Title: Decomposition of torsion theories
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Title: Decomposition of torsion theories
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Creator: Bronowitz, Richard, 1946-
Publication Date: 1972
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DECOMPOSITION OF TORSION THEORIES


By

RICHARD BRONOWITZ










A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY



UNIVERSITY OF FLORIDA


1972





















ACKNOWLEDGEMENTS


No words can adequately express the author's thanks to

his wife, Carol, and his parents. Without their faith and

encouragement this dissertation would not have been written.

The author would like to express his sincere appreciation to

his supervisory committee. Special recognition is due to his

dissertation director, Professor Mark L. Teply, whose patience,

suggestions and constant interest made this work possible.

The author would also like to thank Professor Jonathan S. Golan

for his interest and advice during the preparation of the dis-

sertation. Finally, the author would like to acknowledge an

excellent job of typing the manuscript by Jean Sheffield.
























TABLE OF CONTENTS


Page


Acknowledgments

Abstract

Chapter


1.

2.

3.

4.

5.

Bibliography


Introduction

Preliminaries

The Lattice of Torsion Classes

Decomposition of Torsion Classes

Completely Reducible Torsion Classes








Abstract of Dissertation Presented to the Graduate
Council of the University of Florida in Partial Fulfillment
of the Requirements for the Degree of Doctor of Philosophy

DECOMPOSITION OF TORSION THEORIES

By

Richard Bronowitz

August, 1972

Chairman: Mark L. Teply
Major Department: Department of Mathematics

The purpose of this paper is to extend the concept of

Primary Decomposition of modules in the simple torsion class

to modules in an arbitrary torsion class. The main theorem

gives necessary and sufficient conditions on a certain sub-

class of cyclic torsion modules for the torsion class to ad-

mit a decomposition. In order to obtain this, results con-

cerning torsion preradicals and the lattice of torsion classes

are proved. Results of Dickson, Alin and Albu are obtained

as corollaries to the main theorem. Finally, those rings,

for which all torsion classes are central splitting, are

determined to be finite direct sums of full matrix rings over

local semi-artinian rings.









CHAPTER 1

Introduction

An element of an Abelian group is said to be a torsion

element if it has finite order. Since Abelian groups may be

regarded as modules over the integers, the definition may be

stated in module-theoretic terms. That is to say, an element

of an Abelian group is a torsion element if it has a non-zero

annihilator. An Abelian group is called a torsion Abelian

group if each of its elements is a torsion element. Then any

Abelian group has a maximal torsion subgroup; namely the sub-

group consisting of all the torsion elements of the group.

This concept is easily generalized to modules over any

commutative integral domain. However, when one moves to an

even slightly more general ring, the collection of elements

which have non-zero annihilators does not form a submodule.

An example of this occurs when the ring under consideration

is the'direct product of two fields. However, there is at

least one reasonable way of generalizing the concept of tor-

sion.

This generalization was introduced by Dickson in 1966 [5].

His definition holds for certain Abelian categories. For the

purposes of this dissertation, the definition will be re-

stricted to R-mod, the category of left unital R-modules,






2

where R is any ring with identity. All modules in this work

will be members of R-mod. According to Dickson, a torsion

theory is a pair, (3,3), of classes of R-modules satisfying

the following axioms:

(1) n 3 = (0);

(2) 3 is closed under homomorphic images;

(3) 3 is closed under submodules;

(4) For all R-modules, M, there is a submodule, 3(M),

of M such that 3(M) E 3 and M/3(M) E 3.

A class of R-modules, 3, is called a torsion (torsion-

free) class if it is the first (second) coordinate of some

torsion theory. A class of R-modules, 3, is a torsion class

if 3 is closed under direct sums, homomorphic images, and

extensions of one module in 3 by another. Recall that 3 is

closed under extensions if for all exact sequences,

0 A B C 0,

with A and C in 3, then B is also in 3. Dually, a class of

modules, 3, is a torsion-free class if it is closed under

submodules, direct products, and extensions of one module in

3 by another. In a torsion theory, (3,3), the modules in 3

are called torsion, and the modules in 3 are called torsion-

free.

Once a torsion class, 3, is specified, the torsion theory

having 3 as its torsion class is uniquely determined. In this

situation, 3 = (M E R-modIHom(T,M) = 0 for all T E 3). Simi-

larly, if a torsion-free class, 3, is specified, then the

torsion theory having 3 as its torsion-free class is uniquely





3

determined. In this instance, 3 = (M E R-modjHom(M,F) = 0

for all F E 5). The above results are due to Dickson [5].

A torsion theory is called hereditary if its torsion

class is closed under submodules. In this event, the torsion

class is called hereditary. This concept has appeared in

the literature under such names as torsion radical [10],

idempotent radical [15], and strongly complete Serre class

[20]. Dickson [5] shows that a torsion class is hereditary

if and only if the corresponding torsion-free class is closed

under injective envelopes. If M is an R-module, then the

injective envelope of M, written E(M), is a minimal injective

extension of M [9]. Unless otherwise mentioned all torsion

theories will be hereditary, and the term "hereditary" will

be omitted.


Definition 1.1 [11]. A collection of left ideals of R, F(5),

is called a topologizing and idempotent filter if it satisfies

the following conditions:

(1) If I E F(3) and I c J, then J E F(3);

(2) If I E F(3) and J E F(3), then I n J E F(3);

(3) If I E F(3) and r E R, then (I:r) = (x E RIxr E I)

E F(3);

(4) If (J:r) E F(3) for all r E I and I E F(3), then

J E F(J).


Gabriel then proved that there is a 1-1 correspondence

between hereditary torsion classes, 3, and topologizing and

idempotent filters, F(3), given by the mapping

3 F(U) = (I g RIR/I E 3).






4

These definitions can be made for classes of right R-modules

and filters of right ideals in an entirely analogous manner.

It is in order to look at some examples of torsion

theories. This will be done by specifying their torsion

classes. By a previous remark, this means that the entire

torsion theory is specified. Two torsion classes of a some-

what trivial nature are (0) and R-mod. For the sake of

completeness, the next example will be an example of a non-

hereditary torsion class. If R is any ring with identity,

then 3 = (M E R-modlHom(M,R) = 0) is a torsion class, which,

as in the case of Abelian groups, is not necessarily heredi-

tary. A torsion class defined in a similar manner is Jans'

E(R)-class [13]. In this theory, j = (MIHom(M,E(R)) = 0)

is the torsion class. This class is hereditary, because E(R)

is an injective R-module.


Definition 1.2 [12]. If N is a submodule of M, then N is

called an essential submodule of M if N n N' 0 for all non-

zero submodules, N', of M.



Following Goldie [12], define Z(M) = (m E M (0:m) is an

essential left ideal of R). If nT is the natural map from M
-1
onto M/Z(M), then define Z2(M) = 1 (Z(M/Z(M))). Let

Q = (M E R-modjZ2(M) = M). Q is a torsion class, which is

known as the Goldie torsion class. The filter corresponding

to the Goldie torsion class is the smallest filter containing

all the essential left ideals of R. The torsion-free modules






5

are those modules, M, such that Z(M) = 0. If Z(R) = 0, then

the Goldie torsion class is equal to the E(R)-torsion class.

The final examples are due to Dickson and are called

the torsion theories of simple type [5]. If G is any collec-

tion of non-isomorphic simple R-modules, then JG = (MIEvery

non-zero homomorphic image of M has a submodule isomorphic

to some member of G) is a torsion class. The filter for this

torsion class is the smallest filter containing all the maxi-

mal left ideals, M, of R such that R/M is isomorphic to a

member of G. The torsion-free class is precisely the class

of modules which have no submodules isomorphic to an element

of G.

There are two torsion classes of simple type which deserve

special recognition. If G = (S), then 3G is denoted by JS

and is called the S-primary torsion class. If G is a complete

representative set of non-isomorphic simple R-modules, the

torsion class is called the simple torsion class and is de-

noted by 8. The torsion-free modules in this torsion theory

are precisely those modules with no simple submodules.

A ring, R, is said to satisfy the Primary Decomposition

if g(M) = SES8 US(M) for all M E R-mod. In this case, the

simple theory is said to have the Primary Decomposition.

If 3G is a torsion theory of simple type, then 39 is said

to have the Primary Decomposition if 3 (M) = ESE S 5(M), for

all M E R-mod. The question of when the simple theory has

the Primary Decomposition has been investigated in recent

papers; e.g. [2], [6], [7], and [18].














6

The purpose of this dissertation is to generalize the

Primary Decomposition in the following way. Let 3 be an

arbitrary torsion class, and let (U aaEG be a collection of

subtorsion classes of 3, i.e. 3' 3 and J is a torsion
a a
class for all a E G. When is it true that 3(M) = E U (M)
aEG a
for all M E R-mod?

Toward this end, Chapter 2 develops some tools which will

be used in later chapters. These tools involve a generaliza-

tion of bounded Abelian groups due to Walker and Walker [20]

and some results concerning torsion preradicals. Chapter 3

investigates the lattice of torsion classes. Chapter 4 leads

to a theorem providing necessary and sufficient conditions on

a certain subclass of torsion cyclic R-modules for a decomposi-

tion to occur. Separating examples are given to show that the

theorems are meaningful. Finally, Chapter 5 studies completely

reducible torsion classes and eventually yields two additional

characterizations of rings studied by Dlab [8].

All homological concepts and basic facts concerning them

may be found in any standard text; e.g. [14].









CHAPTER 2

Preliminaries

Every left ideal of R is an element of the power set of R.

Then every topologizing and idempotent filter of left ideals

of R is a subset of the power set of R, and is an element of

the power set of the power set of R. The collection of all

such filters is a subset of the power set of the power set of

R, and is consequently a set. Since the collection of torsion

classes is equipotent to this set, it must be a set also.

Chapter 3 will determine a lattice structure on the set

of torsion classes on R-mod for a fixed ring, R. Before

attempting this, several indispensable tools will be developed

in this chapter. Section 1 is devoted to developing some

properties of uniformly negligible modules, and Section 2

investigates some aspects of torsion preradicals and the

torsion classes they generate.



Section 1. Uniformly Negligible Modules.

Following Teply [19], make the following definition:


Definition 2.1. Let I be a left ideal of R, and let M be an

R-module. M is called I-uniformly negligible (I-bounded) if,

for all m E M, there exists (rl,...,rn(m) c R such that
n(m)
A (I:ri)m = 0(Im = 0). If 5 is a torsion class, then an
i=l







R-module, M, is called 7-uniformly negligible (a-bounded)

if there is some I E F(3) such that M is I-uniformly negli-

gible (I-bounded).


The prefixes, I and a, in the above definition will be

omitted whenever no confusion is possible. The following

proposition was mentioned without proof in [19].


Proposition 2.2. If I is a left ideal of R, then the class of

I-uniformly negligible R-modules is closed under submodules,

homomorphic images, and arbitrary direct sums.


Proof. Let I be a left ideal of R, and let M be an I-uniformly

negligible module. Assume that N C M, and let x N. Then

x E M, and there exists (rl,...,rn) c R such that
n
n (I:ri)x = 0. Thus, N is I-uniformly negligible. Moreover,
i=l
if x E M, then there exists (r ,...,r) c R such that
m m
n (I:ri)x = 0. Then n (I:r.)x c N. Since x was arbitrary,
i=l i=l
M/N is I-uniformly negligible.

Now let (Ma aEG be an arbitrary collection of I-uniformly

negligible modules. Let M = e M and let x E M. Then
aEGa
x = mI + m2 + ... + mk, where m. E M For each m., there
1 ai 1
exists (rl,...,r ) = A. c R such that n (I:r)mi = 0.
k rEAi
If B = U Ai, then n (I:r)x = 0, and M is I-uniformly
i=l rEB
negligible. C


Proposition 2.3. If I is a two-sided ideal, then a module, M,

is I-bounded if and only if M is I-uniformly negligible.








Proof. The proof follows immedi-ately from the fact that

I c (I:r) whenever I is a two-sided ideal and r E R. rI


It follows that, for any torsion class over a left duo

ring, the concepts of uniformly negligible and bounded are

equivalent. Also notice that both of the above concepts

generalize the concept of a bounded Abelian group.


Definition 2.4 [13]. A torsion class, 3, is called a TTF class

if it is closed under arbitrary direct products. In this case,

the torsion theory with 9 as its torsion class is also called

TTF.


Proposition 2.5 [13, Theorem 2.1]. A torsion class, 5, is TTF

if and only if there exists a minimal ideal in F(3). In this

event the minimal left ideal in F(3) is an idempotent, two-

sided ideal of R.


Proposition 2.6. Let 5 be a torsion class. Then 9 is a TTF

class if and only if all torsion modules are bounded.


Proof. The proof is immediate and will be omitted.


Example. The following is an example of a torsion class

which is not TTF, but in which all torsion modules are uni-

formly negligible. The ring was studied by Cozzens [4].

Let F2 denote an algebraic closure of Z/(2), and let
22
p be the automorphism of F2 defined by p(z) = z Let

F2[t,p] denote the ring of "twisted" polynomials in t over F2.

The elements in F2[t,p] are precisely the elements of F2[t]







10

The addition is the same as in the polynomial ring, and the

multiplication is determined by the identity ta = p(a)t for

all a E F2 together with the distributive laws. The ring,

F2[tp], is a principal left and right ideal domain [4].

Let M = (t kk is a non-negative integer]. Let
a a b
RM = a E L2[t,p] and m E M). Make the identity k k+i
m k k a+i
if b = t a. Addition is defined by a/t + b/t = a b and
J tbj
multiplication is defined by p3(a)b where p (a)
t ti t +
th
is the element of F2[t,p] obtained by applying the j iterate

of p to each coefficient of a. Cozzens [4] showed that RM is

a simple principal right and left ideal domain, and there is

a unique simple R -module up to isomorphism, which is injective.
n
Furthermore, every right ideal is of the form n (t-a.)RM,
i=l
where ai E F2[t,p]\(0).


Lemma 2.7. If I is a non-zero right ideal of RM, then RM/I is

isomorphic to a finite direct sum of simple modules.

n
Proof. If I is a non-zero right ideal of R then I= n (t-a.)R
i=l
where ai f 0 for all i. The proof of the lemma will follow by

induction on n.

If n = 1, then the assertion is true by Lemma 2.2 of [4].

Assume that the assertion is true for all right ideals of RM
n n+l
of the form n (t a.)RM. Let I = n (t a.)RM. Recall that
i=l M i=l

for any module, P, the socle of P, written soc(P), is the sum





11

of the simple submodules of P. Consider the exact sequence
n
n (t-ai)RM R
i=l M RM
0 --- -
n+l n+l n
n (t-ai)RM n (t-a i) (t-ai)RM
i=l i=l i=l


The left side is (t an+l)RM bounded and is therefore its own

socle. Since every simple module is injective and RM is

Noetherian, the left side is injective. Hence the sequence

splits. Since the right side is a direct sum of simple

modules, the assertion is verified. 0


Consider the simple torsion class, 8, on right RM modules.

By Lemma 2.7 every proper right ideal is in F(S). Since RM is

not a division ring, soc(RM) = 0. Therefore, 0 f F(S). Then 8

cannot be a TTF class, because RM is a simple ring and because

the minimal right ideal in the filter for a TTF class is a two-

sided ideal by Proposition 2.5.

Let N be any maximal right ideal of RM. If T is any module

in 8 and x E T, then x E socT. Therefore, T = socT. Since each

simple module is isomorphic to RM/N, and N-uniformly negligible

module, Proposition 2.2 yields that T is an N-uniformly negli-

gible module. [


The final lemma in this section gives the structure of M-

uniformly negligible modules, where M is a maximal left ideal

of R.


LEMMA 2.8. Let M be a maximal left ideal of R. Then every M-

uniformly negligible module is equal to a direct sum of cyclic

modules, each of which is isomorphic to R/M.








12

Proof. It is sufficient to prove that every cyclic M-uni-

formly negligible module is isomorphic to a direct sum of

copies of R/M. Let rl,r2,.. .,rn ) R. Claim that
n
R/ n (M:ri) is isomorphic to a direct sum of copies of R/M.
i=l
The proof will follow by induction on n. If n = 1, then

the assertion is immediate. Assume that the assertion is

true for all subsets of R of cardinality less than n. Let

(rl,...,rn) c R. Without loss of generality, one may assume

that none of the r. is in M. Call the subset irredundant if
1
n (M:r ) i (M:r ) for j = 1,2,...,n. Let J = (rl,r2,...,rn)

and let K = (r2,r3,...,rn). If J is not irredundant, then

the assertion is true by the induction hypothesis. Assume

that J is irredundant. Then


(M:r) + n (M:ri (M: n (M:ri)
R iEK 1 r iEK R
n (M:ri) n- (M:ri) n (M:ri) n (M:ri) n (M:ri)
iEJ iEJ iEJ iEJ iEK

R
(M:r)


However, R/ n (M:ri) satisfies the conditions of the induction
iEK
hypothesis and is therefore isomorphic to a direct sum of copies

of R/M. R/(M:ri) is isomorphic to R/M so that the induction is

complete. The proof is completed by noting that every cyclic

M-uniformly negligible module is a homomorphic image of a module

of the above form. O






13

Section 2. Torsion Preradicals..


Definition 2.9 [10]. A subfunctor, P, of the identity functor

from R-mod to R-mod is said to be a torsion preradical if and

only if the following conditions are satisfied:

(1) If a:M N is a homomorphism, then a(P(M)) c P(N);

(2) If N c M, then P(N) = N N P(M).

If M = P(M), then M is called a P-module.


Proposition 2.10. A subfunctor, P, of the identity functor on

R-mod is a torsion preradical if and only if the following con-

ditions are satisfied:

(1) 2 = ;

(2) N c M implies that P(N) g P(M);

(3) The class of P-modules is closed under both submodules

and homomorphic images.


Proof. ( ). Let M be a P-module, and let N g M. Then

P(N) = N n P(M) = N n M = N. Hence N is also a P-module. Let

r be the natural map from M onto M/N. Since M/N = n(M) =

n(P(M)) c P(M/N), M/N is a P-module. Therefore, the class of

P-modules is closed under submodules and homomorphic images.

In [10], Fuchs showed that P2 = P. If N c M, then condition (2)

of the definition implies that P(N) a P(M).

Conversely, assume that P satisfies the three conditions

of this proposition. Let M be an R-module, and let N Q M.

P(N) c P(M) n N. Since P is idempotent and the class of P-

modules is closed under submodules, P(M) n N = P(P(M) n N).






14

Consequently, P(N) = P(M) n N, because every P-module contained

in M is also contained in P(M). Condition (1) of the defini-

tion follows from the fact that the class of P-modules is closed

under homomorphic images. [


Let P be a torsion preradical. For any module, M, define

Po(M) = 0. Let a be an ordinal number, and assume that for all

ordinal numbers, 0 < a, P (M) is defined. If P (M) is defined,

then let rB be the natural map from M onto M/P (M). Suppose

that a = 0 + 1. Define Pa(M) = (n )- (P (M/P (M))). If a is a

limit ordinal, then define P (M) = U P (M). For every module,
a M, there is a smallest ordinal, X(M), such P ()M) = P,(M)+(1 ).

Let U = (MIPI ) (M) = M). The two following propositions con-

cerning this construction appear in [10] without proof. The

proofs will be included here for completeness.


Proposition 2.11. If P is defined as above, then P is a
a a
torsion preradical.

Proof. If a = 1, then Pa is a torsion preradical by the defini-

tion of P Assume that P is a torsion preradical for all

0 < a. Assume that a = 0 + 1. Let f:M N be a homomorphism.

Then a(P (M)) g P (N). Consider the induced homomorphism

f:M/P (M) N/P (N). Then f(P(M/P (M)) 9 P(N/P (N)). There-

fore, f(P (M)) g P (N). Assume that N c M. P (N) =N n P (M).
aoa f (3M)
Let x E N n P (M). Then N+ (M) which
a x + P () P (M)

implies that x + (N n P (M)) E P(N/N n P (M)) = P(N/P (N));






15

therefore x E P (N). Hence Pa is a torsion preradical. If a

is a limit ordinal and f:M N, then f(P (M)) = f( U P (M)) r
a U P (N) = P (N). If x E N P (M), then x EN P (PM) for
P3 some 0 < a. Thus x E Pp(N) r P (N). O


Proposition 2.12. Let P be a torsion preradical, and let P

be defined as above, where a is an ordinal number. Define

(M) = P(M) (M). If 3 = (MP ) (M) =M), then is the smallest

torsion class containing the class of P-modules.


Proof. The first objective is to show that 3 is a torsion class.

Let M E U, and let N c M. Since PX(M) is a torsion preradical,

P (M)(N) = P (M) n N = M N = N. Since M is a (M)-module,

Proposition 2.10 implies that M/N is aP (M)-module also. There-

fore 3 is closed under submodules and homomorphic images. Let

(Ma ) E be a collection of U-modules. For each a E G,

card(M ) s X(M ). Therefore, X(M ) a card( e M ) = y for all
a a a aG a
aEG
a E G. By [10, Lemma 1], P ( N M ) = P Y(M) = e M .
aEG aEG aEG

Therefore, 3( N M ) = e M and 3 is closed under arbitrary
aEG aEG
direct sums.

Finally, it is necessary to show that 3 is closed under

extensions. Toward this goal, let

0 N M M/N 0

be an exact sequence such that N and M/N are in U. If a is the

smallest ordinal such that m + N E Pa(M/N), then m E P ()+aM).

The proof of this assertion will follow by transfinite induction

on a. If a = 1, then















-1 -1 M
m E6 1 (P(M/N)) H P (N) (M) = (N)+1(M)


where 1 and n0 are the appropriate natural maps. Assume that

the assertion is true for all ordinals 0 < a and all N c M.

Let a = 0 + 1 be the smallest ordinal such that m + N E P (M/N)

-l M/N
(n ) (P (M/N) )), where u is the natural map from M/N onto

M/N) Let P (M/N) = L/N. By the induction hypothesis,
PA (M/N)

L c (N) + (M). But

-1 -1 M
m+NE (n )(P(M/L)) c (nT)P( (M)) P M)
0 (N) +(M (N) +S+I
((N)+P

P (N)+a (M)

where n0 and nl are the natural maps from M/N onto M/L and

M/P (N)+p (M) respectively. Since a cannot be a limit ordinal,

the induction is complete. This completes the original asser-

tion, and 3 is a torsion class.

Let 5' be a torsion class which contains the class of P-

modules. If M E 3, then M E 7'. The proof follows by a routine

transfinite induction on X(M) and will be omitted. E








CHAPTER 3

The Lattice of Torsion Classes

Let R be a fixed ring with identity. Consider the set,

Z, of torsion classes, and order by inclusion. Define two

binary operations, V and A, and as follows: 1 V 72 is the

smallest torsion class containing the set theoretic union,

51 U 72' and 1 A J2 is the largest torsion class contained in
the intersection a1 n ;2. Under these operations Z becomes

a lattice, which will be denoted by (S,v,A). (,V,A) has a zero

(namely (0)) and a 1 (namely R-mod). It is necessary to examine

the meet and join operations a little more closely.

The meet operation is more accessible and consequently will

be treated first. The following proposition yields the result

that the meet of two torsion classes is precisely the set

theoretic intersection of the classes. In fact, the result

shows that any collection of torsion classes has a greatest

lower bound and consequently a meet.


Proposition 3.1. If ( aaEG is any non-empty collection of

torsion classes, then n 3a is also a torsion class.
aEG

Proof. It is sufficient to verify that the class, f a is
aEG
closed under submodules, homomorphic images, arbitrary direct

sums, and extensions. Since each 7a is closed under these

properties, the proof is routine and will be omitted.r








The join of two torsion classes may be constructed in

two ways, internally and externally. The internal construc-

tion is somewhat more complicated than the external one, but

it has the advantage of being much more instructive. Since

it is simpler, the external construction of the join will be

made first.


Proposition 3.2. The join of two torsion classes, 31 and 32'

is the intersection of all torsion classes which contain

U1 U 72'

Proof. R-mod is a torsion class which contains 1 n 2. Hence

G = ([ E (,V,A)~1 2 1 U 23) is a non-empty set. By Proposi-

tion 3.1, U = n 3 is a torsion class. Clearly 1 U 52 m U.
JEG
If V is any torsion class containing 31 U 52, then V E G.

Hence 1 g Vi. Hence U is the smallest torsion class containing

71 U ,2' which means u is the join of 1 and 32' 0

Before the internal construction of 31 v U2 can be made,

a preliminary result is required.


Lemma 3.3. Let 3 and 1 be torsion classes. For all M E R-mod,
-1
define P(M) = (nM ) ((M/31(M))), where nM is the natural map

from M onto M/3r(M). Then P is a torsion preradical.


Proof. P is a subfunctor of the identity functor, because

U and 1 are torsion classes. Let M = P(M). Then

(-)) 1 M/ M) (M)
P~(P(M)) = (nM )(3(P(M)/51 P M))])= rM (5[M/51 M)]) =P(M).






19
Hence P is idempotent. Let N c-M. If x E P(N), then

x + 3Z(N) E 3(N/~I(N)). Therefore, (~r(N):x) E F(3).
However, (3~(M):x) = (3~(N):x), because 1 is a torsion
preradical. Thus, x + 3l(M) E Z(M/3 (M)), which implies

that x E P(M). Since this is true for all x E P(N),

P(N) c p(M).

By Proposition 2.10, it suffices to show that the class

of P-modules is closed under both submodules and homomorphic

images. Let M = P(M), and let N c M. If x E N, then

(3U(N):x) = (3 (M):x) E F(3). Thus, x E P(N), and N is a
P-module. Furthermore, if x E M, then (3U (M):x) E F(3).

Since ( 1(M) + N)/N C 1(M/N), x E P(M/N). C

By constructing the sequence of torsion preradicals, Pa,

one obtains a torsion class, U, which contains the class of

p-modules. In fact, by Proposition 2.10, U is the smallest

torsion class containing the class of p-modules.

Proposition 3.4. If l is defined as above, then U = 5 v 3 '

Proof. Since all modules in 3 U 31 are P-modules, 3 v 31 c U.

Let V be any torsion class containing 3 U 31' and let M = P(M).
Consider the exact sequence

0 31(M) M/ 1(M) 0.

By hypothesis M/31(M) E J. Consequently, since M is the ex-
tension of two V-modules, M E V. By the remark preceding this

proposition, u c V. Since a V a1 contains 3 U 31, the proposi-

tion is completed. O






20

An interesting aspect of this construction is that, in

some sense, it is not commutative. That is, if 3 and a1 are

torsion classes, and for any module, M, nM is the natural map

from M onto M/j(M), and pM is the natural map from M onto
-1
M/31(M), then (n M) ([M/a1(M)]) does not necessarily equal
-1
(PM ) (~1[M/a(M)]). However, after the transfinite construction,

the torsion classes turn out to be the same. An example of the
C0
noncommutativity occurs when R = n F., where each F. is an
i=l
isomorphic copy of some fixed field, F. Let 8 be the simple

torsion class, and let Q be the Goldie torsion class.
-1 -1
(n) (gR/Q(R)]) = 9 F and (pR) (Q[R/8(R)]) = R. It is
i=l
essential for studying decompositions of torsion classes to

decide when the construction in commutative. In this regard,

consider the following two results.


Lemma 3.5. Let 3 and a1 be torsion classes on R-mod. Then

the following statements are equivalent:

(1) 3 n = o().

(2) If I E F(3) and J E F(U1), then I + J = R.

(3) There are no non-zero simple modules in a n 1'.


Proof. (1) = (2). Let I E F(3), and let J E F(U1). Then

R/I+J is a homomorphic image of both R/I and R/J. Thus,

R/I +J E a n 3 = (0). Consequently, R = I + J.

(2) = (3). Let S be a non-zero simple module in u n a1.

S 2 R/M, for some maximal left ideal, M. Then M E F(3) nF( 1).

By condition (2), R = M + M = M, which contradicts M being

proper.






21

(3) = (1). Let ME a n 3l.. If M contains a non-zero

cyclic submodule, Rm, then Rm E 3 n 31 also. Rm a R/L, where

L is a proper left ideal of R. By a Zorn's Lemma argument, L

is contained is some maximal left ideal N. R/N is a simple

homomorphic image of R/L. Thus, R/N E 3a which contra-

dicts condition (3). F1

Proposition 3.6. Let 3 and 31 be torsion classes such that

S0n 3l = (0). For all modules, M, let rM be the natural map

from M onto M/3(M), and let pM be the natural map from M onto

M/31(M). Then the following statements are equivalent
-1 -1
(1) (p ) ([M/3l(M)D) = (r1 )(31[M/3(M)]), for all

M E R-mod.
-1 -1
(2) (pMI )(3[M/31(M)]) = (M) e 31(M) = (n1 )(u [M/3(M)]),
for all M E R-mod.

(3) 3(M/[3(M) e31(M)]) = (0) = 31(M/[3(M) 31(M)]), for

all M E R-mod.

(4) (3 V 31) (M) = 3(M) 3'1(M), for all M E R-mod.

-1
Proof. (1) = (2). If m E (pM )(3[M/, (M)]), where M is an

R-module, then there exists I E F(3) such that Im c 31(M).
-1
By condition (1), m E (nM1) ( 1[M/3(M)]). There exists J E F(3)

such that Jm c U(M). Since 7 n 3l = (0), two facts are immediate

One fact is that 3(M) + 31(M) is always a direct sum. The second

fact, which follows from Lemma 3.5, is that I + J = R. Thus,

Rm = (I + J)m c Im + Jm c 3J(M) E 3(M). Since m was arbitrary,
(-1) ([M/(M) (M)
(p, )(3[M/31(M)]) c a(M) e 31(M).






22

Conversely, let m = mi + m2- E (M) e ~1(M). Then

(0:ml)m c 31(M). By the way mI was chosen (0:ml) E F(Z).

Therefore, m + .7(M) E 7(M/U1(M)). The other equality is

proved in an entirely analagous manner.

(2) = (3). Let M be an R-module. For all m such that

m + (U(M) e ~1(M)) E 3(M/[5(M) e .1(M)]), consider the exact

sequence

U (M) e 3l(M) Rm+ (U(M)GU1 (M)) Rm + (3(M) e~ (M))
0 1(M) 1(M) (M) 0.

Since both ends of the sequence are in 3, the middle must be a
member of 3 also. By condition (2), m E U(M) E 31(M). The

other equality may be obtained in a similar manner.

(3) = (1). Let M be an R-module. If m + U (M) E 3(M/ U(M)),

then m + (U (M) e 3 (M))) E 3(M/[3(M) e U1(M)]). By condition (3),
-1
m E 9(M) E ~1(M). Consequently, (pM1) (3[M/31(M)]) D U(M) 9~ (M).
The reverse inclusion follows from the proof of (1) = (2). An
-1
entirely analagous argument yields that (TTM ) (5[M/3(M)]) =

S(M) e 3 (M).
(3) = (4). In this case conditions (1) and (2) may also

be assumed. Let P(M) = (p1) ()([M/1 (M)]) for all M E R-mod.

By Lemma 3.3 and Proposition 3.4, P is a torsion preradical

which generates 3 V U1. By condition (2), P(M) = (M) e U(M).

However,
-1 -1
P2(M) = (aM )(P[M/P(M)]) = (M ) (0) = U(M) e (M),
where oM is the natural map from M onto M/P(M). Since P2(M)

P(M) for all M E R-mod, U(M) 1(M) = (3 v i) (M).






23

(4) = (1). This is immediate from the construction of

Sv 3av D


It was noted earlier that any arbitrary collection of

torsion classes has a greatest lower bound; namely their inter-

section. Any collection of torsion classes also has a least

upper bound; namely the intersection of all torsion classes

containing their set theoretic union. Hence the lattice of

torsion classes is a complete lattice. Once again, this con-

struction is not very instructive, and an internal construction

will be given. Toward this end we have the following result.


Proposition 3.7. Let (3Yal E be a collection of torsion classes

such that any two elements of the collection are comparable.

Define P(M) = a3 (M) for all M E R-mod. Then p is a torsion
aEG
preradical. If, in addition, for all I E F(3 ), there exists a

finitely generated J E F(3p) such that J c I, then the class of

P-modules is a torsion class which is the join of (a aEG.


Proof. Since any two elements of ([a ) E are comparable,P is

a subfunctor of the identity functor on R-mod. Let M be an R-

module. Then

P(P(M)) = S 3 (P(M)) = ((3 (M) nP(M)) = 3 (M) =P(M).
aEG aEG aEG

If N c M, then

P(N) = E 3 (N) = E 3a (M) n N C E 39 (M) = P(M).
aECG aEG a aEG

By Proposition 2.10, it is sufficient to show that the collection

of P-modules is closed under submodules and homomorphic images.






24

Suppose that M = p(M), and that N E M. Let m E N. From

the definition of P, m = ml + ... mk, where each m. E Uai

Since any two of these torsion classes are comparable, m E 53

for some j. Consequently, m E 3a (M) n N = 3 (N) c P(N).

If m E M, then m E 3a (M) for some a E G. Thus,
a0

m + N E 3 (M/N) C 3a (M/N) = P(M/N).
SaEG a

For the second part of the proposition it suffices to show

that the class of P-modules is closed under extensions because

all torsion preradicals preserve direct sums [10]. Assume that

0 A B B/A 0

is an exact sequence with the ends being p-modules. If b E B,

then b + A E 9 (B/A) for some a E G. By hypothesis, there

exists a finitely generated left ideal, J E F(J ), such that

Jb a A. Assume that J = (rl,...,r ). Then each r.b E ai(A)

for some a. E G. Let 7 be the maximum torsion class of
i .y
(al ,...,U an ). Then Jb U (A). Consider the exact
a1 "n P Y
sequence
Rb
0 Jb Rb 0.
Jb
The ends are in 3 which implies that Rb E P(B). Since B was

arbitrary, B = P(B), and the proposition is complete. O


Notice that even if the extra hypothesis does not hold, one

could construct the join of any collection of comparable torsion

classes by first forming the torsion preradical, P, of Propo-

sition 3.7, and by then observing that the join of the collec-

tion is the torsion class generated by P. What happens if the






25

collection, (7 aaEG' is not comparable? First, well-order G,

so that C might be thought of as a cardinal number. Re-index

each 3a by letting a '= a+l Let 7 = 0 if a is a limit

ordinal less than G. The new indices are certainly contained

in G + 1, so that they, too, form a set. Call this set 8.

V 3 = V '. Define = (0], and let U2 = 2 Assume
aEG a PBE3
that 1b has been defined for all p < a. If a = p + 1, then

define a = 1A V 3a if a' exists. If 3a' does not exist,

then let Ua = U. If a is a limit ordinal, then let ha be the

torsion class generated by the torsion preradical, Z Ui(M).
3 This is a torsion preradical, because the (U )3

parable collection of torsion classes. Then (b )oBS form a

comparable class, so that P(M) = 1 U (M) is a torsion pre-

radical which generates V I It is clear that v 3 =
i3ER aEG
v q.
V U.
AER

Since every torsion class is determined by the cyclic

modules which are in it, each torsion class, 3, is

V 3rR/I where R/ is the smallest torsion class containing
IEF(3) R/I

R/I. It would be interesting to construct JR/I for an arbitrary

left ideal I, for this would lead to a way of characterizing

torsion classes.


Lemma 3.8. Let I be a left ideal of R. For any R-module, M,

define p (M) = (m E MI There exist [rl,...,r) n R such that
n
n (I:ri)m = 0). Then I is a torsion preradical, and if p
i=l







26

is a torsion preradical such that P(R/I) = R/I, then

P (M) c P(M) for all M E R-mod.


Proof. Let M be an R-module, and let mi and m2 be elements

of PI (M). Then there exist (rl,...,rn) c R and (s1,...,sn) CR
n m
such that n (I:r.)m = 0 and n (I:si)m2 = 0, so that
i=l i=l
n m
[( n (I:r.)) n ( n (I:s.))][mI + m2] = 0. If r E R, then
i=l i=l

n (I:rr.)m = 0. Hence P (M) is a submodule of M, for all
i=l

M E R-mod. Proposition 2.2 yields that P -modules are closed

under submodules and homomorphic images. That P is idempotent

follows immediately. Finally, assume that N c M, and that

x E I (N). It is clear that x E P (M). By Proposition 2.10,

I is a torsion preradical.

Let p be a torsion preradical such that P(R/I) = R/I. It

is sufficient to show that every pI-module is also a p-module.

If r E R, then the fact that Rr + I R/I implies that
IRr + I
R/(I:r) Rr+ I is a P-module. If frl,...,rn) g R, then
I
n n n
e R/(I:r.) is a p-module. Map R/ n (I:r ) into e (R/(I:r.)
i=l i=l i=l
by

n
1 + n (I:ri) (1 + (I:rl), 1 + (I:r2),..., 1 + (I:r n))
i=l
n
Since this map is a monomomorphism, R/ n (I:ri) is a P-module.
i=l
Since every p -module is a homomorphic image of a direct sum

of modules of this form, every P -module is a p-module. O






27

Definition 3.9. If a is a torsion class, then a collection, P,

of left ideals of R is called a subbasis for 3 if 3 = v R/I.
IEP
A subbasis, P, of 3 is called a separating subbasis if

ar n \ V R/J/ = (0) for all I E P.
R/I JEP\[I) R/J

Every torsion class, 3, has a subbasis, namely F(3).

However, a torsion class may have a much smaller subbasis.

For example, the simple torsion class has a subbasis consisting

of any collection, P, of maximal left ideals such that every

simple R-module is isomorphic to R/M for some M E P. There are

torsion classes which do not have a separating subbasis as the

following example shows.

Example. Let R = F[xl,...,x n,... be the polynomial ring in

countably many commuting indeterminants over a field, F.

Localize R by the maximal ideal M = (xl,x2J...,x ,...), and

call the localization RM. Recall that R = I a E R and

b E R\M) with multiplication and addition defined in the ob-

vious manner. RM is a local ring with unique maximal I
X1 2 xn n xn+l n+2
1 ,1 1 Let In 1 1 ') for

all n = 1, 2, ... .The collection of In formsa decreasing

chain I 12 D ... D In .... Consequently, R/I 7
1 + 2 + + n + R/I "R/I2

... C R/in ... Let 5 = V R/i By Proposition 3.7,
R/In n=l R/In

P(M) = a RI (M) is a torsion preradical which generates 3.

Since aR/I (R) = 0 for all n, .(R) = 0. Since n In = 0, 3
n n=l
is not a TTF class. Assume that a has a separating subbasis





28

of one element, J. P (R/In) = 0 unless J c In, because I

is a prime ideal. But R/J (R/In) = R/In, so that J E In for

all n. Thus, J = 0, which contradicts the fact that J(R) = 0.

Since RM is a local ring, Lemma 3.5 implies no separating

subbasis could have more than one element in it.


Proposition 3.10. If P is a separating subbasis for a torsion

class, 5, then 3 = (Mj every non-zero homomorphic image of M

has a non-zero I-uniformly negligible module for some I E P ).


Proof. Let C = (M E R-modjM is I-uniformly negligible for

some I E Pi]. G is closed under submodules and homomorphic

images. Proposition 3.10 follows from [17, Proposition 2.9]. O


The final results in this chapter show that (,v,A) is

distributive and determines how complements in this lattice

act.


Proposition 3.11. (,v,A) is distributive.


Proof. It suffices to show that, for any torsion classes J,

U, and VI, A (L V b) c (3 A u) V (3 A V), because the reverse
-1
inclusion holds in any lattice. Let P(M) = (nM ) ([M/V(M)]),

where nM is the natural map from M onto M/L(M). Define P in

the canonical manner used in constructing U V V. Let

M E6 A (U V I) = n (iU v s). The proof will follow by a

transfinite induction on X(M), where X(M) is the smallest

ordinal such that p (M) = P(M)+ (M) If M = X(M) (M), then

say that M has P-length X(M). Assume that M E 5 and that M
1. I is cas
has n-length i. In this case, M = (rM )(u[M/i(M)]). Consider








the exact sequence

0 V(M) M M/i(M) 0.
The left side is in 3 A V, and the right side is in 3 A U.

Hence M must be in (3 A V) V (U A u).

Assume that M E (3 A u) V (3 A I) for all M E 3 A (U v I),

where M has P-length less than a. Suppose that M has P-length

a and that a = 0 + 1. P (M) E (U A 1) V (;3 A I) by the induc-

tion hypothesis, and M/p (M) E (3 A u) V (3 A I) by the case

a = 1. Therefore, M E (U A u) V (3 A i), because M is the

extension of one element in (3 A u) v (3 A V) by another. If

a is a limit ordinal and M has p-length a, then RmE (3A h) V

(3 A b) for all m E M. This is true because m E P (M) implies

m E P (M) for some 0 < a. Since m was an arbitrary element

of M, the proof of the proposition is complete. n


Since (,V,A) is distributive, a torsion class has at most

one lattice complement. Recall that 3 has a lattice complement,

U, if and only if 3 A U = (0) and 3; V U = R-mod. When does a
torsion class have a complement?

Proposition 3.12. Let 3 and 31 be torsion classes on R-mod,

and let 3 and 51 be the torsion-free classes determined by

3 and 1', respectively. Then the following statements are

equivalent:

(1) 3 and l3 are lattice complements.

(2) 3 n 31 = [0) and if 3(M) + 31(M) = 0, then M = 0.

(3) 5 n 31 = (0) and 3 n 3n = (0).






30

Proof. Since U and 51 are lattice complements, 3 3n = (0).
-1
Define P(M) = (nM ) ([M/13(M)]), where M is the natural map

from M onto M/'1 (M). Define 6P in the canonical manner used

in constructing 3 V 3U1 If 3(M) + 3'(M) = 0, then P(M) = 0.

Since 3 v 3' = R-mod, this implies that M = 0.

(2) = (3). a n 3a = (0) by hypothesis. Let M E 3 n 51.

3(M) + 3'(M) = 0, which implies that M = 0.
(3) = (1). A 3'1 = 3 n 3' = (0). Let M be any R-module

W(3 V 35) (M) E 5 n 51, which implies that M = (7 V U1) (M).

Since M was arbitrary, 5 V 51 = R-mod. r3


There are rings (such as the integers) for which no non-

trivial torsion class has a complement. But what rings have

the property that ([,v,A) is complemented?


Proposition 3.13. Let I be a principal lattice ideal (and

consequently a lattice in its own right) of (,v,A) generated

by 3. Then I is complemented if and only if 3 is of simple

type.


Proof. (=). The underlying set for I is the collection of all

torsion classes contained in 3. Let G be a complete represen-

tative set of non-isomorphic simple modules of 3. By hypo-

thesis, a3 has a complement, u, in 3. But U must be zero,

because it cannot contain any simple modules. Therefore,

G = 7, and 3 is of simple type.

(4). Let U C 3. h is also of simple type. Let G be a

complete representative set of non-isomorphic simple modules



















31

in U. Define V = V 5S Then U and V are clearly


lattice complements in I. O


Definition 3.14. A ring, R, is semi-artinian if 8 = R-mod.


Corollary 3.15. (,v,A) is complemented if and only if R is

semi-artinian.


Proof. (,V,A) is the principal lattice ideal generated by

R-mod. By hypothesis, R-mod is of simple type. By Proposition

3.13, [.,V,A) is complemented. O


Corollary 3.16. The lattice ideal of (,V,A) generated by

the simple torsion class is a complemented distributive

lattice and hence is a Boolean Algebra.









CHAPTER 4

Decomposition of Torsion Classes

Definition 4.1. Let 3 be a torsion class. 3 admits a decom-

position if there exist proper subtorsion classes, 31 and 32'

such that 3(M) = 53 (M) 9 3'2(M), for all M E R-mod.


If a ring satisfies the Primary Decomposition, and if the

ring has at least two non-isomorphic simple R-modules, then

the simple torsion class admits a decomposition. For the con-

cept of decomposition of torsion classes to be an interesting

generalization of the Primary Decomposition, there must be

torsion classes which admits decompositions but are not of

simple type. Proposition 4.4 leads to just such an example.


Definition 4.2 [3]. A torsion class, 3, is central splitting

if there is a torsion class, u, such that M = 3(M) e U(M),

for all M E R-mod.


Proposition 4.3 [13, Theorem 2.4]. A torsion class, 5, is

central splitting if and only if there exists a torsion class,

U, such that R = 3(R) E U(R).


Proposition 4.4. Let 3 be any TTF class. Then the following

statements are equivalent:

(1) 3 admits a decomposition.






33

(2) If I is the minimal ideal in F(3), then R/I is the

direct product of two non-zero rings.

(3) If I is the minimal ideal in F(y), then there is a

central splitting torsion class on R/I-mod.


Proof. (1) = (2). Assume that 3(M) = g1(M) e 32(M) for all

M E R-mod. Furthermore, assume that neither U1 nor 32 is zero.

That is, assume that the decomposition is non-trivial. Since

R/I E J, R/I = 31(R/I) E 32(R/I) = J/I K/I, where J and K

are left ideals of R. If x E J and r E R, then (I:x)xr r I,

because I is a two-sided ideal of R. Therefore, (I:x) (xr+ I)= 0,

which implies that xr + I E 39(R/I) = J/I. Hence, J is a two-

sided ideal of R. Similarly, one obtains that K is a two-sided

ideal of R. Thus, J/I and K/I are two-sided ideals of R/I, and

their sum is a ring direct sum.

(2) = (3). Suppose that R/I = J/I + K/I. Since J/I and

K/I are idempotent two-sided ideals of R/I, the collection of

J/I-bounded (K/I-bounded) modules forms a torsion class, 31 (2)

[13, Corollary 2.3]. However, R/I = 3 (R/I) 32(R/I), which

implies that R/I is central splitting by Proposition 4.3.

(3) = (1). Assume that R/I = (R/I) E 32(R/I) = J/IeK/I,

where J/I and K/I are idempotent two-sided ideals of R/I. Then

J and K are idempotent two-sided ideals of R. Let u1 be the

class of J-bounded modules, and let U2 be the class of K-bounded

modules. Ul and U2 are TTF-classes with minimal filter ideals

J and K respectively [13, Corollary 2.3]. Let M E J. Then

M = RM = (J + K)M g JM + KM.








However, KJ = JK J n K = I. Therefore, JM E U2 and IM E Li'

So, for any module, M, M = C (M) + U2(M). All that remains

is to show that the sum is direct. If M E Ui n h2, then JM =

0 = KM, which implies that (J + K)M = RM = 0. D


Example. The following is an example of a torsion class which

admits a decomposition but which is not of simple type. Let

R = n F., where each F. is an isomorphic copy of a fixed
i=l
field, F. Let Il = ((xn) E RIx2n+l = 0 for all positive inte-

gers, n), and let 12 = ((xn) E RIx2n = 0 for all positive inte-

gers, n, not of the form 2m for any positive integer, m). If

I = I1 n I2, then I, Ii, and 12 are idempotent two-sided ideals

of R. Consequently, the class of I-bounded modules is a TTF

class with I as its minimal filter ideal. R/I = Ii/I 9 I2/I.

By Proposition 4.4, the class of I-bounded modules, which will

be denoted by 5, admits a decomposition. All that remains is

to show that 3 is not of simple type.

Suppose that a is of simple type. Then since R/I is

torsion, R/I + ( e Fi) is also torsion. However, R/I + F.
i=l i=l
has zero socle. This implies that R = I + E Fi, which is
i=l
not true. Hence, is not of simple type.


The largest part of the remainder of this section is

dedicated to proving a theorem which gives necessary and

sufficient conditions on a separating subbasis of a torsion

class for the torsion class to admit a decomposition. Several

lemmas will be needed for the proof of the theorem.






35

Lemma 4.5. Let y and 31 be torsion classes such that

3 0 3u = (0). If M/N = (A/N) E (B/N) = (M/N) e 31(M/N),

and if B = N e N1 = 1(B) 3a1(B), then M = N1 E n- (A/N),

where r is the natural map from M onto M/N.

-1
Proof. -1 (A/N) is an extension of N by A/N, both of which
-1
are in 5. Thus, -1 (A/N) is also in 3. By hypothesis,

N1 n TT (A/N) = 0. Since TT(N) = B/N, M = N1 n1 (A/N)

[5, Lemma 2.5]. [

Definition 4.6. Let M be an R-module. If G is an ordinal

number, and if (M )aEG is a collection of submodules of M,

then the collection is called a C-chain for M if the following

conditions hold:

(1) M = U M
aEG

(2) If a < P, then M a Mg Moreover, Ma = MO if and

only if M = M.

(3) Ma+/Ma is a direct sum of cyclic modules for all

a + 1 E G.


Proposition 4.7. 'Every R-module has at least o:.. C-chain.


Proof. For any non-zero R-module, M, let M1 be any non-zero

cyclic submodule of M. Such a module must exist as long as

M 1 0. Assume that M1 is defined for all p < a, where a is

some ordinal number. Whenever M is defined, let TT be the

natural map from M onto M If a = P + 1, then define Ma to

be the inverse image under iu of any non-zero cyclic submodule






36

of M/M If a is a limit ordinal, then define Ma to be U M
P There exists an ordinal number a such that M = M. Let a be
a o
the smallest such ordinal number. It is clear that (M )

is a C-chain for M. O


The manner of constructing a C-chain is far from unique.

However, there is some way of associating modules and their

C-chains.


Definition 4.8. For any module, M, let a(M) be the smallest

ordinal number such that there exists a collection of submodules

of M, (M ))} a(M) which is a C-chain for M. Then M is said to

have C-length a(M).


Lemma 4.9. Let j and 31 be torsion classes on R-mod such that

3 n 51 = (0). If 0 N M M/N 0 is an exact sequence with

N E 3 and M/N E '1, then N = 3(M).


Proof. Since N E 3, N c J(M). For all m E U(M), (N:m) E F(51).

By Lemma 3.5, (N:m) = (N:m) + (0:m) = R. Therefore, Rm c N

and the proof is complete. [


Lemma 4.10. Let G be any set, and let (p aaEG be a collection

of torsion preradicals on R-mod. For all a E G, let 3a be the

torsion class generated by P Suppose that E a(M) is a
aEG
direct sum for all M E R-mod. Then P(M) = P (M) is a
aEG a
torsion preradical. Moreover, the smallest class containing

the class of P-modules is v 3
aEG






37

Proof. First note that Z P (M) is, indeed, a direct sum.
aEGa
Let M be an R-module. Then

P(P (M)) = P (P (M)) = e P (M) = p(M).
aEG aEG

Assume that M is an R-module and that N E M. Then


P(N) = e P (N) = E ( (M) n N) E P (M) = P(M).
aEG aEG aEG

By Proposition 2.10, it suffices to show that the class of P-

modules is closed under submodules and homomorphic images.

Suppose that M = P(M), and let N e M. If m E N, then

m = mI = m2 + ... + mk, where mi E Pi (M) and the a. are dis-

tinct. A proof by induction on K will show that mI E N.

Notice that when this claim is established, the fact that the

class of P-modules is closed under submodules will also be

established. If K = 1, then the assertion is trivial.

Assume that whenever mi + m2 + ... + mk E N, where each

mi E Pa(M), then mI E N. Let ml + m2 + ... + mk+1 E N, where

each m i Pi(M). There exist I E F(3a ) and J E F(a+ )
1i ai a1 ak+l
such that Im, = 0 = Jmk+1. Lemma 3.5 yields that I + J = R.

Let 1 = a + ak+1, where al E I and ak+l E J. Then
k+1 k
a k+l( mi) = + m. E N. By the induction hypothesis,
i=l i=l
ak+l ml E N. However, mi = (al + ak+l)ml = ak+l mi. Hence,
the class of p-modules is closed under submodules.

Suppose that M = P(M), and let N g M. Since torsion pre-

radicals are closed under homomorphic images, (p (M) + N)/N g

p (M/N), for all ordinal numbers a. Since M = P a(M), the
aEG
assertion follows.






38

Finally, let U be the torsion class generated by p.

Since every P-module is contained in v a', a c v a'
aEG aEG
Conversely, since 5 cg for all a E G, v 5 s 3a O
aE a

Lemma 4.11. Let a be a torsion class with a separating sub-

basis, P. If [P1,P2) is a partition of P, and if 3 =

v aR/I for i = 1,2, then 3' D a2 = (0).
IEP^ R/I 1 2
IEP


Proof. Define P(M) = G P I(M), where I is the I-uniformly
IEP2

negligible submodule of M. Lemma 3.8 implies that each I is

a torsion preradical. Lemma 4.10 and the fact that P is a

separating subbasis imply that P is a torsion preradical. 52

is the torsion class generated by P. For all R-modules, M,

and left ideals, I, of P2, P (M) n 3'(M) = (0). Therefore,

P(M) n jI(M) = (0), which implies that 3' n 32 = [0).


Lemma 4.12. Let P be a torsion preradical, and let 32 be the

torsion class generated by p. Let 31 be a torsion class such

that P(M) n 71(M) = 0 for all M E R-mod. If Ext(M,T) = 0 for

all M = P(M) and T = 3 (T), then Ext(M,T) = 0 for all M = 52(M)

and T = 31(T)


Proof. The proof will follow by induction on the P-length of

M. Recall [10] that a '2-module has P-length a if a is the

smallest ordinal number such that M = P (M). If M has P-length

1, then the assertion follows from the hypothesis. Assume that

Ext(M,T) = 0 for all T E 31 and M = P (M), where 0 < a.






39

Suppose that a = 3 + 1, and M = P (M). Consider any

exact sequence of the form

0 T X M O,

where T E 31. This sequence induces the exact sequence
-1 f
0 T f ( (M)) p (M) 0.

The induction hypothesis applies in this case; so the sequence

splits. Therefore, f ((p(M)) = T e N, where N 2 P (M). Now

consider the exact sequence
f'
0 T N/N X/N M/p (M) 0,

where f'(m + N) = f(m) + P (M). This sequence satisfies the

conditions of the induction when a = 1. Thus, the sequence

splits and X/N = (T e N)/N e K/N, where K/N M/P (M). The

hypotheses for Lemma 4.5 are satisfied. If n is the natural
-i -i
map from X onto X/N, then X = T e nT (K/N). Since M a n (K/N),

the original sequence splits. Therefore, Ext(M,T) = 0.

Suppose that a is a limit ordinal. Consider an exact

sequence of the form

0 T X f M 0,

where M has p-length a and T E 751 For any ordinal number

3 < a, there is an induced exact sequence
-1
0 T f (P (M)) P0(M) 0.

This sequence splits by the induction hypothesis. Therefore,

f- (P (M)) c T 9 U2(X). Since X = U f- (P (M)), X = Te 2 (X).
3 Thus, the original sequence splits and the lemma is established.[]


Lemma 4.13. Let j be a torsion class with separating subbasis,

P, and let (P1,P2) be a partition of P. Let i = V 9R/I-
IEP






40

where i = 1, 2, and let Ci = [(JR/J is I-uniformly negligible

for some I E P.). Then the following statements are equiva-

lent:

(1) 3(M) = (M) e 32 (M) for all M E R-mod.

(2) Ext(R/J, Mi) = 0, whenever J E P. and M. E ,i'

where i f j.


Proof. (1) = (2). Consider an exact sequence of the form

0 M1 X R/J 0,

where M1 E 31 and J E P2. X is the extension on one member

of 3 by another. Therefore, X is in a, and the hypothesis

implies that X = 31(X) e 32(X). Lemma 4.9 implies that

M1 = a1(X). Hence, the sequence splits and Ext(R/J, M ) = 0.

A similar argument yields the second part of condition (2).

(2) = (1). Let P(M) = e P(M), where P (M) is the
IEP2

I-uniformly negligible submodule of M. The sum is direct

because P is a separating subbasis of 3. Lemma 4.10 implies

that P is a torsion preradical which generates a2.

Assume that T = 0 (T) for some I E P2, and that M E 31.

Consider any exact sequence of the form
f
0 M X T 0.

For all x E T, there is an induced exact sequence
-1 f
0 M f (Rx) f Rx 0.

Since Rx a R/J for some J i C2, condition (2) implies that the

sequence splits. Thus, f-1(Rx) c a1(X) e 32(X), because
-1
Lemma 4.9 implies that M = 3 (X). Since X = U (Rx),
xET
X = a3(X) e 2 (X) = M A 32(X). Hence Ext(T,M) = 0.






41

Now let T = P(T) and let M E 51. Ext(T,M) =

= Ext( E P (T),M) n Ext(P (T),M) = 0. Hence, by
IEP2 IEP2

Lemma 4.12, Ext(T,M) = 0 for all T E 32 and M E 3 '
-1
To complete the proof, let m E nrX (2 [X/1 1(X)]) where nX
is the natural map from X onto X/1 (X), and X is any module.

There exists an exact sequence

Rm + 3 (X)
0 r1(X) Rm + 1((X) (X) -

-1
This sequence splits, so that X 1(32[X/l.(X)]) = 31(X) e 2(X).

A similar argument yields the analagous equality with the

roles of I1 and a2 inter-changed. Proposition 3.6 implies

that (75 V 32) (M) = 93(M) e 2 (M) for all M E R-mod. Since

a = 31 U2' the lemma is established. 0

Lemma 4.14. Let 3 be a torsion class with a separating subbasis,

P. Let ([P,P2) be a partition of P with 5i and Ci defined as

in Lemma 4.13. If 32([rr R/J]/ R/J)jECl = 0, then


32 ( [R/ (M)]/e 3R/I (M)) IP = 0, for all M E R-mod.


Proof. If M is an R-module, then let M* = ([ n 3R/I(M)]/
IEP1

e UR/I(M)). Define P(M) = e I (M). Then Lemma 4.10
IEP1 IEP2
implies that P is a torsion preradical which generates 32.

Let M be an R-module, and let x = (xI) + e /I(M) E
IEP 1R/

2 (M*). Without loss of generality, assume that x E P I(M*)






42

for some I E P2. Rx -a R/J, where J E C2o For all K E Pl'

define 0(K) to be the smallest ordinal number such that

xK E (PK (K) (M). For all K, 0 (K) = a(K) + 1 is not a limit
ordinal.

For each K E P1, define the submodule of M, BK = (PK)() (M).

For all K E PL, xK + BK E PKR/K(M)/BK). Whenever xK / 0,

there exists a homomorphism fK E Hom([Rx + BK]/BK, R/JK), where

R/JK is K-uniformly negligible, and fK(xK + BK) / 0. Let

Y = (fK(XK + BK)) + e R/J E (7 R/J/e R/J) JEC Let r E J.
JEC1 1

Since rx = 0, r annihilates all but a finite number of xK.

Consequently, Jy = 0. This implies that y = 0, because

J E F(32). Consequently, x = 0 and the lemma is established. O

Theorem 4.15. Let U be a torsion class with a separating sub-

basis, P. Let [PP 2) be a partition of P, and define 3i and

Ci as in Lemma 4.13 for i = 1, 2. Then the following conditions

are equivalent:


(1)

(2)


J(M) = 31(M) E a2(M) = 0, for all M E R-mod.

(a) l(n R/J/9 R/J) = 0 = U2(n R/J/e R/J), where

the product and sum run over C2 and C1 res-

pectively.

(b) Ext(R/I, R/J) = 0 = Ext(R/J, R/I) for all I E C

and J E C2.

(c) If G is any set, then i (n R/Ja / R/Ja) = 0,

where a E G and each R/Ja is I-uniformly negli-

gible for a fixed I E P., i 7 j.


1





43

(d) Whenever 0 3 (Rx) Rx R/J 0 is an exact

sequence with J E Cj, i f j, then i (Rx) has

non-limit ordinal P -length, where P (M) =

D P (M).
IEPi
i

(e) Let 0 B Rx R/J 0 be an exact sequence

with B I-uniformly negligible for some I E P.,

and J E C., where i f j. Then B has a C-chain

of finite length.


Proof. Define pl and 2 as in condition (d) of the hypothesis.

Lemma 4.10 and the fact that P is a separating subbasis for 3
1 2
imply that pl and 2 are torsion preradicals which generate 31

and 32' respectively. These definition will hold throughout

the remainder of this proof.

(1) = (2). (a) Assume that there exists a submodule, M,

of n R/J such that (M + R/J)/ e R/J E 1. Without
JEC2 JEC2 JEC2

loss of generality, one may obtain a submodule, N/ e R/J, of
JEC2

(M + e R/J)/ e R/J which is isomorphic to R/K, for some
JEC2 JEC2

K E C1. Consider the exact sequence

0 E- R/J N R/K 0.
JEC2

Since the left end of the sequence is in 2', Lemma 4.13 implies

that the sequence splits, and that R/K is isomorphic to a sub-

module of 7 R/J. However, this is a contradiction because
JEC2
Hom(R/K, TT R/J) a n Hom(R/K,R/J) = 0. The argument for the other
JEC2 JeC2
equality is entirely analagous.







44

(b) This condition follows immediately from Lemma 4.13.

(c) Let G be any set, and let I E P2. Let Mo= 1 R/J ,
aEG
and let N = e R/J where R/J is I-uniformly negligible,
0 a a
aEG
for all a E G. Let L E C1. Then the exact sequence

0 N M M /N 0

yields the exact sequence

Hom(R/L,Mo) Hom(R/L,Mo/N ) Ext(R/L,N ).

However, Hom(R/L,M ) a Hom(R/L, R/J ) = 0. By Lemma 4.13,
aEG a
Ext(R/L, N) =0. Therefore, P (M /N ) = 0, which implies that

3l(Mo/No) = 0. The argument for the other equality is analagous.
(d) Let 0 U1(Rx) Rx R/J 0, where J E C2. By

Lemma 4.13, the sequence splits. Therefore, 51(Rx) is cyclic

and cannot have limit ordinal p -length.

(e) Let 0 B Rx R/J 0 be an exact sequence, where

B is I-uniformly negligible for some I E Pl, and where J E C2'

By Lemma 4.13, this sequence splits and B is cyclic. Therefore,

B has C-length 1.

(2) = (1). Let M E J. By Lemmas 4.9 and 4.13 it is

sufficient to show that Ext(R/J, 7U(M)) = 0, for all J E C2.

First, assume that a (M) has pl-length 1. Then 3~ = e P (M).
I IEP1
Let N = n P (M). For all J E C2, the exact sequence
IEP1
0 71 ( N- N/(M) 0

induces an exact sequence

Hom(R/J, N/ 1(M)) Ext(R/J, Gl(M)) Ext(R/J, N).







45

Condition (a) and Lemma 4.14 imply that Hom(R/J, N/31(M)) = 0.

On the other hand, Ext(R/J, N) Tn Ext(R/J, P (M)). The case
IEP1
1P1
when 31(M) has p l-length 1 will will be complete if Ext(R/J,B)=0,

where J E C2, and B = P (B) for some I E P1. The proof will

follow by induction on the C-length of B.

Suppose that B has C-length 1; i.e. B is a direct sum of

cyclic modules. Then B a e R/J, where each R/J is I-uniformly
6E8
negligible for some fixed I E P1. Let J E C2, and let

M = r R/J The exact sequence
6Eb
0 B M M/B 0

induces an exact sequence

Hom(R/J, M/B) Ext(R/J, B) Ext(R/J, M).

Condition (c) implies that Hom(R/J, M/B) = 0. Ext(R/J, M) a

nr (R/J, R/J ) = 0, by condition (b). Thus, Ext(R/J, B) = 0,

and the first step in the induction is complete. Assume that

whenever B has C-length less than a, Ext(R/J, B) = 0, for all

J E !2 and B I-uniformly negligible for some I E Pl. Assume

that B has C-length a.

If a = 0 + 1, then let [B 6} be a C-chain for B. For

any exact sequence

0 B X R/J 0,

where J E C2, induces an exact sequence
f'
0 B/BB X/BB f R/J 0,

where f'(x + B ) = f(x). Since B/B has C-length 1, the in-

duction hypothesis implies that this sequence splits. Therefore,






46

X/B3 = B/B 9 H/B where H/B R/J. Since B has C-length

3, the induction hypothesis implies that the exact sequence

0 B) H R/J 0

splits. H = B3 8 H1, where H1 R/J. The hypotheses of

Lemma 4.5 are satisfied. Hence, X = B 9 HI, and the original

sequence splits.

Now assume that a is a limit ordinal, and that B has C-

length a. Assume that

0 B X R/J 0

is an exact sequence with B having C-length a. For x E X\B,

there is an induced exact sequence

0 B n Rx Rx R/(B:x) 0.

Condition (e) implies that B n Rx has finite C-length. By the

induction hypothesis, this sequence splits. Therefore,

Rx g a5(X) 9 52(X) for all x E X\B. However, B = 9 (X) by

Lemma 4.9. Hence, X = 31(X) 9e 2(X) = B e 72(X), and the

original sequence splits. This completes the induction

hypothesis, and consequently, the case for 31(N) having P-

length 1 is finished. Assume that whenever al(M) has P-

length less than a, then Ext(R/J, a (M)) = 0 for all J E C2.

Suppose that 3 (M) has yl-length a.

If a = 0 + 1, then the induction hypothesis yields that

Ext(R/J, (1) (M)) = 0 = Ext(R/J, a1(M)/(Pl) (M)). Conse-

quently, Ext(R/J, M) = 0.

Finally, assume that a is a limit ordinal, and that 9 (M)

has l-length a. Suppose that there is an exact sequence

0 M X R/J 0.





47

Let x E X such that f(x) = r + J d 0. There is an exact

sequence

0 M n Rx Rx R/(J:r) 0.

Lemma 4.9 implies that M n Rx = 7 (Rx). Condition (d) implies

that M n Rx does not have limit ordinal P -length. Since (p )
a
is a torsion preradical, the p -length of M o Rx is strictly

less than a. The induction hypothesis applies and Rx c ~ (X)

a2(X). If f(x) = 0, then x E Ms 1(X). Consequently,
X = 3;(X) e 32(X). Lemma 4.9 implies that M = 1(X). The

original sequence splits and Ext(R/J, 31(M)) = 0 for all J C2

and M E R-mod. Lemma 4.13 implies that the theorem is estab-

lished. r3


The next two propositions reduce the problem of when

U(M) = e 3 (M), where (a a) is a collection of subtorsion
aEG
classes of 3, to the problem handled in Theorem 4.15. The

case where G is a finite set leads to a particularly nice

result and will be done first.


Proposition 4.16. Let be a torsion class with a separating

subbasis, P. Let K = V 1,R/I, where (Pl'" ,P ) is a
IF Pk
partition of P. Then the following statements are equivalent:
n
(1) 3(M) = 3'.(M) for all M E R-mod.
i=l

(2) If i f j, then (3i V 3j)(M) = (M) 3j(M) for all

M E R-mod.


Proof. (1) = (2). Let M E 3i V j., where i / j. Condition (1)
n
implies that M = e 3 (M). If k f i,j, then k n (3i V a) =0
i=l 1





48

because P is a separating subbasis of U. Hence, condition (2)

is established.
n
(2) = (1). Define P(M) = e i (M) for all M E R-mod.
i=l1
Lemma 4.10 implies that P is a torsion preradical which generates
n
V = 3. Let M = P2(M), and let m E P2(M). If m P (M),
i=l
then one may assume, without loss of generality, that

m + Pl(M) E 3i(M/P(M)) for some i = l,...,n. Consider the

exact sequence

(*) 0 e (M) Rm + (M) [Rm + +p(M)]/ j (M) 0.
j i ji
Consider the exact sequence

0 P(M)/ 3. (M) Rm+ (M)/e 3. (M) Rm+ P(M)/P(M) 0.
j i jii 3

Both ends of this sequence are in 3i. Hence, Rm+P(M)/e j (M)
jfi
E 3.i Notice that Ext(Rm+ (M)/g9 j (M), 3.(M)) -
j5i ji i
9 Ext(Rm+P(M)/ 3 k(M), 3.(M)), because the direct sum is
jfi jfi
finite. However, condition (2) implies that Ext(Rm+ P(M)/e 3.(M),
j i
e 3.(M)) = 0, and that (*) splits. Therefore, Rm + P(M) e
j i J
n
e6 (M) = P(M). Thus, P2(M) = p(M) = 3(M) for all MER-mod. ]
i=l

Proposition 4.17. Let 7 be a torsion class with a separating

subbasis, P. Let (P )EG be a partition of P. Let 3 =

V UR/I. Then the following statements are equivalent:
IP R/I
a
(1) j(M) = e (M), for all M E R-mod.
aEG

(2) If a E G, then 3(M) = 3a (M) e ( V ) (M), for
all M E R-m aa
all M 6 R-mod.






49

Proof. (1) = (2). This part is immediate, because condition

(1) implies that 3(M) c 3; (M) e ( V a )(M), for all
o afoo
M E R-mod.

(2) = (1). Let p(M) = ,e 7 (M). Lemma 4.10 implies
aEG
that P is a torsion preradical which generates U. Let M E 3,
-i
and let m E n P([M/P(M)]), where n is the natural map from M

onto M/P(M). If m g P(M), then one may assume that there is
-1
an a E G such that m E n (3o [M/P(M)]). There exists J E

F(a ) such that Jm c P(M). By hypothesis, m = mi + m2, where

m E 3a and m2 E V U (M). Without loss of generality,
o o aa

assume that J c (0:ml). Therefore, Jm2 c P(M). Lemma 3.5

implies that J + (0:m2) = R. Consequently, m2 E p(M). Since

mi E P(M), m E P(M). Therefore, U(M) = P2(M) = P(M), for all

M E R-mod. F]


We are now prepared to obtain some corollaries of these

results where 3 is the simple torsion class.


Corollary 4.18 [2, Theorem 3.1]. Let 8 be the simple torsion

class. Then the following statements are equivalent:

(1) R satisfies the Primary Decomposition.

(2) (a) Soc( T S / E S ) = 0, where (S 5) is a
aEG aECa

complete representative set of non-isomorphic

simple R-modules.

(b) Ext(U,V) = 0 whenever U and V are non-isomor-

phic simple R-modules.





50

(c) If G is any set, then U ( n Va/ e V) = 0
aG aEG

whenever U and V are non-isomorphic simple

R-modules and V a V for all a E G.
a
(d) If L c M are left ideals of R such that M/L E 3V

and R/M a- q then the P-length of M/L is not a

limit ordinal where P(M) = Soc(M).


Proof. Let P = [Ma )E, where [R/M 3aEG is a complete repre-

sentative set of non-isomorphic simple R-module. Then P is a

separating subbasis for g.

(1) = (2). Proposition 4.17 implies that g(M) = R/Ma(M) e


V /M (M) for all M E R-mod and a E G. Since each S is M -
30 R/M a a
uniformly negligible, condition (a) follows from condition (a)

of Theorem 4.15. Lemma 2.8 implies that every M -uniformly

negligible module is a direct sum of simples each of which is

isomorphic to R/Ma. Hence conditions (b), (c), and (d) follow

immediately from conditions (b), (c), and (d) of Theorem 4.15.

(2) = (1). Let P1 = (Ma 1) and let P2 = (M]aE\(al


By Proposition 4.17, it suffices to show that S(M) = 3R/M (M) e
/ a1

v 5R/M (M), for all M E R-mod. Lemma 2.8 implies that every
a/a1 a

M -uniformly negligible module is a direct sum of cyclic R-

modules, and hence condition (e) of Theorem 4.15 is satisfied.

By Lemma 2.8 and conditions (a) (d) of this corollary, con-

ditions (a) (d) of Theorem 4.15 are satisfied. Therefore,

the corollary is established. O






51

Corollary 4.19. For a left Noetherian ring, R, let 7 be a

torsion class with separating subbasis, P. Let (P ,P2) be

a partition of P, and let i = v R/I for i = 1,2. Then
iSP. R/I
EPi
the following statements are equivalent:

(1) U(M) = U1(M) e 32(M), for all M E R-mod.
(2) Ext(R/I, R/J) = 0 = Ext(R/J, R/I), for all J E C1

and I E C2 where C1 and C2 are defined as in

Lemma 4.13.


Proof. (1) = (2). This follows immediately from Theorem 4.15.

(2) = (1). It is sufficient to show that conditions (a) -

(e) of Theorem 4.15 are satisfied. Conditions (d) and (e)

follow directly from the fact that R is Noetherian. Assume

aI(M/N) 1 0, where M = n R/J and N = e R/J. There exist
JEC2 JEC2

I E F(3O) and (x ) + N E M/N such that I((xj) + N) = 0, or

I(xj) c N. Since R is Noetherian, I is finitely generated.

Each generator annihilates all but a finite number of x There-

fore, I annihilates all but a finite number of x However,

since Rxj E 32 and I E F( r1), the only xj that I could anni-

hilate were zero already. Hence condition (a) is established.

Let G be any set. Consider a collection, (R/Ja)aEG of

cyclic R-modules which are all I-uniformly negligible for some

fixed I E 92. Assume that 3 (M/N) f 0, where M = n R/Ja
aEG
and N = e R/J There exist I E F(31) and (x ) + N E M/N
aEG
such that I((xj) + N) = 0, or Ixj c N. Since R is Noetherian,








52

I is finitely generated. Each generator annihilates all but

finite number of xj. However, since Rxj E 32, the only x that

I could annihilate were zero already. Hence, condition (c) is

established. Condition (b) is condition (2) of the corollary,

and the corollary is established by Theorem 4.15. 0


Corollary 4.20 [6, Theorem 1.1]. For a left Noetherian ring,

R, the following statements are equivalent:

(1) R satisfies the Primary Decomposition.

(2) Ext(U,V) = 0 whenever U and V are non-isomorphic

simple R-modules.


Proof. Let P = [M) aEG be a collection of left ideals of R

such that (R/Ma aEG is a complete representative set of non-

isomorphic simple R-modules. Then P is a separating subbasis

for g. Let Mo E P. Consider the partition of P, ((M)oP\(Mo)).

(1) = (2). This follows immediately from Corollary 4.18.

(2) = (1). Corollary 4.19 and the discussion above imply

that S(M) = R/Mo (M) V R /M (M) for all M E P.
S M EP\(Mo) a

Proposition 4.17 implies that R satisfies the Primary Decom-

position. ]


Let 3 be a torsion class of simple type. If 5 admits a

decomposition, then does it necessarily follow that a has the

Primary Decomposition? The following example answers this

question in the negative.









Example. Before the example can be stated several lemmas are

required.


Lemma 4.21. If R is a Noetherian ring, then the upper tri-

angular matrix ring over R is also Noetherian.


Proof. If R is a Noetherian ring, then the ring S = ( )
0 R
is also a Noetherian ring. Since the upper triangular matrix

ring is a finitely generated module over S, it must be a

Noetherian module, and hence a Noetherian ring. 0


Lemma 4.22 [2, Corollary 4.3]. Let U s R/M and V R/N be non-

isomorphic simple R-modules. If M and N are two-sided ideals

such that NM = N n M, then Ext(U,V) = 0.


Now the example can be demonstrated. Consider the Abelian

group G = Z/(2) Z/(3). Make this group into a zero ring, i.e.

define xy = 0, for all x,y E G. Following McCoy [16, page 8],

imbed G into a ring, R, with identity. In this construction,

R = [(a,b)la E G and b E Z). Addition is defined co-ordinate

wise, and (a,b)(x,y) = (bx + ya,by). Let S1 = ((a,O)a E Z/(2) e0]

and let S2 = ((a,0)]a E 0 e Z/(3)). Then S1 and S2 are non-iso-

morphic simple submodules of R such that Ext(S1,S2) = 0 =

Ext(S2,S1). If Mi is the annihilator of Si, then MM2 = M2M =

M1 D M2. Hence, R is a commutative Noetherian ring with two

non-isomorphic simple R-modules in Soc(R). Furthermore, the

product of the annihilators of these simple R-modules equals

their intersection.




54

Let R be as above, and let T be the upper triangular

matrix ring over R. Let A = ( 1 ), B ( 2 R ), C = (
0 R 0 B O R 2
RR
and D = ( ) be maximal left ideals of T which are also two-
O M2

sided ideals of T. Let P1 = (A,C), and let P2 = (B,D). For

all I E P1 and J E P2, IJ = JI = I n J. Lemma 4.22 implies

that Ext(R/I, R/J) = Ext(R/J, R/I) = 0. However, Ext(R/A, R/C)0O.

To see this, consider the exact sequence

0 S 0 S 0 S





The left side is isomorphic to R/A, and the right side is iso-

morphic to R/C. If this sequence split, then the middle term

of the sequence would be annihilated by A n C, which it is not.
Therefore, Ext(R/A, R/C) 0.

Let 3 be the torsion class of simple type with separating

subbasis P = (A,B,C,D). By Corollary 4.20, a does not satisfy

the Primary Decomposition. Let Pl = (A,C] and P2 = (B,D).

Corollary 4.19 implies that 3 admits a decomposition.


The results developed in this paper are now used to gener-

alize a theorem due to Albu (1, Theorem 5].


Definition 4.23. Let 3 be a torsion class, and let b = (a5raEG

be a collection of torsion subclasses of 3 such that J= V 3'
aEG a
ForM E3', define the s-support of M = (a E G 3' (M) d 0). M is
said to haviie ini -support if the B-support of M is a finite

set. Finally, 3is called finitely O-decomposable if, whenever
ME o such t hat M has finite -supporlt, then M = mi e (M).
aEG
aE~






55

Theorem 4.24. Let 3 be a torsion class with subbasis, P.

Let (PaE ) be a partition of P, and let 3a be the torsion

class with subbasis, P for all a E G. Let C = (JsRIR/J

is I-uniformly negligible for some I E P ). If for all

a E G, 3a n v = 0, then the following statements are
a ^
equivalent:

(1) 3 is finitely S-decomposable.

(2) Ext(R/J, M) = 0, for all J E C and M E 3 where
a a a
a f a'.


Proof. (1) = (2). Consider an exact sequence of the form

0 M X R/J 0,

where J E a', and M E 'a,, and a a'. Then claim that 3 (X)= 0

for all P g (a,a'). Let x E 3' (X). (M:x) E F( a), and

(0:x) E F(3'). Lemma 3.5 implies that (M:x = (M:x) + (0:x) = R.

Therefore, x E M. This contradicts the fact that M n 3-(X) = 0,

unless x = 0. Thus X has finite &-support. Condition (1) implies

that the original sequence splits and (2) is established.

(2) = (1). Let P(M) = 3a (M). Lemma 4.10 implies that
aEG a
P is a torsion preradical which generates 3. Assume that M

is a torsion module which has finite i-support. Let X E P2(M)

such that X + p(M) E a' (M/p(M)). By the construction of ao,

one may assume, without loss of generality, that X + p(M) E

I (M/p(M)), for some I E Pa Assume that a 0 -support of M.
0
By condition (2) the exact sequence
n n n
0 a 3 (M) Rx + E 3 (M) R/(E 3a (M):x) 0,
i=l i i=l i i=l i












56

where al,...,a' n is the D-support of M, splits. This con-

tradicts the fact that 3a (M) = 0.
a

Now assume that a E s-support of M. Consider the exact

sequence

0 9e S(M) Rx + P(M) (Rx+ P(M)/ e a(M) 0.
aao aa0o a

The right side of this sequence is in 3T However, condition
a
(2), Lemma 4.13, and the fact that M has finite b-support imply

that this sequence splits. Therefore, x E (a (M) c P(M). Con-

sequently, P(M) = P2(M) = N(M) and 3 is finitely s-decomposable.


Corollary 4.25 [1, Theorem 5]. Let S be the simple torsion

class, and let P = [M aaEG be a separating subbasis for S,

where each M is a maximal left ideal of R. For all a E G,

let 3a = 3R/M If = [a)aE G then the following statements

are equivalent:

(1) 8 is finitely O-decomposable.

(2) Ext(R/M M) = 0, for all M E 3a ,, where a o a'.


Proof. (1) = (2). This follows immediately from Theorem 4.24.

(2) = (1). This follows from Lemma 2.8 and Theorem 4.24. O








CHAPTER 5

Completely Reducible Torsion Classes

Definition 51l. Let 3 be a torsion class on R-mod. Then 5

is called completely reducible if whenever 31 is a torsion

class in 3, then there exists a torsion class, j2, contained

in 3 such that 3(M) = 1 (M) e 32(M) for all M E R-mod.


The primary goal of this chapter is to characterize those

rings for which all torsion classes are completely reducible.

These rings have been studied by Dlab [8]. A sequence of

lemmas will lead to the main result.


Lemma 5.2. If 3 is a torsion class and if F is a complete

representative set of non-isomorphic simple modules in 5,

then the following statements are equivalent:

(1) 7 is completely reducible.

(2) For each S0 E F, u(M) = S (M) e VSEr\(S S (M).


Proof. (1) = (2). Let 3 be a completely reducible torsion

class, and let r be a complete representative set of non-iso-

morphic simple R-modules in 3. Let 3 be the smallest torsion

class containing all the simple R-modules in 3. Therefore,

3p e 3. By hypothesis, there exists a torsion class, U, such

that J(M) = 3p(M) e h(M) for all M E R-mod. If U were not

the zero class, then there would be a non-zero simple R-module,







S, in U. Since U s 3, S E 3. By the definition of F, S is

isomorphic to an element of r. Therefore, S E 9p, which con-

tradicts the fact that p n f = (0). Consequently, U = [0),

and 3 is of simple type.

Let S E F. By hypothesis, there exists a torsion class,

U, such that U(M) = h(M) 9 ;S (M) for all M E R-mod. Since 5

is of simple type, u is of simple type also. Hence, h= V 3S
SEF\{S )

By Lemma 4.17, 3 satisfies the Primary Decomposition.

(2) = (1). Let i be a torsion class contained in 3, where

a satisfies condition (2). Let V = VSEr-\ 3S. 3(M) = U(M) e

U(M) for all M E R-mod, because 3 satisfies the Primary Decom-

position. []


Lemma 5.3. Let R and S be rings with identity, and let f:R S

be a ring homomorphism. Any S-module, M1 may be made an R-

module via the definition rm = f(r)m for all r E R and m 1 M.

If 3 is a torsion class on R-mod, then 3l = (M E S-modjM E 5,

if M is considered as an R-module) is a torsion class on S-mod.


Proof. Recall [5, Theorem 2.3] that a class of modules, 3, is

a torsion class if 3 is closed under submodules, homomorphic

images, direct sums, and extensions of one module in 3 by

another. Noting this fact, the result follows routinely and

will be omitted. C]


Lemma 5.4. Let R = S + T (ring direct sum), and let U be a

torsion class on S-mod (T-mod). If M is an R-module, then

SM(TM) is an S-module (T-module) with the multiplication





59

defined in the obvious way. Then )' = (M E R-modJSM E u)

and h" = [M E R-modjTM E I) are torsion classes on R-mod.


Proof. The proof is routine and will be omitted.


Lemma 5.5. Let R = R + R2 + ... + Rn (ring direct sum).

Then the following statements are equivalent:

(1) All torsion classes on R-mod are central splitting.

(2) All torsion classes on R.-mod are central splitting
1
for i = 1, 2, ...,n.


Proof. (1) = (2). It suffices to show that if 3 is a torsion

class on R -mod, then 3 is central splitting. By Lemma 5.3,

a = (M E R-modjR1M E 3) is a torsion class on R-mod. Since

condition (1) implies that 31 is central splitting, there

exists a torsion class, U on R-mod such that M= U (M) 9 U (M)

for all M E R-mod. In particular, R = (R) e U (R) and
31 = (M E R-modul(R)M = 0). R1 = (R ) U1 (R) and

l (R ) = R1 1(R1) E 3, because (Ri) C 3 Since l(R) F(3),

(M E R -mod ll(R )M = 0) is contained in 3. Conversely, let

M E 3. Make M an R-module via the definition rm= (rl+ ...+ rn)m=

r m for all m E M and r E R. Since RM =M M 3J, M E 3 There-
1 1
fore, 1, (R)M = 0, and consequently, (R )M = 0. Therefore, a

is central splitting.

(2) = (1). Let 3 be a torsion class on R-mod, and let f.

be the canonical map from R onto R. for i = 1,2, ...,n. For

each i = 12, 2...,n, let 3i be the torsion class induced on R -

mod by f., as described in Lemma 5.4. By condition (2) each

3i is central splitting. Then each Ri = 3i(Ri) + Ki (ring





60

direct sum), where K. is the minimal ideal in F(i.). Then
n n n n
R = ( e (R.)) + K.. Since e .i(R.) E J, K K. EF(5).
i=l 1 1 i=l 1 i=l 1 i=l 1
n
Therefore, (M E R-modl( e K.)M = 0) is contained in 3. Con-
i=l
n
versely, let M E 5. Then M = e R.M. If one makes R.M into
1 1
i=l

an R.-module in the obvious way, then R.M. E J. for all i =

1, 2, ..., n. Therefore, R.M is annihilated by e R. E K.
i j
for all i. M is annihilated by the intersection of annihilators,
n
which is 9 K.. Hence, 3 is central splitting. F
i=l 1


We are now ready for the main theorem of this chapter.


Theorem 5.6. For any ring, R, the following statements are

equivalent:

(1) All torsion classes on R-mod are central splitting.

(2) R is a finite direct sum (ring direct sum) of full


(3)

(4)


matrix rings over local semi-artinian rings.

All torsion classes on R-mod are completely reducible.

R is semi-artinian and satisfies the Primary Decom-

position.


Proof. (2) o (4). This is known [8, Theorem 2].

(1) = (2). Let 8 be the simple torsion class. Condition

(1) implies that S is central splitting. Hence, there exists

a torsion class, U, such that M = g(M) e u(M) for all M E R-mod.

However, u cannot contain any simple R-modules, because

S n u = (0) and all simple R-modules are in g. Therefore,

1 = (0), and g = R-mod. -Thus, R is semi-artinian.






61

Let S be a non-zero simple R-module, and let 3S be the
o
S -primary. Condition (1) implies that JS is central split-

ting. Hence, there exists a torsion class, h, such that

M = UJ(M) 3S (M) for all M E R-mod. In particular, R =

U (R) 3US (R) If S (R) = 0, then R = U (R) and S = (0).
o o o
Since So 0, this is impossible. Therefore, an isomorphic

copy of So appears in soc(R).

hl(R) and US (R) are both two-sided ideals of R, because
0
they are the torsion submodules of R for some torsion class.

Hence, their sum is a ring direct sum. First, a short examina-

tion of 3S (R) is in order. Lemma 5.5 implies that all torsion
0
classes in So(R)-mod are central splitting. From the first

paragraph of this proof, it follows that aS (R) is a semi-

artinian ring. From the second paragraph one deduces that

there is a unique simple 3S (R)-module up to isomorphism. Con-

sequently, there are precisely two torsion classes on 7 (R)-

mod.

Now proceed to the other summand of R; namely u1(R). By

Lemma 5.5, every torsion class on uA(R)-mod is central split-

ting. Let S1 be a simple Ul(R)-module. Then ul(R) contains

an isomorphic copy of S Observe that S1 can be made into

an R-module in a canonical manner, and that S is not iso-

morphic to S1 as R-modules. By repeating the process described

in the preceding paragraph, hl(R) = 3S (R) U 2(R). By making

an inductive construction, one obtains that R = 35 (R)

SI (R) G ... 9 JS (R) U1n(R), where Sii=1,2,...,n is a set
1 i L J-J, . 1







62

of non-isomorphic simple R-modules, and Un is a torsion class
n
on R-mod. Let In = S. (R), and consider the chain of
1=1 i
ideals I 12 C ... C I c ... If this chain terminates
n
in a finite number of steps, then R = 9 R. (ring direct sum)
i=l 1
where each R. is a ring with only two torsion classes on R.-
1 1
mod. However, each R. is isomorphic to a full matrix ring

over a local semi-artinian ring [8, Theorem 2], and condition

(2) is established.

Hence, it suffices to show that the chain IL c 12 .

c In ... terminates in a finite number of steps. Observe
n
that the chain may be assumed to be strictly increasing. Thus,

J=U In is an idempotent two-sided ideal of R. Consequently,
n=l
UR/J = (M E R-modlJM = 0) is a torsion class of R-mod with J

being the minimal ideal in F(UR/J) [13, Corollary 2.3]. By

hypothesis, aR/J is central splitting. There exists a torsion

class, U, such that R = U(R) 9 UR/J(R), where J = U(R). Let

1 = e + f, where e E J and f E R/J(R). Since e E J, e E In

for some integer n. In this case, eSn+p(R) = 0 for all p > 1.
n+p
Since e is the identity for J, this implies that the chain

terminates in a finite number of steps. This contradicts the

assumption that the chain is infinite, and condition (2) is

established.

(4) = (3). Let 3 be a torsion class on R-mod. Since R

is semi-artinian, U is of simple type. Let T be a complete

representative set of non-isomorphic simple R-modules in 3.


















63

Then a = v aS* Since R satisfies the Primary Decomposition,
SEP
J(M) = e 9S(M) for all M E R-mod. If 3i is a torsion class
SEP
contained in 5, then define U = v 1 3. 3(M) = a 1(M) 9U(M)
SEF\\a "
for all M E R-mod. Thus, 5 is completely reducible.

(3) = (4). This follows immediately from Lemma 5.2 by

letting 3 = R-mod.

(4) = (1). There exist a finite number of simple R-modules,
n
S1,...,Sn, such that R = 3S. (R). If S is any simple R-
i=l 1
module, then S Si for some i = 1, 2, ..., n. Let 7 be any

torsion class on R-mod. Then 5 = V aS. 3(R) = e (3S (R)).
SiEJ i SiEU i

Let h = V S Then R = 3(R) hU(R), and a is central


splitting. n








BIBLIOGRAPHY


1. M. T. Albu, Modules de Torsion A Support Fini, C. R.
Acad. Sc. 273 (1971), 335-338.

2. J. S. Alin, Primary Decomposition of Modules, Math.
Zeit. 107 (1968), 319-325.

3. R. L. Bernhardt, Splitting Hereditary Torsion Theories
Over Semiperfect Rings, Proc. Amer. Math. Soc. 22
(1969), 681-687.

4. J. H. Cozzens, Homological Properties of the Ring of
Differentiable Polynomials, Bull. Amer. Math. Soc.
76 (1970), 75-79.

5. S. E. Dickson, A Torsion Theory for Abelian Categories,
Trans. Amer. Math. Soc. 121 (1966), 223-235.

6. S. E. Dickson, Decomposition of Modules I: Classical Rings,
Math. Zeit. 90 (1965), 9-13.

7. S. E. Dickson, Decomposition of Modules II: Rings Without
Chain Conditions, Math. Zeit. 104 (1968), 349-357.

8. V. Dlab, On a Class of Perfect Rings, Can. J. Math. XXII,
4 (1970), 822-826.

9. B. Eckmann and A. Schopf, Uber Injective Modular, Archiv
der Math. 4 (1953), 75-78.

10. L. Fuchs, Torsion Preradicals and Ascending Loewy Series
of Modules, J. Reine Agnew. Math. 239/240 (1970),
169-179.

11. P. Gabriel, Des Categories Abeliennes, Bull. Soc. Math.
France, 90 (1962), 323-448.

12. A. W. Goldie, Torsion Free Modules and Rings, J. Algebra
1 (1964), 268-287.

13. J. P. Jans, Some Aspects of Torsion, Pacific J. Math. 15
(1965), 1249-1259.

























14. S. MacLane, Homology, Berlin-G6ttingen-Heidelberg, Springer,
1963.

15. J. M. Maranda, Injective Structures, Trans. Amer. Math.
Soc. 110 (1964), 98-135.

16. N. H. McCoy, The Theory of Rings, New York, Macmillan,
1964.

17. B. Stenstr6m, Rings and Modules of Quotients, Lecture Notes
in Mathematics, 237, Springer-Verlag Berlin (1970).

18. M. L. Teply, Direct Decomposition of Modules, Math. Japon.
15 (1970), 85-90.

19. M. L. Teply, Uniformly Negligible Modules, J. London Math.
44 (1969), 424-428.

20. E. A. Walker and C. Walker, Quotient Categories and Rings
of Quotients, preprint, 1963.
























BIOGRAPHICAL SKETCH



Richard Bronowitz was born in New York City on

April 13, 1946. Upon graduating from Miami Norland

Senior High School in 1963, he attended the Massachusetts

Institute of Technology where he received his B.S. in

mathematics in 1967. He then entered the University of

Florida Graduate School, Department of Mathematics. On

June 14, 1970, he married Carol Linda Stein. They have

a daughter, Amy Michelle, born on May 29, 1971.










I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




Mark L. Teply, Chai man
Assistant Professor of Math atics






I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




Alexander R. Bednarek
Professor and Chairman of Mathematics






I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




I OPr1^AA A /. Ai
Thomas T. Bowman
Assistant Professor of Mathematics












I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.





Thomas H. Hanson
Assistant Professor of Mathematics





I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.


Professor of Education


This dissertation was submitted to the Department of
Mathematics in the College of Arts and Sciences and to the
Graduate Council, and was accepted as partial fulfillment
of the requirements for the degree of Philosophy.

August, 1972


Dean, Graduate School




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