Vienam Journal of Mathematics 25:l (1997) 33-39
Vietnarn
Journal
of
MATHEMATICS
q Springer-Verlag 1997
The Bass-PappTheoremand SomeRelatedResults
A. Harmanci,P. F. Smith, A. Tercan,and Y. Tiras
Department of Mathematics, Hacettepe Uniuersity, Beytepe, Ankara, Turkey
Received
July8, 1995
Absfract.In the spirit of the Bass-Papptheorem,this paperis concerned
with ringsfor
whichany(finite)directsumof quasi-continuous
modulesis quasi-continuous
or for which
any (finite)directsumof CS-modules
is CS.
l. Introduction
we show that a ring R is right QI if and only if the direct sum of any two quasiinjective right R-modulesis quasi-injective.on the other hand, a ring R is semiprime Artinian if and only if the direct sum of any two quasi-continuousright
R-modulesis quasi-continuous.A right nonsingularring R is right Noetherianif
and only if everydirect sum of injectiveright R-modulesis cs. we also show that
if R is a ring with finitely generatedright soclesuchthat the direct sum of any two
cs right R-module is cs, then every right R-modulesis cs and R is right and
left Artinian.
All rings are associativeand have identity elementsand all modulesare unital
right modules.Let R be any ring. A right R-module M is called a cs-module if
every submoduleis essentialin a direct summand("c,s" standsfor complements
are summands).Recall also that the module M is called quasi-continuous
if M is
cs and for all direct summandsK andz with K a L: 0, the submoduleK @ z is
also a direct summandof M. The module M is called continuousif M is cs and
for eachdirect summandN of M and eachmonomorphisme; N - M, the submodule p(N) is also a direct summandof M.Finally, the module M is quasiinjectiueif M is M-injective, i.e., for every submodule 11 of M, any homomorphism 0: H - M can be lifted to M.It is well known that the following
implicationshold for M:
M is injective + M is quasi-injective+ M is continuous
> M is quasi-continuous
+ M ls CS.
For thesefacts and a good accountofthis area,see[3] or [7].
The Bass-Papptheoremstatesthat a ring R is ring Noetherianif and only if
everydirect sum of (a countablenumber of) injectiveright R-modulesis injective
34
A. Harmanci, P. F. Smith, A. Tercan, and Y. Tiras
(see,for example,[1, Proposition18.13]or [1], Theorem4.lD. It is naturalto raise
the following generalquestion:for which rings R is every direct sum of quasiinjective (respectively,continuous,quasi-continuous,CS) right R-modulesquasiinjective(continuous,quasi-continuous,CS)?The purposeof this note is to try to
answerthesequestions.
For any unexplainedterminology,pleasesee[1], [3], or [7].
Modules
2. Quasi-Iniective,Continuousand Quasi-Continuous
In this section,we are concernedwith the problem of finding which rings R have
the property that the classesof quasi-injectiveor continuousor quasi-continuous
modulesare closedunder taking direct sums(coproducts).lf M is an R-module,
then E(M) will denotethe injectivehull of M, Soc(M) the socleof M, Z(M) the
singularsubmoduleof M,i.e.,
right ideal,4 of R}
Z(M) : {m e M : mA :0 for someessential
and Z2(M) the secondsingularsubmoduleof M, r.e.,Zz(M) is the submoduleof
M, containin9Z(M) suchthat Z2(M)lZ(M) : Z(M lZ(M)). A ring R is called
a right Ql-ing if everyquasi-injectiveright R-moduleis injective.
Our first result givesa characterizationof right Q/-rings.
Propositionl. Thefollowing statementsare equiualent
for a ring R:
(i) R ls a right Ql-ring.
(11)The direct sum of any two quasi-injectiue
right R-modulesis quasi-injectiue.
(111)Thedirectsumof anyfamily of quasi-injectiue
right R-modulesis quasi-injectiue.
Proof. (i) = (iii). Let R be a right Ql-ri'ng. Every semisimpleR-module is quasiinjectiveand henceinjective.By [], Theorem4.1], R is right Noetherian.Now
(iii) follows by the Bass-Papptheorem.
(iii) + (ii). Clear.
(ii) > (i). Let M be any quasi-injectiveright R-module. Then E(Ra) @ M is
quasi-injectiveby hypothesis,and henceM is an injective R-moduleby [7, Proposition1.31.
I
Proposition2. Thefollowing statementsare equiualent
for a ring R:
(i) Euerycontinuousright R-moduleis injectiue.
(ii) The direct sum of any two continuousright R-modulesis continuous.
Qll fhe direct sum of anyfamily of continuousright R-modulesis continuous.
Proof. (1)+ (iii). In particular, R is a ight Ql-nng and henceis right Noetherian.
Now (iii) follows by the Bass-Papptheorem.
(iii) - (ii). Clear.
(ii) + (i). Let M be a continuous R-module. The E(Rn @ M) is continuous
by hypothesis.Hence, M is an injective R-module by [7, Propositions 1.3 and
2.101.
r
The Bass-Papp Theorem and Some Related Results
35
Propositions I and 2 raise the following natural question: if R is a right
Ql-ing, is everycontinuousright R-moduleinjective?
Theorem3. Thefollowing statementsare equiualent
for a ring R:
(i) R r semiprimeArtinian.
(ri) Eueryquasi-continuous
right R-moduleis injectiue.
(iii) Thedirectsumof any two quasi-continuous
right R-modulesis quasi-continuous.
(iv) The direct sum of any family of quasi-continuous
right R-modulesis quasicontinuous.
Proof. (i) > (iv) + (iii). Clear.
(iii) + (ii). By the proof of Proposition2 (ii) + (i).
(iD - G). Suppose every quasi-continuousR-module is injective. Then R
is a right Ql-ing and hence is right Noetherian. By [ll, Theorem 4.4],
E(Rn) : (Di. r E;, whereE; is indecomposableinjective for each i e I. Let i e I.
Let 0 * u e Ei. Then zR is uniform, whencequasi-continuous.By hypothesis,aR
is injective.Hence,rzRis a direct summandof E; and we haveEi:uR. Thus,Ei
is simple.It followsthat.E(Ra) is semisimpleand henceso too is Ra.
I
It is not the case that every right Ql-ing is semiprimeArtinian (see for
example,14,19.491).
3. Direct Sumsof Iniectives
A ring R is right Noetherian if and only if the direct sum of every (countable)
collectionof injectiveright R-modulesis quasi-continuous.First, supposethat the
direct sum of every countable collection of injective right R-modulesis quasicontinuous.Let E;(i e 1) be any countablecollectionof injectiveR-modules.Then
(@rE,)@E(RR) is quasi-continuous
and hence,by [7, Proposition2.10],@7E;
is an injectiveR-module.By [1, Theorem4.1], R is right Noetherian.The conversealsoclearlyfollowsfrom [1], Theorem4.1].
We shall show that a right nonsingularring R is right Noetherianif and only if
everydirect sum of injectiveright R-modulesis CS.
Lemma4. Let R be a ring with right socleS.
(1) If the directsumof euery'countablecollectionof injectiuehulls of singularsimple
right R-modulesis injectiue,thenR/^S,sright Noetherian.
(ii) If RIS is right Noetherian,then euery direct sum of singular injectiueright
R-modules is injectiue.
Proof. (l) By a result of Goodearl [5, Proposition3.6],it is sufficientto prove that
the R-moduleRIE is Noetherianfor any essentialright ideal -E of R. This is done
by adaptingthe proof of [11, Theorem4. I ].
(ii) By [9, Corollary 1l].
Lemma5. Let R be a right nonsingularring right socleS suchthat the direct sumof
A. Harmanci, P. F. Smith, A. Tercan,and Y. Tiras
euerycountablecollectionof injectiueright R-modulesis CS. Thenthe right RIS is
right Noetherian.
Proof. Let & (l e 1) be singular simple right R-modules. Let E E ( S r ) @ E ( S z ) @ E ( S r ) @ . . . a n dl e t M : E ( R n ) @ E . B y h y p o t h e s iM
s, is a
CS-module.By [6, Theorem 1], becauseE(Rn) is nonsingularand E : Zz(M),
we deduce that E is E(Rp)-injective. Hence, E is an injective R-module. By
Lemma 4, RIS is right Noetherian.
I
Recall that a ring R hasfinite right unifurm dimensionif it containsno infinite
direct sumsof nonzeroright ideals.
Corollary 6. A right nonsingularring R is right Noetherianif and only if R hasfinite
right unifurm dimensionand the direct sum of euery( countable
) collectionof indecomposable
injectiueright R-mo&tlesis CS.
Proof. By Lemma 5.
Let R be a right nonsingularring and let Q denotethe maximal right quotient
ring of R. Recall that Q is a von Neumann regular right self-injectivering, R is a
subringof Q, and the right R-module Q is the injectivehull of the right R-module
R. Let X be a right Q-modulesuch that as a right R-module, X is nonsingular
and CS. Let Y be any submoduleof Xg. Thereexist submodulesZ, Zt of Xp such
that X : Z @ Z' and I is essentialin Z. SinceXa is nonsingular,it is easy to
checkthat X : (ZQ)@Q'Q).Moreover, Z is essential
in (ZQ)^ so that I is
essentialin (ZQ) ^. But this implies that Y is essentialin (ZQ) o.It follows that the
p-module X is CS.
I
We haveproved the following.
Lemma7. Let R be a right nonsingularring with maximal right quotientring Q. Let
X be a right Q-modulesuchthat the right R-moduleX is nonsingularCS. Thenthe
right Q-moduleX is CS.
Corollary 8. Let R be a right nonsingularring with maximal right quotientring Q.
ThenR hasfinite right uniform dimensionif and only if the right R-moduleQV) tt
CSfor any index set I.
Proof. Sttpposethat R has finite right uniform dimension.Clearly, Q is a nonsingular right R-module. Let A be a right ideal of R and let g : A ---+Q0 6s ^
holomorphism. There exists a finitely generatedright ideal ,B of R such that
-B is an essentialsubmoduleof Ap. There exists a finite subsetJ of I such that
a@) = 0(4. Sitrcepa is nonsingular,it follows that q(A) c. Q@ also.But Q@ is
injective,so that there existsq e Q@ c. 90 with 9@) - qa for all a in.4. Thus,
0(1) ir injective and hence CS. Conversely,if the right R-module 20 is C^Sfor
any index set 1, then the right Q-moduleQ0 is CS (Lemma 7). Thus, Q is right
Artinian by [8, TheoremII] (or see[3, 11.13D.It follows that R has finite right
uniform dimension.
I
The Bass-Papp Theorem and Some Related Results
Combining Lemma 5 and Corollary 8 we have at once:
Theorem9. A right nonsingularring R is right Noetherianif and only if the direct
sum of euerycollectionof injectiueright R-modulesis CS.
4. Direct Sumsof C,S-Modules
In this section,our concernis with rings R havingthe property that any direct sum
of CS-modulesis CS. It turns out that if the direct sum of any two CS-modulesis
CS, then in many casesthe direct sum of any collection of CS-modulesis CS.
Recall that any uniform module is CS. Note that if R is any ring, then the following statementsare equivalent:
(i) the direct sum of any two uniform right R-modulesis CS;
(ii) the direct sum of any collection of uniform right R-modules is CS (see
[3, l3.l]).
A ring R is a right V-ring if every simpleright R-moduleis injective.It is well
known that a commutativering R is a Z-ring if and only if R is von Neumann
regular.A ring R is calledright semi-Artinianif everynonzeroright R-modulehas
nonzerosocle.For examplesof suchrings, see[2].
Proposition10. Let R be a ring which is either (a) a commutatiueuon Neumann
regular ring or (b) a right semi-Artinion,right V-ring. Then eueryunifurm right
R-moduleis simpleand euerydirect sum of unifurm right R-modulesis CS.
Proof. First, supposethat R is a commutativevon Neumann regular ring. Let U
be a uniform R-module.Let 0 * u e U. Let r eR with ur + 0. Then rR : eR for
s o m ei d e m p o t e net i n R . A l s o ,0 * u r R : u e R c U e .N o w U : U e @ U ( l - e )
givesU(l - e) :0 and henceu(l - e) : 0. Thus,tt : ue e urR.It follows that uR
is simpleand henceinjective.Thus, [/: aR. Hence,U is simple.
Now supposethat R is a right semi-artinianright Z-ring. Let U be any uniform right R-module.Then U containsa simplesubmoduleS and S is injective,so
that U: S.
Thus in either caseuniform modules are simple and clearly direct sums of
uniform modulesare CS, becausesemisimplemodulesare CS.
I
Lemma ll. Let R be any ring, M a semisimpleright R-module,and M2 a right Rmodulewith zero soclesuchthat M : Mr @ M2 is a CS-module.ThenM1 is M2injectiue.
Proof. Clearly,Mr : Soc(M). Let N be any submoduleof M2 and let rp: N -+ M1
be a homomorphism.Let L: {* - q(*): x e N}. Then Z is a submoduleof M
a n d L a M r : 0 . T h e r e e x i s t s u b m o d uKl,e K
s ' o f M s u c ht h a t M : K @ K t a n d
Z is an essentialsubmodule of K. Note that Soc(K) : K a Mt :0, so that
Mt :Soc(M) = K' . Thus, K/ : Mr @ (K' a M2) and M : K @ Mr @ (K' n 1',Iz).
Let n ; M ---+M1 denotethe projection with kernel K @ (K' n TrIz).Let 0 denote
the restriction of g to Mz. Then 0 : M2---+M1 and 0(x) : 9Q) for all x in N. It
follows that Ml is M2-injective.
38
A. Harmanci, P. F. Smith, A. Tercan, and Y. Tiras
Lemma 12. Let R be a ring suchthat the direct sum of any two CS right R-module
is CS. ThenR is right semi-Artinian.
Proof. Stppose first that R has zero right socle.Let X be any semisimpleright
R-module and let E : E(Rn). By hypothesis,M : X @ n is CS and hence,by
Lemma 11, X is E-injective. It follows that every semisimpleright R-module is
injective.By [1], Theorem4.1], R is right Noetherian.Now [3, 13.3]givesthat
everyright R-moduleis CS and [3, 13.5]givesR right Artinian.
I n g e n e r a ll,e t 0 : . S e ( R ) = S 1 ( R )c . . - & ( R ) c S " * 1 ( R )c . . . d e n o t e
the right Loewy seriesof R, i.e., for each ordinal a ) 0, S'11(R)/S-(R) is the
right socleof the ring R/&(R) and S,(R) : (Jl<p<oSp(R)
if a is a limit ordinal.
There existsan ordinal p > 0 suchthat Sr(R) : Sp+r(R),i.e., the ring R/S'(R)
has zero right socle. By the above argument, R/S,(R) is right Artinian
and hence ,Sp(R): R. It follows that R is semi-Artinian (see, for example,
[2, Propositionl]).
Theorem13. Thefollowing statementsare equiualentfor a ring R with Jacobson
radical J.
(1) R hasfinite right uniformdimensionand the directsum of any two uniform right
R-modulesis CS.
(ii) R hasfinite left uniform dimensionand the direct sum of any two unifurm left
R-modulesis CS.
(iiD R hasfinitely generatedright socleand the direct sum of any two CS right
R-modulesis CS.
(iv) R has finitely generatedleft socle and the direct sum of any two CS left
R-modulesis CS.
(v) Eueryright R-moduleis CS.
(vi) Eueryleft R-moduleis CS.
(vii) R is (right and left) Artinian serialand Jz :0.
Proof.(v)+(vi)+(vii). By [3, 13.5].
(v) + (i). Clear.
(i) = (v). There exist a positive integer n and indecomposableinjective right
R - m o d u l eEsi ( l < i < n ) s u c ht h a t E ( R a ): E r @ . . . @ 8 , . B y [ 3 , 1 3 . 1 E
] ; is
Artinian for each I and hence,R is right Artinian. Clearly, every cyclic right
R-modulehasfinite uniform dimension.By [3, 13.3],everyright R-moduleis CS.
(ii)e(v).
(v) + (iii).
(iii) + (i).
(iv)e(v).
As for (i)e(v).
Clear.
By Lemma 12.
Similar to (iii)+(v).
T
Two questionsspring to mind at this point. First of all, if R is a ring suchthat
the direct sum of any two CS-modulesis CS, then is any direct sum of CS-modulesCS? Relatedto this, we also ask: if R is a ring suchthat the direct sum of any
two CS-modulesis CS, is any R-module CS?
The Bass-Papp Theorem and Some Related Results
Finally, we give an example of a commutative ring for which the direct sum of
every collection of uniform modules is CS but not every direct sum of CS-modules
is CS.
Example. Let R denote the group algebra K[G] where K is a field of characteristic
O and G is the direct product of an infinite family of cyclic groups. Then
(i) R is a commutative von Neumann regular ring (and hence is nonsingular).
(ii) R has zero socle (and hence R is not semi-Artinian).
(iii) Every direct sum of uniform R-modules is CS.
(iv) There exist CS R-modules X and I such that X @ I is not CS.
Proof. (i) by [10, Theorem 3.1.5].
(ii) SupposeR has nonzero socle. Let U be a minimal ideal. Let0*ueU.
Then there exist a finite subgroup G1 and a nontrivial subgroup Gz with
G : Gr x G2 and u e KlGl]. Let | * g e Gz. Then z(l - S) +0 and hence, u :
u ( I - g ) r f o r s o m e r e R . S u p p o s eg h a s o r d e r n . T h e n z ( l - t s * . . . * g n - t ) : 0
a n d h e n c e ,u + u g + . . . + u g n - t - 0 . I t f o l l o w s t h a t u : 0 , a c o n t r a d i c t i o n .
(iii) By Proposition 10.
(iv) By (ii) and Lemma 12.
I
Acknowledgement.
This work was made possibleby TUBITAK Grank number TBAG1200 and was carried out during visits of the secondauthor to HacettepeUniversity in
September1993and April 1994.The authors wish to thank both organisationsfor their
support.
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