim Hom~(Ui,C ) - lime.. H°m~(S(Ai)'C) = ~om£(S(~Ai),C) adjoint of
= ~om~(S(M),C)
M Ai
is the colimit is
A A , we get
= *---limHomA(Ai,Hom~(U,C)) " Hom,(lim~._~ S(Ai),C ) , and t h ~ s
S
i~ a left
T .
Before we go o n a n d prove the exactness of following observation:
S , we m a k e the
=
52
Proposition 10. 4.
Let
a strongest topology ~-closed. A module if and only if and
M cM
M ~
L
be any class of modules. There exists such that all modules in
M
are
is a torsion module for this topology
Hom(A,M)
~ P Hom(Ann(x),M)
for every
x c L
.
Proof. If
~
is any topology, then a module
if and only if
M
is
_E-torsion-free and
is equivalent to requiring
M
and
M
is
_E-closed
_E-injective, which
E(M)/M
to be
_E-torsion-
free(Proposition 6.3). It follows that the torsion theory cogenerated by
~ E(M) ~ E(E(M)/M)
I M C M__~, which is hereditary
by § 2, Exercise 4, defines the strongest topology for which all modules in
M
are closed.
It remains to determine the torsion modules for this strongest topology each
F_~ . If
x ¢ L
L
is a torsion module, then
and hence
Ann(x) e ~
Hom(A,M) ~ Hom(Ann(x),M)
Conversely, if a module
L
for all
for M 6 M .
satisfies this later condition, we
may restate this as Hom(C,M) = ExtI(c,M) = 0
for every cyclic submodule
C
of
L ,
as one sees from the exact sequence
O-*Hom(A/Ann(x),M)--~Hom(A,M).-~Hom(Ann(x),M)-~Extl(A/Ann(x),M)-'tO But if
Hom(C,M) = 0 , then
Consequently we have submodule
C
of
L
ExtI(c~M) ~ Hom(C,E(M)/M)
Hom(C,M~E(M)/M) and
M
e M u
.
= 0
.
for every cyclic
This implies that
L
is
a
torsion module, because of the following easily verified fact: Lemma IO.~.
If
L
and
M
and only if
Hom(C,M) = 0
are modules, then
Hom(L,E(M)) = O
for every cyclic suhmodule
C
of
if L .
53
We continue the proof of Theorem 10.3, where it remains to show that
S
is exact.
all modules
Let
F
be the strongest topology for which
T(C) , C c ~ , are
corresponding
category of
F_-closed. Let
~
be the
F - c l o s e d m o d u l e s . We h a v e a d i a g r a m
of f u n o t o r s
i
Im T ~ a'i
~ Mod-A
i' D
where
T'
is an equivalence. We have
definition of
F . It now suffices to show that
is an equivalence, because adjoint of
S
i'
of
i'a'i ~ i
T
T = i'-a'T
is the composition of the left adjoint
both these two adjoints are exact, i'a'T
= T
and i t
r e m a i n s f o r u s t o show t h a t
S
is full an faithful, also
Thus let
M
M
is
be an
will he exact. Since a'T
a'T(C)
isomorphic to
a'
a'T ! since
is full and faithful,
e v e r y module i n
i s o m o r p h i c t o a m o d u l e o f t h e form module
a'iT' = a'T
implies that the left
and the left adjoint of the equivalence
F-closed
by the
, i.e.
D that
is every
T(C).
F-closed module. Choose an exact sequence m
(~)
eA
("i j)
I in
and t h e r e f o r e
~M
~ 0
J
Mod-A . Since the functor
oolimits
~A
carries
S
(~)
has a right adjoint, it preserves
into
an e x a c t sequence
54
s(alj)
(.~) e u
,ou
I in
~s(M)
~0
J
C .
Lemma 10.6.
The functor
a'T,~-~
is exact and preserves
direct su~s. We conclude the proof
of
the theorem before we prove the lemma.
By applying the Lemma to ( ~ )
and noting that
A - T(U)
is
~-closed, we obtain the upper exact row of the following diagram in
D :
a'Ts(~ij) OA
.......
a,TS(N)
~@A
I
,o
J
fl ei A
u .........
a,faij~,
~
M
ej A
The lower row is obtained by applying exact. The diagram commutes because has the form
a'
0
~
to ( ~ ) ,
A - T(U)
~iJ = i(~ij) ' and one has
and is also
implies that
a'TSi = a'iT'Si =
= a'iai = a'i . We conclude from the diagram that Proof of Laama 10.6, We already know the funotor
M ~ a'TS(M) . a'T
to
be
exact, so to prove exactness it will suffice to show that it preserves epimorphisms. This means that if epimorphism in an
C , then we should show that
f,C'~
C"
is an
Coker T(f)
is
F-torsion module. By Proposition 10.4 this is equivalent
to showing that for each
x ¢ T(C")
we have
Hom(A,T(C)) -~ Hom((Im T(f) ,x),T(C)) for all
C s _C . Define
maps
into
i
hsU-~C"
such that
x . From the pullback in
aij
C
T(h).A-~T(C")
left
55
0 --~ K
0--*
K
k
p
~
g ~
U --* 0
C ' ------e C " - - t O f
we get a pullback diagram with exact rows in 0 ---~T(K)
T(k)
~ T(P)
1
....T(g)
~
l
(Im T(f) :x) = Im T(g) . Note that
T(k)
in the subcategory
equivalence.
A
1 T(C")
o---, and
Mod-A
Im T , because
A
is the cokernel of T',~-~Im
T
is an
Y% follows that if we have a homomorphism
Im T ( g ) - * T ( C )
, for some
C c ~ , then it factors uniquely over
A = Coker T(k) , and this is precisely what we wanted to show. It remains to show that
a'T
preserves direct sums. Actually
we prove a little more, namely that ~ions.
Let
(Ca)
a'T
preserves directed
be a directed family of subobjects of
C c ~ .
We must show that the cokernel of the monomorphism
f,U (ca) a
is a torsion module. By Proposition 10.4 this means that for each x C T(U T(C'))
Ca )
we shall show that
for all
T(h):A-~T(~Ca)
C' 6 ~ . Define maps
1
Pa ~
Ca
~
to
Hom(A,T(C')) ~ Hom((Im f :x), hzU--~UC a
such that
x . From the pullback diagram U
U Ca
56
we obtain a commutative diagram
UT( )
- g
.- A t T(h) T(U C=)
UT(C.) f
which is a pullback diagram because pullbacks are preserved both by
T
have
and when taking direct limits in (Im f :x) = Im g ~ U T ( P a )
Mod-A . Therefore we
. Now
Hom(~JT(Pa),T(C'))
lim HomCT(P~),TCC')) ~ lim HomCP ,C') : H o m C ~ P a , C' ) = 4-= Hom(U,C') ~ Hom(A,T(C')) , where we have utilized the fact that exactness of
lim --~
Example: Let
A
implies
~Pa
= U .
be any ring. Proposition 10.4 provides
a strongest topology
~
for which
A
is
logy is called the canonical topology on
A
with
F_-closed. This top@A .
Exercises: I.
Let
~
be a topology on
Mod-A
and let
an exact functor into an abelian category T(M) = 0
for all
T:Mod-A-P~ ~ , such that
F-torsion modules. Show that
unique factorization
T = T'a
modules. (The category of
be
T
has a
over the category of
F_-olosed
F_-closed modules is thus a
solution of a universal problem). 2. Let
~
be a Giraud subcategory of
Mod-A
left adjoint of the inclusion functor (i)
If
E
is an injective object in
an injective module. (ii)
i
preserves injective envelopes.
and let
a
be the
i . Show that: ~ , then ~ i(E)
is
57
(iii)
I f the t o r s i o n theory corresponding to
(§ 4) and
E
injeetive in
C is stable
is an inJeotive module, then
a(E)
is
C .
Referencess Bucur-Deleanu [12] (oh. 6 , written by N. Popescu),
[~] (§ 4), P o ~ o u - G ~ b ~ l [62], (oh. I), ~akeuohi [112].
G~h~i.1 [31] (oh. 3), ~ b ~ Roos [66]
Chapter 3.
General properties of rings of cuotients
§ II. Lattices of
M
~-p~e
submodules
We will assume
F
to be a topology on
and submodule
L
of
M , we define
={,. The operation
}.
L ~ Lc
all submodules of were called
Note that
of
is a closure operation on the lattice of
M . Those submodules
F-submodules of
L° = L
particular,
if
for which
M/L
is torsion-free!
then
CF(M )
L~L
°
is a complete lattice with intersection
It remains to verify modularity.
= (K~(H
is the family
is a closure operation it
as meet ([18], Ch. 2.1). The join is given by
C~(M)
with
Let
H C K . Then
H , K
V L i = (~ Li)C . and
K c ~(M)
i~omorphism $ ( M ) - ~ ( K ) m
Proof. If
L ~ $(M)
~ven
be
, using the
modularity of the lattice of all submodules of If
L
L) = K ° O (H + L) c =
Kn(H~
+ L)) ° = (H + ( K ~ L ) ) ° = K V ( K ~ L )
Propqsition 11.2.
in
is a complete modular lattice.
Proof. From the fact that
members of
Le = M
M (§ 6, Exercise).
C_F(M)
C~(M)
L
in § 6. On the other hand, we
is torsion-free,
~-pure submodules of
follows that
M
if and only if
M
Proposition II.i.
A . For each module
M .
, then there is a lattice
by
L~
~n K .
m
, then clearly
L~ K ~ $(K)
. The inverse
m
map is defined as
L~L
c , where the closure is taken in
M .
59
For if
if
L C C_~(K) , then
L e ~(M)
, then
LO~K
= { x c K 1 (L:x) ¢ ~ =
(L~K) ° = LONK c = LnM
= L .
In the following we will mainly be concerned with is torsion-free. Proposition 11.3.
In case Let
M
M be
is F-closed if and only if Proof. If is
L CM
, then
is also
~(M)
when
F_-injective, we have:
F_-closed. A submodule
L c ~(M)
M/L
L , while
L
of
M
.
is torsion-free
if and only if
F_-closed, as is seen from the exact sequence
0 = Hom(A/I,M)-~ E x t I ( A / I ,Hom(A/I,M/L)-~ L ) . . . ~~E x t.I ( A / I , M. )-. . . . . Recall that a submodule L
L
is maximal with respect to
Proposition 11.4. module
M
Proof. If
L° = L
M
L~K
= O
KC
M (§ 4).
submodule of a torsion-free
C_F(M ) .
M
L~K
= 0 , then
is torsion-free.
LC~K c =
We must then
The following properties of a torsion-free
are equivalent:
(a)
C_F(M )
(b)
C_~(M) consists of the complemented
(o)
Every essential
Proof.
for some
if
by maximality.
Proposition 11.~. module
since
- = 0 .
is called oomp!emented
is maximal such that
= ( L ~ K) ° = 0 c = 0 have
M
Every complemented
is a member of L
of
- -
(a)~
is a complemented
lattice.
submodule of
M
(b): Every complemented
11.4. Suppose conversely
L c Ca(M) -L
submodules of
is an
M .
F_-submodule.
submodule is in • By hypothesis
~(M)
by
there exists
60
K CM
such that
maximal xlcK
KNL
with respect + L
= 0 to
for some
L C_~F(M )
implies
(b)=~ (c): Let
L
x c L
L
an
xANK
and hence
be an essential
x ¢ L'
= 0 , so
L = L'
submodule
and essential
L C ~(M)
= 0 . K + L
in
Proposition
, choose
K
be
we have
xI¢
K .
is complemented. of
M . Lc
M . Hence
maximal
is then an essential
F-submodule.
C_F(M )
= 0 . For each
L' D L
is
Lc = M ,
is an F_-submodule.
(C) =~ (a): If KNL
L'NK
(K + L) c = M . Let
I ~ F . But
then both complemented and
and
Thus we have
11.6.
Let
M
is complemented.
(i)
Every
(ii)
The endomorphism
K~L
be an
with respect
submodule
= M
and
of
KNL
to
M , hence i 0 .
F-closed module
such that
Then:
F_-closed submodule ring of
of
M
M
is a direct
is regular
smnmand.
(in the sense
of yon Nemnann). Proof.
(i). If
L CM
Hence there exists essential,
hence an
is
F_-closed~
then
M
such that
K~L
KC
F-submodule,
we may extend the canonical morphism (ii): Let
L
= 0
and
by II.3. K + L
M . BY Proposition
projection
M - ~ L . This makes f:M-, M
in
L c ~_F(M )
K + L--~L
6.2
to a homo-
into a direct su~mand of
be an endomorphism.
Then
is
Kerf
M .
c ~(M) m
because if f(x)l = 0
x e M and
M
and
xlC
torsion-free
thus a direct su~,nand of an isomorphism
Kerf
f~K:K-*Im
for some
implies
M . Write
I c F_ , then
x c Kerf
M = Ker f~
f , so also
Im f
. Kerf K . f
is an
is
induces
F_-closed
61
module. Hence
Im f
therefore extend
is a direct s,nmand of
(fJK) -I
to a homomorphism
M , and we may h:M-PM
. Then
f = fhf , and we have established the regularity of the endomorphism ring. The lattice
C_~(A) has an interesting description as the D
set of annihilators of subsets of an injective module. Recall that
.CF(A) = [ 1% A/I
is torsion-free ~ .
three propositions we let
E
In the following
be an injective module which
cogenerates the torsion theory corresponding t o F = { I~ Hom(A/I,E) = 0 } . we p u t
Ann(S) = ~ a
If
S
is a subset of any module,
S A i Sa ,, 0 7 ,
Proposition 11. 7 . CF(A ) = ~Ann(S) ~ subsets Proof. Suppose
F , i.e.
S C E ~.
S C E . Then
Ann(S) = N Ann(x) , where each xsS A/Ann(x) C E is torsion-free. But C_F(A)
AnnCx) ¢ CF(A ) since m
is closed under intersections,
so also
Suppose on the other hand that
s o{x
Elxl - 0h
a C Ann(S) i.e. that
Then
belongs to
I e ~(A)
and put
Ann(S)= I . To show that every
I , it suffices to show that
Hom(A/(I:a),E)=
monomorphism
Ann(S) e CF(A ) .
0.
Let
a(~) = a-~ . For every
a:A/(l:a)-~A/I f:A/(I:a)-@ E
(l:a) 8 _F , be the we get a
commutative diagram A/(I:a) ~
f ~
A/I
~ g
E ~ where
g(~) = xb
for some
x c E . Then necessarily
xl = 0 ,
62
so
x c S . But then also
xa = 0 , so
g~ = 0 . Hence
f = 0 .
Before the statement of the nex% result we need to make two definitions. A lattice is called noetherian if every ascending chain is stationary. A n inJective module is called
~-in~ective
if every direct sum of copies of the module is injective. .Prgl0osition 11.8. (a)
The following assertions are equivalent:
Every direct sum of torsion-free injective modules is injective.
(b)
E
(o)
C_F(A )
Proof.
is Z - i n j e c t i v e . is a noetherian lattice.
(a) ~ ( b )
(b) =~(c):
By 11.7 it suffices to show that every strictly
ascending chain E
is trivial.
IlCl
must be finite.
for each
n
2 C
....
of annihilators of subsets of
If the chain were not finite, we could choose
an element
xn c E
such that
_.xnI n = 0
but
w
.x• In+ 1 $ 0 . Put
I = U I n and define f : I - ~ Q E as f(a) = i i = (xla , x2a,... ) . Note that f is well-defined! Since E is -injective,
f
has the form
f(a) = ya
Y = (Yl ~ Y2 ''''~ Ym' 0~...) e ~ E I choice of the elements x .
for some
, which contradicts the
n
(C)~
(a): Let
modules.
(Ea)
Suppose we are given a homomorphism
suffices to show that many
be a family of torsion-free injective
f
maps
I
f:I-~QE=
. It
into the sum of finitely
E a . Suppose on the contrary that there exists an infinite
sequence of indices
= , which we write as
a = I, 2,...
, such
63
that
Im f
has non-zero coordinates in each
I n = f-l(E I @
E
. Put
n
.... ~) En) . The ascending chain
IICC12
is by hypothesis finite, which implies that for some I k C In c
have which
f(a)
Since
a e In
for all
k
Let
a e Ik
has non-zero coordinate o
, we have
aJ C I n
the assumption that
Proposition II.9:
Ek
If
n
...
we
be some element for
xk
in
for some
f(a)J C E 1 ~ .... • E n • But then
c C
~
, for some
k> n .
J ¢ ~ . This gives
xkJ = 0 , which contradicts
is torsion-free.
C_F(A )
is noetherian,
then
[
contains
a cofinal family of finitely generated right ideals. Proof.
For each right ideal
xI = 0 } . Note that operations
Ann
I 8 ~
and
A
I
of
A
we set
if and only if
E
I A = 0 . The
define an order-inverting bijsction
between the set of right ideals of the form set of submodules of
IA = { x c E
of the form
Ann(S)
and the
I A . Since we have ACC
on the former set (by 11.7), we must have DCC on the submodules family J
I A . Let { jA
IJ
I
be any right ideal in
F . Consider the
finitely generated right ideal
be a minimal member of this family. For each
right ideal
Jl = J + aA
¢ I~
and let
a e I , the
is also finitely generated
C I
and
satisfies
Jl~ C j A . By minimality we must have
Jl~ = J~ so in
particular
JAa = 0 . Since this holds for all
a e I , we have
JAI = 0 and
JeF
and thus .
J ~ C l A , which implies that also
jA = 0
64
We will now show that
C~(A)
lattice of right ideals in
is isomorphic to a corresponding
A F , assuming for simplicity that m
A
is torsion-free, so that ZF
= [right ideals
Proposition I!.I0.
Proof.
Suppose
J
AF I J N A
and
q c $ ~ Z,
because
I
J e ZF
q ¢ J . Then
and for each
S F
Jf~A ¢ F
Proposition II.II.
Jn X ~ F_ and
A
be a right ideal of
for which
={ b s A F ~ab s I } G A and hence
(J,q)n A --{a ~ A l is
AF , so we may use Lemma 6.1. Thus T I is
such that there exists
I•A
AF .
. Then
satisfied. Next we verify T 2. Let
ah C I h A ~
A F. Define
S [}.
is a topology on
F_e
J ~ F_~
F_-submodule of
is a subring of
of
qa ~ J} ~ {a ~ A iqa ~ J - A ~ an
A
(l:q) e F e
a e Jf~A
S F . T 2 for
for all
we have F
AF
{b e A
implies
I ¢ Fe .
AF
is equal to its ring of quotients with
m
respect to
Fe .
Proof. The ring of quotients of ~e-injective envelope of = E(A)
~
is isomorphic to the
A F , and is a submodule of
(Propositions 8.1 and 8.2). But the
envelope of
AF
coincides with its
E(AF) =
F_e-injective
F--injective envelope, as
m
one immediately verifies, and
Proposition 11.!2.
If
M
AF
is
F_-olosed.
is a torsion-free
A-module, then
CF(M ) =~ C_Fe(MF) Proof. Since
M
~= CF(MF)_ _
11.2. It remains to see that
by
is an
F-submodule of
MF , we have
CF(M ) =~
~(MF)_ _ = .CFe(MF) . D
65
Suppose
L e ~F(MF) . L
is therefore an xJC
L
AF-mOdule. If
for some
Consequently
is then an
F_-closed module by ii.3 and
x e MF
J e F_e , then
has the property that
x(JNA) C L
and hence
x e L .
L e _CFe(MF_)
Suppose conversely that
L e _CFe(MF_) . If
x e MF_ and
xIC
m
for some
I ¢ ~ , then
xlA F C L
and hence
IAF ¢ F.. e
implies
m
. It follows that
.
Examples, I.
Let
~
be the Goldie topology, i.e the topology generated
by the family of essential right ideals (§ 3, Example I). From Propositions 11.5 and 11.6 we obtain:
Proposition I!.13.
The endomorphism ring of a non-singular
injective module is regular. For the Goldie topology one can prove the converse of Proposition 11.9, namelys Proposition !!.!4.
The following assertions are equivalent for
the Goldie topology (a)
F :
Every direct sum of non-singular injective modules is injeutive. lattice
(h)
$(A)
of
complemented right ideals is
noetherian. (c)
~
contains a oofinal family of finitely generated right
ideals. Proof. It remains to prove (c) ~ (a). Let
(Ea)
be a family of
L
66
no~-singular injective modules, and let morphism,
for an arbitrary right ideal
ideal
such that
J
l~J
= 0
and
f : l - * ~ ) E a he a homoa I . Choose a right
I + J
is essential in
By (c) there exists a finitely generated right ideal contained An
I + J . Extend
f
to
and then restrict to a homomorphism
finitely
generated~
g
I ~ J-'~ g:K-e6
E~
K e ~
by
f~J = 0
E ~ . Since
maps i n t o t h e sum o f f i n i t e l y
K
is
maw
E a , and is therefore extendable to
h:A--~6 E a • Since a the usual argument shows that hll = f
is torsion-free,
A .
(~E a a (of.
the proof of Lemma 8.3).
2.
Taking
11.8 that
~= A
{ A} , we otain as a special case of Proposition is right noetherian if and only if every direct
s~n of injeutive modules is injective.
3.
Let
D
be the family of dense right ideals of
Example 2). Then
D_e
A
(§ 3,
is the family of dense right ideals of
Qm ' as one easily verifies by means of Proposition 3.8. Exercises: i. Let
A
be a regular ring and let
~
be the family of
essential right ideals. Show that: (i)
A
is non-singular.
(ii)
~(A)
is noetherian if and only if
A
is semi-simple.
2. Show that the following two properties of a topology
~
equivalent: (a) If
I I C 12 C ....
is a countable ascending chain of
are
67
righ~ ideals such that U I n e ~ , then some (b) If
I I C 1 2 C ....
In e ~ .
is a oountahle ascending chain in
~F(A) , then also t J I n S C_~(A) . Show that these properties are satisfied if every contains a finitely generated
I ¢
J e ~ .
3. Show that the following properties of a right self-injective ring
A
are equivalent:
(a)
A
satisfies ACC on right annihilators of subsets of
(b)
AA
(c)
Every projective module is injective.
is
4. The ring
A
~-injective.
is called right finite-dimensional ' if no right
ideal can be written as a direct sum of infinitely many non-zero right ideals of
(i)
A
A . Show that:
is right finite-dimensional if and only if every
right ideal is an essential extension of a finitely generated right ideal.
(ii)
Every right finite-dimensional ring satisfies the conditions of Proposition 11.14. (Hint: use 3.6 to
verify ll.14(c) ). 5. Show that if
E
is an injective module and
M
is non-
singular, then every exact sequence
O-*K-*E-*M--~O
splits. Using this, show that if
and
E
E'
submodules of a non-singular module, then also inJective.
are injective E + E'
is
A .
68
6. Show that the conditions of Proposition 11.14 are equivalent to:
(d) Every non-singular module contains a unique maximal injective submodule. 7. Let
~
be a topology on
topology
~'
on
AF
A . Show that
such that all
F_e
is the strongest
F_-closed modules
i
(considered as
AF-mOdules ) are
~'-closed.
~eferencss, A=endaris [85], Faith [27], C28] (§'7 ~d 8), JTohnson [40], Teply [75],[76], utumi [8o]. § 12. Finiteness conditions on to~ologies
In this §
we will consider two kinds of finiteness conditions
on the topology
F . The first one is introduced in the next
proposition, where modules and
Proposition 12.1.
(b)
If
as usual denotes the category of
i:C-~ Mod-A
q:Mod-A--PMod-A F
(a)
C
F-closed
is the inclusion functor~ while
is the functor
M~-~M F .
The following assertions are equivalent:
I I C 12 C...
is a countable ascending chain~ of right
ideals such that
n .
t ~ I k ¢ F , then In ¢ F_ for some I Every direct sum of F--closed modules is F_-closed.
(b') Every direct sum of coumtably many
F_-closed modules is
F--closed. (c)
The functor
i
commutes with direct sums.
(d)
The functor
q
commutes with direct sums.
Proof. (a) ~ (b): Let { M }
be a family of
F_-closed modules.
69
~M~ is
is of course torsion-free, ~-injective.
Considering
Let
$ Ma
f:l-~$
Ma
be any homomorphism with
as a suhmodule of the
Mm , there exists all
and we must show that it also
x = (xa) e ~ M a
F_-injective module
such that
a e I . We only have to show that
for
~
for which
i~n~.
Since
then implies that so
x
I = 0
xai @ 0 . Put
f(1) C ~ M m
implies
Let
(Ma)
xa
e i(M=)
is
(b')~(a):
F_-closed, then
.
m o d u l e s . The
i(~)Ma)
~
of
F_-closed
oposition 9 . 2 . So i f =(Bi(M=)
.
are rather o b v i o u s . be an a s c e n d i n g c h a i n w i t h
There is a well-defined
canonical
map
I--tO A/I n .
I e F , one obtains a commutative diagram -
~
(D A/I n -
f(a)
such that
- xa
f o r some
xn = 0
for
then lies in the kernel of
torsion
S-closed
hy
I 1 C 12 C . . .
I
m
e I ~ xaia = 0
in the category
and (b) ~ ( b ' )
Let
I = WI n c ~
Ma
a(ei(Ma))
(o) ~ ( d ) = ~ ( b )
where
In = { a
of
, we have
be a family of
sum o f t h e m o d u l e s
modules i s
Since
(al' m2'''" )
n
(h)~(c): direct
for
I = U In . But (a) n for some n . Now M a is torsion-free, n = 0 , which is a contradiction.
I = In
n
f(a) = xa
x e ~ M a . If this were
not true, there would exist an infinite set indices
I e ~ .
A
' @ (A/In)F_
x = (Xn) ~ e n)m
( A / I n ) ~ • There e x i s t s
. The image of
A/Im--@(A/Im) F , so
m o d u l e . The e x a c t s e q u e n c e
I
in
A/I m
I/I m
is a
70
0 ---* I/I m ....~ A/I m .....~ A/I ~ where also hence
I
m
A/I
0
is torsion, shows that
A/I m
is torsion, and
c F . --
We strengthen the finiteness condition somewhat by considering topologies with the following properties:
P ~ p o s i t i o n 12.2. a topology (a)
~
F
The following assertions are equivalent for
:
contains a cofinal family of finitely generated right
ideals. (b)
Every directed union of
(c)
The functor
(d)
The torsion radical
Proof.
(a)=~(b):
i
F_-closed modules is
F_-closed.
commutes with directed unions.
Let
t
commutes with direct limits.
(Ma)
submodules of some module.
be a directed family of ~.~Ma
F_-closed
is then torsion-free,
so it
a
remains to verify that it is morphiem generated
f:l--~tJM J c ~ . f
F_-injeetivity of = xa
for
f(a) = xa
Ma
F_-injective.
where maps
I c [ . I J
a c J . Since also for all
~M= a c I
contains a finitely
into some
there exists
Consider any homo-
x c M=
M a , so by the such that
is torsion-free,
f(a) =
one then has
(by the same argument as in the
proof of Proposition 7.8). (b)4~(c) (b)~(a):
similarly to the preceding Proposition. Write
I c ~
generated right ideals
as the directed union of finitely I a . Then
ia(A) = ia(1) = ia(t)la) =
7!
=Uia(la)
,
and so the canonical homomorphism
as A
= . Let
Uia(Ia) g:ia(l=)-~ia(A)
want to show that Ia ¢ ~ . g
g
is an isomorphism,
, so
gf = ~
~ - c l o s e d submodule of that
g
be the canonical map. We
is obviously a monomorphism,
~ g f = jf = ~
factors
ia(A)
""'
ia(I a) ¢ for some
~:A-~ AF
because this would imply since
. It follows that
A?
containing
Im~
,
~g
= j . We have
Im g
is an
and we conclude
is an epimorphism~
(a) ~ ( d ) :
Let
t(Mm)-+M a
(Ma)
be a direct system of modules. The inclusions
i n d u c e i n t h e l i m i t an i n c l u s i o n
lim t(Ma)--+lim M..
The class of torsion modules is closed under direct limits, since it is closed under direct sums and quotients, is therefore a submodule of have equality,
suppose
finitely generated
t(l~
x C t(l~
still
xaI = 0 . Then
(d)~(a):
Write
I e ~
a e A choose
comes from some and
J e ~
a
so that
xl = 0
for some
is finitely generated,
by some
xa e
Ma
it
such that
x e lim --@ t(Ma)
-
as the directed union of finitely I= . A/I = lim A/I a
A/I = t(A/I) = lim t(A/la)
T e A/I
x
I
x a c t(M ) , and
generated right ideals so
. To show that we actually
M ) . Then
I e mF . Since
is clear that we may represent
Ma)
lim t(Ma)
. In particular the generator
t(A/la)
such that
is a torsion module,
aJC
, which means that there exist Ia
l-a e I a , and then
and
l-a e I . We may
J C I a . Hence
Ia e ~ .
72 It is clear that every topology satisfying Proposition 12.2 also satisfies 12.1 (cf. Exercise 2 of § Ii). The converse holds e.g. when all right ideals in
A
are countably generated.
References; Goldman t331, Roos [661 (oh. l ) . § 13. Fl,at epimorphisms of rings In many examples of rings of quotients one obtains the module ME
of quotients as
M F = M ~ A AF . Here we will prove that this
is equivalent to several other nice properties of the localization, e.g.
that
AF
is obtained by
a
kind of generalized calculus of
fractions. Let
~
be a topology on
A
and let ~ :A--~AF
be the canonical
m
ring homomorphism. We have the diagram of functors (of. § 7):
Mod-k
- 0
and so
sicij o
J . We now make use of condition (ii), and J
elements
tjl,...,tjr c A
and
bBl,..,h'jr 6 B
such that
sicijtjk = 0 i 'e ( t ) ~ , Jk
Jk
= 1
"
We then have hi = ~ b . a
and E sicijtjk = 0 i relation in
Remark I:
j,k , i.e. the given relation comes from a
A , as was to be shown.
Note that condition (c) of the theorem implies in
particular that each h = ~b~(si)h
b e B
i =~(ai)h
i
may be wTitten as with
~(si)b
i = I . B
is thus
obtained by a sort of generalized calculus of fractions (of. §15).
8i
Remark 2.
As was noted in the proof of ( c ) ~
(a) , condition
(i) of (o) may be restated as: (i)'
For every family
bl,..~b r c B
' b'n c B bl'''''
and
such that
there exist
Sl,..,s n c A
hj%O(si) c ~ ( A )
and
Z ~(si)b ~ : 1 i Corol!ar ~ 13.12.
There is a I-I correspondence between perfect
topologies on
and equivalence classes of ring epimorphisms
A--~B
A
such that
AB
is flat.
Exercise s: I. Show that a topology
F
is perfect if and only if all
AF-modules are torsion-free as 2. Suppose
F
A-modules.
is a perfect topology. Show that:
m
(i)
MA
isatorsio, module if and only if M ® A ~ - - 0
(ii) t(M) -- TOrl(M,~JA) A 3. Let
for ~ l mod~es
torsion-free.
A C B
be commutative rings. Show that
epimorphism if and only if (Hint: note that if ideals
I= , then
References: Spircu
~ , ass~ing
[I04~
(A:b)B = B
laB = B (~la)B = B
A c~ B
for every
is a flat b c B .
for a finite family of ).
Goldman [33], Lambek [48] (§ 2), Popesou and , Roos [66] (ch. I), Silver [70], Walker and
Walker [81]. For commutative rings: Akiba [84], Nastasescu and Popescu [59], S@m. Samuel [67] (Expos@ 6 by Olivier).
82
§ 14. Maximal flat epimor~hic extension of a ring
Theorem 14.1.
For every ring
A
there exists a ring
a ring homomorphism ~ : A - ~ M ( A )
M(A)
and
such that
is an injective epimorphism and
M(A)
is a flat left
A-module. (ii)
For every injective epimorphism that
AB
a:A--~B
of rings such
is fiat, there exists ~ unique ring homomorphism
~:B-*M(A)
such that
~a = ~
! moreover,
~
is also
injective. Proof. Every flat epimorphism is obtained as the canonical homomorphism ~ : A - * ~
for a perfect topology
is injective if and ordy if and only if
F CD
A
has no
~
(Theorem 13.10).
~-torsion, i.e. if
(the family of dense right ideals). 8o if m
M(A)
exists, it should be a subring of the maximal ring of
quotients
AD = ~
of all subrings
. This leads us to consider the family B
of
~
such that
A C B
and
A~B
is
a flat epimorphism. Note that inclusion of subrings in corresponds to inclusion of the corresponding perfect topologies.
Lemma 14.2.
The family
P
is directed under inclusion.
w
Proof. We will show that if then the smallest subring is also a member of
B D
and of
Qm
C
are members of containing both
P . Every element of
D
d = blC I .... bnC n
with
b i c B , ci c C
B
P , and
is a sum of
elements of the form (~)
m
,
83
and we may assume that each same length given and
d
appearing in a given sum has the
n . We will verify condition (i) of 13.10(c), i.e.:
dl,...~d r
of length
Sl,...,s m c A
d l' ' ' ' " d'm
r~ , there exist
such that
dis k C A
We do this by induction on the length
for all
n . For
i , k
D
and
n = 0 , i.e.
d = I , the condition is clearly satisfied. Suppose the condition has been verified for
in
n> I
and
n-1 . To simplify the
notation somewhat, we only consider the case
r = 1 ! it is
easy to see that our argument extends immediately to the case of a family form ( ~ ) . in
D
and
dl,...,d r c D . So suppose we are given By the induction hypothesis there exist t l,...,t m
in
x i = b2c 2 .... bnCnt i c A
A
C
and
s l,...,sp
xij = ClXiS j ¢ A
in
for all
A
4'''"
d' m
i , and
~tid i
C , there exist
t
t
Cl,..,c p
such that
for all
Similarly there exist
of the
such that
By the remark 2 of § 13~ applied to in
d
i , j , and
hl,...,bq
in
B
~sjc~ and
= I .
rl,...,rq
in
A
such that blXijr k ~ A
for all
i , j , k , and
We have thus got elements
t
t
t
bkCjdii ¢ D
dtisjr k = blClXiSjrk c A
and
~rk~ and
= i
tisjr k c A
such that
~- t.s.rkb~Ct.d.~ = I . i,j,k I J J
This finishes the proof of the Lemma, and we may continue the proof of the Theorem. Define
M(A)
as the union of all rings
84
in
~ . It is obvious that
M(A)
is flat as a left
~:A--pM(A)
A-module
is an epimorphism, and
since it is a direct limit of
flat A-modules. Suppose
m:A-~B
is any other injective flat epimorphism. There
is a corresponding perfect topology we have
F C D
[ , and as we noticed above,
and hence a commutative diagram A t
~
Qm
B
But also
~:B-@Q m
must be injective, because
mean that there exists a homomorphism
~(b) = 0
f:l--~A
with
such that
f~J = 0
for some
(J:a) c ~
and
D-torsion-free, so it follows that
is
We conclude that the image of obviously is unique ( a
J C ~ ! for each
~
lies within
a ¢ I
would
I c F
we have f = 0 .
M(A) . Since
is an epimorphism), we have proved the
Theorem. We will investigate the properties of Propositi0n 14.3.
If
AB
J = f-l(j)B
is flat, then
Proof. Put by
J=IB
is a ring epimorphism such that for every right ideal
I = f-l(j) . Tensoring the inclusion
B , we obtain
B/IB-*B/J
f:A--~B
M(A).
B/IB ~--~B/JeA B = B/J
J
of
B .
A/I c.@ B/J
by 13.7(o). Since
obviously is surjective~ it is an isomorphism and
.
Corollar~ !4.4. ~opositi..on 14.~.
If
A If
is right noetherian% then so is also M(A) . A
is von Neumann regular, then
Proof. By Lemma 13.11 we have
M(A) = A .
M(A)/A ® A M(A) = 0 . But since
85
A
is regular,
A c_~ M(A)
MCA)/A® A A~
induces a monomorphism
~(A)/Ae A ~(A) , so
Further results on methods. When
A--~ B
M(A)
~(A)/A -- 0
may be obtained by using homolcgical
is a ring homomcrphism such that
flat, one has the following formulas
AB
is
([13], ch. 6~ § 4):
(1)
~t~CM,~) "= ~ t ~ ( M ~ A B,~)
for
MA , N B ,
(2)
TornA(M,N) ~ Torn(M, A B,N)
for
MA , ~
.
Combining these formulas with 13.7(o) we obtain: Proposition 14.6. that
AB
If
~ :A--~ B
is a ring epimorphism such
is flat, then
~t~(M,~) = ~ ( M , ~
for ~ , ~
•or~ "=Tor~(M,~>
for
Corollar[ 14.[.
where
r.gl.dim
5
' ~
, "
r.gl.dim
M(A)@ r.gl.dim A
w.gl.dim
M(A)6 w.gl.dim A
and ,
is the right global dimension and
w.gl.dim
is the weak global dimension. The ring
A
of flat left
is called right coherent if every direct product A-modules is flat [17].
Proposition 14.8. Proof. Let
(N=)
If
A
is right coherent, then so is also M(A) .
be a family of flat left
each
N
is flat as an
over
A , but is then flat also over
M(A)-modules. Then
A-module by (2). Thus M(A)
~ N~ a by 14.6.
is flat
86
Example: It follows from Corollary 13.4 that when hereditary, then
A
is right noetherian
M(A) = Qm " This result will be improved in
§ 20 (Theorem 20.2). Exercises: I. Let if
~
be the Goldie topology (§ 3, Example i). Show that
~
is perfect, then
AF
is right noetherian.
n
2. Suppose
A
is a ring for which
Qcl
exists and such that
every finitely generated right ideal of Show that
A
is principal.
Qcl = M(A) .
References: Knight [991, Popescu and Spircu [1041. For the commutative case: Akiba [841, Lazard [I00] (ch. 4).
§ 15. A
l--topologies and rings of fractions l-topolog~ on
A
is a topology containing a cofinal family
of principal right ideals. A 1-topology
set
ZC_F) = { s
Proposition I~.I.
between I-topologies on SI.
I eS
$2.
s, t e S
$3.
If
If
~
defines a I-I correspondence
and subsets
S
of
A
satisfying:
.
s e S
such that $4.
A
is determined by the
}.
A l sA The map
F
implies and
st e S .
a ¢ A , then there exist
sb = at
ab e S , then
( S
t e S
is right permutable).
a e S .
and
b e A
87
Proof. Let
F
be a 1-topology.
~ (F)
obviously satisfies S I.
S 2 follows from axiom T 2 for topologies, sa s s A
we have
(stA:sa) D
from T I, because we have
(tA:a) ¢ F
(sA:a) D
tA
because for each
by T I. S 3 is clear for some
tA S F
. S 4
is immediate from T 3. Conversely, for some
if
S
satisfies S I-3 and one sets
s e S t , then
F
F ={ I I IDsA
is easily verified to be a 1-topology.
S 4 is a saturation axiom which makes the correspondence
F_,@S_
one-to-one. For a 1-topology one may describe the modules of quotients in a rather explicit way. Proposition 15.2.
-{ (x,s)
If
F
is a 1-topology and
s L sa.
0
in
A
implies
M c Mod-A , then
zat
0
f o r some
where ~0 i s the e q u i v a l e n c e r e l a t i o n given by
(x,s)~ (y,t)
there exist
and
Proof.
a , b ¢ A
such that
Recall that we have
sa = t b e
MF = ~ m
S
if
xa = yb .
HomA(SA,M/t(M))
with
m
s e S . A homomorphism element i.e.
x e M
x~t = xbt
same element of if and only if
c~:sA--~M/t(M)
such that for some MF cp
sa = sb
in
~
and
to the relation
~J.
implies ~0
xa-xb e t(M) , gives the
, determined by
coincide on some
u G S , i.e. if and only if there exist u = sa = tb c S
A
t e S . In the limit,
as ~ : t A - @ M / t ( M ) and
is dtermined by an
uA c sAf~tA
a , b G A
y e M , with
such that
xa-yb c t(M) . This clearly corresponds
88
One easily verifies the Proposition,
that under the isomorphism
the module operations
(x,s) + (y,t) = (xa+yb,u) (x,s)
• (a,t) = (xb, tv)
where
in
~
described
in
take the form:
sa = tb = u c S !
for some
b ¢ A , v c S
such that
a V = sb . Proposition S = ~ (~) $5'.
I~.3.
A 1-topology
F o r every
s e S
Proof.
Perfectness
exists
q c ~
(s,l)(a,t)
of
F
a
for some
if for each
as in S 5' , then
saa' = l.b'
s 8 S
(a, sa)
s 6 S
q = (a,t)
u c S
Suppose . Then such aa'
and if
and
(a,t) c A F
(by 15.2).
there exists
represents
there
a', b' ~ A
saa' = b' 6 S
S
u c S .
. We may then take
ta'b = b'h = saa'b = 0
aa'bu = 0
Conversely,
and
sac
for some
(Theorem 13.1(g)).
, so there exist
in S 5', for
saa'b = 0 , then
ahu = O
Using 15.2, we write
8 S
such that
that for every
sq = 1
= (sa, t ) ~ (I,I)
as the element
a ~ A
implies
means
such that
ta' = l.b'
one has
there exists
sab = 0
that so is the case.
implies
is perfect if and only if
satisfies:
and such that
that
~
a 8 A
an element in
such that ~
and
(s,l)(a, sa) = 1
The axiom S 5' is a weakened
form of the perhaps more well-
known condition: $5.
If ( S
sa = 0
with
s c S , then
is right reversible).
at = 0
for some
t C S
89
The most important examples of 1-topologies are those given by the rings of fractions. Let subset of tO
S
S
be a multiplioatively closed
A . A right ring of fractions of
is a ring
A[S -I]
A
with respect
and a ring homomorphism ~ : A - * A[S -I]
satisfying: FI.
~(s)
is invertible for every
F2.
every element in A[S-1] with
s e S .
~(a)
= 0
F3.
if and only if
s C S .
has the form
as = 0
~(a)~(s) -1
for some
Similarly one defines the left ring of fractions with respect to
s e S . [S-I]A
of
A
S . It is not immediately clear that the axioms f
F 1-3 dte~ine
~
ALs-~J
follows from the fact
uniquely, but that so is the case AIS -I]
is a solution of a universal
A[S-I~
exists, it has the following
problem: Proposition I~. 4.
If
property: for every ring homomorphism is invertible in
B
for every
homomorphi~ ~ , B ~ A [ S Proof.
We define
~
-1]
~:A-*B
such that ~(s)
s ¢ S , there exists a unique
such t h a t
,~-~
as ~ ( c ~ ( a ) C ~ ) ( s ) - 1 )
=
Ut(a)~(s)
-1
. We
then have to verify that this is well-defined. So suppose C~(a)~(s) -I = ~ ( b ) ~ ( t ) -I . Then = ~
(b)~(c)~(u)
@~(a)$(u)
= bcv (V)
=
~
f o r some
-1
f o r some
(h)~(c)
~(a)
= cp(b)%O(t)-l~(s) =
c 9 A , u c S , by F 2. So
, and by F 3 this implies that
v e S . Then
~(a)~(u)
= ?,(b) 7(c)
is invertible, and we may go backwards to obtain
auv
=
since
90
~ ( a )- ~ C "sI)
that
""
"
~
=
~ ' ( b -) ~ ( t")I
""
"
.
We leave
the reader
%0
is a homomorphism. It is clear that
verify
to
~c~ . ~
and that
is unique. Corollar~ I~.~.
A[S -I]
is unique up to isomorphism.
The unicity of the solution of a universal problem also implies: Corollary
1~.6.
both
If
A[S -I]
and
[S-I]A
exist, then they
are isomorphic. Since there are examples of but
A[S -I]
A , S
such that
[S-I]A
exists
does not exist ([I0], p. 163), a ring may satisfy
the universal property of 15.4 without being a right ring of fractions with respect to the existence of
A[S -I] .
Proposition i~.7. A[S -1]
S . We now turn to the question of
Let
S
be multiplicatively closed in
exists if and only if
right reversible. If
A[S-~
S
A .
is both right permutable and
exists, then
s ~ S}
where
the topology
~ = {I~ I D sA
Proof. If
is right permutable and right reversible, ~ e n
S
for some
A[S -I] = A F
is perfect.
is a perfect topology by Proposition 15.3. Let
~:A--~A F
the canonical homomorphism, and use the description of Proposition 14.2. S 5 guarantees that element of of
~(s)
AF
whenever
s e S . (l,s)
. Every element
= ~ ( a ) ~ ( s ) -I. Finally, if some
s c S . Hence
Suppose conversely
~
(l,s)
(a,s) c A F
A?
be in
represents an
will be an inverse has the form
(a,l)~(O,l)
, then
(a,l)(l,s) =
as = 0
for
satisfies F I-3.
A[S -I]
exists. S
is then right permutable,
91
for
if
a
by F 2 .
¢
A
i.e.
for some
,
s
c
S
~p(at)
are
-- T ( s b )
u c S . Since
s c S , then
given,
~(a)
then
~ ( s ) - l ~ "( a ) "
. ByF
3 this
"
by F I and
at = 0
= ~ ( b )-~ ( t ")I
means t h a t
tu c S , we have S 3. If
= 0
"
atu--
sa = 0
for some
""
sbu
with
t c S
by
F 3~ so we have S 5. Corollary I~.8. There is a I-i correspondence between right rings of fractions of
For
AL$-I]
A
and subsets of
A
satisfying S 1-5.
one may simplify the formula of Proposition 15.2
somewhat, in that MF = M~ S/~
, where
Of course one has
N
is defined as before.
M F = M ~ A A[S -I]
.
Examples: I.
When
A
is commutative~
satisfied. The theory
S 3 and S 5 are automatically
of rings of fractions is well-known
in that case [I0]. 2.
Let
S
be the set of non-zero-divisors
of
called the classical right ring of quotients of be denoted by of quotients
A . A[S-II
is
A , and will
Qcl " It is a subring of the maximal right ring Qm
of
A , and is in fact a suhring of
M(A)
.
From Proposition 15.7 we get: Pro~osition I~. 9.
The classical right ring of quotients of
exists if and only if for
a c A
A
satisfies the "Ore condition",
and a non-zero-divisor
a non-zero-divisor
t
such that
s
there exist
sb = at .
b ¢ A
"
i.e. and
A
92
Note that if
A
has both a classical right ring of quotients
and a classical left ring of quotients, then these two rings coincide, by Corollary 15.6. Exercises: I. Show that if
A
has no nilpotent elements ~ 0 , then S 5 is
always satisfied. 2. Let M
AtS -I] is
F-closed for the corresponding topology
only if
M
for every 3. Let
be a ring of fractions. Show that an
A
b = 0~.
F
is torsion-free and divisible (i.e.
if and M = Ms
s ¢ S ).
be a regular ring and let Show that
S
S =~a
e A~ ba = 0
satisfies S 1-5' and that
S 5 only if all elements in A
A-module
S
are invertible.
S
implies satisfies
(llluetration:
is the endomorphi~n ring o£ an infinite-dimensional vector
space). References.- Als,~vist [3], Bourbaki [I0] (p. 162-163), Eriksson
t261, G a b r i e l [31],Gabriel-Zism
[89].
Chapter 4.
§ 16.
Self-in~ective rings
The endomorphism rin~ of an in~ective module
A ring
A
for every
is called regular (in the sense of von Neumann) if a s A
there exists
x s A
such that
axa = a . We
recall the following alternative characterizations of regular
(of. [I0], p. 64):
rings
Pro~gsition 16.!.
The following properties of
A
are equi-
valent: (a)
A
is regular.
(b)
Every finitely generated right ideal of
A
is generated
by an idempotent element. (c)
Every right
A-module is flat.
It follows from (b) that every right noetherian regular ring is semi-simple. More generally, (b) implies that if
A
is
regular and has no infinite family of orthogonal idempotents, then
A
is semi-simple.
Theorem 16.2.
Let
morphism ring
H
E
be an injective
and let
J
A-module with endo-
be the Jacobson radical of
H .
Then:
(i) (ii)
is re.nat. Idempotents may be lifted modulo
The last assertion means that if
e
J .
is an idempotent in
H/J ~ then there exists an idempotent in
H
mapping canonically
9~
onto
e . It is well-known
that if idempotents
then one may lift any countable potents in
H/J
orthogonal
so that orthogonality
may be lifted,
family of idem-
is preserved
([47],
§ 3.6). The proof of the Theorem will be broken up into several steps. We define N =
~h
c H ~ Ker h
Lemma 16.3.
N
is an essential
is a two-sided
submodule
ideal of
of
E}.
H , and
H/N
is a
regular ring. Proof.
If
f, g e N , then
Ker f C~ Ker g . If
f e N
Ker fh = h-l(Ker f) , and thus a two-sided Let
f+g e N
ideal.
and
essential
that
submodule
is a monomorphism, g:E-*E
of
K~Ker
h~K
Ker hf D K e r f
K
of
E
h = 0 . K + Ker h
E
is injective, for
since . N
of
is
H/N .
which is maximal is then an
E . Since the restriction
gh(x) = x
K
since
fh c N
It remains to show regularity
and since
such that
Ker(f+g)
h c H , then
hf e N
h ~ H . Choose a submodule
with the property
since
of
h
to
K
there exists
x e K .
.~ E
c~n,1 E If
y e K + Ker h , we write
Then
hgh(y)
shows that Lemma 16.4. Proof.
: hgh( H/N
) : h(x)
y = x+z h(y)
,
with
x e K
hgh-h
c
and
h(z) -- 0 .
. This
is regular.
N = J .
Suppose
h c N o Since
K e r ( 1 - h ) ~ Ker h = 0
and
Ker h
95
is essential Im(l-h)
in
E
must then be a direct
injective.
If
Im(l-h)
and so
= E . l-h
is a monomorphism. E , because
x = (l-h)(x) essential
, so
in
is thus invertible
E
is
Im(l-h)
E , and we must
for every
h e N ,
N C J .
intersection radical
of
J/N = 0
H/N
and
idempotents
is
. But
H/N
it is clear that the
is regular by 16.3,
it remains
may be lifted mod J . Assume
h c H
L = Ker(h-h 2)
envelope
on
J/N
is defined as the
the proof of the Theorem,
of
thus has the form eh = h
radical
so
J = N .
which means that injective
since the Jacobson
of all maximal right ideals,
To conclude
f = ~
h(L)
Im e
L , so
f = eh
for some idempotent
If
A
Im(l-e)
on
L' , so
e
radical
(i)
J
is the right singular
(ii)
A/J
h-h 2 e J ,
E . The E
and
H . Then
f = e + eh(l-e)
and note
and note that
L'
+ Im e = E . f-eh = e-ehe
is ,
f-eh e N . Since we already f-h e N = J .
is a right self-injective
Jacobson
J , then: ideal of
A .
is regular.
(iii) Idempotents
and
in
+ h(L)
of
eh-h c N , it follows that
C o r o l l a r ~ ! 6 . ~.
in
to show that
is a direct st~mand of
L' = Im(l-e)
submodule
and therefore
is essential
eh-h e N . Put
. Put
an essential
have
l-h
stmmand of
is therefore
On the other hand,
that
that
h(x) = 0 , then
Ker h . Im(l-h) have
it follows
,
may be lifted modulo
J .
ring with
96
We remark that the ring
H/J
of the Theorem may be shown to
be right self-injective (Osofsky [61~, Renault [107], Roos [109]), but this is a fact which we will not need. Instead we are interested in the case when
H/J
Definition. The ring
is semi-perfect if
where
J
lifted
A
is semi-simple.
is the Jacobson radical of
A/J
is semi-simple,
A , and idempotents may be
mod J .
It is well-known that one can always lift idempotents when
J
is a nilideal ([47], § 3.6), and therefore every right or left artinian ring is semi-perfect. From 16.5 follows immediately: PropQsition 16.6.
A right self-injective ring is semi-perfect
if and only if it has no infinite family of orthogonal idempotents. A module
M
is called finite-dimensional if it does not
contain any infinite family of non-zero submodules that their s~n
[M.
M.
such
i
is direct. It is clear that an injective
I
module is finite-dimensional if and only if its endomorphism ring has no infinite family of orthogonal idempotents. Hence Theorem 16.2 gives the following generalization of 16.6: Pro~qsition 16.7. E
The endomorphism ring of an injective module
is semi-perfect if and only if
E
is finite-dimensional.
Examples: I. If
E
is an indecomposable injective module, then
a division ring ([52]).
H/J
is
97
2. The ring
A
is called right finite-dimensional if
AA
is
a finite-dimensional module (cf. § II, Exercise 4). Such a ring cannot have any infinite family of orthogonal idempotents. Exercises: I. Show that a commutative ring is semi-perfect if and only if it is a product of finitely many local rings. 2. Show that a semi-perfect ring is a.direct st,n of indecomposable right ideals. 3. Show that a module E(M)
M
is finite-dimensional if and only if
is finite-dimensional.
References: Faith [28~, § 5 , Lambek [471, § 4.4.
§ 17.
Coperfect rin~s
As a natural generalization of the class of artinian rings we define: Definition. J
A
is a semi-primary ring if the Jacobson radical
is nilpotent and If
A
A/J
is semi-simple.
is semi-primary and right noetherian, then
A
is right
artinian~ by a classical argument ([9], § 6, Prop. 12). A further generalization leads to: Definition.
A
is a right coperfect ring if it satisfies DCC
on finitely generated right ideals. The usual Zorn's lemma argument shows that the DCC is equivalent to the condition that every non-empty family of finitely
98
generated right ideals contains a minimal member. It is known that a ring is right coperfect if and only if it is left
p e r f e c t in the sense cf Bass [7]
(Bj rk [86]),
but this fact
we do not have tc use.
Proposition 17.1.
Every
semi-primary ring is right and left
coperfect. Proof. We use inducticn ~on the smallest integer jn = 0 . When
n -- 1 , A
n>~ I
such that
is semi-simple and obvicusly right and
left coperfect. Suppose the assertion has been proved for all semi-primary rings such that
jm : 0
be semi-primary with
but
I I D 12 D
....
~-I
m < n . Let
A
@ 0 . Suppose
is a descending chain of finitely generated
right ideals of
A . The canonical map
chain into a chain
ideals of
jn = 0
for some
~i D T 2 3
...
A - * A/J n-I
takes this
of finitely generated right
A/J n-l. But the radical cf
A/Jn-z
is
j/jn-I , so
this later chain is stationary by the induction hypothesis. Hence we have Since
Ik C I r + jn-I
jn = 0 , this gives
consider the chain
s
we have
we may assume Prcp. 6) gives
IkJ = IrJ
Ii/IkJ D 12/IkJ D
the semi-simple ring some
for all
r)k
, for some
for all ..
r>k
. Now
of right ideals over
A/J . This chain is stationary,
I s C I t + IkJ = I t + IsJ
k .
(all
sc for
t) s ) since
s~ k . The Nakayama lemma ([9], § 6, Cot. 2 of Is = It
for
t>,s .
Some basic properties of ccperfect rings are summarized in the following statement:
99
Propositi0n !~.2.
If
A
is right coperfect, then:
(i)
A
is right semi-artinian.
(ii)
J
is a nilideal.
(iii) A Proof. A/I
is semi-perfect. (i): It suffices to show that every cyclic right module
contains a simple submodule. Let
generated right ideal of
~I
. !+I'/I
I'
be a minimal finitely
is then a simple submodule
A/I .
(ii): Let
s
be the preradical associating to each module its
socle. From § 3, Example 5, we recall that J = s0(J ) o(a)
for some ordinal
as the smallest
0
a limit ordinal, for if some
0 • For each
such that a s~sa(J
a s J
hilated by
) , then
a, b c J
s +l(J)/sa(J ) = s(J/sa(J)) J , so
we have
the sequence
is never
a s sa(J )
for
for some ordinal
a .
and is therefore anni-
s + I ( J ) . J C sa(J ) . This shows that for any h(ab)(h(a)
h(a n)
we define
a c s0(J ) . o(a)
a~ 0 a . Hence we may write o(a) = a+l
Recall that
J = ~(J) , i.e.
. If
a c J
were not nilpotent,
would be an infinite strictly decreasing
sequence of ordinals, which is impossible. (iii): Since
J
is a nilideal, we may lift idempotents. It
remains to see that A/J
is semi-simple. It is easy to see that
also is right semi-artinian. If
ideal of that I
A/J
I
is a minimal right
A/J , then there exists a maximal ideal
I~I'
= 0
since the Jacobson radical of
A/J
I'
such is zero.
is therefore generated by an idempotent. Since we can lift
idempotents modulo
J , and since a right coperfect ring
100
obviously cannot have any infinite family of orthogonal idempotents, the right eocle of
A/J
is a finite direct stun of
minimal right ideals, and is therefore itself a direct summand of
A/J . Its complementary summand must then have zero socle
and is therefore zero. Hence
A/J
is semi-simple.
It should be remarked here that it can be shown that conditions (i) and (iii) of 17.2 imply conversely that
A
is left perfect
(Bass ~7]) and hence right coperfect (BJGrk ~86]). PropQsition 17.3. coperfect, then Proof. If
A
If A
A
is right noetherian and right or left
is right artinian.
is right noetherian and right coperfect, then
is obviously right artinian. Suppose left coperfect. J
A
A
is right noetheriau and
is a nilideal, but every nilideal in a
noetherian ring is nilpotent ([47], P. 70). Since semi-simple by 17.2, A
A/J
is
is a semi-primary ring and hence right
artinian by a previous remark.
This result may be generalized somewhat. For this we introduce s o m e terminology. For each subset
r(S)
= { a ¢ A I
sa
= o
}
,
l(S)
S
of
A
= { a ¢ A I
we put aS = 0 } .
A right ideal is said to be a right annihilator if it has the form
r(S)
for some
Proposition 17.4.
If
S CA A
. satisfies ACC on right annihilators
and is left coperfect, then Proof.
A
is semi-primary.
Consider the ascending chain of right annihilators
I01
r(J) C r ( j 2 ) C for some
....
n . If
A/r(J n)
By hypothesis we have
r(J n) = r(J n+l)
r(J n) ¢ A , then the left
A-module
has non-zero soole, which is of the form
for some left ideal I/r(J n)
I D r(J n) . But the semi-simple module
is annihilated by
J . so
I C r ( J n+l) = r(J n) . Then and hence
I/r(J n)
r(J n) = A , so
Jl C r(J n) , which gives
I = r(J n) , which is a contradiction, jn = 0 .
References: Bass /.7~, Faith t27J.
§ 18.
Quasi-Frobenius rings
In this § we discuss three classes of rings: S-rings, PF-rings and @Y-rings (in order of d e c r e a s i ~ generality). An injective module M @ 0 E
E
is a co6enerator if
Hom(M~E) ~ 0
for every
module
(of course it suffices to consider cyclic modules
M , so
is a cogenerator if and only if for every right ideal
there exists
x @ 0
Proposition we let
Proposition 18.1.
in E(A)
E
with
I @ A
xl = 0 ). In the following
denote the injective envelope of
The following properties of
A
AA .
are equivalent:
(a)
E(A)
is an injective oogensrator.
(b)
Every simple right module is isomorphic to a minimal right ideal of
A .
(c)
Hom(C,A) @ 0
(d)
l(I) ~ 0
(e)
A
for every cyclic module
for every right ideal
C @ 0 .
I ~ A .
has no proper dense right ideals.
Proof. (a) ~ (b): If
S
is a simple module, there exists a non-
102
zero of
f : S - @ E(A) . Since f
must lie in
(b) ~ ( c ) :
(c)
A
is essential in
E(A) , the image
A .
Clear, because
C
has a simple quotient module.
(d): Ob ous.
(d) ~ ( e ) :
Immediate from Proposition 3.8.
(e) ~=~(a): A right ideal only if
I
is by definition dense
if and
Hom(A/I,E(A)) = 0 .
Definition.
A
is called a right
S-ring if it has the properties
of 18.1. (We have reversed the terminology of Morita [56], who calls these rings "left S-rings" and furthermore asstunes minimum conditions on Proposition 18.2.
A ).
When
A
is a right self-injective ring, the
following conditions are equivalent: (a)
A
is a right S-ring.
(b)
AA
(c)
r(l(1)) = I
is an injeotive cogenerator. for every right ideal
Proof. (a) ~ ( b ) : (b) ~ ( c ) :
Clear, since
Suppose
homomorphism
f:r(l(1))/l-,A
have
g(a) = ba
is an injeotive module.
r(l(I)) @ I . Then there exists a non-zero
composed homomorphism the form
AA
I .
. Since
A
is injective, the
g:r(l(1))-~ r(l(I))/l-* A for some
b s I(I) . But then
b C A • Since
g(a) = 0
for every
must be of
g(1) = O , we a ¢ r(l(1))
which is impossible. (c) =>(a): Obvious by 18.1(d). Definition. cogenerator.
A
is a right PF-ring if
AA
is an injective
103
Proposition 18.3.
The following properties of
A
are equivalent:
(a)
A
is a right PF-rin~.
(b)
A
is right self-injective and an essential extension of
its right socle~ and (c)
A/J
is semi-simple.
Every faithful right module is a generator for
Mod-A .
Proof. We will only prove the implication ( b ) ~ ( a ) ,
since this
is the only one we will need in the following. For the proof of (a)~(b)
and ( b ) ~ ( c )
(cf. also
Kato
(b) ~ ( a ) :
[42]).
We write
right ideals
A/J
I' G J
Ik
of
and hence
minimal right ideal
morphic to some
I i ~ Ik
Jk
A
so that
I' = 0 . Each
Ik
to an indecompos... ~ I n $ I' .
contains a unique
A-module
S
may also
A/J-module, and is therefore iso-
Yk " In particular, each of the minimal right
must be isomorphic to some
since
~k
A = Ii@
Jk " Every simple
be considered as a simple
ideals
as a direct sum of indecomposable
YI,..,Yn . We can lift each
able right ideal Then
we refer to Azumaya [61 and Osofsk~ [60]
Ik
. If
is an injective envelope of
that the number of non-isomorphic right ideals the number of non-isomorphic modules
Ji = Jk ' then Jk " It follows Jk
is equal to
~k " Hence each
then every simple module, is isomorphic to some
~
Jk ' and
, and A
is thus a right S-ring.
Ne next prove a very usefu~ property of self-injective rings:
104
Proposition 18. 4. (a)
The following properties of
Every homomorphism
f:l-@A
, where
I
generated right ideal, has the form (b)
A
satisfies
l(IlnI2)
(i)
A
are equivalent:
is a finitely
f(a) = ca
= 1(Ii)
+
I(I2)
for some for
all
finitely generated right ideals
(ii) Proof.
(a) =~ (b): If
ideals,
l(r(a)) = Aa II
then obviously
the other hand that morphism
and
have
such that
cb = b , i.e.
as
are finitely generated right
a(b) =
I
a e A
we have
Aa C l(r(a))
with some
. If
as
b e l(r(a))
xa,-~xb
f:I--~A
I = alA + . . . + a n A .
being obvious.
. It must be
c ¢ A , so in particular
where
I
is finitely generated,
We use induction on
8o we may assume there exist
f(a) =~oa
n , the case c
and
for
a e I' = alA +...+an_iA
for
a e ariA
c'
n = 0 in
!
~c'a For
,
b e Aa .
(b) = ~ (a): Consider
such that
we thus
, which proves (i).
given by left multiplication
say
b e II
(c-l)b = 0 . As a consequence we may write
A a - ~ Ab
and
b e II b e 12
I i ~ I 2 . By (a) there
=(b) = cb . For
we can define a homomorphism
b = ca
b (l+a)b
coincide on
a = (o-I) + (l+a-o) e l(Ii) + 1(12) For every
a e A .
a e I(Ii(~ 12) . We can define a homo-
a-I I + 12--~A
c e A
Ii, 12 .
i(II) + 1(12) C l(Ii(~ 12) . Suppose on
since these two expressions exists
12
for every
c e A.
a e I'~anA
we must have
e l(I'(~anA ) = l ( I ' )
+ l(anA )
(o-c')a = 0 , so
o-c' e
by ( i ) . We accordingly write
A
105
c-c' = b-b'
with
plication by
and
c-b = c'-h'
Proposition 18.~. (i)
l(lln12)
(ii)
l(r(1)) = I
Proof.
bl' = 0
If
A
b'a
= 0 . Then left multi-
n
coincides with
f
on
I = I' + a n A
is right self-injective,
= I(Ii) + 1(12)
.
then:
for all right ideals
I I , 12 .
for every finitely generated left ideal
I .
(i): The argument used in the preceding proof works
equally well in the present situation. (ii): Write
I = Aa I + ... +Aa n . Then
= l(r(Aal)a...nr(Aan)) Aa I + ...
Definition. both right
+ Aa n
= l(r(Aal)) by (i)
A
+ ...
+ l(r(Aan))
=
=
and 18.4(ii).
A i s a QF-ring ( o r q u a s i - F r o b e n i u s r i p ~ ) i f i t and l e f t
artinian
We will show, however, make
l(r(Aa I +...+Aan))
and
r i g h t and l e f t
is
self-injective.
that much weaker conditions suffice to
a QF-ring. Our first step is to show that the conditions
may be made one-sided. For this we need the following reformulation of the definition: Lemma 18.6.
A right and left artinian ring is a QF-ring if and
only if it satisfies r(l(1)) = I for all right ideals Proof.
and
l(r(I')) = I' I
and left ideals
I' .
A QF-ring satisfies the double annihilator conditions
by 18.5, Conversely, r
,
1
these conditions imply that the operations
define an anti-isomorphlsm between the ordered sets
of left resp. right ideals. They therefore reverse the lattice operations,
so conditions (i) of 18.4 are satisfied, and hence
the ring is right and left self-injective.
106
Proposition 18.[.
If
A
is right or left artinian and is right
or left self-injective, then
A
is a QF-ring.
Proof. There are two cases we must consider: I)
A
is right artinian and right self-injective!
2)
A
is left artinian and right self-injective.
We can reduce case I) to 2): If
IIC
I2C
...
is an ascending
chain of finitely generated left ideals, then the descending chain
r(l I) D r(12)D
... is stationary since
A
is right
artinian. The ascending chain is then also stationary by 18.5, and
A
is thus left noetherian. But a ring which is both left
noetherian and right artinian, is also left artinian. We now consider case 2). Since
A
is left artinian, it is
right coperfect and therefore right semi-artinian (Propositions 17.1 and 17.2). It then follows from 18.3 that PF-ring, and this implies by 18.2. We also have
r(l(1)) = I
l(r(I)) = I
A
is a right
for every right ideal
for every left ideal
I
I , by
18.5. By an argument similar to the previous reduction of I) to 2), we finally have that
A
is right artinian, and by applying
Lemma 18.6 we may conclude that
A
We may weaken the conditions for Proposition 18.8.
If
A
is QF.
A
to be QF quite a bit more:
is right or left noetherian and is
right or left self-injective, then
A
is a QF-ring.
Proof. There are two cases to be considered: I)
A
is left noetherian and right self-injective!
2)
A
is right noetherian and right self-injective.
107
Case I): To show that that A/J
A
A
is left artinian, it suffices to show
is semi-primary (cf. remark at the beginning of § 17).
is a left noetherian regular ring by
simple. It remains to show that chain of two-sided ideals since
A
J
follows that
...
is stationary
r(Jn) = r(J n+l) . From 18.5
jn = / n + l which implies
§6, Cor.2 de
lsmma
is nilpotent. The ascending
r(J) C r(j2) C
is left noetherian. Let
16.5, hence semi-
jn = 0
by the Nakayama
op.6).
Case 2): We prove a slightly stronger statement: Theorem ! 8 . 9 .
If
A
satisfies ACC o n right or o n left annihi-
lators and is right or left self-injective, then
A
is a
QF-ri~g. Proof. Case I)- Suppose
A
satisfies ACC on left annihilators
and is right self-injective. Every finitely generated left ideal is a left annihilator by 18.5, so
A
is left noetherian. A is
then QF by case I) of 18.8. Case 2): Suppose
A
satisfies ACC on right annihilators and is
right self-injective. If
I I ~ 12 ~ ... is a descending chain
of finitely generated left ideals, then stationary by hypothesis, and so semi-primary by 17.4.
A
A
r(l I) C r(12)C
... is
is left coperfect and even
is then a right PF-ring by 18.3, which
implies that every right ideal is a right annihilator (18.2). A
is therefore right noetherian. But since we have just proved
that
A
is semi-primary,
conclude from 18.7 that
A A
must then be right artinian. We is QF.
108
An interesting property of QF-rings is: Proposition 18.10.
Every module over a @y-ring is a submodule
of a free module. Proof. It suffices to show that every injective module is projective when
A
is QF. Since
A
is noetherian,
every
injective module is a direct sum of indecomposable injective modules,
so it clearly suffices to show that
projective, where
S
is any simple module. But
morphic to a right ideal of E(S)
E(S)
is a direct summand of
A , and since
A
S
is is iso-
is self--injeotive,
A .
Combining this result with § II, Exercise 3, we find that over a QF-rin~, a module is injective if and only if it is projective. Example: ~-ring
The group ring of a finite group over a field is a ([201, p. 402).
Exercises: I. Show that an injective module is a cogenerator if and only if it contains a copy of each simple module. 2. Show that a right PF-ring with zero Jacobson radical cannot have essential right ideals @ A
and must therefore be semi-
simple.(This is a special case of the implication (a) ~ (b) of 18.3, which we did not prove). 3. Show that if
A
right ideal, then
is a right PF-rin~ and l(I)
I
is a maximal
is a minimal left ideal.
109
References;
Azumaya
Faith [27], ~ e a ~ x ~
[6],
BjSrk
[8],
Eilenberg-Nakayama
[37], Kato t421.
[23~,
Cha~_ter 5-
Maximal and classical rings of qu_qtients
§ 19. The maximal rin~ of quotients
We recall that a right ideal
I
of
has no left annihilators for any
A
is dense if
(l:a)
a c A , and that the family
of dense right ideals is a topology, corresponding to the hereditary torsion theory cogenerated by is the maximal right rin G of quotients of
E(A) . The ring
AD
A , and is denoted
by % . P~PPqsiti°n 19"1"
Qm
is its own maximal right ring of quotients.
Proof. This follows from Proposition Ii.ii since family of dense right ideals of Let
H
~
De
is the
.
be the endomorphism ring of
E(A) . We recall from § 8
that there is a commutative diagram A
~
~ HomH(E(A ) ,E(A))
E(A) where
k
is a ring isomorphism,
(Proposition 8.2) and
Proposition 19.2.
~
is the canonical inclusion
£(f) = f(1) .
The following assertions are equivalent:
(a)
~
is a right self-injeotive ring.
(b)
I,Qm_-~E(A )
(c)
The right ideal
(d)
There is a ring isomorphism
is an isomorphism. (A:x)
is dense for every H-@Q m
x ¢ E(A) .
such that the diagram
111
commutes, where
e(h) = h(1) .
~oof. The equivalence of (b) and (c) is clear from Proposition 8.1, and the equivalence of (a) and (b) from Proposition 8.2. (b)@(d):
~
(d)~(b):
•
Hom%(E(A),E(A)) = HomA(E(A),E(A)) . is ob~ously s~jective, so if ~
exists, then
is also s~Jective. Lemma 19.3. is also
If
~
A
is a r i ~ t finit~dimensional ring, then so
.
Proof. If a r i ~ t
ideal
J
r i ~ t ideals
of
, then each
i d e ~ of
J=
~
of
~
A . It follows that ~ e
Proposition 19.4.
Suppose
a semi-perfect ring if ~ d
~
is a direct s t JaVA
family
of non-zero
is a non-zero right
( ~)
m~t
be fibre.
is r i ~ t self-injective. ~
only if
A
is
is r i ~ t finite-
dimensional. Proof. ~d
~
~ HomA(E(A),E(A)) by 19.2, so
only if
E(A)
E(A)
~
is semi-perfect if
is a finite-dimensional mod~e, by 16.7. But
is finite-dimensional if ~ d
Proposition 19.~.
only if
A
is so.
The following assertions are ~ u i ~ l e n t :
(a) % = M(A) (h)
~
is f l a t
as a left
epimorphism. (c)
~
is a right
S-ring.
A-mod~e and
A ~
is a ring
112
Proof. (a)@~p(b) is obvious since one always has (a) ~ ( c ) :
If
ideals of
A
then C
J
Qm = M(A) , then the topology is perfect. If
JN A c ~ , and
(c)~(a):
of dense right
is a dense right ideal of
by Example 3 of § Ii. Therefore
J = Qm
Zf
J
~
M(A) C Qm .
Qm '
Qm = (J~A)Qm C.
"
I C2
Thus there exists
and
I%
~ %
O # q e Qm
impossible, because
Qm
is
Corollar~ 19.6 ,. Suppose
Qm
, then
such that
1(I~)
%0
by 18.1.
ql = 0 . But this is
_D-torsion-free.
is right self-injective.
~
is
a right PF-ring if and only if
Qm = M(A) .
PropQsition I~. T.
is right self-injective. The
Suppose
~
following assertions are equivalent: (a)
Qm
is a quasi-Frobenius ring.
(b)
Qm
is a ~-injective right
(c)
Every direct sum of
A-module.
D-torsion-free injective
A-modules
is injective. (d)
The lattice
(e)
A
~D(A)
of
_D-pure right ideals is noetherian.
satisfies ACC on right annihilators of subsets of
E(A)
Proof. The equivalence of (b), (c), (d) and (e) was proved in Propositions 11.7 and 11.8. Since every QF-ring is noetherian,
(a) ~l~(d) by Proposition 11.12. (e)=)(a): By Theorem 18.9 it suffices to show that the ring Qm
satisfies ACC on right annihilator ideals. In view of
113
condition (e), it will be enough to show that if are subsets of
Qm ' then
if and only if
{a s A~ Sla = 0~ C I a
SI
~q c Qm I Slq = 0 } C I q
and
S2
s Qml S2q = 0
c A ~ S2a = 0 } .
The first
inclusion trivially implies the second one. Suppose the second inclusion holds, and such that S2q = 0
qlCA since
References:
§ 20.
Slq = 0
for some
. Slql = 0 Qm
Lambek
implies
q s Qm " Choose
I c
S2ql = 0 , which gives
is torsion-free. [47], Mewborn-Winton
[54], StenstrGm [72].
The maximal rin 6 of quotients of a non-singular rin~
For non-singular rings there are more precise results on the structure of
Qm
than we were able to obtain in the general
case. Recall that when
~
is right non-singular,
the t o p o l o ~
of dense right ideals coincides with the family of essential right ideals, which we denote by (§ 3, Example I) reduces to Proposition 20.1.
E
E . Also the Goldie t o p o l o ~ in this case.
The following assertions are equivalent:
(a)
A
is right non-singular.
(b)
Qm
is a regular ring.
(c)
Qm
is a right self-injective regular ring.
(d)
The Jacobsen radical of
Proof. Suppose (A:x)
A
HomA(E(A),E(A))
is right non-singular.
If
is zero. x c E(A) , then
is essential. From Proposition 19.2 fellows that
right self-injective and that (a) @ ( d ) ,
then also (a)@(c)
Qm = H°mA(E(A)'E(A))
Qm
is
" If we prove
will follow, as a consequence of
114
Theorem 16.2. (a)~
(d): Recall that the Jacobson radical of
consists of those ~ :E(A)--~E(A) in
E(A)
we have
~(x)I
hence
~=
(d)~
Ker~
essential
(lemma 16.4). But if so is the case, then for each
x ¢ E(A)
and
which have
Hom(E(A),E(A))
= 0 0
xl C K e r ~
for some essential right ideal
implies ~(x)
= 0
since
E(A)
is non-singular!
.
(a): Suppose
aI = 0
where
a c A
and
I
is an essential
right ideal. The homomorphism
A-*E(A)
extends to an endomorphism
~
of
essential in
is in the Jacobson radical of the
E(A) , so
~
endomorphism ring, and hence (b) ~
(a): Suppose again
that
and
I
dicts
b~ab
Ker~
~ = 0 . This implies where
a = aqa
is essential in
0 % qab e I . But then
given by
E(A) . Then
al = 0
essential. We may write qa @ 0
0 # a ¢ A
for some ~
D I
is
a = 0 . and
I
is
q c Qm " Since
, there exists
ab = aqab C al = O
b ~ A
such
which contra-
qab # 0 .
We see in particular from this Proposition that if semi-simple ring, then Theorem 20.2.
Let
A
A
is~on-singular.
~
is a
Conversely we have:
be right non-singular.
The following
assertions are equivalent: (a)
qm
is semi-simple.
(o)
Qm
is flat as a left A-module and
A-~
is a ring epi-
morphism. (d)
I ,
Every essential right ideal contains a finitely generated
115
essential right ideal. (e)
A
is right finite-dimensional.
(f)
The lattice
~(A)
of complemented right ideals is
noetherian. (g)
Every direct sum of non-singular injective modules is injective.
(h)
E(A)
is a
-injeotlve module.
Proof. We know to begin with that
Qm
is right self-injective
and regular. Therefore the equivalence of (a), (f), (g) and (h) is immediate from Proposition 19.7. We have (a) =~(b) by 19.6, (b)@~(c) trivially, (c) $ ( d )
by Theorem IB.l.
Ne get (d)~=p(e) from Propositions 19.4 and 19.3, and (d)4=~(f) from Proposition 11.14. Corollar~ 20.3.
If
(i)
evelv right
(ii)
every left
Qm
is semi-simple, then:
Qm-module is injective as an Qm-module is flat as an
A-module.
A-module.
Proof. Follows from the formulas (i) and (2) of § 14. Corol!ar~ 20.4.
If
Qm
is semi-simple and
A-module, then
M ® A Qm
Proof. We have
ME-- E(M)
Proposition 20.~. singular. Qm A
Qm
is a non-singular
is an injective envelope of
A
by 8.1 , and
M~ = M S A Qm
M . by 13.1.
be right finite-dimenslonal non-
is then also a maximal left ring of quotients of
if and only if
Proof. If
Let
M
Qm
is flat as a right
A-module.
is a maximal left ring of quotients of
A , then
116
it is right flat over if
Qm
A
by Theorem 20.2. On the other hand,
is right flat, then
of rings. This implies that
A--~Q m Qm
is a right flat epimorphism
is a subring of
MI(A)
the maximal right flat epimorphic extension ring of Theorem 14.1. But of quotients of
MI(A )
(i.e.
A ) by
is a subring of the maximal left ring
A . Since
Qm
is left self-injective, it must
then coincide with the maximal left ring of quotients of
A
(of. Proposition 19.1). We will take a closer look at the lattices submodules. We assume
A
C_E(M)
of
_E-pure
to be right non-singular. As a special
case of Proposition 11.5 we have: Proposition 20.6.
If
M
is a non-singular module, then
~E(M)
is a complemented modular lattice consisting of the complemented submodules of
M .
Ne may describe
~E(A)
to a certain extent, because this
lattice is isomorphic to the lattice of complemented right ideals of
Qm
(Proposition 11.12). Since
Qm
is right self-injective
regular, this second lattice consists of the direct summands of
qm " We thus have:
Proposition 20.7.
If
A
is right non-singular,
then
~E(A)
is isomorphic to the lattice of principal right ideals of Corollary 20.8. then
~E(A)
If
A
Qm "
is right finite-dimensional non-singular,
is a modular lattice of finite length.
117
Examples: I. If
A
is a Boolean ring (i.e. a commutative regular ring
with all elements idempotent), of
A
Qm
is the "completion"
in the sense of Boolean algebra theory.
2. It may happen if
then
Qm
that
Qm
is flat as a left
is not semi-simple.
A-module even
In fact, one can just take
A
to
be a regular ring whieh is not semi-simple. More generally: Proposition 20. 9 .
If
A
is right semi-hereditary,
is right non-singular and Proof. If module,
0 $ a c A , then
so
r(a)
To show tha%
Qm
A
is flat over
generated right ideal
right
A/r(a) ~ aA
A . Hence
I@Qm--~AQQm
a projective
is flat as a left
A
and can therefore
is non-singular. A , it suffices to show that
I . Since
I
is projective,
IQQ m
is
Qm-module and is therefore non-singular as a
A-module.
$uppose
i , and for each
= 0 . l®Qm
A-module.
is a monomorphism for every finitely
~ai@qi
There is an essential right ideal all
A
is a projective
is a direct summand of
not be essential in
the map
Qm
then
a s J
~ I®Qm J
such that
we get
non-singular implies
and
~ aiq i = 0 . qiJCA
(~ a i @ q i ) a
for
=~ai(qia)®l
~ a i ~ qi = 0 .
E x e r c i se s :
I. Show that if A-module,
then
free, induces
A
is regular and
Qm
is projective as a right
A = Qm " (Hint: an imbedding AC-~F'
, F'
E(rA)C-pF , F
finitely generated free! then
use the fact that a finitely presented flat module is projective).
=
118
2. Let
K
be a skew-field and
2 ~ 2-matrices over
(i)
A
the ring of upper triangular
K . Show that:
The matrices of the form
constitute a minimal essential right ideal of (ii)
The ring
M2(K )
of all 2.2-matrices
right (and left) ring of quotients of (iii) ~ ( K )
~ .
is the maximal A .
is projective as a right (and as a left)
A-modul e.
References: Cateforis [15],[16], Faith [28], Lambek [47], Sandomierski [681, [69], Utumi [80], Walker and Walker [81].
§ 21.
The maximal tin6 of quotients of a reduced ring
A ring is called reduced if there are no nilpotent elements % 0 . In this § we want to find out when the maximal ring of quotients of a reduced ring also is reduced. For a commutative ring, this is always the case, because if 0 ~ aq e A
for some A
a e A
q c ~
and also
Lemma 21.1.
If
is reduced and
(i)
r(S)
is a two-sided ideal.
(ii)
S~r(S)
(iii)
Every idempotent is central.
(iv)
A
= 0 .
is non-singular.
and
qn = 0 , then
(aq) n = 0 .
S CA
, then:
119
Proof.
(i): It suffices to show that
implies
aS = 0
aS = 0
because
implies
= 0 , so (iv):
When
e
and
(ea(l-e)) 2
ae = eae , so
cannot be essential in = 0
A
S~r('S) = 0 .
a c A , then
ea = eae . Similarly
= aA~r(aA)
Sa = 0
Sa = 0 .
e2
r(a)
. But
(aS) 2 = aSaS = 0 , and similarly
(if): ( S n r ( S ) ) 2 C Sr(S) = 0 , so (iii): If
r(S) = l(S)
A
since
ea(l-a)ea(l-e)
ea = ae . a A n r(a) =
by (i) and (ii).
is commutative,
the converse of (iv) holds, because
a non-singular ring is a subring of a commutative regular ring, which obviously cannot have nilpotent elements % 0 . In the noncommutative case, this is no longer true (consider e.g. matrix rings). Since
A
reduced implies
Qm
regular, we should first
determine under what conditions a regular ring is reduced.
Proposition 21.2. The following properties of a ring
A
are
equival en%: (a)
A
is a reduced regular ring.
(b)
Every principal right ideal is generated by a central idempotent.
(c)
A
is regular and every right ideal is two-sided.
(d)
A
is strongly regular,
x c A Proof.
such that
(a)~
(b)~(c):
i.e. for every
a c A
there exists
2 a -- a x .
(b): Clear from Lemma 21.1(iii).
Since every principal right ideal is two-sided, all
right ideals are two-sided.
120
(c)$(d): Since a
2
F o r every
axA
also is a left ideal,
A
n n-I = a x
is o b v i o u s l y
for each
If
Qm
such that
axa = bax
reduced
a = axa
for some
a = axa
2 a = a x
since
n . We also have
2 2 a x a - axa = 0 , so
x
b
.
. Then
2 a = a x .
= a x a . a = baxa = ba , so
(d)~(a):
-
a e A there exists
and
is a s t r o n g l y r e g u l a r ring,
2
(a-axa) 2 A
implies
=
a
+
a =
2 axa xa
-
is regular.
then
A
m u s t be reduced.
C O n v e r s e l y we have: Proposition
21.3.
The f o l l o w i n g
properties
of a r e d u c e d r i n g
A
are equivalent: (a)
Qm
(b)
Every complemented
(c)
aA~bA
(d)
InJ of
is s t r o n g l y regular.
= 0 = 0
implies implies
right
ideal of
A
is a t w o - s i d e d
ab = 0 , for all
ideal.
a , b s A .
IJ = 0 , for all r i g h t ideals
I , J
A .
F o r the p r o o f w e need: L e m m a 21.4.
If
A
is any r i n g w i t h a t o p o l o g y
A C A F , then for each Proof.
Suppose
Choose
J c [
=ZaiqiJCI
P r o o f of 21.3: has the form
a =
~aiq i C A
such that
, and
I s~(A)
I = I~NA
with
qi J C A
Z c$(A)
(a)@(b):
one has
Each
implies
and qi
.
qi c A F .
" Then
aJ =
a C Z .
complemented
by the Lemma.
such that
I = IAFNA
ai 6 1
for all
~
right ideal of
Since
I~
A
is a two-
121
sided ideal of (b)$(d):
If
~
, I
l~J
with respect to
is a two-sided ideal in
= 0 , let INK
K D J
= 0 . K
A .
be a right ideal maximal
is then two-sided,
so
IJC
INK
= O
(d)~ (c): Trivial. (c)=~(a): We know that
Qm
is the endomorphism ring of
(Proposition 19.2). Suppose f ~ 0 ring
but Qm
f:E(A)-@ E(A)
E(A)
is nilpotent,
say
f2 = 0 . Since the Jacobson radical of the regular is zero,
So there exists
Ker f
is not an essential submodule of
O # a s A
such that
f(a) s Ker f , so (c) implies that and we must have
a
2
= 0 . Since
aA~Ker
a 2 S Ker f ,
is reduced, this is a contra-
diction.
Propositio n 21.~.
Let
A
be reduced with
Qm
strongly regular.
Then:
(i)
C_F(A )
is the family of right annihilator ideals.
m
(ii)
~F(A)
is
a
complete Boolean algebra with
l~-~r(1)
as a
w
complementation. Proof.
(i): Every ideal of the form
r(S) , S C A
by Proposition 11.7. Suppose conversely I
is maximal with respect to
21.3(d), and since also z
=
r(J)
I
= 0
, is complemented
is complemented,
I n J = 0 . Then
r(J)~J
fOr(J)
say
by
by 21.i, we must have
.
(ii): A Boolean algebra is the same as lattice [18]. That
r(1)
21.1 and 21.3(d). Since
a
complemented distributive
is a complement of CF(A ) m
.
f = 0 . But
f(a)a = 0 . Hence A
E(A)
I
follows from
is isomorphic to the lattice of
122 idempotents in
Qm ' it suffices to show that this second lattice
is distributive, i.e. = ef
and
Example: A
eA(fvg)
= (eA f) v ( e A g )
e ~ f = e+f-ef , and one has
Let
A
A . B
(e.g. by 21.3(b)). For each such that
B
be the maximal
is then also strongly regular
b ¢ B
there exists an idempotent
0 @ eb e A . Sirrce every idempotent in
central, we also have
equal
=
be strongly regular and right self-injective,
left ring of quotients of
submodule of
eaf
e(f+g-fg) = ef+eg-efg .
is then also left self-injective. For let
e s A
. But
B . Since
0 ~ be c A , so A
A
~
is
is an essential right
is right self-injective,
B
must
A .
Exercises: I. Show that every strongly regular ring is a subring of a product of skew-fields. 2. Let
K
be a skew-field and
the subring of
11K i
I
an arbitrary set. Let
consisting of
all (i.e. all except a finite number) (i)
A
(ii)
~K. i I N K. I i
(iii)
Referenoes:
(xi) xi
are equal. Show :
is a minimal essential ideal of
A .
is the maximal ring of quotients of
[113].
be
such that almos~
is strongly regular.
Renault [106], Utt~i
A
A .
§ 22.
The classical rin~ of ~uotients
When the classical ring of quotients subring of
Qm ~ in fact a subring of
self-injective,
we have
Proposition 22.1. (a)
Qcl
(b)
A
Every
. If
is right finite-dimensional
If
Qcl
q c Qcl
zero-divisor.
for each
and
x S E(A)
may be written in
Qcl
finite-dimensional
furthermore
Qcl = E(A)
where
a s A
.
and
b
contains a nonthen
A
is
by Proposition 19.4. A
is right finite-dimensional
(A:x)
is then dense, because for any = (A:xa)
for every a c A
and semi-perfect
19.2 and 19.4. We will show that
~
A . Since for each
for some non-zero-divisor
a
of
every non-zero-divisor
of
A
be the map
. Every
we have
a
q~-~aq
Im~a
by Propositions
is a classical right ring q s Qm
we have
qa s A
A , it suffices to show that is invertible in
~
. Since the kernel of
zero intersection with the essential is a monomorphism.
x s E(A)
and
, which has no non-zero left annihilators.
is right self-injective
~a:Qm--*Qm
ring.
cor~tains a non-
is semi-perfect,
contains a non-zero-divisor
of
are equivalent:
semi-perfect
then
(A:q)
(A:x)
of quotients
is right
.
q = ab -I
A . H~nce
Suppose corr~ersely that
((A:x):a)
Qcl
A
(A:x)
is right self-injective,
If
it is a
Qcl = N(A) = Qm "
exists and is a right self-injective
is a non-zero-divisor
Qm
N(A)
exists,
The following properties of
zero-divisor Proof.
Qcl
submodule
A
of
. Let ~a
has
Qm ' ~ a
is thus an injective submodule of
124
Qm ' so
Qm = ~ a ( Q m ) ~ K
~a(K)
~ K , using the fact that
generally~ Since
A
for each
n
one gets
is finite-dimensional
fore necessarily have so
a
. Iterating this, one gets ~a
is a monomorphism.
Qm = ~ ( Q m and
Qm = ~ ( Q m
)(~
More
) ~ @ P ~ ~I(K) ~ ' " ~
K .
Qm = E(A) , we must there-
K = (0) . ~ a
is thus also an epimorphism,
has a right inverse. To conclude the proof, we use:
Lemma 22.2.
Suppose
idempotents.
If
A
a s A
has no infinite family of orthogonal has a right inverse,
then
a
is
invertible. Proof. Suppose ei = Each
ei
ab = I
bia i
-
but
bi+lai+l
for
i = I, 2, ....
is different from zero, because
would give
bia i
=
ba = I
bia i = bi+lai+l
abi+lai+ib = ablalb = bi-la i-I , using
repeatedly the fact that to
ba # 1 . Define elements
ab = i , and this would finally lead
One verifies immediately that the
gonal idempotents,
Lemma 22.3.
If
A
e. i
are ortho-
contrary to our assumption.
satisfies ACC on right annihilators,
then
the singular right ideal is nilpotent. Proof. Ne will show that the ascending chain would be strictly ascending if Z n ~ 0 , choose an element
Z
a s Z
were not nilpotent. with
possible right annihilator. For each since that
r(b) ac¢
is essential in 0
but
r(Z) C r(Z 2) C
zn-la # 0
b s Z
bac = 0 , which means that
If
and largest
we have
A . So there exists
...
r ( b ) n aA = 0
c c A
r(ba)
such
is strictly
125
larger than have
r(a) , and by the choice of
zn-lba = 0 . Since
and hence
r(~: -I)
b G Z
a
is arbitrary, we get
is strictly oon%ained in
Proposition 22. 4.
Suppose
A
K
is right finite-dimensional!
2)
A
satisfies ACC on right annihilators!
3)
(A:x) Qcl
Zna = 0 ,
r(Z n) .
satisfies:
I)
Then
we must therefore
contains a non-zero-divisor for each
x c E(A) .
exists and is a semi-primary right self-injective
ring. Proof. In view of Proposition 22.1 it remains to show that the Jacobsor~radical know that that
J J
of
Qcl
is nilpotent. From Lemma 16.4 we
is the singular submodule of the right
A-module
Qcl ' and
for some
n , and from this it follows that also
fact, if
ql,..,qn c J , write
are in
A
and
bi
Z(A) = J ~ A
. Lemma 22.3 says that
qi = aibi -I
are non-zero-divisors.
We may write
ql....-qn
Z(A)
a non-zero-divisor,
and
b
condition. Hence
in the form
Zn = 0
jn = 0 . In
where Note that
ai
and
bi
a i c Z(A) .
t al.. .a~b -I , with
a~I ¢
by repeated use of the Ore
ql...qn = 0 .
By strengthening the condition (2) slightly we obtain a quasiFrobenius classical ring of quotients.
126
Propositign ' 22.~. The following two assertions are equivalent: (a)
A
satisfies
I)
A
is right finite-dimensional!
2)
A
satisfies ACC on right annihilators of subsets of
~(A) , 3)
(b)
(A:x)
Qcl
Proof.
contains
exists
a non-zero-aivisor
for each
x e E(A) .
aria i s a O~-ring.
(a)@(b):
Qcl
exists and coincides
with
~
by 2 2 . 1 .
~t i s QF by 19.7. (b)=~(a): This follows also from Propositions 19.7 and 22.1.
We may now characterize those rings for which
Qcl
is semi-
simple ("Goldie's theorem"). Theorem 22.6. (a)
Qcl
(b)
~
The following properties of
A
~re equivalent:
exists and is a semi-simple ring. is right finite-dimensional and right non-singular, and
has no nilpotent two-sided ideals @ 0 . (c)
A
is right finite-dimensional, satisfies A C C on right
annihilators, and has no nilpotent two-sided ideals # 0 . (d)
A
is right finite-dimensional and every essential right
ideal contains a non-zero-divisor. Proof. (a)@(d):
Since
Qcl = Qm
is semi-simple, we know from
Theorem 20.2 that for every essential right ideal Qcl = IQcl " Write ql = blCl Then
-I
where
I = ~aiq i
with
I
one has
al,...,a n c Y . Write
b I , cI ~ A and cI is a non-zero-divisor. n c I = alb I + ~ aiqic I . Write q2cl = b2c2 -I with b 2 , 2
127
02 ¢ A and 02 a non-zero-divisor. Then ClC 2 = alblC 2 + a2b 2 + n + ~ aiqiclc 2 . Continuing in this manner, we get non-zero-divisors
3 o l,...,c n
such that
c I .... .cn ¢ ~ a i A C I . Thus
I
contains
a non-zero-divisor. (d)~(a):
Qcl
22.1. Since
exists and is a right self-injective ring by
A
is right non-singular and right finite-dimensional,
it follows from Theorem 20.2 that (a)~(o).
Qc~ = Qm
From Proposition 22.5 follows that we only have to
show that every nilpotent two-sided ideal an essential right ideal, for if let
n
exists (a)~
is semi-simple.
a @ 0
be the smallest integer such that b ¢ I n-I
such that
ab @ 0
and
I
is zero.
l(I)
is any element of
is A ,
al n = 0 ! then there abc
i(I) . But since
(d), every essential right ideal contains a non-zero-
divisor, and therefore (c)~(b):
Z(A)
I
must be zero.
is a nilpotent two-sided ideal by Lemma 22.3.
By hypothesis it must be zero, so (b)=~(c):
Qm
A
is right non-singular.
is semi-simple by Theorem 20.2 and this implies
ACC on right annihilators by Proposition 19.7. (c)-~(d): We first prove: Lemma 22.7.
If
A
satisfies ACC on right annihilators and has
no nilpotent two-sided ideals @ 0 , then every right or left nilideal is zero. Proof.
Since
Aa
is nil if and only if
to consider a nilideal elements of
Aa . Assume
Aa , choose one
b c Aa
aA
is nil, it suffices
Aa # 0 . Among the non-zero with maximal right annihi-
128
lator. For each r(b) C
c c A , let
r((cb) k-l)
cb c r(b)
and
(cb) k = 0 , (cb) k-I @ 0 . Since
, we must have equality by maximality. Hence
bAb = O , which gives
nilpotent ideal is zero, we get diction,
and so
b = 0 . This is a contra-
Aa = 0 .
We return to the proof of ( c ) ~ ( d ) : right ideal. Since not a nilideal that
A
l~r(al) a2 @ 0
and
% 0 , we go on and get
akb = a l b l + . . . + S k _ l ~ _ l
! since for each
right finite-dimensional,
alA ~
and hence
c = al+...+a k C I , then
Lemma 22.8.
If
Proof.
cA
alA ~
such
r(a2)
~ and
... @ akA . This
... (~ ak_iA
is direct and
i ~ k we have aia k = 0 , k-I ~ a.b. = 0 . But A is i I z
= 0 . Then
-- 0 , so if
then
I
is
so the process must stop at some stage,
l~r(al)~...~r(ak)
A
in
a 3 ¢ Inr(al)N
suppose
b i c r(ai 2) = r(ai)
aI @ 0
I
r(a2) = r(a22 ) . Yf then
is proved by induction:
where we have
be an essential
@ 0 , we continue and choose
so on. At each step we obtain a direct sum
we get
I
(by the Lemma), there exists
such that
l~r(al) ~r(a2)
Let
has ACC' on right annihilators and
r(al) = r(al2 ) . If
a 2 c I n r ( a I)
(AbA) 2 = 0 . Since every
r(al)~...~r(a
k) =
r(c) = 0 . We need:
is right finite-dimensional and
r(c) = 0 ,
is an essential right ideal. If
I
is any right ideal and
easy to see that we get a direct sum
cA(~l = 0 , then it is I ( ~ cI ~ ... ~ cnI (~ . . . .
We may now conclude the proof of the Theorem.
Since
A
is right
129
non-singular,
it follows from the Lemma that
is thus a non-zero-divisor belonging to
1(o) = 0 , and
c
I .
Exercises Let
A
be the ring of matrioes
0 with
a , b e ~
b and
q c Q . Show that the ring of upper tri-
angular matrices over quotients of
Q
A , while
ring of quotients of
is a two-sided classical ring of M2(Q)
is the maximal right (and left)
A . (~ote that
A
has nilpotent two-
sided ideals ~ 0 ).
~sferenoes, Q~briel [31], Goldie [ ~ ] ,
Jans [391, ~ b o =
Winton !>41, P~ocesi and Small [ 6 4 , S~domier~i [68].
~d
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