# Planar, outerplanar and ring graph

of the intersection graph
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## Abstract

Let be two integers, and be a -module. Let be the set of all non- zero proper ideals of . The -intersection graph of , denoted by is a graph with the vertex set , and two distinct vertices and are adjacent if and only if . In this paper, we determine the values of and for which is planar, outerplanar or ring graph.

## 1 Introduction

Let be a commutative ring, and be the set of all non-zero proper ideals of . There are many papers on assigning a graph to a ring , for instance see [1], [2], [5], [6] and [13]. Also the intersection graphs of some algebraic structures such as groups, rings and modules have been studied by several authors, see [3, 4, 8, 9]. In [8], 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 [4], the intersection graph of submodules of an -module , denoted by , is defined to be the graph whose vertices are the non-zero proper submodules of and two distinct vertices are adjacent if and only if they have non-zero intersection. In [11], the intersection graph of ideals of denoted by , is defined to be the graph with the vertex set , and two distinct vertices and are adjacent if and only if . Also, the intersection graph of , was studied in [12], where are integers and is a -module. Clearly, if , then is exactly the same as the intersection graph of ideals of . This implies that is a generalization of . As usual, denotes the integers modulo .

Now, we recall some definitions and notations on graphs. Let be a graph with the vertex set and the edge set . If , we say is adjacent to and write — . If , then a path from to is a series of adjacent vertices — — — — — . For with , denotes the length of a shortest path from to . If there is no such path, then we define . We say that is connected if there is a path between any two distinct vertices of . A cycle is a path that begins and ends at the same vertex in which no edge is repeated and all vertices other than the starting and ending vertex are distinct. We denote the complete graph of order by . A graph is bipartite if its vertices can be partitioned into two disjoint subsets and such that each edge connects a vertex from to one from . A bipartite graph is a complete bipartite graph if every vertex in is adjacent to every vertex in . We denote the complete bipartite graph, with part sizes and by . The disjoint union of graphs and , which is denoted by , where and are two vertex-disjoint graphs, is a graph with and . A graph may be expressed uniquely as a disjoint union of connected graphs. These graphs are called the connected components, or simply the components, of . Recall that a graph is said to be planar if it can be drawn in the plane so that its edges intersect only at their ends. A subdivision of a graph is any graph that can be obtained from the original graph by replacing edges by paths.

In [12], the authors were mainly interested in the study of intersection graph of ideals of . For instance, they determined the values of for which is connected, complete, Eulerian or has a cycle. In this article, we determine all integer numbers and for which is planar, outerplanar or ring graph.

## 2 Results

Let be integers and be a -module. Clearly, is a -module if and only if divides . Throughout the paper, 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 , . The cardinality of is denoted by . Also, we denote the least common multiple of integers and by . We write () if divides ( does not divide ). It is easy to see that divides and . If , then is an isolated vertex of . Obviously, and are adjacent in if and only if .

In this section, we study the planarity of . A remarkable simple characterization of the planar graphs was given by Kuratowski in 1930.

###### Theorem A

. [7, Theorem 10.30] (Kuratowski’s Theorem) A graph is planar if and only if it contains no subdivision of either or .

###### Theorem 1

. Let be a -module. Then is planar if and only if one of the following holds:

and .

and .

.

.

and .

.

and .

.

###### Proof.

One side is obvious. For the other side assume that is planar. With no loss of generality assume that , where and , for . We note that . Since otherwise, forms a , a contradiction. Consider three following cases:

Case 1. . If , then forms a , a contradiction. Hence . If , then forms a , a contradiction. Therefore . There are three following subcases:

Subcase 1. . Clearly, is planar, where . Therefore holds.

Subcase 2. . We note that . Otherwise, forms a , a contradiction. If , then forms a , a contradiction. This implies that . If , then it is easy to check that is planar, where . Therefore holds. Now, let . If , then forms a , a contradiction. Therefore . Obviously, is planar and holds.

Subcase 3. . If , then forms a , a contradiction. Hence . Similarly, we find that . If , then forms a , a contradiction. It is easy to see that is planar and holds.

Case 2. . If , then forms a , a contradiction. Hence with no loss of generality we may assume that .

First assume that . If , then forms a , a contradiction. Therefore . If , then forms a , a contradiction. Thus . If , then forms a , a contradiction. Therefore . If , then forms a , a contradiction. Therefore . It is easy to see that is planar, where and holds.

Now, assume that . Let . If , then forms a , a contradiction. Hence . If , then forms a , a contradiction. Therefore and similarly, . It is clear that is planar and holds. Now, assume that . Then we find that . Also, one can easily check that is planar, where . Therefore holds.

Case 3. . If , then forms a , a contradiction. Therefore . If , then forms a , a contradiction. Therefore and similarly . If , then forms a , a contradiction. Therefore and similarly we have . If , then forms a , a contradiction. Therefore or . With no loss of generality, we may assume that . By The same argument as we saw we find that or . Thus we have or and . It is clear that if , then is planar. Therefore holds. Also, is not planar because forms a which is impossible.

###### Corollary 1

. Let be a positive integer number. Then is planar if and only if .

Let be a graph with vertices and edges. We recall that a chord is any edge of joining two nonadjacent vertices in a cycle of . Let be a cycle of . We say is a primitive cycle if it has no chords. Also, a graph has the primitive cycle property (PCP) if any two primitive cycles intersect in at most one edge. The number is called the free rank of and it is the number of primitive cycles of . Also, the number is called the cycle rank of , where is the number of connected components of . The cycle rank of G can be expressed as the dimension of the cycle space of . By [10, Proposition 2.2], we have . A graph is called a ring graph if it satisfies in one of the following equivalent conditions (see [10, Theorem 2.13]).

,

satisfies the PCP and does not contain a subdivision of as a subgraph. Also, an undirected graph is an outerplanar graph if it can be drawn in the plane without crossings in such a way that all of the vertices belong to the unbounded face of the drawing. There is a characterization for outerplanar graphs that says a graph is outerplanar if and only if it does not contain a subdivision of or . By [10, Proposition 2.17], we find that every outerplanar graph is a ring graph and every ring graph is a planar graph.

Example 1. In Fig.1, let . Also, in Fig.2, assume that

{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=3,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]5 \Vertex[x=0,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=0,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]2 \Vertex[x=1.5,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6 \Vertex[x=3,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3 \Vertex[x=1.5,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]1 \Edges(3,5) \Edges(5,6) \Edges(3,6) \Edges(6,4) \Edges(2,4) \Edges(2,6)

Fig.1 Fig.2

Example 2. In Fig.3, let . In Fig.4, let .

[vstyle=Classic] \Vertex[x=3,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]5 \Vertex[x=0,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=0,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]2 \Vertex[x=1.5,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6 \Vertex[x=1.5,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]1 \Vertex[x=3,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3 \Edges(3,5) \Edges(5,6) \Edges(3,6) \Edges(6,4) \Edges(2,4) \Edges(2,6)

{tikzpicture}\GraphInit

[vstyle=Classic] \Vertex[x=2,y=0,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]1 \Vertex[x=3,y=1.5,style=black,minimum size=3pt,LabelOut=true,Lpos=0,L=]2 \Vertex[x=4,y=3,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]3 \Vertex[x=1,y=1.5,style=black,minimum size=3pt,LabelOut=true,Lpos=180,L=]4 \Vertex[x=2,y=3,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]5 \Vertex[x=0,y=3,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]6 \Edges(1,2) \Edges(2,3) \Edges(4,5) \Edges(2,5) \Edges(1,4) \Edges(4,6) \Edges(3,5) \Edges(5,6) \Edges(4,2)

Fig.3 Fig.4

Example 3. In Fig.5, let .

[vstyle=Classic] \Vertex[x=2,y=1,style=black,minimum size=3pt,LabelOut=true,Lpos=90,L=]1 \Vertex[x=1,y=-1,style=black,minimum size=3pt,LabelOut=true,Lpos=180,L=]2 \Vertex[x=3,y=-1,style=black,minimum size=3pt,LabelOut=true,Lpos=0,L=]3 \Vertex[x=0,y=-2,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]4 \Vertex[x=2,y=-2,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]5 \Vertex[x=4,y=-2,style=black,minimum size=3pt,LabelOut=true,Lpos=270,L=]6

\Edges(1,2) \Edges(2,3) \Edges(3,1)

Fig.5

###### Theorem 2

. Let be a -module. Then is a ring graph if and only if one of the following holds:

and .

and .

.

.

and .

.

and .

.

###### Proof.

One side is obvious. For the other side assume that is a ring graph. Then it is planar and by Theorem 1, we have eight following cases:

Case 1. and . If , then forms a , a contradiction. Therefore . It is easy to check that is a ring graph and holds.

Case 2. and . Clearly, if , then forms a , a contradiction. Hence . Now, has at most three non-isolated vertices and so it is a ring graph and holds.

Case 3. . In this case, is the set of all non-isolated vertices of . This yields that is a ring graph and holds.

Case 4. . It is easy to see that is a ring graph and holds.

Case 5. and . If , then forms a , a contradiction. Therefore . By Fig.1 and Fig.2, and are ring graphs and holds.

Case 6. . In this case, by Fig.3, we conclude that is a ring graph and holds.

Case 7. and . If , then forms a , a contradiction. Hence and similarly, . Now, by [12, Theorem 11], we have , where is the set of all non-isolated vertices of . Therefore holds.(see Fig.5)

Case 8. . By Fig.4, we find that is a ring graph and holds.

###### Corollary 2

. Let be a positive integer number. Then is a ring graph if and only if .

Using theorems in this section and the proof of pervious theorem, we have the following result.

###### Theorem 3

. Let be a -module. Then is outerplanar if and only if one of the following holds:

and .

and .

.

.

and .

.

and .

.

Also, we have the following corollary.

###### Corollary 3

. Let be a positive integer number. Then is outerplanar if and only if .

### Footnotes

- thanks: Key words: intersection graph, ring graph, planarity, outerplanarity.

2010 Mathematics Subject Classification: 05c10, 05c25, 13a15.

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