On Semiannahilator Supplement Submodules

Let R be associative; ring; with an identity and let D be unitary left Rmodule; . In this work we present semiannihilator; supplement submodule as a generalization of R-asupplement submodule, Let U and V be submodules of an R-module D if D=U+V and whenever Y≤ V and D=U+Y, then annY≪R;. We also introduce the the concept of semiannihilator -supplemented ;modules and semiannihilator weak; supplemented modules, and we give some basic properties of this conseptes. KeywordssaSmall Submodule, sa-Hollow And sa -Lifting Modules;, sa supplemented Modules هلمكملا هفلاتلا هبش هيئزجلا تاساقملا للاه مشاه ىليل 1 * ، ساي دهمحم هرهاس ني 2 1 مدق يندطلا ةسدظهلا ة ، سدظهلا ةيلك ة ، قا رعلا ، دادغب ، دادغب ةعماج 2 تايضايرلا مدق ، دادغب ةعماج ، مهلعلا ةيلك ، دادغب , قا رعلا هصلاخلا نكتل R و دياحم رصظع تاذ هيعيطجت هقلح D دياحم ساقم رديا ئزجلا ساقطلا . ي N نم D هبش ىعدي هلاح يفريغص فلات فلات K يف ريغص R ثيح K نم ىئزج ساقم قطلا سا D امدظع D = K+K نكتل. K و U نم هيئزج تاساقم D نا لهقن. K لطكم فلات هبش يئزجلا ساقطلل U يف ساقطلا D اذا D = U+K كلذك و امدظع Y > K و D = U+Y هكي ن فلات هبش Y يف ريغص R ريهطت هه ثحبلا نم يساسلاا ضرغلا . ساقطلاو هلطكطلا هفلاتلا هبش تاساقطلل هيساسلاا صاهخلا تلا هبش تا هفيعضلا هلطكطلا هفلا . Introduction " Throughout this paper; all rings are associative; ring with identity and modules are unitary left modules, D is named a "hollow module" if every proper submodule; is small; in D, where a submodule B of R-module D is named small in D (N << M) if B +K ≠ D for each proper submodule K of D . A proper submodule; B of R-module D is named an essential; in D, if for every non-zero submodule K of M then N∩K ≠0 [1]. The concept of "small submodule" has been generalized by some researchers, for this see [2,3]. the authors; in [4] introduced; the concept of R-annihilator; small submodules, that is; a submodule B of an R-module D is called R-annihilator small, if whenever B+K=D, where K a submodule of D; ,implies that ann(K)=0, where ann(K) ={r R: r.K=0}. In [5] sahira introduce the concept of semiannihilator; small submodules, in case ann(K) << R where K is a submodule; of D whenever; B+ K =D. Clear that every :R-annihilator small submodule is semiannihilator; small, but the convers is not true [5] . Recall that a submodule V of M is called a supplement of U in M. If V is a minimal element in the set of submodules L of M with U+L=M. Let ISSN: 0067-2904 Helal and Yaseen Iraqi Journal of Science, 2020, Vol. 61, Special Issue, pp: 16-20 17 M be an R-module, M is called a supplemented module if every submodule of M has a supplement in M. Let U, V be submodules of an R-module M. If M=U+V and U∩V≪ M, then V is called a weak supplement of U in M. Let M be an R-module if every submodule of M has a weak supplement in M, then M is called a weakly supplemented module[6]. In this work we present semiannihilator;supplement submodule , Let U and V be submodules of an R-module D if D=U+V and whenever Y≤ V and D=U+Y, then annY≪R;. We also introduce the the concept of semiannihilator supplemented ;modules and semiannihilator weak; supplemented modules, and we give some basic properties of this conseptes. In section two we introduce the notion of semiannihilator lifting modules and discus some characteristics ; of this kind; of modules.. In part three; we introduce the concept of semiannihilator supplement submodule and basic properties. We show that;. If D and D′ be R-modules and let f: D→D′ be an epimorphism, if D′ is semiannihilator supplemented module, then D is semiannihilator supplemented; module. In part four , the concept of semiannihilator weak supplement; submodules" with some examples and basic properties was introduced .The below lemma award the characteristics of semiannihilator; small submodules. Lemma[5]: 1Let D be an R-module with submodules A,N.such that A⊆N. If N ≪ sa D then A ≪sa D. 2Let D be an R-module with submodules A,N.such that A⊆N, if A ≪sa N then A ≪sa D. 3-Let D1 ,D2 be an Rmodules . If N1≪sa D1and N2 ≪sa D1thus N1⨁N2 ≪sa D1 ⨁D2. 4let D and S be an Rmodules and f : D→S be an epimorphism . If H ≪sa S, then f (H) ≪sa D. 1. Semiannihilator lifting modules. An R-module Dis called lifting; if for any submodule N of D there exist submodule K of N such that D=K⨁ K′ with K′≤ M and N∩ K′≪ K. In this part we introduce the notion of semiannihilator lifting modules as generalization of R-Annihilator Lifting modules and discus some properties of this kind of modules. Definition 1.1 : An R1-module D is named semiannihilator lifting; (salifting ) if for any submodule B of D , there exist submodules K , K ′ of D such that D= K ⨁ K ′ with K ⊆ B and B⋂ K ′ ≪sa K The below theorem given a characterization of following semiannihilator lifting; modules. Theorem 1. For an R-module; D the statement are equivalent : 1) M is salifting; 2) Every submodule N of D, N can be written as N= A⨁ B where A is direct summand of D and B ≪sa D. 3) For every submodule; N of D, there exist a direct summand; K of D s.t K ≤ N and N/K≪ sa D/ K. Proof : See proof of lemma2 in [7]. A nontrivial R-module D is called semiannihilator –hollow;(sa-hollow) if every proper submodule of D is sa-small in M [5]. Examples 1.2 : 1Z as Zmodule is salifting; module but it is not lifting. 2Z6 and Z4 as Zmodule are not salifting module. Not :Every sahollow is sa-lifting Proof: For a submodule B of D if B ≠ D, B≪sa D, B = (0) ⨁B the result go after, directly by theorem 1 Proposition 1. 3 : Let D be indecomposable; module then D is sahollow module if and only if D is salifting. Proof: Let D be sa-hollow then D is salifting ,Conversely suppose that D is sa-lifting and A a proper submodule of D, by Theorem.1,We have A= N⨁K where N is a direct summand in D and K≪sa D, but D is indecomposable . Then either N = (0) or N=D then D=N⊆ A which attend that A=D which is contradiction, if N= (0), so A=K ≪sa D, and D is sahollow. Proposition 1. 4 :Let; D=H1⨁H2 be duo module .If H1 and H2 are ;sa-lifting; modules, then; D is sa – lifting module. Proof: Let H1 and H2 be sa-lifting modules, and N submodule of D ,then N= (N∩ H1 ) ⨁ (N∩ H2) .For each i∈ {1,2},there exists a direct summand; Ki of Hi ,such that Hi= Ki⨁Li with Di ⊆ N∩ Hi and N∩ Helal and Yaseen Iraqi Journal of Science, 2020, Vol. 61, Special Issue, pp: 16-20 18 Li ≪sa Li then , D= (K1⨁ L1) ⨁ (K1⨁ L2) = (K1⨁ K2) ⨁ (L1 ⨁ L2) ,we have (K1⨁ K2) ⊆N ,and N∩(L1⨁ L2) ≪sa (L1⨁ L2) thus D is salifting; module. Corollary 1.5: Let D= H1⨁ H2 be a module such that R;= ann(H1);+ann(H2).If H1 and H2 are salifting; modules ,then D is sa-lifting; module. Proposition 1.6 : Let; D be a "multiplication" R –module, if D ;sa-lifting; module , then R is ;salifting ring; . Proof: Assume; that D is sa-lifting; module;. Where I is an ideal in R. ; D is "multiplication Then N = ID; is a submodule; of D, thus there exist submodules K; and K' in M with ;K ⊆ N, D = K⨁ K' and (N ∩ K') ) ≪sa D . D is a multiplication ;Rmodule, so there are ideals J and J' of R such that K = JD and K' = J'D. Since K ⊆ N then J ⊆ I. We have ;D =K ⨁K' = JD ⨁J'D = (J ⨁J')D implies that R = J⨁ J '.Now N ∩ K' = (ID ∩ J'D) ≪sa D and since; (J∩J')D ⊆ ID ∩J'D; it follows that (J∩J')D ≪sa D [5] and according to [3] we get; [(J∩J')D:D] ≪sa R. But;[(J∩J')D:D] = I ∩ J', then (I∩J') ≪sa R and R is sa-lifting; ring . 2. Semiannihilator supplemente submodule. In this section we present the definition of semiannihilator (sa-supplement) supplement class. Then, some basic properties of this class are presented. In addition, several examples are given to illustrate the results. Definition 2.1: Let V and U be submodules of an R-module D. We say that V is "semiannihilator supplement" (sa-supplement) of U in D if D=U+V and whenever Y≤ V and D=U+Y, then annY ≪ R. Let D be an R-module. We say that D is semiannihilator supplemented(((sa-supplemented) module if every proper submodule of D has sa-supplement. Let R be a commutative ring and let I be an ideal of R. We say that R is sa-supplemented; ring of R is sa-supplemented as an R-module. The bellow proposition gives a charactrization of sa-supplement submodule Proposition 2.2: For submodules U and V of an R-module D. Then V is sa-supplment of U if and only if D=U+V and U∩V ≪saV. Proof: Let V be sa-supplement; of U. To show that U∩V sasmall in V, let V= (U∩V) +Y. Now D= U+ V = U+ (U∩V) +Y= U+Y. But Y≤V, therefore annY ≪R1 and U∩V ≪saV. Conversely, Let D=U+V and U∩V ≪sa V . We want to show that V sa-supplement of U. Let Y≤V such that D=U+Y. By (Modular law);, V= (U∩V )+Y. But U∩V ≪sa V , therefore annY ≪ R. Then V is sa-supplement; of D. Examples and Remarks 2.3 1sa-supplement;; submodule not supplement submodule to see; that consider; Z as Z-module. For every proper submodule nZ of Z , Z= nZ+Z and nZ ∩ Z = nZ≪saZ. Then Z is sa-supplement; of nZ. Thus every proper submodule of Z has sa-supplement. But it is known that every non trivial submodule of Z has no supplement in Z. Where Z is indecomposable; and {0} is the only small; submodule of Z. 2A supplement;; submodule; need not be sa-supplement submodule. For example, Let Z4 as Zmodule. Z4 is a supplement; of {0 ,2 }. And Z4 is not sa-supplement;;of {0 ,2 },where {0 ,2 }∩ Z4={0 ,2}is not sa-small in Z4, since Z4={0 ,2 } + Z4 and ann Z4={n ∈ Z ; n. Z4 = 0}=4Z not small in Z. 3Let D be an R-module. Then every sa-small; submodule of D has sa-supplement; in D. That is if , N be sa-small submodule of D. Then D =N+D and N∩D =N is sa-small submodule of D. Thus D is sasupplement of N in D. 4-Let U and V be two submodules in an R-module D such that V sa-supplement in U. If D= W+V, where W a submodule in U, then V sa-supplement in W. Proof: Since V is sa-supplment of U, then D=U+V and U∩V ≪sa V. Since W≤ U. Then W∩V ≪sa V, by prop. (2.4) in [5]. Thus V is sasupplement of W in D. Let D be an R1-module its known that every direct summand of D has a supplment in D. But this is not true for sa-supplement; as the below examples shows : Example2.4: Let Z6 as Z-module. and U={0 ,2 ,4 } ,V={0 ,3 }, Z6= U⨁V. U and V are supplement; of each others. But each of U and V has no sa-supplement in Z6, where ann Z6={n∈Z ; n. Z6=0}=6Z not small in Z.. Hence Z6 has no sa-small submodule [5]. Thus every submodule of Z6 has no sasupplement in Z6. Helal and Yaseen Iraqi Journal of Science, 2020, Vol. 61, Special Issue, pp: 16-20 19 Proposition2.5 : For a finitely generated D in R-module and For submodules U and V in D such that V sa-supplment of U in D. Thus there exists a finitly generated sa-supplement W in U s.t W≤ V. Proof: Let M=Rx1+ Rx2+ . . . +Rxn, where xi∈D, ∀i=1,2,...,n. Since D=U+V, then xi = ui+ vi, where ui∈U, vi∈V, ∀i=1,2,...,n. Now let W=Rv1+Rv2+ . . . +Rvn. Clearly that D=U+W. Since V sasupplment of U and W≤V, then annW ≪ R. is clear that W is sa-supplement of U. Proposition 2.6: Let D and N be R1-modules and let f: D→N an epimorphism,; if N is sasupplemented module, then D is sa-supplmented; module. Proof: For a submodule K of D, then f(K) submodule in N. Since N is sa-supplemented; module. Then there exists a submodul L in N s.t N= f(K)+L and f(K)∩L is sa-small in L. M=f -1 (N)=f -1 (f(K)+L)= f 1 (f(K))+f -1 (L)=K+ker f+f -1 (L)=K+f -1 (L).Claim that K∩f -1 (L) is sa-small submodule of f -1 (L). Since f(K) ∩L≪sa L, then by prop (2-7) in [3] , f(f(K) ∩L)≪sa f(L) .But f(f(K) ∩L)=f(f(K)) ∩f 1 (L)=(K+Kerf) ∩ f -1 (L) =kerf+(K∩f -1 (L)), by( Modular Law) . ker f +(K∩f -1 (L)) ≪sa f(L). By Prop (2-4)in[5], K∩ f -1 (L) ≪sa f (L).So f(L) is sa-supplment of K in D. Then M is sasupplemented module. Proposition 2.7: Let D be a finitely generated; faithful multiplication; modul over a commutative ring R and let I be an ideal of R. if ID has sa-supplment in D, so I has sa-supplment in R. Proof: For an ideal I in R1 such that ID has sa-supplement in D. Then there exists a submodul N in D s.t D=ID+N and ID∩N ≪saN. D a multiplication modul, thus N=JD, for some ideal J of D. Now D=R1 D=ID+JD=(I+J)D. But D is finitely generated; faithful multiplication modul and hence R=I+J[8]. ID∩N=ID∩JD=(I∩J)D ≪sa JD. To show that I∩J≪saJ. Let J=(I∩J)+L, where L an ideal of R. Then JD=((I∩J)+L)D=(I∩J)D+LD. Therefore ann (LD)≪R. ann L≤ ann LD. Thus ann L≪R and I∩J≪sa J. Then J is sasupplement of I. 3. Semiannihilator weakly supplmented modules;. In this part, we introduce; the definition of semiannihilator; weakly supplmented modules. And we introduce some basic characterization; of this modules:. Definition 3.1: For submodules U and V of an R-module D. We say that V semiannihilator -weak supplment (sa-weak supplement) of U in D if D=U+V and U∩V ≪sa D. We say that D is semiannihilator weakly (sa-weakly) supplemented module if every submodule of D has sa-weak supplment in D. Remarks and examples; 3.4: 1. semiannihilator weak supplement submodule not be weak supplement submodule. For example, consider Z as Z module. Cleary that Z=2Z+3Z and 2Z∩3Z=6Z≪sa Z . Thus 3Z is sa-weak supplement of 2Z. But {0} is the only small submodule of Z and hence 3Z is not weak supplement of 2Z. 2. Every sa-supplemented module is sa-weakly supplemented module. To show that, let D be an sasupplemented module and let U be a proper submodule of D , then there exists a submodule V of D, such that D=U+V and U∩V ≪saV. By prop (2-3) in [8] , U∩V ≪sa D. Hence V sa-weak supplement of U. Clearly that D=D+0 and D∩{0}=0≪sa D. So {0}is sa-weak supplement of D . Thus D is saweak-supplemented module. 3. sa-weak supplement submodule need not be sa-supplement; submodule. For example, let D be a faithful R-module. Then D=D+0 and D∩{0}=0≪saD. Thus {0} is sa-weak supplement of D. Now D∩{0}=0 is not sa-small in 0, when 0=0+0 ann0=R not small in R. 4. Let X and Y be submodules of R1-module D if X is sa-weak supplement of Y, then Y is sa-weak supplement of X, where D=X+Y and X∩Y ≪sa D. Proposition 3.5:-For N , K and L submodules in an R-module D such that L≤ N. If K is sa-weak supplement of N and D=L+K, there K is sa-weak supplment of L in D. Proof: Since K sa-weak supplement in N, then D=N+K and N∩K ≪sa D. Now D=L+K and L∩K≤ N∩K ≪sa D. Hence L∩K ≪sa D by prop (2-4) in [5]. Thus K is sa-weak supplement in L is D. Proposition 3.6: For submodules N and L of a "finitely generated R-module" D. if L sa-weak supplment of N, then L contains a finitly; generated sa-weak supplment of N. Proof:Let D =Rx1+Rx2+. . . +Rxn, xi ∈ D, for some xi ∈ D, ∀ i=1,2,...,n. Since M=N+L, then xi=ai+bi , where ai ∈ N, bi ∈ K, ∀ i=1,2,..., n. Now let L'=Rb1 +Rb2+ . . . +Rbn , D=N+K′. Clearly that K′ ≤ K. But N∩ L′=N∩L ≪sa D, therefore N∩ L'≪sa D, by prop (2-4) in [5].Thus L′ is a finitely generated; sa-weak supplement of N. Helal and Yaseen Iraqi Journal of Science, 2020, Vol. 61, Special Issue, pp: 16-20 20 References 1. Wisbauer; R. 1991. Foundation of Module and Ring Theory. Gordon and Breach, Philadelphia. 2. Beyranvand, R. and Moradi, F. 2015. Small submodules with respect to an arbitrary submodule. Journal of Algebra and Related Topics, 3(2): 43-51. 3. Mehdi Sadiq A. and Hiba Ali S. 2017. Fully Annihilator Small Stable Modules. International Mathematical Forum, 12(19): 915 927 4. Al-Hurmuzy, H. and Al-Bahrany, B. 2016. RAnnihilator – Small Submodules. M.Sc theses, College of science ,Baghdad University . 5. Yaseen, S. M. 2018. Semiannihilator Small Submodules, International Journal of Science and Research (IJSR)), 7(1): 955-958. 6. Yaseen, S. M. 2018. RAnnihilator – Hollow and R-Annihilator Lifting modules, Sci. Int.(Lahore), 30(2): 204-207 7. Truong C. Q. , Phan H. T. 2013. Some properties of e-supplemented and e-lifting modules",vitnam journa lof math., 41(3): 303312. 8. Abdelkader, B.H. 2001. On Lifting Modules, M.SC, Thesies, College; of Science, University of Baghdad,

Introduction " Throughout this paper; all rings are associative; ring with identity and modules are unitary left modules, D is named a "hollow module" if every proper submodule; is small; in D, where a submodule B of R-module D is named small in D (N << M) if B +K ≠ D for each proper submodule K of D . A proper submodule; B of R-module D is named an essential; in D, if for every non-zero submodule K of M then N∩K ≠0 [1]. The concept of "small submodule" has been generalized by some researchers, for this see [2,3]. the authors; in [4] introduced ; the concept of R-annihilator; small submodules, that is; a submodule B of an R-module D is called R-annihilator small, if whenever B+K=D, where K a submodule of D; ,implies that ann(K)=0, where ann(K) ={r R: r.K=0}. In [5] sahira introduce the concept of semiannihilator; small submodules, in case ann(K) << R where K is a submodule; of D whenever; B+ K =D. Clear that every :R-annihilator small submodule is semiannihilator; small, but the convers is not true [5] . Recall that a submodule V of M is called a supplement of U in M. If V is a minimal element in the set of submodules L of M with U+L=M. Let

ISSN: 0067-2904
M be an R-module, M is called a supplemented module if every submodule of M has a supplement in M. Let U, V be submodules of an R-module M. If M=U+V and U∩V≪ M, then V is called a weak supplement of U in M. Let M be an R-module if every submodule of M has a weak supplement in M, then M is called a weakly supplemented module [6]. In this work we present semiannihilator;supplement submodule , Let U and V be submodules of an R-module D if D=U+V and whenever Y≤ V and D=U+Y, then annY≪R;. We also introduce the the concept of semiannihilator supplemented ;modules and semiannihilator weak; supplemented modules, and we give some basic properties of this conseptes.
In section two we introduce the notion of semiannihilator lifting modules and discus some characteristics ; of this kind; of modules.. In part three; we introduce the concept of semiannihilator supplement submodule and basic properties. We show that;. If D and D′ be R-modules and let f: D→D′ be an epimorphism, if D′ is semiannihilator supplemented module, then D is semiannihilator supplemented; module. In part four , the concept of semiannihilator weak supplement; submodules" with some examples and basic properties was introduced .The below lemma award the characteristics of semiannihilator; small submodules. Lemma [5]: 3-Let D 1 ,D 2 be an R-modules . If N 1 ≪sa D 1 and N 2 ≪sa D 1 thus N 1 ⨁N 2 ≪sa D 1 ⨁D 2 . 4-let D and S be an R-modules and f : D→S be an epimorphism . If H ≪sa S, then f −1 (H) ≪sa D.

Semiannihilator lifting modules.
An R-module Dis called lifting; if for any submodule N of D there exist submodule K of N such that D=K⨁ K′ with K′≤ M and N∩ K′≪ K. In this part we introduce the notion of semiannihilator lifting modules as generalization of R-Annihilator Lifting modules and discus some properties of this kind of modules. Definition 1.1 : An R 1 -module D is named semiannihilator lifting; (sa-lifting ) if for any submodule B of D , there exist submodules K , K ′ of D such that D= K ⨁ K ′ with K ⊆ B and B⋂ K ′ ≪sa K The below theorem given a characterization of following semiannihilator lifting; modules. Li ≪sa Li then , D= (K 1 ⨁ L 1 ) ⨁ (K 1 ⨁ L 2 ) = (K 1 ⨁ K 2 ) ⨁ (L 1 ⨁ L 2 ) ,we have (K 1 ⨁ K 2 ) ⊆N ,and N∩(L 1 ⨁ L 2 ) ≪sa (L 1 ⨁ L 2 ) thus D is sa-lifting; module. Corollary 1.5: Let D= H 1 ⨁ H 2 be a module such that R;= ann(H 1 );+ann(H 2 ).If H 1 and H 2 are salifting; modules ,then D is sa-lifting; module. Proposition 1.6 : Let; D be a "multiplication" R -module, if D ;sa-lifting; module , then R is ;salifting ring; . Proof: Assume; that D is sa-lifting; module;. Where I is an ideal in R. ; D is "multiplication Then

Semiannihilator supplemente submodule.
In this section we present the definition of semiannihilator (sa-supplement) supplement class. Then, some basic properties of this class are presented. In addition, several examples are given to illustrate the results. Definition 2.1: Let V and U be submodules of an R-module D. We say that V is "semiannihilator supplement" (sa-supplement) of U in D if D=U+V and whenever Y≤ V and D=U+Y, then annY ≪ R.
Let D be an R-module. We say that D is semiannihilator supplemented(((sa-supplemented) module if every proper submodule of D has sa-supplement. Let R be a commutative ring and let I be an ideal of R. We say that R is sa-supplemented; ring of R is sa-supplemented as an R-module. The bellow proposition gives a charactrization of sa-supplement submodule Proposition 2.2: For submodules U and V of an R-module D. Then V is sa-supplment of U if and only if D=U+V and U∩V ≪saV. Proof: Let V be sa-supplement; of U. To show that U∩V sa-small in V, let V= (U∩V) +Y. Now D= U+ V = U+ (U∩V) +Y= U+Y. But Y≤V, therefore annY ≪R 1 and U∩V ≪saV. Conversely, Let D=U+V and U∩V ≪sa V . We want to show that V sa-supplement of U. Let Y≤V such that D=U+Y. By (Modular law);, V= (U∩V )+Y. But U∩V ≪sa V , therefore annY ≪ R. Then V is sa-supplement; of D. Examples and Remarks 2.3 1-sa-supplement;; submodule not supplement submodule to see; that consider; Z as Z-module. For every proper submodule nZ of Z , Z= nZ+Z and nZ ∩ Z = nZ≪saZ. Then Z is sa-supplement; of nZ. Thus every proper submodule of Z has sa-supplement. But it is known that every non trivial submodule of Z has no supplement in Z. Where Z is indecomposable; and {0} is the only small; submodule of Z. 2-A supplement;; submodule; need not be sa-supplement submodule. For example, Let Z 4 as Zmodule. Z 4 is a supplement; of {0 ,2 }. And Z 4 is not sa-supplement;;of {0 ,2 },where {0 ,2 }∩ Z 4 ={0 ,2}is not sa-small in Z 4 , since Z 4 ={0 ,2 } + Z 4 and ann Z 4 ={n ∈ Z ; n. Z 4 = 0}=4Z not small in Z. 3-Let D be an R-module. Then every sa-small; submodule of D has sa-supplement; in D. That is if , N be sa-small submodule of D. Then D =N+D and N∩D =N is sa-small submodule of D. Thus D is sasupplement of N in D. 4-Let U and V be two submodules in an R-module D such that V sa-supplement in U. If D= W+V, where W a submodule in U, then V sa-supplement in W. Proof: Since V is sa-supplment of U, then D=U+V and U∩V ≪sa V. Since W≤ U. Then W∩V ≪sa V, by prop. (2.4) in [5]. Thus V is sa-supplement of W in D.
Let D be an R 1 -module its known that every direct summand of D has a supplment in D. But this is not true for sa-supplement; as the below examples shows : Example2.4: Let Z 6 as Z-module. and U={0 ,2 ,4 } ,V={0 ,3 }, Z 6 = U⨁V. U and V are supplement; of each others. But each of U and V has no sa-supplement in Z 6 , where ann Z 6 ={n∈Z ; n. Z 6 =0}=6Z not small in Z.. Hence Z 6 has no sa-small submodule [5]. Thus every submodule of Z 6 has no sasupplement in Z 6 .