#### Abstract

Let be a commutative ring with identity. In this paper we classify rings such that the complement of comaximal graph of is planar. We also consider the subgraph of the complement of comaximal graph of induced on the set of all nonunits of with the property that each element of is not in the Jacobson radical of and classify rings such that this subgraph is planar.

#### 1. Introduction

All rings considered in this paper are commutative with identity . The graphs considered here are undirected and simple. Sharma and Bhatwadekar in [1] introduced a graph on a commutative ring , whose vertices are the elements of and two distinct vertices and are adjacent if and only if . In [1, Theorem 2.3], the authors showed that if and only if the ring is finite and in this case , where and denote, respectively, the number of maximal ideals of and number of units of . In [2], Maimani et al. named the graph studied by Sharma and Bhatwadekar as the comaximal graph of . Also in [2], the authors studied the subgraphs , and , where is the subgraph of induced on the set of units of , is the subgraph of induced on the set of nonunits of , and is the subgraph of induced on the set of nonunits of which are not in = the Jacobson radical of . Several other researchers also investigated the comaximal graphs of rings [35]. In [6], Gaur and Sharma studied the graph whose vertices are the elements of a ring and two distinct vertices and are adjacent if and only if there exists a maximal ideal of containing both and . They called this graph the maximal graph of the ring .

Let be a simple graph. Recall from [7, Definition 1.1.13] that the complement of is defined by taking and two distinct vertices and are adjacent in if and only if they are not adjacent in . Thus the maximal graph of a ring studied in [6] is the complement of the comaximal graph of a ring studied by Sharma and Bhatwadekar in [1].

For any set denotes the cardinality of . For any ring , and denote, respectively, the set of units of and the set of nonunits of . We denote by and the set of all maximal ideals of by Max(). For the sake of convenience, for any ring with at least two maximal ideals we denote by . For any , denotes the ring of integers modulo . For any prime number and , denotes the finite field with exactly elements.

We first recall the following definitions and results from graph theory. A graph is said to be complete if every pair of distinct vertices of are adjacent in . A complete graph on vertices is denoted by [7, Definition 1.1.11]. A graph is said to be bipartite if the vertex set can be partitioned into two nonempty subsets and such that each edge of has one end in and another one in . The pair () is called a bipartition of . A bipartite graph with bipartition () is denoted by . A bipartite graph is said to be complete if each vertex of is adjacent to all the vertices of . If is a complete bipartite graph with and , then it is denoted by [7, Definition 1.1.12]. Let be a graph. By a clique of , we mean a complete subgraph of [7, Definition 1.2.2]. We say that the clique number of equals if is the largest positive integer such that is a subgraph of [7, page 185]. The clique number of a graph is denoted by the notation . If contains as a subgraph for all , then we set .

A graph is said to be planar if it can be drawn in a plane in such a way that no two edges of intersect in a point other than a vertex of [7, Definition 8.1.1]. Two adjacent edges of a graph are said to be in series if their common vertex is of degree two [8, page 9]. Two graphs are said to be homeomorphic if one graph can be obtained from the other graph by the creation of edges in series (i.e., by insertion of vertices of degree two) or by the merger of edges in series [8, page 100]. Recall from [8, page 93] that is referred to as Kuratowski's first graph and is referred to as Kuratowski's second graph. A celebrated theorem of Kuratowski says that a necessary and sufficient condition for a graph to be planar is that does not contain either of Kuratowski's two graphs or any graph homeomorphic to either of them [8, Theorem 5.9].

Motivated by the interesting theorems proved on comaximal graphs of rings in [15] and also the theorems proved on its complement in [6], in this paper we attempt to characterize commutative rings such that is planar, where is the comaximal graph of a ring . Since any subgraph of a planar graph is planar, it follows that (= ) is planar if is planar. Hence we first focus our study on investigating when is planar and at the end of this paper we classify the rings for which is planar. While investigating when is planar, we assume that has at least two maximal ideals.

In view of Kuratowski's Theorem, [8, Theorem 5.9] we introduce the following definitions.

Let be a graph.(i)We say that satisfies if does not contain as a subgraph.(ii)We say that satisfies if does not contain as a subgraph.(iii)We say that satisfies if satisfies and, moreover, does not contain any subgraph homeomorphic to .(iv)We say that satisfies if satisfies and, moreover, does not contain any subgraph homeomorphic to .

If a graph is planar, then it follows from Kuratowski's theorem [8, Theorem 5.9] that satisfies both and . Hence satisfies both and . It is interesting to note that a graph may be nonplanar even if it satisfies both and . For example of this type refer to [8, Figure 5.9(a), page 101] and the graph in this example does not satisfy . It is not difficult to construct an example of a graph such that satisfies but does not satisfy .

Let be a ring with at least two maximal ideals. We now give a brief summary of the results proved in this paper. It is shown in Lemma 8 that if satisfies or , then . For a ring with exactly two maximal ideals, it is proved in Theorem 19 that satisfies if and only if satisfies if and only if is planar. Moreover, the statement (2) of Theorem 19 describes such rings up to isomorphism. If is a ring with exactly three maximal ideals, then it is shown in Theorem 24 that satisfies if and only if satisfies if and only if is planar if and only if as rings.

Let be a ring with at least two maximal ideals. As is a subgraph of , we obtain from Lemma 8 that if satisfies or , then . It is proved in Theorem 32 that, for a ring with exactly two maximal ideals, satisfies and if and only if satisfies if and only if is planar. Also, in statement (3) of Theorem 32, rings with such that is planar are listed up to isomorphism of rings. In Remark 34, an example of a ring with is provided to illustrate that which satisfies need not imply that it satisfies . Moreover, in Theorem 35, rings with such that satisfies are listed up to isomorphism of rings. For a ring with , it is proved in Theorem 36 that satisfies if only if satisfies if and only if is planar if and only if as rings.

In Section 6, quasilocal rings which are not fields are considered and necessary and sufficient conditions are determined for to be planar. It is proved in Theorem 37 that, for a quasilocal ring which is not a field, satisfies and if and only if satisfies if and only if , , and if and only if is planar. It is observed in Lemma 38 that for a quasilocal ring which is not a field, satisfies if and only if , , and . In Remark 40, up to isomorphism of rings, all local rings which are not fields of order 4,8,9,25 and all local rings with and are mentioned with the help of the theorems proved in [9] and from the results proved in [10].

#### 2. Preliminaries

As is mentioned in the introduction, the rings considered in this paper are commutative with identity . In this section we state and prove several lemmas which are needed for proving the main results of this paper.

Lemma 1. Let be a ring which admits at least one nonzero nonunit. If the set of nonunits of is finite, then the ring is finite.

Proof. Suppose that has an infinite number of units. Let “” be any nonzero nonunit of . Then the set of all nonunits of . As is finite, there exist such that for So for which implies that are distinct nonunits of , for Hence we obtain that is not finite. This is a contradiction. Thus must be finite. Therefore, is a finite ring.

Lemma 2. Let . If the ring is such that , then .

Proof. Suppose that has more than maximal ideals. Let . Let for and . We claim that the subgraph of induced on is a clique. Let . Now but . So and are distinct. Note that for and but . Hence . Let , . Since , . Hence there exists . Note that both and are in . It is clear that and are in for . Observe that both and belong to . Hence the subgraph of induced on is a clique. This is in contradiction to the assumption that . Therefore, .

Lemma 3. Let be a semiquasilocal ring which is not quasilocal. If there exists such that for all , then is finite.

Proof. Let be the set of all maximal ideals of . Note that . In view of Lemma 1, to prove is finite, it is enough to prove that is finite. Observe that . By hypothesis, for . Thus that is finite will follow if we show that is finite. We claim that . Let . Note that and so . Hence . This proves that is finite and so is a finite ring.

Corollary 4. Let . Let be a ring with at least two maximal ideals. If , then is finite.

Proof. By assumption, . Hence by Lemma 2, . Moreover, since for any maximal ideal of , the subgraph of induced on is a clique, it follows that for any . Therefore, we obtain from Lemma 3 that is finite.

Lemma 5. Let be the set of all maximal ideals of a ring R. Then has exactly two components and if for , then is planar.

Proof. Let (resp., ) be the subgraph of induced on (resp., ). Let and . Observe that , and . For any and , . Hence there is no edge of whose one end vertex is in and the other is in . Therefore, there is no path in between any and . Moreover, it is clear that and are complete. This proves that and are the only components of . It is well known that any complete graph on at most four vertices is planar. Hence if for , it follows that is planar.

Lemma 6. Let be a quasilocal ring with . If with is such that , then . In particular, if , then .

Proof. Now . We assert that . Suppose that . Let . Let . Note that for any quasilocal ring , if with , then for any , . Therefore, it follows that . Hence . This is a contradiction to the hypothesis that . Therefore, .
Let . Now . Hence for some , . If , then we obtain and so it follows from the previous paragraph that . Hence . This is a contradiction. Therefore, .

Remark 7. Let be a finite local ring which is not a field. If , then .

Proof. Now . As and , it follows that .

Observation 1. Let be a graph. If contains as a subgraph, then neither satisfies nor satisfies .

Proof. It is obvious.

Lemma 8. Let R be a ring with at least four maximal ideals. Then neither satisfies nor satisfies .

Proof. By hypothesis, has at least four maximal ideals. Let . Consider the elements , , , , , defined by.
It is clear that the above elements , , , , , are distinct. As they all belong to , it follows that the subgraph of induced on is a clique. Hence contains as a subgraph. Therefore, by Observation 1, we obtain that neither satisfies nor satisfies .

Lemma 9. Let R be a ring with at least two maximal ideals. If satisfies or , then is finite.

Proof. Assume that satisfies either or . Then it is clear that . Therefore, we obtain from Corollary 4 that is a finite ring.

Lemma 10. Let be a ring with at least two maximal ideals. If satisfies or , then R is isomorphic to the direct product of finite local rings with .

Proof. We know from Lemma 8 that and from Lemma 9 that is finite. Since any finite ring is isomorphic to the direct product of finite local rings and as , it follows that either or as rings, where is a finite local ring for each .

#### 3. Characterization of Rings with Exactly Two Maximal Ideals Such That Is Planar

The aim of this section is to characterize rings with such that is planar.

Lemma 11. Let , where and are fields. If satisfies or , then for .

Proof. Now . Note that has exactly two maximal ideals and , where and . Observe that . As satisfies either or , for . Hence for . Therefore, for . As there is no field with exactly 6 elements, it follows that for .

Lemma 12. Let , where and are fields. If for , then is planar.

Proof. Note that and are the only maximal ideals of . Now , and . So we obtain from Lemma 5 that is planar.

Lemma 13. Let , where is a field and is a quasilocal ring which is not a field. If satisfies or , then and . And in the case , must be a local ring with .

Proof. Note that and are the only maximal ideals of . As satisfies or , and . Now implies that . Hence . Therefore, and so . Now implies that . As , we obtain that . Therefore, and this implies that . Now is a local ring which is not a field and . And as the order of any finite local ring is a power of a prime, it follows that . Suppose that and . Then by Remark 7. Hence . Since the subgraph of induced on is , it follows that neither satisfies nor satisfies . Hence is not possible. Suppose that and . Then by Lemma 6. Note that . In this case the subgraph of induced on is . This is impossible as satisfies or . This proves that if satisfies or , then either and or and .

Lemma 14. Let , where is a field with and is a quasilocal ring which is not a field with . Then is planar.

Proof. Note that and are the only maximal ideals of . Now and . Hence by Lemma 5, we obtain that is planar.

Lemma 15. Let , where is a field with and is a quasilocal ring which is not a field with . Then is planar.

Proof. Note that . Now and . So by Lemma 5, it follows that is planar.

Lemma 16. Let , where is a field with and is a quasilocal ring which is not a field with .Then is planar.

Proof. Note that . Now and . So by Lemma 5, we obtain that is planar.

Lemma 17. Let , where is a quasilocal ring which is not a field for . If satisfies or , then .

Proof. Note that and are the only maximal ideals of . As satisfies or , for . Now, and as and , we obtain that . This implies that . Hence and so . Similarly from , it follows that .

Lemma 18. Let , where is a quasilocal ring which is not a field with for . Then is planar.

Proof. Note that . Now and . Hence it follows from Lemma 5 that is planar.

The following theorem determines necessary and sufficient conditions for to be planar, where is a ring with .

Theorem 19. Let be a ring with exactly two maximal ideals. Then the following statements are equivalent.(1) satisfies .(2) is isomorphic to one of the following rings.(a), where and are fields with for .(b), where is a field with and is a local ring which is not a field and and if , then must be a local ring with .(c), where both and are local rings which are not fields with for .(3) satisfies .(4) is planar.

Proof. (resp., )
Now has exactly two maximal ideals and satisfies (resp., ). Hence from Lemma 10 we obtain that as rings, where is a quasilocal ring for . If both and are fields, then Lemma 11 implies that is isomorphic to the ring given in (a). If is a field and is not a field, then from Lemma 13, we obtain that is isomorphic to the ring described in (b). If both and are not fields, then it follows from Lemma 17 that is isomorphic to the ring given in (c).
.
This follows immediately from Lemmas 12, 14, 15, 16, and 18:
and .
Indeed it follows from Kuratowski's theorem [8, Theorem 5.9] that satisfies both and .

#### 4. Characterization of Rings with Exactly Three Maximal Ideals Such That Is Planar

The main result of this section is Theorem 24 which characterizes rings with such that is planar.

Lemma 20. Let , where is a quasilocal ring for . If at least one is not a field, then neither satisfies nor satisfies .

Proof. Without loss of generality we may assume that is not a field. Let . Then the subgraph of induced on is a clique. Thus contains as a subgraph. Therefore, neither satisfies nor satisfies .

Lemma 21. Let , where is a field for . If satisfies , then for .

Proof. Note that and . As satisfies , . This implies that . So . It follows that . Hence for . Similarly we obtain that . Therefore, as rings.

Lemma 22. Let , where is a field for . If satisfies , then for .

Proof. Assume that satisfies . We assert that for each . Suppose that for some . Without loss of generality we may assume that . Let . Note that . Let and . Note that and each vertex of is adjacent to every vertex of in . Hence the subgraph of induced on contains as a subgraph. This is in contradiction to the assumption that satisfies . Therefore, for .

Lemma 23. If , then is planar.

Proof. Note that is the union of cycles:: ,: . Observe that is a cycle of length 6 and it can be represented by means of a hexagon and is a cycle of length 3. Note that each edge of is a diagonal of the hexagon representing and hence it can be represented by means of a triangle which can be drawn inside the hexagon without any crossing over of the edges. This shows that is planar.

Theorem 24. Let be a ring with exactly three maximal ideals. Then the following statements are equivalent.(1) satisfies .(2) as rings.(3) satisfies .(4) is planar.

Proof. (resp., ).
Now by hypothesis has exactly three maximal ideals and satisfies (resp., ). Hence we obtain from Lemma 10 that as rings, where is a quasilocal ring for . Now it follows from Lemmas 20 and 21 (resp., Lemmas 20 and 22) that for . Therefore, as rings.

This follows from Lemma 23.
and .
It follows from Kuratowski's theorem [8, Theorem 5.9] that satisfies both and .

#### 5. Characterization of Rings with at Least Two Maximal Ideals Such That Is Planar

The aim of this section is to characterize rings with at least two maximal ideals such that is planar. If is planar, then being a subgraph of must also be planar. Hence we obtain from Lemma 10 that is isomorphic to the direct product of either two or three finite local rings.

Lemma 25. Let , where and are fields. If at least one contains five elements for some , then does not satisfy .

Proof. Note that and are the only maximal ideals of . Without loss of generality we may assume that . Now . It is clear that the subgraph of induced on is a clique. So it follows that contains as a subgraph. Therefore, does not satisfy .

Lemma 26. Let be the set of all maximal ideals of a ring with for . If , then is planar.

Proof. Note that the vertex set of . Let for , , . It is clear that for all distinct . Observe that the subgraph of induced on is a complete graph on at most three vertices for and each vertex of is an isolated vertex. The vertex 0 is adjacent to every vertex of and . Now it is clear that can be drawn in a plane in such a way that there is no crossing over of the edges. This proves that is planar.

Corollary 27. Let , where and are fields with for . Then is planar.

Proof. Observe that and are the only maximal ideals of with for and . Hence it follows immediately from Lemma 26 that is planar.

Lemma 28. Let , where is a quasilocal ring for and is not a field. If satisfies or , then and .

Proof. Note that is the set of all maximal ideals of . If satisfies or , then for . Since is not a field, . Thus if , then . This is impossible. Therefore, . Thus and so . We next verify that . Since is not a field, it follows that . As , we obtain that . Therefore, .

Lemma 29. Let , where is a field with and is a quasilocal ring which is not a field with . Then is planar.

Proof. Now . As , it follows that . Let . Note that . Observe that the vertex set of , where , , and . Note that the subgraph of induced on is complete for . Each vertex of is an isolated vertex. Thus is the union of , and two isolated vertices. Since is planar and as the edge is the only edge which is common to both and , can be drawn in a plane in such a way that there is no crossing over of the edges. This proves that is planar.

Lemma 30. Let , where is a quasilocal ring for . If satisfies or , then for .

Proof. Note that is a ring with exactly three maximal ideals. If satisfies (resp., ), then satisfies (resp., ). Hence it follows from (resp., ) of Theorem 24 that for .

Lemma 31. If , then is planar.

Proof. Observe that . It was already noted in the proof of Lemma 23 that is the union of the cycles : and : . Note that is a cycle of length 6 and it can be represented by means of a hexagon. The vertex is adjacent to all the six vertices of in . The vertex can be plotted inside the hexagon representing and it can be joined to all the vertices of without any crossing over of the edges. The edges of , each of which is a diagonal of the hexagon representing , can be drawn outside the hexagon in such a way that there is no crossing over of the edges. The vertex is an isolated vertex of . This proves that is planar.

Theorem 32. Let be a ring with exactly two maximal ideals. Then the following statements are equivalent. (1) satisfies and .(2) satisfies .(3) is isomorphic to one of the following rings:(a), where and are fields with for ;(b), where is a field with and is a local ring which is not a field and .(4) is planar.

Proof. This is clear.

Now by hypothesis has exactly two maximal ideals and satisfies . As is a subgraph of , it follows that satisfies . Hence from Lemma 10 it follows that as rings, where is a local ring for . Now if both and are fields, then it follows from Lemma 25 that is isomorphic to the ring given in (a). If either or is not a field, then it follows from Lemma 28 that is isomorphic to the ring given in (b).
.
This follows from Corollary 27 and Lemma 29:
.
It follows from Kuratowski's theorem [8, Theorem 5.9] that satisfies both and .

Lemma 33. Let , where and are fields. Then the following statements are equivalent.(1) satisfies .(2) for .

Proof.
As is a subgraph of , we obtain that satisfies . So, by Lemma 11, it follows that for .
.
Now , where for . Note that is the set of all maximal ideals of with for . Let , , , and . Let be the subgraph of induced on . Observe that is a complete graph on at most four vertices for and each vertex of is an isolated vertex, and is adjacent to each vertex of . Note that the vertex set of and for all distinct . It is clear from the above description of that it satisfies .

Remark 34. Let be a ring with exactly two maximal ideals. If satisfies , then we obtain from of Theorem 32 that satisfies . We mention below an example of a ring with exactly two maximal ideals such that satisfies but it does not satisfy .
Let , where is a field with and is a field with . Note that is the set of all maximal ideals of with and . As the subgraph of induced on is a clique on five vertices, it follows that does not satisfy but by Lemma 33, we obtain that satisfies .

Theorem 35. Let be a ring with exactly two maximal ideals. Then the following statements are equivalent.(1) satisfies .(2) is isomorphic to one of the following rings.(a), where and are fields with for .(b), where is a field with and is a local ring which is not a field and .

Proof.
As is a subgraph of , it follows that satisfies . Hence from Lemma 10 we obtain that as rings, where is a quasilocal ring for . If both and are fields, then Lemma 33 implies that is isomorphic to the ring given in (a). If is not a field, then from Lemma 28, we obtain that is isomorphic to the ring described in (b).

follows immediately from Theorem 32 and Lemma 33.

Theorem 36. Let be a ring with exactly three maximal ideals. Then the following statements are equivalent.(1) satisfies .(2) as rings.(3) satisfies .(4) is planar.

Proof. and .
As is a subgraph of , and follow from Theorem 24.

follows immediately from Lemma 31.
and .
Indeed it follows from Kuratowski's theorem [8, Theorem 5.9] that satisfies both and .

#### 6. Characterization of Quasilocal Rings Which Are Not Fields Such That Is Planar

Let be a quasilocal ring. If , then is a field. In such a case, each vertex of is an isolated vertex and hence it is clear that is planar. So, we assume that . The main result of this section is Theorem 37 which characterizes quasilocal rings which are not fields such that is planar.

Theorem 37. Let be a quasilocal ring which is not a field. Then the following statements are equivalent.(1) satisfies and .(2) satisfies .(3), , and .(4) is planar.

Proof. This is clear.
.
As satisfies , it follows that . Now follows from Lemma 6. Hence :
.
Now, , , and . Note that is a graph with vertex set . Each unit of is an isolated vertex of and as , it is clear that is planar:

It follows from Kuratowski's theorem [8, Theorem 5.9] that satisfies both and .

Lemma 38. Let be a quasilocal ring which is not a field. Then the following statements are equivalent.(1) satisfies .(2), , and .

Proof.
As satisfies , it follows that . Now follows from Lemma 6. Hence .

Now and . As each unit of is an isolated vertex of and any graph on at most five vertices satisfies , it follows that satisfies .

Example 39. Let . Note that is a local ring with unique maximal ideal . As , it is clear that does not satisfy . Moreover, from Lemma 38, it follows that satisfies .

Remark 40. With the help of theorems proved by Corbas and Williams in [9], Belshoff and Chapman in [10] have listed up to isomorphism, the local rings of order and all the local rings of order 16. From the above list, we mention below up to isomorphism of rings, all local rings which are not fields of order 4, 8, 9, 25 and all local rings with and .(1)Local rings of order 4 which are not fields are and .(2)Local rings of order 8 which are not fields are , , , and .(3)Local rings of order 9 which are not fields are and .(4)Local rings with and are and .

#### Conflict of Interests

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

#### Acknowledgment

The authors are very much thankful to the Academic Editors for their suggestions and support.