80 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 βŠ•-s-extending modules Department of Mathematics, College of Science, Mustansiriyah University, Baghdad, Iraq. ayaadnan994@gmail.com saalsaadi@uomustansiriyah.edu.iq π€π›π¬π­π«πšπœπ­ The concept βŠ•-s-extending modules will be purpose of this paper, a module M is βŠ•- s-extending if each submodule in M is essential in submodule has a supplement that is direct summand. Initially, we give relation between this concept with weakly supplement extending modules and βŠ•-supplemented modules. In fact, we gives the following implications: Lifting modules ⟹ βŠ•-supplemented modules ⟹ βŠ•-s-extending modules ⟹ weakly supplement extending modules. It is also we give examples show that, the converse of this result is not true. Moreover, we study when the converse of this result is true. Keyword: Extending modules, weakly supplement extending modules, βŠ•-supplemented modules, βŠ•-s-extending modules. 1. Introduction In this paper, motivated by the concept of closed ⨁-supplemented that is, a module M is called closed ⨁-supplemented, if each closed submodule in M has a supplement which is direct summand, and weakly supplement extending module we give concept that is βŠ•- s-extending module, a module M is βŠ•-s-extending if each submodule in M is essential in submodule has a supplement which is direct summand. We can answer the question: When is the βŠ•-s-extending module inherited by direct summand. Following [1]. Studied a weakly supplement extending modules, a module M is called weakly supplement extending if each submodule in M is essential in weakly supplement submodule in M. Also, βŠ•-supplemented module were studied by [2]. A module M is ⨁-supplemented, if for any submodule N in M has a supplement which is Article history: Received 31 October 2019, Accepted 16 December 2019, Published in July 2020. Doi: 10.30526/33.3.2475 Aya A. Al-Rubaye Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Saad A. Al-Saadi mailto:ayaadnan994@gmail.com2 mailto:saalsaadi@uomustansiriyah.edu.iq 81 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 Direct summand. In this work, R is associative rings with identity and M is left R- module. A submodule W of M is said to be essential, if W∩V=0 then V=0[2]. A submodule W of M is called small (denoted by Wβ‰ͺM) if for each submodule L in M such that Mβ‰ W+L implies Lβ‰ M [2]. A submodule N of M is called closed, if N has no proper essential extension in M [2]. A submodule N of M is weakly supplement, if there exists a submodule W in M such that M=W+N and N∩ Wβ‰ͺM. A module M is called weakly supplemented if each submodule W in M is weakly supplement of M [2]. A submodule V of M is supplement, if there is a submodule W of M such that M=V+W and V∩ Wβ‰ͺV [2]. A module M is called supplemented if each submodule V of M has a supplement submodule in M [2]. A module M is said to be uniform if each nonzero submodule in M is essential submodule in M [3]. A module M is called non-singular if Z(M)=0 where Z(M)={m∈M | Xm=0 for some essential left ideal X of R} , and M is singular if Z(M)=M [3]. A module 𝑀 is called lifting, if each submodule V in M there is a direct summand W in M with WβŠ† V such that M=WβŠ•W' and Wβ€² ∩ 𝑉 β‰ͺW' [2]. If f(N)βŠ†(N) for each R-endomorphism f of M, then a submodule N of a module M is said to be fully invariant [4]. A module M is semi-simple, if every submodule is direct summand [3]. A module M is called extending if each submodule in M is essential in direct summand of M [3]. The extending property and their generalization are studied by different authors such as [1, 5,6]. 2. βŠ•-s-extending modules In this work, we will present the following concept that is stronger than of weakly supplement extending modules: Definition (1) A module M is called βŠ•-s-extending if each submodule in M is essential in submodule has a supplement which is direct summand. Proposition (2) A module M is called βŠ•-s-extending if and only if each submodule in M is essential in submodule has a weakly supplement which is direct summand. Proof Let N be a submodule of βŠ•-s-extending module, then N is essential in submodule K in M has a supplement which is direct summand. Since every supplement submodule is weakly supplement, then we have K has a weakly supplement which is direct summand. Conversely, let N be a submodule of a module M. By hypothesis, N is essential in submodule K has a weakly supplement L which is direct summand (i.e) M=K+L and K∩Lβ‰ͺM where L is direct summand in M. Since K∩LβŠ†LβŠ†M and L is direct summand in M, thus we have K∩Lβ‰ͺL. Then N is essential in submodule K has a supplement L which is direct summand. Hence M is βŠ•-s-extending. Proposition (3) A module M is βŠ•-s-extending if and only if each closed submodule in M has a (weakly) supplement which is direct summand. 82 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 Proof (⟹) Let B be a closed submodule in M. Since M is βŠ•-s-extending, so B is essential in submodule K has a supplement which is direct summand. But B is closed, so we have B has a supplement which is direct summand. (⟸) Let B be submodule in M. Then by using Zorn's lemma, so there is a closed submodule D in M such that B is essential in D. By hypothesis, D has a supplement which is direct summand. Following [1]. A module M is closed ⨁-supplemented if each closed submodule in M has a weakly supplement which is direct summand. The next result is directly by proposition (2). Corollary (4) A module M is βŠ•-s-extending if and only if M is closed ⨁-supplemented. Remarks and Examples (5) 1. Each extending module is βŠ•-s-extending, while it is not conversely. For example, M=𝑍8⨁𝑍2 as Z-module is βŠ•-s-extending which it is not extending [1]. 2. Every βŠ•-s-extending module is weakly supplement extending, while the other direction is not true in general. 3. Every ⨁-supplemented module is βŠ•-s-extending, while it is not conversely. In fact, Z is βŠ•-s-extending Z-module which it is not ⨁-supplemented (because a submodule 2Z in Z has no a supplement submodule in Z). 4. Every uniform module is βŠ•-s-extending, while the converse is not true. For example, 𝑍10 is βŠ•-s-extending Z-module which it is not uniform. 5. In [1]. every lifting module is weakly supplement extending. This result can be generalized to βŠ•-s-extending (i.e), every lifting module is βŠ•-s-extending (since every lifting module is βŠ•-supplemented), while the converse is not true. In fact, Z is βŠ•-s- extending Z-module which is not lifting. 6. Every weakly (supplemented) module is weakly supplement extending [1]. This result is not still valid for βŠ•-s-extending. Not every weakly supplemented module is βŠ•-s- extending. Moreover, not every βŠ•-s-extending module is weakly supplemented, Z is βŠ•-s- extending Z-module which it is not weakly supplemented. 7. Following [7]. Every semi-simple module is βŠ•-supplemented. So every semi-simple is βŠ•-s-extending, while the converse is not true. Q is βŠ•-s-extending Z-module which it is not semi-simple. Recall that, a module M is called refinable if for every submodule W, V in M with W+V=M, then there is a direct summand W' in M such that W'βŠ†W and W'+V=M [2]. Following [1]. Every closed ⨁-supplemented module is weakly supplement extending. Also, from [1]. Studied when the converse is true. Moreover, we have this corollary: Corollary (6) Let 𝑀 be a refinable module. Then the following are equivalent: 1. 𝑀 is weakly supplement extending module. 2. 𝑀 is βŠ•-s-extending module. 83 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 Recall that, a module M is said to be wd-module if each weakly supplement submodule is direct summand [1]. Proposition (7) Let 𝑀 be a wd-module. Then the following statement are equivalent: 1. M is βŠ•-s-extending module. 2. M is weakly supplement extending module. Proof (1⟹2) Directly by (Remarks and Examples (5)). (2⟹1) Let W be a closed submodule of weakly supplement extending module M, so W is a weakly supplement submodule of V in M. Also V is weakly supplements of W in M. But M is wd-module. Hence we have V is direct summand of M and so M is βŠ•-s- extending module. Proposition (8) Let M be a wd-module. Then the following statement are equivalent: 1. M is extending module. 2. M is βŠ•-s-extending module. Proof (1⟹2) Directly by (Remarks and Examples (5)). (2⟹1) Let A be a closed submodule of βŠ•-s-extending module M, so A has a weakly supplement submodule K in M which is direct summand. Also K is weakly supplement of A in M. But M is wd-module. Then we have A is direct summand of M. Hence M is βŠ•-s-extending module. The following lemma helps us in the next results Lemma (9) Let M be a βŠ•-s-extending module. Suppose that W be a closed submodule in M and B is small in M. t Then there is a submodule C in M such that M=W+C=W+C+B and W∩Cβ‰ͺM, so C∩(W+B) β‰ͺM. Proof Let W be a closed submodule of βŠ•-s-extending module M, then W has a weakly supplements C in M that is direct summand (i.e) M=W+C and W∩Cβ‰ͺM. Now let h: Mβ†’(Mβˆ• π‘Š)⨁(Mβˆ•C) is defined by h(d)=(d+W, d+C) and let j: (Mβˆ• π‘Š)⨁(Mβˆ•C) β†’ (Mβˆ• π‘Š + 𝐡)⨁(Mβˆ•C) is defined by j(d+W, m'+C)=(d+W+B, m'+C). Since W∩Cβ‰ͺM, then h is epimorphism and Kerh= W∩Cβ‰ͺM, and since Kerj=((W+Bβˆ•W)βŠ•0 and (W+B)βˆ•W=(B)β‰ͺMβˆ•W when the canonical epimorphism 𝜈: Mβ†’Mβˆ•W. So we have j is a small epimorphism and jh is small epimorphism since Kerjh= C∩(B+W)β‰ͺM. We noticed that, every βŠ•-supplemented module is βŠ•-s-extending see (Remarks and Example (5)). Next we explained when the convers is true. Proposition (10) Let M be a module in which each submodule L in M there exists a closed submodule W (depending on L) of M such that L=W+C or W=L+C for some C small in M. Then M is βŠ•-s-extending module if and only if M βŠ•-supplemented. 84 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 Proof Let L be a submodule in βŠ•-s-extending module M, so there exists a closed submodule W in M such that L=W+C where Cβ‰ͺM. But M is βŠ•-s-extending, thus W has a supplements D in M that is direct summand (i.e) M=W+D=W+C+D=L+D where Cβ‰ͺM and W∩Dβ‰ͺD, then L∩D βŠ† (W+C)∩D. Since W∩Dβ‰ͺD, then W∩Dβ‰ͺM So by lemma (9) (W+C)∩Dβ‰ͺM. Thus L∩Dβ‰ͺM. Now since L∩D βŠ†DβŠ†M and D direct summand, so we have L∩Dβ‰ͺD and hence M is βŠ•-supplemented. Or, W=L+C where Cβ‰ͺM. Since M is βŠ•- s-extending, thus W has a supplement D in M that is direct summand (i.e) M=W+D=L+C+D=L+D where C β‰ͺM and L∩DβŠ†W∩Dβ‰ͺD, so L∩Dβ‰ͺD. thus M is βŠ•- supplemented. Conversely see (Remarks and Examples (5)). Recall that, a module M is called injective hull of a module Q if it is both an essential extension of Q and an injective module [3]. The following proposition gives another characterization of βŠ•-s-extending module. Proposition (11) For any a module M. t The following statement are equivalent: 1. M is βŠ•-s-extending module. 2. The intersection of M with any direct summand of injective hull of M, has a weakly supplements submodule in M that is direct summand. Proof (1) ⟹ (2) Let W be a direct summand of injective hull of M, i.e E(M)=W⨁V, where V is a submodule of injective hull of M. It is easy to show that W∩M is closed of M. So by proposition (3) W∩M has a weakly supplement submodule of M which is direct summand. (2) ⟹ (1) Let W be a submodule of M. and let V be a relative complement of W in M, then W⨁V is essential in M, but M is essential in injective hull of M, Therefore W⨁V is essential in injective hull of M, Then E(M) =E(W⨁V) =E(W) ⨁E(V). Since E(W) is direct summand of E(M) then E(W)∩M has a weakly supplement submodule in M that is direct summand. But W is essential in E(W) and M is essential in M, So W=W∩M is essential in E(W)∩M which has a weakly supplement in M that is direct summand. Hence M is βŠ•-s-extending module. Recall that, if each submodule A of M, there is an ideal J of R such that A=JM, then a module M is said to be multiplication [8]. Following [4]. let K be a submodule of a module M and R be a ring. The ideal {X∈R| XMβŠ†K}will be denoted by [K:M] and the annihilater of M denoted by annR (M) is annR(M)=[0:M]. Also, a module M is called faithful if annR(M)=0 Proposition (12) Let R be a commutative ring and M a finitely generated faithful multiplication module. Then R is βŠ•-s-extending if and only if M is βŠ•-s-extending. Proof Let N be a closed submodule in M, then by [9]. There exists a closed ideal I in R such that IM=N. Since R is βŠ•-s-extending, so I has a weakly supplement J which is direct 85 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 summand in R (i.e) R= I+J and I∩Jβ‰ͺR. Let K=MJ where K is submodule in M. Now since M is multiplication then M=RM= (I+J)M=IM+JM=N+K and by [10, lemma (4.11)]. N∩Kβ‰ͺM. Also, since J is direct summand, so we have J+S =R and J∩S=0 where S is ideal in R and such that SM=F. Now since M is multiplication, then M=RM=(J+S)M=JM+SM=K+F and K∩F=JM∩SM=(J∩S)M=0. Thus we have K is direct summand in M. Hence M is βŠ•-s-extending. Conversely, let I be a closed ideal in R, then by [10, lemma (4.10)]. There is a closed submodule N be in M such that N=IM. Since M is βŠ•-s-extending, so N has a weakly supplement K in M which is direct summand (i.e) M=K+N and K∩Nβ‰ͺM. Let K=JM where J is ideal in R. Since M is multiplication then M=N+K=IM+JM=(I+J)M=RM Thus we have R=I+J and by [10, lemma (4.11)]. I∩Jβ‰ͺR and J is direct summand in R. Then R is βŠ•-s-extending. Recall that, let f :R⟢T be a ring hommorphism and M a right T-module. On can be defined to as a right R-module by mr=m f (r) for all m∈M and r∈R. Moreover, if f is an epimorphism and M is a right R-module such that kerf βŠ† r(M), so also can also be define to be a right T-module by mt=mr, where f (r)=t. We denote by 𝑀𝑇 , 𝑀𝑅 then M is a right T-module, right R-module [10]. Proposition (13) Let f: R⟢T be a ring epimorphism and M a right R-module with kerf βŠ† r(M). Then 𝑀𝑅 is βŠ•-s-extending if and only if 𝑀𝑇 is βŠ•-s-extending. Proof Let 𝐴𝑇 be a closed submodule of 𝑀𝑇, then 𝐴𝑅 is closed submodule in 𝑀𝑅, since 𝑀𝑅 is βŠ•-s-extending, so 𝐴𝑅 has a weakly supplements 𝐡𝑅 that is direct summand. So 𝐡𝑅 we can define to be T-module by mt=mr, where f (r)=t. Thus 𝐴𝑇 has a weakly supplements 𝐡𝑇which is direct summand in 𝑀𝑇. Then 𝑀𝑇 is βŠ•-s-extending. Conversely, Let 𝐴𝑅 be a closed submodule of 𝑀𝑅, then 𝐴𝑇 is closed submodule in 𝑀𝑇, since 𝑀𝑇 is βŠ•-s-extending, so 𝐴𝑇 has a weakly suplement 𝐡𝑇 that is direct summand. So 𝐡𝑇 we can define to be R- module by mr=mf (r) for each r∈R and m∈M. Thus 𝐴𝑅 has a weakly supplement 𝐡𝑅which is direct summand in 𝑀𝑅. Hence 𝑀𝑅 is βŠ•-s-extending. It is known that, a factor module of βŠ•-supplemented need not necessary βŠ•- supplemented [7]. Also, in βŠ•-s-extending module is not verified. Thus in the following result we obtain a condition under it a factor module of βŠ•-s-extending module is βŠ•-s- extending. Proposition (14) Let M be a βŠ•-s-extending module. Then any nonsingular (epimorphic) image of M is βŠ•-s-extending module. Proof Let f: Mβ†’N be an epimorphism mapping and let K be a closed submodule of N, so H=π‘“βˆ’1(K) is a closed submodule of M. since M is βŠ•-s-extending module, so H has a weakly supplement submodule W of M which is direct summand. Then M=H+W and H∩Wβ‰ͺM. Now N= f (M)=f (H+W)=f (H)+f (W)=K+ f (W) and (since f is epimorphism and ker f βŠ†H) f (H∩W)= f (H) ∩f (W) β‰ͺf (M), so we have K∩ f (W) β‰ͺ 𝑓 (𝑀) =N. Then f (W) is weakly supplement of K in N. Now to show that f (W) is direct summand of N. Let M =W+F and W∩F=0 where F is submodule of M, so N= f (M)= f (F+W)=f (F)+f (W) and 86 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 f (F∩W)= f (F) ∩f (W)=0. Then f (W) is direct summand of N and thus N is βŠ•-s-extending module. Proposition (15) If a module 𝑀1 is βŠ•-s-extending module and 𝑀1 β‰… 𝑀2, then 𝑀2 is βŠ•-s-extending module. Proof Let f: 𝑀1→𝑀2 be an isomorphic image and 𝑀1is βŠ•-s-extending module. Let W is a submodule of 𝑀2 then we have 𝑓 βˆ’1(W) is a submodule of 𝑀1. Since 𝑀1 is βŠ•-s-extending module, so π‘“βˆ’1 (W) is essential in V where V has a weakly supplement submodule of 𝑀1 which is direct summand, so W= f (π‘“βˆ’1 (W)) is essential in f (V). Since V has a weakly supplement submodule in 𝑀1 then there exists H is a submodule of 𝑀1 such that 𝑀1=V+H and V∩H β‰ͺ 𝑀1. Then f (𝑀1)= f (V)+ f (H), so 𝑀2= f (V)+ f (H) and (since f is monomorphism then kerf=0) so we have f (V∩H)= f (V)∩ f (H) β‰ͺ 𝑀2. Then f (V) has a weakly supplement submodule in 𝑀2, since H is direct summand so, we have 𝑀1=H+L and H∩L=0 where L is submodule of 𝑀1 then f (𝑀1)= f (H)+ f (L) and f (H) ∩ f (L)=0. Then 𝑀2 βŠ•-s-extending module. The next result gives a condition under which a direct summand of βŠ•-s-extending module is βŠ•-s-extending. Proposition (16) Every direct summand A of βŠ•-s-extending module such that the intersection of two direct summand in M is direct summand in N is βŠ•-s-extending module. Proof Let A be a direct summand of βŠ•-s-extending module M and let W be a closed submodule in A, so W is a submodule of M. Since A is a direct summand of M, so A is closed submodule of M and we have W is closed submodule of M, but M is βŠ•-s- extending module. Thus by using proposition (2.3), W has a weakly supplement submodule of M which is direct summand then M=W+H and W∩H << M, where H is a direct summand of M. So A∩M=A∩(W+H) then (by modular law) A=W+(A∩H) and W∩(A ∩M)=( W∩H)∩A. Since ( W∩H)∩A be a submodule of W∩H and W∩Hβ‰ͺ M. So, W∩(A∩H) β‰ͺ M and since W∩(A∩H) be a submodule in A and A is a submodule in M. Hence W∩(A∩H ) β‰ͺ A, then W has a weakly supplement submodule in A, then by hypothesis (A∩H ) is direct summand in A, then by proposition (3), A is βŠ•-s-extending module. Proposition (17) Every fully invariant direct summand A of βŠ•-s-extending module M is βŠ•-s- extending. Proof Let A be a direct summand fully invariant of βŠ•-s-extending module M and let L be a closed submodule in A, so L is closed in M. But M isβŠ•-s-extending. Then L has a weakly 87 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 supplement W which is direct summand in M, so M=WβŠ•W', M=W+L and W∩Lβ‰ͺM. Now A=A∩M=A∩(W+L) by (modular law) A=L+(A∩W), but A=(A∩ π‘Š) βŠ• (𝐴 ∩ π‘Šβ€²). So A∩ π‘Š is direct summand in A and L∩ (𝐴 ∩ π‘Š)= W∩Lβ‰ͺM, since A is direct summand in M. Hence A∩ π‘Š is weakly supplement of L in N. Thus A is βŠ•-s-extending. Recall that, if for each closed submodules F, G and S in a module M such that F∩(G+S)= F∩G+F∩S, then a module M is said to be local distributive [10]. A direct sum of βŠ•-s-extending module need not necessary βŠ•-s-extending. In fact, for example. M=Z⨁Z2 as Z-module is not βŠ•-s-extending (since it is not weakly supplement extending module, while Z and Z2 are βŠ•-s-extending. We give the condition so we get the proposition: Proposition (18) Let M=𝑀1⨁ 𝑀2 where 𝑀1and 𝑀2 are βŠ•-s-extending such that M is local distributive module. Then M is βŠ•-s-extending module. Proof Let F be a closed submodule in M. To prove Fβˆ©π‘€π‘– is closed in 𝑀𝑖, since M is local distributive module, then we have F=((Fβˆ©π‘€1) ⨁ (Fβˆ©π‘€2)). Hence Fβˆ©π‘€1 is closed in 𝑀1 and Fβˆ©π‘€2 is closed in 𝑀2. But 𝑀1 and 𝑀2 are βŠ•-s-extending module. Then there exists a weakly supplement submodule 𝐺1 of 𝑀1, 𝐺2 of 𝑀2 and 𝐺1, 𝐺2 are direct summand such that 𝐺1+(Fβˆ©π‘€1)= 𝑀1 π‘Žπ‘›π‘‘ 𝐺2 + (𝐹 ∩ 𝑀2) = 𝑀2 , 𝐺1 ∩ (𝐹 ∩ 𝑀1) = ( 𝐺1∩F) β‰ͺ 𝑀1 and 𝐺2∩(Fβˆ©π‘€2)=( 𝐺2∩F)β‰ͺ 𝑀2. Now M= 𝑀1⨁𝑀2 = (𝐺1+(Fβˆ©π‘€1)) ⨁ (𝐺2+(Fβˆ©π‘€2))=(𝐺1 ⨁𝐺2)+F. Then M = (𝐺1⨁ 𝐺2) + F and (𝐺1⨁𝐺2)∩F = ( 𝐺1∩F)⨁( 𝐺2∩F)β‰ͺ(Fβˆ©π‘€1) ⨁(Fβˆ©π‘€2) β‰ͺ 𝑀1⨁ 𝑀2 β‰ͺ M. We have 𝐺1 ⨁𝐺2 is direct summand of M. Then M is βŠ•-s-extending module. Recall that, a module M is distributive if for any submodule F, G and S in M such that F∩(G+S)= F∩G+F∩S. Following [10]. Every distributive module is local distributive. Thus, we have this corollary: Corollary (19) Let M=𝑀1⨁ 𝑀2 where 𝑀1and 𝑀2 are βŠ•-s-extending module such that M is a distributive module. Then M is βŠ•-s-extending. 2. Conclusions We proved the following results: 1. A module M is βŠ•-s-extending if and only if each submodule in M is essential in submodule has a weakly supplement which is direct summand. 2. A module M is βŠ•-s-extending if and only if each closed submodule in M has a (weakly) supplement which is direct summand. 3. A module M is βŠ•-s-extending if and only if M is closed ⨁-supplemented. 4. We obtain the following result: Extending modules β†šβƒ—βƒ—βƒ— βŠ•-s-extending modules β†šβƒ—βƒ—βƒ— weakly supplement extending modules. 88 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (3) 2020 References 1. Al-Rubaye, A.A.; Al-Saadi, S.A. Weakly supplement extending modules, to appear. 2. Clark, J.; Lamp, C.; Vanaja, N.; Wisbauer, R. Lifting modules supplements and projectivity in module theory, Birkhauser Verlag, Basel- Boston-Berlin.2006, ISBN 3-7643-7572-8. 3. Dung, N.V.; Huynh, D.V.; Smith, P.F.; Wisbauer, R. Extending modules, Pitman Research Notes in Math. 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