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 are introduced. Several characterizations of them are furnished.

1. Introduction

Γ-semiring was introduced by Rao in [1] as a generalization of a ring, a Γ-ring, and a semiring. Ideals in semirings were characterized by Ahsan in [2], Iséki in [3, 4], and Shabir and Iqbal 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 in semirings known as k-ideals. Some more characterizations of 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] and in [6] Dutta and Sardar gave some of its properties. Author studied k-ideals and full k-ideals of Γ-semirings in [10]. The concept of a bi-ideal of a Γ-semiring was given by author in [11].

In this paper efforts are made to introduce 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. Discuss some characterizations of a k-idempotent Γ-semiring, a right k-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 Dutta and Sardar [6].

Definition 1.

Let S and Γ be two additive commutative semigroups. 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:

aαb+c=aαb+(aαc);

b+cαa=bαa+(cαa);

a(α+β)c=aαc+(aβc);

aαbβc=aαbβc, 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 an absorbing zero if 0αa=0=aα0, a+0=0+a=a, for all a∈S and α∈Γ.

Definition 3.

A nonempty subset T of a Γ-semiring S is said to be 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 called a 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-semiring S, then T is known as an ideal of S.

Definition 6.

A right ideal I of a Γ-semiring S is said to be a right 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 an ideal I is both right and left k-ideal, then I is known as a k-ideal of S.

Example 7.

Let N0 denote the set of all positive integers with zero. S=N0 is 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. In={1,2,3,…,n} is a k-ideal for any n∈I.

Definition 9.

For a nonempty I of a Γ-semiring S,
(1)I¯=a∈S∣a+x∈I,forsomex∈I.I¯ is called k-closure of I.

Now we give a definition of a bi-ideal.

Definition 10 (see [<xref ref-type="bibr" rid="B7">11</xref>]).

A nonempty subset B of a Γ-semiring S is said to be a bi-ideal of S if B is a sub-Γ-semiring of S and BΓSΓB⊆B.

Example 11.

Let N be the set of natural numbers and Γ=2N. Then both N and Γ are additive commutative semigroups. An image of a mapping N×Γ×N→N is denoted by aαb and defined as aαb= product of a,α,b, for all a,b∈N and α∈Γ. Then N forms a Γ-semiring. B=4N is a bi-ideal of N.

Example 12.

Consider a Γ-semiring S=M2×2(N0), where N0 denotes the set of natural numbers with zero and Γ=S. Define AαB = usual matrix product of A, α and B, for A,α,B∈S. (2)P=0x0y∣x,y∈N0
is a bi-ideal of a Γ-semiring S.

Definition 13.

An element 1 in a Γ-semiring S is said to be an unit 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 if aαb=bαa, for all a,b∈S and for all α∈Γ.

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

Lemma 15.

For nonempty subsets A and B of S, we have the following.

If A⊆B, then A¯⊆B¯.

A¯ is the smallest (left k-ideal, right k-ideal) k-ideal containing (left k-ideal, right k-ideal) k-ideal A of S.

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

A¯¯=A¯, where A is a (left k-ideal, right k-ideal) k-ideal of S.

A¯ΓB¯¯=AΓB¯, where A and B are (left k-ideals, right k-ideals) k-ideals of S.

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

Result 1.

For each nonempty subset X of S, the following statements hold.

SΓX is a left ideal of S.

XΓS is a right ideal of S.

SΓXΓS is an ideal of S.

Result 2.

For a∈S, the following statements hold.

SΓa is a left ideal of S.

aΓS is a right ideal of S.

SΓaΓS is an ideal of S.

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

In this section we introduce and characterize the notion of a k-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 if every k-ideal of S is k-idempotent.

Theorem 18.

In S the following statements are equivalent.

S is k-idempotent.

For any a∈S, a∈SΓaΓSΓaΓS¯.

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+N0a. Then a∈a¯=aΓS+SΓa+SΓaΓS+N0a¯. Hence by assumption a¯=a¯Γa¯¯=aΓS+SΓa+SΓaΓS+N0a¯ΓaΓS+SΓa+SΓaΓS+N0a¯¯.

2⇒(3). Let A⊆S and a∈A. Hence by assumption we have a∈SΓaΓSΓaΓS¯. Therefore a∈SΓAΓSΓAΓS¯. Thus we get A⊆SΓAΓSΓAΓS¯.

3⇒(1). Let A be any k-ideal of S. Then by assumption A⊆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 of S if SΓIΓS¯⊆I and if a∈I and x∈S such that a+x∈I, then x∈I.

Theorem 20.

If S is a k-idempotent Γ-semiring, then a subset of S is a k-ideal if and only if it is a k-interior ideal.

Proof.

Let S be a k-idempotent Γ-semiring. As every k-ideal is 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 is a k-ideal of S, let x∈I and t∈S. As S is a k-idempotent Γ-semiring, x∈SΓxΓSΓxΓS¯ (see Theorem 18). Therefore, for any α∈Γ, xαt∈SΓxΓSΓxΓS¯ΓS⊆SΓxΓSΓxΓSΓS¯⊆SΓIΓSΓIΓSΓS¯⊆SΓIΓS¯⊆I. Similarly we can show that tαx∈I. Therefore I is a k-ideal of S.

Theorem 21.

S is k-idempotent if and only if AΓB¯=A∩B, for any k-interior ideals A and B of S.

Proof.

Suppose a Γ-semiring S is k-idempotent. Let A and B be any two k-interior ideals of S. Then, by Theorem 20, A and 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∩BΓ(A∩B)¯⊆AΓB¯. Therefore AΓB¯=A∩B. Conversely, let A be any k-ideal of S. As every k-ideal of S is a k-interior ideal of S, by assumption, AΓA¯=A∩A=A. Therefore S is a k-idempotent Γ-semiring.

A Γ-semiring S is said to be right k-weakly regular if, for any a∈S, a∈(aΓS)2¯.

Theorem 23.

In S, the following statements are equivalent.

S is right k-weakly regular.

R2¯=R, for each right k-ideal R of S.

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¯=aΓSΓaΓS¯⊆RΓSΓRΓS¯⊆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 right k-ideal of S, then by assumption (aΓS)2¯=aΓS. Therefore a∈(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∩I2¯=R∩I. Consider R∩I=R∩I2¯=R∩IΓ(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 of S. Therefore RΓI¯⊆R∩I. Hence R∩I=RΓI¯.

3⇒(2). Let R be a right k-ideal of S and let (R) be two sided ideal generated by R. Then R=SΓRΓS. By assumption R∩(R)¯=RΓ(R)¯. Hence R=RΓSΓRΓS¯⊆RΓR¯=R2¯. Therefore R2¯=R.

Theorem 24.

S is right k-weakly regular if and only if every right k-ideal of S is semiprime.

Proof.

Suppose S is right k-weakly regular. Let R be a right k-ideal of S such that AΓA¯⊆R, for any right k-ideal A of S. A=AΓA¯. Then AΓA¯⊆R¯=R. Therefore A⊆R. Hence R is a semiprime right k-ideal of S. Conversely, suppose every right k-ideal of S is semiprime. Let R be a right k-ideal of S. RΓR¯ is also a right k-ideal of S. By assumption RΓR¯ is a semiprime right k-ideal of S. RΓR¯⊆RΓR¯ implies R⊆RΓR¯. Therefore R=R¯⊆RΓR¯=R2¯. Therefore R2¯=R. Hence S is right k-weakly regular by Theorem 23.

Definition 25 (see [<xref ref-type="bibr" rid="B2">12</xref>]).

A lattice L is said to be Brouwerian if, for any a,b∈L, the set of all x∈L satisfying the condition a∧x≤b contains the greatest element.

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

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

Let LS denote the family of all k-ideals of S. Then LS,⊆ is a partially ordered set. As 0,S∈LS and ⋂α∈ΔIα∈LS, for all Iα∈LS and Δ is an indexing set, we have LS is a complete lattice under ∧ and ∨ defined by I∨J=I+J¯ and I∧J=I∩J. Further we have the following.

Theorem 26.

If S is a right k-weakly regular Γ-semiring, then LS is a Brouwerian lattice.

Proof.

Let B and C be any two k-ideals of S. Consider the family of k-ideals K=I∈LS∣I∩B⊆C. Then by Zorn’s lemma there exists a maximal element M in K. Select I∈LS such that B∩I⊆C. By Theorem 23, we have BΓI¯⊆C. To show that BΓI+M¯¯⊆C. Let x∈BΓI+M¯. Then x=∑i=1nbiαixi, where bi∈B, αi∈Γ, and xi∈I+M¯. Therefore ai+xi∈I+M for some ai∈I+M (see Definition 9).

x=∑i=1nbiαi(ai+xi)∈BΓI+M=BΓI+BΓM⊆C, as BΓI⊆C and BΓM⊆C. Hence BΓI+M¯⊆C implies BΓI+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 LS is Brouwerian.

As LS satisfies infinite meet distributive property property of lattice, we have the following.

Corollary 27.

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

Theorem 28.

If S is a right k-weakly regular Γ-semiring, then a 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 P be a k-ideal of S. If P is a prime k-ideal of S, then clearly P is an irreducible k-ideal. Suppose P is an irreducible k-ideal of S. To show that P is a prime k-ideal. Let A and B be any two k-ideals of S such that AΓB¯⊆P. Then, by Theorem 23, we have A∩B⊆P. Hence A∩B+P¯=P. As LS is 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 by Shabir and Iqbal in [5], we define a fully k-prime Γ-semiring in [10] as follows.

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

Theorem 29.

A Γ-semiring S is a fully k-prime Γ-semiring if and only if S is right k-weakly regular and the set of k-ideals of S 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. As every prime k-ideal is a semiprime k-ideal, we have S which is a right k-weakly regular Γ-semiring by Theorem 24. For any two k-ideals A and B of S, AΓB¯⊆A∩B. By assumption A∩B is a prime k-ideal and hence we have A⊆A∩B or B⊆A∩B. But then A∩B=A or A∩B=B. Hence either A⊆B or B⊆A. This shows that the set of k-ideals of S is a totally ordered set by the set inclusion. Conversely, assume S is right k-weakly regular and the set of k-ideals of S is a totally ordered set by set inclusion. To show that S is a fully k-prime Γ-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 or B⊆A. Therefore A∩B=A or A∩B=B. Hence either A⊆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 said to be a k-bi-ideal of S if B 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. Let B be a k-bi-ideal and let I be a k-ideal of S. Let a∈B∩I. Then a∈(aΓS)2¯=aΓSΓaΓS¯⊆BΓ(SΓIΓS)¯⊆BΓI¯. Therefore B∩I⊆BΓI¯. Conversely, let R be a right k-ideal of S. Then R itself is a k-bi-ideal of S. Hence by assumption R=R∩R¯⊆RΓR¯¯=RΓSΓRΓS¯=RΓSΓ(RΓS)¯⊆RΓR¯. Therefore R=RΓR¯=R2¯. Hence, by Theorem 23, S is a right k-weakly regular Γ-semiring.

Theorem 32.

S is right k-weakly regular if and only if B∩I∩R⊆BΓIΓR¯, for any k-bi-ideal B, k-ideal I, and a right k-ideal R of S.

Proof.

Suppose S is a right k-weakly regular Γ-semiring. Let B be a k-bi-ideal, let I be a k-ideal, and let R be a right k-ideal of S. Let a∈B∩I∩R. Then a∈(aΓS)2¯=aΓSΓaΓS¯⊆aΓSΓaΓSΓaΓS¯ΓS¯⊆BΓSΓIΓSΓ(RΓS¯)⊆BΓIΓR¯. Therefore B∩I∩R⊆BΓIΓR¯. Conversely, for a right k-ideal R of S, R itself is being a k-bi-ideal and S itself is being a k-ideal of S. Then by assumption R∩S∩R⊆RΓSΓR¯⊆RΓR¯. Therefore R⊆RΓR¯. Therefore R=RΓR¯=R2¯. Then, by Theorem 23, S is right k-weakly regular.

5. Right Pure <inline-formula>
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M713">
<mml:mrow>
<mml:mi>k</mml:mi></mml:mrow>
</mml:math></inline-formula>-Ideals

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

Definition 33.

A k-ideal I of a Γ-semiring S is said to be a right pure 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-ideal of S. Then clearly RΓI¯⊆R∩I. Now let a∈R∩I. As I is a right pure k-ideal, a∈aΓI¯⊆RΓI¯. This gives R∩I⊆RΓI¯. By combining both the inclusions we get R∩I=RΓI¯. Conversely, assume the given statement holds. Let I be k-ideal of S and a∈I. (a)r¯ denotes a right k-ideal generated by a and (a)r¯=N0a+aΓI¯, where N0 denotes the set of nonnegative integers. Then a∈(a)r¯∩I=(a)r¯ΓI¯=(N0a+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 a right pure k-ideal of S.

Proof.

Let A and B be right pure k-ideals of S. Then for any right k-ideal R of S we have R∩A=RΓA¯ and R∩B=RΓB¯. We consider R∩A∩B=R∩A∩B=RΓA¯∩B=RΓA¯ΓB¯=RΓAΓB¯=RΓ(A∩B)¯. Therefore A∩B is a right pure k-ideal of S.

Characterization of a right k-weakly regular Γ-semiring in terms of right pure k-ideals is given in the following theorem.

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, by Theorem 23, R∩I=RΓI¯. Hence, by Theorem 34, any k-ideal I of S is right pure. Conversely, suppose that any k-ideal of S is right pure. Then, from Theorems 34 and 23, we get that S is a right k-weakly regular Γ-semiring.

6. Space of Prime <inline-formula>
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M797">
<mml:mrow>
<mml:mi>k</mml:mi></mml:mrow>
</mml:math></inline-formula>-Ideals

Let S be a Γ-semiring and let ℘S be the set of all prime k-ideals of S. For each k-ideal I of S define ΘI=J∈℘S∣I⊈J and
(3)ζ℘S=ΘI∣Iisak-idealofS.

Theorem 37.

If S is a right k-weakly regular Γ-semiring, then ζ℘S forms a topology on the set ℘S. There is an isomorphism between lattice of k-ideals LS and 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:

Θ0=J∈℘s∣0⊈J=Φ. Therefore Φ∈ζ℘S. As S itself is a k-ideal, ΘS=J∈℘S∣S⊈J=℘S imply ℘S∈ζ℘S. Now let ΘIk∈ζ℘S for k∈Λ; Λ is an indexing set, and Ik is a k-ideal of S. Therefore ΘIk=J∈℘S∣Ik⊈J. As ⋃kΘIk=J∈℘S∣∑kIk¯⊈J, ∑kIk¯ is a k-ideal of S. Therefore ⋃kΘIk=Θ∑kIk¯∈ζ℘S. Further let ΘA, ΘB∈ζ℘S.

Let J∈ΘA⋂ΘB; J is a prime k-ideal of S. Hence A⊈J and B⊈J. Suppose that A∩B⊆J. In a right weakly regular Γ-semiring, prime k-ideals and strongly irreducible k-ideals coincide. Therefore J is a strongly irreducible k-ideal of S. As J is a strongly irreducible k-ideal of S, A⊆J or B⊆J, which is a contradiction to A⊈J and B⊈J. Hence A∩B⊈J implies J∈ΘA∩B. Therefore ΘA⋂ΘB⊆ΘA∩B. Now let J∈ΘA∩B. Then A∩B⊈J implies A⊈J and B⊈J. Therefore J∈ΘA and J∈ΘB imply J∈ΘA⋂ΘB. Thus ΘA∩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 ϕI=ΘI, for all I∈LS. Let I,K∈LS. Consider ϕI∩K=ΘI∩K=ΘI⋂ΘK=ϕI∩ϕK.

Consider ϕI+K¯=ΘI+K¯=ΘI∪ΘK=ϕI∪ϕK. Therefore ϕ is a lattice homomorphism. Now consider ϕI=ϕK. Then ΘI=ΘK.

Suppose that I≠K. Then there exists a∈I such that a∉K. As K is a proper k-ideal of S, there exists an irreducible k-ideal J of S such that K⊆J and a∉J (see Theorem 6 in [10]). Hence I⊈J. As S is a right k-weakly regular Γ-semiring, J is a prime k-ideal of S by Theorem 28. Therefore J∈ΘK=ΘI implies I⊆J, which is a contradiction. Therefore I=K. Thus ϕI=ϕK implies I=K and hence ϕ is one-one. As ϕ is onto the result follows.

Conflict of Interests

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

Acknowledgment

The author is thankful for the learned referee for his valuable suggestions.

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