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arXiv:2110.14639v1 [math.AC] 27 Oct 2021
ON WEAKLY S-PRIME SUBMODULES
HANI A. KHASHAN AND ECE YETKIN CELIKEL
Abstract. Let Rbe a commutative ring with a non-zero identity, Sbe a
multiplicatively closed subset of Rand Mbe a unital R-module. In this
paper, we define a submodule Nof Mwith (N:RM)∩S=φto be weakly
S-prime if there exists s∈Ssuch that whenever a∈Rand m∈Mwith
06=am ∈N, then either sa ∈(N:RM) or sm ∈N. Many properties,
examples and characterizations of weakly S-prime submodules are introduced,
especially in multiplication modules. Moreover, we investigate the behavior of
this structure under module homomorphisms, localizations, quotient modules,
cartesian product and idealizations. Finally, we define two kinds of submodules
of the amalgamation module along an ideal and investigate conditions under
which they are weakly S-prime.
1. Introduction
Throughout this paper, unless otherwise stated, Rdenotes a commutative ring
with non-zero identity and Mis a unital R-module. It is well-known that a proper
submodule Nof Mis called prime if rm ∈Nfor r∈Rand m∈Mimplies
r∈(N:RM) or m∈Nwhere (N:RM) = {r∈R|rM ⊆N}. Since prime ideals
and submodules have a vital role in ring and module theory, several generalizations
of these concepts have been studied extensively by many authors (see, for example,
[3], [6], [13], [16], [18], [19]).
In 2007, Atani and Farzalipour introduced the concept of weakly prime submod-
ules as a generalization of prime submodules. Following [7], a proper submodule N
of Mis said to be weakly prime if for r∈Rand m∈M, whenever 0 6=rm ∈N,
then r∈(N:RM) or m∈N. In 2019 a new kind of generalizations of prime
submodules has been introduced and studied by S¸engelen Sevim et. al. [18]. For
a multiplicatively closed subset Sof R, they called a proper submodule Nof an
R-module Mwith (N:RM)∩S=∅an S-prime if there exists s∈Ssuch that for
r∈Rand m∈M, whenever rm ∈N, then either sr ∈(N:RM) or sm ∈N. In
particular, an ideal Iof Ris called an S-prime ideal if Iis an S-prime submodule
of an R-module R, [13]. Recently, Almahdi et. al. generalized S-prime ideals by
defining the notion of weakly S-prime ideals. A proper ideal Iof Rdisjoint with
Sis said to be weakly S-prime if there exists s∈Ssuch that for a, b ∈Rand
06=ab ∈I, then either sa ∈Ior sb ∈I, [3].
Our objective in this paper is to define and study the concept of weakly S-prime
submodules as an extension of the above concepts. Let Sbe a multiplicatively closed
subset of R. We call a submodule Nof an R-module Mwith (N:RM)∩S=∅a
2010 Mathematics Subject Classification. Primary 13A15, 16P40, Secondary 16D60.
Key words and phrases. S–prime ideal, weakly S-prime ideal, S-prime submodule, weakly
S-prime submodule, amalgamated algebra.
This paper is in final form and no version of it will be submitted for publication elsewhere.
1
2 HANI A. KHASHAN AND ECE YETKIN CELIKEL
weakly S-prime submodule if there exists s∈Ssuch that for a∈Rand m∈M,
whenever 0 6=am ∈Nthen either sa ∈(N:RM) or sm ∈N. In the second section,
we obtain many equivalent statements to characterize this class of submodules (see
Theorems 1 and 2), particularly in multiplication modules (Theorem 4). Moreover,
various properties of weakly S-prime submodules are considered and many examples
are given for supporting the results (see for example Theorem 3, Propositions 2, 1
and Examples 1, 3). We investigate the behavior of this structure under module
homomorphisms, localizations, quotient modules, cartesian product of modules (see
Propositions 4, 8, Theorem 5 and Corollary 3). Let Sbe a multiplicatively closed
subset of R,Mbe an R-module and consider the idealization ring R⋉M. For any
submodule Kof M, the set S⋉K={(s, k) : s∈S,k∈K}is a multiplicatively
closed subset of R⋉M. In Theorem 7, we justify the relation among weakly S-
prime ideals of R, weakly S-prime submodules of Mand weakly S⋉K-prime ideals
of the the idealization ring R⋉M.
Let f:R1→R2be a ring homomorphism, Jbe an ideal of R2,M1be an
R1-module, M2be an R2-module (which is an R1-module induced naturally by f)
and ϕ:M1→M2be an R1-module homomorphism. The subring R1⋊⋉fJ=
{(r, f(r) + j) : r∈R1,j∈J}of R1×R2is called the amalgamation of R1and R2
along Jwith respect to f, [11]. The amalgamation of M1and M2along Jwith
respect to ϕis defined in [12] as
M1⋊⋉ϕJM2={(m1, ϕ(m1) + m2) : m1∈M1and m2∈J M2}
which is an (R1⋊⋉fJ)-module with the scaler product defined as
(r, f(r) + j)(m1, ϕ(m1) + m2) = (rm1, ϕ(rm1) + f(r)m2+jϕ(m1) + jm2)
For submodules N1and N2of M1and M2, respectively, the sets
N1⋊⋉ϕJM2={(m1, ϕ(m1) + m2)∈M1⋊⋉ϕJM2:m1∈N1}
and
N2
ϕ={(m1, ϕ(m1) + m2)∈M1⋊⋉ϕJM2:ϕ(m1) + m2∈N2}
are submodules of M1⋊⋉ϕJM2. Section 3 is devoted for studying several conditions
under which the submodules N1⋊⋉ϕJM2and N2
ϕof M1⋊⋉ϕJM2are (weakly) S-
prime submodules, (see Theorems 8, 10). Furthermore, we conclude some particular
results for the duplication of a module along an ideal (see Corollaries 4-6, 7-9 and
Theorem 9).
For the sake of completeness, we start with some definitions and notations which
will be used in the sequel. A non-empty subset Sof a ring Ris said to be a
multiplicatively closed set if Sis a subsemigroup of Runder multiplication. An
R-module Mis called multiplication provided for each submodule Nof M, there
exists an ideal Iof Rsuch that N=IM . In this case, Iis said to be a presentation
ideal of N. In particular, for every submodule Nof a multiplication module M,
ann(M/N) = (N:RM) is a presentation for N. The product of two submodules
Nand Kof a multiplication module Mis defined as N K = (N:RM)(K:RM)M.
For m1, m2∈M, by m1m2,we mean the product of Rm1and Rm2which is equal
to IJM for presentation ideals Iand Jof m1and m2, respectively [4]. Let Nbe
a proper submodule of an R-module M. The radical of N(denoted by M-rad(N))
is defined in [9] to be the intersection of all prime submodules of Mcontaining N.
If Mis multiplication, then M-rad(N) = {m∈M|mk⊆Nfor some k≥0}.As
usual, Z,Znand Qdenotes the ring of integers, the ring of integers modulo nand
ON WEAKLY S-PRIME SUBMODULES 3
the field of rational numbers, respectively. For more details and terminology, one
may refer to [1], [2], [8], [14], [15].
2. Characterizations of Weakly S-prime Submodules
We begin with the definitions and relationships of the main concepts of the paper.
Definition 1. Let Sbe a multiplicatively closed subset of a ring Rand Nbe a
submodule of an R-module Mwith (N:RM)∩S=∅. We call Na weakly S-prime
submodule if there exists (a fixed) s∈Ssuch that for a∈Rand m∈M, whenever
06=am ∈Nthen either sa ∈(N:RM)or sm ∈N. The fixed element s∈Sis
said to be a weakly S-element of N.
It is clear that every S-prime submodule is a weakly S-prime submodule. Since
the zero submodule is (by definition) a weakly S-prime submodule of any R-module,
then the converse is not true in general. For a less trivial example, let Mbe a non-
zero local multiplication R-module with the unique maximal submodule Ksuch
that (K:RM)K= 0. If we consider S={1R}, then every proper submodule of
Mis weakly S-prime, [1]. Hence, there is a weakly S-prime submodule in Mthat
is not S-prime.
Also, every weakly prime submodule Nof an R-module Msatisfying (N:R
M)∩S=∅is a weakly S-prime submodule of Mand the two concepts coincide if
S⊆U(R) where U(R) denotes the set of units in R. The following example shows
that the converse need not be true.
Example 1. Consider the Z-module M=Z×Z6and let N= 2Z× h¯
3i. Then
Nis a (weakly) S-prime submodule of Mwhere S={2n:n∈N∪{0}}. Indeed,
let (0,¯
0) 6=r.(r′, m)∈Nfor r, r′∈Zand m∈Z6such that 2r /∈(N:M) = 6Z.
Then r.m ∈ h¯
3iwith r /∈3Zand so m∈ h¯
3i. Thus, 2.(r′, m)∈Nas needed. On
the other hand, Nis not a weakly prime submodule since (0,¯
0) 6= 2.(1,¯
0) ∈Nbut
2/∈(N:M)and (1,¯
0) /∈N.
Let Nbe a submodule of an R-module Mand Ibe an ideal of R. The residual
of Nby Iis the set (N:MI) = {m∈M:Im ⊆N}. It is clear that (N:MI) is
a submodule of Mcontaining N. More generally, for any subset A⊆R, (N:MA)
is a submodule of Mcontaining N.
Theorem 1. Let Sbe a multiplicatively closed subset of a ring R. For a submodule
Nof an R-module Mwith (N:RM)∩S=∅, the following conditions are equivalent.
(1) Nis a weakly S-prime submodule of M.
(2) There exists s∈Ssuch that (N:Ma) = (0 :Ma) or (N:Ma)⊆(N:Ms)
for each a /∈(N:RsM).
(3) There exists s∈Ssuch that for any a∈Rand for any submodule Kof
M, if 0 6=aK ⊆N, then sa ⊆(N:RM) or sK ⊆N.
(4) There exists s∈Ssuch that for any ideal Iof Rand a submodule Kof
M, if 0 6=I K ⊆N, then sI ⊆(N:RM) or sK ⊆N.
Proof. (1)⇒(2). Let s∈Sbe a weakly S-element of Nand a /∈(N:MsM).
Let m∈(N:Ma). If am = 0, then clearly m∈(0 :Ma). If 0 6=am ∈N,
then, we conclude sm ∈Nas sa /∈(N:RM) and Nis weakly S-prime in M.
Thus, m∈(N:Ms) and so (N:Ma)⊆(0 :Ma)∪(N:Ms). Therefore,
(N:Ma)⊆(0 :Ma) (which implies (N:Ma) = (0 :Ma)) or (N:Ma)⊆(N:Ms).
4 HANI A. KHASHAN AND ECE YETKIN CELIKEL
(2)⇒(3). Choose s∈Sas in (2) and suppose 0 6=aK ⊆Nand sa /∈(N:RM)
for some a∈Rand a submodule Kof M. Then K⊆(N:Ma)\(0 :Ma) and by
(2) we get K⊆(N:Ma)⊆(N:Ms). Thus, sK ⊆Nas required.
(3)⇒(4). Choose s∈Sas in (3) and suppose 0 6=I K ⊆Nand sI *(N:RM)
for some ideal Iof Rand a submodule Kof M. Then there exists a∈Iwith
sa /∈(N:RM). If aK 6= 0, then by (3), we have sK ⊆Nas needed. Assume that
aK = 0. Since IK 6= 0, there is some b∈Iwith bK 6= 0. If sb /∈(N:RM),then
from (3), we have sK ⊆N. Now, assume that sb ∈(N:RM).Since sa /∈(N:RM),
we have s(a+b)/∈(N:RM). Hence, 0 6= (a+b)K⊆Nimplies sK ⊆Iagain by
(3) and we are done.
(4)⇒(1). Let a∈R, m ∈Mwith 0 6=am ∈N. The result follows directly by
taking I=aR and K=< m > in (4).
Theorem 2. Let Mbe a faithful multiplication R-module and Sbe a multiplica-
tively closed subset of R. The following are equivalent.
(1) Nis a weakly S-prime submodule of M.
(2) N∩SM =φand there exists s∈Ssuch that whenever K, L are submodules
of Mand 0 6=KL ⊆N, then sK ⊆Nor sL ⊆N.
Proof. Clearly, we have N∩SM =φif and only if (N:RM)∩S=∅.
(1)⇒(2). Let Ibe a presentation ideal of Kand sbe a weakly S-element of N.
Then 0 6=IL ⊆Nyields that either sI ⊆(N:RM) or sL ⊆Nby Theorem 1.
Hence, sK =sIM ⊆Nor sL ⊆N, as needed.
(2)⇒(1). Let s∈Sbe as in (2) and suppose 0 6=IL ⊆Nfor some ideal I
of Rand submodule Lof M. Put K=IM and assume that sL *N. Then
06=KL ⊆Nwhich implies sK ⊆N. Therefore, sI ⊆(N:RM) and the result
follows by Theorem 1.
Theorem 3. Let Nbe a submodule of an R-module Mand Sbe a multiplicatively
closed subset of R. The following statements hold.
(1) If Nis a weakly S-prime submodule of M, then for every submodule Kwith
(N:RK)∩S=∅and Ann(K) = 0,(N:RK) is a weakly S-prime ideal of
R. In particular, if Mis faithful, then (N:RM) is a weakly S-prime ideal
of R.
(2) If Mis multiplication and (N:RM) is a weakly S-prime ideal of R, then
Nis a weakly S-prime submodule of M.
(3) If Mis faithful multiplication and Iis an ideal of R, then Iis weakly
S-prime in Rif and only if IM is a weakly S-prime submodule of M.
(4) If Nis a weakly S-prime submodule of Mand Ais a subset of Rsuch that
(0 :MA) = 0 and Z(N:RM)(R)∩A=φ, then (N:MA) is a weakly S-prime
submodule of M.
Proof. (1) Suppose s∈Sis a weakly S-element of Nand let a, b ∈Rwith 0 6=ab ∈
(N:RK). Since Ann(K) = 0, we have 0 6=abK ⊆Nwhich implies sa ∈(N:RM)
or sbK ⊆Nby Theorem 1. Hence, sa ∈(N:RK) or sb ∈(N:RK). Thus,
(N:RK) is a weakly S-prime ideal associated with the same s∈S. The ”in
particular” part is clear.
(2) Suppose Mis multiplication and (N:RM) is a weakly S-prime ideal of R.
Let Ibe an ideal of Rand Kbe a submodule of Mwith 0 6=I K ⊆N. Since M
ON WEAKLY S-PRIME SUBMODULES 5
is multiplication, we may write K=JM for some ideal Jof R. Thus, 0 6=I J ⊆
(N:RM), and by [13, Theorem 1], there exists an s∈Ssuch that sI ⊆(N:RM)
or sJ ⊆(N:RM). Thus, sI ⊆(N:RM) or sK =sJ M ⊆(N:RM)M=N.
Therefore, Nis a weakly S-prime submodule of Mby (4) of Theorem 1.
(3) Suppose Mis faithful multiplication and Iis an ideal of R. Since (IM :R
M) = I, the result follows from (1) and (2).
(4) Let s∈Sbe a weakly S-element of N. We firstly note that ((N:MA) :R
M)∩S=φ. Indeed, if t∈((N:MA) :RM)∩S, then tA ⊆(N:RM) and so
t∈(N:RM) as Z(N:RM)(R)∩A=φ, a contradiction. Let r∈Rand m∈M
such that 0 6=rm ∈(N:MA). Then 0 6=Arm ⊆Nsince (0 :MA) = 0. By
assumption, either sr ∈(N:RM) or sAm ⊆N. Thus, sr ∈((N:MA) :RM) or
sm ∈(N:MA) as needed.
We show by the following example that the condition ”faithful module” in The-
orem 3 (1) is crucial.
Example 2. Let p1, p2and p3be distinct prime integers. Consider the non-
faithful Z-module M=Zp1p2×Zp1p2and the multiplicatively closed subset S=
{pn
3:n∈N∪ {0}} of Z. While N=0×0is a weakly S-prime submodule of M, we
have clearly (N:ZM) = hp1p2iis not a weakly S-prime ideal of Z.
Let Nbe a proper submodule of an R-module M. Then Nis said to be a maximal
weakly S-prime submodule if there is no weakly S-prime submodule which contains
Nproperly. In the following corollary, by Z(M), we denote the set {r∈R:rm = 0
for some m∈M\{0M}}.
Corollary 1. Let Nbe a submodule of Msuch that Z(N:RM)(R)∪Z(M)⊆(N:R
M). If Nis a maximal weakly S-prime submodule of M, then Nis an S-prime
submodule of M.
Proof. Let s∈Sbe a weakly S-element of N. Suppose that am ∈Nand sa /∈
(N:RM) for some a∈Rand m∈M. Since a /∈(N:RM), then by assumption,
a /∈Z(N:RM)(R) and (0 :Ma) = 0. It follows by Theorem 3 (4) that (N:Ma)
is a weakly S-prime submodule of M. Therefore, sm ∈(N:Ma) = Nby the
maximality of Nand so Nis an S-prime submodule of M.
As N= (N:M)Mfor any submodule Nof a multiplication R-module M, we
have the following consequence of Theorem 3.
Theorem 4. Let Mbe a faithful multiplication R-module and Nbe a submodule
of M. The following are equivalent.
(1) Nis a weakly S-prime submodule of M.
(2) (N:RM) is a weakly S-prime ideal of R.
(3) N=IM for some weakly S-prime ideal Iof R.
For a next result, we need to recall the following lemma.
Lemma 1. [2] For an ideal Iof a ring Rand a submodule Nof a finitely generated
faithful multiplication R-module M, the following hold.
(1) (IN :RM) = I(N:RM).
(2) If Iis finitely generated faithful multiplication, then
(a) (IN :MI) = N.
6 HANI A. KHASHAN AND ECE YETKIN CELIKEL
(b) Whenever N⊆IM , then (JN :MI) = J(N:MI)for any ideal Jof
R.
Proposition 1. Let Ibe a finitely generated faithful multiplication ideal of a ring
R, S a multiplicatively closed subset of Rand Na submodule of a finitely generated
faithful multiplication R-module M.
(1) If I N is a weakly S-prime submodule of Mand (N:RM)∩S=φ, then
either Iis a weakly S-prime ideal of Ror Nis a weakly S-prime submodule
of M.
(2) Nis a weakly S-prime submodule of I M if and only if (N:MI) is a weakly
S-prime submodule of M.
Proof. (1) Let s∈Sbe weakly S-element of IN . Suppose N=M. In this case,
I=I(N:RM) = (IN :RM) is a weakly S-prime ideal of Rby Theorem 4. Now,
suppose that Nis proper. Hence, Lemma 1 implies N= (I N :MI) and so we
conclude that (N:RM) = ((IN :MI) :RM) = (I(N:RM) :MI). Suppose
a∈R, m ∈Msuch that 0 6=am ∈Nand sa /∈(N:RM).Since Iis faithful, then
(0 :MI) = AnnR(I)M= 0, [2] and so 0 6=Iam ⊆IN. Since clearly sa /∈(IN :RM)
and IN is a weakly S-prime submodule, sI m ⊆I N by Theorem 1. By Lemma 1
(2), we have sm ∈(IN :MI) = N, and thus Nis a weakly S-prime submodule of
M.
(2) Suppose Nis a weakly S-prime submodule of IM with a weakly S-element
s′∈S. Then ((N:MI) :RM)∩S= (N:RI M )∩S=φ. Let a∈Rand m∈M
with 0 6=am ∈(N:MI) and s′a /∈((N:MI) :RM) = (N:RIM ). If amI = 0,
then am ∈(0M:I) = AnnR(I)M= 0, a contradiction. Thus, 0 6=amI ⊆N.
Since Nis a weakly S-prime submodule of I M, Theorem 1 yields that s′mI ⊆N,
and so s′m∈(N:MI), as required.
Conversely, suppose (N:MI) is a weakly S-prime submodule of Mwith a weakly
S-element s′∈S. Then (N:RI M )∩S= ((N:MI) :RM)∩S=φ. Now, let a∈R
and m′∈IM such that 0 6=am′∈Nand s′a /∈(N:RIM ) = ((N:MI) :RM).
Then a(hm′i:MI) = (ham′i:MI)⊆(N:MI). If a(hm′i:MI) = 0, then by (2)
of Lemma 1, we have am′∈a(Im′:MI)⊆a(hm′i:MI) = 0, a contradiction.
Thus, 0 6=a(hm′i:MI)⊆(N:MI) and so s′(hm′i:MI)⊆(N:MI) as s′a /∈
((N:MI) :RM). Again, by Lemma 1, we conclude that s′m′∈(Ihs′m′i:MI) =
Is′(hm′i:MI)⊆I(N:MI) = (IN :MI) = N. Therefore, Nis a weakly S-prime
submodule of IM.
Proposition 2. Let Sbe a multiplicatively closed subset of a ring Rand Nbe a
submodule of an R-module Msuch that (N:RM)∩S=φ. If (N:Ms)is a weakly
prime submodule of Mfor some s∈S, then Nis a weakly S-prime submodule of
M. The converse holds for non-zero submodules Nif S∩Z(M) = ∅.
Proof. Suppose (N:Ms) is a weakly prime submodule of Mfor some s∈Sand let
a∈R, m ∈Msuch that 0 6=am ∈N⊆(N:Ms). Then either a∈((N:Ms) :R
M) = ((N:RM) :Rs) or m∈(N:Ms) and so either sa ∈(N:RM) or sm ∈N
as required. Conversely, suppose N6= 0Mis weakly S-prime submodule of Mwith
weakly S-element s∈S. Let a∈Rand m∈Msuch that 0 6=am ∈(N:Ms).
Since S∩Z(M) = ∅, we have 0 6=sam ∈Nwhich implies either s2a∈(N:RM) or
sm ∈N. If sm ∈N, then m∈(N:Ms) and we are done. Suppose s2a∈(N:RM).
If s2aM = 0, then s2∈S∩Z(M), a contradiction. Hence, 0 6=s2aM ⊆Nimplies
ON WEAKLY S-PRIME SUBMODULES 7
either s3∈(N:RM) or saM ⊆N. But (N:RM)∩S=φimplies saM ⊆N
and so a∈(N:RsM ) = ((N:Ms) :RM). Therefore, (N:Ms) is a weakly prime
submodule of M.
If S∩Z(M)6=∅, then the converse of Proposition 2 need not be true as we can
see in the following example.
Example 3. Consider the Z-module M=Z×Z6and let N=h0i × h¯
0i. Then N
is a weakly S-prime submodule of Mfor S={3n:n∈N}. Now, for each n∈N,
we have clearly (N:M3n)= h0i×h¯
2iwhich is not a weakly prime submodule of M.
Indeed, 2.(0,¯
1) ∈(N:M3n)but 2/∈((N:M3n) :RM) = h0iand (0,¯
1) /∈(N:M
3n). We note that S∩Z(M) = S6=∅.
Proposition 3. Let Mbe a faithful multiplication R-module and Sbe a multiplica-
tively closed subset of R.
(1) If Nis a weakly S-prime submodule of Mthat is not S-prime, then
s√0RN= 0Mfor some s∈S.
(2) If Nand Kare two weakly S-prime submodules of Mthat are not S-prime,
then sNK = 0Mfor some s∈S.
Proof. (1) Let Nbe a weakly S-prime submodule of Mwhich is not S-prime.
Then by (1) of Theorem 3 and [18, Proposition 2.9 (ii)], (N:RM) is a weakly
S-prime ideal of Rthat is not S-prime. Hence, we get s(N:RM)√0R= 0Rby [3,
Proposition 9] and thus, sN√0R=s(N:RM)M√0R= 0RM= 0M.
(2) Since Nand Kare two weakly S-prime submodules that are not S-prime,
(N:RM) and (K:RM) are weakly S-prime ideals of Rthat are not S-prime by
Theorem 3 and [18, Proposition 2.9 (ii)]. Hence, there exists some s∈Ssuch that
s(N:RM)(K:RM) = 0Rby [3, Corollary 11] and so sN K = 0.
Corollary 2. Let Mbe a faithful multiplication R-module, Sbe a multiplicatively
closed subset of a ring R. If Nis a weakly S-prime submodule of M, then either
N⊆√0RMor s√0RM⊆Nfor some s∈S. Additionally, if Ris a reduced ring,
then N= 0Mor Nis S-prime.
Proof. Suppose that Nis a weakly S-prime submodule of M. Then from Theorem
3 (1), (N:RM) is a weakly S-prime ideal of Rand by [3, Corollary 6], we conclude
either (N:RM)⊆√0Ror s√0R⊆(N:RM).Since N= (N:RM)M, we are
done.
Proposition 4. Let Nbe a submodule of an R-module Mand Sbe a multiplica-
tively closed subset of Rwith Z(M)∩S=∅.
(1) If Nis a weakly S-prime submodule of M, then S−1Nis a weakly prime
submodule of S−1Mand there exists an s∈Ssuch that (N:Mt)⊆(N:M
s) for all t∈S.
(2) If Mis finitely generated, then the converse of (1) holds.
Proof. (1) Suppose s∈Sis a weakly S-element of N. In proving that S−1Nis a
weakly prime submodule of S−1Mwe do not need the assumption Z(M)∩S=∅.
Let 0S−1M6=r
s1
m
s2∈S−1Nfor some r
s1∈S−1Rand m
s2∈S−1M. Then urm ∈N
for some u∈S. If urm = 0, then r m
s1s2=urm
us1s2= 0S−1M, a contradiction. Hence,
06=urm ∈Nyields either sur ∈(N:RM) or sm ∈N. Thus, r
s1=sur
sus1∈
8 HANI A. KHASHAN AND ECE YETKIN CELIKEL
S−1(N:RM)⊆(S−1N:S−1RS−1M) or m
s2=sm
ss2∈S−1Nand so S−1Nis
a weakly prime submodule of S−1M. Now, let t∈Sand m∈(N:Mt).Then
06=tm ∈Nas Z(M)∩S=∅and so st ∈(N:MM)∩Sor sm ∈N. Since the first
one gives a contradiction, we have m∈(N:Ms).Thus, (N:Mt)⊆(N:Ms) for
all t∈S.
(2) Suppose Mis finitely generated choose s∈Sas in (1). If (N:RM)∩S6=∅,
then clearly S−1N=S−1M, a contradiction. Let 0 6=am ∈Nfor some a∈R
and m∈M. Since Z(M)∩S=∅,we have 0 6=a
1
m
1∈S−1N. By assumption,
either a
1∈(S−1N:S−1RS−1M) = S−1(N:RM) as Mis finitely generated or
m
1∈S−1N. Hence, va ∈(N:RM) for some v∈Sor wm ∈Nfor some w∈S. If
va ∈(N:RM), then our hypothesis implies aM ⊆(N:Mv)⊆(N:Ms) and so
sa ∈(N:RM).If wm ∈N, then again m∈(N:Mw)⊆(N:Ms), and so sm ∈N.
Therefore, Nis a weakly S-prime submodule of M.
However, S−1Nis a weakly prime submodule of S−1Mdoes not imply that N
is a weakly prime submodule of M. For example, it was shown in [18, Example
2.4] that N=Z×{0}is not a (weakly) S-prime submodule of the Z-module Q×Q
where S=Z\ {0}. But S−1Nis a weakly prime submodule of the vector space
(over S−1Z=Q)S−1(Q×Q).
Remark 1. Let Mbe an R-module and S,Tbe two multiplicatively closed subsets
of Rwith S⊆T. If Nis a weakly S-prime submodule of Mand (N:RM)∩T=∅,
then Nis a weakly T-prime submodule of M.
Let Sbe a multiplicatively closed subset of a ring R. The saturation of Sis
the set S∗={x∈R:xy ∈Sfor some y∈R}, see [14]. It is clear that S∗is a
multiplicatively closed subset of Rand that S⊆S∗.
Proposition 5. Let Sbe a multiplicatively closed subset of a ring Rand Nbe a
submodule of an R-module Msuch that (N:RM)∩S=∅. Then Nis a weakly
S-prime submodule of Mif and only if Nis a weakly S∗-prime submodule of M.
Proof. Let Nbe a weakly S∗-prime submodule of Mwith a weakly S-element
s∗∈S∗. Choose r∈Rsuch that s=s∗r∈S. Suppose 0 6=am ∈Nfor some
a∈Rand m∈M. Then either s∗a∈(N:RM) or s∗m∈N. Thus, sa ∈(N:RM)
or sm ∈Nand we are done. Conversely, suppose Nis weakly S∗-prime. By using
Remark 1, it is enough to prove that (N:RM)∩S∗=∅. Suppose there exists
s∗∈(N:RM)∩S∗. Then there is r∈Rsuch that s=s∗r∈(N:RM)∩S, a
contradiction.
Lemma 2. Let Sbe a multiplicatively closed subset of a ring R. If Iis a weakly
S-prime ideal of Rand {0R}is an S-prime ideal of R, then √Iis an S-prime ideal
of R.
Proof. Suppose Iis weakly S-prime associated to s1and {0R}is S-prime associated
with s2. Since I∩S=∅, we have √I∩S=∅. Let a, b ∈Rwith ab ∈√I. Then
anbn∈Ifor some positive integer n. If anbn6= 0, then we have s1an∈Ior
s1bn∈Ithat is s1a∈√Ior s1b∈√I. If anbn= 0, then by assumption, either
s2an= 0 or s2bn= 0 and so s2a∈√Ior s2b∈√I . Thus, √Iis an S-prime ideal
of Rassociated with s=s1s2.
Proposition 6. Let Mbe a finitely generated faithful multiplication R-module and
Sbe a multiplicatively closed subset of R. If Nis a weakly S-prime submodule of
ON WEAKLY S-PRIME SUBMODULES 9
Mand {0R}is an S-prime ideal of R, then M-rad(N)is an S-prime submodule
of R.
Proof. By [16, Lemma 2.4], we have (M-rad(N) : M) = p(N:RM). Since Nis
a weakly S-prime submodule of M , (N:RM) is so by Theorem 3. By Lemma 2,
p(N:RM) is an S-prime ideal of R. Thus, the claim follows from [18, Proposition
2.9 (ii)].
Proposition 7. Let Sbe a multiplicatively closed subset of a ring R. If Nis a
weakly S-prime submodule of an R-module M, then for any submodule Kof M
with (K:RM)∩S6=∅,N∩Kis a weakly S-prime submodule of M. Additionally,
if Mis multiplication, then N K is a weakly S-prime submodule of M.
Proof. Note that (N∩K:RM)∩S=∅as (N:RM)∩S=∅. Let s∈Sbe
a weakly S-element of Nand let 0 6=am ∈N∩K⊆N. Then sa ∈(N:RM)
or sm ∈N. Choose s′∈(K:RM)∩S. Then ss′a∈(N:RM)∩(K:RM) =
(N∩K:RM) or ss′m∈N∩(K:RM)M=N∩K. Thus, N∩Kis a weakly
S-prime submodule of Mwith a weakly S-element t=ss′.Putting in mind that
NK = (N:RM)(K:RM)M, the rest of the proof is very similar.
Notice that if Nis weakly prime and Kis as above, then N∩Kneed not be
weakly prime. For instance, consider the Z12 -module Z12 ,S={¯
1,¯
3,9},N=h¯
2i
and K=h¯
3i. Then N∩K=h¯
6iis not a weakly prime submodule of Z12 .
Proposition 8. Let f:M1→M2be a module homomorphism where M1and
M2are two R-modules and Sbe a multiplicatively closed subset of R. Then the
following statements hold.
(1) If fis an epimorphism and Nis a weakly S-prime submodule of M1con-
taining Ker(f), then f(N) is a weakly S-prime submodule of M2.
(2) If fis a monomorphism and Kis a weakly S-prime submodule of M2with
(f−1(K) :RM1)∩S=∅, then f−1(K) is a weakly S-prime submodule of
M1.
Proof. (1) First, observe that (f(N) :R2M2)∩S=∅. Indeed, assume that t∈
(f(N) :R2M2)∩S. Then f(tM1) = tf (M1) = tM2⊆f(N), and so tM1⊆Nas
Kerf ⊆N. It follows that t∈(N:M1)∩S, a contradiction. Let sbe a weakly
S-element of Nand a∈R,m2∈M2with 0 6=am2∈f(N). Then m2=f(m1) for
some m1∈M1and 0 6=af (m1) = f(am1)∈f(N) and since Kerf ⊆N, we have
06=am1∈N. This yields either sa ∈(N:RM1) or sm1∈N. Thus, clearly we
have either sa ∈(f(N) :RM2) or sm2=f(sm1)∈f(N) as required.
(2) Let sbe a weakly S-element of Kand let a∈R, m ∈M1with 0 6=am ∈
f−1(K). Then 0 6=f(am) = af (m)∈Kas fis a monomorphism. Since Kis a
weakly S-prime submodule of M2, we have sa ∈(K:RM2) or sf (m)∈K. Thus,
clearly we have sa ∈(f−1(K) :RM1) or sm ∈f−1(K) as needed.
Corollary 3. Let Sbe a multiplicatively closed subset of a ring Rand N,Kare
two submodules of an R-module Mwith K⊆N.
(1) If Nis a weakly S-prime submodule of M, then N/K is a weakly S-prime
submodule of M/K.
(2) If K′is a weakly S-prime submodule of Mwith (K′:RN)∩S=∅, then
K′∩Nis a weakly S-prime submodule of N.
10 HANI A. KHASHAN AND ECE YETKIN CELIKEL
(3) If N/K is a weakly S-prime submodule of M/K and Kis an S-prime
submodule of M, then Nis an S-prime submodule of M.
(4) If N/K is a weakly S-prime submodule of M/K and Kis a weakly S-prime
submodule of M, then Nis a weakly S-prime submodule of M.
Proof. Note that (N/K :RM/K )∩S=∅if and only if (N:RM)∩S=∅.
(1). Consider the canonical epimorphism π:M→M/K defined by π(m) =
m+K. Then π(N) = N/K is a weakly S-prime submodule of M/K by (1) of
Proposition 8.
(2). Let K′be a weakly S-prime submodule of Mand consider the natural
injection i:N→Mdefined by i(m) = mfor all m∈N. Then (i−1(K′) :R
N)∩S=∅. Indeed, if s∈(i−1(K′) :RN)∩S, then sN ⊆i−1(K′) = K′∩N⊆K′
and so s∈(K′:RN)∩S, a contradiction. Thus i−1(K′) = K′∩Nis a weakly
S-prime submodule of Mby (2) of Proposition 8. .
(3). Let s1be a weakly S-element of N/K and suppose Kis an S-prime sub-
module of Massociated with s2∈S. Let a∈Rand m∈Msuch that am ∈N.
If am ∈K, then s2a∈(K:RM)⊆(N:RM) or s2m∈K⊆N. If am /∈K,
then K6=a(m+K)∈N/K which implies either s1a∈(N/K :RM/K) or
s1(m+K)∈N/K. Thus, s1a∈(N:RM) or s1m∈N. It follows that Nis an
S-prime submodule of Massociated with s=s1s2∈S.
(4). Similar to (3).
The next example shows that the converse of Corollary 3 (1) is not valid in
general.
Example 4. Consider the submodules N=K=h6iof the Z-module Zand the
multiplicatively closed subset S={5n:n∈N∪{0}} of Z.It is clear that N/K is
a weakly S-prime submodule of Z/K but Nis not a weakly S-prime submodule of
Zas 06= 2 ·3∈Nbut neither 2s∈(N:ZZ)nor 3s∈Nfor all s∈S.
Proposition 9. Let Sbe a multiplicatively closed subset of a ring Rand N,Kbe
two weakly S-prime submodules of an R-module Msuch that ((N+K) :RM)∩S=
∅. Then N+Kis a weakly S-prime submodule of M.
Proof. Suppose Nand Kare two weakly S-prime submodules of M. By Corollary
3 (1), N/(N∩K) is a weakly S-prime submodule of M/(N∩K). Now, from the
module isomorphism N/(N∩K)∼
=(N+K)/K, we conclude that (N+K)/K is a
weakly S-prime submodule of M/K. Thus, N+Kis a weakly S-prime submodule
of Mby Corollary 3 (4).
Theorem 5. Let S1, S2be multiplicatively closed subsets of rings R1,R2respec-
tively and N1,N2be non-zero submodules of an R1-module M1and an R2-module
M2, respectively. Consider M=M1×M2as an (R1×R2)-module, S=S1×S2
and N=N1×N2. Then the following are equivalent.
(1) Nis a weakly S-prime submodule of M.
(2) N1is an S1-prime submodule of M1and (N2:R2M2)∩S26=∅or N2is an
S2-prime submodule of M2and (N1:R1M1)∩S16=∅
(3) Nis a S-prime submodule of M.
Proof. (1)⇒(2). Suppose Nis a weakly S-prime submodule of Mwith weakly
S-element s= (s1, s2)∈S. Assume that (N1:R1M1)∩S1and (N2:R2M2)∩S2
are both empty. Choose 0 6=m∈N1.Then (0M1,0M2)6= (1,0)(m, 1M2)∈Nwhich
ON WEAKLY S-PRIME SUBMODULES 11
implies (s1, s2)(1,0) ∈(N:RM) = (N1:R1M1)×(N2:R2M2) or (s1, s2)(m, 1M2)∈
N1×N2.Hence, we have either s1∈(N1:R1M1)∩S1or s2∈N2∩S2⊆(N2:R2
M2)∩S2, a contradiction. Now, we may assume that (N1:R1M1)∩S16=∅and
we show that N2is an S2-prime submodule of M2.Suppose am′∈N2for some
a∈R2and m′∈M2. Then (0M1,0M2)6= (1R1, a)(m, m′)∈Nimplies either
(s1, s2)(1R1, a)∈(N1:R1M1)×(N2:R2M2) or (s1, s2)(m, m′)∈N1×N2. Thus,
s2a∈(N2:R2M2) or s2m′∈N2and so N2is an S2-prime submodule of M2.
(2)⇒(3). Follows from [18, Theorem 2.14]
(3)⇒(1). Straightforward.
Theorem 6. Let M=M1×M2×··· ×Mnbe an R1×R2×· ··×Rn-module and
S=S1×S2×· ··×Snwhere Ri’s are rings, Siis a multiplicatively closed subset of
Riand Niis a non-zero submodule of Mifor each i= 1,2, ..., n. Then the following
assertions are equivalent.
(1) N=N1×N2× · ·· × Nnis a weakly S-prime submodule of M.
(2) For i= 1,2, ..., n,Niis an S-prime submodule of Miand (Nj:RjMj)∩Sj6=
∅for all j6=i.
Proof. We prove the claim by using mathematical induction on n. The claim follows
by Theorem 5 for n= 2.Now, we assume that the claim holds for all k < n and prove
it for k=n. Suppose N=N1×N2×···×Nnis a weakly S-prime submodule of M.
Then Theorem 5 implies that N=N′×Nnwhere, say, N′=N1×N2×··· ×Nn−1
is a weakly S-prime submodule of M′=M1×M2× · ·· × Mn−1and Sn∩(Nn:Rn
Mn)6=∅. Thus, the result follows by the induction hypothesis.
Let Mbe an R-module and Sbe a multiplicatively closed subset of Rwith
S∩AnnR(M) = φ. Following [18], Mis called S-torsion-free if there is s∈Ssuch
that whenever rm = 0 for r∈Rand m∈M, then sr = 0 or sm = 0. Compare
with [18, Proposition 2.24], we have the following result.
Proposition 10. Let Sbe a multiplicatively closed subset of a ring Rand Nbe
a submodule of an S-torsion-free R-module M. If η:R→R/(N:RM)is the
canonical homomorphism, then Nis weakly S-prime in Mif and only if M/N is a
η(S)-torsion-free R/(N:RM)-module.
Proof. First, we clearly note that s∈S∩(N:RM) if and only if ¯s∈η(S)∩
AnnR/(N:RM)(M/N).
⇒) Suppose Nis weakly S-prime in Mwith weakly S-element s1∈S. Let
¯r∈R/(N:RM), ¯m∈M/N such that ¯r¯m=¯
0. Then rm ∈Nand we have
two cases. If rm = 0, then by assumption there is s2∈Ssuch that s2r= 0 or
s2m= 0. Thus ¯s2¯r=¯
0 or ¯s2¯m=¯
0 where ¯s2∈η(S) as needed. If rm 6= 0, then
s1r∈(N:RM) or s1m∈Nand so ¯s1¯r=¯
0R/(N:RM)or ¯s1¯m=¯
0M/N where
¯s1∈η(S). Therefore, M/N is a η(S)-torsion-free R/(N:RM)-module associated
to ¯s1¯s2∈η(S).
⇐) Follows directly by [18, Proposition 2.24].
Let Rbe a ring and Mbe an R-module. Recall that the idealization of Min R
denoted by R⋉Mis the commutative ring R⊕Mwith coordinate-wise addition
and multiplication defined as (r1, m1)(r2, m2) = (r1r2, r1m2+r2m1) [17]. For an
ideal Iof Rand a submodule Nof M , The set I⋉N=I⊕Nis not always an
ideal of R⋉Mand it is an ideal if and only if IM ⊆N[5, Theorem 3.1].Among
12 HANI A. KHASHAN AND ECE YETKIN CELIKEL
many other properties of an ideal I⋉Nof R⋉M, we have √I⋉N=√I⋉M
and in particular, √0⋉0 = √0⋉M, [5, Theorem 3.2]. It is clear that if Sis a
multiplicatively closed subset of Rand Na submodule of M, then S⋉K={(s, k) :
s∈S,k∈K}is a multiplicatively closed subset of R⋉M. In [3, Proposition 27],
it is proved that if I⋉Mis a weakly S⋉M-prime (or weakly S⋉0-prime ideal of
R⋉Mwhere Iis an ideal of Rdisjoint with S, then Iis a weakly S-prime ideal of
R. In general, we have:
Theorem 7. Let Sbe a multiplicatively closed subset of a ring R, I be an ideal of
Rand K⊆Nbe submodules of an R-module Mwith IM ⊆N. Let I⋉Nbe a
weakly S⋉K-prime ideal of R⋉M. Then
(1) Iis a weakly S-prime ideal of Rand Na weakly S-prime submodule of M
whenever (N:RM)∩S=φ.
(2) There exists s∈Ssuch that for all a, b ∈R,ab = 0, sa /∈I,sb /∈Iimplies
a, b ∈ann(N) and for all c∈R,m∈M,cm = 0, sc /∈(N:RM), sm /∈N
implies c∈ann(I) and m∈(0 :MI).
(3) If I⋉Nis not S⋉K-prime, then (s, k)(I⋉N) = (sI ⋉0) ⊕(0 ⋉sN +Ik)
for some (s, k)∈S⋉K.
(4) If I⋉Nis S⋉K-prime then sM ⊆Nfor some s∈S.
Proof. Let (s, k)∈S⋉Kbe a weakly S⋉K-element of I⋉N.
(1) Note that clearly (S⋉K)∩(I⋉N) = φif and only if I∩S=φ. Suppose
that a, b ∈Rwith 0 6=ab ∈I. Then (0,0) 6= (a, 0)(b, 0) ∈I⋉Nimplies that
either (s, k)(a, 0) ∈I⋉Nor (s, k)(b, 0) ∈I⋉N. Thus, either sa ∈Ior sb ∈I
and Iis weakly S-prime in R. Now, let 0 6=rm ∈Nfor r∈R,m∈M. Then
(0,0) 6= (r, 0)(0, m)∈I⋉Nand so (sr, rk) = (s, k)(r, 0) ∈I⋉Nor (0, sm) =
(s, k)(0, m)∈I⋉N. In the first case, we get sr ∈I⊆(N:RM) and the second
case implies sm ∈N. Therefore, Nis a weakly S-prime submodule of M.
(2) Let a, b ∈Rsuch that ab = 0 and sa /∈I,sb /∈I. Suppose a /∈ann(N) so that
there exists n∈Nsuch that an 6= 0. Thus, (0,0) 6= (a, 0)(b, n) = (0, an)∈I⋉N
and so either (s, k)(a, 0) ∈I⋉Nor (s, k )(b, n)∈I⋉N. Hence, sa ∈Ior sb ∈I,
a contradiction. Similarly, if b /∈ann(N),then we get a contradiction. Therefore,
a, b ∈ann(N) as needed. Next, we assume cm = 0 for c∈R,m∈Mand
sc /∈(N:RM), sm /∈N. We have two cases.
Case 1. If c /∈ann(I), then there exists a∈Isuch that ca 6= 0. Hence, (0,0) 6=
(c, 0)(a, m) = (ca, 0) ∈I⋉Nand so (s, k)(c, 0) ∈I⋉Nor (s, k )(a, m)∈I⋉N.
Therefore, sc ∈I⊆(N:RM) or sm +ka ∈N(and so sm ∈Nas K⊆N) which
contradicts the assumption.
Case 2. If m /∈(0 :MI), then there exists a∈Isuch that am 6= 0. Thus,
(0,0) 6= (a, m)(c, m) = (ac, am)∈I⋉Nimplies either (s, k)(a, m)∈I⋉Nor
(s, k)(c, m)∈I⋉N. It follows that either sc ∈I⊆(N:RM) or sm ∈Nwhich is
also a contradiction.
(3) If I⋉Nis not S⋉K-prime, then (s, k)(I⋉N)(√0⋉M) = (0,0) for some
(s, k)∈S⋉Kby [3, Proposition 9]. Thus, by [5, Theorem 3.3] s√0I⋉(sIM +
s√0N+√0Ik) = (0,0). Then clearly sI M = 0 and so sI ⋉0 is an ideal of R⋉M.
Now, (s, k)(I⋉N) = sI ⋉(sN +Ik) = (sI ⋉0) ⊕(0 ⋉sN +I k) as required.
(4) If I⋉Nis S⋉K-prime in R⋉M, then (s, k)(√0⋉M)⊆(I⋉N) for some
(s, k)∈S⋉Kby [3, Corollary 6]. Thus, s√0⋉(sM +√0k)⊆(I⋉N) and so
clearly, sM ⊆Nas needed.
ON WEAKLY S-PRIME SUBMODULES 13
In general if Iis a (weakly) S-prime ideal of a ring Rand Na (weakly) S-prime
submodule of an R-module M, then I⋉Nneed not be a (weakly) S⋉K-prime
ideal of R⋉M.
Example 5. Consider the multiplicatively closed subset S={3n:n∈N}of Z.
While clearly 0is (weakly) S-prime in Zand h¯
2iis (weakly) S-prime in the Z-
module Z6, the ideal 0⋉h¯
2iis not (weakly) S⋉0-prime in Z ⋉ Z6. Indeed, (0,0) 6=
(0,¯
1)(2,¯
1) = (0,¯
2) ∈0⋉h¯
2ibut (s, ¯
0)(0,¯
1) /∈0⋉h¯
2iand (s, ¯
0)(2,¯
1) /∈0⋉h¯
2ifor
all s∈S.
3. (Weakly) S-prime Submodules of Amalgamation Modules
Let Rbe a ring, Jan ideal of Rand Man R-module. We recall that the set
R⋊⋉ J={(r, r +j) : r∈R,j∈J}
is a subring of R×Rcalled the the amalgamated duplication of Ralong J, see [11].
Recently, in [10], the duplication of the R-module Malong the ideal Jdenoted by
M⋊⋉ Jis defined as
M⋊⋉ J={(m, m′)∈M×M:m−m′∈JM }
which is an (R⋊⋉ J)-module with scalar multiplication defined by (r, r+j).(m, m′) =
(rm, (r+j)m′) for r∈R,j∈Jand (m, m′)∈M⋊⋉ J. Many properties and results
concerning this kind of modules can be found in [10].
Let Nbe a submodule of an R-module Mand Jbe an ideal of R. Then clearly
N⋊⋉ J={(n, m)∈N×M:n−m∈JM }
and
¯
N={(m, n)∈M×N:m−n∈JM}
are submodules of M⋊⋉ J. If Sis a multiplicatively closed subset of R, then ob-
viously, the sets S⋊⋉ J={(s, s +j) : s∈S,j∈J}and ¯
S={(r, r +j) : r+j∈S}
are multiplicatively closed subsets of R⋊⋉ J.
In general, let f:R1→R2be a ring homomorphism, Jbe an ideal of R2,M1
be an R1-module, M2be an R2-module (which is an R1-module induced naturally
by f) and ϕ:M1→M2be an R1-module homomorphism. The subring
R1⋊⋉fJ={(r, f(r) + j) : r∈R1,j∈J}
of R1×R2is called the amalgamation of R1and R2along Jwith respect to f. In
[12], the amalgamation of M1and M2along Jwith respect to ϕis defined as
M1⋊⋉ϕJM2={(m1, ϕ(m1) + m2) : m1∈M1and m2∈J M2}
which is an (R1⋊⋉fJ)-module with the scalar product defined as
(r, f(r) + j)(m1, ϕ(m1) + m2) = (rm1, ϕ(rm1) + f(r)m2+jϕ(m1) + jm2)
For submodules N1and N2of M1and M2, respectively, clearly the sets
N1⋊⋉ϕJM2={(m1, ϕ(m1) + m2)∈M1⋊⋉ϕJM2:m1∈N1}
and
N2
ϕ={(m1, ϕ(m1) + m2)∈M1⋊⋉ϕJM2:ϕ(m1) + m2∈N2}
14 HANI A. KHASHAN AND ECE YETKIN CELIKEL
are submodules of M1⋊⋉ϕJ M2. Moreover if S1and S2are multiplicatively closed
subsets of R1and R2, respectively, then
S1⋊⋉fJ={(s1, f (s1) + j) : s∈S1,j∈J}
and
S2
ϕ={(r, f(r) + j) : r∈R1,f(r) + j∈S2}
are clearly multiplicatively closed subsets of M1⋊⋉ϕJM2.
Note that if R=R1=R2,M=M1=M2,f=IdRand ϕ=IdM, then the
amalgamation of M1and M2along Jwith respect to ϕis exactly the duplication of
the R-module Malong the ideal J. Moreover, in this case, we have N1⋊⋉ϕJM2=
N⋊⋉ J,N2
ϕ=¯
N,S1⋊⋉fJ=S⋊⋉ Jand S2
ϕ=¯
S.
Theorem 8. Consider the (R1⋊⋉fJ)-module M1⋊⋉ϕJM2defined as above. Let
Sbe a multiplicatively closed subsets of R1and N1be submodule of M1. Then
(1) N1⋊⋉ϕJM2is an S⋊⋉fJ-prime submodule of M1⋊⋉ϕJ M2if and only if
N1is an S-prime submodule of M1.
(2) N1⋊⋉ϕJM2is a weakly S⋊⋉fJ-prime submodule of M1⋊⋉ϕJ M2if and
only if N1is a weakly S-prime submodule of M1and for r1∈R1,m1∈M1
with r1m1= 0 but s1r1/∈(N1:R1M1) and s1m1/∈N1for all s1∈S, then
f(r1)m2+jφ(m1) + jm2= 0 for every j∈Jand m2∈J M2.
Proof. We clearly note that (N1⋊⋉ϕJM2:R1⋊⋉fJM1⋊⋉ϕJM2)∩S⋊⋉fJ=φif and
only if (N1:R1M1)∩S=φ.
(1) Suppose (s, f (s) + j) is an S⋊⋉fJ-element of N1⋊⋉ϕJM2and let r1m1∈N1
for r1∈R1and m1∈M1. Then (r1, f (r1)) ∈R1⋊⋉fJand (m1, ϕ(m1)) ∈M1⋊⋉ϕ
JM2with (r1, f (r1))(m1, ϕ(m1)) = (r1m1, ϕ(r1m1)) ∈N1⋊⋉ϕJM2. Thus, either
(s, f (s) + j)(r1, f(r1)) ∈(N1⋊⋉ϕJ M2:R1⋊⋉fJM1⋊⋉ϕJM2)
or
(s, f (s) + j)(m1, ϕ(m1)) ∈N1⋊⋉ϕJM2
In the first case, for all m∈M1, (s, f (s)+j)(r1, f (r1))(m, ϕ(m)) ∈N1⋊⋉ϕJM2and
so sr1M1⊆N1. In the second case, sm1∈N1and so N1is an S-prime submodule
of M1. Conversely, let sbe an S-element of N1. Let (r1, f (r1) + j1)∈R1⋊⋉fJand
(m1, ϕ(m1) + m2)∈M1⋊⋉ϕJM2such that
(r1m1, ϕ(r1m1) + f(r1)m2+j1ϕ(m1) + j1m2)
= (r1, f (r1) + j1)(m1, ϕ(m1) + m2)∈N1⋊⋉ϕJM2
Then r1m1∈N1and hence either sr1M1⊆N1or sm1∈N1. If sr1M1⊆N1,
then clearly (s, f (s))(r1, f(r1) + j1)∈(N1⋊⋉ϕJM2:R1⋊⋉fJM1⋊⋉ϕJM2) and if
sm1∈N1, then (s, f (s))(m1, ϕ(m1) + m2)∈N1⋊⋉ϕJM2. Therefore, N1⋊⋉ϕJM2
is an S⋊⋉fJ-prime submodule of M1⋊⋉ϕJM2associated to (s, f (s)) ∈S⋊⋉fJ.
(2) Suppose (s, f (s)+j) is a weakly S⋊⋉fJ-element of N1⋊⋉ϕJ M2. Let r1∈R1
and m1∈M1such that 0 6=r1m1∈N1so that (0,0) 6= (r1, f (r1))(m1, ϕ(m1)) =
(r1m1, ϕ(r1m1)) ∈N1⋊⋉ϕJM2. By assumption, either (s, f (s) + j)(r1, f(r1)) ∈
(N1⋊⋉ϕJM2:R1⋊⋉fJM1⋊⋉ϕJM2) or (s, f (s)+j)(m1, ϕ(m1)) ∈N1⋊⋉ϕJM2and so
N1is S-prime in M1as in the proof of (1). Now, we use the contrapositive to prove
ON WEAKLY S-PRIME SUBMODULES 15
the other part. Let r1∈R1,m1∈M1with r1m1= 0 and f(r1)m2+jφ(m1)+jm26=
0 for some j∈Jand some m2∈J M2. Then
(0,0) 6= (r1, f (r1) + j)(m1, ϕ(m1) + m2)
= (0, f (r1)m2+jϕ(m1) + jm2)∈N1⋊⋉ϕJ M2
By assumption, either (s, f (s) + j)(r1, f (r1) + j)∈(N1⋊⋉ϕJM2:R1⋊⋉fJM1⋊⋉ϕ
JM2) or (s, f (s) + j)(m1, ϕ(m1) + m2)∈N1⋊⋉ϕJM2and so again sr1∈(N1:R1
M1) or sm1∈N1as needed. Conversely, let sbe a weakly S-element of N1and let
(r1, f (r1) + j)∈R1⋊⋉fJand (m1, ϕ(m1) + m2)∈M1⋊⋉ϕJM2such that
(0,0) 6= (r1m1, ϕ(r1m1) + f(r1)m2+jϕ(m1) + jm2)
= (r1, f (r1) + j)(m1, ϕ(m1) + m2)∈N1⋊⋉ϕJM2
If 0 6=r1m1, then the proof is similar to that of (1). Suppose r1m1= 0. Then
f(r1)m2+jϕ(m1) + jm26= 0 and so by assumption there exists s′∈Ssuch that
either s′r1∈(N1:R1M1) or s′m1∈N1. Thus, (s′, f (s′))(r1, f (r1) + j)∈(N1⋊⋉ϕ
JM2:R1⋊⋉fJM1⋊⋉ϕJM2) or (s′, f (s′))(m1, ϕ(m1)+ m2)∈N1⋊⋉ϕJ M2. Therefore,
N1⋊⋉ϕJM2is a weakly S⋊⋉fJ-prime submodule of M1⋊⋉ϕJM2associated to
(ss′, f (ss′)) ∈S⋊⋉fJ.
In particular, if Sis a multiplicatively closed subset of R1, then S×f(S) is
a multiplicatively closed subset of R1⋊⋉fJ. Moreover, one can similarly prove
Theorem 8 if we consider S×f(S) instead of S⋊⋉fJ.
Corollary 4. Consider the (R1⋊⋉fJ)-module M1⋊⋉ϕJM2defined as in Theorem
8 and let N1be a submodule of M1. Then
(1) N1⋊⋉ϕJM2is a prime submodule of M1⋊⋉ϕJ M2if and only if N1is a
prime submodule of M1.
(2) N1⋊⋉ϕJM2is a weakly prime submodule of M1⋊⋉ϕJ M2if and only if N1is
a weakly prime submodule of M1and for r1∈R1,m1∈M1with r1m1= 0
but r1/∈(N1:R1M1) and m1/∈N1, then f(r1)m2+jφ(m1) + jm2= 0 for
every j∈Jand m2∈JM2.
Proof. We just take S={1R1}(and so S×f(S) = {(1R1,1R2)}) and use Theorem
8.
Theorem 9. Consider the (R1⋊⋉fJ)-module M1⋊⋉ϕJM2defined as in Theorem
8 where fand ϕare epimorphisms. Let Sbe a multiplicatively closed subsets of R2
and N2be a submodule of M2. Then
(1) N2is an S-prime submodule of M2if and only if N2
ϕis an Sϕ-prime
submodule of M1⋊⋉ϕJ M2.
(2) If N2
ϕis an Sϕ-prime submodule of M1⋊⋉ϕJ M2, and (N2:R2JM2)∩S=
φ, then (N2:M2J) is an S-prime submodule of M2.
Proof. (1). We note that (N2
ϕ:R1⋊⋉fJM1⋊⋉ϕJM2)∩Sϕ=φif and only if
(N2:R2M2)∩S=φ. Indeed if (t, f (t) + j) = (t, s)∈Sϕsuch that (t, s)(M1⋊⋉ϕ
JM2)⊆N2
ϕ, then for each m2=ϕ(m1)∈M2, we have (t, s)(m1, m2)∈N2
ϕ.
Therefore, sm2∈N2and s∈(N2:R2M2). The converse is similar.
16 HANI A. KHASHAN AND ECE YETKIN CELIKEL
Suppose N2is an S-prime submodule of M2associated to s=f(t)∈S. Let
(r1, f (r1) + j)∈R1⋊⋉fJand (m1, ϕ(m1) + m2)∈M1⋊⋉ JM2such that
(r1, f (r1) + j)(m1, ϕ(m1) + m2)∈N2
ϕ
Then (f(r1) + j)(ϕ(m1) + m2)∈N2and so s(f(r1) + j)∈(N2:R2M2) or
s(ϕ(m1) + m2)∈N2. If s(f(r1) + j)∈(N2:R2M2), then for all (m1, ϕ(m1) +
m2)∈M1⋊⋉ϕJM2, clearly (t, s)(r1, f (r1) + j)(m1, ϕ(m1) + m2)∈N2
ϕand so
(t, s)(r1, f (r1) + j)∈(N2
ϕ:R1⋊⋉fJM1⋊⋉ϕJM2). If s(ϕ(m1) + m2)∈N2, then
(t, s)(m1, ϕ(m1) + m2)∈N2
ϕand the result follows. Conversely, suppose N2
ϕis
an Sϕ-prime submodule of M1⋊⋉ϕJM2associated to (t, f (t) + j) = (t, s)∈Sϕ.
Let r2=f(r1)∈R2and m2=ϕ(m1)∈M2such that r2m2∈N2. Then (r1, r2)∈
R1⋊⋉fJand (m1, m2)∈M1⋊⋉ϕJ M2with (r1, r2)(m1, m2)∈N2
ϕ. Thus,
(t, s)(r1, r2)(M1⋊⋉ϕJM2)⊆N2
ϕor (t, s)(m1, m2)∈N2
ϕ. If (t, s)(r1, r2)(M1⋊⋉ϕ
JM2)⊆N2
ϕ, then for all m=ϕ(m′)∈M2, we have (t, s)(r1, r2)(m′, m)∈N2
ϕ
and so sr2M2⊆N2. If (t, s)(m1, m2)∈N2
ϕ, then sm2∈N2and we are done.
(2). Suppose N2
ϕis an Sϕ-prime submodule of M1⋊⋉ϕJM2associated to
(t, f (t) + j′) = (t, s)∈Sϕ. Let r2∈R2,m2∈M2such that r2m2∈(N2:M2
J). Then r2Jm2⊆N2and so for all j∈J, we have (r1, f(r1))(0, j m2)∈N2
ϕ
where f(r1) = r2. By assumption, (t, s)(r1, r2)∈(N2
ϕ:R1⋊⋉fJM1⋊⋉ϕJM2) or
(t, s)(0, jm2)∈N2
ϕ. If (t, s)(r1, r2)∈(N2
ϕ:R1⋊⋉fJM1⋊⋉ϕJM2), then for all
m2∈M2and all j∈J, we have (t, s)(r1, r2)(0, jm2)∈N2
ϕand so sr2jm2∈N2.
Thus, sr2∈(N2:R2JM2) = ((N2:M2J) :R2M2). If (t, s)(r1, r2)/∈(N2
ϕ:R1⋊⋉fJ
M1⋊⋉ϕJM2), then (t, s)(0, jm2)∈N2
ϕfor all j∈Jand so sm2∈(N2:M2J) as
required.
In particular, if we consider S={1R2}and take T={(1R1,1R2)}instead of Sϕ
in Theorem 9, then we get the following corollary.
Corollary 5. Consider the (R1⋊⋉fJ)-module M1⋊⋉ϕJM2defined as in Theorem
8 where fand ϕare epimorphisms and let N2be a submodule of M2. Then
(1) N2is a prime submodule of M2if and only if N2
ϕis a prime submodule of
M1⋊⋉ϕJM2.
(2) If N2
ϕis a prime submodule of M1⋊⋉ϕJM2and J*(N2:R2M2), then
(N2:M2J) is a prime submodule of M2.
Theorem 10. Consider the (R1⋊⋉fJ)-module M1⋊⋉ϕJM2defined as in Theorem
8 where fand ϕare epimorphisms. Let Sbe a multiplicatively closed subsets of R2
and N2be a submodule of M2. Then
(1) N2
ϕis a weakly Sϕ-prime submodule of M1⋊⋉ϕJ M2if and only if N2is a
weakly S-prime submodule of M2and for r1∈R1,m1∈M1,m2∈J M2,
j∈Jwith (f(r1) + j)(ϕ(m1) + m2) = 0 but s(f(r1) + j)/∈(N2:R2M2)
and s(ϕ(m1) + m2)/∈N2for all s∈S, then r1m1= 0.
(2) If N2
ϕis a weakly Sϕ-prime submodule of M1⋊⋉ϕJM2, (N2:R2JM2)∩S=
φand ZR2(M2)∩J={0}, then (N2:M2J) is a weakly S-prime submodule
of M2.
ON WEAKLY S-PRIME SUBMODULES 17
Proof. (1). Suppose s=f(t)∈Sis a weakly S-element of N2. Let (r1, f (r1)+ j)∈
R1⋊⋉fJand (m1, ϕ(m1) + m2)∈M1⋊⋉ϕJ M2such that
(0,0) 6= (r1, f (r1) + j)(m1, ϕ(m1) + m2)∈N2
ϕ
Then (f(r1) + j)(ϕ(m1) + m2)∈N2. If (f(r1) + j)(ϕ(m1) + m2)6= 0, then the
result follows as in the proof of (1) in Theorem 9. Suppose (f(r1)+j)(ϕ(m1)+m2) =
0 so that r1m16= 0. Then by assumption, there exists s′=f(t′)∈Ssuch that
s′(f(r1) + j)∈(N2:R2M2) or s′(ϕ(m1) + m2)∈N2. It follows clearly that
(t′, s′)(r1, f (r1)+j)∈(N2
ϕ:R1⋊⋉fJM1⋊⋉ϕJM2) or (t′, s′)(m1, ϕ(m1)+m2)∈N2
ϕ.
Hence, (tt′, ss′) is a weakly Sϕ-element of N2
ϕ. Conversely, let (t, f (t) + j) = (t, s)
be a weakly Sϕ-element of N2
ϕ. Let r2=f(r1)∈R2and m2=f(m1)∈M2
such that 0 6=r2m2∈N2. Then (r1, r2)∈R1⋊⋉fJand (m1, m2)∈M1⋊⋉ϕJ M2
with (0.0) 6= (r1, r2)(m1, m2)∈N2
ϕ. Hence, either (t, s)(r1, r2)∈(N2
ϕ:R1⋊⋉fJ
M1⋊⋉ϕJM2) or (t, s)(m1, m2)∈N2
ϕ. In the first case, for all m=ϕ(m′)∈M2,
(tr1, sr2)(m′, m)∈N2
ϕ. Hence, sr2m∈N2and then sr2∈(N2:R2M2). In the
second case, we have sm2∈N2and so sis a weakly S-element of N2. Now, let
r1∈R1,m1∈M1,m2∈JM2,j∈Jwith (f(r1) + j)(ϕ(m1) + m2) = 0 and
suppose r1m16= 0. Then (0,0) 6= (r1, f (r1) + j)(m1, ϕ(m1) + m2)∈N2
ϕand so
(t, s)(r1, f (r1) + j)∈(N2
ϕ:R1⋊⋉fJM1⋊⋉ JM2) or (t, s)(m1, ϕ(m1) + m2)∈N2
ϕ.
Hence, clearly, either s(f(r1) + j)∈(N2:R2M2) or s(ϕ(m1) + m2)∈N1and the
result follows by contrapositive.
(2) Suppose (t, f (t)+ j) = (t, s) is a weakly Sϕ-element of N2
ϕ. Let r2=f(r1)∈
R2,m2∈M2such that 0 6=r2m2∈(N2:M2J). Then r2Jm2⊆N2and so for all
j∈J, we have (r1, r2)(0, jm2)∈N2
ϕ. If j6= 0 and (r1, r2)(0, j m2) = (0,0), then
r2jm2= 0 and so r2m2= 0 as ZR2(N2)∩J={0}, a contradiction. Thus, for all
j6= 0, (r1, r2)(0, jm2)6= (0,0). By assumption and similar to the proof of (2) of
Theorem 9, we have for all j6= 0, either sr2jm2∈N2or (t, s)(0, j m2)∈N2
ϕfor all
m2∈M2. Thus, sr2∈(N2:R2J M2) = ((N2:M2J) :R2M2) or sm2∈(N2:M2J)
and we are done.
Corollary 6. Consider the (R1⋊⋉fJ)-module M1⋊⋉ϕJM2defined as in Theorem
8 where fand ϕare epimorphisms. If N2is a submodule of M2, then
(1) N2
ϕis a weakly prime submodule of M1⋊⋉ϕJM2if and only if N2is a
weakly prime submodule of M2and for r1∈R1,m1∈M1,m2∈J M2,
j∈Jwith (f(r1) + j)(ϕ(m1) + m2) = 0 but (f(r1) + j)/∈(N2:R2M2) and
(ϕ(m1) + m2)/∈N2, then r1m1= 0.
(2) If N2
ϕis a weakly prime submodule of M1⋊⋉ϕJM2,J*(N2:R2M2) and
ZR2(N2)∩J={0}, then (N2:M2J) is a weakly prime submodule of M2.
Corollary 7. Let Nbe a submodule of an R-module M,Jan ideal of Rand Sa
multiplicatively closed subset of R.
(1) N⋊⋉ Jis an (S⋊⋉ J)-prime submodule of M⋊⋉ Jif and only if Nis an
S-prime submodule of M.
(2) N⋊⋉ Jis a weakly (S⋊⋉ J)-prime submodule of M⋊⋉ Jif and only if N
is a weakly S-prime submodule of Mand for r∈R,m∈Mwith rm = 0
but sr /∈(N:R1M) and sm /∈Nfor all s∈S, then (r+j)m′= 0 for every
j∈Jand m′∈Mwhere (m, m′)∈M⋊⋉ J.
18 HANI A. KHASHAN AND ECE YETKIN CELIKEL
Corollary 8. Let Nbe a submodule of an R-module M,Jan ideal of Rand Sa
multiplicatively closed subset of R.
(1) Nis an S-prime submodule of Mif and only if Nis an S-prime submodule
of M⋊⋉ J.
(2) If Nis an S-prime submodule of M⋊⋉ Jand (N:RJ M )∩S=φ, then
(N:MJ) is an S-prime submodule of M.
Corollary 9. Let Nbe a submodule of an R-module M,Jan ideal of Rand Sa
multiplicatively closed subset of R.
(1) Nis a weakly S-prime submodule of M⋊⋉ Jif and only if Nis a weakly
S-prime submodule of Mand for r∈R,m∈M,m′∈JM,j∈Jwith
(r+j)(m+m′) = 0 but s(r+j)/∈(N:RM) and s(m+m′)/∈Nfor all
s∈S, then rm = 0.
(2) If Nis a weakly S-prime submodule of M⋊⋉ J, (N:MJ)∩S=φand
ZR(N)∩J={0}, then (N:MJ) is a weakly S-prime submodule of M.
In the following example, we show that in general Nis a weakly S-prime sub-
module of Mdoes not imply N⋊⋉ Jis a weakly (S⋊⋉ J)-prime submodule of
M⋊⋉ J.
Example 6. Consider the Z-submodule N= 0×h¯
0iof M=Z×Z6and let J= 2Z.
Then Nis a weakly prime submodule of M. Now
M⋊⋉ J={(m, m′)∈M×M:m−m′∈JM = 2Z× h¯
2i}
and
N⋊⋉ J={(n, m)∈N×M:n−m∈2Z× h¯
2i}
If we consider (2,4) ∈Z ⋊⋉ Jand ((0,¯
3),(0,¯
1)) ∈M⋊⋉ J, then we have (2,4).((0,¯
3),(0,¯
1)) =
((0,¯
0),(0,¯
4)) ∈N⋊⋉ J. But we have (2,4) /∈((N⋊⋉ J) :Z⋊⋉I(M⋊⋉ J)) as for ex-
ample (2,4)((2,¯
2),(0,¯
0)) /∈N⋊⋉ Jand ((0,¯
3),(0,¯
1)) /∈N⋊⋉ J. Thus, N⋊⋉ Jis
not a weakly prime submodule of M⋊⋉ J.
We note that the condition in the reverse implication of Corollary 7 (2) does not
hold in example 6. For example, if we take r= 2 and m= (0,¯
3) ∈M, then clearly,
rm = 0, r /∈(N:RM) = 0 and m /∈Nbut for m′= (0,¯
2) ∈JM = 2Z× h¯
2i, we
have (r+ 0)m′6= 0.
Also, if the condition in the reverse implication of Corollary 9 (1) does not hold,
then we may find a weakly S-prime submodule Nof Msuch that Nis not a weakly
S-prime submodule of M⋊⋉ J.
Example 7. Consider N,Mand Jas in Example 6. If we consider (2,4) ∈
Z ⋊⋉ Jand ((0,¯
1),(0,¯
3)) ∈M⋊⋉ J, then we have (2,4).((0,¯
1),(0,¯
3)) = N. But
(2,4) /∈(¯
N:Z⋊⋉I(M⋊⋉ J)) and ((0,¯
1),(0,¯
3)) /∈N. Thus, ¯
Nis not a weakly prime
submodule of M⋊⋉ J.
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Department of Mathematics, Faculty of Science, Al al-Bayt University, Al Mafraq,
Jordan.
Email address:hakhashan@aabu.edu.jo.
Department of Basic Sciences, Faculty of Engineering, Hasan Kalyoncu University,
Gaziantep, Turkey.
Email address:ece.celikel@hku.edu.tr, yetkinece@gmail.com.
