In this section, we introduce the concept of a φprime 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 φprime semimodule if r ∈ R, x ∈ M and rx ∈ N \ φ(N), then r ∈ (N :_{R}M) or x ∈ N.
Since N \ φ(N) = N \ (N ∩ φ(N)), we will assume throughout this article that φ(N) ⊆ N. In the rest of the article we use the following functions .
$$\begin{array}{l}{\phi}_{\varnothing}(N)=\varnothing ,\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\forall N\in \mathcal{S}(M),\\ {\phi}_{0}(N)=\{0\},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\forall N\in \mathcal{S}(M),\\ {\phi}_{1}(N)=(N{:}_{R}M)N,\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\forall N\in \mathcal{S}(M),\\ {\phi}_{2}(N)={(N{:}_{R}M)}^{2}N,\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\forall N\in \mathcal{S}(M),\\ {\phi}_{w}(N)={\cap}_{i=1}^{\infty}{(N{:}_{R}M)}^{i}N,\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\forall N\in \mathcal{S}(M).\end{array}$$
Then it is clear that φ_{∅︀} and φ_{0}prime subsemimodules are prime and weakly prime subsemimodules respectively. Evidently for any subsemimodule and every positive integer n ⩾ 2, we have the following implications:
$$prime\Rightarrow {\phi}_{w}prime\Rightarrow {\phi}_{n}prime\Rightarrow {\phi}_{n1}prime.$$
For functions , we write φ ⩽ ψ if φ(N) ⊆ ψ(N) for each . So whenever φ ⩽ ψ, any φprime subsemimodule is ψprime.
Definition 2.2
A proper subsemimodule N of M is called Msubtractive if N and φ(N) are subtractive subsemimodules of M.
The following propositions asserts that under some conditions φprime subsemimodules are prime.
Proposition 2.3
If N is a φprime subsemimodule of M and φ(N) is a prime subsemimodule, then N is a prime subsemimodule of M.
Proof
Let r ∈ R, x ∈ M and rx ∈ N. If rx ∉ φ(N), since N is a φprime subsemimodule, then x ∈ N or r ∈ (N :_{R}M), and if rx ∈ φ(N), then x ∈ φ(N) ⊆ N or r ∈ (φ(N) :_{R}M) ⊆ (N :_{R}M).
Proposition 2.4
Let R be a semiring and M be an Rsemimodule. Letbe a function and N be a φprime Msubtractive subsemimodule of M such that (N :_{R}M)N ⊈ φ(N). Then N is a prime subsemimodule of M.
Proof
Let a ∈ R and x ∈ M be such that ax ∈ N. If ax ∉ φ(N), then since N is φprime, we have a ∈ (N :_{R}M) or x ∈ N.
So let ax ∈ φ(N). In this case we may assume that aN ⊆ φ(N). For, let aN ⊈ φ(N). Then there exists p ∈ N such that ap ∉ φ(N), so that a(x + p) ∈ N\φ(N). Therefore a ∈ (N :_{R}M) or x + p ∈ N and hence a ∈ (N :_{R}M) or x ∈ N. Second we may assume that (N :_{R}M)x ⊆ φ(N). If this is not the case, there exists u ∈ (N :_{R}M) such that ux ∉ φ(N) and so (a + u)x ∈ N\φ(N). Since N is a φprime subsemimodule, we have a+u ∈ (N :_{R}M) or x ∈ N. By [7, Lemma 1.2], (N :_{R}M) is a subtractive ideal, and hence either a ∈ (N :_{R}M) or x ∈ N. Now since (N :_{R}M)N ⊈ φ(N), there exist r ∈ (N :_{R}M) and p ∈ N such that rp ∉ φ(N). So (a+r)(x+p) ∈ N\φ(N), and hence a+r ∈ (N :_{R}M) or x+p ∈ N. Therefore a ∈ (N :_{R}M) or x ∈ N, since N is a subtractive subsemimodule and by [7, Lemma 1.2], (N :_{R}M) 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 :_{R}M)P ≠ 0. Then P is a prime subsemimodule of M.
Proof
In Proposition 2.4 set φ = φ_{0}.
Corollary 2.6
Let M be an Rsemimodule and P be a φprime Msubtractive subsemimodule of M. Then

(i) $(P{:}_{R}M)\subseteq \sqrt{(\phi (P){:}_{R}M)}\hspace{0.17em}$or$\sqrt{(\phi (P){:}_{R}M)}\subseteq (P{:}_{R}M)$.

(ii) If$(P{:}_{R}M)\u2288\sqrt{(\phi (P){:}_{R}M)}$then P is not a prime subsemimodule of M.

(iii) If$\sqrt{(\phi (P){:}_{R}M)}\u2288(P{:}_{R}M)$, then P is a prime subsemimodule of M.

(iv) If φ(P) is a radical subsemimodule of M, either (P :_{R}M) = (φ(P) :_{R}M) 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 :_{R}M)P ⊆ φ(P). Hence
$$\sqrt{{(P{:}_{R}M)}^{2}}\subseteq \sqrt{((P{:}_{R}M)P{:}_{R}M)}\subseteq \sqrt{(\phi (P){:}_{R}M)}.$$
So $(P{:}_{R}M)\subseteq \sqrt{(\phi (P){:}_{R}M)}$. If P is a prime subsemimodule of M, then by [12, Lemma 3.3], $\sqrt{(\phi (P){:}_{R}M)}\subseteq \sqrt{(P{:}_{R}M)}=(P{:}_{R}M)$(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, $\sqrt{(\phi (P){:}_{R}M)}\subseteq (P{:}_{R}M)$. Also by assumption $(P{:}_{R}M)\u2288\sqrt{(\phi (P){:}_{R}M)}$. Thus $\sqrt{(\phi (P){:}_{R}M)}\subseteq (P{:}_{R}M)\u2288\sqrt{(\phi (P){:}_{R}M)}$, which is a contradiction.
(iii) Suppose P is not a prime subsemimodule of M. Then by (i) we have, $(P{:}_{R}M)\subseteq \sqrt{(\phi (P){:}_{R}M)}$. Also by assumption $\sqrt{(\phi (P){:}_{R}M)}\u2288(P{:}_{R}M)$. Thus $(P{:}_{R}M)\subseteq \sqrt{(\phi (P){:}_{R}M)}\u2288(P{:}_{R}M)$, which is a contradiction.
(iv) Suppose that P is not prime subsemimodule. Then by (i) and definition of radical subsemimodule we have
$$(P{:}_{R}M)\subseteq \sqrt{(\phi (P){:}_{R}M)}=(\phi (P){:}_{R}M)\subseteq (P{:}_{R}M),$$
so (P :_{R}M) = (φ (P) :_{R}M).
Let M be an Rsemimodule and N a subsemimodule of M. For every r ∈ R, {m ∈ M rm ∈ N} is denoted by (N :_{M}r). It is easy to see that (N :_{M}r) is a subsemimodule of M containing N.
Theorem 2.7
Let N be a Msubtractive subsemimodule of M and letbe a function. Then the following are equivalent:

(i) N is a φprime subsemimodule of M;

(ii) for x ∈ M \ N, (N :_{R}x) = (N :_{R}M) ∪ (φ(N) :_{R}x);

(iii) for x ∈ M \ N, (N :_{R}x) = (N :_{R}M) or (N :_{R}x) = (φ(N) :_{R}x);

(iv) for any ideal I of R and any subsemimodule L of M, if IL ⊆ N and IL ⊈ φ(N), then I ⊆ (N :_{R}M) or L ⊆ N;

(v) for any r ∈ R \ (N :_{R}M), (N :_{M}r) = N ∪ (φ(N) :_{M}r);

(vi) for any r ∈ R \ (N :_{R}M), (N :_{M}r) = N or (N :_{M}r) = (φ(N) :_{M}r).
Proof
(i) ⇒ (ii). Let x ∈ M \N, a ∈ (N :_{R}x) \ (φ(N) :_{R}x). Then ax ∈ N \ φ(N). Since N is a φprime subsemimodule of M, So a ∈ (N :_{R}M). 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 IL ⊆ N. Suppose I ⊈ (N :_{R}M) and L ⊈ N. We show that IL ⊆ φ(N). Let a ∈ I and x ∈ L. First let a ∉ (N :_{R}M). Then, since ax ∈ N, we have (N :_{R}x) ≠ (N :_{R}M). Hence by our assumption (N :_{R}x) = (φ(N) :_{R}x). So ax ∈ φ(N). Now assume that a ∈ I ∩ (N :_{R}M). Let u ∈ I \ (N :_{R}M). Then a + u ∈ I \ (N :_{R}M). So by the first case, for each x ∈ L we have ux ∈ φ(N) and (a+u)x ∈ φ(N). This gives that ax ∈ φ(N). Thus in any case ax ∈ φ(N), since N is Msubtractive. Therefore IL ⊆ φ(N).
(iv) ⇒ (i). Let ax ∈ N \ 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 φprime subsemimodule such that r ∉ (N :_{R}M). Let m ∈ (N :_{M}r). If rm ∉ φ(N), then N is φprime implies m ∈ N, if rm ∈ φ(N), then m ∈ (φ(N) :_{M}r). Hence (N :_{M}r) ⊆ N ∪ (φ(N) :_{M}r). The other containment holds trivially.
(v) ⇒ (vi) Let (N :_{M}r) = N ∪ (φ(N) :_{M}r) for r ∈ R \ (N :_{R}M). Then by [10, Lemma 3.13] and [5, Theorem 6], either (N :_{M}r) = N or (N :_{M}r) = (φ(N) :_{M}r). Therefore, the result follows.
(vi) ⇒ (i) Let r ∈ R\(N :_{R}M), m ∈ M such that rm ∈ N \φ(N). By assumption, either (N :_{M}r) = N or (N :_{M}r) = (φ(N) :_{M}r). As rm ∉ φ(N), then m ∉ (φ(N) :_{M}r) and as m ∈ (N :_{M}r), we have (N :_{M}r) ≠ (φ(N) :_{M}r). Hence (N :_{M}r) = N and so m ∈ N as required.
Recall from [13, Definition 2] that, an Rsubsemimodule N of M is said to be a strong subsemimodule if for each x ∈ N, there exists y ∈ N such that x + y = 0.
Theorem 2.8
Let f : M → M′ be an epimorphism of Rsemimodules with f(0) = 0 and letandbe two functions. Then the following statements hold:

(i) If N′ is a φ′prime subsemimodule of M′ and φ(f^{−1}(N′)) = f^{−1}(φ′ (N′)), then f^{−1}(N′) is a φprime subsemimodule of M.

(ii) If N is a strong φprime 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 a ∈ R and m ∈ M such that am ∈ f^{−1}(N′) \ φ(f^{−1}(N′)). Since am ∈ f^{−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 m ∈ f^{−1}(N′). Therefore either af^{−1}(M′) ⊆ f^{−1}(N′), or m ∈ f^{−1}(N′). Thus, either a ∈ (f^{−1}(N′) : M) or m ∈ f^{−1}(N′). It follows that f^{−1}(N′) is a φprime subsemimodule of M.
(ii) Let ax ∈ f(N) \ φ′ (f(N)) for some a ∈ R and x ∈ M′. Since ax ∈ f(N), there exists an element x′ ∈ N such that ax = f(x′). Since f is epimorphism and x ∈ M′, then there exists y ∈ M 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″ ∈ Kerf ⊆ N. Thus ay ∈ N, since N is a substractive subsemimodule. Since ax ∉ φ′ (f(N)), one can see easily that ay ∉ φ(N). So ay ∈ N \ φ(N), and hence either a ∈ (N : M) or y ∈ N, as N is φprime. Thus, either aM ⊆ N, or f(y) ∈ f(N), and hence either af(M) ⊆ f(N), or x ∈ f(N). Thus, a ∈ (f(N) : M′) or x ∈ f(N). It follows that f(N) is a φprime subsemimodule of M′.
The notion of fractions of semimodule M was defined by Atani in [11]. Also it show that there is a onetoone correspondence between the set of all prime subsemimodules N of M with (N : M) ∩ S = ∅︀ and the set of all prime subsemimodules of the R_{P}semimodule M_{P} , see [11, Corollary 2]. In the next theorem we want to generalize this fact for φprime subsemimodules. Let S be a multiplicatively closed subset of R and N(S) = {x ∈ M : ∃s ∈ S, sx ∈ N}. We know that N(S) is a subsemimodule of M containing N and S^{−1}(N(S)) = S^{−1}N. Let be a function and define by φ_{S}(S^{−1}N) = S^{−1}(φ(N(S)) if φ(N(S)) ≠ ∅︀, and φ_{S}(S^{−1}N) = ∅︀ if φ(N(S)) = ∅︀. Since φ(N) ⊆ N, φ_{S}(S^{−1}N) ⊆ S^{−1}N.
Lemma 2.9
If N is a subtractive subsemimodule of M, so is S^{−1}N ∩M.
Proof
Let a, a + b ∈ S^{−1}N ∩ M. Then a/1 = c_{1}/s_{1} and (a + b)/1 = c_{2}/s_{2}, such that c_{1}, c_{2} ∈ N and s_{1}, s_{2} ∈ S. So t_{2}(a + b)s_{2} = t_{2}c_{2} ∈ N and t_{2}t_{1}as_{1} = t_{2}t_{1}c_{1} ∈ N for some t_{1}, t_{2} ∈ S. Consequently t_{1}t_{2}as_{1}s_{2} + t_{1}t_{2}bs_{1}s_{2} = t_{1}t_{2}s_{1}c_{2} ∈ N. Since N is subtractive, therefore t_{1}t_{2}bs_{1}s_{2} ∈ N. Thus b/1 = (t_{1}t_{2}s_{1}s_{2}b)/(t_{1}t_{2}s_{1}s_{2}) ∈ S^{−1}N, and so b ∈ S^{−1}N ∩M. Hence S^{−1}N ∩M is subtractive.
Theorem 2.10
Let N be a φprime subsemimodule of M. Let S be a multiplicatively closed subset of R such that S^{−1}N ≠ S^{−1}M and S^{−1}(φ(N)) ⊆ φ_{S}(S^{−1}N). Then S^{−1}N is a φ_{S}prime subsemimodule of S^{−1}M. Furthermore if N is a Msubtractive subsemimodules of M and S^{−1}N ≠ S^{−1}(φ(N)), then N = N(S) = S^{−1}N ∩M.
Proof
First suppose that there is an element s which is common to (N :_{R}M) and S. So, sM ⊆ N. Suppose now that m/s′ ∈ S^{−1}M. Then m/s′ = sm/ss′ ∈ S^{−1}N. This show that S^{−1}N = S^{−1}M and the first assertion follows. From here on we assume that (N :_{R}M) ∩ S = ∅︀ and N be a φprime subsemimodule of M. Let (a_{1}/s_{1})(a_{2}/s_{2}) ∈ S^{−1}N \ φ_{S}(S^{−1}N) for some a_{1} ∈ R, a_{2} ∈ M, s_{1}, s_{2} ∈ S. Then there exist c ∈ N and s′ ∈ S such that a_{1}a_{2}/s_{1}s_{2} = c/s′. Then us′ a_{1}a_{2} = us_{1}s_{2}c ∈ N for some u ∈ S. If us′ a_{1}a_{2} ∈ φ(N), then (a_{1}/s_{1})(a_{2}/s_{2}) = (us′ a_{1}a_{2})/(us′ s_{1}s_{2}) ∈ S^{−1}(φ(N)) ⊆ φ_{S}(S^{−1}N), a contradiction. Hence us′ a_{1}a_{2} ∈ N − φ(N). As N is a φprime subsemimodule, we get us′ a_{1} ∈ (N :_{R}M) or a_{2} ∈ N. It follows that a_{1}/s_{1} = (us′ a_{1})/(us′ s_{1}) ∈ S^{−1}(N :_{R}M) ⊆ (S^{−1}N :_{S−1R}S^{−1}M), or a_{2}/s_{2} ∈ S^{−1}N. Consequently S^{−1}N is a φ_{S}prime subsemimodule of S^{−1}M.
To prove the last part of the theorem, assume that N be Msubtractive subsemimodules of M and S^{−1}N ≠ S^{−1}(φ(N)). Clearly, N ⊆ S^{−1}N ∩ M. For the reverse containment, pick an element a ∈ S^{−1}N ∩M. Then there exist c ∈ N and s ∈ S with a/1 = c/s. Therefore, tsa = tc for some t ∈ S. If (ts)a ∉ φ(N), then (ts)a ∈ N \φ(N) and (N :_{R}M)∩S = ∅︀ gives a ∈ N. So assume that (ts)a ∈ φ(N). In this case a ∈ S^{−1}(φ(N)) ∩M. Therefore, S^{−1}N ∩M ⊆ N ∪ (S^{−1}(φ(N)) ∩M). It follows that either S^{−1}N ∩ M = S^{−1}(φ(N)) ∩ M or S^{−1}N ∩ M = N by Lemma 2.9 and [5, Theorem 6]. If the former case holds, then by [14, Lemma 2.5], S^{−1}N = S^{−1}(φ(N)) which is a contradiction. So the result follows.
We say that r ∈ R is a zerodivisor of a semimodule M if rm = 0 for some nonzero element m of M. The set of all zerodivisors of M is denoted by Z_{R}(M) (see [11, Definition1 (f)]).
Proposition 2.11
Let S^{−1}N be a φ_{S}prime subsemimodule of S^{−1}M such that S^{−1}(φ(N)) = φ_{S}(S^{−1}N), S ∩ Z_{R}(M/N) = ∅︀ and S ∩ Z_{R}(N/φ(N)) = ∅︀. Then N is a φprime subsemimodule of M.
Proof
Let a ∈ R, m ∈ M, am ∈ N\φ(N) and m ∉ N. Then (a/1)(m/1) = am/1 ∈ S^{−1}N. If (a/1)(m/1) ∈ φ_{S}(S^{−1}N) = S^{−1}(φ(N)), then there exists s ∈ S such that sam ∈ φ(N) which contradicts S ∩ Z_{R}(N/φ(N)) = ∅︀. Therefore (a/1)(m/1) ∈ S^{−1}N \ φ_{S}(S^{−1}N), and so either a/1 ∈ (S^{−1}N : S^{−1}M) or m/1 ∈ S^{−1}N. If m/1 ∈ S^{−1}N, then sm ∈ N for some s ∈ S, and so s ∈ S ∩ Z_{R}(M/N) which is a contradiction. Thus a/1 ∈ (S^{−1}N : S^{−1}M). Hence for every c ∈ M, sac ∈ N and since s ∈ S, we have s ∈ Z_{R}(M/N). Since S ∩ Z_{R}(M/N) = ∅︀, it follows that ac ∈ N, and so a ∈ (N :_{R}M).
A subsemimodule N of an Rsemimodule M is called partitioning subsemimodule (= Qsubsemimodule) if there exists a subset Q of M such that

(1) M = ∪{q + N : q ∈ Q}, and

(2) (q_{1} + N) ∩ (q_{2} + N) ≠ ∅︀, for any q_{1}, q_{2} ∈ Q if and only if q_{1} = q_{2}.
Let N be a Qsubsemimodule of an Rsemimodule M, and M/N_{(Q)} = {q +N : q ∈ Q}. Then M/N_{(Q)} forms an Rsemimodule under the binary operations ⊕ and ⊙ , which are defined as follows: (q_{1} + N) ⊕ (q_{2} + N) = q_{3} + N where q_{3} ∈ Q is the unique element such that q_{1} + q_{2} + N ⊆ q_{3} + N and r ⊙ (q_{1} + N) = q_{4} + N, where r ∈ R and q_{4} ∈ Q is the unique element such that rq_{1} + N ⊆ q_{4} + N. Then, this Rsemimodule M/N_{(Q)} is called the quotient semimodule of M by N.
Moreover, if N is a Qsubsemimodule of an Rsemimodule M and L is a ksubsemimodule of M containing N, then N is a Q ∩ Lsubsemimodule of L and L/N = {q +N : q ∈ Q ∩ L} is a subtractive subsemimodule of M/N_{(Q)} as given by Chaudhari and Bonde in 2010, see [8].
Let N be a Qsubsemimodule of M and be a function. We define by φ_{N}(P/N) = (φ(P))/N for every with N ⊆ φ(P) where P is a Msubtractive subsemimodule (and φ_{N}(P/N) = ∅︀ if φ(P) = ∅︀).
Theorem 2.12
Let N be a Qsubsemimodule of an Rsemimodule M, P a M  subtractive subsemimodule and N a φprime subsemimodule of M with N ⊆ φ(P). Then P is a φprime subsemimodule of M if and only if P/N_{(Q∩P)}is a φprime subsemimodule of M/N_{(Q)}.
Proof
Let P be a φprime subsemimodule of M. Let r ∈ R, q_{1} + N ∈ M/N_{(Q)} be such that r ⊙ (q_{1} +N) ∈ P/N_{(Q∩P)} \ (φ(P))/N. By [8, Theorem 3.5], there exists a unique q_{2} ∈ Q ∩ P such that r ⊙ (q_{1} +N) = q_{2} +N where rq_{1} +N ⊆ q_{2} +N. Since N ⊆ P, rq_{1} ∈ P.
If rq_{1} ∈ φ(P), then rq_{1} ∈ (q_{2} +N) ∩ (q_{0} +N) and q_{0} ∈ φ(P). So q_{2} = q_{0} and hence q_{0}+N = q_{2}+N a contradiction. Thus rq_{1} ∉ φ(P). As P is φprime subsemimodule, either rM ⊆ P or q_{1} ∈ P. If q_{1} ∈ P, then q_{1} ∈ Q ∩P and hence q_{1}+N ∈ P/N_{(Q∩P)}. Suppose rM ⊆ P. For q + N ∈ M/N_{(Q)}, let r ⊙ (q + N) = q_{3} + N where q_{3} is a unique element of Q such that rq + N ⊆ q_{3} + N. Since N ⊆ P and P is a subtractive subsemimodule of M, q_{3} ∈ P. Hence q_{3} ∈ Q ∩ P. Now r ⊙ (q + N) = q_{3}+N ∈ P/N_{(Q∩P)} and hence r ⊙ M/N_{(Q)} ⊆ P/N_{(Q∩P)}. So P/N_{(Q∩P)} is a φprime subsemimodule of M/N_{(Q)}.
Conversely, suppose that N,P/N_{(Q∩P)} are φprime subsemimodules of M, M/N_{(Q)} respectively. Let rm ∈ P \ φ(P) where r ∈ R,m ∈ M. If rm ∈ N, then we are through, since rm ∈ N \ φ(N) and N is a φprime subsemimodule of M. So suppose rm ∈ P \ N. By using [7, Lemma 3.6], there exists a unique q_{1} ∈ Q such that m ∈ (q_{1} + N) and rm ∈ r ⊙ (q_{1} + N) = q_{2} + N where q_{2} is a unique element of Q such that rq_{1} + N ⊆ q_{2} + N. Now rm ∈ P,rm ∈ q_{2} + N implies q_{2} ∈ P, as P is a subtractive subsemimodule and N ⊆ P. Hence r ⊙ (q_{1} + N) = q_{2} + N ∈ P/N_{(Q∩P)} \ (φ(P))/N. As P/N_{(Q∩P)} is a φprime subsemimodule, either r ⊙ M/N_{(Q)} ⊆ P/N_{(Q∩P)} or q_{1} + N ∈ P/N_{(Q∩P)}. If q_{1} + N ∈ P/N_{(Q∩P)}, then q_{1} ∈ P. Hence m ∈ q_{1} + N ⊆ P. So assume that r ⊙ M/N_{(Q)} ⊆ P/N_{(Q∩P)}. Let x ∈ M. By using [7, Lemma 3.6], there exists a unique q_{3} ∈ Q such that x ∈ q_{3} + N and rx ∈ r ⊙ (q_{3} + N) = q_{4} + N where q_{4} is a unique element of Q such that rq_{3} +N ⊆ q_{4} +N. Now q_{4} +N = r ⊙ (q_{3} +N) ∈ P/N_{(Q∩P)} and hence q_{4} ∈ P. As rx ∈ q_{4} + N and N ⊆ P implies rx ∈ P. So rM ⊆ P.
Theorem 2.13
Let R be semiring, I a Qideal of R and P a subtractive ideal of R with I ⊆ P. Then P is a prime (weakly prime) ideal of R if and only if P/I_{(Q∩P)}is a prime (weakly prime) ideal of R/I_{(Q)}.
Proof
The proof is similar as in the proof of Theorem 2.12.