Research ArticleRight 𝑘-Weakly Regular Γ-Semirings

R. D. Jagatap

Y. C. College of Science, Karad, India

Correspondence should be addressed to R. D. Jagatap; ravindrajagatap@yahoo.co.in

Received 20 July 2014; Accepted 11 November 2014; Published 15 December 2014

Academic Editor: K. C. Sivakumar

Copyright © 2014 R. D. Jagatap.This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The concepts of a 𝑘-idempotent Γ-semiring, a right 𝑘-weakly regular Γ-semiring, and a right pure 𝑘-ideal of a Γ-semiring areintroduced. Several characterizations of them are furnished.

1. Introduction

Γ-semiring was introduced by Rao in [1] as a generalizationof a ring, a Γ-ring, and a semiring. Ideals in semirings werecharacterized by Ahsan in [2], Iseki in [3, 4], and Shabir andIqbal in [5]. Properties of prime and semiprime ideals in Γ-semirings were discussed in detail by Dutta and Sardar [6].Henriksen in [7] defined more restricted class of ideals insemirings known as 𝑘-ideals. Some more characterizationsof 𝑘-ideals of semirings were studied by Sen and Adhikari in[8, 9]. 𝑘-ideal in a Γ-semiring was defined by Rao in [1] andin [6] Dutta and Sardar gave some of its properties. Authorstudied 𝑘-ideals and full 𝑘-ideals of Γ-semirings in [10]. Theconcept of a bi-ideal of a Γ-semiring was given by author in[11].

In this paper efforts are made to introduce the conceptsof a 𝑘-idempotent Γ-semiring, a right 𝑘-weakly regular Γ-semiring, and a right pure 𝑘-ideal of a Γ-semiring. Discusssome characterizations of a 𝑘-idempotent Γ-semiring, a right𝑘-weakly regular Γ-semiring, and a right pure 𝑘-ideal of a Γ-semiring.

2. Preliminaries

First we recall some definitions of the basic concepts of Γ-semirings that we need in sequel. For this we follow Duttaand Sardar [6].

Definition 1. Let 𝑆 and Γ be two additive commutativesemigroups. 𝑆 is called a Γ-semiring if there exists a mapping

𝑆 × Γ × 𝑆 → 𝑆 denoted by 𝑎𝛼𝑏, for all 𝑎, 𝑏 ∈ 𝑆 and 𝛼 ∈ Γ

satisfying the following conditions:

(i) 𝑎𝛼(𝑏 + 𝑐) = (𝑎𝛼𝑏) + (𝑎𝛼𝑐);

(ii) (𝑏 + 𝑐)𝛼𝑎 = (𝑏𝛼𝑎) + (𝑐𝛼𝑎);

(iii) 𝑎(𝛼 + 𝛽)𝑐 = (𝑎𝛼𝑐) + (𝑎𝛽𝑐);

(iv) 𝑎𝛼(𝑏𝛽𝑐) = (𝑎𝛼𝑏)𝛽𝑐, for all 𝑎, 𝑏, 𝑐 ∈ 𝑆 and for all 𝛼, 𝛽 ∈Γ.

Definition 2. An element 0 in a Γ-semiring 𝑆 is said to be anabsorbing zero if 0𝛼𝑎 = 0 = 𝑎𝛼0, 𝑎 + 0 = 0 + 𝑎 = 𝑎, for all𝑎 ∈ 𝑆 and 𝛼 ∈ Γ.

Definition 3. Anonempty subset𝑇 of a Γ-semiring 𝑆 is said tobe a sub-Γ-semiring of 𝑆 if (𝑇, +) is a subsemigroup of (𝑆, +)and 𝑎𝛼𝑏 ∈ 𝑇, for all 𝑎, 𝑏 ∈ 𝑇 and 𝛼 ∈ Γ.

Definition 4. A nonempty subset 𝑇 of a Γ-semiring 𝑆 is calleda left (resp., right) ideal of 𝑆 if 𝑇 is a subsemigroup of (𝑆, +)and 𝑥𝛼𝑎 ∈ 𝑇 (resp., 𝑎𝛼𝑥 ∈ 𝑇) for all 𝑎 ∈ 𝑇, 𝑥 ∈ 𝑆 and 𝛼 ∈ Γ.

Definition 5. If 𝑇 is both left and right ideal of a Γ-semiring𝑆, then 𝑇 is known as an ideal of 𝑆.

Definition 6. A right ideal 𝐼 of a Γ-semiring 𝑆 is said to be aright 𝑘-ideal if 𝑎 ∈ 𝐼 and 𝑥 ∈ 𝑆 such that 𝑎+𝑥 ∈ 𝐼; then 𝑥 ∈ 𝐼.

Similarly we define a left 𝑘-ideal of a Γ-semiring 𝑆. If anideal 𝐼 is both right and left 𝑘-ideal, then 𝐼 is known as a 𝑘-ideal of 𝑆.

Hindawi Publishing CorporationAlgebraVolume 2014, Article ID 948032, 5 pageshttp://dx.doi.org/10.1155/2014/948032

2 Algebra

Example 7. Let𝑁0denote the set of all positive integers with

zero. 𝑆 = 𝑁0is a semiring and with Γ = 𝑆, 𝑆 forms a Γ-

semiring. A subset 𝐼 = 3𝑁0\ {3} of 𝑆 is an ideal of 𝑆 but not a

𝑘-ideal. Since 6 and 9 = 3 + 6 ∈ 𝐼 but 3 �∈ 𝐼.

Example 8. If 𝑆 = 𝑁 is the set of all positive integers, then(𝑆, max., min.) is a semiring and with Γ = 𝑆, 𝑆 forms a Γ-semiring. 𝐼

𝑛= {1, 2, 3, . . . , 𝑛} is a 𝑘-ideal for any 𝑛 ∈ 𝐼.

Definition 9. For a nonempty 𝐼 of a Γ-semiring 𝑆,

𝐼 = {𝑎 ∈ 𝑆 | 𝑎 + 𝑥 ∈ 𝐼, for some 𝑥 ∈ 𝐼} . (1)

𝐼 is called 𝑘-closure of 𝐼.

Now we give a definition of a bi-ideal.

Definition 10 (see [11]). A nonempty subset 𝐵 of a Γ-semiring𝑆 is said to be a bi-ideal of 𝑆 if 𝐵 is a sub-Γ-semiring of 𝑆 and𝐵Γ𝑆Γ𝐵 ⊆ 𝐵.

Example 11. Let𝑁 be the set of natural numbers and Γ = 2𝑁.Then both𝑁 and Γ are additive commutative semigroups. Animage of a mapping𝑁 × Γ × 𝑁 → 𝑁 is denoted by 𝑎𝛼𝑏 anddefined as 𝑎𝛼𝑏 = product of 𝑎, 𝛼, 𝑏, for all 𝑎, 𝑏 ∈ 𝑁 and 𝛼 ∈ Γ.Then𝑁 forms a Γ-semiring. 𝐵 = 4𝑁 is a bi-ideal of𝑁.

Example 12. Consider a Γ-semiring 𝑆 = 𝑀2×2(𝑁0), where

𝑁0denotes the set of natural numbers with zero and Γ =

𝑆. Define 𝐴𝛼𝐵 = usual matrix product of 𝐴, 𝛼 and 𝐵, for𝐴, 𝛼, 𝐵 ∈ 𝑆.

𝑃 = {(0 𝑥

0 𝑦) | 𝑥, 𝑦 ∈ 𝑁

0} (2)

is a bi-ideal of a Γ-semiring 𝑆.

Definition 13. An element 1 in a Γ-semiring 𝑆 is said to be anunit element if 1𝛼𝑎 = 𝑎 = 𝑎𝛼1, for all 𝑎 ∈ 𝑆 and for all 𝛼 ∈ Γ.

Definition 14. A Γ-semiring 𝑆 is said to be commutative if𝑎𝛼𝑏 = 𝑏𝛼𝑎, for all 𝑎, 𝑏 ∈ 𝑆 and for all 𝛼 ∈ Γ.

Some basic properties of 𝑘-closure are given in thefollowing lemma.

Lemma 15. For nonempty subsets 𝐴 and 𝐵 of 𝑆, we have thefollowing.

(1) If 𝐴 ⊆ 𝐵, then 𝐴 ⊆ 𝐵.

(2) 𝐴 is the smallest (left 𝑘-ideal, right 𝑘-ideal) 𝑘-idealcontaining (left 𝑘-ideal, right 𝑘-ideal) 𝑘-ideal 𝐴 of 𝑆.

(3) 𝐴 = 𝐴 if and only if 𝐴 is a (left 𝑘-ideal, right 𝑘-ideal)𝑘-ideal of 𝑆.

(4) 𝐴 = 𝐴, where 𝐴 is a (left 𝑘-ideal, right 𝑘-ideal) 𝑘-idealof 𝑆.

(5) 𝐴Γ𝐵 = 𝐴Γ𝐵, where 𝐴 and 𝐵 are (left 𝑘-ideals, right𝑘-ideals) 𝑘-ideals of 𝑆.

Some results from [11] are stated which are useful forfurther discussion.

Result 1. For each nonempty subset 𝑋 of 𝑆, the followingstatements hold.

(i) 𝑆Γ𝑋 is a left ideal of 𝑆.(ii) 𝑋Γ𝑆 is a right ideal of 𝑆.(iii) 𝑆Γ𝑋Γ𝑆 is an ideal of 𝑆.

Result 2. For 𝑎 ∈ 𝑆, the following statements hold.(i) 𝑆Γ𝑎 is a left ideal of 𝑆.(ii) 𝑎Γ𝑆 is a right ideal of 𝑆.(iii) 𝑆Γ𝑎Γ𝑆 is an ideal of 𝑆.

Now onwards 𝑆 denotes a Γ-semiring with an absorbingzero and an unit element unless otherwise stated.

3. 𝑘-Idempotent Γ-Semiring

In this section we introduce and characterize the notion of a𝑘-idempotent Γ-semiring.

Definition 16. A subset 𝐼 of a Γ-semiring 𝑆 is said to be 𝑘-idempotent if 𝐼 = 𝐼Γ𝐼.

Definition 17. A Γ-semiring 𝑆 is said to be 𝑘-idempotent ifevery 𝑘-ideal of 𝑆 is 𝑘-idempotent.

Theorem 18. In 𝑆 the following statements are equivalent.(1) 𝑆 is 𝑘-idempotent.(2) For any 𝑎 ∈ 𝑆, 𝑎 ∈ 𝑆Γ𝑎Γ𝑆Γ𝑎Γ𝑆.(3) For every 𝐴 ⊆ 𝑆, 𝐴 ⊆ 𝑆Γ𝐴Γ𝑆Γ𝐴Γ𝑆.

Proof. (1) ⇒ (2). Suppose that 𝑆 is a 𝑘-idempotent Γ-semiring. For any 𝑎 ∈ 𝑆, (𝑎) = 𝑎Γ𝑆 + 𝑆Γ𝑎 + 𝑆Γ𝑎Γ𝑆 + 𝑁

0𝑎.

Then 𝑎 ∈ (𝑎) = 𝑎Γ𝑆 + 𝑆Γ𝑎 + 𝑆Γ𝑎Γ𝑆 + 𝑁0𝑎.

Hence by assumption (𝑎) = (𝑎)Γ(𝑎) =

(𝑎Γ𝑆 + 𝑆Γ𝑎 + 𝑆Γ𝑎Γ𝑆 + 𝑁0𝑎)Γ(𝑎Γ𝑆 + 𝑆Γ𝑎 + 𝑆Γ𝑎Γ𝑆 + 𝑁

0𝑎).

Therefore (𝑎) = (𝑎Γ𝑆 + 𝑆Γ𝑎 + 𝑆Γ𝑎Γ𝑆 + 𝑁0𝑎)Γ

(𝑎Γ𝑆 + 𝑆Γ𝑎 + 𝑆Γ𝑎Γ𝑆 + 𝑁0𝑎).

Hence (𝑎) ⊆ 𝑆Γ𝑎Γ𝑆Γ𝑎Γ𝑆. Therefore 𝑎 ∈ 𝑆Γ𝑎Γ𝑆Γ𝑎Γ𝑆.(2) ⇒ (3). Let𝐴 ⊆ 𝑆 and 𝑎 ∈ 𝐴. Hence by assumption we

have 𝑎 ∈ 𝑆Γ𝑎Γ𝑆Γ𝑎Γ𝑆. Therefore 𝑎 ∈ 𝑆Γ𝐴Γ𝑆Γ𝐴Γ𝑆. Thus we get𝐴 ⊆ 𝑆Γ𝐴Γ𝑆Γ𝐴Γ𝑆.

(3) ⇒ (1). Let 𝐴 be any 𝑘-ideal of 𝑆. Then by assumption𝐴 ⊆ 𝑆Γ𝐴Γ𝑆Γ𝐴Γ𝑆 ⊆ 𝐴Γ𝑆Γ𝐴 ⊆ 𝐴Γ𝐴. As 𝐴 is a 𝑘-ideal of 𝑆,𝐴Γ𝐴 ⊆ 𝐴. Therefore 𝐴Γ𝐴 = 𝐴, which shows that 𝑆 is a 𝑘-idempotent Γ-semiring.

Definition 19. A sub-Γ-semiring 𝐼 of 𝑆 is a 𝑘-interior ideal of𝑆 if 𝑆Γ𝐼Γ𝑆 ⊆ 𝐼 and if 𝑎 ∈ 𝐼 and 𝑥 ∈ 𝑆 such that 𝑎 + 𝑥 ∈ 𝐼, then𝑥 ∈ 𝐼.

Theorem 20. If 𝑆 is a 𝑘-idempotent Γ-semiring, then a subsetof 𝑆 is a 𝑘-ideal if and only if it is a 𝑘-interior ideal.

Algebra 3

Proof. Let 𝑆 be a 𝑘-idempotent Γ-semiring. As every 𝑘-idealis a 𝑘-interior ideal, one part of theorem holds. Conversely,suppose a subset 𝐼 of 𝑆 is a 𝑘-interior ideal of 𝑆. To show 𝐼 isa 𝑘-ideal of 𝑆, let 𝑥 ∈ 𝐼 and 𝑡 ∈ 𝑆. As 𝑆 is a 𝑘-idempotentΓ-semiring, 𝑥 ∈ 𝑆Γ𝑥Γ𝑆Γ𝑥Γ𝑆 (see Theorem 18). Therefore,for any 𝛼 ∈ Γ, 𝑥𝛼𝑡 ∈ 𝑆Γ𝑥Γ𝑆Γ𝑥Γ𝑆Γ𝑆 ⊆ 𝑆Γ𝑥Γ𝑆Γ𝑥Γ𝑆Γ𝑆 ⊆

𝑆Γ𝐼Γ𝑆Γ𝐼Γ𝑆Γ𝑆 ⊆ 𝑆Γ𝐼Γ𝑆 ⊆ 𝐼. Similarly we can show that𝑡𝛼𝑥 ∈ 𝐼. Therefore 𝐼 is a 𝑘-ideal of 𝑆.

Theorem 21. 𝑆 is 𝑘-idempotent if and only if𝐴Γ𝐵 = 𝐴∩𝐵, forany 𝑘-interior ideals 𝐴 and 𝐵 of 𝑆.

Proof. Suppose a Γ-semiring 𝑆 is 𝑘-idempotent. Let 𝐴 and 𝐵be any two 𝑘-interior ideals of 𝑆. Then, by Theorem 20, 𝐴and 𝐵 are two 𝑘-ideals of 𝑆. Hence 𝐴Γ𝐵 ⊆ 𝐴 and 𝐴Γ𝐵 ⊆ 𝐵.Therefore 𝐴Γ𝐵 ⊆ 𝐴 ∩ 𝐵. By assumption 𝐴 ∩ 𝐵 = (𝐴 ∩ 𝐵)

2=

(𝐴 ∩ 𝐵)Γ(𝐴 ∩ 𝐵) ⊆ 𝐴Γ𝐵.Therefore𝐴Γ𝐵 = 𝐴∩𝐵. Conversely,let 𝐴 be any 𝑘-ideal of 𝑆. As every 𝑘-ideal of 𝑆 is a 𝑘-interiorideal of 𝑆, by assumption, 𝐴Γ𝐴 = 𝐴 ∩𝐴 = 𝐴. Therefore 𝑆 is a𝑘-idempotent Γ-semiring.

4. Right 𝑘-Weakly Regular Γ-Semiring

Definition 22. A Γ-semiring 𝑆 is said to be right 𝑘-weaklyregular if, for any 𝑎 ∈ 𝑆, 𝑎 ∈ (𝑎Γ𝑆)2.

Theorem 23. In 𝑆, the following statements are equivalent.

(1) 𝑆 is right 𝑘-weakly regular.

(2) 𝑅2 = 𝑅, for each right 𝑘-ideal 𝑅 of 𝑆.

(3) 𝑅 ∩ 𝐼 = 𝑅Γ𝐼, for a right 𝑘-ideal 𝑅 and a 𝑘-ideal 𝐼 of 𝑆.

Proof. (1) ⇒ (2). For any right 𝑘-ideal 𝑅 of 𝑆, 𝑅2 = 𝑅Γ𝑅 ⊆

𝑅Γ𝑆 ⊆ 𝑅. Hence 𝑅2 = 𝑅Γ𝑅 ⊆ 𝑅. For the reverse inclusion,let 𝑎 ∈ 𝑅. As 𝑆 is right 𝑘-weakly regular, 𝑎 ∈ (𝑎Γ𝑆)

2=

(𝑎Γ𝑆)Γ(𝑎Γ𝑆) ⊆ (𝑅Γ𝑆)Γ(𝑅Γ𝑆) ⊆ 𝑅Γ𝑅 = 𝑅2. Thus 𝑅2 = 𝑅,for each right 𝑘-ideal 𝑅 of 𝑆.

(2) ⇒ (1). For any 𝑎 ∈ 𝑆, 𝑎 ∈ 𝑎Γ𝑆 ⊆ 𝑎Γ𝑆 and 𝑎Γ𝑆 is a right𝑘-ideal of 𝑆, then by assumption (𝑎Γ𝑆)2 = (𝑎Γ𝑆). Therefore𝑎 ∈ (𝑎Γ𝑆)

2. Hence 𝑆 is right 𝑘-weakly regular.(2) ⇒ (3). Let 𝑅 be a right 𝑘-ideal and 𝐼 be a 𝑘-ideal of 𝑆.

Then 𝑅 ∩ 𝐼 is a right 𝑘-ideal of 𝑆. By assumption (𝑅 ∩ 𝐼)2 =𝑅 ∩ 𝐼. Consider 𝑅 ∩ 𝐼 = (𝑅 ∩ 𝐼)2 = (𝑅 ∩ 𝐼)Γ(𝑅 ∩ 𝐼) ⊆ 𝑅Γ𝐼.Clearly 𝑅Γ𝐼 ⊆ 𝑅 and 𝑅Γ𝐼 ⊆ 𝐼. Then 𝑅Γ𝐼 ⊆ 𝑅 = 𝑅 and 𝑅Γ𝐼 ⊆𝐼 = 𝐼, since 𝑅 is a right 𝑘-ideal and 𝐼 is a two sided 𝑘-ideal of𝑆. Therefore 𝑅Γ𝐼 ⊆ 𝑅 ∩ 𝐼. Hence 𝑅 ∩ 𝐼 = 𝑅Γ𝐼.

(3) ⇒ (2). Let 𝑅 be a right 𝑘-ideal of 𝑆 and let (𝑅) be twosided ideal generated by𝑅.Then (𝑅) = 𝑆Γ𝑅Γ𝑆. By assumption𝑅 ∩ (𝑅) = 𝑅Γ(𝑅). Hence 𝑅 = (𝑅Γ𝑆)Γ(𝑅Γ𝑆) ⊆ 𝑅Γ𝑅 = 𝑅2.Therefore 𝑅2 = 𝑅.

Theorem 24. 𝑆 is right 𝑘-weakly regular if and only if everyright 𝑘-ideal of 𝑆 is semiprime.

Proof. Suppose 𝑆 is right 𝑘-weakly regular. Let 𝑅 be a right𝑘-ideal of 𝑆 such that 𝐴Γ𝐴 ⊆ 𝑅, for any right 𝑘-ideal 𝐴 of 𝑆.𝐴 = 𝐴Γ𝐴. Then𝐴Γ𝐴 ⊆ 𝑅 = 𝑅. Therefore𝐴 ⊆ 𝑅. Hence 𝑅 is asemiprime right 𝑘-ideal of 𝑆. Conversely, suppose every right𝑘-ideal of 𝑆 is semiprime. Let 𝑅 be a right 𝑘-ideal of 𝑆. 𝑅Γ𝑅 isalso a right 𝑘-ideal of 𝑆. By assumption 𝑅Γ𝑅 is a semiprimeright 𝑘-ideal of 𝑆. 𝑅Γ𝑅 ⊆ 𝑅Γ𝑅 implies 𝑅 ⊆ 𝑅Γ𝑅. Therefore𝑅 = 𝑅 ⊆ 𝑅Γ𝑅 = 𝑅2. Therefore 𝑅2 = 𝑅. Hence 𝑆 is right𝑘-weakly regular byTheorem 23.

Definition 25 (see [12]). A latticeL is said to be Brouwerianif, for any 𝑎, 𝑏 ∈ L, the set of all 𝑥 ∈ L satisfying thecondition 𝑎 ∧ 𝑥 ≤ 𝑏 contains the greatest element.

If 𝑐 is the greatest element in this set, then the element 𝑐is known as the pseudocomplement of 𝑎 relative to 𝑏 and isdenoted by 𝑎 : 𝑏.

Thus a latticeL is a Brouwerian if 𝑎 : 𝑏 exists for all 𝑎, 𝑏 ∈L.

LetL𝑆denote the family of all 𝑘-ideals of 𝑆.Then ⟨L

𝑆, ⊆⟩

is a partially ordered set. As {0}, 𝑆 ∈ L𝑆and ⋂

𝛼∈Δ𝐼𝛼∈ L𝑆,

for all 𝐼𝛼∈ L𝑆and Δ is an indexing set, we have L

𝑆is a

complete lattice under ∧ and ∨ defined by 𝐼 ∨ 𝐽 = 𝐼 + 𝐽 and𝐼 ∧ 𝐽 = 𝐼 ∩ 𝐽. Further we have the following.

Theorem 26. If 𝑆 is a right 𝑘-weakly regular Γ-semiring, thenL𝑆is a Brouwerian lattice.

Proof. Let 𝐵 and 𝐶 be any two 𝑘-ideals of 𝑆. Consider thefamily of 𝑘-ideals 𝐾 = {𝐼 ∈ L

𝑆| 𝐼 ∩ 𝐵 ⊆ 𝐶}. Then by

Zorn’s lemma there exists a maximal element𝑀 in 𝐾. Select𝐼 ∈ L

𝑆such that 𝐵∩𝐼 ⊆ 𝐶. ByTheorem 23, we have 𝐵Γ𝐼 ⊆ 𝐶.

To show that 𝐵Γ(𝐼 +𝑀) ⊆ 𝐶. Let 𝑥 ∈ 𝐵Γ(𝐼 +𝑀). Then𝑥 = ∑

𝑛

𝑖=1𝑏𝑖𝛼𝑖𝑥𝑖, where 𝑏

𝑖∈ 𝐵, 𝛼

𝑖∈ Γ, and 𝑥

𝑖∈ 𝐼 +𝑀.

Therefore 𝑎𝑖+𝑥𝑖∈ 𝐼+𝑀 for some 𝑎

𝑖∈ 𝐼+𝑀 (seeDefinition 9).

𝑥 = ∑𝑛

𝑖=1𝑏𝑖𝛼𝑖(𝑎𝑖+ 𝑥𝑖) ∈ 𝐵Γ(𝐼 + 𝑀) = (𝐵Γ𝐼) + (𝐵Γ𝑀) ⊆

𝐶, as 𝐵Γ𝐼 ⊆ 𝐶 and 𝐵Γ𝑀 ⊆ 𝐶. Hence 𝐵Γ(𝐼 +𝑀) ⊆ 𝐶

implies 𝐵Γ(𝐼 +𝑀) ⊆ 𝐶 = 𝐶, since 𝐶 is a 𝑘-ideal. Therefore,by Theorem 23, 𝐵 ∩ (𝐼 +𝑀) ⊆ 𝐶. But, by the maximality,we have 𝐼 +𝑀 ⊆ 𝑀 which implies 𝐼 ⊆ 𝑀. Hence L

𝑆is

Brouwerian.

As L𝑆satisfies infinite meet distributive property prop-

erty of lattice, we have the following.

Corollary 27. If 𝑆 is a right 𝑘-weakly regular Γ-semiring, thenL𝑆is a distributive lattice (see Birkoff [12]).

Theorem 28. If 𝑆 is a right 𝑘-weakly regular Γ-semiring, thena 𝑘-ideal 𝑃 of 𝑆 is prime if and only if 𝑃 is irreducible.

Proof. Let 𝑆 be a right 𝑘-weakly regular Γ-semiring and let 𝑃be a 𝑘-ideal of 𝑆. If𝑃 is a prime 𝑘-ideal of 𝑆, then clearly𝑃 is anirreducible 𝑘-ideal. Suppose𝑃 is an irreducible 𝑘-ideal of 𝑆. Toshow that𝑃 is a prime 𝑘-ideal. Let𝐴 and𝐵 be any two 𝑘-idealsof 𝑆 such that𝐴Γ𝐵 ⊆ 𝑃.Then, byTheorem 23, we have𝐴∩𝐵 ⊆

4 Algebra

𝑃. Hence (𝐴 ∩ 𝐵) + 𝑃 = 𝑃. AsL𝑆is a distributive lattice, we

have (𝐴 + 𝑃) ∩ (𝐵 + 𝑃) = 𝑃. Therefore 𝑃 is an irreducible 𝑘-ideal implies 𝐴 + 𝑃 = 𝑃 or 𝐵 + 𝑃 = 𝑃. Then 𝐴 ⊆ 𝑃 or 𝐵 ⊆ 𝑃.Therefore 𝑃 is a prime 𝑘-ideal of 𝑆.

As a generalization of a fully prime semiring defined byShabir and Iqbal in [5], we define a fully 𝑘-prime Γ-semiringin [10] as follows.

A Γ-semiring 𝑆 is said to be a fully 𝑘-prime Γ-semiring ifeach 𝑘-ideal of 𝑆 is a prime 𝑘-ideal.

Theorem 29. A Γ-semiring 𝑆 is a fully 𝑘-prime Γ-semiring ifand only if 𝑆 is right 𝑘-weakly regular and the set of 𝑘-ideals of𝑆 is a totally ordered set by the set inclusion.

Proof. Suppose that a Γ-semiring 𝑆 is a fully 𝑘-prime Γ-semiring. Therefore every 𝑘-ideal of 𝑆 is a prime 𝑘-ideal. Asevery prime 𝑘-ideal is a semiprime 𝑘-ideal, we have 𝑆which isa right 𝑘-weakly regular Γ-semiring by Theorem 24. For anytwo 𝑘-ideals 𝐴 and 𝐵 of 𝑆, 𝐴Γ𝐵 ⊆ 𝐴 ∩ 𝐵. By assumption𝐴 ∩ 𝐵 is a prime 𝑘-ideal and hence we have 𝐴 ⊆ 𝐴 ∩ 𝐵 or𝐵 ⊆ 𝐴 ∩ 𝐵. But then 𝐴 ∩ 𝐵 = 𝐴 or 𝐴 ∩ 𝐵 = 𝐵. Hence either𝐴 ⊆ 𝐵 or 𝐵 ⊆ 𝐴. This shows that the set of 𝑘-ideals of 𝑆 is atotally ordered set by the set inclusion. Conversely, assume 𝑆is right 𝑘-weakly regular and the set of 𝑘-ideals of 𝑆 is a totallyordered set by set inclusion. To show that 𝑆 is a fully 𝑘-primeΓ-semiring. Let 𝑃 be a 𝑘-ideal of 𝑆 and 𝐴Γ𝐵 ⊆ 𝑃, for any 𝑘-ideals 𝐴 and 𝐵 of 𝑆. Then 𝐴Γ𝐵 ⊆ 𝑃. Hence by Theorem 23,we have 𝐴 ∩ 𝐵 = 𝐴Γ𝐵 ⊆ 𝑃. By assumption either 𝐴 ⊆ 𝐵 or𝐵 ⊆ 𝐴.Therefore𝐴∩𝐵 = 𝐴 or𝐴∩𝐵 = 𝐵. Hence either𝐴 ⊆ 𝑃

or 𝐵 ⊆ 𝑃. This shows that 𝑃 is a prime 𝑘-ideal of 𝑆.Therefore 𝑆 is a fully 𝑘-prime Γ-semiring.

Now we define a 𝑘-bi-ideal of a Γ-semiring.

Definition 30. A nonempty subset 𝐵 of a Γ-semiring 𝑆 is saidto be a 𝑘-bi-ideal of 𝑆 if𝐵 is a sub-Γ-semiring of 𝑆,𝐵Γ𝑆Γ𝐵 ⊆ 𝐵,and for 𝑎 ∈ 𝐵 and 𝑥 ∈ 𝑆 such that 𝑎 + 𝑥 ∈ 𝐵; then 𝑥 ∈ 𝐵.

Theorem 31. 𝑆 is right 𝑘-weakly regular if and only if 𝐵 ∩ 𝐼 ⊆𝐵Γ𝐼, for any 𝑘-bi-ideal 𝐵 and 𝑘-ideal 𝐼 of 𝑆.

Proof. Suppose 𝑆 is a right 𝑘-weakly regular Γ-semiring. Let𝐵 be a 𝑘-bi-ideal and let 𝐼 be a 𝑘-ideal of 𝑆. Let 𝑎 ∈ 𝐵∩𝐼.Then𝑎 ∈ (𝑎Γ𝑆)

2= (𝑎Γ𝑆)Γ(𝑎Γ𝑆) ⊆ 𝐵Γ(𝑆Γ𝐼Γ𝑆) ⊆ 𝐵Γ𝐼. Therefore

𝐵 ∩ 𝐼 ⊆ 𝐵Γ𝐼. Conversely, let 𝑅 be a right 𝑘-ideal of 𝑆. Then𝑅 itself is a 𝑘-bi-ideal of 𝑆. Hence by assumption 𝑅 = 𝑅 ∩

(𝑅) ⊆ 𝑅Γ(𝑅) = 𝑅Γ(𝑆Γ𝑅Γ𝑆) = (𝑅Γ𝑆)Γ(𝑅Γ𝑆) ⊆ 𝑅Γ𝑅. Therefore𝑅 = 𝑅Γ𝑅 = 𝑅2. Hence, by Theorem 23, 𝑆 is a right 𝑘-weaklyregular Γ-semiring.

Theorem32. 𝑆 is right 𝑘-weakly regular if and only if𝐵∩𝐼∩𝑅 ⊆

𝐵Γ𝐼Γ𝑅, for any 𝑘-bi-ideal 𝐵, 𝑘-ideal 𝐼, and a right 𝑘-ideal 𝑅 of𝑆.

Proof. Suppose 𝑆 is a right 𝑘-weakly regular Γ-semiring. Let𝐵 be a 𝑘-bi-ideal, let 𝐼 be a 𝑘-ideal, and let 𝑅 be a right 𝑘-idealof 𝑆. Let 𝑎 ∈ 𝐵 ∩ 𝐼 ∩ 𝑅. Then 𝑎 ∈ (𝑎Γ𝑆)

2= (𝑎Γ𝑆)Γ(𝑎Γ𝑆) ⊆

(𝑎Γ𝑆)Γ(𝑎Γ𝑆)Γ(𝑎Γ𝑆)Γ𝑆 ⊆ 𝐵Γ(𝑆Γ𝐼Γ𝑆)Γ(𝑅Γ𝑆) ⊆ 𝐵Γ𝐼Γ𝑅.Therefore 𝐵∩𝐼∩𝑅 ⊆ 𝐵Γ𝐼Γ𝑅. Conversely, for a right 𝑘-ideal 𝑅of 𝑆, 𝑅 itself is being a 𝑘-bi-ideal and 𝑆 itself is being a 𝑘-idealof 𝑆.Then by assumption𝑅∩𝑆∩𝑅 ⊆ 𝑅Γ𝑆Γ𝑅 ⊆ 𝑅Γ𝑅.Therefore𝑅 ⊆ 𝑅Γ𝑅. Therefore 𝑅 = 𝑅Γ𝑅 = 𝑅2. Then, by Theorem 23, 𝑆is right 𝑘-weakly regular.

5. Right Pure 𝑘-Ideals

In this section we define a right pure 𝑘-ideal of a Γ-semiring𝑆 and furnish some of its characterizations.

Definition 33. A 𝑘-ideal 𝐼 of a Γ-semiring 𝑆 is said to be a rightpure 𝑘-ideal if, for any 𝑥 ∈ 𝐼, 𝑥 ∈ 𝑥Γ𝐼.

Theorem 34. A 𝑘-ideal 𝐼 of 𝑆 is right pure if and only if 𝑅∩𝐼 =𝑅Γ𝐼, for any right 𝑘-ideal 𝑅 of 𝑆.

Proof. Let 𝐼 be a right pure 𝑘-ideal and let 𝑅 be a right 𝑘-idealof 𝑆. Then clearly 𝑅Γ𝐼 ⊆ 𝑅 ∩ 𝐼. Now let 𝑎 ∈ 𝑅 ∩ 𝐼. As 𝐼 is aright pure 𝑘-ideal, 𝑎 ∈ 𝑎Γ𝐼 ⊆ 𝑅Γ𝐼. This gives 𝑅 ∩ 𝐼 ⊆ 𝑅Γ𝐼.By combining both the inclusions we get 𝑅 ∩ 𝐼 = 𝑅Γ𝐼. Con-versely, assume the given statement holds. Let 𝐼 be 𝑘-ideal of𝑆 and 𝑎 ∈ 𝐼. (𝑎)

𝑟denotes a right 𝑘-ideal generated by 𝑎 and

(𝑎)𝑟= 𝑁0𝑎 + 𝑎Γ𝐼, where 𝑁

0denotes the set of nonnegative

integers. Then 𝑎 ∈ (𝑎)𝑟∩ 𝐼 = (𝑎)

𝑟Γ𝐼 = (𝑁

0𝑎 + 𝑎Γ𝐼)Γ𝐼 =

(𝑁0𝑎 + 𝑎Γ𝐼)Γ𝐼 ⊆ 𝑎Γ𝐼 (see Result 2).Therefore 𝐼 is a right pure

𝑘-ideal of 𝑆.

Theorem 35. The intersection of right pure 𝑘-ideals of 𝑆 is aright pure 𝑘-ideal of 𝑆.

Proof. Let 𝐴 and 𝐵 be right pure 𝑘-ideals of 𝑆. Then for anyright 𝑘-ideal𝑅 of 𝑆we have𝑅∩𝐴 = 𝑅Γ𝐴 and𝑅∩𝐵 = 𝑅Γ𝐵.Weconsider 𝑅∩(𝐴∩𝐵) = (𝑅∩𝐴)∩𝐵 = (𝑅Γ𝐴)∩𝐵 = (𝑅Γ𝐴)Γ𝐵 =𝑅Γ(𝐴Γ𝐵) = 𝑅Γ(𝐴 ∩ 𝐵).Therefore𝐴∩𝐵 is a right pure 𝑘-idealof 𝑆.

Characterization of a right 𝑘-weakly regular Γ-semiringin terms of right pure 𝑘-ideals is given in the followingtheorem.

Theorem 36. 𝑆 is right 𝑘-weakly regular if and only if any 𝑘-ideal of 𝑆 is right pure.

Proof. Suppose that 𝑆 is a right 𝑘-weakly regular Γ-semiring.Let 𝐼 be a 𝑘-ideal and let 𝑅 be a right 𝑘-ideal of 𝑆. Then, byTheorem 23, 𝑅∩ 𝐼 = 𝑅Γ𝐼. Hence, byTheorem 34, any 𝑘-ideal𝐼 of 𝑆 is right pure. Conversely, suppose that any 𝑘-ideal of 𝑆is right pure. Then, from Theorems 34 and 23, we get that 𝑆is a right 𝑘-weakly regular Γ-semiring.

Algebra 5

6. Space of Prime 𝑘-Ideals

Let 𝑆 be a Γ-semiring and let℘𝑆be the set of all prime 𝑘-ideals

of 𝑆. For each 𝑘-ideal 𝐼 of 𝑆 define Θ𝐼= {𝐽 ∈ ℘

𝑆| 𝐼 ⊆ 𝐽} and

𝜁 (℘𝑆) = {Θ

𝐼| 𝐼 is a 𝑘-ideal of 𝑆} . (3)

Theorem 37. If 𝑆 is a right 𝑘-weakly regular Γ-semiring, then𝜁(℘𝑆) forms a topology on the set ℘

𝑆. There is an isomorphism

between lattice of 𝑘-idealsL𝑆and 𝜁(℘

𝑆) (lattice of open subsets

of ℘𝑆).

Proof. As {0} is a 𝑘-ideal of 𝑆 and each 𝑘-ideal of 𝑆 contains{0}, we have the following:

Θ{0}

= {𝐽 ∈ ℘𝑠| {0} ⊆ 𝐽} = Φ. Therefore Φ ∈ 𝜁(℘

𝑆).

As 𝑆 itself is a 𝑘-ideal, Θ𝑆= {𝐽 ∈ ℘

𝑆| 𝑆 ⊆ 𝐽} = ℘

𝑆imply

℘𝑆∈ 𝜁(℘

𝑆). Now let Θ

𝐼𝑘∈ 𝜁(℘

𝑆) for 𝑘 ∈ Λ; Λ is an indexing

set, and 𝐼𝑘is a 𝑘-ideal of 𝑆. Therefore Θ

𝐼𝑘= {𝐽 ∈ ℘

𝑆| 𝐼𝑘

⊆ 𝐽}.As ⋃𝑘Θ𝐼𝑘= {𝐽 ∈ ℘

𝑆| ∑𝑘𝐼𝑘

⊆ 𝐽}, ∑𝑘𝐼𝑘is a 𝑘-ideal of 𝑆.

Therefore ⋃𝑘Θ𝐼𝑘= Θ∑𝑘𝐼𝑘∈ 𝜁(℘

𝑆). Further let Θ

𝐴, Θ𝐵∈

𝜁(℘𝑆).Let 𝐽 ∈ Θ

𝐴⋂Θ𝐵; 𝐽 is a prime 𝑘-ideal of 𝑆. Hence 𝐴 ⊆ 𝐽

and 𝐵 ⊆ 𝐽. Suppose that 𝐴 ∩ 𝐵 ⊆ 𝐽. In a right weakly regularΓ-semiring, prime 𝑘-ideals and strongly irreducible 𝑘-idealscoincide. Therefore 𝐽 is a strongly irreducible 𝑘-ideal of 𝑆. As𝐽 is a strongly irreducible 𝑘-ideal of 𝑆,𝐴 ⊆ 𝐽 or𝐵 ⊆ 𝐽, which isa contradiction to 𝐴 ⊆ 𝐽 and 𝐵 ⊆ 𝐽. Hence 𝐴 ∩ 𝐵 ⊆ 𝐽 implies𝐽 ∈ Θ

𝐴∩𝐵. Therefore Θ

𝐴⋂Θ𝐵⊆ Θ𝐴∩𝐵

. Now let 𝐽 ∈ Θ𝐴∩𝐵

.Then𝐴∩𝐵 ⊆ 𝐽 implies𝐴 ⊆ 𝐽 and𝐵 ⊆ 𝐽.Therefore 𝐽 ∈ Θ

𝐴and

𝐽 ∈ Θ𝐵imply 𝐽 ∈ Θ

𝐴⋂Θ𝐵. Thus Θ

𝐴∩𝐵⊆ Θ𝐴⋂Θ𝐵. Hence

Θ𝐴⋂Θ𝐵= Θ𝐴∩𝐵

∈ 𝜁(℘𝑆). Thus 𝜁(℘

𝑠) forms a topology on

the set ℘𝑆.

Now we define a function 𝜙 : L𝑆→ 𝜁(℘

𝑆) by 𝜙(𝐼) = Θ

𝐼,

for all 𝐼 ∈ L𝑆. Let 𝐼, 𝐾 ∈ L

𝑆. Consider 𝜙(𝐼 ∩ 𝐾) = Θ

𝐼∩𝐾=

Θ𝐼⋂Θ𝐾= 𝜙(𝐼) ∩ 𝜙(𝐾).

Consider 𝜙(𝐼 + 𝐾) = Θ𝐼+𝐾

= Θ𝐼∪ Θ𝐾= 𝜙(𝐼) ∪ 𝜙(𝐾).

Therefore 𝜙 is a lattice homomorphism. Now consider 𝜙(𝐼) =𝜙(𝐾). Then Θ

𝐼= Θ𝐾.

Suppose that 𝐼 = 𝐾. Then there exists 𝑎 ∈ 𝐼 such that 𝑎 ∉𝐾. As𝐾 is a proper 𝑘-ideal of 𝑆, there exists an irreducible 𝑘-ideal 𝐽 of 𝑆 such that𝐾 ⊆ 𝐽 and 𝑎 ∉ 𝐽 (seeTheorem 6 in [10]).Hence 𝐼 ⊆ 𝐽. As 𝑆 is a right 𝑘-weakly regular Γ-semiring, 𝐽 isa prime 𝑘-ideal of 𝑆 by Theorem 28. Therefore 𝐽 ∈ Θ

𝐾= Θ𝐼

implies 𝐼 ⊆ 𝐽, which is a contradiction.Therefore 𝐼 = 𝐾.Thus𝜙(𝐼) = 𝜙(𝐾) implies 𝐼 = 𝐾 and hence 𝜙 is one-one. As 𝜙 isonto the result follows.

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper.

Acknowledgment

The author is thankful for the learned referee for his valuablesuggestions.

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[8] M. K. Sen andM. R. Adhikari, “On k-ideals of semirings,” Inter-national Journal of Mathematics andMathematical Sciences, vol.15, no. 2, pp. 347–350, 1992.

[9] M. K. Sen and M. R. Adhikari, “On maximal 𝑘-ideals ofsemirings,” Proceedings of the American Mathematical Society,vol. 118, no. 3, pp. 699–703, 1993.

[10] R. D. Jagatap and Y. S. Pawar, “k-ideals in Γ-semirings,” Bulletinof Pure and Applied Mathematics, vol. 6, no. 1, pp. 122–131, 2012.

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