aa r X i v : . [ m a t h . A C ] A ug FULLY S -COIDEMPOTENT MODULES F. FARSHADIFAR* AND H. ANSARI-TOROGHY**
Abstract.
Let R be a commutative ring with identity, S be a multiplica-tively closed subset of R , and M be an R -module. A submodule N of M is called coidempotent if N = ((0 : M Ann R ( N )). Also, M is called fullycoidempotent if every submodule of M is coidempotent. In this article, we in-troduce the concepts of S -coidempotent submodules and fully S -coidempotent R -modules as generalizations of coidempotent submodules and fully coidem-potent R -modules. We explore some basic properties of these classes of R -modules. Introduction
Throughout this paper R will denote a commutative ring with identity and S will denote a multiplicatively closed subset of R . Also, Z will denote the ring ofintegers. Definitions 1.1.
Let M be an R -module.(a) M is said to be a multiplication module if for every submodule N of M ,there exists an ideal I of R such that N = IM [8].(b) M is said to be a comultiplication module if for every submodule N of M ,there exists an ideal I of R such that N = (0 : M I ). It is easy to see that M is a comultiplication module if and only if N = (0 : M Ann R ( N )) foreach submodule N of M [5].(c) A submodule N of M is said to be pure if IN = N ∩ IM for every ideal I of R [2]. M is said to be fully pure if every submodule of M is pure [6].(d) A submodule N of M is said to be copure if ( N : M I ) = N + (0 : M I ) forevery ideal I of R [7]. M is said to be fully copure if every submodule of M is copure [6].(e) 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 [6].(f) A submodule N of M is said to be coidempotent if N = ((0 : M Ann R ( N )).Also, M is said to be fully coidempotent if every submodule of M is coidem-potent [6].Recently the notions such as of S -Noetherian rings, S -Noetherian modules, S -prime submodules, S -multiplication modules, S -2-absorbing submodules, S -secondsubmodules, S -comultiplication modules, classical S -2-absorbing submodules, S -pure submodules, S -copure submodules, fully S -idempotent module, etc. intro-duced and investigated [3, 1, 9, 18, 4, 19, 10, 20, 17, 11, 12, 13]. Mathematics Subject Classification.
Key words and phrases.
Coidempotent submodule, fully coidempotent module, multiplica-tively closed subset, S -coidempotent submodule, fully S -coidempotent module. Definitions 1.2.
Let M be an R -module.(a) A multiplicatively closed subset S of R is said to satisfy the maximal mul-tiple condition if there exists an s ∈ S such that t | s for each t ∈ S . Forexample, if S is finite or S ⊆ U ( R ), then S satisfying the maximal multiplecondition [4].(b) A submodule N of M is said to be an S -finite submodule if there exist afinitely generated submodule K of M and s ∈ S such that sN ⊆ K ⊆ N .Also, M is said to be an S -Noetherian module if every submodule of M is S -finite. In particular, R is said to be an S -Neotherian ring if it is an S -Noetherian R -module [3].(c) A submodule N of M is said to be an S -direct summand of M if thereexist a submodule K of M and s ∈ S such that sM = N + K (d.s.). M issaid to be an S -semisimple module if every submodule of M is an S -directsummand of M [12].(d) M is said to be an S -comultiplication module if for each submodule N of M ,there exist an s ∈ S and an ideal I of R such that s (0 : M I ) ⊆ N ⊆ (0 : M I )[20].(e) A submodule N of M is said to be S -pure if there exists an s ∈ S such that s ( N ∩ IM ) ⊆ IN for every ideal I of R . Also, M is said to be fully S -pure if every submodule of M is S -pure [11].(f) A submodule N of M is said to be S -copure if there exists an s ∈ S suchthat s ( N : M I ) ⊆ N + (0 : M I ) for every ideal I of R . Also, M is said tobe fully S -copure if every submodule of M is S -copure [12].(g) A submodule N of M is said to be an S -idempotent submodule if thereexists an s ∈ S such that sN ⊆ ( N : R M ) M ⊆ N . Also, M is said toa fully S -idempotent module if every submodule of M is an S -idempotentsubmodule [13].In this paper, we introduce the notions of S -coidempotent submodules and fully S -coidempotent modules as a generalization of coidempotent submodules and fullycoidempotent modules. Also, these notions can be regarded as a dual notions of S -idempotent submodules and fully S -idempotent modules. We consider variousfundamental properties of fully S -coidempotent R -modules.2. Main resultsDefinition 2.1.
We say that a submodule N of an R -module M is an S -coidempotentsubmodule if there exists an s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . Definition 2.2.
We say that an R -module M is a fully S -coidempotent module ifevery submodule of M is an S -coidempotent submodule. Example 2.3.
Let M be an R -module with Ann R ( M ) ∩ S = ∅ . Then clearly, M is a fully S -coidempotent R -module. Proposition 2.4.
Every fully coidempotent R -module is a fully S -coidempotent 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 examples show that the converse of Proposition 2.4 is not true ingeneral.
ULLY S -COIDEMPOTENT MODULES 3 Example 2.5.
Consider the Z -module Z . Then for each positive integer t , (0 : Z Ann Z ( t Z )) = Z implies that each submodule of Z is not coidempotent and so Z isnot a fully coidempotent Z -module. Now, take the multiplicatively closed subset S = Z \ { } of Z . Then for each positive integer t , t (0 : Z Ann Z ( t Z ) ) = t Z . Thuseach submodule of Z is S -coidempotent and hence Z is a fully S -coidempotent Z -module. Example 2.6.
Consider the Z -module M = Z ⊕ Z . Take the multiplicativelyclosed subset S = { n : n ∈ N ∪ { }} of Z , where N denotes the set of positiveintegers. Then M as a Z -module is fully S -coidempotent, while M is not fullycoidempotent.Lemma 2.7 and Example 2.8 show that the notion of S -comultiplication R -module is a generalization of fully S -coidempotent R -module. Lemma 2.7.
Let M be a fully S -coidempotent R -module. Then M is an S -comultiplication R -module. Proof.
Let N be a submodule of M . There exists an s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . This implies that s (0 : M Ann R ( N )) ⊆ s ( N : M Ann R ( N )) ⊆ s (0 : M Ann R ( N )) ⊆ N. (cid:3) The following example shows that the converse of Lemma 2.7 is not true ingeneral.
Example 2.8.
Take the multiplicatively closed subset S = Z \ Z of Z . Then Z isan S -comultiplication Z -module. But Z is not a fully S -coidempotent Z -module.A proper submodule N of an R -module M is said to be completely irreducible if N = T i ∈ I N i , where { N i } i ∈ I is a family of submodules of M , implies that N = N i for some i ∈ I . It is easy to see that every submodule of M is an intersection ofcompletely irreducible submodules of M [14].In the following theorem, we characterize the fully S -coidempotent R -modules,where S satisfying the maximal multiple condition. Theorem 2.9.
Let S satisfying the maximal multiple condition and let M be an R -module. Then the following statements are equivalent: (a) M is a fully S -coidempotent module; (b) Every completely irreducible submodule of M is S -coidempotent; (c) For all submodules N and K of M , we have s (0 : M Ann R ( N ) Ann R ( K )) ⊆ N + K for some s ∈ S .Proof. ( a ) ⇒ ( b ) This is clear.( b ) ⇒ ( a ) As S satisfying the maximal multiple condition, there exists an s ∈ S such that for each completely irreducible submodule L of M we have s (0 : M Ann R ( L )) ⊆ L . Now, let N be a submodule of M and N = T i ∈ I L i , where each F. FARSHADIFAR* AND H. ANSARI-TOROGHY** L i is a completely irreducible submodule of M . Then we have s (0 : M Ann R ( N )) = s (0 : M Ann R ( \ i ∈ I L i )) ⊆ \ i ∈ I s (0 : M Ann R ( L i )) ⊆ \ i ∈ I L i = N. ( a ) ⇒ ( c ) Let N and K be two submodules of M . Then there exists an s ∈ S such that s (0 : M Ann R ( N + K )) ⊆ N + K . Hence we have s (0 : M Ann R ( N ) Ann R ( K )) ⊆ s (0 : M Ann R ( N + K )) ⊆ N + K. ( c ) ⇒ ( a ) For a submodule N of M for some s ∈ S , we have s (0 : M Ann R ( N )) = s (0 : M Ann R ( N ) Ann R ( N )) ⊆ N + N = N. Thus M is a fully S -coidempotent module (cid:3) Theorem 2.10.
Let S satisfying the maximal multiple condition and M be an R -module. Then we have the following. (a) If M is an S -comultiplication module such that every completely irreduciblesubmodule of M is an S -direct summand of M , then M is a fully S -coidempotent module. (b) If M is an S -semisimple S -comultiplication module, then M is a fully S -coidempotent module.Proof. (a) By Theorem 2.9, it is enough to show that every completely irreduciblesubmodule of M is S -coidempotent. So, let L be a completely irreducible submoduleof M . By hypothesis, there exists an s ∈ S such that sM = L + K (d.s.), where K is a submodule of M . Thus s (0 : M Ann R ( L )) ⊆ (0 : sM Ann R ( L )) = (0 : L Ann R ( L )) + (0 : K Ann R ( L )) = L. (b) This follows from part (a). (cid:3) The saturation S ∗ of S is defined as S ∗ = { x ∈ R : x/ is a unit of S − R } . Itis obvious that S ∗ is a multiplicatively closed subset of R containing S [15]. Proposition 2.11.
Let M be an R-module. Then we have the following.(a) If S ⊆ S are multiplicatively closed subsets of R and M is a fully S -coidempotent R -module, then M is a fully S -coidempotent R -module.(b) M is a fully S -coidempotent R -module if and only if M is a fully S ∗ -coidempotent R -module.(c) If M is a fully S -coidempotent R -module, then every homomorphic imageof M is a fully S -coidempotent R -module. Proof. (a) This is clear.(b) Let M be a fully S -coidempotent R -module. Since S ⊆ S ∗ , M is a fully S ∗ -coidempotent R -module by part (a). For the converse, assume that M is afully S ∗ -coidempotent module and N is a submodule of M . Then there exists an x ∈ S ∗ such that x (0 : M Ann R ( N )) ⊆ N . As x ∈ S ∗ , x/ S − R and ULLY S -COIDEMPOTENT MODULES 5 so ( x/ a/s ) = 1 for some a ∈ R and s ∈ S . This implies that us = uxa for some u ∈ S . Thus we have us (0 : M Ann R ( N )) = uxa (0 : M Ann R ( N )) ⊆ x (0 : M Ann R ( N )) ⊆ N. Therefore, M is a fully S -coidempotent R -module.(c) Let N be a submodule of M and K/N a submodule of
M/N . By assumption,there exists an s ∈ S such that s (0 : M Ann R ( K )) ⊆ K . Hence(0 : M Ann R ( K )) = ((0 : M Ann R ( K )) : M Ann R ( K )) ⊆ ( K : M sAnn R ( K )) ⊆ (0 : M sAnn R ( K )) . It follows that s (0 : M Ann R ( K )) ⊆ K . Thus s (0 : M/N
Ann ( K/N )) ⊆ s (0 : M Ann R ( N ) Ann R ( K/N )) /N ⊆ s (0 : M Ann R ( K )) /N ⊆ K/N, as needed. (cid:3)
The following theorem provides some characterizations for fully coidempotent R -modules. Theorem 2.12.
Let M be an R-module. Then the following statements are equiv-alent: (a) M is a fully coidempotent R -module; (b) M is a fully ( R − P ) -coidempotent R -module for each prime ideal P of R ; (c) M is a fully ( R − m ) -coidempotent R -module for each maximal ideal m of R ; (d) M is a fully ( R − m ) -coidempotent R -module for each maximal ideal m of R with M m = 0 m .Proof. ( a ) ⇒ ( b ) Let M be a fully coidempotent R -module and P be a primeideal of R . Then R − P is multiplicatively closed set of R and so M is a fully( R − P )-coidempotent R -module by Proposition 2.4.( b ) ⇒ ( c ) Since every maximal ideal is a prime ideal, the result follows from thepart (b).( c ) ⇒ ( d ) This is clear.( 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 )-coidempotent module, there exists an s m such that s (0 : M Ann R ( N )) ⊆ N . This implies that(0 : M Ann R ( N )) m = ( s (0 : M Ann R ( N ))) m ⊆ N m . If M m = 0 m , then clearly (0 : M Ann R ( N )) m ⊆ N m . So, we have (0 : M Ann R ( N )) m ⊆ N m for each maximal ideal m of R . It follows that (0 : M Ann R ( N )) ⊆ N . Thus N = (0 : M Ann R ( N )) because the inverse inclusion is clear. (cid:3) Proposition 2.13.
Let f : M → ´ M be an R -monomorphism of R -modules. Thenwe have the following.(a) If ´ M is a fully S -coidempotent module, then M is a fully S -coidempotentmodule.(b) If M is a fully S -coidempotent module and t ´ M ⊆ f ( M ) for some t ∈ S ,then ´ M is a fully S -coidempotent module. F. FARSHADIFAR* AND H. ANSARI-TOROGHY**
Proof. (a) Let N be a submodule of M . Then as ´ M is a fully S -coidempotentmodule, there exists an s ∈ S such that s (0 : ´ M Ann R ( f ( N ))) ⊆ f ( N ). This impliesthat sf − ((0 : ´ M Ann R ( f ( N )))) ⊆ f − ( f ( N )). It follows that s (0 : M Ann R ( N )) ⊆ N .(b) Let ´ N be a submodule of ´ M and t ´ M ⊆ f ( M ) for some t ∈ S . Since M is a fully S -coidempotent module, there exists an s ∈ S such that s (0 : M Ann R ( f − ( ´ N ))) ⊆ f − ( ´ N ). Now as f is monomorphism, we have that s (0 : f ( M ) Ann R ( f − ( ´ N ))) ⊆ f ( f − ( ´ N )) = f ( M ) ∩ ´ N ⊆ ´ N .
One can see that
Ann R ( f − ( ´ N )) = Ann R ( ´ N ). Therefore, s (0 : f ( M ) Ann R ( ´ N )) ⊆ ´ N . Thus ts (0 : ´ M Ann R ( ´ N )) ⊆ ´ N , as needed. (cid:3) Corollary 2.14.
Every submodule of a fully S -coidempotent R -module is a fully S -coidempotent R -module. Proof.
Let N be a submodule of a fully S -coidempotent R -module M . Then theresult follows from Proposition 2.13 by using the inclusion homomorphism f : N ֒ → M . (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 an R -module with componentwise addition and scalarmultiplication. 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 2.15.
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 -coidempotent R -module if and only if M is a fully S -coidempotent R -module and M is a fully S -coidempotent R -module.Proof. First assume that M is a fully S -coidempotent R -module, without loss ofgenerality we show that M is a fully S -coidempotent R -module. Let N be a sub-module of M . Then N ×{ } is a submodule of M . As M is a fully S -coidempotent R -module, there exists an s = ( s , s ) ∈ S × S such that ( s , s )(0 : M Ann R ( N ×{ } )) ⊆ N × { } . This in turn implies that s (0 : M Ann R ( N )) ⊆ N . Thus M is a fully S -coidempotent R -module. Now assume that M is a fully S -coidempotent R -module and M is a fully S -coidempotent R -module. Let N be a submodule of M . Then N must be in the form of N × N , where N is asubmodule of M and N is a submodule of M . As M is a fully S -coidempotent R -module, there exists an s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . Simi-larly, there exists an element s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . Set s = ( s , s ) ∈ S . Then we have( s , s )(0 : M Ann R ( N )) ⊆ s (0 : M Ann R ( N )) × s (0 : M Ann R ( N )) ⊆ N × N = N. So, M is a fully S -coidempotent R -module. (cid:3) Theorem 2.16.
Let M i be an R i -module and S i be a multiplicatively closed subsetof R i for i = 1 , , .., n . Assume that M = M × ... × M n , R = R × ... × R n and S = S × ... × S n . Then the following statements are equivalent: ULLY S -COIDEMPOTENT MODULES 7 (a) M is a fully S -coidempotent R -module; (b) M i is a fully S i -coidempotent R i -module for each i ∈ { , , ..., n } .Proof. We use induction. If n = 1, the claim is trivial. If n = 2, the claim followsfrom Theorem 2.15. Assume that the claim is true for n < k and we show that it isalso 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 Theorem 2.15, M is fully S -coidempotent R -module if and only if M × ... × M n − is a fully ( S × ... × S n − )-coidempotent( R × R × ... × R n − )-module and M n is a fully S n -coidempotent R n -module. Nowthe rest follows from the induction hypothesis. (cid:3) The following lemma is known, but we write its proof here for the sake of refer-ences.
Lemma 2.17.
Let M be an R -module. Then we have the following.(a) If S satisfying the maximal multiple condition, then S − (0 : M I ) = (0 : S − M S − I ) and S − ( Ann R ( N )) = Ann S − R ( S − N ) for each ideal I of R andsubmodule N of M .(b) If I is an S -finite ideal of R and N is an S -finite submodule of M , then S − (0 : M I ) = (0 : S − M S − I ) and S − ( Ann R ( N )) = Ann S − R ( S − N ). Proof. (a) Let x/h ∈ (0 : S − M S − I ) and a ∈ I . Then ( x/h )( a/
1) = 0. Hence thereexists an u ∈ S such that uxa = 0. As S satisfying the maximal multiple condition,there exists an s ∈ S such that t | s for each t ∈ S . Thus s = uv for some v ∈ S .Therefore, sxI = 0. So, sx ∈ (0 : M I ). Hence, x/h = ( xs ) / ( sh ) ∈ S − (0 : M I ) andso (0 : S − M S − I ) ⊆ S − (0 : M I ). The reverse inclusion is clear. Similarly, onecan see that S − ( Ann R ( N )) = Ann S − R ( S − N ).(b) Let I be an S -finite ideal of R . Then there exist t ∈ S and a , a , ..., a n ∈ I such that tI ⊆ Ra + ... + Ra n ⊆ I . Now let x/h ∈ (0 : S − M S − I ). Then( x/h )( a i /
1) = 0 for i = 1 , , .., n . Hence there exists an u i ∈ S such that u i xa i = 0for i = 1 , , .., n . Set U = u u ...u n . Then we have uxa i = 0 for i = 1 , , .., n .This implies that utxI = 0. Hence, utx ∈ (0 : M I ). Thus x/h = ( xut ) / ( uth ) ∈ S − (0 : M I ) and so (0 : S − M S − I ) ⊆ S − (0 : M I ). The reverse inclusion is clear.Similarly, if N is an S -finite submodule of M one can see that S − ( Ann R ( N )) = Ann S − R ( S − N ). (cid:3) Theorem 2.18.
Let S satisfying the maximal multiple condition and M be an R -module. Then M is a fully S -coidempotent R -module if and only if S − M is a fullycoidempotent S − R -module.Proof. First note that as S satisfying the maximal multiple condition, S − (0 : M Ann R ( N )) = (0 : S − M Ann S − R ( S − N )) for each submodule N of M by Lemma2.17 (a). Now suppose that M is a fully S -coidempotent R -module and S − N is asubmodule of S − M . Then there exists an t ∈ S such that t (0 : M Ann R ( N )) ⊆ N .Let x/h ∈ (0 : S − M Ann S − R ( S − N )). Then x/h ∈ S − (0 : M Ann R ( N ))and so x/h = y/h , where y ∈ (0 : M Ann R ( N )) and h ∈ S . Thus there ex-ists u ∈ S such that xh u = yh u . So, txh u = yh ut ∈ N . It follows that x/h = ( xth u ) / ( h th u ) ∈ S − N . This in turn implies that S − N = (0 : S − M Ann S − R ( S − N )) and S − M is a fully coidempotent S − R -module. Conversely,assume that S − M is a fully coidempotent S − R -module and N is a submodule of M . Then S − N = (0 : S − M Ann S − R ( S − N )). Let m ∈ (0 : M Ann R ( N )). Then F. FARSHADIFAR* AND H. ANSARI-TOROGHY** m/ n/s for some n ∈ N and s ∈ S . Thus ms s = ns for some s ∈ S . As S satisfying the maximal multiple condition, there exists an s ∈ S such that t | s foreach t ∈ S . Therefore, ms = nsv ∈ N for some v ∈ S . Thus x ∈ ( N : M s ). Thus s (0 : M Ann R ( N )) ⊆ N , as needed. (cid:3) Proposition 2.19.
Let R be an S -Noetherian ring. If M is a fully S -coidempotent R -module, then S − M is a fully coidempotent S − R -module. Proof.
This is straightforward by using the technique of Theorem 2.18 and Lemma2.17 (b). (cid:3)
Theorem 2.20.
Let M be an S -comultiplication R -module and N be a submoduleof M . Then the following statements are equivalent: (a) N is an S -copure submodule of M ; (b) M/N is an S -comultiplication R -module and N is an S -coidempotent sub-module of M ; (c) M/N is an S -comultiplication R -module and there exists an s ∈ S suchthat s ( N : M Ann R ( K )) ⊆ K , where K is a submodule of M with N ⊆ K ; (d) M/N is an S -comultiplication R -module and there exists an s ∈ S suchthat s ( N : M Ann R ( K )) ⊆ ( N : M ( N : R K )) , where K is a submodule of M .Proof. ( a ) ⇒ ( b ) Assume that K/N is a submodule of
M/N . Since N is an S -copure submodule of M , there exists an t ∈ S such that t ( N : M Ann R ( K/N )) ⊆ N + (0 : M Ann R ( K/N )). Also, as M is an S -comultiplication module, there existsan s ∈ S such that s (0 : M Ann R ( K )) ⊆ K . Now we have st (0 : M/N
Ann R ( K/N )) = st (( N : M Ann R ( K/N )) /N )= s (( t ( N : M Ann R ( K/N )) + N ) /N ) ⊆ s ( N + (0 : M Ann R ( K/N )) /N ) ⊆ ( N + s (0 : M Ann R ( K )) /N ⊆ ( N + K ) /N = K/N.
Thus
M/N is an S -comultiplication R -module. Now we show that N is an S -coidempotent submodule of M . As M is an S -comultiplication module, there existsan u ∈ S such that u (0 : M Ann R ( N )) ⊆ N . As N is an S -copure submodule of M , there exists an h ∈ S such that h ( N : M uAnn R ( N )) ⊆ N + (0 : M uAnn R ( N )).Now we have h (0 : M Ann R ( N )) ⊆ h (( N : M u ) : M Ann R ( N ))= h ( N : M uAnn R ( N )) ⊆ N + (0 : M uAnn R ( N ))= ((0 : M An R ( N ) : M u ) . This implies that uh (0 : M Ann R ( N )) ⊆ (0 : M An R ( N ). Thus u h (0 : M Ann R ( N )) ⊆ u (0 : M An R ( N ) ⊆ N , as desired.( b ) ⇒ ( c ) Let K be a submodule of M with N ⊆ K . Since M/N is an S -comultiplication R -module, there exists an t ∈ S such that t (0 : M/N
Ann R ( K/N )) ⊆ K/N . It follows that t ( N : M Ann R ( K/N ))+ N ⊆ K and so t ( N : M Ann R ( K/N )) ⊆ ULLY S -COIDEMPOTENT MODULES 9 K . As N is an S -coidempotent, there exists an s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . Now we have( N : M Ann R ( K )) /N ⊆ ( N : M Ann R ( N ) Ann R ( K/N )) /N ⊆ ((0 : M Ann R ( N )) : M Ann R ( K/N )) N ⊆ (( N : M sAnn R ( K/N )) /N. It follows that s ( N : M Ann R ( K )) ⊆ ( N : M Ann R ( K/N )). Thus st ( N : M Ann R ( K )) ⊆ t ( N : M Ann R ( K/N )) ⊆ K. ( c ) ⇒ ( a ) Let I be an ideal of R . Since N ⊆ (0 : M I ) + N , there exists an s ∈ S such that s ( N : M Ann R ((0 : M I ) + N )) ⊆ (0 : M I ) + N by part (c). As M is an S -comultiplication module, there exists an t ∈ S such that t (0 : M Ann R ( N )) ⊆ N .Now we have( N : M I ) ⊆ ((0 : M Ann R ( N )) : M I ) = ((0 : M I ) : M Ann R ( N )) ⊆ ( N + (0 : M I ) : M Ann R ( N )) ⊆ ((0 : M Ann R ( N + (0 : M I )) : M Ann R ( N )) ⊆ ( N : M tAnn R ( N + (0 : M I ))) . This implies that t ( N : M I ) ⊆ ( N : M Ann R ( N + (0 : M I ))). Thus st ( N : M I ) ⊆ s ( N : M Ann R ( N + (0 : M I ))) ⊆ N + (0 : M I ) . ( b ) ⇒ ( d ) Let K be a submodule of M . As N is S -coidempotent, there exists an s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . Now we have( N : M Ann R ( K )) ⊆ ( N : M Ann R ( N )( N : R K ))= (( N : M Ann R ( N )) : M ( N : R K )) ⊆ ((0 : M Ann R ( N )) : M ( N : R K )) ⊆ ( N : M s ( N : R K )) . This implies that s ( N : M Ann R ( K )) ⊆ ( N : M ( N : R K )).( d ) ⇒ ( b ) Take K = N . (cid:3) Corollary 2.21.
Let M be an R -module. Then we have the following.(a) If M is a fully S -coidempotent module, then M is fully S -copure.(b) If M is an S -comultiplication fully S -copure module, then M is fully S -coidempotent. Proof. (a) By Proposition 2.11, every homomorphic image of M is a fully S -coidempotent module and so every homomorphic image of M is an S -comultiplicationmodule by Lemma 2.7. Now the result follows from Theorem 2.20 ( b ) ⇒ ( a ).(b) This follows from Theorem 2.20 ( a ) ⇒ ( b ). (cid:3) The following example shows that in part (b) of the Corollary 2.21, the condition M is an S -comultiplication R -module can not be omitted. Example 2.22.
Set M = Z p ⊕ Z p for some prime number p . Take the multiplica-tively closed subset S = Z \ p Z . Then M as a Z -module is a fully S -copure module,while M is not a fully S -coidempotent module because the submodule 0 ⊕ Z p of M is not S -coidempotent. Proposition 2.23.
Let S satisfying the maximal multiple condition and M be afully S -coidempotent R -module. Then for each submodule K of M and each col-lection { N λ } λ ∈ Λ of submodules of M , there exists an s ∈ S such that s ( T λ ∈ Λ ( N λ + K )) ⊆ T λ ∈ Λ N λ + K. Proof.
Let K be a submodule of M and { N λ } λ ∈ Λ be a collection of submodules of M . By Lemma 2.7, M is an S -comultiplication R -module. So, there exists an s ∈ S such that s (0 : M Ann R ( N λ ) ⊆ N λ for each λ ∈ Λ since S satisfying the maximalmultiple condition. Now by using the fact that M is fully S -copure by Corollary2.21 (a), we have s ( \ λ ∈ Λ ( N λ + K )) ⊆ \ λ ∈ Λ s ( N λ + K ) ⊆ \ λ ∈ Λ s ((0 : M Ann R ( N λ )) + (0 : M Ann R ( K )) ⊆ \ λ ∈ Λ s ((0 : M Ann R ( N λ ) Ann R ( K )) ⊆ \ λ ∈ Λ s ( N λ : M sAnn R ( K )) ⊆ \ λ ∈ Λ s ( N λ + (0 : M sAnn R ( K )) ⊆ \ λ ∈ Λ s ( N λ : M Ann R ( K )) ⊆ \ λ ∈ Λ s ( N λ + (0 : M Ann R ( K )) ⊆ \ λ ∈ Λ ( sN λ + s (0 : M Ann R ( K )) ⊆ \ λ ∈ Λ ( N λ + K ) . (cid:3) Proposition 2.24.
Let M be an S -comultiplication R -module and N be an S -puresubmodule of M . Then N is an S -coidempotent submodule of M . Proof. As N is an S -pure submodule of M , there exists an s ∈ S such that s ( N ∩ Ann R ( N ) M ) ⊆ Ann R ( N ) N = 0 . Hence, N ∩ Ann R ( N ) M ) ⊆ (0 : M s ). Since M is an S -comultiplication module,there exists an t ∈ S such that t (0 : M Ann R ( N )) ⊆ N . Hence we have(0 : M Ann R ( N )) = ((0 : M Ann R ( N )) : M Ann R ( N )) ⊆ (( N : M t ) : M Ann R ( N ))= ( N : M tAnn R ( N ))= ( N ∩ Ann R ( N ) M : M tAnn R ( N )) ⊆ ((0 : M s ) : M tAnn R ( N ))= ((0 : M Ann R ( N )) : M st ) ⊆ ( N : M st ) . Therefore, st (0 : M Ann R ( N )) ⊆ N . (cid:3) 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 [4]. Corollary 2.25.
Let M be an R -module. Then we have the following.(a) If M is an S -multiplication fully S -copure module, then M is fully S -pure.(b) If M is an S -comultiplication fully S -pure module, then M is fully S -copure. ULLY S -COIDEMPOTENT MODULES 11 (c) If M is an S -multiplication fully S -coidempotent module, then M is fully S -idempotent.(d) If M is an S -comultiplication fully S -idempotent module, then M is fully S -coidempotent. Proof. (a) By [13, Proposition 2.23], every submodule of M is S -idempotent. Hencethe result follows from [13, Corollary 2.22].(b) By Proposition 2.24, every submodule of M is S -coidempotent. Hence theresult follows from Corollary 2.21 (a).(c) This follows from Corollary 2.21 (a) and [13, Proposition 2.23].(d) By [13, Corollary 2.22], M is fully S -pure. Thus by part (b), M is fully S -copure. So the result follows from Corollary 2.21 (b). (cid:3) Lemma 2.26.
Let M be an S -comultiplication R -module. If I and J are ideals of R such that (0 : M I ) ⊆ (0 : M J ). Then there exists an s ∈ S such that sJM ⊆ IM . Proof.
As (0 : M I ) ⊆ (0 : M J ), we have Ann R ( IM ) = ((0 : M I ) : R M ) ⊆ ((0 : M J ) : R M ) = Ann R ( JM ) . Now by [20, Lemma 1], there exists an s ∈ S such that sJM ⊆ IM . (cid:3) In the following theorem we see that if M is an S -finite R -module, then M is afully S -idempotent module iff M is a fully S -coidempotent module. Theorem 2.27.
Let M be an R -module. Then we have the following. (a) If M is an S -finite fully S -coidempotent R -module, then M is a fully S -idempotent module. (b) If M is an S -Noetherian fully S -idempotent module, then M is a fully S -coidempotent module.Proof. (a) Let N be a submodule of M . Since M is a fully S -coidempotent module,there exists an s ∈ S such that s (0 : M Ann R ( N )) ⊆ N . Hence s (0 : M/N
Ann R ( N )) = s (( N : M Ann R ( N )) /N ) ⊆ s ((0 : M Ann R ( N )) : M Ann R ( N )) /N )= s ((0 : M Ann R ( N )) /N )= ( s (0 : M Ann R ( N )) + N ) /N = 0 . This implies that (0 :
M/N
Ann R ( N )) ⊆ (0 : M/N s ). By Proposition 2.11 (c), M/N is an S -coidempotent R -module. Thus M/N is an S -comultiplication module byLemma 2.7. Hence by Lemma 2.26, there exists t ∈ S such that ts ( M/N ) ⊆ Ann R ( N )( M/N ). As M is an S -finite R -module, one can see that M/N is an S -finite R -module. Therefore, there exist h ∈ S and a ∈ Ann R ( N ) such that( h + a )( M/N ) = 0 by [16, Lemma 2.1]. So, h + a ∈ ( N : R M ). Hence, ( h + a ) ∈ ( N : R M ) . This implies that Rh ⊆ Ann R ( N ) + ( N : R M ) . Thus we have h N = Rh N ⊆ Ann R ( N ) N + ( N : R M ) N ⊆ ( N : R M ) N ⊆ ( N : R M ) M. Therefore, M is fully S -idempotent. (b) Let N be a submodule of M . Since M is fully S -idempotent, there exists an s ∈ S such that sN ⊆ ( N : R M ) M ⊆ ( N : R M ) N . Since M is S -Noetherian, N is S -finite. Now by [16, Lemma 2.1], there exist h ∈ S and a ∈ ( N : R M ) such that( h + a ) N = 0. So, h + a ∈ Ann R ( N ). Hence, ( h + a ) ∈ Ann R ( N ). This yieldsthat Rh ⊆ ( N : R M ) + Ann R ( N ). Thus h (0 : M Ann R ( N )) = Rh (0 : M Ann R ( N )) ⊆ ( N : R M )(0 : M Ann R ( N )) + Ann R ( N )(0 : M Ann R ( N ))= ( N : R M )(0 : M Ann R ( N )) ⊆ ( N : R M ) M ⊆ N, as needed. (cid:3) References
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