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Chiang Mai J. Sci. 2013; 40(1) 77

Chiang Mai J. Sci. 2013; 40(1) : 77-82 http://it.science.cmu.ac.th/ejournal/ Contributed Paper

Ideals in Quotient Shahabaddin Ebrahimi Atani and Ameneh Gholamalipour Garfami Faculty of Mathematical Sciences, University of Guilan, P.O. Box 1914, Rasht, Iran. *Author for correspondence; e-mail: [email protected]

Received: 10 November 2010 Accepted: 5 October 2011

ABSTRACT Since the class of quotient rings is contained in the class of quotient semirings, in this paper, we will make an intensive study of the properties of quotient semirings as compared to similar properties of quotient rings. The main aim of this paper is that of extending some well-known theorems in the theory of quotient rings to the theory quotient semirings.

Keywords: quotient semirings, weakly prime ideals, weakly primal , semidomain like semirings

1. INTRODUCTION Semirings are natural topic in algebra ideal as in the case. There are many to study because they are the algebraic different definitions of a quotient structure of the of natural numbers. appearing in the literature. P.J. Allen ([2]) Semirings also appear naturally in many introduced the notion of Q-ideal and a . For example, semirings construction process was presented by are useful in the area of theoretical computer which one can build the quotient structure science as well as in the solution of of a semiring a Q-ideal (also see problem in graph theory and optimiza- the results listed in [3-5,7,8]). If I is an tion. In structure, semirings lie between ideal of a semiring R, then Golan has and rings. The class of rings is presented the notion quotient semiring contained in the class of semirings [7,8]. R/I, but this definition is different from Therefore, all of the properties given here the definition of Allen (see Section 2). apply to rings. Here we follow the definition of Golan. This paper generalizes some well- The main part of this paper is devoted to known result on quotient rings in stating and proving analogues to several commutative rings to commutative well-known theorems in the theory of semirings. The maindifficulty is figuring quotient rings (see Section 2). out what additional hypotheses the In order to make this paper easier to semiring or ideal must satisfy to get similar follow, we recall here various notions results. Quotient semirings are determined from semiring theory which will be used by equivalence relations rather than by in the sequel. A commutative semiring R 78 Chiang Mai J. Sci. 2013; 40(1)

is defined as an algebraic system (R,+,.) (r + I) + (s + I) = r + s + I and (r + I) + such that (R,+) and (R,.) are commutative (s + I) = rs + I [7]. Our starting point is semigroups, connected by a(b + c) = ab + ac the following lemma. for all a, b, c ∈ R, and there exists 0 ∈ R such that r + 0 = r and r0 = 0r = 0 for each Lemma 2.1 Let I be an ideal of a semiring R. r ∈ R. In this paper all semirings considered Then the following hold: will be assumed to be commutative with (1) If a ∈ I, then a + I = I. non-zero identity. A subset I of a semiring (2) If I is a k-ideal of R and a ∈ I, then R will be called an ideal if a, b ∈ I and a + I = b + I for every b ∈ R if and only if r ∈ R implies a + b ∈ I and ra ∈ I. b ∈ I. In particular, c + I = I if and only if A subtractive ideal (= k-ideal) J is an ideal c ∈ I. such that if x, x + y ∈ J then y ∈ J (so {0} is a k-ideal of R). A prim ideal of R is a Proof. (1) Since a + 0 = 0 + a, we conclude proper ideal P of R in which x ∈ P or that a ~ 0; hence a + I = 0 + 1. y ∈ P whenever xy ∈ P. A (2) Let a + I = b + I for every b ∈ R. P of R is a proper ideal of R such that, if Then a + u = b + v for some u, v ∈ I; so xy ∈ P and x ∉ P, then y ∈ P = {r ∈ R : rn ∈ P b ∈ I since I is a k-ideal. The other for some positive n}. A semiring R implication follows from (1) and the fact I is said to be a semidomain if ab = 0 (a, b ∈ R), is a k-ideal of R. then either a = 0 or b = 0. A is a semiring in which non-zero elements from Lemma 2.2 Let I and J be ideals of a a under multiplication. An element semiring R with I ⊆ J. Then the following a of a semiring R is called zero-divisor of hold: R if there exists 0 ≠ b ∈ R such that ab = 0 (1) J/I ={a + I: a ∈ J} is an ideal of (not here that we include 0 in the set of R/I. In particuar, if J is a k-ideal of R, then zero-divisors of semiring). The collection J/I is a k-ideal of R/I. of all zero-divisors of R will be denoted by (2) If 1+I ∈ J/I, then R/I= J/I. Z(R). Furthermore, the subset {a ∈ R: (3) If a+I is a invertible element of R/I there exists a positive integer such that with a+I ∈ J/I, then R/I=J/I. an = 0} of Z(R) consisting of the elements of R will be denoted by nil (R), Proof. (1)Cleary, 0 + I ∈ J/I. Let a + I, b + the nilradical of R. I ∈ J/I and r + I ∈ R)/I. It is easy to see that (a + I) + (b + I) = a + b + I ∈ J/I and (r + I) 2. RESULTS (a + I) = ra + I ∈ J/I. Thus J/I is an ideal Quotients semirings are determined of R/I. Finally, assume that u + I ∈ J/I and by equivalence relations rather than by (u + I) + (v + I) = u + v + I ∈ J/I, where ideals as in the ring case. If I is an ideal of u ∈ J and v ∈ R. It then follows that semiring R, we define a on R, given u + v + t1 = c + t2 for some t1, t2 ∈ I and c ∈ J; by r1 ~ r2 if and only if there exist a1, a2 ∈ I hence v ∈ J since J is a k-ideal. Thus v + I satisfying r1 + a1 = r2 + a2. Then is an ∈ J/I, and the proof is complete. equivalence relation on R, and we denote (2) Let x + I ∈ R/I. Then (x + I) (1 + I) the equivalence class of r by r + I and these = x + I ∈ J/I; hence R/I ⊆ J/I, as required. collection of all equivalence classes by R/I. (3) Follows from (2). Golan shows that R/I is a semiring with Chiang Mai J. Sci. 2013; 40(1) 79

Theorem 2.3 Let I be an ideal of a semiring Theorem 2.4 Let P be a proper k-ideal of R. Then the following hold: a semiring R. Then the following hold: (1) If L is an ideal of R/I, then L = J/I for (1) P is a maximal k-ideal of R if and some ideal J of R. only if R/I is a semifield. (2) If P is a k-ideal of R with I ⊆ P, then (2) If I is an ideal of R with I ⊆ P, then P is a of R if and only if P/I is a P is a maximal k-ideal of R if and only if prime ideal of R/I. P/I is a of R/I. (3) I is a prime k-ideal of R if and only R/I (0) if is a semidomain. In particular, is Proof. (1) Let P be a maximal ideal of R. prime if and onlyif only R is a semidomain. It suffices to show that every non-zero Proof. (1) Assume that J = {r ∈ R : r +I ∈ L} element a + P of R/P is invertible. By and let a ∈ I. Then by Lemma 2.1, a + I = Lemma 2.1, a ∉ P; hence P + Ra = R by 0 + I ∈ L; hence I ⊆ J. Let a,b ∈ J and r ∈ R. maximality of P. There exist r ∈ R and p ∈ P Then (a + I)+(b + I) = a + b + I ∈ L; so such that ra + p = 1. It then follows from a + b ∈ J. Similarly, ra ∈ J. Thus J is an Lemma 2.1 that (r + P)(a + P) = 1 + P. ideal of R. Finally, it is easy to see that Thus a + P is invertible. Conversely, L = J/I. assume that R/P is a semifield and P J (2) Let P be a prime ideal of R. Suppose for some k-ideal J of R; we show that J = R. that r + I, s + I ∈ P/I are such that (r + I) Then there is an element b ∈ J-P such that (s + I) = rs + I ∈ P/I, where r, s ∈ R. Then b + P is invertible in R/P, so (b+P)(c+P) = rs + I = P + I for some p ∈ P. This implies bc + P = 1+P forsome c + P ∈ R/P. that rs ∈ P since P is a k-ideal. Then P Since J is a k-ideal, we conclude that 1 ∈ J, prime gives either r ∈ P or s ∈ P; so either as needed. r + I ∈ P/I or s + I ∈ P/I by Lemma 2.2. (2) Suppose that P is a maximal k-ideal Conversely, suppose that P/I is prime. of R and let L be a k-ideal of R/I such that Let a, b ∈ R such that ab ∈ P. Then by P/I U. There exists a k-ideal J of R such Lemma 2.1, (a + I)(b + I) = ab + I = 0 + I that P/I L = J/I by Theorem 2.3 (1), ∈ P/I; thus either a + I ∈ P/I (so a ∈ P) or so P J; hence J = R. Thus L = R/I. The b + I ∈ P/I (so b ∈ P), as required. other implication is similar. (3) Let I be a prime ideal of R and let If R is a semiring, then R is Noetherian a + I and b + I be elements of R/I such (resp. Artinian) if any non-empty set of that (a + I)(b + I)=ab + I = 0 + I, where a, k-ideals of R has a maximal member (resp. b ∈ R. By Lemma 2.1 (2), ab ∈ I; so either minimal member) with respect to set a ∈ I or b ∈ I. Therefore, by Lemma 2.1 inclusion. This definition is equivalent to (1), either a + I = I or b + I = I. Thus a + the ascending chain condition (resp. R/I is semidomain. The proof of the other descending condition) on k-ideals. It is easy implication is similar. to see that If I and J are k-ideals of R, I + J is a k-ideal of R, and an intersection of a A proper ideal I in a semiring R is said family of k-ideals of R is k-ideal. to be maximal (resp. k-maximal)if J is an ideal (resp. a k-ideal) in R such that I ⊆ J, Theorem 2.5 Let I be a k-ideal of a then I = R. Moreover, it is clear that if I, J semiring R. R is Noetherian (resp. Artinian) and L are ideals of R with I ⊆ J, I ⊆ L and if and only if both I and R/I are Noetherian J/I = L/I, then J = L. (resp. Artinian). 80 Chiang Mai J. Sci. 2013; 40(1)

Proof. Let J1 ⊆ J2 ⊆ … ⊆ Jn ⊆ Jn+1 ⊆ … be show that J ∩ K = L. Let x ∈ J ∩ K. Then an ascending chain of k-ideals of R. Then x + I ∈ (J/I) ∩ (K/I) = L/I so x + a1 = y + a2

J1 ∩ I ⊆ J2 ∩ I ⊆ … ⊆ Jn ∩ I ⊆ Jn+1∩ I ⊆ … is for some y ∈ L and a1, a2 ∈ I; thus x ∈ L an ascending chain of k-ideals of I, and so since L is a k-ideal. Thus J ∩ K ⊆ L. For the there is a positive integer s such that Js ∩ I reverse inclusion, suppose that z ∈ L. Then

= Js+i ∩ I for all positive integer i. So (J1 + I)/ z + I ∈ (J/I) ∩ (K/I); so z + b1 = u + b2 and

I ⊆ (J2 + I)/I ⊆ … ⊆ (Jn+ I)/I ⊆ (Jn+1+ I)/I ⊆ z + c1 = v + c2 for some b1, b2, c1, c2 ∈ I, u ∈ k

… is a chain of k-ideals of R/I. Thus there and v ∈ J. Thus z ∈ J ∩ K since K and J are exists a positive integer t such that (Jt + I)/I k-ideals, and so we have equality. The other

= (Jt+i + I)/I for all positive integer i, so that implication is similar.

I + Jt = I +Jt+i for all i. Let u = max{s,t}. We show that, for each positive integer i i, Theorem 2.7 Let R be a semiring, I an ideal

Ju = Ju+i. Since the inclusion Ju ⊆ Ju+i is of R and J a strongly irreducible k-ideal of trivial, we will prove the reverse inclusion. R with I ⊆ J. Then J/I is a strongly irreducible

Let x ∈ Ju+i. Since x ∈ I + Ju+i = I + Ju, ideal of R/I. we must have x=a+b for some a ∈ I and b ∈ Ju ⊆ Ju+i. Hence a ∈ Ju+i, since it is a Proof. Let N and M be k-ideals of R/I such k-ideal. It follows that a ∈ I ∩ Ju+i = I ∩ Ju; that N ∩ M ⊆ J/I. Then there are k-ideals hence both a and b belong to Ju and x ∈ Ju. K,H of R such that N = K/I and M = H/I Thus R is Noetherian. Convesely, assume by Theorem 2.3; hence Lemma 2.7 gives that R is Noetherian. By Theorem 2.3, is K ∩ H ⊆ J. Since J is strongly irreducible it an ascending chain of k-ideals of R/I must follows that either K ⊆ J or H ⊆ J; hence have the form J1/I ⊆ J2/I ⊆ … ⊆ Jn/I ⊆ either N = K/I ⊆ J/I or M = H/I ⊆ J/I. So

Jn+1/I ⊆ …, where J1 ⊆ J2 ⊆ … ⊆ Jn ⊆ Jn+1 J/I is strongly irreducible. ⊆ … is an ascending chain of k-ideals of A semiring R is called semidomainlike R all of which contain I. Since the latter semiring, if Z(R) ⊆ nil (R). A classic result chain must eventually become stationary, of commutative semiring theory is that an so must the former. Thus R/I is Noetherian. ideal P is prime if and only if R/P is a Since every subideal of I is an ideal of R, it semidomain (see Theorem 2.3 (3)). The is clear from the definition of Noetherian following theorem is a parallel result for semiring that I is Noetherian. The Artinian semidomainlike semirings. case can be proved in a very similar manner tothe way in which was proved above, and Theorem 2.8 Let P be a proper k-ideal of a we omit it. semiring R.Then P is primary if and only if An ideal I of a semiring R is strongly R/P is a semidomainlike semiring. irreducible if for ideals J and K of J, the inclusion J ∩ K ⊆ I implies that either J Proof. Assume that P is a primary ideal of ⊆ I or K ⊆ I. R and let a + P ∈ Z(R/P). Then there exists a non- b + P of R/P such that Lemma 2.6 Let I be an ideal of a semiring (a + P)(b + P) = 0 + P; so ab ∈ P by R. If J,K and L are k-ideals of R containing Lemma 2.1. If b ∈ P, then b + P = 0 + P I, then (J/I) ∩ (K/I)=L/I if and only if J ∩ K. (see Lemma 2.1),which is a contradiction. Then P primary gives (a + P)n = an+ P = 0 + P Proof. Suppose that (J/I) ∩ (K/I) = L/I; we for somen by Lemma 2.1. Thus a + P ∈ Chiang Mai J. Sci. 2013; 40(1) 81

nil R/P and so R/P is a semidomainlike Theorem 2.10 Let R be a semiring, I an semiring. Conversely, let ab ∈ P, where ideal of R and P a k-ideal of R with I ⊆ P. a, b ∈ R. Then (a + P)(b + P) = ab + P = Then the following hold: 0 + P by Lemma 2.1. If a + P = 0 + P, (1) If P is a weakly primary ideal (resp. then a ∈ P. Similarly, for b + P = 0 + P. weakly prime ideal) of R, then P/I is So we may assume that a + P ≠ 0 + P and aweakly primary ideal (resp. weakly prime b + P ≠ 0 + P. Therefore, by assumption ideal ) of R/I. (a + P)n = an + P = 0 + P for some n; hence (2) If both I and P/I are weakly primary an ∈ P. Thus P is primary. (resp. weakly prime ideal ) ideal, then P is In general, a semidomainlike semirings weakly primary ideal (resp. weakly prime is not necessarily a semidomain, but as the ideal ). next result shows for semirings of the form R/ I, where I is anideal of R, the two Proof. (1) Assume that P is weakly prime concepts are equivalent. and let a + I and b + I be elements of R/I such that (0 + I) ≠ (a + I)(b + I) = ab + I ∈ Theorem 2.9 Let I be an ideal of a semiring P/I, so 0 ≠ ab ∈ P by Lemma 2.1. Then P n R. Then R/ I is semidomainlike if and only weakly primary gives either a ∈ P or b ∈ P for somen. If a ∈ P, then a + I ∈ P/I by if R/ I is a semidomain. In particular, Lemma 2.2. So suppose that bn ∈ P. It R/nil(R) is semidomailike if and only if follows that (b + I)n = bn + I ∈ P/I. Thus R/nil(R) is a semidomain. P/I is weakly primary. (2) Let 0 ≠ ab ∈ P, where a, b ∈ R. If Proof. Let R/ I be semidomainlike, and ab ∈ I, then I weakly primary gives either let (a + I)(b + I) = ab + I = 0 + I in a ∈ I ⊆ P or bs ∈ I ⊆ P for some s. So we R/ I with a + I ≠ 0 + I . Then ab ∈ I may suppose that ab ∈ I. By Lemma 2.1, by Lemma 2.1 and a ∉ I. Since R/ I we must have (0 + I) ≠ (ab + I) = (a + I) (b + I) P/I a + I P/I (b + is semidomainlike, I is primary by ∈ ; so either ∈ or I)m = bm + I ∈ P/I for some m since P/I is Theorem 2.7. Therefore, bm ∈ I for some weakly primary. If a + I ∈ P/I, then a ∈ P by positive integer m, whence b ∈ I, Lemma 2.1. If bm + I ∈ P/I, then bm ∈ P. b + I = 0 + I, and R/ I is a semidomain. Thus P is weakly primary. The other implication is similar. Let I be an ideal of a semiring R. An Let R be a semiring. We define a proper element a ∈ R is called weakly prime to I if ideal I of R to be weakly primary (resp. 0 ≠ ra ∈ (r ∈ R) implies that r ∈ I, and let weakly prime) if 0 ≠ ab ∈ I implies a ∈ I p(I) be the set of elements of R that are not or bm ∈ I for some positive integerm (resp. weakly prime to I. 0 is always weakly a ∈ I or b ∈ I) [1, 6]. So a primary ideal prime to I. A proper ideal I of R is called (resp. prime ideal) is a weakly primary weakly primal if the set P = p(I) ∪ {0} form (resp. weakly prime). However, since 0 is an ideal: this ideal is called the weakly always weakly primary (resp. weakly adjoint ideal P of I. Let R be a commutative prime) by definition, a weakly primary semiring which is not a semidomain. Then ideal (a weakly prime ideal) need not be 0 is a 0-weakly primal ideal of R (by primary (resp. prime). Clearly, every definition). Let I be an ideal of R and A a weakly prime is weakly primary. subset of R. We say that A satisfies (*) if A 82 Chiang Mai J. Sci. 2013; 40(1)

is exactly the set of elements of R that are that 0 ≠ (a + J)(r + J) = ar + J ∈ I/J; hence not weakly prime to I. We use the notation 0 ≠ ar ∈ I with r ∉ J. Thus a is not weakly A* to refer to the non-zeroelements of A. prime to I. Now assume that a is not weakly prime to I (so a ≠ 0); we show that Theorem 2.11 Let J be a weakly prime a ∈ P. We can assume that a ∉ I. Then k-ideal of a semiring R and I a proper there is an element r ∈ R − I such that 0 ≠ k-ideal of R with J ⊆ I. Then I is a weakly ar ∈ I. Therefore, 0 ≠ (r + J)(a + J) = ra + J primal ideal of R if and only if I/J is a weakly ∈ I/J with r + J ∉ I/J; hence a + J ∈ (P/J)* primal ideal of R/J. In particular, there is a since I/J is P/J-weakly primal. Thus a ∈ P, bijective correspondence between the weakly as required. primal ideals of R containing J and the weakly primal ideals of R/J. REFERENCES [1] Anderson D.D., Weakly prime ideals, Proof. First suppose that I is a P-weakly Houston J. Math., 2003; 29: 831-840. primal ideal of R with J ⊆ I. Then by [3, [2] Allen P.J., A fundamental theorem of Remark 3.2 and Theorem 3.4], J ⊆ P and for semirings, Proc. P is a weakly prime ideal of R; hence P/J is Am. Math. Soc., 1969; 21: 412-416. a weakly prime ideal of R/J by Theorem 2.10. It suffices to show that (P/J)* satisfies [3] Ebrahimi Atani S., On primal and (* ). Let a + J ∈ (P/J)*, where a ∈ P. Since weakly primal ideals over commu- by Lemma 2.1, 0 + J ≠ a + J, we must tative semirings, Glas. Math., 2008; 43: 13-23. have a ≠ 0 and a is not weakly prime to I; hence there exists r ∈ R − I such that 0 ≠ [4] Ebrahimi Atani S., The zero-divisor ra ∈ I. If 0 ≠ ra ∈ J, then J weakly prime graph with respect to ideals of a gives r ∈ J, which is a contradiction since commutative semiring, Glas. Math., r ∉ I. So we may assume that 0 ≠ ra ∉ J. It 2008; 43: 309-320. follows that 0 ≠ (r + J)(a + J) ∈ I/J with [5] Ebrahimi Atani R. and Ebrahimi r + J ∉ I/J, so a + J is not weakly prime to Atani S., in commutative I/J. Now assume that b + J ≠ 0 + J is not semirings, Bul. Acad. Stiinte Repub. weakly prime to I/J, where b ∈ I. Then Mold. Mat., 2008; 2: 14-23. there exists c + J ∈ R/J − I/J such that 0 ≠ [6] Ebrahimi Atani S., On k-weakly (c + J)(b + J) = cb + J ∈ I/J, so cb ∈ I with primary ideals over semirings, c ∉ I by Lemma 2.1. Thus b ≠ 0 is not Sarajevo J. Math., 2007; 3(15): 9-13. * weakly prime to I. Therefore, b + J ∈ (P/J) [7] Golan J.S., The theory of semirings and the proof is complete. with applications in mathematics and Conversely, suppose that I/J is a P/J- theoretical computer Science, Pitman weakly primal ideal of R/J; we show that I Monographs and Surveys in Pure and is a P-weakly primal ideal of R. By Applied Mathematics, Longman [3, Theorem 3.4] and Theorem 2.10, P is a Scientific and Technical, Harlow UK, weakly prime ideal of R. It is enough to 1992. * * show that P satisfies (* ). Let a ∈ P . By [8] Hebisch U. and Weinert U.J. [3, Remark 3.2], we can assume that a ∉ J. Halbringe - Algebraische Theorie und As J is a weakly prime ideal and 0 ≠ a + J Anwendungen in der Informatik, ∈ P/J, there exists r + J ∈ R/J−I/J such Teubner, Stuttgart 1993.