검색
Article Search

JMB Journal of Microbiolog and Biotechnology

OPEN ACCESS eISSN 0454-8124
pISSN 1225-6951
QR Code

Article

Kyungpook Mathematical Journal 2020; 60(4): 673-682

Published online December 31, 2020

Copyright © Kyungpook Mathematical Journal.

Direct Sums of Strongly Lifting Modules

Shahabaddin Ebrahimi Atani, Mehdi Khoramdel* and Saboura Dolati Pishhesari

Department of Mathematics, University of Guilan, P. O. Box 1914, Rasht, Iran
e-mail : ebrahimi@guilan.ac.ir, mehdikhoramdel@gmail.com and saboura_dolati@yahoo.com

Received: June 20, 2019; Revised: August 5, 2020; Accepted: August 18, 2020

For the recently defined notion of strongly lifting modules, it has been shown that a direct sum is not, in general, strongly lifting. In this paper we investigate the question: When are the direct sums of strongly lifting modules, also strongly lifting? We introduce the notion of a relatively strongly projective module and use it to show if M=M1⊕ M2 is amply supplemented, then M is strongly lifting if and only if M1 and M2 are relatively strongly projective and strongly lifting. Also, we consider when an arbitrary direct sum of hollow (resp. local) modules is strongly lifting.

Keywords: lifting modules, strongly lifting modules, coclosed submodules, hollow modules, relatively strongly projective modules

Supplemented and lifting modules are worthy of study in module theory since they are the duals of complemented and extending modules. A number of results concerning lifting modules have appeared in the literature in recent years. Lifting modules were first introduced by Takeuchi [11] but under the name codirect modules. An R-module M is called lifting if every submodule of M lies above a direct summand. The notion of strongly extending modules was introduced in [5]. In this paper, we study modules with properties that are dual to strongly extending modules. The notion of strongly lifting modules was introduced in [8, 15]. An R-module M is called a strongly lifting module if for any submodule N of M, there exists a fully invariant direct summand K of M such that KN and N/K is a small submodule of M/K.

It is of natural interest to investigate whether or not an algebraic notion for modules is inherited by direct summands and direct sums. The purpose of this paper is to study the direct sum of strongly lifting modules. The direct sum of two strongly lifting modules need not be strongly lifting. We look at when direct sums of finitely many strongly lifting modules are strongly lifting. It is shown that every strongly lifting module is a direct sum of hollow modules and every strongly lifting module is π-projective and a direct sum of hollow modules. We introduce the notion of relatively strongly projective modules and use it to show if M=M1 ⊕ M2 is amply supplemented, then M is strongly lifting if and only if M1 and M2 are relatively strongly projective and strongly lifting. Also, we consider when an arbitrary direct sum of hollow (resp. local) modules is strongly lifting.

Throughout, all rings (not necessarily commutative rings) have identity and all modules are unital right modules. For completeness, we now state some definitions and notations used in this paper. Let M be a module over a ring R. For submodules N and K of M, N ≤ K means that N is a submodule of K and End(M) denotes the ring of right R-module endomorphisms of M. We denote module direct summands by "". The symbols Z, n and Q stand for the ring of integers, the ring of residues modulo n and ring of rational numbers, respectively. Let M be a module. Let N and L be submodules of M. N is called a supplement of L if it is minimal with the property M = N + L, equivalently, M = N + L and N ∩ L ≪ N. N is called a supplement submodule if N is a supplement of some submodule of M. A module M is called a supplemented module if every submodule of M has a supplement. A module M is called amply supplemented if, for any submodules A, B of M with M = A + B there exists a supplement P of A such that P ⊆ B. Let M be a module and K≤ N≤ M. If N/K≪ M/K, then N is called coessential submodule of M and denoted by KceN. Also N is called coessential extension of K. A submodule K of M is called coclosed if K has no proper coessential submodule; this is denoted by NccM (every supplement submodule is coclosed). Furthermore, N is called the s-closure of K in M, if KceN and KccM. An idempotent e ∈ R is called left (resp. right) semicentral if re=ere (resp. er=ere), for each r ∈ R, equivalently, eR (resp. Re) is an ideal of R. The set of all left (resp. right) semicentral idempotents of R will be denoted by Sl(R) (resp. Sr(R)). If e2 = e ∈ End(M), then e∈ Sl(End(M)) if and only if eM is a fully invariant direct summand. Also e∈ Sl(R) if and only if 1-e∈ Sr(R) [2, 3]. A module M is said to have the strong summand sum property (SSSP), if the sum of any family of direct summands is a direct summand of M [14]. A module M is called a C3-module, if M1 and M2 are direct summands of M with M1M2={0}, then M1 ⊕ M2 is also a direct summand of M [5]. In [12], Talebi and Vanaja defined

Z¯(M)={Ker(φ):φHom(M,N),NE(N)}

and Z¯2(M)=Z¯(Z¯(M)), where E(N) is the injective hull of N. A module M is called a cosingular (resp. noncosingular) module, if Z(M) = 0 (resp. Z(M) = M [12]. A module M is called a 𝒯 -non-cosingular module if, for every non-zero endomorphism f of M, Im(f) is not small in M [14]. A submodule N of a module M is called t-small in M, denoted by NtM, if for every submodule K of M, Z¯2(M)N+K implies that Z¯2(M)K [1]. A module M is called t-lifting if every submodule N of M contains a direct summand K of M such that N/K ≪ t M/K [1]. A module M is called dual Rickart (resp. t-dual Rickart) if for each φEnd(M), ⊎(M) (resp. φ(Z¯2(M))) is a direct summand of M [9] ([6]).

The following are used in the sequel.

Proposition 1.1.

  • (i)([13, Proposition 1.5]) Let M be an amply supplemented module. Then every submodule of M has an s-closure.

  • (ii)([4, 3.7(6)]) Let M be an R-module and K≤ L≤ M. If KccM, then KccL and the converse is true if LccM.

  • (iii)([13, 31.4(2)]) Let M be an amply supplemented module and B ≤ C submodules of M such that C/B is co-closed in M/B and B is co-closed in M. Then C is co-closed in M.

  • (iv)([2, Lemma 1.1]) If M=⊕i∈ IMi and N is a fully invariant submodule of M, then N=⊕i∈ I(N∩ Mi).

Theorem 1.2. ([8, 15])

The following are equivalent for an R-module M with S=End(M).

  • (1) M is strongly lifting;

  • (2) For each N ≤ M, there exists e∈ Sl(S) such that eM⊆ N and (1-e)M∩ N≪ (1-e)M.

  • (3) M is lifting and each direct summand of M is fully invariant.

  • (4) M is lifting and S is Abelian.

  • (5) M is amply supplemented and each coclosed submodule is a fully invariant direct summand in M.

Proposition 1.3. ([8, 15])

Let M be a strongly lifting module. Then each direct summand of M is strongly lifting.

Theorem 1.4. ([8, 15])

If M is strongly lifting with S=End(M), then M has SSSP.

Lemma 1.5. ([8, 15])

Let M=M1 ⊕ M2 be an amply supplemented module. Suppose that for every co-closed submodule N of M such that either M = N + M1 or M = N + M2, N is a fully invariant direct summand of M. Then M is strongly lifting.

This section is devoted to investigate when direct sums of strongly lifting modules are strongly lifting. While it was shown in [8, 15] that every direct summand of a strongly lifting module is always strongly lifting, the following examples show that in general, the direct sum of strongly lifting modules is not a strongly lifting module.

Example 2.1.

  • (1) Let R=2202, M1=2200 and M2=0002. Then M1 and M2 are strongly lifting R-modules, however R=M1 ⊕ M2 is not strongly lifting, by Theorem 1.2.

  • (2) Let R be a uniserial ring and I a nonzero proper right ideal of R. Then R and I are strongly lifting R-modules. If R ⊕ I is strongly lifting, then R ⊕ I has SSSP. So it is a C3-module. This implies that I≤ R, by [7, Corollary 3.2], which is a contradiction.

Proposition 2.2.

  • (i) Let M=⊕i∈ IMi. If M is a strongly lifting module, then Hom(Mi,Mj)=0 for each i≠ j of I.

  • (ii) A free R-module F is strongly lifting if and only if RR is strongly lifting and rank(F)=1.

Proof. (i) As M is strongly lifting, every direct summand of M is fully invariant in M. Hence the result is clear.

(ii) Assume that F is a strongly lifting free R-module. If rank(F)≥ 2, then (i) gives Hom(R,R)=0, a contradiction. Thus rank(F)=1, and R is strongly lifting. The converse is clear.

Theorem 2.3.

  • (i) Let M be an R-module. The following are equivalent:

    • (1) M is strongly lifting;

    • (2) If M=N+K (N,K ≤ M), then there exists a fully invariant direct summand X say M=X ⊕ Y such that X ⊆ N, Y ⊆ K and Y ∩ N ≪ Y.

  • (ii) Let M be a strongly lifting module. Then M is π-projective.

Proof. (i) (1) ⇒ (2) Let M=N+K. As M is amply supplemented, there exists YK such that M=Y+N and YNY. Also, there exists XN such that M=X+Y and YXX. Therefore XYM. As X is supplement of Y and Y is supplement of X, X and Y are coclosed submodules of M. Hence X and Y are fully invariant direct summands of M by Theorem 1.2. Let X=eM and Y=fM for some e,fSl(End(M)). It can be seen XY=efMM. As XYM, XY=0. Therefore M=X ⊕ Y and XN and YK and YNY.

(2)(1) is clear.

(ii) is clear from (i).

Theorem 2.4.

Let M be a strongly lifting module. Then there is a decomposition M=iIMi with hollow modules Mi, and, for every direct summand N of M, there exists a subset J of I with M=(iJMi)N.

Proof. Let M be a strongly lifting module. Then M has SSSP by Theorem 1.4. So by [10 Theorem 2.17], M=iIMi, where Mi is indecomposable for each i ∈ I. By Proposition 1.3, Mi is strongly lifting for each i ∈ I. An inspection shows that Mi is hollow for each i ∈ I.

Now, let N be a direct summand of M. Then by Theorem 1.2, N is fully invariant in M. Hence by Proposition 1.1(iv), N=iI(NMi). As NM, NMiMi for each i ∈ I. Since Mi is indecomposable, either NMi=Mi or NMi=0. Thus N=iKMi for some KI and so M=(iJMi)N for some JI.

Corollary 2.5.

Let R be a ring. Then R is strongly lifting as an R-module if and only if R is a direct product of finite local rings.

Proof. Let R be strongly lifting as an R-module. By Theorem 2.4, R=i=1nHi where n and Hi is a hollow module, for each 1in. By Theorem 1.2, every Hi is fully invariant. It is clear that Hi is finitely generated, for each 1in. Therefore R=R1×R2×...×Rn, where each Ri is a local ring. Conversely, assume that R is a direct product of finite local rings. This implies that R is a semiperfect ring by [16, 42.6]. Hence R is lifting as an R-module. Moreover, R is Abelian. Hence by Theorem 1.2, R is strongly lifting.

By Corollary 2.5, we can see that every strongly lifting ring (considered as a module over itself) is semiperfect, the following example shows that the converse is not true, in general.

Example 2.6.

Let R=FFFF, where F is a field. Then R is a semiperfect ring, which is not strongly lifting as an R-module.

We now define a relative version of particular projective condition which is useful in our main theorems.

Definition 2.7.

Let M and N be two R-modules. Then M is called N-strongly projective (or strongly projective relative to N) if Hom(M,T)=0 for each factor module T of N.

This is obviously true if and only if M is N-projective and Hom(M,N)=0. R-modules {Mi:iI} are called relatively strongly projective if Mi is Mj-strongly projective for all distinct i,j ∈ I.

Lemma 2.8.

  • (i) Let M and N be two modules. If M is N-strongly projective and NN, then M is N-strongly projective and N/N-strongly projective.

  • (ii) iIMi is N-strongly projective if and only if Mi is N-strongly projective for each i ∈ I.

  • (iii) M is i=1nNi-strongly projective if and only if M is Ni-strongly projective for each 1 ≤ i ≤ n.

  • (iv) Let M be a finitely generated module. Then M is i ∈ INi-strongly projective if and only if M is Ni-strongly projective for each i∈ I.

Proof. (i) Suppose that Hom(M,T)=0 for each factor module T of N and NN. Then Hom(M,T)=0 for each factor module T of N, because every factor module of N is a submodule of a factor module of N. Now, since every factor module of N/N is a factor module of N, Hom(M,T)=0 for each factor module T of N/N.

(ii) Let T be a factor module of N. Then Hom(iIMi,T)=0 if and only if Hom(Mi,T)=0 for each iI. Hence iIMi is N-strongly projective if and only if Mi is $N$-strongly projective for each iI.

(iii) The necessity is clear by (i). For the sufficiency, it is sufficient to show that if M is relative strongly projective to N1 and N2, then M is N1⊕ N2-strongly projective. Let fHom(M,(N1N2)/L) where LN1N2. Let π:(N1N2)/L(N1N2)/(N1+L) be natural homomorphism. As (N1N2)/(N1+L) is a factor module of N2 and M is N2-strongly projective, πf=0. Hence f(M)Ker(π)=(N1+L)/L. Since M is N1-strongly projective, f=0. Hence M is N1⊕ N2-strongly projective.

(iv) Assume that M is iINi-strongly projective. Then M is Ni-strongly projective by (i). Conversely, let M be Ni-strongly projective for each i∈ I and fHom(M,(iINi)/L) where LiINi. Since M is finitely generated, f(M)(iFNi+L)/L for some finite subset F of I. By (iii), M is iFNi-strongly projective; hence f=0. Thus M is iINi-strongly projective.

The result of Lemma 2.8(iii) does not extend to infinite direct sums, as the following example shows.

Example 2.9.

It is clear that Q is Z-strongly projective because Hom(,n)=0 for each integer n. As Q is a homomorphic image of a free Z-module (I) for some infinite set I, Hom(,T)0 for some factor module T of (I). Hence Q is not (I)-strongly projective.

Lemma 2.10.

Let M=M1M2. Then the following are equivalent:

  • (1) M2 is M1-strongly projective;

  • (2) If M=K+M1 for some K ≤ M, then M2 ⊆ K.

Proof. (1)(2) Let K be a submodule of M such that M=K+M1. Hence we have M2M/M1=(K+M1)/M1K/(M1K). We show K=(KM1)(KM2). Define f:K/(M1K)M1/(M1K) so that f(k+M1K)=k1+M1K where k=k1+k2, k1M1 and k2M2. Clearly, f is a homomorphism. Since Hom(M2,M1/(M1K))=0 and M2K/(KM1), we have f=0. This implies that for each k=k1+k2 where k1M1 and k2M2, k1M1K. Therefore k2M2K. Hence K=(KM1)(KM2). As M=K+M1, M=M1(KM2). By using modular law we have M2=KM2. Thus M2K.

(2)(1) Let f:M2L be a homomorphism and L a factor module of M1. Let π:M1L be natural homomorphism. Set K={m2m1:m2M2, m1M1 and f(m2)=π(m1)}. It can be seen that K is a submodule of M and M=K+M1. By (2), M2K. This implies that f(m2)=0 for each m2M2. Thus f=0, as desired.

In the following theorem, we present necessary and sufficient conditions under which direct sum of finite strongly lifting modules is strongly lifting.

Theorem 2.11.

Let M=M1 ⊕ M2 and M be an amply supplemented module. Then M is strongly lifting if and only if

  • (i) M1 and M2 are relatively strongly projective.

  • (ii) M1 and M2 are strongly lifting.

Proof. We show M2 is M1-strongly projective. Let M=K+M1 for some K ≤ M. By Theorem 2.3, there exists a fully invariant direct summand X of M say M=XY such that XM1 and YK and YM1M. As M is strongly lifting, every direct summand of M is fully invariant by Theorem 1.2, this implies that M1YM. Hence M1Y=0 and M=YM1. Since M2 is fully invariant, M2=(M2Y)(M2M1)=M2Y, by Proposition 1.1(iv). Thus M2YK. Therefore, by Lemma 2.10, M2 is M1-strongly projective. Similarly M1 is M2-strongly projective.

Conversely, let $K$ be a coclosed submodule of $M$ such that $K+M_1=M$. By Lemma 2.10, M2K and so K=M2(KM1). Since KM1K, KM1ccK. As KM1KM and KM1ccK and KccM, We have KM1ccM by Proposition 1.1(ii). Since KM1M1M, and KM1ccM, we have KM1ccM1 by Proposition 1.1(ii). Since M1 is strongly lifting, KM1 is a fully invariant direct summand of M1. Therefore K=KM1M2M. We show K is fully invariant in M. Let fEnd(M). Since Hom(M1,M2)=0 and Hom(M2,M1)=0, f=f1f2 where f1End(M1) and f2End(M2). Thus f(K)=f1(KM1)f2(M2). Since KM1 is fully invariant in M1, f1(KM1)KM1. This implies that f(K)K and so K is fully invariant. Similarly, if K is a coclosed submodule of M with K+M2=M, then K is a fully invariant submodule of M. Thus by Lemma 1.5, M is strongly lifting.

Corollary 2.12.

Let M=M1...Mn be a finite direct sum of modules Mi. Then M is strongly lifting if and only if M is amply supplemented, for each 1in, Mi is strongly lifting and modules {Mi}i=1n are relatively strongly projective.

Proof. Let M be strongly lifting. Then by Theorem 2.11, Mi is jiMi-strongly projective. Hence Mi is Mj-strongly projective by Lemma 2.8. Hence {Mi}1in are relatively strongly projective. The other implication is clear from Theorem 1.2 and Proposition 1.3.

Conversely, assume that M is amply supplemented and Mi is strongly lifting and relatively strongly projective. Then by induction on n, it is enough to prove that M is strongly lifting when n = 2. This follows from Theorem 2.11 and Lemma 2.8.

In [5], it is shown an R-module M is strongly extending if and only if M=Z2(M)N for some N≤ M where Z2(M) ( second singular submodule of M) and N are both strongly extending and Hom(K,Z2(M))=0 for each submodule K of N. The next theorem is dual of this result and similar to [12, Theorem 4.1].

Theorem 2.13.

Let M be an R-module. Then M is strongly lifting if and only if M=Z¯2(M)N for some submodule N of M, where Z¯2(M) and N are both strongly lifting and N is Z¯2(M)-strongly projective and M is amply supplemented.

Proof. The necessity is clear by Theorem 2.11 and Theorem 1.2 (since Z¯2(M) is coclosed by [12, Corollary 3.4], Z¯2(M)M). For the sufficiency, it suffices to show that Z¯2(M) is N-strongly projective. Let L≤ N and N/L be a factor module of N and fHom(Z¯2(M),N/L). By [12, Theorem 3.5], Z¯2(N/L)=(Z¯2(N)+L)/L. As Z¯2(N)=0, Z¯2(N/L)=0. Since f(Z¯2(M))Z¯2(N/L)=0, f=0. Thus  Hom(Z¯2(M),N/L)=0. Hence by Theorem 2.11, M is strongly lifting.

Corollary 2.14.

Let M be a strongly lifting module. Then M is t-dual Rickart.

Proof. By Theorem 2.13, M=Z¯2(M)N for some submodule N of M, where Z¯2(M) and N are both strongly lifting and M is amply supplemented. Therefore by [1, Theorem 1], M is t-lifting. As Z¯2(M) is noncosingular (and so 𝒯-noncosingular), Z¯2(M) is dual Rickart. Therefore by [6, Theorem 3.2], M is t-dual Rickart.

In the next theorem, it is considered when direct sum of arbitrary hollow modules is strongly lifting.

Theorem 2.15.

Let M=iIMi with Mi hollow. Then the following are equivalent:

  • (1) M is strongly lifting;

  • (2) Mi is j ∈ I, j ≠ iMj-strongly projective.

Proof. (1) ⇒ (2) is from Theorem 2.11.

(2)⇒(1) For each i ∈ I, Mi is hollow, hence End(Mi) is Abelian. Also for each i,jI with $i\neq j $, Hom(Mi,Mj)=0 by Lemma 2.8(i). This follows that End(M) is Abelian. It suffices to show that M is lifting. Let N≤ M. Let {Nα}αΓ be all direct summands of M that are contained in N. Let K= αΓNα. We will show KM and K=iTMi for some TI. Since End(M) is Abelian, Nα is fully invariant for each α ∈ Γ. Hence Nα=iI(NαMi) by Proposition 1.1(iv). Since NαM, NαMiMi for each i∈ I. Therefore Mi is hollow gives either NαMi=Mi or NαMi=0. This implies that Nα=iTαMi for some TαI. Set T=αΓTα. Then it can be easily seen K=iTMi and so KM say M=KK. We show KNM. Assume that KN+S=M for some submodule S of M. Let Q= SM, SSS. Then QM and Q=iJMj for some JI by argument mentioned before. Let jJ and π:MMj be natural projection. As KN+S=M, Mj=π(KN)+π(S). Since Mj is hollow, π(KN)=Mj or π(S)=Mj. Hence M=Ker(π)+KN or M=Ker(π)+S. By (2) and Lemma 2.10, MjKN or MjS. If MjKN, then MjN and MjK. Hence MjK and so Mj=0. If MjS, then MjQ and so Mj=0. Hence M=Q. This gives S=M, as desired.

In the following, it is considered when direct sum of an arbitrary local modules is strongly lifting.

Corollary 2.16.

Let M=iIMi where each Mi is local. Then M is strongly lifting if and only if Mi is Mj-strongly projective where j ≠ i.

Proof. Let M be strongly lifting. Then by Theorem 2.11, Mi is ijIMj-strongly projective. Thus by Lemma 2.8(i), Mi is Mj-strongly projective for each i≠ j∈ I.

Conversely, let Mi be Mj-strongly projective for each i≠ j of I. By Lemma 2.8(iv), Mi is iIMj-strongly projective. Hence M is strongly lifting by Theorem 2.15.

In the next theorem, we use Theorem 2.15 to show that each factor module by a fully invariant submodule of a strongly lifting module is strongly lifting.

Theorem 2.17

Let M be a strongly lifting module and N a fully invariant submodule of M. Then M/N is strongly lifting.

Proof. Let M be a strongly lifting module and N a fully invariant submodule of M. Then by [4, 22.2(4)], M/N is lifting. We show End(M/N) is Abelian. By Theorem 2.4, M=iIHi where Hi is hollow for each i∈ I. Since N is fully invariant, N=iI(NHi) by Proposition 1.1(iv). Hence M/N=iIHi/(NHi). Since Hi/(NHi) is indecomposable for each i∈ I, it suffices to show that Hom(Hi/(NHi),Hj/(NHj))=0 for each distinct i,j of I. Letf:Hi/(NHi)Hj/(NHj) be a homomorphism and i ≠ j ∈ I. Let π:HiHi/(NHi) be natural homomorphism. By Theorem 2.15, $H_i$ is jI,ijHj-strongly projective, so Hi is Hj-strongly projective for each distinct i,j of I by Lemma 2.8(i). Hence fπ=0. Thus f=0, as desired.

We would like to thank the referees for valuable comments.

  1. T. Amouzegar, D. K. Tütüncü and Y. Talebi, t-dual Baer modules and t-lifting mod-ules, Vietnam J. Math., 42(2)(2014), 159-169.
    CrossRef
  2. G. F. Birkenmeier, B. J. Müller and S. T. Rizvi, Modules in which every fully invariant submodule is essential in a direct summand, Comm. Algebra, 30(2002), 1395-1415.
    CrossRef
  3. G. F. Birkenmeier, J. K. Park and S. T. Rizvi, Modules with fully invariant submodules essential in fully invariant summands, Comm. Algebra, 30(2002), 1833-1852.
    CrossRef
  4. J. Clark, C. Lomp, N. Vanaja and R. Wisbauer, Lifting modules, supplements and projectivity in module theory, Frontiers in Mathematics, Birkh¨auser Verlag, 2006.
  5. S. Ebrahimi Atani, M. Khoramdel and S. Dolati Pish Hesari, Strongly extending modules, Kyungpook Math. J., 54(2014), 237-247.
    CrossRef
  6. S. Ebrahimi Atani, M. Khoramdel and S. Dolati Pish Hesari, t-dual Rickart modules, Bull. Iranian Math. Soc., 42(3)(2016), 627-642.
  7. S. Ebrahimi Atani, M. Khoramdel and S. Dolati Pish Hesari, C3-modules, Demonstr. Math., 49(3)(2016), 282-292.
    CrossRef
  8. M. Khoramdel, On Baer and Rickart modules and some results on semirings, Ph.D thesis, University of Guilan, 2015.
  9. G. Lee, S. T. Rizvi, and C. S. Roman, Dual Rickart modules, Comm. Algebra, 39(11)(2011), 4036-4058.
    CrossRef
  10. S. H. Mohamed and B. J. Müller, Continuous and discrete modules, London Math-ematical Society Lecture Note Series 147, Cambridge University Press, Cambridge, 1990.
  11. T. Takeuchi, On cofinite-dimensional modules, Hokkaido Math J., 5(1976), 1-43.
    CrossRef
  12. Y. Talebi and N. Vanaja, The torsion theory cogenerated by M-small modules, Comm. Algebra, 30(2002), 1449-1460.
    CrossRef
  13. D. K. Tütüncü, On lifting modules, Comm. Algebra, 28(2000), 3427-3440.
    CrossRef
  14. D. K. Tütüncü and R. Tribak, On dual Baer modules, Glasg. Math. J., 52(2010), 261-269.
    CrossRef
  15. Y. Wang, Strongly lifting modules and strongly dual Rickart modules, Front. Math. China, 12(1)(2017), 219-229.
    CrossRef
  16. R. Wisbauer, Foundations of module and ring theory, Gordon and Breach Science Publishers, Philadelphia, 1991.