#### Abstract

Let be a commutative ring with identity. The zero-divisor graph of , denoted , is the simple graph whose vertices are the nonzero zero-divisors of , and two distinct vertices and are linked by an edge if and only if . The genus of a simple graph is the smallest integer such that can be embedded into an orientable surface . In this paper, we determine that the genus of the zero-divisor graph of , the ring of integers modulo , is two or three.

#### 1. Introduction

This paper concerns the zero-divisor graphs of rings. For a commutative ring , define a simple graph called* zero-divisor graph*, denoted by , whose vertices are the nonzero zero-divisors of , and two distinct vertices and are adjacent if and only if in . This definition was first introduced by Beck in [1]. However, he let all elements of be the vertices of the graph and mainly considered the coloring of this graph. Here our definition is the same as in [2], where some basic properties of are established. The zero-divisor graph, as well as other graphs of rings, is an active research topic in the last two decades (see, e.g., [3–11]).

Let us first recall some needed notions in graph theory. Let be a simple graph, that is, no loops and no multiedges. The* degree* of a vertex , denoted by , is the number of edges of incident with . If , then is the subgraph of obtained by deleting the vertices in and all edges incident with them. If or , then we use for the subgraph and call it the* reduction* of . A* complete bipartite graph* is a bipartite graph (i.e., a set of graph vertices decomposed into two disjoint sets such that no two vertices within the same set are adjacent) such that every pair of vertices in the two sets are adjacent. The complete bipartite graph with partitions of sizes and is denoted by . The* complete graph *on vertices, denoted , is the graph in which every pair of distinct vertices is joined by an edge. A surface is said to be of* genus * if it is topologically homomorphic to a sphere with handles. A graph that can be drawn without crossings on a compact surface of genus , but not on one of genus , is called a graph of* genus *. We write for the genus of the graph . It is clear that , where is the reduction of , and for any subgraph of .

Determining the genus of a graph is one of the most fundamental problems in topological graph theory. It has been shown to be NP-complete by Thomassen in [12]. Several papers focus on the genera of zero-divisor graphs. For instance, in [6, 7, 13, 14], the authors studied the planar zero-divisor graphs (genus equals to 0); Wang et al. investigated the genus one zero-divisor graphs in [11, 15, 16], respectively; and Bloomfield and Wickham determined all local rings whose zero-divisor graphs have genus two in [8]. In this paper, we study the zero-divisor graph of , the ring of integers modulo . In particular, we determine when or . Here we first summarize the results about the genus of from [5, Theorem 5.1(a)], [8, Theorem 1], and [16, Section 5].

Theorem 1. *Let be not empty. Then the following hold. *(1)* if and only if , where is prime.*(2)* if and only if .*(3)* if and only if .**All rings considered in this paper will be commutative rings with identity. Let be an element of a ring . Then the principal ideal generated by is denoted by . For a set , means the order of .*

#### 2. The Genus of

The following two lemmas are frequently used in the proofs of our main results.

Lemma 2 ([17, Theorem 6.38]). *, where is the least integer that is greater than or equal to .*

Lemma 3 ([17, Theorem 6.37]). *, where is the least integer that is greater than or equal to .*

Lemma 4 ([17, Corollary 6.15]). *Suppose a simple graph is connected with vertices and edges. If has no triangles, then .*

Lemma 5. *Let be a graph with vertex set and the edge set . Then .*

*Proof. *Note that there are no triangles in . As has 40 edges and 15 vertices, by Lemma 4.

We first consider the case that has only one prime divisor.

Theorem 6. *Let , where is prime and . Then if and only if .*

*Proof. *(). Let . Then . As any two vertices in are adjacent, there exists a complete subgraph in . It follows that if , then by Lemma 2. Therefore, . We proceed with three cases.*Case **1* ( or 7). By Theorem 1, and . So we can further assume . Let and . Then is empty and , . As each vertex of is adjacent to every vertex of , there exists a complete bipartite subgraph in , which implies that by Lemma 3. *Case **2* (). From Theorem 1, we have , , and . So we may assume . Let and . Then is empty and , . Since each vertex of is adjacent to every vertex of , contains a complete bipartite subgraph , which implies that by Lemma 3. *Case **3* (). By Theorem 1, if , and .

If , we can assume . We let , . Then , . Note that each vertex of is adjacent to each vertex of and is empty, so there exists a complete bipartite subgraph in , which implies that by Lemma 3. Therefore, .

(). For , let and . Then , , and is empty. Since each vertex in is adjacent to each vertex in , there exists a complete bipartite subgraph in , which implies that by Lemma 3. On the other hand, we can embed the reduction of into as shown in Figure 1. Therefore, .

This completes our proof.

We now consider the case that has exactly two prime divisors.

Theorem 7. *Let , where are primes and . Then if and only if , and if and only if .*

*Proof. *We first prove that if then . We then determine for each . We proceed with four cases to get the result.*Case **1* (). It is clear that is a complete bipartite graph . In this case, if and , then is a subgraph of ; it follows that by Lemma 3. If or , by Theorem 1, we have . So , ; that is, . *Case **2* ( and ). If , set and . Then is empty and each vertex in is adjacent to each vertex in . If , then , , which implies that contains a complete bipartite subgraph . Therefore by Lemma 3. If and , then , , which implies that there exists a complete bipartite subgraph in . Therefore by Lemma 3. Hence, in this situation, we have .

Consider now ; that is, . Let and . Then is empty and , . Notice that each vertex in is adjacent to each vertex in , so there exists a complete bipartite subgraph in . If , then . If and then . By Theorem 1, we know . So . *Case **3* ( and ). Let and . Then , , and is empty. Since each vertex in is adjacent to each vertex in , there exists a complete bipartite subgraph in . Therefore . Simply checking, we can see that there are only seven cases satisfying the inequality ; . By Theorem 1, we have and . So .

For , we set , , and . Note that each vertex in is adjacent to each vertex in , and the vertices 10, 20, and 30 are adjacent to each vertex in , so by Lemma 5, .

For the case , by Theorem 1, we have and . If , that is, , we set , , and . Note that each vertex in is adjacent to each vertex in , and the vertices 12, 24, and 36 are adjacent to each vertex in , so by Lemma 5, . If , we set and . Then , and is empty. Since each vertex in is adjacent to each vertex in , there is a complete bipartite subgraph in . It then follows that by Lemma 3.*Case **4* ( and ). We set . Then . Since any two vertices in are adjacent, there exists a complete subgraph in . If , then by Lemma 2. So or . If , set and ; then , , and is empty. Since each vertex in is adjacent to each vertex in , there exists a complete bipartite subgraph in . If or , then . So , ; that is . If , similarly, there exists a complete bipartite subgraph in . Therefore, by Lemma 3.

Now we have proved that if , then .

In the following, we determine for each .

It is easy to see that . So by Lemma 3.

For , as mentioned in Case 4 above, contains a subgraph . So . can be embedded into as shown in Figure 2. So .

Since the reductions of the graphs and are and , respectively, we have and , respectively.

For , as mentioned in Case 2 above, we have by Lemma 3. We can embed the reduction of the graph into as shown in Figure 3. Thus, .

For , we set and . Then , , and is empty. Since each vertex in is adjacent to each vertex in , there is a complete bipartite subgraph in . It then follows that . On the other hand, we can embed the reduction of the graph into as shown in Figure 4. Thus, .

For , let and . Then , , and is empty. Since each vertex in is adjacent to each vertex in , there exists a complete bipartite subgraph in . Therefore . On the other hand, we can embed the graph into as shown in Figure 5.

This completes the proof.

The final case is that has more than two prime divisors.

Theorem 8. *Let (), where are primes. Then if and only if , and if and only if .*

*Proof. *Let and . Then , and is empty; moreover, every vertex in is adjacent to each vertex in . Thus, there exists a complete bipartite subgraph in . It then follows that as . So either or .

For the former case, set and ; then , and is empty. Note that each vertex in is adjacent to each vertex in , so there exists a complete bipartite subgraph in . If , then by Lemma 3. So ; that is, .

Let and . Then , , and is empty. Since each vertex in is adjacent to each vertex in , there exists a complete bipartite subgraph in . It then follows that . On the other hand, we can embed into as shown in Figure 6, so .

For the latter case, with a similar argument above, we have . Let and . Then , , and is empty. Since each vertex in is adjacent to each vertex in , there exists a complete bipartite subgraph in , which implies that . On the other hand, we can embed into as shown in Figure 7, so .

This completes our proof.

Now we have completely determined when or 3. We summarize the result by the following theorem.

Theorem 9. *
(1) if and only if .**
(2) if and only if .*

#### Conflict of Interests

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

#### Acknowledgments

The authors thank the anonymous referees for their very careful reading of the paper and for their many valuable comments which improved the paper. This work was supported by the National Natural Science Foundation of China (11161006) and the Guangxi Education Committee Research Foundation (LX2014223).