On some powerful ideals of semigroups

Gulf Journal of Mathematics Vol 5, Issue 1 (2017) 102-107 ON SOME POWERFUL IDEALS OF SEMIGROUPS ABUL BASAR1∗ AND MOHAMMAD YAHYA ABBASI2 Abstract. Let S be a non trivial additive cancellation commutative semigroup having a zero 0 with the quotient group G = {s1 − s2 | s1 , s2 ∈ S}. A prime ideal P of S is called strongly prime if for a, b ∈ G, whenever a + b ∈ P , we get a ∈ P or b ∈ P . We show that a prime ideal of S is strongly prime ⇔ it is powerful. Finally, we prove that if O is an oversemigroup of S and S and O have the nonzero ideal I where I is powerful in O, then 3I is a powerful ideal of S. 1. Introduction and preliminaries Let S be an additive commutative cancellation semigroups with a zero 0 and S 6= 0. The set G = {s1 − s2 | s1 , s2 ∈ S} is a torsion-free commutative group which is called the quotient group of S and S is a subsemigroup of G. The nonempty set O is called an oversemigroup of S if O is a subsemigroup of G which contains S. Suppose G is the quotient group of S. For each a ∈ G, a is called an integral element over S if na ∈ S for some positive integer n. An oversemigroup O is called an integral semigroup over S if every element a of O is an integral ′ ′ element over S. If S is the set of all integral elements over S in G, we term S as the integral closure of S. A nonempty subset I of a semigroup S is called an ideal of S if I + S ⊂ I, that is, x + s ∈ I for each x ∈ I and s ∈ S. A proper ideal P of S is called a prime ideal of S if a + b ∈ P with a, b ∈ S implies either a ∈ P or b ∈ P . Let N be the set of positive integers. The radical of an ideal I is rad(I) = {s ∈ S | ns ∈ I for some n ∈ N }. A semigroup S is a valuation semigroup if s ∈ G, then s ∈ S or −s ∈ S. We call S a seminormal semigroup [3] if 2s, 3s ∈ S for s ∈ G, we have s ∈ S. A commutative semigroup (with unit) which has only finitely many maximal ideals is called quasi-local semigroup and is denoted by (S, X) where X is a maximal ideal. An element a ∈ S is called a unit if a + b = 0 for some b ∈ S. Hedstrom and Houston[2] defined and studied strongly prime ideals of integral domains. Ayman Badawi and Evan Houston[1] introduced and investigated powerful ideals of integral domains as a generalization of strongly prime ideals of integral domains. Date: Received: Sep 10, 2016; Accepted: Dec 21, 2016. ∗ Corresponding author. 2010 Mathematics Subject Classification. 16D25, 20M12, 20M14. Key words and phrases. semigroups, powerful ideals, strongly prime ideals. 102 SEMIGROUPS AND POWERFUL IDEALS 103 2. Semigroups and Powerful Ideals We start with introducing the following definitions. Definition 2.1. A prime ideal L of S is called strongly prime if for any a, b ∈ G, whenever a + b ∈ L, we get a ∈ L or b ∈ L. Definition 2.2. A semigroup S is called a pseudo-valuation semigroup if every prime ideal of S is strongly prime. Definition 2.3. A non-zero ideal T of S is called powerful if for any a, b ∈ G, we have a ∈ S or b ∈ S, whenever a + b ∈ T . Theorem 2.4. Every valuation semigroup is a pseudo-valuation semigroup. Proof. Suppose S is a valuation semigroup, and suppose P is a prime ideal of S. Let a + b ∈ P where a, b ∈ G, the quotient group of S. If a,b ∈ S, there is nothing to prove. Let a ∈ / S. We obtain −a ∈ S since S is a valuation semigroup. Hence b = (a + b) − a ∈ P .  To prove Theorem 2.6, we need the following Lemma. Lemma 2.5. Let T be an ideal of S. For each a ∈ G\S, we have that T +(−a) ⊆ S if and only if T is powerful. Proof. Let T + (−a) ⊆ S, for each a ∈ G \ S. Assume b + c ∈ T , b; c ∈ G. Let b∈ / S. Then c = (−b) + (b + c) ∈ (−b) + T ⊆ T. Conversely, let T be a powerful ideal of S, and assume a ∈ G \ S. Then we have [a + (−a)] + x = x ∈ T , for x ∈ T and so (−a) + x ∈ S.  Theorem 2.6. Let P be a proper subset of T where P is a prime ideal of S and T is a powerful ideal of S. Then T /P is powerful in S/P . Proof. Suppose ϕ : S → S/P be the canonical homomorphism. Let a = ϕ(b) − ϕ(c) is a member of the quotient group of S/P and a ∈ / S/P . Therefore, b−c ∈ / S. So if x ∈ T , we obtain (c−b)+x ∈ S and then it implies that (ϕ(c)−[ϕ(b)+ϕ(x)] ∈ S/P . Hence S/P ⊇ (−a) + (T /P ) by Lemma 2.5.  Theorem 2.7. A prime ideal of S is strongly prime if and only if it is powerful. Proof. ⇒ It is trivial. ⇐ Let T be a powerful prime ideal of S. Suppose a + b ∈ T for some a, b ∈ G. So (2a) + (2b) ∈ T . Therefore, we may now consider a ∈ / S and b ∈ S. If 2a ∈ S, then (2a) + (2b) ∈ T and since a ∈ / S, 2a ∈ / T , it follows that 2b ∈ T , so b ∈ T . Again, since [2b − (a + b)] + 2a ∈ T , so if 2a ∈ / S then we have 2b − (a + b) ∈ S. Hence 2b = [2b − (a + b)] + (a + b) ∈ T , and we again obtain b ∈ T .  Definition 2.8. Let L be a strongly prime ideal of S and I be an ideal of S. We say L and I are comparable if L ⊂ I or I ⊂ L. Theorem 2.9. Suppose that T is a powerful ideal of S. Then: (i) Assume that I is an ideal of S. Then I ⊆ T or T + T ⊆ I. 104 A. BASAR AND M. Y. ABBASI (ii) Assume that I is a prime ideal of S. Then T and I are comparable. (iii) The prime ideals of S which are contained in Rad(T ) are linearly ordered. Proof. (i) Let I be an ideal of S and I T . We choose x ∈ I \ T , and suppose y, z ∈ T . Then [(y + z) − x] + (x − y) ∈ T , and because T is powerful such that (x − y) ∈ / S, we have (y + z) − x ∈ S. Hence (y + z) ∈ x + S ⊆ I. (ii) Easily follows from (i) (iii) Let X, Y be prime ideals which are properly contained in Rad(T ). Then X ⊂ T and Y ⊂ T and so both X and Y are powerful. Hence they are comparable by (ii).  Theorem 2.10. A semigroup S contains a unique greatest powerful ideal if it contains a powerful ideal. Proof. Let {Tβ } be a class of powerful ideals of S. Then P by Lemma 2.5, (−a) + Tβ ⊆ S for each β if a ∈ G \ S. Therefore (−a) + β Tβ ⊆ S, and we obtain P  β Tβ is powerful by the same Lemma 2.5. Theorem 2.11. Suppose T is a powerful ideal of S. Let a, b ∈ G and a + b ∈ Rad(T ), then there exists a positive integer n such that either na ∈ T or nb ∈ T . Proof. Obviously, n(a + b) ∈ T for some n > 0. Therefore (3n)a − (na + nb) + (3n)b − (na + nb) = na + nb ∈ T . Now we have either [3na − (na + nb)] ∈ S or [3nb − (na + nb)] ∈ S since T is powerful. Hence either 3na ∈ T or 3nb ∈ T .  The radical of a powerful ideal need not be powerful. We illustrate it by the following example. Example 2.12. Suppose E = G + I is a rank one discrete valuation semigroup, where G is a group and I = s + E is the maximal ideal of E. Further suppose a + b ∈ 3I, with a, b ∈ G. Express a = x + ns, b = y + ms, where x, y are units of E and n, m are integers. Since a + b ∈ 3I, we obtain n + m ≥ 3. Therefore either n ≥ 2 or m ≥ 2. Let n ≥ 2. Then a = x + ns ∈ 2I ⊆ S. Hence 3I is a powerful ideal of S. However, Rad(3I) = I + I = 2I is not powerful in S since 2s ∈ 2I but s ∈ / S. We now investigate a seminormal semigroup in which the radical of a powerful ideal is powerful. For this, first we introduce the following definition. Definition 2.13. A radical ideal R of S is called strongly radical if a ∈ G and na ∈ R for some n > 0 implies that a ∈ R. Theorem 2.14. Suppose T is a proper powerful ideal of S. Then Rad(T ) is powerful if and only if Rad(T ) is strongly radical. Proof. Obviously a powerful radical ideal is a strongly radical. Let Rad(T ) be strongly radical, and suppose that (a + b) ∈ Rad(T ) where a, b ∈ G. Then using Theorem 2.11. we obtain pa ∈ t or pb ∈ T for some p > 0. So now we may let pa ∈ T . Then pa ∈ Rad(T ) and so a ∈ S.  Lemma 2.15. Let T be a powerful ideal of S. If a ∈ G and na ∈ T for some n > 0, then (n + m)a ∈ S for each m ≥ 0. SEMIGROUPS AND POWERFUL IDEALS 105 Proof. Suppose l =min{p ≥ 1 | pa ∈ S}. Further, suppose that m is a positive integer. Express m = r + l + s where 0 ≤ s ≤ l. If s = 0, then it is obvious to realize that (n + m)a ∈ S. Let s > 0 and we obtain (l − s)a + (r + l + n + s)a = na + [(r + 1) + l]a ∈ T . We obtain (n + m)a = (r + l + n + s)a ∈ S since (l − s)a ∈ / S.  Theorem 2.16. Suppose T is a powerful ideal of S and O is an oversemigroup of S. Then we have that T + O is a powerful ideal of O. If T + O = O, then O is a valuation semigroup. Proof. Suppose a ∈ G \ O. Then a ∈ / S, and so by Lemma 2.5, (−a) + T ⊆ S. This shows that (−a) + T + O ⊆ O. Therefore, again by the same Lemma 2.5, T + O is powerful in O. Now it is straightforward to infer that O is powerful as an ideal of O ⇔ a semigroup O is a valuation semigroup.  Lemma 2.17. Suppose P is a prime ideal of a semigroup S with the quotient group G. Then (−a) + P ⊂ P if and only if P is strongly prime for a ∈ G − S. Proof. Let P be strongly prime. We have p = p + (−a) + a ∈ P if a ∈ G − S and p ∈ P , and so p + (−a) ∈ P or a ∈ P . We obtain p + (−a) ∈ P because a ∈ / S. Hence (−a) + P ⊂ P . Conversely, let (−a) + P ⊂ P if a ∈ G − S, and suppose x + y ∈ P . If x, y ∈ S, it is trivial. Therefore, we suppose that x ∈ / S and so (−x) + P ⊂ P and y = (−x) + (x + y) ∈ P .  Theorem 2.18. Suppose (S, X) is a quasi-local semigroup. Then the following assertions are equivalent: (i) S is a pseudo-valuation semigroup. (ii) For each pair P , Q of ideals of S, either P ⊂ Q or X + Q ⊂ X + P. (iii) For each pair of ideals P , Q of S, either P ⊂ Q or X + Q ⊂ P . (iv) X is strongly prime. Proof. (i) ⇒ (ii). Let P ( Q and choose x contained in P −Q. We have x−y ∈ /S for each y ∈ Q such that X + y ⊂ X + x ⊂ X + P and (y − x) + X ⊂ X. Hence X + P ⊃ X + Q. (ii) ⇒ (iii). This is straightforward. (iii) ⇒ (iv). Suppose x, y ∈ S and (x − y) ∈ / S. We have (x) ( (y) since (x−y) ∈ / S and hence X +y ⊂ (x) and (X +y)−a ⊂ S. We obtain X = (S +x)−y and (x − y) ∈ S if (X + y) − a = S. This leads to a contradiction. Therefore X + (y − a) ⊂ X. (iv) ⇒ (i). Suppose a is a member of the quotient group G of S, a ∈ / S, and assume that P is a prime ideal. So, by Lemma 2.17, it suffices to prove that (−a)+P ⊂ P . Suppose p ∈ P and because P ⊂ X, we obtain (−a) + p ∈ X. Therefore (−a) + p + (−a) ∈ X, and so [(−a) + p] + [(−a) + p] = (−a) + p + (−a) + p ∈ P . Thus we obtain (−a) + p ∈ P because P is prime and (−a) + p ∈ S.  Theorem 2.19. Let T be a powerful ideal of S and X = Rad(T ) is a maximal ideal of S. Then: (i) S is quasilocal having a maximal ideal X. ′ ′ (ii) T + S ⊆ X, and so T + S is an ideal of S. 106 A. BASAR AND M. Y. ABBASI ′ Proof. (i) Immediately follows from Theorem 2.9. For (ii), suppose a ∈ S \ S. ′ Recall that (−a) ∈ / S since S is integral over S. Therefore, by Lemma 2.5, a + T ⊆ S. If a + T = S, then −a ∈ S. So, indeed, a + T ⊆ X. This implies that ′ T + S ⊆ X.  Theorem 2.20. Suppose T is a powerful ideal of S, and suppose O 6= G is an oversemigroup of S. Then both S and O have a common ideal that is powerful in both S and O. We have the following statements: (i) Let X be the maximal ideal of O. If T + O = O, then Q = X ∩ S is a shared ideal which is powerful in both semigroups. (ii) If T + O 6= O, then 2T + O is a common ideal, and 3T + O is powerful in both semigroups. Proof. (i) By Theorem 2.16, O is a valuation semigroup and by Theorem 2.9, T is comparable to Q. It follows that Q ( T , and so Q is powerful and hence strongly prime in S. We recall that Q + O is powerful in O by Theorem 2.16. Suppose a ∈ O \ S. Obviously, (−a) ∈ / Q. Therefore we obtain a + Q ⊆ Q since (−a) + a + Q ⊆ Q and Q is strongly prime. (ii) Suppose a ∈ O \ S. If −a ∈ / S, then a + 2T ⊆ a + T ⊆ S by Lemma 2.5. If −a ∈ S, then by the assumption, −a ∈ / T , and so 2T ⊆ (−a) + S by Theorem 2.9. Therefore, a + 2T ⊆ S. Hence 2T + O is an ideal of S. Now 3T + O is powerful in S since 3T + O ⊆ T and by Theorem 2.16, 3T + O is powerful in O.  Theorem 2.21. Let O be an oversemigroup of S and that S and O have the nonzero ideal I. If I is powerful in O, then 3I = I + I + I is a powerful ideal of S. / O. Proof. Suppose a ∈ G \ S. We have that (−a) + I ⊆ O by Lemma 2.5 if a ∈ So we obtain (−a) + 3I ⊆ 2I + O ⊆ S. Let a ∈ O. By Theorem 2.9, we obtain 2I ⊆ a + O since a ∈ / I. Hence (−a) + 3I ⊆ I + O = I ⊆ S.  If O is a valuation semigroup, then 2I = I + I the second multiple of I is powerful in S. But, for general O, the third multiple is best possible. We support it by the following example. √ √ Example 2.22. Suppose G is the quotient group of the form Q( 2) = {a+b 2 | a, b ∈ Q}, where Q is a set of rational number and suppose that B = G[[Y ]] = G+X, X = Y +G[[Y ]]. Now suppose that O = Q+X, I = Y +O, and S = Q+I. Then S and O have the ideal I, and since O is a pseudo I is √ semigroup, √ √ valuation powerful √ in O. However, 2I is not powerful in S, since 2Y + 2Y = 2 2Y ∈ 2I, but 2Y ∈ / S. Acknowledgement:The first author is thankful to the National Board of Higher Mathematics, Department of Atomic Energy, Government of India for its financial assistance provided through Post-Doctoral Fellowship under Grant No: 2/40(30)/2015/R&D-II/9473. SEMIGROUPS AND POWERFUL IDEALS 107 References 1. A. Badawi and E. Houston, Powerful ideals, Strongly Primary Ideals, Almost Pseudo- Valuation Domains, and Conductive Domains, Communications in Algebra, 30(4) (2002), 1591 - 1606. 2. John R. Hedstrom and E. G. Houston, Pseudo-Valuation Domains, Pacific Journal of Mathematics,75(1) (1978), 138 - 147. 3. R. Matsuda and M. Kanemitsu, On seminormal semigroups, Archiv der Mathematic,(69) (1997), 279 - 285. 1 Department of Mathematics, Jamia Millia Islamia, New Delhi-110 025, India. E-mail address: [email protected] 2 Department of Mathematics, Jamia Millia Islamia, New Delhi-110 025, India E-mail address: yahya [email protected]