# Classification of Rings with Toroidal and Projective Coannihilator Graph

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

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## Abstract

LetS be a commutative ring with unity, and a set of nonunit elements is denoted by WS. The coannihilator graph of S, denoted by AG′S, is an undirected graph with vertex set WS∗ (set of all nonzero nonunit elements of S), and α∼β is an edge of AG′S⇔α∉αβS or β∉αβS, where δS denotes the principal ideal generated by δ∈S. In this study, we first classify finite ring S, for which AG′S is isomorphic to some well-known graph. Then, we characterized the finite ring S, for which AG′S is toroidal or projective.


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## Body

## 1. Introduction

Throughout this study, all ringsS are commutative ring with unity which is not a field. For α∈S, annihilator of α, which is denoted by annα, is defined as annα=β∈S:αβ=0. The set of zero-divisor of S, the set of nonunit elements of S, and the set of unit elements of S is denoted by ZS, WS, and US, respectively. For a finite ring S, ZS=WS. For A⊆S, we denote A∗=A∖0. For any undefined notation or terminology in ring theory, please refer to [1].Recall that two adjacent verticesα,β∈V in a graph GV,E are denoted by α∼β. A graph GV,E is said to be complete if α∼β for each α,β∈V. Also, if V=n, then G is denoted by Kn. A graph GV,E is said to be bipartite if V=A∪B, such that for α,β∈A (or B), α≁β. Also, if α∼β, for each α∈A and β∈B, then GV,E is said to be complete bipartite graph. If A=n and B=m, then G is denoted by Kn,m. A graph G, which contains a unique cycle, is called a unicycle graph. A graph G, which contains no cycle, is called acyclic. A connected acyclic graph is said to be a tree. A graph GV,E is called a split graph if V=A∪B, such that A∩B=∅, where A is an independent set and B is a clique. For more details on graph theory, see [2, 3].The concept of zero-divisor graph was first introduced by Beck [4], where he was mainly interested in the coloring of a commutative ring. In [5], Anderson and Livingston modified the definition of zero-divisor graph and denoted it by ΓS, whose vertex set is ZS∗, and α∼β is an edge of ΓS⇔αβ=0; for more details, see [6–11]. After that, Badawi [12] introduced the annihilator graph of a commutative ring S, denoted by AGS, whose vertex set is ZS∗, and α∼β is an edge of AGS⇔annαβ≠annα∪annβ. Recently, Afkhami and Khashyarmanesh [13] introduced and studied the cozero-divisor graph of a commutative ring S, denoted by Γ′S with vertex set WS∗, and α∼β is edge of Γ′S⇔α∉βS and β∉αS. For more details about cozero-divisor graph, see [14–16].In this study, we study about the coannihilator graph of a commutative ringS, which was introduced by Afkhami et al. [17]. It is an undirected graph denoted by AG′S with vertex set WS∗, and α∼β is an edge of AG′S⇔α∉αβS or β∉αβS. First, we classify all the finite rings S, for which AG′S is isomorphic to some well-known graphs. Then, we characterized all the finite rings S, for which AG′S is toroidal or a projective graph.We recall some results from [17], which are useful for the proof of the subsequent sections.Lemma 1 (see [17], Lemma 2.1).
LetS be a commutative ring. Then, the following hold:(1)
For some distinctα,β∈WS∗, α∼β is not an edge of AG′S⇔αS=αβS=βS(2)
Ifα∼β is an edge of ΓS, for some distinct α,β∈ZS∗, then α∼β is an edge of AG′S.(3)
Ifα∼β is an edge of Γ′S, for some distinct α,β∈WS∗, then α∼β is an edge of AG′S.(4)
Ifα∼β is an edge of AGS, for some distinct α,β∈ZS∗, then α∼β is an edge of AG′S.Theorem 1 (see [17], Theorem 5).
LetS be a commutative ring. Then, AG′S is connected and diamAG′S≤2.Theorem 2 (see [17], Theorem 6).
LetS be a local commutative ring. Then, AG′S is complete.


## 2. Basic Properties ofAG′S

In this section, we classify all the finite ringsS, whose coannihilator graph AG′S is isomorphic to some well-known graphs such as complete graph, unicycle graph, tree, or split graph.Theorem 3.
LetS be a finite commutative ring. Then, AG′S is a complete graph, ⇔S is local, or S≅ℤ2n for n≥2.Proof.
Assume thatAG′S is complete. Since S is finite, S≅S1×S2×⋯×Sn, where Si is local ring for each i. If n=1, we are done. Suppose n≥2. Suppose that Sj is not a field for some j. Consider α=0,0,…,0,1jthplace,0,…,0 and β=0,0,…,0,1−ajthplace,0,…,0∈WS∗, where a∈WSj∗. Since αS=αβS=βS, α≁β in AG′S, which contradicts our assumption. Hence, Si is a field for each i.
SupposeSk≥3. Consider δ=0,0,…,0,xkthplace,0,…,0 and ω=0,0,…,0,ykthplace,0,…,0∈WS∗, where x,y∈Sk∗. Since δS=δωS=ωS, δ≁ω in AG′S, again contradicts our assumption. Hence, Si=2 for each i, which implies that Si≅ℤ2 for each 2≤i≤n.
Converse is clear.Corollary 1.
LetS be a finite local commutative ring. Then, the following hold:(1)
AG′S≅K1⇔S≅ℤ4, ℤ2x/x2(2)
AG′S≅K2⇔S≅ℤ9, ℤ3x/x2(3)
AG′S≅K3⇔S≅ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, ℤ4x/x2+x+1(4)
AG′S≅K4⇔S≅ℤ25, ℤ5x/x2(5)
AG′S≅K6⇔S≅ℤ49, ℤ7x/x2(6)
AG′S≅K7⇔S≅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, ℤ2x,y,z/x,y,z2Theorem 4.
LetS be a finite commutative ring. Then, AG′S is unicycle, ⇔S is one of the following rings:(1)ℤ8,ℤ2xx3,ℤ4xx3,x2−2,ℤ4x2x,x2,ℤ2x,yx2,xy,y2,F4xx2,ℤ4xx2+x+1orℤ3×ℤ3.Proof.
Assume thatAG′S is unicycle. Since S is finite, S≅S1×S2×⋯×Sn, where Si is the local for each i. Suppose n≥3. Consider α1=1,0,0,…,0, α2=0,1,0,…,0, α3=0,0,1,0,…,0, and α4=0,1,1,0,…,0∈WS∗. Since α1α2=α1α3=α1α4=α2α3=0 and α4∉α2α4S, α1∼α2∼α3∼α1 and α1∼α2∼α4∼α1 are two different cycles in AG′S, which contradict our assumption. Hence, n≤2.
Supposen=2 and S1 is not a field with maximal ideal m1≠0. Then, there is a∈m1∗, such that anna=m1. Consider β1=1,0, β2=0,1, β3=a,1, and β4=a,0∈WS∗. Since β1β2=β2β4=0, β1∉β1β4S, β3∉β1β3S, and β3∉β3β4S, β1∼β2∼β4∼β1 and β1∼β3∼β4∼β1 are two distinct cycles in AG′S, which contradict our assumption. Hence, S1 and S2 both are fields, which show that AG′S≅KS1−1,S2−1. Since AG′S is a unicycle, S1=3 and S2=3. Hence, S≅ℤ3×ℤ3.
Supposen=1. Then, S is a local ring. Thus, AG′S is complete by Theorem 2. Since AG′S is a unicycle, WS∗=3. Hence, S≅ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, or ℤ4x/x2+x+1.
The converse is clear.Theorem 5.
LetS be a finite commutative ring. Then, AG′S is a tree, ⇔S is one of the following rings:(2)ℤ4,ℤ2xx2,ℤ9,ℤ3xx2orℤ2×F,where F is the finite field.Proof.
Assume thatAG′S is a tree. Since S is finite, S≅S1×S2×⋯×Sn, where Si is the local ring for each i. Suppose n≥3. Consider α1=1,0,0,…,0, α2=0,1,0,…,0, and α3=0,0,1,0,…,0∈WS∗. Since α1α2=α2α3=α3α1=0, α1∼α2∼α3∼α1 is a cycle in AG′S, which contradicts our assumption. Hence, n≤2.
Supposen=2 and S1 is not a field with maximal ideal m1≠0. Then, there is a∈m1∗, such that anna=m1. Consider β1=1,0, β2=a,0, and β3=0,1∈WS∗. Since β1β3=β2β3=0 and β1∉β1β2S, β1∼β2∼β3∼β1 is a cycle in AG′S, which contradicts our assumption. Hence, S1 and S2 both are fields, which show that AG′S≅KS1−1,S2−1. Since AG′S is a tree, S1=2 or S2=2. Hence, S≅ℤ2×F, where F is the finite field.
Supposen=1. Then, S is a local ring. Thus, AG′S is complete by Theorem 2. Since AG′S is a tree, WS∗≤2. This implies that S≅ℤ4, ℤ2x/x2, ℤ9, or ℤ3x/x2.
The converse is clear.Theorem 6 (see [2]).
LetG be a connected graph. Then, G is a split graph, ⇔G contains no induced subgraph isomorphic to 2K2, C4, and C5.Theorem 7.
LetS be a finite commutative ring with WS∗≥2. Then, AG′S is split graph, ⇔S is one of the following rings:(3)ℤ9,ℤ3xx2,ℤ8,ℤ2xx3,ℤ4xx3,x2−2,ℤ4x2x,x2,ℤ2x,yx2,xy,y2,F4xx2,ℤ4xx2+x+1orℤ2×F,where F is the finite field.Proof.
Assume thatAG′S is a split graph. Since S is finite, S≅S1×S2×⋯×Sn, where Si is the local ring for each i. Suppose n≥3. Consider α1=1,0,0,…,0, α2=0,1,0,0,…,0, α3=0,0,1,0,…,0, and α4=1,1,0,0,…,0∈WS∗. Since α1α2=α2α3=α3α4=0 and α4∉α1α4S, α1∼α2∼α3∼α4∼α1 is C4 in AG′S, which contradicts our assumption by Theorem 6. Hence, n≤2.
Supposen=2 and S1 is not a field with maximal ideal m1≠0. Then, there is a∈m1∗, such that anna=m1. Consider δ1=1,0, δ2=a,0, δ3=u,0, and δ4=0,1∈WS∗, where 1≠u∈US1. Since δ1∉δ1δ2S, δ3∉δ2δ3S, and δ3δ4=δ4δ1=0, δ1∼δ2∼δ3∼δ4∼δ1 is C4 in AG′S, which contradicts our assumption by Theorem 6. Hence, S1 and S2 both are fields, which show that AG′S≅KS1−1,S2−1. Since AG′S is a split graph, S1=2 or S2=2. Hence, S≅ℤ2×F, where F is the finite field.
Supposen=1. Then, S is a local ring. Thus, AG′S is complete by Theorem 2. Since AG′S is split, WS∗|=2 or 3. Hence, S≅ℤ8, ℤ2x/x3, ℤ4x/x3,x2−2, ℤ4x/2x,x2, ℤ2x,y/x2,xy,y2, F4x/x2, or ℤ4x/x2+x+1.
The converse is clear.


## 3. Genus ofAG′S

If a graphG can be embedded on a sphere Sk with k handles without crossing itself, then the minimality of such positive integer k is called the genus of the graph G, and it is denoted by γG. Also, if γG=1, then the graph G is called toroidal.In this section, we classify all the finite commutative ringsS for which the coannihilator graph AG′S is toroidal.In order to classify the rings with genus one coannihilator graph, we need the following results, which deal with genus properties of graphs.Lemma 2.
(see [3])(1)
Letn≥3. Then,(4)γKn=112n−3n−4.In particular,γKn=1 if n=5,6,7.(2)
Letn,m≥2. Then,(5)γKm,n=14m−2n−2.
In particular,γK4,4=γK3,n=1 if n=3,4,5,6. Also, γK5,4=γK6,4=γK3,m=2, if m=7,8,9,10.Theorem 8.
LetS be a finite local commutative ring. Then, γAG′S=1⇔S is one of the following rings:(6)ℤ49,ℤ7xx2,ℤ16,ℤ2xx4,ℤ4xx2−2,x4,ℤ2xx3−2,x4,ℤ4xx3+x2−2,x4,ℤ2x,yx3,xy,y2−x2,ℤ2xx3,x2−2x,ℤ8xx2−4,2x,ℤ4x,yx3,x2−2,xy,y2−2,y3,ℤ4xx2,ℤ4x,yx2,y2,xy−2,ℤ2x,yx2,y2,ℤ2x,yx2,y2,xy,ℤ4xx3,2x,ℤ4x,yx3,x2−2,xy,y2,ℤ8xx2,2x,F8xx2,ℤ4xx3+x+1,ℤ4x,y2x,2y,x2,y2,xyorℤ2x,y,zx,y,z2.Proof.
SinceS is local ring, then by Theorem 2, AG′S is a complete graph. Hence, γAG′S=1⇔5≤WS∗≤7. This completes the proof.Theorem 9.
LetS be a finite nonlocal reduced commutative ring. Then, γAG′S=1⇔S is one of the following rings:(7)F4×F4,F4×ℤ5,ℤ5×ℤ5,F4×ℤ7orℤ2×ℤ2×ℤ2.Proof.
SinceAG′F4×F4=K3,3, AG′F4×ℤ5=K3,4, AG′F4×ℤ7=K3,6, AG′ℤ5×ℤ5=K4,4, and AG′ℤ2×ℤ2×ℤ2=K6, γAG′S=1 by Lemma 2 and Figure 1.
Now, conversely assume thatγAG′S=1. Since S is a finite nonlocal reduced ring, S≅F1×F2×⋯×Fn, where Fi is a field for each i and n≥2. Suppose n≥4. Consider α1=1,0,0,…,0, α2=0,1,0,…,0, α3=0,0,1,0,…,0, α4=0,0,0,1,0,…,0, α5=0,1,1,0,…,0, α6=0,1,0,1,0,…,0, α7=0,0,1,1,0,…,0, and α8=0,1,1,1,0,…,0∈WS∗. Since α1α2=α1α3=α1α4=α1α5=α1α6=α1α7=α1α8=α2α3=α2α4=0, α5∉α2α5S, α6∉α2α6S, α2α7=0, α8∉α2α8S, α3α4=0, α5∉α3α5S, α3α6=0, α7∉α3α7S, α8∉α3α8S, α4α5=0, α6∉α4α6S, α7∉α4α7S, α8∉α4α8S, α5∉α5α6S, α5∉α5α7S, α8∉α5α8S, α6∉α6α7S, α8∉α6α8S, and α8∉α7α8S, the set α1,α2,…,α8 induces K8 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which contradicts our assumption. Hence, n≤3.
Supposen=3 and F1≥3. Consider β1=1,0,0, β2=a,0,0, β3=0,1,0, β4=0,0,1, δ1=1,1,0, δ2=1,0,1, δ3=0,1,1, δ4=a,0,1, and δ5=a,1,0∈WS∗, where 1≠a∈F1∗. Since β1δ3=β2δ3=β3δ3=β3δ1=β3δ4=β4δ1=β4δ5=0, δ1∉β1δ1S, δ2∉β1δ2S, δ4∉β1δ4S, δ1∉β2δ1S, δ5∉β1δ5S, δ2∉β2δ2S, δ4∉β2δ4S, δ5∉β2δ5S, δ1∉β3δ1S, δ3∉β3δ3S, δ5∉β3δ5S, δ2∉β4δ2S, δ3∉β4δ3S, and δ4∉β4δ4S, the set β1,β2,β3,β4∪δ1,δ2,δ3,δ4,δ5 induces K4,5 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which contradicts our assumption. Hence, Fi=2, for each i=1,2,3, which shows that S≅ℤ2×ℤ2×ℤ2.
Supposen=2. Then, AG′S is a complete bipartite graph of order F1−1,F2−1. Since γAG′S=1, by Lemma 2, S≅F4×F4, F4×ℤ5, ℤ5×ℤ5, or F4×ℤ7.Figure 1
Toroidal embedding ofAG′ℤ2×ℤ2×ℤ2.Theorem 10.
LetS be a finite nonlocal nonreduced commutative ring. Then, γAG′S=1⇔S is one of the following rings:(8)ℤ4×ℤ3orℤ2xx2×ℤ3.Proof.
SupposeS≅ℤ4×ℤ3 or S≅ℤ2x/x2×ℤ3, then the toroidal embedding of AG′S is shown in Figure 2.
Conversely, assume thatγAG′S=1. Since S is nonreduced nonlocal finite ring, then the following two cases occur:Case (i):S≅S1×S2×⋯×Sn, where Si,mi is the local ring with mi≠0 for each i and n≥2. Then, there is ai∈mi∗, such that annai=mi for each i. Consider α1=1,0,0,…,0, α2=u,0,0,…,0, α3=0,1,0,…,0, α4=0,v,0,…,0, β1=a1,0,0,…,0, β2=0,a2,0,…,0, β3=a1,a2,0,…,0, β4=a1,1,0,…,0, and β5=1,a2,0,…,0∈WS∗, where 1≠u∈US1 and 1≠v∈US2. Since α1∉α1β1S, α1β2=0, β3∉α1β3S, β4∉α1β4S, β5∉α1β5S, α2∉α2β1S, α2β2=0, β3=α2β3S, β4∉α2β4S, β5∉α2β5S, α3β1=0, α3∉α3β2S, β3∉α3β3S, β4∉α3β4S, β5∉α3β5S, α4β1=0, α4∉α4β2S, β3∉α4β3S, β4∉α4β4S, and β5∉α4β5S, the set α1,α2,α3,α4∪β1,β2,β3,β4,β5 induces K4,5 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which contradicts our assumption.Case (ii):S≅S1×S2×⋯×Sn×F1×F2×⋯×Fm, where Si,mi is the local ring with mi≠0 for each i, Fj is a field for each j, and n,m≥1. If n≥2, then we get a contradiction as above. Hence, n=1.
Supposem≥2 and a∈m1∗, such that anna=m1. Consider δ1=1,0,0,…,0, δ2=a,0,0,…,0, δ3=0,1,0,…,0, δ4=0,0,1,0,…,0, ω1=1,1,0,…,0, ω2=1,0,1,0,…,0, ω3=0,1,1,0,…,0, ω4=a,0,1,0,…,0, and ω5=a,1,0,…,0∈WS∗. Since ω1∉δ1ω1S, ω2∉δ1ω2S, δ1ω3=0, ω4∉δ1ω4S, ω5∉δ1ω5S, ω1∉δ2ω1S, ω2∉δ2ω2S, δ2ω3=0, ω4∉δ2ω4S, ω5∉δ2ω5S, ω1∉δ3ω1S, δ3ω2=0, ω3∉δ3ω3S, δ3ω4=0, ω5∉δ3ω5S, δ4ω1=0, ω2∉δ4ω2S, ω3∉δ4ω3S, ω4∉δ4ω4S, and δ4ω5=0, the set δ1,δ2,δ3,δ4∪ω1,ω2,ω3,ω4,ω5 induces K4,5 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which contradicts our assumption. Hence, m=1, which shows that S≅S1×F1.
Supposem1∗=2, then S1≅ℤ9 and US1=6. Let a,b∈m1∗, such that ab=0 and anna=m1. Consider θ1=u1,0, θ2=u2,0, θ3=u3,0, θ4=u4,0, ϕ1=0,1, ϕ2=a,0, ϕ3=b,0, ϕ4=a,1, and ϕ5=b,1∈WS∗, where ui∈US1. Since θ1ϕ1=0, θ1∉θ1ϕ2S, θ1∉θ1ϕ3S, ϕ4∉θ1ϕ4S, ϕ5∉θ1ϕ5S, θ2ϕ1=0, θ2∉θ2ϕ2S, θ2∉θ2ϕ3S, ϕ4∉θ2ϕ4S, ϕ5∉θ2ϕ5S, θ3ϕ1=0, θ3∉θ3ϕ2S, θ3∉θ3ϕ3S, ϕ4∉θ3ϕ4S, ϕ5∉θ3ϕ5S, θ4ϕ1=0, θ4∉θ4ϕ2S, θ4∉θ4ϕ3S, ϕ4∉θ4ϕ4S, and ϕ5∉θ4ϕ5S, the set θ1,θ2,θ3,θ4∪ϕ1,ϕ2,ϕ3,ϕ4,ϕ5 induces K4,5 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which contradicts our assumption. Hence, m1∗≠2.
Supposem1∗≥3, then US1≥4. Let x,y,z∈m1∗, such that xy=xz=0 and annx=m1. Consider π1=v1,0, π2=v2,0, π3=v3,0, π4=v4,0, σ1=0,1, σ2=x,0, σ3=y,0, σ4=z,0, and σ5=x,1∈WS∗, where vi∈US1. Since π1σ1=0 and πi∉πiσjR, for j=2,3,4 and σ5∉πiσ5S for each i=1,2,3,4, the set π1,π2,π3,π4∪σ1,σ2,σ3,σ4,σ5 induces K4,5 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which contradicts our assumption. Hence, m1∗=1, which shows that S1≅ℤ4 or ℤ2x/x2.
SupposeF1≥4 and a∈m1∗, such that a2=0. Consider μ1=1,0, μ2=0,1, μ3=0,x, μ4=0,y, μ5=u,0, ν1=a,0, ν2=a,1, ν3=a,x, and ν4=a,y∈WS∗, where u∈US1 and 1≠x,y∈F1∗. Since μ1∉μ1ν1S, ν2∉μ1ν2S, ν3∉μ1ν3S, ν4∉μ1ν4S, μ2ν1=0, ν2∉μ2ν2S, ν3∉μ2ν3S, ν4∉μ2ν4S, μ3ν1=0, ν2∉μ3ν2S, ν3∉μ3ν3S, ν4∉μ3ν4S, μ4ν1=0, ν2∉μ4ν2S, ν3∉μ4ν3S, ν4∉μ4ν4S, μ5∉μ5ν1S, ν2∉μ5ν2S, ν3∉μ5ν3S, and ν4∉μ5ν4S, the set μ1,μ2,μ3,μ4,μ5∪ν1,ν2,ν3,ν4 induces K4,5 as a subgraph of AG′S. Thus, γAG′S>1 by Lemma 2, which is a contradiction to our assumption. Hence. F1≤3. From ([17], Theorem 3.1), it is clear that F1≇ℤ2. Hence, F1≅ℤ3.Figure 2
Toroidal embedding ofAG′ℤ5×ℤ5≅AG′ℤ2x/x2×ℤ3.


## 4. Crosscap ofAG′S

The crosscap of a graphG, denoted by γG, is the minimum integer k, such that G can be embedded in Nk, where Nk denotes a sphere with k crosscap attached to it. It is clear that γ¯G≥γ¯G˜, where G˜ is the reduction of G, and γ¯H≤γ¯G for any subgraph H of G. A graph G is said to be projective if γ¯G=1.In this section, we classify all the finite commutative ringsS, for which the coannihilator graph AG′S is projective. The following results are useful for the proof of subsequent results of this section.Lemma 3.
(see [3])(1)
Letn≥3. Then,(9)γ¯Kn=n−3n−46,ifn≥3andn≠7;3,ifn=7.(2)
Letn,m≥2. Then,(10)γ¯Km,n=m−2n−22.Lemma 4 (see [3]).
LetG be a connected graph with n≥3 vertices and q edges. Then,(11)γ¯G≥q3−n+2.Theorem 11.
LetS be a finite commutative ring. Then, γ¯AG′S=1⇔S is one of the following rings: ℤ49, ℤ7x/x2, F4×F4, F4×ℤ5, or ℤ2×ℤ2×ℤ2.Proof.
SinceAG′ℤ49=AG′ℤ7x/x2=AG′ℤ2×ℤ2×ℤ2=K6, AG′F4×F4=K3,3, and AG′F4×ℤ5=K3,4, by Lemma 3 and Figure 3, γ¯AG′S=1.
Conversely, assume thatγ¯AG′S=1. Since S is finite, S≅S1×S2×⋯×Sn, where Si is the local ring for each i. Suppose n≥4. Consider α1=1,0,0,…,0, α2=0,1,0,…,0, α3=0,0,1,0,…,0, α4=0,0,0,1,0,…,0, α5=0,1,1,0,…,0, β1=0,1,0,1,0,…,0, β2=0,0,1,1,0,…,0, and β3=0,1,1,1,0,…,0∈WS∗. Since α1β1=α1β2=α1β3=α2β2=α3β1=0, β1∉α2β1S, β3∉α2β3S, β2∉α3β2S, β3∉α3β3S, β1∉α4β1S, β2∉α4β2S, β3∉α4β3S, α5∉α5β1S, α5∉α5β2S, and β3∉α5β3S, the set α1,α2,α3,α4,α5∪β1,β2,β3 induces K5,3 as a subgraph of AG′S. Thus, γ¯AG′S>1 by Lemma 3, which is a contradiction to our assumption. Hence, n≤3. The following three cases occur:Case (i):n=3. Suppose that S1 is not a field with maximal ideal m1≠0, then there is a∈m1∗, such that anna=m1. Consider δ1=1,0,0, δ2=a,0,0, δ3=0,1,0, ω1=1,1,0, ω2=1,0,1, ω3=0,1,1, ω4=a,0,1, and ω5=a,1,0∈WS∗. Since ω1∉δ1ω1S, ω2∉δ1ω2S, δ1ω3=0, ω4∉δ1ω4S, ω5∉δ1ω5S, ω1∉δ2ω1S, ω2∉δ2ω2S, δ2ω3=0, ω4∉δ2ω4S, ω5∉δ2ω5S, ω1∉δ3ω1S, δ3ω2=0, ω3∉δ3ω3S, δ3ω4=0, and ω5∉δ3ω5S, the set δ1,δ2,δ3∪ω1,ω2,ω3,ω4,ω5 induces K3,5 as a subgraph of AG′S. Thus, γ¯AG′S>1 by Lemma 3, which is a contradiction to our assumption. Hence, Si is a field for each i=1,2,3.SupposeS1≥3. Consider θ1=1,0,0, θ2=a,0,0, θ3=0,1,0, ϕ1=1,1,0, ϕ2=1,0,1, ϕ3=0,1,1, ϕ4=a,0,1, and ϕ5=a,1,0∈WS∗, where 1≠a∈S1∗. Since θ1ϕ3=θ2ϕ3=θ3ϕ3=θ3ϕ1=θ3ϕ4=0, ϕ1∉θ1ϕ1S, ϕ2∉θ1ϕ2S, ϕ4∉θ1ϕ4S, ϕ1∉θ2ϕ1S, ϕ5∉θ1ϕ5S, ϕ2∉θ2ϕ2S, ϕ4∉θ2ϕ4S, ϕ5∉θ2ϕ5S, ϕ1∉θ3ϕ1S, ϕ3∉θ3ϕ3S, and ϕ5∉θ3ϕ5S, the set θ1,θ2,θ3∪ϕ1,ϕ2,ϕ3,ϕ4,ϕ5 induces K3,5 as a subgraph of AG′S. Thus, γ¯AG′S>1 by Lemma 3, which contradicts our assumption. Hence, Si=2 for each i=1,2,3, which shows that S≅ℤ2×ℤ2×ℤ2.Case (ii):n=2. If S1 and S2 both are not fields, then by Theorem 10, γ¯AG′S>1, a contradiction. Hence, at least one of S1 or S2 must be a field. Suppose S1 is not a field with maximal ideal m1≠0. Since γAG′S≤γ¯AG′S=1, by Theorem 10, S≅ℤ4×ℤ2, ℤ2x/x2×ℤ2, ℤ4×ℤ3, or ℤ2x/x2×ℤ3.IfS≅ℤ4×ℤ2 or ℤ2x/x2×ℤ2, then AG′S is planar by ([17], Theorem 8), a contradiction to our assumption.Also, ifS≅ℤ4×ℤ3 or ℤ2x/x2×ℤ3, then γ¯AG′S>1 by Lemma 4 and Figure 2, again a contradiction. Hence, S1 and S2 both are fields, which show that AG′S is a complete bipartite graph of order S1−1,S2−1. Since γ¯AG′S=1, by Lemma 3, S≅F4×F4 or F4×ℤ5.Case (iii):n=1. Then, S is a local ring. Thus, AG′S is complete by Theorem 2. Since γ¯AG′S=1, by Lemma 3, 5≤WS∗≤6. Hence, S≅ℤ49 or ℤ7x/x2.Figure 3
Projective embedding ofAG′ℤ2×ℤ2×ℤ2.



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