6_xxx_tekir.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 3, 2011, 251-265 ISSN 1307-5543 – www.ejpam.com A Zariski Topology For Semimodules Shahabaddin Ebrahimi Atani1, Reza Ebrahimi Atani2, Ünsal Tekir3,∗ 1 Faculty of Mathematical Sciences, University of Guilan, P. O. Box 1914, Rasht, Iran 2 Department of Computer Engineering, University of Guilan, P.O. Box 3756, Rasht, Iran 3 Department of Mathematics, The University of Marmara, Goztepe, Istanbul, Turkey Abstract. Given a very strong multiplication semimidule M over a commutative semiring R, a Zariski topology is defined on the spectrum Speck(M) of prime k-subsemimodules of M . The properties and possible structures of this topology are studied. 2000 Mathematics Subject Classifications: 16Y60 Key Words and Phrases: Prime subsemimodule, Strong multiplication semimodule, Very strong mul- tiplication semimodule, Prime spectrum 1. Introduction As a generalization of rings, semirings have been found useful for solving problems in dif- ferent areas of applied mathematics and information sciences, since the structure of a semiring provides an algebraic framework for modelling and studying the key factors in these applied areas. They play an important role in studying optimization theory, graph theory, theory of discrete event dynamical systems, generalized fuzzy computation, automata theory, coding theory, cryptography theory, and so on (see Golan [16], Glazek [17], Hebisch and Weinert [18], Simon [23]). Ideals of semirings play a central role in the structure theory and are useful for many purposes. However, they do not in general coincide with the usual ring ideals and, for this reason, their use is somewhat limited in trying to obtain analogues of ring theo- rems for semirings. Indeed, many results in rings apparently have no analogues in semirings using only ideals. In order to overcome this deficiency, the present authors [14] defined a more resttricted class of ideals in semirings, which is called the class of “strong ideals”, with the property that they are k-ideals. Let M be a module over a commutative ring R with identity. The prime spectrum Spec(R) and the topological space obtained by introducing Zariski topology on the set of prime ideals of R play an important role in the fields of commutative algebra, algebraic geometry and lattice ∗Corresponding author. Email address: utekir�marmara.edu.tr (Ü. Tekir) http://www.ejpam.com 251 c© 2011 EJPAM All rights reserved. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 252 theory. Also, recently the notion of prime submodules and Zariski topology on Spec(M), the set of all prime R-submodules of M , are studied by many authors (for example see Lu[19], Lu [20], Lu [21], MacCasland, Moore and Smith [22]). The main dificulty is figuring out what additional hypotheses the semimodule must satisfy to get similar results of modules. The two new key notions are that a “strong subsemimodule” and a “very strong subsemimodule”. In this paper, Section 3, we list some basic properties concerning very strong multiplication semimodules. For example, we give some results to characterize the prime k-subsemimodules of very strong multiplication semimodules. In Section 4, we concentrate on Zariski topology of very strong multiplication semimodules M over commutative semirings R with identity and generalize the some well known results of Zariski topoloy on the sets of prime ideals of a commutative ring to Spec(M), the sets of prime k-subsemimodules of M and investigate the basic properties of this topology. In this regard we strongly use the notion of product of subsemimodules of very strong multiplication semimodules. For example, we prove that an open set X = Speck(M) is compact if and only if it is a finite union of basic open sets.. 2. Preliminaries Throughout this paper R is a commutative semiring with identity. In order to make this paper easier to follow, we recall in this section various notions from semimodule theory which will be used in the sequel. For the definitions of monoid, semirings, semimodules and subsemi- modules we refer [16, 18, 13, 7]. All semiring in this paper are commutative with non-zero identity. Definition 1. (1) 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 + y ∈ N, then y ∈ N (so {0M} is a k-subsemimodule of M). (2) A prime subsemimodule of M is a proper subsemimodule N of M in which x ∈ N or rM ⊆ N whenever r x ∈ N. The collection of all prime k-subsemimodules of M is called the k-spectrum of M and denoted by Speck(M). We define k-ideals and prime k-ideals of a semiring R in a similar fashion. (3) A subsemimodule L of M is said to be semiprime if L is an intersection of prime k- subsemimodules of M. (4) Let T be a proper subsemimodule of an R-semimodule M. Then the prime radical rad(T ) of T (in M) is the intersection of all prime k-subsemimodules of M containing T or, in case there are no such prime k-subsemimodules, rad(T ) is M. Note that T ⊆ rad(T ) and that rad(T ) = M or rad(T ) is semiprime k-subsemimodule of M. (5) A prime subsemimodule N of M is called extraordinary if whenever A and B are semiprime k-subsemimodules of M with A∩ B ⊆ N, then A⊆ N or B ⊆ N. (6) An R-semimodule M is called multiplication semimodule provided that for every subsemi- module N of M there exists an ideal I of R such that N = IM. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 253 (7) A proper ideal I of a semiring R is said to be strong ideal (or strongly zero-sum ideal), if for each a ∈ I there exists b ∈ I such that a+ b = 0 (see [14, Example 2.3] and [11]). Quotient semirings are determined by equivalence relations rather than by ideals as in the ring case. Allen [1] has presented the notion of a partitioning ideal (= Q-ideal) I in the semir- ing R and constructed the quotient semiring R/I . If I is an ideal of a semiring R, we define a relation ∼ on R, given by r1 ∼ r2 if and only if there exist a1, a2 ∈ I satisfying r1+a1 = r2+a2. Then ∼ is an equivalence relation on R, and we denote the equivalence class of r by r+ I and these collection of all equivalence classes by R/I . Golan shows that R/I is a semiring with (r + I)+ (s+ I) = r + s+ I and (r + I)(s+ I) = rs+ I . The semiring R/I has additive identity 0+ I and multiplicative identity 1+ I (see Golan [16]). In this paper, we will follow Golan’s terminology for quotient semirings. Quotient semimodules over a semiring R have already been introduced and studied by present authors in [13]. Chaudhari and Bonde [7] extended the definition of QM -subsemimodule of a semimodule to a more general case: A subsemimodule N of a semimodule M over a semir- ing R is called a partitioning subsemimodule (= QM -subsemimodule) if there exists a subset QM of M such that M = ∪{q+ N : q ∈ QM} and if q1,q2 ∈ QM then (q1 + N)∩ (q2+ N) 6= ; if and only if q1 = q2. Let N be a QM -subsemimodule of M and let M/N = {q + N : q ∈ QM}. Then M/N forms an R-semimodule under the operations ⊕ and ⊙ defined as follows: (q1 + N)⊕ (q2 + N) = q3 + N , where q3 ∈ QM is the unique element such that q1 + q2 + N ⊆ q3 + N and r ⊙ (q1 + N) = q4 + I , where r ∈ R and q4 ∈ QM is the unique element such that rq1 + N ⊆ q4 + N . This R-semimodule M/N is called the quotient semimodule of M by N [7]. By [7, Lemma 2.3], there exists a unique element q0 ∈ QM such that q0 + N = N . Thus q0 + N is the zero element of M/N . Also, [7, Theorem 2.4] show that the structure (M/N ,⊕,⊙) is essentially independent of QM (see [7, Example 2.6]). Lemma 1. Let M be a semimodule over a semiring R. If {Mi}i∈Λ is a collection of subsemimodules of M, then ∑ i∈ΛMi and ⋂ i∈ΛMi are subsemimodules of M. 3. Properties of Strong Multiplication Semimodules In this section, we list some basic properties concerning very strong multiplication semi- modules. Our starting point is the following lemma. Lemma 2. Let N be a QM -subsemimodule of a semimodule M over a semiring R. If T is a k-subsemimodule of M containing N, then (T :R M) = (T/N :R M/N). Proof. Let r ∈ (T : M). If q + N ∈ M/N , then there exists a unique element q′ of QM such that r(q + N) = q′ + N , where rq + N ⊆ q′ + N ; so q′ ∈ T ∩QM since rq ∈ T and T is a k-subsemimodule. Thus (T : M) ⊆ (T/N : M/N). For the other inclusion, assume that a ∈ (T/N : M/N) and m ∈ M . Then m = q1 + n for some q1 ∈ QM and n ∈ N ; so there is a unique element q2 of QM with a(q1 + N) = q2 + N ∈ T/N , where aq1 + N ⊆ q2 + N . Thus T being a k-subsemimodule gives aq1 ∈ T . As am = aq1 + an ∈ T , we have a ∈ (T : M). S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 254 Theorem 1. Let R be a semiring with identity, M an R-semimodule and N an QM -subsemimodule of M. Then there is a one-to-one correspondence between prime k-subsemimodules of R-semimodule M/N and prime k-subsemimodules of M containing N. Proof. Let T be a prime k-subsemimodule of M containing N . It then follows from [7, Theorem 3.6] that T/N is a proper k-susemimodule of M/N . Let a(q1+N) = q2+N ∈ T/N , where q2 ∈ QM ∩ T and aq1 + N ⊆ q2 + N , so aq1 ∈ T since T is a k-subsemimodule. Then T prime gives either q1 ∈ T (so q1 + N ∈ T/N) or a ∈ (T : M) = (T/N : M/N) by Lemma 2. Thus, T/N is a prime k-subsemimodule of M/N . Conversely, assume that T/N is a prime k-subsemimodule of M/N . To show that T is a prime k-subsemimodule of M , suppose that rm ∈ T , where r ∈ R and m ∈ M . We may assume that r 6= 0. There are elements q ∈ QM and n ∈ N such that m = q+ n, so rm = rq+ rn ∈ T ; hence rq ∈ T since T is a k- subsemimodule. Therefore, there exists a unique element q′ ∈QM such that r(q+N) = q′+N , where rq+ N ⊆ q′ + N ; hence q′ ∈ T . Thus r(q+ N) ∈ T/N . Then T/N prime gives either q+ N ∈ T/N (so m ∈ T) or r ∈ (T/N : M/N) = (T : M), and the proof is complete. Corollary 1. Let R be a semiring with identity, M an R-semimodule and N an QM -subsemimodule of M. Then there is a one-to-one correspondence between semiprime k-subsemimodules of R- semimodule M/N and semiprime k-subsemimodules of M containing N. Proof. Apply Theorem 1. Definition 2. Let M be a semimodule over a semiring R. A 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. Example 1. (1) Clearly, every submodule of a module over a ring R is a strong subsemimodule. (2) Let R denote the semiring of non-negative integers with the usual operations of addition and multiplication, and let M = Z6 denote the monoid of integers modulo 6. Then M is a semimodule over R by [16, p. 151], and an inspection will show that N = {0̄, 2̄, 4̄} and M are strong subsemimodules of M. Definition 3. A semimodule M over a semiring R is called a strong multiplication semimodule whenever N is a k-subsemimodule of M, then there exists a strong ideal I of R such that N = IM. Definition 4. A semimodule M over a semiring R is called a very strong semimodule if I is an ideal of R and m ∈ M, then the ideal {r ∈ R : rm ∈ IM} is a strong k-ideal of R. Definition 5. A very strong semimodule M over a semiring R is called a very strong multiplication semimodule whenever N is a k-subsemimodule of M, then there exists a strong ideal I of R such that N = IM. Proposition 1. Let M be a semimodule over a semirig R. Then the following statements hold: (i) If N is a strong subsemimodule of M, then N is a k-subsemimodule. (ii) If I is a strong ideal of R, then IM is a strong k-subsemimodule. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 255 (iii) If N is a strong subsemimodule of M, then N + L is a strong k-subsemimodule of M for every strong subsemimodule L of M. (iv) If I is a strong ideal of R and N is a strong QM -subsemimodule of M, then I(M/N) = (IM + N)/N. Proof. (i) Let a, a+ b ∈ N for some a, b ∈ M . Then a+ a′ = 0 for some a′ ∈ N ; hence b = a+ a′ + b ∈ N . (ii) Let z = ∑n i=1 rimi ∈ IM . Then there exists si ∈ I such that ri + si = 0 for every i = 1, . . . , n; so z + ∑n i=1 simi = 0. Now the assertion follows from (i). (iii) Let a + b ∈ N + L, where a ∈ N and b ∈ L. Then a + a′ = 0 for some a′ ∈ N and b+ b′ = 0 for some b′ ∈ L; hence (a+ b) + (a′ + b′) = 0. Thus N + L is a strong ideal. Now the assertion follows from (i). (iv) First we show that I(M/N) ⊆ (IM + N)/N . It is enough to show that for each r ∈ I and for each q+ N ∈ M/N we have r.(q + N) ∈ (IM + N)/N . Let r.(q+ N) = q′ + N , where q′ ∈QM is the unique element such that rq+ N ⊆ q′+ N , so rq+ n= q′ + n′ for some n, n′ ∈ N . It follows that q′ ∈QM ∩ (IM + N) since IM + N is a k-subsemimodule by (ii) and (iii). Thus r.(q+ N) ∈ (IM + N)/N . For the reverse inclusion, assume that q1 + N ∈ (IM + N)/N , where q1 ∈ QM ∩ (IM + N). Then there are elements q′ i ∈ QM , t i ∈ N , ri ∈ R and n ∈ N such that q1 = ∑s i=1 ri(q ′ i + ni) + n; hence q1 = ∑s i=1 riq ′ i + x , where x ∈ N . Let q0 + N is the zero in M/N . Clearly x + N ⊆ N . Assume that y ∈ N . Since N = q0+N by [7, Lemma 2.3], there exist a, b, c ∈ N with y = q0+ a, x = q0+ b and b+ c = 0; so y = q0+ b+a+ c = x+a+ c ∈ x+N ; hence x+N = N . An inspection will show that q1+ N = ∑s i=1 ri.(q ′ i +N) ∈ I(M/N). Thus (IM +N)/N ⊆ I(M/N), and so we have equality. Theorem 2. Let N be a strong QM -subsemimodule of a strong multiplication semimodule M over a semiring R. Then M/N is a strong multiplication R-semimodule. Proof. Let L be a k-subsemimodule of M/N . Then by [7, Theorem 3.6], L = T/N for some k-subsemimodule T of M with N ⊆ T , so there exists a strong ideal I of R such that T = IM . Therefore I(M/N) = (IM + N)/N = T/N = L by Proposition 1 (iv), as needed. Theorem 3. Let N be a strong QM -subsemimodule of a very strong multiplication semimodule M over a semiring R. Then M/N is a very strong multiplication R-semimodule. Proof. By Theorem 2 and Definition 5, it suffices to show that M/N is a very strong semimodule. Let I be an ideal of R and q+ N ∈ M/N , where q ∈ QM and set J = {r ∈ R : r.(q+ N) ∈ I(M/N)}; we show that J is a strong k-ideal of R. Let r, r ′ ∈ J . There are unique elements q1,q2 ∈QM such that r.(q+ N) = q1+ N ∈ I(M/N) and S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 256 r ′.(q+N) = q2+N ∈ I(M/N), where rq+N ⊆ q1+N , r ′q+N ⊆ q2+N . Then there exists a unique element q3 ∈QM such that (q1+N)⊕ (q2+N) = q3+N , where q1+ q2+N ⊆ q3+N , so (r + r ′)q + N ⊆ q1 + q2 + N ⊆ q3 + N ∈ I(M/N); hence r + r ′ ∈ J . Similarly, if s ∈ R, then sr ∈ J . Thus J is an ideal of R. By assumption, there must exists a strong ideal I ′ of R such that N = I ′M , and the set J1 = {r ∈ R : rq ∈ (I + I ′)M} is a strong k-ideal of R since M is a very strong semimodule. Let t ∈ J . Then tq + N ⊆ q′ + N ∈ I(M/N) for some q′ ∈ (IM + I ′M)∩QM , so tq ∈ (I + I ′)M since it is a k- subsemimodule by Proposition 1 (iii); hence t ∈ J1. Thus J ⊆ J1. It follows that J is a strong k-ideal of R, and this completes the proof. Proposition 2. Let M be a non-strong semimodule over a semiring R with M 6= 0. Then M has at least one strong maximal QM -subsemimodule. Proof. Since {0M} is a proper strong QM -subsemimodule of M with respect to the set QM = M − {0M}, the set ∆ of all proper strong QM -subsemimodules of M is not empty. Of course, the relation of inclusion, ⊆, is a partial order on ∆. If {Mi}i∈I is a chain of strong QM - subsemimidules of M , then N = ⋃ i∈I Mi is a strong QM -subsemimodule of M . Furthermore, N is proper since M is not strong. So by Zorn’s Lemma ∆ has a maximal element, i.e., M has a strong maximal QM -subsemimodule. Theorem 4. Let M be a non-strong semimodule over a semiring R with M 6= 0. Then Speck(M) 6= ;. Proof. By Proposition 2, there exists a strong maximal QM -subsemimodule N of M ; so it is a k-prime subsemimodule of M by [13, Theorem 14] (since every QM -subsemimodule is a k-subsemimodule by [7, Theorem 3.2]), as required. Recall that we follows Golan’s terminology for quotient semirings in the following lemma. Lemma 3. Let I be an ideal of a semiring R with 1 6= 0. Then the following hold: (i) If L is a k-ideal of R/I , then L = J/I for some k-ideal J of R. (ii) If I ⊆ P, then P is a maximal k-ideal of R if and only P/I is a maximal k-ideal of R/I . (iii) R has at least one strong maximal k-ideal. (iv) If J is a proper strong k-ideal of R, then J ⊆ P for some strong maximal k-ideal P of R. Proof. (i) Assume that J = {r ∈ R : r+ I ∈ L} and let a ∈ I . Since a+ I = 0+ I ∈ L, we have I ⊆ J . Let a, b ∈ J and r ∈ R. Then (a+ I) + (b+ I) = a+ b+ I ∈ L; so a+ b ∈ J . Similarly, ra ∈ J . Thus J is a k-ideal of R. Finally, it is easy to see that L = J/I . (ii) Suppose that P is a maximal k-ideal of R and let L be a k-ideal of R/I such that P/I $ L. There exists a k-ideal J of R such that P/I $ L = J/I by (i), so P $ J ; hence J = R. Thus L = R/I . The other implication is similar. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 257 (iii) Since {0} is a proper strong k-ideal of R, the set ∆ of all proper strong k-ideals of R is not empty. So by Zorn’s Lemma ∆ has a maximal element (with respect to ⊆), i.e., R has a proper strong maximal k-ideal. (iv) Since R/I is non-trivial, and so, by (iii), has a strong maximal k-ideal L, which, by (i), will have to have the form P/I for some k-ideal P of R with I ⊆ P. It now follows from (ii) that P is a maximal k-ideal of R. It remains to show that P is a strong ideal of R. Let a ∈ P. Then a+ I ∈ P/I . By assumption, (a+ I) + (b+ I) = a+ b+ I = 0+ I = I for some b+ I ∈ P/I , so a+ b ∈ I . Then there is an element c ∈ I such that a+ b+ c = 0, as required. Theorem 5. Let M be a non-zero very strong multiplication semimodule over a semiring R. Then every proper strong QM -subsemimodule of M is contained in a strong maximal k-subsemimodule of M. Proof. Assume that N is a proper strong QM -subsemimodule of M and let q0 + N is the zero in M/N . Then there exists x ∈ M − N such that x = q+ n for some q ∈ QM with q /∈ N (since every QM -subsemimodule is a k-subsemimodule by [7, Theorem 3.2]) and n ∈ N ; hence q0+N 6= q+N ∈ M/N . Then M/N is a non-zero very strong multiplication semimodule. Thus it is sufficient to prove that any non-zero very strong multiplication semimodule contains a maximal k-subsemimodule. Let 0 6= m ∈ M . If I = {0}, then the ideal J = {r ∈ R : rm = 0} is a proper strong k-ideal of R since M is a very strong semimodule and hence J ⊆ P for some strong maximal k-ideal P of R by Lemma 3 (iv). If M = PM , then Rm = T M for some ideal T of R, so Rm = T PM = PRm = Pm and hence m = pm for some p ∈ P. There exists p′ ∈ P such that p+ p′ = 0. But this implies (1+ p′)m = 0; so 1+ p′ ∈ I ⊆ P, a contradiction. Thus M 6= PM . Since P is a strong ideal of R, we have PM is a strong subsemimodule of M by Proposition 1 (i). Let PM $ N = LM ⊆ M for some ideal L of R. It follws that there is an element a ∈ L with a /∈ P; so P + Ra = R. Therefore there exist t ∈ P and r ∈ R such that t + ra = 1; whence m = tm+ ram ∈ PM + LM = N . Thus M = N and hence PM is a strong maximal k-subsemimodule of M . Definition 6. Let M be a non-zero semimodule over a semiring R. An element u of M is said to be unit provided that u is not contained in any strong maximal k-subsemimodule of M. Theorem 6. Let M be a non-zero very strong multiplication semimodule over a semiring R. Then u ∈ M is unit if and only if M = Ru. Proof. The sufficiency is clear. Conversely, suppose that u is an unit element of M . Then Ru is not contained in any strong maximal k-subsemimodule of M ; hence M = Ru by Theorem 5. Assume that P is a strong maximal k-ideal of a semiring R and let M be a semimodule over R. We say that M is a P-cyclic provided there exist p ∈ P and m ∈ M such that (1+p)M ⊆ Rm. We say that a subset TP(M) of M is P-torsion precisely when TP(M) = {m ∈ M : (1+ p)m = 0 for some p ∈ P}. The Definition is the same as that intro- duced by Z. El-Bast and P. F. Smith in [6]. It is easy to see that TP(M) is a subsemimodule of M . S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 258 Proposition 3. (i) If M is a strong multiplication semimodule over a semiring R, then for every strong maximal k-ideal P of R either M = TP (M) or M is P-cyclic. (ii) If M is a faithful very strong multiplication semimodule over a semiring R, then⋂ i∈Λ(Ii M) = ( ⋂ i∈Λ Ii)M for any non-empty collection of strong ideals Ii (i ∈ Λ) of R. (iii) Let P be a strong prime k-ideal of a semiring R and M a faithful very strong multiplication semimodule over R. Let a ∈ R, x ∈ M satisfy ax ∈ PM. Then a ∈ P or x ∈ PM. In particular, if M 6= PM, then PM is a strong prime subsemimodule of M. Proof. (i) Let P be a strong maximal k-ideal of R. Suppose M = PM . Let m ∈ M . Then Rm = IM for some strong ideal I of R by [15, Proposition 2.4]. Hence Rm = IM = I PM = Pm and m= pm for some p ∈ P. By assumption, there exists p′ ∈ P such that pm+ p′m = (1+ p′)m = 0 and m ∈ TP(M). It follows that TP (M) = M . Now suppose that PM 6= M . There exists y ∈ M such that y /∈ PM . There is a strong ideal J of R such that Ry = J M . Clearly, J * P. Since, J + P is a strong ideal of R, we must have J+P = R, so 1= e+q for some e ∈ J and q ∈ P. There exists q′ ∈ P such that q+q′ = 0; hence 1+ q′ ∈ J . It follows that (1+ q′)M ⊆ Ry and M is P-cyclic. (ii) Let Ii (i ∈ Λ) be any non-empty collection of strong ideals of R. Set I = ⋂ i∈Λ Ii . Clearly, IM ⊆ ⋂ i∈Λ(Ii M). For the reverse inclusion, assume that x ∈ ⋂ i∈Λ(Ii M). Then K = {r ∈ R : r x ∈ IM} is a strong k-ideal of R. Suppose K 6= R. Then by Lemma 3 (iv), there exists a strong maximal k-ideal P of R such that K ⊆ P. Clearly, x /∈ TP(M). For if x ∈ TP(M), then (1+ p)x = 0 ∈ IM for some p ∈ P; hence (1+ p) ∈ K ⊆ P, a contradiction. Therefore, M is P-cyclic by (i). There exist p ∈ P and m ∈ M such that (1+ p)M ⊆ Rm. Then (1+ p)x ∈ ⋂ i∈Λ(Iim). For each i ∈ Λ, there is an element ai ∈ Ii such that (1+ p)x = aim. Choose j ∈ Λ. Then for each i ∈ Λ, a jm = aim. By assumption, ai + a′i = 0 for some a′i ∈ Ii; hence a jm+ a′im = 0. Now (1+ p)(a j + a′i)M ⊆ (a j + a′i)Rm = 0 implies (1+ p)(a j + a′i) = 0 since M is faithful. Therefore, a j + pa j + a′i + pa′i = 0, so (1+ p)a j = (1+ p)ai ∈ Ii; hence (1+ p)a j ∈ I . Thus (1+ p)2 x = (1+ p)(aim) ∈ IM . It follows that (1+ p)2 ∈ K ⊆ P, a contradiction. So K = R; hence x ∈ IM , and (ii) is proved. (iii) Let a /∈ P. Then the ideal K = {r ∈ R : r x ∈ PM} is a strong k-ideal of R. Suppose K 6= R. Then there exists a strong maximal k-ideal P ′ of R such that K ⊆ P ′. Clearly, x /∈ TP ′(M). By (i), M is P ′-cyclic, that is, there exist m ∈ M and q ∈ P ′ such that (1 + q)M ⊆ Rm. In particular, (1 + q)x = sm for some s ∈ R. Therefore we have (1+ q)ax ∈ (1+ q)PM ⊆ PRm = Pm; hence asm = pm for some p ∈ P. By assumption, p+ p′ = 0 for some p ∈ P ′; hence (as+ p′)m= 0. Since (1+ q)ann(m)M ⊆ Rann(m)m = 0, we must have (1 + q)(as + p′) = 0, (1 + q)as = (1+ q)p ∈ P. But P ⊆ K ⊆ P ′ so that s ∈ P and (1+ q)x = sm ∈ PM . Thus (1+ q) ∈ K ⊆ P ′, which is a contradiction. It follows that K = R and x ∈ PM , as required. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 259 Remark 1. (i) (Change of semirings) Assume that I is an ideal of a semiring R with I ⊆ (0 : M) and let M be an R-semimodule. We show now how M can be given a natural structure as a semimodule over R/I . Let r, s ∈ R such that r + I = s + I , and let m ∈ M. Then r + a = s+ b for some a, b ∈ I , and rm = sm. Hence we can unambiguously define a mapping R/I ×M into M (sending (r + I , m) to rm) and it is routine to check that this turns the commutative additive semigroup with a zero element M into an R/I-semimodule. It should be noted that a subset of M is an R-subsemimodule if and only if it is an R/I- subsemimodule. (ii) Assume that N is a proper k-subsemimodule of a semimodule M over a semiring R and let I be an ideal of R with I ⊆ (0 : M). Then N is a prime R-subsemimodule of M if and only if N is a prime subsemimodule of M as an R/I-semimodule. Theorem 7. The following statements are equivalent for a proper k-subsemimodule N of a very strong multiplication semimodule M over a semiring R. (i) N is a strong prime subsemimodule of M. (ii) (N : M) is a strong prime k-ideal of R. item N = PM for some strong prime ideal P of R with (0 : M) ⊆ P. Proof. (i)⇒ (ii). By [13, Lemma 4], (N : M) is a prime ideal of R. If 0 6= m ∈ M , then J = {r ∈ R : rm ∈ (N : M)M} is a proper strong k-ideal of R (since M is very strong multipli- cation) with (N : M) ⊆ J ; hence (N : M) is a strong prime k-ideal of R. (ii)⇒ (iii) is clear. (iii) ⇒ (i). Since N = PM 6= M and as an R/(0 : M)-semimodule, N is a strong prime subsemimodule by Proposition 3 (iii), so is a strong prime as an R-subsemimodule of M by Remark 1. Theorem 8. Let R be a semiring, N a proper subsemimodule of a very strong multiplication R-semimodule M and A= (N : M). Then rad(N) = rad(A)M. Proof. Without loss of generality M is a faithful R-semimodule. LetB denote the collection of all strong prime ideals P of R such that A ⊆ P and C denote the collection of all prime ideals P of R such that A ⊆ P. Clearly, C ⊆ B . If B = rad(A), then B = ⋂ P∈C P [see, 1], and hence by Proposition 3 (ii), BM = ⋂ P∈C (PM) ⊆ ⋂ P∈B PM . Let P ∈ B . If M = PM , then rad(N) ⊆ PM . If M 6= PM , then N = AM ⊆ PM implies rad(N) ⊆ PM by Theorem 7. It follows that rad(N)⊆ BM . Conversely, suppose that L is a strong prime subsemimodule of M containing N . By Theorem 7, there exists a strong prime ideal P ′ of R such that L = P ′M . Since AM = N ⊆ L = P ′M 6= M it follows that A⊆ P ′ by [13, Theorem 7], and hence B ⊆ P ′. Thus BM ⊆ L. It follows that BM ⊆ rad(N), and so we have equality. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 260 4. Prime spectrum Assume that R is a semiring and let M be an R-semimodule and N be a subsemimodule of M such that N = IM for some ideal I of R. Then we say that I is a presentation ideal of N . Clearly, every subsemimodule of M has a presentation ideal if and only if M is a multiplication semimodule. Let N and K be subsemimodules of a multiplication R-semimodule M with N = I1M and K = I2M for some ideals I1 and I2 of R. The product N and K denoted by N K is defined by N K = I1 I2M . Let N = I1M = I2M = N ′ and K = J1M = J2M = K ′ for some ideals I1, I2, J1 and J2 of R. It is easy to show that N K = N ′K ′, that is, N K is independent of presentation ideals of N and K (The proof is similar to Ameri [4, Theorem 3.4]). It is easy to see that N K is a subsemimodule of M and N K ⊆ N ∩ K . For m, m′ ∈ M by mm′, we mean the product of Rm and Rm′, that is, mm′ = IJ M , where I and J are presentations for m and m′, respectively. Lemma 4. Let N be a proper subsemimodule of a multiplication semimodule M over a semiring R. Then the follwing statements hold: (i) N is prime if and only if whenever UV ⊆ N for some subsemimodules U and V of M, then U ⊆ N or V ⊆ N. (ii) N is prime if and only if whenever m.m′ ⊆ N for some m, m′ ∈ M, then Rm ⊆ N or Rm′ ⊆ N. Proof. The proofs are straightforward (the proofs are similar to Ameri [4, Theorem 3.16 and Corollary 3.17]). Let M be a non-strong semimodule over a semiring R with M 6= 0. Then by Theorem 4, the k-spectrum X = Speck(M) is non-empty. For any subsemimodule N of a semimodule M by V (N) we mean the set of all prime k-subsemimodules of M containing N . Clearly, V (M) = ; and V ({0}) = Speck(M) = X . Throughout this section we may assume that Speck(M) is non-empty. Lemma 5. Let M be a semimodule over a semiring R. Then the following statements hold: (i) If N is a subsemimodule of M, then V (N) = V (rad(N)). (ii) If {Ni}i∈I is a family of subsemimodules of M, then V ( ∑ i∈I Ni) = ⋂ i∈I V (Ni). Proof. (i) Since N ⊆ rad(N), we have V (rad(N)) ⊆ V (N). For the reverse inclusion, assume that P ∈ V (N). Then N ⊆ P; hence rad(N)⊆ P, and so we have equality. (ii) Let P ∈ ⋂ i∈I V (Ni). Then Ni ⊆ P for every i ∈ I , so ∑ i∈I Ni ⊆ P, which implies that⋂ i∈I V (Ni)⊆ V ( ∑ i∈I Ni). The reverse inclusion is similar. S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 261 If ζ(M) denotes the collection of all subsets V (N) of Speck(M), then ζ(M) contains the empty set and Spec(M) and is closed under arbitrary intersection by Lemma 5 (ii). If also ζ(M) is closed under finite union, that is, for every subsemimodules N and L of M such that V (N) ∪ V (L) = V (T ) for some subsemimodule T of M , for in this case ζ(M) satisfies the axioms of closed subsetes of a topological spaces, which is called Zariski topology. In MacCasland, Moore and Smith [22] a module with Zariski topology is called a top module. Lemma 6. The following statements are equivalent for a semimodule M over a semiring R. (i) M is a top semimodule. (ii) Every prime k-subsemimodule of M is extraordinary. (iii) V (T )∪ V (L) = V (T ∩ L) for any semiprime subsemimodules T and L of M. Proof. (i)⇒ (ii). Let N be any prime k-subsemimodule of M and let T and L be semiprime subsemimodules of M such that T ∩ L ⊆ N . By (i), there exists a subsemimodule U of M such that V (T )∪ V (L) = V (U). Now set T = ⋂ i∈I Ni, where Ni is a prime k-subsemimodule of M (i ∈ I). For each i ∈ I , Ni ∈ V (T ) ⊆ V (U), so that U ⊆ Ni. Thus U ⊆ T . Similarly, U ⊆ L. Thus U ⊆ T ∩ L. Now we have V (T ) ∪ V (L) ⊆ V (T ∩ L) ⊆ V (U) = V (T ) ∪ V (L), that is, V (T )∪ V (L) = V (T ∩ L). Now N ∈ V (T ∩ L) gives T ⊆ N or L ⊆ N by Lemma 5. (ii)⇒ (iii) is clear. (iii)⇒ (i). Let A and B be any k-subsemimodules of M . If V (A) = ;, we are done. So we may assume that V (A) and V (B) are both non-empty. Then V (A)∪V (B) = V (rad(A))∪V (rad(B)) = V (rad(A)∩ rad(B)) by Lemma 5 and (iii), as required. Theorem 9. If N is a QM -subsemimodule of a top semimodule M over a semiring R, then M/N is a top semimodule. Proof. Note that any semiprime k-subsemimodule of M/N has the form U/N where U is a semiprime k-subsemimodule of M containing N by Corollary 1. Let T/N be any prime k-subsemimodule of M/N and let U/N and L/N be semiprime k-subsemimodules of M/N such that (L/N)∩ (U/N)⊆ T/N . Then (L ∩ U)/N ⊆ (L/N)∩ (U/N)⊆ T/N , so U ∩ L ⊆ T ; hence either U ⊆ T or L ⊆ T since T is extraordinary by Lemma 6. Thus either U/N ⊆ T/N or L/N ⊆ T/N . Now the assertion follows from Lemma 6 (ii). Theorem 10. Let N , L be k-subsemimodules of a semimodule M over a semiring R. Then the following statements hold: (i) If S is a subset of M, then V (S) = V (< S >). (ii) V (N)∪ V (IM) = V (IN) = V (N ∩ IM) for every strong ideal I of R . (iii) V (IM)∪ V (J M) = V (IJ M) = V (IM ∩ J M) for every strong ideals I and J of R. (iv) If V (N)⊆ V (L), then L ⊆ rad(N). S. Atani, R. Atrani, Ü. Tekir / Eur. J. Pure Appl. Math, 4 (2011), 251-265 262 (v) V (N) = V (L) if and only if rad(N) = rad(L) (vi) If M is a strong multiplication semimodule, then V (N)∪ V (L) = V (N L) = V (N ∩ L). (vii) Every strong multiplication semimodule is a top module. (viii) If M is a strong multiplication semimodule, then every prime k-subsemimodule of M is extraordinary. Proof. (i) Obvious. (ii) It is clear that V (N) ∪ V (IM) ⊆ V (N ∩ IM) ⊆ V (IN). Let P ∈ V (IN). Then IN ⊆ P and hence N ⊆ P or IM ⊆ P by [13, Theorem 7]. Thus P ∈ V (N) or P ∈ V (IM), i.e. P ∈ V (N)∪ V (IM). Hence V (IN)⊆ V (N)∪ V (IM). (iii) Follows from (ii). (iv) Obvious. (v) Let V (N) = V (L). By Lemma 5, we have V (N) ⊆ V (rad(L)); hence rad(L) ⊆ rad(N) by (iv). Similarly, rad(N) ⊆ rad(L), and so we have equality. The other implication is similar. (vi) Apply (iii). (vii) Follows from (vi). (viii) Follows from (vii) and Lemma 6. Remark 2. Assume that M is a semimodule over a semiring R and let X = Speck(M). For each subset S of M, by XS we mean X − V (S) = {P ∈ X : S * P}. If S = {m}, we denote by Xm = {P ∈ X : Rm * P} = {P ∈ X : m * P}. Clearly, the sets Xm are open, and they are called basic open sets. Lemma 7. Let M be a strong multiplication semimodule over a semiring R. Then the following statements hold: (i) X I M ∩ X J M = X I J M for every strong ideals I and J of R. (ii) The setA = {Xm : m ∈ M} forms a base for the Zariski topology on X . Proof. (i) Immediately follows from Theorem 10 (iii) (taking complements). REFERENCES 263 (ii) Suppose that U is an open set in X . Then U = X − V (N) for some k-subsemimodule N of M . Let N =< {mi : i ∈ I} >, where {mi : i ∈ I} is a generator set of N . Then V (N) = V ( ∑ i∈I Rmi) = ⋂ i∈I V (Rmi) by Lemma 5 (ii). It follows that U = X−V (N) = X− ⋂ i∈I V (Rmi) = ⋃ i∈I Xmi . ThusA is a base for the Zariski topology on X . Proposition 4. Let M be a very strong multiplication semimodule over a semiring R. Then every basic open set of X is compact. Proof. By Lemma 7 (ii), it suffices to show that every cover of basic open sets has a finite subcover. Suppose that Xm ⊆ ⋃ t∈I Xmt , and let N be the subsemimodule of M generated by {mt : t ∈ I}. It follows that ⋂ t∈I V (Rmt) = V (N)⊆ V (Rm), so V (rad(N)) ⊆ V (rad(< m >)) by Lemma 5 (i); hence rad(< m >) ⊆ rad(N) by Theo- rem 10 (iv). Moreover, by Theorem 8, rad(N) = rad(A)M , where A = (N : M). By as- sumption, there exists a finite subset J of I and ri ∈ rad(A) (i ∈ J) such that m = ∑ t∈J rt mt . For ri ∈ rad(A), there is a positive integer si such that r si i ∈ A. If s = ∑ i∈J si , then r s i ∈ A for every i ∈ J . For each i ∈ J , there exists a strong ideal Ii of R such that Rmi = Ii M ; so by Proposition 3 (ii), m ∈ ∑ i∈J ri Ii M = ( ∑ i∈J (ri Ii))M . Thus ms ⊆ ( ∑ i∈J (ri Ii) s)M ⊆ AM . Therefore Theorem 10 gives V (N) = ⋂ i∈I V (Rmi)⊆ ⋂ i∈J V (Rmi)⊆ V (m) = V (Rm) = V (ms). Taking complements, we have Xm ⊆ ⋃ i∈J Xmi , and so the proof is complete. Theorem 11. Let M be a very strong multiplication semimodule over a semiring R. Then an open set of X is compact if and only if it is a finite union of basic open sets. Proof. Apply Lemma 7 and Proposition 4. Corollary 2. 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