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 r ∈ R, x ∈ M and rx ∈ N \ φ(N), then r ∈ (N :RM) 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
.
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:
For functions
, we write φ ⩽ ψ if φ(N) ⊆ ψ(N) for each
. So whenever φ ⩽ ψ, 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 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).
Proposition 2.4
Let R be a semiring and M be an R-semimodule. Let
be a function and N be a φ-prime M-subtractive subsemimodule of M such that (N :RM)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 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.
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) or.
-
(ii) Ifthen P is not a prime subsemimodule of M.
-
(iii) If, 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)P ⊆ φ(P). Hence
So . If P is a prime subsemimodule of M, then by [12, Lemma 3.3], (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, . Also by assumption . Thus , which is a contradiction.
(iii) Suppose P is not a prime subsemimodule of M. Then by (i) we have, . Also by assumption . Thus , 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.
Theorem 2.7
Let N be a M-subtractive subsemimodule of M and let
be a function. Then the following are equivalent:
-
(i) N is a φ-prime subsemimodule of M;
-
(ii) for x ∈ M \ N, (N :Rx) = (N :RM) ∪ (φ(N) :Rx);
-
(iii) for x ∈ M \ 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 IL ⊆ N and IL ⊈ φ(N), then I ⊆ (N :RM) or L ⊆ N;
-
(v) for any r ∈ R \ (N :RM), (N :Mr) = N ∪ (φ(N) :Mr);
-
(vi) for any r ∈ R \ (N :RM), (N :Mr) = N or (N :Mr) = (φ(N) :Mr).
Proof
(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.
Theorem 2.8
Let f : M → M′ be an epimorphism of R-semimodules with f(0) = 0 and let
and
be 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 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 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 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) = {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 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−1N ∩M.
Proof
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.
Theorem 2.10
Let N be a φ-prime subsemimodule of M. Let S be a multiplicatively closed subset of R such that S−1N ≠ S−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−1N ≠ S−1(φ(N)), then N = N(S) = S−1N ∩M.
Proof
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 \ φS(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)) ⊆ φS(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 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−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 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), S ∩ ZR(M/N) = ∅︀ and S ∩ ZR(N/φ(N)) = ∅︀. Then N is a φ-prime subsemimodule of M.
Proof
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) ∈ φS(S−1N) = S−1(φ(N)), then there exists s ∈ S such that sam ∈ φ(N) which contradicts S ∩ ZR(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 sm ∈ N for some s ∈ S, and so s ∈ S ∩ ZR(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 ∈ ZR(M/N). Since S ∩ ZR(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 φN(P/N) = (φ(P))/N for every
with N ⊆ φ(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 N ⊆ φ(P). Then P is a φ-prime subsemimodule of M if and only if P/N(Q∩P)is a φ-prime subsemimodule of M/N(Q).
Proof
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.
Theorem 2.13
Let R be semiring, I a Q-ideal 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.