Zero Divisor Graph for the Ring of Eisenstein Integers Modulo n

Let ω be a primitive third root of unity. Then the set of complex numbers a + bω, where a, b are integers, is called the set of Eisenstein integers and is denoted by E. Since E is a subring of the field of complex numbers, it is an integral domain. Moreover, the mappingN(a + bω) = a2 + b2 − ab is a Euclidean norm on E. Thus E is a principal ideal domain. The units of E are ±1, ±ω, and ±ω. The primes of E (up to a unit multiple) are the usual prime integers that are congruent to 2modulo 3 and Eisenstein integers whose norm is a usual prime integer. It is easily seen that, for any positive integer n, the factor ring E/nE is isomorphic to the ring E n = {a + bω |


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 ( + ) =  2 +  2 −  is a Euclidean norm on .Thus  is a principal ideal domain.The units of  are ±1, ±, and ±.The primes of  (up to a unit multiple) are the usual prime integers that are congruent to 2 modulo 3 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   = { +  | ,  ∈ Z  }.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 Z  .Recall that a ring is local if it has a unique maximal ideal.The following are sample results of [1].
We deduce the following.
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 .
The (undirected) zero divisor graph Γ() 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 0. 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][6][7]).[4] that, for any commutative ring with identity, there is a path between any two vertices of Γ() and that diam(Γ()) ≤ 3. It is shown in [9] that the girth of the zero divisor graph of a commutative ring with identity is either 3, 4, or ∞.Throughout this paper we will use  to denote a usual prime integer that is congruent to 2 modulo 3 and use  to denote other prime integers.

Diameter and Girth of the Zero Divisor Graph of 𝐸 𝑞 𝑘
In this section we find the diameter and girth of the zero divisor graphs of   where  = 3 or  is a usual prime integer congruent to 1 modulo 3.
Recall that a graph Γ is bipartite if the set of vertices of Γ can be split into two disjoint sets  and  so that each edge of Γ 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.
Proof.Let  be a prime integer congruent to 2 modulo 3.
Then  is an Eisenstein prime integer.Hence   is a field.
This completes the proof.

Zero Divisor Graph of 𝐸 𝑛
We start with finding the diameter and girth of the graph Γ(  ); then we characterize  for which Γ(  ) is complete, complete bipartite, bipartite, regular, Eulerian, Hamiltonian, or chordal.

Lemma 8.
Let  be a product of at least three commutative rings with identities.Then (Γ()) = 3.
Theorem 9. Let  > 1.Then (i) (Γ(  )) = 3 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 3 distinct prime divisors; (3)  is a product of two primes, with at least one of them congruent to 1 modulo 3.
(ii) (Γ(  )) = 4 if  is a prime that is congruent to 1 modulo 3 or a product of two distinct primes congruent to 2 modulo 3 or  is 3 times a prime congruent to 2 modulo 3.
(2) In this case,   can be written as a product of at least three commutative rings with identity.Thus, by Lemma 8, (Γ(  )) = 3.
(ii) The case  =  is treated in the second part of Theorem 5.If  is a product of two distinct primes congruent to 2 modulo 3, then   is a product of two fields.Thus, by Lemma 3, Γ(  ) is complete bipartite.If  is 3 times a prime congruent to 2 modulo 3, then   ≅  3 ×   .One can easily show that Γ( 3 ×   ) is a bipartite graph that contains a cycle.Therefore, (Γ(  )) = 4. Thus by Theorem 7,  = 3  for some  > 1 or  =   for some  > 2 or  =  or  =  1  2 .If  = 3  for some  > 1 or  =   for some  > 2, then by Theorems 2 and 6 the girth is not 4.
Lemma 12.If  is a local ring with maximal ideal  such that || > 4, then Γ() contains a triangle and hence cannot be bipartite.
Proof.Let  be the maximal ideal of .If  2 = {0}, then any three nonzero elements in  form a triangle in Γ().Assume Then,  is adjacent to every other vertex in Γ().If there is no other adjacency, then  ≃ Z 2 × , where  is a field (see [4]).This contradicts the fact that  is local.Thus, there exist ,  ∈  \ {0, } such that {, , } form a triangle in Γ(), and so Γ() cannot be bipartite.
, where  2 is a ring with at least two nonzero zero divisors, then Γ() cannot be bipartite.In particular, if  is a product of more than two rings, then Γ() cannot be bipartite.
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 Γ(  ) is, is there a graph of the form Γ(  ) 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 Γ() is regular.
Proof.Assume that  =  1 ×  2 .Then, Γ() is a complete bipartite graph by Lemma 3. So, every vertex has the same number of neighbors.Thus, Γ() is regular.
Lemma 16.Let  =  1 ×  2 , where  1 and  2 are commutative rings with identity and zero divisors.If (without loss of generality) the graph Γ( 1 ) is not regular, then the graph Γ() is not regular.
(3) For any  > 2 and any prime , the graph Γ(   ) is not regular.
(4) If  is divisible by the product of two distinct primes, then the graph Γ(  ) is not regular.
(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, 0, , 2, . . ., ( − 1).Thus, there are  2 − 1 neighbors of .Similarly, there are  2 − 1 neighbors of .Since  ̸ = ,  2 − 1 ̸ =  2 − 1; that is, the number of neighbors of  and  is different.
Theorem 18.The graph Γ(  ) is regular if and only if  = 3 and  =  2 for some prime  such that  ≡ 2 ( mod 3) or  =  for some prime  such that  ≡ 1 (mod 3).