# Characterizations of Regular Ordered Semirings by Ordered Quasi-Ideals

Authors: Pakorn Palakawong na Ayutthaya; Bundit Pibaljommee
Journal: International Journal of Mathematics and Mathematical Sciences
 (2016)

---

## Abstract

We introduce the notion of an ordered quasi-ideal of an ordered semiring and show that ordered quasi-ideals and ordered bi-ideals coincide in regular ordered semirings. Then we give characterizations of regular ordered semirings, regular ordered duo-semirings, and left (right) regular ordered semirings by their ordered quasi-ideals.


---

## Body

## 1. Introduction

The concept of a quasi-ideal was defined first by Steinfeld for semigroups and for rings [1–3] as a generalization of a right ideal and a left ideal. Then Iséki [4] introduced the notion of a quasi-ideal in a semiring without zero and investigated some of its properties. In 1994, Dönges [5] studied quasi-ideals of a semiring with zero, investigated connections between left (right) ideals, bi-ideals, and quasi-ideals and characterized regular semirings using their quasi-ideals. Later, Shabir et al. [6] have studied some properties of quasi-ideals, using quasi-ideals to characterize regular and intraregular semirings and regular duo-semirings. As a generalization of quasi-ideals of semirings the quasi-ideals of Γ-semirings were investigated by many authors; see, for example, [7–9].In 2011, the notion of an ordered semiring was introduced by Gan and Jiang [10] as a semiring with a partially ordered relation on the semiring such that the relation is compatible to the operations of the semiring. In the paper, the concept of a left (right) ordered ideal, a minimal ordered ideal, and a maximal ordered ideal was defined. Then Mandal [11] studied fuzzy ideals in an ordered semiring with the least element zero and gave a characterization of regular ordered semirings by their fuzzy ideals.In this paper, we introduce the notion of an ordered quasi-ideal of an ordered semiring and show that ordered quasi-ideals and ordered bi-ideals coincide in regular ordered semirings. Then characterizations of regular ordered semirings, regular ordered duo-semirings, and left (right) regular ordered semirings by their ordered quasi-ideals have been investigated.


## 2. Preliminaries

An ordered semiring is a system(
S
,
+
,
·
,
≤
) consisting of a nonempty set S such that (
S
,
+
,
·
) is a semiring, (
S
,
≤
) is a partially ordered set, and for any a
,
b
,
x
∈
S the following conditions are satisfied:(i)
ifa
≤
b then a
+
x
≤
b
+
x and x
+
a
≤
x
+
b;
(ii)
ifa
≤
b then a
x
≤
b
x and x
a
≤
x
b.An ordered semiringS is said to beadditively commutative if a
+
b
=
b
+
a for all a
,
b
∈
S. An element 0
∈
S is said to be anabsorbing zero if 0
a
=
0
=
a
0 and a
+
0
=
a
=
0
+
a for all a
∈
S. In this paper we assume that S is an additively commutative ordered semiring with an absorbing zero 0.For any subsetsA
,
B of S and a
∈
S, we denote (1)
A
=
x
∈
S
∣
x
≤
a
for  some
a
∈
A
,
A
B
=
a
b
∈
S
∣
a
∈
A
,
b
∈
B
,
Σ
A
=
∑
i
∈
I
a
i
∈
S
∣
a
i
∈
A
and
I
is  a  finite  subset  of
N
,
Σ
A
B
=
∑
i
∈
I
a
i
b
i
∈
S
∣
a
i
∈
A
,
b
i
∈
B
and
I
is  a  finite  subset  of
N
,
N
a
=
Σ
a
.Now, we mention some properties of finite sums on an ordered semiring.Remark 1.
For any subsetsA
,
B of S, the following statements hold:(i)
Σ
(
A
]
⊆
(
Σ
A
];
(ii)
Σ
(
Σ
A
)
=
Σ
A;
(iii)
A
(
Σ
B
)
⊆
Σ
A
B and (
Σ
A
)
B
⊆
Σ
A
B;
(iv)
Σ
(
A
Σ
B
)
⊆
Σ
A
B and Σ
(
Σ
A
)
B
⊆
Σ
A
B;
(v)
Σ
(
A
+
B
)
=
Σ
A
+
Σ
B.We note that, for anyA
⊆
S, Σ
A
=
A if and only if A
+
A
⊆
A ((
A
,
+
) is a subsemigroup of (
S
,
+
)).Now, we give the basic properties of the operator(
] which are not difficult to verify.Lemma 2.
LetA
,
B
,
C be subsets of an ordered semiring S. Then the following statements hold: (i)
A
⊆
(
A
] and (
(
A
]
]
=
(
A
];
(ii)
IfA
⊆
B then (
A
]
⊆
(
B
];
(iii)
A
(
B
]
⊆
(
A
]
(
B
]
⊆
(
A
B
] and (
A
]
B
⊆
(
A
]
(
B
]
⊆
(
A
B
];
(iv)
A
+
(
B
]
⊆
(
A
]
+
(
B
]
⊆
(
A
+
B
] and (
A
]
+
B
⊆
(
A
]
+
(
B
]
⊆
(
A
+
B
];
(v)
A
(
B
+
C
]
⊆
(
A
B
+
A
C
] and (
A
+
B
]
C
⊆
(
A
C
+
B
C
];
(vi)
(
A
∪
B
]
=
(
A
]
∪
(
B
];
(vii)
(
A
∩
B
]
⊆
(
A
]
∩
(
B
].In (vii) of the above lemma, we have(
A
∩
B
]
=
(
A
]
∩
(
B
] when (
A
]
=
A and (
B
]
=
B.Lemma 3.
LetS be an ordered semiring and ∅
≠
A
⊆
S. If A
⊆
(
Σ
A
2
+
Σ
A
S
A
] then Σ
A
2
⊆
(
Σ
A
S
A
].Proof.
Assume thatA
⊆
(
Σ
A
2
+
Σ
A
S
A
]. Then (2)
Σ
A
2
⊆
Σ
Σ
A
2
+
Σ
A
S
A
A
⊆
Σ
Σ
A
2
A
+
Σ
A
S
A
A
⊆
Σ
Σ
A
3
+
Σ
A
S
A
=
Σ
Σ
A
3
+
Σ
Σ
A
S
A
=
Σ
A
3
+
Σ
A
S
A
=
Σ
A
A
A
+
Σ
A
S
A
⊆
Σ
A
S
A
+
Σ
A
S
A
=
Σ
A
S
A
.Definition 4 (see [10]).
LetS be an ordered semiring and ∅
≠
A
⊆
S. Then A is said to be aleft ordered ideal (right ordered ideal) if the following conditions are satisfied. (1)
A is a left ideal (right ideal) of S.
(2)
Ifx
≤
a for some a
∈
A then x
∈
A (i.e., A
=
(
A
]).
We callA anordered ideal if it is both left ordered ideal and right ordered ideal of S.Example 5 (see [10]).
Let[
0,1
] be the unit interval of real numbers. Define binary operations ⊕ and ⊙ on [
0,1
] by letting a
,
b
∈
0,1,(3)
a
⊕
b
=
max
⁡
a
,
b
,
a
⊙
b
=
max
⁡
a
+
b
-
1,0
,and an ordered relation ≤ is the natural order on real numbers. It is easy to show that L
=
(
[
0,1
]
,
⊕
,
⊙
,
≤
) is an ordered semiring. Let I
=
[
0
,
1
/
2
]. Then we can prove that I is an ordered ideal of L.Lemma 6.
LetA be a nonempty subset of an ordered semiring S. Then (i)
(
Σ
S
A
] is a left ordered ideal of S;
(ii)
(
Σ
A
S
] is a right ordered ideal of S;
(iii)
(
Σ
S
A
S
] is an ordered ideal of S.Proof.
(i) Letx
,
y
∈
(
Σ
S
A
]. Then x
≤
x
′ and y
≤
y
′ for some x
′
,
y
′
∈
Σ
S
A. It is clear that x
+
y
≤
x
′
+
y
′
∈
Σ
S
A, and so x
+
y
∈
(
Σ
S
A
]. By Remark 1 and Lemma 2, we obtain S
(
Σ
S
A
]
⊆
(
S
Σ
S
A
]
⊆
(
Σ
S
S
A
]
⊆
(
Σ
S
A
]. We have (
(
Σ
S
A
]
]
=
(
Σ
S
A
]. Hence, (
Σ
S
A
] is a left ordered ideal of S.
(ii) and (iii) can be proved similar to (i).Corollary 7.
LetS be an ordered semiring. Then, for any a
∈
S, (i)
(
S
a
] is a left ordered ideal of S;
(ii)
(
a
S
] is a right ordered ideal of S;
(iii)
(
Σ
S
a
S
] is an ordered ideal of S.LetA be a nonempty subset of an ordered semiring S. We denote L
(
A
)
,
R
(
A
) and I
(
A
) as the smallest left ordered ideal, right ordered ideal, and ordered ideal of S containing A, respectively. In particular, we can show that if A is a left ideal (right ideal, ideal) of S then (
A
] is the smallest left ordered ideal (resp., right ordered ideal and ordered ideal) of S containing A.Lemma 8.
LetA be a nonempty subset of an ordered semiring S. Then (i)
L
(
A
)
=
(
Σ
A
+
Σ
S
A
];
(ii)
R
(
A
)
=
(
Σ
A
+
Σ
A
S
];
(iii)
I
(
A
)
=
(
Σ
A
+
Σ
S
A
+
Σ
A
S
+
Σ
S
A
S
].Proof.
(i) SinceS has an absorbing zero, we have, for every a
∈
A, a
=
a
+
0
∈
Σ
A
+
Σ
S
A
⊆
(
Σ
A
+
Σ
S
A
]. Hence, A
⊆
(
Σ
A
+
Σ
S
A
]. Let x
,
y
∈
(
Σ
A
+
Σ
S
A
]. Then x
≤
x
′ and y
≤
y
′ for some x
′
,
y
′
∈
Σ
A
+
Σ
S
A. Thus x
′
=
a
1
+
b
1 and y
′
=
a
2
+
b
2 for some a
1
,
a
2
∈
Σ
A and b
1
,
b
2
∈
Σ
S
A. It is easy to show that a
1
+
a
2
∈
Σ
A and b
1
+
b
2
∈
Σ
S
A. It follows that x
+
y
≤
x
′
+
y
′
∈
Σ
A
+
Σ
S
A, and so x
+
y
∈
(
Σ
A
+
Σ
S
A
]. By Remark 1 and Lemma 2, we obtain (4)
S
Σ
A
+
Σ
S
A
⊆
S
Σ
A
+
Σ
S
A
⊆
S
Σ
A
+
S
Σ
S
A
⊆
Σ
S
A
+
Σ
S
S
A
⊆
Σ
S
A
+
Σ
S
A
=
Σ
S
A
⊆
Σ
A
+
Σ
S
A
.Since (
(
Σ
A
+
Σ
S
A
]
]
=
(
Σ
A
+
Σ
S
A
], L is a left ordered ideal of S. Let K be any left ordered ideal of S containing A. It turns out Σ
A
⊆
K and Σ
S
A
⊆
K, so Σ
A
+
Σ
S
A
⊆
K. It follows that (
Σ
A
+
Σ
S
A
]
⊆
(
K
]
=
K. Therefore, (
Σ
A
+
Σ
S
A
] is the smallest left ordered ideal of S containing A.
(ii) and (iii) can be proved similar to (i).As a special case of Lemma8, if A
=
{
a
} then we have the following corollary.Corollary 9.
LetS be an ordered semiring. Then, for any a
∈
S,(i)
L
(
a
)
=
(
N
a
+
S
a
];
(ii)
R
(
a
)
=
(
N
a
+
a
S
];
(iii)
I
(
a
)
=
(
N
a
+
S
a
+
a
S
+
Σ
S
a
S
].An elemente of an ordered semiring S is said to be anidentity if e
a
=
a
=
a
e for all a
∈
S. If S has an identity, then we denote 1 as the identity of S.It is not difficult to show that ifS has an identity, then L
(
A
)
=
(
Σ
S
A
]
,
R
(
A
)
=
(
Σ
A
S
] and I
(
A
)
=
(
Σ
S
A
S
] for any A
⊆
S. In particular case, we have L
(
a
)
=
(
S
a
]
,
R
(
a
)
=
(
a
S
] and I
(
a
)
=
(
Σ
S
a
S
] for any a
∈
S.


## 3. Ordered Quasi-Ideals in Ordered Semirings

Here, we present a notion of an ordered quasi-ideal of an ordered semiring. Then, in ordered semiring with an identity, we show that every ordered quasi-ideal can be expressed as an intersection of an ordered left ideal and an ordered right ideal.Definition 10.
Let(
S
,
+
,
·
,
≤
) be an ordered semiring and let (
Q
,
+
) be a subsemigroup of (
S
,
+
). Then Q is said to be anordered quasi-ideal of S if the following conditions are satisfied: (1)
(
Σ
S
Q
]
∩
(
Σ
Q
S
]
⊆
Q;
(2)
ifx
≤
q for some q
∈
Q then x
∈
Q (i.e., Q
=
(
Q
]).It is clear that every left ordered ideal (right ordered ideal and ordered ideal) of an ordered semiringS is an ordered quasi-ideal of S. Moreover, each ordered quasi-ideal of S is a subsemiring of S; indeed, Q
Q
⊆
(
Q
Q
]
⊆
(
S
Q
]
∩
(
Q
S
]
⊆
(
Σ
S
Q
]
∩
(
Σ
Q
S
]
⊆
Q.Example 11.
LetS
=
{
a
,
b
,
c
,
d
}. Define binary operations + and · on S by the following equations: (5)
+
a
b
c
d
a
a
b
c
d
b
b
b
b
b
c
c
b
c
d
d
d
b
d
d
,
·
a
b
c
d
a
a
a
a
a
b
a
b
b
b
c
a
c
c
c
d
a
b
b
b
.Then (
S
,
+
,
·
) is an additively commutative semiring with an absorbing zero a. Define a binary relation ≤ on S by (6)
≤
≔
a
,
a
,
b
,
b
,
c
,
c
,
d
,
d
,
b
,
d
.We give the covering relation “≺” and the figure of S: (7)
≺
≔
b
,
d
.
Now,(
S
,
+
,
·
,
≤
) is an ordered semiring. Let Q
=
{
a
,
b
}. We have (
Σ
S
Q
]
∩
(
Σ
Q
S
]
=
{
a
,
b
,
c
}
∩
{
a
,
b
}
=
Q and (
Q
]
=
Q. Hence, Q is an ordered quasi-ideal of S but is not a left ordered ideal of S, since S
Q
=
{
a
,
b
,
c
}
⊈
Q.Lemma 12.
LetS be an ordered semiring and let {
Q
i
∣
i
∈
I
} be a family of ordered quasi-ideals of S. Then ⋂
i
∈
I
‍
Q
i is an ordered quasi-ideal of S.LetA be a nonempty subset of an ordered semiring S. We denote Q
(
A
) the smallest ordered quasi-ideal of S containing A.Theorem 13.
LetS be an ordered semiring and let A be a nonempty subset of S. Then Q
(
A
)
=
(
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
)
].Proof.
LetQ
=
(
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
)
]. Since S has an absorbing zero, we have a
=
a
+
0
∈
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
)
⊆
Q for every a
∈
A. Hence, A
⊆
Q. Let x
,
y
∈
Q. Then x
≤
x
′ and y
≤
y
′ for some x
′
,
y
′
∈
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
). Thus x
′
=
a
1
+
b
1 and y
′
=
a
2
+
b
2 for some a
1
,
a
2
∈
Σ
A and b
1
,
b
2
∈
(
Σ
S
A
]
∩
(
Σ
A
S
]. Clearly, a
1
+
a
2
∈
Σ
A and b
1
+
b
2
∈
(
Σ
S
A
]
∩
(
Σ
A
S
]. It follows that x
+
y
≤
x
′
+
y
′
∈
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
), and so x
+
y
∈
Q. By Remark 1 and Lemma 2, we obtain (8)
Σ
S
Q
∩
Σ
Q
S
⊆
Σ
S
Q
=
Σ
S
Σ
A
+
Σ
S
A
∩
Σ
A
S
⊆
Σ
S
Σ
A
+
Σ
S
A
⊆
Σ
S
Σ
A
+
S
Σ
S
A
⊆
Σ
Σ
S
A
+
Σ
S
S
A
⊆
Σ
Σ
S
A
+
Σ
S
S
A
⊆
Σ
Σ
S
A
+
Σ
S
A
⊆
Σ
Σ
S
A
⊆
Σ
S
A
=
Σ
S
A
.Similarly, we can show that (
Σ
S
Q
]
∩
(
Σ
Q
S
]
⊆
(
Σ
A
S
]. Thus (
Σ
S
Q
]
∩
(
Σ
Q
S
]
⊆
(
Σ
S
A
]
∩
(
Σ
A
S
]
⊆
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
)
⊆
Q. Since (
Q
]
=
Q, we obtain that Q is an ordered quasi-ideal of S containing A. Let K be any ordered quasi-ideal of S containing A. It follows that (
Σ
S
A
]
∩
(
Σ
A
S
]
⊆
(
Σ
S
K
]
∩
(
Σ
K
S
]
⊆
K. So Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
)
⊆
K. Hence, Q
=
(
Σ
A
+
(
(
Σ
S
A
]
∩
(
Σ
A
S
]
)
]
⊆
(
K
]
=
K. Therefore, Q is the smallest ordered quasi-ideal of S containing A.As a special case of Theorem13, if A
=
{
a
} then we have the following corollary.Corollary 14.
LetS be an ordered semiring. Then Q
(
a
)
=
(
N
a
+
(
(
S
a
]
∩
(
a
S
]
)
] for any a
∈
S.IfS has an identity, then it is easy to check that Q
(
A
)
=
(
Σ
S
A
]
∩
(
Σ
A
S
] for any A
⊆
S. In particular case, we have Q
(
a
)
=
(
S
a
]
∩
(
a
S
] for any a
∈
S.LetQ
(
S
) be the set of all ordered quasi-ideals of an ordered semiring S. Using Lemma 12, we define the operations ∧ and ∨ on Q
(
S
) by letting P
1
,
P
2
∈
Q
(
S
), (9)
P
1
∧
P
2
=
P
1
∩
P
2
,
P
1
∨
P
2
=
Q
P
1
∪
P
2
.Then we obtain the following theorem.Theorem 15.
LetS be an ordered semiring. Then (
Q
(
S
)
,
∧
,
∨
) is a complete lattice.Theorem 16.
The intersection of a left ordered idealL and a right ordered ideal R of an ordered semiring S is an ordered quasi-ideal of S.Proof.
It is easy to show thatL
∩
R is a subsemigroup of (
S
,
+
). By Remark 1 and Lemma 2, we obtain (10)
Σ
S
L
∩
R
∩
Σ
L
∩
R
S
⊆
Σ
S
L
∩
R
=
Σ
S
L
∩
S
R
⊆
Σ
S
L
⊆
L
,
Σ
S
L
∩
R
∩
Σ
L
∩
R
S
⊆
Σ
L
∩
R
S
=
Σ
L
S
∩
R
S
⊆
Σ
R
S
⊆
R
.Hence, (
Σ
S
(
L
∩
R
)
]
∩
(
Σ
(
L
∩
R
)
S
]
⊆
L
∩
R. Let s
∈
S such that s
≤
x for some x
∈
L
∩
R. Then s
∈
(
L
∩
R
]
⊆
(
L
]
∩
(
R
]
=
L
∩
R.The converse of Theorem16 is not true as Example 2.1 page 8 in [2] given by A. H. Clifford.Corollary 17.
LetS be an ordered semiring. Then the following statements hold. (i)
(
Σ
S
A
]
∩
(
Σ
A
S
] is an ordered quasi-ideal of S, for any A
⊆
S.
(ii)
(
S
a
]
∩
(
a
S
] is an ordered quasi-ideal of S, for any a
∈
S.Proof.
(i) By Lemma6, we have (
Σ
S
A
] and (
Σ
A
S
] a left and a right ordered ideal of S, respectively. Then by Theorem 16, we have that (
Σ
S
A
]
∩
(
Σ
A
S
] is an ordered quasi-ideal of S.
(ii) It is a particular case of (i).Now, we will show that the converse of Theorem16 is true if S contains an identity as the following theorem.Theorem 18.
LetS be an ordered semiring with identity. Then every ordered quasi-ideal Q of S can be written in the form Q
=
R
∩
L for some right ordered ideal R and left ordered ideal L of S.Proof.
Assume thatS has an identity. Let Q be an ordered quasi-ideal of S. Then R
(
Q
)
=
(
Σ
Q
S
] and L
(
Q
)
=
(
Σ
S
Q
]. We obtain Q
⊆
R
(
Q
)
∩
L
(
Q
) and R
(
Q
)
∩
L
(
Q
)
=
(
Σ
Q
S
]
∩
(
Σ
S
Q
]
⊆
Q. Hence, Q
=
R
(
Q
)
∩
L
(
Q
).


## 4. Regular Ordered Semirings

In this section, we show that in regular ordered semirings the converse of Theorem16 is true and ordered quasi-ideals coincide with ordered bi-ideals. Then we give characterizations of regular ordered semirings, regular ordered duo-semirings, and left regular and right regular ordered semirings by their ordered quasi-ideals.Definition 19 (see [11]).
An elementa of an ordered semiring S is said to beregular if a
≤
a
x
a for some x
∈
S. An ordered semiring S is said to beregular if every element a
∈
S is regular.The following lemma is characterizations of regular ordered semiring which directly follows Definition19.Lemma 20.
LetS be an ordered semiring. Then the following statements are equivalent: (i)
S is regular;
(ii)
A
⊆
(
Σ
A
S
A
] for each A
⊆
S;
(iii)
a
∈
(
a
S
a
] for any a
∈
S.Now, we will show that the converse of Theorem16 is true in regular ordered semirings.Theorem 21.
Every ordered quasi-ideal of a regular ordered semiringS can be written in the form Q
=
R
∩
L for some right ordered ideal R and left ordered ideal L of S.Proof.
LetQ be an ordered quasi-ideal of S. By Lemma 8, we have R
(
Q
)
=
(
Σ
Q
+
Σ
Q
S
] and L
(
Q
)
=
(
Σ
Q
+
Σ
S
Q
]. Now, Q
⊆
R
(
Q
)
∩
L
(
Q
). Let q
∈
Q. Since S is regular, there exists x
∈
S such that q
≤
q
x
q
∈
Σ
Q
S. So Q
⊆
(
Σ
Q
S
]. Since Q
+
Q
⊆
Q, Σ
Q
=
Q. It follows that (11)
Σ
Q
S
⊆
Σ
Q
+
Σ
Q
S
=
Q
+
Σ
Q
S
⊆
Σ
Q
S
+
Σ
Q
S
⊆
Σ
Q
S
.This implies that R
(
Q
)
=
(
Σ
Q
S
]. Similarly, we can show that L
(
Q
)
=
(
Σ
S
Q
]. Hence, R
(
Q
)
∩
L
(
Q
)
=
(
Σ
Q
S
]
∩
(
Σ
S
Q
]
⊆
Q. Therefore, Q
=
R
(
Q
)
∩
L
(
Q
).Definition 22.
Let(
S
,
+
,
·
,
≤
) be an ordered semiring. A subsemigroup (
B
,
+
) of (
S
,
+
) is said to be anordered bi-ideal of S if the following conditions hold: (1)
B
S
B
⊆
B;
(2)
ifx
≤
b for some b
∈
B, then x
∈
B (i.e., B
=
(
B
]).We note that condition (1) of Definition22 is equivalent to Σ
B
S
B
⊆
B.Theorem 23.
Every ordered quasi-ideal of an ordered semiringS is an ordered bi-ideal of S.Proof.
LetQ be an ordered quasi-ideal of S. Then Σ
Q
S
Q
⊆
Σ
Q
S
⊆
(
Σ
Q
S
] and Q
S
Q
⊆
Σ
S
Q
⊆
(
Σ
S
Q
]. So, Σ
Q
S
Q
⊆
(
Σ
S
Q
]
∩
(
Σ
Q
S
]
⊆
Q. Hence, Q is an ordered bi-ideal of S.The converse of Theorem23 is not generally true as the following example.Example 24.
LetS
=
{
a
,
b
,
c
,
d
,
e
}. Define binary operations + and · by the following equations: (12)
+
a
b
c
d
e
a
a
b
c
d
e
b
b
b
d
d
d
c
c
d
d
d
d
d
d
d
d
d
d
e
e
d
d
d
e
,
·
a
b
c
d
e
a
a
a
a
a
a
b
a
a
a
a
a
c
a
a
b
b
b
d
a
a
b
b
b
e
a
a
b
b
b
.Then (
S
,
+
,
·
) is an additively commutative semiring with an absorbing zero a. Define a binary relation ≤ on S by (13)
≤
≔
a
,
a
,
b
,
b
,
c
,
c
,
d
,
d
,
e
,
e
,
a
,
b
,
a
,
c
,
a
,
e
,
a
,
d
,
b
,
d
,
c
,
d
,
e
,
d
.We give the covering relation “≺” and the figure of S: (14)
≺
≔
a
,
b
,
a
,
c
,
a
,
e
,
b
,
d
,
c
,
d
,
e
,
d
.
Then(
S
,
+
,
·
,
≤
) is an ordered semiring but not regular, since d
≰
d
x
d for any x
∈
S. Let B
=
{
a
,
e
}. It is easy to show that B is an ordered bi-ideal but not an ordered quasi-ideal of S, since (
Σ
S
B
]
∩
(
Σ
B
S
]
=
{
a
,
b
}
⊈
B
.Now, we show that in regular ordered semirings, ordered bi-ideals and ordered quasi-ideals coincide as the following theorem.Theorem 25.
LetS be a regular ordered semiring. Then ordered bi-ideals and ordered quasi-ideals coincide in S.Proof.
By Theorem23, we have that every ordered quasi-ideal of S is an ordered bi-ideal of S. Now, we show that every ordered bi-ideal of S is an ordered quasi-ideal of S. Let B be an ordered bi-ideal of S. Let a
∈
(
Σ
S
B
]
∩
(
Σ
B
S
]. By Lemma 20, Remark 1, and Lemma 2, we obtain a
∈
(
a
S
a
]
⊆
(
(
Σ
B
S
]
S
(
Σ
S
B
]
]
⊆
(
(
Σ
B
S
S
]
(
Σ
S
B
]
]
⊆
(
(
Σ
B
S
)
(
Σ
S
B
)
]
⊆
(
Σ
(
B
S
(
Σ
S
B
)
)
]
⊆
(
Σ
(
Σ
B
S
S
B
)
]
⊆
(
Σ
B
S
B
]
⊆
B. Hence, B is an ordered quasi-ideal of S.Theorem 26.
LetS be an ordered semiring. Then the following statements are equivalent: (i)
S is regular;
(ii)
(
Σ
R
L
]
=
R
∩
L for every right ordered ideal R and left ordered ideal L of S;
(iii)
B
=
(
Σ
B
S
B
] for each ordered bi-ideal B of S;
(iv)
Q
=
(
Σ
Q
S
Q
] for each ordered quasi-ideal Q of S.Proof.
(
i
)
⇒
(
ii
): assume that S is regular and let R and L be a right ordered ideal and a left ordered ideal of S, respectively. So, (
Σ
R
L
]
⊆
(
Σ
R
]
=
R and (
Σ
R
L
]
⊆
(
Σ
L
]
=
L. Hence, (
Σ
R
L
]
⊆
R
∩
L. Let a
∈
R
∩
L. Since S is regular, a
≤
a
x
a for some x
∈
S. Since a
∈
R, x
a
∈
L. It follows that a
≤
a
(
x
a
)
∈
R
L. This means a
∈
(
R
L
]
⊆
(
Σ
R
L
]. Therefore, (
Σ
R
L
]
=
R
∩
L.
(
ii
)
⇒
(
iii
): assume that (ii) holds. Let B be an ordered bi-ideal of S. It is clear that (
Σ
B
S
B
]
⊆
B. By assumption, B
⊆
R
(
B
)
∩
L
(
B
)
=
(
Σ
R
(
B
)
L
(
B
)
]. By Lemma 8, Remark 1, and Lemmas 2 and 3, we have (15)
B
⊆
Σ
B
+
Σ
B
S
∩
Σ
B
+
Σ
S
B
=
Σ
Σ
B
+
Σ
B
S
Σ
B
+
Σ
S
B
⊆
Σ
Σ
B
+
Σ
B
S
Σ
B
+
Σ
S
B
⊆
Σ
Σ
B
Σ
B
+
Σ
S
B
+
Σ
B
S
Σ
B
+
Σ
S
B
⊆
Σ
Σ
B
2
+
Σ
B
S
B
+
Σ
B
S
B
+
Σ
B
S
S
B
⊆
Σ
Σ
B
2
+
Σ
B
S
B
=
Σ
Σ
B
2
+
Σ
Σ
B
S
B
=
Σ
B
2
+
Σ
B
S
B
⊆
Σ
B
S
B
+
Σ
B
S
B
⊆
Σ
B
S
B
.
(
iii
)
⇒
(
iv
): it follows from Theorem 23.
(
iv
)
⇒
(
i
): let a
∈
S. Then Q
(
a
)
=
(
Q
(
a
)
S
Q
(
a
)
]. By Corollary 14, Remark 1, and Lemma 2, we have (16)
a
∈
N
a
+
S
a
∩
a
S
=
Σ
N
a
+
S
a
∩
a
S
S
N
a
+
S
a
∩
a
S
⊆
Σ
N
a
+
a
S
S
N
a
+
S
a
⊆
Σ
N
a
+
a
S
S
N
a
+
S
a
⊆
Σ
N
a
+
a
S
S
N
a
+
S
a
⊆
Σ
a
S
N
a
+
S
a
⊆
Σ
a
S
N
a
+
S
a
⊆
Σ
a
S
a
=
a
S
a
.By Lemma 20, S is regular.Theorem 27.
LetS be a regular ordered semiring. Then the following statements hold: (i)
every ordered quasi-idealQ of S can be written in the form Q
=
R
∩
L
=
(
R
L
] for some right ordered ideal R and left ordered ideal L of S;
(ii)
(
Q
2
]
=
(
Q
3
] for each ordered quasi-ideal Q of S.Proof.
(i) It is obvious by Theorems21 and 26.
(ii) LetQ be an ordered quasi-ideal of S. Clearly, (
(
Q
Q
)
Q
]
⊆
(
Q
Q
]. Let x
∈
(
Q
Q
]. Then x
≤
q
1
q
2 for some q
1
,
q
2
∈
Q. Since S is regular, there exists s
∈
S such that x
≤
q
1
q
2
≤
(
q
1
q
2
)
s
(
q
1
q
2
)
∈
Q
Q
S
Q
Q. Hence, x
∈
(
Q
(
Q
S
Q
)
Q
]
⊆
(
Q
Q
Q
]
. Therefore, (
Q
2
]
=
(
Q
3
]
.Theorem 28.
LetS be an ordered semiring. Then S is regular if and only if B
∩
I
∩
L
⊆
(
B
I
L
] for every ordered bi-ideal B, every ordered ideal I, and every left ordered ideal L of S.Proof.
LetB
,
I, and L be an ordered bi-ideal, an ordered ideal, and a left ordered ideal of S, respectively. Let a
∈
B
∩
I
∩
L. Since S is regular, a
≤
a
x
a
≤
a
x
a
x
a
x
a
∈
B
I
L. Hence, B
∩
I
∩
L
⊆
(
B
I
L
].
Conversely, assume thatB
∩
I
∩
L
⊆
(
B
I
L
] for every ordered bi-ideal B, every ordered ideal I, and every left ordered ideal L of S. Then we obtain R
∩
L
=
R
∩
S
∩
L
⊆
(
R
S
L
]
⊆
(
R
L
]
⊆
(
Σ
R
L
] for every right ordered ideal R and left ordered ideal L of S. On the other hand, we have (
Σ
R
L
]
⊆
R
∩
L. Hence, (
Σ
R
L
]
=
R
∩
L. By Theorem 26, S is regular.Definition 29.
An ordered semiringS is said to be anordered duo-semiring if every one-sided (right or left) ordered ideal of S is an ordered ideal of S.We note that every multiplicatively commutative ordered semiring is an ordered duo-semiring, but the converse is not generally true. Now, we give an example of a multiplicatively noncommutative ordered semiring which is an ordered duo-semiring.Example 30.
LetS
=
{
a
,
b
,
c
,
d
,
e
}. Define binary operations + and · by the following equations: (17)
+
a
b
c
d
e
a
a
b
c
d
e
b
b
c
c
c
c
c
c
c
c
c
c
d
d
c
c
c
c
e
e
c
c
c
c
,
·
a
b
c
d
e
a
a
a
a
a
a
b
a
e
c
e
c
c
a
c
c
c
c
d
a
c
c
e
c
e
a
c
c
c
c
.Then (
S
,
+
,
·
) is an additively commutative semiring with an absorbing zero a. Define a binary relation ≤ on S by (18)
≤
≔
a
,
a
,
b
,
b
,
c
,
c
,
d
,
d
,
e
,
e
,
e
,
c
.We give the covering relation “≺” and the figure of S: (19)
≺
≔
e
,
c
.
Then(
S
,
+
,
·
,
≤
) is an ordered semiring which is not multiplicatively commutative, since b
d
≠
d
b. We have all one-sided ordered ideals of S which are as follows: (20)
a
,
a
,
c
,
a
,
c
,
e
,
a
,
b
,
c
,
e
,
a
,
c
,
d
,
e
,
S
.It is not difficult to check that all of them are ordered ideals of S. This shows that S is an ordered duo-semiring.Lemma 31.
LetS be an ordered semiring. Then the following conditions are equivalent: (i)
S is an ordered duo-semiring;
(ii)
R
(
A
)
=
L
(
A
) for each A
⊆
S;
(iii)
R
(
a
)
=
L
(
a
) for each a
∈
S.Proof.
(
i
)
⇒
(
ii
) and (
ii
)
⇒
(
iii
) are obvious.
(
iii
)
⇒
(
i
): let L be a left ordered ideal of S and let x
∈
L
,
s
∈
S. By assumption, we have x
s
∈
R
(
x
)
S
⊆
R
(
x
)
=
L
(
x
)
⊆
L
(
L
)
=
L. It follows that L is a right ordered ideal of S. Similarly, we have that every right ordered ideal of S is a left ordered ideal of S. Hence, S is an ordered duo-semiring.Theorem 32.
LetS be an ordered duo-semiring. Then S is regular if and only if (
Σ
Q
1
Q
2
]
=
Q
1
∩
Q
2 for each two ordered quasi-ideals Q
1 and Q
2 of S.Proof.
Assume thatS is a regular ordered semiring. Let Q
1 and Q
2 be ordered quasi-ideals of S. By Theorem 21, Q
1 and Q
2 can be written in the forms (21)
Q
1
=
R
1
∩
L
1
,
Q
2
=
R
2
∩
L
2for some R
1
,
R
2 and L
1
,
L
2 which are right ordered ideals and left ordered ideals of S, respectively. Since S is an ordered duo-semiring, R
1
,
R
2
,
L
1, and L
2 are ordered ideals of S. It follows that Q
1 and Q
2 are ordered ideals of S. By Theorem 26, we have (
Σ
Q
1
Q
2
]
=
Q
1
∩
Q
2.
Conversely, assume that(
Σ
Q
1
Q
2
]
=
Q
1
∩
Q
2 for each two ordered quasi-ideals Q
1 and Q
2 of S. Let A
⊆
S. By assumption, A
⊆
Q
(
A
)
∩
Q
(
A
)
=
(
Σ
Q
(
A
)
Q
(
A
)
]. By Theorem 13, Remark 1, and Lemmas 2 and 3, we have (22)
A
⊆
Σ
Σ
A
+
Σ
S
A
∩
Σ
A
S
Σ
A
+
Σ
S
A
∩
Σ
A
S
⊆
Σ
Σ
A
+
Σ
A
S
Σ
A
+
Σ
S
A
⊆
Σ
Σ
A
+
Σ
A
S
Σ
A
+
Σ
S
A
⊆
Σ
Σ
A
+
Σ
A
S
Σ
A
+
Σ
S
A
⊆
Σ
Σ
A
Σ
A
+
Σ
S
A
+
Σ
A
S
Σ
A
+
Σ
S
A
⊆
Σ
Σ
A
2
+
Σ
A
S
A
+
Σ
A
S
A
+
Σ
A
S
S
A
⊆
Σ
Σ
A
2
+
Σ
A
S
A
=
Σ
Σ
A
2
+
Σ
Σ
A
S
A
=
Σ
A
2
+
Σ
A
S
A
⊆
Σ
A
S
A
+
Σ
A
S
A
⊆
Σ
A
S
A
.By Lemma 20, S is a regular ordered semiring.Theorem 33.
LetS be an ordered duo-semiring. Then the following conditions are equivalent: (i)
S is regular;
(ii)
(
Σ
L
1
L
2
]
=
L
1
∩
L
2 and (
Σ
R
1
R
2
]
=
R
1
∩
R
2 for each two left ordered ideals L
1
,
L
2 and right ordered ideals R
1
,
R
2 of S;
(iii)
(
Σ
R
L
]
=
R
∩
L
=
(
Σ
L
R
], for each right ordered ideal R and left ordered ideal L of S.Proof.
It is obvious by Theorem26.Definition 34.
LetS be an ordered semiring. Then an element a
∈
S is said to beleft regular (right regular) if a
≤
x
a
2
(
a
≤
a
2
x
) for some x
∈
S. An ordered semiring S is said to beleft regular (right regular) if every element a
∈
S is left regular (right regular).Example 35.
LetS
=
{
a
,
b
,
c
,
d
,
e
,
f
}. Define binary operations + and · on S by the following equations: (23)
+
a
b
c
d
e
f
a
a
b
c
d
e
f
b
b
b
b
b
b
b
c
c
c
c
c
c
c
d
d
d
d
d
d
d
e
e
e
e
e
e
e
f
f
f
f
f
f
f
,
·
a
b
c
d
e
f
a
a
a
a
a
a
a
b
a
b
b
b
b
b
c
a
b
b
b
b
b
d
a
b
b
d
b
d
e
a
e
e
e
e
e
f
a
e
e
f
e
f
.Then (
S
,
+
,
·
) is a semiring with an absorbing zero a. Define a binary relation ≤ on S by (24)
≤
≔
a
,
a
,
b
,
b
,
c
,
c
,
d
,
d
,
e
,
e
,
f
,
f
,
b
,
c
,
b
,
d
,
b
,
e
,
b
,
f
,
c
,
d
,
c
,
e
,
c
,
f
,
d
,
f
,
e
,
f
.We give the covering relation “≺” and the figure of S: (25)
≺
≔
b
,
c
,
c
,
d
,
c
,
e
,
d
,
f
,
e
,
f
.
Now,(
S
,
+
,
·
,
≤
) is an ordered semiring. Clearly, a
,
b
,
d
,
e, and f are left regular. We consider c
≤
e
c
2
=
e
b
=
e. This implies that S is left regular. Since there does not exist x
∈
S such that c
≤
c
x
c, S is not regular.Example 36.
Consider the ordered semiringS
=
(
N
∪
{
0
}
,
m
a
x
,
m
i
n
,
≤
) where ≤ is the natural order relation on numbers. Since n
≤
m
i
n
{
n
,
n
,
n
} for any n
∈
N, we get S a regular, left regular, and right regular ordered semiring.The following lemmas can be easily proved using Definition34.Lemma 37.
LetS be an ordered semiring. Then the following statements are equivalent: (i)
S is left regular;
(ii)
A
⊆
(
Σ
S
A
2
] for each A
⊆
S;
(iii)
a
∈
(
S
a
2
] for each a
∈
S.Lemma 38.
LetS be an ordered semiring. Then the following statements are equivalent: (i)
S is right regular;
(ii)
A
⊆
(
Σ
A
2
S
] for each A
⊆
S;
(iii)
a
∈
(
a
2
S
] for each a
∈
S.Definition 39.
LetT be a nonempty subset of an ordered semiring S. Then T is said to besemiprime if for any a
∈
S, if a
2
∈
T, then a
∈
T.We note that a nonempty subsetT of S is semiprime if and only if, for any ∅
≠
A
⊆
S, A
2
⊆
T implies A
⊆
T. Because, if T is semiprime, ∅
≠
A
⊆
T, and a
∈
A then a
2
∈
T and so a
∈
T; that is, A
⊆
T.Theorem 40.
LetS be an ordered semiring. Then S is left regular and right regular if and only if every ordered quasi-ideal of S is semiprime.Proof.
LetQ be an ordered quasi-ideal of S. Let A be a nonempty subset of S such that A
2
⊆
Q. Since S is left regular and by Lemma 37, we have A
⊆
(
Σ
S
A
2
]. Since S is right regular and by Lemma 38, we have A
⊆
(
Σ
A
2
S
]. Hence, A
⊆
(
Σ
S
A
2
]
∩
(
Σ
A
2
S
]
⊆
(
Σ
S
Q
]
∩
(
Σ
Q
S
]
⊆
Q. Therefore, Q is semiprime.
Conversely, assume that every ordered quasi-ideal ofS is semiprime. Let A
⊆
S. By Theorem 13, we have Q
(
A
2
)
=
(
Σ
A
2
+
(
(
Σ
S
A
2
]
∩
(
Σ
A
2
S
]
)
]. Since A
2
⊆
Q
(
A
2
) is semiprime, A
⊆
Q
(
A
2
)
=
(
Σ
A
2
+
(
(
Σ
S
A
2
]
∩
(
Σ
A
2
S
]
)
]. Then we obtain (26)
A
⊆
Σ
A
2
+
Σ
S
A
2
∩
Σ
A
2
S
⊆
Σ
A
2
+
Σ
S
A
2
⊆
Σ
A
2
+
Σ
S
A
2
,
A
⊆
Σ
A
2
+
Σ
S
A
2
∩
Σ
A
2
S
⊆
Σ
A
2
+
Σ
A
2
S
⊆
Σ
A
2
+
Σ
A
2
S
.In case A
⊆
(
Σ
A
2
+
Σ
S
A
2
], we get (27)
Σ
A
2
⊆
Σ
A
Σ
A
2
+
Σ
S
A
2
⊆
Σ
A
Σ
A
2
+
A
Σ
S
A
2
⊆
Σ
Σ
A
3
+
Σ
A
S
A
2
⊆
Σ
Σ
A
3
+
Σ
S
A
2
⊆
Σ
Σ
A
3
+
Σ
Σ
S
A
2
⊆
Σ
A
3
+
Σ
S
A
2
⊆
Σ
S
A
2
+
Σ
S
A
2
=
Σ
S
A
2and so A
⊆
(
Σ
A
2
+
Σ
S
A
2
]
⊆
(
(
Σ
S
A
2
]
+
Σ
S
A
2
]
⊆
(
Σ
S
A
2
]. By Lemma 37, S is left regular. Similarly, in case A
⊆
(
Σ
A
2
+
Σ
A
2
S
], we get A
⊆
(
A
2
S
]. By Lemma 38, S is right regular.



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*Source: 4272451-2016-01-18.xml*