ISRN Discrete MathematicsVolume 2014 (2014), Article ID 896270, 7 pageshttp://dx.doi.org/10.1155/2014/896270

Research Article

The Graph of Equivalence Classes of Zero Divisors

Vinayak Joshi, B. N. Waphare, and H. Y. Pourali

Department of Mathematics, University of Pune, Pune 411007, India

Received 13 December 2013; Accepted 5 February 2014; Published 8 May 2014

Academic Editors: G. Isaak, H.-J. Kreowski, and J. A. Rodriguez Velazquez

Copyright © 2014 Vinayak Joshi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

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

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 with 0. The main aspect of the graph of equivalence classes of zero divisors of is the connection to the associated primes of . In general, every vertex of either corresponds to an associated prime or is adjacent to an associated prime of . Further, we consider the equivalence relation on a meet semilattice with 0, if and only if for and construct the meet semilattice . It is proved that is an meet semilattice with the property that , the zero divisor graph of . Hence it is clear that the study of is nothing, but the study of zero divisor graph of meet semilattices. This observation helps us to prove that for bounded meet semilattices and , if and only if . This extends the result of LaGrange [16] for Boolean algebra. It is proved that for an meet semilattice , is connected and either it contains a cycle of length 3 or . In the class of a finite semicomplemented meet semilattice , is isomorphic to the zero divisor graph if and only if is .

2. Properties of the Graph

We begin with the necessary definitions and terminology.

A nonempty subset of a meet semilattice is said to be semi-ideal, if implies that . A semi-ideal of a meet semi-lattice is said to be an ideal, if implies that . Dually we have the concept of a filter. A proper semi-ideal (ideal) is said to be prime, if implies either or .

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

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

A meet semilattice with is said to be semicomplemented if, for , there exists such that and is said to be section semicomplemented  (in brief ) if for with , there exists nonzero such that and . It is clear that every lattice is a semicomplemented lattice, but not conversely. More details about poset can be found in Thakare et al. [18, 19], Joshi [20].

Let be a meet semilattice with 0. The set of all zero divisors of is denoted by and the set of all nonzero zero divisors is denoted by . Clearly, .

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

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

Definition 1 (Joshi [6], Lu and Wu [12]). Let be a semilattice with . We associate a simple undirected graph with , called the zero divisor graph of with respect to , denoted by , with the set of vertices is = and two distinct vertices are adjacent if and only if .For any , we say that if and only if , where . Note that is an equivalence relation on . Furthermore, if and , then and hence . We define , the equivalence class of , as follows: .

Lemma 2. Let be a meet semilattice with 0 and , where . Then is a meet semilattice under the partial order if and only if .

Proof. Since , we have . We claim that . Let be another lower bound of and . Then . Now, we claim that . Let . Then . This gives ; that is, . Hence ; that is, . This proves that . Thus and this yields . Clearly, if is a finite lattice, then is also a lattice.

Lemma 3. is an meet semilattice.

Proof. Let . Then . Hence there exists such that . Thus . But then and . Thus is an meet semilattice.

Now, we introduce the graph of equivalence classes of zero divisors of a meet semilattice with , denoted by .

Definition 4. Let be a meet semilattice with 0. We associate a simple undirected graph with whose vertices are the equivalence classes of elements in and with each pair of distinct classes and joined by an edge if and only if .

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

Figure 1: The graph of equivalence classes of zero divisors.

Consider the following two infinite lattices and .

We can see that the zero divisor graph of lattices , is infinite, but the graph of equivalence classes of zero divisors of the lattice , namely, , is finite. Also for the lattices , , and both are infinite, see Figure 2.

Figure 2: The graph is finite, whereas is infinite.

Corollary 5. Let be a meet semilattice with 0. Then there exists an meet semilattice (as constructed in Lemma 2) such that .

Proof. From the construction of , it is clear that .

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

Remark 6. Recall that a prime semi-ideal of a meet semilattice is an associated prime if it is the annihilator of some elements of ; that is, for some nonzero . The set of associated primes is denoted by . If , then 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 . Clearly, the reverse inclusion is trivial. Hence ; that is, the set of zero-divisors of equals the union of all associated primes of . It is obvious to prove that a map defined by , where 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 as an associated prime if . It will be clear from context whether refers to an equivalence class, a vertex, or a specific annihilator.

Theorem 7. Let be a meet semilattice with , and let . If and are distinct prime semi-ideals of , then and are adjacent in . Furthermore, every vertex of is either an associated prime semi-ideal or adjacent to a maximal semi-ideal in .

Proof . Since and are distinct, we have . This further gives that . Thus and are adjacent. Furthermore, suppose that is not an associated prime semi-ideal. Then there exists such that . Hence . We claim that is maximal. Otherwise, since , has a maximal element, say containing , then . It is clear that every maximal element of is prime. Therefore . Thus there is an edge between and . This completes the proof.

Remark 8. For a Boolean lattice, for every , and hence in this case, and are isomorphic. This fact is illustrated in Figure 3.

Figure 3: The isomorphism of and .

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

Question 1. Find a class of meet semilattices such that for .

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

Lemma 9. Let be a meet semilattice with . Then is if and only if for , implies .

Proof. Suppose that is . Further, assume that there exist such that and . Since is , there exist such that and . This gives , a contradiction to the fact that . Hence .

Theorem 10. Let be a finite semicomplemented meet semilattice. Then is a graph isomorphism if and only if is .

Proof . Let be a meet semilattice. Then follows from Lemma 9.Conversely, assume that . First, we prove that , . Let for . Then we have for , a contradiction to . Now, let . Since is semicomplemented, . This gives . Otherwise, if , then , a contradiction. From the proof of Lemma 9, it is clear that is .

If we consider that the is a graph isomorphism defined by , then we do not need even the finiteness of .

Theorem 11. Let be a semicomplemented meet semilattice. Then is a graph isomorphism defined by if and only if is .

Proof. Let be a meet semilattice. Then follows from Lemma 9.Conversely, assume that is a graph isomorphism defined by . This gives , for all . Let . Since is semicomplemented, . This gives . From the proof of Lemma 9, it is clear that is .

From Figure 4, it is clear that for the posets , . 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 such that if and only if for the posets .

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 and be Boolean posets. Then if and only if .

It is well known that there is a 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 and be Boolean algebras. Then if and only if .

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

Theorem 15. Let and be bounded meet semilattices. Then the following statements are equivalent: (1);(2);(3).

Proof. The equivalence of statements and follows from Lemma 9. Since and are semicomplemented, for . Let be a graph isomorphism. Define a map such that , for all along with and , where is the smallest (largest) element of . Since is bijective, is bijective. We claim that is a biorder preserving map.Let . First, we show that . If any of is either or , then we are through. Hence assume that . Suppose that ; that is, . Since is , there exists such that and . Since , we have . This gives that and are adjacent in . Since is a graph isomorphism, we have that and are adjacent which further imply that . Hence, in . Thus and are adjacent in . Hence, we have . But gives , a contradiction. Hence in .Conversely, assume that in . We claim that . If any of or is or , then we are through. Let us assume that . This yields . Now, suppose on the contrary that for . Since is , there exists such that and . Thus and are adjacent in . Then and are adjacent in . Then in . Replacing the role of in the above proof by , respectively, we have . This implies that . But this gives , a contradiction. Hence . Thus . Obvious.

Lemma 16. Let be a bounded meet semilattice, and be a bounded semicomplemented meet semilattice such that . Then is also .

Proof. Since and are semicomplemented, for . Let be a graph isomorphism. Define a map such that , for all along with and , where () is the smallest (largest) element of . Since is a bijective map, is bijective. Let in . Since is semicomplemented, and if or , then we are through. Hence assume that and . Then clearly, and . Since is and , there exists and . Since and are adjacent in , we have and are adjacent in ; that is, . Further and are nonadjacent in . Then and are nonadjacent in . Consider say. Clearly, and (as implies ). Thus is .

Theorem 17. Let and be bounded semicomplemented meet semilattices. If , then is .

Proof. For the given lattice , we can construct an meet semilattice (as given in Lemma 2), such that . This gives . By Lemma 16, is .

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

Theorem 18 (Anderson and LaGrange [14, Theorem 2.6]). Let and be commutative reduced rings with and . If , then 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 be a meet semilattice with 0. Then the graph is connected and .

Theorem 20. Let be a meet semilattice with 0. Then the graph is connected and .

Proof. Follows from Corollary 5, Theorem 19, and the fact that adjacency in implies adjacency in . Hence .

Theorem 21. Let be an lattice. Then is either or contains a cycle of length . Hence .

Proof. If , then , by Theorem 19. Now assume that . Choose such that is a path with and . Since is and without loss of generality assume that , there exists such that and . Clearly, . Since , we have . Thus we have a cycle of length .

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

Corollary 22. Let be a meet semilattice with 0; then is either or contains a cycle of length . Hence, if , then .

Proof . It follows from Corollary 5 and Theorem 21.

Remark 23. In the above result, the condition of existence of a cycle in is necessary. In Figure 1, we can observe that has a cycle but has no cycle.

3. Correlation between Diameter of and

The following result gives the correlation between diameter of and .

Theorem 24. The following statements are true for a meet semilattice with 0. (a)If , then .(b)If , then or .(c)If , if and only if .(d)If , then or .(e)If , then .

Proof. It is clear that . Hence the result follows from Theorem 20. Suppose that . So we have a path in . We have two possibilities.Case  1. If are distinct vertices of , we get the path . Hence by Theorem 20, .Case  2. The only possibility in this case is , and hence, . Let . Then there exists a path . Since gives ; otherwise , a contradiction. Now assume that in . Then there exists such that . This yields that , again a contradiction. Thus by Theorem 20, . The converse follows from Theorem 20. Suppose that . Then , by . Hence the result. Suppose that ; then there exists a path in . This yields a path in . Thus . The result follows from and .

Remark 25. The assertions and of Theorem 24 are justified in Figures 1 and 5, whereas the assertions and are justified in Figures 5 and 6. Figure 3 verifies the assertion .

Figure 5: .

Figure 6: .

Corollary 26. Let be a meet semilattice with . Then if and only if .

4. Cut Vertices in

In this section, we examine the properties of cut vertices of .

Definition 27. A lattice with is said to be 0-distributive, if implies that ; 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 be a -distributive lattice. If is a cut vertex of , then forms an ideal of .

Proof. Let be a cut vertex of . Let and be nonzero elements of . Since , . Now, it is enough to show that . Clearly, .Let . This gives . Since is -distributive, we have ; that is, . Thus . This implies that .Now suppose that and , to show that . If or , then . Let and . Then . Since is a cut vertex of , therefore for any two arbitrary vertices , we have a path . Therefore ; that is, , which further yields . Similarly, . Therefore, we get a path . Since is cut vertex, . Thus is an ideal.

Remark 29. It is clear from the proof of the above theorem that in a general lattice, if is a cut vertex of , then is a semi-ideal.

Lemma 30. Let be a meet semilattice with 0. If is a cut vertex of , then is maximal in . Hence is a prime semi-ideal.

Proof. Let be a cut vertex of , and let and be mutually separated subgraphs of with . Let and , therefore we have the path . Suppose that for some . Since is an edge, . Then . Similarly, . Thus we have another path passing through of . Since is a cut vertex of , then . Thus . It is easy to prove that is a prime semi-ideal.

Lemma 31. If is a cut vertex of , then all other associated primes of are contained in only one component of .

Proof. Suppose that and are two mutually separated connected components of , and each contains an associated prime. It is easy to observe that the associated primes are adjacent, and hence , are connected, a contradiction.

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

Problem 32. Characterize those posets for which .

In view of Corollary 5 and the following result, it is clear that for an meet semilattice , .

Theorem 33. Let be a meet semilattice with 0. If and have at least cut vertices, then .

Proof. Let and be cut vertices of . Since is a cut vertex of , there is some such that any path connecting and must include . Similarly, since is a cut vertex, there is some such that any path connecting and must include . Clearly, and any path from to must include and and so . By Theorem 20, . Thus , by Corollary 22.

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

5. Beck’s Conjecture for

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 be a meet semilattice with 0. If , then , where is the number of minimal prime semi-ideals of . From Theorem 34 and Corollary 5, it is clear that Beck’s Conjecture is true for the graph . In the sequel, we calculate the chromatic number of . As a preparation, we need the following easy lemma.

Lemma 35. Let be a minimal prime semi-ideal of a meet semilattice with 0. Then for any , there exists such that .

Proof. It follows from the fact that is a maximal filter of and .

Lemma 36. Let be a meet semilattice with 0, and let be a meet semilattice with 0 (as constructed in Lemma 2). If is a minimal prime semi-ideal of , then is a minimal prime semi-ideal of .

Proof. First, we prove that is a semi-ideal. Let . On the contrary, assume that . Hence . Further, gives which yields . Since , by Lemma 35, there exists such that ; that is, . But then , a contradiction. Thus is a semi-ideal.Let . Then . By primeness of , either or . This gives either or . Thus is prime. Let be a prime semi-ideal of such that . Hence there exists such that . Again by Lemma 35, there exists such that . This gives . But then , a contradiction. Thus is a minimal prime semi-ideal of .

On similar lines, we can prove the following result.

Lemma 37. Let be a meet semilattice with 0, and let be a meet semilattice with 0 (as constructed in Lemma 2). If is a minimal prime semi-ideal of , then is a minimal prime semi-ideal of .

Let us denote the set of all minimal prime semi-ideals of by .

Theorem 38. Let be a meet semilattice with 0, and let be a meet semilattice with 0 (as constructed in Lemma 2). Let be a map defined by . Then is bijective.

Proof. Let be a map defined by . By Lemma 36, is a minimal prime semi-ideal. First, we show that is one-to-one. Let . Then for , we have . This gives . Hence . Similarly, . Thus is one-to-one. Let be any minimal prime semi-ideal in . Then is a minimal prime semi-ideal of , by Lemma 37. It is clear that . Thus is onto.

With this preparation, we now prove Beck’s Conjecture for .

Theorem 39. Let be a meet semilattice with 0. If , then , where is the number of all minimal prime semi-ideals of .

Proof. Let . Then , by Corollary 5, where is the semilattice constructed as in Lemma 2. By Theorem 34, , where is the number of minimal prime semi-ideals of . Now, by Theorem 38, the number of minimal prime semi-ideals of is also . Thus we have .

Conflict of Interests

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

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