Rings and Modules of Quotients

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

(
F_(S) = { right

s ~ S

with

as e ]~}

is easily verified to he a topology. 5. Let

C

be the class of semi-simple modules, which is closed ~

under submodules, quotients and direct sums. The corresponding pretopology consists of all finite intersections of maximal right ideals, and the corresponding left exact preradical to each module

M

its socle

s

associates

s(M) . The corresponding torsion

radical is as usual denoted by

~ . Note that the torsion-free

modules are those modules which have zero socle. Pr0position 3.1.!,. When A

is a right noetherian ring,

the sum of all artinian submodules of

~(M)

is

M .

Proof. A module is called artinian if it satisfies DCC (descending chain condition) on submodules. submodules of

M . t

Put

t(M) = sum of all artinian

is clearly a left exact preradical. It is

not quite so obvious that it is a radical, and here we need the noetherian hypothesis. We write

t(M/t(M)) = L/t(M)

to show that every finitely generated submodule

artinian.

We have

CLIt(M)

and h e n c e

,.'It(I, ')

-- I , ' / L ' n

L'It(L')

t(M)

is a r t i n i a n .

, and want

L'

of

= Z,' + t ( I ~ ) / t ( l ~ ) Since

A

T. is

C

is n o e t h e r i a n ,

39

also

t(L')

is finitely generated and hence artinian. Since the

class of artinian modules is closed under extensions, it follows that t

L'

is artinian.

thus defines a torsion theory, and this torsion theory is the

same as the one given by

~ , because the torsion-free modules

are clearly the same in both cases.

The ring

A

is called right semi-artinian if every non-zero

module has non-zero socle. In that case we have no torsion-free modules

# 0 ~ and thus every module is a torsion module, i.e.

~(M) = M

for all

M . In particular we see from 3.11 that a

right noetherian and semi-artinian ring is right artinian. Proposition 3.12.

A

is right semi-artinian if and only if

every hereditary torsion theory is generated by a class of simple modules. Proof. If

A

is right semi-artinian, then the torsion class

generated by all simple modules is

Mod-A . It follows from § 2,

Exercise 3 that any hereditary torsion theory is generated by a class of simple modules. The converse follows from the trivial fact that

Mod-A

is a

hereditary torsion class. Exercises: I. Show that if II~

F

is a topology and

I , J c F , then also

F .

B

2. Suppose the ring

A

is commutative and

such that every ideal in

F

F_. is a pretopology

contains a finitely generated

w

ideal in

F . Show that if the product of two ideals in

F

20

always is in

F , then

F

m

3. (i) Show that

A

is a topology.

(as a right

module if and only if

Z(A)

(ii) Show that the ring

A-module) is a Goldie torsion is an essential right ideal in

Z_/p2~

(where

p

A .

is a prime number)

is a Goldie torsion module over itself. 4. (i) Let

I

be a right ideal. Show that the family of right

ideals containing

I

two-sided ideal and

is a topology if and onl~ if

I

is a

12 = I .

(ii) Show that the following assertions are equivalent for a hereditary torsion class

T : m

(a)

T

is closed under direct products.

(b)

The corresponding topology is of the form for some right ideal

(C)

~ ={M~

References, Alin

Xar

da

,

MI = O ~

I .

for some ideal

Ill , Bourhaki

ishina and sko

{J~ J D I

I •

~o] (p. 157), Gabriel [Zl] (p. 411),

S

ov

[Sb], Roos

l).

§ 4. S~able torsion theories In this and the following sections of this chapter, we will consider some more particular aspects of torsion. The results of these sections are not necessary for the understanding of the following chapters. A hereditary torsion theory

(5' ~)

is called stable if

is closed under injective envelopes. Alternative characterizations of stable torsion may be given by using the concept of "complemented" (or "closed") submodules. A submodule

L

of

M

is

21

complemented if there exists a suhmodule L

K

of

M

such that

is a maximal submodule having the property that

In such a case, Lemma 4.1. (i.e.

L

L

K + L

M

L

is maximal such that

E(M) • E(K) • E(L)

K + L . Therefore

and by maximality of

L

M ).

which reduces to the identity and

it follows that L

E(L)~ M ~ K

L~K

K~L

= 0 . Then also

= 0 , for if

L' D L

essential submodule of in

L'

and hence

is "essentially closed" in

L

K

M .

of

M

such

is maximal with respect to

and

L'~K

- 0 , then

L' + K , which implies that

L+K L

is an is essential

L = L' .

It should be remarked here that imbedding in

= O ,

E(L)(~M = L .

By Zorn's lemma there exists a maximal submodule that

L = E(L)~IM

L ~ K - 0 . There is an

E(L)f~MDL

Suppose conversely that

M .

if and only if

has no essential extension within

isomorphism map .on

is an essential submodule of

is complemented in

Proof. Suppose

K f~ L = 0 .

E(L)

has in general no unique

E(M) .

Proposition 4.2.

The following properties of a hereditary torsion

theory are equivalents (a)

The theory is stable.

(b)

The torsion submodule every module

(o)

t(M)

is complemented in

M , for

M .

Every injective module is the direct sum of a torsion module and a torsion-free

module.

22

Proof. ( a ) @ E(t(M))C

(b): Since

t(E(M))

submodule of

implies ~ves

t(E(M))

, for if LNM

t(M)

is an essential

0 @ L C t(E(M)) , then

~ 0 , so

E(t(M)) = t(E(M))

t(M) = t ( Z ( M ) ) n M

L

is

L ~ t(M) ~ 0 . Therefore

. Left exactness of

= Z~t(~))nM

t

now

, and ~em~a ~ 4.1

(h).

(b) ~ ( O ) :

If

E

is an injective module, then

injective by Lemma 4.1, so (c) ~ ( a ) : T

is a torsiom module, we have

by stability. But

a torsion module and we must have

t(M)

If

M

t(E)

we must have

F

is torsion-free.

Let

r

~

is stable. Then

the unique maximal essential extension of

r(M)

~(M).

If

L~(M).

there exists ~(L)NK

r(M)

in

~(M)

is

M , for

is an essential submodule of L~r(M)=

left exactness. This implies r(M)

T D M ,

M .

Proof. We first show that

Suppose

Since

where

be a left exact preradical such that

the torsion theory associated to

Suppose

E(M) = T ( D F

F = 0 .

Proposition 4.3.

every module

is also

splits off.

is a torsion module, then

is a torsion module and

t(E)

0,

then

such that

= O . But then

Hence

K = 0

and

&,ives

Ln~(M) = ~ ( ~ )

~(L)

L

=L,

by

of

M . By 4.2

is maximal with respect to

r(K) = 0 so we must have

~(L) = L

0

0 = ~(L) = L .

is essential in some submodule K CL

r(L)=

KNr(M)

= O .

by the maximality condition. This

and so

~C~(M)

.

23

An i m p o r t a n t example of s t a b l e t o r s i o n is g i v e n byt Proposition 4.4. Proof. If

M

is a torsion module, consider the exact sequence

0 where also

The G o l d i e t o r s i o n t h e o r y i s s t a b l e .

N

E(K) --. E(.)/M

o

is a torsion module by definition of the

E(M)/M

Goldie theory. It follows that

Reference, Gabriel

E(M)

is a torsion module.

[~I] (P. 426).

§ 5. T o p o l o g i e s f o r a commutative n o e t h e r i a n r i n ~ Our purpose i s t o d e t e r m i n e a l l n o e t h e r i a n r i n g . The f i r s t

t o p o l o g i e s f o r a commutative

s t e p i n t h a t d i r e c t i o n i s t a k e n by

d e t e r m i n i n g a l l i n J e o t i v e modules. Proposition ~.1. If

A

is a right noetherian ring,

then every

d i r e c t l i m i t o f i n j e o t i v e modules i s i n J e c t i v e . Proof. Let let

(Ea)

fzl-~ l~m E~

Since

I

that

f(I)

be a d i r e c t f a m i l y o f i n j e c t i v e modules and be a homomorphimm from a right ideal

is finitely generated, there exists an index lies in the image of

finitely generated submodule The kernel Since K

K

K

of

f(I)

goes to zero in

goes to zero already in

inJeotive module to

L

A

L

E of

in E~

I . 8

such

lim..~E= . Choose a which maps onto

f(I) .

is then also finitely generated. lira E= , there exists

~~ ~

E~ . We may then factor

E~ , and it follows that

f

f

such that over the

may be extended

24

Diagram for 5. I:

I

K

~

L

~

r E E

Coro!!ar~ }.2.

If

A

f(I) /

'

/ i

I ~: E

C.@

P

lira E -"*

"

Y

is right noetherian,

then every direct sum

of injective modules is inJective.

Prcpositicn ~. 3.

If

A

is right ncetherian,

then every injective

module is a direct sum cf indecomposable injective modules. Proof. Let

E

be an inJective mcdule. Consider all families

(Ei)

of indecomposable injective submodules such that the su~

~E i

in

E

is direct. By Zorn,s lemma there exists a maximal

such family

(Ei) I . From 5.2 we know that the sum

injective module,

(D E i is an I E = (v~Ei) (D E' . We want to

sc we can write

~M

show that

E' = 0 , and fcr this it suffices tc show that every

injeotive module Let

x

E' $ 0

be any element

modules

E"

of

E

contains an indecomposable direct summand. • 0

in

such that

E' . Consider all injective subx ~ E" . By Zcrn's lemma and 5.1

there exists a maximal such submodule and we assert that with

L'

and

of the modules

L

is indecomposable.

L" $ 0 , then E"(~L'

E" . We have

or

For if

(E"~L')n (E"eL") E"~L"

contradicts the maximality of

E" .

E' = E" ~ L ,

L = L' (~ L" = ~'

does nct contain

sc one x , which

25

It is possible to prove the converse statement, namely that if every inJeotive module is a direct sum of indecomposable modules, then

A

must be right noetherian (see e.g. Faith-Walker

Lemma ~.4.

An injective module

E = E(A/I)

where

I

E

[30])? )

is indeoomposable if and only

is an irreducible right ideal.

Proof.Easy exercise.

For the res% of this § we assume noetherian rin 6. We write of

A . For each module

A

Spec(A) M

for the family of prime ideals

we define

Ass(M) - { 2 ~ Speo(A) I ~ = A ~ ( x ) Lemma ~.~.

If

M % 0 , then

Proof. The family of ideals member

Ann(x)

since

A

for some x ,

Ann(y)

for

b ~ Ann(x) . Then

and since

and

Ann(x) = Ann(bx) . NOW

M~.

y , 0

has a maximal

with

abe

Ann(bx)~

a e Ann(bx)

Ann(x)

Ann(x) but

Ann(x) , we mus~

implies

a e Ann(x) ,

is p r i m e .

Ann(x)

Proposition ~.6. ideals of

in

is noetherian. We assert that a, b ¢ A

have

o

Ass(M) , ~ .

is a prime ideal. Suppose bx , O

t o be a commutative

A

There is a i-I correspondence between prime

and isomorphism classes of indeoomposable inJeotive

A-modules, g i v e n by

2 ~ E(A/2) •

~)Our "proof' of this result in a paper in Arkiv fSr Matematik, vol. 7 (Theorem 2 on p. 428) is insufficient! one has to use some kind of "counting' argument, as in

[~O].

26

Proof. If E(A/~) such

~

is a prime ideal, then

~

is irreducible and

is indecomposable by 5.4. If

~

and

E(A/~) ~ E(A/~)

that

we consider both

is indecomposable. But

x ¢ A/~

are prime ideals A/~

and likewise

Ann(x) = ~

Ann(y) = ~

A/~

and

E(A/2 ) , and we then have A / 2 n A / a ~

submodules of E(A/~)

,

E

0

as

since

for every non-zero

for non-zero

y e A/~ , so

it follows that

~ = E .

Finally, let

E

be an indecomposable injective module. By 5.5

there exists

~ e Ass(E) , and since

must

Let

have

Z

E

=

E

is indecomposable we

E(A/~) .

be a topology on

A , and let

(~, ~)

be the oorreB-

ponding torsion theory. Proposition ~.7. the form

A/~

(~, ~)

where

is generated by the cyclic modules of

~ e Spec(A)~

.

Proof. We must show that if a module Hom (A/~, M) = 0

M

for all prime ideals

has the property that ~

in

~ , then

torsion-free. We decompose the injeotive envelope of E(M) = ~ E ( A ~ )

. For each stsnmand we have

for each non-zero

x e A/~

we have

is prime. By the hypothesis on hence each

A/~

M

M

~ #~

, and is

is torsion-free.

a sequence of submodules

M = Mn~Mn_

such that

where

M

as

since

E(M) = e Z ( A / £ )

Lemma 5.8. For each finitely generated module

M i/Mi_ 1 ~= A/~ i

M

is

@ 0 , and

Ann(x) = ~

this implies

is torsion-free. Then

torsion-free by heredity, and so

A/~M

M

M

there exists

1 D ..... ~ M 0 = (0)

~i e Spec(A) .

is a torsion module if and only if each

~i e ~ .

27

Proof. Suppose we have constructed Mi_ I 4 M , we have H iC

M

M o , ..... , Mi_ I . If

Ass(M/Mi_l) % ~

such that

Mi/Mi_ I =" A / ~

by 5.5, so there exists

. Since

M

is noetherian, the

chain will eventually stop. It is clear that module if and only if each The0rem ' ~.~.

A family

F

Mi/Mi_ I

is a torsion

is a torsion module.

of ideals of

and only if there exists

M

P_ C Spec(A)

A

is a topology if

such that

_• . { I ! I~ 20"..-.'20 where ~ c P}. Proof. Suppose

F

is a topology. If

5.8 that there exist ~o" "'" "~n if

I

I c F , it follows from

~o ''''' ~n c S p e o ( A ) N F

annihilates

A/I , i.e.

such that

20-....-20 C I . Conversely,

contains a product of ideals in

Spec(A)N_F , then

I ¢ F

by § 3, Exercise I. On t h e o t h e r hand,

I ~ 20......20

if

with

PC ~

Speo(A)

i s g i v e n and

c _P } , then

F = { I

I

_r i s c l e a r l y a p r ~

t o p o l o g y and i s i n f a o t a t o p o l o g y by § 3, E x e r c i s e 2. LmRa Ann(x)

p.lO.

If

P 9 Spec(A)

is an irreducible

Proof. Since

E(A/~)

envelope of

A/Ann(x)

and

x ¢ E(A/~) , x # 0 , then

l~-primary ideal.

is indeooaposahle,

it is an injeotive

. Lemma 5.4 gives that

Ann(x)

is an

irreducible ideal, hence primary (~g~], p. 209). Let the prime ideal associated to integer s ~ h

that

as submodules of there exists

Ann(x) . Let

~¢A~(x).

Considering

E(A/2 ) , we have

a e E n'l

such that

n

in

be

be the smallest AI2

and

A/pN~n-I/Ann(x) E % 0

~

AIA~(x)

~ 0 . Hence

A/Ann(x)

and

28

Ann(Z) = 2 • We have

bE = 0 , then

other hand, if Ann(x)

is

~aCAnn(x)

~-primary. Hence

Corollary ~.iI. Proof. Since

If

Ann(~) D ~ • But on the

ba e Ann(x)

is

and

b ~ ~

since

~ = Ann(F) = ~ .

x c E(A/~) , then

Ann(x)

Propositiqn 5.12.

, so

~-primary,

j~nx = 0

n

for some

C Ann(x)

n>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

<

q ,~

:

Mod-A F

C

where in ~artio~ar

q(~) . ME

ana

a natural transformation G : ~ ~-~ q defined as 9 M ( x @ q )

ool~utes.

~*(M) = M ® A ~

• ~here i~

where O M : M @ A F-~ NF

= ~M(x)q . The diagram

is

Y3

Theorem 13.I.

The following properties of

F

are equivalent:

(a)

The functor

j:~-~Mod-A F

is an equivalence.

(b)

The functor

i:~-*Mod-A

has a right adjoint.

(c)

The functor

i:~-~Mod-A

is exact and commutes with direct

sums.

(d)

~

contains a cofinal family of finitely generated right

ideals and

i

is exact.

(e)

@ :~-~q

is a natural equivalence of functors.

(f)

KerCM~M®A ~ 3 = t(~)

for a l l

(~)

,I,(I)~=~

Le~.

for every

Proof. (a) =~(b): hypothesis~ and ~

i = ~4 j , where

MA .

j

has a right adjoint by

has the right adjoint

(b) ~ ( c ) :

Clear, since

(c)~(d):

Follows from Proposition 12.2.

(c) =~(e): We have

i

HOmA(AF,. ) .

always is left exact.

~a q = ia , which by hypothesis preserves

arbitrary colimits. Since the forgetful functor serves colimits, it follows that

q

T~ also pre-

preserves colimits. A well-

known argumen't ([I021, p.157) then shows that (~ equivalence between (e) ~

" ~A

(f): K e r ( M - * M O A F )

AF

and

is a natural

q .

-- Ker(M-~MF)

= t(M) .

u

(f) ~

(g): If

the map

I e F , then

A/I

is a torsion module and hence

A/I--~A/IQA F = A F / ~ ( 1 ) A F

is zero. This implies

~(~), (g) =~ (a): We must show that every right as an

A-module. M

for some

AF-module M

is torsion-free, for if

I c F , then

x~

= xlA F = 0

x e M

and hence

and

is

F_-closed

xl = 0

x = 0 .

74

We next verify that homomorphism with i =~T(ai)qi

M

is

F_-injective.

I c ~ . Write

with

qi c AF

Let

I c AF

and

f:I--FM

be a

in the form

a i c I . Then

J =

m

=

N ( ~ ( A ) : q i) ~ ~ and I ~ J c ~ . Since M is torsion-free, i we may factor f over ~ : ~ ( I ) - , M . For every a 6 I~ J we get

f(a) = ~ ( ~ (a)) = ~(~ ?(ai)qia ) = ~ ~ ( ~ (ai))qia =

= ~ f(ai)qia . Thus the restriction of extension to have

h:A-~M

h~l = f

f

to

I~ J

has an

, and as usual we must then necessarily

(cf. proof of Proposition 7.8).

Definition. A topology is called ~erfect if it has the properties listed in the Theorem. Co rollar ~ !3.2. flat as a left

If

~

is a perfect topology,

then

~

is

A-module.

Proof. The functor

• ~A F = q

is left exact.

The main drawback of Theorem 13.1 is that none of the conditions (a) - (g) are internal, i.e. they do not give conditions for to be perfect solely in terms of of

F

and the ideal structure

A . So e.g. when one is applying (d), it is usually difficult

to tell from

F

m

whether

i

is exact. In one special case this

is however possible: Pr0~ositi0 ~ 13.3.

If

A

for all topologies on Proof. Exactness of between Since

is right hereditary,

is

i

is exact

A . i

means that if

F-closed modules~ Ker f

then

~-closed,

then also

f:L--~M Coker f

is a homomorphism is

F-closed.

it clearly suffices to consider

75

monomorphisms

f:L-~M

. M/L

is then torsion-free,

as one sees

from the exact sequence HomCAII,M)---~ Let

g:I-~M/L

Hom(AII,MIL)

~ ExtI(A/I,L)

be any homomorphlsm with

a projective module, we may lift ~-closed. we may extend A-~M-~M/L

h

then extends

Corollar[ 13.4. If

A

to

g

to

A-~M

g , and

.

I c [ . Since h:l-@M

I

. Since

is M

is

. The composition

M/L

is

F_-injective.

is a right noetherian hereditary ring,

then every topology is perfect.

There is an interesting abstract characterization of those ring hcmomorphisms which are of the form topology

A--~A F

for a perfect

~ , namely they are precisely the "flat epimorphisms"

in the category of rings. As a preparation for this result, we are going to review some facts about epimorphisms of rings. Let ~ : A - - ~ B

be a ring homomcrphism.

Recall that

$

is an

e~imorphism (in the category of rings) if for any ring homomorphisms

a , ~:B--~C , a S = ~

generally we say that always implies dominated by ~

b ¢ B

implies

if

~(b) = ~(b) . The set of elements of

~? = ~ B

B , called the dominion of ~ .

is an epimorphism if and only if its dominion equals Proposition 13.~. ~ : A - , B h®l

= l®b

in

Proof. Suppose

dominates

b ~ B

morphisms such that

= l~b aS = ~.

and let

B .

if and only if

B® A B . b@l

and

a = ~ . More

is dominated by $

is a subring of

C

a ,~IB-~C

be homo-

Define a homomorphism of

76

A-A-bimodules

= l@b

T:B~) A B-~C

implies

as

a(b) = W ( b ® l )

Suppose o n the other hand that

T(beb')

= a(b)~(b')

=r(l®b)

= ~(b) .

@~

dominates

Let

B-B-bimodule. for all

~ :A--)B

If

x c M

Proof. We make

BX M

x = I@I

be a ring homomorphism and has the property that

a c A , then

bx = xb

for all

b

=

b . The assertion

now follows by applying the following lemma to

Lemma 13.6.

. b@l

c B@ A B .

M

a

T(a)x = x T(a)

in the dominion of ~ .

into a ring by defining

(b,y) + (b,,y,) = (b + b',y + y~) (b,y)

• (b',y') = (bb',by' + yb')

The ring axioms are easily verified, an identity, namely as

=(b) = (b,O)

(I,0)

and

ring homomorphisms, , then

in particular

. Define two maps

~(b) = (b,bx-xb)

and

a(b) = o(b)

.

a~ = ~ and thus

. So if

has

a ~ ~ :B-~B~

. Both b

BK M

a

and

~

M are

is dominated by

bx = xb .

If one is interested in dominions in the category of commutative rings, one may prove an analogue of

Proposition 13.5. The proof

in the commutative case has to be done separately (for Lemma 13.6 would introduce non-commutative rings), but is easy since B~ A B

is a ring

whenever

A

is commutative.

We will elaborate on the consequences of Proposition 13.5. Let Mod-A be the funotors a left adjoint of

4

~*(M) = M ~

.

Mod-B A ~ ' ~,(N) = N . Recall that ~% is

77

Proposition 13.7~ morphism

The following properties of a ring homo-

~:A--~B

are equivalent:

(a)

~

is an epimorphism.

(b)

The canonical map

(c)

The adjunction transformation

B@ A B-*B

is bijectiva. ~mT~-@id.

is a natural

equivalence of functors. (d)

The fm~ctor

~:Mod-B-@Mod-A

Proof. ( a ) ~ (d), Suppose =:M--~N

is

M

and

A-linear. For each

~ : B ® A B--~N

is full. N

x e M

a(xb) = a(x)b , and therefore

given by The map

B-modules and

oonsider the map

@~iven by~ ~(b@b,) = a(xb)b, . Note that this

really is a welldefined map. Since

(d)~(c):

ar~

a

l®b is

= b®l

, we have

B-linear.

We have to show that the homomorphism ~ : M O A B - @ M x@b~xb

, is an isomorphism for every

M--,MQ A B

given by

and by hypothesis therefore

x~xQl

B-module

is clearly

M .

A-linear,

B-linear. It is the desired inverse

of ~ . (c)~(b)

is clear, while ( b ) ~ ( a )

follows from Proposition 13.5.

An immediate consequence of (d) is: Corollar~ 13.8. right

Let ~ :A--~B

B-module such that

MA

be an epimorphism. If is injective, then

MB

M

is a is

injective. For the proof of the next theorem we need the following result:

78

Lemma 13. 9 .

Let there be given modules

LA

and

A M . The

following assertions a~e equivalent: (a)

xA@M

= 0

(b)

For every

for every y e M

x s L .

and

X1,...,X"

e L , there exist m

yl,...,y m e M

and

al,...,a m e A

such that

y =5

aiY i I

and (c)

x.a. = 0 i S

For every and

for all

y s M

and

j .

and

a.,..i"'am s A

for all

i

x e L , there exist yl,...,ym e M m such that y = ~ a_v_ivl and xa. = 0 I J

j .

Proof. The equivalence of (a) and (c) is well-known § 2, Lemma I0). (a) implies that C

of

C®M

= 0

for every submodule

L , and one easily proves that this implies

for every submodule

C

of a direct sum

([i0], oh. I,

Ln

C~M

= 0

of copies of

(b) is then obtained from (a) by considering (Xl,..,xn) element of

Ln

Theorem 13.10.

L . as an

.

Let

~ :A-~ B

be a ring homomorphism.

The

following assertions are equivalent: (a)

~

is an epimorphism and makes

(b)

The family

F

of right ideals

B

into a flat left

I

of

A

such that

is a topology, and there exists a ring isomorphism such that (c)

G~=

~A

A-module. ~(1)B

= B

~ :B--*A F

"

The following two conditions are satisfied: (i)

for every

b s B

bl,...,b n e B (ii) if $ ( a )

there exist

such that

Sl,...,s n c A

b~(si)

= 0 , then there exist

bl,...,b n e B

such that

as i = 0

e~(A)

and Z $ ( s i ) b i = 1 !

Sl...,s n e A and

and

~(si)b

and i = i .

79

Proof.

(a) :=~ (b): The f o r g e t f u l

makes

Mod-B

functor

~:~t:Mod-B--,Mod-A

equivalent to a full suboategory of

Proposition 13.7(d).

~

is exact, since

is flat. We may thus consider

AB

a Giraud subcategory of

has a left adjoint

Mod-A

~*-- " ~ A

I

such that

bl C ~ ( A )

Mod-B

as

~

of right

AF = B .

b s B = A F , there exists

I c ~

. By Theorem 13.1 we may assume

say generated by

which

~m(1) ~ ~4(A) , i.e. such that ~o (1)B = B ,

~s a t o p o l o g y and that (b),~ (c): If

B

Mod-A . From Theorem 10.2, combined

with Theorem 13.1, we may conclude that the family ideals

by

Sl,..,s n . Then

I

such that

finitely generated,

I =~(si)b

i

for some

b i s B , and (i) is verified. If ~ ( a ) such that

= 0 , then

i =~

generated by

(si)b i . Thus also (ii) is verified.

If

~ :A-@B

condition (i) of (c), then of

I

I c

First of all we note:

Lemma 13.11.

C

, so there exists

al = 0 . As before one may assume

Sl, .... ,sn , and (c)~(a):

a S Ker(A-~AF)

is a ring homomorphism satisfying C~ A B = 0

for every submodule

B/~O(A)

Proof. (i) implies that 13.9(c) is satisfied. We now prove (c)~(a). Since we have just seen that B/~(A)~

A B = 0 , we o b t a i n

B ~ A® A B

) ~(A)®A

The canonical map

B@ A B-~B

epimorphisms

B

, B(Z)A B

.

must then be a bijection, and

80

thi;s shows that

~

is an epimorphism (Proposition 13.7).

It remains to see

AB

is flat. For this we use Proposition 1 9

of [101, oh. I, § 2. Suppose we have bl,...,hn ¢ B tO

1 ~ B

and

such that

and

~(si)b

and

i = 0 . Applying Lemma 13.9(b)

~l,...,~n c B / ~ ( A )

h~,..,h'm c B

Sl,...,s n c A

, we obtain

al,..,am • A

such that

a

h i ~ ( % ) =~(oiS) hen

for some cij c A

Z (sici ) = Z?(si)bi

8 Ker T

for each

obtain for each

(aj> - 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~<M,~> "=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

REFERENCES: 1.

J . S . A l i n , S t r u c t u r e o f t o r s i o n modules, Ph.D. T h e s i s , U. of Nebraska 1967.

2.

J.S. Alin and S.E. Diokson, Goldie's torsion theory and its derived functor, Pacific J. Math. 24, 1968, 195-203.

3.

G. Almkvist, Fractional categories, Arkiv f. Mat. 7, 1968, 449-476.

4.

S.A. Amitsur, General theory of radicals II, American J. Math. 76, 1954, 100-125.

5-

K. Asano, Uber die Quotientenbildung yon Sohiefringen, J. Math. Soo. Japan I, 1949, 73-78.

6.

G. Azumaya, Completely faithful modules and self-injeotive rings, Nagoya Math. J. 27, 1966, 697-708.

7.

H. Bass, Finitistio dimension and a homologioal generalization of semi-primary rings, Trans. Amer. Math. Soo. 95, 1960,

466-488. 8.

J.-E. Bj~rk, Rings satisfying certain chain conditions, to appes~

9.

N. Bourbaki, Algbbre, oh. 8, Hermann 1958.

I0.

- , -

, Alg6bre commutative, oh. I and 2, Hermann 1961.

II. B. Brainerd and J. Lambek, On the ring of quotients of a Boolean ring, Canad. Math. Bull. 2, 1959, 25-29. 12. I. Buour and A. Deleanu, Introduction to the theory of oategorie~ and functors, J. Wiley 1968. 13. H. Cartan and $. Eilenberg, Homological algebra, Princeton 1956. 14. V.C. Cateforis, Flat regular quotient rings, Trans. Amer. Math. Soc. 138, 1969, 241-250.

131

15. V.C. Cateforis, On regular self-inJeotive rings, Pacific J.

Math. 30, 1969, 39-45. 16.

- , -

, Two-sided semisimple maximal quotient rings~

Trans. Amer. Math. Soo. 149, 1970, 339-349. 17. S.U. Chase, Direct products of modules, Trans. Amer. Math. Soo.

97, 1960, 457-473.

18. P.M. Cohn, Universal a l g e b r a , Harper and Row 1965. 19. R.S. C~ningham, E.A. R u t t e r and D.R. Ttu~nidge, Rings o f q u o t i e n t s o f endomorphism r i n g s o f p r o j e c t i v e modules, to appear. ~O. C.W. C u r t i s and I . Reiner, R e p r e s e n t a t i o n t h e o r y o f f i n i t e groups and a s s o c i a t i v e a l g e b r a s , J . Wiley 1962. 21. S.E. Diokson, A t o r s i o n t h e o r y f o r a b e l i a n c a t e g o r i e s , Trans. Amer. Math. Soo. 121, 1966, 223-235. 22. M. D J a b a l i , Annenu de f r a c t i o n s d,un J-anneau, Canad. J . Math. 17, 1965, 1041-1052. 23. S. Eilenberg and T. Nakayama, O~ the dimension of modules and algebras II, Nagoya Math. J. 9, 1955, 1-16. 24. V.P. E1izarov, On quotient rings of associative rings, Izv. Akad. Nauk SSSR 24, 1960, 153-170. (Aaer. Math. Soo. Transl. 52, 1966).

2~5,

- . -

, Flat extensions of rings, Soviet Math. Dokl. 8,

1967, 905-907. 26. H. Eriksson, Fraktionkategorier (mimeographed),

och Grothendieoktopologier

Stockholm 1967.

27. C. Faith, Rings with ascending condition on annihilators, Nagoya Math. J. 27, 1966, 179-191. 28.

- " - t Lectures on inJeotive modules and quotient rings, Springer Lecture Notes 49, 1967.

30. C. Faith and E.A. Walker, Direot-sm, representations

of inJectiw

modules, J. Algebra 5, 1967, 203-221. 31. P. Gabriel, Des oat6gories ab@liennes, ~ 1 .

Soc. Math. France

90, 1962, 323-448. 32. A.W. Goldie, Some aspects of ring theory, Bull. London Math. Sos. 1, 1969, 129-154.

132

33. O. Goldman, Rings and modules o f q u o t i e n t s , J. Algebra 13,

1969, 10-47. 34. R.N. Gupta, Self-inJeotive quotient rings and inJectlve quotient modules, Osaka J. Math. 5, 1968, 69-87. 35. M. Hacque, Loealisations exaotes et localisations plates, Pub1. D6p. Math. Lyon 6, 1969, 97-117. 36. G. Helzer, On divisibility and inJectivit¥,

Canad. J .

Math.

18, 1966, 901-919. 37. M. Iked~ and T. Nakayama, On some characteristic properties of quasi-FTobenius and regular rings, Prec. Amer. Math. See. 5, 1954, 15-18. 38. J.P. Jans, Some aspects of torsion, Pacific J. Math. 15, 1965,

1249-1259. 39.

- ,, --

~ On. o r d e r s

in quasi-Frobenius r i n g s ,

J. Algebra 7~

1967, 35-43. 40. R.E. Johnson, Structure theory of faithful rings, I and II, Trans. Amer. Math. Soc. 84, 1957, 508-544. 41. A.I.

Ka~u,

Closed

classes

of left

A-modules

and closed

sets

of left ideals of the ring A, Mat. Zametki 5, 1969,

381-39o. 42. T. Kate, Self-injective rings, Tohoku Math. J. 19, 1967, 485--495. 43.

- , - , Torsionless modules, Tohoku Math. J. 20, 1968, 234-243.

44. A.G. Kurosoh, Radicals in rings and algebras, Mat. Sb. 33, 1953, 13-26. 45. J. Lambek, On the structure of semi-prime rings and their rings of quotients, 46.

-,-

Canad, J. Math. 13, 1961, 392-417.

, On Utuli's ring of quotients,

Canad. J. Math. 15,

1963, 363-370. 47.

- . -

, Lectures on rings and modules, Blaisdell 1966.

48.

- . -

, Torsion theories, additive semantics and rings of quotients,

49.

- , -

Springer Lecture Notes 177, 1971.

, Bioommutators of nice inJectives,

to appear.

50. L. Levy, Torsion-free and divisible modules over non-integral domains, Canad. J. Math. 15, 1963, 132-151.

133

51. J.-M. Ma~anda,

InJeotive structures,

Trans. Amer. Math. Soo.

llO, 1964, 98-135. 52. E. Matlis, InJeotive modules over noetherian rings, Pacific J. Math. 8, 1958, 511-528. 53.

- , -

, Modules with descending chain condition, Trans. Amer. Math. Soo. 97, 1960, 495-508.

54. A.C. Mewborn and C.N. Winton, Orders in self-inJective semiperfect rings, J. Algebra 13, 1969, 5-9. 55. A.P. Mishina and L.A. SkornJakov, Abelian groups and modules (in R u s s i a n ) , Moscow 1969. 56. K. Morita, On S-rings, Nagoya Math. J. 27, 1966, 687-695. 57.

- ,, -

, Localization in oategories of modules I, Math. Z. 114, 1970, 121-144.

58.

- " -

, Localization in categories of modules II, J. Reine Angew. Math. 242, 1970, 163-169.

59. C. N~st~sescu and N. Popescu, On the localization ring of a ring, J. Algebra 15, 1970, 41-56. 60. B.L. Osofsky, A generalization of quasi-Frobenius rings, J. Algebra 4, 1966, 373-387. 61.

- , -

, Endomorphism rings of quasi-injective modules,

Canad. J . Math. 20, 1968, 859-903. 62. N. Popescu and P. Gabriel, Caraot6risation des oat6gories ab@lienmes avee g@n@rateurs et limites inductives exactes, C.R. Aoad. Soi. Paris 258, 1964, 4188-4190. 63. N. Popescu and T. Spircu, Sur les @pimorphismes plats d, anneaux, C.R. Aoad. Soi. Paris 268, 1969, 376-379. 64. C. Procesi and L. Small, On a theorem of Goldie, J. Algebra 2, 1965, 80-84. 65. J.-E. Roos, Sur l,anneau maximal de fractions des AW~-alg~bres et des anneaux de Baer, C.R. Acad. Sci. Paris 266, 1968,

449-452. 66.

- ,, -

, On the structure of abelian categories with generators and exact direct limits. Applications,

to appear.

134

67. S@minaire P. Samuel, Les @pimorphismes d, anneaux, Paris 1968. 68. F.L. Sandomierski,

Semisimple maximal quotient rings, Trans.

Amer. Math. Soc. 128, 1967, 112-120. 69.

-., -

, Non-singular rings, Proo. Amer. Math. Soc.

19, 1968~ 225-230. 70. L. Silver, Non-commutative

localization and applications,

J. Algebra 7, 1967, 44-76. 71. L.A. Skornjakov, Elizarov's quotient ring and the localization principle, Mat. Zametki I, 1967, 263-268. 72. B. Stenstram, On the completion of modules in an additive topology, 73.

- ,, -

J.Algebra 16, 1970, 523-540.

, Flatness and localization over monoids, to

appear in Math. Naohrichten. 74. H. Tachikawa, Double centralizers and dominant dimensions, Math. Z. 116, 1970, 79-88. 75. M.L. Teply, Torsion-free injective modules, Pacific J. Math. 28, 1969, 441-453. 76.

- , -

, Some aspects of Goldie's torsion theory, Pacific

J. Math. 29, 1969, 447-459.

77. M.L. Teply and J.D. Fuelberth, The torsion submodule splits off, to appear. 78. D.R. Turnidge, Torsion theories and semihereditary rings, Prom. Amer. Math. Soo. 24, 1970~ 137-143. 79.

- , -

, Torsion theories and rings of quotients of

Morita equivelent rings, to appear. 80. y. Utuli, On quotient rings, Osaka Math. J. 8, 1956, 1-18. 81. C.L. Walker and E.A. Walker, Quotient categories and rings of quotients, mimeographed,

1963.

82. L.E.T. Wut H.Y. Moohizuki and J.P. Jans, A characterization of QF-3 rings, Nagoya Math. J. 27~ 1966, 7-13. 83. O. Zariski and P. Samuel, Commutative algebra I, van Nostrand

1958.

I35

Additional referen~esz 84. T. Akiba, Remarks on generalized rings of quotients,

III,

J. of Math. Kyoto 9, 1969, 205-212. 85. E.P. Armendariz,

On finite-dimensional

torsion-free modules

and rings, Proc. Amer. Math. Soc. 24, 1970, 566-571. 86. J.E. BJ~rk, Rings satisfying a minimum condition on principal ideals, J. Reine Angew. Math. 236, 1969, 112-119. 87. A. Cailleau and G. Renault, Sur l,enveloppe inJective des anneaux semi-premiers & l,id6al singulier nul, J. Algebra 15, 1970, 133-141. 88. V.P. Elizarov, Rings of quotients, Algebra i Logika Seminar 8, 1969, 381-424. 89. P. Gabriel and M. Zisman, Calculus of fractions and homotopy theory, Springer 1967. 90. A.W. Goldie, The structure of prime rings under ascending chain conditions,

Proo. London Math. Soc. 8, 1958,

589-608.

91.

- - -

, Semi-prime rings with maximum condition, Proo. London Math. Soc. 10, 1960, 201-220.

92. R.N. Gupta, Self-inJective quotient rings and injective quotient modules, Osaka J. Math. 5, 1968, 69-87. 93. N.

Jaoobson, Structure of rings, Amer. Math. Soc. Coll. Pub1. 37, revised ed. 1964.

94. R.E. Johnson, Prime rings, Duke Math. J. 18, 1951, 799-809. 95.

- , -

, Extended centralizer of a ring over a module, Proc. Amer. Math. Soc. 2, 1951, 891-895.

96.

- , -

, Quotient rings of rings with zero singular ideal, Pacific Math. J. II, 1961, 1385-1392.

97.

- ,, -

, Rings with zero right and left singular ideals, Trans. Amer. Math. Soo. 118, 1965, 150-157.

98. R.E. Johnson and E.T. Wong, Quasi-injective modules and irreducible rings, I. London Math. Soo. 36, 1961, 260-268.

136

99.

J.T. Knight, On epimorphisms of non-commutative rings, Proc. Cambr. Phil. Soc. 68, 1970, 589-601.

I00. D. Lazard, Autour de la platitude, Bull. Soo. Math. France 97, 1969, 81-128. I01. A.C. Mewborn, Some conditions on commutative semiprime rings, J. Algebra 13, 1969, 422-431. 102. B. Mitchell, Theory of categories, Academic Press 1965. 103. K. Morita, Localization in categories of modules IIY, Math. Z. 119, 1971, 313-320. 104. N. Popescu and T. Spircu, Quelques observations sur les @pimorphismes plats (8 gauche) d'anneaux, J. Algebra 16, 1970, 40-59. 105. N. Popescu and D. Spulber, Sur les quasi-ordres ('~ gauche) dans un anneau, .J. Algebra 17, 1971, 474-481. 106. G. Renault, Anneaux r@duits non commutatifs, J. Math. Pures Appl. 46, 1967, 203-214. 107.

- ,, -

, Anneau associ@ ~ un module injectif, Bull. Sci.

Math. 92, 1968, 53-58. 108. J.C. Robson, Artinian quotient rings, Proc. London Math. Soc. 17, 1967, 600-616. 109. J.E. Roos, Locally distributive spectral categories and strongly regular rings, Reports Midwest Category Seminar, Springer Lecture Notes 47, 1967, 156-181. II0. L. Small, Orders in artinian rings, J. Algebra 4, 1966, 13-41 and 505-507. Iii. H. Tachikawa, Localization and artinian quotient rin~s, Math. Z. 119, 1971, 239-253. 112. M. Takeuchi, A simple proof of Gabriel and Popescu's theorem, J. Algebra 18, 1971, 112-113. 113. Y. Ut~i, On rings of which any one-sided quotient rings are two-sided, Proc. Amer. Math. Soc. 14, 1963, 141-147. 114. E.T. Wong and R.E. Johnson, Self-injective rings, Canad. Math. Bull. 2, 1951, 167-173.