EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 2, 2022, 486-495 ISSN 1307-5543 – ejpam.com Published by New York Business Global Closed Ideals and Annihilators of Distributive Dual Weakly Complemented Lattice Eman Ghareeb Rezk1,2 1 Department of Mathematics, Faculty of Science, Tanta University, Egypt 2 Princess Nourah bint Abdulrahman University, P.O.Box 84428, Riyadh 11671, Saudi Arabia Abstract. The goal of this paper is to study closed ideals and annihilators over the class of distributive dual weakly complemented lattices (DDWCLs). The algebraic structure of ideals, closed ideals, and dense ideals are shown. The connection between closed ideals and annihilators in this class is obtained. 2020 Mathematics Subject Classifications: 06C15, 06B10, 06D15, 06E75 Key Words and Phrases: Dual Weakly Complemented Lattices, Distributive Lattices, Ideals, Annihilators 1. Introduction A dual weakly complemented lattice was introduced by Wille and Kwuida in[9] and [18]. It is a bounded lattice equipped with a unary operation called a dual weak comple- mentation. M.Mandelker introduced the concept of annihilator in lattices in[11]. Cornish defined an annihilator on a distributive lattice in [3] and [4] and discussed its properties. Later many authors discussed the concept of annihilator in different algebraic structures and classes e.g.[1],[2],[5],[7], [8], [10], [12],[13],[14],[15], [16] and [17]. This contribution connected the notion of annihilators of DDWCL with a certain type of ideals is called closed ideals. The given definition of closed ideal depends on the dual weak complementation operation on the lattice of all ideals I(L) of L. Some im- portant properties of closed and dense ideals are proved. A new entry to the concept of annihilator in the class of DDWCLs is displayed. After preliminaries in section 2, the definition of closed and dense ideals are given and some properties are discussed in section 3. In addition, algebraic structures of them are obtained. In section 4, the definition of a closed annihilator is given, and we prove that DOI: https://doi.org/10.29020/nybg.ejpam.v15i2.4329 Email address: eman.rezk@science.tanta.edu.eg (E. G. Rezk) https://www.ejpam.com 486 © 2022 EJPAM All rights reserved. E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 487 the set of all closed annihilators forms a maximal Boolean algebra which is contained in the ortho lattice S(I(L)) of closed ideals of a DDWCL L. In the special case, if the unary operation is a pseudocomplementation, there is a one-to-one corresponding between the annihilators and the closed ideals of L. 2. Preliminaries In this section, we recall some basic definitions and results that are needed in the remaining parts. Definition 1. [9] A dual weakly complemented lattice is a bounded lattice L equipped with one unary operation H called dual weak complementation, and satisfied the following conditions for all a, b ∈ L (1) a ≤ aHH, (2) If a ≤ b implies aH ≥ bH, (3) (a ∨ b) ∧ (a ∨ bH) = a. In a distributive lattice, the condition (3) becomes a ∧ aH = 0. The trivial dual weakly complemented lattice is the lattice in which every dual weak complementation of any non-zero element is zero. A distributive pseudocomplemented lattice (p-algebra) is an algebra< L;∧,∨,∗ , 0, 1 >, where < L;∧,∨, 0, 1 > is a bounded distributive lattice and the unary operation ∗ is de- fined by: x ≤ a∗ iff x ∧ a = 0, a ∈ L. The operation ∗ is called pseudocomplementation on L. The distributive pseudocomple- mented lattice is a perfect example of the distributive dual weakly complemented lattice. Some properties of the operation H are listed in the following theorem. Theorem 1. [9] For any a, b of dual weakly complemented lattices L, we have the following: (1) 0H = 1, 1H = 0, (2) aHHH = aH, (3) (a ∨ aH)H = 0, (4) aH ≥ b iff bH ≥ a, (5) (a ∨ b)H = aH ∧ bH, (6) (a ∨ b)HH ≥ aHH ∨ bHH, (7) a ∨ bH ≥ b iff a ≥ b, (8) If aH ≥ b then a ∧ b = 0, (9) Tf a ∨ b = 1 then bH ≤ a, (10) If a ∨ aH = 1 then a = aHH, (11) a ∧ (a ∧ b)H ≥ a ∧ bH. E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 488 Definition 2. [6] An orthocomplemented lattice (ortho lattice) is a bounded lattice < L;∧,∨ > equipped with one unary operation ⊥ called orthocomplementation on L and satisfied the following conditions, for all a, b ∈ L (i) a ≤ b implies a⊥ ≥ b⊥, (ii) a⊥⊥ = a, (iii) a ∧ a⊥ = 0 and a ∨ a⊥ = 1. A distributive ortho lattice is a Boolean algebra. The skeleton of dual weakly com- plemented lattice L is defined by S(L) = {a ∈ L : a = aHH}. It forms an ortho lattice with the same meet operation of L and join ”Y ”operation defined as: a Y b = (aH ∧ bH)H, for any a, b ∈ S(L). The dual weak complemented of an element a ∈ S(L) is its orthocomple- mented in S(L). The set D(L) of dense element of L defined as D(L) = {x ∈ L : xH = 0}. refer to [9]. Definition 3. [6] A non-empty subset I of a lattice L is called an ideal of L if (i) a, b ∈ I implies a ∨ b ∈ I, (ii) a ∈ L, b ∈ I and a ≤ b implies a ∈ I. The ideal (a] = {x ∈ L : x ≤ a} is called the principal ideal generated by a ∈ L. Let I(L) be the set of all ideals of L under set inclusion forms a complete lattice with the smallest element (0] = {0} and the largest element (1] = L. For any I,K ∈ I(L), the infimum I ∧K = I ∩K, and the supremum I ∨K = {x ∈ L : x ≤ i∨ k for some i ∈ I and some k ∈ K}. The lattice of all ideals I(L) of a distributive lattice L is distributive. For a, b ∈ L, meet and join operations of two principal ideals are given as: (a] ∧ (b] = (a ∧ b] and (a] ∨ (b] = (a ∨ b], see[6]. Definition 4. [4] For a non-empty subset A of a lattice L. Define the set A∗ = {x ∈ L : x ∧ a = 0, for all a ∈ A} is called the annihilator ideal of A in L. The set of annihilator ideals A(L) of a distributive lattice L is a complete Boolean algebra with the smallest element (0], the largest element L, set-theoretic intersection as the infimum, and the map I → I∗ as complementation. The supremum of I and J in A(L) is given by I Y J = (I∗ ∩ J∗)∗. If I∗ = (0] then I is called dense ideal. The set of all dense ideals of a distributive lattice L is denoted by D∗(L) and it forms a distributive lattice too. An ideal of form (a]∗ is called an annulet of a ∈ L. Each annulet is an annihilator ideal, where (a]∗ = (a∗], a∗ is the pseudocomplemented of a in L, see[4]. From now on, L stands for non-trivial distributive dual weakly complemented lat- tice. E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 489 3. Closed and dense Ideals of DDWCLs In the present section, the dual weak complementation operation is defined on the lattice of all ideals I(L) of L. Concepts of closed and dense ideals are introduced and it’s basic properties are proved. Definition 5. Let I be an ideal. Define the set IO = {x ∈ L : x ≤ aH, for all a ∈ I}. Note that, If I is an ideal and x ∈ IO then x ∧ a = 0, for all a ∈ I. Consequently x ∈ I∗ and so IO ⊆ I∗. Proposition 1. A set IO is an ideal, moreover is the dual weak complemented of the ideal I in I(L) . Proof. Assume that x, y ∈ IO, then x, y ≤ aH for all a ∈ I. Thus x ∨ y ≤ aH. Conse- quently, x ∨ y ∈ IO. Let z ∈ L and z ≤ x for some x ∈ IO. Then z ≤ aH for all a ∈ I. Therefore IO is an ideal of L. Now we show that IO satisfies the conditions of dual weak complementation: (1) If a ∈ I and b ∈ IO then b ≤ aH. From (4) in Theorem 1, this is equivalent that a ≤ bH. Thus a ∈ IOO. (2) Let I ≤ K and x ∈ KO. Then x ≤ bH for all b ∈ K. So x ≤ bH for all b ∈ I. It implies x ∈ IO. So KO ≤ IO, (3) Clearly I ∧ I∗ = (0], then I ∧ IO = (0]. If L is a distributive pseuodocomplemented lattice, then IO = I∗, and the lattice I(L) of its ideals become a distributive pseuodocomplemented lattice. Lemma 1. Let I,K be two ideals and a ∈ L. Then,: (1) (a]O = (aH], (2) IO = ∩a∈I(a] O, (3) (0]O = (1] and (1]O = (0], (4) IO = IOOO, (5) IO ≥ K iff KO ≥ I, (6) IO ⊆ KO iff KOO ⊆ IOO, (7) I ⊆ KO implies I ∩K = (0]. Proof. (1) Suppose x ∈ (aH], then x ≤ yH for all y ∈ (a].So x ∈ (a]O. Conversely, let y ∈ (a]O and aH ≤ y. Thus yHH ≥ aH and by using (7) in Theorem 1 we get yHH∨aHH ≥ aH. Hence, aH ∧ (yHH ∨ aHH) = (aH ∧ yHH) ∨ (aH ∧ aHH) ≥ aH, which is a contradiction. So, y ≤ aH.Therefore, (a]O = (aH]. E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 490 (2) Let x ∈ ∩a∈I(a] O i.e., x ∈ (a]O, for all a ∈ I. Then x ≤ aH and x ∈ IO. Conversely, if x ∈ IO then x ≤ aH for all a ∈ I. It means x ∈ ∩a∈I(a] O. Therefore IO = ∩a∈I(a] O. (3) We get (0]O = {x ∈ L : x ≤ 0H = 1} = (1], (1]O = {x ∈ L : x ≤ 1H = 0} = (0]. (4) By using Proposition1, we get IOOO ≤ IO. Conversely, if b ∈ IOOO, then b ≤ aH for all a ∈ IOO. But I ⊆ IOO so x ≤ cH for all c ∈ I.Thus x ∈ IO. (5) Suppose IO ≥ K then I ≤ IOO ≤ KO and vice versa. (6) Let IO ≤ KO. Then, from Proposition1, IOO ≥ KOO. The opposite direction can be get by using (4). (7) Assume I ⊆ KO then we get I ∧KO = (0], since K ∧KO = (0]. Proposition 2. Let I,K ∈ I(L). Then: (1) I ∨KO ≥ K iff I ≥ K, (2) I ∨K = (1] implies KO ≤ I, (3) (I ∨K)O = IO ∩KO, (4) (I ∩K)OO ≤ IOO ∩KOO, (5) (I ∨ IO)O = (0], (6) (I ∨K)OO ≥ IOO ∨KOO, (7) I ∩ (I ∩K)O ≥ I ∩KO. Proof. (1) Suppose that I ∨ KO ≥ K, we get K = K ∧ (I ∨ KO) = (K ∧ I) ∨ (K ∧ KO) = K ∧ I, which means I ≥ K. Conversely, if I ≥ K then IO ≤ KO. So I ∨KO ≥ I ∨ IO ≥ I ≥ K. (2) Assume I ∨K = (1]. Meet each side by KO to get KO ∧ I = KO. Thus KO ≤ I. (3) Let x ∈ IO ∩ KO and z = a ∨ b ∈ I ∨ K, where a ∈ I and b ∈ K. Then x ≤ aH and x ≤ bH. It implies x ≤ aH ∧ bH = (a ∨ b)H = zH. Hence, x ∈ (I ∨ K)O and so IO ∩KO ⊆ (I ∨K)O. The converse is trivial. (4) From (3) we get (I ∩K)OO ≤ IOO ∩KOO. (5) Using (3) we get (I ∨ IO)O = IO ∩ IOO = (0]. (6) Since K, I ≤ I ∨K, then IOO,KOO ≤ (I ∨K)OO. Hence IOO ∨KOO ≤ (I ∨K)OO. (7) Since (I ∩K)O ≥ KO. Meeting both sides by I we get I ∩ (I ∩K)O ≥ I ∩KO. The skeleton S(I(L)) and the set of dense elements of I(L) are defined as: S(I(L)) = {I ∈ I(L) : I = IOO}, and D(I(L)) = {I ∈ I(L) : IO = (0]}. The elements of S(I(L)) and D(I(L)) are called closed and dense ideals, respectively. E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 491 Lemma 2. Let I and K be two ideals. Then: (1) (a]OO = (a] iff a ∈ S(L), (2) I ∩K = (IO ∨KO)O, for all I,K ∈ S(I(L)), (3) If I ∨ IO = (1] then I ∈ S(I(L)), (4) If I is an annihilator then I ∈ S(I(L)). Proof. (1) Immediately from (1) in Lemma 1. (2) By using (3) in Proposition 2, I ∩K = IOO ∩KOO = (IO ∨KO)O. (3) Assume I ∨ IO = (1], meet both sides by IOO to get IOO = IOO ∧ (I ∨ IO) = (IOO ∧ I) ∨ (IOO ∧ IO) = IOO ∧ I = I. Therefore I = IOO. (4) Let I be an annihilator and I∗O be a proper subset of I = I∗∗. So there exists x ∈ I∗∗ such that x � aH for all a ∈ I∗. Thus x ≤ a∗ and x � a∗, at the same time, which is a contradiction. Therefore I = I∗O ∈ S(I(L)). Theorem 2. The skeleton S(I(L)) of the lattice I(L) forms an ortho lattice. Proof. Two binary operations ∩ and t will be defined for I,K ∈ S(I(L)) by: I ∩K = (IO ∨KO)O and I tK = (IO ∩KO)O. It is clear that the meet operation ∩ is the usual intersection of I(L). Now we prove that the supremum I tK of any two ideals I and K of S(I(L)) equals (IO ∩KO)O: Since IO ∩KO ≤ IO,KO, we get (IO ∩KO)O ≥ IOO = I,KOO = K. If J ∈ S(I(L)) and J ≥ I,K, then JO ≤ IO,KO. It implies JO ≤ IO ∩KO i.e., JOO = J ≥ (IO ∩KO)O. If I ∈ S(I(L)),then IO is the orthcomplemented of I. especially, (0]O = (1] and (1]O = (0]. Therefore < S(I(L));∩,t,O , (0], (1] > forms an ortho lattice. Theorem 3. The skeleton S(L) of the lattice L is embedded in the skeleton S(I(L)) of its lattice of ideals I(L). Proof. Consider the map ψ from S(L) into S(I(L)) which mapping the element a ∈ S(L) to (a]OO ∈ S(I(L)) is a well defined and satisfies the following: ψ(a ∧ b) = (a ∧ b]OO = (a ∧ b] = (a] ∩ (b] = (a]OO ∧ (b]OO = ψ(a) ∩ ψ(b), ψ(a ∨ b) = ψ((aH ∧ bH)H) = ((aH ∧ bH)H]OO = ((aH ∧ bH)H] = (aH ∧ bH]O = ((aH] ∧ (bH])O = ((a]O ∧ (b]O)O = (a] t (b] = (a]OO t (b]OO = ψ(a) t ψ(b), ψ(aH) = (aH]OO = (a]OOO = (ψ(a))O, ψ(0) = (0]OO = (0], and ψ(1) = (1]OO = L. It is easy to show that ψ is an injective map and this completed the proof. Theorem 4. The following statements are equivalent: (1) I is closed ideal, (2) If a ∈ L and a ≤ bH, for all b ∈ IO then a ∈ I, (3) I = KO for some ideal K ∈ I(L). E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 492 Proof. Let I be a closed ideal and a ∈ L such that a ≤ bH for all b ∈ IO, then a ∈ IOO = I. Let (2) be satisfied and let x ∈ IOO. Then x ≤ bH for all b ∈ IO. Therefore x ∈ I. Assume I = IOO, set K = IO then KO = I ∈ S(I(L)). Let I = KO for some K ∈ I(L). Thus IOO = KOOO = KO = I. Lemma 3. Let D(I(L)) be the set of all dense ideals of L. Then: (1) (1] ∈ D(I(L)), (2) If I,K ∈ I(L) such that I ⊆ K and I ∈ D(I(L)) then K ∈ D(I(L)), (3)If I,K ∈ D(I(L)), then I ∨K ∈ D(I(L)), (4)If I ∈ D(I(L)), then I ∨ IO ∈ D(I(L)), (5)If I ∩D(L) 6= φ then I ∈ D(I(L)). (6) D∗(L) ⊆ D(I(L)). Proof. (1) Since (1]O = (0], then (1] ∈ D(I(L)). (2) Let I,K ∈ I(L), I ⊆ K and IO = (0]. This implies (0] = IO ⊇ KO i.e., KO = (0] and K ∈ D(I(L)). (3) Assume I,K ∈ D(I(L)), from (3) in Proposition 2, we get (I ∨K)O = IO ∩KO = (0]. Thus I ∨K ∈ D(I(L)). (4) If I ∈ D(I(L)) then I ⊆ I ∨ IO. It implies (0] = IO ⊇ (I ∨ IO)O. Therefore I ∨ IO ∈ D(I(L)). (5) Assume I ∩D(L) 6= φ, then there exists a non-zero element a ∈ I such that aH = 0. Hence IO = (0] and I ∈ D(I(L)). (6) Assume I ∈ D∗(L). Since IO ⊆ I∗ = (0], then I ∈ D(I(L)). Theorem 5. D(I(L)) forms a join semilattice with one. 4. Closed Annihilators of DDWCLs This section introduces the concept of a closed annihilator. The structure of the set of all closed annihilators of DDWCL L and the connection with the set of all closed ideals S(I(L))are investegated. Definition 6. An ideal I is called a closed annihilator iff I = IO∗. The set of all annihilators is denoted by IO∗(L). The only dense ideal belongs to IO∗(L) is (1]. Lemma 4. (1) If I is an ideal then I∗O ⊆ IOO ⊆ IO∗, E. G. Rezk / Eur. J. Pure Appl. Math, 15 (2) (2022), 486-495 493 (2) If I is a closed annihilator then I∗ = IO, (3) If I and K are closed annihilators then I ∩K and I tK are too, (4) I ∈ A(L) iff I ∈ IO∗(L). Proof. (1) We have that IO ⊆ I∗. then IOO ⊇ I∗O. Suppose x ∈ IOO i.e., x ≤ aH for all a ∈ IO.It implies x ∧ a = 0. Therefore I∗O ⊆ IOO ⊆ IO∗. (2) If I is a closed annihilator then I = IO∗ = I∗O. The dual weak complementation of I is given by IO = IO∗O = I∗OO ⊇ I∗. Therefore I∗ = IO. (3) For meet operation we have I ∩K ≤ I,K. Thus (I ∩K)O∗ ≤ IO∗,KO∗. Accordingly, (I∩K)O∗ ≤ IO∗∩KO∗ = I∩K, and I∩K ⊆ (I∩K)OO ⊆ (I∩K)O∗. Then I∩K ∈ IO∗(L).For join operation we have [I tK]O∗ = (IO ∩KO)OO∗ = (IO ∩KO)O = I tK. (4) If I ∈ A(L) then I = I∗O = I∗∗ and I∗ ∈ A(L). Hence IO∗ = I∗OO∗ = I∗∗ = I. Conversely, if I ∈ IO∗(L) then I = IO∗ = I∗∗ and I∗ = IO ∈ IO∗. So I∗O = IO∗∗O = IOO = I. The subset of ortho lattice L which forms a Boolean algebra under the same opera- tions of L is called a Boolean algebra induced from L. Theorem 6. The set IO∗(L), of all closed annihilators forms a maximal Boolean algebra indued from S(I(L)). Proof. To prove IO∗(L) forms a Boolean algebra it is enough to prove the distributivity of it. Initially, we prove the inequality (1): for I, J,K,H ∈ IO∗(L) (I t J) ∩H ≤ I t (J ∩H) ...(1) Since, H ∩ IO ∩ (J ∩H)O ≤ IO...(2) and, J ∩ H ∩ (IO ∩ (J ∩ H)O) = IO ∩ (J ∩ H) ∩ (J ∩ H)O = (0]. It implies that, H ∩ IO ∩ (J ∩H)O ≤ JO Thus, H ∩ IO ∩ (J ∩H)O ≤ JO...(3) From (2) and (3), H ∩ IO ∩ (J ∩H)O ≤ IO ∩JO and from the properties of ”O”, we get (H∩IO∩(J∩H)O)∩(IO∩JO)O = (0], implies that, ((IO∩JO)O∩H)∩(IO∩(J∩H)O) = (0]. So, H ∩ (IO ∩ JO)O ≤ (IO ∩ (J ∩H)O)O. Therefore, (IO ∩ JO)O ∩H ≤ (IO ∩ (J ∩H)O)O. By putting H = I t K in inq.(1), we get (I t J) ∩ (I t K) ≤ I t [(J ∩ I t K)] ≤ I t [I t (J ∩K)] = I t (J ∩K). Also, J ∩K ≤ (I t J)∩ (I tK) and I ≤ (I t J)∩ (I tK). Then, I t (J ∩K) ≤ (I t J) ∩ (I tK). Consequently, I t (J ∩K) = (I t J) ∩ (I tK). Therefore, < IO∗(L);t,∩,O , (0], (1] > forms a Boolean algebra. Next, we prove the maximality of IO∗(L). Let B be a Boolean algebra induced from S(I(L)) such that IO∗(L) ⊂ B. Then there exist ideal H ∈ B and H 6∈ IO∗(L), then H ⊂ HO∗. Since HO ∩HO∗ = (0] then. HO∗ ⊆ HOO = H. So that H = HO∗, which is a contradiction. REFERENCES 494 Theorem 7. (1) A(L) isomorphic to IO∗(L), (2) If IO = KO, for any closed annihilator I and any ideal K such that I ⊂ K, then S(I(L)) is a Boolean algebra, (3) If there exists closed annihilator I and ideal K such that I ⊂ K and KO ⊂ IO then S(I(L)) is not Boolean algebra. Proof. (1) We defined a map α : A(L) → IO∗(L) as: α(I) = IO∗. Assume I, J ∈ A(L) then: α(I ∩ J) = (I ∩ J)O∗ = I ∩ J = IO∗ ∩ JO∗, α(I Y J) = (I Y J)O∗ = (I∗ ∩ J∗)∗O∗ = (IO ∩ JO)O = I t J , α(I∗) = I∗O∗ = I∗ = IO, Clearly that α is bijective map. When for any closed annihilator I and any ideal K such that I ⊂ K we get KO = IO = I∗. Then the set of all closed ideals S(I(L)) coincides with the set IO∗(L) of all closed annihilators. Consequently, S(I(L)) becomes a Boolean algebra. While if there exists closed annihilator I and ideal K such that I ⊂ K and KO ⊂ IO then KO and its dual weak complementation are added to S(I(L)). 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