A TORSION THEORY FOR MODULES OVER

RINGS WITHOUT IDENTITIES

By

JOHN MICHAEL KELLETT

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

1968

To my wife

ACKNOWLEDGEMENTS

I would like to thank my chairman, Dr. W. Edwin

Clark, for his instruction and guidance during my graduate

study and the writing of this dissertation. The prepa-

ration of this paper has been a pleasant and rewarding

experience due, in large measure, to Dr. Clark's patience,

encouragement, and aid.

I am grateful to Dr. John Maxfield and Dr. Alex R.

Bednarek for their timely encouragement and for their aid

in procuring for me generous financial assistance, without

which I would have been unable to continue my work.

To the members of my committee, to the Mathematics

faculty, and especially to Dr. T. O. Moore and Dr. Y. L.

Lee I would like to express my appreciation for their

guidance and friendship.

Finally, I am deeply grateful to my wife, Christine,

for her understanding, help, and patience during these

years of study.

TABLE OF CONTENTS

Page

ACKNOWLEDGEMENTS . . . . . . . . iii

INTRODUCTION . . .. . . . . . . . 1

SECTION

I. THE CLASSES '., RJ, AND Ri (R-MODULES,

TORSION R-MODULES, AND TORSION FREE

R-MODULES) . . . . . . . . 3

II. THE COUPLE (R a, R,) AS A TORSION THEORY

R R

AND THE CLASS OF TORSION RINGS . . .. 20

III. THE EXTENSION OF THE STRUCTURE THEOREM

FOR SEMI-SIMPLE ARTINIAN RINGS TO THE

CATEGORY R . . . . . . .. 35

IV. A CLOSURE OPERATOR ON SUBMODULES AND

A CLASS OF RINGS WHICH SATISFIES A

CERTAIN CLOSURE PROPERTY . . . .. 57

V. A LIST OF OPEN QUESTIONS . . . . 68

BIBLIOGRAPHY . . . . . . . . . . 70

BIOGRAPHICAL SKETCH . . . . . . . .. 71

INTRODUCTION

The initial objective of this dissertation was to

consider the possibility of developing homological ring

theory for a ring R without identity. As a possible

generalization of the unital case, we consider the full

subcategory RR of 2, the category of all R-modules, con-

sisting of those modules M such that R does not annihilate

any nonzero element of M. We call the modules in R torsion

free. This category has the desired property that if R has

an identity, then RR is the category of unital modules.

The category RR is shown to be closed under taking sub-

modules, products, and extensions and is the largest sub-

category of R5 which has these properties and contains no

nonzero modules M such that RM = 0.

A basic theorem in homological ring theory is that

if R has an identity, then the following are equivalent:

(1) All unital modules are projective;

(2) All unital modules are injective; and

(3) R is semi-simple Artinian.

One of our main results is the extension of the above theo-

rem to rings without identities by replacing unital modules

with torsion free modules.

The main difficulty encountered in working with R ,

as compared to the category of unital modules, is that RR

is not, in general, closed under taking quotients. In

fact, a basic result is that if the right annihilator of R

is zero, then RR is closed under taking quotients if and

only if R has an identity. We say that a submodule N of

an R-module M is closed if M/N is in RR. As with normal

subgroups in the category of groups and ideals in the cate-

gory of rings, closed submodules are of particular im-

portance. We show that the closed submodules of an R-

module form a complete modular lattice and give necessary

and sufficient conditions for R such that each such lattice

is complemented.

R is shown to be a torsion free class for a torsion

theory (in the sense of S. E. Dickson [2]) on 1. A. W.

R

Goldie in [5] and J. P. Jans in [8] have considered other

torsion theories on R-, and we have found necessary and

sufficient conditions such that their torsion theories

coincide with the torsion theory determined by R' for

rings with zero right annihilator ideals.

SECTION I

THE CLASSES i, R, AND R (R-MODULES,

RR R

TORSION R-MODULES, AND TORSION FREE R-MODULES)

In this section we define and discuss the elementary

properties of torsion modules and torsion free modules. We

prove the expected result that every module contains a unique

maximal torsion submodule whose quotient is torsion free.

It turns out that if the right annihilator of a ring R is 0,

then R has an identity if and only if the class RR of torsion

free R-modules is closed under homomorphic images. A de-

scription of free modules relative to RR is given that is

very useful in the later sections.

1.1 Definition. We shall call an R-module M torsion

free if, for each nonzero element m in M, Rm / 0.

Of course, if R is the ring of integers and M is a

unital R-module, then the above definition differs from the

usual notion of torsion freeness. See Section II for a

justification of this terminology.

Notation. R2 will be used to designate the cate-

gory of torsion free R-modules. It is a full subcategory

of R.

1.2 Proposition. R is closed under taking submodules.

Proof. The proof is clear from the definition.

1.3 Proposition. If R has an identity, then RR is the

category of unital modules.

Proof. The category of unital modules is clearly a

subcategory of R .

Let M E RR and 0 9 m E M. Then r(lm m) = rl(lm m)

= 0 for all r E R. Hence R(lm m) = 0, which implies that

Im = m, where 1 is the identity of R. Therefore M is a

unital module.

1.4 Proposition. ,R is closed under taking factors if

and only if for each torsion free module M and each m E M,

m E Rm.

Proof. Assume R is closed under taking factors.

Let M E RR with m M; then b/Rm is torsion free by

ass Tumption.

Hence m E Rn since R(m + R.) = Rm = 0.

Conversely, assume for each torsion free module X4

and each m E M, m E Rm.

Let L be a submodule of the torsion free module M.

If R(m + L) = L, then Rm < L and m E L by assumption.

Therefore M/L is torsion free.

1.5 Proposition. If R is nilpotent, then RR contains

only the zero module.

Proof. Assume R = 0 for k, a positive integer.

Note that if M is a nonzero torsion free module, then RM

is a nonzero torsion free module.

Inductively, RkM is a nonzero torsion free module if

M is nonzero torsion free. However, RM = 0 for all R-

modules since Rk = 0. Therefore R2 contains only the zero

module.

Definition. Let R = R X Z(Z the ring of integers)

where addition is defined by (r,k) + (r',k') =

(r + r', k + k') and multiplication by (r,k)(r',k') =

(rr' + kr' + k'r,kk'). As usual, we identify R with

R [0); then R1 is considered an R-module where the module

multiplication is defined in terms of the ring multipli-

cation.

1.6 Proposition. R is a torsion free R-module if and

only if for each nonzero element (r,-n) of R1 there exists

a E R such that ar / na.

Proof. The proof is immediate from the definition.

1.7 Proposition. If R is a ring with a right identity e,

then I = [r E Rier = 01 is an ideal of R and R/I has the

identity (e + I).

Note. I = r(e) = (1 e)R where for x E R, r(x) =

(y E Rlxy = 0] and (1 e)R = fr erir E R}.

1.8 Proposition. If R is a ring with a right identity e

and I = r(e), then an R-module M is torsion free if and

only if M is a unital R/I module (under the definition

(r + I)m = rm).

Proof. If M is a unital R/I module, then M is

clearly a torsion free R-module.

If M is a torsion free R-module, then for m E M and

r E I, Rrm = (Re)rm = R(er)m = 0. This implies rm = 0,

since M is torsion free. Hence IM = 0. Therefore M can

be considered an R/I module. It follows, from Proposition

1.3, that M is a unital R/I module.

Note that if e is a right identity for R, then r(e)

= r(R) = (x E RIRx = 0}. Hence we have

1.9 Proposition. If R is a torsion free R-module and

R has a right identity, then R has an identity.

1.10 Definition. A submodule L of an R-module M will be

said to be closed in M if M/L is torsion free.

1.11 Proposition. L, a submodule of M, is closed in M

if and only if for each m E M, Rm c L = m E L.

Proof. Immediate from the definition.

1.12 Proposition. The intersection of closed submodules

of M is closed in M.

Proof. This proposition follows from 1.11.

1.13 Prooosition. If L is a closed submodule in N and N

is a closed submodule in M, then L is a closed submodule in

M.

Proof. Let m E M and Rm n L; then Rm is contained

in N. By (1.11) m E N, which implies m E L.

1.14 Proposition. If L is closed in M and H is a sub-

module of M, then L n H is closed in H.

Notation. For an R-module N, let TR(N) be the inter-

section of all closed submodules of N. TR(N) will be written

T(N) when there is no possibility of confusion.

1.15 Definition. An R-module K will be called a torsion

module if K is the only closed submodule of K. That is,

K is torsion if its only torsion free homomorphic image

is 0.

Example. Let R = 2Z (Z the integers) and L1, L2,

L3, and L4 be the submodules of RR, 4Z, 6Z, 8Z, and 12Z

respectively. It is easily seen that R/L1 and R/L3 are

torsion modules and R/L2 is torsion free. R/L4 is not

torsion free since R(6 L4) = 0 and it is not torsion

since R/L2 is a homomorphic image of R/L which implies

R/L4 contains a proper closed submodule.

1.16 Proposition. If N is an R-module, then N/T(N) is

a torsion free module.

Proof. The proof follows directly from Proposition

1.12 and the definition of T(N).

1.17 Proposition. If N is an R-module, then T(N) is a

torsion module.

Proof. If H is closed in T(N), then H is closed in

N since T(N) is closed in N. Therefore H closed in T(N)

implies that H = T(N), since T(N) is the intersection of

all closed submodules of N.

1.18 Prooosition. If L is a torsion module and L is a

submodule of N, then L is a submodule of T(N).

Proof. Assume L is a submodule of N. Then T(N)) L

is a closed submodule in L since T(N) is closed in N. Hence

T(N) n L = L and so L T(N).

Comment. From 1.17 and 1.18 we note that T(M) is

the union of all torsion submodules of M for any R-module M.

1.19 Proposition. (1) M is a torsion free module if and

only if T(M) = [0}. (2) M is a torsion module if and only

if T(M) = M.

Proof. If M is torsion free, then [0] is a closed

submodule. Hence T(M) = (0). If T(M) = [0], then M/(0) O M

is torsion free by 1.16. Part (2) follows from 1.17 and

1.18.

1.20 Proposition. If N is an R-module and M is a torsion

free R-module, with g a homomorphism from N into M, then

N/ker g is a torsion free module.

1.21 Definition. A left ideal K of R will be said to

be closed if it is closed as a submodule of RR.

Notation. TR(R) is the intersection of all closed

left ideals of R.

1.22 Pronosition. If M is a torsion free module, then

T(R)M = 0.

Proof. Let m o M and r : R such that rm / 0. It

is sufficient to show that r V T(R). Define a mapping

g from R into M by g(r') = r'm for r' R. Clearly g is

an R-module homomorphism, and r is not an element of the

kernel of g since rm 0.

By Proposition 1.20, R/ker g is torsion free. There-

fore ker g is a closed submodule of R which implies T(R) is

a submodule of ker g. Hence r is not an element of T(R).

Comment. From 1.22 we have that T(R)Mc T(M).

1.23 Proposition. T(R) is an ideal of R. (This is noted

by Jans in [8] in a more general context.)

Proof. T(R) is a left ideal since it is a submodule

of R. R/T(R) is a torsion free module by 1.16. Hence

T(R)(R/T(R)) = 0 by 1.22. Therefore T(R) is a right ideal.

1.24 Proposition. If M is a torsion free module, then M

is a R/T(R) torsion free module.

Proof. This follows from 1.22 and 1.23.

1.25 Definition. We say that a ring is torsion free if

R is torsion free, i.e., if T(R) = 0.

R

1.26 Definition. Let M be an R-module and define:

Ml =[ m M Rm = 0O

For a any ordinal number with the predecessor a 1

define:

M = m C MIRm InC M

For a a limit ordinal define:

M = 'j M.

1.27 Prooosition. (a) M is a submodule of M for any

ordinal a.

(b) M is a submodule of M if a is less than or

equal to 3.

(c) There exists an ordinal number such that

M = M for all a greater than or equal to z.

(d) If M= M then M = T(M).

Proof. (a) and (b) follow easily from the definition.

To prove (c), it is clear from (b) that there exists

an ordinal number 5 such that M = M+ Let a be the least

ordinal such that M = M Assume that M = M tor all

o oi+l o

o < a < v. If v is a limit ordinal, then by the definition

of M it is clear that M = M If v has a predecessor

v v o

v 1, then M = ({mRm c M = m|Rm M ] = M = M .

Hence M = Mo for all a > o where a is the least ordinal

such that M = M This proves part (c).

To prove (d) assume M = M Let m M and

a a+l

Rm c M then m E M since M +1= [m[Rm c M i. Hence M

a o+ +l O o

is a closed submodule of M. This implies T(M) is contained

in M From the definition of torsion module, it is easily

seen that M1 is a torsion module; hence M1 is contained in

T(M). Assume MI is contained in T(M) for all a < 3. If g

is a limit ordinal, then clearly M, is contained in T(M).

If 5 has a predecessor - 1, then m E M. implies Rm is

contained in M- which is contained in R(M). Since T(M)

is closed in M, Rm T(M), which implies m E T(M). This

proves M. is contained in T(M) for all ordinals 3. There-

p

fore if M = M then M = T(M).

o o+l 0

1.28 Proposition. If M is an R-module and 0 7 m E Rm,

then m r T(M)

Proof. The proof follows from 1.27 if we note that

m M M for any a by transfinite induction.

1.29 Proposition. If M is an R-module and there exist s,

r 8 R, and m a M such that srm = rm / 0, then m t T(M).

Proof. By 1.28, rm T(M). Hence m T(M) since

T(M) is a submodule.

Notation. For any ordinal number a define R = (RR)a

as in Definition 1.26.

1.30 Proposition. For any ring R,

(a) R is an ideal of R for any ordinal a.

(b) R c R if and only if a < p.

(c) If R Ra then R = T(R).

a a+l' a

Proof. (b) and (c) are immediate from 1.27 and (a)

is easily verified.

Example. Let A be the set of all ordinal numbers

less than or equal to a given ordinal ao and V be a vector

space with a basis {v )EA. Let R be the ring of all

linear transformations y on V of finite rank such that for

any B E A, y(v ) / 0 implies that y(v ) E [v ,], It is

easily seen that Rp is properly contained in R if 0 is less

than ao and R = R if P is greater than a where R, is as

in 1.26. Hence for any ordinal o there exists a torsion

module M with M properly contained in M.

1.31 Proposition. (a) If R has a left identity, then

T(R) = 0.

(b) If R has a right identity or, more generally,

2

if R = R, then R1 = T(R).

Recall that I is a nil ideal of a ring R if I is

an ideal of R and for each x E I, there exists n such that

x = 0.

The nil radical of a ring R is the union of all nil ideals

of R. The nil radical is a nil ideal of R (see, e.g.,

[3, p. 19,].).

1.32 Proposition. The nil radical N of R is closed in R,

i.e., it is closed as a left ideal of R.

Proof. Suppose r E R and Rr C N. Note (RrR + Rr

+ rR + Zr) is contained in RrR. Since N is an ideal, RrR

is contained in N which implies (RrR + Rr + rR + Zr) is

contained in N. It follows that (RrR + Rr + rR + Zn) is

a nil ideal. Therefore (RrR + Rr + rR + Zr) is contained

in N and r E N. By Proposition 1.11, N is a closed left

ideal.

1.33 Corollary. T(R) is contained in the nil radical of

R.

Example. Let R = Rk where R = (2)/(2k).

kEN

It is easily checked that Rpk is a torsion R-module

for k E N. However, RR is not a torsion module since it is

not a nil ring. Hence the direct product of torsion R-

modules is not necessarily a torsion module. Clearly we

could obtain similar examples by replacing Rk by any nil-

potent ring whose index of nilpotency is k.

Recall that an element x of R is called left quasi-

regular if there exists y in R such that x + y + yx = 0.

If I is a left ideal of a ring R and if every element in I

is left quasi-regular, then I is called a left auasi-regular

left ideal. Let J be the union of all left quasi-regular

left ideals of a ring R, then J is a left quasi-regular

two-sided ideal. J is called the Jacobson Radical of the

ring R, (see, e.g., [3, p. 92]).

1.34 Proposition. J, the Jacobson Radical of R, is closed

in R.

Proof. As is well known, the nil radical of R is

contained in the Jacobson Radical of R ([3, p. 931). Hence

0 is the nil radical of R/J. By 1.32, 0 is a closed left

ideal of R/J, i.e. R/J is a torsion free R/J-module. It is

easily seen that this implies R/J is a torsion free R-module.

1.35 Proposition. If R is an ideal of S and S has an

identity, then TR(RS) = [s E SIRs C TR(RR)3.

Proof. In this proof T always means TR. Let

L = (s E SIRs c T(RR)}. By 1.16,we have T(R)(S/T(S)) = 0,

i.e., T(R)S C T(S). Since S has an identity element,

T(R) c R(S). For s E S and Rs a subset of T(R), we note

that s 6 T(S) by Proposition 1.11. Hence L is contained

in T(RS).

Let s E S L. Then there exists r such that

rs j T(R). This implies that Rrs is not a subset of T(R)

by Proposition 1.11 since T(R) is closed in R. It follows

that rs is not an element of L, which implies that L is

closed in S. Hence T(RS) is contained in L.

Therefore T(S) = L.

1.36 Corollary. TR(R1) = ((r,k)Isr + ks E TR(R), for all

s ~ R1.

1.37 Corollary. If T(R) = 0, then T(RI) = ((r,k)Isr = -ks

for all s E Rj.

1.38 Proposition. r E T(R) if and only if rM = 0 for all

torsion free modules M.

Proof. If r E T(R), then rM = 0 for all torsion free

modules by Proposition 1.22.

Assume r E R\T(R). Note that r(0,1) + T(R)) =

(r,0) + T(R') and Rr 2 T(R); hence (r,0) ) T(R1) by 1.36.

Therefore r((0,l) + T(R )) / O, i.e., r(R /T(RI)) o 0.

Comment. It is obvious from the above argument

that ((0,1) + T(R')) 5 0.

1.39 Proposition. If R is torsion free, then the follow-

ing are equivalent:

(1) R has an identity.

(2) RR is closed under taking factors.

Proof. If R has an identity, then (2) follows from

Proposition 1.3.

Conversely, we assume Rn is closed under taking fac-

tors. By Proposition 1.4, we note that (0,1) + T(R ) E

R((0,1) + T(R )). Hence (0,1) + T(RI) = (r,0) + T(R1) for

some r E R. This implies that (r,-l) 6 T(R1) for some

r E R. Therefore, by Proposition 1.37, sr = s for all

s E R, i.e., R has a right identity. It follows from Pro-

position 1.9, that R has an identity since R has a right

identity.

1.40 Corollary. The following are equivalent:

(1) R/T(R) has an identity.

(2) RR is closed under taking factors.

R

Proof. Note that since T(R)M = 0 for all M E R

it is clear that the category RR is isomorphic to the cate-

gory R/T(R).

1.41 Proposition. If the additive group R of R is di-

visible and if T(R) = 0, then T(RR') / 0 if and only if R

has an identity.

Proof. Assume that R is divisible, that R is a

torsion free R-module, and that T(R ) # 0. Then there

exists b E R and k E Z such that (b,k) / 0 and rb = kr for

all r E R by Proposition 1.37. Note that if b / 0, then

k Z 0 since Rb / 0, and k / 0 implies b 4 0 since kR = R.

It follows then that b / 0 / k, and b = kc for some c E R,

since R = kR.

Now let r E R and let r = kr'. Then re = kr'c =

r'kc = r'b = kr' = r. Hence c is a right identity for R

and, since R is torsion free, R has an identity by Propo-

sition 1.9. Conversely, we note if R has an identity, e,

then R(e,-l) = 0, which implies (e,-l) E T(R'). This

proves the proposition.

1.42 Proposition. If M is an R-module, N is a torsion

free module, and g a homomorphism from M into N, then g can

be factored through M/T(M), i.e., there exists a unique

homomorphism h from M/T(M) into N such that h o f = g

where f is the canonical homomorphism onto M/T(M).

Proof. Let g be a homomorphism from M into N,

where N is torsion free. Then M/ker g is torsion free.

It follows that T(M) is contained in ker g. Therefore

g can be factored through M/T(M).

1.43 Proposition. The direct product of modules is

torsion free if and only if the component modules are

torsion free.

Note. It is easily seen that H is an R-module if

and only if H is a (unital) R -module. For H an R-module,

(r,k) E R and h E H, (r,k)h is defined to equal rh + kh.

If H is an R -module, then rh is defined to equal (r,O)h

for r E R and h E H. It is also easy to see that the map,

a, is an R homomorphism from RA into RB if and only if a

is an RI homomorphism from 1A into RB. Thus, the cate-

gory of all R-modules is isomorphic to the category of all

unital R -modules.

1.44 Proposition. @ Mi/T' K M.) is isomorphic to

icl iEI

S M./T(M) .

L 1 1

iEI

Proof. Since, in general, if N. is a submodule of

M., we have (G M.)/(B N.) a (M./N.), it suffices to prove

i i

that T(@ M.) = G T(M.). (For simplicity assume all direct

i i

1 1

sums are internal.) Since a direct sum of torsion free

modules is torsion free, 0 (M./T(M.)) 2 (G M./(@ T(M.)) is

i 1 1

torsion free. Hence D T(M.) is closed and therefore contains

1

T(G M.). On the other hand, each T(M.) is torsion and a sub-

i 1

module of D M., and so each T(M.) is contained in T(i M.).

1

It follows that G T(M.) c T(@ ).

i i

1.45 Definition. We call F an R-free module if F E R

and F has a generating set, A, such that any mapping from A

to a torsion free module, M, can be extended uniquely to a

homomorphism from F into M. A is said to be a free basis

for F in the category RR

1.46 Proposition. Let F be a free R -module with basis A.

Then RF/T(RF) is an R-free R-module with basis [a + T(RF)I

a E A).

Proof. Let n : F F/T(R) be the natural projection.

Let g : n(A) M where M is any torsion free R-module.

Then gr yields a mapping from A to M which has a unique

extension f : F M. Here we consider M as an R -module

and so f is an R -homomorphism and hence also an R-

homomorphism. By 1.20, T(F) c ker f and so f induces a

unique mapping g*, making the following diagram commutative.

F ------ F/T(F)

Then for a E A, g*(n(a)) = f(a) = f(r(a)). That g* is

unique follows from the fact that n(A) generates F/T(F).

1.47 Proposition. An r-module F is R-free if and only if

it is isomorphic to a direct sum of copies of RRI/T(RR ).

Proof. It is immediate from the definition that

R-free modules whose bases have the same cardinality are

isomorphic. Hence, by 1.46, each R-free module is iso-

morphic to F/T(F) where F is a free R'-module. Since F is

a direct sum of copies of R we deduce immediately from 1.44

that F/T(F) is a direct sum of copies of RR /r(RR ).

1.48 Definition. A torsion free R-module, P, will be said

to be 9-projective if for A and B, any torsion free modules,

with f a homomorphism from A onto B and g a homomorphism

from P into B, there exists h, a homomorphism from P into A,

such that f o h = g.

The reader should note that this definition differs

from the usual definition of a projective object in a cate-

gory since, as we shall soon show, epimorphisms in the

category RR need not be onto.

A torsion free R-module Q, is said to be R-injective

if for C and D, any torsion free modules with i a monomor-

phism from C into D and k a homomorphism from C into Q there

exists q, a homomorphism from D into Q such that g o i = k.

As usual, one shows:

1.49 Pronosition. An R-free module is R-projective.

1.50 ProDosition. Every torsion free module is a homo-

morphic image of an q-projective module.

1.51 Proposition. Every R-projective module is a direct

summand of an R-free module.

1.52 Definition. Recall that a morphism f in a category P

is said to be an epimorohism if whenever g and h are mor-

phisms in P such that gf = hf, then g = h.

1.53 Prooosition. All epimorphisms in R are onto if and

only if R/T(R) has an identity.

Proof. Assume all epimorphisms in R are onto and

suppose R/T(R) does not have an identity. By 1.40, there

exists M E RR with L a submodule of RR such that M/L f R.

R B R

Let H equal the intersection of all closed submodules in M

containing L (M is such a submodule) and note H is closed

in M. Let f be the inclusion homomorphism from L into H

and suppose gf = hf where g and h are morphisms in R. It

follows that the ker(g-h) is a closed submodule of H con-

taining L. Hence ker(g-h) = H since H by construction con-

tains no proper closed submodules which contain L. This

implies that f is an epimorphism which contradicts the

assumption. Therefore R/T(R) has an identity.

The converse follows easily from 1.40.

Comment. It follows from 1.53 that if R does not

have an identity and T(R) = 0, then there exists an R-free

module which is not projective in the category Ri in the

categorical sense of projectivity. (See [10, p. 69 ].)

Hence, if T(R) = 0 and all torsion free modules are assumed

projective in R, then R has an identity.

B'

SECTION II

THE COUPLE (R R) AS A TORSION THEORY

R R

AND THE CLASS OF TORSION RINGS

In this section it is shown that the couple (R R)

satisfies a set of axioms given by S. E. Dickson for torsion

theories in Abelian categories. It is proven that Ra is the

smallest torsion class for a torsion theory on Dr which con-

R

tains the class of R-modules M such that RM = 0. The torsion

theory (R ,R9) is shown to coincide with torsion theories of

A. W. Goldie and J. P. Jans if R satisfies certain conditions.

The properties of the class of rings such that R9 is trivial,

(i.e., that it only contains the zero module) are discussed.

2.1 Definition (S. E. Dickson). A torsion theory for R

consists of a couple (5,R) of classes of modules of -

satisfying the following axioms. j is called the torsion

class and 2 the torsion free class with M $ j called j-

torsion and F E R called 2-torsion free.

(1) 5 and 9 have only zero in common.

(2) If T E j and T A 0 is exact, then A 3.

(3) If F E R and 0 H F is exact, then H 2 R.

(4) For each M in R2, there is an exact sequence

S- T X F with T and F

0 T X F 0 with T E U and F E 2.

2.2 Definition (S. E. Dickson). A torsion theory (7,R)

is said to be closed under taking submodules if 5 is closed

under taking submodules.

Recall that a class of modules, T, of the category

Di is said to be closed under taking extensions if

O A B C 0 is an exact sequence with A and C elements

of Z implies B is an element of D.

2.3 The following are results proven by S. E. Dickson.

In the statements, U will designate a torsion class and R

a torsion free class for an arbitrary torsion theory on R.3

(1) 2 is closed under direct products and U is

closed under direct sums.

(2) Both 5 and R are closed under extensions.

(3) J is closed under taking subobjects if and only

if R is closed under taking minimal injectives.

(4) Given U closed under taking factors, extensions,

and direct sums, there exists a unique ;? = fFHom(T,F) = 0,

for all T C} such that (U,P) is a torsion theory.

(5) Given R closed under taking subobjects, exten-

sions, and direct products, there exists a unique class

5 = (TIHom(T,F) = 0, for all F E RS

such that (U,2) is a torsion theory.

(6) For M E R-2 there exists a unique largest sub-

module Mt of M which is a member of j and M/Mt E F. The

T of axiom (4) equals this unique submodule

Mt = T i MIT S or Mt = '[S c MM/S 6 R).

t t

(7) The correspondence M Mt defines a functor

t : D t having the properties:

R R

(a) Given f : A B, t(f) : At Bt is the

restriction of f,

(b) t(A/A ) = 0,

(c) t2 = t.

2.4 Proposition. (R RR) is a torsion theory where RU,

R are the class of torsion and torsion free modules as

defined in 1.15 and 1.1.

Proof. R and R have only the zero module in com-

mon by Propositions 1.19, 1.16, 1.10, and 1.15.

Let B be a submodule of A and A S RU. The canonical

homomorphism from A onto A/B/T(A/B) must have ker A since

it would be closed in A. This implies T(A/B) = A/B. Hence

A/B is a torsion module. Therefore R3 is closed under

taking factors.

R is closed under submodules by Proposition 1.2.

Let M be an R-module. Then we have the following

exact sequence: 0 T(M) M M/T(M) 0 with T(M) an

element of R3 by Proposition 1.17 and M/T(M) E RV by Propo-

sition 1.16.

Comment. In general, we will use the symbols Rj and

R to represent the full subcategories of RS with torsion

and torsion free modules, respectively,

2.5 Prooosition. If L is a submodule of M, then

T(L) = L r T(M).

Proof. Using notation as defined in 1.26, we note

L = M1 L. Assume L = MC n L for all a < . If 3 is

a limit ordinal, we have

L = U L = U (M n L) = ( U M ) n L = M. n L.

a< L a a a

If 5 has the predecessor p-1, then

L = (m E MIRm C M _l] n L = M1 n L.

Hence L = M n L for all ordinal numbers a. It follows

a a

from Proposition 1.27(d) that T(L) = T(M) n L.

2.6 Proposition. RV is closed under taking submodules.

Proof. If H E Rj and A c H, then T(A) = T(H) n A

= H q A = A. Therefore A is a torsion module.

2.7 Proposition. R is closed under taking minimal in-

jectives.

Proof. The proof follows from 2.3(3) and 2.6.

2.8 Proposition. If (C,2) is a torsion theory on R

such that RM!RM = 0 'C then < .

Proof. Suppose M RV by Proposition 2.1(3).

There exist T : a and F 2 ? such that 0 T M F O

is exact. Since M j V, T is a proper submodule of M by

2.3(6). Let H = [m E MR c T3. T is a proper submodule

of H since M contains no proper closed submodules for it

was assumed to be an element of R. This implies that H/T

R

is a nonzero submodule of F. Hence H/T is an element of t,

since ? is closed under taking submodules. H/T is an ele-

ment of V since R(H/T) = 0. This is a contradiction.

Therefore RV is contained in V.

H\

We will now discuss some other torsion theories on

R and find conditions on R such that the torsion theories

R

mentioned coincide with (R ,R ).

2.9 Let M be any R-module. For m E M, define (m)L

= [r E Rlrm = 0].

2.10 After Jans [8], we let 52 = (M E REIHom(M,E(R)) = 0]

2. R

where E(R) denotes the injective hull of R.

In [8], Jans proves:

2.11 Proposition. j2 is a torsion class for a torsion

theory (2', 2) on R.

2.12 We recall that L, a left ideal of R, is called an

essential left ideal of R if L intersects any nonzero left

ideal of R nontrivially. We also recall that the singular

submodule Z(M) of an R-module M is the set of elements of

M which are annihilated by an essential left ideal of R.

Z(R) is called the singular ideal.

2.13 After Goldie r5, we let Z2(M) = _m E MIEm c Z(M)

for some essential left ideal E of R}. Then, as noted by

Alin and Dickson -1, if we let 53 = iM R Z2(M) = M

R3 2

and R3 = [M E RTIZ2(M) = 0], then we have:

2.14 Proposition. (33',3) is a torsion theory in RV.

2.15 Prooosition. If (;,5) and (5', ') are torsion

theories on Rt, then 5 properly contained In a' implies

2' properly contained in R, and properly contained in V'

implies a' properly contained in Z.

Proof. The proof follows from Proposition 2.3,

(1), (2), (4), and (5).

2.16 Proposition. R is contained in 3'.

Proof. Let M E nT such that R(M) = 0. Then M = Z(M)

R

and hence M = Z2(M). Hence M is an element of 3. The

conclusion follows from Proposition 2.8.

2.17 Proposition. If (3,R) is a torsion theory on R and

2 is contained in 2', then R E R.

Proof. If R R, then there exists T, a left ideal

of R, which is contained in 2; but Hom(T,E(R)) # 0. There-

fore T S 2.

2.18 Proposition. If (Z,R) is a torsion theory on R 9

which is closed under taking submodules and R E 2, then

J is contained in 2'.

Proof. If R E R and (2,q) is closed under taking

submodules, we have E(R) E R by Proposition 2.7. Let M E

and a Hom(M,E(R)). Then Im a E 2 since E(R) E R and R

is closed under taking submodules. Hence Im a = 0 and so

a = 0. This implies that M E j2 Since M was an arbitrary

element of 5, we have that T is contained in 2'

2.19 Proposition. R- is contained in 52 if and only if

T(R) = 0.

Proof. The proof follows from Propositions 2.17

and 2.18.

Note that if Z (R) 0, then Z2(R) E 23\ 2 since

Hom(Z2(R),E(R)) = 0 if and only if Z2(R) = 0. In fact,

we have the following:

2.20 Prooosition. 3, = j2 if and only if Z(R) = 0.

Proof. See Gentile [4].

Examples.

(1) If R is the ring of integers, then Z(R) = 0;

hence 53 = 52. Note R(R/(n)) E 33 for R(R/(n)) E RR for

any n E R.

(2) If R is the ring of integers modulo 12, then

Z(R) 3 0. However, T(R) = 0. Hence 0 7 Z2(R) E 33 Rn 2 .

2.21 Definition. For L a submodule of M, we define the

closure of L in M, cl LM, to be the intersection of all

closed submodules of M containing L. (Recall that N c M is

closed if M/N E RR.)

2.22 Proposition. If L is a submodule of M, then cl L

is a closed submodule of M.

2.23 Definition. We call a ring R full if the closure

in R of each essential left ideal of R is R. That is, R

is full if and only if it does not contain a proper es-

sential closed left ideal.

2.24 Proposition. R is a full ring if and only if for

each nonzero element m of a torsion free module M, (m)

is not an essential left ideal of R.

Proof. Assume that R is a full ring, Note that

(m)L is a closed proper submodule of R if m is a nonzero

element of a torsion free module M since (m)L = ker am

where am E Hom(R,M) defined by a (r) = rm. Therefore

m m

(m)L is not an essential submodule of R, since it is not

equal to R and it is closed.

Conversely we assume for each nonzero element m

of a torsion free module M, (m)L is not an essential left

L

ideal of R. Suppose R is not a full ring, then R con-

tains a proper closed essential left ideal L. Let

r' E R\L and H = [r E Rlrr' E L}. Then H is easily

seen to be an essential left ideal of R. However,

H c (r + L)L where (r + L) is an element of the torsion

-- L

free module R/L. This contradicts the assumption. Hence

we see that R is a full ring.

2.25 Prooosition. R is a full ring if and only if 3 = R.

Proof. Assume R is a full ring and let F E R .

Note that for 0 / m E F, (m)L is a proper closed left ideal

of R. From this we see that (m), is not an essential left

ideal of R. Therefore, Z2(F) = 0, which implies F E Z3.

It follows, from Propositions 2.3 (6) and 2.15, that

R 3'

Conversely, we assume 3 = R and let L be an

3 R

essential closed left ideal of R. Then each nonzero ele-

ment of R/L is annihilated by an essential left ideal of R

(as noted in the proof of Proposition 2.24). It follows

that R/L is an element of j3 and RR. Hence R/L = 0 since

R = R3. Therefore L = R which implies that R is a full

ring.

2.26 If T(R) = 0 and R is a full ring, then R = 2'

Proof. By Proposition 2.19, R5 is contained in

32 if T(R) = 0.

Assume T(R) = 0 and R is a full ring.

Suppose M 4 Rj, then M/T(M) is a nonzero torsion free

module. Let m + T(M) be a nonzero element of M/T(M).

Then there exists r e R such that r(m + T(M)) / 0 and

Rr n(m + T(M))L = 0. Hence the homomorphism a from Rr

onto Rr(m + T(M)) defined by a(r'r) = r'r(m + T(M)) for

r'r E Rr is an isomorphism. Consider the following diagram

of homomorphisms.

j

0----Rr (m + T(M))--->M/T(M)

-1 I/

a'

E(R)

where j is the injection. There exists a from M/T(M) into

-1

E(R) such that j. = a since E(R) is an injective module.

-1

a is a nonzero homomorphism since a is nonzero. There-

fore M is not an element of 52. Hence 52 is contained in

R. Therefore 2 = R.

2.27 Proposition. If Z(R) = 0, then R is a full ring if

and only if R3 = 2'

Proof. This follows from Propositions 2.20 and 2.25.

We will now consider the function T as an operator

on the category of rings. Note that while T is a subfunctor

of the identity functor on R1, it is not on the category of

rings. Let, e.g., S be a nilpotent subring of a ring R with

an identity, then TS(S) = S, but TR(R) = 0. Hence, to em-

phasize this distinction, when we think of T as an operator

on the category of rings, we write

H(R) = TR(RR).

Recall that H(R) is a two-sided ideal of R.

Obviously, we have

2.28 Proposition. H(R) = 0 if and only if r(R) = 0.

2.29 Proposition. If R and S are rings and f is a ring

homomorphism from R onto S, then f(H(R)) c H(S).

Proof. Consider R --> S -s-> S/H(S) where g is

the natural ring homomorphism. Note S/H(S) is a torsion

free S-module. It is easily seen that S/H(S) is a torsion

free R-module where r(S + H(S)) = f(r)(S + H(S)). This

implies that the ker fg contains H(R). (Note f and g are

now being considered R-module homomorphisms and S, as an

R-module.) Hence f(H(R)) is contained in H(S).

Comment. If f is not onto,we can conclude that

f(H(R)) C H(f(R)). However, as we see from the example

mentioned above, we cannot conclude that f(H(R)) c H(S).

2.30 Corollary. If H(R) = R and S is a (ring) homomor-

phic image of R, then H(S) = S.

2.31 Definition. R will be said to be H-torsion if

H(R) = R. The class of H-torsion rings will be designated

by y.

2.32 Proposition. Y is closed under taking extensions.

Proof. Consider 0 --> J -> S --> K --> 0,

an exact sequence of rings with H(J) = J t 0 and

H(K) = K S/J / 0. If either end is zero,the proof is

immediate. We want to show that H(S) = S. If

(s + J) = (S/J) then sr E S1 for all r E Jl since s's

is an element of J for all s' E S. Suppose (s + J) E (S/J)

implies sr E S for all r E Jl and all a < P. If 0 is a

limit ordinal and (s + J) E (S/J) then (s + J) E (S/J)

for some a < p, which implies sr E S_ < S for all r e J1.

If 8 has a predecessor B 1, then if s + J E (S/J) ,

then S/J(s + J) c (S/J)-1, which implies s's + J E (S/J)-_1

for all s' E S. Hence s'sr E S,-1 for all r E J1 and all

s' E S by the inductive assumption. Therefore, sr E S

for all r E Jl. It follows from the above induction that

SJ1 is contained in Ts(S) since H(S/J) = S/J and JJl = 0.

If SJ1 = 0, then J1, which is not equal to zero, is

contained in S If SJ2 0, then Ts (S) / 0, which implies

S1 is not equal to zero. Hence S1 is not equal to zero.

Suppose S = S+l and S is properly contained in S

for some ordinal p. By 1.27, S = H(S), which is an ideal

of S by 1.23. Getting (J + S )/S equal to J', we have

H(J') = J' by Proposition 2.29 since J' is isomorphic to

J/J n S and H(J) = J. Writing S' for S/S we note that J'

is an ideal of S' and S'/J' is isomorphic to S/J+S Hence

H(S'/J') = S'/J' since S'/J' is a homomorphic image of S/J.

Hence we have 0 ---> J' --> S' --> S'/J' -- 0 is an

exact sequence of rings with the ends H-torsion. If either

end is zero, then H(S') = S' / 0 which implies that Si / 0.

If the ends are not zero, then it follows from the above

induction that S{ / 0. Noting that

Si = [s + S IS/S (O + S ) = 01 / 0

it follows that S -1 S This contradiction implies that

B +-'-

T s(S) = H(S) = S.

s

2.33 Proposition. M is closed under taking direct sums.

Proof. Let [R aaEA be a collection of rings such

that H(R ) = R for all aEA and consider R = a X R It

aEA

is easily seen by transfinite induction that (R' ) is

contained in R for any ordinal a and for all aEA where

(R' ) = a (Ra) Ta the injection from R into R. Hence

aa a at a a

R' is contained in H(R) for all aEA. Therefore H(R) = R.

a

Example. If R = T (2)/(2"), then H(R) / R since R

n=2

is not a nil ring, but H(2)/(2n) = (2)/(2n) for all n.

Therefore direct products of H-torsion rings are not neces-

sarily H-torsion.

2.34 Proposition. V is closed under taking subobjects.

Proof. Let K be a subring R and H(R) = R. From

Proposition 2.5 we note that TR(K) = K n TR(RR) = K. It

is easily shown that K when considered an R-module is

contained in K when considered a K-module for all ordinal

numbers a. Hence TR(K) is contained in TK(K). Therefore

H(K) = K.

We will now show that M is not a class of radical

rings for any radical property. First we recall some defi-

nitions. (See [3, Chapter 1].)

Let S be a property a ring may possess. A ring R is

called an S-ring if it possesses the property S. An ideal

J of R is an S-ideal if J is an S-ring. A ring which does

not contain any nonzero S-ideals is called S-semi-simple.

S is called a radical property if the following con-

ditions hold:

(1) A homomorphic image of an S-ring is an S-ring.

(2) Every ring contains an S-ideal K which contains

every other S-ideal of the ring.

(3) The factor ring R/K is S-semi-simple.

For a given radical property S, the class of S-rings

is called the S-radical class of rings.

Let ) be any nonempty class of rings. A ring is said

to be of first degree over Z if it is a homomorphic image of

some ring in Z.

Assume that rings of degree a over T have been de-

fined for every a < p. A ring R is said to be of degree 3

over if every nonzero homomorphic image of R contains a

nonzero ideal which is of degree a over I, for some a < .

Let Z be the class of all rings which are of any

degree over T. Define a ring to be an S- ring if R is an

element of '.

2.35 Prooosition. S, is a radical property and '

is contained in !. ([3, p. 12.3)S3 is called the lower

radical property determined by 2.

2.36 Pronosition. If J is a radical property and every

ring in Z is a U-radical ring, then the lower radical pro-

perty determined by Z is contained in 3, i.e., if R E '

then R is a U-radical ring. ([3, p. 13.])

2.37 We recall that if is the class of nilpotent rings

then Sg is called the Baer Lower radical property

(frequently called the Baer-McCoy radical), (see e.g.,

[3], p. 59).

2.38 Proposition. If H(R) = R,then R is of second degree

over 2 where Z is the class of all zero rings. Hence all

H-torsion rings are Baer-McCoy radical rings.

Proof. Let H(R) = R and let S be a homomorphic

image of R. Then H(S) = S by 2.30. Then S1 is the desired

nilpotent ideal of S.

2.39 Proposition. The class of Baer-McCoy radical rings

is the smallest class of radical rings which contains i.

Proof. This proposition follows from 2.36 and 2.38

when one notes that M contains all nilpotent rings.

Remarks. An example of a Baer-McCoy radical ring

which is not an H-torsion ring is given by any Baer-McCoy

radical ring of degree greater than two over the nilpotent

rings.

However, there are even Baer-McCoy radical rings of

degree 2 over the nilpotent rings which are H-torsion free.

For example, the commutative algebra A over the field of

real numbers with basis < x = x if a + < 1 and

x x, = 0 if z + > 1 (see 3, p. 19]).

p p

Since the ring A is commutative and nil, each ele-

ment generates a nilpotent ideal. It follows that every

homomorphic image of A also has this property; so A is of

degree 2 over the class of nilpotent rings. Further, the

right annihilator of A is 0 and therefore H(A) = 0.

34

It is also of interest to note that every nonzero

(left) ideal of A is essential and that the left annihi-

lator of every element is not zero. Hence A = Z(S) and

A is a Goldie-torsion ring.

Since the Baer-McCoy radical of a ring R is the

intersection of those ideals Q of R such that R/Q has no

nilpotent ideals, ([3, p. 561), it is clear that the Baer-

McCoy radical is closed in R.

SECTION III

THE EXTENSION OF THE STRUCTURE THEOREM

FOR SEMI-SIMPLE ARTINIAN RINGS TO THE CATEGORY R

R

The principal objective of this section is to prove

that the following conditions are equivalent for Jacobson

semi-simple rings:

(a) All torsion free R-modules are projective in R;

(b) All torsion free R-modules are injective in R3;

(c) R has d.c.c. on closed left ideals;

(d) R is semi-simple Artinian.

In general, we show that if R is any ring such that

T(R) = 0 and all torsion free R-modules are projective, then

RR = J R S, where J is the Jacobson radical of R and S is

a semi-simple Artinian subring of R. It is an open question

whether or not J = 0 under these conditions. If in the

above theorem R is assumed nil semi-simple, then R = J G S

is a ring direct sum. The above theorems are shown to

apply to the torsion theories of A. W. Goldie and J. P. Jans

under suitable conditions on R.

3.1 Recall that a ring R is said to be a dense ring of

linear transformations on V such that for any finite linearly

independent set x1, x2, ..., xn in V and any finite arbitrary

set yl' Y2' ...' n in V, there exists an element r E R such

that r(x.) = y. for i = 1, 2, 3, ..., n. Such a ring is

called primitive.

For the purposes of this paper, when we speak of R

as a primitive ring it will be understood that R is a dense

subring of the full ring of linear transformations on the

vector space V over the division ring D. The rank of r E R

is understood to mean the dimension of r(V).

3.2 Proposition. If R is a primitive ring, r E R, and

rank r < m, then there exists r' E R such that r'r = r.

3.3 Proposition. If R is a primitive ring and r 5 R,

then rank r'r < rank r.

3.4 Proposition. If R is a primitive ring, then L is a

simple left ideal of R if and only if L = Rg where rank

g = 1.

3.5 Proposition. If L is a simple left ideal of the

primitive ring R, then L is a closed left ideal of R.

Proof. Assume L is a simple left ideal of R; then,

by Proposition 3.4, L = Rg where rank of g = 1. Note that

if r is a nonzero element of L, then rank r = 1 by Propo-

sition 3.3. Let 0 / h E R such that Rh is contained in L.

It is easily seen that rank h = 1. Hence, by Proposition

3.2, h E Rh which is contained in L. Therefore, L is a

closed left ideal of R by Proposition 1.11.

3.6 Recall that the Socle (Soc M) of RM is defined to be

the sun of all simple submodules of M.

3.7 Proposition. If Soc R 5 0, R is a primitive ring,

and U is an infinite dimensional subspace of V, then there

exists [e.] c R such that e.e. = 6. .e. for all e, j E I,

I is infinite, and e.(V) = [w.] where w. E U for all e E I.

Proof. Assume Soc R / 0 and U is an infinite dimen-

sional subspace of V. By 3.4, there exists g E R such that

rank g = 1 since R contains simple left ideals. Let vl and

v2 be nonzero vectors such that g(v ) = v2 and wl a nonzero

vector in U. Since R is a dense ring of linear transfor-

mations, there exists f, h E R such that f(wl) = v and

h(v2) = w1. Since the rank g = 1, we have hgf(V) = [w1]

2

and (hgf) = hgf. Ker hgf 0 U is an infinite dimensional

subspace since U = ker hgf n U G hgf(U) and hgf(U) is a

one-dimensional space. We have shown that there exists

(el R such that elel = el, ker el n U is infinite dimen-

sional and el(V) = w l] where wl E U.

Assume there exists [el, .... en} c R such that

n

e.e. = 6 ..e., i, j E 1, .. ., n}. A ker e. U is an

1 1 1 j=1

infinite dimensional space, e (V) = [w.], 0 / w. E U,

i = 1, ...,n. It follows from the orthogonality of idem-

potent elements [el, .... en] that (wI, .. w ] is a

linearly independent set of vectors. Let 0 w n+ E M

n

= A ker e. f U. Now (w w2, ..., w w n+ is a linear

j=l 2 1 2 n n+1

independent set since MD [w ... w] = 0. As in the

first paragraph of the proof, it can be shown that there

exists en1 an idempotent element of rank 1 in R such that

e n(w.) = 6 w., j = 1, ..., n + 1. This implies

n+1 j n,3 3

e e n+ en+1 e. =6 n+,j e j = 1, ..., n + i.

Ker en+1 n M is an infinite dimensional subspace since

M = ker en+1 n M N e n+(M) and e n+(M) is a one-dimensional

space. Hence, by induction, we can form a chain

(el l E el, 0232 [fel, e2, e33 3 ... where [el, ..., en

satisfies the conclusion of the proposition with the ex-

ception that the set is finite. Let B = U el, .... e }m

m=l

Clearly, B satisfies the conclusion of the proposition.

3.8 Proposition. Lee R be any ring such that T(R) = 0.

Let (e.i i E I be an infinite family of pairwise orthogonal

a-

idempotents in R. If the sets Pr = [ei 0 e. r E Zei}

and P = (e. 0 7 re. E Ze.i are finite for all r E R,

then RR contains a module which is not R-projective.

Proof. Let fe 1 be an infinite set of orthogonal

---- 3 jI

idempotent elements in R satisfying the hypothesis of the

proposition. Let In be the set of j in I such that nej = 0

where n is a fixed element of Z. Partition I into I1 and 12

such that II and I are infinite and, if In is infinite,

then In n I1 and I n 12 are infinite.

Define a : R/T(R) --> Re. by a(0,1) = h,((O,1)

jEI 3

= (0,1) + T(RI)) where h(j) = e. if j I,, and h(j) = 0 if

j E 12, and extend a to a homomorphism.

We shall show that M = Im a is not 2-projective. If

it were, there would exist 9 : M -> RI/T(R1) such that

a 9 = 1. Let 9(h) = (r,n). Then h = a 9(h) = a(r,n) =

a(r(0,l) + n(0,1)) = rh + nh. Thus rh = (l-n)h, and this

implies that re. = (l-n)e. for j E I. It follows from the

hypothesis of the proposition that (l-n)e. = 0 for all but

a finite number of the j's in II. Hence (l-n)e. = 0 for an

infinite number of j's in II. By construction of I1 and I2,

it follows that In- I1 and In-l 12 are infinite sets.

Note e.h = 0 if j E 12 For j E I 1n- 12, we have the fol-

32

lowing qualities:

0 = 6(e.h) = e.e(h) = e.(r,n) = (e r + ne .,0). Hence,

3 m 3

by Corollary 1.37, R(e.r + ne.) = 0. Hence, for

n-l

j E I n 12, 0 = e.(e.r + ne.) = e.r + ne. = e.r + e. since

2 3 3 3 J 3

n-l

(n-l)e. = 0 for j E I n 12. This implies e.r = -e. for

3 2

j I I n 12, which is impossible since In- 12 is an in-

finite set, and this is contrary to the hypothesis of the

proposition. Therefore M is non R-projective.

3.9 Pronosition. The following are equivalent:

(1) All torsion free modules are R-projective.

(2) If L is a closed submodule of a torsion free

module M, then L is a direct summand of M.

(3) If M, N E RR and a is a homomorphism from M onto

N, then M = ker a G N' where N' is isomorphic to N.

3.10 Proposition. If all R-modules in RR are 9-projective

and R is a primitive ring, then Soc R 0.O

Proof. Assume that all R-modules in RR are

R-projective and that R is a primitive ring. Note that for

nonzero v in V, V = Rv is a simple torsion free R-module.

Let v be any nonzero element of V. Define a:R Rv by

a(r) = rv for r E R. Ker a is a closed submodule of R

since Rv is torsion free. Hence, by Proposition 3.9,

R = ker a @ L where L is a simple R-module since L is

isomorphic to Rv. Therefore R contains a simple left ideal

which implies that Soc R / 0.

3.11 Proposition. The Socle of R is the set of linear

transformations of finite rank contained in R.

Proof. (See [7], p 75.)

3.12 Proposition. If all torsion free modules are

projective and R is a primitive ring, then Soc R is a

closed left ideal of R.

Proof. Assume R is a primitive ring and all torsion

free modules are R-projective. Suppose that Soc R is not

closed in R. Then there exists h E R\Soc R such that

rh E Soc R for all r E R. It follows from Proposition 3.11

that rank h = and rank rh < > for all r R. Let h(V) = U.

Then, from Propositions 3.10 and 3.7, there exists an infinite

set of orthogonal idempotent elements e} contained in R

where the image space of e. is a one-dimensional subspace

of U. Since e.}jEI is an infinite set of orthogonal idem-

potent elements of rank 1, there exists a set of linearly

independent vectors H whose elements can be indexed by I

such that for w. E H, e.(w.) = w..

3 3 3 J

Let r E R and define P = [ej.re. = cjej. 0, c. E ZT.

r 3 3 2 2 3

Note if e rP, then r(w.) = r(e.(wj)) = re.(w.) = c.e.(w.)

Sc .w. 0. Hence (c.wj e. P] is a subset of rh(V) and

therefore must be finite since rh(V) is finite dimensional.

Therefore P is finite for all r in R.

r

Let P = [e.e .r = c.e 0 for some c. in Z1. Note

r j j 3 3 3

that for e. E P we have e.r(w.) = cej.(wj) = cj(w.) / 0.

It follows that fr(w.j)e. E P} is linearly independent.

j r

Suppose Za ekr(w ) = a jckek(wj) = akck k = 0. Since

ckWk / 0, we have ak = 0.

Now, since w. is in h(V), r(w.) is in rh(V), which,

as noted above, is finite dimensional. It follows that P

r

is finite.

Thus P and P are finite for all r in R and so, by

r r

3.9, there exists M E RS which is not R-projective. This

contradicts the assumption. Hence Soc R is a closed left

ideal of R.

3.13 Proposition. If all torsion free modules are pro-

jective and R is a primitive ring, then R is simple

Artinian.

Proof. Assume all torsion free modules are pro-

jective and R is a primitive ring. By Propositions 3.10

and 3.12, Soc R is a nonempty closed left ideal of R.

From Proposition 3.9, R = RSoc R @ K. Since Soc R is an

R R

ideal of R, and Soc R / 0, this implies that K = 0. Hence

R = Soc R.

Suppose V is infinite dimensional. Then R contains

an infinite set of orthogonal idempotent elements fe.j]j

of rank 1 by Proposition 3.7. For r E R, r(V) is finite

dimensional since Soc R = R. We arrive at a contradiction

by essentially repeating the second half of the proof of

Proposition 3.12. Therefore V is finite dimensional.

The conclusion follows since R was assumed to be a dense

ring of linear transformations.

3.14 Recall that a ring R is said to be a subdirect sum

of a family of rings [Ri.i E I if R is a subring of i R.

iEI

and, for each i E I, the projection u. restricted to R is

1

onto R..

1

3.15 Proposition. If all M E RB are R-projective and R

is a subdirect sum of a family [R.Ji E K} of simple Artinian

rings, then R is a semi-simple Artinian ring.

Proof. Assume that all M E R are j-projective and

R is a subdirect sum of the family [R ii E K} of simple

Artinian rings. Let r. be the projection homomorphism from

1

R onto R. for each i E K. Note n. can be considered a ring

1 1

homomorphism or an R-module homomorphism with R and R. c R 9

By the assumption that all torsion free modules are

projective, we have RR = R(ker -) R R where R R RR

R R J

for each j E K.

We now show that RR = R(ker ) + RR is, i.e.,

ker -. and R3 are ideals of R. Ker -. is an ideal of R

3 3

since 7. can be considered a ring homomorphism.

2

(Ker )R3 = O since ker -. is an ideal, RK is a left ideal,

and ker R = This implies (R(ker r)) 0. Note

that R3(ker .) is an ideal. Hence R (ker ) = 0 since R

is Jacobson semi-simple and, hence, also semi-simple. It

follows that Rj is an ideal of R. Therefore R = (ker n.)

+ R2 as a ring direct sum.

We note that Rj is ring isomorphic to R. since 7.

2 3

restricted to R3 is an isomorphism onto R.. Hence Rj is

also a simple Artinian ring. Let e. denote the identity of

3

R3. Since R3 is an ideal of R, we have e Re. = e.R = Re.

= R Further, each e. is in the center of R. Since each

3

R is simple, if Ri R3, then RiR3 = 0. It follows that

ee.e = 0. Let N denote the set [e.j of all such idem-

potents. One verifies immediately that J = LejRej = ZRi

is a direct sum. Furthermore, J is an ideal since each R3

is an ideal. Note that it is possible that, for i j, we

might have Ri = R; but, nevertheless, there is a one-to-

one correspondence between the set of ideals Rj and N.

We now observe that an element r in R is contained

in J if and only if the set [ej 6 M Ie r / 03 is finite.

If r E J, r = ejlrjl + ... + ejnrjn, and so it is clear

that e.r = 0 except when j is one of j ..., in

Conversely, assume that e.r Z 0 if and only if

e E [ejl, ..., ejm}. Suppose 0 $ s = ejlr + ... + ejmr r.

Since s = 0, there exists 7ksuch that k(s) 0. Hence

there exists rk R such that (rk) = (s). Note that

0 nk(rk = k(ekrk = k(ek) (rk) = ((ek)k( s) = (k(eks)

= nk(ekejlr + ... + ekejmr ekr) = -k(0) = 0. This con-

tradiction proves that if e.r / 0 for only finitely many

e.'s in N, then r E J.

3

It follows from the above observation that, if ; is

a finite set, then R is semi-simple Artinian.

We now assume that 9 is an infinite set. We wish to

show that J is closed in R. If J = R, then it is closed in

R. Assume J / R and let r be an arbitrary element in R

which is not in J. It is enough to prove that Rr is not

contained in J. Note that I = [e je r / 0; e. E M3 is not

a finite set since r 4 J. For p, a positive prime number,

let Kp = ([e E Ilpej = 0}.

Let S denote the set of primes p such that K is

finite and non-empty. Note that since e. is the identity

3

of a semi-simple Artinian ring, its characteristic is prime.

Hence I = U K is a disjoint union. We now partition I into

P

Il and 12 so that

(i) If K is finite, then K c I or K 12.

p p 1 p 2

(ii) If S is infinite, both II and I2 contain infi-

nitely many K .

(iii) If K is infinite, then I q K and 12 n K are

P 1 P 2 p

both infinite.

Now, as in the proof of 3.8, we define

c:R /T(R ) --> "Re. by 2(0,1) = h where h(j) = e. for

ej E I and h(j) = 0 for j c 12. By hypothesis, M = Im

is projective and so, as in 3.8, we obtain 0 / s in R and

k in Z such that

(1) se = (l-k)e ej 6 I1

(2) e.s = -ke., e. E 12

We now claim that, for infinitely many e. in I,

se. = k .e O. First, if k = O, this is clear from (1);

so we may assume k / 0. There are two cases. First, sup-

pose S is infinite. In this case, by (i) and (ii), there

are infinitely many primes p such that Kp 12 and p > k.

If e. E K for such p, since k 0 we have e.s = -ke. / 0.

3 p 3 3

Next, suppose S is finite. Then there is a q such that K

q

is infinite and, by (iii), K f Il and K n I2 are both in-

finite. If qlk, e. = se. for e. I fl K by (1). If

3 D j 1 q

(q,k) = 1, e.s = -ke. $ 0 for e I 12 K by (2).

Now it follows from the above paragraph and the fact

that each e. is in the center of R that e.sr = se.r = k.e.r

3 3 3 3 3

S0 for infinitely many e. in I. Hence sr is not in J and

so J is closed.

It follows that R = J RH since by assumption all

torsion free modules are R-projective. Suppose H 0 O and

0 / r c H, then there exists i E K such that -.(r) / 0,

from which it follows that e.r / 0. However, e.r E J n H

1 1

since J is an ideal and H is a left ideal. This is a contra-

diction. Hence H = 0 and R = J.

It follows from the above characterization of J that

7 is a set of idempotent elements of R satisfying the hypo-

theses of Proposition 3.7. This implies that there exists

a torsion free R-module M which is not 2-projective. It

follows from this contradiction that 1 is finite. There-

fore R is Artinian semi-simple.

3.16 Prooosition. If all torsion free R-modules are

projective and S is a homomorphic image or R, then all

torsion free S modules are R-projective.

Proof. Any S-module M can be considered a torsion

free R-module by defining rm = f(r)m where r E R, m E M,

and f is the homomorphism from R onto S. It is easily

seen that, for M and N torsion free S-modules,

a:M N is an S-module homomorphism and a is an R-module

homomorphism. The proposition follows immediately from

this observation.

3.17 Proposition. If R is a Jacobson semi-simple ring

and all torsion free modules are R-projective, then R is

Artinian semi-simple.

Proof. Assume R is a Jacobson semi-simple ring and

all torsion free modules are R-projective. From[page 102

of 3], R is a subdirect sum of a family (Rili E I of

dense rings of linear transformations on vector spaces V.

1

over division rings D.. Since the projections 7. are onto

R. for j E I, we have that all torsion free R.-modules are

D 3

RR-projective for j E I. By 3.13, R. is a simple Artinian

Rj 3

ring for each j in I. From Proposition 3.15, R is a semi-

simple Artinian ring.

3.18 Proposition. The following are equivalent if R is

a Jacobson semi-simple ring:

(1) M E R' implies M is --projective.

(2) M E RR implies M is R-injective.

(3) The closed submodules of R have d.c.c.

(4) R is semi-simple Artinian.

Proof. Clearly (4) implies (1), (2), (3), (4).

(1) implies (4) is Proposition 3.16.

Assume (2) and R is a Jacobson semi-simple ring.

Note that T(R) = 0 by Proposition 1.34. Consider the fol-

lowing diagram of homomorphisms:

0 --> R R /T(R

i ---

R

j(r) = (r,0) + T(R ) which is a monomorphism by 1.38,

i(r) = r, and a is the homomorphism whose existence is gua-

ranteed by the 9-injective property of R. For r E R,

r i(r) = a .(r) = a((r,0) + T(R1)) = rc((0,1) + T(R1))

= rh where h = a((0,1) + T(R )). Hence h is a right identi-

ty for R. By Proposition 1.9, R has an identity. Hence R

is the category of unital R-modules by Proposition 1.3, and

all unital modules are projective since all torsion free

modules are R-injective. Therefore, by the Artin-

Wedderburn theorem, R is semi-simple Artinian.

Assume (3) and that R is a Jacobson semi-simple ring.

Since R is Jacobson semi-simple, it is a subdirect sum of a

set of rings [R j]jI where Rj is a dense ring of linear

transformations on a vector space V. over a division ring.

The projection homonorphism from R to R. is onto R. and,

3 3 3

if L is a closed left ideal of R., then it is easily seen

-1

that nj. (L) is a closed left ideal of R. It follows that

R. has d.c.c. on closed left ideals since R has d.c.c. on

3

closed left ideals for each j E I.

Suppose V. is an infinite dimensional vector space

for some j E I. Let IvI, v2 ... be a basis for V. and

define:

M1 = ker al where al:R --> V ; a (r) = rv1

M2 = ker a2 where a2:MI -> Vj ; 2(r) = rv2

M = ker a where a :M ----> V a (r) = rvn'

n n n n-i

It is easily seen that Mi+1 is a left ideal of R

properly contained in .. Mi+l is closed in M. since V.

-i i+1 ]

is a torsion free R-module. It follows from Proposition

1.13 that M is a closed left ideal of R for all positive

n

integers n. This contradicts the assumption of d.c.c. on

closed left ideals of R. Hence V. is a finite vector space

3

for j E I. Therefore R. is a simple Artinian ring for all

3

j f I.

We assume that the family (R.}. E I has been indexed

by an initial set of ordinal numbers and define L = R,

L1 = (r E R r(l) = O}. If a has the predecessor 1 1, then

define L = (r e L a-_r() = 0 and, if a is a limit ordinal,

define L = (r E n L r(a) = 0. This clearly defines L

a
for each a 6 I, and it is easily seen that L is a closed

left ideal of R with L < L Since we are assuming d.c.c.

on closed left ideals of R, we have that H = (a E IIL is

properly contained in L, for all 3 < a] is a finite set,

i.e., H = (al, .... a n

1 n' n

We define f from R to G 7 Ra by f(r)a. = r(a.).

iEl

It is easily checked that f is a ring homomorphism onto

n

SRai by f(r) i = r(a ). It is easily checked that f is

iEl

n

a ring homomorphism onto j Ri.. We now want to show that

i=l

f is a monomorphism. Suppose there exists r E R such that

r / 0 and f(r) = 0. Let P be the least ordinal such that

r() /- 0. $ is not an element of H since 0 = f(r).. =

1

= r(a.). Note that r E L for a < since r(p) = 0 for all

1 0

o < 3 and r 4 L since r(3) / 0. This implies that L, is

properly contained in L for all c < 3. However, this im-

plies that $ E H,which is a contradiction. Hence 0 / r E R

implies f(r) 0. Therefore, f is a ring monomorphism.

If for each r E R, r(a.) / 0 implies r(a ) / 0 for

some j 4 i, then R is a subdirect sum of a G R k.

akEH [i]

Hence we can assume that for each a. E H, there exists r E R,

1

such that r(a ) / 0 and r(a.) / 0 for j i. From the above

3 J

observation and noting that Rai is a simple ring with 7,

restricted to R onto Ri., it is seen that f is onto

n

S@ Rai. Therefore f is an isomorphism, which proves that

i=l

R is a semi-simple Artinian ring.

Comment. It is not necessary in 3.18 to assume R is

Jacobson semi-simple; it is enough to assume T(R) = 0.

3.19 Prooosition. If R E R5 and all torsion free R-

modules are 2-projective, then R = N G K RS where N + K

R R R R

is the Jacobson radical of R, N is the nil radical of R, and

S is a semi-simple Artinian subring of R.

Proof. Assume R E RR and all torsion free R-modules

are R-projective. Let J be the Jacobson radical of R. Then

R/J is a semi-simple Artinian ring by Propositions 3.17 and

3.18, and RR = J @ S where S R/J by Propositions 1.34 and

R K

3.9. Divinsky, in [3], points out that the nil radical N

of R is contained in the Jacobson radical of R. Hence, by

Propositions 1.14, 1.32, and 3.9, R = N RK. Therefore,

RR = RN 9 RK G RS.

Comment. If R is not assumed to be an element of RK

in the above proposition, then the conclusion holds for the

ring R/T(R).

3.20 Proposition. If R is a nil semi-simple ring and all

torsion free R-modules are R-projective, then R = J 0 S

(a ring direct sum) where J is the Jacobson radical of R and

S is semi-simple Artinian.

Proof. Assume R is nil semi-simole and all torsion

free R-modules are 9-projective. By Proposition 1.32, RR

is torsion free. Hence, by Proposition 3.19, RR = RJ S

where J is the Jacobson radical of R and S is semi-simple

Artinian. Since J is an ideal of R and S is a left ideal

of R, we note, as in the proof of Proposition 3.15, that

JS = 0 since R is a nil semi-simple ring. Therefore J and

S are ideals of R, i.e., R = J G S, a ring direct sum.

3.21 Proposition. If R is a ring whose only closed left

ideals are 0 and R, then R is a division ring or a Jacobson

radical ring with no zero divisors.

Proof. Assume the only closed left ideals of R are

0 and R. Then T(R) = 0 since 0 is closed. Hence R does not

annihilate any elements of R, which implies that R does not

have any zero divisors since a zero divisor would determine

an R homomorphism from R into R whose kernel would be a

proper closed left ideal of R. Since the Jacobson radical

or R is a closed left ideal, we have that R is Jacobson semi-

simple or a radical ring. If R is Jacobson semi-simple, then

it follows from 3.18 and the comment following 3.19 that R is

a division ring since R has d.c.c. on closed left ideals.

Using the relationships between the torsion theories

(R,R R), (2, R2), and ( 3,R3) established in Section II, we

now relate the results of this section to the torsion theo-

ries ( 2, 2) and ( 3, 3).

3.22 Definition. For (3,R) a torsion theory on RT, an R-

torsion free R-module M will be called R-injective if it is

injective in the full subcategory R of RS, and R-projective

if it is projective in the full subcategory R with respect

to the onto epimorphisms in R, as in 1.48.

3.23 Proposition. If (3,R) is a torsion theory on R !

closed under taking submodules, then all R-torsion free

modules are 9-injective if and only if all R-torsion free

modules are injective, i.e., injective in the category 3R.

Proof. Assume (3,R) is a torsion theory on RD

closed under taking submodules and all R-torsion free mod-

ules are R-injective. For a submodule L of M, t(L) is

contained in t(M) by 2.3(7) where, for any R-module N, t(N)

is the sum of the 3-torsion submodules of N. Hence t(L) is

contained in t(M) n L. Note that t(M) n L E 3 since t(M) is

torsion and 3 is closed under taking submodules. Therefore,

t(L) = t(M) n L.

Suppose Q is R-injective and consider the following

diagram of R-homomorphisms:

0 ----)M ----->H

Q

Note that it gives rise to the following diagram of homo-

morphisms since a

0 --- M------ >H

0 --- M/t(M) --->H/t (H)

Q-

restricted to t(M) = 0 by 2.3(5) and i is a monomorphism

since j(t(M)) = t(j(M)) = t(H) q j(M) = j(t(M)) from the

above argument. 3 is the R-homomorphism guaranteed by the

R-injectivity of Q. Hence it is easily seen that Q is in-

jective in R(.

The converse is immediate.

Alin and Dickson, in [1%, proved that if R is a ring

with identity, then all RM E ?3 are injective if and only if

3\

R = T @ S where T has an essential singular ideal and S is

semi-simple Artinian. The following proposition is very

similar to Alin and Dickson's result.

3.24 Proposition. If the image of every essential left

ideal of R is an essential left ideal in the ring R/Z2(R),

then all RM E 23 are injective if and only if R = K @ S

where Z2(K) = K and S is a semi-simple Artinian ring.

Proof. Assume the image of every essential left

ideal of R is an essential left ideal in the ring R/Z2(R)

and all R E 3 are injective. From 2.3(7) and 2.13, we

note that T3(R) = Z2(R). It is easily seen that T3(R)M = 0

for M E 2 We will write R/Z (R) as R. It follows from

the assumption on R that RM is R -torsion free if and only

if RM is R3-torsion free where we define rm=(r+T3(R))m.

Hence all 3 -torsion free R-modules are 3-injective. This

implies that R does not contain an essential left ideal

since submodules of -R are 9 -torsion free. Hence R is a

R 3

full ring which implies that -M 3 if and only if -M c -?.

This implies that all torsion free R-modules are injective.

Noting that T(R) = 0 since Z(R) = 0, we have R is a semi-

simple Artinian ring by 3.18. Hence

n n

R R = Re where

i=l i=l

ee. = S., amd RRei is a simple R-module for i = 1, ..., n.

Consider the sequence of homomorphisms

n Sj

R = --> Re > Re where

i=l

p is the projection homomorphism and a is the canonical

homomorphism from R onto R/T3(R). Note that .j RRej and

e .e. = = a(s ) = a(sj)a(s ) = a(ss ) = s.a(s ), where

-1

s. C a (e.). Ker a@. is not an essential left ideal of R

since Re is 3 -torsion free, being a submodule of R. Hence

there exists H. a left ideal of R such that H. f ker ap. = 0,

which implies that a restricted to RH. is an isomorphis-. onto

Re. for j = 1, ..., n. Suppose H. = H. where i / j. Then

R 3 j 1

there exist r., r. E H. such that a(r.) = e. and a(r.) = Z..

3 1 1 3 3 1 1

However, a(r.r.) = a(r.)a(r.) = e = 0, which implies that

r.r. = 0. This implies that ker a. is a proper submodule of

3 1 1

H. where c.(r) = rr, since r.r.i 0 and r.r. = 0. This is

R 3 i 1 1i 3 1

impossible since H. is a simple module. Hence H. / H. if

i 7 j. We now note that, if h + ... + h = 0, where h. c H.

1 n 3 3

for j = 1, ... ..n, then a(h )e. + ... + a(h )e. = 0, and

this implies that a(h.)ej = 0, from which it follows that

hj = 0. Hence it is easily seen that RR = RH 9 RZ2(R) where

n

a restricted to H = H an isomorohism onto R. Note that

j=l

R3 is closed under homomorphic images since all 3-torsion

free modules are injective. Therefore, since H is 5 -

R 3

torsion, we have that R = H 5 Z2(R) is a ring direct sum

since right multiplications in R are left R-homomorphisms.

Conversely, assume that R = K G S where Z2(K) = K

and S is semi-simple Artinian. Clearly, Z2(R) = K and RN

is 3-torsion free if and only if R/T (R)M is 3 -torsion

free. Since R/T3(R) is a ring isomorphic to S and S is a

full ring, we have R = 2 = s where sZ is the category of

3 s s s

unital S-modules. Hence all R -torsion free modules are R -

injective.

3.25 Proposition. If Z (R) = 0 and all R3-torsion free

modules are injective, then R is semi-simple Artinian.

Proof. This is a corollary of the previous propo-

sition. However, it also follows very quickly from Propo-

sition 3.18.

3.26 Proposition. If T(R) = 0, then all 2-torsion free

modules are R2-injective if and only if R is semi-simple

Artinian.

Proof. Assume T(R) = 0 and all 2 -torsion free mod-

ules are 2 -injective. From the definition of 2 it im-

mediately follows that R is 2 -torsion free. Hence R con-

tains no essential left ideals since any left ideal of R is

92-injective. By 2.26 and 2.15, R = 2 since R is a full

ring and T(R) = 0. This implies that all torsion free mod-

ules are 2-injective. By 3.18, R is semi-simple Artinian.

The converse follows easily from the theory of modules for

semi-simple Artinian rings. It also can be seen to follow

easily from Propositions 2.26, 2.15, and 1.3.

3.27 Proposition. If R is a torsion free full ring and

all (2) )3-torsion free modules are (2 ) 3-projective, then

(a) RR = J RS where J is the Jacobson radical of

R and S is a semi-simple Artinian subring of R,

(b) R = J G S (a ring direct sum) if R is nil semi-

simple,

56

(c) R = S if R is Jacobson semi-simple (J and S

as in part (a)).

Proof. The proof for the R3 case follows from 2.26,

3.18, 3.19 and 3.20.

SECTION IV

A CLOSURE OPERATOR ON SUBMODULES

AND A CLASS OF RINGS

WHICH SATISFIES A CERTAIN CLOSURE PROPERTY

In this section the properties of a closure opera-

tor on the submodules of an R-module M are discussed. It

is shown that the lattice of closed submodules of an R-

module is a complete modular lattice which is complemented

if and only if R is a full ring. The properties of full

rings are also investigated.

We will first consider the closure of L in M. To

do this, we generalize the definition of closure as given

in 2.21.

Let L be a submodule of M and (Mf,9), a torsion theory

on 2r, with torsion functor t as discussed in 2.3(6) and (7).

L, a submodule of M, will be called closed in M with respect

to (Z,9) if M/L E 2. We define the closure of L in M with

respect to (7,9) to be the intersection of all closed sub-

modules in M with respect to (J,9) which contain L. It

will be written cl LM. We shall omit the j when discussing

Rj and the M when it will not lead to confusion.

4.1 Proposition. If L is a submodule of M and (5,9) is

a torsion theory on R, then M/cl _L E ?.

K J

Proof. Let M be the set of all closed submodules of

M which contain L. Define an R-module homomorphism from M

onto @ M/L where g(m)a, = in + L Clearly

L Ea

ker g = 0 L and q N M/L 6 R. It follows that

a

M/cl L E R.

4.2 Proposition. If L is a submodule of M, then

-i

o (t(M/L)) = cl L where o is the canonical f homomorphism

from M onto M/L and t is the torsion functor for (J,q).

-l

Proof. -a (t(M/L)) is closed in M since

-1

M/L/t(M/L) 6 9. Hence clL is contained in (t(M/L)).

Suppose L' is a closed submodule of M containing L. Then

ker is closed in M/L where P is the canonical homomorphism

from M/L onto M/L'. This implies, by 2.3(6), that r(M/L) is

contained in ker 3. It follows that L' is contained in

a- (t(M/L)). Hence 7-(t(M/L)) is contained in cl L.

Therefore cl.L = -l (t(M/L)).

4.3 Prooosition. If (3', ') and ('",Q") are torsion

theories on R- with j' contained in j", then, for L a sub-

module of M, cl.,(L) is contained in cl ,,(L) and

dM M

ci, (cl,C(L)M)M = cl, (L)M

Proof. The proof follows from 2.3(6) and 4.2.

Let L C M be submodules of a module N, then we have

4.4 Definition. For a submodule L of M we define L. = L,

L1 = [m E MlRni L} and for a, an ordinal number with pre-

decessor a 1, we define L = (m 6 MIRm C L -1; for a,

a limit ordinal, we define L = U L

S

4.5 Proposition. If L is a submodule of M, then clR LM

= L for some ordinal a.

Proof. This follows from 1.27 and 4.2.

4.6 Proposition. Let L c M be submodule of a module N,

then

(a) cl LM = cl LN n M.

(b) M E RR implies that cl LM is an essential ex-

tension of L.

(c) M a closed submodule of N implies cl LM = cl LN

(d) L is closed in M if and only if L = ker a where

a is a R-homomorphism from M into a torsion free module.

Proof. (a) By 4.2 and 2.5, we have the following

qualities where a is the canonical homomorphism from N to

-1 -1

N/L and co = aIM:cl LM = ao (T(M/L)) = ao (T(N/L)nM/L)

-1 -l -1

= a (T(N/L)fna (M/L) = a (T(N/L))OM = cl LN M.

(b) L a submodule of a torsion free module M implies

that E(M) is a torsion free module and E(L) can be taken to

be a submodule of E(M) where, for N an R-module, E(N) is the

injective hull of N. E(L) is closed in E(M) since it is a

direct summand of E(M). Hence E(L) 2 M is closed in M,

which implies that cl L is contained in E(L) n M. There-

fore cl L is an essential extension of L since E(L) is an

essential extension of L.

Parts (c) and (d) are easily seen.

4.7 Proposition. If L and H are submodules of M, then

cl LM n cl HM = cl(L n H)M and cl LM + cl M C cl(L + H)M-

Proof. The only part of this proposition which is

not immediate is cl LM n cl Hc c ci(L n H)M. Using the

description of cl LM given in 4.5, we will show cl LM n H

cl(L n H)M. Suppose m E L1 n H; then Rm C L and m E H

which implies that Rm C L n H. This implies that

m E cl(L 0 H)M. Assume La H c cl(L n H) for all a < 8.

Let m E L 0 H. Then if P has a predecessor, p 1, then

Rm L-1 n H c cl(L n H) which implies m E cl(L n H); if

p is a limit ordinal, then m E L n H for some a < 3 which

implies that m E cl(L n H). It follows, by 4.5, that

cl LM n H H cl(L A H)M. Similarly, it follows that

cl LM n cl Hi 1 cl(cl LM A H)M c cl(L n H)M. This proves

the proposition.

Examples. Let R = 2Z (Z the integers) and consider

the family of ideals of fL kkEN (N the counting numbers),

where Lk = 2 Z for k E N. It is easily observed that

(cl L)R = R for all k E N. Hence cl( L)R kN (cl LkR.

Therefore the closure of the intersection of an infinite

family of left ideals is not necessarily equal to the inter-

section of the closure of the same family of left ideals.

Let V be a countably infinite dimensional vector

space with basis {vj.]N, and let R be the subring of the

ring of linear transformations which contains all linear

transformations of finite rank and the transformation a

where a(V3j) = V3j+1, C(V3j+l) = 0, and c(V3j+2) = 0 for

j E N'(N' = N U(O}). Clearly, ra or ar is of finite rank

for r E R. Let I, = Cr E Rir / a and r(v ) = 0 for all

j E N'). One easily notes that II and 12 are closed left

ideals of R and that cl I1 + cl I2 cl(I + 2) = R. There-

fore the sum of the closure of left ideals is not necessarily

equal of the closure of the sum of left ideals.

4.8 Definition. An element L of RR is said to be pure

if L is closed in any torsion free module containing it.

The following proposition is noted by Alin and Dickson

in [1].

4.9 Proposition. L E Q is pure if and only if E(L)/L is

torsion free.

Proof. Assume L is a closed submodule of E(L) and

E(L) QR. If L is contained in a torsion free module M,

then E(L) is clearly closed in E(M). (E(L) can be considered

a submodule of E(M)), which implies that E(L) n M is a

closed submodule of M. Hence L is closed in M.

The converse is immediate.

4.10 Definition. For L and H closed submodules of M, we

define L v H = cl(L + H)M and L \ H = cl LM I cl H .

4.11 Pronosition. The set of closed submodules of an R-

module M, with binary operations v and A defined on it as

above, forms a complete modular lattice.

Proof. The only part of the statement which is not

immediate is that it is a modular lattice.

Suppose A, B, and C are closed submodules of RM with

B C and A C = 0. Claim (A V B) A C = B. Note A + B

n C = B. Assume (A + B) n C = B if a < 3,using the nota-

tion of 4.4. If $ is a limit ordinal, then clearly

(A + B) n C = B. If p has 0 1 as a predecessor and

m E (A + B) n C, then Rm c (A + B)1 and m E C. This im-

plies Rm is contained in B since (A + B)_1 n C = B by the

inductive assumptions,which implies that m E B since B is

a closed submodule of M. Hence (A + B) n C = B. From

Proposition 4.5, cl(A + B)M q C = (A V B) A C = B. It fol-

lows from [9 p. 167, that the lattice of closed sub-

modules is modular.

4.12 Proposition. A submodule H of M is closed in M with

respect to R3 if and only if for m E MNH there exists r E R

such that Rr + r n(m + H)L = 0 where (m + H) E M/H.

4.13 Proposition. The following are equivalent: (L is a

submodule of an arbitrary R-module M.)

(a) cl23 LM = clRq LM;

(b) R is a full ring;

(c) The lattice of closed submodules of an R-module

is complemented.

Proof. Noting that for an arbitrary torsion theory

(3,)) on Rt,t(M) = cl q[01 where t is the torsion functor,

we have that (a) implies R = j3. Hence by 2.15 and 2.25

(a) implies (b). From 2.25, one easily sees that (b) implies

(a).

Assume (b) and suppose that H is a closed submodule

of N. Let n E N H. Since (b) and (c) are equivalent, H is

closed in N with respect to )3. Hence there exists r E R

such that rn K H and Rr n(n + H) = 0. This implies that

Rrn n H = 0 and Rrn / 0 since r(n + H) / 0 and (n + H) is

an element of the torsion free module N/H. It follows that

there exists a nonzero submodule K of N such that H n K = 0,

and any submodule of N that properly contains K intersects H

nontrivially. It is easily seen that K is closed in N and

that K + H is an essential submodule of N. Hence cl(H + K)N

= N. Therefore, the lattice of closed submodules of a module

is complemented.

That (c) = (b) is immediate. This proves the pro-

position.

4.14 Proposition. If T(R) = 0, the following are equiva-

lent:

(a) R is a full ring;

(b) For a torsion free R-module M and a submodule L

of M, cl LM = E(L) n M:

(c) For N a torsion free R-module N and a submodule

H of N, cl H = (n n Nlrn 5 0 implies Rrn n H / 0 for all

r c R).

Proof. Assume (a) and that M is a torsion free R-

module with L a submodule of M. Take E(L) to be a submodule

of E(M). Then cl LE(M) is contained in E(L) by 4.6(c).

From 4.6(a), it follows that cl LM is contained in E(L) A M.

Let m E E(L) N M and suppose that rm 4 L. Then

Rrm / 0 since M is torsion free. This implies that Rrm r L

/ 0 since E(L) is an essential extension of L. Hence

m E cl LM by 4.12 and 4.13. Therefore cl LM = E(L) q M

where E(L) is contained in E(M).

Assume (b) and suppose K is an essential left ideal

of R. Then, if E(K) is taken as a submodule of E(R),

E(K) ( R = R since R is an essential extension of K. Hence

(b) implies cl KR = R, i.e., R is a full ring. Therefore,

(a) and (b) are equivalent, assuming T(R) = 0.

Assume (a) and let N be a torsion free R-module with

H a submodule. Define D = (n E Njrn / 0 implies Rrn n H / 0

for all r e R}. Since cl HN = cl 3HN, it follows by 4.14

that D is contained in cl HN. It is easily seen by trans-

finite induction that cl HN is contained in D by 4.4 and 4.5.

Hence cl HN = D.

It is obvious that (c) implies (a). This completes

the proof.

4.15 Proposition. If R is a full ring, then the lattice

of closed submodules of an R-module has a.c.c. on closed

submodules if and only if it has d.c.c. on closed submodules.

Proof. Assume R is a full ring and M is an R-module

with d.c.c. on closed submodules. Suppose C1 and C2 are

proper closed submodules in M with C1 properly contained in

C2. There exists a closed submodule B1 in M such that

cl(B1 + C ) = M and Bl n C1 = 0 by Proposition 4.13.

If C2 B1 is an essential submodule of B1, then C2

contains B1 since C2 B1 is a closed submodule of B Hence

C2 B1 is a closed submodule of BI. We note that C2 2 B

is not an essential submodule of B1 since cl(C2 + B ) = M

and C1 is properly contained in M. Since the lattice of

closed submodules of M is modular, C2 Bi 1 0. Let B2 be

a submodule of B1 maximal with respect to the property that

B n C2 = 0. It is easily checked that B2 is a closed sub-

module of B1. Therefore B2 is a closed submodule of M prop-

erly contained in B1 such that B2 v C2 = M and B2 A C2 = 0.

It follows that M has a.c.c. on closed submodules.

The proof that a.c.c. on closed submodules of an R-

module M implies d.c.c., assuming R is a full ring, is simi-

lar to the above proof.

4.16 Definition. N and H submodules of M are said to be

related if N q X = 0 if and only if H n X = 0 for any sub-

module X of M.

4.17 Proposition. If N and H are submodules of M, then:

(1) cl 3(N)M ~ N + cl,3(0)M.

(2) P ~ N P c cl 3NM

This is proven by Goldie. (See 16, p. 169].)

4.18 Recall that a module M is said to be finite dimen-

sional if it does not contain an infinite direct sum of non-

zero submodules.

4.19 Recall that a submodule U of M is said to be uniform

if U 0 and every pair of nonzero submodules of U has non-

zero intersections. (See [6, Chapter 3].)

4.20 Definition. A nonzero submodule L of M will be

called a minimal closed submodule if it is closed and does

not contain a proper closed submodule.

4.21 Pronosition. If M is a finite dimensional torsion

free R-module, then every nonzero submodule contains a uni-

form submodule.

Proof. See [6, Chapter 3 ].

4.22 Proposition. If U is a uniform submodule of M and

M E RR, then cl UM is a uniform submodule of M.

Proof. This follows directly from 4.6(b).

4.23 Proposition. A minimal closed submodule of a torsion

free module is a uniform submodule.

Proof. The proof follows immediately from 4.6(b).

4.24 Proposition. If R is a full ring, then L is a mini-

mal closed submodule of a torsion free module M if and only

if L is a maximal uniform submodule of M, i.e., there does

not exist a uniform submodule of M which properly contains L.

4.25 Proposition. If M is a finite dimensional torsion

free R-module, then there exists an integer n > 0 such that:

(1) any direct sum of uniform submodules of M having

maximal length has n terms;

(2) every direct sum of nonzero submodules of M has

at most n terms;

(3) a submodule N of M is essential if and only if

it contains a direct sum of n uniform submodules.

Proof. See [6, Chapter 3].

4.26 Prooosition. If R is a full ring and M is a torsion

free module of dimension n, then M = cl(U1 ... U ) where

1 n

U. is a minimal closed submodule of M for j = 1 ...., n.

4.27 Prooosition. If R is a torsion free full ring and R

has dimension n, then RR = cl(U1 @ ... 9 Un) where U. is a

R I n j

minimal closed submodule of R and, as a subring of R, U. is

a zero ring or has no divisors of zero for j = 1, ..., n.

4.28 Proposition. The ring direct sum of full rings is a

full ring.

4.29 Proposition. The homomorphic image of a full ring

is a full ring.

4.30 Proposition. If R is a dense ring of linear trans-

formations on a vector space V over a division ring D with

Soc R / 0, then R is a full ring if and only if RR = cl Soc R.

4.31 Proposition. If R is a full ring, then R has d.c.c.

on closed left ideals if and only if R is finite dimensional.

Comment. If R is not assumed to be full in 4.31,

then the proposition does not hold since R = 2Z is a finite

dimensional ring which does not have d.c.c. on closed left

ideals.

SECTION V

A LIST OF OPEN QUESTIONS

1. We noted in Section I that T(R)M C T(M). Under what

conditions does equality hold? (W. E. Clark)

2. If R has an identity, then it is easily seen that RR

is isomorphic to R(R /T(R )). Does the converse of

this theorem hold?

3. If R is a torsion free Jacobson radical ring, is it

possible that all torsion free modules are R-projective?

(W. E. Clark)

4. Let R be the subring of the 2 x 2 matrix ring with

rational entries such that the elements of R are of the

form Are all torsion free R-modules

0 0

R-projective? (W. E. Clark)

5. If R is Jacobson semi-simple and all (92)93-torsion

free R-modules are (2 )3 -projective, is R semi-simple

Artinian?

6. What conditions on R are necessary to guarantee that

n (La) = ( n L )M where L a}aA is a set of

aEA A a

submodules of a torsion free module M?

7. Give a ring-theoretical or module-theoretical

characterization of rings of the form S. where

iI 1

I is possibly infinite and S. is a semi-simple

Artinian ring for each i. (W. E. Clark)

8. If R has no identity, is it true that the categories

!2 and .V, where R is the n X n matrix ring with

R R n

n

entries from R, are equivalent? This holds for unital

modules over rings with identity. However,

(R )n ; (Rn) so this is apparently not helpful.

(W. E. Clark)

9. If R is nil semi-simple and has d.c.c. on closed left

ideals, does R = J 0 S where J is the Jacobson radical

and S is Artinian semi-simple?

10. Is it possible to characterize those rings R which are

of the form I1 for some ideal I in R by considering the

properties of the torsion theory ( 3I, R)? (W. E. Clark)

11. Is it possible to find necessary and sufficient con-

ditions on R such that cl LM + cl H = cl(L + H)M

for all R-modules M?

BIBLIOGRAPHY

[1] J. S. Alin and S. E. Dickson. "Goldie's Torsion

Theory and its Derived Functor." To be

published.

[2] S. E. Dickson. "A Torsion Theory for Abelian

Categories," Trans. Am. Math. Soc. 121 (1966),

223-235.

[3] N. J. Divinsky. Rings and Radicals, University of

Toronto Press (1965).

[4] E. Gentile. "Singular submodule and injective hull,"

Indag. Math. 24 (1962), 426-433.

[5] A. W. Goldie. "Torsion free modules and rings,"

J. Algebra 1 (1964), 268-287.

F6] A. W. Goldie. "Rings with Maximum Condition," Yale

University Notes, 1964.

E7] N. Jacobson. "Structure of rings," Am. Math. Colloq.

Pub., Vol. 37 (1956).

"8 J. P. Jans. "Some aspects of torsion," Pac. Jour.

Math., Vol. 15, No. 4, (1965).

F9] A. G. Kurosh. Lectures on General Algebra, Chelsea

Pub. Co., New York (1963).

[10] B. Mitchell. Theory of Catecories, Academic Press,

New York and London (1965).

BIOGRAPHICAL SKETCH

John Michael Kellett was born December 19, 1935,

in Milford, Massachusetts. He was graduated from St. Mary's

High School, Milford, in June, 1953, and received his

Bachelor of Science degree from Worcester State College in

June, 1957. He taught high school mathematics for a year

and a half before entering the United States Navy. While

serving as a naval officer, he attended the University of

New Mexico as a part-time student.

In 1965, the author received his Master of Science

degree in mathematics from Rutgers The State University.

In September of that year, he came to the University of

Florida where he is presently an interim instructor and is

working toward the degree of Doctor of Philosophy.

Mr. Kellett is married to the former Christine Hunter

of Gainesville. They have four children, William Hunter,

John Michael, Jr., Jane Therese, and Paul Cyril.

This dissertation was prepared under the direction

of the chairman of the candidate's supervisory committee

and has been approved by all members of that committee.

It was submitted to the Dean of the College of Arts and

Sciences and to the Graduate Council, and was approved as

partial fulfillment of the requirements for the degree of

Doctor of Philosophy.

August, 1968

Dean, Coll e of A-ts & Sciences

Dean, The Graduate School

Supervisory Committee:

Chairman a.

Chairman

J

J, 2., 7-/7 &f<