The Graph of Equivalence Classes of Zero Divisors

Abstract

We introduce a graph 𝐺_𝐸(𝐿) of equivalence classes of zero divisors of a meet semilattice 𝐿 with 0. The set of vertices of 𝐺_𝐸(𝐿) are the equivalence classes of nonzero zero divisors of 𝐿 and two vertices [𝑥] and [𝑦] are adjacent if and only if [𝑥]∧[𝑦]=[0]. It is proved that 𝐺_𝐸(𝐿) is connected and either it contains a cycle of length 3 or 𝐺_𝐸(𝐿)≅𝐾₂. It is known that two Boolean lattices 𝐿₁ and 𝐿₂ have isomorphic zero divisor graphs if and only if 𝐿₁≅𝐿₂. This result is extended to the class of SSC meet semilattices. Finally, we show that Beck's Conjecture is true for 𝐺_𝐸(𝐿) .

1. Introduction

The idea of a zero divisor graph was introduced by Beck in [1] to investigate the interplay between ring theoretic properties and graph theoretic properties. The concept of zero divisor graph is also well studied in ordered structures; see Alizadeh et al. [2, 3], Halaš and Jukl [4], Halaš and Länger [5], Joshi [6], Joshi et al. [7–11], and Lu and Wu [12].

The graph of equivalence classes of zero divisors of commutative rings is well studied in Allen et al. [13], Anderson and LaGrange [14], and Spiroff and Wickham [15].

In this paper, we extend this concept to a meet semilattice 𝐿 with 0. The main aspect of the graph 𝐺_𝐸(𝐿) of equivalence classes of zero divisors of 𝐿 is the connection to the associated primes of 𝐿. In general, every vertex of 𝐺_𝐸(𝐿) either corresponds to an associated prime or is adjacent to an associated prime of 𝐿. Further, we consider the equivalence relation ∼ on a meet semilattice 𝐿 with 0, 𝑥∼𝑦 if and only if ann(𝑥)=ann(𝑦) for 𝑥,𝑦∈𝐿 and construct the meet semilattice 𝐿^′={[𝑥]∣𝑥∈𝑍_{0}(𝐿)^∗}. It is proved that 𝐿^′ is an SSC meet semilattice with the property that 𝐺_𝐸(𝐿)≅𝐺_{0}(𝐿^′), the zero divisor graph of 𝐿^′. Hence it is clear that the study of 𝐺_𝐸(𝐿) is nothing, but the study of zero divisor graph of SSC meet semilattices. This observation helps us to prove that for bounded SSC meet semilattices 𝐿₁ and 𝐿₂, 𝐺_{0}(𝐿₁)≅𝐺_{0}(𝐿₂) if and only if 𝐿₁≅𝐿₂. This extends the result of LaGrange [16] for Boolean algebra. It is proved that for an SSC meet semilattice 𝐿, 𝐺_{0}(𝐿) is connected and either it contains a cycle of length 3 or 𝐺_{0}(𝐿)≅𝐾₂. In the class of a finite semicomplemented meet semilattice 𝐿, 𝐺_𝐸(𝐿) is isomorphic to the zero divisor graph 𝐺_{0}(𝐿) if and only if 𝐿 is SSC.

2. Properties of the Graph 𝐺_𝐸(𝐿)

We begin with the necessary definitions and terminology.

A nonempty subset 𝐼 of a meet semilattice 𝐿 is said to be semi-ideal, if 𝑦⩽𝑥∈𝐼 implies that 𝑦∈𝐼. A semi-ideal 𝐼 of a meet semi-lattice 𝐿 is said to be an ideal, if 𝑎,𝑏∈𝐼 implies that 𝑎∨𝑏∈𝐼. Dually we have the concept of a filter. A proper semi-ideal (ideal) is said to be prime, if 𝑎∧𝑏∈𝐼 implies either 𝑎∈𝐼 or 𝑏∈𝐼.

A prime semi-ideal (ideal) 𝑃 of a meet semilattice 𝐿 is said to be minimal prime semi-ideal (ideal), if there is no prime ideal 𝑄 such that 𝑄⫋𝑃.

The concept of SSC lattices is introduced by Janowitz [17]. We write this definition for a meet semilattice with 0.

A meet semilattice 𝐿 with 0 is said to be semicomplemented if, for 𝑎≠1, there exists 𝑏≠0 such that 𝑎∧𝑏=0 and 𝐿 is said to be section semicomplemented  (in brief SSC) if for 𝑎,𝑏∈𝐿 with 𝑎<𝑏, there exists nonzero 𝑐 such that 0<𝑐≤𝑏 and 𝑎∧𝑐=0. It is clear that every SSC lattice is a semicomplemented lattice, but not conversely. More details about SSC poset can be found in Thakare et al. [18, 19], Joshi [20].

Let 𝐿 be a meet semilattice with 0. The set of all zero divisors of 𝐿 is denoted by 𝑍_{0}(𝐿)={𝑥∈𝐿∣𝑥∧𝑦=0forsome𝑦∈𝐿⧵{0}} and the set of all nonzero zero divisors is denoted by 𝑍_{0}(𝐿)^∗. Clearly, 𝑍_{0}(𝐿)^∗∪{0}=𝑍_{0}(𝐿).

Let 𝐺 be a graph, and let 𝑥,𝑦 be distinct vertices in 𝐺. We denote by 𝑑(𝑥,𝑦) the length of a shortest path from 𝑥 to 𝑦 if it exists and put 𝑑(𝑥,𝑦)=∞ if no such path exists. The diameter of 𝐺, denoted by diam(𝐺), is zero if 𝐺 is the graph on one vertex and is diam(𝐺)=sup{𝑑(𝑥,𝑦)where𝑥and𝑦aredistinctverticesof𝐺} otherwise. A cycle in a graph 𝐺 is a path that begins and ends at the same vertex. The girth of 𝐺, written girth(𝐺), is the length of the shortest cycle in 𝐺 (and girth(𝐺)=∞ if 𝐺 has no cycles). The chromatic number of 𝐺 is denoted by 𝜒(𝐺). Thus, 𝜒(𝐺) is the minimum number of colors which can be assigned to the elements of 𝐿 such that adjacent elements receive different colors. If this number is not finite, write 𝜒(𝐺)=∞. A subset 𝐶 of 𝐺 is a clique if any two distinct vertices of 𝐶 are adjacent. If 𝐺 contains a clique with 𝑛 elements and every clique has at most 𝑛 elements then the clique number of 𝐺 is Clique(𝐺)=𝑛. If the sizes of the cliques are not bounded, then Clique(𝐺)=∞. We always have 𝜒(𝐺)⩾Clique(𝐺).

Joshi [6] introduced the concept of the zero divisor graph of a poset 𝑃 having the smallest element 0 with respect to an ideal 𝐼 of 𝑃. We mention below this definition when the corresponding poset is a meet semilattice and 𝐼={0}. Note that this definition coincides with the definition given by Lu and Wu [12] when 𝐼={0}.

Definition 1 (Joshi [6], Lu and Wu [12]). Let 𝐿 be a semilattice with 0. We associate a simple undirected graph with 𝐿, called the zero divisor graph of 𝐿 with respect to {0}, denoted by 𝐺_{0}(𝐿), with the set of vertices is 𝑉(𝐺_{0}(𝐿)) = 𝑍_{0}(𝐿)^∗={𝑥∈𝐿⧵{0}∣𝑥∧𝑦=0forsome𝑦∈𝐿⧵{0}} and two distinct vertices 𝑥,𝑦 are adjacent if and only if 𝑥∧𝑦=0.
For any 𝑥,𝑦∈𝐿, we say that 𝑥∼𝑦 if and only if ann(𝑥)=ann(𝑦), where ann(𝑎)={𝑏∣𝑎∧𝑏=0}. Note that ∼ is an equivalence relation on 𝐿. Furthermore, if 𝑥₁∼𝑥₂ and 𝑥₁∧𝑦=0, then 𝑦∈ann(𝑥₁)=ann(𝑥₂) and hence 𝑥₂∧𝑦=0. We define [𝑥], the equivalence class of 𝑥, as follows: [𝑥]={𝑧∈𝐿∣ann(𝑧)=ann(𝑥)}.

Lemma 2. Let 𝐿 be a meet semilattice with 0 and 𝐿^′={[𝑥]∣𝑥∈𝐿}, where [𝑥]={𝑧∈𝐿∣𝑎𝑛𝑛(𝑧)=𝑎𝑛𝑛(𝑥)}. Then 𝐿^′ is a meet semilattice under the partial order [𝑥]⩽[𝑦] if and only if 𝑎𝑛𝑛(𝑦)⊆𝑎𝑛𝑛(𝑥).

Proof. Since ann(𝑥),ann(𝑦)⊆ann(𝑥∧𝑦), we have [𝑥∧𝑦]⩽[𝑥],[𝑦]. We claim that [𝑥]∧[𝑦]=[𝑥∧𝑦]. Let [𝑡] be another lower bound of [𝑥] and [𝑦]. Then ann(𝑥),ann(𝑦)⊆ann(𝑡). Now, we claim that ann(𝑥∧𝑦)⊆ann(𝑡). Let 𝑧∈ann(𝑥∧𝑦). Then 𝑧∧𝑥∈ann(𝑦)⊆ann(𝑡). This gives 𝑥∧𝑧∧𝑡=0; that is, 𝑧∧𝑡∈ann(𝑥)⊆ann(𝑡). Hence 𝑧∧𝑡=0; that is, 𝑧∈ann(𝑡). This proves that ann(𝑥∧𝑦)⊆ann(𝑡). Thus [𝑡]⩽[𝑥∧𝑦] and this yields [𝑥]∧[𝑦]=[𝑥∧𝑦]. Clearly, if 𝐿 is a finite lattice, then 𝐿^′ is also a lattice.

Lemma 3. 𝐿^′ is an 𝑆𝑆𝐶 meet semilattice.

Proof. Let 0<[𝑥]≨[𝑦]. Then ann(𝑦)⫋ann(𝑥). Hence there exists 𝑧∈ann(𝑥) such that 𝑧∉ann(𝑦). Thus 0≠𝑧∧𝑦=𝑡. But then [0]<[𝑡]⩽[𝑦] and [𝑡]∧[𝑥]=[0]. Thus 𝐿^′ is an SSC meet semilattice.

Now, we introduce the graph 𝐺_𝐸(𝐿) of equivalence classes of zero divisors of a meet semilattice 𝐿 with 0, denoted by 𝐺_𝐸(𝐿).

Definition 4. Let 𝐿 be a meet semilattice with 0. We associate a simple undirected graph with 𝐿 whose vertices are the equivalence classes of elements in 𝑍_{0}(𝐿)^∗ and with each pair of distinct classes [𝑥] and [𝑦] joined by an edge if and only if [𝑥]∧[𝑦]={0}=[0].

We illustrate the concept of the graph of equivalence classes of zero divisor with an example, see Figure 1.

Figure 1 

The graph 𝐺𝐸(𝐿) of equivalence classes of zero divisors.

Consider the following two infinite lattices 𝐿 and 𝐿₁.

We can see that the zero divisor graph of lattices 𝐿, 𝐺_{0}(𝐿) is infinite, but the graph of equivalence classes of zero divisors of the lattice 𝐿, namely, 𝐺_𝐸(𝐿), is finite. Also for the lattices 𝐿₁, 𝐺_{0}(𝐿₁), and 𝐺_𝐸(𝐿₁) both are infinite, see Figure 2.

Figure 2 

The graph 𝐺𝐸(𝐿) is finite, whereas 𝐺𝐸(𝐿1) is infinite.

Corollary 5. Let 𝐿 be a meet semilattice with 0. Then there exists an 𝑆𝑆𝐶 meet semilattice 𝐿^′ (as constructed in Lemma 2) such that 𝐺_𝐸(𝐿)=𝐺_{0}(𝐿^′).

Proof. From the construction of 𝐿^′, it is clear that 𝐺_𝐸(𝐿)=𝐺_{0}(𝐿^′).

Hence from Corollary 5, it is clear that the study of the graph of equivalence classes of zero divisors of 𝐿 is nothing, but the study of zero divisor graphs of SSC meet semilattices.

Remark 6. Recall that a prime semi-ideal 𝑃 of a meet semilattice 𝐿 is an associated prime if it is the annihilator of some elements of 𝐿; that is, 𝑃=ann(𝑥) for some nonzero 𝑥∈𝐿. The set of associated primes is denoted by Ass(𝐿). If Clique(𝐺_{0}(𝐿))<∞, then Ass(𝐿) is nonempty and finite. Further, any maximal element of the family of annihilator semi-ideals is prime; see Halaš and Jukl [4, Lemma 2.4, 2.6]. Note that every zero divisor is contained in an annihilator ideal and maximal annihilators are associated primes; we have 𝑍_{0}(𝐿)⊆⋃_{𝑃∈Ass(𝐿)}𝑃. Clearly, the reverse inclusion is trivial. Hence 𝑍_{0}(𝐿)=⋃_{𝑃∈Ass(𝐿)}𝑃; that is, the set of zero-divisors of 𝐿 equals the union of all associated primes of 𝐿. It is obvious to prove that a map 𝜙∶Ass(𝐿)→𝑉(𝐺_𝐸(𝐿)) defined by 𝜙(𝑃)=[𝑥], where 𝑃=ann(𝑥) is an injective map. We adopt the conventions of Spiroff and Wickham [15], and by a slight abuse of terminology we will refer to the vertex [𝑦] as an associated prime if ann(𝑦)∈Ass(𝐿). It will be clear from context whether [𝑦] refers to an equivalence class, a vertex, or a specific annihilator.

Theorem 7. Let 𝐿 be a meet semilattice with 0, and let 𝐶𝑙𝑖𝑞𝑢𝑒(𝐺_{0}(𝐿))<∞. If 𝑎𝑛𝑛(𝑥) and 𝑎𝑛𝑛(𝑦) are distinct prime semi-ideals of 𝐿, then [𝑥] and [𝑦] are adjacent in 𝐺_𝐸(𝐿). Furthermore, every vertex of 𝐺_𝐸(𝐿) is either an associated prime semi-ideal or adjacent to a maximal semi-ideal in ℒ={𝑎𝑛𝑛(𝑥)∣(0≠)𝑥∈𝐿}.

Proof . Since ann(𝑥) and ann(𝑦) are distinct, we have 𝑥∧𝑦=0. This further gives that [𝑥]∧[𝑦]=[𝑥∧𝑦]=[0]. Thus [𝑥] and [𝑦] are adjacent. Furthermore, suppose that [𝑥]∈𝑉(𝐺_𝐸(𝐿)) is not an associated prime semi-ideal. Then there exists [𝑧]∈𝑉(𝐺_𝐸(𝐿)) such that [𝑥]∧[𝑧]=[0]={0}. Hence 𝑥∈ann(𝑧). We claim that ann(𝑧) is maximal. Otherwise, since Clique(𝐺_{0}(𝐿))<∞, ℒ has a maximal element, say ann(𝑧^′)(𝑧^′≠0) containing ann(𝑧), then 𝑥∈ann(𝑧)⊆ann(𝑧^′). It is clear that every maximal element of ℒ is prime. Therefore 𝑥∧𝑧^′=0. Thus there is an edge between [𝑥] and [𝑧^′]. This completes the proof.

Remark 8. For a Boolean lattice, [𝑎]={𝑎} for every 𝑎∈𝑍_{0}(𝐿), and hence in this case, 𝐺_𝐸(𝐿) and 𝐺_{0}(𝐿) are isomorphic. This fact is illustrated in Figure 3.

Figure 3 

The isomorphism of 𝐺𝐸(𝐿) and 𝐺{0}(𝐿).

In view of Remark 8, we raise the following problem.

Question 1. Find a class of meet semilattices ℒ such that 𝐺_{0}(𝐿)≅𝐺_𝐸(𝐿) for 𝐿∈ℒ.

Before we answer this question, we need the following result which follows from the definition of SSC meet semilattice.

Lemma 9. Let 𝐿 be a meet semilattice with 0. Then 𝐿 is 𝑆𝑆𝐶 if and only if for 𝑎,𝑏∈𝐿, 𝑎𝑛𝑛(𝑎)=𝑎𝑛𝑛(𝑏) implies 𝑎=𝑏.

Proof. Suppose that 𝐿 is SSC. Further, assume that there exist 𝑎,𝑏∈𝐿 such that ann(𝑎)=ann(𝑏) and 𝑎≠𝑏. Since 𝐿 is SSC, there exist 𝑐∈𝐿 such that 0<𝑐⩽𝑎 and 𝑐∧𝑏=0. This gives 𝑐∈ann(𝑏)=ann(𝑎), a contradiction to the fact that 𝑐≠0. Hence 𝑎=𝑏.

Theorem 10. Let 𝐿 be a finite semicomplemented meet semilattice. Then 𝜙∶𝐺_{0}(𝐿)→𝐺_𝐸(𝐿) is a graph isomorphism if and only if 𝐿 is 𝑆𝑆𝐶.

Proof . Let 𝐿 be a SSC meet semilattice. Then 𝐺_𝐸(𝐿)≅𝐺_{0}(𝐿) follows from Lemma 9.
Conversely, assume that 𝐺_𝐸(𝐿)≅𝐺_{0}(𝐿). First, we prove that [𝑎]={𝑎}, ∀𝑎∈𝑍_{0}(𝐿)^∗. Let 𝑏∈[𝑎] for 𝑎≠𝑏∈𝑍_{0}(𝐿)^∗. Then we have [𝑎]=[𝑏] for 𝑎≠𝑏, a contradiction to |𝑉(𝐺_𝐸(𝐿))|=|𝑉(𝐺_{0}(𝐿))|. Now, let 𝑎<𝑏(𝑎≠0). Since 𝐿 is semicomplemented, 𝑎,𝑏∈𝑍_{0}(𝐿)^∗. This gives ann(𝑏)⫋ann(𝑎). Otherwise, if ann(𝑎)=ann(𝑏), then 𝑏∈[𝑎]={𝑎}, a contradiction. From the proof of Lemma 9, it is clear that 𝐿 is SSC.

If we consider that the 𝜙∶𝐺_{0}(𝐿)→𝐺_𝐸(𝐿) is a graph isomorphism defined by 𝜙(𝑎)=[𝑎], then we do not need even the finiteness of 𝐿.

Theorem 11. Let 𝐿 be a semicomplemented meet semilattice. Then 𝜙∶𝐺_{0}(𝐿)→𝐺_𝐸(𝐿) is a graph isomorphism defined by 𝜙(𝑎)=[𝑎] if and only if 𝐿 is 𝑆𝑆𝐶.

Proof. Let 𝐿 be a SSC meet semilattice. Then 𝐺_𝐸(𝐿)≅𝐺_{0}(𝐿) follows from Lemma 9.
Conversely, assume that 𝜙∶𝐺_{0}(𝐿)→𝐺_𝐸(𝐿) is a graph isomorphism defined by 𝜙(𝑎)=[𝑎]. This gives [𝑎]={𝑎}, for all 𝑎∈𝑍_{0}(𝐿)^∗. Let 𝑎<𝑏(𝑎≠0). Since 𝐿 is semicomplemented, 𝑎,𝑏∈𝑍_{0}(𝐿)^∗. This gives ann(𝑏)⫋ann(𝑎). From the proof of Lemma 9, it is clear that 𝐿 is SSC.

From Figure 4, it is clear that for the posets 𝑃₁,𝑃₂, 𝐺_{0}(𝑃₁)≅𝐺_{0}(𝑃₂)⇏𝑃₁≅𝑃₂. Hence, it is worth to study the following problem.

Figure 4 

Nonisomorphic lattices with isomorphic zero divisor graphs.

Problem 12. Find a class of posets 𝒫 such that 𝐺_{0}(𝑃₁)≅𝐺_{0}(𝑃₂) if and only if 𝑃₁≅𝑃₂ for the posets 𝑃₁,𝑃₂∈𝒫.

In [8], Joshi and Khiste answered this problem in the case of Boolean posets. More details about Boolean posets can be found in Waphare and Joshi [21].

Theorem 13 (Joshi and Khiste [8, Theorem 2.11]). Let 𝑃₁ and 𝑃₂ be Boolean posets. Then 𝐺_{0}(𝑃₁)≅𝐺_{0}(𝑃₂) if and only if 𝑃₁≅𝑃₂.

It is well known that there is a 1−1 correspondence between Boolean algebras and Boolean rings.

A consequence of the above theorem is the following result.

Corollary 14 (LaGrange [18,  Theorem 4.1], Mohammadian [22,Theorem 7]). Let 𝑃₁ and 𝑃₂ be Boolean algebras. Then 𝐺_{0}(𝑃₁)≅𝐺_{0}(𝑃₂) if and only if 𝑃₁≅𝑃₂.

Now, we are ready to extend this result to the class of SSC meet semilattices. Note that every Boolean lattice is SSC but not conversely. Consider the lattice 𝐿₁ depicted in Figure 2 is SSC but not a Boolean lattice.

Theorem 15. Let 𝐿₁ and 𝐿₂ be bounded 𝑆𝑆𝐶 meet semilattices. Then the following statements are equivalent: (1)𝐺_𝐸(𝐿₁)≅𝐺_𝐸(𝐿₂);(2)𝐺_{0}(𝐿₁)≅𝐺_{0}(𝐿₂);(3)𝐿₁≅𝐿₂.

Proof. The equivalence of statements (1) and (2) follows from Lemma 9.
(2)⇒(3)∶ Since 𝐿₁ and 𝐿₂ are semicomplemented, 𝑉(𝐺_{0}(𝐿_𝑖))=𝐿_𝑖⧵{0_𝐿𝑖,1_𝐿𝑖} for 𝑖=1,2. Let 𝜙∶𝑉(𝐺_{0}(𝐿₁))→𝑉(𝐺_{0}(𝐿₂)) be a graph isomorphism. Define a map 𝑓∶𝐿₁→𝐿₂ such that 𝑓(𝑎)=𝜙(𝑎), for all 𝑎∈𝑉(𝐺_{0}(𝐿₁)) along with 𝑓(0_𝐿1)=0_𝐿2 and 𝑓(1_𝐿1)=1_𝐿2, where 0_𝐿𝑖(1_𝐿𝑖) is the smallest (largest) element of 𝐿_𝑖. Since 𝜙 is bijective, 𝑓 is bijective. We claim that 𝑓 is a biorder preserving map.
Let 𝑎⩽𝑏. First, we show that 𝑓(𝑎)⩽𝑓(𝑏). If any of 𝑎,𝑏 is either 0_𝐿1 or 1_𝐿1, then we are through. Hence assume that 𝑎,𝑏∉{0_𝐿1,1_𝐿1}. Suppose that 𝑓(𝑎)⩽^/𝑓(𝑏); that is, 𝜙(𝑎)⩽^/𝜙(𝑏). Since 𝐿₂ is SSC, there exists 𝜙(𝑐)∈𝐿₂ such that 0_𝐿2<𝜙(𝑐)⩽𝜙(𝑎) and 𝜙(𝑐)∧𝜙(𝑏)=0_𝐿2. Since 𝑏≠0_𝐿1, we have 𝜙(𝑏)≠0_𝐿2. This gives that 𝜙(𝑐) and 𝜙(𝑏) are adjacent in 𝐺_{0}(𝐿₂). Since 𝜙 is a graph isomorphism, we have that 𝑐 and 𝑏 are adjacent which further imply that 𝑐∧𝑏=0_𝐿1. Hence, 𝑐∧𝑎=0_𝐿1 in 𝐿₁. Thus 𝑎 and 𝑐 are adjacent in 𝐺_{0}(𝐿₁). Hence, we have 𝜙(𝑐)∧𝜙(𝑎)=0_𝐿2. But 𝜙(𝑐)⩽𝜙(𝑎) gives 𝜙(𝑐)=0_𝐿2, a contradiction. Hence 𝑓(𝑎)⩽𝑓(𝑏) in 𝐿₂.
Conversely, assume that 𝑓(𝑎)⩽𝑓(𝑏) in 𝐿₂. We claim that 𝑎⩽𝑏. If any of 𝑓(𝑎) or 𝑓(𝑏) is 0_𝐿2 or 1_𝐿2, then we are through. Let us assume that 𝑓(𝑎),𝑓(𝑏)∈𝐿₂⧵{0_𝐿2,1_𝐿2}. This yields 𝑎,𝑏∉{0_𝐿1,1_𝐿1}. Now, suppose on the contrary that 𝑎⩽^/𝑏 for 𝑎,𝑏∉{0_𝐿1,1_𝐿1}. Since 𝐿₁ is SSC, there exists 𝑐∈𝐿₁ such that 0_𝐿1<𝑐⩽𝑎 and 𝑐∧𝑏=0_𝐿1. Thus 𝑐 and 𝑏 are adjacent in 𝐺_{0}(𝐿₁). Then 𝜙(𝑐) and 𝜙(𝑏) are adjacent in 𝐺_{0}(𝐿₂). Then 𝜙(𝑐)∧𝜙(𝑏)=0_𝐿2 in 𝐿₂. Replacing the role of 𝑎,𝑏 in the above proof by 𝑐,𝑎, respectively, we have 𝜙(𝑐)⩽𝜙(𝑎). This implies that 𝜙(𝑐)⩽𝜙(𝑎)⩽𝜙(𝑏). But this gives 𝜙(𝑐)=𝜙(𝑐)∧𝜙(𝑏)=0_𝐿2, a contradiction. Hence 𝑎⩽𝑏. Thus 𝐿₁≅𝐿₂.
(3)⇒(2)∶ Obvious.

Lemma 16. Let 𝐿₁ be a bounded 𝑆𝑆𝐶 meet semilattice, and 𝐿₂ be a bounded semicomplemented meet semilattice such that 𝐺_{0}(𝐿₁)≅𝐺_{0}(𝐿₂). Then 𝐿₂ is also 𝑆𝑆𝐶.

Proof. Since 𝐿₁ and 𝐿₂ are semicomplemented, 𝑉(𝐺_{0}(𝐿_𝑖))=𝐿_𝑖⧵{0_𝐿𝑖,1_𝐿𝑖} for 𝑖=1,2. Let 𝜙∶𝑉(𝐺_{0}(𝐿₁))→𝑉(𝐺_{0}(𝐿₂)) be a graph isomorphism. Define a map 𝑓∶𝐿₁→𝐿₂ such that 𝑓(𝑎)=𝜙(𝑎), for all 𝑎∈𝑉(𝐺_{0}(𝐿₁)) along with 𝑓(0_𝐿1)=0_𝐿2 and 𝑓(1_𝐿1)=1_𝐿2, where 0_𝐿𝑖(1_𝐿𝑖) is the smallest (largest) element of 𝐿_𝑖. Since 𝜙 is a bijective map, 𝑓 is bijective. Let 𝑓(𝑎)<𝑓(𝑏) in 𝐿₂. Since 𝐿₂ is semicomplemented, and if 𝑓(𝑎)=0_𝐿2 or 𝑓(𝑏)=1_𝐿2, then we are through. Hence assume that 𝑓(𝑎)≠0_𝐿2 and 𝑓(𝑏)≠1_𝐿2. Then clearly, 𝑎≠𝑏 and 𝑏≠1_𝐿1. Since 𝐿₁ is SSC and 𝑎∧𝑏<𝑏, there exists 0_𝐿1<𝑐⩽𝑏 and 𝑐∧𝑎=0_𝐿1. Since 𝑐 and 𝑎 are adjacent in 𝐺_{0}(𝐿₁), we have 𝜙(𝑐) and 𝜙(𝑎) are adjacent in 𝐺_{0}(𝐿₂); that is, 𝜙(𝑐)∧𝜙(𝑎)=0_𝐿2. Further 𝑏 and 𝑐 are nonadjacent in 𝐺_{0}(𝐿₁). Then 𝜙(𝑏) and 𝜙(𝑐) are nonadjacent in 𝐺_{0}(𝐿₂). Consider 0_𝐿2≠𝜙(𝑏)∧𝜙(𝑐)=𝑓(𝑏)∧𝑓(𝑐)=𝑓(𝑑) say. Clearly, 0_𝐿2<𝑓(𝑑)⩽𝑓(𝑏) and 𝑓(𝑑)∧𝑓(𝑎)=0_𝐿2 (as 𝜙(𝑐)∧𝜙(𝑎)=0_𝐿2 implies 𝑓(𝑐)∧𝑓(𝑎)=0_𝐿2). Thus 𝐿₂ is SSC.

Theorem 17. Let 𝐿₁ and 𝐿₂ be bounded semicomplemented meet semilattices. If 𝐺_𝐸(𝐿₁)≅𝐺_{0}(𝐿₂), then 𝐿₂ is 𝑆𝑆𝐶.

Proof. For the given lattice 𝐿₁, we can construct an SSC meet semilattice 𝐿^′₁ (as given in Lemma 2), such that 𝐺_𝐸(𝐿₁)≅𝐺_{0}(𝐿^′₁). This gives 𝐺_{0}(𝐿^′₁)≅𝐺_{0}(𝐿₂). By Lemma 16, 𝐿₂ is SSC.

The above result is analogues to the following result of Anderson and LaGrange [14].

Theorem 18 (Anderson and LaGrange [14, Theorem 2.6]). Let 𝑅 and 𝑆 be commutative reduced rings with 1≠0 and 𝑍(𝑆)≠{0}. If Γ_𝐸(𝑅)≅Γ(𝑆), then 𝑆 is a Boolean ring.

We quote the result of Joshi [6, Theorem 2.4] when the corresponding poset is a meet semilattice.

Theorem 19 (Joshi [6, Theorem 2.4]). Let 𝐿 be a meet semilattice with 0. Then the graph 𝐺_{0}(𝐿) is connected and 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))⩽3.

Theorem 20. Let 𝐿 be a meet semilattice with 0. Then the graph 𝐺_𝐸(𝐿) is connected and 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))⩽𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))⩽3.

Proof. Follows from Corollary 5, Theorem 19, and the fact that adjacency in 𝐺_𝐸(𝐿) implies adjacency in 𝐺_{0}(𝐿). Hence diam(𝐺_𝐸(𝐿))⩽diam(𝐺_{0}(𝐿))⩽3.

Theorem 21. Let 𝐿 be an 𝑆𝑆𝐶 lattice. Then 𝐺_{0}(𝐿) is either 𝐾₂ or contains a cycle of length 3. Hence 𝑔𝑖𝑟𝑡ℎ(𝐺_{0}(𝐿))=3.

Proof. If 𝑉(𝐺_{0}(𝐿))=2, then 𝐺_{0}(𝐿)≅𝐾₂, by Theorem 19. Now assume that |𝑉(𝐺_{0}(𝐿))|⩾3. Choose 𝑎,𝑏,𝑐∈𝑉(𝐺_{0}(𝐿)) such that 𝑎−𝑏−𝑐 is a path with 𝑎≠𝑐 and 𝑎∧𝑐≠0. Since 𝐿 is SSC and without loss of generality assume that 𝑎∧𝑐<𝑎, there exists 𝑥 such that 0<𝑥⩽𝑎 and 𝑐∧𝑥=0. Clearly, 𝑥∈𝑉(𝐺_{0}(𝐿)). Since 𝑎∧𝑏=0, we have 𝑥∧𝑏=0. Thus we have a cycle 𝑏−𝑥−𝑐−𝑏 of length 3.

An immediate consequence of the above theorem is the following corollary.

Corollary 22. Let 𝐿 be a meet semilattice with 0; then 𝐺_𝐸(𝐿) is either 𝐾₂ or contains a cycle of length 3. Hence, if 𝐺_𝐸(𝐿)≇𝐾₂, then 𝑔𝑖𝑟𝑡ℎ(𝐺_𝐸(𝐿))=3=𝑔𝑖𝑟𝑡ℎ(𝐺_{0}(𝐿)).

Proof . It follows from Corollary 5 and Theorem 21.

Remark 23. In the above result, the condition of existence of a cycle in 𝐺_𝐸(𝐿) is necessary. In Figure 1, we can observe that 𝐺_{0}(𝐿) has a cycle but 𝐺_𝐸(𝐿) has no cycle.

3. Correlation between Diameter of 𝐺_{0}(𝐿) and 𝐺_𝐸(𝐿)

The following result gives the correlation between diameter of 𝐺_{0}(𝐿) and 𝐺_𝐸(𝐿).

Theorem 24. The following statements are true for a meet semilattice 𝐿 with 0.(a)If 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))=1, then 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))=1.(b)If 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))=2, then 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))=1 or 2.(c)If 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))=3, if and only if 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))=3.(d)If 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))=1, then 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))=1 or 2.(e)If 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))=2, then 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))=2.

Proof. (𝑎) It is clear that diam(𝐺_𝐸(𝐿))≠0. Hence the result follows from Theorem 20.
(𝑏) Suppose that diam(𝐺_{0}(𝐿))=2. So we have a path 𝑎−𝑏−𝑐 in 𝐺_{0}(𝐿). We have two possibilities.
Case  1. If [𝑎],[𝑏],and[𝑐] are distinct vertices of 𝐺_𝐸(𝐿), we get the path [𝑎]−[𝑏]−[𝑐]. Hence by Theorem 20, diam(𝐺_𝐸(𝐿))=2.
Case  2. The only possibility in this case is [𝑎]=[𝑐], and hence, diam(𝐺_𝐸(𝐿))=1.
(𝑐) Let diam(𝐺_{0}(𝐿))=3. Then there exists a path 𝑎−𝑐−𝑑−𝑏. Since 𝑐∈ann(𝑎) gives [𝑎]≠[𝑏]; otherwise 𝑑(𝑎,𝑏)=2, a contradiction. Now assume that 𝑑([𝑎],[𝑏])=2 in 𝐺_𝐸(𝐿). Then there exists [𝑥]∈𝑉(𝐺_𝐸(𝐿)) such that [𝑎]−[𝑥]−[𝑏]. This yields that 𝑑(𝑎,𝑏)=2, again a contradiction. Thus by Theorem 20, diam(𝐺_𝐸(𝐿))=3. The converse follows from Theorem 20.
(𝑑) Suppose that diam(𝐺_𝐸(𝐿))=1. Then diam(𝐺_{0}(𝐿))≠3, by (𝑐). Hence the result.
(𝑒) Suppose that diam(𝐺_𝐸(𝐿))=2; then there exists a path [𝑎]−[𝑐]−[𝑏] in 𝐺_𝐸(𝐿). This yields a path 𝑎−𝑐−𝑏 in 𝐺_{0}(𝐿). Thus diam(𝐺_{0}(𝐿))≠1. The result follows from (𝑐) and diam(𝐺_{0}(𝐿))≠3.

Remark 25. The assertions (𝑎) and (𝑑) of Theorem 24 are justified in Figures 1 and 5, whereas the assertions (𝑏) and (𝑒) are justified in Figures 5 and 6. Figure 3 verifies the assertion (𝑐).

Figure 5 

diam(𝐺𝐸(𝐿))=diam(𝐺{0}(𝐿))=1.

Figure 6 

diam(𝐺𝐸(𝐿))=diam(𝐺{0}(𝐿))=2.

Corollary 26. Let 𝐿 be a meet semilattice with 0. Then 𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿))⩽2 if and only if 𝑑𝑖𝑎𝑚(𝐺_{0}(𝐿))⩽2.

4. Cut Vertices in 𝐺_𝐸(𝐿)

In this section, we examine the properties of cut vertices of 𝐺_𝐸(𝐿).

Definition 27. A lattice 𝐿 with 0 is said to be 0-distributive, if 𝑎∧𝑏=𝑎∧𝑐=0 implies that 𝑎∧(𝑏∨𝑐)=0; Varlet [23]. More details about 0-distributive posets can be found in Joshi and Waphare [24]; see also Joshi and Mundlik [25].

Theorem 28. Let 𝐿 be a 0-distributive lattice. If [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), then [𝑎]∪{0} forms an ideal of 𝐿.

Proof. Let [𝑎] be a cut vertex of 𝐺_𝐸(𝐿). Let 𝑎₁ and 𝑎₂ be nonzero elements of [𝑎]. Since 𝑎₁,𝑎₂∈[𝑎], ann(𝑎₁)=ann(𝑎)=ann(𝑎₂). Now, it is enough to show that ann(𝑎₁∨𝑎₂)=ann(𝑎). Clearly, ann(𝑎₁∨𝑎₂)⊆ann(𝑎).
Let 𝑡∈ann(𝑎). This gives 𝑡∧𝑎₁=𝑡∧𝑎₂=𝑡∧𝑎=0. Since 𝐿 is 0-distributive, we have 𝑡∧(𝑎₁∨𝑎₂)=0; that is, 𝑡∈ann(𝑎₁∨𝑎₂). Thus ann(𝑎)⊆ann(𝑎₁∨𝑎₂). This implies that 𝑎₁∨𝑎₂∈[𝑎].
Now suppose that 𝑥∈[𝑎]∪{0} and 𝑦⩽𝑥, to show that 𝑦∈[𝑎]∪{0}. If 𝑥=0 or 𝑦=0, then 𝑦∈[𝑎]∪{0}. Let 𝑦≠0 and 𝑦⩽𝑥. Then ann(𝑥)⊆ann(𝑦). Since [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), therefore for any two arbitrary vertices [𝑏],[𝑐]∈𝑉(𝐺_𝐸(𝐿)), we have a path [𝑏]−[𝑎]−[𝑐]. Therefore 𝑏∈ann(𝑎)=ann(𝑥); that is, 𝑏∧𝑥=0, which further yields 𝑏∧𝑦=0. Similarly, 𝑐∧𝑦=0. Therefore, we get a path [𝑏]−[𝑦]−[𝑐]. Since [𝑎] is cut vertex, [𝑎]=[𝑦]. Thus [𝑎]∪{0} is an ideal.

Remark 29. It is clear from the proof of the above theorem that in a general lattice, if [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), then [𝑎]∪{0} is a semi-ideal.

Lemma 30. Let 𝐿 be a meet semilattice with 0. If [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), then 𝑎𝑛𝑛(𝑎) is maximal in ℒ={𝑎𝑛𝑛(𝑥)∣0≠𝑥∈𝐿}. Hence 𝑎𝑛𝑛(𝑎) is a prime semi-ideal.

Proof. Let [𝑎] be a cut vertex of 𝐺_𝐸(𝐿), and let 𝐺₁ and 𝐺₂ be mutually separated subgraphs of 𝐺_𝐸(𝐿) with 𝑉(𝐺₁∪𝐺₂)=𝑉(𝐺_𝐸(𝐿))⧵[𝑎]. Let [𝑏]∈𝐺₁ and [𝑐]∈𝐺₂, therefore we have the path [𝑏]−[𝑎]−[𝑐]. Suppose that ann(𝑎)⊆ann(𝑥) for some ann(𝑥)∈ℒ. Since [𝑏]−[𝑎] is an edge, 𝑏∈ann(𝑎)⊆ann(𝑥). Then [𝑏]∧[𝑥]={0}. Similarly, [𝑐]∧[𝑥]={0}. Thus we have another path [𝑏]−[𝑥]−[𝑐] passing through of [𝑥]. Since [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), then [𝑥]=[𝑎]. Thus ann(𝑥)=ann(𝑎). It is easy to prove that ann(𝑎) is a prime semi-ideal.

Lemma 31. If [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), then all other associated primes of 𝐺_𝐸(𝐿) are contained in only one component of 𝐺_𝐸(𝐿)⧵[𝑎].

Proof. Suppose that 𝐺₁ and 𝐺₂ are two mutually separated connected components of 𝐺_𝐸(𝐿)⧵[𝑎], and each contains an associated prime. It is easy to observe that the associated primes are adjacent, and hence 𝐺₁, 𝐺₂ are connected, a contradiction.

Alizadeh et al. [2] raised the following problem.

Problem 32. Characterize those posets 𝑃 for which (diam(𝐺_{0}(𝑃)),girth(𝐺_{0}(𝑃)))=(2,3)or(3,3).

In view of Corollary 5 and the following result, it is clear that for an SSC meet semilattice 𝐿, (diam(𝐺_{0}(𝐿)),girth(𝐺_{0}(𝐿)))=(3,3).

Theorem 33. Let 𝐿 be a meet semilattice with 0. If |𝑉(𝐺_𝐸(𝐿))|⩾4 and 𝐺_𝐸(𝐿) have at least 2 cut vertices, then (𝑑𝑖𝑎𝑚(𝐺_𝐸(𝐿)),𝑔𝑖𝑟𝑡ℎ(𝐺_𝐸(𝐿))=(3,3).

Proof. Let [𝑎] and [𝑏] be cut vertices of 𝐺_𝐸(𝐿). Since [𝑎] is a cut vertex of 𝐺_𝐸(𝐿), there is some [𝑎₁] such that any path connecting [𝑎₁] and [𝑏] must include [𝑎]. Similarly, since [𝑏] is a cut vertex, there is some [𝑏₁] such that any path connecting [𝑏₁] and [𝑎] must include [𝑏]. Clearly, [𝑎₁]≠[𝑏₁] and any path from [𝑎₁] to [𝑏₁] must include [𝑎] and [𝑏] and so diam([𝑎],[𝑏])⩾3. By Theorem 20, diam(𝐺_𝐸(𝐿))=3. Thus (diam(𝐺_𝐸(𝐿)),girth(𝐺_𝐸(𝐿))=(3,3), by Corollary 22.

It is easy to see that Figure 3 fulfills the conditions of Theorem 33.

5. Beck’s Conjecture for 𝐺_𝐸(𝐿)

Joshi [6] (see also Halaš and Jukl [4], Nimbhorkar et al. [26], and Lu and Wu [12]) proved that Beck’s Conjecture is true for the zero divisor graphs of posets. We quote this result when the poset is a meet semilattice and ideal is a zero ideal.

Theorem 34 (Joshi [6, Theorem 2.9]). Let 𝐿 be a meet semilattice with 0. If 𝐶𝑙𝑖𝑞𝑢𝑒(𝐺_{0}(𝐿))<∞, then 𝐶𝑙𝑖𝑞𝑢𝑒(𝐺_{0}(𝐿))=𝜒(𝐺_{0}(𝐿))=𝑛, where 𝑛 is the number of minimal prime semi-ideals of 𝐿.
From Theorem 34 and Corollary 5, it is clear that Beck’s Conjecture is true for the graph 𝐺_𝐸(𝐿). In the sequel, we calculate the chromatic number of 𝐺_𝐸(𝐿). As a preparation, we need the following easy lemma.

Lemma 35. Let 𝑃 be a minimal prime semi-ideal of a meet semilattice 𝐿 with 0. Then for any 𝑥∈𝑃, there exists 𝑦∉𝑃 such that 𝑥∧𝑦=0.

Proof. It follows from the fact that 𝐿⧵𝑃 is a maximal filter of 𝐿 and 𝑥∉𝐿⧵𝑃.

Lemma 36. Let 𝐿 be a meet semilattice with 0, and let 𝐿^′ be a meet semilattice with 0 (as constructed in Lemma 2). If 𝑃 is a minimal prime semi-ideal of 𝐿, then 𝑃^′={[𝑥]∣𝑥∈𝑃} is a minimal prime semi-ideal of 𝐿^′.

Proof. First, we prove that 𝑃^′ is a semi-ideal. Let [𝑦]<[𝑥]∈𝑃^′. On the contrary, assume that [𝑦]∉𝑃^′. Hence ann(𝑦)⊆𝑃. Further, [𝑦]<[𝑥] gives ann(𝑥)⫋ann(𝑦) which yields ann(𝑥)⊆𝑃. Since 𝑥∈𝑃, by Lemma 35, there exists 𝑦∉𝑃 such that 𝑥∧𝑦=0; that is, 𝑦∈ann(𝑥). But then 𝑦∈ann(𝑥)⊆𝑃, a contradiction. Thus 𝑃^′ is a semi-ideal.
Let [𝑥]∧[𝑦]∈𝑃^′. Then 𝑥∧𝑦∈𝑃. By primeness of 𝑃, either 𝑥∈𝑃 or 𝑦∈𝑃. This gives either [𝑥]∈𝑃^′ or [𝑦]∈𝑃^′. Thus 𝑃^′ is prime. Let 𝑄^′ be a prime semi-ideal of 𝐿^′ such that 𝑄^′⫋𝑃^′. Hence there exists [𝑥]∈𝑃^′ such that [𝑥]∉𝑄^′. Again by Lemma 35, there exists 𝑦∉𝑃 such that 𝑥∧𝑦=0. This gives [𝑥]∧[𝑦]=[0]∈𝑄^′. But then [𝑦]∈𝑄^′⊆𝑃^′, a contradiction. Thus 𝑃^′ is a minimal prime semi-ideal of 𝐿^′.

On similar lines, we can prove the following result.

Lemma 37. Let 𝐿 be a meet semilattice with 0, and let 𝐿^′ be a meet semilattice with 0 (as constructed in Lemma 2). If 𝑃^′is a minimal prime semi-ideal of 𝐿^′, then 𝑃={𝑥∣[𝑥]∈𝑃^′} is a minimal prime semi-ideal of 𝐿.

Let us denote the set of all minimal prime semi-ideals of 𝐿 by Min_𝑠(𝐿).

Theorem 38. Let 𝐿 be a meet semilattice with 0, and let 𝐿^′ be a meet semilattice with 0 (as constructed in Lemma 2). Let 𝜙∶Min_𝑠(𝐿)→Min_𝑠(𝐿^′) be a map defined by 𝜙(𝑃)=𝑃^′={[𝑥]∣𝑥∈𝑃}. Then 𝜙 is bijective.

Proof. Let 𝜙∶Min_𝑠(𝐿)→Min_𝑠(𝐿^′) be a map defined by 𝜙(𝑃)=𝑃^′={[𝑥]∣𝑥∈𝑃}. By Lemma 36, 𝑃^′ is a minimal prime semi-ideal. First, we show that 𝜙 is one-to-one. Let 𝜙(𝑃)=𝜙(𝑄). Then for 𝑥∈𝑃, we have [𝑥]∈𝑃^′=𝜙(𝑃)=𝜙(𝑄)=𝑄^′. This gives 𝑥∈𝑄. Hence 𝑃⊆𝑄. Similarly, 𝑄⊆𝑃. Thus 𝜙 is one-to-one. Let 𝑄^′ be any minimal prime semi-ideal in 𝐿^′. Then 𝑄={𝑥∣[𝑥]∈𝑄^′} is a minimal prime semi-ideal of 𝐿, by Lemma 37. It is clear that 𝜙(𝑄)=𝑄^′. Thus 𝜙 is onto.

With this preparation, we now prove Beck’s Conjecture for 𝐺_𝐸(𝐿).

Theorem 39. Let 𝐿 be a meet semilattice with 0. If 𝐶𝑙𝑖𝑞𝑢𝑒(𝐺_𝐸(𝐿))<∞, then 𝜒(𝐺_𝐸(𝐿))=𝜒(𝐺_{0}(𝐿))=𝐶𝑙𝑖𝑞𝑢𝑒(𝐺_𝐸(𝐿))=𝐶𝑙𝑖𝑞𝑢𝑒(𝐺_{0}(𝐿))=𝑛, where 𝑛 is the number of all minimal prime semi-ideals of 𝐿.

Proof. Let Clique(𝐺_𝐸(𝐿))<∞. Then Clique(𝐺_{0}(𝐿^′))<∞, by Corollary 5, where 𝐿^′ is the semilattice constructed as in Lemma 2. By Theorem 34, Clique(𝐺_{0}(𝐿^′))=𝜒(𝐺_{0}(𝐿^′))=𝑛, where 𝑛 is the number of minimal prime semi-ideals of 𝐿^′. Now, by Theorem 38, the number of minimal prime semi-ideals of 𝐿 is also 𝑛. Thus we have Clique(𝐺_𝐸(𝐿))=𝜒(𝐺_𝐸(𝐿))=𝑛.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.