`International Journal of Algebra, Vol. 4, 2010, no. 26, 1299 - 1306Prime Subsemimodules of SemimodulesReza Ebrahimi Atani Department of Computer Engineering University of Guilan P.O. Box 3756, Rasht, Iran [email protected]Abstract In this paper we characterize prime subsemimodules of a semimodule over a commutative semiring.Mathematics Subject classification: 16Y60 Keywords: Quotient semimodules, Strong ideals, Prime subsemimodules1IntroductionSemimodules over semirings also appear naturally in many areas of mathematics. For example, semimodules are useful in the area of theoretical computer science as well as in the cryptography [10]. In the present paper, we introduce and investigate the quotient semimodules of a module over a semiring. Moreover, we extend some basic results of Lu [9] to semimodules over semirings. For the definitions of monoid, semirings, semimodules and subsemimodules of a semimodule we refer [8, 6, 1, 2]. All semiring in this paper are commutative with non-zero identity. A semiring R is said to be semidomain whenever a, b  R with ab = 0 implies that either a = 0 or b = 0. A semifield is a semiring in which non-zero elements form a group under multiplication. The semiring R is considered to be also a semimodule over itself. In this case, the subsemimodules of R are called ideals of R. An R-semimodule M is said to be semivector space if R is a semifield. 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). A prime subsemimodule of M is a proper subsemimodule N of M in which x  N or rM  N whenever rx  N. We define k-ideals and prime ideals of a semiring R in a similar fashion. We say that r  R is a zero-divosor for a semimodule M if rm = 0 for some non-zero element m of M. The set of zero-divisors of M is written ZR (M).1300R. Ebrahimi Atani2Quotient semimodulesQuotient semimodules are determined by equivalence relations rather than by subsemimodules as in the module case. If N is a sunsemimodule of a semimodule M over a semiring R, we define a relation  on M, given by m1  m2 if and only if there exist n1 , n2  N satisfying m1 + n1 = m2 + n2 . Then  is an equivalence relation on M, and we denote the equivalence class of m by m + N and these collection of all equivalence classes by M/N. Then M/N forms a commutative additive semigroup which has zero element under the binary operation  defined as follows: (m + N)  (m + N) = m + m + N. Then 0M + N is the zero element of M/N. Now let r  R and suppose that m1 + N, m2 + N  M/N are such that m1 + N = m2 + N in M/N. Then there are elements a, b  N such that rm1 + ra = rm2 + rb, so rm1  rm2 ; hence rm1 + N = rm2 + N. Hence we can unambiguously define a mapping from R × M/N into M/N (sending (r, m + N) to rm + N) and it is routine to check that this turns the commutative semigroup M/N into an R-semimodule. We call this R-semimodule the quotient semimodule or factor semimodule of M modulo N. Lemma 2.1 Let N be a subsemimodule over a semimodule M over a semiring R. Then the following hold: (i) If a  N, then a + N = N. (ii) If N is a k-subsemimodule of M and a  N, then a + N = b + N for every b  M if and only if b  N. In particular, c + N = N if and only if c  N. Proof. (i) Since a + 0M = 0M + a, we conclude that a  0; hence a + N = 0 + N = N. (ii) Let a + N = b + N for every b  M. Then a + u = b + v for some u, v  N; so b  N since N is a k-subsemimodule. The other implication follows from (i) and the fact that N is a k-subsemimodule of M. 2 Proposition 2.2 Let N and K be subsemimodules of a semimodule M over a semiring R with N  K. Then K/N = {m + N : m  K} is a subsemimodule of M/N. In particular, if K is a k-subsemimodule of M, then K/N is a k-subsemimodule of M/N. Proof. Clearly, 0 + N  K/N. Let m + N, m + N  K/N and r  R. It is easy to see that (m + N)  (m + N) = m + m + N  K/N and r(m + N) = rm + N  K/N. Thus K/N is a subsemimodule of M/N. Finally, assume that u + N  K/N and (u + N)  (v + N) = u + v + N  K/N, where u  K and v  M. It then follows that u + v + t1 = c + t2 for some t1 , t2  N and c  K; hence v  K since K is a k-subsemimodule. Thus v + N  K/N, and the proof is complete. 2Prime subsemimodules of semimodules1301Theorem 2.3 Let M be a semimodule over a semiring R, N a subsemimodule of M and L a k-subsemimodule of M/N. Then L = T /N for some ksubsemimodule of T of M. Proof. Assume that T = {m  M : m + N  L} and let n  N. Then by Lemma 2.1, n + N = 0M + N  L; hence N  T . Let a, b  T and r  R. Then (a + N)  (b + N) = a + b + N  L; so a + b  T . Similarly, ra  T . Thus T is a subsemimodule of M. Let a, a + b  T . Then a + N, a + b + N = (a + N)  (b + N)  L; hence b + N  L since L is a k-subsemimodule of M/N. It follows that b  T , whence T is a k-subsemimodule of M. Finally, an inspection will show that L = T /N. 2 Let M and N be semimodules over the semiring R, and let f : M  N be a homomorphism of R-semimodules. It is easy to see that Kerf = {m  M : f (m) = 0N } is a subsemimodule of M and Im(f ) is a subsemimodule of N. Theorem 2.4 Let M and N be semimodules over the semiring R, and let f : M  N be a homomorphism of R-semimodules. Then f induces an ¯ ¯ isomorphism f : M/Kerf  Imf for which f (m + Kerf ) = f (m) for all m  M. In particular, if f is surjective, then M/Kerf  N. = Proof. The proof is straightforward. 2Let R be a semiring. We define the Jacobson radical of R, denoted by Jac(R), to be the intersection of all the maximal k-ideals of R. Then by [3, Lemma 2], the Jacobson radical of R always exists and by [2, Lemma 2.12], it is a k-ideal of R. A non-zero element a of R is said to be semi-unit in R if there exist r, s  R such that 1 + ra = sa. An proper ideal I of a semiring R is said to be a strong ideal, if for each a  I there exists b  I such that a + b = 0 (see [5]) . Now we state and prove a version of Nakayama's lemma. Theorem 2.5 Let M be a finitely generated semimodule over a semiring R and let J be a strong ideal of R contained in the Jacobson radical of R. Then the following hold: (i) If M = JM, then M = 0. (ii) If N is a subsemimodule of M such that JM + N = M, then M = N. Proof. By [6, Lemma 2.2], (1 + t)M = 0 for some t  J. Let x  M. Then (1 + t)x = 0. Since 1 + t is a semi-unit by [7, Lemma 3.4], we must have 1 + (1 + t)t = (1 + t)s for some t, s  R; hence x = 0. (ii) Let m1 , m2 , ..., mk be elements which generate M. Then the R-semimodule generated by m1 + N, m2 + N, ..., mk + N is just M/N. Since J(M/N) = (IM + N)/N = M/N, we must have M/N = {0 + N} by (i). Now the assertion follows from Lemma 2.1. 21302R. Ebrahimi Atani3Basic properties of prime subsemimodulesLet R be a semiring, M an R-semimodule and N a prime subsemimodules of M. Then (N :R M) is a prime ideal of R [4, Lemma 9]. Let M be a semimodule over a semiring R. We say that M is a torsion-free R-semimodule whenever r  R and m  M with rm = 0 implies that either m = 0 or r = 0 (so every semivector space over a semifield R is a torsion-free R-semimodule). Allen [1] has presented the notion of a Q-ideal I in the semiring R and constructed the quotient semiring R/I. The results proven in [1] and [2] will be used in the next results. Remark 3.1 (Change of semirings) Assume that I is a Q-ideal of a semiring R and let N be a subsemimodule of an R-semimodule M. We show now how M/N can be given a natural structure as a semimodule over R/I. Let q1 , q2  Q such that q1 + I = q2 + I, and let m, m  M such that m + N = m + N. Then q1 = q2 , and q1 m + q1 n = q2 m + q1 n for some n, n  N; thus q1 m + N = q2 m + N. Hence we can unambiguously define a mapping R/I × M/N into M/N (sending (q1 + I, m + N) to q1 m + N) and it is routine to check that this turns the commutative additive semigroup with a zero element M/N into an R/I-semimodule. It should be noted that a subset of M/N is an Rsubsemimodule if and only if it is an R/I-subsemimodule. We next give six other characterizations of prime subsemimodules. Theorem 3.2 Let N be a proper k-subsemimodule of a semimodule over a semiring R with (N : M) = P a Q-ideal of R. Then the following statements are equivalent: (1) N is a prime subsemimodule of M; (2) M/N is a torsion-free R/P -semimodule; (3) (N :M &lt; r &gt;) = N for every r  R - P ; (4) (N :M J) = N for every ideal J P ; (5) (N :R &lt; m &gt;) = P for every m  M - N; (6) (N :R L) = P for every subsemimodule L of M properly containing N; (7) ZR (M/N) = P . Proof. (1)  (2) Note that M/N is an R/P -semimodule by Remark 3.1. Let (q + P )(m + N) = qm + N = 0M + N where q  Q and m  M, so qm  N by Lemma 2.1. Therefore, N prime gives either q  P or m  N. If q  P , then q + P is the zero in R/P (otherwise, m + N is the zero in M/N by Lemma 2.1 again). Thus M/N is torsion-free semimodule as an R/P -semimodule. (2)  (3) Assume that q0 + P is the zero element in R/P . It suffices to show that (N :M &lt; r &gt;)  N. Let m  (N :M &lt; r &gt;). Then rm  N andPrime subsemimodules of semimodules1303r = q + a for some q  Q and a  P (so q  P ); hence qm  N since N / is a k-subsemimodule. Since (q + P )(m + N) = 0M + N by Lemma 2.1 and q + P = q0 + P , we must have m + N = 0M + N; hence m  N by Lemma 2.1, and so we have equality. (3)  (4) Clearly, N  (N :M J). For the reverse inclusion, assume that m  (N :M J). By assumption, there exist r  J such that r  R - P and rm  N; so (N :M &lt; r &gt;) = N by (3). This completes the proof. (4)  (5) Since P M  N, we conclude that P  (N :R &lt; m &gt;) for every m  M - N. for the other containment, assume that m  M - N and r  (N :R &lt; m &gt;); we show that r  P . Suppose not. Then J =&lt; r &gt; P , and so m  (N :M J) = N by (4), which is a contradiction, as required. (5)  (6) If a  P , then aL  aM  N; so P  (N :R L). Now suppose that b  (N :R L). By assumption, there exists m  L such that m  M - N. Then b  (N :R &lt; m &gt;) = P by (5), as needed. (6)  (7) Let r  ZR (M/N). Then there exists m  M - N such that r(m + N) = rm + N = 0M + N, so rm  N by Lemma 2.1; hence r  (N :R Rm + N) = P by (6). Thus ZR (M/N)  P . For the reverse conclusion, assume that a  P . By assumption, there is an element m  M - N such that am  N, so a(m + N) = am + N = 0M + N; thus a  ZR (M/N). This completes the proof. (7)  (1) Let rm  N for some r  R and m  M - N; we show that r  P . Then r(m + N) = 0M + N by Lemma 2.1; hence r  ZR (M/N) = P by (7), as required. 2 Proposition 3.3 Let N be a proper k-subsemimodule of a semimodule M over a semiring R with (N : M) = P a maximal Q-ideal of R. Then N is a prime subsemimodule. In particular, P M is a prime subsemimodule of an R-semimodule M for every maximal Q-ideal P of R such that P M = M. Proof. By [2, Theorem 2.10], R/P is a semifield, so M/N is a semivector space over the semifield R/P by Remark 3.1; hence it is a torsion-free R/P semimodule. Thus N is prime by Theorem 3.2. Finally, suppose that (P M : M) = J = R. Then P  J, so J = P since P is maximal, as required. 2 Proposition 3.4 Let N be a proper k-subsemimodule of a semimodule M over a semiring R with (N : M) = P a Q-ideal of R and let P be a maximal ideal of R. Then N is a P -prime if and only if P M  N. In particular, if N is an P -prime subsemimodule of M, then so is every proper subsemimodule of M containing N. Proof. It suffices to show that if P M  M, then N is P -prime. Let p  P . Then p  (N : M), so P = (N : M) by maximality of P . Now apply Proposition 3.3. 21304R. Ebrahimi AtaniProposition 3.5 Let N be a proper maximal k-subsemimodule of a semimodule over a semiring R. Then N is a prime subsemimodule and (N : M) is a maximal ideal of R. Proof. Let rm  N such that r  R and m  M - N. Note that N is a maximal subsemimodule if and only if M/N is a simple R-semimodule. Hence M/N is the cyclic R-semimodule Rm, where m = m+N. By assumption and Lemma ¯ ¯ 2.1, we must have r m = 0M + N, so r(M/N) = r(Rm) = R(r m) = 0M + N; ¯ ¯ ¯ hence rM  N. Thus N is prime. Finally, The mapping f : R  Rm defined ¯ by f (r) = r m for all r  R is a surjective R-semimodule homomorphism, so ¯ R/Kerf  Rm by Theorem 2.4; hence Kerf = ann(m) = ann(M/N) = (N : ¯ = ¯ M) is a maximal ideal of R. 2 Proposition 3.6 Let N1 , N2 , ..., Nn be subsemimodules of an R-semimodule M and let N be a prime subsemimodule of M. If N1  N2  ...  Nn  N, then there exists an i such that either Ni  N or (Ni : M)  (N : M). / Proof. Suppose not. Then there exists m  N1 such that m  N and a / ai  (Ni : M) such that ai  (N : M) for every i = 1. Therefore, ai m  N1 Ni for every i = 1 so that a2 a3 ...an m  N1  N2  ...  Nn  N. However, m  N / / and a2 a3 ...an  (N : M), which is a contradiction. 2 Lemma 3.7 Let R be a semiring, I a strong ideal in R, M an R-semimodule generated by n elements, and x an element of R satisfying xM  IM. Then (xn + y)M = 0 for some y  I. Proof. We use induction on n. Consider first the case in which n = 1. Here we have x &lt; m &gt; I &lt; m &gt;. So xm = sm for some s  I; hence there is an element s  I such that (x + s )m = sm + s m = 0. It follows that (x + s )M = 0. We now turn to the inductive step. Assume, inductively, that n = k + 1, where k  1, and that the result has been proved in the case where n = k. Then we must have (x + a)(xk + b)M = (xk+1 + axk + bx + ab)(&lt; m1 , ..., mk &gt; + &lt; mk+1 &gt;) = 0 for some a, b  I, so (xk+1 + c)M = 0, where axk + bx + ab = c  J. This completes the proof. 2 Proposition 3.8 Let M be a finitely generated semimodule over a semiring R and let I be a strong k-ideal of R such that I = rad(I). Then (IM : M) = I if and only if ann(M)  I. Proof. The necessity is clear. Assume that ann(M)  I and let x  (IM : M). If M generated by n elements, then there exists a y  I such that xn + y  ann(M)  I by Lemma 3.7. Since I is a k-ideal, we must have xn  I and, therefore, (IM : M)  rad(I) = I. Now we can see easily that (IM : M) = I. 2Prime subsemimodules of semimodules1305Theorem 3.9 If M is a finitely generated semimodule over a semiring R and P is a maximal Q-ideal of R containing ann(M), then P M = M so that P M is a prime subsemimodule of M. In particular, If M is a finitely generated faithful R-semimodule, then P M is a prime subsemimodule of M for every maximal Q-ideal P of R. Proof. Apply Proposition 3.3 and Proposition 3.8 (not that every Q-ideal is a k-ideal). 2 Acknowledgements I would like to thank Prof. S. Ebrahimi Atani for several useful suggestions on the first draft of the manuscript.References[1] P. J. Allen, A fundamental theorem of homomorphisms for simirings, Proc. Amer. Math. Soc. 21 (1969), 412-416. [2] S. Ebrahimi Atani, The ideal theory in quotients of commutative semirings, Glasnik Matematicki 42 (2007), 301-308. [3] R. Ebrahimi Atani and S. Ebrahimi Atani, Ideal theory in commutative semirings, Bul. Acad. Stiinte Repub. Mold. Mat 2 (2008), 14-23. [4] R. Ebrahimi Atani and S. Ebrahimi Atani, On subsemimodules of semimodules, Bul. Acad. Stiinte Repub. Mold. Mat. to appear (2010). [5] S. Ebrahimi Atani and R. Ebrahimi Atani, Some remarks on partitioning semirings, An. St. Univ. Ovidius Constanta, 18 (1) (2010), 49-62. [6] S. Ebrahimi Atani and M. Shajari Kohan, A note on finitely generated multiplication semimodules over comutative semirings, International Journal of Algebra 4(8) (2010), 389-396. [7] S. Ebrahimi Atani, The zero-divisor graph with respect to ideals of a commutative semiring, Glas. Math. 43 (2008), 309-320. [8] J. S. Golan, The theory of semirings with applications in mathematics and theoretical computer Science, Pitman Monographs and Surveys in Pure and Applied Mathematics, Longman Scientific and Technical, Harlow UK, 1992. [9] C. P. Lu, Prime submodules of modules. Comment. Univ. St. Pauli 33 (1984), 61-69.1306R. Ebrahimi Atani[10] G. Maze, C. Monico and J. Rosenthal, Public key cryptography based on semigroup actions, Adv. Mathematics of communication, 1(4) (2007), 489-281. Received: July, 2010`

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