# S-Semiprime Submodules and S-Reduced Modules

Authors: Ayten Pekin; Ünsal Tekir; Özge Kılıç
Journal: Journal of Mathematics
 (2020)

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## Abstract

This article introduces the concept ofS-semiprime submodules which are a generalization of semiprime submodules and S-prime submodules. Let M be a nonzero unital R-module, where R is a commutative ring with a nonzero identity. Suppose that S is a multiplicatively closed subset of R. A submodule P of M is said to be an S-semiprime submodule if there exists a fixed s∈S, and whenever rnm∈P for some r∈R,m∈M, and n∈ℕ, then srm∈P. Also, M is said to be an S-reduced module if there exists (fixed) s∈S, and whenever rnm=0 for some r∈R,m∈M, and n∈ℕ, then srm=0. In addition, to give many examples and characterizations of S-semiprime submodules and S-reduced modules, we characterize a certain class of semiprime submodules and reduced modules in terms of these concepts.


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## Body

## 1. Introduction

In this article, all rings are assumed to be commutative with a nonzero identity, and all modules are assumed to be nonzero unital. LetR always denote such a ring and M always denote such an R-module. Recalling from [1], an R-module M is said to be a reduced module if a2m=0 for each a∈R and m∈M implying that am=0. Note that M is a reduced module if and only if anm=0 for some a∈R,m∈M, and n∈ℕ implying that am=0. Let P be a submodule of M.P is said to be a semiprime submodule; if a2m∈P, where a∈R and m∈M, then am∈P [2]. It is easy to see that M is a reduced module if and only if the zero submodule is semiprime. Also, it is clear that a submodule P of M is semiprime if and only if anm∈P for some a∈R,m∈M, and n∈ℕ implying that am∈P. As a generalization of the prime submodule (torsion-free module), the notion of the semiprime submodule (reduced module) has been widely studied in many papers. See, for example, [1–5]. Our aim, in this paper, is to introduce S-semiprime submodules and S-reduced modules which are generalizations of semiprime submodules and reduced modules, respectively. For the sake of completeness, we begin by giving some notions and notations which will be used throughout the paper. SpecR and MaxR denote the set of all prime ideals and maximal ideals of R, respectively. A nonempty subset S of R is said to be a multiplicatively closed set (briefly, m.c.s) if 1∈S and S is a subsemigroup of R under multiplication. Let P be a submodule of M,L be a nonempty subset of M, and J be a nonempty subset of R; the residuals of P by L and J are defined as follows:(1)P:LR=x∈R:xL⊆P,P:JM=m∈M:Jm⊆P.In particular, ifJ=a is the singleton, where a∈R, we use P:aR instead of P:aR, and also, we use annM to denote 0:RM. For each x∈R, the set 0:xM is denoted by annMx, and if annMx≠0,x∈R is called a zero divisor on M. Furthermore, the set of all zero divisors on M is denoted by zM=x∈R:annMx≠0. It is clear that the set of all units uR of R and R−zM are always a m.c.s of R.The concepts of prime ideals/submodules have a distinguished place in commutative algebra. Since certain class of rings and modules are characterized in terms of prime ideals/submodules, they have been widely studied by many authors. See, for example, [6–11]. Recently, Sevim et al., in [12], introduced S-prime submodules and S-torsion-free modules and used them to characterize certain prime submodules and torsion-free modules. Let S be a m.c.s of R. A submodule P of M is said to be an S-prime submodule if there exists a fixed s∈S, and whenever am∈P, then either sa∈P:MR or sm∈P for each a∈R and m∈M. Note that if P:MR∩S≠∅, then P is (trivially) an S-prime submodule, and so, the authors in [12] defined S-prime submodules with the condition that P:RM∩S=∅ to avoid the trivial case. Similarly, an R-module M with annM∩S=∅ is said to be an S-torsion-free module if there exists a fixed s∈S, and whenever am=0 for some a∈R and m∈M, then either sa=0 or sm=0. They showed in [12], Theorem 2.26, that R-module M with annM∩S=∅ is a simple module if and only if its each proper submodule is an S-prime submodule.LetS be a m.c.s of R and P be a submodule of M. Then, we call P an S-semiprime submodule if there exists a fixed s∈S, and whenever anm∈P for some a∈R,m∈M, and n∈ℕ, then sam∈P. Also, M is said to be an S-reduced module if there exists (fixed) s∈S, and whenever anm=0 for some a∈R,m∈M, and n∈ℕ, then sam=0. To avoid the trivial case, we assume that P:RM∩S=∅ for each S-semiprime submodule P of M and annM∩S=∅ for each S-reduced module M. Among other results in this paper, we show that the class of S-semiprime submodules properly contains the class of semiprime submodules and the class of S-prime submodules (see Proposition 1 and Examples 1 and 2). Also, we show that, in Proposition 2, if P is an S-semiprime submodule of M, then S−1P is a semiprime submodule of the quotient module S−1M of M. Also, we investigate the behaviour of S-semiprime submodules under homomorphism, in factor modules, and in Cartesian products of modules (see Proposition 5, Corollary 3, and Theorems 2 and 3). An R-module M is said to be a multiplication module if each submodule P of M has the form P=IM for some ideal I of R. In Theorem 1, we determine all S-semiprime submodules of finitely generated multiplication modules. Also, we characterize certain semiprime submodules of an arbitrary module in terms of S-semiprime submodules (see Theorem 5). Using Theorem 5, we determine all semiprime submodules of modules over quasi-local rings in terms of S-semiprime submodules (see Corollary 4). Finally, we characterize reduced modules in terms of S-reduced modules (see Theorem 6).


## 2. Characterization ofS-Semiprime Submodules

Definition 1.
LetP be a submodule of M with P:RM∩S=∅, where S⊆R is a m.c.s of R. P is said to be an S-semiprime submodule if there exists a fixed s∈S, and whenever rnm∈P for some r∈R,m∈M, and n∈ℕ, then srm∈P.
LetS be a m.c.s of R. If we consider the ring R as a module over itself, then we say that P is an S-semiprime ideal if it is an S-semiprime submodule of R. Note that an ideal P of R with P∩S=∅ is an S-semiprime ideal if and only if there exists (fixed) s∈S, and whenever an∈P for some a∈R and n∈ℕ, then sa∈P.Proposition 1.
LetS⊆R be a m.c.s of R and M be an R-module. The following statements are satisfied:(i)
IfP is a semiprime submodule of M provided that P:RM∩S=∅, then P is an S-semiprime submodule of M(ii)
IfP is an S-semiprime submodule of M and S⊆uR, then P is a semiprime submodule of M(iii)
EveryS-prime submodule is also an S-semiprime submoduleProof.
(i)
LetP be a semiprime submodule of M and P:RM∩S=∅. Now, we will show that P is an S-semiprime submodule of M. To see this, take r∈R and m∈M such that rnm∈P for some n∈ℕ. Since P is a semiprime submodule, we have rm∈P, and this implies that srm∈P for each s∈S. Then, P is an S-semiprime submodule of M.(ii)
LetP be an S-semiprime submodule of M. Then, we know that rnm∈P, where r∈R,m∈M, and n∈ℕ, implies that srm∈P for a fixed s∈S. Since S⊆uR, there is a t∈R such that st=1, and so, rm=tsrm∈P. Therefore, P is a semiprime submodule of M.(iii)
Suppose thatP is an S-semiprime submodule of M. Let rnm∈P for some r∈R,m∈M, and n∈ℕ. Since P is an S-prime submodule, there exists a fixed s∈S such that sr∈P:RM or srn−1m∈P. If sr∈P:RM, then srm∈P, and so, the proof is completed. Now, assume that sr∉P:RM, that is, srn−1m=rsrn−2m∈P. This implies that s2rn−2m∈P. If we continue in this manner, we conclude that snm∈P. As P is an S-prime submodule, we get either sn+1∈P:RM or sm∈P. As P:RM∩S=∅ and sn+1∈S, we have sm∈P, and so, srm∈P.
The converses of Proposition1 (i) and (iii) need not be true. See the following examples.Example 1.
Consider theℤ-module ℤ×ℤ4 and the zero submodule P=0×0¯. First, note that P:ℤ×ℤ4=0 and 220,1¯=0,0¯∈P. Since 20,1¯∉P, P is not a semiprime submodule of ℤ×ℤ4. Now, take the m.c.s S=ℤ∗=ℤ−0 of ℤ, and put s=4∈S. Now, we will show that P is an S-semiprime submodule. To see this, let akx,m¯∈P for some a,x,m∈ℤ and k∈ℕ. Then, we have akx=0. If a=0, then sax,m¯=0,0¯∈P. Otherwise, we have x=0, and so, sax,m¯=0,0¯. Therefore, P is an S-semiprime submodule.Example 2.
LetR=ℤ and M=ℤpqr, where p,q,andr are distinct prime numbers. Consider the multiplicatively closed subset S=pn:n∈ℕ∪1 of R. Take the submodule P=0¯. Then, note that qpr¯=0¯, and also, pnq∉P:RM=pqrℤ and pnpr¯∉P for any pn∈S. Thus, P is not an S-prime submodule. Also, it is clear that P is an S-semiprime submodule of M.
LetS⊆R be a m.c.s of R. Then, S−1R=λ:λ=a/s,∃a∈R,s∈S is called the quotient ring of R. For any m.c.s S of R, the saturation S∗ of S is defined as a∈R:a/1is a unit ofS−1R [13]. Note that S∗ is a m.c.s of R containing S.Proposition 2.
LetS⊆R be a m.c.s of R and M be an R-module. The following statements hold:(i)
IfS1⊆S2 is a m.c.s of R and P is an S1-semiprime submodule of M, then P is an S2-semiprime submodule of M in case P:RM∩S2=∅(ii)
P is an S-semiprime submodule of M if and only if P is an S∗-semiprime submodule of M(iii)
IfP is an S-semiprime submodule of M, then S−1P is a semiprime submodule of S−1MProof.
(i)
It is straightforward.(ii)
LetP:RM∩S=∅. It is clear that S⊆S∗. Then, we need to show that P:RM∩S∗=∅. Assume that P:RM∩S∗≠∅. So, there exists a∈P:RM∩S∗. Since a∈S∗,a/1 is a unit of S−1R, and so, a/1x/y=1 for some x/y∈S−1R. This yields that axu=yu for some u∈S. Now, put s′=yu∈S. Then, s′=uax=yu∈P:RM∩S, a contradiction. Thus, P:RM∩S∗=∅. So, by (i), P is an S∗-semiprime submodule of M. For the converse, let P be an S∗-semiprime submodule of M. Then, P:RM∩S∗=∅, and thus, P:RM∩S=∅. Let rnm∈P for some r∈R,m∈M, and n∈ℕ. Since P is an S∗-semiprime submodule, there is an s′∈S∗ such that s′rm∈P. As s′∈S∗, there exists x/y∈S−1R such that s′/1x/y=1. Then, we conclude that us′x=yu for some u∈S. Now, take s=yu∈S. Then, we get srm=yurm=uxs′rm∈P, and hence, P is an S-semiprime submodule of M.(iii)
Suppose thatP is an S-semiprime submodule of M. Let r/skm/t=rkm/skt∈S−1P for some r/s∈S−1R,m/t∈S−1M, and k∈ℕ. Then, there exists u∈S such that urkm=rkum∈P. As P is an S-semiprime submodule of M, we get us′rm∈P for some s′∈S. This implies that r/sm/t=us′rm/us′st∈S−1P. Hence, S−1P is a semiprime submodule of S−1M.
The following example shows that the converse of Proposition2 (iii) is not true in general.Example 3.
LetR=ℤ and M=ℚ×ℚ, where ℚ is the field of rational numbers. Take the submodule P=ℤ×0 and the m.c.s S=ℤ∗=ℤ−0 of ℤ. It is easy to see that S−1M is a vector space over S−1R, and thus, S−1P is a prime (semiprime) submodule of S−1M. Now, we will show that P is not S-semiprime. Let s be an arbitrary element of S. Choose a prime number p with gcdp,s=1. Then, note that p21/p2,0=1,0∈P and sp1/p2,0=sp/p2,0∉P. Thus, P is not an S-semiprime submodule of M.Proposition 3.
LetM be an R-module and S=s1,s2,…,sn be a finite m.c.s of R. Suppose that P is a submodule of M provided that P:RM∩S=∅. Then, P is an S-semiprime submodule of M if and only if S−1P is a semiprime submodule of S−1M.Proof.
Suppose thatP is an S-semiprime submodule of M. Then, by Proposition 2 (iii), S−1P is a semiprime submodule of S−1M. For the converse, take a semiprime submodule S−1P of S−1M. Let rnm∈P for some r∈R,m∈M, and n∈ℕ. Then, we have r/1nm/1=rnm/1∈S−1P. Since S−1P is a semiprime submodule of S−1M, we conclude that r/1m/1=rm/1∈P, and this yields that sirm∈P for some si∈S. Now, put s=s1s2⋯sn∈S. Then, we conclude that srm∈P, and so, P is an S-semiprime submodule of M.Lemma 1.
SupposeP is a submodule of M and S is a m.c.s of R provided that P:RM∩S=∅. The following statements are equivalent:(i)
P is an S-semiprime submodule of M(ii)
There is a fixeds∈S and JkN⊆P for some k∈ℕ implying that sJN⊆P for each ideal J of R and submodule N of MProof.
i⟹ii: let P be an S-semiprime submodule of M. Suppose that JkN⊆P for some ideal J of R, some submodule N of M, and k∈ℕ. Now, we will show that sJN⊆P. Suppose to the contrary. Then, there exist a∈Jandm∈N such that sam∉P. Since akm∈JkN⊆P and P is an S-semiprime submodule of M, we conclude that sam∈P, a contradiction. Therefore, sJN⊆P.ii⟹i: conversely, let rnm∈P for some r∈R,m∈M, and k∈ℕ. Now, put J=Rr and N=Rm. Then, we have JnN=Rrnm⊆P. Hence, by assumption, sJN=Rsrm⊆P for a fixed s∈S, and so, srm∈P. Then, P is an S-semiprime submodule of M.
As immediate consequences of the previous lemma, we give the following corollary which will be used in the sequel.Corollary 1.
Suppose thatS is a m.c.s of R and P is an ideal of R with P∩S=∅.The following statements are equivalent:(i)
P is an S-semiprime ideal of R(ii)
There is a (fixed)s∈SandIkJ⊆P for some ideals I,J of R and k∈ℕ implying that sIJ⊆PProposition 4.
LetM be an R-module and S be a m.c.s of R. Suppose that P is a submodule of M with P:RM∩S=∅. The following statements hold:(i)
IfP is an S-semiprime submodule of M, then P:RM is an S-semiprime ideal of R(ii)
IfM is a multiplication module and P:RM is an S-semiprime ideal of R, then P is an S-semiprime submodule of MProof.
(i)
Letxk∈P:RM for some x∈R and k∈ℕ. Then, we have xkm∈P for each m∈M. Since P is an S-semiprime submodule, we conclude that sxm∈P, and this yields that sx∈P:RM. Therefore, P:RM is an S-semiprime ideal of R.(ii)
Assume thatM is a multiplication module and P:RM is an S-semiprime ideal of R. Let JkN⊆P for some ideal J of R, submodule N of M, and k∈ℕ. Then, we conclude that JkN:RM⊆JkN:RM⊆P:RM. Also, note that, by Corollary 1, there exists a fixed s∈S such that sJN:RM⊆P:RM. Since M is a multiplication module, we have sJN=sJN:RMM⊆P:RMM=P. Then, by Lemma 1, P is an S-semiprime submodule of M.Corollary 2.
Suppose thatP is a submodule of a multiplication R-module M and S is a m.c.s of R such that P:RM∩S=∅. Then, the following statements are equivalent:(i)
P is an S-semiprime submodule of M(ii)
There exists a fixeds∈S such that LkN⊆P for some submodules L,N of M and k∈ℕ implying sLN⊆PProof.
i⟹ii: suppose that P is an S-semiprime submodule of M. Let LkN⊆P for some submodules L,N of M and k∈ℕ. Since M is a multiplication module, L=IM and K=JM for some ideal I,J of R. Also, note that LkN=IkJM=IkN⊆P. Since P is an S-semiprime submodule, by Lemma 1, there exists s∈S such that sIN⊆P, and this yields that sIJM=sLN⊆P.ii⟹i: suppose that JkN⊆P for some ideal J of R, submodule N of M, and k∈ℕ. Then, we have JkN=JkN:RMM. Now, put L=JM, and note that Lk=JkM. This implies that JkN=JkN:RMM=LkN⊆P. Then, by assumption, there exists a fixed s∈S such that sLN=sJN:MM=sJN⊆P. Then, by Lemma 1, P is an S-semiprime submodule of M.Theorem 1.
LetP be a submodule of a finitely generated multiplication R-module M and S be a m.c.s of R with P:RM∩S=∅. The following statements are equivalent:(i)
P is an S-semiprime submodule of M(ii)
P:RM is an S-semiprime ideal of R(iii)
P=IM for some S-semiprime ideal I of R with annM⊆IProof.
i⟹ii: follows from Proposition 4 (i).ii⟹iii: it is straightforward.iii⟹i: suppose that P=IM for some S-semiprime ideal I of R with annM⊆I. Assume that JkN⊆P for some ideal J of R, some submodule N of M, and k∈ℕ. Then, we obtain JkN:RMM⊆IM. As M is a finitely generated multiplication module, by [14], Theorem 9 Corollary, we have JkN:RM⊆I+annM=I. Since I is an S-semiprime ideal of R, by Corollary 1, for a fixed s∈S such that sJN:RM⊆I, this yields that sJN=sJN:RMM⊆IM=P. Then, by Lemma 1, P is an S-semiprime submodule of M.Proposition 5.
Leth:M⟶M′ be an R-homomorphism. The following statements are satisfied:(i)
IfP′ is an S-semiprime submodule of M′ such that h−1P′:RM∩S=∅, then h−1P′ is an S-semiprime submodule of M(ii)
Ifh is an epimorphism and P is an S-semiprime submodule of M such that Kerh⊆P, then hP is an S-semiprime submodule of M′Proof.
(i)
Letrnm∈h−1P′ for some r∈R,m∈M, and n∈ℕ. Then, we conclude that hrnm=rnhm∈P′. Since P′ is an S-semiprime submodule, we have srhm=hsrm∈P′ for some s∈S. Then, we get srm∈h−1P′, and thus, h−1P′ is an S-semiprime submodule of M.(ii)
Letrnm′∈hP for some r∈R,m′∈M′, and n∈ℕ. As h is an epimorphism, we can write m′=hm for some m∈M, and so, rnm′=rnhm=hrnm∈hP. Since Kerh⊆P, we conclude that rnm∈P. Since P is an S-semiprime submodule, there exists s∈S such that srm∈P, and so, we obtain hsrm=srhm=srm′∈hP. Consequently, hP is an S-semiprime submodule of M′.Corollary 3.
Suppose thatS is a m.c.s of R and L is a submodule of M. Then, the following statements are satisfied:(i)
IfP′ is an S-semiprime submodule of M with P′:RL∩S=∅, then L∩P′ is an S-semiprime submodule of L.(ii)
Suppose thatP is a submodule of M with L⊆P. Then, P is an S-semiprime submodule of M if and only if P╱L is an S-semiprime submodule of M╱L.Proof.
(i)
Consider the injectioni:L⟶M defined by im=m for all m∈L. Then, note that i−1P′=L∩P′. Now, we will show that i−1P′:RL∩S=∅. Assume that s∈i−1P′:RL∩S. Then, we have sL⊆i−1P′=L∩P′⊆P′, and thus, s∈P′:RL∩S, a contradiction. By Proposition 5 (i), we can say that L∩P′ is an S-semiprime submodule of L.(ii)
Assume thatP is an S-semiprime submodule of M. Consider the natural epimorphism π:M⟶M/L, defined by πm=m+L, for all m∈M. By Proposition 5 (ii), P/L is an S-semiprime submodule of M/L. For the converse, let P/L be an S-semiprime submodule of M/L. Take r∈R and m∈M with rkm∈P for some k∈ℕ. Then, we get rkm+L=rkm+L∈P/L. Since P/L is an S-semiprime submodule of M/L, for a fixed s∈S, we conclude that srm+L=srm+L∈P/L. This implies that srm∈P, and hence, P is an S-semiprime submodule of M.
LetMi be an Ri module and Si be a m.c.s of Ri for each i=1,2,…,n, where n∈ℕ. Suppose that M=M1×M2×⋯×Mn,R=R1×R2×⋯×Rn and S=S1×S2×⋯×Sn. Then, M is an R-module with componentwise addition and scalar multiplication, and note that S is a m.c.s of R. Also, each submodule P of M has the form P=P1×P2×⋯×Pn, where Pi is a submodule of Mi for each i=1,2,…,n.Theorem 2.
Suppose thatMi is an Ri-module, Pi is a submodule of Mi, and Si is a m.c.s of Ri for each i=1,2. Let M=M1×M2,R=R1×R2 and S=S1×S2. The following statements are equivalent for P=P1×P2:(i)
P is an S-semiprime submodule of M.(ii)
P1 is an S1-semiprime submodule of M1 and P2:R2M2∩S2≠∅ or P1:R1M1∩S1≠∅, and P2 is an S2-semiprime submodule of M2 or Pi is an Si-semiprime submodule of Mi for each i=1,2.Proof.
i⟹ii: let P be an S-semiprime submodule of M. Then, we have P:RM∩S=∅, and this yields that P1:R1M1∩S1=∅ or P2:R2M2∩S2=∅. Without loss of generality, we may assume that P2:R2M2∩S2=∅ and P1:R1M1∩S1≠∅. We must show that P2 is an S2-semiprime submodule of M2. To prove this, take r∈R2andm∈M2 such that rkm∈P2 for some k∈ℕ. Then, 1,rk0,m=0,rkm∈P. Since P is an S-semiprime submodule of M, there exists a fixed s=s1,s2∈S such that s1,r0,m=0,s2rm∈P. This implies that s2rm∈P2. Hence, P2 is an S2-semiprime submodule of M2. One can similarly show that if P2:R2M2∩S2≠∅, then P1 is an S1-semiprime submodule of M1. Also, if P1:R1M1∩S1=P2:R2M2∩S2=∅, then a similar argument shows that Pi is an Si-semiprime submodule of Mi for each i=1,2.ii⟹i: let P1:R1M1∩S1≠∅ and P2 be an S2-semiprime submodule of M2. Then, we have s1∈P1:R1M1∩S1. Let r1,r2nm1,m2=r1nm1,r2nm2∈P for some ri∈Riandmi∈Mi, where i=1,2. This implies that r2nm2∈P2, and so, s2r2m2∈P2 for a fixed s2∈S2 since P2 is an S2-semiprime submodule of M2. Now, take s=s1,s2∈S. Then, sr1,r2m1,m2=s1r1m1,s2r2m2∈P. Similarly, we can show that P is an S-semiprime submodule of M in other cases.Theorem 3.
Letn≥1,Mi be an Ri module and Si be a m.c.s of Ri for each i=1,2,…,n. Suppose that M=M1×M2×⋯×Mn,R=R1×R2×⋯×Rn, and S=S1×S2×⋯×Sn. Let P=P1×P2×⋯×Pn, where Pi is a submodule of Mi for each i=1,2,…,n. The following statements are equivalent:(i)
P is an S-semiprime submodule of M(ii)
Pi is an Si-semiprime submodule of Mi for each i∈t1,t2,…,tk:1≤k≤n, and Pj:RjMj∩Sj≠∅ for each j∈1,2,…n−t1,t2,…,tkProof.
We use mathematical induction to prove the claimi⟺ii. For n=1, the result is clear. If n=2, the claim i⟺ii follows from Theorem 2. Suppose that i and ii are equivalent for each k<n. Now, we will show that the claim i⟺ii is true for k=n. Let P′=P1×P2×⋯×Pn−1,M′=M1×M2×⋯×Mn−1 and also S′=S1×S2×⋯×Sn−1 and R′=R1×R2×⋯×Rn−1. Note that P=P′×Pn,M=M′×Mn and also S=S′×Sn,R=R′×Rn. Then, by Theorem 2, P is an S-semiprime submodule of M if and only if P′ is an S′-semiprime submodule of M′ and Pn:RnMn∩Sn≠∅ or P′:R′M′∩S′≠∅ and Pn is an Sn-semiprime submodule of Mn or P′ is an S′-semiprime submodule of M′, and Pn is an Sn-semiprime submodule of Mn. The rest follows from induction hypothesis.Theorem 4.
LetP be a submodule of M and S be a m.c.s of R such that P:RM∩S=∅. Then, P is an S-semiprime submodule of M if and only if P:Ms′ is a semiprime submodule of M for some s′∈S.Proof.
LetP be an S-semiprime submodule of M, and put s′=s2∈S. Now, we will show that P:Ms′ is a semiprime submodule of M. Let anm∈P:Ms′ for some a∈R,m∈M and n∈ℕ. Then, we get s′anm=s2anm∈P. If n=1, then we have am∈P:Ms′. So, assume that n≥2. Then, clearly we have snanm=sanm∈P, and this gives ssam=s2am∈P. Then, we conclude that am∈P:Ms2=P:Ms′. Hence, P:Ms′ is a semiprime submodule of M. Conversely, assume that P:Ms′ is a semiprime submodule of M for some s′∈S. Let anm∈P for some a∈R,m∈M, and n∈ℕ. Since P:Ms′ is a semiprime submodule and anm∈P⊆P:Ms′, we conclude that am∈P:Ms′, and hence, s′am∈P. Therefore, P is an S-semiprime submodule of M.Theorem 5.
LetP be a submodule of M such that P:RM⊆JacR, where JacR is the Jacobson radical of R. The following statements are equivalent:(i)
P is a semiprime submodule of M(ii)
P is an Sm=R−m-semiprime submodule of M for each maximal ideal m of RProof.
i⟹ii: suppose that P is a semiprime submodule of M. Then, by Proposition 1, P is an Sm=R−m-semiprime submodule of M for each maximal ideal m of R.ii⟹i: let rnm∈P for some r∈R,m∈M, and n∈ℕ. As P is an Sm=R−m-semiprime submodule of M, there exists an sm∉m such that smrm∈P. Now, consider the set Ω=sm:∃m∈MaxR,sm∉msuch thatsmrm∈P. Now, we will show that Ω=R. Assume that Ω≠R. Then, there exists a maximal ideal m∗ of R containing Ω. By the definition of Ω, there exists sm∗∈Ω such that sm∗∉m∗. Since Ω⊆Ω⊆m∗, we have sm∗∈m∗ which is a contradiction. Thus, Ω=R, and so, there exists sm1,sm2,…,smn∈Ω such that 1=x1sm1+x2sm2+⋯+xnsmn for some x1,x2,…,xn∈R. Since smirm∈P for each i=1,2,…,n, we conclude that rm=x1sm1rm+x2sm1rm+⋯+xnsmnrm∈P. Therefore, P is an S-semiprime submodule of M.
As immediate consequence of the previous theorem, we give the following result.Corollary 4.
LetM be a module over a quasi-local ring R,m. Suppose that P is a submodule of M. The following statements are equivalent:(i)
P is a semiprime submodule of M(ii)
P is an Sm=R−m-semiprime submodule of MDefinition 2.
LetM be an R-module and S be a m.c.s of R.M is said to be an S-reduced module if there exists s∈S, and whenever rnm=0, where r∈R,m∈M, and n∈ℕ, then srm=0.Proposition 6.
Suppose thatM is an R-module and S is a m.c.s of R. The following statements are satisfied:(i)
IfM is a reduced module, then M is an S-reduced module. In particular, the converse holds if S⊆R−zM, where zM=x∈R:annMx≠0M.(ii)
IfM is an S-torsion-free module, then M is an S-reduced module.(iii)
M is an S-reduced module if and only if the zero submodule is an S-semiprime submodule.(iv)
LetP be a submodule of M with P:RM∩S=∅. Then, P is an S-semiprime submodule if and only if R-module M/P is an S-reduced module.(v)
IfM is an S-reduced module, then S−1M is a reduced module.Proof.
(i)
The claim “reduced module implies theS-reduced module” is obvious. Let M be an S-reduced module such that S⊆R−zM. Let anm=0 for some a∈R, m∈M, and n∈ℕ. Since M is an S-reduced module, there exists s∈R−zM such that sam=0. As annMs=0M, we have am∈annMs=0, and so, am=0. Hence, M is a reduced module.(ii)
LetM be an S-torsion-free module and anm=aan−1m=0 for some a∈R,m∈M, and n∈ℕ. Since M is an S-torsion-free module, there exists s∈S such that sa=0 or san−1m=0. If sa=0, then sam=0 which completes the proof. So, assume that sa≠0. Since san−1m=asan−2m=0, we conclude that ssan−2m=s2an−2m=0 since M is an S-torsion-free module. If we continue in the previous way, we conclude that snm=0. Since M is an S-torsion-free module, we get either sn+1=0 or sm=0. If sn+1=0, then sn+1=0∈S, which is a contradiction so that sm=0, and this yields sam=0.(iii)
It follows from Definitions1 and 2.(iv)
It follows from (iii).(v)
LetM be an S-reduced module. Then by (iii), 0 is an S-semiprime submodule of M. Again by Proposition 2 (iii), S−10 is a semiprime submodule of S−1M. Thus, S−1M is a reduced module.
Now, we will characterize reduced modules in terms ofS-reduced modules.Theorem 6.
The following statements are equivalent for anyR-module M:(i)
M is a reduced module(ii)
M is an Sp=R−p-reduced module for each P∈SpecR(iii)
M is an Sm=R−m-reduced module for each m∈MaxRProof.
i⟹ii: it follows from Proposition 6.ii⟹iii: it follows from the fact that MaxR⊆SpecR.iii⟹i: let M be an Sm=R−m-reduced module for each m∈MaxR. Choose a∈R and m∈M such that a2m=0. Since M is an Sm=R−m-reduced module for each m∈MaxR, there exists sm∉m such that smam=0. Now, consider the set Ω=sm:∃m∈MaxR,sm∉m and smam=0. Note that Ω is not empty since M is an Sm=R−m-reduced module. Similar argument in Theorem 5 shows that Ω=R, and so, there exists s1,s2,…,sn∈Ω such that r1s1+r2s2+⋯+rnsn=1 and siam=0. This yields that am=r1s1+r2s2+⋯+rnsnam=r1s1am+r2s2am+⋯+rnsnam=0. Hence, M is a reduced module.


## 3. Conclusion

This paper is mainly concerned withS-semiprime submodules of modules over commutative rings. We first investigate some properties of S-semiprime submodules similar to semiprime submodules. Then, we introduce S-reduced modules and give some new characterizations of semiprime submodules and reduced modules in terms of these concepts.



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*Source: 8824787-2020-10-21.xml*