# Planar, Outerplanar, and Toroidal Graphs of the Generalized Zero-Divisor Graph of Commutative Rings

Authors: Abdulaziz M. Alanazi; Mohd Nazim; Nadeem Ur Rehman
Journal: Journal of Mathematics
 (2021)

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## Abstract

LetA be a commutative ring with unity and let set of all zero divisors of A be denoted by ZA. An ideal ℐ of the ring A is said to be essential if it has a nonzero intersection with every nonzero ideal of A. It is denoted by ℐ≤eA. The generalized zero-divisor graph denoted by ΓgA is an undirected graph with vertex set ZA∗ (set of all nonzero zero-divisors of A) and two distinct vertices x1 and x2 are adjacent if and only if annx1+annx2≤eA. In this paper, first we characterized all the finite commutative rings A for which ΓgA is isomorphic to some well-known graphs. Then, we classify all the finite commutative rings A for which ΓgA is planar, outerplanar, or toroidal. Finally, we discuss about the domination number of ΓgA.


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## Body

## 1. Introduction

In this paper, all ringsA are commutative with unity. For any element x1∈A, the annihilator of x1, which is denoted by annx1, is defined as annx1=x2∈A:x1x2=0. An ideal ℐ1 of a ring A is called essential ideal of A, which is denoted by ℐ1≤eA, if ℐ1∩ℐ2≠0 for every nonzero ideal ℐ2 of A. The set of all nilpotent elements, set of all unit elements, and set of all zero-divisors of a ring A are dented by NA, UA, and ZA, respectively. For any S⊆A, S∗ denotes the set S\0. For any undefined notation or terminology in ring theory, we refer the reader to [1].In 1988, Beck [2] began to investigate the possibility of coloring a commutative ring A by associating to the ring a zero-divisor graph, defined as a simple graph, the vertices of which are the elements of the ring A, with two distinct elements x1 and x2 being adjacent if and only if x1x2=0. While Beck concentrated on the connection between the clique number and the chromatic number of the graph, various works inspired from this construction have concentrated on the interplay of commutative rings and their zero-divisor graphs. Retaining the original definition, the next decade brought little progress. However, in 1999, Anderson and Livingston [3] modified and studied the zero-divisor graph whose vertices are the nonzero zero divisors of the commutative ring (see [4–7]). Further Redmond [8] has extended this concept to any arbitrary ring, whereas the same concept for semigroup was put forward by Demeyer et al. [9]. These graphs turn out to best exhibit the properties of the set of zero divisors of a ring. Actually zero-divisor graphs help us to study the algebraic properties of rings using graph theoretical tools. We can translate some algebraic properties of a ring to graph theory language, and then the geometric properties of graphs help us to explore some interesting results in the algebraic structures of rings like the structure of zero-divisors.In [10], Jahanbakhsh Basharlou et al. introduced and studied the generalized zero-divisor graph of a commutative ring A, which is denoted by ΓgA, whose vertex set is ZA∗=ZA\0 and x1∼x2 is an edge of ΓgA if and only if annx1+annx2≤eA. The authors in [10] discussed some connections between the graph-theoretic properties of ΓgA and some algebraic properties of rings. Moreover, they investigated the affinity between generalized zero-divisor graph and zero-divisor graph associated with a ring and proved that ΓA is a spanning subgraph of ΓgA.In this article, we characterized all the finite commutative ringsA for which the generalized zero-divisor graph ΓgA is a unicycle, a tree, or a split graph. Further, we classify all the finite commutative rings A for which ΓgA is a planar graph, an outerplanar graph, or a toroidal graph. Finally, we discuss about the domination number of ΓgA and prove that the domination number of ΓgA can be any arbitrary number.


## 2. Basic Properties of Generalized Zero-Divisor Graph

In this section, we classify all the finite commutative ringsA for which the generalized zero-divisor graph ΓgA is isomorphic to some well-known graphs such as a unicycle, a tree, or a split graph. We begin our discussion with the following result due to Basharlou et al. [10], which is essential to prove our theorems.Lemma 1 (see [10, Lemma 2.1]).
For any ringA, the following holds:(1)
Ifx∈A, then Axn+annx is an essential ideal of A, for every integer n≥0.(2)
x is adjacent to all other vertices in ΓgA, for every x∈NA∗.(3)
x∼y will be an edge of ΓgA if x∼y is an edge of ΓA.Theorem 1.
For any ringA, the following holds:(1)
IfA,m is a local ring, then ΓgA is a complete graph.(2)
ΓA is a spanning subgraph of ΓgA.Proof.
Clear from parts (2) and (3) of Lemma1.A graphG, which contains a unique cycle, is called a unicycle graph. In the following result, the finite rings A are classified for which ΓgA is unicycle.Theorem 2.
LetA be a finite ring (not a field). Then, ΓgA is unicycle ⟺A is one of these rings:
ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, ℤ4x/x2+x+1, ℤ3×ℤ3, or ℤ2×ℤ2×ℤ2.Proof.
Suppose thatΓgA is unicycle. Since A is finite, A≅A1×A2×⋯×An, where Ai is local ring for each i. Suppose n≥4. Consider four vertices of ΓgA given by α1=1,0,0,…,0, α2=0,1,0,…,0, α3=0,0,1,0,…,0, and α4=0,0,0,1,0,…,0. Since annαi+annαj≤eA for each i,j, then α1∼α2∼α3∼α1 and α1∼α2∼α4∼α1 are two different cycles in ΓgA, which contradicts our assumption. Hence, n≤3.
Supposen=3 with A1≥3. Consider four vertices of ΓgA given as β1=1,0,0, β2=0,1,0, β3=0,0,1, and β4=a,0,0, where 1≠a∈A1∗. Since β1β2=β1β3=β2β3=β2β4=β3β4=0, then β1∼β2∼β3∼β1 and β2∼β3∼β4∼β2 are two different cycles in ΓgA, which contradicts our assumption. Hence, Ai=2 for each i=1,2,3, which implies that A≅ℤ2×ℤ2×ℤ2.
Supposen=2 and A1 is not a field with maximal ideal m≠0; then, there is b∈m∗ such that annb=m. Consider δ1=1,0, δ2=0,1, δ3=b,0, δ4=u,0∈ZA∗, where u∈UA1. One can see that δ1∼δ2∼δ3∼δ1 and δ2∼δ3∼δ4∼δ2 are two different cycles in ΓgA, which contradicts our assumption. Hence, A1 and A2 both are fields, and so ΓgA is a complete bipartite graph. Since ΓgA is unicycle, A1=3 and A2=3, which implies that A≅ℤ3×ℤ3.
Supposen=1. Then, ΓgA is a complete graph by Theorem 1. Since ΓgA is unicycle, ZA∗=3. Hence, A≅ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, or ℤ4x/x2+x+1.A graphG, which contains no cycle, is called acyclic. A connected acyclic graph is said to be a tree. In the following result, we classify all the finite rings A whose generalized zero-divisor graph ΓgA is a tree.Theorem 3.
LetA be a finite ring (not a field). Then, ΓgA is tree ⟺A is one of these rings:
ℤ4, ℤ2x/x2, ℤ9, ℤ3x/x2, or ℤ2×ℱ, where ℱ is a finite field.Proof.
Suppose thatΓgA is a tree. Since A is finite ring, A≅A1×A2×⋯×An, where Ai is local ring for each i. Suppose n≥3. Consider three distinct vertices of ΓgA given by α1=1,0,0,…,0, α2=0,1,0,…,0, and α3=0,0,1,0,…,0. Since annαi+annαj≤eA for each i,j, then α1∼α2∼α3∼α1 is a cycle in ΓgA, which contradicts our assumption. Hence, n≤2.
Supposen=2. We claim that Ai is a field for each i=1,2. Suppose on the contrary that A1 is not a field with nonzero maximal ideal m. Then, there is a∈m∗ such that anna=m. Consider β1=1,0, β2=0,1, β3=a,0∈ZA∗. Since annβi+annβj≤eA for each i,j, then β1∼β2∼β3∼β1 is a cycle in ΓgA, which contradicts our assumption. Hence, A1 and A2 both are fields, which implies that ΓgA is a complete bipartite graph. Since ΓgA is a tree, Ai=2 for some i=1,2. Hence, A≅ℤ2×ℱ, where ℱ is finite field.
Supposen=1. Then, ΓgA is complete by Theorem 1. Since ΓgA is a tree, A has at most two nonzero zero-divisors. Hence, A≅ℤ4, ℤ2x/x2, ℤ9, or ℤ3x/x2.A graphG, whose vertex set may be partitioned into an independent set and a clique, is called a split graph. The following result is useful to classify the finite commutative ring A whose generalized zero-divisor graph ΓgA is a split graph.Theorem 4 (see [11]).
G is a split graph ⟺G contains no induced subgraph which is isomorphic to 2K2, C4, C5, where G is a connected graph.Theorem 5.
LetA be a finite nonlocal ring and ZA∗≥2. Then, ΓgA is split graph ⟺A≅ℤ2×ℱ, where ℱ is a finite field or A≅ℤ2×ℤ2×ℤ2.Proof.
Assume thatΓgA is a split graph. Since A is a finite ring, A≅A1×A2×⋯×An, where Ai is a local ring for each i and n≥2. Suppose n≥4. Consider four distinct vertices of ΓgA defined as α1=1,0,0,…,0, α2=0,1,0,…,0, α3=0,0,1,0,…,0, and α4=0,0,0,1,0,…,0. Since annαi+annαj≤eA for each i,j, then ΓgA contains C4, which contradicts our assumption. Hence, n≤3.
Supposen=3 with A2≥3. Consider distinct vertices of ΓgA given by β1=1,0,0, β2=0,1,0, β3=0,0,1, and β4=0,a,0, where 1≠a∈A2∗. Since β1∼β2∼β3∼β4∼β1, then ΓgA contains C4, which contradicts our assumption. Hence, Ai=2 for each i=1,2,3. Hence, A≅ℤ2×ℤ2×ℤ2.
Supposen=2 and A1 is not a field with maximal ideal m≠0; then, there is b∈m∗ such that annb=m. Consider δ1=1,0, δ2=b,0, δ3=u,0, δ4=0,1∈ZA∗, where 1≠u∈UA1. Since annδ1+annδ2≤eA, annδ2+annδ3≤eA, annδ3+annδ4≤eA, and annδ1+annδ4≤eA, then ΓgA contains a cycle of length four, which contradicts our assumption. Hence, Ai is a field for each i=1,2, which shows that ΓgA is a complete bipartite graph of order A1−1,A2−1. Since ΓgA is a split graph, then A1=2 or A2=2, which implies that A≅ℤ2×ℱ, where ℱ is a finite field.


## 3. Planarity of Generalized Zero-Divisor Graph

If a graphG can be drawn in the plane in such a way that no two edges intersect each other except the end points, then it is called a planar graph. A planar graph G is said to be outerplanar if all the vertices of G lie on the unbounded face of the drawing.In this section, we classify all the finite commutative rings whose generalized zero-divisor graph is planar or outerplanar graph. Before going to our main results, first we state some results which are useful for the proof of the subsequent results.Theorem 6 (Kuratowski’s theorem).
A graphG is planar ⟺ it contains no subdivision of K5 or K3,3.Theorem 7 (see [10, Theorem 3.3]).
LetA be a reduced ring and x,y∈ZA∗. Then, these statements are equivalent:(1)
ΓA=ΓgA.(2)
Ifxy≠0, then annannx+anny≠0.Theorem 8 (see [6, Theorem 3.5.1]).
LetA,m be a finite local ring and ΓA be the zero-divisor graph of A. Then, ΓA is planar ⟺A is one of these rings:
ℤ4, ℤ2x/x2, ℤ9, ℤ3x/x2, ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, ℤ4x/x2+x+1, ℤ25, ℤ5x/x2, ℤ16, ℤ2x/x4, ℤ4x/x2−2,x4, ℤ2x/x3−2,x4, ℤ4x/x3+x2−2,x4, ℤ2x,y/x3,xy,y2−x2, ℤ2x/x3,x2−2x, ℤ8x/x2−4,2x, ℤ4x,y/x3,x2−2,xy,y2−2,y3, ℤ4x/x2, ℤ4x,y/x2,y2,xy−2, ℤ2x,y/x2,y2, ℤ27, ℤ3x/x3, ℤ9x/x2−3,x3, ℤ9x/x2+3,x3.Theorem 9 (see [5, Theorem 3.7]).
LetA=ℱ1×ℱ2×⋯×ℱn be a finite ring, where each ℱi is a field and n≥2. Then, ΓA is planar ⟺A is one of these rings:
ℤ2×ℱ, ℤ3×ℱ, ℤ2×ℤ2×ℤ2, ℤ2×ℤ2×ℤ3, where ℱ is a finite field.The following result characterized all the finite local commutative ringsA for which ΓgA is planar.Theorem 10.
LetA be a finite local ring. Then, ΓgA is planar ⟺A is one of these rings:
ℤ4, ℤ2x/x2, ℤ9, ℤ3x/x2, ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, ℤ4x/x2+x+1, ℤ25 or ℤ5x/x2.Proof.
It is clear from Theorem1 and 6.Now, we are in a position to classify all the finite nonlocal ringsA for which ΓgA is planar.Theorem 11.
LetA be a finite reduced ring (not a field). Then, ΓgA is planar ⟺A is one of these rings:
ℤ2×ℱ, ℤ3×ℱ, ℤ2×ℤ2×ℤ2, ℤ2×ℤ2×ℤ3, where ℱ is a finite field.Proof.
SinceA is a finite reduced ring, A≅ℱ1×ℱ2×⋯×ℱn, where ℱi is a field for each i and n≥2. Thus, by Theorem 7, ΓA=ΓgA. Hence, the result is clear from Theorem 9.Remark 1.
IfA≅A1×A2×⋯×An is a finite commutative ring, where Ai,mi is local ring with mi≠0 for each i and n≥2, then ΓgA contains a copy of K3,3, which shows that ΓgA is not planar. Hence, for ΓgA to be planar, at least one of Ai must be a field.Theorem 12.
LetA≅A1×A2×⋯An×ℱ1×ℱ2×⋯×ℱm be a finite ring, where Ai,mi is local ring with mi≠0 for each i, and ℱj is a field for each j and n,m≥1. Then, ΓgA is planar if and only if A≅ℤ4×ℤ2 or A≅ℤ2x/x2×ℤ2.Proof.
Suppose thatΓgA is planar. If n≥2, then by Remark 1, ΓgA is not planar, which contradicts our assumption. Hence, n=1.
Supposem≥2. Since A1 is local ring with m1≠0, then there is a∈m1∗ such that anna=m1. Consider x1=a,0,0,…,0, x2=1,0,0,…,0, x3=u,0,0,…,0, y1=0,1,0,…,0, y2=0,0,1,0,…,0, y3=0,1,1,0,…,0∈ZA∗, where 1≠u∈UA1. Since annxi+annyj≤eA for each i,j, then ΓgA contains a copy of K3,3, which contradicts our assumption. Hence, m=1.
Supposem1∗≥2, then UA1≥3. Consider a1=u1,0, a2=u2,0, a3=u3,0, b1=0,1, b2=a,0, b3=a,1∈ZA∗, where ui∈UA1 for each i=1,2,3. Since annai+annbj≤eA for each i,j, then ΓgA contains K3,3 as a subgraph, which contradicts our assumption. Hence, m1∗=1, which implies that A1≅ℤ4 or ℤ2x/x2.
Now, suppose thatℱ1≥3 and x∈m1∗ such that x2=0. Consider r1=1,0, r2=x,0, r3=u,0, s1=0,1, s2=0,y, s3=x,y∈ZA∗, where 1≠u∈UA1 and 1≠y∈ℱ1∗. Since annri+annsj≤eA for each i,j, then ΓgA contains a subdivision of K3,3, which contradicts our assumption. Hence, ℱ1=2, which implies that ℱ1≅ℤ2.
Converse is clear from Figure1.Figure 1
Planar embedding ofΓgℤ4×ℤ2≅Γgℤ2x/x2×ℤ2.Theorem 13 (see [12]).
A graphG is outerplanar ⟺ there is no subdivision of K4 or K2,3 in G.We end this section with the following result which characterized all the finite commutative ringsA for which ΓgA is outerplanar.Theorem 14.
LetA be a finite ring (not a field). Then, ΓgA is outerplanar ⟺A is one of these rings:
ℤ4, ℤ2x/x2, ℤ9, ℤ3x/x2, ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, ℤ4x/x2+x+1, ℤ2×ℱ, ℤ3×ℤ3 or ℤ2×ℤ2×ℤ2, where ℱ is a finite field.Proof.
Assume thatΓgA is outerplanar. Since A is finite, A≅A1×A2×⋯×An, where Ai is local ring for each i. Suppose n≥4. Consider v1=1,0,…,0, v2=0,1,0,…,0, w1=0,0,1,0,…,0, w2=0,0,0,1,0,…,0, w3=0,0,1,1,0,…,0∈ZA∗. Since annvi+annwj≤eA for each i,j, then ΓgA contains a copy of K2,3, which contradicts our assumption. Hence, n≤3.
Supposen=3 and A1≥3. Consider a1=1,0,0, a2=x,0,0, b1=0,1,0, b2=0,0,1, b3=0,1,1∈ZA∗, where 1≠x∈A1∗. Since aibj=0 for each i,j, then ΓgA contains K2,3 as a subgraph. Hence, Ai=2 for each i=1,2,3. This implies that A≅ℤ2×ℤ2×ℤ2.
Supposen=2 and A1 is not a field with nonzero maximal ideal m1. Then, there is a∈m1∗ such that anna=m1. Consider u1=1,0, u2=a,0, u3=u,0, v1=0,1, v2=a,1∈ZA∗, where 1≠u∈UA1. Since annui+annvj≤eA for each i,j, then ΓgA contains a copy of K2,3, which contradicts our assumption. Hence, A1 and A2 both are fields, which shows that ΓgA is a complete bipartite graph with order A1−1,A2−1. Since ΓgA is outerplanar, A≅ℤ2×ℱ or ℤ3×ℤ3, where ℱ is a finite field.
Supposen=1. Then, by Theorem 1, ΓgA is a complete graph. Since ΓgA is outerplanar, ZA∗≤3. Hence, A≅ℤ4, ℤ2x/x2, ℤ9, ℤ3x/x2, ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2 or ℤ4x/x2+x+1.
Converse is clear.


## 4. Genus of Generalized Zero-Divisor Graph

If a graphG can be embedded on the sphere with k handles without crossing itself, then the minimality of such integer k is called the genus of G. It is denoted by λG.In this section, we classify all the finite commutative rings for which the generalized zero-divisor graph is toroidal.In order to characterized the rings with toroidal generalized zero-divisors graph, we need the following results, which deal with genus properties of graphs.Lemma 2 (see [12]).
λKn=⌈1/12n−3n−4⌉, where ⌈x⌉ is the least integer, that is, ≥x. In particular, λKn=1 if n=5,6,7.Lemma 3 (see [12]).
λKm,n=⌈1/4m−2n−2⌉, where ⌈x⌉ is the least integer, that is, ≥x. In particular, λK4,4=λK3,n=1 if n=3,4,5,6. Also, λK5,4=λK6,4=λK3,m=2, if m=7,8,9,10.Theorem 15 (see [6], Theorem 3.5.2).
LetA,m be a finite local ring and ΓA be the zero-divisor graph of A. Then, λΓA=1⟺A is one of these rings:
ℤ49, ℤ7x/x2, ℤ2x,y/x3,xy,y2, ℤ4x/x3,2x, ℤ4x,y/x3,x2−2,xy,y2, ℤ8x/x2,2x, F8x/x2, ℤ4x/x3+x+1, ℤ4x,y/2x,2y,x2,xy,y2, ℤ2x,y,z/x,y,z2, ℤ32, ℤ2x/x5, ℤ4x/x3−2,x5, ℤ4x/x4−2,x5, ℤ8x/x2−2,x5, ℤ8x/x2−2x+2,x5, ℤ8x/x2+2x−2,x5.Theorem 16 (see [7, Theorem 3.1]).
LetA=ℱ1×ℱ2×⋯×ℱn be a finite ring, where each ℱi is a field and n≥2. Then, λΓA=1⟺A is one of these rings:
F4×F4, F4×ℤ5, ℤ5×ℤ5, F4×ℤ7, ℤ2×ℤ2×F4, ℤ2×ℤ2×ℤ5, ℤ2×ℤ2×ℤ7, ℤ2×ℤ2×ℤ3, ℤ2×ℤ3×ℤ3, ℤ3×ℤ3×ℤ3, ℤ2×ℤ3×F4 or ℤ2×ℤ2×ℤ2×ℤ2.The following result classifies all the finite local commutative ringsA for which ΓgA is toroidal.Theorem 17.
LetA be a finite local ring (not a field). Then, λΓgA=1⟺A is one of these rings:
ℤ49, ℤ7x/x2, ℤ16, ℤ2x/x4, ℤ4x/x2−2,x4, ℤ2x/x3−2,x4, ℤ4x/x3+x2−2,x4, ℤ2x,y/x3,xy,y2−x2, ℤ2x/x3,x2−2x, ℤ8x/x2−4,2x, ℤ4x,y/x3,x2−2,xy,y2−2,y3, ℤ4x/x2, ℤ4x,y/x2,y2,xy−2, ℤ2x,y/x2,y2, ℤ2x,y/x2,y2,xy, ℤ4x/x3,2x, ℤ4x,y/x3,x2−2,xy,y2, ℤ8x/x2,2x, F8x/x2, ℤ4x/x3+x+1, ℤ4x,y/2x,2y,x2,y2,xy or ℤ2x,y,z/x,y,z2.Proof.
SinceA is finite local ring, ΓgA is complete graph by Theorem 1. Thus, by Lemma 2, λΓgA=1 if and only if A has at least 5 and at most seven nonzero zero-divisors. This completes the proof.Now, we are in a position to characterize all the finite nonlocal commutative ringsA for which ΓgA is toroidal.Theorem 18.
LetA be finite reduced ring (not a field). Then, λΓgA=1⟺A is one of these rings:
F4×F4, F4×ℤ5, ℤ5×ℤ5, F4×ℤ7, ℤ2×ℤ2×F4, ℤ2×ℤ2×ℤ5, ℤ2×ℤ2×ℤ7, ℤ2×ℤ2×ℤ3, ℤ2×ℤ3×ℤ3, ℤ3×ℤ3×ℤ3, ℤ2×ℤ3×F4 or ℤ2×ℤ2×ℤ2×ℤ2.Proof.
SinceA is finite reduced ring, A≅ℱ1×ℱ2×⋯×ℱn, where ℱi is field for each i and n≥2. Thus, by Theorem 7, ΓgA=ΓA. Hence, the proof follows from Theorem 16.Remark 2.
IfA≅A1×A2×⋯×An is a finite commutative ring, where Ai,mi is local ring with mi≠0 for each i and n≥2, then ΓgA contains a copy of K5,5, which shows that λΓgA>1. Hence, for ΓgA to be toroidal, at least one of Ai must be a field.Theorem 19.
LetA≅A1×A2×⋯An×ℱ1×ℱ2×⋯×ℱm be a finite ring, where Ai,mi is local ring with mi≠0 for each i, and ℱj is a field for each j and n,m≥1. Then, λΓgA=1 if and only if A is one of these rings:
ℤ4×ℤ3, ℤ2x/x2×ℤ3, ℤ4×F4, or ℤ2x/x2×F4.Proof.
Assume thatλΓgA=1. If n≥2, then by Remark 2, λΓgA>1, which is a contradiction to our assumption. Hence, n=1.
Supposem≥2 and there is a∈m1∗ such that anna=m1. Consider the set S=a1.a2,…,a10, where a1=1,0,0,…,0, a2=a,0,0,…,0, a3=u,0,0,…,0, a4=0,1,0,…,0, a5=0,0,1,0,…,0, a6=0,1,1,0,…,0, a7=a,1,0,…,0, a8=a,0,1,0,…,0, a9=a,1,1,0,…,0, a10=1,0,1,0,…,0∈ZA∗, where 1≠u∈UA1. Let G1 be the graph induced by the set S. Then, G1 is a subgraph of ΓgA containing G′ as shown in Figure 2. From Figure 2, it is clear that G′ contains K3,6, which implies that λG′≥1.
Assume thatλG′=1. Consider an embedding of K3,6 on torus; then, by Euler’s formula, the embedding of K3,6 has 9 faces, say F1′,F2′,…,F9′. Let fi denote the length of the face Fi′ for each i. Then, ∑t=19fi=2e=36 and fi≥4 for each i. This implies that fi=4 for each i. Let P=a7,a8,a4,a10⊂VG′; then, the subgraph G″ induced by P is K3 such that EG″∩EK3,6=∅. Since K3 cannot be embedded in the torus along with an embedding of G″ on K3,6 together in torus, this implies that λG′≥2. Since G′ is a subgraph of ΓgA, λΓgA≥2, which is a contradiction to our assumption. Hence, m=1.
We claim thatm1∗=1. Suppose on the contrary that m1∗=2; then, A1≅ℤ9 or ℤ3x/x2, and hence U~A~1=6. Let a,b∈m1∗ such that ab=0 and anna=m1. Consider p1=a,0, p2=b,0, p3=u1,0, p4=u2,0, p5=u3,0, p6=u4,0, p7=u5,0, q1=0,1, q2=a,1, q3=b,1∈ZA∗, where ui∈UA1. Since annpi+annqj≤eA for each i,j, then K3,7 is a subgraph of ΓgA. Thus, by Lemma 3, λΓgA>1, which is a contradiction to our assumption. Hence, m1∗≠2.
Supposem1∗≥3; then, UA1≥4. Let a,b,c∈m1∗ such that ab=bc=0 and anna=m1. Consider w1=u1,0, w2=u2,0, w3=u3,0, w4=u4,0, w5=a,0, w6=b,0, w7=c,0, z1=0,1, z2=a,1, z3=b,1∈ZA∗, where ui∈UA1. Since annwi+annzj≤eA for each i,j, then ΓgA contains K3,7 as a subgraph, which is a contradiction by Lemma 3. Hence, m1∗=1, which shows that A1≅ℤ4 or ℤ2x/x2.
Now, supposeℱ1≥5 and a∈m1∗ such that a2=0 and anna=m1. Consider r1=1,0, r2=a,0, r3=u,0, s1=0,x1, s2=0,x2, s3=0,x3, s4=0,x4, s5=a,x1, s6=a,x2, s7=a,x3∈ZA∗, where 1≠u∈UA1 and xi∈ℱ1∗. Since annri+annsj≤eA for each i,j, then K3,7 is a subgraph of ΓgA, which is a contradiction to our assumption. Hence, |ℱ1|≤4. From Theorem 12, it is clear that ℱ1≇ℤ2. Hence, ℱ1≅ℤ3 or F4.
The converse follows from Figures3 and 4.Figure 2
The graph ofG′.Figure 3
Toroidal embedding ofΓgℤ4×ℤ3≅Γgℤ2x/x2×ℤ3.Figure 4
Toroidal embedding ofΓgℤ4×F4≅Γgℤ2x/x2×F4.


## 5. Domination of Generalized Zero-Divisor Graph

LetGV,E be a graph and S⊆V. Then, S is said to be a dominating set of G if for u∈V∖S there exists v∈S such that u∼v. A dominating set with minimum cardinality is said to be γ-set of G. The cardinality of γ-set of G is called the domination number of G and is denoted by γG.In this section, we discuss about the domination number of the generalized zero-divisor graph of a commutative ring. The following result shows that the dominating number ofΓgA may be any arbitrary number.Proposition 1.
LetA=ℱ1×ℱ2×⋯×ℱn be a ring, where ℱi is a field for each i and n≥3. Then, γΓgA=n.Proof.
SinceA is a reduced ring, then by Theorem 7, ΓgA=ΓA. Hence, by [6], (Proposition 8), γΓgA=γΓA=n.The following result gives the necessary and sufficient conditions for a commutative ringA for which γΓgA=1.Theorem 20.
LetA be a ring. Then, γΓgA=1⟺ one of the following holds:(1).
There exists0≠x∈A such that annx is essential ideal in A.(2)
A≅ℤ2×S, where S is a reduced ring.Proof.
Suppose there exists0≠x∈A such that annx≤eA. Then, by Lemma 1, x is adjacent to every other vertex of ΓgA. Hence, γΓgA=1. Also, if A≅ℤ2×S, where S is a reduced ring, then 1,0 is adjacent to every other vertex of ΓgA. Hence, γΓgA=1.
Conversely, assume thatγΓgA=1. The following two cases occur:Case (i):A is nonreduced ring. Let a be the nonzero nilpotent element of A. Then, by Lemma 1, anna≤eA.Case (ii):A is reduced ring. Since γΓgA=1, then let x be the dominating set of ΓgA. If x2≠x, then annx2=annx+annx2≤eA, and we are done. Thus, suppose that x=x2; then, A≅A1×A2 with dominating set 1,0, where Ai is a reduced ring for each i=1,2. Let 1≠a∈A1∗; then, a,0∈ZA∗. Since anna,0+ann1,0=anna×A2 and A is a reduced ring, then anna×A2∩aA1×A2=0, which implies that a,0 is not adjacent to 1,0, a contradiction. Hence, A1≅ℤ2.We immediately get the following corollary from Theorem20, which classifies the domination number of ΓgA for all finite commutative rings A.Corollary 1.
LetA=A1×A2×⋯×An be a finite ring, where Ai is a local ring for each i and n≥1. Then, the following holds:(1)
IfAi is not a field for some i, then γΓgA=1.(2)
IfAi is a field for each i and n≥3, then γΓgA=n.(3)
IfA1 and A2 are fields and minA1,A2≥3, then γΓgA=2.(4)
IfA1 and A2 are fields and minA1,A2=2, then γΓgA=1.Proof.
The proof of (1) and (4) follows from Theorem20 and the proof of (2) is clear from Proposition 1. Now for part (3), the set 1,0,0,1 is a dominating set of ΓgA. Hence, γΓgA=2.



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