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.
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:
if a≤b then a+x≤b+x and x+a≤x+b;
if a≤b then ax≤bx and xa≤xb.
An ordered semiring S is said to be additively commutative if a+b=b+a for all a,b∈S. An element 0∈S is said to be an absorbing zero if 0a=0=a0 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 subsets A,B of S and a∈S, we denote (1)A=x∈S∣x≤a for some a∈A,AB=ab∈S∣a∈A,b∈B,ΣA=∑i∈Iai∈S∣ai∈A and I is a finite subset of N,ΣAB=∑i∈Iaibi∈S∣ai∈A,bi∈B and I is a finite subset of N,Na=Σa.
Now, we mention some properties of finite sums on an ordered semiring.
Remark 1.
For any subsets A,B of S, the following statements hold:
Σ(A]⊆(ΣA];
Σ(ΣA)=ΣA;
A(ΣB)⊆ΣAB and (ΣA)B⊆ΣAB;
Σ(AΣB)⊆ΣAB and Σ(ΣA)B⊆ΣAB;
Σ(A+B)=ΣA+ΣB.
We note that, for any A⊆S, ΣA=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.
Let A,B,C be subsets of an ordered semiring S. Then the following statements hold:
A⊆(A] and ((A]]=(A];
If A⊆B then (A]⊆(B];
A(B]⊆(A](B]⊆(AB] and (A]B⊆(A](B]⊆(AB];
A+(B]⊆(A]+(B]⊆(A+B] and (A]+B⊆(A]+(B]⊆(A+B];
A(B+C]⊆(AB+AC] and (A+B]C⊆(AC+BC];
(A∪B]=(A]∪(B];
(A∩B]⊆(A]∩(B].
In (vii) of the above lemma, we have (A∩B]=(A]∩(B] when (A]=A and (B]=B.
Lemma 3.
Let S be an ordered semiring and ∅≠A⊆S. If A⊆(ΣA2+ΣASA] then ΣA2⊆(ΣASA].
Proof.
Assume that A⊆(ΣA2+ΣASA]. Then (2)ΣA2⊆ΣΣA2+ΣASAA⊆ΣΣA2A+ΣASAA⊆ΣΣA3+ΣASA=ΣΣA3+ΣΣASA=ΣA3+ΣASA=ΣAAA+ΣASA⊆ΣASA+ΣASA=ΣASA.
Definition 4 (see [10]).
Let S be an ordered semiring and ∅≠A⊆S. Then A is said to be a left ordered ideal (right ordered ideal) if the following conditions are satisfied.
A is a left ideal (right ideal) of S.
If x≤a for some a∈A then x∈A (i.e., A=(A]).
We call A an ordered 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=maxa,b,a⊙b=maxa+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.
Let A be a nonempty subset of an ordered semiring S. Then
(ΣSA] is a left ordered ideal of S;
(ΣAS] is a right ordered ideal of S;
(ΣSAS] is an ordered ideal of S.
Proof.
(i) Let x,y∈(ΣSA]. Then x≤x′ and y≤y′ for some x′,y′∈ΣSA. It is clear that x+y≤x′+y′∈ΣSA, and so x+y∈(ΣSA]. By Remark 1 and Lemma 2, we obtain S(ΣSA]⊆(SΣSA]⊆(ΣSSA]⊆(ΣSA]. We have ((ΣSA]]=(ΣSA]. Hence, (ΣSA] is a left ordered ideal of S.
(ii) and (iii) can be proved similar to (i).
Corollary 7.
Let S be an ordered semiring. Then, for any a∈S,
(Sa] is a left ordered ideal of S;
(aS] is a right ordered ideal of S;
(ΣSaS] is an ordered ideal of S.
Let A 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.
Let A be a nonempty subset of an ordered semiring S. Then
L(A)=(ΣA+ΣSA];
R(A)=(ΣA+ΣAS];
I(A)=(ΣA+ΣSA+ΣAS+ΣSAS].
Proof.
(i) Since S has an absorbing zero, we have, for every a∈A, a=a+0∈ΣA+ΣSA⊆(ΣA+ΣSA]. Hence, A⊆(ΣA+ΣSA]. Let x,y∈(ΣA+ΣSA]. Then x≤x′ and y≤y′ for some x′,y′∈ΣA+ΣSA. Thus x′=a1+b1 and y′=a2+b2 for some a1,a2∈ΣA and b1,b2∈ΣSA. It is easy to show that a1+a2∈ΣA and b1+b2∈ΣSA. It follows that x+y≤x′+y′∈ΣA+ΣSA, and so x+y∈(ΣA+ΣSA]. By Remark 1 and Lemma 2, we obtain (4)SΣA+ΣSA⊆SΣA+ΣSA⊆SΣA+SΣSA⊆ΣSA+ΣSSA⊆ΣSA+ΣSA=ΣSA⊆ΣA+ΣSA.Since ((ΣA+ΣSA]]=(ΣA+ΣSA], L is a left ordered ideal of S. Let K be any left ordered ideal of S containing A. It turns out ΣA⊆K and ΣSA⊆K, so ΣA+ΣSA⊆K. It follows that (ΣA+ΣSA]⊆(K]=K. Therefore, (ΣA+ΣSA] is the smallest left ordered ideal of S containing A.
(ii) and (iii) can be proved similar to (i).
As a special case of Lemma 8, if A={a} then we have the following corollary.
Corollary 9.
Let S be an ordered semiring. Then, for any a∈S,
L(a)=(Na+Sa];
R(a)=(Na+aS];
I(a)=(Na+Sa+aS+ΣSaS].
An element e of an ordered semiring S is said to be an identity if ea=a=ae 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 if S has an identity, then L(A)=(ΣSA],R(A)=(ΣAS] and I(A)=(ΣSAS] for any A⊆S. In particular case, we have L(a)=(Sa],R(a)=(aS] and I(a)=(ΣSaS] 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 an ordered quasi-ideal of S if the following conditions are satisfied:
(ΣSQ]∩(ΣQS]⊆Q;
if x≤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 semiring S is an ordered quasi-ideal of S. Moreover, each ordered quasi-ideal of S is a subsemiring of S; indeed, QQ⊆(QQ]⊆(SQ]∩(QS]⊆(ΣSQ]∩(ΣQS]⊆Q.
Example 11.
Let S={a,b,c,d}. Define binary operations + and · on S by the following equations: (5)+abcdaabcdbbbbbccbcdddbdd,·abcdaaaaababbbcacccdabbb.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 (ΣSQ]∩(ΣQS]={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 SQ={a,b,c}⊈Q.
Lemma 12.
Let S be an ordered semiring and let {Qi∣i∈I} be a family of ordered quasi-ideals of S. Then ⋂i∈IQi is an ordered quasi-ideal of S.
Let A be a nonempty subset of an ordered semiring S. We denote Q(A) the smallest ordered quasi-ideal of S containing A.
Theorem 13.
Let S be an ordered semiring and let A be a nonempty subset of S. Then Q(A)=(ΣA+((ΣSA]∩(ΣAS])].
Proof.
Let Q=(ΣA+((ΣSA]∩(ΣAS])]. Since S has an absorbing zero, we have a=a+0∈ΣA+((ΣSA]∩(ΣAS])⊆Q for every a∈A. Hence, A⊆Q. Let x,y∈Q. Then x≤x′ and y≤y′ for some x′,y′∈ΣA+((ΣSA]∩(ΣAS]). Thus x′=a1+b1 and y′=a2+b2 for some a1,a2∈ΣA and b1,b2∈(ΣSA]∩(ΣAS]. Clearly, a1+a2∈ΣA and b1+b2∈(ΣSA]∩(ΣAS]. It follows that x+y≤x′+y′∈ΣA+((ΣSA]∩(ΣAS]), and so x+y∈Q. By Remark 1 and Lemma 2, we obtain (8)ΣSQ∩ΣQS⊆ΣSQ=ΣSΣA+ΣSA∩ΣAS⊆ΣSΣA+ΣSA⊆ΣSΣA+SΣSA⊆ΣΣSA+ΣSSA⊆ΣΣSA+ΣSSA⊆ΣΣSA+ΣSA⊆ΣΣSA⊆ΣSA=ΣSA.Similarly, we can show that (ΣSQ]∩(ΣQS]⊆(ΣAS]. Thus (ΣSQ]∩(ΣQS]⊆(ΣSA]∩(ΣAS]⊆ΣA+((ΣSA]∩(ΣAS])⊆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 (ΣSA]∩(ΣAS]⊆(ΣSK]∩(ΣKS]⊆K. So ΣA+((ΣSA]∩(ΣAS])⊆K. Hence, Q=(ΣA+((ΣSA]∩(ΣAS])]⊆(K]=K. Therefore, Q is the smallest ordered quasi-ideal of S containing A.
As a special case of Theorem 13, if A={a} then we have the following corollary.
Corollary 14.
Let S be an ordered semiring. Then Q(a)=(Na+((Sa]∩(aS])] for any a∈S.
If S has an identity, then it is easy to check that Q(A)=(ΣSA]∩(ΣAS] for any A⊆S. In particular case, we have Q(a)=(Sa]∩(aS] for any a∈S.
Let Q(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 P1,P2∈Q(S), (9)P1∧P2=P1∩P2,P1∨P2=QP1∪P2.Then we obtain the following theorem.
Theorem 15.
Let S be an ordered semiring. Then (Q(S),∧,∨) is a complete lattice.
Theorem 16.
The intersection of a left ordered ideal L and a right ordered ideal R of an ordered semiring S is an ordered quasi-ideal of S.
Proof.
It is easy to show that L∩R is a subsemigroup of (S,+). By Remark 1 and Lemma 2, we obtain (10)ΣSL∩R∩ΣL∩RS⊆ΣSL∩R=ΣSL∩SR⊆ΣSL⊆L,ΣSL∩R∩ΣL∩RS⊆ΣL∩RS=ΣLS∩RS⊆ΣRS⊆R.Hence, (ΣS(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 Theorem 16 is not true as Example 2.1 page 8 in [2] given by A. H. Clifford.
Corollary 17.
Let S be an ordered semiring. Then the following statements hold.
(ΣSA]∩(ΣAS] is an ordered quasi-ideal of S, for any A⊆S.
(Sa]∩(aS] is an ordered quasi-ideal of S, for any a∈S.
Proof.
(i) By Lemma 6, we have (ΣSA] and (ΣAS] a left and a right ordered ideal of S, respectively. Then by Theorem 16, we have that (ΣSA]∩(ΣAS] is an ordered quasi-ideal of S.
(ii) It is a particular case of (i).
Now, we will show that the converse of Theorem 16 is true if S contains an identity as the following theorem.
Theorem 18.
Let S 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 that S has an identity. Let Q be an ordered quasi-ideal of S. Then R(Q)=(ΣQS] and L(Q)=(ΣSQ]. We obtain Q⊆R(Q)∩L(Q) and R(Q)∩L(Q)=(ΣQS]∩(ΣSQ]⊆Q. Hence, Q=R(Q)∩L(Q).
4. Regular Ordered Semirings
In this section, we show that in regular ordered semirings the converse of Theorem 16 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 element a of an ordered semiring S is said to be regular if a≤axa for some x∈S. An ordered semiring S is said to be regular if every element a∈S is regular.
The following lemma is characterizations of regular ordered semiring which directly follows Definition 19.
Lemma 20.
Let S be an ordered semiring. Then the following statements are equivalent:
S is regular;
A⊆(ΣASA] for each A⊆S;
a∈(aSa] for any a∈S.
Now, we will show that the converse of Theorem 16 is true in regular ordered semirings.
Theorem 21.
Every ordered quasi-ideal of a regular ordered semiring S can be written in the form Q=R∩L for some right ordered ideal R and left ordered ideal L of S.
Proof.
Let Q be an ordered quasi-ideal of S. By Lemma 8, we have R(Q)=(ΣQ+ΣQS] and L(Q)=(ΣQ+ΣSQ]. Now, Q⊆R(Q)∩L(Q). Let q∈Q. Since S is regular, there exists x∈S such that q≤qxq∈ΣQS. So Q⊆(ΣQS]. Since Q+Q⊆Q, ΣQ=Q. It follows that (11)ΣQS⊆ΣQ+ΣQS=Q+ΣQS⊆ΣQS+ΣQS⊆ΣQS.This implies that R(Q)=(ΣQS]. Similarly, we can show that L(Q)=(ΣSQ]. Hence, R(Q)∩L(Q)=(ΣQS]∩(ΣSQ]⊆Q. Therefore, Q=R(Q)∩L(Q).
Definition 22.
Let (S,+,·,≤) be an ordered semiring. A subsemigroup (B,+) of (S,+) is said to be an ordered bi-ideal of S if the following conditions hold:
BSB⊆B;
if x≤b for some b∈B, then x∈B (i.e., B=(B]).
We note that condition (1) of Definition 22 is equivalent to ΣBSB⊆B.
Theorem 23.
Every ordered quasi-ideal of an ordered semiring S is an ordered bi-ideal of S.
Proof.
Let Q be an ordered quasi-ideal of S. Then ΣQSQ⊆ΣQS⊆(ΣQS] and QSQ⊆ΣSQ⊆(ΣSQ]. So, ΣQSQ⊆(ΣSQ]∩(ΣQS]⊆Q. Hence, Q is an ordered bi-ideal of S.
The converse of Theorem 23 is not generally true as the following example.
Example 24.
Let S={a,b,c,d,e}. Define binary operations + and · by the following equations: (12)+abcdeaabcdebbbdddccddddddddddeeddde,·abcdeaaaaaabaaaaacaabbbdaabbbeaabbb.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≰dxd 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 (ΣSB]∩(ΣBS]={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.
Let S be a regular ordered semiring. Then ordered bi-ideals and ordered quasi-ideals coincide in S.
Proof.
By Theorem 23, 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∈(ΣSB]∩(ΣBS]. By Lemma 20, Remark 1, and Lemma 2, we obtain a∈(aSa]⊆((ΣBS]S(ΣSB]]⊆((ΣBSS](ΣSB]]⊆((ΣBS)(ΣSB)]⊆(Σ(BS(ΣSB))]⊆(Σ(ΣBSSB)]⊆(ΣBSB]⊆B. Hence, B is an ordered quasi-ideal of S.
Theorem 26.
Let S be an ordered semiring. Then the following statements are equivalent:
S is regular;
(ΣRL]=R∩L for every right ordered ideal R and left ordered ideal L of S;
B=(ΣBSB] for each ordered bi-ideal B of S;
Q=(ΣQSQ] 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, (ΣRL]⊆(ΣR]=R and (ΣRL]⊆(ΣL]=L. Hence, (ΣRL]⊆R∩L. Let a∈R∩L. Since S is regular, a≤axa for some x∈S. Since a∈R, xa∈L. It follows that a≤a(xa)∈RL. This means a∈(RL]⊆(ΣRL]. Therefore, (ΣRL]=R∩L.
(ii)⇒(iii): assume that (ii) holds. Let B be an ordered bi-ideal of S. It is clear that (ΣBSB]⊆B. By assumption, B⊆R(B)∩L(B)=(ΣR(B)L(B)]. By Lemma 8, Remark 1, and Lemmas 2 and 3, we have (15)B⊆ΣB+ΣBS∩ΣB+ΣSB=ΣΣB+ΣBSΣB+ΣSB⊆ΣΣB+ΣBSΣB+ΣSB⊆ΣΣBΣB+ΣSB+ΣBSΣB+ΣSB⊆ΣΣB2+ΣBSB+ΣBSB+ΣBSSB⊆ΣΣB2+ΣBSB=ΣΣB2+ΣΣBSB=ΣB2+ΣBSB⊆ΣBSB+ΣBSB⊆ΣBSB.
(iii)⇒(iv): it follows from Theorem 23.
(iv)⇒(i): let a∈S. Then Q(a)=(Q(a)SQ(a)]. By Corollary 14, Remark 1, and Lemma 2, we have (16)a∈Na+Sa∩aS=ΣNa+Sa∩aSSNa+Sa∩aS⊆ΣNa+aSSNa+Sa⊆ΣNa+aSSNa+Sa⊆ΣNa+aSSNa+Sa⊆ΣaSNa+Sa⊆ΣaSNa+Sa⊆ΣaSa=aSa.By Lemma 20, S is regular.
Theorem 27.
Let S be a regular ordered semiring. Then the following statements hold:
every ordered quasi-ideal Q of S can be written in the form Q=R∩L=(RL] for some right ordered ideal R and left ordered ideal L of S;
(Q2]=(Q3] for each ordered quasi-ideal Q of S.
Proof.
(i) It is obvious by Theorems 21 and 26.
(ii) Let Q be an ordered quasi-ideal of S. Clearly, ((QQ)Q]⊆(QQ]. Let x∈(QQ]. Then x≤q1q2 for some q1,q2∈Q. Since S is regular, there exists s∈S such that x≤q1q2≤(q1q2)s(q1q2)∈QQSQQ. Hence, x∈(Q(QSQ)Q]⊆(QQQ]. Therefore, (Q2]=(Q3].
Theorem 28.
Let S be an ordered semiring. Then S is regular if and only if B∩I∩L⊆(BIL] for every ordered bi-ideal B, every ordered ideal I, and every left ordered ideal L of S.
Proof.
Let B,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≤axa≤axaxaxa∈BIL. Hence, B∩I∩L⊆(BIL].
Conversely, assume that B∩I∩L⊆(BIL] 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⊆(RSL]⊆(RL]⊆(ΣRL] for every right ordered ideal R and left ordered ideal L of S. On the other hand, we have (ΣRL]⊆R∩L. Hence, (ΣRL]=R∩L. By Theorem 26, S is regular.
Definition 29.
An ordered semiring S is said to be an ordered 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.
Let S={a,b,c,d,e}. Define binary operations + and · by the following equations: (17)+abcdeaabcdebbccccccccccddcccceecccc,·abcdeaaaaaabaececcaccccdacceceacccc.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 bd≠db. 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.
Let S be an ordered semiring. Then the following conditions are equivalent:
S is an ordered duo-semiring;
R(A)=L(A) for each A⊆S;
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 xs∈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.
Let S be an ordered duo-semiring. Then S is regular if and only if (ΣQ1Q2]=Q1∩Q2 for each two ordered quasi-ideals Q1 and Q2 of S.
Proof.
Assume that S is a regular ordered semiring. Let Q1 and Q2 be ordered quasi-ideals of S. By Theorem 21, Q1 and Q2 can be written in the forms (21)Q1=R1∩L1,Q2=R2∩L2for some R1,R2 and L1,L2 which are right ordered ideals and left ordered ideals of S, respectively. Since S is an ordered duo-semiring, R1,R2,L1, and L2 are ordered ideals of S. It follows that Q1 and Q2 are ordered ideals of S. By Theorem 26, we have (ΣQ1Q2]=Q1∩Q2.
Conversely, assume that (ΣQ1Q2]=Q1∩Q2 for each two ordered quasi-ideals Q1 and Q2 of S. Let A⊆S. By assumption, A⊆Q(A)∩Q(A)=(ΣQ(A)Q(A)]. By Theorem 13, Remark 1, and Lemmas 2 and 3, we have (22)A⊆ΣΣA+ΣSA∩ΣASΣA+ΣSA∩ΣAS⊆ΣΣA+ΣASΣA+ΣSA⊆ΣΣA+ΣASΣA+ΣSA⊆ΣΣA+ΣASΣA+ΣSA⊆ΣΣAΣA+ΣSA+ΣASΣA+ΣSA⊆ΣΣA2+ΣASA+ΣASA+ΣASSA⊆ΣΣA2+ΣASA=ΣΣA2+ΣΣASA=ΣA2+ΣASA⊆ΣASA+ΣASA⊆ΣASA.By Lemma 20, S is a regular ordered semiring.
Theorem 33.
Let S be an ordered duo-semiring. Then the following conditions are equivalent:
S is regular;
(ΣL1L2]=L1∩L2 and (ΣR1R2]=R1∩R2 for each two left ordered ideals L1,L2 and right ordered ideals R1,R2 of S;
(ΣRL]=R∩L=(ΣLR], for each right ordered ideal R and left ordered ideal L of S.
Proof.
It is obvious by Theorem 26.
Definition 34.
Let S be an ordered semiring. Then an element a∈S is said to be left regular (right regular) if a≤xa2(a≤a2x) for some x∈S. An ordered semiring S is said to be left regular (right regular) if every element a∈S is left regular (right regular).
Example 35.
Let S={a,b,c,d,e,f}. Define binary operations + and · on S by the following equations: (23)+abcdefaabcdefbbbbbbbcccccccdddddddeeeeeeefffffff,·abcdefaaaaaaababbbbbcabbbbbdabbdbdeaeeeeefaeefef.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≤ec2=eb=e. This implies that S is left regular. Since there does not exist x∈S such that c≤cxc, S is not regular.
Example 36.
Consider the ordered semiring S=(N∪{0},max,min,≤) where ≤ is the natural order relation on numbers. Since n≤min{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 Definition 34.
Lemma 37.
Let S be an ordered semiring. Then the following statements are equivalent:
S is left regular;
A⊆(ΣSA2] for each A⊆S;
a∈(Sa2] for each a∈S.
Lemma 38.
Let S be an ordered semiring. Then the following statements are equivalent:
S is right regular;
A⊆(ΣA2S] for each A⊆S;
a∈(a2S] for each a∈S.
Definition 39.
Let T be a nonempty subset of an ordered semiring S. Then T is said to be semiprime if for any a∈S, if a2∈T, then a∈T.
We note that a nonempty subset T of S is semiprime if and only if, for any ∅≠A⊆S, A2⊆T implies A⊆T. Because, if T is semiprime, ∅≠A⊆T, and a∈A then a2∈T and so a∈T; that is, A⊆T.
Theorem 40.
Let S 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.
Let Q be an ordered quasi-ideal of S. Let A be a nonempty subset of S such that A2⊆Q. Since S is left regular and by Lemma 37, we have A⊆(ΣSA2]. Since S is right regular and by Lemma 38, we have A⊆(ΣA2S]. Hence, A⊆(ΣSA2]∩(ΣA2S]⊆(ΣSQ]∩(ΣQS]⊆Q. Therefore, Q is semiprime.
Conversely, assume that every ordered quasi-ideal of S is semiprime. Let A⊆S. By Theorem 13, we have Q(A2)=(ΣA2+((ΣSA2]∩(ΣA2S])]. Since A2⊆Q(A2) is semiprime, A⊆Q(A2)=(ΣA2+((ΣSA2]∩(ΣA2S])]. Then we obtain (26)A⊆ΣA2+ΣSA2∩ΣA2S⊆ΣA2+ΣSA2⊆ΣA2+ΣSA2,A⊆ΣA2+ΣSA2∩ΣA2S⊆ΣA2+ΣA2S⊆ΣA2+ΣA2S.In case A⊆(ΣA2+ΣSA2], we get (27)ΣA2⊆ΣAΣA2+ΣSA2⊆ΣAΣA2+AΣSA2⊆ΣΣA3+ΣASA2⊆ΣΣA3+ΣSA2⊆ΣΣA3+ΣΣSA2⊆ΣA3+ΣSA2⊆ΣSA2+ΣSA2=ΣSA2and so A⊆(ΣA2+ΣSA2]⊆((ΣSA2]+ΣSA2]⊆(ΣSA2]. By Lemma 37, S is left regular. Similarly, in case A⊆(ΣA2+ΣA2S], we get A⊆(A2S]. By Lemma 38, S is right regular.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
This work has been supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
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