aa r X i v : . [ m a t h . A C ] J u l FULLY S -IDEMPOTENT MODULES FARANAK FARSHADIFAR
Abstract.
Let R be a commutative ring with identity and S be a multiplica-tively closed subset of R . The aim of this paper is to introduce the notion offully S -idempotent modules as a generalization of fully idempotent modulesand investigate some properties of this class of modules. Introduction
Throughout this paper R will denote a commutative ring with identity and Z will denote the ring of integers.Let M be an R -module. M is said to be a multiplication module [6] if forevery submodule N of M , there exists an ideal I of R such that N = IM . It iseasy to see that M is a multiplication module if and only if N = ( N : R M ) M for each submodule N of M . A submodule N of M is said to be idempotent if N = ( N : R M ) M . Also, M is said to be fully idempotent if every submodule of M is idempotent [4].Let S be a multiplicatively closed subset of R . In [2], the authors introducedand investigated the concept of S -multiplication modules as a generalization ofmultiplication modules. An R -module M is said to be an S -multiplication module if for each submodule N of M , there exist s ∈ S and an ideal I of R such that sN ⊆ IM ⊆ N [2].Let S be a multiplicatively closed subset of R . In this paper, we introduce thenotions of fully S -idempotent R -modules as a generalization of fully idempotentmodules and provide some useful information concerning theis new class of modules.We say that a submodule N of an R -module M is an S -idempotent submodule ifthere exists an s ∈ S such that sN ⊆ ( N : R M ) M ⊆ N (Definition 2.2). Wesay that an R -module M is a fully S -idempotent module if every submodule of M is an S -idempotent submodule (Definition 2.3). Clearly every fully idempotent R -module is a fully S -idempotent R -module (Proposition 2.5). Example 2.6 showsthat the converse is not true in general. In Theorem 2.10, we characterize the fullyidempotent R -modules. Also, we characterize the fully S -idempotent R -modules,where S satisfying the maximal multiple condition (Theorem 2.13). Let M i be an R i -module for i = 1 , , .., n and let S , ..., S n be multiplicatively closed subsets of R , ..., R n , respectively. Assume that M = M × ... × M n , R = R × ... × R n and S = S × ... × S n . Then we show that the following statements are equivalent.(a) M is a fully S -idempotent module.(b) M i is a fully S i -idempotent module for each i ∈ { , , ..., n } . Mathematics Subject Classification.
Key words and phrases.
Idempotent submodule, fully idempotent module, multiplicativelyclosed subset, S -idempotent submodule, fully S -idempotent module. Also, among other results, it is shown that (Theorem 2.21) if M an S -multiplication R -module and N be a submodule of M , then the following statements are equiva-lent.(a) N is an S -pure submodule of M .(b) N is an S -multiplication R -module and N is an S -idempotent submoduleof M .Finally, we prove that if M is a fully S -idempotent R -module, then M is a fully S -pure R -module. The converse holds if M is an S -multiplication R -module (Corol-lary 2.22). 2. Main resultsDefinition 2.1.
Let S be a multiplicatively closed subset of R . We say that anelement x of an R -module M is an S -idempotent element if there exist s ∈ S and a ∈ ( Rx : R M ) such that sx = ax . Definition 2.2.
Let S be a multiplicatively closed subset of R . We say that asubmodule N of an R -module M is an S -idempotent submodule if there exists an s ∈ S such that sN ⊆ ( N : R M ) M ⊆ N . Definition 2.3.
Let S be a multiplicatively closed subset of R . We say that an R -module M is a fully S -idempotent module if every submodule of M is an S -idempotent submodule. Example 2.4.
Let S be a multiplicatively closed subset of R and M be an R -module with Ann R ( M ) ∩ S = ∅ . Then clearly, M is a fully S -idempotent R -module. Proposition 2.5.
Let S be a multiplicatively closed subset of R . Every fullyidempotent R -module is a fully S -idempotent R -module. The converse is true if S ⊆ U ( R ), where U ( R ) is the set of units in R . Proof.
This is clear. (cid:3)
The following example shows that the converse of Proposition 2.5 is not true ingeneral.
Example 2.6.
Take the Z -module M = Z p ∞ for a prime number p . Then weknow that all proper submodules of M are of the form G t = h /p t + Z i for some t ∈ N ∪ { } and ( G t : Z M ) = 0. Therefore, M is not a fully idempotent Z -module.Now, take the multiplicatively closed subset S = { p n : n ∈ N ∪ { }} of Z . Then p t G t = 0 ⊆ ( G t : Z M ) M ⊆ G t . Hence, G t is an S -idempotent submodule of M for each t ∈ N ∪ { } . So, M is a fully S -idempotent Z -module. Lemma 2.7.
Let S be a multiplicatively closed subset of R . Then every fully S -idempotent R -module is an S -multiplication R -module. Proof.
Let M be a fully S -idempotent R -module and N be a submodule of M .Then there is an s ∈ S such that sN ⊆ ( N : R M ) M ⊆ N . This implies that sN ⊆ ( N : R M ) M = ( N : R M )( N : R M ) M ⊆ ( N : R M ) M ⊆ N, as needed. (cid:3) The following example shows that the converse of Lemma 2.7 is not true ingeneral.
ULLY S -IDEMPOTENT MODULES 3 Example 2.8.
Take the multiplicatively closed subset S = Z \ Z of Z . Then Z is an S -multiplication Z -module. But Z is not a fully S -idempotent Z -module.Let S be a multiplicatively closed subset of R . Recall that the saturation S ∗ of S is defined as S ∗ = { x ∈ R : x/ is a unit of S − R } . It is obvious that S ∗ is amultiplicatively closed subset of R containing S [9]. Proposition 2.9.
Let S be a multiplicatively closed subset of R and M be anR-module. Then we have the following.(a) If S ⊆ S are multiplicatively closed subsets of R and M is a fully S -idempotent R -module, then M is a fully S -idempotent R -module.(b) M is a fully S -idempotent R -module if and only if M is a fully S ∗ -idempotent R -module.(c) If M is a fully S -idempotent R -module, then every submodule of M is afully S -idempotent R -module. Proof. (a) This is clear.(b) Let M be a fully S -idempotent R -module. Since S ⊆ S ∗ , by part (a), M is a fully S ∗ -idempotent R -module. For the converse, assume that M is a fully S ∗ -idempotent module and N is a submodule of M . Then there exists x ∈ S ∗ suchthat xN ⊆ ( N : R M ) M . As x ∈ S ∗ , x/ S − R and so ( x/ a/s ) = 1for some a ∈ R and s ∈ S . This yields that us = uxa for some u ∈ S . Thus wehave usN = uxaN ⊆ xN ⊆ ( N : R M ) M . Therefore, M is a fully S -idempotent R -module.(c) Let N be a submodule of M and K be a submodule of N . By Lemma 2.7, M is an S -multiplication R -module. Hence there exists s ∈ S such that sK ⊆ ( K : R M ) M ⊆ K implies that s K ⊆ s ( K : R M ) M ⊆ s ( K : R M ) K ⊆ ( K : R M )( K : R M ) M ⊆ ( K : R M ) M. Thus s K ⊆ ( K : R M ) M ⊆ ( K : R N ) ( N : R M ) M ⊆ ( K : R N ) N. Therefore, N is fully S -idempotent. (cid:3) In the following theorem, we characterize the fully idempotent R -modules. Theorem 2.10.
Let M be an R-module. Then the following statements are equiv-alent: (a) M is a fully idempotent R -module; (b) M is a fully ( R − p ) -idempotent R -module for each prime ideal p of R ; (c) M is a fully ( R − m ) -idempotent R -module for each maximal ideal m of R ; (d) M is a fully ( R − m ) -idempotent R -module for each maximal ideal m of R with M m = 0 m .Proof. ( a ) ⇒ ( b ) Let M be a fully idempotent R -module and p be a prime idealof R . Then R − p is multiplicatively closed set of R and so M is a fully ( R − p )-idempotent R -module by Proposition 2.5.( b ) ⇒ ( c ) Since every maximal ideal is a prime ideal, the result follows from thepart (b).( c ) ⇒ ( d ) This is clear. FARANAK FARSHADIFAR ( d ) ⇒ ( a ) Let N be a submodule of M . Take a maximal ideal m of R with M m = 0 m . As M is a fully ( R − m )-idempotent module, there exists s m suchthat sN ⊆ ( N : R M ) M . This implies that N m = ( sN ) m ⊆ (( N : R M ) M ) m ⊆ N m . If M m = 0 m , then clearly N m = (( N : R M ) M ) m . Thus we conclude that N m =(( N : R M ) M ) m for each maximal ideal m of R . It follows that N = ( N : R M ) M ,as needed. (cid:3) Proposition 2.11.
Let S be a multiplicatively closed subset of R and N be an S -idempotent submodule of an R -module M . Then there is an s ∈ S such that sN ⊆ Hom R ( M, N ) N = X { ϕ ( N ) : ϕ : M → N } . Proof. As N is an S -idempotent submodule of M , there is an s ∈ S such that sN ⊆ ( N : R M ) M and so sN ⊆ ( N : R M ) N . Let x ∈ N . Then there exists r ∈ ( N : R M ) and y ∈ N such that sx = ry . Now consider the homomorphism f : M → N defined by f ( m ) = rm . Then we have sx = f ( y ) ∈ X { ϕ ( N ) : ϕ : M → N } = Hom R ( M, N ) N. Therefore, sN ⊆ Hom R ( M, N ) N . (cid:3) The following example shows that the converse of Proposition 2.11 is not true ingeneral.
Example 2.12.
Let p be a prime number. Take the multiplicatively closed subset S = Z \ p Z of Z . Then one can see that the submodule N = Z p ⊕ Z -module M = Z p ⊕ Z p is not S -idempotent but sN ⊆ Hom Z ( M, N ) N = N for each s ∈ S .A multiplicatively closed subset S of R is said to satisfy the maximal multiplecondition if there exists an s ∈ S such that t | s for each t ∈ S .In the following theorem, we characterize the fully S -idempotent R -modules,where S is a multiplicatively closed subset of R satisfying the maximal multiplecondition. Theorem 2.13.
Let S be a multiplicatively closed subset of R satisfying the max-imal multiple condition (e.g., S is finite or S ⊆ U ( R ) ) and let M be an R -module.Then the following statements are equivalent: (a) M is a fully S -idempotent module; (b) Every cyclic submodule of M is S -idempotent; (c) Every element of M is S -idempotent; (d) For all submodules N and K of M , we have s ( N ∩ K ) ⊆ ( N : R M )( K : R M ) M for some s ∈ S .Proof. ( a ) ⇒ ( b ) and ( b ) ⇒ ( c ) are clear.( c ) ⇒ ( a ). Let N be a submodule of M and x ∈ N . Then by hypothesis, thereexist s x ∈ S and a ∈ ( Rx : R M ) such that s x x = ax . Hence as x x = a x and so s x x = s x ax = as x x = a x . Thus s x Rx ⊆ ( Rx : R M ) M . Now as S satisfyingthe maximal multiple condition, there exists an s ∈ S such that sRx ⊆ ( Rx : R M ) M ⊆ ( N : R M ) M . Therefore, sN ⊆ ( N : R M ) M , as required.( a ) ⇒ ( d ). Let N and K be two submodules of M . Then for some s ∈ S wehave s ( N ∩ K ) ⊆ ( N ∩ K : R M ) M ⊆ ( N : R M )( K : R M ) M. ULLY S -IDEMPOTENT MODULES 5 ( d ) ⇒ ( a ). For a submodule N of M , we have sN = s ( N ∩ N ) ⊆ ( N : R M )( N : R M ) M = ( N : R M ) M for some s ∈ S . (cid:3) Let R i be a commutative ring with identity, M i be an R i -module for each i =1 , , ..., n , and n ∈ N . Assume that M = M × M × ... × M n and R = R × R × ... × R n . Then M is clearly an R -module with componentwise addition andscalar multiplication. Also, if S i is a multiplicatively closed subset of R i for each i = 1 , , ..., n , then S = S × S × ... × S n is a multiplicatively closed subset of R .Furthermore, each submodule N of M is of the form N = N × N × ... × N n , where N i is a submodule of M i . Theorem 2.14.
Let M i be an R i -module and S i ⊆ R i be a multiplicatively closedsubset for i = 1 , . Assume that M = M × M , R = R × R , and S = S × S .Then M is a fully S -idempotent R -module if and only if M is a fully S -idempotent R -module and M is a fully S -idempotent R -module.Proof. For only if part, without loss of generality we will show M is a fully S -idempotent R -module. Take a submodule N of M . Then N ×{ } is a submoduleof M . Since M is a fully S -idempotent R -module, there exists s = ( s , s ) ∈ S × S such that ( s , s )( N × { } ) ⊆ ( N × { } : R M ) M . By focusing on first coordinate,we have s N ⊆ ( N : R M ) M . So M is a fully S -idempotent R -module. Nowassume that M is a fully S -idempotent module and M is a fully S -idempotentmodule. Take a submodule N of M . Then N must be in the form of N × N ,where N ⊆ M , N ⊆ M . Since M is a fully S -idempotent R -module, thereexists an s ∈ S such that s N ⊆ ( N : R M ) M . Similarly, there exists anelement s ∈ S such that s N ⊆ ( N : R M ) M . Now, put s = ( s , s ) ∈ S .Then we get( s , s ) N ⊆ s N × s N ⊆ ( N : R M ) M × ( N : R M ) M ⊆ ( N : R M ) M. Hence, M is a fully S -idempotent R -module. (cid:3) Theorem 2.15.
Let M i be an R i -module for i = 1 , , .., n and let S , ..., S n bemultiplicatively closed subsets of R , ..., R n , respectively. Assume that M = M × ... × M n , R = R × ... × R n and S = S × ... × S n . Then the following statementsare equivalent. (a) M is a fully S -idempotent module. (b) M i is a fully S i -idempotent module for each i ∈ { , , ..., n } .Proof. We use mathematical induction. If n = 1, the claim is trivial. If n = 2,the claim follows from Theorem 2.14. Assume that the claim is true for n < k andwe will show that it is also true for n = k . Put M = ( M × ... × M n − ) × M n , R = ( R × R × ... × R n − ) × R n , and S = ( S × ... × S n − ) × S n . By Theorem2.14, M is fully S -idempotent module if and only if M × ... × M n − is a fully( S × ... × S n − )-idempotent ( R × R × ... × R n − )-module and M n is a fully S n -idempotent R n -module. Now the rest follows from the induction hypothesis. (cid:3) Let M be an R -module. The idealization or trivial extension R ∝ M = R ⊕ M of M is a commutative ring with componentwise addition and multiplication( a, m )( b, ´ m ) = ( ab, a ´ m + bm ) for each a, b ∈ R , m, ´ m ∈ M [3]. If I is an ideal of R and N is a submodule of M , then I ∝ N is an ideal of R ∝ M if and only if FARANAK FARSHADIFAR IM ⊆ N . In that case, I ∝ N is called a homogeneous ideal of R ∝ M . Also if S ⊆ R is a multiplicatively closed subset, then S ∝ N is a multiplicatively closedsubset of R ∝ M [3, Theorem 3.8].Let I be an ideal of R and S ⊆ R be a multiplicatively closed subset of R . If I is a fully S -idempotent R -module, then we say that I is a fully S -idempotent ideal of R . Theorem 2.16.
Let N be a submodule of an R -module M and S be a multiplica-tively closed subset of R . Then the following statements are equivalent: (a) N is a fully S -idempotent R -module; (b) 0 ∝ N is a fully S ∝ -idempotent ideal of R ∝ M ; (c) 0 ∝ N is a fully S ∝ M -idempotent ideal of R ∝ M .Proof. ( a ) ⇒ ( b ). Suppose that N is a fully S -idempotent R -module. Take an ideal J of R ∝ M contained in 0 ∝ N . Then J = 0 ∝ ´ N for some submodule ´ N of M with ´ N ⊆ N . Since N is a fully S -idempotent module, there exists s ∈ S with s ´ N ⊆ ( ´ N : R N ) N ⊆ ´ N . It follows that( s, J = ( s, ∝ ´ N ) = 0 ∝ s ´ N ⊆ ∝ ( ´ N : R N ) N =(( ´ N : R N ) ∝ M )(0 ∝ N ) ⊆ ∝ ´ N = J. This implies that 0 ∝ N is a fully S ∝ R ∝ M .( b ) ⇒ ( c ). This follows from the fact that S ∝ ⊆ S ∝ M and Proposition 2.9(a).( c ) ⇒ ( a ). Suppose that 0 ∝ N is a fully S ∝ M -idempotent ideal of R ∝ M .Let ´ N be a submodule of N . Then 0 ∝ ´ N ⊆ ∝ N and 0 ∝ ´ N is an idealof R ∝ M . Since 0 ∝ N is a fully S ∝ M -idempotent ideal of R ∝ M , thereexists ( s, m ) ∈ S ∝ M such that ( s, m )(0 ∝ ´ N ) ⊆ ´ J (0 ∝ N ) ⊆ (0 ∝ ´ N ), where´ J = (0 ∝ ´ N : R ∝ M ∝ N ). Clearly, ( ´ N : R N ) ∝ N = (0 ∝ ´ N : R ∝ M ∝ N ). Now,set J = ´ J + 0 ∝ M . As´ J (0 ∝ N ) = ´ J (0 ∝ N ) + (0 ∝ M )(0 ∝ N ) = ( ´ J + 0 ∝ M )(0 ∝ N ) , we may assume that ´ J contains 0 ∝ M . Then ´ J = ( ´ N : R N ) ∝ M . This impliesthat( s, m )(0 ∝ ´ N ) = 0 ∝ s ´ N ⊆ (( ´ N : R N ) ∝ M ) (0 ∝ N ) = 0 ∝ ( ´ N : R N ) N ⊆ ∝ ´ N and so s ´ N ⊆ ( ´ N : R N ) N ⊆ ´ N . Hence, N is a fully S -idempotent R -module (cid:3) Proposition 2.17.
Let M and ´ M be R -modules. Assume that S is a multiplica-tively closed subset of R and f : M → ´ M is an R -epimorphism. If M is a fully S -idempotent module, then ´ M is a fully S -idempotent module. Conversely, sup-pose that ´ M is a fully S -idempotent module and tker ( f ) = 0 for some t ∈ S . Then M is a fully S -idempotent module. Proof.
Let ´ N be a submodule of ´ M . Then N := f − ( ´ N ) is a submodule of M .As M is a fully S -idempotent module, there exists s ∈ S such that sN ⊆ ( N : R M ) M ⊆ N . Hence, f ( sN ) ⊆ f (( N : R M ) M ) ⊆ f ( N ). This yields that s ´ N = sf ( N ) ⊆ ( N : R M ) f ( M ) = ( N : R M ) ´ M ⊆ ´ N. Thus s ´ N ⊆ ( N : R M ) ´ M ⊆ ´ N . Hence, ´ M is a fully S -idempotent module. For theconverse, let N be a submodule of M and tker ( f ) = 0. Since f ( M ) = ´ M is a fully ULLY S -IDEMPOTENT MODULES 7 S -idempotent module, there exists s ∈ S with sf ( N ) ⊆ ( f ( N ) : R f ( M )) f ( M ).Hence sN + ker ( f ) ⊆ ( N : R M ) M + ker ( f ) . Multiplying through by t and noting that tker ( f ) = 0 gives that( st ) N ⊆ t ( N : R M ) M ⊆ ( N : R M ) M ⊆ N. So, M is a fully S -idempotent R -module. (cid:3) Corollary 2.18.
Let S be a multiplicatively closed subset of R , M a fully S -idempotent R -module, and N a submodule of M . Then M/N is a fully S -idempotent R -module. Conversely, if M/N is a fully S -idempotent R -module and there exists t ∈ S with tN = 0, then M is a fully S -idempotent R -module. Proposition 2.19.
Let S and T be multiplicatively closed subsets of R . Put˜ S = { s/ ∈ T − R : s ∈ S } , a multiplicatively closed subset of T − R . Supposethat M is a fully S -idempotent R -module. Then T − M is a fully ˜ S -idempotent T − R -module. Hence if S ⊆ T ∗ , then T − M is a fully idempotent T − R -module.Thus S − M is a fully idempotent S − R -module. Proof.
Let N be a T − R -submodule of T − M , so N = T − ´ N for some submodule´ N of M . Since M is a fully S -idempotent module, there exists s ∈ S with s ´ N ⊆ ( ´ N : R M ) M ⊆ ´ N . Then( s/ N = T − ( s ´ N ) ⊆ ( T − ( ´ N : R M ) )( T − M ) ⊆ T − ´ N = N. So T − M is a fully ˜ S -idempotent T − R -module. If S ⊆ T ∗ , then ˜ S ⊆ U ( T − R ).Hence, T − M is a fully idempotent T − R -module by Proposition 2.5. (cid:3) Corollary 2.20.
Let M be an R -module and S a multiplicatively closed subsetof R satisfying the maximal multiple condition (e.g., S is finite or S ⊆ U ( R )).Then M is a fully S -idempotent module if and only if S − M is a fully idempotent S − R -module. Proof. ( ⇒ ): This follows from Proposition 2.19.( ⇐ ): Assume that S − M is a fully idempotent S − R -module. Take a submodule N of M . Since S − M is a fully idempotent S − R -module, we have S − N = ( S − N : S − R S − M ) ( S − M ) = S − (( N : R M ) M ) . Choose s ∈ S with t | s for each t ∈ S . Note that for each m ∈ N , we have m/ ∈ S − N = S − (( N : R M ) M ) and so there exists t ∈ S such that tm ∈ ( N : R M ) M and hence sm ∈ ( N : R M ) M . Therefore, we obtain s N ⊆ s ( N : R M ) M ⊆ ( N : R M ) M ⊆ N. Hence, M is a fully S -idempotent module. (cid:3) Let S be a multiplicatively closed subset of R and M be an R -module. Asubmodule N of M is said to be pure if IN = N ∩ IM for every ideal I of R [1]. M is said to be fully pure if every submodule of M is pure [4]. A submodule N of M is said to be S -pure if there exists an s ∈ S such that s ( N ∩ IM ) ⊆ IN for everyideal I of R [7]. M is said to be fully S -pure if every submodule of M is S -pure [7]. Theorem 2.21.
Let S be a multiplicatively closed subset of R , M an S -multiplication R -module, and N be a submodule of M . Then the following statements are equiva-lent. FARANAK FARSHADIFAR (a) N is an S -pure submodule of M . (b) N is an S -multiplication R -module and N is an S -idempotent submoduleof M . (c) N is an S -multiplication R -module and there exists an s ∈ S such that sK ⊆ ( N : R M ) K , where K is a submodule of N . (d) N is an S -multiplication R -module and there exists an s ∈ S such that s ( K : R N ) N ⊆ ( K : R M )( N : R M ) M , where K is a submodule of M .Proof. ( a ) ⇒ ( b ). Let K be a submodule of N . As M is an S -multiplication module,there exists an s ∈ S such that sK ⊆ ( K : R M ) M . Now since N is S -pure, thereis an t ∈ S such that ( K : R N ) N ⊇ t ( N ∩ ( K : R N ) M ). Hence,( K : R N ) N ⊇ t ( N ∩ ( K : R N ) M ) ⊇ t ( N ∩ ( K : R M ) M ) ⊇ t ( N ∩ sK ) = tsK. This implies that N is an S -multiplication R -module. Since M is an S -multiplicationmodule, there exists an u ∈ S such that uN ⊆ ( N : R M ) M . Now as N is S -pure,there is an v ∈ S such that ( N : R M ) uN ⊇ v ( N ∩ u ( N : R M ) M ). Therefore,( N : R M ) M = ( N : R M )( N : R M ) M ⊇ ( N : R M ) uN ⊇ v ( N ∩ u ( N : R M ) M ) = vu ( N : R M ) M ⊇ vu N. So, N is an S -idempotent submodule.( b ) ⇒ ( c ). Let K be a submodule of N . Since N is an S -multiplication R -module, there exists an s ∈ S such that sK ⊆ ( K : R N ) N . As N is S -idempotent,there is an t ∈ S such that tN ⊆ ( N : R M ) M . Therefore, tsK ⊆ t ( K : R N ) N = ( K : R N ) tN ⊆ ( K : R N )( N : R M ) M = ( N : R M )( K : R N )( N : R M ) M ⊆ ( N : R M )( K : R N ) N ⊆ ( N : R M ) K. ( c ) ⇒ ( a ). Let I be an ideal of R . Since IM ∩ N ⊆ N , by part (c), there is an s ∈ S such that s ( IM ∩ N ) ⊆ ( N : R M )( IM ∩ N ). Hence,( N ∩ IM )( N : R M ) ⊆ ( N : R M ) IM = IN implies that s ( IM ∩ N ) ⊆ IN . Thus, N is an S -copure submodule of M .( b ) ⇒ ( d ). Let K be a submodule of M . Since N is S -idempotent, there is an s ∈ S such that sN ⊆ ( N : R M ) Ms ( K : R N ) N ⊆ ( K : R N )( N : R M ) M ⊆ ( K : R M )( N : R M ) M. ( d ) ⇒ ( b ). Take K = N . (cid:3) Corollary 2.22.
Let S be a multiplicatively closed subset of R and M be an R -module. Then we have the following.(a) If M is a fully S -idempotent R -module, then M is a fully S -pure R -module.(b) If M is an S -multiplication fully S -pure R -module, then M is a fully S -idempotent R -module. Proof. (a) By Proposition 2.9, every submodule of M is a fully S -idempotent R -module. Hence, by Lemma 2.7, every submodule of M is an S -multiplication R -module. Now the result follows from Theorem 2.21 ( b ) ⇒ ( a ).(b) This follows from Theorem 2.21 ( a ) ⇒ ( b ). (cid:3) ULLY S -IDEMPOTENT MODULES 9 Let S be a multiplicatively closed subset of R and M be an R -module. Asubmodule N of M is said to be copure if ( N : M I ) = N + (0 : M I ) for every ideal I of R [5]. A submodule N of M is said to be S -copure if there exists an s ∈ S such that s ( N : M I ) ⊆ N + (0 : M I ) for every ideal I of R [8]. M is said to be fully S -copure if every submodule of M is S -copure [8]. Proposition 2.23.
Let S a multiplicatively closed subset of R and M be an S -multiplication R -module. If N is an S -copure submodule of M , then N is S -idempotent. Proof.
Suppose that N is an S -copure submodule of M . Then there exists an s ∈ S such that s ( N : M ( N : R M )) ⊆ N + (0 : M ( N : R M )) . This in turn implies that sM ⊆ N + (0 : M ( N : R M )). It follows that s ( N : R M ) M ⊆ ( N : R M ) N. As M is an S -multiplication module, there is an t ∈ S such that tN ⊆ ( N : R M ) M .Hence, we have st N ⊆ st ( N : R M ) M ⊆ ( N : R M ) tN ⊆ ( N : R M ) M, as needed. (cid:3) Acknowledgments.
The author would like to thank Prof. Habibollah Ansari-Toroghy for his helpful suggestions and useful comments.
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Department of Mathematics, Farhangian University, Tehran, Iran.
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