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

Published online December 31, 2020

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 $K⊆N$ 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 $Z$, $ℤn$ 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 $K↪ceN$. 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 $N↪ccM$ (every supplement submodule is coclosed). Furthermore, N is called the s-closure of K in M, if $K↪ceN$ and $K↪ccM$. 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 $M1∩M2={0}$, then M1 ⊕ M2 is also a direct summand of M [5]. In [12], Talebi and Vanaja defined

$Z¯(M)=∩{Ker(φ):φ∈Hom(M,N),N≪E(N)}$

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 $N≪tM$, 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 $φ∈End(M)$, ⊎(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 $K↪ccM$, then $K↪ccL$ and the converse is true if $L↪ccM$.

• (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.

2. Direct Sums of Strongly Lifting Modules

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 $R=ℤ2ℤ20ℤ2$, $M1=ℤ2ℤ200$ and $M2=000ℤ2.$ 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 $Y⊆K$ such that M=Y+N and $Y∩N≪Y$. Also, there exists $X⊆N$ such that M=X+Y and $Y∩X≪X$. Therefore $X∩Y≪M$. 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 $e,f∈Sl(End(M))$. It can be seen $X∩Y=efM≤⊕M$. As $X∩Y≪M$, $X∩Y=0$. Therefore M=X ⊕ Y and $X⊆N$ and $Y⊆K$ and $Y∩N≪Y$.

$(2)⇒(1)$ is clear.

(ii) is clear from (i).

Theorem 2.4.

Let M be a strongly lifting module. Then there is a decomposition $M=⊕i∈IMi$ with hollow modules Mi, and, for every direct summand N of M, there exists a subset J of I with $M=(⊕i∈JMi)⊕N$.

Proof. Let M be a strongly lifting module. Then M has SSSP by Theorem 1.4. So by [10 Theorem 2.17], $M=⊕i∈IMi$, 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), $N=⊕i∈I(N∩Mi)$. As $N≤⊕M$, $N∩Mi≤⊕Mi$ for each i ∈ I. Since Mi is indecomposable, either $N∩Mi=Mi$ or $N∩Mi=0$. Thus $N=⊕i∈KMi$ for some $K⊆I$ and so $M=(⊕i∈JMi)⊕N$ for some $J⊆I$.

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 $n∈ℕ$ and Hi is a hollow module, for each $1≤i≤n$. By Theorem 1.2, every Hi is fully invariant. It is clear that Hi is finitely generated, for each $1≤i≤n$. 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 ${Mi:i∈I}$ 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 $N′≤N$, then M is $N′$-strongly projective and $N/N′$-strongly projective.

• (ii) $⊕i∈IMi$ 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 $N′≤N$. 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 $Hom(⊕i∈IMi,T)=0$ if and only if $Hom(Mi,T)=0$ for each $i∈I$. Hence $⊕i∈IMi$ is N-strongly projective if and only if Mi is $N$-strongly projective for each $i∈I$.

(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 $f∈Hom(M,(N1⊕N2)/L)$ where $L≤N1⊕N2$. Let $π:(N1⊕N2)/L→(N1⊕N2)/(N1+L)$ be natural homomorphism. As $(N1⊕N2)/(N1+L)$ 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 $⊕i∈INi$-strongly projective. Then M is Ni-strongly projective by (i). Conversely, let M be Ni-strongly projective for each i∈ I and $f∈Hom(M,(⊕i∈INi)/L)$ where $L≤⊕i∈INi$. Since M is finitely generated, $f(M)⊆(⊕i∈FNi+L)/L$ for some finite subset F of I. By (iii), M is $⊕i∈FNi$-strongly projective; hence f=0. Thus M is $⊕i∈INi$-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 $Hom(ℚ,ℤn)=0$ for each integer n. As Q is a homomorphic image of a free $Z$-module $ℤ(I)$ for some infinite set I, $Hom(ℚ,T)≠0$ for some factor module T of $ℤ(I)$. Hence $Q$ is not $ℤ(I)$-strongly projective.

Lemma 2.10.

Let $M=M1⊕M2$. 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 $M2≅M/M1=(K+M1)/M1≅K/(M1∩K)$. We show $K=(K∩M1)⊕(K∩M2)$. Define $f:K/(M1∩K)→M1/(M1∩K)$ so that $f(k+M1∩K)=k1+M1∩K$ where $k=k1+k2$, $k1∈M1$ and $k2∈M2$. Clearly, f is a homomorphism. Since $Hom(M2,M1/(M1∩K))=0$ and $M2≅K/(K∩M1)$, we have f=0. This implies that for each $k=k1+k2$ where $k1∈M1$ and $k2∈M2$, $k1∈M1∩K$. Therefore $k2∈M2∩K$. Hence $K=(K∩M1)⊕(K∩M2)$. As $M=K+M1$, $M=M1⊕(K∩M2)$. By using modular law we have $M2=K∩M2$. Thus $M2⊆K$.

$(2)⇒(1)$ Let $f:M2→L$ be a homomorphism and L a factor module of M1. Let $π:M1→L$ be natural homomorphism. Set . It can be seen that K is a submodule of M and M=K+M1. By (2), $M2⊆K$. This implies that $f(m2)=0$ for each $m2∈M2$. 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 $M=X⊕Y$ such that $X⊆M1$ and $Y⊆K$ and $Y∩M1≪M$. As M is strongly lifting, every direct summand of M is fully invariant by Theorem 1.2, this implies that $M1∩Y≤⊕M$. Hence $M1∩Y=0$ and $M=Y⊕M1$. Since M2 is fully invariant, $M2=(M2∩Y)⊕(M2∩M1)=M2∩Y$, by Proposition 1.1(iv). Thus $M2⊆Y⊆K$. 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, $M2⊆K$ and so $K=M2⊕(K∩M1)$. Since $K∩M1≤⊕K$, $K∩M1↪ccK$. As $K∩M1≤K≤M$ and $K∩M1↪ccK$ and $K↪ccM$, We have $K∩M1↪ccM$ by Proposition 1.1(ii). Since $K∩M1≤M1≤M$, and $K∩M1↪ccM$, we have $K∩M1↪ccM1$ by Proposition 1.1(ii). Since M1 is strongly lifting, $K∩M1$ is a fully invariant direct summand of M1. Therefore $K=K∩M1⊕M2≤⊕M$. We show K is fully invariant in M. Let $f∈End(M)$. Since $Hom(M1,M2)=0$ and $Hom(M2,M1)=0$, $f=f1⊕f2$ where $f1∈End(M1)$ and $f2∈End(M2)$. Thus $f(K)=f1(K∩M1)⊕f2(M2)$. Since $K∩M1$ is fully invariant in M1, $f1(K∩M1)⊆K∩M1$. 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 $1≤i≤n$, 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 $⊕j≠iMi$-strongly projective. Hence Mi is Mj-strongly projective by Lemma 2.8. Hence ${Mi}1≤i≤n$ 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 $f∈Hom(Z¯2(M),N/L)$. 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 . 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 $M=⊕i∈IMi$ 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 $i,j∈I$ 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 ${Nα}α∈Γ$ be all direct summands of M that are contained in N. Let $K=∑ α∈ΓNα$. We will show $K≤⊕M$ and $K=⊕i∈TMi$ for some $T⊆I$. Since $End(M)$ is Abelian, Nα is fully invariant for each α ∈ Γ. Hence $Nα=⊕i∈I(Nα∩Mi)$ by Proposition 1.1(iv). Since $Nα≤⊕M$, $Nα∩Mi≤⊕Mi$ for each i∈ I. Therefore Mi is hollow gives either $Nα∩Mi=Mi$ or $Nα∩Mi=0$. This implies that $Nα=⊕i∈TαMi$ for some $Tα⊆I$. Set $T=∪α∈ΓTα$. Then it can be easily seen $K=⊕i∈TMi$ and so $K≤⊕M$ say $M=K⊕K′$. We show $K′∩N≪M$. Assume that $K′∩N+S=M$ for some submodule S of M. Let . Then $Q≤⊕M$ and $Q=⊕i∈JMj$ for some $J⊆I$ by argument mentioned before. Let $j∈J$ and $π:M→Mj$ be natural projection. As $K′∩N+S=M$, $Mj=π(K′∩N)+π(S)$. Since Mj is hollow, $π(K′∩N)=Mj$ or $π(S)=Mj$. Hence $M=Ker(π)+K′∩N$ or $M=Ker(π)+S$. By (2) and Lemma 2.10, $Mj⊆K′∩N$ or $Mj⊆S$. If $Mj⊆K′∩N$, then $Mj⊆N$ and $Mj⊆K′$. Hence $Mj⊆K$ and so $Mj=0$. If $Mj⊆S$, then $Mj⊆Q$ 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 $M=⊕i∈IMi$ 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 $⊕i≠j∈IMj$-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 $⊕i≠∈IMj$-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, $M=⊕i∈IHi$ where Hi is hollow for each i∈ I. Since N is fully invariant, $N=⊕i∈I(N∩Hi)$ by Proposition 1.1(iv). Hence $M/N=⊕i∈IHi/(N∩Hi)$. Since $Hi/(N∩Hi)$ is indecomposable for each i∈ I, it suffices to show that $Hom(Hi/(N∩Hi),Hj/(N∩Hj))=0$ for each distinct i,j of I. Let$f:Hi/(N∩Hi)→Hj/(N∩Hj)$ be a homomorphism and i ≠ j ∈ I. Let $π:Hi→Hi/(N∩Hi)$ be natural homomorphism. By Theorem 2.15, $H_i$ is $⊕j∈I,i≠jHj$-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.

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