Citation: Ze Gu, Xiang-Yun Xie, Jian Tang. On C-ideals and the basis of an ordered semigroup[J]. AIMS Mathematics, 2020, 5(4): 3783-3790. doi: 10.3934/math.2020245
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Ideal theory play an important role in studying ordered semigroups. The concepts of ideals, bi-ideals, quasi-ideals, weakly prime ideals and prime ideals in ordered semigroups have been introduced by N. Kehayopulu in [1,2,3]. Moreover, fuzzy ideals, fuzzy bi-ideals, fuzzy quasi-ideals, weakly prime fuzzy ideals and prime fuzzy ideals in ordered semigroups have been studied in [4,5,6,7,8,9,10].
The concept of C-ideals in semigroups has been introduced by I. Fabrici in [11]. The second author Xie of this paper extended the concept to ordered semigroups in [12]. Xie has studied some basic properties of C-ideals in ordered semigroups in [13] and characterized the ordered semigroups in which every proper ideal is a C-ideal in [12]. Mao, Xu and Lian have further studied some properties of C-ideals in ordered semigroups in [14]. Motivated by the previous work, in this paper, we study ordered semigroups containing the greatest ideal and give the conditions of the greatest ideal to be a C-ideal in an ordered semigroup. Moreover, we introduce the concept of a basis of an ordered semigroup and study the relationship between the greatest C-ideal and the basis in an ordered semigroup. If the order relation is trivial in an ordered semigroup S, then S is a semigroup. Consequently, all the results in this paper are true for semigroups.
We call (S,⋅,≤) an ordered semigroup if (S,⋅) is a semigroup, (S,≤) is an ordered set and it satisfies:
a≤b⇒ax≤bx,xa≤xb(∀a,b,x∈S).([15]) |
We will just use S to denote an ordered semigroup when the operation and order are understood. A nonempty subset I of an ordered semigroup S is called an ideal if I satisfies: 1) IS,SI⊆I; 2) ∀a∈S,a≤b∈I⇒a∈I. An ideal I of S is called proper if I≠S. A proper ideal M of S is called the greatest ideal if every proper ideal is contained in M. A proper ideal M of S is called a maximal ideal if whenever there exists an ideal N of S such that M⊆N, then N=S. An ordered semigroup S is called simple if S contains no proper ideals. Let H be a nonempty subset of S. Denote
(H]:={x∈S∣∃h∈H,x≤h};[H):={x∈S∣∃h∈H,h≤x}. |
If H has only one element a, then we denote (H] and [H) by (a] and [a) respectively. For any x∈S, denote by I(x) the ideal generated by x. Then we have I(x)=(x∪Sx∪xS∪SxS] (see [1,2]).
Green's relation J:={(x,y)∈S×S∣I(x)=I(y)} on an ordered semigroup S was introduced by N. Kehayopulu in [16]. It is easy to see that the relation J is an equivalence on S. For any x∈S, denote by Ix the J-class containing x. We define a relation "≼" on the set of all J-classes in S as below:
Ix≼Iy⇔I(x)⊆I(y)(∀x,y∈S). |
It is routine to verify that "≼" is an order relation. A J-class Ix of S is called maximal if there is no other J-class Iy such that Ix≼Iy. A J-class Ix of S is called the greatest J-class if other J-classes are all contained in Ix. From [12], we know that M is a maximal ideal of S if and only if S−M is a maximal J-class.
Let S be an ordered semigroup. A proper ideal M of S is called a C-ideal if M⊆(S(S−M)S]. We know from [14] the following basic properties of C-ideals.
Lemma 1. (1) (Theorem 5, [14]) If S is not simple, then S contains at least one C-ideal;
(2) (Theorem 1, [14]) If S contains two different proper ideals M1,M2 such that M1∪M2=S, then neither of them is a C-ideal of S;
(3) (Corollary 1, [14]) If S contains more than one maximal ideal, then none of them is a C-ideal of S;
(4) (Theorem 4, [14]) If S contains only one maximal ideal M and M is a C-ideal, then M is the greatest ideal of S;
(5) (Theorem 2, Theorem 3 [14]) If M1 and M2 are two C-ideals of S, then M1∪M2 and M1∩M2 are C-ideals of S.
(6) (Theorem 7, [14]) If S contains the greatest ideal M∗ and M∗ is a C-ideal, then every proper ideal of S is a C-ideal.
However, the following result given in [14] is incorrect, we next improve it.
Result 1. (Theorem 6, [14]) Let S be an ordered semigroup with an identity 1. If S contains the greatest ideal, denoted by M∗, then M∗ is a C-ideal or there exists a∈S−(S3] such that S−[a)⊆M∗.
Remark 1. If S has an identity 1, then S=(S3], i.e. S−(S3]=∅. Consequently, we next improve Result 1 by the following result.
Theorem 1. Let S be an ordered semigroup with an identity 1. Then every proper ideal of S is a C-ideal. In particular, if S contains the greatest ideal, denoted by M∗, then M∗ is a C-ideal.
Proof. Let M be a proper ideal of S. Then 1∉M. Suppose that 1∈M. Then S=S⋅1⊆SM⊆M, i.e. S=M, this is a contradiction. Thus 1∈S−M. It follows that (S(S−M)S]=S. Therefore M⊆(S(S−M)S], i.e. M is a C-ideal.
Since M∗ is a proper ideal of S, we have M∗ is a C-ideal.
In the following, we study ordered semigroups containing the greatest ideal and give the conditions of the greatest ideal to be a C-ideal in an ordered semigroup.
Lemma 2. If S≠(S2], then I(x)=(x] for any x∈S−(S2].
Proof. Let x∈S−(S2]. Suppose that I(x)≠(x]. Then there exists a∈S or b∈S or c,d∈S such that ax>x or xb>x or cxd>x. Since ax,xb,cxd∈S2, we have x∈(S2]. This is a contradiction.
Theorem 2. If S≠(S2], then S contains the greatest ideal if and only if S=(a] for some a∈S−(S2].
Proof. (⇒) Denote by M∗ the greatest ideal of S. Let a∈S−M∗. Since (S2]⊆M∗, a∈S−(S2]. Moreover, Ia=S−M∗ is the greatest J-class. Thus I(x)⊆I(a) for any x∈M∗ and so M∗⊆I(a). In addition, I(a)∪M∗=S. It follows that S=I(a)=(a] from Lemma 2.
(⇐) Let S=(a] for some a∈S−(S2]. Next we prove that S−{a} is an ideal of S. In fact: for any x,y∈S and b∈S−{a}, we have xb,by∈S−{a}. Otherwise, a=xb=by∈S2 which contradicts that a∈S−(S2]. Let c∈S−{a} and z∈S. If z≤c, then z∈S−{a}. Otherwise, z=a which contradicts that c<a. It follows that S−{a} is an ideal of S.
Finally, we show that S−{a} is the greatest. Let I be a proper ideal of S. Then a∉I and thus I⊆S−{a}.
Theorem 3. Let M∗ be the greatest ideal of S.
(1) If S=(S2], then M∗ is a C-ideal.
(2) If S≠(S2], then M∗=S−{a} for some a∈S−(S2].
Proof. (1) Since (S(S−M∗)S] is an ideal of S and M∗ is the greatest ideal of S, we have (S(S−M∗)S]=S or (S(S−M∗)S]⊆M∗. We distinguish three cases.
1) If (S(S−M∗)S]=S, then M∗⊆(S(S−M∗)S]. Hence M∗ is a C-ideal.
2) If (S(S−M∗)S]=M∗, then M∗ is a C-ideal clearly.
3) If (S(S−M∗)S]⊂M∗, then (S3]=(S(S−M∗)S∪SM∗S]⊆(S(S−M∗)S]∪(SM∗S]⊂M∗∪(M∗]=M∗∪M∗=M∗⊂S. Since S=(S2], we have S=(S3] which contradicts (S3]⊂S.
(2) It can be easily obtained from the proof of Theorem 2.
Corollary 1. Let M∗ be the greatest ideal of S. If M∗ is a C-ideal, then (S2]=(S3]. In particular, if S≠(S2], then M∗ is a C-ideal if and only if M∗=(S3].
Proof. If S=(S2], then (S2]=(S3] obviously. Next we consider the case of S≠(S2]. From Theorem 3, we know that M∗=S−{a} for some a∈S−(S2]. Since M∗ is a C-ideal and (S2] is a proper ideal, we have (S2]⊆M∗⊆(S(S−M∗)S]=(SaS]⊆(S3]⊆(S2]. It follows that M∗=(S2]=(S3].
Let S≠(S2]. We need only prove the sufficiency. We know that S=(a] for some a∈S−(S2] from Theorem 2 and M∗=S−{a} from Theorem 3(2). Since M∗=(S3], we have M∗=(S3]=(S(a]S]=(SaS]=(S(S−M∗)S]. Thus M∗ is a C-ideal.
In this section, we introduce the concept of a basis of an ordered semigroup and study the relation between the existence of the greatest C-ideal and the existence of a basis in an ordered semigroup.
Definition 1. A nonempty subset A of an ordered semigroup S is called a basis of S if A satisfies the following conditions:
1) (A∪SA∪AS∪SAS]=S;
2) There is no proper subset B of A such that (B∪SB∪BS∪SBS]=S.
Example 1. We consider the order semigroup S={a,b,c,d} with the multiplication "⋅" and the order relation "≤" below:
![]() |
≤:={(a,a),(a,b),(a,c),(a,d),(b,b),(c,c),(d,d)}. |
Let A={c,d}. Then (A∪SA∪AS∪SAS]=S. However, d∉(c∪Sc∪cS∪ScS] and c∉(d∪Sd∪dS∪SdS]. Thus A is a basis of S.
Theorem 4. Let A be a nonempty subset of S. Then A is a basis of S if and only if A satisfies:
1) For any x∈S, there exists a∈A such that I(x)⊆I(a);
2) For any a1,a2∈A, if Ia1≼Ia2, then a1=a2.
Proof. (⇒) Obviously, condition 1) can be obtained from Definition 1. Next we prove condition 2).
If Ia1≼Ia2, then I(a1)⊆I(a2). Suppose that a1≠a2. Let B=A−{a1}. Then (A∪SA∪AS∪SAS]⊆(B∪SB∪BS∪SBS]. Therefore (B∪SB∪BS∪SBS]=S. This contradicts Definition 1.
(⇐) By the condition 1), we have
S=⋃x∈SI(x)⊆⋃a∈AI(a)=(A∪SA∪AS∪SAS]⊆S. |
Hence (A∪SA∪AS∪SAS]=S. Suppose that there exists B⊂A such that (B∪SB∪BS∪SBS]=S. Let a1∈A−B. Then there exists b1∈B or b2∈B or b3∈B or b4,b5∈B,s∈S such that a1≤b1 or a1≤sb2 or a1≤b3s or a1≤b4sb5. It implies that I(a1)⊆I(b1) or I(a1)⊆I(b2) or I(a1)⊆I(b3) or I(a1)⊆I(b4), i.e. Ia1≼Ib1 or Ia1≼Ib2 or Ia1≼Ib3 or Ia1≼Ib4. This contradicts 2).
Remark 2. By Theorem 4, we can see that if A is a basis of S, then every element of A belongs to some maximal J-class and there is only one element in A for every maximal J-class.
Proposition 1. Let {Mα∣α∈Λ} be a set of all maximal ideals of S, ˆM=⋂α∈ΛMα and ¯Mα=S−Mα. Then we have
1) S=(⋃α∈Λ¯Mα)∪ˆM;
2) If α≠β, then ¯Mα∩¯Mβ=∅;
3) If α≠β, then ¯Mα⊆Mβ;
4) If α≠β, then ¯Mα¯Mβ⊆ˆM, i.e. ˆM≠∅;
5) Let I be an ideal of S and I∩¯Mα≠∅, then ¯Mα⊆I.
Proof. 1) Since
ˆM=⋂α∈ΛMα=⋂α∈Λ(S−¯Mα)=S−⋃α∈Λ¯Mα, |
we have
S=(⋃α∈Λ¯Mα)∪ˆM. |
2) We know that ¯Mα is a maximal J-class for any α∈Λ. Thus ¯Mα∩¯Mβ=∅ when α≠β.
3) It can be obtained from 2) obviously.
4) Let α≠β,μα∈¯Mα,μβ∈¯Mβ and μ=μαμβ∉ˆM. By 1), we have S=(⋃α∈Λ¯Mα)∪ˆM. Thus there exists μγ∈¯Mγ such that μ=μγ.
i) If γ≠α, then ¯Mα⊆Mγ. Thus ¯Mα¯Mβ⊆Mγ. Hence μγ∈Mγ which contradicts μγ∈¯Mγ.
ii) If γ=α, then ¯Mβ⊆Mγ. It follows that ¯Mα¯Mβ⊆Mγ Thus μγ∈Mγ which is also a contradiction.
By i) and ii), we get μ=μαμβ∈ˆM. Hence ¯Mα¯Mβ⊆ˆM, i.e. ˆM≠∅.
5) If I is an ideal of S and I∩¯Mα≠∅, then Mα∪I=S. Thus, ¯Mα⊆I.
Remark 3. 1) If S contains the greatest C-ideal, denote it by Mg, then Mg⊆ˆM. Indeed: If there exists α∈Λ such that Mg⊈Mα, then Mg∪Mα=S. By Lemma 1(2), we can see that Mg is not a C-ideal. This is a contradiction.
2) The greatest C-ideal Mg does not necessarily exist in all ordered semigroups. See the following example.
Example 2. Let S={a,b,c,d,e} with the multiplication "⋅" and the order "≤" below:
![]() |
≤:={(a,a),(a,b),(b,b),(c,c),(d,b),(d,c),(d,d),(e,c),(e,e)}. |
It is easy to check that (S,⋅,≤) is an ordered semigroup and {a},{d},{e} are C-ideals of S. However, there is not the greatest C-ideal.
In the following, we give the relationship between the basis and the greatest C-ideal Mg of S.
Theorem 5. Let S contain a basis A. Then S contains the greatest C-ideal Mg. Moreover, Mg=(S3]∩ˆM where ˆM=⋂α∈ΛMα and {Mα∣α∈Λ} is the family of all maximal ideals of S.
Proof. For any a∈A, Ia is a maximal J-class. Thus S−Ia is a maximal ideal. Hence ˆM≠∅. Since ˆM and (S3] are ideals of S, ˆM∩(S3]≠∅. Denote N=ˆM∩(S3]. Then for any x∈N, there exists c∈S such that x∈(ScS]. If c∉A, then there exists b∈A such that I(c)⊆I(b). Therefore c∈(b∪Sb∪bS∪SbS]. Obviously, c≠b. Next we distinguish two cases.
1) If c<b, then (ScS]⊆(SbS]. Thus x∈(SbS].
2) If c∈(Sb∪bS∪SbS], then (ScS]⊆(S(Sb∪bS∪SbS]S]⊆(S2bS∪SbS2∪S2bS2]⊆(SbS]. Hence x∈(SbS].
By 1) and 2), we have x∈(SbS]⊆(SAS]⊆(S(S−ˆM)S]⊆(S(S−N)S] (Because S−ˆM=S−⋂α∈ΛMα=⋃α∈Λ¯Mα and every element of A is in some maximal J-class). Consequently, N is a C-ideal.
Finally, we prove that N is the greatest. Let T be a C-ideal of S. Then T⊆(S(S−T)S]⊆(S3]. By Lemma 1(4), we have every C-ideal is contained in all maximal ideals. Thus T⊆ˆM. Then T⊆(S3]∩ˆM. It follows that N=(S3]∩ˆM is the greatest C-ideal Mg.
Proposition 2. Let S contain the greatest C-ideal Mg and Mg⫋(S3]. Then every J-class of (S3]−Mg is maximal and I(a)=(SaS] for any a∈(S3]−Mg.
Proof. Since (S3] and Mg are ideals of S, (S3]−Mg is a union of some J-classes of S. Let Mγ be any one of them. Then Mγ⊆(S3]. It implies that there exist x,y,b∈S such that a≤xby for any a∈Mγ. Thus a∈(SbS]. Next we show b∈Mγ. If there exists δ≠γ such that b∈Mδ, then a∈(SbS] which implies that I(a)⊆I(b) and b∉I(a). Otherwise, b∈I(a). Hence I(b)⊆I(a) and so I(a)=I(b) which contradicts δ≠γ. Therefore b∈S−I(a). It follows that a∈(S(S−I(a))S]. Then I(a)⊆(S(S−I(a))S], i.e. I(a) is a C-ideal. Therefore I(a)∪Mg is a C-ideal properly containing Mg, which is impossible. Thus b∈Mγ. This implies I(a)⊆(SbS]⊆I(b)=I(a), hence I(a)=(SbS]=I(b). Next we show that I(a)=(SaS].
1) If a=b, then I(a)=(SaS].
2) If a≠b, then b∈(a∪Sa∪aS∪SaS].
i) If b<a, then I(b)=(SbS]⊆(SaS]⊆I(a)=I(b). Thus I(a)=(SaS].
ii) If b∈(Sa∪aS∪SaS], then I(b)=(SbS]⊆(S(Sa∪aS∪SaS]S]⊆(S2aS∪SaS2∪S2aS2]⊆(SaS]⊆I(a)=I(b). Hence I(a)=(SaS].
Finally, we show that Mγ is maximal. Suppose that I(a)=(SaS]⫋I(c) for a∈Mγ⊆(S3]−Mg and some c∈S. Then a∈I(c)=(c∪Sc∪cS∪ScS]. Obviously, a≠c. We show that I(a)⊆(ScS].
1) If a<c, then I(a)=(SaS]⊆(ScS].
2) If a∈(Sc∪cS∪ScS], then I(a)=(SaS]⊆(S(Sc∪cS∪ScS]S]⊆(S2cS∪ScS2∪S2cS2]⊆(ScS].
Since c∉I(a), we have I(a)⊆(S(S−I(a))S], i.e. I(a) is a C-ideal. Therefore I(a)∪Mg is a C-ideal properly containing Mg, this is a contradiction. It follows that Mγ is a maximal J-class.
Theorem 6. Let S contain the greatest C-ideal Mg. If S≠(S2] and any two elements of S−(S2] are incomparable, then S contains a basis.
Proof. Firstly, we have Mg⊆(S(S−Mg)S]⊆(S3]⊆(S2]⊆S. We denote by Mα a J-class of S−(S2], by Mβ a J-class of (S2]−(S3] and by Mγ a J-class of (S3]−Mg.
1) From Lemma 2, we have
I(x)=I(y)⇔(x]=(y]⇔x=y(∀x,y∈S−(S2]). |
Thus there is only one element in Mα. Since any two elements of S−(S2] are incomparable, Mα is maximal.
2) For any x∈Mβ⊆(S2]−(S3], there exist u,v∈S such that x≤uv. Here u,v∈S−(S2]. Otherwise u,v∈(S2], then x∈(S3] which is a contradiction. Therefore x∈I(u)=(u].
3) From Proposition 2, we get that Mγ is maximal. Since Mg⊆(S(S−Mg)S], there exists z∈S−Mg such that y∈I(z) for any y∈Mg. It follows that y∈Mα or Mβ or Mγ. Next we construct a set A.
i) If z∈Mα or z∈Mγ, then we choose z into A.
ii) If z∈Mβ, then there exists u∈Mα, such that z∈I(u). Hence we can choose u into A.
Now, we have
Mg⊆(A∪SA∪AS∪SAS];(S2]−(S3]⊆(A∪SA∪AS∪SAS];S−(S2]⊆(A∪SA∪AS∪SAS];(S3]−Mg⊆(A∪SA∪AS∪SAS]. |
Therefore S⊆(A∪SA∪AS∪SAS], i.e. (A∪SA∪AS∪SAS]=S.
To prove that A is a basis of S, it remains to show that there is no proper subset B⫋A with the property (B∪SB∪BS∪SBS]=S. This is evident, because A is constructed by means of elements of maximal J-classes of S, and from each maximal J-class just one element is chosen into A.
Ideal theory play an important role in studying ordered semigroups. In this paper, we first study ordered semigroups containing the greatest ideal and give the conditions of the greatest ideal to be a C-ideal in an ordered semigroup. Furthermore, we introduce the concept of a basis of an ordered semigroup and establish the relationship between the greatest C-ideal and the basis in an ordered semigroup.
We thank the referees whose comments led to significant improvements to this paper and the editor for his/her warm work.
This research was supported by the National Natural Science Foundation of China (No. 11701504, 11801081); the Natural Science Foundation of Guangdong Province (No. 2014A030313625, No. 2018A030313063); the Project for Characteristic Innovation of 2018 Guangdong University (No. 2018KTSCX234); the Teaching Team Project in Guangdong Province (Guangdong teach highletter [2018] no. 179); the Demonstration Project of Grass-roots Teaching and Research Section in Anhui Province (No. 2018jyssf053); the University Natural Science Project of Anhui Province (No. KJ2018A0329) and the Innovation Education Project of Graduate Students of Guangdong Province (No. 2016SFKS-40).
The authors declare no conflict of interest.
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