On the intersection graph of ideals of a commutative ring Keywords: Intersection graph, perfect graph, clique number, chromatic number, diameter, girth.   2010 Mathematics Subject Classification: 05C15, 05C17, 05C69, 13A99, 13C99.

# On the intersection graph of ideals of a commutative ring 1

## Abstract

Let be a commutative ring and be an -module, and let be the set of all non-trivial ideals of . The -intersection graph of ideals of , denoted by , is a graph with the vertex set , and two distinct vertices and are adjacent if and only if . For every multiplication -module , the diameter and the girth of are determined. Among other results, we prove that if is a faithful -module and the clique number of is finite, then is a semilocal ring. We denote the -intersection graph of ideals of the ring by , where are integers and is a -module. We determine the values of and for which is perfect. Furthermore, we derive a sufficient condition for to be weakly perfect.

## 1 Introduction

Let be a commutative ring, and be the set of all non-trivial ideals of . There are many papers on assigning a graph to a ring , for instance see [1–4]. Also the intersection graphs of some algebraic structures such as groups, rings and modules have been studied by several authors, see [3, 6, 8]. In [6], the intersection graph of ideals of , denoted by , was introduced as the graph with vertices and for distinct , the vertices and are adjacent if and only if . Also in [3], the intersection graph of submodules of an -module , denoted by , is defined to be the graph whose vertices are the non-trivial submodules of and two distinct vertices are adjacent if and only if they have non-zero intersection. In this paper, we generalize to , the -intersection graph of ideals of , where is an -module.
Throughout the paper, all rings are commutative with non-zero identity and all modules are unitary. A module is called a uniform module if the intersection of any two non-zero submodules is non-zero. An -module is said to be a multiplication module if every submodule of is of the form , for some ideal of . The annihilator of is denoted by . The module is called a faithful -module if . By a non-trivial submodule of , we mean a non-zero proper submodule of . Also, denotes the Jacobson radical of and denotes the ideal of all nilpotent elements of . By , we denote the set of all maximal ideals of . A ring having only finitely many maximal ideals is said to be a semilocal ring. As usual, and will denote the integers and the integers modulo , respectively.
A graph in which any two distinct vertices are adjacent is called a complete graph. We denote the complete graph on vertices by . A null graph is a graph containing no edges. Let be a graph. The complement of is denoted by . The set of vertices and the set of edges of are denoted by and , respectively. A subgraph of is said to be an induced subgraph of if it has exactly the edges that appear in over . Also, a subgraph of is called a spanning subgraph if . Suppose that . We denote by the degree of a vertex in . A regular graph is a graph where each vertex has the same degree. We recall that a walk between and is a sequence  —  —  —  of vertices of such that for every with , the vertices and are adjacent. A path between and is a walk between and without repeated vertices. We say that is connected if there is a path between any two distinct vertices of . For vertices and of , let be the length of a shortest path from to ( and if there is no path between and ). The diameter of , , is the supremum of the set . The girth of , denoted by , is the length of a shortest cycle in ( if contains no cycles). A clique in is a set of pairwise adjacent vertices and the number of vertices in the largest clique of , denoted by , is called the clique number of . The chromatic number of , , is the minimal number of colors which can be assigned to the vertices of in such a way that every two adjacent vertices have different colors. A graph is perfect if for every induced subgraph of , . Also, is called weakly perfect if .
In the next section, we introduce the -intersection graph of ideals of , denoted by , where is a commutative ring and is a non-zero -module. It is shown that for every multiplication -module , and . Among other results, we prove that if is a faithful -module and is finite, then and . In the last section, we consider the -intersection graph of ideals of , denoted by , where are integers and is a -module. We show that is a perfect graph if and only if has at most four distinct prime divisors. Furthermore, we derive a sufficient condition for to be weakly perfect. As a corollary, it is shown that the intersection graph of ideals of is weakly perfect, for every integer .

## 2 The M-intersection graph of ideals of R

In this section, we introduce the -intersection graph of ideals of and study its basic properties.

Definition. Let be a commutative ring and be a non-zero -module. The -intersection graph of ideals of , denoted by , is the graph with vertices and two distinct vertices and are adjacent if and only if .

Clearly, if is regarded as a module over itself, that is, , then the -intersection graph of ideals of is exactly the same as the intersection graph of ideals of . Also, if and are two isomorphic -modules, then is the same as .

###### Example 1

. Let . Then we have the following graphs.

{tikzpicture}\GraphInit

[vstyle=Classic] \Vertex[x=1,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=1,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]2 \Vertex[x=2.3,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6 \Vertex[x=2.3,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3 \Edges(2,3) \Edges(6,3) \Edges(2,6) \Edges(2,4)

{tikzpicture}\GraphInit

[vstyle=Classic] \Vertex[x=1,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=1,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]2 \Vertex[x=2.3,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6 \Vertex[x=2.3,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3

{tikzpicture}\GraphInit

[vstyle=Classic] \Vertex[x=1,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=1,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]2 \Vertex[x=2.3,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6 \Vertex[x=2.3,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3 \Edges(2,4)

{tikzpicture}\GraphInit

[vstyle=Classic] \Vertex[x=1,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=1,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]2 \Vertex[x=2.3,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6 \Vertex[x=2.3,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3 \Edges(2,3) \Edges(6,3) \Edges(2,6)

###### Example 2

. Let be an integer. If is the least common multiple of two distinct integers , then . Thus and are adjacent in if and only if does not divide .

###### Example 3

. Let be a prime number and be two positive integers. If divides , then is an isolated vertex of . Therefore, since is a uniform -module, so is a disjoint union of an infinite complete graph and its complement. Also, (the quasi-cyclic -group), is a uniform -module and . Hence is an infinite complete graph.

###### Remark 1

. Obviously, if is a faithful multiplication -module, then is a complete graph if and only if is a uniform -module.

###### Remark 2

. Let be a commutative ring and let be a non-zero -module.

1. If is a faithful -module, then is a spanning subgraph of . To see this, suppose that and are adjacent vertices of . Then implies that and so . Therefore is adjacent to in .

2. If is a multiplication -module, then is an induced subgraph of . Note that for each non-trivial submodule of , there is a non-trivial ideal of , such that and so we can assign to . Also, is adjacent to in if and only if , that is, if and only if is adjacent to in .

###### Theorem 1

. Let be a commutative ring and let be a faithful -module. If is not connected, then is a direct sum of two -modules.

###### Proof.

Suppose that and are two distinct components of . Let and . Since is a faithful -module, so implies that and . Now if , then  —  —  is a path between and , a contradiction. Thus and so .

The next theorem shows that for every multiplication -module , the diameter of has possibilities.

###### Theorem 2

. Let be a commutative ring and be a multiplication -module. Then .

###### Proof.

Assume that is a connected graph with at least two vertices. So is a faithful module. If there is a non-trivial ideal of such that , then is adjacent to all other vertices. Hence . Otherwise, we claim that is connected. Let and be two distinct vertices of . Since is a multiplication module, so and , for some non-trivial ideals and of . Suppose that  —  —  —  is a path between and in . Therefore,  —  —  —  —  is a walk between and . Thus, we conclude that there is also a path between and in . The claim is proved. So by [3, Theorem 2.4], . Now, suppose that and are two distinct vertices of . If , then and are two distinct vertices of . Hence there exists a non-trivial submodule of which is adjacent to both and in . Since is a multiplication module, so , for some non-trivial ideal of . Thus is adjacent to both and in . Therefore .

###### Theorem 3

. Let be a commutative ring and be a multiplication -module. If is a connected regular graph of finite degree, then is a complete graph.

###### Proof.

Suppose that is a connected regular graph of finite degree. If , then . So assume that . We claim that is an Artinian module. Suppose to the contrary that is not an Artinian module. Then there is a descending chain of submodules of , where ’s are non-trivial ideals of . This implies that is infinite, a contradiction. The claim is proved. Therefore has at least one minimal submodule. To complete the proof, it suffices to show that contains a unique minimal submodule. By contrary, suppose that and are two distinct minimal submodules of . Hence and , where and are two non-trivial ideals of . Since , so and are not adjacent. By Theorem 2, there is a vertex which is adjacent to both and . So both and are contained in . Thus each vertex adjacent to is adjacent to too. This implies that , a contradiction.

Also, the following theorem shows that for every multiplication -module , the girth of has possibilities.

###### Theorem 4

. Let be a commutative ring and be a multiplication -module. Then .

###### Proof.

Suppose that  —  —  —  —  is a cycle of length in . If , we are done. Thus assume that . Since and is a multiplication module, we have , where is a non-zero ideal of . If is a proper ideal of and , then  —  —  —  is a triangle in . Otherwise, we conclude that or . Similarly, we can assume that or , for every , . Without loss of generality suppose that . Now, if , then  —  —  —  is a cycle of length 3 in . Therefore assume that . Since or , so  —  —  —  is a triangle in . Hence if contains a cycle, then .

###### Lemma 1

. Let be a commutative ring and be a non-zero -module. If is an isolated vertex of , then the following hold:

1. is a maximal ideal of or .

2. If , then , for every .

###### Proof.

There is a maximal ideal of such that . Assume that . Then we have , since is an isolated vertex. So .
Suppose that and . Since is an isolated vertex, we have and so , a contradiction. Thus .

###### Theorem 5

. Let be a commutative ring and be a faithful -module. If is a null graph, then it has at most two vertices and is isomorphic to one of the following rings:

1. , where and are fields;

2. , where is a field;

3. , where is a coefficient ring of characteristic , for some prime number .

###### Proof.

By Lemma 1, every non-trivial ideal of is maximal and so by [10, Theorem 1.1], cannot have more than two different non-trivial ideals. Thus has at most two vertices. Also, by [11, Theorem 4], is isomorphic to one of the mentioned rings.

In the next theorem we show that if is a faithful -module and , then is a semilocal ring.

###### Theorem 6

. Let be a commutative ring and be a faithful -module. If is finite then and .

###### Proof.

First we prove that . Let . By contradiction, assume that are distinct maximal ideals of . We know that , for every , . Otherwise, , for some , . So and hence by Prime Avoidance Theorem [5, Proposition 1.11], we have , for some , , which is impossible. This implies that is a clique in , a contradiction. Thus .
Now, we prove that . By contrary, suppose that . Since , for every , and is finite, we conclude that , for some integers . Hence , for some . Since , so is a unit. This yields that , a contradiction. The proof is complete.

## 3 The Zn-intersection graph of ideals of Zm

Let be two integers and be a -module. In this section we study the -intersection graph of ideals of the ring . Also, we generalize some results given in [9]. For abbreviation, we denote by . Clearly, is a -module if and only if divides .
Throughout this section, without loss of generality, we assume that and , where ’s are distinct primes, ’s are positive integers, ’s are non-negative integers, and for . Let and . The cardinality of is denoted by . For two integers and , we write () if divides ( does not divide ).
First we have the following remarks.

###### Remark 3

. It is easy to see that divides and . Let be a -module. If , then is an isolated vertex of . Obviously, and are adjacent if and only if . This implies that is a subgraph of .

###### Remark 4

. Let be a -module and be a divisor of . We set . Clearly, . Suppose that is a clique of . Then is an intersecting family of subsets of . (A family of sets is intersecting if any two of its sets have a non-empty intersection.) Also, if is an intersecting family of subsets of and is non-empty, then is a clique of . (If is a non-empty subset of and , then we will denote by .) Thus we have

Now, we provide a lower bound for the clique number of .

###### Theorem 7

. Let be a -module. Then

###### Proof.

Suppose that . With the notations of the previous remark, let . Then is an intersecting family of subsets of and so is a clique of . Clearly, . Therefore and hence the result holds.

Clearly, if , then equality holds in the previous theorem. Also, if has only two distinct prime divisors, that is, , then again equality holds. So the lower bound is sharp.

###### Example 4

. Let , where are distinct primes. Thus and . It is easy to see that and . Also, . Let , for . Hence , for . If , then . Therefore .

By the strong perfect graph theorem, we determine the values of and for which is a perfect graph.

###### Theorem A

. (The Strong Perfect Graph Theorem [7]) A finite graph is perfect if and only if neither nor contains an induced odd cycle of length at least .

###### Theorem 8

. Let be a -module. Then is perfect if and only if has at most four distinct prime divisors.

###### Proof.

First suppose that and , where ’s are distinct primes and ’s are positive integers. Let , , , , and . Now, assume that , for . Hence  —  —  —  —  —  is an induced cycle of length 5 in . So by Theorem A, is not a perfect graph.
Conversely, suppose that is not a perfect graph. Then by Theorem A, we have the following cases:
Case 1.  —  —  —  —  —  is an induced cycle of length 5 in . Let , for . So and , for . Let and , for . Clearly, are distinct and thus .
Case 2.  —  —  —  —  —  is an induced path of length 5 in . Let , for . So , for . Let , for . Clearly, are distinct and hence .
Case 3. There is an induced cycle of length 5 in . So contains an induced cycle of length 5 and by Case 1, we are done.
Case 4.  —  —  —  —  —  is an induced path of length 5 in . Since , and , we may assume that , where and , for some distinct . Similarly, we find that , for some distinct and also . Now, since and , we deduce that .

###### Corollary 1

. The graph is perfect if and only if has at most four distinct prime divisors.

In the next theorem, we derive a sufficient condition for to be weakly perfect.

###### Theorem 9

. Let be a -module. If for each , then is weakly perfect.

###### Proof.

Let be a non-empty subset of and . As we mentioned in Remark 4, if is non-empty, then is a clique of . Also, the vertices of (if ) are adjacent to all non-isolated vertices. Suppose that and are two non-empty subsets of and . Since for each , so . This implies that and hence .
Let be an intersecting family of subsets of and . Let . We show that or . Assume that . So there is such that . Thus and hence . We claim that , for each . Suppose to the contrary, and . If and , then . So we have . Let . Then is an intersecting family of subsets of and , a contradiction. The claim is proved.
Now, we show that has a proper -vertex coloring. First we color all vertices of with different colors. Next we color each family of vertices out of with colors of vertices of . Note that if , then and . Suppose that and are two adjacent vertices of . Thus . Without loss of generality, one can assume . So we deduce that and . Therefore, and have different colors. Thus and hence .

As an immediate consequence of the previous theorem, we have the next result.

###### Corollary 2

. The graph is weakly perfect, for every integer .

In the case that for each , we determine the exact value of . It is exactly the lower bound obtained in the Theorem 7.

###### Theorem 10

. Let be a -module. If for each , then .

###### Proof.

Let be a proper subset of . Then and hence . Also, the vertices of (if ) are adjacent to all non-isolated vertices and . Clearly if is an intersecting family of subsets of , then . Moreover, if and , then . Thus by Theorem 9, .

###### Corollary 3

. Let , where ’s are distinct primes. Then .

Problem. Let be a -module. Then is it true that is a weakly perfect graph?

### Footnotes

1. thanks: Keywords: Intersection graph, perfect graph, clique number, chromatic number, diameter, girth.
2010 Mathematics Subject Classification: 05c15, 05c17, 05c69, 13a99, 13c99.

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