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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(3): 445-453

Published online September 30, 2020

### ϕ-prime Subsemimodules of Semimodules over Commutative Semirings

Fatemeh Fatahi and Reza Safakish*

Faculty of Mathematical Sciences, Bouali Sina University, Hamadan, P. O. Box 65178-38695, Iran
e-mail : f.fatahi.kntu@gmail.com and re.sa.hamedani@gmail.com

Received: December 9, 2016; Revised: November 26, 2018; Accepted: January 21, 2019

### Abstract

Let R be a commutative semiring with identity and M be a unitary R-semimodule. Let be a function, where is the set of all subsemimodules of M. A proper subsemimodule N of M is called φ-prime subsemimodule, if rR and xM with rxN \ φ(N) implies that r ∈ (N :RM) or xN. So if we take φ(N) = ∅︀ (resp., φ(N) = {0}), a φ-prime subsemimodule is prime (resp., weakly prime). In this article we study the properties of several generalizations of prime subsemimodules.

Keywords: prime subsemimodule, weakly prime subsemimodule, ϕ,-prime subsemimodule

### 1. Introduction

Anderson and Bataineh [3] have introduced the concept of φ-prime ideals in a commutative ring as a generalization of weakly prime ideals in a commutative ring introduced by Anderson and Smith [2]. After that several authors [1, 4, 9, 14, 15], etc. explored this concept in different ways either in commutative ring or semiring theory. In this article, we have extended the results of prime ideals of commutative ring to prime subsemimodules of semimodules.

For the definitions of monoid, semiring, semimodule and subsemimodule of a semimodule we refer [6]. All semirings in this paper are commutative with non-zero identity. The semiring R is to be also a semimodule over itself. In this case, the subsemimodules of R are called ideals of R. Let M be a semimodule over a semiring R. A subtractive subsemimodule (= k-subsemimodule) N is a subsemimodule of M such that if x, x + yN, yM, then yN. If N is a proper subsemimodule of an R-semimodule M, then we denote (N :RM) = {rR : rMN} and $(N:RM)={r∈R:rnM⊆N for some n∈N}$. We say that N is a radical submodule of M if $(N:RM)=(N:RM)$.

### 2. φ-prime Subsemimodules

In this section, we introduce the concept of a φ-prime subsemimodule of a semimodule M over a commutative semiring R and prove some results related to it.

### Definition 2.1

Let be the set of subsemimodule of M and be a function. The proper subsemimodule N of M is called a φ-prime semimodule if rR, xM and rxN \ φ(N), then r ∈ (N :RM) or xN.

Since N \ φ(N) = N \ (Nφ(N)), we will assume throughout this article that φ(N) ⊆ N. In the rest of the article we use the following functions .

$φ∅(N)=∅, ∀N∈S(M),φ0(N)={0}, ∀N∈S(M),φ1(N)=(N:RM)N, ∀N∈S(M),φ2(N)=(N:RM)2N, ∀N∈S(M),φw(N)=∩i=1∞(N:RM)iN, ∀N∈S(M).$

Then it is clear that φ∅︀ and φ0-prime subsemimodules are prime and weakly prime subsemimodules respectively. Evidently for any subsemimodule and every positive integer n ⩾ 2, we have the following implications:

$prime⇒φw-prime⇒φn-prime⇒φn-1-prime.$

For functions , we write φψ if φ(N) ⊆ ψ(N) for each . So whenever φψ, any φ-prime subsemimodule is ψ-prime.

### Definition 2.2

A proper subsemimodule N of M is called M-subtractive if N and φ(N) are subtractive subsemimodules of M.

The following propositions asserts that under some conditions φ-prime subsemimodules are prime.

### Proposition 2.3

If N is a φ-prime subsemimodule of M and φ(N) is a prime subsemimodule, then N is a prime subsemimodule of M.

Proof

Let rR, xM and rxN. If rxφ(N), since N is a φ-prime subsemimodule, then xN or r ∈ (N :RM), and if rxφ(N), then xφ(N) ⊆ N or r ∈ (φ(N) :RM) ⊆ (N :RM).

### Proposition 2.4

Let R be a semiring and M be an R-semimodule. Letbe a function and N be a φ-prime M-subtractive subsemimodule of M such that (N :RM)Nφ(N). Then N is a prime subsemimodule of M.

Proof

Let aR and xM be such that axN. If axφ(N), then since N is φ-prime, we have a ∈ (N :RM) or xN.

So let axφ(N). In this case we may assume that aNφ(N). For, let aNφ(N). Then there exists pN such that apφ(N), so that a(x + p) ∈ N\φ(N). Therefore a ∈ (N :RM) or x + pN and hence a ∈ (N :RM) or xN. Second we may assume that (N :RM)xφ(N). If this is not the case, there exists u ∈ (N :RM) such that uxφ(N) and so (a + u)xN\φ(N). Since N is a φ-prime subsemimodule, we have a+u ∈ (N :RM) or xN. By [7, Lemma 1.2], (N :RM) is a subtractive ideal, and hence either a ∈ (N :RM) or xN. Now since (N :RM)Nφ(N), there exist r ∈ (N :RM) and pN such that rpφ(N). So (a+r)(x+p) ∈ N\φ(N), and hence a+r ∈ (N :RM) or x+pN. Therefore a ∈ (N :RM) or xN, since N is a subtractive subsemimodule and by [7, Lemma 1.2], (N :RM) is a subtractive ideal. Hence N is a prime subsemimodule of M.

### Corollary 2.5

Let P be a weakly prime subtractive subsemimodule of M such that (P :RM)P ≠ 0. Then P is a prime subsemimodule of M.

Proof

In Proposition 2.4 set φ = φ0.

### Corollary 2.6

Let M be an R-semimodule and P be a φ-prime M-subtractive subsemimodule of M. Then

• (i) $(P:RM)⊆(φ(P):RM)$or$(φ(P):RM)⊆(P:RM)$.

• (ii) If$(P:RM)⊈(φ(P):RM)$then P is not a prime subsemimodule of M.

• (iii) If$(φ(P):RM)⊈(P:RM)$, then P is a prime subsemimodule of M.

• (iv) If φ(P) is a radical subsemimodule of M, either (P :RM) = (φ(P) :RM) or P is a prime subsemimodule of M.

Proof

(i) If P is not a prime subsemimodule of M, then by Proposition 2.4, we have (P :RM)Pφ(P). Hence

$(P:RM)2⊆((P:RM)P:RM)⊆(φ(P):RM).$

So $(P:RM)⊆(φ(P):RM)$. If P is a prime subsemimodule of M, then by [12, Lemma 3.3], $(φ(P):RM)⊆(P:RM)=(P:RM)$(note that we may assume that φ(P) ⊆ P), and all the claims of the corollary follows.

(ii) Suppose P is a prime subsemimodule of M. Then by (i) we have, $(φ(P):RM)⊆(P:RM)$. Also by assumption $(P:RM)⊈(φ(P):RM)$. Thus $(φ(P):RM)⊆(P:RM)⊈(φ(P):RM)$, which is a contradiction.

(iii) Suppose P is not a prime subsemimodule of M. Then by (i) we have, $(P:RM)⊆(φ(P):RM)$. Also by assumption $(φ(P):RM)⊈(P:RM)$. Thus $(P:RM)⊆(φ(P):RM)⊈(P:RM)$, which is a contradiction.

(iv) Suppose that P is not prime subsemimodule. Then by (i) and definition of radical subsemimodule we have

$(P:RM)⊆(φ(P):RM)=(φ(P):RM)⊆(P:RM),$

so (P :RM) = (φ (P) :RM).

Let M be an R-semimodule and N a subsemimodule of M. For every rR, {mM| rmN} is denoted by (N :Mr). It is easy to see that (N :Mr) is a subsemimodule of M containing N.

### Theorem 2.7

Let N be a M-subtractive subsemimodule of M and letbe a function. Then the following are equivalent:

• (i) N is a φ-prime subsemimodule of M;

• (ii) for xM \ N, (N :Rx) = (N :RM) ∪ (φ(N) :Rx);

• (iii) for xM \ N, (N :Rx) = (N :RM) or (N :Rx) = (φ(N) :Rx);

• (iv) for any ideal I of R and any subsemimodule L of M, if ILN and ILφ(N), then I ⊆ (N :RM) or LN;

• (v) for any rR \ (N :RM), (N :Mr) = N ∪ (φ(N) :Mr);

• (vi) for any rR \ (N :RM), (N :Mr) = N or (N :Mr) = (φ(N) :Mr).

Proof

(i) ⇒ (ii). Let xM \N, a ∈ (N :Rx) \ (φ(N) :Rx). Then axN \ φ(N). Since N is a φ-prime subsemimodule of M, So a ∈ (N :RM). As we may assume that φ(N) ⊆ N, the other inclusion always holds.

(ii) ⇒ (iii). It is follows by [7, Lemma 1.2] and [16, Lemma 6].

(iii) ⇒ (iv). Let I be an ideal of R and L be a subsemimodule of M such that ILN. Suppose I ⊈ (N :RM) and LN. We show that ILφ(N). Let aI and xL. First let a ∉ (N :RM). Then, since axN, we have (N :Rx) ≠ (N :RM). Hence by our assumption (N :Rx) = (φ(N) :Rx). So axφ(N). Now assume that aI ∩ (N :RM). Let uI \ (N :RM). Then a + uI \ (N :RM). So by the first case, for each xL we have uxφ(N) and (a+u)xφ(N). This gives that axφ(N). Thus in any case axφ(N), since N is M-subtractive. Therefore ILφ(N).

(iv) ⇒ (i). Let axN \ φ(N). Then (a)(x) ⊆ N \ φ(N). By considering the ideal (a) and the submodule (x), the result follows.

(i) ⇒ (v) Suppose N is a φ-prime subsemimodule such that r ∉ (N :RM). Let m ∈ (N :Mr). If rmφ(N), then N is φ-prime implies mN, if rmφ(N), then m ∈ (φ(N) :Mr). Hence (N :Mr) ⊆ N ∪ (φ(N) :Mr). The other containment holds trivially.

(v) ⇒ (vi) Let (N :Mr) = N ∪ (φ(N) :Mr) for rR \ (N :RM). Then by [10, Lemma 3.13] and [5, Theorem 6], either (N :Mr) = N or (N :Mr) = (φ(N) :Mr). Therefore, the result follows.

(vi) ⇒ (i) Let rR\(N :RM), mM such that rmN \φ(N). By assumption, either (N :Mr) = N or (N :Mr) = (φ(N) :Mr). As rmφ(N), then m ∉ (φ(N) :Mr) and as m ∈ (N :Mr), we have (N :Mr) ≠ (φ(N) :Mr). Hence (N :Mr) = N and so mN as required.

Recall from [13, Definition 2] that, an R-subsemimodule N of M is said to be a strong subsemimodule if for each xN, there exists yN such that x + y = 0.

### Theorem 2.8

Let f : MMbe an epimorphism of R-semimodules with f(0) = 0 and letandbe two functions. Then the following statements hold:

• (i) If Nis a φ-prime subsemimodule of Mand φ(f−1(N′)) = f−1(φ′ (N′)), then f−1(N′) is a φ-prime subsemimodule of M.

• (ii) If N is a strong φ-prime subsemimodule of M containing Ker(f) and φ′ (f(N)) = f(φ(N)), then f(N) is a φ-prime subsemimodule of M′.

Proof

(i) Since f is epimorphism, f−1(N′) is a proper subsemimodule of M. Let aR and mM such that amf−1(N′) \ φ(f−1(N′)). Since amf−1(N′), so af(m) ∈ N′. Also, af(m) ∉ φ′ (N′) as φ(f−1(N)) = f−1(φ ′ (N′)). Therefore af(m) ∈ N′ \ φ′ (N′), and so either a ∈ (N′ : M′) or f(m) ∈ N′. Hence either aM′ ⊆ N′, or mf−1(N′). Therefore either af−1(M′) ⊆ f−1(N′), or mf−1(N′). Thus, either a ∈ (f−1(N′) : M) or mf−1(N′). It follows that f−1(N′) is a φ-prime subsemimodule of M.

(ii) Let axf(N) \ φ′ (f(N)) for some aR and xM′. Since axf(N), there exists an element x′ ∈ N such that ax = f(x′). Since f is epimorphism and xM′, then there exists yM such that f(y) = x. As x′ ∈ N and N is a strong subsemimodule of M, there exists x″ ∈ N such that x′ +x″ = 0, which gives f(x′ +x″) = 0. Therefore, ax+f(x′) = 0 or f(ay)+f(x″) = 0 which implies that ay + x″ ∈ KerfN. Thus ayN, since N is a substractive subsemimodule. Since axφ′ (f(N)), one can see easily that ayφ(N). So ayN \ φ(N), and hence either a ∈ (N : M) or yN, as N is φ-prime. Thus, either aMN, or f(y) ∈ f(N), and hence either af(M) ⊆ f(N), or xf(N). Thus, a ∈ (f(N) : M′) or xf(N). It follows that f(N) is a φ-prime subsemimodule of M′.

The notion of fractions of semimodule M was defined by Atani in [11]. Also it show that there is a one-to-one correspondence between the set of all prime subsemimodules N of M with (N : M) ∩ S = ∅︀ and the set of all prime subsemimodules of the RP-semimodule MP , see [11, Corollary 2]. In the next theorem we want to generalize this fact for φ-prime subsemimodules. Let S be a multiplicatively closed subset of R and N(S) = {xM : ∃sS, sxN}. We know that N(S) is a subsemimodule of M containing N and S−1(N(S)) = S−1N. Let be a function and define by φS(S−1N) = S−1(φ(N(S)) if φ(N(S)) ≠ ∅︀, and φS(S−1N) = ∅︀ if φ(N(S)) = ∅︀. Since φ(N) ⊆ N, φS(S−1N) ⊆ S−1N.

### Lemma 2.9

If N is a subtractive subsemimodule of M, so is S−1NM.

Proof

Let a, a + bS−1NM. Then a/1 = c1/s1 and (a + b)/1 = c2/s2, such that c1, c2N and s1, s2S. So t2(a + b)s2 = t2c2N and t2t1as1 = t2t1c1N for some t1, t2S. Consequently t1t2as1s2 + t1t2bs1s2 = t1t2s1c2N. Since N is subtractive, therefore t1t2bs1s2N. Thus b/1 = (t1t2s1s2b)/(t1t2s1s2) ∈ S−1N, and so bS−1NM. Hence S−1NM is subtractive.

### Theorem 2.10

Let N be a φ-prime subsemimodule of M. Let S be a multiplicatively closed subset of R such that S−1NS−1M and S−1(φ(N)) ⊆ φS(S−1N). Then S−1N is a φS-prime subsemimodule of S−1M. Furthermore if N is a Msubtractive subsemimodules of M and S−1NS−1(φ(N)), then N = N(S) = S−1NM.

Proof

First suppose that there is an element s which is common to (N :RM) and S. So, sMN. Suppose now that m/s′ ∈ S−1M. Then m/s′ = sm/ss′ ∈ S−1N. This show that S−1N = S−1M and the first assertion follows. From here on we assume that (N :RM) ∩ S = ∅︀ and N be a φ-prime subsemimodule of M. Let (a1/s1)(a2/s2) ∈ S−1N \ φS(S−1N) for some a1R, a2M, s1, s2S. Then there exist cN and s′ ∈ S such that a1a2/s1s2 = c/s′. Then usa1a2 = us1s2cN for some uS. If usa1a2φ(N), then (a1/s1)(a2/s2) = (usa1a2)/(uss1s2) ∈ S−1(φ(N)) ⊆ φS(S−1N), a contradiction. Hence usa1a2Nφ(N). As N is a φ-prime subsemimodule, we get usa1 ∈ (N :RM) or a2N. It follows that a1/s1 = (usa1)/(uss1) ∈ S−1(N :RM) ⊆ (S−1N :S−1RS−1M), or a2/s2S−1N. Consequently S−1N is a φS-prime subsemimodule of S−1M.

To prove the last part of the theorem, assume that N be M-subtractive subsemimodules of M and S−1NS−1(φ(N)). Clearly, NS−1NM. For the reverse containment, pick an element aS−1NM. Then there exist cN and sS with a/1 = c/s. Therefore, tsa = tc for some tS. If (ts)aφ(N), then (ts)aN \φ(N) and (N :RM)∩S = ∅︀ gives aN. So assume that (ts)aφ(N). In this case aS−1(φ(N)) ∩M. Therefore, S−1NMN ∪ (S−1(φ(N)) ∩M). It follows that either S−1NM = S−1(φ(N)) ∩ M or S−1NM = N by Lemma 2.9 and [5, Theorem 6]. If the former case holds, then by [14, Lemma 2.5], S−1N = S−1(φ(N)) which is a contradiction. So the result follows.

We say that rR is a zero-divisor of a semimodule M if rm = 0 for some non-zero element m of M. The set of all zero-divisors of M is denoted by ZR(M) (see [11, Definition1 (f)]).

### Proposition 2.11

Let S−1N be a φS-prime subsemimodule of S−1M such that S−1(φ(N)) = φS(S−1N), SZR(M/N) = ∅︀ and SZR(N/φ(N)) = ∅︀. Then N is a φ-prime subsemimodule of M.

Proof

Let aR, mM, amN\φ(N) and mN. Then (a/1)(m/1) = am/1 ∈ S−1N. If (a/1)(m/1) ∈ φS(S−1N) = S−1(φ(N)), then there exists sS such that samφ(N) which contradicts SZR(N/φ(N)) = ∅︀. Therefore (a/1)(m/1) ∈ S−1N \ φS(S−1N), and so either a/1 ∈ (S−1N : S−1M) or m/1 ∈ S−1N. If m/1 ∈ S−1N, then smN for some sS, and so sSZR(M/N) which is a contradiction. Thus a/1 ∈ (S−1N : S−1M). Hence for every cM, sacN and since sS, we have sZR(M/N). Since SZR(M/N) = ∅︀, it follows that acN, and so a ∈ (N :RM).

A subsemimodule N of an R-semimodule M is called partitioning subsemimodule (= Q-subsemimodule) if there exists a subset Q of M such that

• (1) M = ∪{q + N : qQ}, and

• (2) (q1 + N) ∩ (q2 + N) ≠ ∅︀, for any q1, q2Q if and only if q1 = q2.

Let N be a Q-subsemimodule of an R-semimodule M, and M/N(Q) = {q +N : qQ}. Then M/N(Q) forms an R-semimodule under the binary operations ⊕ and ⊙ , which are defined as follows: (q1 + N) ⊕ (q2 + N) = q3 + N where q3Q is the unique element such that q1 + q2 + Nq3 + N and r ⊙ (q1 + N) = q4 + N, where rR and q4Q is the unique element such that rq1 + Nq4 + N. Then, this R-semimodule M/N(Q) is called the quotient semimodule of M by N.

Moreover, if N is a Q-subsemimodule of an R-semimodule M and L is a k-subsemimodule of M containing N, then N is a QL-subsemimodule of L and L/N = {q +N : qQL} is a subtractive subsemimodule of M/N(Q) as given by Chaudhari and Bonde in 2010, see [8].

Let N be a Q-subsemimodule of M and be a function. We define by φN(P/N) = (φ(P))/N for every with Nφ(P) where P is a M-subtractive subsemimodule (and φN(P/N) = ∅︀ if φ(P) = ∅︀).

### Theorem 2.12

Let N be a Q-subsemimodule of an R-semimodule M, P a M - subtractive subsemimodule and N a φ-prime subsemimodule of M with Nφ(P). Then P is a φ-prime subsemimodule of M if and only if P/N(QP)is a φ-prime subsemimodule of M/N(Q).

Proof

Let P be a φ-prime subsemimodule of M. Let rR, q1 + NM/N(Q) be such that r ⊙ (q1 +N) ∈ P/N(QP) \ (φ(P))/N. By [8, Theorem 3.5], there exists a unique q2QP such that r ⊙ (q1 +N) = q2 +N where rq1 +Nq2 +N. Since NP, rq1P.

If rq1φ(P), then rq1 ∈ (q2 +N) ∩ (q0 +N) and q0φ(P). So q2 = q0 and hence q0+N = q2+N a contradiction. Thus rq1φ(P). As P is φ-prime subsemimodule, either rMP or q1P. If q1P, then q1QP and hence q1+NP/N(QP). Suppose rMP. For q + NM/N(Q), let r ⊙ (q + N) = q3 + N where q3 is a unique element of Q such that rq + Nq3 + N. Since NP and P is a subtractive subsemimodule of M, q3P. Hence q3QP. Now r ⊙ (q + N) = q3+NP/N(QP) and hence rM/N(Q)P/N(QP). So P/N(QP) is a φ-prime subsemimodule of M/N(Q).

Conversely, suppose that N,P/N(QP) are φ-prime subsemimodules of M, M/N(Q) respectively. Let rmP \ φ(P) where rR,mM. If rmN, then we are through, since rmN \ φ(N) and N is a φ-prime subsemimodule of M. So suppose rmP \ N. By using [7, Lemma 3.6], there exists a unique q1Q such that m ∈ (q1 + N) and rmr ⊙ (q1 + N) = q2 + N where q2 is a unique element of Q such that rq1 + Nq2 + N. Now rmP,rmq2 + N implies q2P, as P is a subtractive subsemimodule and NP. Hence r ⊙ (q1 + N) = q2 + NP/N(QP) \ (φ(P))/N. As P/N(QP) is a φ-prime subsemimodule, either rM/N(Q)P/N(QP) or q1 + NP/N(QP). If q1 + NP/N(QP), then q1P. Hence mq1 + NP. So assume that rM/N(Q)P/N(QP). Let xM. By using [7, Lemma 3.6], there exists a unique q3Q such that xq3 + N and rxr ⊙ (q3 + N) = q4 + N where q4 is a unique element of Q such that rq3 +Nq4 +N. Now q4 +N = r ⊙ (q3 +N) ∈ P/N(QP) and hence q4P. As rxq4 + N and NP implies rxP. So rMP.

### Theorem 2.13

Let R be semiring, I a Q-ideal of R and P a subtractive ideal of R with IP. Then P is a prime (weakly prime) ideal of R if and only if P/I(QP)is a prime (weakly prime) ideal of R/I(Q).

Proof

The proof is similar as in the proof of Theorem 2.12.

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