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Decomposition of torsion theories 

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iv, 65 leaves. : ; 28 cm. 

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Bronowitz, Richard, 1946 

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1972 

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1972 
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Torsion ( lcsh ) Mathematics thesis Ph. D Dissertations, Academic  Mathematics  UF 

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Thesis  University of Florida. 

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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 semiartinian 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 moduletheoretic terms. That is to say, an element
of an Abelian group is a torsion element if it has a nonzero
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 nonzero 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 Rmod, the category of left unital Rmodules,
2
where R is any ring with identity. All modules in this work
will be members of Rmod. According to Dickson, a torsion
theory is a pair, (3,3), of classes of Rmodules satisfying
the following axioms:
(1) n 3 = (0);
(2) 3 is closed under homomorphic images;
(3) 3 is closed under submodules;
(4) For all Rmodules, M, there is a submodule, 3(M),
of M such that 3(M) E 3 and M/3(M) E 3.
A class of Rmodules, 3, is called a torsion (torsion
free) class if it is the first (second) coordinate of some
torsion theory. A class of Rmodules, 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 torsionfree 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 RmodIHom(T,M) = 0 for all T E 3). Simi
larly, if a torsionfree class, 3, is specified, then the
torsion theory having 3 as its torsionfree class is uniquely
3
determined. In this instance, 3 = (M E RmodjHom(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 torsionfree class is closed
under injective envelopes. If M is an Rmodule, 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 11 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 Rmodules
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 Rmod. 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 RmodlHom(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 Rmodule.
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 RmodjZ2(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 torsionfree 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 nonisomorphic simple Rmodules, then JG = (MIEvery
nonzero 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 torsionfree 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 Sprimary torsion class. If G is a complete
representative set of nonisomorphic simple Rmodules, the
torsion class is called the simple torsion class and is de
noted by 8. The torsionfree 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 Rmod. 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 Rmod. 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 Rmod?
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 Rmodules 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 Rmod 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
Rmodule. M is called Iuniformly negligible (Ibounded) 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
Rmodule, M, is called 7uniformly negligible (abounded)
if there is some I E F(3) such that M is Iuniformly negli
gible (Ibounded).
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
Iuniformly negligible Rmodules is closed under submodules,
homomorphic images, and arbitrary direct sums.
Proof. Let I be a left ideal of R, and let M be an Iuniformly
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 Iuniformly 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 Iuniformly negligible.
Now let (Ma aEG be an arbitrary collection of Iuniformly
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 Iuniformly
i=l rEB
negligible. C
Proposition 2.3. If I is a twosided ideal, then a module, M,
is Ibounded if and only if M is Iuniformly negligible.
Proof. The proof follows immediately from the fact that
I c (I:r) whenever I is a twosided 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 nonnegative 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 (ta.)RM,
i=l
where ai E F2[t,p]\(0).
Lemma 2.7. If I is a nonzero right ideal of RM, then RM/I is
isomorphic to a finite direct sum of simple modules.
n
Proof. If I is a nonzero right ideal of R then I= n (ta.)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 (tai)RM R
i=l M RM
0  
n+l n+l n
n (tai)RM n (ta i) (tai)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 Nuniformly negligible
module, Proposition 2.2 yields that T is an Nuniformly 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 Muni
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
Muniformly 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 Rmod to Rmod 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 Pmodule.
Proposition 2.10. A subfunctor, P, of the identity functor on
Rmod 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 Pmodules is closed under both submodules
and homomorphic images.
Proof. ( ). Let M be a Pmodule, and let N g M. Then
P(N) = N n P(M) = N n M = N. Hence N is also a Pmodule. 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 Pmodule. Therefore, the class of
Pmodules 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 Rmodule, 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 Pmodule contained
in M is also contained in P(M). Condition (1) of the defini
tion follows from the fact that the class of Pmodules 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 Pmodules.
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 Umodules. 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 Rmod). 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 nonempty 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. Rmod is a torsion class which contains 1 n 2. Hence
G = ([ E (,V,A)~1 2 1 U 23) is a nonempty 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 Rmod,
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 cM. 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 Pmodules 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
Pmodule. 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
pmodules. In fact, by Proposition 2.10, U is the smallest
torsion class containing the class of pmodules.
Proposition 3.4. If l is defined as above, then U = 5 v 3 '
Proof. Since all modules in 3 U 31 are Pmodules, 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 Vmodules, 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 Rmod. 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 nonzero 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 nonzero 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 nonzero
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 Rmod.
1 1
(2) (pMI )(3[M/31(M)]) = (M) e 31(M) = (n1 )(u [M/3(M)]),
for all M E Rmod.
(3) 3(M/[3(M) e31(M)]) = (0) = 31(M/[3(M) 31(M)]), for
all M E Rmod.
(4) (3 V 31) (M) = 3(M) 3'1(M), for all M E Rmod.
1
Proof. (1) = (2). If m E (pM )(3[M/, (M)]), where M is an
Rmodule, 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 Rmodule. 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 Rmodule. 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 Rmod.
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 Rmod, 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 Rmod. 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
Pmodules 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 Rmod. 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 Pmodules 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 Pmodules 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 pmodules. 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, wellorder G,
so that C might be thought of as a cardinal number. Reindex
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 Rmodule, 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 Rmod.
Proof. Let M be an Rmodule, 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 Rmod. 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 pImodule is also a pmodule.
If r E R, then the fact that Rr + I R/I implies that
IRr + I
R/(I:r) Rr+ I is a Pmodule. If frl,...,rn) g R, then
I
n n n
e R/(I:r.) is a pmodule. 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 Pmodule.
i=l
Since every p module is a homomorphic image of a direct sum
of modules of this form, every P module is a pmodule. 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 Rmodule 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 nonzero homomorphic image of M
has a nonzero Iuniformly negligible module for some I E P ).
Proof. Let C = (M E RmodjM is Iuniformly 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 Plength X(M). Assume that M E 5 and that M
1. I is cas
has nlength 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 Plength less than a. Suppose that M has Plength
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 plength 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 = Rmod. When does a
torsion class have a complement?
Proposition 3.12. Let 3 and 31 be torsion classes on Rmod,
and let 3 and 51 be the torsionfree 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' = Rmod, 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 Rmodule
W(3 V 35) (M) E 5 n 51, which implies that M = (7 V U1) (M).
Since M was arbitrary, 5 V 51 = Rmod. 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 nonisomorphic 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 nonisomorphic 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 semiartinian if 8 = Rmod.
Corollary 3.15. (,v,A) is complemented if and only if R is
semiartinian.
Proof. (,V,A) is the principal lattice ideal generated by
Rmod. By hypothesis, Rmod 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 Rmod.
If a ring satisfies the Primary Decomposition, and if the
ring has at least two nonisomorphic simple Rmodules, 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 Rmod.
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 nonzero rings.
(3) If I is the minimal ideal in F(y), then there is a
central splitting torsion class on R/Imod.
Proof. (1) = (2). Assume that 3(M) = g1(M) e 32(M) for all
M E Rmod. Furthermore, assume that neither U1 nor 32 is zero.
That is, assume that the decomposition is nontrivial. 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 twosided 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 twosided
ideal of R. Thus, J/I and K/I are twosided 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 twosided ideals of R/I, the collection of
J/Ibounded (K/Ibounded) 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 twosided ideals of R/I. Then
J and K are idempotent twosided ideals of R. Let u1 be the
class of Jbounded modules, and let U2 be the class of Kbounded
modules. Ul and U2 are TTFclasses 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 twosided ideals
of R. Consequently, the class of Ibounded 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 Ibounded 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 Rmodule. If G is an ordinal
number, and if (M )aEG is a collection of submodules of M,
then the collection is called a Cchain 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 Rmodule has at least o:.. Cchain.
Proof. For any nonzero Rmodule, M, let M1 be any nonzero
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 nonzero 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 Cchain for M. O
The manner of constructing a Cchain is far from unique.
However, there is some way of associating modules and their
Cchains.
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 Cchain for M. Then M is said to
have Clength a(M).
Lemma 4.9. Let j and 31 be torsion classes on Rmod 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 Rmod. 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 Rmod. Then P(M) = P (M) is a
aEG a
torsion preradical. Moreover, the smallest class containing
the class of Pmodules is v 3
aEG
37
Proof. First note that Z P (M) is, indeed, a direct sum.
aEGa
Let M be an Rmodule. Then
P(P (M)) = P (P (M)) = e P (M) = p(M).
aEG aEG
Assume that M is an Rmodule 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 Pmodules 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 pmodules 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 Pmodule 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 Iuniformly
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 Rmodules, 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 Rmod. 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 Plength of
M. Recall [10] that a '2module has Plength a if a is the
smallest ordinal number such that M = P (M). If M has Plength
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 plength 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 Iuniformly negligible
for some I E P.). Then the following statements are equiva
lent:
(1) 3(M) = (M) e 32 (M) for all M E Rmod.
(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
Iuniformly 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, f1(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 interchanged. Proposition 3.6 implies
that (75 V 32) (M) = 93(M) e 2 (M) for all M E Rmod. 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 Rmod.
Proof. If M is an Rmodule, 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 Rmodule, 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 Kuniformly 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 Rmod.
(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 Iuniformly 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
nonlimit 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 Iuniformly negligible for some I E P.,
and J E C., where i f j. Then B has a Cchain
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 Iuniformly 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 Iuniformly 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 Clength 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 pllength 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 llength 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 Clength of B.
Suppose that B has Clength 1; i.e. B is a direct sum of
cyclic modules. Then B a e R/J, where each R/J is Iuniformly
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 Clength less than a, Ext(R/J, B) = 0, for all
J E !2 and B Iuniformly negligible for some I E Pl. Assume
that B has Clength a.
If a = 0 + 1, then let [B 6} be a Cchain 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 Clength 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 Clength
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 Clength 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 Clength. 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 yllength 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 llength 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 Rmod. 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 Rmod.
i=l
(2) If i f j, then (3i V 3j)(M) = (M) 3j(M) for all
M E Rmod.
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 Rmod.
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 MERmod. ]
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 Rmod.
aEG
(2) If a E G, then 3(M) = 3a (M) e ( V ) (M), for
all M E Rm aa
all M 6 Rmod.
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 Rmod.
(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 Rmod. 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 nonisomorphic
simple Rmodules.
(b) Ext(U,V) = 0 whenever U and V are nonisomor
phic simple Rmodules.
50
(c) If G is any set, then U ( n Va/ e V) = 0
aG aEG
whenever U and V are nonisomorphic simple
Rmodules 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 Plength 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 nonisomorphic simple Rmodule. 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 Rmod 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 Rmod. 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 Rmod.
(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 Rmodules which are all Iuniformly 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 nonisomorphic
simple Rmodules.
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 Rmodules. 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 Rmodules. If M and N are twosided 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 coordinate
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 noniso
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
nonisomorphic simple Rmodules in Soc(R). Furthermore, the
product of the annihilators of these simple Rmodules 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 ssupport of M = (a E G 3' (M) d 0). M is
said to haviie ini support if the Bsupport of M is a finite
set. Finally, 3is called finitely Odecomposable 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 Iuniformly 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 Sdecomposable.
(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 isupport. 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 Dsupport of M, splits. This con
tradicts the fact that 3a (M) = 0.
a
Now assume that a E ssupport 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 bsupport 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 sdecomposable.
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 Odecomposable.
(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 Rmod. 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 Rmod.
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 nonisomorphic 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 noniso
morphic simple Rmodules in 3. Let 3 be the smallest torsion
class containing all the simple Rmodules 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 Rmod. If U were not
the zero class, then there would be a nonzero simple Rmodule,
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 Rmod. 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 Rmod, 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 Smodule, 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 Rmod, then 3l = (M E SmodjM E 5,
if M is considered as an Rmodule) is a torsion class on Smod.
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 Smod (Tmod). If M is an Rmodule, then
SM(TM) is an Smodule (Tmodule) with the multiplication
59
defined in the obvious way. Then )' = (M E RmodJSM E u)
and h" = [M E RmodjTM E I) are torsion classes on Rmod.
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 Rmod 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 RmodjR1M E 3) is a torsion class on Rmod. Since
condition (1) implies that 31 is central splitting, there
exists a torsion class, U on Rmod such that M= U (M) 9 U (M)
for all M E Rmod. In particular, R = (R) e U (R) and
31 = (M E Rmodul(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 Rmodule 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 Rmod, 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 Rmodl( 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 Rmod are central splitting.
(2) R is a finite direct sum (ring direct sum) of full
(3)
(4)
matrix rings over local semiartinian rings.
All torsion classes on Rmod are completely reducible.
R is semiartinian 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 Rmod.
However, u cannot contain any simple Rmodules, because
S n u = (0) and all simple Rmodules are in g. Therefore,
1 = (0), and g = Rmod. Thus, R is semiartinian.
61
Let S be a nonzero simple Rmodule, 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 Rmod. 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 twosided 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 Rmodule in a canonical manner, and that S is not iso
morphic to S1 as Rmodules. 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 JJ, . 1
62
of nonisomorphic simple Rmodules, and Un is a torsion class
n
on Rmod. 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 semiartinian 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 twosided ideal of R. Consequently,
n=l
UR/J = (M E RmodlJM = 0) is a torsion class of Rmod 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 Rmod. Since R
is semiartinian, U is of simple type. Let T be a complete
representative set of nonisomorphic simple Rmodules in 3.
63
Then a = v aS* Since R satisfies the Primary Decomposition,
SEP
J(M) = e 9S(M) for all M E Rmod. 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 Rmod. Thus, 5 is completely reducible.
(3) = (4). This follows immediately from Lemma 5.2 by
letting 3 = Rmod.
(4) = (1). There exist a finite number of simple Rmodules,
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 Rmod. 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.
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3. R. L. Bernhardt, Splitting Hereditary Torsion Theories
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(1969), 681687.
4. J. H. Cozzens, Homological Properties of the Ring of
Differentiable Polynomials, Bull. Amer. Math. Soc.
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5. S. E. Dickson, A Torsion Theory for Abelian Categories,
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6. S. E. Dickson, Decomposition of Modules I: Classical Rings,
Math. Zeit. 90 (1965), 913.
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Chain Conditions, Math. Zeit. 104 (1968), 349357.
8. V. Dlab, On a Class of Perfect Rings, Can. J. Math. XXII,
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10. L. Fuchs, Torsion Preradicals and Ascending Loewy Series
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11. P. Gabriel, Des Categories Abeliennes, Bull. Soc. Math.
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12. A. W. Goldie, Torsion Free Modules and Rings, J. Algebra
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Soc. 110 (1964), 98135.
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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

