EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 3, 2022, 1402-1416 ISSN 1307-5543 – ejpam.com Published by New York Business Global Annihilator Hyperideals in Strong Bounded Dual Distributive Meet-hyperlattice Eman Ghareeb Rezk1, Nabilah Hani Abughazalah2,∗ 1 Department of Mathematics, Faculty of Science, Tanta University, Egypt 1,2 Mathematical Sciences Department, College of Science, Princess Nourah bint Abdulrahman University, P.O.Box 84428, Riyadh 11671, Saudi Arabia Abstract. In this paper, we study the properties of annihilator hyperideals in the class of strong bounded dual distributive meet-hyperlattice. We show that the set of all closed hyperideals forms a Boolean algebra. We introduce the concept of homomorphism, which preserves the annihilator hyperideal. Suitable conditions for preserving annihilator hyperideals are obtained. Representation and characterization theorems of annihilator hyperideals in sub-meet-hyperlattice and product meet-hyperlattice are proved. 2020 Mathematics Subject Classifications: 06B75, 06F99, 08A05 Key Words and Phrases: Annihilator,Boolean algebra, Homomorphism, Hyperideal, Hyperlat- tice 1. Introduction The approach to the theory of hyberlattices was first made by M. Konstantinidou and J. Mittas in 1977, [12]. Modular distributive and complemented classes of hyper- lattices were studied by M. Konstantinidou in [10] and [9]. Ideals of hyperlattices were introduced by Rahnamai-Barghi in [15], where he considered the prime ideal theorem for distributive hyperlattices. M. Amiri Bideshki and et al. defined the notions of hyperideals and hyperfilters in strong meet-hyperlattices in [11]. They also introduced the concept of annihilator hyperideals. Annihilators have been started in ring theory over many classes of rings, as examples refer to [19] and [8]. In that sense, the annihilator of a certain set A means the set of killer elements that make each element of A tends to zero by multiplication operation. The mention of annihilators in lattices was first introduced by M.Mandelker in 1970, [13]. He defined the relative annihilator as a generalization of relative pseudocom- plementation. M.Mandelker introduced the relation between prime ideal conditions and annihilator conditions on distributive lattices. W. H. Cornish investigated the annihilator ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4480 Email addresses: eman.rezk@science.tanta.edu.eg (E.G. Rezk), nhabughazala@pnu.edu.sa (N.H. Abughazalah) https://www.ejpam.com 1402 © 2022 EJPAM All rights reserved. E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1403 properties of distributive lattices in [18]. He defined the annihilator of a set A as a set of all elements whose elements tend to zero by the meet operation. The main result- that Cornish proved- is the normality of lattice equivalents to any two elements with zero meeting have a comaximal annihilator of their principal ideals. Moreover, the normality of the lattice equivalents to the annihilator of the principle ideal of the meeting of any two elements equals the joining of annihilators of their principle ideals. In [1], B. A. Davey and Nieminen studied the annihilators in the class of modular lattices. They proved that the weakly atomic modular lattice has necessary and sufficient conditions for its annihilators. As a generalization of lattices Halaš [6] studied annihilators in ordered sets. This was followed by a lot of studies and research on the concept of annihilator in many algebraic structures and classes, for instance: almost distributive lattices [3],0-almost distributive lattices [4], distributive dual weakly complemented lattice [17], BCK-algebras [7], standard QBCC algebras [14], C-algebra[16], and many other classes. In this paper, the properties of annihilator hyperideals of strong bounded dual distribu- tive meet-hyperlattice are investigated. Main terminologies and properties are recalled in Section 2. Important properties of annihilator hyperideals are proved. Moreover, the structure of the set of all closed hyperideals is investigated in Section 3. In Section 4, the conditions of homomorphism map to preserve Annihilator hyperideals are discussed. The proof of the preservation of annihilator hyperideals under the effects of these condi- tions is given. Finally, in section 5, we show that annihilator hyperideals are inherited for sub-meet-hyperlattice and product meet-hyperlattice. 2. Backgrounds We recall here the basic terminologies and concepts of hyperlattices. The reader must be familiar with lattice theory. For more details about lattice theory, see [2] and [5]. Definition 1. [11] Let L be a nonempty set, P∗(L) is the set of all nonempty subsets of L, ∧̄ : L × L → P∗(L) is a hyperoperation and ∨ : L → L is a binary operation. Then L =< L; ∧̄,∨ > is called a meet-hyperlattice if: H1) a ∈ iai∧̄ia,ia = a ∨ a; H2) ai∧̄ibi = ibi∧̄ia,ia ∨ ibi = ib ∨ ai; H3) ai∧̄i(bi∧̄ic)i = i(ai∧̄ib)∧̄ic, a ∨ (b ∨ c) = (a ∨ b) ∨ c; H4) a ∈ (ai∧̄i(a ∨ b))i ∩ i(a ∨ (ai∧̄ib)),i ifor all a, ib, ic ∈ L. The meet-hyperlatticce L is called strong, if it satisfies that: if a ∈ ai∧̄ib then ai ∨ ib = b , i ifor all a, ib ∈ L. Consider an order relation ≤ on L as: a ≤ b iff a ∨ b = b, for all a, b ∈ L. Accordingly, meet-hyperlattice L is bounded if there exist two elements 0, 1 ∈ L such that 0 ≤ a ≤ 1, for all a ∈ L. E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1404 For subsets A,B ⊆ L: Ai∧̄iB = ∪{ai∧̄ib : ia ∈ A, ib ∈ B}, Ai ∨ iB = {ai ∨ ib : ia ∈ A, ib ∈ B}. Proposition 1. [11] Let L be a boundedistrong meet-hyperlattice. Then theifollowing conditions hold: (i) If a, b ̸= 1 iandiai ∨ ib = 1, ithen ia, b /∈ ai∧̄ib; (ii) If ai∧̄ib = L or a, b ∈ ai∧̄ib, then a = b; (iii) For all a ∈ L: a ∈ ai∧̄i1 and 0 ∈ ai∧̄i0. The meet-hyperlattice L is distributive, if ai ∨ i(bi∧̄ic) = (ai ∨ ib)i∧̄i(ai ∨ ic), for all a, b, c ∈ L. Dually, L is dual distributive if ai∧̄i(b ∨ ic) = (ai∧̄ib)i ∨ i(ai∧̄ic). Definition 2. [11] Let I be a nonempty subset of a strong meet-hyperlattice L. I is called a hyperideal if: i) If a, ib ∈ I, theniai ∨ ib ∈ I; ii) Ifia ∈ Iiand b ∈ L, such that b ≤ a, then b ∈ I. The set of all hyperideals of strong bounded meet-hyperlattice L is denoted by I(L). The meet operation on hyprideals is the usual sets intersection ∩ and the join operation is defined as Ii∨iJ = {x ∈ L : x ≤ ai ∨ ib, a ∈ I, b ∈ J}. Then, we get the following theorem: Theorem 1. [11] The structure (I(L);∩,∨, 0, L) forms a bounded distributive lattice. Definition 3. [11] Let L = (L, ∧̄,∨) be aistrong bounded dualidistributive meet-hyperlatticei and A ⊆ L. Then the hyperideal Ar = {x ∈ L : 0 ∈ x∧̄a, forall a ∈ A}, is called annihilator hyperideal or for brevity (annihilator). When A is a singleton subset {a} of L, its annihilator is defined as: ar = {x ∈ L : 0 ∈ a∧̄x}. Proposition 2. [11] Let L = (L, ∧̄,∨) be a strong bounded dualidistributive meet-hyperlattice. Then: (i) 0r = L; (ii) If a, b ∈ L and a ≤ b,ithen br ⊆ ar; E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1405 (iii) Ar = ∩{ar : a ∈ A}; (iv) Ar ∩Br = (A ∪B)r. Notice that (ii) can be generalized to any subsets A and B of L as: If A,B ⊆ L and A ⊆ B then Br ⊆ Ar. In all of the following, the meet-hyperlattice L =< L; ∧̄,∨, 0, 1 > is considered that strong bounded and dual distributive. 3. Annihilator Hyperideals The core of this section is that we prove the main theory, which states that the set of all closed hyperideals forms a Boolean algebra. Theorem 2. Let I, J be subsets of meet-hyperlattice L. Then (i) I ⊆ Irr; (ii) Irrr = Ir; (iii) I ∩ J ⊆ (Ir ∨ Jr)r; (iv) Ir ∩ Jr = (I ∨ J)r; (v) (I ∩ J)rr ⊆ Irr ∩ Jrr; (vi) 1r ⊆ Ir for all I ⊆ L; (vii) Ir ∩ Irr = 1r; (viii) If I ⊆ Jr then I ∩ Jrr = 1r; (ix) 1rr = L. Proof. (i) Since I ∩ Irr = I ∩ {x ∈ L : 0 ∈ x∧̄i for all i ∈ Ir} = {x ∈ I : 0 ∈ x∧̄i for all i ∈ Ir} = I. Then, I ⊆ Irr. (ii) From (i) we have I ⊆ Irr and Ir ⊆ Irrr. On the other side, we have Irrr ⊆ Ir from Proposition 2. Hence the equality is satisfied. E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1406 (iii) Let x ∈ I ∩ J and y ∈ Ir ∨ Jr. Then y ≤ i ∨ j for some i ∈ Ir and j ∈ Jr such that x∧̄y ⊆ x∧̄(i ∨ j) = (x∧̄i) ∨ (x∧̄j). Since x ∈ I, x ∈ J then 0 ∈ (x∧̄i) and 0 ∈ (x∧̄j). Then 0 ∈ x∧̄(i ∨ j). From Proposition 2, we get 0 ∈ (x∧̄y) .Therefore x ∈ (Ir ∨ Jr)r. (iv) Since I, J ⊆ I ∨ J , then Ir, Jr ⊇ (I ∨ J)r. So, (I ∨ J)r ⊆ Ir ∩ Jr. Conversely, let y ∈ Ir ∩ Jr, thus 0 ∈ y∧̄i for all i ∈ I and 0 ∈ y∧̄j for all j ∈ I which indicates that 0 ∈ (y∧̄i) ∨ (y∧̄j) = y∧̄(i ∨ j). So for all a ≤ i ∨ j meeting each side by y to get y∧̄a ⊆ y∧̄(i ∨ j). It impliesithat 0 ∈ y∧̄a and then y ∈ (I ∨ J)r. Therefore Ir ∩ Jr ⊆ (I ∨ J)r. (v) Since I ∩ J ⊆ I and I ∩ J ⊆ J then (I ∩ J)rr ⊆ Irr and (I ∩ J)rr ⊆ Jrr. Thus (I ∩ J)rr ⊆ Irr ∩ Jrr. (vi) Suppose x ∈ 1r and I is a nonemptyisubset of L, then 0 ∈ x∧̄1. From Proposition 2, we get 0 ∈ x∧̄y for all y ∈ I. Therefore x ∈ Ir. (vii) From (iv) in Proposition 2, we get Ir ∩ Irr = (I ∪ Ir)r. Since a ≤ 1 for all a ∈ I ∪ Ir, then 1r ⊆ ar. It implies that 1r ⊆ ∩{ar : a ∈ I ∪ Ir}. Consequently, 1r ⊆ (I∪Ir)r = Ir∩Irr. The opposite direction is taken immediately from (vi).Therefore, Ir ∩ Irr = 1r. (viii) Let I ⊆ Jr. Intersect both sides by Jrr, we get I ∩ Jrr ⊆ Jr ∩ Jrr = 1r. But, from (vi) we have 1r ⊆ I ∩ Jrr. (ix) 1rr = {x ∈ L : 0 ∈ x∧̄a, a ∈ 1r} = {x ∈ L : 0 ∈ x∧̄a, 0 ∈ a∧̄1} = {x ∈ L : 0 ∈ x∧̄a, 0 ∈ a∧̄y, for all y ∈ L}. Definition 4. A hyperideal I of meet-hyperlattice L is called closed if I = Irr. We denotei the set of all closed hyperideals of ∧- hyperlattice L by H(L). E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1407 Lemma 1. Let I, Jand K be closed hyperideals of meet-hyperlattice. Then: (i) (Ir ∨ Jr)r = I ∩ J ; (ii) (I ∩ J)rr = Irr ∩ Jrr; (iii) If I ∩ Jrr = 1r then I ⊆ Jr; (iv) K ∩ (Ir ∩ Jr)r ⊆ ( Ir ∩ (J ∩K)r )r . Proof. (i) Since Ir, Jr ⊆ Ir ∨ Jr then, from Proposition 2, we have (Ir ∨ Jr)r ⊆ Irr = I, and (Ir ∨ Jr)r ⊆ Jrr = J, Thus (Ir ∨ Jr)r ⊆ I ∩ J. The equalty is obtained from (iii) in Theorem 2. (ii) Irr∩Jrr = I ∩J ⊆ (I ∩J)rr. On the other side, from (v) in Theorem 2. the equality is satisfied. (iii) Let I ∩ Jrr = I ∩ J = 1r. Then I ⊆ Jr (from (vii) in Theorem 2). (iv) It is clear that K ∩ Ir ∩ (J ∩K)r ⊆ Ir. (1) Consequently J ∩K ∩ ( Ir ∩ (J ∩K)r ) = Ir ∩ [(J ∩K) ∩ (J ∩K)r] = Ir ∩ [(Jrr ∩Krr) ∩ (J ∩K)r] = Ir ∩ [(J ∩K)rr ∩ (J ∩K)r], (From ii)) = Ir ∩ 1r (From vii) in Theorem 2.) = 1r, which implies that K ∩ Ir ∩ (J ∩K)r ⊆ Jr (2) Thus, from (1), (2) we get K ∩ Ir ∩ (J ∩K)r ⊆ Ir ∩ Jr. Hence ( K ∩ Ir ∩ (J ∩K)r ) ∩ (Ir ∩ Jr)r = 1r, E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1408 which equivalent, Ir ∩ (J ∩K)r ∩ ( K ∩ (Ir ∩ Jr)r ) = 1r. As a result, we get K ∩ (Ir ∩ Jr)r ⊆ ( Ir ∩ (J ∩K)r )r . For I, J ∈ H(L), we define two binary operations I ∧J = I ∩J and I ⊻J = (Ir ∩Jr)r. We get (I ∩ J)rr = Irr ∩ Jrr = I ∩ J ∈ H(L), and, I ⊻ J = (Ir ∩ Jr)r ∈ H(L). Theorem 3. Let L =< L; ∧̄,∨ > be a meet-hyperlattice. Then < H(L);∩,⊻,r , 1r, L > forms a Boolean algebra. Proof. To demonstrate that (H(L),∧,⊻) forms a lattice, only associative and absorp- tion identities are required, as idempotent and commutative identities are trivial. Let I, J,K ⊆ H(L). Then we have (Ii ⊻ iJ) ⊻ iK = (Iri ∩ iJr)ri ⊻ iK = ( (Iri ∩ iJr)rri ∩ iKr )r = ( (Iri ∩ iJr)i ∩ iKr )r = ( Iri ∩ i(Jri ∩ iKr) )r = ( Iri ∩ i(Jri ∩ iKr)rr )r = Ii ⊻ i(Jri ∩ iKr)r = Ii ⊻ i(Ji ⊻ iK). It is easy to prove the second associative identity, (I ∩ J) ∩K = I ∩ (J ∩K). Now we are going to show the absorption identities. Since Ir ⊆ (I ∩ J)r, then Ii ⊻ i(I ∩ J) = ( Ir ∩ (I ∩ J)r )r = Irr = I. Similarly, since Ir ∩ Jr ⊆ Ir,ithen I = Irr ⊆ ( Ir ∩ Jr )r . Therefore I ∩ (Ii ⊻ iJ) = I ∩ (Ir ∩ Jr)r = I. Notice that I ⊆ 0r = L and 1r ⊆ I ∈ H(L), consequently (H(L),∧,⊻) is a bounded lattice. Clearly, Ir is the complement of I, because I∩Ir = 1r and I⊻Ir = (Ir∩Irr)r = 1rr = L. By using (iv) in Lemma 1, we get that: K ∩ (I ⊻ J) ⊆ I ⊻ (J ∩K) (3) Then the distributivity condition is proved by replacing K in (3) by I ⊻K to get: E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1409 (Ii ⊻ iK) ∩ (I ⊻ J) ⊆ Ii ⊻ i ( J ∩ (I ⊻K) ) ⊆ Ii ⊻ i ( Ii ⊻ i(K ∩ J) ) (by replacing K in (3) by J) = Ii ⊻ i(K ∩ J). On the other hand, J ∩ K ⊆ (Ii ⊻ iJ) ∩ (Ii ⊻ iK) and I ⊆ (Ii ⊻ iJ) ∩ (Ii ⊻ iK), which implies that Ii ⊻ i(J ∩K) ⊆ (Ii ⊻ iJ) ∩ (Ii ⊻ iK). (4) Therefore Ii ⊻ i(J ∩K) = (Ii ⊻ iJ) ∩ (Ii ⊻ iK) from (3) and (4). Example 1. Tables 1 and 2 represent the hyperoperation ∧̄ and operation ∨ of meet- hyberlattice L = {0, α, β, γ, δ, 1}. Figure 1 shows Boolean algebra H(L) of closed hyperide- als of L. ∧̄ 0 α β γ δ 1 0 {0} {0} {0} {0} {0} {0} α {0} {0, α} {0} {0, α} {0, α} {0, α} β {0} {0} {β} {β} {0} {β} γ {0} {0, α} {β} {γ} {0, α} {γ} δ {0} {0, α} {0} {0, α} {δ} {δ} 1 {0} {0, α} {β} {γ} {δ} {1} ∨ 0 α β γ δ 1 0 0 α β γ δ 1 α α α γ γ δ 1 β β γ β γ 1 1 γ γ γ γ γ 1 1 δ δ 0 1 1 δ 1 1 1 1 1 1 1 1 Table 1: Represents the hyperoperation Table 2: Represents operation ∧̄ of the meet-hyberlattice L ∨ of the meet-hyberlattice L L {0, α, β, γ} {0, α, δ} 1r = {0, α} Figure1:Boolean Algebra < H(L);∩,⊻,r , 1r, L > 4. Homomorphic Images of Annihilator Hyperideals In this section, we define the homomorphism that is annihilator hyperideal preserv- ing. Many properties related to the homomorphism of annihilator hyperideals are proven. Moreover, we show that homomorphic images and preimages of annihilator hyperideals are annihilator hyperideals. Definition 5. Let L and L′ be two meet-hyperlattices. Then the map ϕ : L → L′ is called homomorphism if the following conditions hold: ϕ(a ∨ b) = ϕ(a) ∨ ϕ(b), iϕ(a∧̄b) = ϕ(a)∧̄ϕ(b). E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1410 Since for any a ∈ L: ϕ(a∨ 0) = ϕ(a)∨ ϕ(0) = ϕ(a), then ϕ(0) = 0′, where 0 and 0′ are the zero elements of L and L′ respectively. Obviously, if a ≤ b then ϕ(a) ≤ ϕ(b). The kernal of the homomorphism ϕ is given by ker(ϕ) = {a ∈ L : ϕ(a) = 0′}. It is clearithat ker(ϕ) is a hyperideal of L′. The set of images of ϕ is denoted by Im(ϕ). It forms sub-meet-hyperlattice of L′. If ϕ is one-to-one and onto, then L and L′ are isomorphic and denoted by L ∼= L′. Example 2. Tables 3 and 4 represent the hyperoperation ∧̄ and operation ∨ of meet- hyberlattice L′ = {0′, x, y, z, 1′}. ∧̄ 0′ x y z 1′ 0′ {0′} {0′} {0′} {0′} {0′} x {0′} {x} {0′} {x} {x} y {0′} {0′} {y} {y} {y} z {0′} {x} {y} {z} {z} 1′ {0′} {x} {y} {z} {1′} ∨ 0′ x y z 1′ 0′ 0′ x y z 1′ x x x z z 1′ y y z y z 1′ z z z z z 1′ 1′ 1′ 1′ 1′ 1′ 1′ Table 3: Represents the hyperoperation Table 4: Represents the operation ∧̄ of meet-hyberlattice L′ ∨ of the meet-hyberlattice L′ Consider the meet-hyperlattice L in Example 1 and the meet-hyperlattic L′. Define a homomorphism f : L → L′ as: f(0) = 0′, f(β) = x, f(α) = y, f(γ) = z and f(δ) = f(1) = 1′. Proposition 3. Let ϕ : L → L′ be a homomorphism between meet-hyperlattices. Then: (i) If ϕ is onto and I is aihyperideal of L, then ϕ(I) is a hyperideal of L′; (ii) If J is aihyperideal of L′, theniϕ−1(J) is a hyperideal of L containing ker(ϕ); (iii) If A is ainonempty subset of L, then ϕ(Ar) ⊆ (ϕ(A))r. Proof. (i) Let x, y ∈ ϕ(I) then there exist a, b ∈ I such that x = ϕ(a) and y = ϕ(b). Then ϕ(a ∨ b) = ϕ(a) ∨ ϕ(b) = x ∨ y ⊆ ϕ(I). Now, suppose x, y ∈ L′, x ∈ ϕ(I) and y ≤ x. Hence y = ϕ(b) ∈ ϕ(a)∧̄ϕ(b) = ϕ(a∧̄b), which indicates that b ∈ a∧̄b and b ≤ a. Thus b ∈ I and y = ϕ(b) ∈ ϕ(I). Consequently ϕ(I) is a hyperideal. (ii) Let x, y ∈ J . Then thereiexist a, b ∈ L′ such that ϕ−1(x) = a and ϕ−1(y) = b. It implies a ∨ b = ϕ−1(x) ∨ ϕ−1(y). By using the effect of ϕ on both sides we get ϕ(a ∨ b) = ϕ ( ϕ−1(x) ∨ ϕ−1(y) ) = ϕ ( ϕ−1(x) ) ∨ ϕ ( ϕ−1(y) ) = x ∨ y. Thus ϕ−1 ( ϕ(a ∨ b) ) = a ∨ b = ϕ−1(x ∨ y) ∈ ϕ−1(J). Let x ∈ J, ϕ−1(x) = a and y ∈ L such that y ≤ a = ϕ−1(x). Then y ∨ a = a which implies ϕ(y) ∨ ϕ(a) = ϕ(y ∨ a) = ϕ(a). Thus ϕ(y) ≤ ϕ(a) = ϕ ( ϕ−1(x) ) = x. Therefore ϕ(y) ∈ J and y ∈ ϕ−1(J). Clearly, 0′ ∈ J which means ker(ϕ) = ϕ−1(0′) ⊆ ϕ−1(J). E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1411 (iii) Let x ∈ Ar and ϕ(x) = b. Then 0′ = ϕ(0) ∈ ϕ(x ∧ a) = ϕ(x)∧̄ϕ(a) for all a ∈ A. It means that 0′ ∈ ϕ(x)∧̄ϕ(a) for all ϕ(a) ∈ ϕ(A) i.e., ϕ(x) ∈ ( ϕ(a) )r . Therefore ϕ(Ar) ⊆ ( ϕ(A) )r . Definition 6. Let ϕ : L → L′ be a homomorphism. Then ϕ is called annihilator hyperideal preserving if for any subset A of L: ϕ(Ar) = [ϕ(A)]r. The homomrphism f in Example 2 is not an annihilator hyperideal preserving. Example 3. Tables 5 and 6 represent the hyperoperation ∧̄ and operation ∨ of meet- hyberlattice L′′ = {0, α, β, γ, δ, 1}. ∧̄ 0 α β γ δ 1 0 {0} {0} {0} {0} {0} {0} α {0} {α} {0} {α} {α} {α} β {0} {0} {β} {β} {0} {β} γ {0} {α} {β} {γ} {α} {γ} δ {0} {α} {0} {α} {δ} {δ} 1 {0} {α} {β} {γ} {δ} {1} ∨ 0 α β γ δ 1 0 0 α β γ δ 1 α α α γ γ δ 1 β β γ β γ 1 1 γ γ γ γ γ 1 1 δ δ 0 1 1 δ 1 1 1 1 1 1 1 1 Table 5: Represents the hyperoperation Table 6: Represents operation ∧̄ of the meet-hyberlattice L′′ ∨ of the meet-hyberlattice L′′ Define a homomorphism f : L′′ → L′ as: g(0) = 0′, g(β) = g(δ) = y, g(α) = x, g(γ) = z and g(1) = 1′. Where L′ is a meet-hyperlattice in Example 2. f is an annihilator hyperideal preserving. Theorem 4. Let L and L′ be two meet-hyperlattice, ϕ : L → L′ be a homomorphism and ker(ϕ) = {0}. Then: (i) If ϕ is onto then: (a) ϕ is annihilator hyperideal preserving; (b) For any nonemptyisubsets A and B of L Ar = Br if andionly if [ϕ(A)]r = [ϕ(B)]r. (ii) ϕ−1 is annihilator hyperideal preserving. Proof. (i) (a) For a inonempty subset A of L, we have ϕ(Ar) ⊆ [ϕ(A)]r, from Proposition 3. So we just need to prove that [ϕ(A)]r ⊆ ϕ(Ar). To do that, let x ∈ [ϕ(A)]r ⊆ L′ then there is a ∈ L such that ϕ(a) = x, but 0′ ∈ x∧̄ϕ(b) for all b ∈ A. Then 0 ∈ a∧̄b for all b ∈ A. Therefore a ∈ Ar and then x = ϕ(a) ∈ ϕ(Ar). E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1412 (b) Suppose that A and B are nonemptyisubsets of L such that Ar = Br. Then by using a) we get [ϕ(A)]r = ϕ(Ar) = ϕ(Br) = [(ϕ(B)]r. Conversely, Let [ϕ(A)]r = [ϕ(B)]r and x ∈ Ar. Then 0 ∈ a∧̄x for all a ∈ A. So 0′ = ϕ(0) ∈ ϕ(a∧̄x) = ϕ(a)∧̄ϕ(x). It means ϕ(x) ∈ [ϕ(A)]r = [ϕ(B)]r. Therefore 0′ = ϕ(0) ∈ ϕ(x)∧̄ϕ(b) for all b ∈ B, which implies that 0 ∈ x∧̄b for all b ∈ B. Accordingly x ∈ Br and hence Ar ⊆ Br. Similarly we can prove that Br ⊆ Ar. (ii) Let x ∈ [ϕ−1(B)]r which means 0 ∈ x∧̄b for all b ∈ ϕ−1(B). Then 0 ∈ x∧̄b for all ϕ(b) ∈ B. It implies ϕ(0) = 0′ ∈ ϕ(x∧̄b) = ϕ(x)∧̄ϕ(b) for all ϕ(b) ∈ B. Hence ϕ(x) ∈ Br, i.e., x ∈ ϕ−1(Br). As a result, [ϕ−1(B)]r ⊆ ϕ−1(Br). Conversely, let x ∈ ϕ−1(Br) and b ∈ ϕ−1(B). Then ϕ(x) ∈ Br and ϕ(b) ∈ B. Thus ϕ(0) = 0′ ∈ ϕ(x)∧̄ϕ(b) = ϕ(x∧̄b). As a result, 0 ∈ x∧ b for all b ∈ ϕ−1(B). Hence, x ∈ [ϕ−1(B)]r and then ϕ−1(Br) ⊆ [ϕ−1(B)]r. Theorem 5. Let L and L′ be two meet-hyperlattice, ϕ : L → L′ be a homomorphism and ker(ϕ) = {0}. Then: (i) If ϕ is annihilator hyperideal preserving and onto then the homomorphiciimage ϕ(I) of annihilator hyperideal I of L is an annihilator hyperideal of L′; (ii) If ϕ−1 is annihilator hyperideal of L′ preserving then the preimage ϕ−1(J) of anni- hilator hyperideal J is an annihilator hyperideal of L; (iii) If ϕ is annihilator hyperideal preserving and onto then ker(ϕ) is an annihilator hyperideal of L. Proof. (i) Let I be an annihilator hyperideal of L. Then by using i) in Proposition 3, ϕ(I) is a hyperideal. From the assumption of ϕ is annihilator hyperideal preserving we get [ϕ(I)]rr = ϕ(Irr) = ϕ(I). Consequently, ϕ(I) is an annihilator ideal of L′. (ii) Let J be an annihilator hyperideal of L′. Then by using ii) in Proposition 3, ϕ−1(I) is a hyperideal. From the assumption of ϕ−1 is annihilator hyperideal preserving we get [ϕ−1(I)]rr = ϕ−1(Irr) = ϕ−1(I). Therefore, ϕ−1(I) is an annihilator ideal of L. (iii) We know that ker(ϕ) = ϕ−1({0′}) and {0′} = 1r is annihilator hyperideal. Then from ii), ker(ϕ) is annihilator hyperideal. E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1413 Corollary 1. Let L and L′ be two meet-hyperlattice, ϕ : L → L′ be an annihilator hy- perideal preserving. Then H(L) and H(L′) are isomorphic. Symbolically writes H(L) ∼= H(L′). Example 4. The meet-hyperlattice L′ in Example 3 has the Boolean algebra of its closed hyperideals which is given by Figure 2. Clearly, its isomorphic to the Boolean algebra in Figure 3 which represents the closed hyperideals of the meet-hyperlattice in Example 2. L′′ {0, α} {0, β} 1r = {0} Figure2: Boolean Algebra < H(L′′);∩,⊻,r , 1r, L′′ > L′ {0, x} {0, y} {0′} Figure3: Boolean Algebra < H(L′);∩,⊻,r , 1r, L′ > 5. Sub-meet-hyperlattice and Product Representations of annihilator hyperideals in sub-meet-hyperlattice and product meet- hyperlattice, are investigated in the following. Definition 7. A subset S ⊆ L, of meet-hyperlattice L, is a sub-meet-hyperlattice iff it is close under the same operations of L. Clearly, sub-meet-hyperlattice of strong bounded dual distributive meet-hyperlattice is also too. Theorem 6. If I is an annihilator hyperideal of L and S is a sub-meet hyperlattice of L. Then I ∩ S is an annihilator hyperideal of S. Proof. Let I be an annihilator hyperideal of L and S be a sub-meet hyperlattice of L. Then there exist a subset K of L such that I = Kr = {x ∈ L : 0 ∈ x∧̄a for all a ∈ K}. Thus I ∩ S = {x ∈ L : 0 ∈ x∧̄a for all a ∈ K ∩ S} = (K ∩ S)r. E.G. Rezk, N.H. Abughazalah / Eur. J. Pure Appl. Math, 15 (3) (2022), 1402-1416 1414 Corollary 2. If I is a closed hyperideal of L and S is a sub-meet hyperlattice of L. Then I ∩ S is a closed hyperideal of S. Example 5. Tables 7 and 8 represent a sub-meet-hyperlattice S of meet-hyperlattice L in Example 1. Figure 4 shows Boolean algebra H(S) of closed hyperideals of S. ∧̄ 0 α β γ 0 {0} {0} {0} {0} α {0} {0, α} {0} {0, α} β {0} {0} {β} {β} γ {0} {0, α} {β} {γ} ∨ 0 α β γ 0 0 α β γ α α α γ γ β β γ β γ γ γ γ γ γ Table 7: Represents the hyperoperation Table 8: Represents operation ∧̄ of the sub-meet-hyberlattice S ∨ of the sub-meet-hyberlattice S S 1r = {0, α} Figure4: Boolean Algebra < H(S);∩,⊻,r , 1r, L > Definition 8. Let L1 =< L1; ∧̄1,∨1, 01, 11 > and L2 =< L2; ∧̄2,∨2, 02, 12 > be two meet- hyperlattices. Then the product L1 × L2 with respect to the pair-wise operations such that for any (a, b), (a′, b′) ∈ L1 × L2: (a, b)∧̄(a′, b′) = (a∧̄1a ′, b∧̄2b ′), and (a, b) ∨ (a′, b′) = (a ∨1 a ′, b ∨2 b ′), forms meet-hyperlattice called the prouduct meet-hyperlattice of L1 and L2 with zero element 0 = (01, 02), and one element1 = (11, 12). Notice that, if L1 and L2 are strong bounded dual distributive meet-hyperlattices, then L1 × L2 is also too. Theorem 7. For any two hyperideals I1 and I2 of two meet-hyperlattices L1 and L2 respectively. I1 and I2 are annihilator hyperideals iff I1 × I2 is an annihilator hyperideal of the product meet-hyperlattice L1 × L2. Proof. Assume I1 and I2 be annihilator hyperideals of L1 and L2, respectively. It implies that I1 = K1 r and I2 = K2 r for some two sets K1 ⊆ L1 and K2 ⊆ L2. I.e., Ii = {xi ∈ Li : 0i ∈ xi∧̄ai for all ai ∈ Ki}, for i = 1, 2. Accordingly I1 × I2 = {(x1, x2) ∈ L1 × L2 : (01, 02) ∈ (x1, x2)∧̄(a1, a2) for all (a1, a2) ∈ K1 × K2}. Therefore, I1 × I2 is an annihilator hyperideal of L1 × L2. Conversely, let I be an annihilator hyperideal REFERENCES 1415 of L1 × L2. It means there exist subset K ⊆ L1 × L2 such that I = Kr. We define the projections πi : L1 × L2 → Li for i = 1, 2. Let I1 and I2 be the projections of I on L1 and L1 respectively. In addition to K1 and K2 are the projections of K on L1 and L2. So πi(I) = Ii and πi(K) = Ki for i = 1, 2. We get I = {(x1, x2) ∈ L1 × L2 : (01, 02) ∈ (x1, x2)∧̄(a1, a2) for all (a1, a2) ∈ K1 × K2}. Thus πi(I) = {xi ∈ L1 : 0i ∈ xi∧̄iai for all ai ∈ Ki}. Therefore I ∼= π1(I)× π2(I). Corollary 3. For any two hyperideals I1 and I2 of two meet-hyperlattices L1 and L2 respectively. I1 and I2 are closed hyperideals iff I1×I2 is a closed hyperideal of the product meet-hyperlattice L1 × L2. 6. Conclusion In our work, the properties of annihilator hyperideals in the class of strong bounded dual distributive meet-hyperlattice were studied and proved. 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