# Rightk-Weakly Regular Γ-Semirings

Authors: R. D. Jagatap
Journal: Algebra
 (2014)

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## Abstract

The concepts of ak-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.


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## Body

## 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 ak-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 ofΓ-semirings that we need in sequel. For this we follow Dutta and Sardar [6].Definition 1.
LetS and Γ be two additive commutative semigroups. S is called a Γ-semiring if there exists a mapping S
×
Γ
×
S
→
S denoted by a
α
b, for all a
,
b
∈
S and α
∈
Γ satisfying the following conditions: (i)
a
α
b
+
c
=
a
α
b
+
(
a
α
c
);
(ii)
b
+
c
α
a
=
b
α
a
+
(
c
α
a
);
(iii)
a
(
α
+
β
)
c
=
a
α
c
+
(
a
β
c
);
(iv)
a
α
b
β
c
=
a
α
b
β
c, for  all a
,
b
,
c
∈
S and for all α
,
β
∈
Γ.Definition 2.
An element 0 in aΓ-semiring S is said to be an absorbing zero if 0
α
a
=
0
=
a
α
0, a
+
0
=
0
+
a
=
a, for all a
∈
S and α
∈
Γ.Definition 3.
A nonempty subsetT of a Γ-semiring S is said to be a sub-Γ-semiring of S if (
T
,
+
) is a subsemigroup of (
S
,
+
) and a
α
b
∈
T, for all a
,
b
∈
T and α
∈
Γ.Definition 4.
A nonempty subsetT of a Γ-semiring S is called a left (resp., right) ideal of S if T is a subsemigroup of (
S
,
+
) and x
α
a
∈
T
(
resp
.
,
a
α
x
∈
T) for all a
∈
T, x
∈
S and α
∈
Γ.Definition 5.
IfT is both left and right ideal of a Γ-semiring S, then T is known as an ideal of S.Definition 6.
A right idealI 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 leftk-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.
LetN
0 denote the set of all positive integers with zero. S
=
N
0 is a semiring and with Γ
=
S, S forms a Γ-semiring. A subset I
=
3
N
0
∖
{
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.
IfS
=
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 nonemptyI of a Γ-semiring S,
(1)
I
¯
=
a
∈
S
∣
a
+
x
∈
I
,
for
some
x
∈
I
.
I
¯ is called k-closure of I.Now we give a definition of a bi-ideal.Definition 10 (see [11]).
A nonempty subsetB of a Γ-semiring S is said to be a bi-ideal of S if B is a sub-Γ-semiring of S and B
Γ
S
Γ
B
⊆
B.Example 11.
LetN be the set of natural numbers and Γ
=
2
N. Then both N and Γ are additive commutative semigroups. An image of a mapping N
×
Γ
×
N
→
N is denoted by a
α
b and defined as a
α
b
= product of a
,
α
,
b, for all a
,
b
∈
N and α
∈
Γ. Then N forms a Γ-semiring. B
=
4
N is a bi-ideal of N.Example 12.
Consider aΓ-semiring S
=
M
2
×
2
(
N
0
), where N
0 denotes the set of natural numbers with zero and Γ
=
S. Define A
α
B = usual matrix product of A, α and B, for A
,
α
,
B
∈
S. 							(2)
P
=
0
x
0
y
∣
x
,
y
∈
N
0
is a bi-ideal of a Γ-semiring S.Definition 13.
An element 1 in aΓ-semiring S is said to be an unit element if 1
α
a
=
a
=
a
α
1, for all a
∈
S and for all α
∈
Γ.Definition 14.
AΓ-semiring S is said to be commutative if a
α
b
=
b
α
a, for all a
,
b
∈
S and for all α
∈
Γ.Some basic properties ofk-closure are given in the following lemma.Lemma 15.
For nonempty subsetsA and B of S, we have the following.(1)
IfA
⊆
B, then A
¯
⊆
B
¯.
(2)
A
¯ is the smallest (left k-ideal, right k-ideal) k-ideal containing (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-ideal of S.
(5)
A
¯
Γ
B
¯
¯
=
A
Γ
B
¯, 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 subsetX of S, the following statements hold.(i)
S
Γ
X is a left ideal of S.
(ii)
X
Γ
S is a right ideal of S.
(iii)
S
Γ
X
Γ
S is an ideal of S.Result 2.
Fora
∈
S, the following statements hold.(i)
S
Γ
a is a left ideal of S.
(ii)
a
Γ
S is a right ideal of S.
(iii)
S
Γ
a
Γ
S is an ideal of S.Now onwardsS denotes a Γ-semiring with an absorbing zero 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 subsetI of a Γ-semiring S is said to be k-idempotent if I
=
I
Γ
I
¯.Definition 17.
AΓ-semiring S is said to be k-idempotent if every k-ideal of S is k-idempotent.Theorem 18.
InS the following statements are equivalent.(1)
S is k-idempotent.
(2)
For anya
∈
S, a
∈
S
Γ
a
Γ
S
Γ
a
Γ
S
¯.
(3)
For everyA
⊆
S, A
⊆
S
Γ
A
Γ
S
Γ
A
Γ
S
¯.Proof.
1
⇒
(
2
). Suppose that S is a k-idempotent Γ-semiring. For any a
∈
S, a
=
a
Γ
S
+
S
Γ
a
+
S
Γ
a
Γ
S
+
N
0
a. Then a
∈
a
¯
=
a
Γ
S
+
S
Γ
a
+
S
Γ
a
Γ
S
+
N
0
a
¯. Hence by assumption a
¯
=
a
¯
Γ
a
¯
¯
=
a
Γ
S
+
S
Γ
a
+
S
Γ
a
Γ
S
+
N
0
a
¯
Γ
a
Γ
S
+
S
Γ
a
+
S
Γ
a
Γ
S
+
N
0
a
¯
¯.
Thereforea
¯
=
a
Γ
S
+
S
Γ
a
+
S
Γ
a
Γ
S
+
N
0
a
¯
Γ
a
Γ
S
+
S
Γ
a
+
S
Γ
a
Γ
S
+
N
0
a
¯.
Hencea
¯
⊆
S
Γ
a
Γ
S
Γ
a
Γ
S
¯. Therefore a
∈
S
Γ
a
Γ
S
Γ
a
Γ
S
¯.
2
⇒
(
3
). Let A
⊆
S and a
∈
A. Hence by assumption we have a
∈
S
Γ
a
Γ
S
Γ
a
Γ
S
¯. Therefore a
∈
S
Γ
A
Γ
S
Γ
A
Γ
S
¯. Thus we get A
⊆
S
Γ
A
Γ
S
Γ
A
Γ
S
¯.
3
⇒
(
1
). Let A be any k-ideal of S. Then by assumption A
⊆
S
Γ
A
Γ
S
Γ
A
Γ
S
¯
⊆
A
Γ
S
Γ
A
¯
⊆
A
Γ
A
¯. As A is a k-ideal of S, A
Γ
A
¯
⊆
A. Therefore A
Γ
A
¯
=
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 S
Γ
I
Γ
S
¯
⊆
I and if a
∈
I and x
∈
S such that a
+
x
∈
I, then x
∈
I.Theorem 20.
IfS 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.
LetS 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
∈
S
Γ
x
Γ
S
Γ
x
Γ
S
¯ (see Theorem 18). Therefore, for any α
∈
Γ, x
α
t
∈
S
Γ
x
Γ
S
Γ
x
Γ
S
¯
Γ
S
⊆
S
Γ
x
Γ
S
Γ
x
Γ
S
Γ
S
¯
⊆
S
Γ
I
Γ
S
Γ
I
Γ
S
Γ
S
¯
⊆
S
Γ
I
Γ
S
¯
⊆
I. Similarly we can show that t
α
x
∈
I. Therefore I is a k-ideal of S.Theorem 21.
S is k-idempotent if and only if A
Γ
B
¯
=
A
∩
B, for any k-interior ideals A and B of S.Proof.
Suppose aΓ-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 A
Γ
B
⊆
A and A
Γ
B
⊆
B. Therefore A
Γ
B
¯
⊆
A
∩
B. By assumption A
∩
B
=
(
A
∩
B
)
2
¯
=
A
∩
B
Γ
(
A
∩
B
)
¯
⊆
A
Γ
B
¯. Therefore A
Γ
B
¯
=
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, A
Γ
A
¯
=
A
∩
A
=
A. Therefore S is a k-idempotent Γ-semiring.


## 4. Rightk-Weakly Regular Γ-Semiring

Definition 22.
AΓ-semiring S is said to be right k-weakly regular if, for any a
∈
S, a
∈
(
a
Γ
S
)
2
¯.Theorem 23.
InS, the following statements are equivalent. (1)
S is right k-weakly regular.
(2)
R
2
¯
=
R, for each right k-ideal R of S.
(3)
R
∩
I
=
R
Γ
I
¯, for a right k-ideal R and a k-ideal I of S.Proof.
1
⇒
(
2
). For any right k-ideal R of S, R
2
=
R
Γ
R
⊆
R
Γ
S
⊆
R. Hence R
2
¯
=
R
Γ
R
¯
⊆
R. For the reverse inclusion, let a
∈
R. As S is right k-weakly regular, a
∈
(
a
Γ
S
)
2
¯
=
a
Γ
S
Γ
a
Γ
S
¯
⊆
R
Γ
S
Γ
R
Γ
S
¯
⊆
R
Γ
R
¯
=
R
2
¯. Thus R
2
¯
=
R, for each right k-ideal R of S.
2
⇒
(
1
). For any a
∈
S, a
∈
a
Γ
S
⊆
a
Γ
S
¯ and a
Γ
S
¯ is a right k-ideal of S, then by assumption (
a
Γ
S
)
2
¯
=
a
Γ
S. Therefore a
∈
(
a
Γ
S
)
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
)
¯
⊆
R
Γ
I
¯. Clearly R
Γ
I
⊆
R and R
Γ
I
⊆
I. Then R
Γ
I
¯
⊆
R
¯
=
R and R
Γ
I
¯
⊆
I
¯
=
I, since R is a right k-ideal and I is a two sided k-ideal of S. Therefore R
Γ
I
¯
⊆
R
∩
I. Hence R
∩
I
=
R
Γ
I
¯.
3
⇒
(
2
). Let R be a right k-ideal of S and let (
R
) be two sided ideal generated by R. Then R
=
S
Γ
R
Γ
S. By assumption R
∩
(
R
)
¯
=
R
Γ
(
R
)
¯. Hence R
=
R
Γ
S
Γ
R
Γ
S
¯
⊆
R
Γ
R
¯
=
R
2
¯. Therefore R
2
¯
=
R.Theorem 24.
S is right k-weakly regular if and only if every right k-ideal of S is semiprime.Proof.
SupposeS is right k-weakly regular. Let R be a right k-ideal of S such that A
Γ
A
¯
⊆
R, for any right k-ideal A of S. A
=
A
Γ
A
¯. Then A
Γ
A
¯
⊆
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. R
Γ
R
¯ is also a right k-ideal of S. By assumption R
Γ
R
¯ is a semiprime right k-ideal of S. R
Γ
R
¯
⊆
R
Γ
R
¯ implies R
⊆
R
Γ
R
¯. Therefore R
=
R
¯
⊆
R
Γ
R
¯
=
R
2
¯. Therefore R
2
¯
=
R. Hence S is right k-weakly regular by Theorem 23.Definition 25 (see [12]).
A latticeL 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.
Ifc 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 latticeL is a Brouwerian if a
:
b exists for all a
,
b
∈
L.LetL
S denote the family of all k-ideals of S. Then L
S
,
⊆ is a partially ordered set. As 0
,
S
∈
L
S and ⋂
α
∈
Δ
‍
I
α
∈
L
S, for all I
α
∈
L
S and Δ is an indexing set, we have L
S is a complete lattice under ∧ and ∨ defined by I
∨
J
=
I
+
J
¯ and I
∧
J
=
I
∩
J. Further we have the following.Theorem 26.
IfS is a right k-weakly regular Γ-semiring, then L
S is a Brouwerian lattice.Proof.
LetB and C be any two k-ideals of S. Consider the family of k-ideals K
=
I
∈
L
S
∣
I
∩
B
⊆
C. Then by Zorn’s lemma there exists a maximal element M in K. Select I
∈
L
S such that B
∩
I
⊆
C. By Theorem 23, we have B
Γ
I
¯
⊆
C. To show that B
Γ
I
+
M
¯
¯
⊆
C. Let x
∈
B
Γ
I
+
M
¯. Then x
=
∑
i
=
1
n
‍
b
i
α
i
x
i, where b
i
∈
B, α
i
∈
Γ, and x
i
∈
I
+
M
¯. Therefore a
i
+
x
i
∈
I
+
M for some a
i
∈
I
+
M (see Definition 9).
x
=
∑
i
=
1
n
‍
b
i
α
i
(
a
i
+
x
i
)
∈
B
Γ
I
+
M
=
B
Γ
I
+
B
Γ
M
⊆
C, as B
Γ
I
⊆
C and B
Γ
M
⊆
C. Hence B
Γ
I
+
M
¯
⊆
C implies B
Γ
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
S is Brouwerian.AsL
S satisfies infinite meet distributive property property of lattice, we have the following.Corollary 27.
IfS is a right k-weakly regular Γ-semiring, then L
S is a distributive lattice (see Birkoff [12]).Theorem 28.
IfS 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.
LetS 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 A
Γ
B
¯
⊆
P. Then, by Theorem 23, we have A
∩
B
⊆
P. Hence A
∩
B
+
P
¯
=
P. As L
S 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.AΓ-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.
AΓ-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 aΓ-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, A
Γ
B
¯
⊆
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 A
Γ
B
¯
⊆
P, for any k-ideals A and B of S. Then A
Γ
B
¯
⊆
P. Hence by Theorem 23, we have A
∩
B
=
A
Γ
B
¯
⊆
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.
ThereforeS is a fully k-prime Γ-semiring.Now we define ak-bi-ideal of a Γ-semiring.Definition 30.
A nonempty subsetB of a Γ-semiring S is said to be a k-bi-ideal of S if B is a sub-Γ-semiring of S, B
Γ
S
Γ
B
¯
⊆
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
⊆
B
Γ
I
¯, for any k-bi-ideal B and k-ideal I of S.Proof.
SupposeS 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
∈
(
a
Γ
S
)
2
¯
=
a
Γ
S
Γ
a
Γ
S
¯
⊆
B
Γ
(
S
Γ
I
Γ
S
)
¯
⊆
B
Γ
I
¯. Therefore B
∩
I
⊆
B
Γ
I
¯. 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
¯
⊆
R
Γ
R
¯
¯
=
R
Γ
S
Γ
R
Γ
S
¯
=
R
Γ
S
Γ
(
R
Γ
S
)
¯
⊆
R
Γ
R
¯. Therefore R
=
R
Γ
R
¯
=
R
2
¯. 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
⊆
B
Γ
I
Γ
R
¯, for any k-bi-ideal B, k-ideal I, and a right k-ideal R of S.Proof.
SupposeS 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
∈
(
a
Γ
S
)
2
¯
=
a
Γ
S
Γ
a
Γ
S
¯
⊆
a
Γ
S
Γ
a
Γ
S
Γ
a
Γ
S
¯
Γ
S
¯
⊆
B
Γ
S
Γ
I
Γ
S
Γ
(
R
Γ
S
¯
)
⊆
B
Γ
I
Γ
R
¯. Therefore B
∩
I
∩
R
⊆
B
Γ
I
Γ
R
¯. 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
⊆
R
Γ
S
Γ
R
¯
⊆
R
Γ
R
¯. Therefore R
⊆
R
Γ
R
¯. Therefore R
=
R
Γ
R
¯
=
R
2
¯. Then, by Theorem 23, S is right k-weakly regular.


## 5. Right Purek-Ideals

In this section we define a right purek-ideal of a Γ-semiring S and furnish some of its characterizations.Definition 33.
Ak-ideal I of a Γ-semiring S is said to be a right pure k-ideal if, for any x
∈
I, x
∈
x
Γ
I
¯.Theorem 34.
Ak-ideal I of S is right pure if and only if R
∩
I
=
R
Γ
I
¯, for any right k-ideal R of S.Proof.
LetI be a right pure k-ideal and let R be a right k-ideal of S. Then clearly R
Γ
I
¯
⊆
R
∩
I. Now let a
∈
R
∩
I. As I is a right pure k-ideal, a
∈
a
Γ
I
¯
⊆
R
Γ
I
¯. This gives R
∩
I
⊆
R
Γ
I
¯. By combining both the inclusions we get R
∩
I
=
R
Γ
I
¯. 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
¯
=
N
0
a
+
a
Γ
I
¯, where N
0 denotes the set of nonnegative integers. Then a
∈
(
a
)
r
¯
∩
I
=
(
a
)
r
¯
Γ
I
¯
=
(
N
0
a
+
a
Γ
I
)
¯
Γ
I
¯
=
(
N
0
a
+
a
Γ
I
)
Γ
I
¯
⊆
a
Γ
I
¯ (see Result 2). Therefore I is a right pure k-ideal of S.Theorem 35.
The intersection of right purek-ideals of S is a right pure k-ideal of S.Proof.
LetA and B be right pure k-ideals of S. Then for any right k-ideal R of S we have R
∩
A
=
R
Γ
A
¯ and R
∩
B
=
R
Γ
B
¯. We consider R
∩
A
∩
B
=
R
∩
A
∩
B
=
R
Γ
A
¯
∩
B
=
R
Γ
A
¯
Γ
B
¯
=
R
Γ
A
Γ
B
¯
=
R
Γ
(
A
∩
B
)
¯. Therefore A
∩
B is a right pure k-ideal of S.Characterization of a rightk-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 thatS 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
=
R
Γ
I
¯. 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 Primek-Ideals

LetS 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 Θ
I
=
J
∈
℘
S
∣
I
⊈
J and
(3)
ζ
℘
S
=
Θ
I
∣
I
is
a
k
-ideal
of
S
.Theorem 37.
IfS 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 L
S 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:
Θ
0
=
J
∈
℘
s
∣
0
⊈
J
=
Φ. Therefore Φ
∈
ζ
℘
S. As S itself is a k-ideal, Θ
S
=
J
∈
℘
S
∣
S
⊈
J
=
℘
S imply ℘
S
∈
ζ
℘
S. Now let Θ
I
k
∈
ζ
℘
S for k
∈
Λ; Λ is an indexing set, and I
k is a k-ideal of S. Therefore Θ
I
k
=
J
∈
℘
S
∣
I
k
⊈
J. As ⋃
k
‍
Θ
I
k
=
J
∈
℘
S
∣
∑
k
‍
I
k
¯
⊈
J, ∑
k
‍
I
k
¯ is a k-ideal of S. Therefore ⋃
k
‍
Θ
I
k
=
Θ
∑
k
‍
I
k
¯
∈
ζ
℘
S. Further let Θ
A, Θ
B
∈
ζ
℘
S.
LetJ
∈
Θ
A
⋂
‍
Θ
B; 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
∈
Θ
A
∩
B. Therefore Θ
A
⋂
‍
Θ
B
⊆
Θ
A
∩
B. Now let J
∈
Θ
A
∩
B. Then A
∩
B
⊈
J implies A
⊈
J and B
⊈
J. Therefore J
∈
Θ
A and J
∈
Θ
B imply J
∈
Θ
A
⋂
‍
Θ
B. Thus Θ
A
∩
B
⊆
Θ
A
⋂
‍
Θ
B. Hence Θ
A
⋂
‍
Θ
B
=
Θ
A
∩
B
∈
ζ
℘
S. Thus ζ
℘
s forms a topology on the set ℘
S.
Now we define a functionϕ
:
L
S
→
ζ
℘
S by ϕ
I
=
Θ
I, for all I
∈
L
S. Let I
,
K
∈
L
S. Consider  ϕ
I
∩
K
=
Θ
I
∩
K
=
Θ
I
⋂
‍
Θ
K
=
ϕ
I
∩
ϕ
K.
Considerϕ
I
+
K
¯
=
Θ
I
+
K
¯
=
Θ
I
∪
Θ
K
=
ϕ
I
∪
ϕ
K. Therefore ϕ is a lattice homomorphism. Now consider ϕ
I
=
ϕ
K. Then Θ
I
=
Θ
K.
Suppose thatI
≠
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
∈
Θ
K
=
Θ
I implies I
⊆
J, which is a contradiction. Therefore I
=
K. Thus ϕ
I
=
ϕ
K implies I
=
K and hence ϕ is one-one. As ϕ is onto the result follows.



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*Source: 948032-2014-12-15.xml*