On Hoehnke ideal in ordered semigroups EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 11, No. 4, 2018, 911-921 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On Hoehnke ideal in ordered semigroups Niovi Kehayopulu Abstract. For a proper subset A of an ordered semigroup S, we denote by HA(S) the subset of S defined by HA(S) := {h ∈ S such that if s ∈ S\A, then s /∈ (shS]}. We prove, among others, that if A is a right ideal of S and the set HA(S) is nonempty, then HA(S) is an ideal of S; in particular it is a semiprime ideal of S. Moreover, if A is an ideal of S, then A ⊆ HA(S). Finally, we prove that if A and I are right ideals of S, then I ⊆ HA(S) if and only if s /∈ (sI] for every s ∈ S\A. We give some examples that illustrate our results. Our results generalize the Theorem 2.4 in Semigroup Forum 96 (2018), 523–535. 2010 Mathematics Subject Classifications: 06F05, 20M10 Key Words and Phrases: Ordered semigroup, semiprime subset (right ideal), completely semiprime ideal, prime subset (right ideal), completely prime ideal, Hoehnke ideal 1. Introduction and prerequisites Regarding the prime ideals, Clifford uses the term “prime” while Petrich the term “completely prime”. Clifford uses the term “semiprime ideal” and Petrich the term “com- pletely semiprime ideal (subset)”. For ordered semigroups I adopted the terminology due to Clifford; the authors in [2] the terminology by Petrich. Since in the present paper we refer to [2], for the sake of completeness, in particular for this paper, we will use the terms prime, semiprime, completely prime, completely semiprime. For an ordered semigroup S the zero of S, denoted by 0, is an element of S such that 0x = x0 = 0 and 0 ≤ x for every x ∈ S [1, 3]. In an ordered semigroup, the order plays an essential role and a relation between the multiplication and the order is needed. Let us first give the following definitions. Definition 1.1. [5; Definition 2] Let S be an ordered semigroup. A subset A of S is called completely prime if for any subsets B,C of S such that BC ⊆ A, we have B ⊆ A or C ⊆ A. Equivalent Definition: if x, y ∈ S such that xy ∈ A, then x ∈ A of y ∈ A. Definition 1.2. [5; Definition 3] Let S be an ordered semigroup. A subset A of S is called prime if for any ideals B,C of S such that BC ⊆ A, we have B ⊆ A or C ⊆ A. DOI: https://doi.org/10.29020/nybg.ejpam.v11i4.3341 Email address: nkehayop@math.uoa.gr (N. Kehayopulu) http://www.ejpam.com 911 c© 2018 EJPAM All rights reserved. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 912 Definition 1.3. [5; Definition 4] Let S be an ordered semigroup. A subset A of S is called completely semiprime if for any subset B of S such that B2 ⊆ A, we have B ⊆ A. Equivalent Definition: for every x ∈ S such that x2 ∈ A, we have x ∈ A. Definition 1.4. [4; Remark 4] Let S be an ordered semigroup. A subset A of S is called semiprime if for any ideal B of S such that B2 ⊆ A, we have B ⊆ A. Clearly, every completely prime (resp. completely semiprime) subset of S is a prime (resp. semiprime) subset. The authors in [2] call a right ideal I of an ordered semigroup S prime if it is proper and for any right ideals A, B of S, AB ⊆ I implies A ⊆ I or B ⊆ I. They call a right ideal of S semiprime if it is proper and for any right ideal A of S, A2 ⊆ I implies A ⊆ I. Their definition regarding the completely semiprime right ideal is the same with the usual one in rings, semigroups, ordered semigroups (see, for example [8]) with the only difference that they defined it as “proper”. Since every ordered semigroup (ring, semigroup) is itself a completely prime (prime) or completely semiprime (semiprime) subset of itself, these concepts have been never defined as “proper” in the existed bibliography (so in proofs, as well). For an ordered semigroup (S, ·,≤) possessing a zero 0 and an identity e of (S, ·) such that e 6= 0 it has been proved in [2] that if A is proper right ideal of S, then the set HA(S) := {h ∈ S such that if s ∈ S\A then s /∈ (shS]} is an interior ideal of S; since S possess an identity, the interior ideal HA(S) is also an ideal of S, but this should be emphasized in [2] since an interior ideal is not an ideal in general. Then the authors proved that if A is a proper right ideal of S then, for any right ideal I of S we have I ⊆ HA(S) if and only if s /∈ (sI] for all s ∈ S\A (property (2) in [2; Theorem 2.4]) and using this property they proved that if A is a proper ideal of S, then A ⊆ HA(S) (property (3) in [2; Theorem 2.4]) and that HA(S) is a semiprime ideal of S (property (1) in the same theorem) in the sense that HA(S) is a proper ideal of S and for any right ideal I of S such that I2 ⊆ HA(S) we have I ⊆ HA(S). That is, it has been proved that (2)⇒ (1) and (3). The set HA(S) has been called “Hoehnke ideal” in [2]. In the present paper we define the HA(S) for any proper subset A of an ordered semi- group S. We keep the definitions of semiprime and prime subsets of ordered semigroups given above, and we first prove that the set HA(S) is a semiprime subset of S. Then we prove that, if A is a proper ideal of S, then A is a subset of HA(S). We show, among others, that if A is a right ideal of S and the set HA(S) is nonempty, then HA(S) is an ideal of S; and hence a semiprime ideal of S. Finally, we prove that if A and I are right ideals of an ordered semigroup S, then we have I ⊆ HA(S) if and only if s /∈ (sI] for every s ∈ S\A. Unlike in [2], we have not used this last property to prove that HA(S) is semiprime and that A ⊆ HA(S); each of the three properties have been proved independently and the proof of Theorem 2.4 in [2] can be also given in the same way. In [2] only the definition of semiprime right ideal is given, there is no the definition of semiprime ideal in the paper. However, according to the proof of Theorem 2.4, the authors call an ideal A of an ordered semigroup S semiprime if it is proper and for any N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 913 right ideal I of S, I2 ⊆ A implies I ⊆ A. In case of ideals, this definition is equivalent to our Definition 1.4. So the results of the present paper generalize the Theorem 2.4 in [2]. On this occasion some information concerning the associate prime ideal has been also given. We give some examples that illustrate our results. 2. Main results Let (S, ·,≤) be an ordered semigroup. A nonempty subset A of S is called a right (resp. left) ideal of S if (1) AS ⊆ A (resp. SA ⊆ A) and (2) if a ∈ A and S 3 b ≤ a, then b ∈ A [4, 5]. For a subset A of an ordered semigroup (S, ·,≤), we denote by (A] the subset of S defined by (A] := {t ∈ S | t ≤ a for some a ∈ A} [4, 5]. If S is a right ideal, left ideal or an ideal of an ordered semigroup S, then (A] = A. For a proper subset A of S we denote by HA(S) the subset of S defined by HA(S) := {h ∈ S such that if s ∈ S\A, then s /∈ (shS]} [2]. Clearly, HA(S) = ∅ or HA(S) 6= ∅. Let us give an example for which HA(S) = ∅. Example 2.1. For the ordered semigroup S = {a, b, c} defined by Table 2 and Figure 2 and the subset A = {a, b} of S, we have HA(S) = ∅. For A = {b, c} we also have HA(S) = ∅. · a b c a a b a b a b a c a b c Table 2. c b a Figure 2. Proposition 2.2. (see also [2; Theorem 2.4(1)]) If S is an ordered semigroup, then the set HA(S) is a semiprime subset of S. Proof. Let I be an ideal of S such that I2 ⊆ HA(S). Then I ⊆ HA(S). Indeed: Let h ∈ I. We have to prove that h ∈ HA(S) that is, if s ∈ S\A, then s /∈ (shS]. Suppose s ∈ S\A and s ∈ (shS]. Since h ∈ I and I is an ideal of S, we have s ∈ (s(IS)] ⊆ (sI] ⊆ (SI] ⊆ (I] = I, N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 914 then s2 ∈ I2 ⊆ HA(S). Since s2 ∈ HA(S) and s ∈ S\A, we have s /∈ (ss2S] = (S] = S which is impossible. � Proposition 2.3. (see also [2; Theorem 2.4(3)] Let S be an ordered semigroup. Then we have the following: If A is a (proper) ideal of S, then A ⊆ HA(S). Proof. Let h ∈ A. Then h ∈ HA(S). Indeed: First of all, since A ⊆ S, we have h ∈ S. Let now s ∈ S\A. Then s /∈ (shS]. In fact: If s ∈ (shS] then, since A is an ideal of S, we have s ∈ (s(hS)] ⊆ (s(AS)] ⊆ (sA] ⊆ (SA] ⊆ (A] = A which is impossible. Since h ∈ S, s ∈ S\A and s /∈ (shS], we have h ∈ HA(S). � Corollary 2.4. If A is a (proper) ideal of S, then HA(S) 6= ∅. In Proposition 2.3 and Corollary 2.4 is not necessary to assume that the ideal A is a “proper” ideal of S; this is because, by writing HA(S), we already accepted that A is a proper subset of S. Proposition 2.5. Let (S, ·,≤) be an ordered semigroup. Then If HA(S) 6= ∅, then HA(S) is a right ideal of S. Proof. By hypothesis, HA(S) is a nonempty subset of S. Let h ∈ HA(S) and t ∈ S. Then ht ∈ HA(S). In fact: First of all, ht ∈ S. Let now s ∈ S\A. Then s /∈ (shtS]. Indeed: if s ∈ (sh(tS)], then s ∈ (shS]. On the other hand, since h ∈ HA(S) and s ∈ S\A, we have s /∈ (shS], we get a contradiction. Let now h ∈ HA(S) and S 3 g ≤ h. Then g ∈ HA(S). Indeed: Let s ∈ S\A. Since h ∈ HA(S) and s ∈ S\A, we have s /∈ (shS]. Since g ≤ h, we have (sgS] ⊆ (shS]. Then we get s /∈ (sgS]. Since s ∈ S\A and s /∈ (sgS], we have g ∈ HA(S). � Proposition 2.6. Let (S, ·,≤) be an ordered semigroup. Then If A is a right ideal of S and HA(S) 6= ∅, then HA(S) is a left ideal of S. Proof. By hypothesis, HA(S) is a nonempty subset of S. Let t ∈ S and h ∈ HA(S). Then th ∈ HA(S). Indeed: Let s ∈ S\A. We have to prove that s /∈ (sthS]. Suppose s ∈ (sthS]. Then we have st ∈ (sthS](S] ⊆ (sthS2] ⊆ (sthS] (1) On the other hand, since h ∈ HA(S), we have st ∈ A. Indeed: Let st ∈ S\A. Since h ∈ HA(S) and st ∈ S\A, we have st /∈ (sthS] which is impossible by (1). Since s ∈ (sthS] and st ∈ A, we have s ∈ ((st)hS] ⊆ (AhS] ⊆ (AS] ⊆ (A] = A, so s ∈ A which is impossible. Finally, as in Proposition 2.5, h ∈ HA(S) and S 3 g ≤ h imply g ∈ HA(S); thus HA(S) is a left ideal of S. � Corollary 2.7. If S is an ordered semigroup, A a right ideal of S and HA(S) 6= ∅, then HA(S) is an ideal of S. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 915 Proof. Since HA(S) 6= ∅, by Proposition 2.5, HA(S) is a right ideal of S. Since A a right ideal of S and HA(S) 6= ∅, by Proposition 2.6, HA(S) is a left ideal of S; and so HA(S) is an ideal of S. � Corollary 2.8. (see also [2; Theorem 2.4(1)]) If S is an ordered semigroup, A a right ideal of S and HA(S) 6= ∅, then HA(S) is a semiprime ideal of S. Proof. Since A is a right ideal of S and HA(S) 6= ∅, by Corollary 2.7, HA(S) is an ideal of S. On the other hand, by Proposition 2.2, HA(S) is a semiprime subset of S. Thus HA(S) is a semiprime ideal of S. � Proposition 2.9. Let (S, ·,≤) be an ordered semigroup and I a right ideal of the semigroup (S, ·). If s /∈ (sI] for every s ∈ S\A, then I ⊆ HA(S). Proof. Let h ∈ I. Then h ∈ HA(S), that is if s ∈ S\A, then s /∈ (shS]. Indeed: Let s ∈ S\A and s ∈ (shS]. Then we have s ∈ (s(IS)] ⊆ (sI], we get a contradiction. � Proposition 2.10. Let S be an ordered semigroup, HA(S) a left ideal of S and I be a subset of S such that I ⊆ HA(S). Then s /∈ (sI] for every s ∈ S\A. Proof. Let s ∈ S\A such that s ∈ (sI]. Then s ∈ (sI] ⊆ (sHA(S)] ⊆ (SHA(S)] ⊆ (HA(S)] = HA(S) since HA(S) is a left ideal of S. We have s ∈ S\A and s ∈ HA(S), so s /∈ (ssS] = (S] = S which is impossible. � Corollary 2.11. Let S be an ordered semigroup, A a right ideal of S, HA(S) 6= ∅ and I a subset of S such that I ⊆ HA(S). Then s /∈ (sI] for every s ∈ S\A. Proof. Since A is a right ideal of S and HA(S) 6= ∅, by Proposition 2.6, HA(S) is a left ideal of S. Since HA(S) is a left ideal of S and I a subset of S such that I ⊆ HA(S), by Proposition 2.10, s /∈ (sI] for every s ∈ S\A. � Corollary 2.12. (see also [2; Theorem 2.4(2)]) Let (S, ·,≤) be an ordered semigroup and A, I right ideals of (S, ·,≤). Then I ⊆ HA(S) if and only if s /∈ (sI] for every s ∈ S\A. Proof. =⇒. Since I is a right ideal of S and I ⊆ HA(S), we have HA(S) 6= ∅. Since A is a right ideal of S, HA(S) 6= ∅ and I is a subset of S such that I ⊆ HA(S), by Corollary 2.11, s /∈ (sI] for every s ∈ S\A. ⇐=. Since I is a right ideal of the ordered semigroup (S, ·,≤), it is a right ideal of the semigroup (S, ·) as well. Since I is a right ideal of (S, ·) and s /∈ (sI] for every s ∈ S\A, by Proposition 2.9, we have I ⊆ HA(S). � Summarizing, from Proposition 2.3, Corollary 2.8 and Corollary 2.12 we have the following theorem Theorem 2.13. Let (S, ·,≤) be an ordered semigroup. Then we have the following: (1) If A is a (proper) ideal of S, then A ⊆ HA(S). (2) If A a right ideal of S and HA(S) 6= ∅, then HA(S) is a semiprime ideal of S. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 916 (3) If A and I are right ideals of S, then I ⊆ HA(S) if and only if s /∈ (sI] for every s ∈ S\A. Again in property (1) the assumption “proper” can be omitted. Theorem 2.13 generalizes the Theorem 2.4 in [2]. It is enough to observe that if S has a zero and A is a proper right ideal of S, then 0 ∈ HA(S) and so HA(S) 6= ∅. We apply the above results to the following examples. The first two examples are on ordered semigroups in general; the third one is an example of an ordered semigroup (S, ·,≤) that contains a zero. Example 2.14. We consider the ordered semigroup S = {a, b, c, d, e, f} defined by Table 3 and Figure 3. · a b c d e f a a b b b e f b a b b b e f c a b b c e f d a b b d e f e e e e e e f f f f f f f f Table 3. a b e c d f Figure 3. For the subset A = {c, d, e} of S, we have HA(S) = ∅. The sets {f} and {e, f} are proper subsets of S, so the sets H{f}(S) and H{e,f}(S) are defined and, by Proposition 2.2, they are semiprime subsets of S. Independently, let us prove that H{e,f}(S) is semiprime. We first prove that H{e,f}(S) = {e, f}. Let now I be an ideal of S such that I2 ⊆ {e, f}. Then I ⊆ {e, f}. Indeed: if x ∈ I, then x2 ∈ I2 ⊆ {e, f}, so x2 = e or x2 = f . If x2 = e then, by Table 3, we have x = e and so x ∈ {e, f}. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 917 If x2 = f , then x = f and again x ∈ {e, f}. Independently, the set H{f}(S) is also a semiprime subset of S. Indeed, we have H{f}(S) = {f}; and if I is an ideal of S such that I2 ⊆ {f} and x ∈ I, then x2 = f and, by Table 3, x = f ; so I ⊆ {f}. The sets {f} and {e, f} are ideals of S. We have already seen that {f} ⊆ H{f}(S) and {e, f} ⊆ H{e,f}; that is a consequence of Proposition 2.3 as well. Since H{f}(S) 6= ∅, by Proposition 2.5, H{f}(S) is a right ideal of S; which is true. In a similar way all of the above results can be applied to this example. Example 2.15. We consider the ordered semigroup S = {a, b, c, d, e, f} defined by Table 4 and Figure 4. · a b c d e f a a a a d a a b a b b d b b c a b c d e e d a a d d d d e a b c d e e f a b c d e f Table 4. e a d b c f Figure 4. The proper ideals of S are the sets {a, d} and {a, b, d}. Moreover we have H{a,d}(S) = {a, d} and H{a,b,d}(S) = {a, b, d}. Since {a, d} (resp. {a, b, d}) is a right ideal of S and H{a,d}(S) 6= ∅ (resp. H{a,b,d}(S) 6= ∅), by Corollary 2.8, the sets H{a,d}(S) and H{a,b,d}(S) are semiprime ideals of S. Independently we can prove that the set H{a,d}(S) is a semiprime subset (and thus a semiprime ideal) of S, by showing that the set I = {a, d} is the only ideal of S such that I2 ⊆ {a, d} or in the way indicated in Example 2.14. The sets {a, b, d} and {a, d} are right ideals of S and {a, d} ⊆ H{a,b,d}(S). So, by the ⇒-part of Corollary 2.12, we have s /∈ (s{a, d}] for every s ∈ S\{a, b, d}. Independently, if s ∈ S\{a, b, d}, then s = c or s = e or s = f ; c /∈ (a, d] = (c{a, d}], e /∈ (a, d] = (e{a, d}] and f /∈ (a, d] = (f{a, d}]. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 918 In addition, since {a, b, d} and {a, d} are right ideals of S and s /∈ (sa, sd] = (s{a, d}] for every s ∈ S\{a, b, d}, by the ⇐-part of Corollary 2.12, we have {a, d} ⊆ H{a,b,d}(S); independently, we can check that this is true. All of the results given above in a similar way can be applied. Example 2.16. We consider the ordered semigroup S = {a, b, c, d, e} defined by Table 5 and Figure 5. This is an ordered semigroup with zero; the element a is the zero element of S; that is ax = xa = a and a ≤ x for every x ∈ S. · a b c d e a a a a a a b a a a a a c a a c c a d a a c c a e a a e e a Table 5. a b c e d Figure 5. The set A = {a, b, c, d} is a right ideal of S, S\A = {e} and HA(S) = {a, b, e}. Since HA(S) 6= ∅, by Corollary 2.8, HA(S) is a semiprime ideal of S. Independently, by looking at Table 5 and Figure 5 we can see that this is indeed an ideal of S. Moreover, it is a semiprime subset of S (and so a semiprime ideal of S as Corollary 2.8 shows). Indeed, if I is an ideal of S such that I2 ⊆ {a, b, e} and x ∈ I, then x2 = a or x2 = b or x2 = e. As there is no element x of S such that x2 = b or x2 = e, we have x2 = a and so x = a or x = b or x = e; that is x ∈ {a, b, e}. Thus we have I ⊆ {a, b, e} and {a, b, e} is semiprime. As one can see, A = {a, b, c, e} is an ideal of S and HA(S) = S. We can check that H{a,b,e}(S) = {a, b, e}. Since H{a,b,e}(S) is a left ideal of S and {a, b} is a subset of S such that {a, b} ⊆ H{a,b,e}(S), by Proposition 2.10, for every s ∈ S\{a, b, e}, we have s /∈ (s{a, b}) = (sa, sb], that is c /∈ (ca, cb] = (a] and d /∈ (da, db] = (a]; independently we can check that this is indeed so as c � a and d � b. All the above results can be applied to this example. N. Kehayopulu / Eur. J. Pure Appl. Math, 11 (4) (2018), 911-921 919 For a subset A of an ordered semigroup (S, ·,≤), we denote by Pr(A) the subset of S defined by Pr(A) := {p ∈ S | ∃ s ∈ S\A such that sp ∈ A} (cf. also [2]). Clearly Pr(A) = ∅ or Pr(A) 6= ∅. For the ordered semigroup S defined in Example 2.1 and the subset A = {c} of S, we have Pr(A) = ∅. If A is a proper left ideal of (S, ·), then A ⊆ Pr(A), and thus Pr(A) 6= ∅. Indeed: Let p ∈ A. Take an element s ∈ S\A (A is proper). We have sp ∈ (S\A)A ⊆ SA ⊆ A and so sp ∈ A. Since p ∈ S, s ∈ S\A and sp ∈ A, we have p ∈ Pr(A). Example 2.17. Let us consider the ordered semigroup of the Example 2.14. For the subset {c, e} of S, we have Pr({c, e}) = {e}. For the subset {c, d, e} of S, we also have Pr({c, d, e}) = {e}. On the other hand, the sets {f} and {e, f} are proper left ideals of S, and we have {f} = Pr({f}) ⊆ Pr({f}) and {e, f} = Pr({e, f}) ⊆ Pr({e, f}). In a semigroup (S, ·) containing an identity e, if A is a proper right ideal of S, then A ⊆ Pr(A). Indeed, as A is proper, we have e ∈ S\A; and if p ∈ A, then ep = p ∈ A, thus p ∈ Pr(A) (see also [2]). According to [2; Proposition 2.5], if (S, ·,≤) is an ordered semigroup, e an identity of (S, ·) and A a proper right ideal of (S, ·,≤), then the set Pr(A) is a completely prime right ideal of S and A ⊆ Pr(A). The first part of this proposition can be also obtained as a corollary to the following proposition. Proposition 2.18. Let A be a proper right of an ordered semigroup S. If Pr(A) is nonempty, then it is a completely prime right ideal of S. In contrast to semigroups containing identity, if S is an ordered semigroup and A is a proper right ideal of S, then the property A ⊆ Pr(A) does not hold in general. Let us show it by the following Example 2.19. Consider the ordered semigroup of the Example 2.14. As we have already seen in Example 2.17, for the subset A = {c, d, e} of S, we have Pr(A) = {e} and so A * Pr(A). We observe here that the set {c, d, e} is not an ideal of S. In this respect, we have the following Proposition 2.20. Let A be a proper ideal of an ordered semigroup (S, ·,≤). Then Pr(A) is a completely prime right ideal of S containing A. Proof. Since A is a proper left ideal of (S, ·), we have A ⊆ Pr(A) and so Pr(A) 6= ∅. Since A is a proper right ideal of (S, ·,≤) and Pr(A) 6= ∅, by Proposition 2.18, Pr(A) is a completely prime right ideal of S containing A. � We apply Proposition 2.20 to the following example Example 2.21. Consider the ordered semigroup S of the Example 2.14. The sets {f} and {e, f} are the only proper ideals of S; and as we have seen in Example 2.17, for the set A = {e, f}, we have Pr(A) = {e, f}. By Proposition 2.20, Pr(A) is a completely prime ideal of S. Independently, we can check that if C,D are subsets of S such that CD ⊆ {e, f}, then C ⊆ {e, f} or D ⊆ {e, f} (or we can check that if x, y ∈ S such that xy ∈ {e, f}, then x ∈ {e, f} or y ∈ {e, f}) which means that {e, f} is a completely REFERENCES 920 prime ideal of S. Similarly, the set Pr({f} (= {f}) is a completely prime ideal of S. This being so, we add to Example 2.14 a second proof that the sets {e, f} and {f} are indeed semiprime ideals of S; as every completely prime ideal is a prime ideal and every prime ideal is a semiprime ideal. According [2; p. 526, l. –9 to –7], if I and J are right ideals of an ordered semigroup S, then (IJ ] is a right ideal of S. The authors assume that each ordered semigroup has an identity and a zero [see p. 525, l. 22–23]. It might be noted that, more generally, if I is a nonempty subset of an ordered semigroup S and J a right ideal of S, then (IJ ] is a right ideal of S. If I is a left ideal of S and J a nonempty subset of S, then (IJ ] is a left ideal of S. As a consequence, if I and J are ideals of S, then (IJ ] is an ideal of S. The finite intersection of right (resp. left, two-sided) ideals of an ordered semigroup S, if it is nonempty, is a right (resp. left, two-sided) ideal of S; this generalizes the corresponding result in [2; Corollary 2.2]. Finally, it might be mentioned that the Proposition 2.1 in [2], actually a lemma used throughout the paper, is not new (see, for example [4; the Lemma] or [5; Lemma 1]). The Proposition 2.3 in [2] is also not new, it is a special case of the Proposition in [5], where has been shown that if an ideal of an ordered semigroup is completely semiprime and prime, then it is completely prime (without using the identity considered in [2]). The fact that every semigroup endowed with the order ≤= {x, y) | x = y} is an ordered semigroup and, as a consequence, the notion of a right chain ordered semigroup generalizes the notion of a right chain semigroup (p. 525, l. 11–19; p. 524, l. –10 to –7] in [2]) is well known as it is known for any type of ordered semigroups –see, for example [4–7]. I would like to thank the two anonymous referees for their time to read the paper carefully, their interest on my work and their prompt reply –something lately not very usual. References [1] G. Birkhoff, Lattice theory. Corrected reprint of the 1967 third edition. American Mathematical Society Colloquium Publications, 25. American Mathematical Society, Providence, R.I., 1979 vi+418 pp. [2] T. Changphas, P. Luangchaisri, R. Mazurek. On right chain ordered semigroups. Semigroup Forum 96(3):523–535, 2018. [3] L. Fuchs. Partially ordered algebraic systems. Pergamon Press, Oxford-London-New York-Paris; Addison-Wesley Publishing Co., Inc., Reading, Mass.-Palo Alto, Calif.- London 1963 ix+229 pp. [4] N. Kehayopulu. On weakly prime ideals of ordered semigroups. Math. Japon. 35(6):1051–1056, 1990. [5] N. Kehayopulu. On prime, weakly prime ideals in ordered semigroups. Semigroup Forum 44(3):341–346, 1992. REFERENCES 921 [6] N. Kehayopulu. Ordered semigroups whose elements are separated by prime ideals. Math. Slovaca 62(3):417–424, 2012. [7] N. Kehayopulu, M. Tsingelis. Archimedean ordered semigroups as ideal extensions. Semigroup Forum 78(2):343–348, 2009. [8] M. Petrich. Introduction to semigroups. Merrill Research and Lecture Series. Charles E. Merrill Publishing Co., Columbus, Ohio, 1973 viii+198 pp.