Let r ∈ R, x ∈ M and rx ∈ N. If rx ∉ φ(N), since N is a φ-prime subsemimodule, then x ∈ N or r ∈ (N :_RM), and if rx ∈ φ(N), then x ∈ φ(N) ⊆ N or r ∈ (φ(N) :_RM) ⊆ (N :_RM).

Let a ∈ R and x ∈ M be such that ax ∈ N. If ax ∉ φ(N), then since N is φ-prime, we have a ∈ (N :_RM) 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 :_RM) or x + p ∈ N and hence a ∈ (N :_RM) or x ∈ N. Second we may assume that (N :_RM)x ⊆ φ(N). If this is not the case, there exists u ∈ (N :_RM) such that ux ∉ φ(N) and so (a + u)x ∈ N\φ(N). Since N is a φ-prime subsemimodule, we have a+u ∈ (N :_RM) or x ∈ N. By [7, Lemma 1.2], (N :_RM) is a subtractive ideal, and hence either a ∈ (N :_RM) or x ∈ N. Now since (N :_RM)N ⊈ φ(N), there exist r ∈ (N :_RM) and p ∈ N such that rp ∉ φ(N). So (a+r)(x+p) ∈ N\φ(N), and hence a+r ∈ (N :_RM) or x+p ∈ N. Therefore a ∈ (N :_RM) or x ∈ N, 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.

In Proposition 2.4 set φ = φ₀.

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

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 $(P{:}_{R}M)⊈\sqrt{(\phi (P){:}_{R}M)}$. Thus $\sqrt{(\phi (P){:}_{R}M)}\subseteq (P{:}_{R}M)⊈\sqrt{(\phi (P){:}_{R}M)}$, 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 $\sqrt{(\phi (P){:}_{R}M)}⊈(P{:}_{R}M)$. Thus $(P{:}_{R}M)\subseteq \sqrt{(\phi (P){:}_{R}M)}⊈(P{:}_{R}M)$, which is a contradiction.

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

so (P :_RM) = (φ (P) :_RM).

Let M be an R-semimodule and N a subsemimodule of M. For every r ∈ R, {m ∈ M| rm ∈ N} is denoted by (N :_Mr). It is easy to see that (N :_Mr) is a subsemimodule of M containing N.

(i) ⇒ (ii). Let x ∈ M \N, a ∈ (N :_Rx) \ (φ(N) :_Rx). Then ax ∈ N \ 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 IL ⊆ N. Suppose I ⊈ (N :_RM) and L ⊈ N. We show that IL ⊆ φ(N). Let a ∈ I and x ∈ L. First let a ∉ (N :_RM). Then, since ax ∈ N, we have (N :_Rx) ≠ (N :_RM). Hence by our assumption (N :_Rx) = (φ(N) :_Rx). So ax ∈ φ(N). Now assume that a ∈ I ∩ (N :_RM). Let u ∈ I \ (N :_RM). Then a + u ∈ I \ (N :_RM). 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 M-subtractive. 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 φ-prime subsemimodule such that r ∉ (N :_RM). Let m ∈ (N :_Mr). If rm ∉ φ(N), then N is φ-prime implies m ∈ N, if rm ∈ φ(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 r ∈ R \ (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 r ∈ R\(N :_RM), m ∈ M such that rm ∈ N \φ(N). By assumption, either (N :_Mr) = N or (N :_Mr) = (φ(N) :_Mr). As rm ∉ φ(N), then m ∉ (φ(N) :_Mr) and as m ∈ (N :_Mr), we have (N :_Mr) ≠ (φ(N) :_Mr). Hence (N :_Mr) = N and so m ∈ N as required.

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

(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 φ-prime 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 φ-prime. 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 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 R_P-semimodule M_P , 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) = {x ∈ M : ∃s ∈ S, sx ∈ N}. 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 φ_S(S⁻¹N) = S⁻¹(φ(N(S)) if φ(N(S)) ≠ ∅︀, and φ_S(S⁻¹N) = ∅︀ if φ(N(S)) = ∅︀. Since φ(N) ⊆ N, φ_S(S⁻¹N) ⊆ S⁻¹N.

Let a, a + b ∈ S⁻¹N ∩ M. Then a/1 = c₁/s₁ and (a + b)/1 = c₂/s₂, such that c₁, c₂ ∈ N and s₁, s₂ ∈ S. So t₂(a + b)s₂ = t₂c₂ ∈ N and t₂t₁as₁ = t₂t₁c₁ ∈ N for some t₁, t₂ ∈ S. Consequently t₁t₂as₁s₂ + t₁t₂bs₁s₂ = t₁t₂s₁c₂ ∈ N. Since N is subtractive, therefore t₁t₂bs₁s₂ ∈ N. Thus b/1 = (t₁t₂s₁s₂b)/(t₁t₂s₁s₂) ∈ S⁻¹N, and so b ∈ S⁻¹N ∩M. Hence S⁻¹N ∩M is subtractive.

First suppose that there is an element s which is common to (N :_RM) and S. So, sM ⊆ N. 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 \ φ_S(S⁻¹N) for some a₁ ∈ R, a₂ ∈ M, s₁, s₂ ∈ S. Then there exist c ∈ N and s′ ∈ S such that a₁a₂/s₁s₂ = c/s′. Then us′ a₁a₂ = us₁s₂c ∈ N for some u ∈ S. If us′ a₁a₂ ∈ φ(N), then (a₁/s₁)(a₂/s₂) = (us′ a₁a₂)/(us′ s₁s₂) ∈ S⁻¹(φ(N)) ⊆ φ_S(S⁻¹N), a contradiction. Hence us′ a₁a₂ ∈ N − φ(N). As N is a φ-prime subsemimodule, we get us′ a₁ ∈ (N :_RM) or a₂ ∈ N. It follows that a₁/s₁ = (us′ a₁)/(us′ s₁) ∈ S⁻¹(N :_RM) ⊆ (S⁻¹N :_S⁻¹RS⁻¹M), or a₂/s₂ ∈ S⁻¹N. Consequently S⁻¹N is a φ_S-prime subsemimodule of S⁻¹M.

To prove the last part of the theorem, assume that N be M-subtractive subsemimodules of M and S⁻¹N ≠ S⁻¹(φ(N)). Clearly, N ⊆ S⁻¹N ∩ M. For the reverse containment, pick an element a ∈ S⁻¹N ∩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 :_RM)∩S = ∅︀ gives a ∈ N. So assume that (ts)a ∈ φ(N). In this case a ∈ S⁻¹(φ(N)) ∩M. Therefore, S⁻¹N ∩M ⊆ N ∪ (S⁻¹(φ(N)) ∩M). It follows that either S⁻¹N ∩ M = S⁻¹(φ(N)) ∩ M or S⁻¹N ∩ M = 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 r ∈ R 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 Z_R(M) (see [11, Definition1 (f)]).

Let a ∈ R, m ∈ M, am ∈ N\φ(N) and m ∉ N. Then (a/1)(m/1) = am/1 ∈ S⁻¹N. If (a/1)(m/1) ∈ φ_S(S⁻¹N) = 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⁻¹N \ φ_S(S⁻¹N), and so either a/1 ∈ (S⁻¹N : S⁻¹M) or m/1 ∈ S⁻¹N. If m/1 ∈ S⁻¹N, then sm ∈ N for some s ∈ S, and so s ∈ S ∩ Z_R(M/N) which is a contradiction. Thus a/1 ∈ (S⁻¹N : S⁻¹M). 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 :_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 : q ∈ Q}, and

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

Let N be a Q-subsemimodule of an R-semimodule M, and M/N_(Q) = {q +N : q ∈ Q}. 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 q₃ ∈ Q is the unique element such that q₁ + q₂ + N ⊆ q₃ + N and r ⊙ (q₁ + N) = q₄ + N, where r ∈ R and q₄ ∈ Q is the unique element such that rq₁ + N ⊆ q₄ + 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 Q ∩ L-subsemimodule 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 Q-subsemimodule 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 M-subtractive subsemimodule (and φ_N(P/N) = ∅︀ if φ(P) = ∅︀).

Let P be a φ-prime subsemimodule of M. Let r ∈ R, q₁ + N ∈ M/N_(Q) be such that r ⊙ (q₁ +N) ∈ P/N_(Q∩P) \ (φ(P))/N. By [8, Theorem 3.5], there exists a unique q₂ ∈ Q ∩ P such that r ⊙ (q₁ +N) = q₂ +N where rq₁ +N ⊆ q₂ +N. Since N ⊆ P, rq₁ ∈ P.

If rq₁ ∈ φ(P), then rq₁ ∈ (q₂ +N) ∩ (q₀ +N) and q₀ ∈ φ(P). So q₂ = q₀ and hence q₀+N = q₂+N a contradiction. Thus rq₁ ∉ φ(P). As P is φ-prime subsemimodule, either rM ⊆ P or q₁ ∈ P. If q₁ ∈ P, then q₁ ∈ Q ∩P and hence q₁+N ∈ P/N_(Q∩P). Suppose rM ⊆ P. For q + N ∈ M/N_(Q), let r ⊙ (q + N) = q₃ + N where q₃ is a unique element of Q such that rq + N ⊆ q₃ + N. Since N ⊆ P and P is a subtractive subsemimodule of M, q₃ ∈ P. Hence q₃ ∈ Q ∩ P. Now r ⊙ (q + N) = q₃+N ∈ P/N_(Q∩P) and hence r ⊙ M/N_(Q) ⊆ P/N_(Q∩P). So P/N_(Q∩P) is a φ-prime subsemimodule of M/N_(Q).

Conversely, suppose that N,P/N_(Q∩P) are φ-prime 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 φ-prime subsemimodule of M. So suppose rm ∈ P \ N. By using [7, Lemma 3.6], there exists a unique q₁ ∈ Q such that m ∈ (q₁ + N) and rm ∈ r ⊙ (q₁ + N) = q₂ + N where q₂ is a unique element of Q such that rq₁ + N ⊆ q₂ + N. Now rm ∈ P,rm ∈ q₂ + N implies q₂ ∈ P, as P is a subtractive subsemimodule and N ⊆ P. Hence r ⊙ (q₁ + N) = q₂ + N ∈ P/N_(Q∩P) \ (φ(P))/N. As P/N_(Q∩P) is a φ-prime subsemimodule, either r ⊙ M/N_(Q) ⊆ P/N_(Q∩P) or q₁ + N ∈ P/N_(Q∩P). If q₁ + N ∈ P/N_(Q∩P), then q₁ ∈ P. Hence m ∈ q₁ + 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 q₃ ∈ Q such that x ∈ q₃ + N and rx ∈ r ⊙ (q₃ + N) = q₄ + N where q₄ is a unique element of Q such that rq₃ +N ⊆ q₄ +N. Now q₄ +N = r ⊙ (q₃ +N) ∈ P/N_(Q∩P) and hence q₄ ∈ P. As rx ∈ q₄ + N and N ⊆ P implies rx ∈ P. So rM ⊆ P.

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