Algebra
Volume 2014 (2014), Article ID 146873, 6 pages
http://dx.doi.org/10.1155/2014/146873

Research Article

Zero Divisor Graph for the Ring of Eisenstein Integers Modulo

Osama Alkam and Emad Abu Osba

Department of Mathematics, Faculty of Science, The University of Jordan, Amman 11942, Jordan

Received 26 March 2014; Revised 19 October 2014; Accepted 24 October 2014; Published 15 December 2014

Academic Editor: Ricardo L. Soto Montero

Copyright © 2014 Osama Alkam and Emad Abu Osba. 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.

Let be the ring of Eisenstein integers modulo . In this paper we study the zero divisor graph . We find the diameters and girths for such zero divisor graphs and characterize for which the graph is complete, complete bipartite, bipartite, regular, Eulerian, Hamiltonian, or chordal.

1. Introduction

Let be a primitive third root of unity. Then the set of complex numbers , where are integers, is called the set of Eisenstein integers and is denoted by . Since is a subring of the field of complex numbers, it is an integral domain. Moreover, the mapping is a Euclidean norm on . Thus is a principal ideal domain. The units of are , , and . The primes of (up to a unit multiple) are the usual prime integers that are congruent to modulo and Eisenstein integers whose norm is a usual prime integer. It is easily seen that, for any positive integer , the factor ring is isomorphic to the ring . Thus is a principal ideal ring. This ring is called the ring of Eisenstein integers modulo . In [1] this ring is studied and its properties are investigated; its units are characterized and counted. Thus, its zero divisors are completely characterized and counted. This characterization uses the fact that is a unit in if and only if is a unit in . Recall that a ring is local if it has a unique maximal ideal. The following are sample results of [1].

(1)If is a prime integer, then the ring is local if and only if or .(2)Let denote the number of units in a ring ; then

(i);(ii).

We deduce the following.

Proposition 1. Let , where , , , , and are primes such that , , for each . Then

Proof. If , then . Thus, .

Since is a finite commutative ring with identity, every element of is a unit or a zero divisor. Let denote the number of nonzero zero divisors of a ring . Then

(1); (2) if ;(3) if ;(4)if , then

The (undirected) zero divisor graph of a commutative ring with identity that has finitely many zero divisors is the graph in which the vertices are the nonzero zero divisors of . Two vertices are adjacent if they are distinct and their product is . The concept of a zero divisor graph was introduced by Beck in [2] and then studied by Anderson and Naseer in [3] in the context of coloring. The definition of zero divisor graphs in its present form was given by Anderson and Livingston in [4]. Numerous results about zero divisor graphs were obtained by Akbari et al. (see [5–7]). The zero divisor graph is studied to get a better understanding of the algebraic structure of the ring . The interplay of the algebraic properties of , graph theoretic properties of , and its relation with is studied. An earlier study was carried out for the zero divisor graph of the ring of Gaussian integers modulo (see [8]). For distinct in , we use to denote the length of the shortest path from to . The diameter of the graph is defined by are distinct elements of . The girth of the graph is if the graph contains no cycles; otherwise it is the length of the shortest cycle. It is shown in [4] that, for any commutative ring with identity, there is a path between any two vertices of and that . It is shown in [9] that the girth of the zero divisor graph of a commutative ring with identity is either , , or . Throughout this paper we will use to denote a usual prime integer that is congruent to modulo and use to denote other prime integers.

2. Diameter and Girth of the Zero Divisor Graph of

In this section we find the diameter and girth of the zero divisor graphs of where or is a usual prime integer congruent to modulo . If and , then . In fact, . These two zero divisors are adjacent. Thus, . Hence, . If and , then the ring is local with the unique maximal ideal . This ideal constitutes the zero divisors of . By Corollary  2.7 in [4], there is a vertex of adjacent to every other vertex. Hence, . Moreover, is not the complete graph because the vertices and are not adjacent. This implies that . Hence, . The following are cycles of length in :

   if  ;   if  .

Thus, if , then . These results are summarized in the following.

Theorem 2. (1) .
(2) If , then .

Now we find the diameter and the girth of , where is a prime congruent to modulo . Let be such a prime. Then, there exist nonassociate Eisenstein primes and such that . The ring is the product of the two rings and (see [1], page 5, Section 3) and the ideals and are the only maximal ideals of . These two (maximal) ideals constitute the zero divisors of .

Recall that a graph is bipartite if the set of vertices of can be split into two disjoint sets and so that each edge of joins a vertex of to a vertex of . A bipartite graph is complete bipartite if each element in is adjacent to every element in and conversely. It is well known that a simple graph is bipartite if and only if it contains no odd cycles (see [10]). Thus, if a simple bipartite graph contains a cycle, then its girth is 4.

Lemma 3. If a ring is a product of two integral domains, then the graph is complete bipartite.

Proof. See Example 2.1 (c) in [11].

Lemma 4. Let , where and are commutative rings with identity. Then .

Proof. See [12].

Theorem 5. Let be a prime integer congruent to 1 modulo 3. Then

(1);(2).

Proof. The ring is a product of two fields. Therefore, the graph is complete bipartite. Hence, and . Let . Since , there exist integers and such that . The ring is isomorphic to the product of the local rings and . These local rings are commutative with identity and, since , each of them contains nonzero zero divisors. Therefore, by Lemma 4, . To show that whenever , it suffices to find a cycle of length . The following are cycles of length for and , respectively.

 . .

3. Diameter and Girth of the Zero Divisor Graph of ;

Theorem 6. Let be a prime integer congruent to modulo . Then the zero divisor graph is the empty graph. If , then

(1);(2).

Proof. Let be a prime integer congruent to modulo . Then is an Eisenstein prime integer. Hence is a field. Thus, the zero divisor graph is the empty graph. If , then the ring is local with unique maximal ideal . This ideal constitutes the zero divisors of . By Corollary  2.7 in [4], there is a vertex of adjacent to every other vertex. Hence, . Clearly when , is a complete graph. Thus, and . Let . Then, and are not adjacent. Hence, is not a complete graph. This implies that . Hence, . The following are cycles of length in :

  if ;  if . Thus, if , then .

This completes the proof.

4. Zero Divisor Graph of

We start with finding the diameter and girth of the graph ); then we characterize for which ) is complete, complete bipartite, bipartite, regular, Eulerian, Hamiltonian, or chordal.

4.1. Diameter and Girth

Theorem 7. equals if and only if or and equals if and only if for some , for some , , or . Otherwise equals .

Proof. Let us say that a positive integer is of type if or , type if for some or for some or or , and type otherwise. If , then is isomorphic to a product of two fields. Thus, by Lemma 3, the graph is complete bipartite. Hence, . Therefore, by the first parts of Theorems 2, 5, and 6, we only need to show that if is a type integer, then . If is such an integer, then there exist two commutative rings and with identities such that contains a nonzero zero divisor and . Therefore, by Lemma 4, .

Lemma 8. Let be a product of at least three commutative rings with identities. Then .

Proof. Let and . Then , , and are zero divisors in this product. Moreover, is a cycle of length .

Theorem 9. Let . Then

(i) in each of the following cases:

(1) is not a square free integer; that is, is divisible by the square of a prime;(2) has at least distinct prime divisors;(3) is a product of two primes, with at least one of them congruent to modulo .

(ii) if is a prime that is congruent to modulo or a product of two distinct primes congruent to modulo or is times a prime congruent to modulo .

Proof. Let .
(i) (1) If is divisible by the square of a prime , then is a cycle of length .
(2) In this case, can be written as a product of at least three commutative rings with identity. Thus, by Lemma 8, .
(3) Let , where is a prime congruent to modulo . If , then this case is treated in the second part of Theorem 5. If , then . There exist integers and such that . The following is a cycle in of length :
.
(ii) The case is treated in the second part of Theorem 5. If is a product of two distinct primes congruent to modulo , then is a product of two fields. Thus, by Lemma 3, is complete bipartite. If is times a prime congruent to modulo , then . One can easily show that is a bipartite graph that contains a cycle. Therefore, .

4.2. Characterization of Complete, Complete Bipartite, Bipartite, Regular, Eulerian, Hamiltonian, or Chordal Graphs

4.2.1. When Is Complete, Complete Bipartite, Bipartite, or Regular?

Theorem 10. The graph is complete if and only if or for some prime integer such that .

Proof. If , then . If , then the zero divisors of are given by the principal ideal . Thus, any two zero divisors are adjacent. Hence, the graph is complete. Conversely, assume that the graph is complete. Then . Thus by Theorem 9, or for some prime integer such that .

Theorem 11. The graph is complete bipartite if and only if or for some primes , , and , where and .

Proof. If or , then is a product of two fields, thus complete bipartite by Lemma 3. Conversely, if is complete bipartite, then and . Thus by Theorem 7, for some or for some or or . If for some or for some , then by Theorems 2 and 6 the girth is not . Hence, or .

Lemma 12. If is a local ring with maximal ideal such that , then contains a triangle and hence cannot be bipartite.

Proof. Let be the maximal ideal of . If , then any three nonzero elements in form a triangle in . Assume such that . If , then clearly contains a triangle. Assume that and let . Then, is adjacent to every other vertex in . If there is no other adjacency, then , where is a field (see [4]). This contradicts the fact that is local. Thus, there exist such that form a triangle in , and so cannot be bipartite.

Lemma 13. If , where is a ring with at least two nonzero zero divisors, then cannot be bipartite. In particular, if is a product of more than two rings, then cannot be bipartite.

Proof. Let such that and . Then the vertices , , and form a triangle in . Hence cannot be bipartite. If is a product of more than two rings, then we may write . The ring contains at least two nonzero zero divisors, namely, and .

Therefore, we deduce the following.

Theorem 14. The graph is bipartite if and only if it is complete bipartite (if and only if or for some primes , , and , where and ).

Recall that a graph is regular if each vertex has the same number of neighbors. An example of a regular graph is the complete graph. A good question about is, is there a graph of the form that is regular and incomplete? Before we give an answer, we prove the following.

Lemma 15. If a ring is a product of two finite integral domains that have the same number of elements, then the graph is regular.

Proof. Assume that . Then, is a complete bipartite graph by Lemma 3. So, every vertex has the same number of neighbors. Thus, is regular.

Lemma 16. Let , where and are commutative rings with identity and zero divisors. If (without loss of generality) the graph is not regular, then the graph is not regular.

Proof. Since the graph is not regular, the ring contains zero divisors with different degrees. Clearly, for and have the same degrees. Thus, and have different degrees. Therefore, the graph is not regular.

Theorem 17. (1) For any , the graph is not regular.
(2) For any prime , the graph is not regular.
(3) For any and any prime , the graph is not regular.
(4) If is divisible by the product of two distinct primes, then the graph is not regular.

Proof. (1) If , then the ring is local with the unique maximal ideal . This ideal constitutes the zero divisors of . By Corollary  2.7 in [4], there is a vertex of that is adjacent to every other vertex. Hence, there is a vertex with neighbors. The vertex has neighbors, namely, not both equal zero and . Since , . Therefore, the graph is not regular.
(2) Let be a prime congruent to modulo . Then there exist integers such that . We will see that and have different number of neighbors. If is a neighbor of , then (in ). So there exists in such that . Since is an integral domain, . Hence, the neighbors of are contained in the ideal . Conversely, every element in this ideal is a neighbor of . Similarly, one can show that the ideal constitutes the neighbors of . Since the ideals and are finite and is strictly contained in , and have different number of neighbors.
(3) Let and . Since , . Moreover, and are zero divisors in because they are not units. A nonzero zero divisor is a neighbor of if and only if if and only if if and only if because is a unit in . There are choices for and the same number of choices for , namely, . Thus, there are neighbors of . One can follow the same technique to show that there are neighbors of . Since , ; that is, the number of neighbors of and is different.
(4) Let be two distinct prime divisors of . Then and are two nonzero zero divisors of . A zero divisor is a neighbor of if and only if and are multiples of . There are choices for and the same number of choices for , namely, . Thus, there are neighbors of . Similarly, there are neighbors of . Since , ; that is, the number of neighbors of and is different.

Theorem 18. The graph is regular if and only if and for some prime such that or for some prime such that .

Proof. If or for some prime such that , then the graph is complete and hence regular. If is a prime such that , then is the product of two finite fields that have the same number of elements. Therefore, by Lemma 15, the graph is regular. The converse follows from Lemma 16 and Theorem 17.

4.2.2. When Is Eulerian, Hamiltonian, or Chordal?

A graph is called Eulerian if there exists a closed trail containing every edge of . We now characterize for which the graph is Eulerian. We will do this in two steps.

Step 1 (when is local). It is shown in [1] that is a local ring if and only if or , where and is a prime integer such that . In this case . Also it is shown in [13] that, for a finite local ring with a maximal ideal , the graph is Eulerian if and only if is even and for each . Hence, we have the following result.

Theorem 19. If is local, then is Eulerian if and only if .

Step 2 (when is nonlocal). If where and are prime integers such that and , then . Moreover, , where and are integers such that . It is shown in [13] that, for a finite nonlocal ring , the graph is Eulerian if and only if is a direct product of fields, each of which is of odd order. See also the result of Akbari and Mohammadian (Proposition  1 in [6]). Thus we have the following theorem.

Theorem 20. If is nonlocal, then is Eulerian if and only if is a prime integer such that or is a square free composite integer not divisible by nor .

Combining the two theorems gives the following.

Theorem 21. The graph is Eulerian if and only if or is a prime integer such that or is a square free composite integer not divisible by nor .

We now characterize for which is Hamiltonian. Recall that a Hamiltonian cycle of a graph is a cycle that contains every vertex of . A graph is Hamiltonian if it contains a Hamiltonian cycle. It is shown in [5] that, for a finite principal ideal ring , if is Hamiltonian, then it is either a complete graph or a complete bipartite graph.

Note that a complete bipartite graph with two parts of different orders cannot be Hamiltonian. Thus we have the following result.

Theorem 22. The graph is Hamiltonian if and only if or for some prime integers and , where and .

We conclude this paper by characterizing for which is a chordal graph. Recall that a graph is chordal if it has no induced cycle of length greater than 3.

Lemma 23. If is a local principal ideal ring, then is chordal.

Proof. Let be the maximal ideal of and let such that . Any element in can be written in the form , where and is a unit in . Since is finite, there exists such that . Assume that is a path in a cycle in . Then , and , and so , which implies that or . Thus, there is a chord in joining ( and ) or ( and ).

Lemma 24. If with for , then is not chordal.

Proof. Let and . Then, is an induced cycle in with no chords.

Therefore, we deduce the following.

Theorem 25. The graph is chordal if and only if is local (if and only if or with and ).

Disclosure

This work has been carried out during sabbatical leave granted to the author Osama Alkam from the University of Jordan during the academic year 2011/2012.

Conflict of Interests

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

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