Home Mathematics The hull-kernel topology on prime ideals in ordered semigroups
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The hull-kernel topology on prime ideals in ordered semigroups

  • Huanrong Wu EMAIL logo and Huarong Zhang
Published/Copyright: August 23, 2024

Abstract

The aim of this study is to develop the theory of prime ideals in ordered semigroups. First, to ensure the existence of prime ideals, we study a class of ordered semigroups which will be denoted by S I P . And then we introduce the hull-kernel topology for the prime ideals P ( S ) and the topological properties like separation axioms, compactness and connectedness are studied. Finally, we focus on the subspace ( S , I ) , minimal prime ideals containing the ideal I in an ordered semigroup S . We investigate topological properties of this subspace and connections between this subspace and the ordered semigroup S .

MSC 2010: 06F05

1 Introduction

An ordered semigroup is a typical combination of algebraic structures and ordered structures. Due to its various applications, many scholars researched such ordered semigroups under different restrictions. For a comprehensive survey, readers can refer to Petrich’s book [1].

In 1937, Stone [2] established the co-equivalence of the category of bounded distributive lattices and certain category of topological spaces. It is widely known that a Boolean algebra can be made into a topological space in a natural manner, the open sets of which correspond to ideals of the Boolean algebra. In detail, an ideal I is associated with the open set which consists of all prime ideals which do not include I . Later, the space of prime ideals in the context of commutative semigroups with 0 was discussed by Kist [3]. And then Henriksen and Jerison [4] studied Stone’s topology on the set of minimal prime ideals of a commutative ring. Speed [5] made use of minimal prime ideals to study distributive lattices with 0. Soon afterwards, a topological representation for the distributive join-semilattices was given by Grätzer [6], which extends the known result obtained by Stone for distributive lattices. During the past few years, the hull-kernel topology of prime ideals, minimal prime ideals of an algebraic structure (poset, lattice, frame, ring, etc.) were studied by many researchers [711]. It is discovered that research works of the hull-kernel topology (also called Zariski topology) on the set of prime ideals play a very important role in the areas of commutative algebra, order theory, lattice theory, algebraic geometry and topology.

Recently, the theory of prime ideals and minimal prime ideals of an ordered semigroup has been studied by lot of scholars [1216]. This motivates us to consider how to define a suitable topology on the set of prime ideals, minimal prime ideals of an ordered semigroup. In this study, we not only develop the theory of prime ideals but also investigate lots of topological properties on the space of prime ideals in ordered semigroups. At first, to ensure the existence of prime ideals, a class of ordered semigroups which will be denoted by S I P is studied, and we show that if S is a commutative or an intra-regular ordered semigroup, then S S I P . And then, the hull-kernel topology for prime ideals P ( S ) is defined and some topological properties like separation axioms, compactness, connectedness are studied. Further, the subspace ( S , I ) , minimal prime ideals containing the ideal I in an ordered semigroup is investigated. Moreover, if S S I P , then ( S , I ) is Hausdorff, totally disconnected, completely regular and has a base L ( S , I ) consisting of clopen subsets. Finally, a result is obtained that if S is a pseudocomplemented commutative ordered semigroup, then the minimal prime ideal space ( S ) is a Boolean space. As an application of the results of this study, the corresponding results in semigroups – without the order structure – can be obtained and one can easily prove that some results on semigroups given in [4] can be obtained as an application of this study.

In Section 2, we shall first briefly introduce ordered semigroups, related basic definitions and study a class of ordered semigroups which will be denoted by S I P . In Section 3, we introduce the hull-kernel topology of prime ideals in an ordered semigroup with 0 and give some topological properties. In Section 4, the subspace for minimal prime ideals containing an ideal I in an ordered semigroup with 0 is investigated. Finally, some conclusions are presented in Section 5.

2 Preliminaries

We begin with the necessary definitions and terminologies used in the study.

2.1 Topology

First, we will give some basic definitions in topology, and for more information, one can refer to [17,18]. Let X be a topological space. X is said to be a T 0 space if for any two distinct points of X , there is an open set containing just one of them. X is said to be a T 1 space, if { x } is a closed set for any x X . X is said to be a Hausdorff space if for each pair x , y of distinct points of X , there exist neighborhoods U 1 and U 2 of x and y , respectively, which are disjoint.

Definition 2.1

[17] Let X be a topological space. A closed subset V of X is said to be reducible if V = V 1 V 2 , where V 1 and V 2 are closed sets in X , and V 1 , V 2 are proper subsets of V . Otherwise, V is called irreducible.

A topological space X is said to be a sober space if every irreducible closed subset of X is the closure of exactly one point of X .

Definition 2.2

[17] Let X be a topological space and A X . A is said to be compact if for every family of open set { U i : i I } with A { U i : i I } , there are i 1 , i 2 , , i n I such that A { U i j : j = 1 , 2 , , n } .

Definition 2.3

[19] Let X be a topological space. A subset A X is called super-compact if for any open subset U i ( i I ) with A { U i i I } , there exists i 0 I such that A U i 0 . The notation S O ( X ) denotes the set of all super-compact open sets of X .

Definition 2.4

[17] A subset A of a topological space X is said to be clopen if it is both closed and open. X is called connected if there is not a clopen set except and X .

A topological space X is said to be totally disconnected if given distinct x , y X , there exists a clopen subset U of X such that x U and y U .

Definition 2.5

[17] A topological space X is said to be completely regular, if it is a T 1 space and for each x 0 X and each closed set A of X not containing x 0 , there exists a continuous function f : X [ 0 , 1 ] such that f ( x 0 ) = 1 and f ( A ) = { 0 } .

Definition 2.6

[20]

  1. If a topological space is compact, Hausdorff and totally disconnected, then it is said to be a Boolean space.

  2. For Boolean spaces, as for every topological space, it is true that the class of all clopen sets is a field. The field of all clopen sets in a Boolean space of a Boolean space is called the dual algebra of a Boolean space.

Definition 2.7

[17] A topological space X is called zero-dimensional if the clopen subsets of X form a basis for the topology.

Definition 2.8

[17] Topological spaces X and Y are called homeomorphic if there exists a map f : X Y which satisfies

  1. f is a bijection;

  2. f is continuous;

  3. the inverse function f 1 is also continuous.

2.2 Ordered semigroup

An ordered semigroup is a semigroup ( S , ) with a partial order on S such that x y implies x z y z and z x z y for all x , y , z S . If ( S , , ) is an ordered semigroup, it is customary to write x y as x y for all x , y S . Sometimes, we denote an ordered semigroup ( S , , ) by S . For subsets A , B of S , let ( A ] = { x S : x a f o r s o m e a A } , [ A ) = { x S : x a f o r s o m e a A } and A B = { a b : a A , b B } . For any x , y S , ( x , y ) = { z S : z x , z y } , x = { z S : z x } , x = { z S : z x } . An identity for any ordered semigroup S is an element e such that s e = s and e s = s for all s in S . A zero element for an ordered semigroup S is an element 0 such that 0 s and s 0 = 0 s = 0 for all s in S . For an ordered semigroup S , S is called idempotent, if for any s S , s s = s . A nonempty subset A of S is called a subsemigroup of S if for any a , b A , a b A . In this study, N + denotes the positive integers and N = N + { 0 } .

The following definitions are from the study by Kehayopulu.

Definition 2.9

[2123] Let ( S , , ) be an ordered semigroup. A nonempty subset I of S is called a left (right) ideal of S if the following holds:

  1. S I I ( I S I ) ;

  2. for x I and y S , y x implies y I .

I is called an ideal of S if it is a left and right ideal of S . An ideal I of S is called a proper if I S .

Let A be a nonempty subset of the ordered semigroup S . We denote I ( A ) the ideal of S generated by A , i.e., the smallest (under inclusion relation) ideal of S containing A and I ( A ) = ( A S A A S S A S ] . For any x S , I ( x ) = ( x S x x S S x S ] . If S is commutative, then I ( A ) = ( A S A ] . We denote I ( A , x ) = I ( A { x } ) . In an ordered semigroup S , the intersection of arbitrary ideals of S is an ideal and the union of arbitrary ideals is also an ideal.

Example 2.10

( N , , ) with the standard multiply operation is an ordered semigroup, where for any m , n N , m n if there exists k N such that m = k n . Moreover, 0 is the zero element of ( N , , ) . The collection of all ideals of ( N , , ) is ( N ) = { A N : A N } . For any A 1 , A 2 N , ( A 1 N ) ( A 2 N ) = A 1 A 2 N and ( A 1 N ) ( A 2 N ) = ( A 1 A 2 ) N . So, the intersection of arbitrary ideals of ( N , , ) is an ideal and the union of arbitrary ideals is also an ideal.

Definition 2.11

[21] Let ( S , , ) be an ordered semigroup. A nonempty subset F of S is called a filter of S if the following holds:

  1. for any x , y S , x y F if and only if x , y F ;

  2. for x S and y F , y x implies x F .

Definition 2.12

[23] Let S be an ordered semigroup. A proper ideal I of S is said to be prime if A , B S , A B I implies A I or B I .

Equivalently, a , b S , a b I implies a I or b I .

In this study, the set of all prime ideals of S is denoted by P ( S ) .

Definition 2.13

[23] Let S be an ordered semigroup. An (a left, a right) ideal I of S is said to be semiprime if A S , A 2 I implies A I .

Equivalently, a S and a 2 I , which implies a I .

Definition 2.14

[24] An ordered semigroup S is called regular if for any s S , there exists x in S such that s s x s .

Definition 2.15

[25,26] If I is an ideal in an ordered semigroup S , then the radical of I , denoted by I , is defined by

I = { s S : n N + , s n I } .

It is easily seen that I I and ( I ] = I .

Notation 2.16

We denote by S I P , the class of ordered semigroups in which for any proper ideal I and a I , there exists a prime ideal containing I and not containing a .

To ensure the existence of prime ideals in an ordered semigroup, we introduce this new class of ordered semigroups. It is easy to see that if a prime ideal P containing an ideal I , then I P . So, in the above notation, we require a I . This class of ordered semigroups is not strict as you can see in Theorem 2.18 and any commutative ordered semigroup is an example.

In order to facilitate the writing of the article, we introduce the following definition for the first time.

Definition 2.17

Let S be an ordered semigroup and F S . F is called a weak filter if F satisfies:

  1. for any x , y F , x y F ;

  2. for any y F , and x S , y x implies x F .

By the above definition, it is easily seen that if F is a filter of S , then F is a weak filter. Moreover, if A is a subsemigroup of S , then [ A ) is a weak filter of S .

In the following theorem, we give a characterization of an ordered semigroup which belongs to S I P and show that if S is a commutative ordered semigroup or an intra-regular ordered semigroup, then S S I P . For any ideal I and weak filter with A I = , define M ( A , I ) = { I : I i s a n i d e a l o f S w i t h I A = } . We show that M ( A , I ) is an ideal. For any x M ( A , I ) and y S , there exists an ideal I 0 with I 0 A = such that x I 0 . Since I 0 is an ideal, we have x y I 0 M ( A , I ) and y x I 0 M ( A , I ) . So, M ( A , I ) S M ( A , I ) and S M ( A , I ) M ( A , I ) . For any x M ( A , I ) , y S with y x , there exists an ideal I 0 with I 0 A = such that x I 0 . Since I 0 is an ideal, we have y M ( A , I ) . Therefore, M ( A , I ) is an ideal which is maximal with respect to the property of not meeting A . In Theorem 2.18, we show that for any S S I P , the ideal M ( A , I ) exists and M ( A , I ) is prime.

Theorem 2.18

Let S be an ordered semigroup. Then, the following conditions are equivalent.

  1. S S I P .

  2. For any ideal I of S, I is the intersection of prime ideals containing I.

  3. For any x , y S , I ( x ) I ( y ) I ( x y ) .

  4. If A is a weak filter of S, which satisfies A I = , then the ideal M ( A , I ) exists; moreover M ( A , I ) is a prime.

Proof

( 1 ) ( 2 ) : If s I , and P is a prime ideal containing I , then for some n N + , we have s n I P , and so s P . Suppose that s I . Since S S I P , there exists a prime ideal M containing I and not containing s . Therefore, I is the intersection of all prime ideals containing I .

( 2 ) ( 3 ) : Suppose that (2) holds. Assume that there exist x , y S such that I ( x ) I ( y ) I ( x y ) . Then, there exist a I ( x ) , b I ( y ) such that a b I ( x y ) . So, there exists a prime ideal P containing I ( x y ) and not containing a b . Since x y I ( x y ) P , we have x P or y P . Without loss of generality, assume that x P . Then, a I ( x ) P and so a b P , which contradicts a b P . Therefore, for any x , y S , I ( x ) I ( y ) I ( x y ) .

( 3 ) ( 4 ) : Suppose that A is a weak filter of S , which satisfies A I = . Let M ( A , I ) = { I : I b e a n i d e a l o f S w i t h I A = } . Then, M ( A , I ) is a maximal ideal with respect to the property of not meeting A . Assume that x y M ( A , I ) but x , y M ( A , I ) . Then, there are elements s , t S such that s A I ( M ( A , I ) , x ) and t A I ( M ( A , I ) , y ) . Since A M ( A , I ) = , we have s , t A , s I ( x ) and t I ( y ) . Thus, s t A ( I ( x ) I ( y ) ) . By the hypothesis, s t A I ( x y ) . So, there exists n N such that ( s t ) n I ( x y ) . Since A is a weak filter, we have ( s t ) n A . Therefore, ( s t ) n A I ( x y ) A M ( A , I ) , which contradicts A M ( A , I ) = . Therefore, M ( A , I ) is a prime.

( 4 ) ( 1 ) : Suppose that I is a proper ideal of S and a I . Let A = { [ a n ) : n N + } . Hence, A is a weak filter. Then, M = { J i s a n i d e a l o f S : A J = } is prime. So, there exists a prime ideal containing I and not containing a . Thus, S S I P .□

The above theorem essentially extends the result of [25, Theorem 3.2], where only the sufficient condition of (2) is discussed.

Definition 2.19

[27] Let S be an ordered semigroup. S is called intra-regular if

a S , x , y S : a x a 2 y ,

i.e., a ( S a 2 S ] , a S or A ( S A 2 S ] , A S .

In fact, if S is intra-regular, then for any A S and a A , there exist x , y S such that a x a 2 y . So, A ( S A 2 S ] . Conversely, suppose that A ( S A 2 S ] , A S . For any a S , let A = { a } . We have { a } ( S { a } 2 S ] . Hence, S is intra-regular.

Proposition 2.20

Let S be an ordered semigroup. Then, the following statements are equivalent:

  1. S is intra-regular;

  2. for any ideal I of S, I is semiprime;

  3. for any ideal I of S, I is the intersection of prime ideals containing I;

  4. for any x , y S , I ( x ) I ( y ) = I ( x y ) .

Proof

( 1 ) ( 2 ) : It follows from [28, Lemma 6].

( 2 ) ( 3 ) : It follows from [25, Theorem 2.4].

( 3 ) ( 4 ) : For any x , y S , let a S with a I ( x y ) . Then, by (3) a P for some prime ideal P containing x y . Since I ( x ) P or I ( y ) P by the primeness of P , we have a I ( x ) or a I ( y ) . Thus, a I ( x ) I ( y ) . So, I ( x ) I ( y ) I ( x y ) . The reverse inclusion is obvious. Hence, I ( x ) I ( y ) = I ( x y ) .

( 4 ) ( 1 ) : Let a S . Since a I ( a ) , by the hypothesis, we have a I ( a ) I ( a ) = I ( a 2 ) = ( a 2 S a 2 a 2 S S a 2 S ] . Then, we have

a 2 ( a ] ( a 2 S a 2 a 2 S S a 2 S ] ( a 3 a S a 2 a 3 S a S a 2 S ] ( S a 2 S a 2 S ] .

Since a ( a 2 S a 2 a 2 S S a 2 S ] , replacing the appearance of a 2 by ( S a 2 S a 2 S ] , we have a ( ( S a 2 S a 2 S ] S a 2 ( S a 2 S a 2 S ] S S a 2 S ] ( S a 2 S a 2 S ] . Thus,

a 2 ( S a 2 S a 2 S ] ( a ] ( S a 3 S a 2 S a ] ( S a 2 S ] .

Since a ( S a 2 S a 2 S ] , replacing the appearance of a 2 by ( S a 2 S ] , we have a ( S ( S a 2 S ] S a 2 S ] ( S a 2 S ] . Therefore, a ( S a 2 S ] , and S is intra-regular.□

Theorem 2.21

Let S be an ordered semigroup. If S is commutative or intraregular, then S S I P .

Proof

If S is commutative, then it is easy to see that I ( x ) I ( y ) I ( x y ) I ( x y ) for any x , y S . If S is intraregular, then I ( x ) I ( y ) I ( x ) I ( y ) = I ( x y ) I ( x y ) holds by Proposition 2.20.□

Now, we study the relationship between P MFP ( P MFP ) in [9] and S I P . We give an ordered semigroup in S I P but not in P MFP ( P MFP ) as a poset.

Example 2.22

Consider the Boolean poset ( P , ) depicted in Figure 1. By Remark 1.5 in [9], ( P , ) is a Boolean poset in which a maximal filter is not prime, that is, P P MFP ( P MFP ).

Consider the ⋅ operation which is defined as: x y = 0 , for any x , y P . Then, ( P , , ) is an ordered semigroup with zero element 0. So, I ( 0 ) = { 0 } , I ( a ) = { a , 0 } , I ( b ) = { b , 0 } , I ( c ) = { c , 0 } , I ( d ) = { d , 0 } , I ( a ) = { a , b , c , d , 0 } , I ( b ) = { b , a , c , d , 0 } , I ( c ) = { c , a , b , d , 0 } , I ( d ) = { d , a , b , c , 0 } . And { 0 } = P . So for any x , y P , I ( x ) I ( y ) = { 0 } { 0 } = I ( x y ) . Therefore, S S I P by Theorem 2.18.

Figure 1 
                  A Boolean poset in which a maximal filter is not prime.
Figure 1

A Boolean poset in which a maximal filter is not prime.

3 Space of prime ideals

In this section, we introduce the hull-kernel topology for prime ideals P ( S ) and some topological properties like separation axioms, compactness, and connectedness are studied.

Let S be an ordered semigroup. For any A S , denote { P P ( S ) : A P } by h ( A ) and call it the hull of A . For any X P ( S ) , denote { P : P X } by k ( X ) , and call it the kernel of X . Without conflicting with the notation, we denote k ( ) = S . Let h ( x ) = { P P ( S ) : x P } for any x S and p ( x ) = P ( S ) \ h ( x ) = { P P ( S ) : x P } .

By the definitions of ideals and prime ideals of an ordered semigroup, we have the following lemma.

Lemma 3.1

Let S be an ordered semigroup and x , y S , A , B S , X 1 , X 2 P ( S ) . Then,

  1. h ( x ) = h ( I ( x ) ) and h ( A ) = h ( I ( A ) ) ;

  2. h ( x ) = h ( x n ) for any n N + ;

  3. x y implies h ( y ) h ( x ) ;

  4. x s y , for some s S , implies h ( y ) h ( x ) ;

  5. A B implies h ( B ) h ( A ) .

Proposition 3.2

Let P be a prime ideal in an ordered semigroup S. If I J P , where I , J are ideals in S, then I P or J P .

Proof

Because I , J are ideals in S , we have I J I and I J J . So, I J I J P . By Definition 2.12, we have I P or J P .□

Proposition 3.3

Let S be an ordered semigroup with 0 and I , J , I α be ideals of S, α Λ . Then,

  1. h ( S ) = , h ( { 0 } ) = P ( S ) ;

  2. h ( I ) h ( J ) = h ( I J ) ;

  3. { h ( I α ) : α Λ } = h ( { I α : α Λ } ) .

Proof

(1) There is no prime ideal containing S , hence h ( S ) = . Also, every prime ideal contains 0, which gives h ( ( 0 ] ) = P ( S ) .

(2) Let P h ( I ) h ( J ) , that is P h ( I ) or P h ( J ) , and without loss of generality, assume that P h ( I ) . This implies that I J I P . Thus, h ( I ) h ( J ) h ( I J ) .

To prove the converse, let P h ( I J ) . Assume on the contrary that P h ( I ) h ( J ) . Then, choose x I \ P and y J \ P . This implies that x y I J P . By primeness of P , we have x P or y P , which is a contradiction. Hence, P h ( I ) h ( J ) .

(3) Assume P { h ( I α ) : α Λ } . Then, I α P for each α Λ , which gives { I α : α Λ } P and P h ( { I α : α Λ } ) . Hence, { h ( I α ) : α Λ } h ( { I α : α Λ } ) .

Conversely, let P h ( { I α : α Λ } ) . Then, { I α : α Λ } P . Hence, I α P for each α Λ . Consequently, P h ( I α ) for each α Λ and thus, P { h ( I α ) : α Λ } .□

Due to Proposition 3.3, the sets h ( I ) for all ideals I of S can be considered as closed sets in a topological space on P ( S ) . So we give the following definition:

Definition 3.4

For an ordered semigroup S with 0, the topological space P ( S ) having { h ( I ) : I is an ideal of S } as the collection of closed sets is called the hull-kernel topology of S . We shall denote P ( S ) with the hull-kernel topology again by P ( S ) . The topological space P ( S ) is called the prime ideal space of S .

For any subset X of P ( S ) , h ( k ( X ) ) = { P P ( S ) : { P : P X } P } . For any subset X of P ( S ) , denote the closure of X in the space P ( S ) by X ¯ . We show that X ¯ = h ( k ( X ) ) . Clearly, h ( k ( X ) ) is a closed set of P ( S ) and X h ( k ( X ) ) . We show that any closed set Y containing X also contains h ( k ( X ) ) . Indeed, by Definition 3.4, Y = h ( I ) for some ideal I of S . As X h ( I ) , we have I P for every P X . Therefore I k ( X ) and thus h ( k ( X ) ) h ( I ) = Y . Hence, X ¯ = h ( k ( X ) ) .

Lemma 3.1 ensures that the correspondence X h ( k ( X ) ) is a closure operator on P ( S ) , which makes the latter set into a topological space. We will show that the topology so defined on P ( S ) is equivalent to the hull-kernel topology on P ( S ) . In fact, since h ( I ) = h ( k ( h ( I ) ) ) , we have h ( I ) is a closed set in the hull-kernel topology. And for any X P ( S ) , k ( X ) is an ideal in S . So h ( k ( X ) ) is a closet set. Thus, the topology defined by Proposition 3.3 is equivalent to the hull-kernel topology defined on P ( S ) .

Example 3.5

( N , , ) with the standard multiply operation is an ordered semigroup, where for any m , n N , m n if there exists k N such that m = k n . Moreover, 0 is the zero element of ( N , , ) . The collection of all prime ideals of ( N , , ) is P ( N ) = { p N : p = 0 o r p i s a p r i m e e l e m e n t o f N } . U is a basis for open sets of the hull-kernel topology defined for prime ideals of ( N , , ) if and only if U = p ( x ) = { p N P ( N ) : x p N } .

Definition 3.6

[29] Let S be an ordered semigroup. S is called a negatively ordered semigroup, if for any x , y S , x y x and x y y .

One can see that if S has the greatest element 1 which is also a multiplicative identity, then S is a negatively ordered semigroup.

Definition 3.7

[30] Let ( P , ) be a poset and I P .

  1. A nonempty subset I of P is called a semi-ideal of ( P , ) if for x I , y P , y x implies y I .

  2. An ideal I of P is called prime if for any x , y P , ( x ] ( y ] I implies x I or y I .

Proposition 3.8

[29] Let S be an ordered semigroup. Then, the following conditions are equivalent.

  1. S is negatively ordered.

  2. The ideals of ( S , , ) coincide with the semi-ideals of ( S , ) .

Lemma 3.9

Let S be a negatively ordered semigroup. Then, a prime ideal of ( S , , ) is a prime semi-ideal of ( S , ) .

Proof

By Proposition 3.8, the ideals of ( S , , ) coincide with the semi-ideals of ( S , ) . So a prime ideal of ( S , , ) is a prime semi-ideal of ( S , ) .□

Lemma 3.9 shows that the prime ideal space of a negatively ordered semigroup is a subspace of the prime semi-ideal space of the poset. The distinction between the prime ideals of an ordered semigroup and the prime semi-ideals of a poset is illustrated in the following theorem.

Theorem 3.10

Let S be an ordered semigroup andis the meet operator. Then, the followings are equivalent.

  1. ( S , , ) is a meet-semilattice.

  2. S is negatively ordered and idempotent.

  3. The prime ideals of ( S , , ) coincide with the prime semi-ideals of ( S , ) .

Proof

( 1 ) ( 2 ) : Obviously.

( 2 ) ( 1 ) : Because for any x , y S , x y x and x y y . So, x y is a lower bound of x and y . Assume that z is a lower bound of x and y . So, z x and z y . Hence, z z x y . Since S is idempotent, we have z x y . That is x y is the largest lower bound of x and y . Therefore, ( S , , ) is a meet-semilattice.

( 2 ) ( 3 ) : Suppose that I is a prime semi-ideal of ( S , ) . Let x , y S and x y I . For any c ( x , y ) , we have c x and c y . Hence, c c x y . Since S is idempotent, we have c x y . Because x y ( x , y ) , we have ( x , y ) = ( x y ) I . Since I is a prime semi-ideal of ( S , ) , we have x I or y I . Therefore, I is a prime ideal of ( S , , ) .

( 3 ) ( 2 ) : First, we show that for any z S , S \ z is a prime semi-ideal of ( S , ) . Obviously, S \ z is a semi-ideal of ( S , ) . Let x , y S and ( x , y ) S \ z . Assume that x , y S \ z , i.e., x z and y z . So, z ( x , y ) S \ z . This is a contradiction. Therefore, S \ z is a prime semi-ideal of ( S , ) . For any x , y S , we have S \ ( x y ) is a prime semi-ideal of ( S , ) and hence S \ ( x y ) is a prime ideal of ( S , , ) . Since x y S \ ( x y ) and S \ ( x y ) is an ideal of ( S , , ) , we have x S \ ( x y ) and y S \ ( x y ) . Hence, x x y and y x y . Therefore, S is negatively ordered.

For any z S , we have z 2 z . Suppose that z 2 z . Then, z 2 S \ z . Since S \ z is a prime semi-ideal of ( S , ) , so S \ z is a prime ideal of ( S , , ) . Hence, z S \ z . This is a contradiction. Hence, z 2 = z . Therefore, S is idempotent.□

Next we give a concrete example of an ordered semigroup that does not satisfy the condition of Theorem 3.10.

Example 3.11

The set S = { a , b } with the multiplication x y = x for any x , y S is a semigroup. S (with the above multiplication) is an ordered semigroup with respect to the order x y x = y . Since is not the meet operator, we have ( S , , ) is not a meet-semilattice. S is idempotent but not negatively ordered. The ideal of S is { x , y } and there is no prime ideal in S . The semi-ideals of S are { x } , { y } , { x , y } and the prime semi-ideals are { x } , { y } .

Mundlik et al. [9] proves similar propositions for posets and the proofs of the following propositions often resemble their proofs.

Proposition 3.12

Let S S I P with 0. The collection { p ( x ) : x S } = { P ( S ) \ h ( x ) : x S } forms a basis for the open sets of P ( S ) .

Theorem 3.13

Let S S I P with 0. Then, the space P ( S ) is always a T 0 space.

Proof

The proof is similar to Theorem 23 in [30].□

Proposition 3.14

Let S S I P with 0. Then, the following statements are equivalent:

  1. P ( S ) is a T 1 space.

  2. { P } = h ( P ) for any P P ( S ) .

  3. ( P ( S ) , ) is an antichain.

Proof

( 1 ) ( 2 ) For every P P ( S ) , k ( { P } ) = P and hence, h ( k ( { P } ) ) = h ( P ) . Since h ( k ( { P } ) ) is the closure of { P } , we have ( 1 ) ( 2 ) .

( 2 ) ( 3 ) Suppose there exist P 1 , P 2 P ( S ) such that P 1 P 2 . Then, P 2 h ( P ) , which is a contradiction.

( 3 ) ( 1 ) If ( P ( S ) , ) is an antichain, then { P } = h ( P ) for any P P ( S ) . Thus, P ( S ) is a T 1 -space.□

Although the prime ideals of an ordered semigroup and the prime semi-ideals of the poset are not the same, the proof of the following proposition is similar to that of Lemma 2.15 in [9]. So we omit it.

Proposition 3.15

Let S S I P with 0 and V be a nonempty closed set in P ( S ) . Then, V is irreducible if and only if I = { P : P V } is a prime ideal of S.

Theorem 3.16

Let S S I P with 0. Then, P ( S ) is a sober space.

Proof

Let V be an irreducible closed set of P ( S ) and I = k ( V ) = { P : P V } . Then, V = h ( k ( V ) ) = h ( I ) . Since I P ( S ) by Proposition 3.15, we have { I } ¯ = h ( k ( { I } ) ) = h ( I ) = V . If there exists another point P 0 P ( S ) such that { P 0 } ¯ = V , then { P 0 } ¯ = h ( k ( P 0 ) ) = h ( P 0 ) = V . Thus, k ( h ( P 0 ) ) = k ( V ) = I . Since k ( h ( P 0 ) ) = P 0 , we have P 0 = I . Therefore, every irreducible closed subset of P ( S ) is the closure of exactly one point of P ( S ) , i.e., P ( S ) is a sober space.□

Proposition 3.17

Let S S I P with 0. Then, a subset U P ( S ) is super-compact open if and only if there exists a S such that U = p ( a ) .

Proof

Let U be super-compact open. If U = , then U = p ( 0 ) . If U , there exists a subset A of S such that U = { p ( x ) x A } . Because U is super-compact, there exists a A such that U p ( a ) , which indicates U = p ( a ) .

Conversely, suppose that x S and { U α : α Λ } is a family of open sets with p ( x ) { U α : α Λ } . By Proposition 3.12, there exists A S such that p ( x ) { U α : α Λ } = { p ( a ) : a A } . Since { P ( S ) \ p ( a ) : a A } = { h ( a ) : a A } = h ( A ) P ( S ) \ p ( x ) , we have h ( A ) p ( x ) = . Thus, h ( A ) h ( x ) . By Theorem 2.18, { P : P h ( A ) } = I ( A ) and { P : P h ( x ) } = I ( x ) . Since h ( A ) h ( x ) , we have I ( x ) I ( A ) . Because x I ( x ) I ( x ) , we have x I ( A ) , i.e., x n I ( A ) for some n N + . Since I ( A ) = ( A ] ( S A ] ( A S ] ( S A S ] , four cases are possible: (1) x n a for some a A ; (2) x n s a for some s S and a A ; (3) x n a s for some s S and a A ; (4) x n s 1 a s 2 for some s 1 , s 2 S and a A . Of these, we only consider the fourth case and the other cases are similar. If the fourth case holds, then h ( a ) h ( s 1 a s 2 ) h ( x n ) = h ( x ) by Lemma 3.1 and h ( a ) p ( x ) h ( x ) p ( x ) = . Therefore, there exists a A such that h ( a ) p ( x ) = , i.e., p ( x ) p ( a ) . This shows that p ( x ) is super-compact.□

Proposition 3.18

Let S S I P with 0. For any closed set V in P ( S ) , if is a subset of { p ( x ) x S } which satisfies (i) is closed under finite intersection and ( i i ) for any U , V U , then V ( U U ) .

Proof

First, we show that for any x , y S , p ( x ) p ( y ) = p ( x y ) . For any P p ( x ) p ( y ) , we have x P and y P . Then, by the primeness of P , x y P , i.e., P p ( x y ) . Hence, p ( x ) p ( y ) p ( x y ) . For any P p ( x y ) , we have x y P . Assume that x P or y P . Since P is an ideal, we have x y P , which is a contradiction. So, x P and y P , i.e., P p ( x ) p ( y ) . Hence, p ( x y ) p ( x ) p ( y ) . Therefore, p ( x ) p ( y ) = p ( x y ) .

Suppose that V is a closed set and is a subset of { p ( x ) x S } , which satisfies ( i ) and ( i i ) . Then, there exists an ideal I such that V = h ( I ) . Let H = { x p ( x ) } . Since satisfies ( i ) , we have that H is a subsemigroup of S . We claim that I [ H ) = . In fact: If there exists a I [ H ) , then h ( I ) h ( a ) and there exists b H such that b a . Hence, h ( I ) h ( a ) h ( b ) and so, h ( I ) p ( b ) = , which contradicts with V U for any U . Therefore, by Theorem 2.18, there exists a prime ideal P such that I P but [ H ) P = , i.e., P V ( U U ) .□

Theorem 3.19

Let X be topological space. Then, the following statements are equivalent.

  1. X is homeomorphic to the space P ( S ) of some ordered semigroup S S I P with 0.

  2. X satisfies that

    1. X is T 0 ,

    2. The super-compact open subsets are closed under finite intersection and form a basis for X,

    3. For any closed set A, if is a subset of S O ( X ) , which satisfies is closed under finite intersection and for any U , A U , then A ( U U ) .

Proof

( 1 ) ( 2 ) : It follows by Propositions 3.17 and 3.12.

( 2 ) ( 1 ) : By hypothesis, ( S O ( X ) , ) is a commutative ordered semigroup with zero element and we will show that X is homeomorphic to P ( S O ( X ) ) . Define a function Δ : X P ( S O ( X ) ) as follows: for any x in X ,

Δ ( x ) = { B S O ( X ) x B } .

Let x X and B S O ( X ) , B Δ ( x ) . Then, x B and so x B B , i.e., B B Δ ( x ) . Moreover, if B 1 , B 2 Δ ( x ) , then x B 1 B 2 , and thus B 1 B 2 Δ ( x ) . Hence, Δ ( x ) P ( S O ( X ) ) . So the map Δ is well-defined. Since X is a T 0 space, we have Δ is injective.

Now, we show it is surjective. Let A P ( S O ( X ) ) and { B S O ( X ) B A } . Then, is a subset of S O ( X ) , which is closed under finite intersection. And consider the set V = { X \ C C A } . Then, V is closed. We claim that V B for any B . If there exists B 1 such that V B 1 = , then B 1 X \ V = { C C A } . Because B 1 is super-compact, there exists C 1 A such that B 1 C 1 , which implies B 1 A , which is a contradiction. So V ( B B ) by (iii), i.e., there exists x 0 X such that x 0 B for any B A and x 0 X \ C for any C A . This indicates x 0 B if and only if B A . So, we can obtain that Δ ( x 0 ) = A .

For any B S O ( X ) , we have

Δ 1 ( { P P ( S O ( X ) ) B P } ) = { x X B Δ ( x ) } = { x X x B } = B

and

Δ ( B ) = { Δ ( x ) x B } = { Δ ( x ) B Δ ( x ) } = { P P ( S O ( X ) ) B P } .

Since S O ( X ) and { { P P ( S O ( X ) ) B P } B S O ( X ) } are open bases for X and P ( S O ( X ) ) , respectively, we can conclude that Δ and Δ 1 are continuous. This completes the proof.□

4 Space of minimal prime ideals

A prime ideal P of an ordered semigroup S is said to be a minimal prime ideal containing the ideal I if I P , and there is no other prime ideal containing I and properly contained in P . In case the ordered semigroup S has 0, then a minimal prime ideal containing 0 is called a minimal prime ideal. In this section, we focus on the subspace ( S , I ) , minimal prime ideals containing the ideal I in an ordered semigroup S .

Let S be an ordered semigroup with 0, I be an ideal of S and denote the set of all minimal prime ideals containing I by ( S , I ) . For any subset A of S , we denote by h M ( A , I ) the set h ( A ) ( S , I ) = { P ( S , I ) : A P } . Further, for x S , let m ( x , I ) = { P ( S , I ) : x P } = ( S , I ) \ h M ( x , I ) . If I = { 0 } , we denote h M ( A ) h M ( A , I ) , m ( x ) m ( x , I ) . We study the subspace of P ( S ) on the set ( S , I ) and denote this subspace again by ( S , I ) . The topological space ( S , I ) is called the minimal prime ideal space containing the ideal I of S. Let ( S ) denote the minimal prime ideal of S and also denote the topological space ( S ) with the hull-kernel topology, i.e., ( S ) = ( S , { 0 } ) . The topological space ( S ) is called the minimal prime ideal space of S.

Propositions 4.1 and 4.2 in the following are consequences of the aforementioned notations.

Proposition 4.1

Let I be any ideal in an ordered semigroup S S I P with 0. The collection { m ( x , I ) : x S } forms a basis for the open sets of ( S , I ) .

Proposition 4.2

Let I be any ideal in an ordered semigroup S S I P with 0 and X be any non-empty subset of ( S , I ) . Then, X ¯ = h M ( k ( X ) , I ) .

Definition 4.3

[31] Let P be a poset. D P is called a directed subset of P if for any x , y D , there exists z D such that x z , y z . P is called a dcpo if for any directed subset of P , there exists a least upper bound in P .

Lemma 4.4

[31] If P is a dcpo, then there exists a maximal element in P.

Lemma 4.5

Let S be an ordered semigroup and I be an ideal of S. If A is a weak filter such that A I = , then A is contained in a weak filter which is maximal with respect to the property of not meeting I.

Proof

Let C = { T is a weak filter of S : I T = } . Then, C is a poset with respect to the inclusion order. Assume that D C is a directed set. Then, for any b S , a D D D with a b , there exists D 0 D such that a D 0 . So b D 0 D D D . For any x , y D D D , there exists D 1 , D 2 D such that x D 1 , y D 2 . Since D is directed, there exists D 3 D such that D 1 D 3 , D 2 D 3 . So x , y D 3 and x y D 3 D D D . Thus, D D D C . And so C is a dcpo. Therefore, A is contained in a weak filter T , which is maximal with respect to the property of not meeting I by Lemma 4.4.□

Theorem 4.6

Let S S I P , I be an ideal of S and P S . Then, P is a minimal prime ideal containing I if and only if S \ P is a weak filter of S, which is maximal with respect to the property of not meeting I.

Proof

Let P be a subset of S such that S \ P is a weak filter of S , which is maximal with respect to the property of not meeting I . By Theorem 2.18, there exists a prime ideal Q such that I Q and Q ( S \ P ) = . So ( S \ Q ) I = and S \ P S \ Q . Because Q is a prime ideal, we have S \ Q is a weak filter. By the maximal property of S \ P , we have S \ P = S \ Q . This shows that P is a prime ideal containing I . Since S \ P is a weak filter of S , which is maximal with respect to the property of not meeting I , we have P is a minimal prime ideal containing I .

Conversely, let P be a minimal prime ideal containing the ideal I . Then S \ P is a weak filter of S , which does not meet I , and by Lemma 4.5, it is contained in a weak filter T , which is maximal with respect to the property of not meeting I . Hence, I S \ T P . By the case just proved, S \ T is a prime ideal containing I . Since P is a minimal prime ideal containing the ideal I , we have S \ T = P , i.e., S \ P = T . Therefore, S \ P is a weak filter of S , which is maximal with respect to the property of not meeting I .□

Corollary 4.7

Let S S I P , I be an ideal of S. Then, there is a minimal prime ideal containing I.

Proof

Assume that P is a prime ideal containing I . Then, S \ P is a weak filter of S , which does not meet I . Thus, S \ P is contained in a weak filter T , which is maximal with respect to the property of not meeting I by Lemma 4.5. By Theorem 4.6, S \ T is a minimal prime ideal containing I , and S \ T P .□

Characteristic of minimal prime ideals of an ordered semigroup is given in the following theorem.

Theorem 4.8

Let S S I P , I be an ideal of S and P be a prime ideal of S containing I. Then, P is a minimal prime ideal containing I if and only if for any x P , there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P such that [ y 1 x n 1 y 2 x n 2 y j x n j ) I , where, for any y S \ P , y x 0 = y .

Proof

Let P be a minimal prime ideal containing I . By Theorem 4.6, S \ P is a weak filter of S , which is maximal with respect to the property of not meeting I . Let x P and

T = n 1 , n 2 n j N , y 1 , y 2 y j S \ P [ y 1 x n 1 y 2 x n 2 y j x n j ) .

Then, T is a weak filter of S and S \ P T . Since for any y S \ P , y x T but y x P , i.e., y x S \ P , we have T properly contains S \ P . Being S \ P maximal with respect to disjoint from I , we have [ y 1 x n 1 y 2 x n 2 y j x n j ) I for some n 1 , n 2 n j N , y 1 , y 2 y j S \ P .

Conversely, assume that P is a prime ideal of S containing I and for any x P there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P such that [ y 1 x n 1 y 2 x n 2 y j x n j ) I , where, for any y S \ P , y x 0 = y . Let I Q P , where Q is a prime ideal of S . Choose x P such that x Q . Then, there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P such that

[ y 1 x n 1 y 2 x n 2 y j x n j ) I .

Because I is an ideal, we have y 1 x n 1 y 2 x n 2 y j x n j I Q . Since Q is prime, there exists i j such that y i Q or x Q , which contradicts y i Q and x Q . Therefore, P is a minimal prime ideal containing I .□

In Lemma 3.1 of [3], the ordered semigroup is commutative. By Theorem 2.21, if S is a commutative ordered semigroup, then S S I P . By using the previous theorem, the following strengthening of Lemma 3.1 of [3] can be obtained.

Theorem 4.9

Let S S I P with 0 and P be a prime ideal of S. Then, P is a minimal prime ideal if and only if for any x P , there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P such that y 1 x n 1 y 2 x n 2 y j x n j = 0 .

Proof

In Theorem 4.8, consider I = { 0 } and then we can obtain the result.□

Proposition 4.10

Let S S I P with 0 and I be an ideal of S. For any x , y S , m ( x , I ) m ( y , I ) = m ( x y , I ) .

Proof

Let x , y S and P m ( x , I ) m ( y , I ) , which implies x P and y P . By primeness of P , we have x y P . Hence, P m ( x y , I ) and m ( x , I ) m ( y , I ) m ( x y , I ) . Let P m ( x y , I ) . Then, x y P . Since P is an ideal, we have x P and y P . Therefore, m ( x , I ) m ( y , I ) = m ( x y , I ) .□

Theorem 4.11

Let S S I P with 0 and I be an ideal of S. Then,

  1. m ( x , I ) is clopen in ( S , I ) for any x S ;

  2. ( S , I ) is Hausdorff;

  3. ( S , I ) is totally disconnected;

  4. ( S , I ) is completely regular.

Proof

(1) We have already observed that m ( x , I ) is a basic open set. Let P ( S , I ) but P m ( x , I ) . Then, x P , and so by Theorem 4.8, there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P such that [ y 1 x n 1 y 2 x n 2 y j x n j ) I . Hence, y 1 x n 1 y 2 x n 2 y j x n j I and m ( y 1 x n 1 y 2 x n 2 y j x n j , I ) = . By Proposition 4.10, m ( y 1 x n 1 y 2 x n 2 y j x n j , I ) = m ( y 1 , I ) m ( y 2 , I ) m ( y j , I ) m ( x , I ) = m ( y 1 y 2 y j , I ) m ( x , I ) . So, m ( y 1 y 2 y j , I ) ( S , I ) \ m ( x , I ) . Because y 1 , y 2 y j S \ P , we have P m ( y 1 y 2 y j , I ) by the primeness of P . So, P m ( y 1 y 2 y j , I ) ( S , I ) \ m ( x , I ) . This shows that m ( x , I ) is closed.

(2) Let P 1 , P 2 ( S , I ) with P 1 P 2 . Without loss of generality, assume that P 2 P 1 . Then, there exists x P 2 but x P 1 . Thus, P 1 m ( x , I ) and P 2 ( S , I ) \ m ( x , I ) . By the item (1), ( S , I ) is Hausdorff.

(3) and (4) By Proposition 4.1 and (1), ( S , I ) is zero-dimensional. Since every zero-dimensional Hausdorff space is totally disconnected and completely regular, (3) and (4) are obvious.□

Theorem 4.12

Let S S I P with 0. If { 0 } = { 0 } and the collection { m ( x ) : x S } is a basis for the closed subsets of ( S ) , then ( S ) is compact.

Proof

Let { U α : α Λ } be a family of open sets with ( S ) { U α : α Λ } . Then, { ( S ) \ U α : α Λ } = . By the hypothesis, for any α Λ , ( S ) \ U α = { m ( b ) : b B α } . Let B = { B α : α Λ } . Then { ( S ) \ U α : α Λ } = { m ( b ) : b B } . Thus, { m ( b ) : b B } = . Assume that for every finite subset { b 1 , b 2 , , b n } of B , m ( b 1 b 2 b n ) = m ( b 1 ) m ( b 2 ) m ( b n ) . Then, b 1 b 2 b n 0 , for every finite subset { b 1 , b 2 , , b n } of B . Let A denote the subsemigroup of S , which is generated by B . Then, [ A ) is a weak filter and 0 [ A ) , i.e., [ A ) { 0 } = . By Lemma 4.5, [ A ) is contained in a weak filter M of S , which is maximal with respect to the property of not meeting { 0 } . Since B [ A ) M , we have B ( S \ M ) = . By Theorem 4.6, S \ M is a minimal prime ideal of S . So S \ M { m ( b ) : b B } , which is impossible. Therefore, there exists a finite subset { b 1 , b 2 , , b n } of B such that m ( b 1 b 2 b n ) = m ( b 1 ) m ( b 2 ) m ( b n ) = . And so, { ( S ) \ U α : α Λ } = m ( b 1 ) m ( b 2 ) m ( b n ) and { U α : α Λ } = ( ( S ) \ m ( b 1 ) ) ( ( S ) \ m ( b 2 ) ) ( ( S ) \ m ( b n ) ) . Since for any b B , there exists α Λ such that b B α and ( S ) \ U α m ( b ) , so there exists α i Λ such that ( S ) \ U α i m ( b ) , i.e., ( S ) \ m ( b ) U α . Therefore, there exists a finite subset { α 1 , α 2 , , α n } of Λ such that ( S ) { U α i : i = 1 , 2 , , n } . Hence, ( S ) is compact.□

We define the pseudocomplement in an ordered semigroup, which generalizes the concept of pseudocomplements in groupoids and semigroups [32] as follows:

Definition 4.13

Let S be an ordered semigroup with 0 and s S . An element s * is said to be a pseudocomplement if

  1. s * s = s s * = 0 ;

  2. for any t S , s t = t s = 0 implies that t s * t = t s * .

    S is called pseudocomplemented if for any s S , there exists a pseudocomplement in S .

A pseudocomplement of an element in an ordered semigroup does not necessarily exists. If it exists, it may not be unique, which is shown in the following example.

Example 4.14

(1) ( N , , ) with the usual ordering and standard multiply operation is an ordered semigroup. We show that for any x N + , x is a pseudocomplement of 0. In fact, for any m , n N + , m 0 = 0 m = n 0 = 0 n = 0 and m m n = n m , n m n = n m . So for any x N + , x is a pseudocomplement of 0. For any m N + , 0 is a pseudocomplement of m .

(2) For any n N , let S = ( [ 0 , 1 ] n , , ) , where for any ( x 1 , x 2 , , x n ) , ( y 1 , y 2 , , y n ) [ 0 , 1 ] n , ( x 1 , x 2 , , x n ) ( y 1 , y 2 , , y n ) if x i y i for 1 i n and ( x 1 , x 2 , , x n ) ( y 1 , y 2 , , y n ) = ( min ( x 1 , y 1 ) , min ( x 2 , y 2 ) , , min ( x n , y n ) ) . Then, S is an ordered semigroup with a zero element ( 0 , 0 , , 0 ) . For any ( x 1 , x 2 , , x n ) [ 0 , 1 ] n , if x i 0 for any 1 i n , then ( 0 , 0 , , 0 ) is a pseudocomplement of ( x 1 , x 2 , , x n ) ; if there exists I { 1 , 2 , , n } such that x i = 0 , i I , then ( y 1 , y 2 , , y n ) is a pseudocomplement of ( x 1 , x 2 , , x n ) , where y i = 1 , i I and y i = 0 , i I . Therefore, S is pseudocomplemented.

In the following example, we study the relationship between S I P and the class of pseudocomplemented ordered semigroups. We will show that if S S I P , then S may not be pseudocomplemented. It is not known whether every pseudocomplemented ordered semigroup is S I P .

Example 4.15

Consider the ordered semigroup ( T , , ) whose Cayley table is shown in Figure 2. The order is defined as: x y x = 0 , for any x , y T . By Theorem 2.18, it is easy to show that T S I P . Since a b b a , we have that 0 * does not exist. So T is not pseudocomplemented.

Figure 2 
               An ordered semigroup in 
                     
                        
                        
                           
                              
                                 S
                              
                              
                                 I
                                 P
                              
                           
                        
                        {{\mathbb{S}}}_{IP}
                     
                  , which is not pseudocomplemented.
Figure 2

An ordered semigroup in S I P , which is not pseudocomplemented.

Lemma 4.16

Let S be an ordered semigroup with 0. If S is pseudocomplemented, then { 0 } = { 0 } .

Proof

Let s { 0 } and s 0 . Suppose that n is the smallest positive integer for which s n = 0 . Put t = s n 1 ; then s t = t s = 0 . So t s * t . If n = 2 , s * t = s * s = 0 . If n > 2 , then s * t = ( s * s ) s n 2 = 0 . Hence, t = 0 , which contradicts the choice of n . So, s = 0 and { 0 } = { 0 } .□

Lemma 4.17

Let S be a pseudocomplemented commutative ordered semigroup. Consider the topological space ( S ) . Then, for any s S , k ( m ( s ) ) = { x S : s x = x s = 0 } .

Proof

Let s S . In case m ( s ) = , then k ( m ( s ) ) = S and for any P ( S ) , s P . So s { 0 } by Theorem 2.18. By Lemma 4.16, we have s = 0 . Hence, { x S : s x = x s = 0 } = S and k ( m ( s ) ) = { x S : x s = x s = 0 } . Otherwise, let x k ( m ( s ) ) . If P m ( s ) , then x k ( m ( s ) ) P and x s , s x P . If P h M ( s ) , then s P and x s , s x P . Hence, for any P ( S ) = m ( s ) h M ( s ) , x s , s x P . So, x s , s x { 0 } by Theorem 2.18 and x s = s x = 0 by Lemma 4.16. Thus, k ( m ( s ) ) { x S : s x = x s = 0 } . Conversely, let x S with s x = x s = 0 and P m ( s ) . Then, x P from the fact that P is a prime ideal. Hence, x k ( m ( s ) ) . So { x S : s x = x s = 0 } k ( m ( s ) ) . Thus, for any s S , we have { x S : s x = x s = 0 } = k ( m ( s ) ) .□

Let I be any ideal in an ordered semigroup S with 0. It should be clear that ( { m ( x , I ) ; x S } , , ) is a semilattice, which will be called the dual semilattice of S , and will be denoted by L ( S , I ) . If I = { 0 } , we denote L ( S , I ) by L S .

Theorem 4.18

Let S be a pseudocomplemented commutative ordered semigroup. Then,

  1. ( S ) is a Boolean space;

  2. the dual semilattice L S coincides with the dual algebra of ( S ) .

Proof

Since S is commutative, by Theorem 2.21, S S I P .

(1) Let s S . By Lemma 4.17, we have k ( m ( s ) ) = { x S : s x = x s = 0 } . By Proposition 4.2, we have m ( s ) = h M ( k ( m ( s ) ) ) = h M ( { x S : s x = x s = 0 } ) . For any x S with s x = x s = 0 , we have x s * x and so h M ( { x S : s x = x s = 0 } ) = h M ( s * ) . Hence, m ( s ) = h M ( s * ) = ( S ) \ m ( s * ) . Let V be a closed subset of ( S ) . By Proposition 4.1, there exists A S such that ( S ) \ V = { m ( a ) : a A } = { ( S ) \ m ( a * ) : a A } . So V = { m ( a * ) : a A } . Thus, L S is a basis for the closed subsets of ( S ) . Therefore, ( S ) is compact by Theorem 4.12. Since S S I P , by Theorem 4.11, we have ( S ) is totally disconnected and Hausdorff. Hence, ( S ) is a Boolean space.

(2) Let U be a subset of ( S ) , which is both open and closed. Since ( S ) is compact, there exist finitely many elements s 1 , s 2 , , s n S such that U = m ( s 1 ) m ( s n ) . By the proof of (1), there exists s S such that U = m ( s ) . Thus, the dual semilattice L S coincides with the dual algebra of ( S ) .□

Theorem 4.19

Let X be a T 0 zero-dimensional space. Then, X is homeomorphic to a dense subspace of ( S ) for some ordered semigroup S.

Proof

Let be the basis, which consists of all clopen subsets of X . Consider = ( , , ) as an ordered semigroup. Define f : X ( ) as follows: for any x in X , let

f ( x ) = { U : x U } .

It is clear that f ( x ) is a prime ideal of . For any U f ( x ) , there exists V = X \ U such that V f ( x ) and V U = . By Theorem 4.9, f ( x ) ( ) . Since X is a zero-dimensional space, we have f is one-to-one. For any U , we have f ( U ) = f ( X ) { P ( ) : U P } , which shows that f and f 1 are continuous. So X is homeomorphic to f ( X ) and f ( X ) is dense in ( ) .□

The following theorem relates the space of minimal prime ideals of an ordered semigroup to the space of minimal prime ideals of the dual semilattice of S .

Theorem 4.20

Let S S I P with 0. Then, ( S ) and ( L S ) are homeomorphic.

Proof

Consider L S = ( { m ( x ) ; x S } , , ) as an ordered semigroup. Define f : ( S ) ( L S ) as follows: for any P in ( S ) , let

f ( P ) = { m ( x ) : x P } .

Then, f ( P ) L S = L S f ( P ) = f ( P ) and for any y S , x P with m ( y ) m ( x ) , since P m ( x ) , we have P m ( y ) . Hence, y P and m ( y ) f ( P ) . So, f ( P ) is an ideal of L S . Because for any x , y S with m ( x ) m ( y ) f ( P ) , we have m ( x y ) f ( P ) by Proposition 4.10. So x y P , and hence x P or y P by the primeness of P . Hence, m ( x ) f ( P ) or m ( y ) f ( P ) . Therefore, f ( P ) is a prime ideal of L S . Since P is a minimal prime ideal of S , by Theorem 4.9, we have for any x P , there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P such that y 1 x n 1 y 2 x n 2 y j x n j = 0 . So for any m ( x ) f ( P ) , i.e., x P , there exist n 1 , n 2 n j N , y 1 , y 2 y j S \ P , i.e., m ( y 1 ) , m ( y 2 ) , , m ( y j ) L S \ f ( P ) such that m ( y 1 x n 1 y 2 x n 2 y j x n j ) = m ( y 1 ) m ( y 2 ) m ( x ) = . Since L S is commutative, we have L S S I P . So f ( P ) is a minimal prime ideal of L S by Theorem 4.9. Therefore, f is well defined.

By Theorem 4.11, ( S ) is a Hausdorff space, which ensures that the mapping f is one-to-one. Let Q be a minimal prime ideal of L S , i.e., Q ( L S ) , and put P = { x S : m ( x ) Q } . It is easy to see that P is a prime ideal in S . So, there exists a minimal prime ideal P 1 in S such that P 1 P by Corollary 4.7. Hence, f ( P 1 ) Q . Since f ( P 1 ) is a prime ideal and Q is a minimal prime ideal of L S , we have f ( P 1 ) = Q . Therefore, f is a bijection. By Proposition 4.1, { { Q ( L S ) : m ( x ) Q } : m ( x ) L S } is a basis for ( L S ) . Since f is a bijection, for any x S , we have { Q ( L S ) : m ( x ) Q } = { f ( P ) ( L S ) : m ( x ) f ( P ) , P ( S ) } = { f ( P ) : x P , P ( S ) } . So, f 1 ( { Q ( L S ) : m ( x ) Q } ) = { P ( S ) : x P } = m ( x ) and f ( m ( x ) ) = { f ( P ) : P m ( x ) } = { f ( P ) : x P , P ( S ) } = { Q ( L S ) : m ( x ) Q } . Because { m ( x ) ; x S } is a basis for ( S ) , we have f and f 1 are continuous. Therefore, ( S ) and ( L S ) are homeomorphic.□

5 Conclusion

Ordered semigroups are very important algebraic structures due to their various applications. The present study has addressed a connection among three research topics, i.e., ordered semigroups, spectra theory and topological spaces, each of them has applications across a wide variety of fields. We exhibited the hull-kernel topology on prime ideals of an ordered semigroup and discussed many topological properties about this topological space. Furthermore we discussed the minimal prime ideal space of an ordered semigroup. In future, we can consider how to apply the hull-kernel topology to other ordered structures, such as quantales, ordered groups and ordered rings.

Acknowledgement

The authors wish to thank the referees for their useful comments.

  1. Funding information: This work was supported by National Natural Science Foundation of China (No. 11701540).

  2. Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2022-02-22
Revised: 2024-05-30
Accepted: 2024-07-28
Published Online: 2024-08-23

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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