# Weakly 2-Absorbing Ideals in Almost Distributive Lattices

Authors: Natnael Teshale Amare
Journal: Journal of Mathematics
 (2022)

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## Abstract

The concepts of weakly 2-absorbing ideal and weakly 1-absorbing prime ideal in an almost distributive lattice (ADL) are introduced, and the necessary conditions for a weakly 1-absorbing prime ideal to become a weakly 2-absorbing ideal in algebraic form are proved. Also, weakly 2-absorbing ideals are characterized in terms of weakly prime ideals and 2-absorbing ideals. Finally, the lattice epimorphic images and inverse images of the weakly 2-absorbing ideal and weakly 1-absorbing prime ideal are discussed.


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## Body

## 1. Introduction

Badawi [1] was introduced the concept of 2-absorbing ideals on a commutative ring and assume that all rings are commutative with 1≠0, which is a generalization of prime ideals and some properties of these were studied. Subsequently, several researchers worked on 2-absorbing ideals in different cases (refer to [2–5]). Later in the paper [6], Yassine et al. obtained on 1-absorbing prime ideals of a commutative ring. In the papers [7–10] (Figures 1and 2), the concept of weakly prime ideals were introduced and some of its properties are investigated. The concept of an almost distributive lattice (ADL) was introduced by Swamy and Rao [11] (Figures 3 and 4) as a common abstraction to most of the existing ring theoretic generalizations of a Boolean algebra and which is an algebra A,∧,∨,0 of type 2,2,0 satisfying all the axioms of a distributive lattice with zero except ∧ commutative, ∨ commutative, and right distributivity of ∨ over ∧. The concept of prime ideal is a vital role in the study of structure theory of distributive lattices in general and of Boolean algebras in particular, see [11].Figure 1
The complete lattice diagram.Figure 2
The Complete Lattice diagram in Example5.Figure 3
The Boolean lattice diagram.Figure 4
The chain 4 diagram.In this paper, we introduce the concept of 2-absorbing ideal in an almost distributive lattice which is a generalization of prime ideals in an ADL. A proper idealP of an ADL A is called a 2-absorbing ideal (2-AI for short) of A if whenever x,y,z∈A and x∧y∧z∈P, then x∧y∈P or y∧z∈P or x∧z∈P. It is shown that a proper ideal P of A is a 2-AI of A if and only if whenever P1∩P2∩P3⊆P for some ideals Pi of A, for 1≤i≤3, then P1∩P2⊆P or P1∩P3⊆P or P2∩P3⊆P. In addition to this, it is observed that the lattice epimorphic image and inverse image of 2-AI and n-AI are also 2-AI and n-AI, respectively. Next, we introduce the concepts of weakly 2-AI of an ADL which is weaker than that of weakly prime ideal and 2-AI of an ADL. The Cartesian product of weakly 2-AIs is also discussed here, and some equivalent conditions for the set of all weakly 2-AIs to become 2-AIs under the Cartesian product are derived. Mainly, we have proved that all prime ideals are 2-AIs and weakly prime ideals, that weakly prime ideals are weakly 2-AIs, and also that all 2-AIs are weakly 2-AIs and vice versa is not true; examples were given to shown these. It is further demonstrated that all prime ideals are 1-absorbing prime ideals, that 1-absorbing prime ideals are 2-AIs, and that all weakly prime ideals and 1-absorbing prime ideals are weakly 1-absorbing prime ideals and that weakly 1-absorbing prime ideals are weakly 2-AIs, and there are examples that show that the converse of these is not true. Finally, it is also shown that the image and inverse image of weakly 2-AI, 1-absorbing prime ideal, and weakly 1-absorbing prime ideal under lattice epimorphism are again weakly 2-AI, 1-absorbing prime ideal, and weakly 1-absorbing prime ideal, respectively.Throughout this paper,A stands for an ADL A,∧,∨,0 with a maximal element and L stands for a complete lattice L,∧,∨,0,1 satisfying the infinite meet distributive law, that is,
(1)a∧∨b∈Sb=∨b∈Sa∧b,for any S⊆L and a∈L.


## 2. Preliminaries

In this section, we recall certain definitions, results, and notations which will be needed later on are presented, see [1, 11, 12].Definition 1.
An algebraA=A,∧,∨,0 of type 2,2,0 is called an ADL if it satisfies the following conditions for all a,b and c∈A.(1)
0∧a=0(2)
a∨0=a(3)
a∧b∨c=a∧b∨a∧c(4)
a∨b∧c=a∨b∧a∨c(5)
a∨b∧c=a∧c∨b∧c(6)
a∨b∧b=bEach of the axioms (1) through (6) above is independent from the others. The element0 is called the zero element.Any bounded below distributive lattice is an ADL.Example 1.
LetX be a nonempty set. Fix an arbitrary element x0∈X. For any x,y∈X, define ∧ and ∨ on X by,
(2)x∧y=y,ifx≠x0,x0,ifx=x0,x∨y=y,ifx≠x0,x0,ifx=x0.ThenX,∧,∨,x0 is an ADL with x0 as its zero element. This ADL is called the discrete ADL.Several ring theoretic generalizations of Boolean algebras (other than Boolean rings which are precisely Boolean algebras) can be made as an ADL. The following example is one such.Example 2.
LetR be a commutative regular ring with identity (that is, R is a commutative ring with unity in which, for each a∈R, there exists an (unique) idempotent a0∈R such that aR=a0R). For any a,b∈R, define
(3)a∧b=a0b,a∨b=a+b−a0b.Then,R,∧,∨,0 is an ADL with the additive identity 0 as the zero element.Theorem 2.
LetA,∧,∨,0 be an ADL. For any a and b∈A, we have
(1)
a∧0=0=0∧a and a∨0=a=0∨a(2)
a∧a=a=a∨a(3)
a∧b∨b=b(4)
a∨b∧a=a(5)
a∧a∨b=a(6)
a∧b=a⇔a∨b=b(7)
a∧b=b⇔a∨b=a(8)
a∨b∨a=a∨bDefinition 3.
LetA,∧,∨,0 be an ADL. For any a and b∈A, define a≤b if a=a∧b (equivalently a∨b=b).
Then≤ is a partial order on A.Theorem 4.
The following hold good for any elementsa,b,c, and d of an ADL A,∧,∨,0.
(1)
a∧b≤b≤b∨a(2)
a≤b⇒a∧b=a=b∧a and a∨b=b=b∨a(3)
a∨b∧c=b∨a∧c(4)
a∧b∧c=a∧b∧c (i.e., ∧ is associative on A)(5)
a∧b∧c=b∧a∧c(6)
The setx∧a:x∈A=y∈A:y≤a is a bounded distributive lattice under the induced operations ∧ and ∨ with 0 as the smallest element and a as the largest element(7)
a∨b=b∨a whenever a∧b=0(8)
a∧b=0⇔b∧a=0(9)
a≤b⇒a∧c≤b∧c and c∧a≤c∧b(10)
a≤b⇒c∨a≤c∨b(11)
a∨b∨c∧d=a∨b∨c∧dTheorem 5.
For any elementsa and b of an ADL A,∧,∨,0, the following are equivalent to each other.
(1)
a∧b∨a=a(2)
a∧b∨a=a(3)
a∧b=b∧a(4)
a∨b=b∨a(5)
Supa,b exists in A,≤ and is equal to a∨b(6)
There existsx∈A such that a≤x and b≤x(7)
infa,b exists in A,≤ is equal to a∧bTheorem 6.
The following statements are equivalent for any ADLA.
(1)
a∧b=b∧a for all a,b∈A(2)
a∨b=b∨a for all a,b∈A(3)
A,∧,∨ is a distributive lattice bounded below(4)
a∧b∨c=a∨c∧b∨c for all a,b,c∈A(5)
b∧a∨b=b(i.e., b≤a∨b) for all a,b∈A(6)
a∧b∨a=a (i.e., a∧b≤a) for all a,b∈A(7)
For anya,b,c∈A,a≤b⇒a∨c≤b∨cAs a consequence, for any idealI of A,x∧a∈I for all a∈I and x∈A. An element m∈A is said to be maximal if, for any x∈A,m≤x implies m=x. It can be easily observed that m is maximal if and only if m∧x=x for all x∈A.Definition 7.
LetI be a nonempty subset of an ADL A. Then, I is called an ideal of A if a,b∈I⇒a∨b∈I and a∧x∈I for all x∈A.Definition 8.
LetA=A,∧,∨,0 be an ADL and for any subset S of A, let
(4)S=∩I∈IA:S⊆I.Then,S is the smallest ideal of A containing S and is called the ideal generated by S in A. Also,
(5)S=∨i=1nsi∧a:n≥0,si∈Sanda∈A.WhenS=x, then we simply write x for x and call this the principal ideal generated by x in A. The principal ideal generated by x in A is given by
(6)x=a∈A:x∧a=a=x∧a:a∈A.Theorem 9.
LetA=A,∧,∨,0 be an ADL and a and b∈A. Then, the following holds good.
(1)
a∩b=a∧b(2)
a∨b=a∨b(3)
a∧b=b∧a and a∨b=b∨aTheorem 10.
LetA=A,∧,∨,0 be an ADL and I and J be ideals of A. Then, in the lattice IA,⊆, I∧J=I∩J, and I∨J=s∨t:s∈Iandt∈J. Also, the lattice IA,⊆ is distributive.Definition 11.
A nonzero proper idealI of R is called a 2-absorbing ideal of R if for any a,b,c∈R and abc∈I, then ab∈I or ac∈I or bc∈I.


## 3. 2-Absorbing Ideal

In this section, we introduce the notion of 2-absorbing ideal (2-AI) andn-absorbing (n-AI) of a given almost distributive lattice (ADL) A=A,∧,∨,0 and prove several structural properties of these.Definition 12.
LetA=A,∧,∨,0 be an ADL. A proper ideal P of A is said to be a 2-absorbing ideal of A and denoted by 2-AI if for any x,y,z∈A(7)x∧y∧z∈P⇒x∧y∈Pory∧z∈Porx∧z∈P.Theorem 13.
LetP be a 2-AI of A. For x,y,z∈A such that x∧y∧z∈P, we have
(8)y∧x∈Porz∧y∈Porz∧x∈P.Proof.
Forx,y,z∈A such that x∧y∧z∈P. Then y∧x=y∧x∧x.
=y∧x∧x (by 2.7(4))
=x∧y∧x. (again by 2.7(4))
SinceP is a 2-AI of A and x∧y∈P, hence x∧y∧x∈P. Thus y∧x∈P. Similarly, z∧y∈P or z∧x∈P.Lemma 14.
LetP1 and P2 be ideals of A and P be a 2-AI of A. The following assertions hold for any x,y∈A.
(1)
x∧y∩P1⊆P⇒x∧y⊆P or x∩P1⊆P or y∩P1⊆P(2)
x∩P1∩P2⊆P⇒x∩P1⊆P or x∩P2⊆P or P1∩P2⊆PProof.
LetP be a 2-AI of A and P1 and P2 ideals of A.
(1)
Supposex∧y∩P1⊆P. Let t∈x∧y. Then, t=x∧y∧t. Since P is a 2-AI of A, if whenever x,y,z∈A and x∧y∧z∈P, then x∧y∈P or y∧z∈P or x∧z∈P. From this, we have t=x∧y∧t∈P. Hence, x∧y⊆P. Suppose x∩P1ÚP and y∩P1ÚP. Now, x∩P1∩y∩P1=x∩y∩P1=x∧y∩P1⊆P (by given); this a gives a contradiction. Thus, x∩P1⊆P or y∩P1⊆P(2)
Supposex∩P1∩P2⊆P. Let x∈A such that x∩P1ÚP and x∩P2ÚP. It follows that x∩P1∩x∩P2ÚP; that is, x∩P1∩P2ÚP, a contradiction. Therefore, x∩P1⊆P or x∩P2⊆PTheorem 15.
LetP be a proper ideal of A. The following statements are equivalent:
(1)
P is a 2-AI of A(2)
For idealsP1,P2,P3 of A, P1∩P2∩P3⊆P⇒P1∩P2⊆P or P1∩P3⊆P or P2∩P3⊆P(3)
For idealsP1,P2,P3 of A, P=P1∩P2∩P3⇒P=P1∩P2 or P=P1∩P3 or P=P2∩P3Proof.
1⇔2: suppose P is a 2-absorbing ideal of A. Let P1∩P2∩P3⊆P, for some proper ideals P1,P2, and P3 of A. Let x,y∈A such that x∩P2ÚP and y∩P3ÚP and put P1=x. It follows that, x∩P3⊆P and y∩P2⊆P. By Lemma 14(2), x∧y∩P2∩P3⊆P, we get either x∧y∩P2⊆P or x∧y∩P3⊆P. If x∧y∩P2⊆P, then x∧y∧z=y∧x∧z (by Theorem 5(4)), for all z∈P2, which implies that x∧z∈P, so x∩P2⊆P, contradiction. Similarly, if x∧y∩P3⊆P, then x∧y∧t=y∧x∧t (by 2.7(4)), for all t∈P3, implies that x∧t∈P, so x∩P3⊆P, contradiction. Thus, either P1∩P2⊆P or P1∩P3⊆P. Conversely suppose P1∩P2∩P3⊆P implies that either P1∩P2⊆P or P1∩P3⊆P or P2∩P3⊆P, for any ideals P1,P2,P3 of A. Let x,y,z∈A, we have x∧y∧z∈P. Suppose also that x∧y∉P and y∧z∉P. Let P1=x,P2=y and P3=z. Since P1∩P2∩P3=x∩y∩z=x∧y∩z⊆P, it follows that P1∩P2ÚP and P2∩P3ÚP. Then by the above lemma, P1∩P3=x∩z=x∧z⊆P; that is, x∧z∈P. Thus, P is a 2-absorbing ideal of A.
(2)⇔3 and 1⇔3 are clearDefinition 16.
LetA1,∧,∨,0 and A2,∧,∨,0 be ADLs and form the set A1×A2 by A1×A2=a,b:a∈A1andb∈A2. Define ∧ and ∨ in A1×A2 by,
(9)a,b∧c,d=a∧c,b∧d,and a,b∨c,d=a∨c,b∨d, for any a,b,c,d∈A1×A2.Then,A1×A2,∧,∨,0 is an ADL under the pointwise operations and 0=0,0 is the zero element in A1×A2.Let us recall from [11] that a proper ideal P of A is said to be a prime ideal if, for any x and y∈A,x∧y∈P⇒ either x∈P or y∈P. Now, we have the following.Theorem 17.
Every prime ideal ofA is a 2-AI of A.Proof.
Assume thatP is a prime ideal of A. Let x,y,z∈A, x∧y∧z∈P. Then, either x∧y∈P or z∈P, or y∧z∈P or x∈P, and hence x∧z∈P (since P is an ideal and by 2.10). If x∧y∈P, then it is obvious and if z∈P, x∧z∈P and y∧z∈P. Thus, P is a 2-AI of A.The following example show that the converse of Theorem44 is not true.Example 3.
LetD=0,x,y be a discrete ADL with 0 as its zero element defined in Example1andL=0,a,b,c,1 be the lattice represented by the Hasse diagram given below:ConsiderD×L=t,s:t∈Dands∈L. Then, D×L,∧,∨,0 is an ADL (note that D×L is not a lattice) under the pointwise operations ∧ and ∨ on D×L and 0=0,0, the zero element in D×L. Then, P=0,0 is a 2-AI of D×L but P is an ideal which is not prime, since 0,a∧x,b=0,0, for all 0,a,x,b∈D×L.Theorem 18.
LetP and Q be prime ideals of A. Then, P∩Q is a 2-AI of A.Proof.
Leta,b,c∈A and a∧b∧c∈P∩Q. Then, a∧b∧c∈P and a∧b∧c∈Q. Since P and Q are prime ideals of A, we have either a∧b∈P or c∈P, or b∧c∈P or a∈P, or a∧c∈P or b∈P (and a∧b∈Q or c∈Q, or b∧c∈Q or a∈Q, or a∧c∈Q or b∈Q). Suppose that a∧b∈P or c∈P and a∧b∈Q or c∈Q. If a∧b∈P and a∧b∈Q, then a∧b∈P∩Q. If c∈P and c∈Q, then either a∧c∈P and a∧c∈Q or b∧c∈P and b∧c∈Q. Hence the theorem.Definition 19.
LetA1 and A2 be ADLs. A mapping f:A1⟶A2 is called a lattice homomorphism if the following are satisfied, for any x,y,z∈A1.
(1)
fx∧y∧z=fx∧fy∧fz(2)
fx∨y∨z=fx∨fy∨fz(3)
f0=0Theorem 20.
LetA1 and A2 be ADLs and f:A1⟶A2 be a lattice homomorphism. Then, the following holds.
(1)
Iff is an epimorphism and Q is a 2-AI of A2, then f−1Q is a 2−AI of A1(2)
Iff is an isomorphism and P is a 2-AI of A1, then fP is a 2−AI of A2Proof.
LetA1 and A2 be ADLs and f:A1⟶A2 be a lattice homomorphism.
(1)
LetQ be a 2-AI of A2 and f−1Q=a∈A1:fa∈Q⊆A2. Let a,b,c∈f−1Q. Then, fa,fb,fc∈Q. Since Q is a 2-AI of A2, fa∧fb∧fc=fa∧b∧c∈Q. Thus, a∧b∧c∈f−1Q. Now if a∈f−1Q and x∈A1, then fa∈Q and fx∈A2 and hence fa∧fx=fa∧x∈Q. Thus, a∧x∈f−1Q. Therefore, f−1Q is a 2-AI of A1(2)
Leta,b,c∈A1 such that fa=x,fb=y and fc=z, for some x,y,z∈A2. As x∧y∧z∈fP, we have fa∧b∧c∈fP. Since P is a 2-AI of A1, a∧b∧c∈P implies that either a∧b∈P or b∧c∈P or a∧c∈P. That is either fa∧b=fa∧fb=x∧y∈fP or y∧z∈fP or x∧z∈fP. Thus, fP is a 2-AI of A2Theorem 21.
LetA1 and A2 be ADLs. If P is a 2-AI of A1, then P×A2 is a 2-AI of A1×A2. Also, if Q is a 2-AI of A2, then A1×Q is a 2-AI of A1×A2.Proof.
LetP be a 2-AI of A1 and a,b,c∈A1 such that a,t∧b,t∧c,t∈P×A2, for every t∈A2. Then, a,t∧b,t∧c,t=a∧b∧c,t∈P×A2. Since P is a 2-AI of A1, we have either a∧b∈P or a∧c∈P or b∧c∈P. So that, either a∧b,t∈P×A2 or a∧c,t∈P×A2 or b∧c,t∈P×A2, for every t∈A2. Thus, P×A2 is a 2-AI of A1×A2. Similarly, A1×Q is a 2-AI of A1×A2.Definition 22.
A proper idealP of A is a weakly prime ideal of A if for any x,y∈A,
(10)0≠x∧y∈P⇒eitherx∈Pory∈P.Lemma 23.
Every prime ideal ofA is a weakly prime ideal of A.Proof.
It is clear.The converse of the above lemma is not true; consider the following example.Example 4.
LetD×L=t,s:t∈Dands∈L be an ADL discussed in Example 3. Let R=0,0. Clearly, R is a weakly prime ideal of A, while R is not a prime ideal of D×L, since 0,a∧x,b∈R implies that 0,a∉R and x,b∉R, for all 0,a,x,b∈D×L. Thus, every weakly prime ideal of D×L is not a prime ideal of D×L.Definition 24.
A proper idealP of A is a weakly 2-AI of A if for any x,y,z∈A,
(11)0≠x∧y∧z∈P⇒eitherx∧y∈Pory∧z∈Porx∧z∈P.Lemma 25.
Every weakly prime ideal ofA is a weakly 2-AI of A.Proof.
It is clear.The following example show that the converse of Lemma25 is not true.Example 5.
LetD=0,x,y be a discrete ADL with 0 as its zero element defined in Example1andL=0,a,b,c,1 be the lattice represented by the Hasse diagram given below.ConsiderD×L=t,st∈Dands∈L. Then, D×L,∧,∨,0 is an ADL (which is not a lattice) under the pointwise operations ∧ and ∨ on D×L and 0=0,0, the zero element in D×L. Let P=x,0,y,c. Then, 0,0≠x,a∧y,b∧y,c∈P implies x,a∧y,b∈P,y,b∧y,c∈P and x,a∧y,c∈P, for all x,a,y,b,y,c∈D×L. Thus, P is a weakly 2-AI of D×L. But P is neither prime ideal nor weakly prime ideal of D×L, since x,a∧y,b∈P0,0≠x,a∧y,b∈P⇒x,a∉P and y,b∉P.Theorem 26.
Every2-AI of A is a weakly 2-AI of A.Proof.
It is clear.The following example show that the converse of Theorem26 is not true.Example 6.
LetD=0,x,y be a discrete ADL with 0 as its zero element defined in Example1 and L=0,a,b,c,d,e,f,1 be a lattice whose Hasse diagram is given below.ConsiderD×L=t,st∈Dands∈L. Then, D×L,∧,∨,0 is an ADL (which is not a lattice) under the pointwise operations ∧ and ∨ on D×L and 0=0,0, the zero element in D×L. Let P=0,0. Clearly P is a weakly 2-AI of D×L. On the other hand, consider 0,d∧x,e∧y,f=0,0∈P which implies that 0,d∧x,e=0,a∉P, x,e∧y,f=y,c∉P and 0,d∧y,f=0,b∉P, for all 0,d,x,e,y,f∈D×L. Thus P is not a 2-AI of D×L. Therefore, every weakly 2-AI of D×L is not a 2-AI of D×L.As a consequence of Theorem18 and Lemmas 23 and 25, we have the following.Theorem 27.
LetP and Q be weakly prime ideals of A. Then, the intersection of P and Q is also a weakly 2-AI of A.Theorem 28.
LetP be a proper ideal of A and P≠0 in A. Then, P is a 2-AI of A if and only if P is a weakly 2-AI of A.Proof.
It is clear.As a consequence of Theorems20 and 26, we have the following.Theorem 29.
Letf:A1⟶A2 be a lattice homomorphism. Then, the following holds.
(1)
Iff is an epimorphism and Q is a weakly 2−AI of A2, then f−1Q is a weakly 2-AI of A1(2)
Iff is an isomorphism and P is a weakly 2−AI of A1, then fP is a weakly 2-AI of A2Theorem 30.
LetA=A1×A2 be an ADL, where A1 and A2 are ADLs. Let P and Q be proper ideal of A1 and A2, respectively. If P×Q is a weakly 2-AI of A, then P and Q are weakly 2-AI of A1 and A2, respectively.Proof.
Suppose thatP×Q is a weakly 2-AI of A. Let a,b,c∈A1 and x,y,z∈A2 such that 0≠x∧y∧z∈Q. Then, 0≠a,x∧y∧z∈P×Q implies that either a,x∧y∈P×Q or a,x∧z∈P×Q or a,y∧z∈P×Q. From this, either x∧y∈Q or x∧z∈Q or y∧z∈Q. Thus, Q is a weakly 2-AI of A2. Similarly, P is a weakly 2-AI of A1.The converse of the above theorem is not true, consider the following example.Example 7.
LetA1=0,a,b,c,1 be the lattice discussed in Example 5 and A2=0,x,y,1 be a chain represented by the diagram given below.ConsiderA1×A2=u,vu∈A1andv∈A2. Let P=c and Q=0 be ideals of A1 and A2, respectively. Then, P×Q=c,0. We note that, for all 1,0,a,x,b,x∈A1×A2, 1,0∧a,x∧b,x=1∧a∧b,0∧x=c,0∈P×Q. Now, 1,0∧a,x=a,0∉P×Q, a,x∧b,x=a∧b,x=c,x∉P×Q and 1,0∧b,x=b,0∉P×Q. It follows that P×Q is not a weakly 2-AI of A1×A2.Theorem 31.
LetA=A1×A2 be ADL, where A1 and A2 be ADLs and P≠0 be a proper ideal of A1. Then, the following are equivalent.
(1)
P×A2 is a weakly 2-AI of A(2)
P×A2 is a 2-AI of A(3)
P is a 2-AI of A1Proof.
1⇒2: assume (1). Let a,b,c∈A1 such that a,x∧b,x∧c,x∈P×A2, for every x∈A2. Then, a,x∧b,x∧c,x=a∧b∧c,x∈P×A2, implies either a∧b,x∈P×A2 or a∧c,x∈P×A2 or b∧c,x∈P×A2, for every x∈A2 (since P×A2 is a weakly 2-AI of A). Thus, P×A2 is a 2-AI of A.
2⇒3: assume (2). Let a,b,c∈A1 such that a∧b∧c∈P. Since P×A2 is a 2-AI of A, a∧b∧c,x∈P×A2, for every x∈A2, which implies that either a∧b,x∈P×A2 or a∧c,x∈P×A2 or b∧c,x∈P×A2. From this, we have that either a∧b∈P or a∧c∈P or b∧c∈P. Therefore, P is a 2-AI of A1.
3⇒1. Suppose P is a 2-AI of A1 and 0≠a,x∧b,x∧c,x∈P×A2, for every x∈A2 and a,b,c∈A1. Then, a,x∧b,x∧c,x=a∧b∧c,x∈P×A2, implies either a∧b,x∈P×A2 or a∧c,x∈P×A2 or b∧c,x∈P×A2, for every x∈A2 (since either a∧b∈P or a∧c∈P or b∧c∈P). Thus, P×A2 is a weakly 2-AI of A.Theorem 32.
LetA=A1×A2 be ADL, where A1 and A2 be ADLs. Let P≠0 and Q≠0 be proper ideal of A1 and A2, respectively. Then, the following are equivalent.
(1)
P×Q is a weakly 2-AI of A(2)
Q=A2 and P is a 2-AI of A1 or Q is a prime ideal of A2 and P is a prime ideal of A1(3)
P×Q is a 2-AI of AProof.
1⇒2: assume (1). If Q=A2, then P is a 2-AI of A1 (by the above theorem). Suppose that Q≠A2. Let x,y∈A2 such that x∧y∈Q and let 1≠t∈P. Then, t,1∧1,x∧1,y=t,x∧y∈P×Q−0,0. Since 1,x∧1,y=1,x∧y∉P×Q, we conclude that either t,1∧1,x=t,x∈P×Q or t,1∧1,y=t,y∈P×Q and hence either x∈Q or y∈Q. Thus, Q is a prime ideal of A2. Similarly, P is a prime ideal of A1.
2⇒3: assume (2). Then, by the above theorem, P×A2 is a 2-AI of A. Suppose that P is a prime ideal of A1 and Q is a prime ideal of A2. Then, clearly P×Q is a prime ideal of A. Let x,y,z,t,a,b∈A such that x,y∧z,t∧a,b∈P×Q. Then, either x,y∧z,t∈P×Q or a,b∈P×Q, or x,y∧a,b∈P×Q or z,t∈P×Q, or x,y∈P×Q or z,t∧a,b∈P×Q. Thus, P×Q is a 2-AI of A.
(3)⇒1 is clear by Theorem 26In the following, we introduce the concept ofn-AI of an ADL A.Definition 33.
LetP be a proper ideal of A and n∈Z+. Then, P is an n-AI of A if whenever x1∧x2∧⋯∧xn+1∈P, for xi∈A, 1≤i≤n+1, then there are n of the xi′s whose meet is in P.Corollary 34.
LetP be a proper ideal of A and n,m∈Z+. Then,
(1)
P is n-AI if and only if whenever x1∧x2∧⋯∧xm∈P, for x1,⋯,xm∈A with m>n, then there are n of the xi′s whose meet is in P(2)
IfP is n-AI, then P is an m-AI, for all m≥nCorollary 35.
Letg:A⟶B be a lattice homomorphism. Then, the following holds.
(1)
Ifg is an epimorphism and Q is an n-AI of B, then g−1Q is an n-AI of A(2)
Ifg is an isomorphism and P is an n-AI of A, then gP is an n-AI of BTheorem 36.
IfPαα∈Δ is a nonempty chain of n-AI of A, then ∧α∈ΔPα is an n-AI of A.Proof.
LetP=∧α∈ΔPα and x1,x2,⋯,xn+1∈A such that ∧i=1n+1xi∈P. Let xi=∧j≠ixj and xi∉P, for all 1≤i≤n. Then, for each 1≤i≤n, there exist an n-AI Pαi such that xi∉Pαi. Assume that Pα1⊆Pα2⊆⋯⊆Pαn. Let β∈Δ. If Pβ⊆Pα1⊆⋯⊆Pαn, then xi∉Pβ, for each 1≤i≤n. Since x1∧x2∧⋯∧xn+1∈P and Pβ is n-AI of A, we have xn+1∈Pβ. Again, x1∧x2∧⋯∧xn+1∈Pα1 and Pα1 is n-AI of A, then xn+1∈Pα1. So, xn+1∈Pβ, for every β∈Δ. Thus, xn+1∈P. Hence the theorem.


## 4. 1-Absorbing Prime Ideals

In this section, we introduce the 1-absorbing prime ideals of ADLs.Definition 37.
LetA be an ADL. A proper ideal P of A is a 1-absorbing prime ideal if for x,y,z∈A such that x∧y∧z∈P, then either x∧y∈P or z∈P.Theorem 38.
LetP be a proper ideal of A. Then, every prime ideal of A is a 1-absorbing prime ideal and every 1-absorbing prime ideal of A is a 2-absorbing ideal of A.The following example show that every 2-AI ofA is not 1-absorbing prime ideal of A.Example 8.
LetD×L=t,st∈Dands∈L be an ADL discussed in Example 6. Let Q=0,0,x,b,y,c. Then, for all 0,d,x,e,y,f∈D×L,
(12)0,d∧x,e∧y,f∈Q⇒x,e∧y,f=y,c∈Q.Thus,Q is a 2-AI of D×L. On the other hand, 0,d∧x,e∧y,f=0,0∈Q implies 0,d∧x,e=0,a∉Q and y,f∉Q. From this, Q is not a 1-absorbing prime ideal of D×L. Therefore, every 2-AI of D×L is not 1-absorbing prime ideal of D×L.Next, we have the following result.Theorem 39.
LetA=A1×A2 be an ADL where A1 and A2 be ADLs with 0, proper ideal P of A and Q a proper ideal of A2. Then, P is a 1-absorbing prime ideal of A if and only if P=R×A2 or P=A1×Q, where R and Q are prime ideals of A1 and A2, respectively.Lemma 40.
LetP be a 1-absorbing prime ideal of A. If x∧y∩Q⊆P, for all proper ideal Q of A and for x,y∈A, then x∧y⊆P or Q⊆P.Theorem 41.
LetP be a proper ideal of A. Then, the following are equivalent.
(1)
P is a 1-absorbing prime ideal of A(2)
IfP1∩P2∩P3⊆P for some proper ideals P1,P2, and P3 of A, then either P1∩P2⊆P or P3⊆PProof.
1⇒2: suppose P is a 1-absorbing prime ideal of A. Let P1∩P2∩P3⊆P for some proper ideals P1,P2, and P3 of A. Let P1∩P2ÚP. Then, there exists x∈P1 and y∈P2 such that x∧y∉P and hence x∧yÚP. Since x∩y∩P3=x∧y∩P3⊆P, it follows that P3⊆P (by the above lemma).
2⇒1: assume (2) hold. Suppose that x∧y∧z∈P, for x,y,z∈A and let x∧y∉P. Suppose also that P1=x,P2=y and P3=z. Then P1∩P2∩P3=x∩y∩z=x∧y∩z⊆P and P1∩P2ÚP (by assumption, x∧y∉P). Thus by the above lemma, z⊆P, that is, P3⊆P and thus z∈P. Therefore, P is a 1-absorbing prime ideal of A.Theorem 42.
LetA1 and A2 be ADLs and f:A1⟶A2 be a lattice homomorphism. Then, the following hold.
(1)
IfP is a 1-absorbing prime ideal of A2, then f−1P is a 1-absorbing prime ideal of A1(2)
Iff is onto and P is a 1-absorbing prime ideal of A1 with kerf⊆P, then fP is a 1-absorbing prime ideal of A2Proof.
(1)
LetP be a 1-absorbing prime ideal of A2 and x∧y∧z∈f−1P for some x,y,z∈A1. Then fx∧y∧z=fx∧fy∧fz∈P which implies that fx∧fy∈P or fz∈P. It follows that x∧y∈f−1P or z∈f−1P. Hence f−1P is a 1-absorbing prime ideal of A1(2)
LetP be a 1-absorbing prime ideal of A1 with kerf⊆P, f be onto and x∧y∧z∈fP for some x,y,z∈A2. Since f is onto, then there exists a,b,c∈A2 such that a=fx,b=fy and c=fz. Therefore, fx∧y∧z=fx∧fy∧fz=a∧b∧c∈fP. Since kerf⊆P, we conclude that x∧y∧z∈P. Thus, x∧y∈P or z∈P and so, a∧b∈fP or c∈fP. Therefore, fP is a 1-absorbing prime ideal of A2Definition 43.
LetP be a proper ideal of A. Then, P is said to be a weakly 1-absorbing prime ideal of A if 0≠x∧y∧z∈P for some x,y,z∈A, then either x∧y∈P or z∈P.Theorem 44.
Every weakly prime ideal ofA is a weakly 1-absorbing prime ideal of A.Proof.
It is clear.The following example show that the converse of Theorem44 is not true.Example 9.
LetD×L=t,st∈Dands∈L be an ADL discussed in Example 5. Let P=0,c,x,0,y,c. Let 0,a,x,b,y,c∈D×L. Then,
(13)0,0≠0,a∧x,b∧y,c∈P⇒0,a∧x,b∈Pandy,c∈P.Thus,P is a weakly 1-absorbing prime ideal of D×L, but P is not weakly prime ideal of D×L, since 0,0≠0,a∧x,b∈P implies that 0,a∉P and x,b∉P. From this, we conclude that, every weakly 1-absorbing prime ideal of D×L is not weakly prime ideal of D×L.Theorem 45.
Every weakly 1-absorbing prime ideal ofA is a weakly 2-absorbing ideal of A.Proof.
It is clear.The following example show that every weakly 1-absorbing prime ideal ofA is not 1-absorbing prime ideal of A.Example 10.
LetD×L=t,st∈Dands∈L be an ADL discussed in Example 6. Let Q=0,0. Clearly Q is a weakly 1-absorbing prime ideal of D×L. But Q is not 1-absorbing prime ideal of D×L, since 0,d∧x,e∧y,f∈P implies 0,d∧x,e=0,a∉P and y,f∉P, for all 0,d,x,e,y,f∈D×L. Thus, every weakly 1-absorbing prime ideal of D×L is not 1-absorbing prime ideal of D×L.Theorem 46.
LetA and B be ADLs and f:A⟶B be a lattice homomorphism. Then, the following holds.
(1)
Iff is a monomorphism and P is a weakly 1-absorbing prime ideal of B, then f−1P is a weakly 1-absorbing prime ideal of A(2)
Iff is an epimorphism and Q is a weakly 1-absorbing prime ideal of A such that kerf⊆Q, then fQ is a weakly 1-absorbing prime ideal of BProof.
(1)
Let0≠x∧y∧z∈f−1P for some x,y,z∈A. Then by assumption, fx∧y∧z=fx∧fy∧fz∈P, for some fx,fy,fz∈B. Since f is monomorphism, we have fx∧y∧z≠0. Also, since P is a weakly 1-absorbing prime ideal of B, we conclude either fx∧fy=fx∧y∈P or fz∈P which implies that x∧y∈f−1P or z∈f−1P. Thus, f−1P is a weakly 1-absorbing prime ideal of A.(2)
Assume that0≠a∧b∧c∈fQ for some a,b,c∈B. Since f is an epimorphism, then there exists x,y,z∈A such that a=fx,b=fy and c=fz. Then, 0≠fx∧fy∧fz=fx∧y∧z∈fQ. Since kerf⊆Q, then we get that 0≠x∧y∧z∈Q. As Q is a weakly 1-absorbing prime ideal of A, we have either x∧y∈Q or z∈Q and which implies that fx∧y=a∧b∈fQ or fz=c∈fQ. Therefore, fQ is a weakly 1-absorbing prime ideal of B


## 5. Conclusion

In this paper, the concepts of 2-absorbing ideal, 1-absorbing prime ideal, weakly 1-absorbing prime ideal, and weakly 2-absorbing ideal of an almost distributive lattice are introduced and obtain certain results of these.



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