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Research Article Right -Weakly Regular -Semiringsdownloads.hindawi.com/archive/2014/948032.pdfย ยท...
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Research ArticleRight ๐-Weakly Regular ฮ-Semirings
R. D. Jagatap
Y. C. College of Science, Karad, India
Correspondence should be addressed to R. D. Jagatap; [email protected]
Received 20 July 2014; Accepted 11 November 2014; Published 15 December 2014
Academic Editor: K. C. Sivakumar
Copyright ยฉ 2014 R. D. Jagatap.This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The concepts of a ๐-idempotent ฮ-semiring, a right ๐-weakly regular ฮ-semiring, and a right pure ๐-ideal of a ฮ-semiring areintroduced. Several characterizations of them are furnished.
1. Introduction
ฮ-semiring was introduced by Rao in [1] as a generalizationof a ring, a ฮ-ring, and a semiring. Ideals in semirings werecharacterized by Ahsan in [2], Iseki in [3, 4], and Shabir andIqbal in [5]. Properties of prime and semiprime ideals in ฮ-semirings were discussed in detail by Dutta and Sardar [6].Henriksen in [7] defined more restricted class of ideals insemirings known as ๐-ideals. Some more characterizationsof ๐-ideals of semirings were studied by Sen and Adhikari in[8, 9]. ๐-ideal in a ฮ-semiring was defined by Rao in [1] andin [6] Dutta and Sardar gave some of its properties. Authorstudied ๐-ideals and full ๐-ideals of ฮ-semirings in [10]. Theconcept of a bi-ideal of a ฮ-semiring was given by author in[11].
In this paper efforts are made to introduce the conceptsof a ๐-idempotent ฮ-semiring, a right ๐-weakly regular ฮ-semiring, and a right pure ๐-ideal of a ฮ-semiring. Discusssome characterizations of a ๐-idempotent ฮ-semiring, a right๐-weakly regular ฮ-semiring, and a right pure ๐-ideal of a ฮ-semiring.
2. Preliminaries
First we recall some definitions of the basic concepts of ฮ-semirings that we need in sequel. For this we follow Duttaand Sardar [6].
Definition 1. Let ๐ and ฮ be two additive commutativesemigroups. ๐ is called a ฮ-semiring if there exists a mapping
๐ ร ฮ ร ๐ โ ๐ denoted by ๐๐ผ๐, for all ๐, ๐ โ ๐ and ๐ผ โ ฮ
satisfying the following conditions:
(i) ๐๐ผ(๐ + ๐) = (๐๐ผ๐) + (๐๐ผ๐);
(ii) (๐ + ๐)๐ผ๐ = (๐๐ผ๐) + (๐๐ผ๐);
(iii) ๐(๐ผ + ๐ฝ)๐ = (๐๐ผ๐) + (๐๐ฝ๐);
(iv) ๐๐ผ(๐๐ฝ๐) = (๐๐ผ๐)๐ฝ๐, for all ๐, ๐, ๐ โ ๐ and for all ๐ผ, ๐ฝ โฮ.
Definition 2. An element 0 in a ฮ-semiring ๐ is said to be anabsorbing zero if 0๐ผ๐ = 0 = ๐๐ผ0, ๐ + 0 = 0 + ๐ = ๐, for all๐ โ ๐ and ๐ผ โ ฮ.
Definition 3. Anonempty subset๐ of a ฮ-semiring ๐ is said tobe a sub-ฮ-semiring of ๐ if (๐, +) is a subsemigroup of (๐, +)and ๐๐ผ๐ โ ๐, for all ๐, ๐ โ ๐ and ๐ผ โ ฮ.
Definition 4. A nonempty subset ๐ of a ฮ-semiring ๐ is calleda left (resp., right) ideal of ๐ if ๐ is a subsemigroup of (๐, +)and ๐ฅ๐ผ๐ โ ๐ (resp., ๐๐ผ๐ฅ โ ๐) for all ๐ โ ๐, ๐ฅ โ ๐ and ๐ผ โ ฮ.
Definition 5. If ๐ is both left and right ideal of a ฮ-semiring๐, then ๐ is known as an ideal of ๐.
Definition 6. A right ideal ๐ผ of a ฮ-semiring ๐ is said to be aright ๐-ideal if ๐ โ ๐ผ and ๐ฅ โ ๐ such that ๐+๐ฅ โ ๐ผ; then ๐ฅ โ ๐ผ.
Similarly we define a left ๐-ideal of a ฮ-semiring ๐. If anideal ๐ผ is both right and left ๐-ideal, then ๐ผ is known as a ๐-ideal of ๐.
Hindawi Publishing CorporationAlgebraVolume 2014, Article ID 948032, 5 pageshttp://dx.doi.org/10.1155/2014/948032
2 Algebra
Example 7. Let๐0denote the set of all positive integers with
zero. ๐ = ๐0is a semiring and with ฮ = ๐, ๐ forms a ฮ-
semiring. A subset ๐ผ = 3๐0\ {3} of ๐ is an ideal of ๐ but not a
๐-ideal. Since 6 and 9 = 3 + 6 โ ๐ผ but 3 ๏ฟฝโ ๐ผ.
Example 8. If ๐ = ๐ is the set of all positive integers, then(๐, max., min.) is a semiring and with ฮ = ๐, ๐ forms a ฮ-semiring. ๐ผ
๐= {1, 2, 3, . . . , ๐} is a ๐-ideal for any ๐ โ ๐ผ.
Definition 9. For a nonempty ๐ผ of a ฮ-semiring ๐,
๐ผ = {๐ โ ๐ | ๐ + ๐ฅ โ ๐ผ, for some ๐ฅ โ ๐ผ} . (1)
๐ผ is called ๐-closure of ๐ผ.
Now we give a definition of a bi-ideal.
Definition 10 (see [11]). A nonempty subset ๐ต of a ฮ-semiring๐ is said to be a bi-ideal of ๐ if ๐ต is a sub-ฮ-semiring of ๐ and๐ตฮ๐ฮ๐ต โ ๐ต.
Example 11. Let๐ be the set of natural numbers and ฮ = 2๐.Then both๐ and ฮ are additive commutative semigroups. Animage of a mapping๐ ร ฮ ร ๐ โ ๐ is denoted by ๐๐ผ๐ anddefined as ๐๐ผ๐ = product of ๐, ๐ผ, ๐, for all ๐, ๐ โ ๐ and ๐ผ โ ฮ.Then๐ forms a ฮ-semiring. ๐ต = 4๐ is a bi-ideal of๐.
Example 12. Consider a ฮ-semiring ๐ = ๐2ร2(๐0), where
๐0denotes the set of natural numbers with zero and ฮ =
๐. Define ๐ด๐ผ๐ต = usual matrix product of ๐ด, ๐ผ and ๐ต, for๐ด, ๐ผ, ๐ต โ ๐.
๐ = {(0 ๐ฅ
0 ๐ฆ) | ๐ฅ, ๐ฆ โ ๐
0} (2)
is a bi-ideal of a ฮ-semiring ๐.
Definition 13. An element 1 in a ฮ-semiring ๐ is said to be anunit element if 1๐ผ๐ = ๐ = ๐๐ผ1, for all ๐ โ ๐ and for all ๐ผ โ ฮ.
Definition 14. A ฮ-semiring ๐ is said to be commutative if๐๐ผ๐ = ๐๐ผ๐, for all ๐, ๐ โ ๐ and for all ๐ผ โ ฮ.
Some basic properties of ๐-closure are given in thefollowing lemma.
Lemma 15. For nonempty subsets ๐ด and ๐ต of ๐, we have thefollowing.
(1) If ๐ด โ ๐ต, then ๐ด โ ๐ต.
(2) ๐ด is the smallest (left ๐-ideal, right ๐-ideal) ๐-idealcontaining (left ๐-ideal, right ๐-ideal) ๐-ideal ๐ด of ๐.
(3) ๐ด = ๐ด if and only if ๐ด is a (left ๐-ideal, right ๐-ideal)๐-ideal of ๐.
(4) ๐ด = ๐ด, where ๐ด is a (left ๐-ideal, right ๐-ideal) ๐-idealof ๐.
(5) ๐ดฮ๐ต = ๐ดฮ๐ต, where ๐ด and ๐ต are (left ๐-ideals, right๐-ideals) ๐-ideals of ๐.
Some results from [11] are stated which are useful forfurther discussion.
Result 1. For each nonempty subset ๐ of ๐, the followingstatements hold.
(i) ๐ฮ๐ is a left ideal of ๐.(ii) ๐ฮ๐ is a right ideal of ๐.(iii) ๐ฮ๐ฮ๐ is an ideal of ๐.
Result 2. For ๐ โ ๐, the following statements hold.(i) ๐ฮ๐ is a left ideal of ๐.(ii) ๐ฮ๐ is a right ideal of ๐.(iii) ๐ฮ๐ฮ๐ is an ideal of ๐.
Now onwards ๐ denotes a ฮ-semiring with an absorbingzero and an unit element unless otherwise stated.
3. ๐-Idempotent ฮ-Semiring
In this section we introduce and characterize the notion of a๐-idempotent ฮ-semiring.
Definition 16. A subset ๐ผ of a ฮ-semiring ๐ is said to be ๐-idempotent if ๐ผ = ๐ผฮ๐ผ.
Definition 17. A ฮ-semiring ๐ is said to be ๐-idempotent ifevery ๐-ideal of ๐ is ๐-idempotent.
Theorem 18. In ๐ the following statements are equivalent.(1) ๐ is ๐-idempotent.(2) For any ๐ โ ๐, ๐ โ ๐ฮ๐ฮ๐ฮ๐ฮ๐.(3) For every ๐ด โ ๐, ๐ด โ ๐ฮ๐ดฮ๐ฮ๐ดฮ๐.
Proof. (1) โ (2). Suppose that ๐ is a ๐-idempotent ฮ-semiring. For any ๐ โ ๐, (๐) = ๐ฮ๐ + ๐ฮ๐ + ๐ฮ๐ฮ๐ + ๐
0๐.
Then ๐ โ (๐) = ๐ฮ๐ + ๐ฮ๐ + ๐ฮ๐ฮ๐ + ๐0๐.
Hence by assumption (๐) = (๐)ฮ(๐) =
(๐ฮ๐ + ๐ฮ๐ + ๐ฮ๐ฮ๐ + ๐0๐)ฮ(๐ฮ๐ + ๐ฮ๐ + ๐ฮ๐ฮ๐ + ๐
0๐).
Therefore (๐) = (๐ฮ๐ + ๐ฮ๐ + ๐ฮ๐ฮ๐ + ๐0๐)ฮ
(๐ฮ๐ + ๐ฮ๐ + ๐ฮ๐ฮ๐ + ๐0๐).
Hence (๐) โ ๐ฮ๐ฮ๐ฮ๐ฮ๐. Therefore ๐ โ ๐ฮ๐ฮ๐ฮ๐ฮ๐.(2) โ (3). Let๐ด โ ๐ and ๐ โ ๐ด. Hence by assumption we
have ๐ โ ๐ฮ๐ฮ๐ฮ๐ฮ๐. Therefore ๐ โ ๐ฮ๐ดฮ๐ฮ๐ดฮ๐. Thus we get๐ด โ ๐ฮ๐ดฮ๐ฮ๐ดฮ๐.
(3) โ (1). Let ๐ด be any ๐-ideal of ๐. Then by assumption๐ด โ ๐ฮ๐ดฮ๐ฮ๐ดฮ๐ โ ๐ดฮ๐ฮ๐ด โ ๐ดฮ๐ด. As ๐ด is a ๐-ideal of ๐,๐ดฮ๐ด โ ๐ด. Therefore ๐ดฮ๐ด = ๐ด, which shows that ๐ is a ๐-idempotent ฮ-semiring.
Definition 19. A sub-ฮ-semiring ๐ผ of ๐ is a ๐-interior ideal of๐ if ๐ฮ๐ผฮ๐ โ ๐ผ and if ๐ โ ๐ผ and ๐ฅ โ ๐ such that ๐ + ๐ฅ โ ๐ผ, then๐ฅ โ ๐ผ.
Theorem 20. If ๐ is a ๐-idempotent ฮ-semiring, then a subsetof ๐ is a ๐-ideal if and only if it is a ๐-interior ideal.
Algebra 3
Proof. Let ๐ be a ๐-idempotent ฮ-semiring. As every ๐-idealis a ๐-interior ideal, one part of theorem holds. Conversely,suppose a subset ๐ผ of ๐ is a ๐-interior ideal of ๐. To show ๐ผ isa ๐-ideal of ๐, let ๐ฅ โ ๐ผ and ๐ก โ ๐. As ๐ is a ๐-idempotentฮ-semiring, ๐ฅ โ ๐ฮ๐ฅฮ๐ฮ๐ฅฮ๐ (see Theorem 18). Therefore,for any ๐ผ โ ฮ, ๐ฅ๐ผ๐ก โ ๐ฮ๐ฅฮ๐ฮ๐ฅฮ๐ฮ๐ โ ๐ฮ๐ฅฮ๐ฮ๐ฅฮ๐ฮ๐ โ
๐ฮ๐ผฮ๐ฮ๐ผฮ๐ฮ๐ โ ๐ฮ๐ผฮ๐ โ ๐ผ. Similarly we can show that๐ก๐ผ๐ฅ โ ๐ผ. Therefore ๐ผ is a ๐-ideal of ๐.
Theorem 21. ๐ is ๐-idempotent if and only if๐ดฮ๐ต = ๐ดโฉ๐ต, forany ๐-interior ideals ๐ด and ๐ต of ๐.
Proof. Suppose a ฮ-semiring ๐ is ๐-idempotent. Let ๐ด and ๐ตbe any two ๐-interior ideals of ๐. Then, by Theorem 20, ๐ดand ๐ต are two ๐-ideals of ๐. Hence ๐ดฮ๐ต โ ๐ด and ๐ดฮ๐ต โ ๐ต.Therefore ๐ดฮ๐ต โ ๐ด โฉ ๐ต. By assumption ๐ด โฉ ๐ต = (๐ด โฉ ๐ต)
2=
(๐ด โฉ ๐ต)ฮ(๐ด โฉ ๐ต) โ ๐ดฮ๐ต.Therefore๐ดฮ๐ต = ๐ดโฉ๐ต. Conversely,let ๐ด be any ๐-ideal of ๐. As every ๐-ideal of ๐ is a ๐-interiorideal of ๐, by assumption, ๐ดฮ๐ด = ๐ด โฉ๐ด = ๐ด. Therefore ๐ is a๐-idempotent ฮ-semiring.
4. Right ๐-Weakly Regular ฮ-Semiring
Definition 22. A ฮ-semiring ๐ is said to be right ๐-weaklyregular if, for any ๐ โ ๐, ๐ โ (๐ฮ๐)2.
Theorem 23. In ๐, the following statements are equivalent.
(1) ๐ is right ๐-weakly regular.
(2) ๐ 2 = ๐ , for each right ๐-ideal ๐ of ๐.
(3) ๐ โฉ ๐ผ = ๐ ฮ๐ผ, for a right ๐-ideal ๐ and a ๐-ideal ๐ผ of ๐.
Proof. (1) โ (2). For any right ๐-ideal ๐ of ๐, ๐ 2 = ๐ ฮ๐ โ
๐ ฮ๐ โ ๐ . Hence ๐ 2 = ๐ ฮ๐ โ ๐ . For the reverse inclusion,let ๐ โ ๐ . As ๐ is right ๐-weakly regular, ๐ โ (๐ฮ๐)
2=
(๐ฮ๐)ฮ(๐ฮ๐) โ (๐ ฮ๐)ฮ(๐ ฮ๐) โ ๐ ฮ๐ = ๐ 2. Thus ๐ 2 = ๐ ,for each right ๐-ideal ๐ of ๐.
(2) โ (1). For any ๐ โ ๐, ๐ โ ๐ฮ๐ โ ๐ฮ๐ and ๐ฮ๐ is a right๐-ideal of ๐, then by assumption (๐ฮ๐)2 = (๐ฮ๐). Therefore๐ โ (๐ฮ๐)
2. Hence ๐ is right ๐-weakly regular.(2) โ (3). Let ๐ be a right ๐-ideal and ๐ผ be a ๐-ideal of ๐.
Then ๐ โฉ ๐ผ is a right ๐-ideal of ๐. By assumption (๐ โฉ ๐ผ)2 =๐ โฉ ๐ผ. Consider ๐ โฉ ๐ผ = (๐ โฉ ๐ผ)2 = (๐ โฉ ๐ผ)ฮ(๐ โฉ ๐ผ) โ ๐ ฮ๐ผ.Clearly ๐ ฮ๐ผ โ ๐ and ๐ ฮ๐ผ โ ๐ผ. Then ๐ ฮ๐ผ โ ๐ = ๐ and ๐ ฮ๐ผ โ๐ผ = ๐ผ, since ๐ is a right ๐-ideal and ๐ผ is a two sided ๐-ideal of๐. Therefore ๐ ฮ๐ผ โ ๐ โฉ ๐ผ. Hence ๐ โฉ ๐ผ = ๐ ฮ๐ผ.
(3) โ (2). Let ๐ be a right ๐-ideal of ๐ and let (๐ ) be twosided ideal generated by๐ .Then (๐ ) = ๐ฮ๐ ฮ๐. By assumption๐ โฉ (๐ ) = ๐ ฮ(๐ ). Hence ๐ = (๐ ฮ๐)ฮ(๐ ฮ๐) โ ๐ ฮ๐ = ๐ 2.Therefore ๐ 2 = ๐ .
Theorem 24. ๐ is right ๐-weakly regular if and only if everyright ๐-ideal of ๐ is semiprime.
Proof. Suppose ๐ is right ๐-weakly regular. Let ๐ be a right๐-ideal of ๐ such that ๐ดฮ๐ด โ ๐ , for any right ๐-ideal ๐ด of ๐.๐ด = ๐ดฮ๐ด. Then๐ดฮ๐ด โ ๐ = ๐ . Therefore๐ด โ ๐ . Hence ๐ is asemiprime right ๐-ideal of ๐. Conversely, suppose every right๐-ideal of ๐ is semiprime. Let ๐ be a right ๐-ideal of ๐. ๐ ฮ๐ isalso a right ๐-ideal of ๐. By assumption ๐ ฮ๐ is a semiprimeright ๐-ideal of ๐. ๐ ฮ๐ โ ๐ ฮ๐ implies ๐ โ ๐ ฮ๐ . Therefore๐ = ๐ โ ๐ ฮ๐ = ๐ 2. Therefore ๐ 2 = ๐ . Hence ๐ is right๐-weakly regular byTheorem 23.
Definition 25 (see [12]). A latticeL is said to be Brouwerianif, for any ๐, ๐ โ L, the set of all ๐ฅ โ L satisfying thecondition ๐ โง ๐ฅ โค ๐ contains the greatest element.
If ๐ is the greatest element in this set, then the element ๐is known as the pseudocomplement of ๐ relative to ๐ and isdenoted by ๐ : ๐.
Thus a latticeL is a Brouwerian if ๐ : ๐ exists for all ๐, ๐ โL.
LetL๐denote the family of all ๐-ideals of ๐.Then โจL
๐, โโฉ
is a partially ordered set. As {0}, ๐ โ L๐and โ
๐ผโฮ๐ผ๐ผโ L๐,
for all ๐ผ๐ผโ L๐and ฮ is an indexing set, we have L
๐is a
complete lattice under โง and โจ defined by ๐ผ โจ ๐ฝ = ๐ผ + ๐ฝ and๐ผ โง ๐ฝ = ๐ผ โฉ ๐ฝ. Further we have the following.
Theorem 26. If ๐ is a right ๐-weakly regular ฮ-semiring, thenL๐is a Brouwerian lattice.
Proof. Let ๐ต and ๐ถ be any two ๐-ideals of ๐. Consider thefamily of ๐-ideals ๐พ = {๐ผ โ L
๐| ๐ผ โฉ ๐ต โ ๐ถ}. Then by
Zornโs lemma there exists a maximal element๐ in ๐พ. Select๐ผ โ L
๐such that ๐ตโฉ๐ผ โ ๐ถ. ByTheorem 23, we have ๐ตฮ๐ผ โ ๐ถ.
To show that ๐ตฮ(๐ผ +๐) โ ๐ถ. Let ๐ฅ โ ๐ตฮ(๐ผ +๐). Then๐ฅ = โ
๐
๐=1๐๐๐ผ๐๐ฅ๐, where ๐
๐โ ๐ต, ๐ผ
๐โ ฮ, and ๐ฅ
๐โ ๐ผ +๐.
Therefore ๐๐+๐ฅ๐โ ๐ผ+๐ for some ๐
๐โ ๐ผ+๐ (seeDefinition 9).
๐ฅ = โ๐
๐=1๐๐๐ผ๐(๐๐+ ๐ฅ๐) โ ๐ตฮ(๐ผ + ๐) = (๐ตฮ๐ผ) + (๐ตฮ๐) โ
๐ถ, as ๐ตฮ๐ผ โ ๐ถ and ๐ตฮ๐ โ ๐ถ. Hence ๐ตฮ(๐ผ +๐) โ ๐ถ
implies ๐ตฮ(๐ผ +๐) โ ๐ถ = ๐ถ, since ๐ถ is a ๐-ideal. Therefore,by Theorem 23, ๐ต โฉ (๐ผ +๐) โ ๐ถ. But, by the maximality,we have ๐ผ +๐ โ ๐ which implies ๐ผ โ ๐. Hence L
๐is
Brouwerian.
As L๐satisfies infinite meet distributive property prop-
erty of lattice, we have the following.
Corollary 27. If ๐ is a right ๐-weakly regular ฮ-semiring, thenL๐is a distributive lattice (see Birkoff [12]).
Theorem 28. If ๐ is a right ๐-weakly regular ฮ-semiring, thena ๐-ideal ๐ of ๐ is prime if and only if ๐ is irreducible.
Proof. Let ๐ be a right ๐-weakly regular ฮ-semiring and let ๐be a ๐-ideal of ๐. If๐ is a prime ๐-ideal of ๐, then clearly๐ is anirreducible ๐-ideal. Suppose๐ is an irreducible ๐-ideal of ๐. Toshow that๐ is a prime ๐-ideal. Let๐ด and๐ต be any two ๐-idealsof ๐ such that๐ดฮ๐ต โ ๐.Then, byTheorem 23, we have๐ดโฉ๐ต โ
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๐. Hence (๐ด โฉ ๐ต) + ๐ = ๐. AsL๐is a distributive lattice, we
have (๐ด + ๐) โฉ (๐ต + ๐) = ๐. Therefore ๐ is an irreducible ๐-ideal implies ๐ด + ๐ = ๐ or ๐ต + ๐ = ๐. Then ๐ด โ ๐ or ๐ต โ ๐.Therefore ๐ is a prime ๐-ideal of ๐.
As a generalization of a fully prime semiring defined byShabir and Iqbal in [5], we define a fully ๐-prime ฮ-semiringin [10] as follows.
A ฮ-semiring ๐ is said to be a fully ๐-prime ฮ-semiring ifeach ๐-ideal of ๐ is a prime ๐-ideal.
Theorem 29. A ฮ-semiring ๐ is a fully ๐-prime ฮ-semiring ifand only if ๐ is right ๐-weakly regular and the set of ๐-ideals of๐ is a totally ordered set by the set inclusion.
Proof. Suppose that a ฮ-semiring ๐ is a fully ๐-prime ฮ-semiring. Therefore every ๐-ideal of ๐ is a prime ๐-ideal. Asevery prime ๐-ideal is a semiprime ๐-ideal, we have ๐which isa right ๐-weakly regular ฮ-semiring by Theorem 24. For anytwo ๐-ideals ๐ด and ๐ต of ๐, ๐ดฮ๐ต โ ๐ด โฉ ๐ต. By assumption๐ด โฉ ๐ต is a prime ๐-ideal and hence we have ๐ด โ ๐ด โฉ ๐ต or๐ต โ ๐ด โฉ ๐ต. But then ๐ด โฉ ๐ต = ๐ด or ๐ด โฉ ๐ต = ๐ต. Hence either๐ด โ ๐ต or ๐ต โ ๐ด. This shows that the set of ๐-ideals of ๐ is atotally ordered set by the set inclusion. Conversely, assume ๐is right ๐-weakly regular and the set of ๐-ideals of ๐ is a totallyordered set by set inclusion. To show that ๐ is a fully ๐-primeฮ-semiring. Let ๐ be a ๐-ideal of ๐ and ๐ดฮ๐ต โ ๐, for any ๐-ideals ๐ด and ๐ต of ๐. Then ๐ดฮ๐ต โ ๐. Hence by Theorem 23,we have ๐ด โฉ ๐ต = ๐ดฮ๐ต โ ๐. By assumption either ๐ด โ ๐ต or๐ต โ ๐ด.Therefore๐ดโฉ๐ต = ๐ด or๐ดโฉ๐ต = ๐ต. Hence either๐ด โ ๐
or ๐ต โ ๐. This shows that ๐ is a prime ๐-ideal of ๐.Therefore ๐ is a fully ๐-prime ฮ-semiring.
Now we define a ๐-bi-ideal of a ฮ-semiring.
Definition 30. A nonempty subset ๐ต of a ฮ-semiring ๐ is saidto be a ๐-bi-ideal of ๐ if๐ต is a sub-ฮ-semiring of ๐,๐ตฮ๐ฮ๐ต โ ๐ต,and for ๐ โ ๐ต and ๐ฅ โ ๐ such that ๐ + ๐ฅ โ ๐ต; then ๐ฅ โ ๐ต.
Theorem 31. ๐ is right ๐-weakly regular if and only if ๐ต โฉ ๐ผ โ๐ตฮ๐ผ, for any ๐-bi-ideal ๐ต and ๐-ideal ๐ผ of ๐.
Proof. Suppose ๐ is a right ๐-weakly regular ฮ-semiring. Let๐ต be a ๐-bi-ideal and let ๐ผ be a ๐-ideal of ๐. Let ๐ โ ๐ตโฉ๐ผ.Then๐ โ (๐ฮ๐)
2= (๐ฮ๐)ฮ(๐ฮ๐) โ ๐ตฮ(๐ฮ๐ผฮ๐) โ ๐ตฮ๐ผ. Therefore
๐ต โฉ ๐ผ โ ๐ตฮ๐ผ. Conversely, let ๐ be a right ๐-ideal of ๐. Then๐ itself is a ๐-bi-ideal of ๐. Hence by assumption ๐ = ๐ โฉ
(๐ ) โ ๐ ฮ(๐ ) = ๐ ฮ(๐ฮ๐ ฮ๐) = (๐ ฮ๐)ฮ(๐ ฮ๐) โ ๐ ฮ๐ . Therefore๐ = ๐ ฮ๐ = ๐ 2. Hence, by Theorem 23, ๐ is a right ๐-weaklyregular ฮ-semiring.
Theorem32. ๐ is right ๐-weakly regular if and only if๐ตโฉ๐ผโฉ๐ โ
๐ตฮ๐ผฮ๐ , for any ๐-bi-ideal ๐ต, ๐-ideal ๐ผ, and a right ๐-ideal ๐ of๐.
Proof. Suppose ๐ is a right ๐-weakly regular ฮ-semiring. Let๐ต be a ๐-bi-ideal, let ๐ผ be a ๐-ideal, and let ๐ be a right ๐-idealof ๐. Let ๐ โ ๐ต โฉ ๐ผ โฉ ๐ . Then ๐ โ (๐ฮ๐)
2= (๐ฮ๐)ฮ(๐ฮ๐) โ
(๐ฮ๐)ฮ(๐ฮ๐)ฮ(๐ฮ๐)ฮ๐ โ ๐ตฮ(๐ฮ๐ผฮ๐)ฮ(๐ ฮ๐) โ ๐ตฮ๐ผฮ๐ .Therefore ๐ตโฉ๐ผโฉ๐ โ ๐ตฮ๐ผฮ๐ . Conversely, for a right ๐-ideal ๐ of ๐, ๐ itself is being a ๐-bi-ideal and ๐ itself is being a ๐-idealof ๐.Then by assumption๐ โฉ๐โฉ๐ โ ๐ ฮ๐ฮ๐ โ ๐ ฮ๐ .Therefore๐ โ ๐ ฮ๐ . Therefore ๐ = ๐ ฮ๐ = ๐ 2. Then, by Theorem 23, ๐is right ๐-weakly regular.
5. Right Pure ๐-Ideals
In this section we define a right pure ๐-ideal of a ฮ-semiring๐ and furnish some of its characterizations.
Definition 33. A ๐-ideal ๐ผ of a ฮ-semiring ๐ is said to be a rightpure ๐-ideal if, for any ๐ฅ โ ๐ผ, ๐ฅ โ ๐ฅฮ๐ผ.
Theorem 34. A ๐-ideal ๐ผ of ๐ is right pure if and only if ๐ โฉ๐ผ =๐ ฮ๐ผ, for any right ๐-ideal ๐ of ๐.
Proof. Let ๐ผ be a right pure ๐-ideal and let ๐ be a right ๐-idealof ๐. Then clearly ๐ ฮ๐ผ โ ๐ โฉ ๐ผ. Now let ๐ โ ๐ โฉ ๐ผ. As ๐ผ is aright pure ๐-ideal, ๐ โ ๐ฮ๐ผ โ ๐ ฮ๐ผ. This gives ๐ โฉ ๐ผ โ ๐ ฮ๐ผ.By combining both the inclusions we get ๐ โฉ ๐ผ = ๐ ฮ๐ผ. Con-versely, assume the given statement holds. Let ๐ผ be ๐-ideal of๐ and ๐ โ ๐ผ. (๐)
๐denotes a right ๐-ideal generated by ๐ and
(๐)๐= ๐0๐ + ๐ฮ๐ผ, where ๐
0denotes the set of nonnegative
integers. Then ๐ โ (๐)๐โฉ ๐ผ = (๐)
๐ฮ๐ผ = (๐
0๐ + ๐ฮ๐ผ)ฮ๐ผ =
(๐0๐ + ๐ฮ๐ผ)ฮ๐ผ โ ๐ฮ๐ผ (see Result 2).Therefore ๐ผ is a right pure
๐-ideal of ๐.
Theorem 35. The intersection of right pure ๐-ideals of ๐ is aright pure ๐-ideal of ๐.
Proof. Let ๐ด and ๐ต be right pure ๐-ideals of ๐. Then for anyright ๐-ideal๐ of ๐we have๐ โฉ๐ด = ๐ ฮ๐ด and๐ โฉ๐ต = ๐ ฮ๐ต.Weconsider ๐ โฉ(๐ดโฉ๐ต) = (๐ โฉ๐ด)โฉ๐ต = (๐ ฮ๐ด)โฉ๐ต = (๐ ฮ๐ด)ฮ๐ต =๐ ฮ(๐ดฮ๐ต) = ๐ ฮ(๐ด โฉ ๐ต).Therefore๐ดโฉ๐ต is a right pure ๐-idealof ๐.
Characterization of a right ๐-weakly regular ฮ-semiringin terms of right pure ๐-ideals is given in the followingtheorem.
Theorem 36. ๐ is right ๐-weakly regular if and only if any ๐-ideal of ๐ is right pure.
Proof. Suppose that ๐ is a right ๐-weakly regular ฮ-semiring.Let ๐ผ be a ๐-ideal and let ๐ be a right ๐-ideal of ๐. Then, byTheorem 23, ๐ โฉ ๐ผ = ๐ ฮ๐ผ. Hence, byTheorem 34, any ๐-ideal๐ผ of ๐ is right pure. Conversely, suppose that any ๐-ideal of ๐is right pure. Then, from Theorems 34 and 23, we get that ๐is a right ๐-weakly regular ฮ-semiring.
Algebra 5
6. Space of Prime ๐-Ideals
Let ๐ be a ฮ-semiring and letโ๐be the set of all prime ๐-ideals
of ๐. For each ๐-ideal ๐ผ of ๐ define ฮ๐ผ= {๐ฝ โ โ
๐| ๐ผ โ ๐ฝ} and
๐ (โ๐) = {ฮ
๐ผ| ๐ผ is a ๐-ideal of ๐} . (3)
Theorem 37. If ๐ is a right ๐-weakly regular ฮ-semiring, then๐(โ๐) forms a topology on the set โ
๐. There is an isomorphism
between lattice of ๐-idealsL๐and ๐(โ
๐) (lattice of open subsets
of โ๐).
Proof. As {0} is a ๐-ideal of ๐ and each ๐-ideal of ๐ contains{0}, we have the following:
ฮ{0}
= {๐ฝ โ โ๐ | {0} โ ๐ฝ} = ฮฆ. Therefore ฮฆ โ ๐(โ
๐).
As ๐ itself is a ๐-ideal, ฮ๐= {๐ฝ โ โ
๐| ๐ โ ๐ฝ} = โ
๐imply
โ๐โ ๐(โ
๐). Now let ฮ
๐ผ๐โ ๐(โ
๐) for ๐ โ ฮ; ฮ is an indexing
set, and ๐ผ๐is a ๐-ideal of ๐. Therefore ฮ
๐ผ๐= {๐ฝ โ โ
๐| ๐ผ๐
โ ๐ฝ}.As โ๐ฮ๐ผ๐= {๐ฝ โ โ
๐| โ๐๐ผ๐
โ ๐ฝ}, โ๐๐ผ๐is a ๐-ideal of ๐.
Therefore โ๐ฮ๐ผ๐= ฮโ๐๐ผ๐โ ๐(โ
๐). Further let ฮ
๐ด, ฮ๐ตโ
๐(โ๐).Let ๐ฝ โ ฮ
๐ดโฮ๐ต; ๐ฝ is a prime ๐-ideal of ๐. Hence ๐ด โ ๐ฝ
and ๐ต โ ๐ฝ. Suppose that ๐ด โฉ ๐ต โ ๐ฝ. In a right weakly regularฮ-semiring, prime ๐-ideals and strongly irreducible ๐-idealscoincide. Therefore ๐ฝ is a strongly irreducible ๐-ideal of ๐. As๐ฝ is a strongly irreducible ๐-ideal of ๐,๐ด โ ๐ฝ or๐ต โ ๐ฝ, which isa contradiction to ๐ด โ ๐ฝ and ๐ต โ ๐ฝ. Hence ๐ด โฉ ๐ต โ ๐ฝ implies๐ฝ โ ฮ
๐ดโฉ๐ต. Therefore ฮ
๐ดโฮ๐ตโ ฮ๐ดโฉ๐ต
. Now let ๐ฝ โ ฮ๐ดโฉ๐ต
.Then๐ดโฉ๐ต โ ๐ฝ implies๐ด โ ๐ฝ and๐ต โ ๐ฝ.Therefore ๐ฝ โ ฮ
๐ดand
๐ฝ โ ฮ๐ตimply ๐ฝ โ ฮ
๐ดโฮ๐ต. Thus ฮ
๐ดโฉ๐ตโ ฮ๐ดโฮ๐ต. Hence
ฮ๐ดโฮ๐ต= ฮ๐ดโฉ๐ต
โ ๐(โ๐). Thus ๐(โ
๐ ) forms a topology on
the set โ๐.
Now we define a function ๐ : L๐โ ๐(โ
๐) by ๐(๐ผ) = ฮ
๐ผ,
for all ๐ผ โ L๐. Let ๐ผ, ๐พ โ L
๐. Consider ๐(๐ผ โฉ ๐พ) = ฮ
๐ผโฉ๐พ=
ฮ๐ผโฮ๐พ= ๐(๐ผ) โฉ ๐(๐พ).
Consider ๐(๐ผ + ๐พ) = ฮ๐ผ+๐พ
= ฮ๐ผโช ฮ๐พ= ๐(๐ผ) โช ๐(๐พ).
Therefore ๐ is a lattice homomorphism. Now consider ๐(๐ผ) =๐(๐พ). Then ฮ
๐ผ= ฮ๐พ.
Suppose that ๐ผ = ๐พ. Then there exists ๐ โ ๐ผ such that ๐ โ๐พ. As๐พ is a proper ๐-ideal of ๐, there exists an irreducible ๐-ideal ๐ฝ of ๐ such that๐พ โ ๐ฝ and ๐ โ ๐ฝ (seeTheorem 6 in [10]).Hence ๐ผ โ ๐ฝ. As ๐ is a right ๐-weakly regular ฮ-semiring, ๐ฝ isa prime ๐-ideal of ๐ by Theorem 28. Therefore ๐ฝ โ ฮ
๐พ= ฮ๐ผ
implies ๐ผ โ ๐ฝ, which is a contradiction.Therefore ๐ผ = ๐พ.Thus๐(๐ผ) = ๐(๐พ) implies ๐ผ = ๐พ and hence ๐ is one-one. As ๐ isonto the result follows.
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper.
Acknowledgment
The author is thankful for the learned referee for his valuablesuggestions.
References
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