Volume 2007, Article ID 58132,8pages doi:10.1155/2007/58132
Research Article
δ
-Small Submodules and
δ
-Supplemented Modules
Yongduo Wang
Received 9 January 2007; Accepted 18 June 2007
Recommended by Akbar Rhemtulla
LetRbe a ring andMa rightR-module. It is shown that (1)δ(M) is Noetherian if and only ifM satisfies ACC onδ-small submodules; (2)δ(M) is Artinian if and only ifM satisfies DCC onδ-small submodules; (3)Mis Artinian if and only ifM is an amplyδ -supplemented module and satisfies DCC onδ-supplement submodules and onδ-small submodules.
Copyright © 2007 Yongduo Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction and preliminaries
In this note, all rings are associative with identity and all modules are unital right modules unless otherwise specified.
LetRbe a ring andMa module. The concept ofδ-small submodules was introduced by Zhou in [1]. Motivated by [2–4], we study modules with ACC (resp., DCC) onδ-small submodules and prove thatδ(M) is Noetherian (resp., Artinian) if and only ifMsatisfies ACC (resp., DCC) onδ-small submodules inSection 2. InSection 3, we give the con-cepts of (amply)δ-supplemented modules viaδ-small submodules. It is shown thatMis Artinian if and only ifM is an amplyδ-supplemented module and satisfies DCC onδ -supplement submodules and onδ-small submodules. InSection 4, we introduce the con-cept ofδ-semiperfect modules and investigate the connections betweenδ-supplemented modules andδ-semiperfect modules.
a submoduleBofM such that M=A+BandA∩BB. The notions which are not explained here will be found in [6].
Lemma 1.1 (see [7, Proposition 5.20]). Suppose thatK1≤M1≤M,K2≤M2≤M, and
M=M1⊕M2. ThenK1⊕K2≤eM1⊕M2if and only ifK1≤eM1andK2≤eM2.
2. Modules with chain conditions onδ-small submodules
In this section, we study modules with chain conditions onδ-small submodules and prove that δ(M) is Noetherian (resp., Artinian) if and only ifM satisfies ACC (resp., DCC) onδ-small submodules. Let us start with the following.
Lemma 2.1 (see [1, Lemma 1.3]). LetMbe a module. (i) For submodulesN,K,LofMwithK≤N,
(1)NδMif and only ifKδMandN/KδM/K; (2)N+LδMif and only ifNδMandLδM.
(ii) IfKδMand f :M→Nis a homomorphism, then f(K)δN. In particular, if KδM≤N, thenKδN.
(iii) LetK1≤M1≤M,K2≤M2≤M, andM=M1⊕M2. ThenK1⊕K2δM1⊕M2
if and only ifK1δM1andK2δM2.
Lemma 2.2 (see [1, Lemma 1.5]). LetMandNbe modules. (1)δ(M)=Σ{L≤M|Lis aδ-small submodule ofM}. (2) If f :M→Nis a homomorphism, then f(δ(M))≤δ(N). (3) IfM=i∈IMi, thenδ(M)=i∈Iδ(Mi).
(4) If every proper submodule ofM is contained in a maximal submodule ofM, then δ(M) is the unique largestδ-small submodule ofM.
Theorem 2.3. LetMbe a module. Thenδ(M) is Noetherian if and only ifMsatisfies ACC onδ-small submodules.
Proof. “⇒” It is clear byLemma 2.2.
“⇐” Suppose thatδ(M) is not Noetherian. LetA1≤A2≤ ··· be an infinite ascending
chain of submodules ofδ(M). Leta1∈A1 andaj∈Aj−Aj−1 for each j >1. For any
k≥1, letNk=Σkj=1ajR. ThenNkis finitely generated andNk≤δ(M). HenceNkδM.
It is clear thatN1≤N2≤ ···and soMfails to satisfy ACC onδ-small submodules. This
completes the proof.
Recall that a moduleMhas finite uniform dimensionk, for some nonnegativek, ifM does not contain any infinite direct sum of nonzero submodules andk is the maximal number of summands in a direct sum of nonzero submodules ofM. In this case, we call kthe uniform dimension ofM, and write udimM=k.
Proposition 2.4. LetMbe a module. Then the following statements are equivalent. (1)δ(M) has finite uniform dimension.
(2) Everyδ-small submodule ofMhas finite uniform dimension and there exists a pos-itive integerksuch that udimN≤kfor anyNδM.
Proof. “(1)⇒(2)” It is obvious because udimN≤udimδ(M) for anyNδM.
“(2)⇒(3)” LetN1⊕N2⊕ ···be an infinite direct sum of nonzeroδ-small submodules
ofM. ThenN1⊕ ··· ⊕Nk+1is aδ-small submodule ofMand udim(N1⊕ ··· ⊕Nk+1)≥
k+ 1. This is a contradiction.
“(3)⇒(1)” Let N1⊕N2⊕ ··· be an infinite direct sum of nonzero submodules of
δ(M). For everyi≥1, letni be a nonzero element ofNi. ThenniRδM. Thusn1R+
n2R+··· is an infinite direct sum of nonzeroδ-small submodules ofM. This is a
con-tradiction and soδ(M) has finite uniform dimension.
Theorem 2.5. LetMbe a module. Then the following statements are equivalent. (1)δ(M) is Artinian.
(2) Everyδ-small submodule ofMis Artinian. (3)Msatisfies DCC onδ-small submodules.
Proof. “(1)⇒(2)⇒(3)” They are clear.
“(3)⇒(1)” It suffices to prove that any factor module ofδ(M) is finitely cogenerated. If there exists a factor module ofδ(M) that is not finitely cogenerated, then the setΩ of submodules ofδ(M), such thatδ(M)/Lis not finitely cogenerated, is nonempty. Let
{Lλ:λ∈Λ}be any chain of submodules inΩ. LetL=λ∈ΛLλ. IfL∈Ω, thenδ(M)/Lis
finitely cogenerated and henceL=Lλfor someλ∈Λ. ThusL∈Ω. By Zorn’s lemma,Ω
has a minimal memberA.
LetNbe a finitely generated submodule ofδ(M). ThenNis aδ-small submodule ofM and hence Artinian by hypothesis. Thusδ(M) is locally Artinian. Now letx∈δ(M),x∈A. ThenxRis Artinian and (xR+A)/AxR/(xR∩A). So (xR+A)/Ais a nonzero Artinian module and henceδ(M)/Ahas essential socle. LetSdenote the submodule ofδ(M), con-tainingA, such thatS/Ais the socle ofδ(M)/A. ThusS/Ais not finitely generated by [7, Proposition 10.7].
Next we show thatAδM. IfM=A+Bfor someB≤MandM/Bis singular, then S=A+ (S∩B). Suppose thatA∩B=A. Thenδ(M)/(A∩B) is finitely cogenerated by the choice ofA. ButS/A=(A+ (S∩B))/A(S∩B)/(A∩B)≤Soc(δ(M)/(A∩B)) and henceS/Ais finitely generated. This is a contradiction. ThusA=A∩B≤Band we have M=A+B=B. SoAδM.
Now suppose thatM=S+Vof some submoduleV ofMandM/Vis singular. Then M/(A+V)=(S+V)/(A+V)S/(A+ (S∩V)). ThusM/(A+V) is semisimple. IfM=
A+V, then there exists a maximal submoduleWofM such thatA+V≤W. ButS≤ δ(M)≤WsinceM/Wis a singular simple module and this gives the contradictionM=
W. ThusM=A+V, henceM=VsinceAδM. ThusSδMand henceSis Artinian by hypothesis. It follows thatS/Ais Artinian, and, in particular,S/Ais finitely generated. This is a contradiction. Thusδ(M) is Artinian.
Example 2.6. Let R=Z, p is a prime and M=Z(p∞), the Pr ¨ufer p-group, then every
Corollary 2.7. LetRbe a ring which satisfies DCC onδ-small right ideals. ThenRsatisfies ACC onδ-small right ideals.
LetN≤M.Nis called aδ-semimaximal submodule ofM ifN=ni=1LiwithM/Li
singular simple for anyi=1,...,n.
Proposition 2.8. LetMbe a module. Then the following statements are equivalent. (1)Mis Artinian.
(2)Msatisfies DCC onδ-small submodules and onδ-semimaximal submodules. (3)Msatisfies DCC onδ-small submodules andδ(M) is aδ-semimaximal submodule.
Proof. “(1)⇒(2)” It is clear.
“(2)⇒(3)” Suppose thatMsatisfies DCC onδ-semimaximal submodules. LetNbe a minimalδ-semimaximal submodule ofM. Clearlyδ(M)≤N. IfM=δ(M), thenδ(M)=
N. Suppose thatM=δ(M). IfPis a maximal submodule ofM withM/Psingular, then N∩P is aδ-semimaximal submodule of M and hence N=N∩P, so that N≤P. It follows thatN≤δ(M). HenceN=δ(M). Thusδ(M) is aδ-semimaximal submodule of M.
“(3)⇒(1)” It is clearδ(M) is Artinian. If M=δ(M), thenM is Artinian. Suppose thatM=δ(M). Thenδ(M)=P1∩P2∩ ··· ∩Pn, whereM/Piis singular simple for any
i=1,...,n. It follows thatM/δ(M) embeds in the finitely generated semisimple module
M/P1⊕ ··· ⊕M/Pn. HenceM/δ(M) is Artinian and soMis Artinian.
3.δ-supplemented modules
Let M be a module. Let N and L be submodules of M. N is called a δ-supplement of Lif M=N+Land N∩LδN.N is called aδ-supplement submodule if N is a δ-supplement of some submodule ofM.M is called a δ-supplemented module if ev-ery submodule ofM has aδ-supplement. On the other hand,M is called an amply δ -supplemented module if for any submodulesA,BofMwithM=A+Bthere exists aδ -supplementPofAsuch thatP≤B. Clearly, supplemented modules areδ-supplemented modules and every amplyδ-supplemented module isδ-supplemented. But the converses are not true.
Lemma 3.1. LetMbe aδ-supplemented module. Then (1)M/δ(M) is semisimple;
(2)La submodule ofMwithL∩δ(M)=0, thenLis semisimple.
Proof. (1) LetNbe any submodule ofMcontainingδ(M). Then there exists aδ -supple-mentK ofN in M, that is,M=N+K and N∩KδK. ThusM/δ(M)=N/δ(M)⊕
(K+δ(M))/δ(M), and so every submodule ofM/δ(M) is a direct summand. Therefore M/δ(M) is semisimple.
(2) It is clear by (1), sinceL∼=L⊕δ(M)/δ(M)≤M/δ(M).
Proposition 3.2. LetMbe an amplyδ-supplemented module. Then homomorphic images are amplyδ-supplemented modules.
Since M is amply δ-supplemented, there exists a submodule X of M such that M=
f−1(A) +X,f−1(A)∩XX≤f−1(B). Now,N=A+f(X) andA∩f(X)=f(f−1(A)∩
X)δ f(X). Clearly f(X)≤B.
Proposition 3.3. LetMbe aδ-supplemented module. ThenM=N⊕Lfor some semisim-ple moduleNand some moduleLwithδ(L)≤eL.
Proof. Forδ(M), there existsN≤Msuch thatN∩δ(M)=0 andN⊕δ(M)≤eM. Since Mis aδ-supplemented module, there existsL≤Msuch thatN+L=MandN∩LδL. SinceN∩L=N∩(N∩L)≤N∩δ(L)≤N∩δ(M)=0,M=N⊕L. ByLemma 3.1,N is semisimple. Thusδ(M)=δ(N)⊕δ(L). SinceN⊕δ(L)≤eM=N⊕L,δ(L)≤eLby
Lemma 1.1. This completes the proof.
Lemma 3.4. LetM1,U≤M and letM1be aδ-supplemented module. IfM1+U has aδ
-supplement inM, then so doesU.
Proof. SinceM1+U has aδ-supplement inM, there existsX≤M such thatX+ (M1+
U)=MandX∩(M1+U)δX. For (X+U)∩M1, sinceM1is aδ-supplemented
mod-ule, there existsY ≤M1such that (X+U)∩M1+Y=M1and (X+U)∩Y δY. Thus
we haveX+U+Y=Mand (X+U)∩YδY, that is,Y is aδ-supplement ofX+U inM. Next, we will show thatX+Y is aδ-supplement ofU inM. It is clear that (X+ Y) +U=M, so it suffices to show that (X+Y)∩UδX+Y. SinceY+U≤M1+U,
X∩(Y+U)≤X∩(M1+U)δX. Thus (X+Y)∩U≤X∩(Y+U) +Y∩(X+U)δ
X+Y byLemma 2.1, as required.
Proposition 3.5. LetM1andM2beδ-supplemented modules. IfM=M1+M2, thenMis
aδ-supplemented module.
Proof. LetUbe a submodule ofM. SinceM1+M2+U=Mtrivially has aδ-supplement
inM,M2+Uhas aδ-supplement inMbyLemma 3.4. ThusUhas aδ-supplement inM
byLemma 3.4again. SoMis aδ-supplemented module.
Proposition 3.6. IfMis aδ-supplemented module, then every finitelyM-generated mod-ule is aδ-supplemented module.
Proof. From Proposition 3.5, we know that every finite sum ofδ-supplemented mod-ules is aδ-supplemented module. Next we will show that every factor module of a δ -supplemented module is again aδ-supplemented module.
LetM be aδ-supplemented module andM/Nany factor module ofM. For any sub-moduleLofMcontainingN, sinceMis aδ-supplemented module, there existsK≤M such thatL+K=MandL∩KδK. ThusM/N=L/N+ (N+K)/Nand (L/N)∩((N+ K)/N)=(N+ (L∩K))/Nδ(N+K)/N, that is, (N+K)/N is aδ-supplement ofL/Nin
M/N, as required.
Proposition 3.7. LetMbe a module. If every submodule ofMis aδ-supplemented module, thenMis an amplyδ-supplemented module.
Proof. LetL,N≤MandM=N+L. By assumption, there isH≤Lsuch that (L∩N) + H=Land (L∩N)∩H=N∩HδH. ThusH+N≥H+ (L∩N)=Land henceH+
Corollary 3.8. LetRbe any ring. Then the following statements are equivalent. (1) Every module is an amplyδ-supplemented module.
(2) Every module is aδ-supplemented module.
A moduleM is said to beπ-projective if for every two submodulesU,V ofM with U+V=Mthere exists f ∈End(M) with Imf ≤Uand Im(1−f)≤V.
Theorem 3.9. LetM be a module. IfMis aπ-projectiveδ-supplemented module, thenM is an amplyδ-supplemented module.
Proof. LetA,Bbe submodules ofMsuch thatM=A+B. SinceMisπ-projective, there exists an endomorphismeofMsuch thate(M)≤Aand (1−e)(M)≤B. Note that (1− e)(A)≤A. LetCbe aδ-supplement ofAinM. ThenM=e(M) + (1−e)(M)=e(M) + (1−e)(A+C)≤A+ (1−e)(C)≤M, so thatM=A+ (1−e)(C). Note that (1−e)(C) is a submodule ofB. Lety∈A∩(1−e)(C). Theny∈Aandy=(1−e)(x)=x−e(x) for somex∈C. Nextx=y+e(x)∈A, so thaty∈(1−e)(A∩C). ButA∩CδCgives that A∩(1−e)(C)=(1−e)(A∩C)δ(1−e)(C). Thus (1−e)(C) is aδ-supplement ofA
inM. It follows thatMis an amplyδ-supplemented module.
Theorem 3.10. LetM be a module. Then M is Artinian if and only ifM is an amply δ-supplemented module and satisfies DCC on δ-supplement submodules and onδ-small submodules.
Proof. The necessity is clear. Conversely, suppose thatM is an amplyδ-supplemented module which satisfies DCC onδ-supplement submodules and onδ-small submodules. Thenδ(M) is Artinian byTheorem 2.5. Next, it suffices to show thatM/δ(M) is Artinian. It is clear thatM/δ(M) is semisimple byLemma 3.1.
Now suppose thatδ(M)≤N1≤N2≤N3≤ ···is an ascending chain of submodules
ofM. BecauseM is an amplyδ-supplemented module, there exists a descending chain of submodulesK1≥K2≥ ···such thatKiis aδ-supplement ofNiinMfor eachi≥1.
By hypothesis, there exists a positive integertsuch thatKt=Kt+1=Kt+2= ···. Because
M/δ(M)=Ni/δ(M)⊕(Ki+δ(M))/δ(M) for all i≥t, it follows that Nt=Nt+1= ···.
ThusM/δ(M) is Noetherian, and hence finitely generated. SoM/δ(M) is Artinian, as
desired.
Example 3.11. ForZZ, the onlyδ-supplement submodules are 0 andZand the onlyδ
-small submodule is 0, butZZis not Artinian.
Corollary 3.12. LetMbe a finitely generatedδ-supplemented module. ThenMis Artinian if and only ifMsatisfies DCC onδ-small submodules.
Proof. “⇐” SinceM/δ(M) is semisimple andMis finitely generated,M/δ(M) is Artinian. Now that M satisfies DCC onδ-small submodules,δ(M) is Artinian by Theorem 2.5. ThusMis Artinian.
“⇒” It is clear.
Remark 3.13. LetRbe a ring. IfRRis an amplyδ-supplemented module, thenRis a right
Let us end this section with the following.
Proposition 3.14. IfM is aδ-supplemented module and satisfies DCC onδ-small sub-modules, then so doesM/Afor any submoduleAofM.
Proof. LetAbe any submodule ofMandB1/A≤B2/A≤ · · ·where eachBi/AδM/A.
LetC be aδ-supplement of Ain M. ThenM/A=(A+C)/AC/A∩C. SinceBi/A is δ-small inM/A,Bi/ADi/A∩CC/A∩Cfor someDi. Next we prove thatDiδM.
Let Di+E=M withM/E singular. Then (Di+ (E+A∩C))/A∩C=M/A∩C. Hence
E+A∩C=M andE=M. Thus we haveD1≤D2≤ · · ·. SinceM satisfies ACC onδ
-small submodules, there existsnsuch thatDk=Dk+1for allk≥n. ThusBk/A=Bk+1/A
for allk≥n. ThereforeM/Asatisfies ACC onδ-small submodules, as required.
4.δ-semiperfect modules
In this section, we introduce the concept ofδ-semiperfect modules and investigate the interconnections betweenδ-supplemented modules andδ-semiperfect modules. Let P andM be modules, we call an epimorphism f :P→Maδ-cover in case Kerf δP. A δ-cover f :P→Mis called a projectiveδ-cover in casePis a projective module.
Definition 4.1. A moduleMis called aδ-semiperfect module if any homomorphic image ofMhas a projectiveδ-cover.
Proposition 4.2. If f :M→N is an epimorphism with Kerf ≤δ(M), then δ(N)=
f(δ(M)).
Proof. It follows from [7, Corollary 8.17].
Lemma 4.3. If both f :P→Mandg:M→Nareδ-covers, theng f :P→Nis aδ-cover.
Proof. If both f :P→Mandg:M→N areδ-covers, then Kerf δPand KergδM. We want to show that Kerg f δP. Let P=Kerg f +L withP/L singular. ThenM=
Kerg+ f(L). SinceM/ f(L) is singular, M= f(L). This implies thatP=LsinceP/L is
singular and Kerf δP, as desired.
Lemma 4.4. If eachfi:Pi→Mi(i=1, 2,...,n) is aδ-cover, thenni=1fi:
n i=1Pi→
n i=1Mi
is aδ-cover.
Proof. It is straightforward.
Theorem 4.5. LetMbe a module andU≤M. Then the following statements are equivalent. (1)M/Uhas a projectiveδ-cover.
(2) IfV≤MandM=U+V, thenUhas aδ-supplementU≤V such thatUhas a projectiveδ-cover.
(3)Uhas aδ-supplementUwhich has a projectiveδ-cover.
Proof. “(1)⇒(2)” Let f :P→M/U be a projectiveδ-cover. SinceM=U+V,g:V →
“(2)⇒(3)” It is obvious.
“(3)⇒(1)” Let f :P→U be a projectiveδ-cover. SinceUis aδ-supplement ofU, the natural epimorphism g:U→U/U∩UU+U/U=M/U is a δ-cover. Hence hg f :P→M/Uis a projectiveδ-cover byLemma 4.3, whereh:U/U∩UU+U/Uis
an isomorphism
Theorem 4.6. LetMbe a module. Then the following statements are equivalent. (1)Misδ-semiperfect.
(2)Mis amplyδ-supplemented byδ-supplements which have projectiveδ-covers. (3)Misδ-supplemented byδ-supplements which have projectiveδ-covers.
Proof. It is clear fromTheorem 4.5.
Example 4.7. A δ-semiperfect module is not necessarily semiperfect. Let Q=Π∞
i=1Fi,
where eachFi=Z2. LetR be the subring ofQ generated by∞i=1Fi and 1Q. Then RR
isδ-semiperfect but not semiperfect. It is also seen thatRR is aδ-supplemented module
but not a supplemented module (see [1, Example 4.1]).
References
[1] Y. Zhou, “Generalizations of perfect, semiperfect, and semiregular rings,” Algebra Colloquium, vol. 7, no. 3, pp. 305–318, 2000.
[2] E. P. Armendariz, “Rings with DCC on essential left ideals,” Communications in Algebra, vol. 8, no. 3, pp. 299–308, 1980.
[3] I. Al-Khazzi and P. F. Smith, “Modules with chain conditions on superfluous submodules,”
Com-munications in Algebra, vol. 19, no. 8, pp. 2331–2351, 1991.
[4] V. Camillo and M. F. Yousif, “CS-modules with ACC or DCC on essential submodules,”
Com-munications in Algebra, vol. 19, no. 2, pp. 655–662, 1991.
[5] Y. Wang and N. Ding, “Generalized supplemented modules,” Taiwanese Journal of Mathematics, vol. 10, no. 6, pp. 1589–1601, 2006.
[6] R. Wisbauer, Foundations of Module and Ring Theory, vol. 3 of Algebra, Logic and Applications, Gordon and Breach Science, Philadelphia, Pa, USA, German edition, 1991.
[7] F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, vol. 13 of Graduate Texts in
Mathematics, Springer, New York, NY, USA, 1974.
Yongduo Wang: Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou 730050, China
