ON P-EXTENDING MODULES
M. A. KAMAL and O. A. ELMNOPHY
Abstract. LetRbe a ring. A rightR-moduleMis calledquasi-principally injectiveif it isM-principally injective.In this paper, we give some characterizations and properties of principally injective modules, which generalize results of Nicholson and Yousif. For a quasi-principally injective moduleM, we show: 1. For isomorphic submodulesH,K of M, we haveSH =SK, where S is the endomorphism ring ofM. 2. M has (P C2), and consequently has (P C3).
We characterize when a direct sum of P-extending modules is P-extending, and when a direct sum of a P-extending module and a semi-simple module is P-extending. We also characterize when a direct sum of FP-extending modules is FP-extending. Finally, we discuss when a direct sum of P-extending modules with relatively EC-injective is P-extending.
1. Introduction
In [7], Nicholson and Yousif have introduced and studied the structure of principally injective rings, and have given some characterizations of such rings in terms of the internal properties of these rings. In fact, they defined principally injective modules in the following sense: A right moduleM over a ringRis calledprincipally injective (for shortP-injective) if every R-homomorphism from a principal right ideal ofR to M can be extended to R.
In [8], Wongwai extended this notion to modules by making use ofM-cyclic submodules of M.
Here, we adopt the extension of the concept of principally injective rings, which is given in [7], to modules.
The fact that, cyclic andM-cyclic submodules of a moduleM are not the same (e.g., asZ-modules, the integers Z is cyclic submodule, but not a Q-cyclic submodule, of Q, andQ is Q-cyclic but not cyclic, of Q), gives the independence of the concepts of N-principally injective by Wongwai and the one we are dealing with.
Received August 19, 2004.
2000Mathematics Subject Classification. Primary 16D50, 16D70, 16D80.
Key words and phrases. Principally injective modules, extending modules.
We also introduce the definitions of principally extending, (for short P-extending), and P-(quasi-)continuous modules as follows: For a rightR-moduleM,
1. M is called aP-extending module if every cyclic submodule of M is essential in a direct summand of M or, equivalently, every EC-closed submodule of M is a summand. M is called anFP-extending module if every finite uniform dimension EC-closed submodule ofM is a summand.
2. M is called a P-quasi-continuous module if it is P-extending, and the following condition holds: (P C3) For each a,b∈M, ifaR andbR≤⊕ M withaR∩bR= 0, thenaR⊕bR≤⊕ M.
3. M is called aP-continuous module if it is P-extending, and the following condition holds: (P C2) For each a,b∈M, ifaR∼=bRandbR≤⊕M, thenaR≤⊕ M.
It is known that in regular rings the condition (C2) is satisfied, and so such rings are continous if and only if they are extending. Consequently, every regular ring is P-continuous as a module over itself. It is also clear that regular rings are P-injective rings. This allows us to find P-injective modules, which are not injective.
Direct sums of extending modules have been investigated in great detail, in a long series of papers, by Dung and Smith [3], and by Kamal and Muller [4], [5]. The present paper studies direct sums of P-extending modules, and we investigate when such direct sums are P-extending.
It is known thatM isN-injective if and only if for every submoduleAofN⊕M withA∩M = 0, there exists a submodule B ofN ⊕M such that A≤B, and N⊕M =B⊕M. In analogue, we introduce the concept of N-EC-injectivity, and give a characterization of such modules different from the diagram description. This helps us to build up blocks of P-extending modules, which are relatively EC-injective to obtain P-extending modules.
We prove that, if M =M1⊕M2, then Mi is P-extending and isMj-EC-injective (i 6=j = 1,2) if and only if M =C⊕Mi0⊕Mj, whereMi0 ≤Mi, for every EC-closed submoduleC ofM withC∩Mj= 0 (i6=j= 1,2).
All modules here are right modules over a ringR. The right (respectively, left) annahilator of a subsetX of a module is denoted byrR(X) (resp. lR(X)). A submoduleA of a moduleM is calledessential inM orM is an essential extension ofA(denoted byA≤eM), if every non-zero submodule ofM has non-zero intersection with A. X ≤⊕M signifies that X is a direct summand ofM.
A submoduleAof M is calledM-cyclic submodule ofM if it is isomorphic toM/X, for some submoduleX ofM. The injective hull and the uniform dimension of a moduleM will be denoted byE(M) andU−dim (M) respectively. The endomorphism ring of a module M is denoted by End (M). A submodule is closed in M if it has no proper essential extensions in M. The graph of a homomorphism f : N → M is the submodule hfi={n−f(n) :n∈N}ofN⊕M.
A module M is extending (n-extending) if every closed submodule A (with U −dim (A) ≤ n) is a direct summand ofM, or equivalently to the requirement that every submoduleA (withU−dim (A)≤n) is essential in a direct summand ofM.
A moduleM is calledquasi-continuous if it is extending module, and the following condition holds: (C3) For allX, andY ≤⊕ M,with X∩Y = 0, one has X⊕Y ≤⊕ M. M is calledcontinuous if it is extending module, and the following condition holds: (C2) If a submoduleAofM is isomorphic to a direct summand ofM, thenA is a direct summand ofM.
2. Principally Injective Modules
LetR be a ring andM,N be R-modules. M is calledN-principally injective (for shortN-P-injective) if every R-homomorphism from a cyclic submodule ofN toM can be extended toN. Equivalently, for eachm∈M and n∈N withrR(n)⊆rR(m), there exists f ∈HomR(N, M) such thatm=f(n).
Within the proof of [2, Proposition 1.1], it was observed thatM isN-injective if and only ifN⊕M =C⊕M, for every complement C of M in N ⊕M. The condition 3. in the next Proposition is analogous with such observation.
Proposition 2.1. Let M andN beR-modules, andS= End (M). Then the following are equivalent:
1. MisN P-injective;
2. For each m∈M andn∈N withrR(n)⊆rR(m), we haveSm⊆HomR(N, M)n;
3. For eachm∈Mandn∈N withrR(n)⊆rR(m), there is a complementC ofM inN⊕M withn−m∈C andN⊕M =C⊕M;
4. For each n∈N,lMrR(n) = HomR(N, M)n;
5. For each n∈N anda∈R,lM[aR∩rR(n)] =lM(a) + HomR(N, M)n.
Proof. 1. ⇒ 2.: Let m ∈ M and n ∈ N with rR(n) ⊆rR(m). Since M is N-P-injective, then there exists a homomorphism f : N → M such that m = f(n). Let φ ∈ S, then φ(m) ∈ HomR(N, M)n. Therefore, Sm⊆HomR(N, M)n.
2.⇒3.: Letm∈M andn∈N with rR(n)⊆rR(m), then by 2. there exists a homomorphismf : N →M such thatm=f(n). HenceN⊕M =hfi⊕M, wherehfiis the graph of a homomorphismf :N →M. Therefore, C=hfiis a complement ofM inN⊕M with N⊕M =C⊕M andn−m∈C.
3.⇒4.: Letn∈N andx∈lMrR(n), thenrR(n)⊆rR(x). By 3. there is a complementC ofM in N⊕M withn−x∈C and N⊕M =C⊕M. So, there exists a homomorphismf :N →M such thatC =hfi. Since n−x∈C, thenn−x=n0−f(n0), for somen0∈N. So,n=n0andx=f(n0) =f(n). Hencex∈HomR(N, M)n, andlMrR(n)⊆HomR(N, M)n. The other conclusion is obvious.
4.⇒5.: Letn∈N,a∈R, and x∈lM[aR∩rR(n)], then x(aR∩rR(n)) = 0 and sorR(na)⊆rR(xa). Hence lMrR(xa)⊆lMrR(na) = HomR(N, M)na, by 4. Therefore, xa =f(na) = f(n)a, for some f ∈HomR(N, M).
So (x−f(n))a = 0 and x−f(n) ∈ lM(a). Thus x ∈ lM(a) + HomR(N, M)n, and so lM[aR∩rR(n)] ⊆ lM(a)+HomR(N, M)n. On the other hand, letx∈lM(a)+HomR(N, M)n, thenx=m+f(n) for somem∈lM(a) andf ∈HomR(N, M). Soxa=ma+f(n)a=f(na). Letar∈aR∩rR(n), thenx(ar) =f(na)r=f(nar) = 0, and sox∈lM[aR∩rR(n)]. ThuslM(a) + HomR(N, M)n⊆lM[aR∩rR(n)].
5. ⇒1.: Letm ∈M and n∈N with rR(n) ⊆rR(m), then lMrR(m)⊆lMrR(n). By 5. we getlMrR(n) = HomR(N, M)n, and so there is a homomorphismf :N →M such thatf(n) =m. ThusM isN-P-injective.
Proposition 2.2. Let M be N-P-injective, then M is X-P-injective, for every submodule X of N. If, in addition,X is a direct summand ofN, thenM isN/X-P-injective.
Proof. It is clear.
Lemma 2.3. Let M beN-P-injective andK≤⊕M, then K is N-P-injective.
Proof. It is obvious.
Lemma 2.4. Let{Mi}i∈I be a family of modules. Then the direct product Q
i∈I
Mi isN-P-injective if and only ifMi isN-P-injective, for everyi∈I.
Proof. It is clear.
Proposition 2.5. If Mis a quasi-principally injective module, andS = End(M), then SH =SK, for any iso- morphicR-submodulesH,K of M.
Proof. SinceH ∼=K, then there is a right R-isomorphism σ :H →K. For each k∈ K, k=σ(h) for some h ∈ H and rR(h) = rR(k). Since M is quasi-principally injective, then Sh = Sk by Proposition 2.1, and so Sk⊆SH, for eachk∈K. ThenSK⊆SH. Similarly, we getSH ⊆SK, and so the result.
Corollary 2.6. Let R be a P-injective ring and H,K be two-sided ideals of R. If H ∼=K, as right ideals of R, thenH =K.
Remark. In Corollary2.6, the condition P-injective for the ring R is not avoided. In fact, there are rings which do not satisfy the result in2.6, for example, the ring Zof integers.
Theorem 2.7. Let M be a quasi-principally injective module, then M has (P C2).
Proof. Leta, b∈M withaR ∼=bRand bR≤⊕ M. Then bR=eM for some idempotent e∈End(M). Since aR ∼= bR , then there is an isomorphism σ : bR → aR. Let σe = h, then aR = hM and σ−1h = e. Since bR≤⊕ M, then by Lemma2.3,bRisM-P-injective, and so there exists a homomorphismφ:M →bRsuch that φ(a) =σ−1(a). Thenφis an epimorphism,φh=e, and sof =hφis an idempotent endomorphism ofM. Hence f M =hφM =h(bR) =heM =hM, and soaR≤⊕ M.
Remark. It is known that every summand right ideal of a ringRis generated by an idempotent element inR.
Then every summand right ideal ofRis cyclic and so,Rhas (P Ci) if and only ifRhas (Ci),i= 2,3. Therefore by [6, Proposition 2.2.], ifR has (P C2), thenRhas (C3).
Corollary 2.8([7], Theorem 2.3.). If R is a P-injective ring, thenR has(C2).
Lemma 2.9. Let M be an R-module. IfM has(P C2), thenM has (P C3).
Proof. LetaR≤⊕ M andbR≤⊕ M withaR∩bR= 0, then aR=eM = Ime, for somee2 =e∈ End(M), and so aR⊕bR = eM ⊕(1−e)bR. Since (1−e)bR ∼= bR ≤⊕ M and M has (P C2), then (1−e)bR = f M for some f2 = f ∈ End(M). Then ef = 0, and h = e+f −f e is an idempotent in End(M). Therefore, aR⊕bR=eM ⊕f M = (e+f−f e)M =hM≤⊕M.
Corollary 2.10. If M is a quasi-principally injective module, then M has (P C3).
Definition 2.1. By an EC-(closed) submodule C of a module M, we mean a (closed) submodule C which contains essentially a cyclic submodule; i.e. there existsc∈C such thatcR≤eC.
Lemma 2.11. Every summand of an EC- submodule of M is EC-submodule.
Proof. LetcR≤eCbe an EC-submodule ofM, andC1≤⊕ C, thenC=C1⊕C2, for some submoduleC2inC.
Letc=c1+c2, wherec1∈C1 andc2∈C2. It is easy to see thatc1R≤eC1. Therefore,C1is an EC-submodule
ofM.
Corollary 2.12. Every summand of an EC-closed submodule of M is EC-closed.
Lemma 2.13. Every summand of a P-(quasi-)continuous module is P-(quasi-) continuous.
Proof. It is obvious by Corollary2.12
Lemma 2.14. For an indecomposable moduleM, the following are equivalent:
1. M is extending;
2. M is P-extending;
3. M is uniform.
Lemma 2.15. A module M over a right noetherian ringR, is 1-extending if and only if it is P-extending.
Proof. Let M be a 1-extending module, and cR ≤e C be an EC-closed submodule of M. Since R is a noetherian ring, thenC has a finite uniform dimension. Since M is 1-extending, then by Proposition (4) in [4], M is n-extending. HenceC is a summand, and soM is P-extending. For the converse, it is obvious.
Corollary 2.16. Let M be a module with finite uniform dimension, then the following are equivalent:
1. M is extending;
2. M is 1-extending;
3. M is P-extending.
Proposition 2.17. Let M = M1 ⊕M2, and let C ∩M1 be an EC-submodule of M, for every EC-closed submodule C of M. Then M is P-extending if and only if every EC-closed submodule C, with C∩M1 = 0, or C∩M2= 0, is a summand.
Proof. The necessary condition is obvious. For the sufficient condition, letcR≤eCbe an EC-closed submodule ofM. IfC∩M1 = 0, then we are done. Otherwise,C∩M1 is an EC-submodule ofM, by assumption. LetC1 be a maximal essential extension ofC∩M1 in C, thenC1 is an EC-closed submodule ofM, with C1∩M2= 0.
Hence by the assumption,C1is a summand ofM. WriteM =C1⊕C2, by the modular law,C=C1⊕(C∩C2).
By Corollary2.12, C∩C2 is an EC-closed submodule ofM with (C∩C2)∩M1= 0, and therefore,C∩C2 is a summand ofM. ThusCis a summand ofM, and therefore,M is P-extending.
Proposition 2.18. Let M =M1⊕M2, where M1 is of finite uniform dimension. Then M is P-extending if and only if every EC-closed submodule C of M, with C∩M1 = 0, or C is of finite uniform dimension, is a summand.
Proof. The necessary condition is obvious. For the sufficient condition, letmR≤eCbe an EC-closed submod- ule ofM. IfC∩M1= 0, then we are done. Now let 06=c∈C∩M1, andC1 be a maximal essential extension of cR in C. SinceM1 is of finite uniform dimension, so is C1. By the given assumption, C1 is a summand of M. Write M =C1⊕K. Hence C =C1⊕C∗, where C∗ := K∩C is closed inM. Let m =c1+c∗, where c1 ∈ C1 and c∗ ∈ C∗. Since C∗ is a summand of an EC-closed submodule C, then by Corollary 2.12, C∗ is EC-closed. If C∗∩M1 = 0, then by assumption C∗ is a summand, and hence C is a summand of M. On the other hand, ifC∗∩M16= 0, then by repeating the previous steps, we haveC∗=C2⊕C3, whereC2is a summand and has a nonzero intersection with M1. Continuing in this manner, we should stop after a finite steps (due to M1 a finite uniform dimensional module) and end with C=C1⊕C2⊕. . .⊕Cn, where Ci is a summand ofM (i= 1,2, . . . , n−1), andCn contains an essential cyclic submodule withCn∩M1= 0. HenceCn is a summand
ofM, by assumption, and thereforeC is a summand ofM.
Corollary 2.19. Let M =M1⊕M2, where M1 is of finite uniform dimension. ThenM is P-extending if and only if every EC-closed submoduleC ofM, withC∩M1= 0, orC∩M2= 0, is a summand.
Proposition 2.20. LetM =M1⊕M2. ThenM is FP-extending if and only if every EC-closed submoduleC ofM with finite uniform dimensional such thatC∩M1= 0, orC∩M2= 0, is a summand.
Proof. Is similar to the proof of Proposition2.18
Proposition 2.21. Let M =M1⊕M2, where M1 is a semisimple module. Then M is P-extending if and only if every EC-closed submoduleC ofM withC∩M1= 0, is a summand.
Proof. The necessary condition is obvious. For the sufficient condition, letC be an EC-closed submodule of M. IfC∩M1 = 0, then we are done. On the other hand, sinceM1 is a semisimple, we getC∩M1 ≤⊕ M1 and soC=C∩M1⊕C∗. SinceC∗ is an EC-closed submodule ofM andC∗∩M1= 0, thenC∗is a summand ofM.
Therefore,Cis a summand ofM.
Proposition 2.22. LetM =M1⊕M2, whereM1 is P-extending andM2 isM1-P-injective. IfM2 is nonsin- gular, then every EC-closed submoduleC ofM, with C∩M2= 0, is a summand ofM.
Proof. LetcR≤eC be an EC-closed submodule ofM withC∩M2= 0, and writec=c1+c2, wherec1∈M1
andc2∈M2. SinceM2is M1-P-injective, then by Lemma 5 in [4],
cR= (c1R)∗={c1r+φ(c1)r:r∈R} ⊆(M1)∗:={m1+φ(m1) :m1∈M1} ∼=M1
and that M = (M1)∗⊕M2, where φ ∈HomR(M1, M2). Let x∈ C and write x =y+m2, where y ∈ (M1)∗ and m2 ∈ M2. Since cR≤e C, then there exists an essential right ideal I of R such that m2I = 0. Since M2 is nonsingular, thenm2 = 0. It follows that C⊆(M1)∗. Since (M1)∗ is P-extending, we haveC≤⊕ (M1)∗≤⊕
M.
Definition 2.2. Let M = M1⊕M2 be a module. The module M2 is called M1-EC-injective, if for every EC-(closed) submoduleN ofM1, and every homomorphism fromN toM2 can be extended toM1.
This is equivalent to for every EC-(closed) submoduleN of M such that N∩M2 = 0, there exists N0 ≤M such thatN ≤N0, andM =N0⊕M2.
Observe that every module over a regular ringRisR-EC-injective.
Lemma 2.23. Let M =M1⊕M2 andM2 beM1-EC-injective. Then:
1. M2 isK-EC-injective, for allK≤M1. 2. H isM1-EC-injective, for allH ≤⊕ M2.
3. H isK-EC-injective, for all K≤⊕M1, andH≤⊕M2.
Proof. LetKbe a submodule ofM1, andN be an EC-submodule ofK⊕M2 withN∩M2= 0. ThenN is an EC-submodule ofM. SinceM2isM1-EC-injective, then there isN0 ≤M such thatN ≤N0, andM =N0⊕M2 . ThenK⊕M2 = (K⊕M2)∩(N0 ⊕M2) = (N0 ∩(K⊕M2))⊕M2 and N ≤ N0 ∩(K⊕M2). Hence M2 is K-EC-injective.
2. LetH be a summand of M2, and N be an EC-submodule ofM1⊕H with N ∩H = 0. Then N is an EC-submodule ofM and N∩M2 = 0. SinceM2 isM1-EC-injective, then there is N0 ≤M such thatN ≤N0, and M =N0 ⊕M2. Since H ≤⊕ M2, then M2 =H ⊕H0, and so M1⊕H = (M1⊕H)∩(N0⊕H ⊕H0) = H⊕(M1⊕H)∩(N0⊕H0) . SinceN ≤N0, thenN ≤(M1⊕H)∩(N0⊕H0). Therefore,H isM1-EC-injective.
3. Follows from 1. and 2.
Proposition 2.24. Let M =M1⊕M2 , where M1 is P-extending and M2 isM1-EC-injective. Then M = C⊕M10⊕M2; where M10 ≤M1, for every EC-closed submodule C ofM, withC∩M2= 0.
Proof. LetcR≤eCbe an EC-closed submodule of M withC∩M2= 0. DefineX :=M1∩(C⊕M2). Then c1R≤eX , wherec=c1+c2, wherec1∈M1andc2∈M2. LetN1be a maximal essential extension ofX inM1. ThenN1 is an EC-closed submodule ofM1. SinceM1is P-extending, we haveN1≤⊕ M1. WriteM1=N1⊕M10, where M10 ≤M1. Now C⊕M2 =X⊕M2 ≤eN1⊕M2; i.e. C ≤N1⊕M2, and C ≤c N1⊕M2. Then C is a complement ofM2 inN1⊕M2. SinceM2 isM1-EC-injective, andN1is a summand ofM1, then by Lemma2.23 1., M2 is N1-EC-injective, and so there exists N0 ≤ N1⊕M2 such that C ≤ N0, and N1⊕M2 = N0 ⊕M2. HenceN0 is a complement ofM2 inN1⊕M2, butCis a complement ofM2 inN1⊕M2. Therefore,N0 =Cand M =M1⊕M2=N1⊕M10⊕M2=C⊕M10 ⊕M2. Corollary 2.25. Let M = M1⊕M2 , where Mi is P-extending and is Mj-EC-injective (i 6= j = 1,2) if and only if M = C⊕Mi0 ⊕Mj; where Mi0 ≤ Mi, for every EC-closed submodule C of M, with C∩Mj = 0 (i6=j= 1,2).
Proposition 2.26. LetM =M1⊕M2 , whereM1 andM2 are relatively EC-injective, and either M1 orM2 is of finite uniform dimension. ThenM is P-extending if and only ifM1 andM2 are P-extending.
Proof. It is follows by Corollaries2.25, and2.19.
Proposition 2.27. Let M =L
i∈I
Mi be an R-module, whereM(F)is P-extending andM(I\F)isM(F)-EC- injective, for all finite subsetF ofI. ThenM is P-extending.
Proof. Let Letc∈M andC be a maximal essential extension ofcRinM. ThencR≤M(F) andcR∩M(I\ F) = 0, for a finite subset F of I. Since cR ≤e C , then C∩M(I\F) = 0. Since M(I\F) is M(F)-EC- injective andC is EC-closed submodule of M, then by Proposition 2.24, C is a summand ofM. Hence M is
P-extending.
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M. A. Kamal, Department of Mathematics, Faculty of Education, Ain Shams University, Cairo, Egypt., e-mail:[email protected]
O. A. Elmnophy, Department of Mathematics, Faculty of Women, Ain Shams University, Cairo, Egypt., e-mail:[email protected]
