Classification of Genus Three Zero-Divisor Graphs

Abstract

In this paper, we consider the problem of classifying commutative rings according to the genus number of its associating zero-divisor graphs. The zero-divisor graph of R, where R is a commutative ring with nonzero identity, denoted by $\mathrm{\Gamma }\left(R\right)$, is the undirected graph whose vertices are the nonzero zero-divisors of R, and the distinct vertices x and y are adjacent if and only if $xy=0$. Here, we classify the local rings with genus three zero-divisor graphs.

1. Introduction

Studying algebraic structures using graph theory properties has been an exciting research topic in the last two decades. Beck [1], who created the concept of a zero-divisor graph of a commutative ring with identity, established the connection between algebra and graph theory by linking a graph with a ring in 1988. In Beck’s definition, the vertices of the graph are the elements of the ring and two distinct vertices x and y are adjacent if and only if $xy=0$. Beck’s graph had edges between zero-divisors to emphasize them, but it also had an edge connecting each element to zero. Later, Anderson and Livingston [2] in 1999 made a slight modification to this concept by only taking into account the nonzero zero-divisors of a ring as vertices of the graph with the same adjacency criterion. For more details on zero-divisor graphs, one may refer to [3].

In recent decades, many geometrical properties of graphs were widely investigated. More precisely, the study focused on the oriented and non-oriented genus of graphs that originate from algebraic ones. The question of which commutative rings have planar zero-divisor graphs was raised by Anderson et al. [4] in 2001. Since then, the literature has been keenly interested in zero-divisor graph embedding. Numerous academics have so far classified all commutative rings with zero-divisor graphs having a genus less than or equal to two in several research articles. The planar zero-divisor graphs were explicitly characterized by Smith [5,6] and Belshoff [7], while the characterization of commutative rings with genus one zero-divisor graphs was obtained by Chiang-Hsieh et al. in [8] and local rings with genus two zero-divisor graphs were characterized by Bloomfied et al. in [9]. Furthermore, Asir et al. [10] have determined all non-local commutative rings with genus two zero-divisor graphs. All commutative rings with projective (non-oriented genus one) zero-divisor graphs were characterized by Chiang-Hsieh [11]. Moreover, Asir et al. [12] determined all commutative rings with crosscap two zero-divisor graphs. One may refer to Chapter 4 of [3] for the complete literature evaluation.

This paper aims to investigate the problem of embedding zero-divisors on higher-genus orientable surfaces. This paper is structured as follows: In Section 2, definitions and tools are developed to aid classification. In Section 3 and Section 4, all commutative rings with genus three zero-divisor graphs are determined.

Let R be a commutative ring with nonzero identity and $Z\left(R\right)$ be its set of zero-divisors for the course of this paper. We use ${\mathbb{Z}}_n$ to represent the ring of integers modulo n, ${\mathbb{F}}_q$ to represent the finite field with q elements, $\left|A\right|$ to represent the cardinality of the set A and $A^{*}$ to represent the set of all non-zero elements of A. We denote the complete bipartite graph with parts of size m and n by $K_{m,n}$, and the complete graph with n vertices by $K_n$. $\overline{K_n}$ is the complement of the complete graph $K_n$. The notation $deg\left(v\right)$ and $\delta \left(G\right)$ indicate the degree of a vertex v in a graph G and the minimum degree of G, respectively. The Cartesian product $G=G_1\square G_2$ is a graph with vertex set $V\left(G\right)=V\left(G_1\right)\times V\left(G_2\right)$, and two distinct vertices $(u_1,u_2)$ and $(v_1,v_2)$ of G are adjacent if and only if either $u_1=v_1$ and $u_2{v}_2\in E\left(G_2\right)$ or $u_2=v_2$ and $u_1{v}_1\in E\left(G_1\right)$. The join $G=G_1\vee G_2$ of graphs $G_1$ and $G_2$ with disjoint vertex sets $V_1$ and $V_2$ and edge sets $E_1$ and $E_2$ is the graph union $G_1$ and union $G_2$ together with all the edges joining $V_1$ and $V_2$.

For non-negative integers ℓ and k, let $S_{\mathsf{\ell }}$ denote the sphere with ℓ handles and $N_k$ denote a sphere with k crosscaps attached to it. For a simple graph G, the minimum ℓ such that G can be embedded in $S_{\mathsf{\ell }}$ is its genus which is denoted by $\gamma \left(G\right)$. Similarly, the minimum k such that G can be embedded in $N_k$ is $\tilde{\gamma }\left(G\right)$, or crosscap number (non-orientable genus). Planar, toroidal and double toroidal graphs are the terms used to refer to the graphs of genus 0, 1 and 2, respectively. The graph that is obtained from $G-xy$ by identifying vertices x and y to create a new vertex z incident with all edges of G that were incident with either x or y is known as the contraction of e in G, denoted as $[x,y]$. This is the case if $e=xy\in E\left(G\right)$. If H can be obtained from G by deleting vertices, edges and contracting edges, then H is a minor of G, written $H\le G$. The graph $\tilde{G}$ denotes the induced subgraph of G induced by the vertices $G-V^{\prime }$ where $V^{\prime }=\{v\in V:deg\left(v\right)\le 2\}$ and we call this graph the reduction in G. It is clear that $\gamma \left(\tilde{G}\right)=\gamma \left(G\right)$ for any graph G. For details on the notion of embedding of graphs on a surface, we refer to White [13].

For convenience, we denote the elements of the ring ${\mathbb{Z}}_n\left[x\right]/\left(x^2\right)$ simply by $\{a_0+a_{1}x:a_0,a_1\in {\mathbb{Z}}_n\}$. In other words, we avoid writing the fixed last term $\left(x^2\right)$ in each element of the ring. Similarly, the ring ${\mathbb{Z}}_n[x,y]/(x^3,xy,y^2)$ simply by $\{a_0+a_{1}x+a_2{x}^2+a_{3}y:a_0,a_1,a_2,a_3\in {\mathbb{Z}}_n\}$.

2. Genus Properties of Graphs

We now state a few results regarding the genus of the graph which will be used to prove the main results of this paper.

Proposition 1 

([14]). Let $n\ge 3$ be a positive integer and, for any real number x, the ceiling function $⌈x⌉$ is the least integer that is greater than or equal to x. Then, $\gamma \left(K_n\right)=⌈\frac{(n-3)(n-4)}{12}⌉$.

Proposition 2 

([14]). Let $m,n$ be positive integers. Then, $\gamma \left(K_{m,n}\right)=⌈\frac{(m-2)(n-2)}{4}⌉$ if $m,n\ge 2$.

Proposition 3 

([14] (Euler formula)). If G is a finite connected graph with n vertices and m edges; then,

$$n-m+f=2-2\gamma$$

where the graph is embedded upon a surface $S_{\gamma }$ with genus γ and f is the number of faces created when G is embedded on $S_{\gamma }$.

Remark 1 

([15]). For any embedded graph G with v vertices, e edges and f faces, we have $2e\ge g\left(G\right)·f$ where $g\left(G\right)$ denotes the girth of G (i.e., the length of the shortest cycle in $G)$.

By Proposition 3 and Remark 3, one can obtain the following result.

Remark 2.

If G is a triangle-free graph with n vertices and m edges, then

$$\gamma \left(G\right)\ge ⌈\frac{m}{4}-\frac{n}{2}+1⌉.$$

Proposition 4 

([16], Theorem 2). The complete bipartite graph $K_{m,n}$ has a two-cell embedding on a compact orientable two-manifold $S_{\gamma }$ of genus γ if and only if

$$\frac{(m-2)(n-2)}{4}\le \gamma \le \frac{(m-1)(n-1)}{2}.$$

Proposition 5 

([13], Proposition 2.1). If G is a graph with n vertices and genus γ, then

$$\delta \left(G\right)\le 6+\frac{12\gamma -12}{n}.$$

Proposition 6 

([9], Lemma 4). Let $(R,\mathfrak{m})$ be a local ring with $\mathfrak{m}$ principal and with an index of nilpotency n, and let $1\le k<n$. For every $z\in M^k-M^{k+1}$, $Ann\left(z\right)=Ann\left(M^k\right)$.

Proposition 7 

([9], Lemma 5). Let $(R,\mathfrak{m})$ be a local ring such that $\mathfrak{m}=(x,y_1,\dots ,y_{k-2})$ and ${\mathfrak{m}}^2=\left(x^2\right)$ for some $k\ge 2$. Then, there exist elements $y_1^{\prime },\dots ,y_{k-2}^{\prime }\in \mathfrak{m}$ such that $\mathfrak{m}=(x,y_1^{\prime },\dots ,y_{k-2}^{\prime })$ and $x{y}_i^{\prime }=0$.

Proposition 8 

([9], Lemma 6). Let $(R,\mathfrak{m})$ be a local ring such that $\mathfrak{m}=(x,y,z_1,\dots ,z_{k-2})$ and ${\mathfrak{m}}^2=\left(xy\right)$ for some $k\ge 3$. Then, there exist elements $z_1^{\prime },\dots ,z_{k-2}^{\prime }\in \mathfrak{m}$ such that $\mathfrak{m}=(x,y,z_1^{\prime },\dots ,z_{k-2}^{\prime })$ and $x{z}_i^{\prime }=0$.

3. Classification of Genus Three Zero-Divisor Graphs

We will now state a few results regarding the genus properties of a zero-divisor graph of a commutative ring which will be further used to prove the main result.

Theorem 1 

([8], Theorem 3.5.1). Let $(R,\mathfrak{m})$ be a finite local ring which is not a field. Then, $\mathrm{\Gamma }\left(R\right)$ is planar if and only if R is isomorphic to one of the following 29 rings:

$$\begin{array}{c}{\mathbb{Z}}_4,{\mathbb{Z}}_8,{\mathbb{Z}}_9,{\mathbb{Z}}_{16},{\mathbb{Z}}_{25},{\mathbb{Z}}_{27},{\mathbb{Z}}_2\left[x\right]/\left(x^2\right),{\mathbb{Z}}_2\left[x\right]/\left(x^3\right),{\mathbb{Z}}_2\left[x\right]/\left(x^4\right),{\mathbb{Z}}_2[x,y]/(x^2,y^2),\\ {\mathbb{Z}}_2[x,y]/(x^2,xy,y^2),{\mathbb{Z}}_2[x,y]/(x^3,xy,y^2-x^2),{\mathbb{F}}_4\left[x\right]/\left(x^2\right),{\mathbb{Z}}_3\left[x\right]/\left(x^2\right),\\ {\mathbb{Z}}_3\left[x\right]/\left(x^3\right),{\mathbb{Z}}_4\left[x\right]/\left(x^2\right),{\mathbb{Z}}_4\left[x\right]/(x^2+x+1),{\mathbb{Z}}_4\left[x\right]/(2x,x^2),{\mathbb{Z}}_4\left[x\right]/(x^2-2,x^4),\\ {\mathbb{Z}}_4\left[x\right]/(x^3-2,x^4),{\mathbb{Z}}_4\left[x\right]/(x^2-2,x^3),{\mathbb{Z}}_4\left[x\right]/(x^3,x^2-2x),\\ {\mathbb{Z}}_4\left[x\right]/(x^3+x^2-2,x^4),{\mathbb{Z}}_4[x,y]/(x^2,y^2,xy-2),{\mathbb{Z}}_4[x,y]/(x^3,x^2-2,xy,y^2-2,y^3),\\ {\mathbb{Z}}_5\left[x\right]/\left(x^2\right),{\mathbb{Z}}_8\left[x\right]/(x^2-4,2x),{\mathbb{Z}}_9\left[x\right]/(x^2-3,x^3),{\mathbb{Z}}_9\left[x\right]/(x^2+3,x^3).\end{array}$$

Theorem 2 

([8], Theorem 3.5.2). Let ($R,\mathfrak{m})$ be a finite local ring which is not a field. Then, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=1$ if and only if R is isomorphic to one of the following 17 rings:

$$\begin{array}{c}{\mathbb{Z}}_{32},{\mathbb{Z}}_{49},{\mathbb{Z}}_2\left[x\right]/\left(x^5\right),{\mathbb{F}}_8\left[x\right]/\left(x^2\right),{\mathbb{Z}}_2[x,y]/(x^3,xy,y^2),{\mathbb{Z}}_2[x,y,z]/{(x,y,z)}^2,\\ {\mathbb{Z}}_4\left[x\right]/(x^3+x+1),{\mathbb{Z}}_4\left[x\right]/(x^3-2,x^5),{\mathbb{Z}}_4\left[x\right]/(x^4-2,x^5),{\mathbb{Z}}_4[x,y]/(x^3,x^2-2,xy,y^2),\\ {\mathbb{Z}}_4\left[x\right]/(x^3,2x),{\mathbb{Z}}_4[x,y]/(2x,2y,x^2,xy,y^2),{\mathbb{Z}}_7\left[x\right]/\left(x^2\right),{\mathbb{Z}}_8\left[x\right]/(x^2,2x),\\ {\mathbb{Z}}_8\left[x\right]/(x^2-2,x^5),{\mathbb{Z}}_8\left[x\right]/(x^2+2x-2,x^5),{\mathbb{Z}}_8\left[x\right]/(x^2-2x+2,x^5).\end{array}$$

As mentioned earlier, Bloomfield and Wickham [9] enumerated the local commutative rings whose zero-divisor graphs have genus two. The relevant result is stated below.

Theorem 3 

([9], Theorem 1). Let ($R,\mathfrak{m})$ be a finite local ring which is not a field. Then, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=2$ if and only if R is isomorphic to one of the following 14 local rings.

$$\begin{array}{c}{\mathbb{Z}}_{81},{\mathbb{F}}_3\left[x\right]/\left(x^4\right),{\mathbb{Z}}_9\left[x\right]/(3x,x^3-3),{\mathbb{Z}}_9\left[x\right]/(x^2-3),{\mathbb{Z}}_9\left[x\right]/(x^2-6),{\mathbb{Z}}_9[x,y]/{(x,y)}^2,\\ {\mathbb{Z}}_9\left[x\right]/(3x,x^2),{\mathbb{F}}_9\left[x\right]/\left(x^2\right),{\mathbb{F}}_2[x,y]/(x^4,xy,y^2-x^3),{\mathbb{Z}}_4[x,y]/(x^2-2,xy,y^2-2x,2y),\\ {\mathbb{Z}}_9\left[x\right]/(x^2+1),{\mathbb{Z}}_4[x,y]/(x^3-2,xy,2x,y^2-2),{\mathbb{Z}}_8\left[x\right]/(2x,x^3-4),{\mathbb{Z}}_{16}\left[x\right]/(2x,x^2-8).\end{array}$$

First, we provide some results needed to classify the genus three zero-divisor graphs.

Theorem 4.

Let $(R,\mathfrak{m})$ be a finite local ring. If $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$, then $|R/\mathfrak{m}|\le 4$; and the following holds:

(i)

If $|R/\mathfrak{m}|=4$, then ${\mathfrak{m}}^3=0$ and $\left|R\right|=64$;

(ii)

If $|R/\mathfrak{m}|=3$, then ${\mathfrak{m}}^4=0$ and $\left|R\right|=81$;

(iii)

If $|R/\mathfrak{m}|=2$, then ${\mathfrak{m}}^6=0$ and $\left|R\right|=64$ or 32.

Proof. 

Assume that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$. Since R is local, there is a prime power q and integers $s>r$ such that $\left|R\right|=q^s$, $\left|\mathfrak{m}\right|=q^r$ and $|R/\mathfrak{m}|=q$. Let t be a largest positive integer such that ${\mathfrak{m}}^t\ne 0$. Therefore, ${\mathfrak{m}}^{t+1}=0$.

Suppose that $q>9$. Then, $|{\mathfrak{m}}^t-{\mathfrak{m}}^{t+1}| \ge 10$. Since ${\mathfrak{m}}^t·{\mathfrak{m}}^t=0$, we have $xy=0$ for every $x,y\in {\mathfrak{m}}^t$. Therefore, the subgraph induced by ${\mathfrak{m}}^t$ in $\mathrm{\Gamma }\left(R\right)$ is complete and so $K_{10}\subseteq \mathrm{\Gamma }\left(R\right)$. Thus, by Proposition 2, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$ is a contradiction. Hence, $|R/\mathfrak{m}|\le 9$.

Case 1: Suppose $5\le q\le 9$. The possibilities for q are $5,7,8$ or 9. If $t\ge 2$, then $|{\mathfrak{m}}^t-{\mathfrak{m}}^{t+1}| \ge 4$ and $|{\mathfrak{m}}^{t-1}-{\mathfrak{m}}^t| \ge 20$. Since ${\mathfrak{m}}^{t-1}·{\mathfrak{m}}^t=0$, every element of ${\mathfrak{m}}^{t-1}$ is adjacent to all the elements of ${\mathfrak{m}}^t$ in $\mathrm{\Gamma }\left(R\right).$ So $K_{4,20}\subseteq \mathrm{\Gamma }\left(R\right)$ and by Proposition 2, we obtain $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 9$. Therefore, $t=1$. That is, ${\mathfrak{m}}^2=0$. If $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2\ge 2$, then $\left|\mathfrak{m}\right|\ge q^2$. Since the collection of non-zero elements of $\mathfrak{m}$ form a complete graph in $\mathrm{\Gamma }\left(R\right)$ and $q^2-1\ge 24$, we have $K_{24}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction. Therefore, $\mathfrak{m}$ is a principal ideal and $\left|\mathfrak{m}\right|=q$. In this case, $\mathrm{\Gamma }\left(R\right)=K_{q-1}$. The assumption $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$ implies that $q=10$, a contradiction.

Case 2: Suppose $q=4$. If $t\ge 3$, then $|{\mathfrak{m}}^t| \ge 4$. This statement implies that $|{\mathfrak{m}}^t-{\mathfrak{m}}^{t+1}| \ge 3$ and $|{\mathfrak{m}}^{t-2}-{\mathfrak{m}}^t| \ge 28$. Since ${\mathfrak{m}}^{t-2}·{\mathfrak{m}}^t=0$, $K_{3,28}\subseteq \mathrm{\Gamma }\left(R\right)$ so that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 7$, a contradiction. Thus, $t\le 2$. That is, ${\mathfrak{m}}^3=0$.

Moreover, suppose that $\left|R\right|=4^{\mathsf{\ell }}$ for some $\mathsf{\ell }\ge 4$. Here, $\left|\mathfrak{m}\right|=4^{\mathsf{\ell }-1}\ge 64$. If ${\mathfrak{m}}^2=0$, then the subgraph induced by $\mathfrak{m}$ forms a complete graph in $\mathrm{\Gamma }\left(R\right)$, which leads to $K_{63}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction. If ${\mathfrak{m}}^2\ne 0$, then $|{\mathfrak{m}}^2| \ge 4$ and $\mathfrak{m}·{\mathfrak{m}}^2=0$. This implies that $K_{3,60}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction. Therefore, $t\le 3$ means that $\left|R\right|\le 64$. But by Theorems 1 and 2, the zero-divisor graph of any local ring of order less than or equal to 16 has a genus less than 2. Thus, $\left|R\right|=64$.

Case 3: Suppose $q=3$. If $t\ge 4$, then $|{\mathfrak{m}}^{t-1}-{\mathfrak{m}}^t| \ge 6$ and $|{\mathfrak{m}}^{t-2}-{\mathfrak{m}}^{t-1}| \ge 18$. Since ${\mathfrak{m}}^{t-1}·{\mathfrak{m}}^{t-2}=0,$ $K_{6,18}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction. Therefore, $t\le 3$. Hence, ${\mathfrak{m}}^4=0$.

Suppose that $\left|R\right|=3^{\mathsf{\ell }}$ for some $\mathsf{\ell }\ge 5$. This implies that $\left|\mathfrak{m}\right|=3^{\mathsf{\ell }-1}\ge 81$.

Case 3.1: If ${\mathfrak{m}}^2=0$, then the subgraph induced by ${\mathfrak{m}}^{*}$ is complete in $\mathrm{\Gamma }\left(R\right)$, leading to $K_{80}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction.

Case 3.2: If ${\mathfrak{m}}^3=0$, then $|{\mathfrak{m}}^2| \ge 3$. Let $|{\mathfrak{m}}^2| =k\ge 9$. Since ${\mathfrak{m}}^2·\mathfrak{m}=0$, we have $K_{k-1,81-k}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction. Therefore, $|{\mathfrak{m}}^2| =3$ and so $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=\mathsf{\ell }-2\ge 3$. Now, by Proposition 2.8 [17], $\mathfrak{m}=(x,y_1,\dots ,y_{\mathsf{\ell }-3})$ for some $x,y_i\in \mathfrak{m}-{\mathfrak{m}}^2$. Without loss of generality, we may assume that either ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(x{y}_1\right)$.

Assume that ${\mathfrak{m}}^2=\left(x^2\right)$. Then by Proposition 7, $x{y}_i=0$ for $1\le i\le \mathsf{\ell }-3$. Since $\{y_1+{\mathfrak{m}}^2\}\cap \{2y_1+{\mathfrak{m}}^2\}=\varnothing$ and $\{x+{\mathfrak{m}}^2\}\cap \{2x+{\mathfrak{m}}^2\}\cap \left\{{\mathfrak{m}}^2\right\}=\varnothing$, we have the sets $y_1+{\mathfrak{m}}^2\cup 2y_1+{\mathfrak{m}}^2$ and $x+{\mathfrak{m}}^2\cup 2x+{\mathfrak{m}}^2\cup {{\mathfrak{m}}^2}^{*}$ which form $K_{6,8}$ in $\mathrm{\Gamma }\left(R\right)$. Therefore $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 6$.

Assume that ${\mathfrak{m}}^2=\left(x{y}_1\right)$. Since any element of ${\mathfrak{m}}^2-{\mathfrak{m}}^3$ generates ${\mathfrak{m}}^2$, we may assume that $x^2,y_i^2\in {\mathfrak{m}}^3=0$. Now, the cosets ${\mathfrak{m}}^2,x+{\mathfrak{m}}^2,2x+{\mathfrak{m}}^2,y_1+{\mathfrak{m}}^2,2y_1+{\mathfrak{m}}^2,y_2+{\mathfrak{m}}^2$ and $2y_2+{\mathfrak{m}}^2$ are mutually disjoint. In particular, $2x\ne 0$ and $2y\ne 0$. Let $A={{\mathfrak{m}}^2}^{*}\cup x+{\mathfrak{m}}^2\cup 2x+{\mathfrak{m}}^2$, $B=y_1+{\mathfrak{m}}^2\cup 2y_1+{\mathfrak{m}}^2$ and $C=y_2+{\mathfrak{m}}^2\cup 2y_2+{\mathfrak{m}}^2$ so that $\left|A\right|=8$, $\left|B\right|=6$ and $\left|C\right|=6$. Therefore, in $\mathrm{\Gamma }\left(R\right)$, the subgraph induced by the set A is isomorphic to $K_8$, and the subgraph induced by the sets B and C is isomorphic to $K_6$. Thus, $2K_6\cup K_8\subseteq \mathrm{\Gamma }\left(R\right)$ so that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

Case 3.3: If ${\mathfrak{m}}^4=0$, then $|{\mathfrak{m}}^3| \ge 3$ and $|{\mathfrak{m}}^2| \ge 9$. Let $|{\mathfrak{m}}^3| =k\ge 9$. Then, $\mathrm{\Gamma }\left(R\right)$ contains $K_{k-1,81-k}$ as a subgraph, a contradiction. Therefore, $|{\mathfrak{m}}^3| =3$ and $|{\mathfrak{m}}^2| =9$. Consequently, $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=\mathsf{\ell }-3\ge 2$. Then, by Proposition 2.8 [17], $\mathfrak{m}=(x,y_1,\dots ,y_{\mathsf{\ell }-4})$ for some $x,y_i\in \mathfrak{m}-{\mathfrak{m}}^2$. So we may assume that ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(x{y}_1\right)$.

Assume that ${\mathfrak{m}}^2=\left(x^2\right)$. Then, by Proposition 7, $x{y}_i=0$ for every i. Notice that the cosets ${\mathfrak{m}}^2$ and $y_1+{\mathfrak{m}}^2$ are mutually disjoint and every element in ${\mathfrak{m}}^2\backslash \left\{0\right\}$ is adjacent to all the elements of $y_1+{\mathfrak{m}}^2$ in $\mathrm{\Gamma }\left(R\right)$. So $\mathrm{\Gamma }\left(R\right)$ contains $K_{8,9}$ as a subgraph and thus $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 11$.

Assume that ${\mathfrak{m}}^2=\left(x{y}_1\right)$. Since any element of ${\mathfrak{m}}^2-{\mathfrak{m}}^3$ generates ${\mathfrak{m}}^2$, we may assume that $x^2,y_i^2\in {\mathfrak{m}}^3$. Clearly, ${\mathfrak{m}}^2\cap y_1+{\mathfrak{m}}^2=\varnothing$. If $a\in y_1+{\mathfrak{m}}^2$ and $b\in {\mathfrak{m}}^2$, then $ab=sx{y}_1^2+rs{x}^2{y}_1^2$ for some $r,s\in R$. Since $y_1^2\in {\mathfrak{m}}^3$ and $x\in \mathfrak{m}$, we have $ab=0$. Therefore, the subgraph induced by ${\mathfrak{m}}^2\backslash \left\{0\right\}\cup y_1+{\mathfrak{m}}^2$ contains $K_{8,9}$ in $\mathrm{\Gamma }\left(R\right)$ and so $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 11$.

Thus, $\left|R\right|=3^{\mathsf{\ell }}$ with $\mathsf{\ell }\le 4$. But by Theorems 1 and 3, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\le 2$ for all local rings R of order less than or equal to 27. Hence, $\left|R\right|=81$.

Case 4: Suppose $q=2$. If $t\ge 6$, then $|{\mathfrak{m}}^{t-1}-{\mathfrak{m}}^{t+1}| \ge 3$ and $|{\mathfrak{m}}^{t-4}-{\mathfrak{m}}^{t-1}| \ge 28$. Since ${\mathfrak{m}}^{t-4}·{\mathfrak{m}}^{t-1}=0$, we have $K_{3,28}\subseteq \mathrm{\Gamma }\left(R\right)$ so that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 7$, a contradiction. Therefore, $t\le 5$, which leads to ${\mathfrak{m}}^6=0$.

Suppose that $\left|R\right|=2^{\mathsf{\ell }}$ for some $\mathsf{\ell }\ge 7$. Notice that $\left|\mathfrak{m}\right|=2^{\mathsf{\ell }-1}\ge 64$.

Case 4.1: If ${\mathfrak{m}}^2=0$, then $K_{63}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction.

Case 4.2: If ${\mathfrak{m}}^3=0$, then $|{\mathfrak{m}}^2| \ge 2$. Let $|{\mathfrak{m}}^2| =k\ge 4$. Then, $\mathrm{\Gamma }\left(R\right)$ contains $K_{k-1,64-k}$ as a subgraph, a contradiction. Therefore, $|{\mathfrak{m}}^2| =2$ and so $di{m}_{R/\mathfrak{m}}{\mathfrak{m}}^2/\mathfrak{m}=\mathsf{\ell }-2\ge 5.$ This implies that $\mathfrak{m}=(x,y_1,\dots ,y_{\mathsf{\ell }-3})$ for $x,y_i\in \mathfrak{m}-{\mathfrak{m}}^2$ and without loss of generality, we may assume ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(x{y}_1\right)$.

Assume that ${\mathfrak{m}}^2=\left(x^2\right)$. Then, by Proposition 7, we have $x{y}_i=0$. Let $u_1=x,u_2=x^2,u_3=x+x^2,v_1=y_1,v_2=y_1+x^2,v_3=y_2,v_4=y_2+x^2,v_5=y_3,v_6=y_3+x^2,v_7=y_4+x^2,v_8=y_1+y_2,v_9=y_1+y_3,v_{10}=y_1+y_4,v_{11}=y_2+y_3,v_{12}=y_2+y_4,v_{13}=y_3+y_4,v_{14}=y_1+y_2+y_3,v_{15}=y_2+y_3+y_4$ and $v_{16}=y_1+y_2+y_3+y_4$. Then, $u_i·v_j=0$ for every $i,j$, so that $K_{3,16}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction.

Assume ${\mathfrak{m}}^2=\left(x{y}_1\right)$. Then, by Proposition 8, $x{y}_i=0$ for $i\ge 2$. Also, any element of ${\mathfrak{m}}^2-{\mathfrak{m}}^3$ generates ${\mathfrak{m}}^2$ and so we may assume that $x^2,y_i^2\in {\mathfrak{m}}^3=0$. Now, let $u_1=x,u_2=x+x{y}_1,u_3=y_1,u_4=y_1+x{y}_1,u_5=x+y_1,u_6=x+y_1+x{y}_1,$ and let $v_1=y_2,v_2=y_2+x{y}_1,v_3=y_3,v_4=y_3+x{y}_1,v_5=y_2+y_3,v_6=y_2+y_3+x{y}_1$. Then, $u_i·v_j=0$ for every $i,j$ and therefore $K_{6,6}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction.

Case 4.3: If ${\mathfrak{m}}^4=0$, then $|{\mathfrak{m}}^3| \ge 2$ and $|{\mathfrak{m}}^2| \ge 4$. Let $|{\mathfrak{m}}^3| =k\ge 4$. Then, $\mathrm{\Gamma }\left(R\right)$ contains $K_{k-1,64-k}$ as a subgraph, a contradiction. Therefore, $|{\mathfrak{m}}^3| =2$ and $|{\mathfrak{m}}^2| =4$. Consequently, $di{m}_{R/\mathfrak{m}}{\mathfrak{m}}^2/\mathfrak{m}=\mathsf{\ell }-3\ge 4.$ Now, by Proposition 2.8 [17], $\mathfrak{m}=(x,y_1,\dots ,y_{\mathsf{\ell }-4})$ for some $x,y_i\in \mathfrak{m}-{\mathfrak{m}}^2$, and without loss of generality, let ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(x{y}_1\right)$. In either case, by Propositions 7 and 8, we have $x{y}_i=0$ for $i\ge 2$. Since $|{\mathfrak{m}}^2| \ge 4$ and ${\mathfrak{m}}^2·{\mathfrak{m}}^2=0$, we can choose two non-zero elements $a,b\in {\mathfrak{m}}^2$ such that $a·b=0$. Let $u_1=x,u_2=x^2,u_3=a,u_4=b,u_5=x+x^2,u_6=x+a,u_7=x+b$ and $v_1=y_2,v_2=y_3,v_3=y_2+x^2,v_4=y_2+a,v_5=y_2+b,v_6=y_3+x^2,v_7=y_3+a,v_8=y_3+b.$ Clearly $u_i·v_j=0$ for every $i,j$ so that $K_{7,8}\subseteq \mathrm{\Gamma }\left(R\right)$. By Proposition 2, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 8$, a contradiction.

A similar proof technique is valid for the cases ${\mathfrak{m}}^5=0$ or ${\mathfrak{m}}^6=0$.

Thus, $\mathsf{\ell }\le 6$. But by Theorems 1 and 2, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\le 1$ for all local rings R of order less than or equal to 16. Hence, $\left|R\right|=32$ or 64. □

Lemma 1 

([8] (Proposition 7.1), [5] (Proposition 5)). Let R be a local ring with maximal ideal $\mathfrak{m}$. Then, there are 54 local commutative rings with order 32.

$$\begin{array}{c}{\mathbb{F}}_{32},{\mathbb{Z}}_{32},{\mathbb{Z}}_2\left[x\right]/\left(x^5\right),{\mathbb{Z}}_2[x,y]/(x^4,xy,y^2-x^3),{\mathbb{Z}}_2[x,y]/(x^4,xy,y^2),{\mathbb{Z}}_2[x,y,z,t]/{(x,y,z,t)}^2,\\ {\mathbb{Z}}_2[x,y]/(x^3,x^{2}y,y^2),{\mathbb{Z}}_2[x,y]/(x^3,y^3,y^2-xy),{\mathbb{Z}}_2[x,y]/(x^3,y^3,y^2-x^2-xy),\\ {\mathbb{Z}}_2[x,y,z]/(x^3,y^2,z^2,xy,xz,yz),{\mathbb{Z}}_2[x,y,z]/(y^2-x^2,z^2,xy,xz,yz),\\ {\mathbb{Z}}_2[x,y,z]/(y^2-x^2,z^2-x^2,xy,xz,yz),{\mathbb{Z}}_2[x,y,z]/(x^2,y^2,z^2,xy,yz),{\mathbb{Z}}_4\left[x\right]/(2x,x^4-2),\\ {\mathbb{Z}}_4\left[x\right]/(2x^2,x^3-2),{\mathbb{Z}}_4[x,y]/(x^3,y^2,xy+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3,x^2+y^2,xy+2,2x,2y),\\ {\mathbb{Z}}_4[x,y]/(x^3,x^2+y^2+2,xy+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3,y^2+2,xy,2x,2y),\\ {\mathbb{Z}}_4[x,y]/(x{y}^2,x^2,y^2+2,xy,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3,x^2+xy,y^2+2,2x,2y),\\ {\mathbb{Z}}_4[x,y]/(x^3-2,y^2,xy,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3-2,y^2-2,xy,2x),{\mathbb{Z}}_4\left[x\right]/(x^4,2x),{\mathbb{Z}}_4\left[x\right]/(x^3,2x^2),\end{array}$$

$$\begin{array}{c}{\mathbb{Z}}_4\left[x\right]/(x^3-2x,2x^2),{\mathbb{Z}}_4[x,y]/(x^2-2,y^2,xy,2y),{\mathbb{Z}}_4[x,y]/(x^2-2,y^2-2x,xy,2y),\\ {\mathbb{Z}}_4[x,y]/(x^2-2x-2,y^2,xy,2y),{\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2,z^2,xy,xz,yz,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2-2,z^2,xy,xz,yz,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2-2,z^2-2,xy,xz,yz,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2,y^2,z^2,xy,xz,yz-2,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2,z^2,xy,xz-2,yz,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2-2,z^2,xy,xz-2,yz,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2,y^2,z^2,xy-2,xz,yz,2x,2y,2z),{\mathbb{Z}}_4[x,y]/(x^2,y^2,xy,2y),\\ {\mathbb{Z}}_4[x,y]/(x^2,y^2-2x,xy,2y),{\mathbb{Z}}_4[x,y]/(x^2-2x,y^2,xy,2y),\\ {\mathbb{Z}}_4[x,y]/{(2,x,y,z)}^2,{\mathbb{Z}}_8\left[x\right]/(2x,x^3),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x-2),\\ {\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x+2),{\mathbb{Z}}_8\left[x\right]/(2x,x^3-4),{\mathbb{Z}}_8\left[x\right]/(2x,x^3),{\mathbb{Z}}_8\left[x\right]/(4x,x^2),\\ {\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x-4),{\mathbb{Z}}_8[x,y]/(x^2,y^2,xy,2x,2y),\\ {\mathbb{Z}}_8[x,y]/(x^2-4,y^2,xy,2x,2y),{\mathbb{Z}}_8[x,y]/(x^2-4,y^2-4,xy,2x,2y),\\ {\mathbb{Z}}_8[x,y]/(x^2,y^2,xy-4,2x,2y),{\mathbb{Z}}_{16}\left[x\right]/(2x,x^2),{\mathbb{Z}}_{16}\left[x\right]/(2x,x^2-8).\end{array}$$

The graph given in Figure 1 plays a vital role in characterizing local rings with genus three zero-divisor graphs. Therefore, we find its genus value in the following result.

Lemma 2.

For a graph H as shown in Figure 1, we have $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$.

Proof. 

In Figure 2, we provide an embedding of the graph H in the surface of $S_3$. Thus, $\gamma \left(\mathrm{\Gamma }\right(H\left)\right)=3$. □

Lemma 3.

Let $(R,\mathfrak{m})$ be a finite local ring and $|R/\mathfrak{m}|=4$. Then, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$ if and only if $\left|R\right|=64$ and $\mathfrak{m}$ has a nilpotency index of 3.

Proof. 

(⇒): Suppose that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3.$ Then, by Theorem 4, we have $\left|R\right|=64$ and ${\mathfrak{m}}^3=0$. It follows that $\left|\mathfrak{m}\right|=16$. If ${\mathfrak{m}}^2=0$, then any two non-zero element of $\mathfrak{m}$ is adjacent in $\mathrm{\Gamma }\left(R\right)$ and so $\mathrm{\Gamma }\left(R\right)$ contains $K_{15}$ as a subgraph. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 10$, a contradiction. Thus, the nilpotency index of $\mathfrak{m}$ is 3.

(⇐): Let $\left|R\right|=64$ and the nilpotency index of $\mathfrak{m}$ be 3. This implies that ${\mathfrak{m}}^2\ne 0$ and $|{\mathfrak{m}}^2|=4$. The facts ${\mathfrak{m}}^2\ne 0$, $\mathfrak{m}·{\mathfrak{m}}^2=0$ and ${\mathfrak{m}}^2·{\mathfrak{m}}^2=0$ imply that, in $\mathrm{\Gamma }\left(R\right)$, the subgraph induced by the set $\mathfrak{m}-{\mathfrak{m}}^2$ is empty, every element in $\mathfrak{m}-{\mathfrak{m}}^2$ is adjacent to all the element of ${\mathfrak{m}}^2$ and the subgraph induced by ${\mathfrak{m}}^2-\left\{0\right\}$ is complete. Therefore, $\mathrm{\Gamma }\left(R\right)$ is isomorphic to $\overline{K_{12}}\vee K_3$. So $\mathrm{\Gamma }\left(R\right)$ is a subgraph of the graph H given in Figure 1. Thus, by Lemma 2, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$. □

Lemma 4.

Let $(R,\mathfrak{m})$ be a finite local ring and $|R/\mathfrak{m}|=3$. Then, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ne 3$.

Proof. 

Assume to the contrary that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$. Then, by Theorem 4, $\left|R\right|=81$ and ${\mathfrak{m}}^4=0$. Note that R is isomorphic to one of the following 22 commutative local rings [18].

$$\begin{array}{c}{\mathbb{F}}_{81},{\mathbb{Z}}_{81},{\mathbb{F}}_9\left[x\right]/\left(x^2\right),{\mathbb{Z}}_3\left[x\right]/\left(x^4\right),{\mathbb{Z}}_3[x,y]/(x^2-y^2,xy),{\mathbb{Z}}_3[x,y]/(x^2-2y^2,xy),\\ {\mathbb{Z}}_3[x,y]/(x^3,xy,y^2),{\mathbb{Z}}_3[x,y,z]/{(x,y,z)}^2,{\mathbb{Z}}_9\left[x\right]/\left(x^2\right),{\mathbb{Z}}_9\left[x\right]/(x^2-3),{\mathbb{Z}}_9\left[x\right]/(x^2-6),\\ {\mathbb{Z}}_9\left[x\right]/(3x,x^3),{\mathbb{Z}}_9\left[x\right]/(x^2-2),{\mathbb{Z}}_9\left[x\right]/(3x,x^3-3),{\mathbb{Z}}_9[x,y]/(x^2,y^2,xy,3x,3y),\\ {\mathbb{Z}}_9[x,y]/(x^2-3,y^2,xy,3x,3y),{\mathbb{Z}}_9[x,y]/(x^2-6,y^2,xy,3x,3y),\\ {\mathbb{Z}}_9[x,y]/(x^2-3,y^2-3,xy,3x,3y),{\mathbb{Z}}_9[x,y]/(x^2-3,y^2-6,xy,3x,3y),\\ {\mathbb{Z}}_{27}\left[x\right]/(3x,x^2),{\mathbb{Z}}_{27}\left[x\right]/(3x,x^2-9),{\mathbb{Z}}_{27}\left[x\right]/(3x,x^2-18).\end{array}$$

(1) If $R\cong {\mathbb{F}}_{81}$, then the zero-divisor graph $\mathrm{\Gamma }\left(R\right)$ is empty, a contradiction.

(2) Let R be isomorphic to any one of the following seven rings:

$$\begin{array}{c}{\mathbb{Z}}_{81},{\mathbb{F}}_9\left[x\right]/\left(x^2\right),{\mathbb{Z}}_3\left[x\right]/\left(x^4\right),{\mathbb{Z}}_9\left[x\right]/(x^2-3),{\mathbb{Z}}_3\left[x\right]/(x^2-6),{\mathbb{Z}}_9\left[x\right]/(3x,x^3-3),\\ {\mathbb{Z}}_9\left[x\right]/(x^2-2).\end{array}$$

Then, by Theorem 3, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=2$, a contradiction.

(3) If $R\cong {\mathbb{Z}}_9\left[x\right]/\left(x^2\right)$, then every vertex of the set $V_1=\{x+3,4x+3,7x+3,2x+6,5x+6,8x+6\}$ is adjacent to all the vertices of the set $V_2=\{2x+3,5x+3,8x+3,x+6,4x+6,7x+6\}$. Therefore, $\mathrm{\Gamma }\left(R\right)$ contains $K_{6,6}$ as a subgraph and so, by Proposition 2, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

(4) If $R\cong {\mathbb{Z}}_3[x,y,z]/{(x,y,z)}^2$, then the subgraph induced by the set $\{x,y,z,2x,2y,2z,x+y,x+z,y+z,x+2y,2x+2y,2x+y,2y+z,y+2z,2y+2z,2x+z,x+2z,2x+2z,x+y+z,2x+y+z,x+2y+z,x+y+2z,2x+2y+z,2x+y+2z,x+2y+2z,2x+2y+2z\}$ forms $K_{26}$ in $\mathrm{\Gamma }\left(R\right)$ and so, by Proposition 1, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 42$.

(5) If $R\cong {\mathbb{Z}}_3[x,y,z]/{(x^3,xy,y^2)}^2$, then the partite sets $V_1=\{x^2,y,2x^2,2y,x^2+y,2x^2+y,x^2+2y,2x^2+2y\}$ and $V_2=\{x,2x,x^2+x,x+y,2x^2+x,x^2+2x,2x^2+2x\}$ form $K_{8,7}$ in $\mathrm{\Gamma }\left(R\right)$, which implies that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

(6) If $R\cong {\mathbb{Z}}_9\left[x\right]/(3x,x^3)$, then the partite sets $V_1=\{3,6,x^2,2x^2\}$ and $V_2=\{x,2x,x^2+x,x^2+2x,2x^2+x,2x^2+2x,x+3,2x+3,x^2+3,x^2+x+3,x^2+2x+3,2x^2+3,2x^2+x+3,2x^2+2x+3,x+6,2x+6,x^2+6,x^2+x+6,x^2+2x+6,2x^2+6,2x^2+x+6,2x^2+2x+6\}$ form $K_{4,22}$ as a subgraph in $\mathrm{\Gamma }\left(R\right)$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 10$.

(7) Let R be isomorphic to any one of the following five rings:

$$\begin{array}{c}{\mathbb{Z}}_9[x,y]/(x^2,y^2,xy,3x,3y),{\mathbb{Z}}_9[x,y]/(x^2-3,y^2,xy,3x,3y),\\ {\mathbb{Z}}_9[x,y]/(x^2-6,y^2,xy,3x,3y),{\mathbb{Z}}_9[x,y]/(x^2-3,y^2-3,xy,3x,3y),\\ {\mathbb{Z}}_9[x,y]/(x^2-3,y^2-6,xy,3x,3y).\end{array}$$

Then, the partite sets $V_1=\{x,2x,x+3,x+6,2x+3,2x+6\}$ and $V_2=\{y,2y,y+3,y+6,2y+3,2y+6\}$ form a subgraph $K_{6,6}$ in $\mathrm{\Gamma }\left(R\right)$. Hence, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

(8) Let $R\cong {\mathbb{Z}}_{27}\left[x\right]/(3x,x^2)$ or ${\mathbb{Z}}_{27}\left[x\right]/(3x,x^2-9)$ or ${\mathbb{Z}}_{27}\left[x\right]/(3x,x^2-18)$. Then, $\mathrm{\Gamma }\left(R\right)$ contains a subgraph $K_{6,8}$, where the partite sets are $V_1=\{3,6,12,15,21,24\}$ and $V_2=\{9,18,x,2x,x+9,x+18,2x+9,2x+18\}$. Thus, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 6$.

(9) Let $R\cong {\mathbb{Z}}_3[x,y]/(x^2-y^2,xy)$ or ${\mathbb{Z}}_3[x,y]/(x^2-2y^2,xy)$. Then, $\mathrm{\Gamma }\left(R\right)$ contains $K_{5,7}$ as a subgraph with the bipartite sets $V_1=\{y,2y,x^2+y,2x^2+y,2x^2+2y\}$ and $V_2=\{x,x^2,2x,2x^2,x^2+x,2x^2+x,2x^2+2x\}$. Thus, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

Hence, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ne 3$ for all the 22 local rings of order 81 with ${\mathfrak{m}}^4=0$, which is in contradiction to our assumption. □

Theorem 5.

Let $(R,\mathfrak{m})$ be a local ring. Then, the genus of the zero-divisor graph of R is 3 if and only if one of the following conditions hold:

(i)

The residue field $|R/\mathfrak{m}|=4$ with $\left|R\right|=64$, $|{\mathfrak{m}}^2|=4$ and ${\mathfrak{m}}^3=0.$

(ii)

The residue field $|R/\mathfrak{m}|=2$ with $\left|R\right|=64$, $|{\mathfrak{m}}^2|=16,|{\mathfrak{m}}^3|=8,|{\mathfrak{m}}^4|=4$, $|{\mathfrak{m}}^5|=2$ and ${\mathfrak{m}}^6=0$.

(iii)

Any one of the local rings of order 32.

$$\begin{array}{c}{\mathbb{Z}}_2[x,y]/(x^4,xy,y^2),{\mathbb{Z}}_2[x,y]/(x^3,x^{2}y,y^2),{\mathbb{Z}}_2[x,y]/(x^3,y^3,y^2-xy),\\ {\mathbb{Z}}_2[x,y]/(x^3,y^3,y^2-x^2-xy),{\mathbb{Z}}_4[x,y]/(x^3,x^2+y^2+2,xy+2,2x,2y),{\mathbb{Z}}_4\left[x\right]/(x^4,2x),\\ {\mathbb{Z}}_4\left[x\right]/(x^3,2x^2),{\mathbb{Z}}_4[x,y]/(x^3,x^2+y^2,xy+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3,y^2,xy+2,2x,2y),\\ {\mathbb{Z}}_8\left[x\right]/(4x,x^2),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x-4),{\mathbb{Z}}_{16}\left[x\right]/(2x,x^2),\\ {\mathbb{Z}}_4[x,y]/(x{y}^2,x^2,y^2+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3-2,y^2,xy,2x,2y),\\ {\mathbb{Z}}_4[x,y]/(x^2-2,y^2,xy,2y),{\mathbb{Z}}_4[x,y]/(x^2-2x-2,y^2,xy,2y),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-z,y^2,z^2,xy,yz,xz-2,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-z,y^2-2,z^2,xy,yz,xz-2,2x,2y,2z).\end{array}$$

Proof. 

Let $(R,\mathfrak{m})$ be a local ring. Suppose $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3.$ Then, by Theorem 4, the number of elements in the residue field $R/\mathfrak{m}$ is 2 or 3 or 4.

If $|R/\mathfrak{m}|=4$, then, by Lemma 3, we have $\left|R\right|=64$ and also $\mathfrak{m}$ has a nilpotency index of 3. This implies that $\left|\mathfrak{m}\right|=16$, $|{\mathfrak{m}}^2|=4$ and ${\mathfrak{m}}^3=0.$

If $|R/\mathfrak{m}|=3$, then, by Lemma 4, we obtain $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ne 3,$ a contradiction.

Let $|R/\mathfrak{m}|=2$. By Theorem 4, we have $\left|R\right|=64$ or 32 with ${\mathfrak{m}}^6=0$.

Case 1: Suppose $\left|R\right|=64$ with ${\mathfrak{m}}^6=0$. Then, $\left|\mathfrak{m}\right|=32$.

We claim that the nilpotency index of $\mathfrak{m}$ is 6. In order to prove this claim, assume to the contrary that ${\mathfrak{m}}^{\mathsf{\ell }}=0$ for some $\mathsf{\ell }\le 5.$

Case 1.1: If ${\mathfrak{m}}^2=0$, then the 31 non-zero elements in $\mathfrak{m}$ together form a complete graph in $\mathrm{\Gamma }\left(R\right)$. So, by Proposition 1, we have $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 63$, a contradiction.

Case 1.2: Suppose ${\mathfrak{m}}^3=0$. If $|{\mathfrak{m}}^2|=k\ge 4$, then $\mathrm{\Gamma }\left(R\right)$ contains $K_{k-1,32-k}$ as a subgraph and so, by Proposition 2, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 7$. Therefore, $|{\mathfrak{m}}^2|=2$. This implies that $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=4$. So we assume $\mathfrak{m}=(x,y_1,y_2,y_3)$ for some $x,y_i\in \mathfrak{m}-{\mathfrak{m}}^2$. Without loss of generality, we may assume that ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(x{y}_1\right)$.

Case 1.2.1: Assume that ${\mathfrak{m}}^2=\left(x^2\right)$. Then, by Proposition 7, we have $x{y}_i=0$.

(i). If $y_i{y}_j=0$ for all $1\le i\ne j\le 3$, then $K_{5,7}\subseteq \mathrm{\Gamma }\left(R\right)$, where the bipartite sets are $\{x,x^2,x+x^2,y_3,y_3+x^2\}$ and $\{y_1,y_1+x^2,y_2,y_2+x^2,y_1+y_2,y_1+y_2+x^2,y_1+y_3\}$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

(ii). Suppose exactly one element of $\{y_1{y}_2,y_1{y}_3,y_2{y}_3\}$ is non-zero, say $y_1{y}_2\ne 0$. That is, $y_1{y}_3=y_2{y}_3=0$. Let $u_1=x,u_2=x^2,u_3=x+x^2,u_4=y_3,u_5=y_3+x^2,v_1=y_1,v_2=y_1+x^2,v_3=y_2,v_4=y_2+x^2,v_5=y_1+y_2,v_6=y_1+y_2+x^2$ and $v_7=y_1+y_3$. Clearly, $u_i{v}_j=0$ for all $i=1,2,\dots ,5$ and $j=1,2,\dots ,7$ and so $K_{5,7}\subseteq \mathrm{\Gamma }\left(R\right)$. Thus $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

(iii). Suppose exactly two elements of $\{y_1{y}_2,y_1{y}_3,y_2{y}_3\}$ are non-zero. If $y_1{y}_2\ne 0$ and $y_1{y}_3\ne 0$, then $y_1{y}_2=x^2=y_1{y}_3$. This implies that $y_1(y_3-y_2)=0$. Let us take $y_3^{\prime }=(y_3-y_2)$. Clearly, $x{y}_3^{\prime }=y_1{y}_3^{\prime }=0$ and so $(x-y_1)y_3^{\prime }=0.$ Therefore, $(x,y_1,y_2,y_3)=(x,y_1,y_2,y_3^{\prime })$. Similarly, if $y_1{y}_2\ne 0$ and $y_2{y}_3\ne 0$, then $y_1{y}_2=x^2=y_2{y}_3$, so that $y_2(y_3-y_1)=0$. Let $y_3^{\prime }=(y_3-y_1)$, and we have $(x,y_1,y_2,y_3)=(x,y_1,y_2,y_3^{\prime })$ and $x{y}_3^{\prime }=y_2{y}_3^{\prime }=0$. So $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

Case 1.2.2: Assume that ${\mathfrak{m}}^2=\left(x{y}_1\right)$. Then, none of $x^2,y_1^2,y_2^2$ and $y_3^2$ are in ${\mathfrak{m}}^2-{\mathfrak{m}}^3$. Thus, $x^2=y_1^2=y_2^2=y_3^2=0$. By Proposition 8, we obtain $x·y_2=0=x·y_3$. If necessary, replacing $y_2$ by $y_2-x$ and $y_3$ by $y_3-x$, we have $y_1·y_2=0=y_1·y_3$. Now, let $u_1=x,u_2=x+x{y}_1,u_3=y_1,u_4=y_1+x{y}_1,u_5=x+y_1,u_6=x+y_1+x{y}_1,v_1=y_2,v_2=y_2+x{y}_1,v_3=y_3,v_4=y_3+x{y}_1,v_5=y_2+y_3$ and $v_6=y_2+y_3+x{y}_1$. Then, $u_i·v_j=0$ for every $i,j$, so that $K_{6,6}\subseteq \mathrm{\Gamma }\left(R\right)$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

Case 1.3: Suppose ${\mathfrak{m}}^4=0$. Then, $|{\mathfrak{m}}^3| \ge 2$ and $|{\mathfrak{m}}^2| \ge 4$. If $|{\mathfrak{m}}^3| \ge 4$, then $\mathrm{\Gamma }\left(R\right)$ contains $K_{3,28}$ as a subgraph and so $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 7$. Therefore, $|{\mathfrak{m}}^3|=2$ and $|{\mathfrak{m}}^2|=4$. Consequently, $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=3$. So, let $\mathfrak{m}=(x,y_1,y_2)$ for some $x,y_i\in \mathfrak{m}-{\mathfrak{m}}^2$, and without loss of generality, assume ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(x{y}_1\right)$.

First, assume that ${\mathfrak{m}}^2=\left(x^2\right)$. Then, $x·y_1=0=x·y_2$. Since $|{\mathfrak{m}}^2| =4$, we let $a,b\in {\mathfrak{m}}^2-\left\{0\right\}$. This implies that $a·b=0$. Let $u_1=x,u_2=x^2,u_3=a,u_4=b,u_5=x+x^2,u_6=x+a,u_7=x+b,v_1=y_1,v_2=y_2,v_3=y_1+x^2,v_4=y_1+a,v_5=y_1+b,v_6=y_2+x^2,v_7=y_2+a$ and $v_8=y_2+b.$ Since $u_i·v_j=0$ for every $i,j$, we have $K_{7,8}\subseteq \mathrm{\Gamma }\left(R\right)$ so that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 8$.

Second, assume that ${\mathfrak{m}}^2=\left(x{y}_1\right)$. Then, by Lemma 8, we have $x{y}_2=0$. Moreover, we may assume that each of $x^2,y_1^2,$ and $y_2^2$ are in ${\mathfrak{m}}^3$. However, ${\mathfrak{m}}^3=\mathfrak{m}·{\mathfrak{m}}^2=(x^2{y}_1,x{y}_1^2)\subseteq {\mathfrak{m}}^4=0$, a contradiction.

Case 1.4: Suppose ${\mathfrak{m}}^5=0$. Then, $|{\mathfrak{m}}^4| \ge 2$, $|{\mathfrak{m}}^3| \ge 4$ and $|{\mathfrak{m}}^2| \ge 8$. If $|{\mathfrak{m}}^3| =8$, then $|{\mathfrak{m}}^2| =16.$ Since ${\mathfrak{m}}^2·{\mathfrak{m}}^3=0$, we have $\mathrm{\Gamma }\left(R\right)$ contains $K_{7,8}$ as a subgraph and so $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 7$, a contradiction. Therefore, $|{\mathfrak{m}}^3| =4$. Consequently, $|{\mathfrak{m}}^4| =2$. Thus, just two cases can occur: either $|{\mathfrak{m}}^2| =8$ or $|{\mathfrak{m}}^2| =16$.

First, suppose that $|{\mathfrak{m}}^2|=8$. Then, $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=2$, and we have $\mathfrak{m}=(x,y)$ for some $x,y\in \mathfrak{m}-{\mathfrak{m}}^2$. Without loss of generality, ${\mathfrak{m}}^2=\left(x^2\right)$ or ${\mathfrak{m}}^2=\left(xy\right)$. Let ${\mathfrak{m}}^2=\left(x^2\right)$. Then, $x·y=0$ and so ${\mathfrak{m}}^3=\left(x^3\right)$. Clearly, $y+\left(x^2\right)\cap \left(x^2\right)$ and $y+\left(x^3\right)\cap \left(x^2\right)$ are mutually disjoint sets. Note that, in $\mathrm{\Gamma }\left(R\right)$, every elements in the set $\left\{x^3\right\}\cup y+{\mathfrak{m}}^3$ is adjacent to all the elements of ${\mathfrak{m}}^2-\left\{0\right\}$ so that $K_{5,7}\subseteq \mathrm{\Gamma }\left(R\right)$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$. Let ${\mathfrak{m}}^2=\left(xy\right)$. Since the cosets $\left(xy\right),x+{\mathfrak{m}}^2$ and $y+{\mathfrak{m}}^2$ are mutually disjoint and since ${\mathfrak{m}}^3\subseteq {\mathfrak{m}}^2$, we have the sets $x+{\mathfrak{m}}^2$, $y+{\mathfrak{m}}^2$ and ${\mathfrak{m}}^3$ which are mutually disjoint. Then, in $\mathrm{\Gamma }\left(R\right)$, every element in $X=x+{\mathfrak{m}}^2\cup y+{\mathfrak{m}}^2$ is adjacent to every element of $Y={\mathfrak{m}}^3-\left\{0\right\}$ so that $K_{3,16}\subseteq \mathrm{\Gamma }\left(R\right)$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

Second, suppose that $|{\mathfrak{m}}^2|=16$. Then, $di{m}_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=1$ and so $\mathfrak{m}$ is principal. Therefore, ${\mathfrak{m}}^i$ is principal for every i. This implies that $di{m}_{R/\mathfrak{m}}{\mathfrak{m}}^i/{\mathfrak{m}}^{i+1}=1$ for all $i=1,2,3$, which contradicts the fact that $di{m}_{R/\mathfrak{m}}{\mathfrak{m}}^2/{\mathfrak{m}}^3=2$.

Case 1.5: Suppose ${\mathfrak{m}}^6=0$. Then, $|{\mathfrak{m}}^5| =2,|{\mathfrak{m}}^4| =4,|{\mathfrak{m}}^3| =8,|{\mathfrak{m}}^2| =16$. Consequently, $di{m}_{R/\mathfrak{m}}{\mathfrak{m}}^i/{\mathfrak{m}}^{i+1}=1$ for all $i=1,\dots ,4$ and ${\mathfrak{m}}^i$’s are the principal ideal for all $i=1,\dots ,5$. Note that, in $\mathrm{\Gamma }\left(R\right)$, the subgraph induced by ${\mathfrak{m}}^3$ is complete, every element in ${\mathfrak{m}}^2-{\mathfrak{m}}^3$ is adjacent to all the vertices of ${\mathfrak{m}}^4$ and every element in $\mathfrak{m}-{\mathfrak{m}}^2$ is adjacent to the single non-zero element of ${\mathfrak{m}}^5.$ By labeling the vertices of $\mathrm{\Gamma }\left(R\right)$ by ${\mathfrak{m}}^4-\left\{0\right\}=\{a,b,c\}$, ${\mathfrak{m}}^3-{\mathfrak{m}}^4=\{1,2,3,4\}$ and ${\mathfrak{m}}^2-{\mathfrak{m}}^3=\{5,6,\dots ,12\}$, it follows that $\widetilde{\mathrm{\Gamma }\left(R\right)}$ is isomorphic to the graph given in Figure 1. Thus, by Lemma 2, we obtain $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$.

Hence, in Case 1, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$ if and only if $\left|R\right|=64$ such that ${\mathfrak{m}}^6=0$, $|{\mathfrak{m}}^5| =2,|{\mathfrak{m}}^4| =4,|{\mathfrak{m}}^3| =8$ and $|{\mathfrak{m}}^2| =16$.

Case 2: Suppose $\left|R\right|=32$ with ${\mathfrak{m}}^6=0$. This implies that $\left|\mathfrak{m}\right|=16$. Then, by Lemma 1, there exist 53 (non-field) local rings.

(1). Assume that R is isomorphic to any one of the following eight rings:

$$\begin{array}{c}{\mathbb{Z}}_{32},{\mathbb{Z}}_2\left[x\right]/\left(x^5\right),{\mathbb{Z}}_4\left[x\right]/(2x,x^4-2),{\mathbb{Z}}_4\left[x\right]/(2x^2,x^3-2),{\mathbb{Z}}_4\left[x\right]/(x^3-2x,2x^2),\\ {\mathbb{Z}}_8\left[x\right]/(4x,x^2-2),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x-2),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x+2).\end{array}$$

Then, by Theorem 2, we have $\mathrm{\Gamma }\left(R\right)$ as toroidal.

(2). Assume that R is isomorphic to one of the following five rings:

$$\begin{array}{c}{\mathbb{Z}}_2[x,y]/(x^4,xy,y^2-x^3),{\mathbb{Z}}_4[x,y]/(x^2-2,y^2-2x,xy,2y),\\ {\mathbb{Z}}_4[x,y]/(x^3-2,y^2-2,xy,2x),{\mathbb{Z}}_8\left[x\right]/(2x,x^3-4),{\mathbb{Z}}_{16}\left[x\right]/(2x,x^2-8).\end{array}$$

Then, by Theorem 3, we have that $\mathrm{\Gamma }\left(R\right)$ is double toroidal.

(3). Assume that R is isomorphic to one of the following 19 rings:

$$\begin{array}{c}{\mathbb{Z}}_2[x,y]/(x^4,xy,y^2),{\mathbb{Z}}_2[x,y]/(x^3,x^{2}y,y^2),{\mathbb{Z}}_2[x,y]/(x^3,y^3,y^2-xy),\\ {\mathbb{Z}}_2[x,y]/(x^3,y^3,y^2-x^2-xy),{\mathbb{Z}}_4[x,y]/(x^3,x^2+y^2+2,xy+2,2x,2y),{\mathbb{Z}}_4\left[x\right]/(x^4,2x),\\ {\mathbb{Z}}_4\left[x\right]/(x^3,2x^2),{\mathbb{Z}}_4[x,y]/(x^3,x^2+y^2,xy+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3,y^2,xy+2,2x,2y),\\ {\mathbb{Z}}_8\left[x\right]/(4x,x^2),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x),{\mathbb{Z}}_8\left[x\right]/(4x,x^2-2x-4),{\mathbb{Z}}_{16}\left[x\right]/(2x,x^2),\\ {\mathbb{Z}}_4[x,y]/(x{y}^2,x^2,y^2+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3-2,y^2,xy,2x,2y),\\ {\mathbb{Z}}_4[x,y]/(x^2-2,y^2,xy,2y),{\mathbb{Z}}_4[x,y]/(x^2-2x-2,y^2,xy,2y),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-z,y^2,z^2,xy,yz,xz-2,2x,2y,2z),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-z,y^2-2,z^2,xy,yz,xz-2,2x,2y,2z).\end{array}$$

Then, the corresponding zero-divisor graphs of R are provided in a tabular format.

Thus, all the zero-divisor graphs displayed in

Table 1. Zero-divisor graphs of certain rings.

Ring	Vertex Partition	Graph
$\mathit{R}$	$\mathit{V}\left(\mathrm{\Gamma }\left(\mathit{R}\right)\right)={\mathit{V}}_{\mathbf{1}}\cup {\mathit{V}}_{\mathbf{2}}$	$\mathrm{\Gamma }\left(\mathit{R}\right)$
$\frac{{\mathbb{Z}}_2[x,y]}{(x^4,xy,y^2)}$	$V_1=\{x^3,x^2+y,y\}$ $V_2=\{x^2+x+y,x^3+x+y,x^3+x^2+x,$ $x^3+x^2+x+y,x+y,x^2+x,x^2,x^3+x^2,$ $x^3+x^2+y,x^2+y,x,x^3+x\}$	Figure 1
$\frac{{\mathbb{Z}}_2[x,y]}{(x^3,x^{2}y,y^2)}$	$V_1=\{x^2,xy,x^2+xy\}$ $V_2=\{y,x^2+y,y+xy,x^2+y+xy,$ $x,x^2+x,x^2+x+y,x+xy,x+y,$ $x^2+x+y+xy,x+y+xy,x^2+x+xy\}$	Figure 1
$\frac{{\mathbb{Z}}_2[x,y]}{(x^3,y^3,y^2-xy)}$	$V_1=\{x^2,xy,x^2+xy\}$ $V_2=\{y,x^2+y,y+xy,x^2+y+xy,$ $x,x^2+x,x^2+x+y,x+xy,x+y,$ $x^2+x+y+xy,x+y+xy,x^2+x+xy\}$	$\overline{K_{12}}\vee K_3$
$\frac{{\mathbb{Z}}_2[x,y]}{(x^3,y^3,y^2-x^2-xy)}$	$V_1=\{x^2,xy,x^2+xy\}$ $V_2=\{y,x^2+y,y+xy,x^2+y+xy,$ $x,x^2+x,x^2+x+y,x+xy,x+y,$ $x^2+x+y+xy,x+y+xy,x^2+x+xy\}$	$\overline{K_{12}}\vee K_3$
$\frac{{\mathbb{Z}}_4[x,y]}{(x^3,x^2+y^2,xy+2,2x,2y)}$	$V_1=\{y^2,2,y^2+2\}$ $V_2=\{x,y,x+y,x+2,y+2,y^2+y,$ $y^2+x,y^2+x+y,y^2+x+2,$ $y^2+y+2,y^2+y+x+2,x+y+2\}$	$\overline{K_{12}}\vee K_3$
$\frac{{\mathbb{Z}}_4[x,y]}{(x^3,x^2+y^2+2,xy+2,2x,2y)}$	$V_1=\{y^2,2,y^2+2\}$ $V_2=\{x,y,x+y,x+2,y+2,y^2+y,$ $y^2+x,y^2+x+y,y^2+x+2,$ $y^2+y+2,y^2+y+x+2,x+y+2\}$	$\overline{K_{12}}\vee K_3$
$\frac{{\mathbb{Z}}_4[x,y]}{(x^3,y^2,xy+2,2x,2y)}$	$V_1=\{x^2,2,x^2+2\}$ $V_2=\{x,y,x+y,x+2,y+2,$ $x^2+x,x^2+y,x^2+x+y,x^2+x+2,$ $x^2+y+2,x^2+x+y+2,x+y+2\}$	Figure 1
$\frac{{\mathbb{Z}}_4[x,y]}{(x{y}^2,x^2,y^2+2,2x,2y)}$	$V_1=\{xy,2,xy+2\}$ $V_2=\{x,x+2,x+xy,x+xy+2,y,$ $y+xy,y+2,x+y,y+xy+2,$ $x+y+xy+2,x+y+2,x+y+xy\}$	Figure 1
$\frac{{\mathbb{Z}}_4[x,y]}{(x^3-2,y^2,xy,2x,2y)}$	$V_1=\{y,2,y+2\}$ $V_2=\{x^2,x^2+2,x^2+y+2,x^2+y,x,$ $x+y,x+2,x^2+x,x^2+x+y,$ $x+y+2,x^2+x+2,x^2+x+y+2\}$	Figure 1
$\frac{{\mathbb{Z}}_4[x,y]}{(x^2-2,y^2,xy,2y)}$	$V_1=\{y,2x,2x+y\}$ $V_2=\{2,2x+2,y+2,2x+y+2,x,3x,x+2,$ $x+y,3x+y,3x+2,3x+y+2,x+y+2\}$	Figure 1
$\frac{{\mathbb{Z}}_4[x,y]}{(x^2-2x-2,y^2,xy,2y)}$	$V_1=\{y,2x,2x+y\}$ $V_2=\{2,2x+2,y+2,2x+y+2,x,3x,x+2,$ $x+y,3x+y,3x+2,3x+y+2,x+y+2\}$	Figure 1
$\frac{{\mathbb{Z}}_4\left[x\right]}{(x^4,2x)}$	$V_1=\{x^3,x^3+2,2\}$ $V_2=\{x,x^2,x^2+2,x^3+x^2,x^3+x+2,$ $x+2,x^3+x+2,x^3+x^2+x+2,$ $x^2+x+2,x^3+x,x^2+x,x^3+x^2+x\}$	Figure 1
$\frac{{\mathbb{Z}}_4\left[x\right]}{(x^3,2x^2)}$	$V_1=\{x^2,2x,x^2+2x\}$ $V_2=\{2,x^2+2,2x+2,x^2+2x+2,x,$ $3x,x^2+x,x+2,x^2+3x,3x+2,$ $x^2+3x+2,x^2+x+2,x^2+2x+2\}$	Figure 1
$\frac{{\mathbb{Z}}_8\left[x\right]}{(4x,x^2)}$	$V_1=\{4,2x,2x+4\}$ $V_2=\{2,6,x,3x,x+4,3x+4,x+2,$ $x+6,3x+2,3x+6,2x+2,2x+6,\}$	Figure 1
$\frac{{\mathbb{Z}}_8\left[x\right]}{(4x,x^2-2x)}$	$V_1=\{4,2x,2x+4\}$ $V_2=\{2,6,x,3x,x+4,3x+4,x+2,$ $x+6,3x+2,3x+6,2x+2,2x+6,\}$	Figure 1
$\frac{{\mathbb{Z}}_8\left[x\right]}{(4x,x^2-2x-4)}$	$V_1=\{4,2x,2x+4\}$ $V_2=\{2,6,x,3x,x+4,3x+4,x+2,x+6,$ $3x+2,3x+6,2x+2,2x+6,\}$	Figure 1
$\frac{{\mathbb{Z}}_{16}\left[x\right]}{(2x,x^2)}$	$V_1=\{8,x,x+8\}$ $V_2=\{2,4,6,10,12,14,x+2,x+4,x+6,$ $x+10,x+12,x+14,3x+6,2x+2,2x+6,\}$	Figure 1

are either the graph H (refer Figure 1) or a subgraph of H. Hence, by Lemma 2, in this case, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$ for all the 17 local rings.

(4). Let $R\cong {\mathbb{Z}}_2[x,y,z,t]/{(x,y,z,t)}^2$ and $S={\mathbb{Z}}_4[x,y,z]/{(2,x,y,z)}^2$. Then, $V\left(\mathrm{\Gamma }\right(R\left)\right)=\{x,y,z,t,x+y,x+z,x+t,x+y+z,x+y+t,x+y+z+t,y+z,y+t,y+z+t,z+t,x+z+t\}$ and $V\left(\mathrm{\Gamma }\right(S\left)\right)=\{x,y,z,2,x+2,y+2,z+2,x+y,x+z,y+z,x+y+z,x+y+2,x+z+2,y+z+2,x=y+z+2\}$. It is clear that $\mathrm{\Gamma }\left(R\right)$ and $\mathrm{\Gamma }\left(S\right)$ are isomorphic to $K_{15}$ and so $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 11$.

(5). Assume that R is isomorphic to one of the following three rings:

$$\begin{array}{c}{\mathbb{Z}}_2[x,y,z]/(x^3,y^2,z^2,xy,xz,yz),{\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2,z^2,xy,xz,yz,2x,2y,2z),\\ {\mathbb{Z}}_8[x,y]/(x^2,y^2,xy,2x,2y).\end{array}$$

If $R\cong {\mathbb{Z}}_2[x,y,z]/(x^3,y^2,z^2,xy,xz,yz)$, then the partite sets $\{x^2,y,z,x^2+y,x^2+z,x^2+y+z,y+z\}$ and $\{x,x^2+x,x^2+x+y,x^2+x+z,x+y,x+z,x+y+z,x^2+x+y+z\}$ forms $K_{7,8}$ in $\mathrm{\Gamma }\left(R\right)$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 8$.

If $R\cong {\mathbb{Z}}_4[x,y]/(x^3,y^2,xy,2x,2y)$, then the partite sets $\{2,2+y,x^2,y,x^2+y,x^2+2,x^2+y+2\}$ and $\{x,x+2,x+y,x^2+x+y,x+y+2,x^2+x+y+2,x^2+x,x^2+x+2\}$ forms $K_{7,8}$ in $\mathrm{\Gamma }\left(R\right)$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 8$.

If $R\cong {\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2,z^2,xy,xz,yz,2x,2y,2z)$, then every vertex in the set $\{2,y,z,y+2,z+2,y+z,y+z+2\}$ is adjacent to all the vertices of $\{x,x+2,x+y,x+z,x+y+2,x+z+2,x+y+z,x+y+z+2\}$ in $\mathrm{\Gamma }\left(R\right)$. So $K_{7,8}\subseteq \mathrm{\Gamma }\left(R\right)$, a contradiction.

If $R\cong {\mathbb{Z}}_8[x,y]/(x^2,y^2,xy,2x,2y)$, then $\mathrm{\Gamma }\left(R\right)$ contains $K_{7,8}$, as a subgraph with bipartite sets $\{4,x,y,4+x,4+y,x+y,x+y+4\}$ and $\{2,6,2+x,2+y,6+x,6+y,x+y+2,x+y+6\}$, a contradiction.

(6). Assume that R is isomorphic to one of the following three rings:

$$\begin{array}{c}{\mathbb{Z}}_2[x,y,z]/(y^2-x^2,z^2,xy,xz,yz),{\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2-2,z^2,xy,xz,yz,2x,2y,2z),\\ {\mathbb{Z}}_8[x,y]/(x^2-4,y^2,xy,2x,2y).\end{array}$$

Note that the number of vertices and number of edges in $\mathrm{\Gamma }\left(R\right)$ is $n=15$ and $m=62$, respectively. If $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$, then, by Proposition 3, there are 43 faces in any $S_3$-embedding of $\mathrm{\Gamma }\left(R\right)$. So the average length of the faces is $\frac{2m}{f}=2.88$, which is less than the girth value(=3) of $\mathrm{\Gamma }\left(R\right)$, a contradiction.

(7). Assume that R is isomorphic to one of the following five rings:

$$\begin{array}{c}{\mathbb{Z}}_4[x,y]/(x^3,x^2+xy,y^2+2,2x,2y),{\mathbb{Z}}_4[x,y]/(x^3,y^2+2,xy,2x,2y),\\ {\mathbb{Z}}_4[x,y]/(x^2-y^2,xy,2x,2y),{\mathbb{Z}}_4[x,y,z]/(x^2-2x,y^2,xy,2y),{\mathbb{Z}}_8\left[x\right]/(2x,x^3)\end{array}$$

Let $R\cong {\mathbb{Z}}_4[x,y]/(x^3,x^2+xy,y^2+2,2x,2y)$. Then, in $\mathrm{\Gamma }\left(R\right)$, every vertex of the set $V_1=\{2,xy,2+xy\}$ is adjacent to all the vertices of $V_2=\{x,y,x+2,y+2,x+y+2,x+xy,y+xy,x+y+xy,x+xy+2,y+xy+2,x+y+xy+2\}$. Therefore, $K_{3,12}$ is a subgraph of $\mathrm{\Gamma }\left(R\right)$. Clearly, $\gamma \left(K_{3,12}\right)=3$. By Euler’s formula, there are 17 faces when drawing $K_{3,12}$ on $S_3$. Fix a representation of $S_3$-embedding of $K_{3,12}$ and denote it as L. Let $f_i$ be the number of i-gons of $K_{3,12}$ corresponding to the representation L. Since $|V_1|=3$, we have $i\le 6$. Therefore, $f_4+f_6=17$ and $4f_4+6f_6=2e=2\times 36$. This implies that there are fifteen rectangular and two hexagonal faces in L. Note that, in $\mathrm{\Gamma }\left(R\right),$ each vertex in $A=\{x+y+2,x+y+xy,x+y+xy+2\}\subseteq V_2$ is adjacent to all the four vertices in $B=\{x,x+2,x+xy,x+xy+2\}\subseteq V_2$. If $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$, then we have to embed all these 12 edges into L. Choose an arbitrary $v\in A$. Then, $de{g}_{K_{3,12}}\left(v\right)=3$ and so v is a part of exactly three faces of L. Notice that any rectangular face can adopt at most one new edge into it and any hexagonal face can adopt at most three edges between the vertices of $V_2$. So to embed the four edges between v and a vertex of B, two rectangular and one hexagonal face are required. Therefore, to embed all the edges between A and B, six rectangular and three hexagonal faces are required, a contradiction to L which has two hexagonal faces.

If $S\cong {\mathbb{Z}}_4[x,y]/(x^3,y^2+2,xy,2x,2y)$, ${\mathbb{Z}}_4[x,y]/(x^2-y^2,xy,2x,2y)$, ${\mathbb{Z}}_4[x,y,z]/(x^2-2x,y^2,xy,2y)$ or ${\mathbb{Z}}_8\left[x\right]/(2x,x^3)$, then it is not hard to verify that the graph $\mathrm{\Gamma }({\mathbb{Z}}_4[x,y]/(x^3,x^2+xy,y^2+2,2x,2y))$ is isomorphic to a subgraph of $\mathrm{\Gamma }\left(S\right)$ so that $\gamma \left(\mathrm{\Gamma }\right(S\left)\right)\ge 4.$

(8). Assume that R is isomorphic to one of the following five rings:

$$\begin{array}{c}{\mathbb{Z}}_2[x,y,z]/(y^2-x^2,z^2-x^2,xy,xz,yz),{\mathbb{Z}}_4[x,y]/(x^2,y^2-2x,xy,2y),\\ {\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2-2,z^2-2,xy,yz,xz,2x,2y,2z),\\ {\mathbb{Z}}_8[x,y]/(x^2-4,y^2-4,xy,2x,2y),{\mathbb{Z}}_8[x,y]/(x^2,y^2,xy-4,2x,2y).\end{array}$$

Let $R\cong {\mathbb{Z}}_2[x,y,z]/(y^2-x^2,z^2-x^2,xy,xz,yz)$. Then, consider the subgraph $H_1=\mathrm{\Gamma }\left(R\right)-\{(x^2,x+y+z),(x^2,x^2+x+y+z),(x^2,z),(x^2,x),(x^2,y),(x^2,x^2+z),(x^2,x^2+x),(x^2,x^2+y),(z,y),(z,x^2+y),(x^2+z,y),(x^2+z,x^2+y),(x+y,x^2+x+y),(x^2+y+z,y+z),(x+z,x^2+x+z)\}$. The diagrammatic representation of $H_1$ is given in Figure 3. Note that $H_1$ is a triangle-free graph and the number vertices and edges of $H_1$ is $n=15$ and $m=38$, respectively. Therefore, by Remark 2, we obtain $\gamma \left(H_1\right)\ge 3$. Suppose that $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)=3$. Then, $\gamma \left(H_1\right)=3$ and also, by Euler’s formula, there are 19 faces in any $S_3$-embedding of $H_1$. Let $f_i$ be the number of i-gons in a $S_3$-embedding of $H_1$. Then, $f_4+f_5+\dots =19$ and $4f_4+5f_5+\dots =2\times 38.$ This implies that all the faces in any $S_3$-embedding of $H_1$ are rectangular. Now, to recover a $S_3$-embedding of $\mathrm{\Gamma }\left(R\right)$ from a $S_3$-embedding of $H_1$, we have to embed all the 15 omitted edges of $\mathrm{\Gamma }\left(R\right)$ into a $S_3$-embedding of $H_1$. Notice that each rectangular face can adapt at most one edge into it. So to embed the eight edges incident with the vertex $x^2$, namely $(x^2,x+y+z),(x^2,x^2+x+y+z),(x^2,z),(x^2,x),(x^2,y),(x^2,x^2+z),(x^2,x^2+x)$ and $(x^2,x^2+y)$, eight rectangular faces which contain $x^2$ are required. But $de{g}_{H_1}\left(x^2\right)=6$. This implies that $x^2$ is a part of exactly six rectangular faces of any $S_3$-embedding of $H_1$, a contradiction. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

If $S\cong {\mathbb{Z}}_4[x,y]/(x^2,y^2-2x,xy,2y)$ or ${\mathbb{Z}}_4[x,y,z]/(x^2-2,y^2-2,z^2-2,xy,yz,xz,2x,2y,2z),$ or ${\mathbb{Z}}_8[x,y]/(x^2-4,y^2-4,xy,2x,2y)$ or ${\mathbb{Z}}_8[x,y]/(x^2,y^2,xy-4,2x,2y)$, then $\mathrm{\Gamma }\left(S\right)\cong \mathrm{\Gamma }\left(R\right)$ and so $\gamma \left(\mathrm{\Gamma }\right(S\left)\right)\ge 4$.

(9). Assume that R is isomorphic to one of the following four rings:

$$\begin{array}{c}{\mathbb{Z}}_4[x,y,z]/(x^2,y^2,z^2,xy,xz,yz-2,2x,2y,2z),{\mathbb{Z}}_4[x,y]/(x^2,y^2,xy,2y),\\ {\mathbb{Z}}_4[x,y]/(x^2,y^2,2x,2y),{\mathbb{Z}}_2[x,y,z]/(x^3,y^2,z^2,xy,yz).\end{array}$$

Let $R\cong {\mathbb{Z}}_4[x,y,z]/(x^2,y^2,z^2,xy,xz,yz-2,2x,2y,2z)$. Take $u_1=2,u_2=x,u_3=2+x$ and $v_1=y,v_2=x+y,v_3=x+y+2,v_4=y+2,v_5=z,v_6=x+z,v_7=z+2,v_8=x+z+2,v_9=y+z,v_{10}=y+z+2,v_{11}=x+y+z,v_{12}=x+y+z+2$. Then, $\mathrm{\Gamma }\left(R\right)$ is isomorphic to Figure 4 and $\gamma \left(\mathrm{\Gamma }\right(R)\ge 3$, since a copy of $K_{3,12}$ is contained in $\mathrm{\Gamma }\left(R\right)$. Suppose that $\gamma \left(\mathrm{\Gamma }\right(R)=3$. Consider the subgraph $H_2=\mathrm{\Gamma }\left(R\right)-\{(v_1,v_2),(v_1,v_3),(v_1,v_4),(v_2,v_3),(v_2,v_4),(v_3,v_4),(v_5,v_6),(v_5,v_7),(v_5,v_8),(v_6,v_7),(v_6,v_8),(v_7,v_8),(v_9,v_{10}),(v_9,v_{11}),(v_9,v_{12}),(v_{10},v_{11}),(v_{10},v_{12}),(v_{11},v_{12}),(u_1,u_2),(u_1,u_3),(u_2,u_3)\}$. Then, $H_2$ is isomorphic to $K_{3,12}$ and clearly $\gamma \left(H_2\right)=3$.

By the Euler formula, the faces of any $S_3$-embedding of $H_2$ have fifteen rectangular and two hexagonal faces. Now, we try to embed the 21 omitted edges of $\mathrm{\Gamma }\left(R\right)$ into an arbitrary $S_3$-embedding of $H_2$.

First, we try to insert the edges $(v_1,v_2),(v_1,v_3),(v_1,v_4),(v_2,v_3),(v_2,v_4),(v_3,v_4)$ into a $S_3$-embedding of $H_2$. To embed edges, we require one hexagonal and three rectangular faces, which we name as $F_1,F_2,F_3,F_4$ (see Figure 5a). Similarly, to embed the next six edges $(v_5,v_6),(v_5,v_7),(v_5,v_8),(v_6,v_7),(v_6,v_8),(v_7,v_8)$, we require another one hexagonal and three rectangular faces, which we name as $F_5,F_6,F_7,F_8$. Here, we notice that the vertices $v_i,v_j$, for $i=1,2,3$ and $j=5,6,7$ are placed in two faces of $F_1,F_2,\dots ,F_8$. Since $de{g}_{K_{3,12}}\left(v_i\right)=de{g}_{K_{3,12}}\left(v_j\right)=3$, one more face should have $v_i,v_j$’s and those which form three rectangular faces, we name as $F_9$, $F_{10}$ and $F_{11}$. The edges $(u_1,u_2),(u_1,u_3)$ and $(u_2,u_3)$ can be embedded in the $F_9,F_{10},F_{11}$, respectively.

Next, to embed the three edges $(v_9,v_{10}),(v_9,v_{11}),(v_9,v_{12})$, we requires three rectangular faces, which we name as $F_{12},F_{13}$ and $F_{14}$, all of which contain $v_9$ (Figure 5b). Furthermore, to embed two edges $(v_{10},v_{11}),(v_{10},v_{12})$, two rectangular faces are required, which we name as $F_{15},F_{16}$, all of which contain $v_{10}$ (Figure 5b). Since $de{g}_{K_{3,12}}\left(v_{11}\right)=de{g}_{K_{3,12}}\left(v_{12}\right)=3$ and 17 faces are in $S_3$-embedding of $H_2$, exactly one more face should have $v_{11},v_{12}$, and we shall denote it as $F_{17}$. Notice that, in any $S_3$-embedding, every edge of a graph is in exactly two faces. Here, the edges $(v_{11},u_1)$ and $(v_{12},u_2)$ are used twice in the faces $F_{12},F_{13},F_{15},F_{16}$. Now, the choices for another one diagonal of $F_{17}$ are in $\{u_1,u_2,u_3\}$. Obviously, we have to select distinct vertices for $F_{17}$ in which one is from $\{u_1,u_2\},\{u_1,u_3\}$ or $\{u_2,u_3\}$. This is a contradiction to the fact that the edges $(v_{11},u_1)$ and $(v_{12},u_2)$ are used twice in the faces $F_{12},F_{13},F_{15},F_{16}$. Therefore, $\gamma \left(\mathrm{\Gamma }\right(R\left)\right)\ge 4$.

Furthermore, if $S\cong {\mathbb{Z}}_4[x,y]/(x^2,y^2,xy,2y),{\mathbb{Z}}_4[x,y,z]/(x^2,y^2,z^2,xy-2,xz,yz,2x,2y,2z)$ or ${\mathbb{Z}}_2[x,y,z]/(x^3,y^2,z^2,xy,yz)$, then $\mathrm{\Gamma }\left(R\right)\cong \mathrm{\Gamma }\left(S\right)$ and so $\gamma \left(\mathrm{\Gamma }\right(S\left)\right)\ge 4$. □

4. Conclusions

Over the last twenty years, a number of research studies have identified all commutative rings with zero-divisor graphs with genus equal to or less than 2. This work lists all isomorphism classes of local commutative rings with identities whose zero-divisor graphs have genus three. Which non-local commutative rings have genus three zero-divisor graphs is, at this time, an open question.