aa r X i v : . [ m a t h . A C ] S e p ON S-PRIMARY SUBMODULES
H. ANSARI-TOROGHY AND S. S. POURMORTAZAVI
Abstract.
Let R be a commutative ring with identity, S a multiplicativelyclosed subset of R , and M be an R -module. In this paper, we study andinvestigate some properties of S -primary submodules of M . Among the otherresults, it is shown that this class of modules contains the family of primary(resp. S -prime) submodules properly. Introduction
Throughout this article, all rings are commutative with identity elements, andall modules are unital left modules. N , Z , and Q will denote respectively the naturalnumbers, the ring of integers and the field of quotients of Z .Consider a non-empty subset S of R . We call S a multiplicatively closed subsetof R if (i) 0 / ∈ S , (ii) 1 ∈ S , and (iii) ss ′ ∈ S for all s, s ′ ∈ S [12]. Note that S = R − p is a multiplicatively closed subset of R for every prime ideal p of R . Let N and K be two submodules of an R -module M and J an ideal of R . Then theresidual N by K and J is defined as follows:( N : R K ) = { r ∈ R | rK ⊆ N } , ( N : M J ) = { m ∈ M | Jm ⊆ N } . Particularly, we use
Ann R ( M ) instead of (0 : R M ), and we use ( N : M s ) insteadof ( N : M Rs ), where Rs is the principal ideal generated by an element s ∈ R . Thesets of prime ideals and maximal ideals of R are denoted by Spec ( R ) and M ax ( R ),respectively.A submodule P of M is called prime if P = M and whenever r ∈ R and e ∈ M satisfy re ∈ P , then r ∈ ( P : R M ) or e ∈ P . The set of all prime submodule of M is denoted by Spec ( M )(See [3, 7]).In [11], the authors introduce the concept of S -prime submodule and investigatesome properties of this class of modules. Let S be a multiplicatively closed subsetof R and P be a submodule of M with ( P : R M ) ∩ S = ∅ . Then P is said to be an S - prime submodule if there exists s ∈ S such that whenever rm ∈ P , where r ∈ R and m ∈ M , then sr ∈ ( P : R M ) or sm ∈ P . Particularly, an ideal I of R is saidto be an S - prime ideal if I is an S -prime submodule of the R -module R .The notion of S -primary submodule was introduced in [5]. Let S be a multiplica-tively closed subset of R and P be a submodule of M with ( P : R M ) ∩ S = ∅ . Then P is said to be an S - primary submodule if there exists s ∈ S such that whenever rm ∈ P , where r ∈ R and m ∈ M , then sr ∈ p ( P : R M ) or sm ∈ P . The authorhas not specified the properties of this family of modules excepts a few results abouta certain R -module. Date : September 22, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Multiplicatively closed subset, prime submodule, S -prime submodule,primary submodule, S -primary submodule. In this paper, we will study the family of S -primary submodules extensively andinvestigate some of their properties. In fact, this family of modules is a generaliza-tion of primary (resp. S -prime) submodules.Among the other results, we provide some notions that each one is equivalent to S -primary (see Theorem 2.2). Examples 2.4 and 2.5 show that these new modulescontain the family of primary and S -prime submodules properly. Further It isproved that if P is an S -primary submodule of M , then S − P is also an S -primarysubmodule of S − M (see Proposition 2.7). Example 2.8 shows that the converse isnot true in general. Also we show that S -primary submodules has a good behaviorwith direct sums (see Theorems 2.20 and 2.21). Moreover, we provide some usefulcharacterization concerning S -primary submodules (see Theorems 2.17, 2.24, and2.25). 2. Main results
Definition 2.1.
Let S be a multiplicatively closed subset of R and P be a sub-module of M with ( P : R M ) ∩ S = ∅ . Then P is said to be an S - primary submodule of M if there exists s ∈ S such that whenever rm ∈ P , where m ∈ M and r ∈ R ,then sr ∈ p ( P : R M ) or sm ∈ P [5, Definition 2.27 ]. In particular, we say thatan ideal I of R is an S -primary ideal if I is an S -primary submodule of R -module R . Theorem 2.2.
Let S be a multiplicatively closed subset of R . For a submodule P of an R -module M with ( P : R M ) ∩ S = ∅ . Then the following are equivalent: (a) P is an S -primary submodule of M ; (b) There exists s ∈ S such that for every r ∈ R , the endomorphism r : s ( M/P ) → s ( M/P ) given by sm = sm + P rsm = rsm + P isinjective or ( rs ) t ( M/P ) = (0) for some t ∈ N ; (c) There exists s ∈ S such that whenever rN ⊆ P , where N is a submodule of M and r ∈ R , then sr ∈ p ( P : R M ) or sN ⊆ P ; (d) There exists s ∈ S such that whenever JN ⊆ P , where N is a submoduleof M and J is an ideal of R , then sJ ⊆ p ( P : R M ) or sN ⊆ P .Proof. (a) ⇒ (b). By hypothesis, there exists s ∈ S such that for every r ∈ R and m ∈ M if rm ∈ P , then we have sm ∈ P or sr ∈ p ( P : R M ). Now for each r ∈ R , we define the endomorphism r : s ( M/P ) → s ( M/P ) by sm + P rsm + P .We show that this endomorphism is injective or rs ∈ p ( P : R M ). Assume rs / ∈ p ( P : R M ). Then we show the other part holds. To see let rsm = rsm + P = P =0. So we have ( rs ) m ∈ P . So by hypothesis, sm ∈ P or s ( rs ) = rs ∈ p ( P : R M ).We conclude sr ∈ p ( P : R M ) which is a contradiction. Hence sm ∈ P , as required.(b) ⇒ (a). It is clear.(a) ⇒ (c). It is clear.(c) ⇒ (d). Let JN ⊆ P , where J is an ideal of R and N is a submodule of M . Wewill show that there exists s ∈ S such that sN ⊆ P or sJ ⊆ p ( P : R M ). Clearly,we have rN ⊆ P for every r ∈ J . So by part (c), there exists s ∈ S such that sN ⊆ P or sr ∈ p ( P : R M ) for every r ∈ J , as desired.(d) ⇒ (a). Take r ∈ R and m ∈ M with rm ∈ P . Now, put J = Rr and N = Rm .Then we can conclude that JN = Rrm ⊆ P . By assumption, there is an s ∈ S so N S-PRIMARY SUBMODULES 3 that sJ = Rrs ⊆ p ( P : R M ) or sN = Rsm ⊆ P and so either sr ∈ p ( P : R M ) or sm ∈ P , as required. (cid:3) Lemma 2.3.
Let M be an R -module and S a multiplicatively closed subset of R .Then we have the following.(a) If P is a primary submodule of M such that ( P : R M ) ∩ S = ∅ , then P isan S -primary submodule of M .(b) If P is an S -primary submodule of M and S ⊆ u ( R ), where u ( R ) denotesthe set of units in R , then P is a primary submodule of M . Proof.
This is clear. (cid:3)
By setting S = { } , we conclude that every primary submodule is an S -primarysubmodule by Lemma 2.3. The following example shows that the converse is nottrue in general. Example 2.4.
Consider the Z -module M = Q ⊕ ( L ni =1 Z p i ), where p i are distinctpositive prime integers. Take the submodule P = (0) and the multiplicativelyclosed subset S = { , p m p m ...p m n n | ∀ i ∈ { , , ..., n } , m i ∈ N ∪ { } } . First note that ( P : Z M ) = (0) and p p ...p n (0 , , , ...,
1) = (0 , , , ..., ∈ P .Since p p ...p n / ∈ p ( P : Z M ) and (0 , , , ..., / ∈ P , P is not a primary submoduleof M . Put s = p p ...p n and let k ( pq , x , x , ..., x n ) = ( kpq , kx , kx , ..., kx n ) ∈ P, where k ∈ Z and ( pq , x , x , ..., x n ) ∈ M . Then kp = 0. This yields that k = 0or p = 0. If k = 0, there is nothing to prove. Thus assume that p = 0. Then s ( pq , x , x , ..., x n ) ∈ P . Therefore, P is an S -primary submodule of M .We recall that a submodule P of an R -module M is S - prime if there exists s ∈ S such that whenever rm ∈ P , where r ∈ R and m ∈ M , then sr ∈ ( P : R M )or sm ∈ P [11]. Clearly, every S -prime submodule is S -primary. The followingexample shows that the converse is not true in general. Example 2.5.
Consider M = Z as a Z -module. Set S = Z \ Z and P = (0). Thuswe have ( P : Z M ) = 4 Z and 2 . ∈ (0). Since for every s ∈ S , 2 s / ∈ ( P : Z M ) and s. / ∈ P , P is not an S -prime submodule of M . Put s = 1 and let ka = 0. If a = 0,there is nothing to prove. Thus assume that a = 0. Then k = 2 k ′ for some k ′ ∈ Z .This implies that k ∈ p ( P : Z M ). Therefore, P is an S -primary submodule of M . Remark 2.6.
Let S be a multiplicatively closed subset of R . Recall that thesaturation S ∗ of S is defined as S ∗ = { x ∈ R | x S − R } . It is obvious that S ∗ is a multiplicatively closed subset of R containing S [6]. Proposition 2.7.
Let S be a multiplicatively closed subset of R and M be an R -module. Then we have the following. H. ANSARI-TOROGHY AND S. S. POURMORTAZAVI (a) If S ⊆ S are multiplicatively closed subsets of R and P is an S -primarysubmodule of M , then P is an S -primary submodule of M in case( P : R M ) ∩ S = ∅ .(b) P is an S -primary submodule of M if and only if P is an S ∗ -primarysubmodule of M .(c) If P is an S -primary submodule of M , then S − P is a primary submoduleof S − R -module S − M . Proof. (a) It is clear.(b) Assume that P is an S -primary submodule of M . We need to prove that( P : R M ) and S ∗ are disjoint. Suppose there exists x ∈ ( P : R M ) ∩ S ∗ . As x ∈ S ∗ , x is a unit of S − R and so ( x )( as ) = 1 for some a ∈ R and s ∈ S .This yields that us = uxa for some u ∈ S . Now we have that us = uxa ∈ ( P : R M ) ∩ S , a contradiction. Thus ( P : R M ) ∩ S ∗ = ∅ . Now as S ⊆ S ∗ , bypart (b), P is an S ∗ -primary submodule of M . Conversely, assume that P is an S ∗ -primary submodule of M . Let rm ∈ P , where r ∈ R and m ∈ M .Then there exists x ∈ S ∗ such that xr ∈ p ( P : R M ) or xm ∈ P . As x isa unit of S − R , there exist u, s ∈ S and a ∈ R such that us = uxa . Put us = s ′ ∈ S . Then note that s ′ r = ( us ) r = uaxr ∈ p ( P : R M ) or s ′ m ∈ P .Therefore, P is an S -primary submodule of M .(c) Let ( rs )( mt ) ∈ S − P , where rs ∈ S − R and mt ∈ S − M . Then urm ∈ P for some u ∈ S . Since P is an S -primary submodule of M , there is an s ′ ∈ S so that s ′ ur ∈ p ( P : R M ) or s ′ m ∈ P . This yields that rs = s ′ urs ′ us ∈ S − p ( P : R M ) ⊆ p ( S − P : S − R S − M ) or mt = s ′ ms ′ t ∈ S − P . Hence, S − P is a primary submodule of S − M . (cid:3) The following example shows that the converse of part (c) of Proposition 2.7 isnot true in general.
Example 2.8.
Consider the Z -module M = Q . Take the submodule N = Z andthe multiplicatively closed subset S = Z − { } of Z . Then ( N : Z M ) = (0). Let s be an arbitrary element of S . Choose a prime number p with gcd ( p, s ) = 1. Thennote that p p = 1 ∈ N . But sp / ∈ p ( N : Z M ) and sp / ∈ N , it follows that N isnot an S -primary submodule of M . Since S − Z = Q is a field, S − ( Q ) is a vectorspace. Therefore the proper submodule S − N is a primary submodule of S − Q . Proposition 2.9.
Suppose f : M → M ′ is an R -homomorphism. Then we havethe following.(a) If P ′ is an S -primary submodule of M ′ provided that ( f − ( P ′ ) : R M ) ∩ S = ∅ , then f − ( P ′ ) is an S -primary submodule of M .(b) If f is an epimorphism and P is an S -primary submodule of M with ker ( f ) ⊆ P , then f ( P ) is an S -primary submodule of M ′ . Proof. (a) Let rm ∈ f − ( P ′ ) for some r ∈ R and m ∈ M . This yields that f ( rm ) = rf ( m ) ∈ P ′ . Since P ′ is an S -primary submodule of M ′ , thereis an s ∈ S so that sr ∈ p ( P ′ : R M ′ ) or sf ( m ) ∈ P ′ . Now we will showthat ( P ′ : R M ′ ) ⊆ ( f − ( P ′ ) : R M ). Take x ∈ ( P ′ : R M ′ ). Then wehave xM ′ ⊆ P ′ . Since f ( M ) ⊆ M ′ , we conclude that f ( xM ) = xf ( M ) ⊆ xM ′ ⊆ P ′ . This implies that xM ⊆ f − ( f ( M )) ⊆ f − ( p ′ ) and thus x ∈ N S-PRIMARY SUBMODULES 5 ( f − ( P ′ ) : R M ). As p ( P ′ : R M ′ ) ⊆ p ( f − ( P ′ ) : R M ), we can concludeeither sa ∈ p ( f − ( P ′ ) : R M ) or sm ∈ f − ( P ′ ). Hence, f − ( P ′ ) is an S -primary submodule of M .(b) First note that ( f ( P ) : R M ) ∩ S = ∅ . Otherwise there would be an s ∈ ( f ( P ) : R M ′ ) ∩ S . Since s ∈ ( f ( P ) : R M ′ ), sM ′ ⊆ f ( P ), but then f ( sM ) = sf ( M ) = sM ′ ⊆ f ( P ). By taking their inverse images under f , we have sM ⊆ sM + ker ( f ) ⊆ f − ( f ( P )) = P + ker ( f ) = P. That means s ∈ ( P : R M ), which contradictions P is an S -primary sub-module of M . Now take r ∈ R and m ′ ∈ M ′ with rm ′ ∈ f ( P ). As f is an epimorphism, there is an m ∈ M such that m ′ = f ( m ). Then rm ′ = rf ( m ) = f ( rm ) ∈ f ( P ). Since Ker ( f ) is a subset of P , we get rm ∈ P . As P is an S -primary submodule of M , there is an s ∈ S so that sr ∈ p ( P : R M ) or sm ∈ P . Since p ( P : R M ) ⊆ p ( f ( P ) : R M ′ ), we have sr ∈ p ( f ( P ) : R M ′ ) or f ( sm ) = sf ( m ) = sm ′ ∈ f ( P ). Accordingly, f ( P )is an S -primary submodule of M ′ . (cid:3) Corollary 2.10.
Let S be a multiplicatively closed subset of R and take a sub-module L of M . Then we have the following.(a) If P ′ is an S -primary submodule of M with ( P ′ : R L ) ∩ S = ∅ , then L ∩ P ′ is an S -primary submodule of L .(b) Suppose that P is a submodule of M with L ⊆ P . Then P is an S -primarysubmodule of M if and only if P/L is an S -primary submodule of M/L . Proof. (a) Consider the injection i : L → M defined by i ( m ) = m for all m ∈ L .Then note that i − ( P ′ ) = L ∩ P ′ . Now we will show that ( i − ( P ′ ) : R L ) ∩ S = ∅ . Assume that s ∈ ( i − ( P ′ ) : R L ) ∩ S . Then we have sL ⊆ i − ( P ′ ) = L ∩ P ′ ⊆ P ′ . This implies that s ∈ ( P ′ : R L ) ∩ S , a contradiction.The rest follows from Proposition 2.9 (a).(b) Assume that P is an S -primary submodule of M . Then consider the canoni-cal homomorphism π : M → M/L defined by π ( m ) = m + L for all m ∈ M .By Proposition 2.9 (b), P/L is an S -primary submodule of M/L . Con-versely, assume that
P/L is an S -primary submodule of M/L . Let rm ∈ P for some r ∈ R and m ∈ M . This yields that r ( m + L ) = rm + L ∈ P/L .As
P/L is an S -primary submodule of M/L , there is an s ∈ S so that sr ∈ p ( P/L : R M/L ) = p ( P : R M ) or s ( m + L ) = sm + L ∈ P/L . There-fore, we have sr ∈ p ( P : R M ) or sm ∈ P . Hence, P is an S -primarysubmodule of M . (cid:3) An R -module 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 [4]. Proposition 2.11.
Let M be an R -module and S be a multiplicatively closedsubset of R . The following statements hold.(a) If P is an S -primary submodule of M , then ( P : R M ) is an S -primary idealof R .(b) If M is multiplication module and ( P : R M ) is an S -primary ideal of R ,then P is an S -primary submodule of M . H. ANSARI-TOROGHY AND S. S. POURMORTAZAVI
Proof. (a) Let xy ∈ ( P : R M ) for some x, y ∈ R . Then xym ∈ P for all m ∈ M . As P is an S -primary submodule, there exists s ∈ S such that sx ∈ p ( P : R M ) or sym ∈ P for all m ∈ M . If sx ∈ p ( P : R M ), there isnothing to prove. Suppose that sx / ∈ p ( P : R M ). Then sym ∈ P for all m ∈ M so that sy ∈ ( P : R M ). Therefore, ( P : R M ) is an S -primary idealof R .(b) Let J be an ideal of R and N a submodule of M with JN ⊆ P . Then wecan conclude that J ( N : R M ) ⊆ ( JN : R M ) ⊆ ( P : R M ). As ( P : R M ) isan S -primary ideal of R , there is an s ∈ S so that s ( N : R M ) ⊆ ( P : R M )or sJ ⊆ p ( P : R M ). Thus, we can conclude that sN = s ( N : R M ) M ⊆ ( P : R M ) M = P or sJ ⊆ p ( P : R M ). Therefore, by Theorem 2.2 (d), P is an S -primary submodule of M . (cid:3) Remark 2.12. (a) Assume that M is a multiplication R -module and K, L aretwo submodules of M . The product of K and L is defined as KL = ( K : R M )( L : R M ) M [1].(b) Let M be an R -module and N a submodule of M . The radical of N , denotedby rad ( N ), is the intersection of all prime submodules of M containing N ;that is, rad ( N ) = T { P | N ⊆ P, P ∈ Spec ( M ) } [8].As an immediate consequence of the Proposition 2.11 and Theorem 2.2 (d), wehave the following explicit result. Corollary 2.13.
Suppose that M is a multiplication R -module and P a submoduleof M provided that ( P : R M ) ∩ S = ∅ , where S is a multiplicatively closed subsetof R . Then the following are equivalent:(a) P is an S -primary submodule of M ;(b) There exists s ∈ S such that whenever LN ⊆ P , where L and N aresubmodules of M , then s ( L : R M ) ⊆ p ( P : R M ) or sN ⊆ P . Corollary 2.14.
Suppose that M is a finitely generated multiplication R -moduleand P is a submodule of M provided that ( P : R M ) ∩ S = ∅ , where S is amultiplicatively closed subset of R . Then the following are equivalent:(a) P is an S -primary submodule of M ;(b) There exists s ∈ S such that whenever LN ⊆ P , where L and N aresubmodules of M , then sL ⊆ rad ( P ) or sN ⊆ P . Proof. (a) ⇒ (b). Assume that LN ⊆ P , where L and N are submodules of M .By Remark 2.12 (a), LN = ( L : R M ) N ⊆ P . Then there exists s ∈ S so that s ( L : R M ) ⊆ p ( P : R M ) or sN ⊆ P by Theorem 2.2 (d). Since M is multiplication,by [4, Theorem 2.12], we have s ( L : R M ) M = sL ⊆ p ( P : R M ) M = rad ( P ) or sN ⊆ P .(b) ⇒ (a). Assume that JN ⊆ P , where N is a submodule of M and J is an idealof R . Set K := JM . As M is a multiplication module, Then we have KN = ( K : R M )( N : R M ) M = J ( N : R M ) M = JN ⊆ P. By assumption, there exists s ∈ S so that sK ⊆ rad ( P ) or sN ⊆ P . As M isfinitely generated, by [9, Thoerem 4.4], sK ⊆ rad ( P ) implies that sJ ⊆ ( sK : R M ) ⊆ ( rad ( P ) : R M ) = p ( P : R M ) . N S-PRIMARY SUBMODULES 7
Therefore P is an S -primary submodule of M by Corollary 2.13. (cid:3) Remark 2.15. (a) Let M be an R -module and p be a maximal ideal of R .In [4], T p ( M ) is defined as follows T p ( M ) = { m ∈ M | (1 − r ) m = 0 for some r ∈ P } . Clearly T p ( M ) is a submodule of M . An R -module M is said to be p - cyclic provided there exist q ∈ p and m ∈ M such that (1 − q ) M ⊆ Rm [4].(b) Let M be an R -module. Then M is a multiplication R -module if and onlyif for every maximal ideal p of R either M = T p ( M ) or M is p -cyclic [4,Theorem 1.2]. Lemma 2.16.
Let S be a multiplicatively closed subset of R , p be an S -primary(resp. S -prime) ideal of R and M be a faithful multiplication R -module. Thenthere exists a fixed s ∈ S and whenever am ∈ pM , where a ∈ R and m ∈ M ,implies that sa ∈ √ p (resp. sa ∈ p ) or sm ∈ pM . Proof.
It is enough to prove it for S -primary submodules. The technique is similarfor S -prime. As p is an S -primary ideal, there exists s ∈ S , whenever rr ′ ∈ p , where r, r ′ ∈ R , then sr ∈ √ p or sr ′ ∈ p . Let a ∈ R and m ∈ M satisfy am ∈ pM . Suppose sa / ∈ √ p . Set K := ( pM : R sm ). Assume that K = R . Then there exists a maximalideal Q of R so that K ⊆ Q . m / ∈ T Q ( M ), since otherwise, there exists q ∈ Q suchthat (1 − q ) m = 0 and so (1 − q ) sm = 0. This implies that (1 − q ) ∈ K ⊆ Q , acontradiction. By [4, Theorem 1.2] M is Q -cyclic, that is there exist m ′ ∈ M and q ∈ Q such that (1 − q ) M ⊆ Rm ′ . In particular, (1 − q ) m = s ′ m ′ , (1 − q ) am = p ′ m ′ for some s ′ ∈ R and p ′ ∈ p . Thus ( as ′ − p ′ ) m ′ = 0. Now (1 − q )( Ann R ( m ′ )) M ⊆ ( Ann R ( m ′ )) Rm ′ = implies (1 − q ) Ann R ( m ) ⊆ Ann R ( M ) = , because M isfaithful, and hence (1 − q ) as ′ = (1 − q ) p ′ ∈ p . As p is an S -primary ideal, ss ′ ∈ p or sa ∈ √ p or s (1 − q ) n ∈ p for some n ∈ N . But p ⊆ K ⊆ Q so that in each case,we have a contradiction. It follows that K = R and sm ∈ pM , as required. (cid:3) In the following Theorem, The Theorem 2.11 in [11] will be extended by removingthe condition “finitely generated”.
Theorem 2.17.
Let M be a multiplication R -module and P a submodule of M provided that ( P : R M ) ∩ S = ∅ , where S is a multiplicatively closed subset of R .Then the following are equivalent: (a) P is an S -primary (resp. S -prime) submodule of M . (b) ( P : R M ) is an S -primary (resp. S -prime) ideal of R . (c) P = IM for some S -primary (resp. S -prime) ideal I of R with Ann ( M ) ⊆ I .Proof. (a) ⇒ (b). It is clear from Proposition 2.11 (a).(b) ⇒ (c). It is clear.(c) ⇒ (a). As M is a faithful multiplication R/Ann R ( M )-module, by Corollary2.10 (b), I/Ann R ( M ) is an S -primary (resp. S -prime) ideal of R/Ann R ( M ). Hence P = IM is an S -primary (resp. S -prime) submodule of R/Ann R ( M )-module M byLemma 2.16. Therefore, P is an S -primary (resp. S -prime) submodule of R -module M , as required. (cid:3) H. ANSARI-TOROGHY AND S. S. POURMORTAZAVI
Proposition 2.18.
Let P be an S -primary submodule of multiplication R -module M . Suppose that N ∩ L ⊆ P for some submodules N and L of M . Then sN ⊆ P or sL ⊆ rad ( P ) for some s ∈ S . Proof.
Since P is an S -primary submodule, there exists s ∈ S such that for every r ∈ R and m ∈ M , if rm ∈ P , then sr ∈ p ( P : R M ) or sm ∈ P . Let sN * P .Then sm ′ / ∈ P for some m ′ ∈ N . Take an element a ∈ ( L : R M ). This yields that am ′ ∈ ( L : R M ) N ⊆ L ∩ N ⊆ P . As P is an S -primary submodule of M and sm ′ / ∈ P , we can conclude that sa ∈ p ( P : R M ) so that s ( L : R M ) ⊆ p ( P : R M ).As M is a multiplication module, by [4, Theorem 2.12] we have sL = s ( L : R M ) M ⊆ p ( P : R M ) M = rad ( P ) . (cid:3) Lemma 2.19.
Let R = R × R and S = S × S where S i is a multiplicativelyclosed subset of R i . Suppose p = p × p is an ideal of R . So the following areequivalent:(a) p is an S -primary ideal of R .(b) p is an S -primary ideal of R and p ∩ S = ∅ or p is an S -primary idealof R and p ∩ S = ∅ . Proof. (a) ⇒ (b). Since (1 , ,
1) = (0 , ∈ p , there exists s = ( s , s ) ∈ S so that s (1 ,
0) = ( s , ∈ √ p or s (0 ,
1) = (0 , s ) ∈ p and thus p ∩ S = ∅ or p ∩ S = ∅ .We may assume that p ∩ S = ∅ . As P ∩ S = ∅ , we have p ∩ S = ∅ . Let xy ∈ p for some x, y ∈ R . Since (0 , x )(0 , y ) ∈ p and p is an S -primary ideal of R . We geteither s (0 , x ) = (0 , s x ) ∈ √ p or s (0 , y ) = (0 , s y ) ∈ p and this yields s x ∈ √ p or s y ∈ p . Therefore, p is an S -primary ideal of R . In the other case, one caneasily show that p is an S -primary ideal of R .(b) ⇒ (a). Assume that p ∩ S = ∅ and p is an S -primary ideal of R . Thenthere exists s ∈ p ∩ S . Let ( a, b )( c, d ) = ( ac, bd ) ∈ p for some a, c ∈ R and b, d ∈ R . This yields that bd ∈ p and thus there exists s ∈ S so that s b ∈ √ p or s d ∈ p . Put s = ( s , s ) ∈ S . Then note that s ( a, b ) = ( s a, s b ) ∈ √ p or s ( c, d ) ∈ p . Therefore, p is an S -primary ideal of R . In other case, one can similarlyprove that p is an S -primary ideal of R . (cid:3) Theorem 2.20.
Suppose that M = M × M and R = R × R -module and S = S × S is a multiplicatively closed subset of R , where M i is a R i -module and S i is a multiplicatively closed subset of R i for each i = 1 , . Assume P = P × P is a submodule of M . Then the following are equivalent: (a) P is an S -primary submodule of M . (b) P is an S -primary submodule of M and ( P : R M ) ∩ S = ∅ or P isan S -primary submodule of M and ( P : R M ) ∩ S = ∅ .Proof. (a) ⇒ (b). By Proposition 2.11, ( P : R M ) = ( P : R M ) × ( P : R M ) isan S -primary ideal of R and so by Lemma 2.19, either ( P : R M ) ∩ S = ∅ or( P : R M ) ∩ S = ∅ . We may assume that ( P : R M ) ∩ S = ∅ . Now we willshow that P is an S -primary submodule of M . Let rm ∈ P for some r ∈ R and m ∈ M . Then (1 , r )(0 , m ) = (0 , rm ) ∈ P . As P is an S -primary, there is an s = ( s , s ) ∈ S so that s (1 , r ) = ( s , s r ) ∈ p ( P : R M ) or s (0 : R m ) = (0 , s m ) ∈ P . This implies that s r ∈ p ( P : R M ) or s m ∈ P . Therefore, P is an S is N S-PRIMARY SUBMODULES 9 an S -primary submodule of M . In the other case, it can be similarly show that P is an S -primary submodule of M .(b) ⇒ (a). Assume that ( P : R M ) ∩ S = ∅ and P is an S -primary sub-module of M . Then there exists s ∈ ( P : R M ) ∩ S . Let ( r , r )( m , m ) =( r m , r m ) ∈ P for some r i ∈ R i and m i ∈ M i , where i = 1 ,
2. Then r m ∈ P . As P is an S -primary submodule of M , there is an s ∈ S so that s r ∈ p ( P : R M ) or s m ∈ P . Now put s = ( s , s ) ∈ S . Then note that s ( r , r ) = ( s r , s r ) ∈ p ( P : R M ) or s ( m , m ) = ( s m , s m ) ∈ P × P = P .Therefore, P is an S -primary submodule of M . Similarly one can show that if P isan S -primary submodule of M and ( P : R M ) ∩ S = ∅ , then P is an S -primarysubmodule of M . (cid:3) Theorem 2.21.
Let M = M × M × · · · M n and R = R × R × · · · R n -module and S = S × S ×· · · S n is a multiplicatively closed subset of R , where M i is a R i -moduleand S i is a multiplicatively closed subset of R i for each i = 1 , , · · · , n . Assume P = P × P × · · · P n is a submodule of M . Then the following are equivalent: (a) P is an S -primary submodule of M . (b) P i is an S i -primary submodule of M i for some i ∈ { , , · · · , n } and ( P j : R j M j ) ∩ S j = ∅ for all j ∈ { , , · · · , n } − { i } .Proof. We apply induction on n . For n = 1, the result is true. If n = 2, then(a) ⇔ (b) follows from Theorem 2.20. Assume that (a) and (b) are equivalent when k < n . Now, we shall prove (a) ⇔ (b) when k = n . Let P = P × P × ... × P n . Put P ′ = P × P × ... × P n − and S ′ = S × S × ... × S n − . Then by Theorem 2.20,the necessary and sufficient condition for P = P ′ × P n is an S -primary submoduleof M is that P ′ is an S -primary submodule of M ′ and ( P n : R n M n ) ∩ S n = ∅ or P n is an S -primary submodule of M n and ( P ′ : R ′ M ′ ) ∩ S ′ = ∅ , where M ′ = M × M × ... × M n − and R ′ = R × R × ... × R n − . The rest follows from theinduction hypothesis. (cid:3) Lemma 2.22.
Suppose that P is an S -primary submodule of M . Then the fol-lowing statements hold for some s ∈ S .(a) ( P : M s ′ ) ⊆ ( P : M s ) for all s ′ ∈ S .(b) (( P : R M ) : R s ′ ) ⊆ (( P : R M ) : R s ) for all s ′ ∈ S Proof. (a) Take an element m ′ ∈ ( P : M s ′ ), where s ′ ∈ S . Then s ′ m ′ ∈ P .Since P is an S -primary submodule of M , there exists s ∈ S such that ss ′ ∈ p ( P : R M ) or sm ′ ∈ P . As ( P : R M ) ∩ S = ∅ , we get sm ′ ∈ P ,namely m ′ ∈ ( P : M s ).(b) Follows from part (a). (cid:3) Proposition 2.23.
Suppose that M is a finitely generated R -module, S is a multi-plicatively closed subset of R , and P is a submodule of M satisfying ( P : R M ) ∩ S = ∅ . Then the following are equivalent:(a) P is an S -primary submodule of M .(b) S − P is a primary submodule of S − M and there is an s ∈ S satisfying( P : M s ′ ) ⊆ ( P : M s ) for all s ′ ∈ S . Proof. (a) ⇒ (b). It is clear from Proposition 2.7 (c) and Lemma 2.22.(b) ⇒ (a). Take a ∈ R and m ∈ M with am ∈ P . Then a . m ∈ S − P . Since S − P is a primary submodule of S − M and M is finitely generated, we can concludethat a ∈ p ( S − P : S − R S − M ) = p S − ( P : R M ) or m ∈ S − P . Then ua ∈ p ( P : R M ) or u ′ m ∈ P for some u, u ′ ∈ S . By assumption, there is an s ∈ S sothat ( P : R s ′ ) ⊆ ( P : R s ) for all s ′ ∈ S . If ua ∈ p ( P : R M ), then a n M ⊆ ( P : M u n ) ⊆ ( P : R s ) for some n ∈ N and thus sa ∈ p ( P : R M ). If u ′ m ∈ P , a similarargument shows that sm ∈ P . Therefore, P is an S -primary submodule of M . (cid:3) Theorem 2.24.
Suppose that P is a submodule of M provided ( P : R M ) ∩ S = ∅ .Then P is an S -primary submodule of M if and only if ( P : M s ) is a primarysubmodule of M for some s ∈ S .Proof. Assume ( P : M s ) is a primary submodule of M for some s ∈ S . Let am ∈ P ,where a ∈ R and m ∈ M . As am ∈ ( P : M s ), we get a ∈ p (( P : M s ) : R M ) or m ∈ ( P : M s ). This yields that as ∈ p ( P : R M ) or sm ∈ P . Conversely, assume that P is an S -primary submodule of M . Then there exists s ∈ S such that whenever am ∈ P , where a ∈ R and m ∈ M , then sa ∈ p ( P : R M ) or sm ∈ P . Now weprove that ( P : M s ) is primary. Take r ∈ R and m ∈ M with rm ∈ ( P : M s ). Then srm ∈ P . As P is S -primary, we get s r ∈ p ( P : R M ) or sm ∈ P . If sm ∈ P ,then there is nothing to show. Assume that sm / ∈ P . Then s r ∈ p ( P : R M ) andhence sr ∈ p ( P : R M ). Thus r n ∈ (( P : R M ) : R s n ) ⊆ (( P : R M ) : R s ) for some n ∈ N , by Lemma 2.22. Thus, we can conclude that r n ∈ (( P : M s ) : R M ), namely r ∈ p (( P : M s ) : R M ). Hence ( P : M s ) is a prime submodule of M . (cid:3) Theorem 2.25.
Suppose that P is a submodule of M provided ( P : R M ) ⊆ Jac ( R ) ,where Jac ( R ) is the Jacobson radical of R . Then the following statements areequivalent: (a) P is a primary submodule of M . (b) ( P : R M ) is a primary ideal of R and P is an ( R − m ) -primary submoduleof M for each m ∈ M ax ( R ) .Proof. (a) ⇒ (b). Since ( P : R M ) ⊆ Jac ( R ), ( P : R M ) ⊆ m for each m ∈ M ax ( R )and hence ( P : R M ) ∩ ( R − m ) = ∅ . The rest follows from Lemma 2.3 (a).(b) ⇒ (a). Let am ∈ P with a / ∈ ( P : R M ) for some a ∈ R and m ∈ M . Let m ∈ M ax ( R ). As P is an ( R − m )-primary submodule of M , there exists s m / ∈ m such that as m ∈ p ( P : R M ) or s m m ∈ P . As ( P : R M ) is a primary ideal of R and s m / ∈ p ( P : R M ), we have as m / ∈ ( P : R M ) and so s m m ∈ P . Now consider the setΩ = { s m | ∃ m ∈ M ax ( R ) , s m / ∈ m and s m m ∈ P } . Then note that (Ω) = R . Tosee this, take any maximal ideal m ′ containing Ω. Then the definition of Ω requiresthat there exists s m ′ ∈ Ω and s m ′ / ∈ m ′ . As Ω ⊆ m ′ , we have s m ′ ∈ Ω ⊆ m ′ , acontradiction. Thus (Ω) = R , and this yields 1 = r s m + r s m + · · · + r n s m n for some r i ∈ R and s m i / ∈ m i with s m i m ∈ P , where m i ∈ M ax ( R ) for each i = 1 , , ..., n . This yields that m = r s m m + r s m m + · · · + r n s m n m ∈ P .Therefore, P is a primary submodule of M . (cid:3) Now we determine all primary submodules of a module over a quasi-local ringin terms of S -primary submodules. N S-PRIMARY SUBMODULES 11
Corollary 2.26.
Suppose M is a module over a quasi-local ring ( R, m ). Then thefollowing statements are equivalent:(a) P is a primary submodule of M .(b) ( P : R M ) is a primary ideal of R and P is an ( R − m )-primary submoduleof M for each m ∈ M ax ( R ). Proof.
It is clear from Theorem 2.25. (cid:3)
Remark 2.27. (a) Suppose that M is an R -module. The idealization R (+) M = { ( a, m ) | a ∈ R, m ∈ M } of M is a commutative ring whose additionis component-wise and whose multiplication is defined as ( a, m )( b, m ′ ) =( ab, am ′ + bm ) for each a, b ∈ R and m, m ′ ∈ M . If S is a multiplicativelyclosed subset of R and P is a submodule of M , then S (+) P = { ( s, p ) | s ∈ S, p ∈ P } is a multiplicatively closed subset of R (+) M [2, 10].(b) Radical ideals of R (+) M have the form I (+) M , where I is a radical idealof R . If J is an ideal of R (+) M , then √ J = √ I (+) M . In particular, if I is an ideal of R and N is a submodule of M , then p I (+) N = √ I (+) M [2,Theorem 3.2 (3)]. Proposition 2.28.
Let M be an R -module and p be an ideal of R such that p ⊆ Ann ( M ). Then the following are equivalent:(a) p is a primary ideal of R .(b) p (+) M is a primary ideal of R (+) M . Proof.
This is straightforward. (cid:3)
Theorem 2.29.
Let S be a multiplicatively closed subset of R , p be an ideal of R provided p ∩ S = ∅ and M be an R -module. Then the following are equivalent: (a) p is an S -primary ideal of R . (b) p (+) M is an S (+)0 -primary ideal of R (+) M . (c) p (+) M is an S (+) M -primary ideal of R (+) M .Proof. (a) ⇒ (b). Let ( x, m )( y, m ′ ) = ( xy, xm ′ + ym ) ∈ p (+) M , where x, y ∈ R and m, m ′ ∈ M . Then we get xy ∈ p . As p is S -primary, there exists s ∈ S such that sx ∈ √ p or sy ∈ p . Now put s ′ = ( s, ∈ S (+)0. Then we have s ′ ( x, m ) = ( sx, sm ) ∈ √ p (+) M = p p (+) M or s ′ ( y, m ′ ) = ( sy, sm ′ ) ∈ p (+) M .Therefore, p (+) M is an S (+)0-primary ideal of R (+) M .(b) ⇒ (c). It is clear from Proposition 2.7.(c) ⇒ (a). Let xy ∈ p for some x, y ∈ R . Then ( x, y, ∈ p (+) M . Since p (+) M is S (+) M -primary, there exists s = ( s , m ) ∈ S (+) M such that s ( x,
0) =( s x, xm ) ∈ p p (+) M = √ p (+) M or s ( y,
0) = ( s y, ym ) ∈ p (+) M and hence weget s x ∈ √ p or s y ∈ p . Therefore p is an S -primary ideal of R . (cid:3) Remark 2.30.
Let M be an R -module and let S be a multiplicatively closed subsetof R such that Ann R ( M ) ∩ S = ∅ . We say that M is an S - torsion-free module inthe case that there is an s ∈ S such that if rm = 0, where r ∈ R and m ∈ M , then sm = 0 or sr = 0 [11, Definition 2.23]. Proposition 2.31.
Let M be an R -module. Assume that P is a submodule of M and S is a multiplicatively closed subset of R such that Ann R ( M ) ∩ S = ∅ . Then P is an S -primary submodule of M if and only if the factor module M/P is a π ( S )-torsion-free R/ p ( P : R M )-module, where π : R → R/ p ( P : R M ) is thecanonical homomorphism. Proof.
Suppose that P is an S -primary submodule of M . Let am = 0 M/P , where a = a + p ( P : R M ) and m = m + P for some a ∈ R and m ∈ M . This yields that am ∈ P . As P is S -primary, there exists s ∈ S such that sa ∈ p ( P : R M ) or sm ∈ P . Then we can conclude that π ( s ) a = 0 R/ √ ( P : R M ) or π ( s ) m = 0 M/P . Therefore,
M/P is a π ( S )-torsion-free R/ p ( P : R M )-module. For the other direction, supposethat M/P is a π ( S )-torsion-free R/ p ( P : R M )-module. Let am ∈ P ,where a ∈ R and m ∈ M . Put a = a + p ( P : R M ) and m = m + P . Then note that am = 0 M/P .As
M/P is a π ( S )-torsion-free R/ p ( P : R M )-module, there exists s ∈ S such that π ( s ) a = 0 R/ √ ( P : R M ) or π ( s ) m = 0 M/P . This yields that sa ∈ p ( P : R M ) or sm ∈ P . Accordingly, P is an S -primary submodule of M . (cid:3) Definition 2.32.
Let M be an R -module and let S be a multiplicatively closedsubset of R such that Ann R ( M ) ∩ S = ∅ . We say that M is a quasi S - torsion-freemodule , if there exists s ∈ S such that whenever rm = 0, where r ∈ R and m ∈ M ,then sm = 0 or ( sr ) t = 0 for some t ∈ N .According to Definition 2.32, Proposition 2.31 can be expressed as follows. Proposition 2.33.
Let M be an R -module. Assume that P is a submodule of M and S is a multiplicatively closed subset of R such that Ann R ( M ) ∩ S = ∅ .Then P is an S -primary submodule of M if and only if the factor module M/P isa quasi π ′ ( S )-torsion-free R/ ( P : R M )-module, where π ′ : R → R/ ( P : R M ) is thecanonical homomorphism. Theorem 2.34.
Let M be a module over an integral domain R . The following areequivalent: (a) M is a torsion-free module; (b) M is a quasi ( R − p ) -torsion-free module for each p ∈ Spec ( R ) ; (c) M is a quasi ( R − m ) -torsion-free module for each m ∈ M ax ( R ) .Proof. (a) ⇒ (b). It is clear.(b) ⇒ (c). It is clear.(c) ⇒ (a). Assume that a = 0. Take m ∈ M ax ( R ). As M is quasi ( R − m )-torsion-free, there exists s m = m so that s m m = 0 or ( s m a ) t = 0 for some t ∈ N . As R is an integral domain, ( s m a ) t = 0. Now, put Ω = { s m ∈ R | ∃ m ∈ M ax ( R ) , s m / ∈ m and s m m = 0 } . A similar argument in the proof of Theorem 2.25 shows thatΩ = R . Then we have ( s m ) + ( s m ) + · · · + ( s m n ) = R for some ( s m i ) ∈ Ω. thisimplies that Rm = P ni =1 ( s m i ) m = (0) and hence m = 0. This means M is atorsion-free module. (cid:3) References [1] R Ameri. On the prime submodules of multiplication modules.
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Department of pure Mathematics, Faculty of mathematical Sciences, University ofGuilan, P. O. Box 41335-19141 Rasht, Iran.
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