Nilpotent graphs with crosscap at most two

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AKCE International Journal of Graphs and Combinatorics ( ) –

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Nilpotent graphs with crosscap at most two

A. Mallika

^∗

, R. Kala

Department of Mathematics, Manonmaniam Sundaranar University, Tirunelveli - 627 012, India

Received 19 April 2017; received in revised form 21 November 2017; accepted 22 November 2017 Available online xxxx

Abstract

LetRbe a commutative ring with identity. The nilpotent graph ofR, denoted byΓ_N(R), is a graph with vertex setZ_N(R)^∗, and two verticesxandyare adjacent if and only ifx yis nilpotent, whereZ_N(R)= {x∈R:x yis nilpotent, for somey∈ R^∗}. In this paper, we characterize finite rings (up to isomorphism) with identity whose nilpotent graphs can be embedded in the projective plane or Klein bottle. Also, we classify finite rings whose nilpotent graphs are ring graph or outerplanarity index 1,2.

c

⃝2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

Keywords:Crosscap; Nilpotent; Planar; Outerplanar; Projective

1. Introduction

Throughout, rings are commutative with 1̸= 0. The zero-divisor of a ring Ris denoted by Z(R) and Z^∗(R) = Z(R)− {0}. Thezero-divisor graphof R, denoted byΓ(R), is an undirected graph whose vertices are elements of Z^∗(R) and two distinct verticesaandbare adjacent if and only ifab=0. The zero-divisor graph of a commutative ring was first investigated by Beck [1]. He was interested in coloring of the graphΓ(R).

Later, it was modified as in present form by Anderson and Livingston [2]. This paper deals with the nilpotent graph of rings. Recall that the nilpotent graph ofR, denoted byΓN(R), is a graph with vertex setZN(R)^∗, and two verticesx andyare adjacent if and only ifx yis nilpotent, whereZN(R)= {x∈ R:x yis nilpotent, for somey∈R^∗}. It was first investigated by Chen [3]. He consoders all the elements of ringRas the vertices of the graph and two verticesxand yare adjacent if and only ifx yis nilpotent. However, in 2010, Li [4] modified and studied the nilpotent graphΓ_N(R) ofR. Note that the usual zerodivisor graphΓ(R) is a subgraph of the graphΓ_N(R). IfR∼=F1×F2× · · · ×Fn,where eachFi is a field andn ≥2, thenΓN(R)∼=Γ(R).

The focus of this paper is on finding non-orientable genus(crosscap) of the graphΓN(R). This work is motivated by the following result of Hsieh [5] who completely determined the finite commutative rings whose zero-divisor graphs are projective planar. This paper is organized as follows: In Section2, we characterize finite rings (up to isomorphism)

Peer review under responsibility of Kalasalingam University.

∗ Corresponding author.

E-mail addresses:mallikamsu2010@gmail.com(A. Mallika),karthipyi91@yahoo.co.in(R. Kala).

https://doi.org/10.1016/j.akcej.2017.11.006

0972-8600/ c⃝2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

with identity whose nilpotent graphs can be embedded in the projective plane or Klein bottle. In Section3, we classify finite rings whose nilpotent graphs are ring graph or outerplanar.

One can refer to [6] for the following: We denote the ring of integers modulonbyZnand the field withqelements byFq. IfS is a subset ofR,we denote S^∗ =S− {0},R^× =U(R)=set of units of Rand Spec(R) is the set of all prime ideals ofR. If Ris a direct product of local rings, then 0R denotes the zero element of R. We useKn for the complete graph withnvertices andKm,nfor the complete bipartite graph.

One can refer to [7] for the following: By a surface, we mean a connected two-dimensional real manifold, i.e., a connected topological space such that each point has a neighborhood homeomorphic to an open disc. Every non- orientable compact surfaceN_k is a connected sum of finite copies ofk projective planes (see [7]). The number of copies of projective planes is called the crosscap number (non-orientable genus) of this surface. Thecrosscap of a graph G, denotedγ¯(G), is the smallest integerk such that the graph can be embedded in Nk. For example, the projective plane is of crosscap one and the Klein bottle is of crosscap two. A non-planar graph is said to beprojective if it can be embedded into a projective plane. The following results are used in the subsequent section.

Lemma 1([7]). Let m,n be integers and for a real number x,let⌈x⌉denote the smallest integer that is greater than or equal to x . Then

(i) γ(Kn)=

{^⌈(n−3)(n−4) 6

⌉

if n≥3and n̸=7

3 if n=7.

(ii) γ(Km,n)=

⌈(m−2)(n−2) 2

⌉, where m,n ≥2.

Theorem 1([8, Kuratowski’s]). The graph G is planar if and only if it contains no subgraph isomorphic to K5or K_3,3or a subdivision of one of these.

Theorem 2([9, Theorem 3.1]). Let(R,m)be a finite local ring with|R| = pⁿfor some prime p and some positive integer n ≥ 2. ThenΓN(R)is planar if and only if R is isomorphic to one of the following rings: Z4,^Z⟨²x^[x]2⟩,Z9or

Z3[x]

⟨x²⟩.

Theorem 3([9, Theorem 3.2]). Let R∼=R1× · · · ×Rn×F1× · · · ×Fmbe a finite commutative ring with identity, where each(Ri,mi)is a local ring and Fjis a field, m,n≥1and m+n≥2. ThenΓN(R)is planar if and only if R is isomorphic to one of the following rings:Z4×Z2or ^Z⟨²x^[x]2⟩ ×Z2.

Theorem 4([9]). Let R ∼= F1×F2× · · · ×Fn be a finite ring, where each Fi is a field and n ≥ 2. ThenΓ(R)is planar if and only if R is isomorphic to one of the following rings:Z2×F,Z3×F,Z2×Z2×Z2orZ2×Z2×Z3, where F is a finite field.

Theorem 5([5]). Let R be a finite ring such that|Spec(R)| ≥5. Thenγ(Γ(R))≥6.

Theorem 6([9]). Let(R,m)be a finite local ring with|R| = pⁿ for some prime p and some positive integer n≥2.

ThenΓN(R)∼=K|m^∗|+K_|R^×|, where R^×=R\mis the set of all units in R.

Theorem 7([5]). Let R be a finite ring such that|Spec(R)| =2. Thenγ(Γ(R))=1if and only if R is isomorphic to one of the following 16 rings: F4×F4,F4×Z5,Z2×^F4[x]

⟨x²⟩,Z2× ^Z4[x]

⟨x²+x+1⟩,Z2× ^Z2[x,y]

⟨x²,x y,y²⟩,Z2× ^Z4[x]

⟨2x,x²⟩,Z3× Z8,Z3×^Z2[x]

⟨x³⟩,Z3× ^Z4[x]

⟨x²−2,x³⟩,Z4×Z5,^Z⟨²x^[x]2⟩ ×Z5,F4×Z4,F4×^Z2[x]

⟨x²⟩,Z4×Z4,Z4×^Z2[x]

⟨x²⟩,^Z⟨²x^[x]2⟩ ×^Z2[x]

⟨x²⟩.

Theorem 8([5]). Let R be a finite ring such that|Spec(R)| =3. Thenγ(Γ(R))=1if and only if R is isomorphic to one of the following 6 rings:Z2×Z2×F4,Z2×Z2×Z4,Z2×Z2×^Z2[x]

⟨x²⟩,Z2×Z2×Z5,Z2×Z3×Z3,Z3×Z3×Z3. Theorem 9([5]). Let R be a finite ring such that|Spec(R)| =4. Thenγ(Γ(R))=1if and only if R is isomorphic to Z2×Z2×Z2×Z2.

Theorem 10([10]). Suppose m≥0,n ≥0, and m≥n−1. The non-orientable genus of K_m+Knis given by γ(K_m+Kn)=

⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

0 if n≤2

⌈(m−2)(n−2) 2

⌉

if n≥2and(m,n)̸=(4,5),

4 if (m,n)=(4,5).

Theorem 11([11]). The complete bipartite graph K3,5 admits the unique embedding into the Klein bottle, up to congruence, as shown inFig.1.

Fig. 1. The unique embedding ofK_3,5on the Klein bottle.

2. Crosscap ofΓN(R)

In this section, we classify all finite rings (up to isomorphism) with identity whose nilpotent graphs can be embedded in a projective plane or Klein bottle.

The following Lemma gives the general formula for the crosscap of the graphΓN(R) in the case of local ring.

Lemma 2. If (R,m)is a finite local ring with|R| =αand|m| =β ≥4, thenγ(Γ_N(R))=

⌈(α−β−2)(β−3) 2

⌉ .

Proof. ByTheorem 6,Γ_N(R)∼=K_|R^×|+K|m^∗|. Also|R^×| ≥ |m^∗| −1≥2 and (|R^×|,|m^∗|)̸=(4,5). ByTheorem 10, γ(Γ_N(R))=γ(K_|R^×|+K|m^∗|)=γ(K_|R^×|,|m^∗|)=⌈₍

α−β−2)(β−3) 2

⌉ .

Lemma 3. If R∼=R₁×R₂× · · · ×R_n(n ≥2)is a finite commutative ring with identity, where R_i’s are local rings which are not fields, thenγ(Γ_N(R))≥5.

Proof. SinceΓN(Z4×Z4) is a subgraph ofΓN(R),ΓN(R) contains a copy ofK4,7which is a subgraph of the graph induced by the set of vertices [(R^×₁×m₂)∪(m₁×R^×₂)∪(m₁×m₂)]\{0R}. ByLemma 1(i i),γ(Γ_N(R))≥γ(K4,7)≥5.

Thus the result follows.

We consider the finite ring R and soR ∼= R1×R2× · · · × Rn where Ri is a local ring. On the observation of Lemma 3, at least oneRi is a field, for somei.

Lemma 4. Let R ∼= F1× · · · ×Fn be a finite commutative ring with identity, where each Fj is a field, n ≥2and

|F1| ≤ |F2| ≤ · · · ≤ |F_n|. Thenγ(Γ_N(R))≥3if one of the following holds:

(i)n≥5

(i i)n =4and|F4| ≥3

(i i i)n=3,|F1| ≥3, and either F2≇ Z3or F3≇ Z3

(iv)n=3,|F1| =2= |F2|and|F3| ≥8 (v)n =3,|F1| =2,|F2| =3and|F3| ≥5 (vi)n=3,|F1| =2and|F2| ≥4.

Proof.

(i) Follows fromTheorem 5.

(ii) Supposen=4 and|F4| ≥3. Then sinceΓN(Z2×Z2×Z2×Z3) is a subgraph ofΓN(R),ΓN(R) contains a copy of K3,5which is a subgraph of a graph induced by set of vertices [(Z2×Z2×0×0)∪(0×0×Z2×Z3)]\{0R}. It then follows fromLemma 1(i i) thatγ(Γ_N(R))≥γ(K_3,5)≥2. We claim thatγ(Γ_N(R))̸=2. ByTheorem 11,K3,5

has a unique embedding inN2. The embedding ofK3,5inN2has six 4-faces and one 6 face. Consider the face f^∗ of length six. LetVi be the partition ofK3,5, fori =1,2. Then f^∗contains exactly 3 vertices from Vi for alli. In order to embed the set of verticesS = {(0,1,1,0),(1,0,0,1),(1,0,0,2)}, we have to choose the face f^∗ of length 6. Clearly N(S) = {(1,0,0,0),(0,1,0,0),(0,0,0,1),(0,0,0,2),(0,0,1,0)}. If f^∗ contains all the vertices from N(S), then the vertices of N(S) that are adjacent to the vertices (0,1,1,0),(1,0,0,1) and (1,0,0,2) make two edge crossings where N(S) denote the set of vertices adjacent to the vertices of S. If f^∗ does not contain all the elements ofN(S), then we cannot insert the vertices ofS without crossing. Thus we cannot embedΓ_N(R) inN₂. Henceγ(Γ_N(R))≥3.

(iii) Suppose n =3,|F1| ≥3, and eitherF2 ≇ Z3or F3 ≇ Z3. ThenΓN(Z3×Z3×F4) is a subgraph ofΓN(R).

SoΓN(R) contains a copy ofK3,8with vertex set [(0×0×F4) ∪ (Z3×Z3×0)]\ {0R}. It then follows from Lemma 1(i i) thatγ(Γ_N(R))≥γ(K3,8)≥3. ByTheorem 8,γ(Γ_N(Z3×Z3×Z3))=1.

(iv) Suppose|F₁| =2 = |F₂|,|F₃| ≥8 andn =3. Then sinceΓ_N(Z2×Z2×F8) is a subgraph ofΓ_N(R),Γ_N(R) contains a copy ofK3,7with vertex set [(Z2×Z2×0) ∪ (0×0×F8)]\ {0R}. It then follows fromLemma 1(i i) thatγ(ΓN(R))≥γ(K3,7)≥3.

(v) Suppose|F1| =2,|F2| =3,|F3| ≥5 andn =3. Then sinceΓ_N(Z2×Z3×Z5) is a subgraph ofΓ_N(R),Γ_N(R) contains a copy ofK_4,5with vertex set [(Z2×Z3×0) ∪ (0×0×Z5)]\ {0_R}. It then follows fromLemma 1(i i) thatγ(ΓN(R))≥γ(K4,5)≥3.

(vi) Suppose|F1| =2,|F2| ≥4 andn =3. Then sinceΓN(Z2×F4×F4) is a subgraph ofΓN(R),ΓN(R) contains a copy ofK3,7with vertex set [(0×0×F4) ∪ (Z2×F4×0)]\ {0_R}. It then follows fromLemma 1(i i) that γ(Γ_N(R))≥γ(K_3,7)≥3.

The following Theorems characterize the ringRfor whichγ(ΓN(R))=1 andγ(ΓN(R))=2.

Theorem 12. Let R be a finite ring. Thenγ(Γ_N(R))=1if and only if R is isomorphic to one of the following rings:

Z8,^Z⟨²x^[x]3⟩,⟨2x^Z,⁴x^[x]2−2⟩,^Z_⟨x^2[x_,y⟩^,y]2 ,_⟨2^Z4_,x⟩^[x]2,Z4×Z3,^Z⟨²x^[x]2⟩ ×Z3,F4×F4,F4×Z5,Z2×Z2×F4,Z2×Z2×Z5,Z2×Z3× Z3,Z3×Z3×Z3orZ2×Z2×Z2×Z2.

Proof. Assume thatγ(Γ_N(R))=1. Then byTheorem 5,|Spec(R)| ≤4. SinceRis finite, R∼=R1×R2× · · · ×R_n, where (Ri,mi) is a local ring for each i. Without loss of generality, assume that|R1| ≤ |R2| ≤ · · · ≤ |Rn|. If

|Spec(R)| =4, then byLemma 4(i i),R∼=Z2×Z2×Z2×Z2. Case 1: |Spec(R)| =3.

IfRi is a field for alli, thenΓN(R)∼=Γ(R) and so byTheorem 8,Ris isomorphic to one of the following rings:

Z2×Z2×F4,Z2×Z2×Z5,Z2×Z3×Z3,Z3×Z3×Z3. If not, without loss of generality, letR1be a local ring but not a field. ThenΓ_N(Z4×Z2×Z2) is a subgraph ofΓ_N(R). SoΓ_N(R) contains a copy ofK3,6which is the subgraph of a graph induced by the set of vertices [(Z4×0×0) ∪ (m₁×Z2×Z2)]\ {0_R}. It then follows fromLemma 1(i i) thatγ(ΓN(R))≥γ(K3,6)≥2, a contradiction.

Case 2:|Spec(R)| =2.

IfRiis a field, thenΓ_N(R)∼=Γ(R) and so byTheorem 7,R∼=F4×F4orF4×Z5. If not, without loss of generality, letR₁be a local ring but not a field. We claim that|m^∗₁| =1 and|R₂| ≤3.

Suppose that |m^∗₁| ≥ 2. Then let z1 = (b1,0),z2 = (b2,0),z3 = (u1,0),z4 = (u2,0),z5 = (u3,0),z6 = (b1,1),z7 = (b2,1) andz8 =(0,1), whereb1,b2 ∈ m^∗₁,ui ∈ R^×₁ for 1 ≤ i ≤ 3. ThenΓN(R) contains a copy of K3,5 which is the subgraph of the graph induced by set of vertices{z1, . . . ,z8}and soγ(ΓN(R)) ≥ γ(K3,5) ≥2, a contradiction. Hence|m^∗₁| =1 and so|R₁| =4.

Suppose that|R2| ≥4. Then letz∈m^∗₁andy1=(z,0),y2=(u1,0),y3=(u2,0),y4=(0, v1),y5=(0, v2),y6= (0, v3),y7=(z, v1),y8=(z, v2),y9 =(z, v3) whereu1,u2∈ R₁^×andvi ∈ R₂^∗for alli. ThenΓN(R) contains a copy ofK3,6which is the subgraph of the graph induced by set of vertices{y1, . . . ,y9}. It then follows fromLemma 1(i i) thatγ(Γ_N(R))≥γ(K_3,6)≥2, a contradiction. Hence|R₂| ≤3.

Fig. 2. Embedding ofΓN(Z4×Z3)∼=ΓN(^Z⟨²x^[x]2⟩ ×Z3) inN₁.

Thus we get R ∼= Z4 × Z2,Z4 ×Z3,^Z⟨²x^[x]2⟩ × Z2,^Z⟨²x^[x]2⟩ ×Z3. By Theorem 3, ΓN(R) is planar for R ∼= Z4×Z2,^Z⟨²x^[x]2⟩ ×Z2. Hence we conclude thatR∼=Z4×Z3or ^Z⟨²x^[x]2⟩ ×Z3.

Case 3:|Spec(R)| =1.

If|m1| ≤3, then byTheorem 2,Γ_N(R) is planar. Hence|m1| ≥4. Now byLemma 2,γ(Γ_N(R))=⌈₍

α−β−2)(β−3) 2

⌉ , where|R| =αand|m| =β. Sinceγ(Γ_N(R))=1,|R| =8 and|m| =4. SoRis isomorphic to one of the following rings: Z8,^Z⟨²x^[x]3⟩,⟨2x^Z,⁴x^[x]2−2⟩,^Z_⟨x^2[x_,y⟩^,y]2 or ^Z4[x]

⟨2,x⟩².

Conversely, supposeRis isomorphic to one of the following rings:Z8,^Z⟨²x^[x]3⟩,⟨2x^Z,⁴x^[x]2−2⟩,^Z_⟨x^2[x_,y⟩^,y]2 or ^Z4[x]

⟨2,x⟩². Then by Lemma 2,γ(Γ_N(R))=1.

If R ∼= Z4 ×Z3 or ^Z⟨²x^[x]2⟩ ×Z3, then byTheorem 3,Γ_N(R) is non-planar. MoreoverFig. 2 shows one explicit embedding ofΓ_N(R) in the projective plane. Thusγ(Γ_N(R))=1.

SupposeRis isomorphic to one of the following rings:F4×F4,F4×Z5,Z2×Z2×F4,Z2×Z2×Z5,Z2×Z3× Z3,Z3×Z3×Z3orZ2×Z2×Z2×Z2. Then byTheorems 7–9,γ(ΓN(R))=1. This completes the proof.

Theorem 13. Let R be a finite ring. Thenγ(Γ_N(R)) = 2 if and only if R is isomorphic to one of the following rings:Z4×F4,^Z⟨²x^[x]2⟩ ×F4,F4×Z7,Z5×Z5,Z2×Z2×Z7orZ2×Z3×F4.

Proof. Assume thatγ(Γ_N(R))=2. Then byTheorem 5,|Spec(R)| ≤4. SinceRis finite, R∼=R1×R2× · · · ×R_n, where (R_i,m_i)’s are local ring for eachi. Without loss of generality, assume that|R₁| ≤ |R₂| ≤ · · · ≤ |R_n|.

Case 1:|Spec(R)| =4.

Then byLemma 4(i i),R ∼= Z2×Z2×Z2×Z2. Also, byTheorem 12,γ(Γ_N(R)) = 1. Thus there is no ring available in this case.

Case 2: |Spec(R)| =3.

If Ri is a field for alli, then byLemma 4(i i i)–(vi), Ris isomorphic to one of the following rings: Z2×Z2× Z2,Z2×Z2×Z3,Z2×Z2×F4,Z2×Z2×Z5,Z2×Z2×Z7,Z2×Z3×Z3,Z2×Z3×F4,Z3×Z3×Z3.

(i) IfRis isomorphic toZ2×Z2×Z2, thenΓ_N(R)∼=Γ(R) and so byTheorem 4,Γ_N(R) is planar.

(i i) IfR is isomorphic to one of the following rings: Z2×Z2×Z3,Z2×Z2×F4,Z2×Z2×Z5,Z2×Z3× Z3,Z3×Z3×Z3, then byTheorem 12,γ(Γ_N(R))=1.

ThusRis isomorphic to the following rings:Z2×Z2×Z7,Z2×Z3×F4.

Suppose that at least one Ri is not a field. Without loss of generality, let R1be a local ring but not a field. Then ΓN(Z4×Z2 ×Z2) is a subgraph ofΓN(R). SoΓN(R) contains a copy of K3,6 which is the subgraph of a graph induced by the set of vertices [(Z4×0×0) ∪ (m×Z2×Z2)]\ {0R}. It then follows from Lemma 1(i i) that γ(ΓN(R))≥γ(K3,6)≥2. We claim thatγ(ΓN(R))̸=2. ByTheorem 11,K3,5has a unique embedding inN2and it has six 4-faces and one 6-face. Hence we can find a unique embedding ofK3,6inN2that contains nine 4-faces. Let

Fig. 3. Embedding ofΓN(Z4×F4)∼=ΓN(^Z⟨²x^[x]2⟩ ×F4) inN₂.

Vi be a partition ofK3,6 fori =1,2. Then each face contains exactly two vertices fromVi. Also eachvfromV2lie in exactly 3 faces. So there may be at most one face that contains the vertices{(0,1,0),(2,0,0),(2,1,0)}. Suppose there exists a face that contains the vertices{(0,1,0),(2,0,0),(2,1,0)}. Since (1,0,1) and (3,0,1) are adjacent to these vertices inΓ_N(R), there must be one edge crossing in the embedding ofΓ_N(R). Suppose not. Then we cannot insert that vertices without crossing. Hence we cannot have the embedding ofΓ_N(R) in the non-orientable surface of crosscap 2. Thusγ(ΓN(R))≥3, a contradiction.

Case 3:|Spec(R)| =2.

If Ri is a field, thenΓ_N(R) ∼= K|F₁^∗|,|F₂^∗|. Sinceγ(Γ_N(R)) =2,|F1| = 4,|F2| = 7 or|F1| = 5 = |F2|. Hence R∼=F4×Z7orZ5×Z5. If not, then at least oneRiis not a field. Without loss of generality, letR1be a local ring but not a field. We claim that|m^∗₁| =1 and|R2| ≤4.

Suppose that|m^∗₁| =2. ThenR1∼=Z9or ^Z⟨³x^[x]2⟩. Letb1,b2∈m^∗₁. Letz1 =(b1,0),z2 =(b2,0),z3 =(u1,0),z4 = (u2,0),z5 = (u3,0),z6 = (b1,1),z7 = (b2,1),z8 = (0,1),z9 = (u4,0),z10 = (u5,0)and z11 = (u6,0), where ui ∈ R₁^×for 1 ≤ i ≤ 6. ThenΓ_N(R) contains a copy ofK5,6 which is the subgraph of the graph induced by set of vertices{z1, . . . ,z11}and soγ(Γ_N(R))≥γ(K5,6)≥6, a contradiction. Hence|m^∗₁| =1 or|m^∗₁| ≥3.

Suppose that |m^∗₁| ≥ 3. Then ΓN(R) contains a copy of K4,7 which consists of vertices of [{(a,0) : a ∈ R₁^×} ∪ {(b,1),(b,0) : b ∈ m^∗₁}]\ {0_R}. It then follows fromLemma 1(i i) that γ(Γ_N(R)) ≥ γ(K4,7) ≥ 5, which contradicts our assumption. Therefore|m^∗₁| =1 and so|R1| =4.

Suppose that|R2| ≥5. Then letz∈m^∗₁andy1=(z,0),y2=(u1,0),y3=(u2,0),y4=(0, v1),y5=(0, v2),y6= (0, v3),y7 = (0, v4),y8 =(z, v1),y9 = (z, v2),y10 =(z, v3),y11 =(z, v4) whereu1,u2 ∈ R₁^×andvi ∈ R^∗₂ for all i. ThenΓ_N(R) contains a copy ofK3,8which is the subgraph of the graph induced by set of vertices{y1, . . . ,y11}. It then follows fromLemma 1(i i) thatγ(Γ_N(R))≥γ(K_3,8)≥3, a contradiction. Hence|R2| ≤4.

Thus we get R is isomorphic to one of the following rings: Z4 ×Z2,Z4 ×Z3,Z4 ×F4,^Z⟨²x^[x]2⟩ ×Z2,^Z⟨²x^[x]2⟩ × Z3,^Z⟨²x^[x]2⟩ ×F4. ByTheorem 2,ΓN(R) is planar for R ∼=Z4×Z2,^Z⟨²x^[x]2⟩ ×Z2. Also, byTheorem 12,γ(ΓN(R))=1 forR∼=Z4×Z3,^Z⟨²x^[x]2⟩ ×Z3. Hence we conclude thatR∼=Z4×F4or ^Z⟨²x^[x]2⟩ ×F4.

Conversely, suppose thatR∼=F4×Z7orZ5×Z5. Then sinceΓN(R)∼=K_|F^∗

1|,|F₂^∗|, byLemma 1(i i),γ(ΓN(R))=2.

Assume thatRis isomorphic to one of the following rings:Z4×F4or ^Z⟨²x^[x]2⟩ ×F4.

Then sinceΓ_N(Z4 ×F4) ∼= Γ_N(^Z⟨²x^[x]2⟩ ×F4),Γ_N(R) contains a copy of K3,6 which is the subgraph of the graph induced by the set of vertices [(Z4×0) ∪ (m×F4)]\ {0R}. ByLemma 1(i i),γ(ΓN(R))≥γ(K3,6)≥2. Moreover Fig. 3shows one explicit embedding ofΓ_N(R) inN2. Thusγ(Γ_N(R))=2.

Assume that R is isomorphic toZ2 ×Z2 ×Z7. Then letri = (0,0,i),r6+i = (0,1,i),r12+i = (1,0,i),r19 = (0,1,0),r20 = (1,0,0) andr21 =(1,1,0) for 1 ≤ i ≤ 6. ThenΓ_N(R) contains a copy of K3,6 which consists of vertices{r1, . . . ,r6,r19,r20,r21}. It then follows fromLemma 1(i i) thatγ(ΓN(R))≥γ(K3,6) ≥2. MoreoverFig. 4 shows the embedding in the non-orientable surface of crosscap 2. Thusγ(Γ_N(R))=2.

Fig. 4. Embedding ofΓ_N(Z2×Z2×Z7) inN2.

Fig. 5. Embedding ofΓN(Z2×Z3×F4) inN2.

Suppose that R ∼= Z2 ×Z3 ×F4. Then let wi = (0,0,a_i), w3+i = (0,1,a_i), w6+i = (0,2,a_i), w9+i = (1,0,a_i), w13 = (0,1,0), w14 = (0,2,0), w15 = (1,0,0), w16 = (1,1,0), w17 = (1,2,0) for 1 ≤ i ≤ 3. Then Γ_N(R) contains a copy ofK3,5which consists of vertices{w1, w2, w3, w13, w14, w15, w16, w17}. It then follows from Lemma 1(i i) thatγ(Γ_N(R)) ≥ γ(K3,5) ≥ 2. MoreoverFig. 5shows one explicit embedding ofΓ_N(R) in the non- orientable surface of crosscap 2. Thusγ(ΓN(R))=2. This completes the proof.

3. Outerplanarity ofΓN(R)

LetGbe a graph withnvertices andqedges. A chord is any edge ofGjoining two nonadjacent vertices in a cycle ofG. LetCbe a cycle ofG. We sayCis a primitive cycle if it has no chords. Also, a graphGhas the primitive cycle property (PCP) if any two primitive cycles intersect in at most one edge. The number f r ank(G) is called the free rank ofGand it is the number of primitive cycles ofG. Also, the numberr ank(G)=q−n+ris called the cycle rank of G, whereris the number of connected components ofG. The cycle rank ofGcan be expressed as the dimension of the cycle space ofG. By [12, Proposition 2.2], we haver ank(G)≤ f r ank(G). A graphGis called a ring graph if it satisfies one of the following equivalent conditions (see [9]).

(i) rank(G) = frank(G),

(i i) Gsatisfies the PCP andGdoes not contain a subdivision ofK4as a subgraph.

The following concepts were defined in [13]. An embeddingφof a planar graph is1-outerplanarif it is outerplanar (i.e.) all vertices are incident on the outerface inφ. An embedding of a planar graph isk-outerplanarif removing all

vertices on the outerface (together with their incident edges) yields a (k−1)-outerplanar embedding. A graph is k-outerplanar if it has ak-outerplanar embedding. Theouterplanarity indexof a graphGis the smallestksuch that Gisk-outerplanar. There is a characterization for outerplanar graphs that says a graph is outerplanar if and only if it does not contain a subdivision ofK4or K2,3. Clearly, every outerplanar graph is a ring graph or every ring graph is a planar graph. In this section, we characterize all rings R such thatΓ_N(R) is a ring graph and outerplanar graph with index 1,2.

Theorem 14. Let R be a finite ring. ThenΓ_N(R)is a ring graph if and only if R is isomorphic to one of the following rings:

Z4,^Z⟨²x^[x]2⟩,Z9,^Z⟨³x^[x]2⟩,Z4×Z2,^Z⟨²x^[x]2⟩ ×Z2,Z2×F,Z3×Z3orZ2×Z2×Z2.

Proof. Assume thatΓ_N(R) is a ring graph. Then since every ring graph is planar, it is enough to check rings in Theorems 2–4, whether it is a ring graph or not. Now the nilpotent graph of rings

Z4,Z2[x]

⟨x²⟩ ,Z9,Z3[x]

⟨x²⟩ ,Z4×Z2,Z2[x]

⟨x²⟩ ×Z2,Z2×F,Z3×Z3,Z2×Z2×Z2,

we have thatr ank(Γ_N(R))= f r ank(Γ_N(R)), and so they are ring graphs. Also for ringsZ2×Z2×Z3,Z3×F, where

|F| ≥4,ΓN(R) contains a subdivision ofK4or does not satisfy PCP (seeFig. 6) and so they are not ring graphs. The converse is obvious.

Fig. 6.

Theorem 15. Let R be a finite ring. Then ΓN(R)is an outerplanar if and only if R is isomorphic to one of the following rings:

Z4,Z2[x]

⟨x²⟩ ,Z2×F,Z3×Z3orZ2×Z2×Z2.

Proof. Assume thatΓ_N(R) is an outerplanar. Then sinceΓ_N(R) is a ring graph, byTheorem 14, we need to check rings

Z4,Z2[x]

⟨x²⟩ ,Z9,Z3[x]

⟨x²⟩ ,Z4×Z2,Z2[x]

⟨x²⟩ ×Z2,Z2×F,Z3×Z3,Z2×Z2×Z2

are outerplanar or not. ByFig. 7, we can find the subdivision ofK2,3inΓ_N(R) for ringsZ9,^Z⟨³x^[x]2⟩,Z4×Z2,^Z⟨²x^[x]2⟩ ×Z2

and so are not outerplanar. Also, the nilpotent graph for ringsZ4,^Z⟨²x^[x]2⟩,Z2×F,Z3×Z3,Z2×Z2×Z2, isP3,K1,q−1,C4

orG(seeFig. 6) respectively and so are outerplanar. The converse is obvious.

Fig. 7.

Theorem 16. Let R be a finite ring. ThenΓ_N(R)has an outerplanarity index 2 if and only if R is isomorphic to one of the following rings:

Z9,^Z⟨³x^[x]2⟩,Z4×Z2,^Z⟨²x^[x]2⟩ ×Z2,Z3×F,Z2×Z2×Z3,|F| ≥4.

Proof. Assume that ΓN(R) has an outerplanarity index 2. Then ΓN(R) is planar, by Theorems 2–4, we need only to check rings:Z4,^Z⟨²x^[x]2⟩,Z9,^Z⟨³x^[x]2⟩,Z4×Z2,^Z⟨²x^[x]2⟩ ×Z2,Z2 ×F,Z3×F,Z2×Z2 ×Z2,Z2 ×Z2×Z3. are 2-outerplanar and is minimum or not. Since Γ_N(R) is not 1-outerplanar, by Theorem 15, R is not isomorphic to the following rings: Z4,^Z⟨²x^[x]2⟩,Z2 ×F,Z3 ×Z3,Z2 ×Z2 ×Z2. By Figs. 6, 7, the nilpotent graph for rings Z9,^Z⟨³x^[x]2⟩,Z4×Z2,^Z⟨²x^[x]2⟩,Z3×F,Z2×Z2×Z3,|F| ≥4, are 2-outerplanar, as the removal of vertices in the outer face remains the graphP3or totally disconnected. The converse is obvious.

Corollary 1. Let R be a finite ring. ThenΓN(R)has an outerplanarity index at most two.

Acknowledgments

The authors are deeply grateful to the referee for careful reading of the paper and helpful comments and many valuable suggestions that has improved the paper a lot. The work reported here is supported by the INSPIRE programme (IF 110684) awarded to the first author by the Department of Science and Technology, Government of India.

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