# Some Properties of the Intersection Graph for Finite Commutative Principal Ideal Rings

Authors: Emad Abu Osba; Salah Al-Addasi; Omar Abughneim
Journal: International Journal of Combinatorics
 (2014)

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## Abstract

LetR be a commutative finite principal ideal ring with unity, and let G(R) be the simple graph consisting of nontrivial proper ideals of R as vertices such that two vertices I and J are adjacent if they have nonzero intersection. In this paper we continue the work done by Abu Osba. We calculate the radius, eccentricity, domination number, independence number, geodetic number, and the hull number for this graph. We also determine when G(R) is chordal. Finally, we study some properties of the complement graph of G(R).


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## Body

## 1. Introduction

All rings are assumed to be finite commutative principal ideal rings with unity1.For each vertexx in a graph Γ, let deg⁡(x) be the number of vertices adjacent to x and let N(x) be the set of vertices adjacent to x in Γ. For any undefined graph theoretical terms, the reader may consult [1].LetF={Aj:j∈J} be a family of nonempty sets. The intersection graph G(F) defined on F is a simple graph whose vertex set is F and two vertices Aj and Ai are adjacent if Aj≠Ai and Aj∩Ai≠ϕ. Many authors worked on the graphs G(F) when the members of Fhave an algebraic structure; see, for example, [2–7].The intersection graphG(R) of ideals of a ring R is a simple graph whose vertices are the nontrivial proper ideals and two vertices I,J are adjacent if I≠J and I∩J≠{0}. Note that if R is a field, then G(R) is the null graph which has no vertices. Extending statements to the null graph would introduce unnecessary distractions, so we ignore the null graph; that is, all rings are assumed to be nonfields except when stated explicitly.In this paper we consider the intersection graphG(R) of nontrivial proper ideals of a finite commutative principal ideal ring R with unity 1. If R is a finite commutative ring, then it can be written as a product of local rings; see [8]. If R=∏‍Rj and R has a unity, then any ideal I of R can be written as ∏‍Ij, where Ij is an ideal in Rj for each j, while if R has no unity, then this needs not to be true; see [9, page 135]. If R is a finite local principal ideal ring with maximal ideal M=aR, then the ideals of R are M=aR, M2=a2R, and M3=a3R,…,Mn-1=an-1R,Mn={0} for some n∈N; see the proof of Proposition 8.8 in [10].This study is a continuation of the study in [2], where the author used the fact that if R is a local ring with maximal ideal M, then there exists n∈N such that Mn={0} but Mn-1≠{0}, to define Nilpotency(R)=n. If R is a nonlocal ring such that R=∏j=1nRj, where Rj is a local ring with Nilpotency(Rj)=nj for j=1,2,…,n, then R has (∏j=1n(nj+1))-2 nontrivial proper ideals. The author characterized when G(R) is Eulerian, Hamiltonian, planar, or bipartite.In this paper we will continue the investigation of properties of the intersection graph and calculate the radius, eccentricity, dominating number, independence number, geodetic number, and the hull number; we also determine whenG(R) is chordal. We conclude this paper by a study of some properties of the complement graph of G(R).


## 2. Then-Cube

Then-cube Qn  (n≥1) is the graph whose vertex set is the set of all binary n-tuples, where two n-tuples are adjacent if and only if they differ in precisely one coordinate.Theorem 1.
LetR=∏j=1nRj, where Rj is a finite local principal ideal ring for each j. Then G(R) contains a subgraph isomorphic to Qn-{(0,0,…,0),(1,1,…,1)}. In particular, if each Rj is a field, then this subgraph is a spanning subgraph of G(R).Proof.
LetA={∏j=1nIj∈G(R):Ij∈{{0},Rj}}. Consider the one-to-one correspondence f:Qn-{(0,0,…,0),(1,1,…,1)}→A, defined by f((b1,b2,…,bn))=∏j=1nIj, where
(1)Ij={Rjifbj=1{0}ifbj=0.
If (b1,b2,…,bn) and (c1,c2,…,cn) are two adjacent vertices of Qn-{(0,0,…,0),(1,1,…,1)}, then there exists a unique j1∈{1,2,…,n} such that {bj1,cj1}={0,1}; say bj1=1 and cj1=0. Since (c1,c2,…,cn)≠(0,0,…,0), we can find j2∈{1,2,…,n}-{j1} such that cj2=1. But bj2=cj2, and hence f((b1,b2,…,bn))=∏j=1nIj and f((c1,c2,…,cn))=∏j=1nTj are adjacent in G(R) because Ij2=Tj2=Rj2. Therefore the subgraph of G(R) induced by A contains a copy of Qn-{(0,0,…,0),(1,1,…,1)}. Note that the set A consists of all vertices of G(R) when R is a product of fields.The previous result shows thatG(R) contains Qn-{(0,0,…,0),(1,1,…,1)} as a subgraph. This subgraph is induced only when n<3. If R=∏j=1nRj, where n≥3 and Rj is a local ring for each j, then I=∏j=1nIj and T=∏j=1nTj, where I1=R1,I2=R2, andIj={0} for j≠1,2 and T1=R1, T3=R3, and Tj={0} for j≠1,3 are two elements of A. Clearly, I and T are adjacent in the subgraph of G(R) induced by A, while f-1(I) and f-1(T) are not adjacent in Qn-{(0,0,…,0),(1,1,…,1)}.


## 3. Dominating Sets and Numbers ofG(R)

In a graphΓ, a dominating set is a set of vertices A such that every vertex outside A is adjacent to at least one vertex in A. Thedomination number of a graph Γ, denoted by γ(Γ), is the smallest number of the form |A|, where A is a dominating set.IfR is a finite local principal ideal ring with maximal ideal M, then G(R) is complete, and so A={M} is a dominating set, and γ(G(R))=1. Assume that R is nonlocal and R=∏j=1nRj, where Rj is local for each j. We have two cases.Case 1.
One factor, sayR1, is not a field, with maximal ideal M. Let I=M×R2×⋯×Rn. Then A={I} is a dominating set, and γ(G(R))=1.Case 2.
R is a product of fields. Let I be a nontrivial proper ideal in R. Then I is of the form ∏j=1nIj, where Ik={0} for at least one k. Let J=∏j=1nJj such that Jk=Rk and Jj={0} for j≠k. Then I∩J={0}, and so, {I} cannot be a dominating set. Thus γ(G(R))>1. Now, let I={0}×R2×⋯×Rn and J=R1×{0}×⋯×{0}. Then A={I,J} is a dominating set, and γ(G(R))=2. Thus we have the following result.Theorem 2.
For any finite principal ideal ringR which is not a field, γ(G(R))=1 except when R is a product of fields; one has γ(G(R))=2.


## 4. Radius and Center ofG(R)

For a graphΓ, the eccentricity of a vertex f in Γ is e(f)=Max⁡{d(f,g):g is a vertex in Γ}. A center of Γ is a vertex f0 with smallest eccentricity. The eccentricity e(f0) is called the radius of Γ and is denoted by ρ(Γ).Now, we calculate the radius ofG(R).Theorem 3.
For any finite principal ideal ringR which is not a field, one has
(2)ρ(G(R))={0ifRislocalwithNilpotency(R)=2∞ifRisaproductoftwofields2ifRisaproductofmorethantwofields1otherwise.Proof.
LetR be a local ring with Nilpotency(R)=2. Then G(R) is K1 and so ρ(G(R))=0.
Assume now thatR is local with Nilpotency(R)>2 or R is not a product of fields. If R is a local ring or R is not a product of fields, then there exists an ideal I in G(R) such that I is adjacent to every other ideal in G(R) and so, e(I)=1=ρ(G(R)). If R=F1×F2 is a product of two fields, then I=F1×{0} and J={0}×F2 are the only vertices in G(R) and they are nonadjacent. So ρ(G(R))=e(J)=e(I)=∞. Let R=∏j=1nFj be a product of fields with n>2, and I,J∈G(R) such that I∩J={0}. If I+J≠R, then I_I+J_J is a path in G(R). If I+J=R, then I=∏j=1nIj, where Ij={0} or Fj and J=∏j=1nJj, where Jj={0} or Fj. Let k be the least element in {1,2,…,n} such that Ik=Fk and let s be the least element in {1,2,…,n} such that Js=Fs. It is clear that k≠s. Define L=∏j=1nLj such that
(3)Lj={Fkj=kFsj=s{0}otherwise.
ThenL is a vertex in G(R) and I_L_J is a path in G(R). Hence d(I,J)=2 and e(I)=2. Since there is no ideal in G(R) that is adjacent to every other ideal, we have ρ(G(R))=2.Note that ifR is a local ring with maximal ideal M, then M is a center for G(R), and if R=∏j=1nRj, where R1 is a local ring with maximal ideal M≠{0}, then M×∏j=2nRj is a center for G(R), while if R=∏j=1nFj, where Fj is a field for each J and n>2, then {0}×∏j=2nFj is a center for G(R).The following result was proved in [2] and will be used in the proof of the next theorem.Lemma 4.
For any finite principal ideal ringR which is not a field, one has
(4)diam⁡(G(R))={0ifRislocalwithNilpotency(R)=21ifRislocalwithNilpotency(R)>2∞ifRisaproductoftwofields2otherwise.The subgraph of a graphΓ induced by the set of centers of Γ is denoted by Cen(Γ). The graph Γ is calledself-centered if Cen(Γ)=Γ.Theorem 5.
LetR be a finite principal ideal ring. The graph G(R) is self-centered if and only if R is local or a product of fields.
Moreover, ifR=∏j=1nRj, where n≥2, Rj is a finite principal ideal local ring for each j, and there is at least one Rj having nilpotency greater than 1, then the vertex set of Cen(G(R)) is {∏j=1nIj:Ij≠{0}foreachj} and Cen(G(R))=K(∏j=1nnj)-1, where nj=Nilpotency(Rj).Proof.
The graphG(R) is self-centered when its diameter and radius are equal. Thus, by Lemma 4 and Theorem 3, G(R) is self-centered when R is local or R is a product of n fields with n≥2. So, suppose that R=∏j=1nRj, where n≥2, Rj is a local ring for each j, and there is at least one Rj having nilpotency greater than 1. Then, by Theorem 3, G(R) has radius 1. Any vertex I=∏j=1nIj with Ij≠{0} for each j is adjacent to every other vertex of G(R). Thus I is a center of G(R). Now let J=∏j=1nJj with Jj0={0} for some j0∈{1,2,…,n} be a vertex of G(R). Take the vertex L=∏j=1nLj with Lj0=Rj0 and Lj={0} for every j≠j0. Obviously, J and L are not adjacent and hence J is not a center of G(R). Thus the vertex set of Cen(G(R)) is {∏j=1nIj:Ij≠{0}foreachj} which has cardinality (∏j=1nnj)-1, where nj=Nilpotency(Rj). The fact that Cen(G(R)) is a complete graph follows directly from the fact that G(R) has radius 1.


## 5. Independence Number ofG(R)

Anindependent vertex set of a graph Γ is a set of vertices such that no two of them are adjacent in Γ. The vertex independence number α(Γ) of a graph, often called simply the independence number, is the cardinality of the largest (vertex) independent set.Theorem 6.
IfR=∏j=1n‍Rj, where Rj is a finite local principal ideal ring for each j, then α(G(R))=n.Proof.
IfR is local, then α(G(R))=1=n. So assume that n>1. Suppose that M is an independent set of G(R). Suppose that N=∏j=1n‍Nj∈M, where Nj1≠{0}, Nj2≠{0}, and 1≤j1≠j2≤n. Define X=∏j=1n‍Xj, where Xj1=Nj1, Xj={0} for j≠j1 and Y=∏j=1n‍Yj, where Yj2=Nj2, Yj={0} for j≠j2. Observe that (M-{N})∪{X,Y} is an independent set whose cardinality is greater than the cardinality of M. So, an independent set with maximum cardinality cannot contain any element N=∏j=1n‍Nj, where Nj≠{0} for two or more indices. Thus I={∏j=1n‍Ij:Ik=RkforexactlyonekandIj={0}forj≠k} is an independent set with maximum cardinality in G(R) and α(G(R))=n.


## 6. Geodetic and Hull Numbers ofG(R)

Letu,v be two vertices of a connected graph Γ. A shortest path between u and v is called a u-vgeodesic. The set of all vertices in Γ that lie on a u-v geodesic is denoted by I[u,v]. For any subset S of G, let I[S]=⋃u,v∈S‍I[u,v]. If I[S]=V(Γ), then S is called a geodetic set of Γ. The minimum cardinality of a geodetic set of Γ is called thegeodetic number of Γ and this number is denoted by g(Γ). A subset S of V(Γ) isconvex if I[S]=S. If A is a subset of V(Γ), then theconvex hull of A (denoted by [A]) is the smallest convex set in Γ containing A. If [A]=V(Γ), then A is called ahull set of Γ. The smallest cardinality of a hull set of Γ is called thehull number of Γ and is denoted by h(Γ). It is clear that h(Γ)≤g(Γ) for any graph Γ. In this section, we find the geodetic and hull numbers of G(R).A vertexv in a graph Γ is called anextreme vertex if the subgraph induced by its neighbors is complete. The set of extreme vertices of Γ is denoted by Ext⁡(Γ). The following lemma can be found in [11] or [12].Lemma 7.
Every geodetic set (resp., hull set) ofΓ contains Ext(Γ).Note that ifR is a local ring with Nilpotency(R)=n, then G(R) is complete, and so g(G(R))=n-1=h(G(R)). The following lemma determines the set of extreme vertices of G(R), where R is a nonlocal ring.Lemma 8.
LetR be a product of finite local principal ideal rings, and T={∏j=1n‍Ij:Ij≠{0}forexactlyonej}. Then T is the set of extreme vertices of G(R).Proof.
LetL=∏j=1n‍Lj and K=∏j=1n‍Kj be two vertices that are adjacent to I=∏j=1n‍Ij, where Ij1≠{0} for j1 only. Then Lj1≠{0} and Kj1≠{0}. Thus L and K are adjacent. Hence the subgraph induced by the neighbors of I is a complete subgraph and I is an extreme vertex. Let N=∏j=1n‍Nj, where Nj2≠{0}, Nj3≠{0}, with 1≤j2≠j3≤n. Define X=∏j=1n‍Xj, where Xj2=Nj2, Xj={0} for j≠j2 and Y=∏j=1n‍Yj, where Yj3=Nj3, Yj={0} for j≠j3. Observe that X and Y are not adjacent, but both are adjacent to N. So, N is not an extreme vertex of G(R). Thus, T is the set of extreme vertices of G(R).Theorem 9.
LetR be a product of finite local principal ideal rings. Then the set T={∏j=1n‍Ij:Ij≠{0}forexactlyonej} is a geodetic set of G(R) and g(G(R))=∑j=1n‍nj, where nj=Nilpotency(Rj).Proof.
LetN=∏j=1nNj be any vertex of G(R)-T. Then there are j1 and j2, where Nj1≠{0}, Nj2≠{0}, with 1≤j1≠j2≤n. Take X,Y∈T, where Xj1=Nj1 and Yj2=Nj2. Then X_N_Y is a geodesic that contains N. Use Lemmas 7 and 8 to get the result.According to Theorem9 and Lemmas 7 and 8, the hull number of G(R) is equal to the geodetic number of G(R). Also the set T is a hull set of G(R). We state that in the following corollary.Corollary 10.
LetR be a product of finite local principal ideal rings. Then the set T={∏j=1n‍Ij:Ij≠{0}forexactlyonej} is a hull set of G(R) and h(G(R))=∑j=1n‍nj, where nj=Nilpotency(Rj).


## 7. When IsG(R) Chordal?

A graphΓ ischordal if it has no induced cycle of length greater than 3.Theorem 11.
LetR be a finite principal ideal ring which is not a field. The graph G(R) is chordal if and only if R is the product of at most three local rings.Proof.
IfR=∏j=1nRj, where n≥4 and Rj is a local ring for each j, then the four vertices R1×R2×∏j=3n{0}, R1×{0}×R3×∏j=4n{0}, {0}×{0}×R3×R4×∏j=5n{0}, and {0}×R2×{0}×R4×∏j=5n{0} induce a 4-cycle G(R). Thus G(R) is not chordal. If R is local with nilpotency n, then G(R)=Kn-1 which is chordal. So, suppose that R=∏j=1nRj, where 2≤n≤3 and Rj is a local ring for each j. Assume to the contrary that I_J_T_M_⋯_I is an induced cycle in G(R) of length greater than 3. Since J=∏j=1nJj is adjacent to the two nonadjacent vertices I and T, we must have two different values j1 and j2 such that Jji≠{0} for i=1,2, Ij1≠{0}=Tj1, and Ij2={0}≠Tj2. Since J is not adjacent to M and we have at most three factors, we must have Jj={0} for j∉{j1,j2}. Similarly, because T is adjacent to the two nonadjacent vertices J and M, we must have n=3, Tj3≠{0} for j3∉{j1,j2}, Mji={0} for i=1,2, and Mj3≠{0}. But now, the other neighbor of M in this cycle must be adjacent to T (they have nontrivial intersection in the factor j3). This contradicts the assumption that this cycle is induced in G(R). Therefore G(R) is chordal.


## 8. The Complement ofG(R)

ThecomplementΓ¯ of a graph Γ is the graph on the same vertex set as Γ but two vertices are adjacent in Γ¯ if and only if they are nonadjacent in Γ.Theorem 12.
LetR be a finite principal ideal ring which is not a field. The graph G(R)¯ is connected if and only if R is either local with nilpotency 2 or a product of fields.Proof.
IfR is local with nilpotency n, then G(R)=Kn-1. Thus G(R)¯ is connected if and only if n=2 (note that, for n=2, G(R)¯=G(R)=K1). If R=∏j=1nRj, where n≥2, Rj is a local ring for each j, and there exists j0∈{1,2,…,n} such that Rj0 has nilpotency greater than 1, then the vertex I=∏j=1nIj with Ij0∉{{0},Rj0} and Ij=Rj for j≠j0 is adjacent to every other vertex of G(R).Thus I is an isolated vertex in G(R)¯. Therefore G(R)¯ is disconnected (note that the order of G(R)¯ is greater than 1). Finally, suppose that R=∏j=1nFj, where n≥2 and Fj is a field for each j. If n=2, then G(R)=2K1 and hence G(R)¯=K2 is connected. So, assume that n≥3. Let I=∏j=1nIj and J=∏j=1nJj be two distinct nonadjacent vertices of G(R)¯. Then I and J are adjacent in G(R). Thus there exists j1 such that Ij1=Jj1=Fj1. But since I≠J, there exists j2≠j1 such that Ij2≠Jj2; say Ij2={0} and Jj2=Fj2. Take the vertex T=∏j=1nTj with Tj2=Fj2 and Tj={0} for j≠j2. Clearly, T is adjacent in G(R)¯ to I but not to J. But there exists j3∉{j1,j2} such that Jj3={0} (since J≠∏j=1nFj). Now take the vertex M=∏j=1nMj with Mj3=Fj3 and Mj={0} for j≠j3. Then M is adjacent in G(R)¯ to both T and J. Thus I_T_M_J is an I-J path in G(R)¯. Therefore G(R)¯ is connected.The next two results determine the diameter and radius ofG(R)¯.The diameter of a graphΓ and its complement are some times related; for instance, if a graph Γ is disconnected, and so diam⁡(Γ)=∞, then its complement Γ¯ is connected with diam⁡(Γ¯)≤2, while if Γ is connected with diam⁡(Γ)≥3, then Γ¯ is connected and diam⁡(Γ¯)≤3. One now can compare Lemma 4 concerning diam⁡(G(R)) with the next theorem.Theorem 13.
LetR be a finite principal ideal ring which is not a field. Then
(5)diam⁡(G(R)¯)={0ifRislocalwithNilpotency(R)=21ifRisaproductoftwofields3ifRisaproductofmorethantwofields∞otherwise.Proof.
IfR is neither local with nilpotency 2 nor a product of fields, then, by Theorem 12, we have that G(R)¯ is disconnected and hence has infinite diameter. If R is local with nilpotency 2 or a product of two fields, then G(R)¯=K1 or K2, respectively. So, let R=∏j=1nFj, where n≥3 and Fj is a field for each j. For any two nonadjacent vertices of G(R)¯ we have found in the proof of Theorem 12 a path of length 3 joining them. Thus diam⁡(G(R)¯)≤3. Take the two nonadjacent vertices I=∏j=1nIj and J=∏j=1nJj of G(R)¯ with In={0}, Ij=Fj for j≠n, J1={0}, and Jj=Fj for j≠1. Then the vertex ∏j=1nMj with Mn=Fn and Mj={0} for j≠n is the unique neighbor of I in G(R)¯. Similarly, the vertex ∏j=1nTj with T1=F1 and Tj={0} for j≠1 is the unique neighbor of J in G(R)¯. Thus I and J have no common neighbor and hence dG(R)¯(I,J)>2. Therefore, since diam⁡(G(R)¯)≤3, we have diam⁡(G(R)¯)=dG(R)¯(I,J)=3.In Theorem3, the radius of G(R) was calculated, and here we calculate ρ(G(R)¯).Theorem 14.
LetR be a finite principal ideal ring which is not a field. Then
(6)ρ(G(R)¯)={0ifRislocalwithNilpotency(R)=21ifRisaproductoftwofields2ifRisaproductofmorethantwofields∞otherwise.Proof.
IfR is neither local with nilpotency 2 nor a product of fields, then, by Theorem 12, we have that G(R)¯ is disconnected and hence has infinite radius. If R is local with nilpotency 2 or a product of two fields, then G(R)¯=K1 or K2, respectively. So, let R=∏j=1nFj, where n≥3 and Fj is a field for each j. Then, since diam⁡(G(R)¯)=3 by Theorem 13, we have ρ(G(R)¯)=2 or 3. To conclude that ρ(G(R)¯)=2, it is enough to find a vertex with eccentricity 2. Take the vertex I=∏j=1nIj with I1=F1 and Ij={0} for j≠1. Let J=∏j=1nJj be another vertex of G(R)¯. If J1={0}, then dG(R)¯(I,J)=1. So, suppose that J1=F1. Then, since J≠I, there exists j1≠1 such that Jj1=Fj1. But there must be j2∉{j1,1} such that Jj2={0}. Now, take the vertex T=∏j=1nTj with Tj2=Fj2 and Tj={0} for j≠j2. Then T is a common neighbor of I and J. Therefore e(I)=2.Now we determineCen(G(R)¯).Theorem 15.
LetR be a finite principal ideal ring which is not a field. Then G(R)¯ is self-centered except when R is a product of n fields with n≥3. If R=∏j=1nFj is a product of n fields with n≥2, then the vertex set of Cen(G(R)¯) is
(7){∏j=1n‍Ij:Ij={0}exceptforonevalueofj},Cen(G(R)¯)=Kn.Proof.
IfR is not a product of fields, then G(R)¯ has equal diameter and radius by Theorems 13 and 14. Thus G(R)¯ is self-centered. If R is a product of two fields, then G(R)¯=K2 which is also self-centered. So, suppose that R=∏j=1nFj, where n≥3 and Fj is a field for each j. By the same process as that in the proof of Theorem 14, we can show that any vertex of the form ∏j=1nIj, with Ij={0} except for one value of j, has eccentricity 2. Thus each such vertex is a center of G(R)¯. Now let I=∏j=1nIj with Ij≠{0} for at least two values j1 and j2 of j. This means that Iji=Fji for i=1,2. Take the vertex J=∏j=1nJj with Jj1={0} and Jj=Fj for j≠j1. Then the unique neighbor of J in G(R)¯ is T=∏j=1nTj with Tj1=Fj1 and Tj={0} for j≠j1. Clearly, T is not adjacent to I in G(R)¯, and hence dG(R)¯(I,J)>2. Thus e(I)>2, which implies that I is not a center of G(R)¯. Therefore the vertex set of Cen(G(R)¯) is {∏j=1nIj:Ij={0}exceptforonevalueofj}. But any two of the n vertices of Cen(G(R)¯) are adjacent in G(R)¯. Thus Cen(G(R)¯)=Kn.


## 9. ColoringG(R)¯

Aclique of a graph Γ is a maximal complete subgraph of Γ. Theclique number of Γ, denoted by ω(Γ), is the largest possible size of a maximum clique in Γ. Aproper coloring of a graph Γ is a function that assigns a color to each vertex such that no two adjacent vertices have the same color. Thechromatic number of Γ, denoted by χ(Γ), is the smallest number of colors necessary to produce a proper coloring.It is clear that ifR is a local ring with Nilpotency(R)=n, then G(R)¯ is (n-1)K1, and so χ(G(R))=1=ω(Γ).Theorem 16.
LetS be a finite principal ideal ring which is not a field, and let R1 be a finite local principal ideal ring. Then χ(G(R1×S)¯)=χ(G(S)¯)+1.Proof.
The vertex set ofG(R1×S)¯ is the disjoint union of the following five sets:
(8)A1={{0}×S},A2={{0}×J:JisapropernontrivialidealinS},A3={I×{0}:IisanontrivialidealinR1},A4={I×J:IisanontrivialidealinR1,JisapropernontrivialidealinS},A5={I×S:IisapropernontrivialidealinR1}.
Let χ(G(S)¯)=k. Then the vertices in A2 can be colored by k colors. The vertex R1×{0} is adjacent to all vertices from A2, so R1×{0} needs necessarily a new color ck+1. This new color ck+1 can be assigned to every other vertex from A3, since A3 is an independent set in G(R1×S)¯. The vertex {0}×S can be colored by one of the previous colors except ck+1, since the neighbors of {0}×Sare precisely the elements of A3. For each proper nontrivial ideal J in S, assign the color of {0}×J to each of the vertices I×J, where I is a nontrivial ideal in R1. Note that the neighbors of I×J in G(R1×S)¯ are precisely the neighbors of {0}×J in A2. This completes coloring the vertices from A4. Finally, since every element of A5 is an isolated vertex in G(R1×S)¯, there is no need for any new color. Therefore, χ(G(R1×S)¯)=k+1.Corollary 17.
LetR=∏j=1nRj, where each Rj is a finite local principal ideal ring. Then χ(G(R)¯)=n.Proof.
Forn=1, we have G(R)¯=K1. Thus χ(G(R)¯)=1. For R=F1×F2, where both F1 and F2 are fields, we have G(R)¯=K2. Thus χ(G(F1×F2)¯)=2. Therefore, the result follows by induction and Theorem 16.Note that forR=∏j=1nRj, where each Rj is a finite local principal ideal ring, we have ω(G(R)¯)=α(G(R))=n by Theorem 6. Thus, by Corollary 17, χ(G(R)¯)=n=ω(G(R)¯).


## 10. BetweenG(R) and the Zero Divisors

After a conversation between the first author and Professor Christian Lomp (Porto University, Portugal), the latter suggested that there may be a relation between the graphG(R)¯ and the zero divisor graph Γ(R), since if (a)∩(b)={0}, then clearly ab=0. In fact we manage to find a very nice result but with the graph ΓE(R).In [13] the zero divisor graph determined by equivalence classes of zero divisors of a commutative Noetherian ring R was introduced as follows.LetZ(R) be the set of zero divisor elements in R and let Z*(R)=Z(R)∖{0}. For x,y∈R, x~y if ann(x)=Ann(y). This relation is an equivalence relation, and a well-defined multiplication was defined on the set of equivalence classes of ~; that is, if [x] denotes the class of x, then the product [x]·[y]=[xy], and they note that [0]={0} and [1]=R∖Z(R) and the other equivalence classes form a partition of Z*(R). The authors defined the graph of equivalence classes ΓE(R) of elements of Z*(R) and with pair of distinct classes [x], [y] joined by an edge if and only if [x]·[y]=[0].Theorem 18.
IfR is a reduced finite principal ideal ring, then ΓE(R)≃G(R)¯.Proof.
SinceR is reduced it is a product of fields and so for any two ideals I and J of R, I∩J=IJ. Thus we have the following for each a,b∈Z*(R).
a
R is adjacent to bR in Z*(R)⇔aR∩bR={0},⇔aRbR={0},⇔ab=0,⇔[a] is adjacent to [b] in ΓE(R).
Thus if we defineφ:G(R)¯→ΓE(R) such that φ(aR)=[a], then φ is an isomorphism.IfR is not reduced in the above theorem, then the result needs not to be true as one can see that ΓE(Z12)≄G(Z12)¯.



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*Source: 952371-2014-09-25.xml*