Int. J. Anal. Appl. (2023), 21:50 On Quasi-ideals and Bi-ideals in AG-Rings Tanaphong Prommai1, Thiti Gaketem2,∗ 1Department of Mathematics School of Science, University of Phayao, Phayao 56000, Thailand 2Fuzzy Algebras and Decision-Making Problems Research Unit, Department of Mathematics School of Science, University of Phayao, Phayao 56000, Thailand ∗Corresponding author: thiti.ga@up.ac.th Abstract. In this paper we study some properties of quasi-ideals and bi-ideals in AG-ring and study some interesting properties of these ideals. 1. Introduction M.A. Kazim and MD. Naseeruddin [2] have introduced the concept of an AG-groupoid. Definition 1.1. A groupoid G is called a left almost semigroup (abbreviated as a LA-semigroup) if its elements satisfy the left invertive law: (ab)c = (cb)a for all a,b,c ∈ G. It is also called an Abel-Grassmann’s groupoid (abbreviated as AG-groupoid). Moreover every AG-groupoids G have a medial law hold (a ·b) · (c ·d) = (a ·c) · (b ·d), ∀a,b,c,d ∈ G. Q. Mushtaq and M. Khan [4, p.322] asserted that, in every AG-groupoids G with left identity (a ·b) · (c ·d) = (d ·c) · (b ·a), ∀a,b,c,d ∈ G. Received: Mar. 14, 2023. 2020 Mathematics Subject Classification. 16Y30, 16Y99. Key words and phrases. ideal; bi-ideal; quasi-ideal. https://doi.org/10.28924/2291-8639-21-2023-50 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-50 2 Int. J. Anal. Appl. (2023), 21:50 Further M. Khan, Faisal, and V. Amjid [3], asserted that, if a AG-groupoid G with left identity the following law holds a · (b ·c) = b · (a ·c), ∀a,b,c ∈ G. M. Sarwar (Kamran) [5, p.112] defined AG-group as the following. Definition 1.2. A groupoid G is called an Abel-Grassmann’s group, abbreviated as AG-group, if (1) there exists e ∈ G such that ea = a for all a ∈ G, (2) for every a ∈ G there exists a′ ∈ G such that, a′a = e, (3) (ab)c = (cb)a for every a,b,c ∈ G. S.M. Yusuf in [11, p.211] introduces the concept of an Abel-Grassmann’s ring (AG-ring). Definition 1.3. An algebraic system 〈R, +, ·〉 is called a Abel-Grassmann’s ring (AG-ring) if (1) 〈R, +〉 is an AG-group, (2) 〈R, ·〉 is an AG-groupoid, (3) a(b + c) = ab + ac and (a + b)c = ac + bc, for all a,b,c ∈ R. Lemma 1.1. In an AG-ring R, (ab)(cd) = (ac)(bd) (1.1) for all a,b,c,d ∈ R. Equation (1.1) is called a medial law in the AG-ring R. Lemma 1.2. If an AG-ring R has a left identity 1, then a(bc) = b(ac) for all a,b,c ∈ R. Lemma 1.3. If an AG-ring R has a left identity 1, then (ab)(cd) = (dc)(ba) (1.2) for all a,b,c,d ∈ R. Equation (1.2) is called a paramedial law in the AG-ring R. Now we have the following property. T. Shah and I. Rehman [11, p.211] asserted that a commutative ring 〈R, +, ·〉, we can always obtain an AG-ring 〈R,⊕, ·〉 by defining, for a,b,c ∈ R, a ⊕b = b −a and a ·b is same as in the ring. We can not assume the addition to be commutative in an AG-ring. Definition 1.4. Let 〈R, +, ·〉 be an LA-ring and S be a non-empty subset of R and S is itself and AG-ring under the binary operation induced by R, the S is called an AG-subring of R, then S is called an LA-subring of 〈R, +, ·〉. Int. J. Anal. Appl. (2023), 21:50 3 Definition 1.5. If S is an AG-subring of an LA-ring 〈R, +, ·〉, then S is called a left ideal of R if RS ⊆ S. Right and two-sided ideals are defined in the usual manner. Lemma 1.4. If an AG-ring R has a left identity 1, then every right ideal is a left ideal. Proof. Let R be an AG-ring with left identity 1 and A is a right ideal of R. Then for a ∈ A,r ∈ R, we have ra = (1r)a = (ar)1 ∈ (AR)R ⊆ AR ⊆ A, where 1 is a left identity, that is ra ∈ A. Therefore A is left ideal of R. � 2. Main Results Definition 2.1. Let R be an AG-ring and Q be a non-empty subset of R. Then Q is said to be a quasi-ideal of R if Q is a AG-subgroup of (R, +) such that RQ∩QR ⊆ Q. Theorem 2.1. Every one-sided ideal or two-sided ideal of an-AG-ring R is a quasi-ideal of R. Proof. Let L be a left ideal of an AG-ring R. Then LR∩RL ⊆ LL ⊆ L. Thus L is a quasi-ideal of an AG-ring R. Similarly let I be a right ideal of R then IR∩RI ⊆ II ⊆ I. Thus I is a quasi-ideal of an AG-ring R. � Theorem 2.2. Let R be an AG-ring. Then the intersection of left ideal L and a right ideal of I of R is a quasi-ideal of R. Proof. Let L be a left ideal and I be a right ideal of R. Then L∩ I is a AG-subgroup of (R, +). Thus R(L∩ I) ∩ (L∩ I)R ⊆ RL∩ IR ⊆ L∩ I. Therefore the intersection of left ideal L and a right ideal of I of R is a quasi-ideal of R. � Theorem 2.3. Arbitrary intersection of quasi-ideal of an AG-ring R is a quasi-ideal of R. Proof. Let T := ⋂ i∈∆{Qi | Qi is a quasi-ideal of R}, where ∆ denotes any indexing set, be a nonempty set. Then T is a AG-subgroup of (R, +). Now RT ∩TR = R (⋂ i∈∆ Q1 ) ∩ (⋂ i∈∆ Q1 ) R ⊆ RQi ∩QiR ⊆ Qi, for all i ∈ ∆. So we see that RT ∩TR ⊆ ⋂ i∈∆ Qi = T . This proof complete. � Definition 2.2. An element e of an AG-ring R is a said idempotent element if e2 = ee = e. 4 Int. J. Anal. Appl. (2023), 21:50 Theorem 2.4. Let R be an AG-ring in which every quasi-ideal is idempotent. Then for left ideal L and right ideal I such that IL = I ∩L ⊆ LI is true. Proof. Let P and Q be two quasi-ideal in R then P ∩Q is also a quasi-ideal. By the idempotent of P ∩Q we have P ∩Q = (P ∩Q)(P ∩Q)(PQ) ∩ (QP ) on other hand (PQ) ∩ (QP ) ⊆ (PR) ∩ (RP ) ⊆ P. Similarly (PQ) ∩ (QP ) ⊆ Q and so P ∩Q = (PQ) ∩ (QP ). Since left and right ideal are always AG-subgroup we have I ∩L = (IL) ∩ (LI) but (IL) ⊆ (R∩L) and so IL = I ∩L ⊆ LI. This proof complete. � Intersection of a quasi-ideal and AG-subring of R is a quasi-ideal of an AG-subring of R. We can prove this in the following theorem. Theorem 2.5. Let R be an AG-ring. If Q is a quasi-ideal and T is an AG-subring of R, then Q∩T is a quasi-ideal of T . Proof. Let Q is a quasi-ideal and T is an AG-subring of R. Then Q∩T is a AG-subgroup of (R, +). Since Q∩T ⊆ T we have Q∩T is a AG-subgroup of (T, +). Then T (Q∩T ) ∩ (Q∩T )T ⊆ TQ∩QT ⊆ RQ∩QR ⊆ Q and T (Q∩T ) ∩ (Q∩T )T ⊆ TT ∩TT ⊆ T ∩T = T. It follows that T (Q∩T ) ∩ (Q∩T )T ⊆ Q∩T. Hence Q∩T is a quasi-ideal of T . � Definition 2.3. Let R be an AG-ring. An additive AG-subgroup B of R is called a bi-ideal of R if (BR)B ⊆ B. Lemma 2.1. Every left (right) ideal of an AG-ring R is a bi-ideal of R. Proof. Let L be a left ideal of R. Then A is an additive AG-subgroup of R. Thus (LR)L ⊆ (RR)L ⊆ RL ⊆ L. This implies that L is a bi-ideal of R. Let I be a right ideal of R. Then I is an additive AG-subgroup of R. Thus (IR)I ⊆ II ⊆ IR ⊆ I. This implies I is a bi-ideal of R. � Corollary 2.1. Every ideal of a Γ-AG-ring R is a bi-ideal of R. Lemma 2.2. Let B be an idempotent bi-ideal of a Γ-AG-ring R with left identity 1. Then B is an ideal of R. Int. J. Anal. Appl. (2023), 21:50 5 Proof. Let B be an idempotent bi-ideal of a Γ-AG-ring R. Then B is an additive AG-subgroup of R. Thus BR = (BB)R = (RB)B = (R(BB))B. By Lemma 1.2 so (R(BB))B = ((BB)R)B = (BR)B ⊆ B. Which implies that B is a right ideal. By Lemma 1.4 so it is left ideal of R. Hence B is an ideal of R. � Theorem 2.6. The product of two bi-ideals of an AG-ring R with left identity 1 is again a bi-ideal of R. Proof. Let H and K be two bi-ideals of R. Then H and K are additive AG-subgroup of R. Thus using medial and RR = R, we get [(HK)R](HK) = [(HK)(RR)](HK), by medial = [(HR)(KR)](HK), by medial = [(HR)H][(KR)K], H,K is a bi-ideal of R ⊆ HK. Hence HK is a bi-ideal of R. � Theorem 2.7. Let B be a bi-ideal of an AG-ring R and A be a left ideal of R with left identity 1, then BA is a bi-ideal of R. Proof. Since A is a left ideal of R and B is a bi-ideal of an AG-ring R, we have BA is an additive AG-subgroup of R. Thus [(BA)R](BA) = [(RA)B](BA) = [(BA)B](RA) ⊆ [(BR)B]A, B is a bi-ideal of R ⊆ BA. It following that BA is a bi-ideal of R. � Theorem 2.8. Let B be a bi-ideal of an AG-ring R and A be a right ideal of R with left identity 1. If A ⊆ B and BB ⊆ B, then AB is a bi-ideal of R. 6 Int. J. Anal. Appl. (2023), 21:50 Proof. Since A is a right ideal of R and B is a bi-ideal of an AG-ring R, we have AB is an additive AG-subgroup of R. Let A ⊆ B and BB ⊆ B. Then using Lemma 1.1, we get [(AB)R](AB) = [(RB)A](AB) = [(AB)A](RB) ⊆ [(AR)A](RB), B ⊆ R ⊆ (AA)(RB), A is a right ideal of R = (AR)(AB), by Γ-medial ⊆ A(AB), A is a right ideal of R ⊆ A(BB), A ⊆ B ⊆ AB, BB ⊆ B. It follows that AB is a bi-ideal of R. � Theorem 2.9. Let R be an AG-ring and A,B be bi-ideals of an AG-ring R. Then A∩B is a bi-ideal of R. Proof. Since A,B is bi-ideals of an AG-ring R, we have A ∩ B is an additive AG-subgroup of R. Thus [(A∩B)R](A∩B) ⊆ (AR)(A∩B) = [(A∩B)R]A ⊆ (AR)A ⊆ A and [(A∩B)R](A∩B) ⊆ (BR)(A∩B) = [(A∩B)R]B ⊆ (BR)B ⊆ B. It following that A∩B is a bi-ideal of R. � Corollary 2.2. Let R be a Γ-AG-ring and Hi is a bi-ideal of R, for all i ∈ I. Then ⋂ i∈I Hi is a bi-ideal of R. Proof. Since 0 ∈ Hi for all i ∈ I, we have 0 ∈ ⋂ i Hi. Then ⋂ i Hi 6= ∅. Since Hi is a bi-ideal of R, we have Hi is an additive AG-subgroup of R Let x,y ∈ Hi then x − y ∈ Hi. Thus x − y ∈ ⋂ i Hi. Let x,y ∈ ⋂ i Hi, r ∈ R. Then (xr)y ∈ (HiR)Hi ⊆ Hi for all i ∈ I Thus (xr)y ∈ Hi. Hence ⋂ i∈I Hi is a bi-ideal of R. � Theorem 2.10. Let I and L be respectively right and left AG-subgroup of R. Then any AG-subgroup B of R such that IL ⊆ B ⊆ I ∩L is a bi-ideal of R. Proof. Since B is a AG-subgroup of (R, +) with IL ⊆ B ⊆ I ∩L we have (BR)B ⊆ ((I ∩L)R)(I ∩L), by B ⊆ I ∩L ⊆ (IR)L, by S ⊆ I ∩L and L ⊆ I ∩L ⊆ IL, by I is a right ideal of R ⊆ B, by IL ⊆ B. Then B is a bi-ideal of R. � Corollary 2.3. Intersection of an arbitrary set of bi-ideal Bλ (λ ∈ ∧) of an AG-ring R is again a bi-ideal of R. Int. J. Anal. Appl. (2023), 21:50 7 Proof. Set B := ⋂ λ∈∧Bλ. Since B is an AG-subgroup of R. From the inclusion (BλR)Bλ ⊆ Bλ and B ⊆ Bλ. This implies that (BR)B ⊆ (BλR)Bλ ⊆ Bλ (∀λ ∈∧). Hence (BR)B ⊆ B. � Theorem 2.11. Every idempotent quasi-ideal is a bi-ideal. Proof. Let Q be an idempotent quasi-ideal of Γ-AG-ring R. Then (QR)Q ⊆ (RR)Q ⊆ RQ and by Lemma 1.2 (QR)Q ⊆ (QR)Γ(QQ) = (QQ)(RQ) ⊆ Q(RQ) ⊆ Q(RR) ⊆ QR which implies that (QR)Q ⊆ QR∩RQ ⊆ Q. � Definition 2.4. An AG-ring R is called a regular AG-ring if for any x ∈ R there exists y ∈ R such that x = (xy)x. Theorem 2.12. Let R be a regular of an AG-ring and B be a bi-ideal of R. Then (BR)B = B. Proof. Since B is a bi-ideal of R we have (BR)B ⊆ B. Let x ∈ B then there exist a ∈ R such that x = (xa)x ∈ (BR)B, since R is a regular of an AG-ring. This implies that B ⊆ (BR)B so (BR)B = B. � Theorem 2.13. For a quasi-ideal Q in a regular AG-ring R, then QR∩RQ = Q. Proof. Let Q be a quasi-ideal in R then QR∩RQ ⊆ Q. Let x ∈ Q then there exist a ∈ R such that x = (xa)x, since R is a regular of a AG-ring. 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