On the intersection graph of gamma sets in the total graph of a commutative ring-I

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February 13, 2013 17:12 WSPC/S0219-4988 171-JAA 1250198

Journal of Algebra and Its ApplicationsVol. 12, No. 4 (2013) 1250198 (18 pages)c© World Scientific Publishing CompanyDOI: 10.1142/S0219498812501988

THE INTERSECTION GRAPH OF GAMMA SETSIN THE TOTAL GRAPH OF A

COMMUTATIVE RING-I

T. TAMIZH CHELVAM∗ and T. ASIR†

Department of MathematicsManonmaniam Sundaranar University

Tirunelveli 627 012, Tamil Nadu, India∗[email protected]

†[email protected]

Received 22 March 2012Accepted 28 August 2012

Published 15 February 2013

Communicated by S. R. Lopez-Permouth

Let R be a commutative ring and Z(R) be its set of all zero-divisors. Anderson andBadawi [The total graph of a commutative ring, J. Algebra 320 (2008) 2706–2719]introduced the total graph of R, denoted by TΓ(R), as the undirected graph with vertexset R, and two distinct vertices x and y are adjacent if and only if x + y ∈ Z(R).Tamizh Chelvam and Asir [Domination in the total graph of a commutative ring, toappear in J. Combin. Math. Combin. Comput.] obtained the domination number of thetotal graph and studied certain other domination parameters of TΓ(R) where R is acommutative Artin ring. The intersection graph of gamma sets in TΓ(R) is denoted byITΓ(R). Tamizh Chelvam and Asir [Intersection graph of gamma sets in the total graph,Discuss. Math. Graph Theory 32 (2012) 339–354, doi:10.7151/dmgt.1611] initiated astudy about the intersection graph ITΓ(Zn) of gamma sets in TΓ(Zn). In this paper, westudy about ITΓ(R), where R is a commutative Artin ring. Actually we investigate theinterplay between graph-theoretic properties of ITΓ(R) and ring-theoretic properties ofR. At the first instance, we prove that diam(ITΓ(R)) ≤ 2 and gr(ITΓ(R)) ≤ 4. Also somecharacterization results regarding completeness, bipartite, cycle and chordal nature ofITΓ(R) are given. Further, we discuss about the vertex-transitive property of ITΓ(R).At last, we obtain all commutative Artin rings R for which ITΓ(R) is either planar ortoroidal or genus two.

Keywords: Artin ring; total graph; gamma sets; intersection graph; vertex-transitive;genus.

Mathematics Subject Classification: 05C25, 05C07, 05C10, 13A15, 13M05, 16P20

1. Introduction

In recent years, the interplay between ring structure and graph structure is studiedby many researchers. For such kind of study, researchers define a graph whose

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T. Tamizh Chelvam & T. Asir

vertices are a set of elements in a ring or a set of ideals in a ring and edges aredefined with respect to a condition on the elements of the vertex set. A graph isdefined out of nonzero zero divisors of a ring and is called zero-divisor graph ofa ring [5]. Interesting variations are also defined like comaximal graph [19], totalgraph [2] and unit graph [6] associated with rings. Also graphs are defined out ofideals of a ring, namely annihilating-ideal graph of a ring [10, 11] and intersectiongraph of ideals of rings [12, 13]. The graphs constructed from rings help us to studythe algebraic properties of rings using graph theoretical tools and vice-versa. LetA be a set and let S be a collection of nonempty subsets of A. The intersectiongraph of S is the graph whose vertices are the elements of S and two vertices areadjacent if the subsets have a nonempty intersection [21]. Let R be a commutativering, Z(R) be its set of zero-divisors. Anderson and Badawi [2] introduced theconcept of the total graph corresponding to a commutative ring. The total graph ofR, denoted by TΓ(R), is the undirected graph with vertex set R, and for distinctx, y ∈ R, the vertices x and y are adjacent if x + y ∈ Z(R). Thereafter variousresearch papers have been published on the total graph of a commutative ring(see [1, 3, 4, 7, 20, 22, 23, 26]). Also a generalized total graph has been studied byBarati et al. in [9].

Now, in this paper, we study about the intersection graph of gamma sets in thetotal graph of a commutative ring R with vertex set as the collection of all γ-setsof TΓ(R) and two distinct γ-sets u and v are adjacent if and only if u ∩ v �= ∅.This graph is denoted by ITΓ(R). We investigate the interplay between the graph-theoretic properties of ITΓ(R) and the ring-theoretic properties of R. Throughoutthis paper, R is a commutative ring with 1 �= 0 and 2 = 1 + 1. For any a ∈ R,Ann(a) = {x ∈ R : ax = 0} is the annihilator ideal of a in R and the idealgenerated by a is denoted by (a). Let I be an annihilator ideal with |R/I| finiteand |R/I| = min{|R/J | : J is an annihilator ideal of R}. Note that such an ideal I

turns out to be a maximal ideal in R. Let us take |I| = λ, |Z(R)| = α, |R/Z(R)| = β

and |R/I| = µ. We denote R1 × R2 for the direct product of two rings R1 and R2.For a general reference on rings, we refer to Kaplansky [17].

Let G = (V, E) be a simple graph with vertex set V and edge set E. A graph inwhich each pair of distinct vertices is joined by an edge is called a complete graph.We use Kn to denote the complete graph with n vertices. For a vertex v ∈ V , deg(v)is the degree of a vertex v, δ(G) = min{deg(v) : v ∈ V (G)}, ∆(G) = max{deg(v) :v ∈ V (G)}. We say that G is connected if there is a path between any two distinctvertices of G. For vertices u and v of G, the length of a shortest path from u to v

is denoted by d(u, v). The diameter of G is diam(G) = sup{d(u, v) : u, v ∈ V (G)}.The eccentricity of a vertex v in G is defined by e(v) = max{d(v, w) : w ∈ V (G)}.A graph G is said to be self-centered if e(v) = rad(G) for all v ∈ V (G). The girthof a graph is the length of the shortest cycle in the graph. A graph is said tobe vertex-transitive if its automorphism group acts transitively on the vertex set.Let G1 and G2 be two graphs. The union of G1 and G2, denoted by G1 ∪ G2, isa graph with vertex set V (G1) ∪ V (G2) and edge-set E(G1) ∪ E(G2). Graphs G

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Intersection Graph of Gamma Sets in the Total Graph-I

and H are said to be isomorphic to one another, written G ∼= H , if there exists aone-to-one correspondence f : V (G) → V (H) such that for each pair x, y of verticesof G, xy ∈ E(G) if and only if f(x)f(y) ∈ E(H). A subset S of V is called adominating set in G, if every vertex in V − S is adjacent to at least one vertexin S. The domination number of G, denoted by γ(G), is the minimum cardinalityof a dominating set in G and such a dominating set is called as a γ-set of G. ForS ⊆ V, 〈S〉 denotes the subgraph induced by S. For basic graph theory parameters,we refer to Chartrand [14] and for the domination parameters, we use Haynes [16].

The purpose of this paper is to study the basic graph theoretical properties ofthe intersection graph of gamma sets in TΓ(R). In Sec. 2, we list certain fundamentalproperties regarding domination number and dominating sets in the total graph of acommutative ring, which are prelude for subsequent sections. In Sec. 3, we begin thestudy about ITΓ(R) and discuss various fundamental properties including degree,diameter and girth of ITΓ(R). In Sec. 4, we discuss about the vertex-transitiveproperty of ITΓ(R). In Sec. 5, we characterize all rings R for which the genus ofITΓ(R) is zero (planar) or one (toroidal) or two. Many of the results proved in thispaper generalize the results proved in [24].

2. Properties of Dominating Sets in TΓ(R)

In this section, we list certain fundamental properties regarding domination numberand dominating sets in TΓ(R) for a commutative Artin ring R. Further, we charac-terize all γ-sets in TΓ(R), which are vital for the study on ITΓ(R). The followingresult proved by Maimani et al. [20, Lemma 1.1] is used frequently and hence givenbelow.

Observation 2.1 ([20, Lemma 1.1]). Let R be a finite commutative ring, Z(R)be its set of all zero-divisors in R and TΓ(R) be the total graph of R. Then thefollowing are true:

(i) If 2 ∈ Z(R), then deg(v) = |Z(R)| − 1 for every v ∈ V (TΓ(R)).(ii) If 2 /∈ Z(R), then deg(v) = |Z(R)| − 1 for every v ∈ Z(R) and deg(v) = |Z(R)|

for every vertex v /∈ Z(R).

The results given below in Lemma 2.1 and Theorem 2.1 are part of a paperaccepted for publication [25]. For the sake of completeness, we present proofs also.

Lemma 2.1 ([25, Lemma 2.2]). Let R be a commutative ring (not necessarilyfinite) with identity, I be an annihilator ideal in R such that |R/I| = µ is finite andminimum. Then γ(TΓ(R)) ≤ µ.

Proof. Let H be the spanning subgraph of TΓ(R) in which two distinct verticesx, y ∈ R are adjacent if x + y ∈ I. By noting that I is an ideal contained in Z(R),one can obtain, in the same way as in [2, Theorem 2.2], H is a subgraph of TΓ(R)

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and further

(i) if 2 ∈ I, then H is the union of µ disjoint complete graphs,(ii) if 2 /∈ I, then H is the union of one complete graph together with µ−1

2 disjointcomplete bipartite graphs.

Note that every connected component of H which is complete corresponds toa coset of I in R. On the other hand, a connected component which is completebipartite corresponds to two cosets of I in R. From this, any set containing exactlyone element from each of the cosets of I in R is a minimum dominating set of H

and so γ(H) = µ. Since H is a spanning subgraph of TΓ(R), γ(TΓ(R)) ≤ µ.

If Z(R) is an ideal of R, then Z(R) is an annihilator ideal and |R/Z(R)| isminimum. In view of this and by Lemma 2.1, one can have the following lemma.

Lemma 2.2. Let R be a commutative ring such that Z(R) is an ideal of R,

|R/Z(R)| = µ and R/Z(R) = {ai +Z(R) : 1 ≤ i ≤ µ}. Then the following are true:

(i) γ(TΓ(R)) = µ.

(ii) The set of vertices {x1, x2, . . . , xµ} is a γ-set of TΓ(R) if and only if xi ∈ai + Z(R) for 1 ≤ i ≤ µ.

Theorem 2.1 ([25, Theorem 2.5]). Let R be a commutative Artin ring which isnot an integral domain, I be an annihilator ideal of R such that |R/I| = µ is finiteand minimum. Then γ(TΓ(R)) = µ.

Proof. Clearly by Lemma 2.1, γ(TΓ(R)) ≤ µ. For the converse, since R is anArtin ring, by [8, Theorem 8.7], one can write R ∼= R1 × R2 × · · · × Rn whereeach Ri is an Artin local ring with maximal annihilator ideal mi = Z(Ri). ByLemma 2.2, γ(TΓ(Ri)) = |Ri/mi| for all i. Let Ii = R1 × · · · × Ri−1 × mi ×Ri+1 × · · · × Rn and µi = |R/Ii|. Note that |R/Ii| = |Ri/mi|. It is easy to seethat for 1 ≤ i ≤ n, Ii’s are the only maximal annihilator ideals of R and soµ = |R/I| = min{µi : 1 ≤ i ≤ n} = min{γ(TΓ(Ri)) : 1 ≤ i ≤ n}. Now, supposeS = {(a11, a12, . . . , a1n), (a21, a22, . . . , a2n), . . . , (a(µ−1)1, a(µ−1)2, . . . , a(µ−1)n)} is adominating set of TΓ(R). Let Sk = {a1k, a2k, . . . , a(µ−1)k} for 1 ≤ k ≤ n. Sinceγ(TΓ(Rj)) > µ − 1, Sj is not a dominating set of TΓ(Rj) and so there exists anelement bj ∈ Rj such that bj is not dominated by Sj for every j = 1, 2, . . . , n.Therefore bj + Sj ∈ Reg(Rj). Since Reg(R) ∼= Reg(R1) × · · · × Reg(Rn), we have(b1, b2, . . . , bn) ∈ V (TΓ(R)) is not dominated by S, a contraction to S is a dominat-ing set of TΓ(R). Hence γ(TΓ(R)) = µ.

Remark 2.1. If R is a finite commutative ring, then there exists at least one anni-hilator ideal I such that |R/I| is finite and minimum. On the other hand, thereare infinite rings with no annihilator ideal I such that |R/I| is finite. For example,consider the ring R = Z × Z[x]

〈x2〉 . Then R is infinite and, for any annihilator ideal I

of R, |R/I| is infinite.

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Just because of the fact mentioned above, throughout this paper, we assume thatR is a commutative Artin ring with at least one annihilator ideal I such that |R/I|is finite. Further, we assume that µ = min{|R/I| : I is an annihilator ideal of R}.Since the number of annihilator ideals I of R with |R/I| finite and minimum is alsofinite, hereafter we assume that Ii’s (i = 1, 2, . . . , t) are annihilator ideals in R suchthat for each i, |R/Ii| is finite and minimum, i.e. |R/Ii| = µ for each i. The resultsgiven below characterize all γ-sets in TΓ(R).

Theorem 2.2. Let R be a commutative Artin ring which is not an integral domain.

Assume that Ii’s (i = 1, 2, . . . , t) are the annihilator ideals of R such that |R/Ii| = µ

is finite and minimum, and R/Ii = {aij + Ii : 1 ≤ j ≤ µ} for each i, 1 ≤ i ≤ t.

Then {x1, x2, . . . , xµ} is a γ-set of TΓ(R) if and only if xj ∈ aij +Ii for j = 1, . . . , µ

and for some fixed i.

Proof. Let S = {x1, . . . , xµ} ⊂ V (TΓ(R)) where xj ∈ aij + Ii for j = 1, . . . , µ andfor some fixed i. For i = 1, 2, . . . , t, let Hi be the spanning subgraph of TΓ(R) inwhich two distinct vertices x, y ∈ R are adjacent if x + y ∈ Ii. By the proof ofLemma 2.1, a set with exactly one element from each of the cosets of Ii in R is adominating set of TΓ(R) and so by Theorem 2.1, S is a γ-set of TΓ(R).

Conversely, assume that S = {x1, . . . , xµ} is a γ-set of TΓ(R). Let R ∼= R1 ×· · · × Rn where each R� is a local ring. Let Z(R�) = m� and |R�/m�| = µ� for1 ≤ � ≤ n. Since each maximum annihilator ideal of R will be of the form R1 ×· · · × Rs−1 × ms × Rs+1 × · · · × Rn for some s, we get that t ≤ n. Assume thatIi = R1 × · · · × Ri−1 × mi × Ri+1 × · · · × Rn for 1 ≤ i ≤ t. Let us take forr = 1, 2, . . . , µ, xr = (br1, br2, . . . , brn) where br� ∈ R� and Sk = {b1k, b2k, . . . , bµk}for 1 ≤ k ≤ n.

Suppose for each i = 1, 2, . . . , t, there exist p and q such that p �= q and xp, xq ∈aij + Ii for some j with 1 ≤ j ≤ µ. Then there exist bpe and bqf such that bpe, bqf ∈aij + Z(Ri) and so by Lemma 2.2(ii), Si is not a dominating set of TΓ(Ri). Thusfor each i = 1, 2, . . . , t, there is an element ci ∈ Ri such that ci is not dominatedby Si. Since µ� > µ for all � ≥ t + 1, S� is not a dominating set of TΓ(R�) andso there exists c� ∈ R� such that c� is not dominated by S�. Since Reg(R) ∼=Reg(R1) × · · · × Reg(Rn), (c1, c2, . . . , cn) ∈ V (TΓ(R)) is not dominated by S, acontradiction.

If {x1, x2, . . . , xµ} is a γ-set of TΓ(R) with xj ∈ aij + Ii for 1 ≤ j ≤ µ and forsome fixed i (1 ≤ i ≤ t), then we denote the same as a γ-set with respect to Ii.By the definition of the total graph, we have the following lemma applicable to anintegral domain.

Lemma 2.3. Let R be an integral domain and k = |{x ∈ R : 2x = 0}|. Then thefollowing are true:

(i) γ(TΓ(R)) = k + |R|−k2 .

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(ii) Every γ-set of TΓ(R) contains 0. Further, S = {0, x1, . . . , xk+ |R|−k2

} ⊂V (TΓ(R)) is a γ-set of TΓ(R) if and only if 2xi = 0 for each i = 1, . . . , k

and xi �= −xj for all k < i �= j ≤ k + |R|−k2 .

3. Basic Properties of ITΓ(R)

As mentioned earlier, the intersection graph of gamma sets in the total graph of acommutative ring R, denoted by ITΓ(R), is the undirected graph with vertex setas collection of all γ-sets of TΓ(R) and two distinct vertices u and v are adjacentif and only if u ∩ v �= ∅. Since we could attempt the domination number of TΓ(R)only for a commutative Artin ring R with at least an annihilator ideal I such thatR/I is finite, we define the intersection graph of gamma sets in the total graphcorresponding to a commutative Artin ring with at least an annihilator ideal I suchthat R/I is finite.

In this section, we study some basic properties of the intersection graph ITΓ(R).In particular, we show that if |R| ≥ 3, then ITΓ(R) is connected and has small diam-eter and girth. Also we characterize when ITΓ(R) is in the class of cycles, bipartitegraphs, complete graphs and star graphs. We first start with some examples.

Example 3.1. (a) Let R be an integral domain. As mentioned in Lemma 2.3, eachγ-set of TΓ(R) contains 0 and so ITΓ(R) is a complete graph.(b) Below are the intersection graphs of gamma sets in TΓ(R) for several rings R.Note that in the following figure, an element (x, y) of Z2 × Z2 is denoted as xy.

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One may notice that even though Z4 is not isomorphic to Z2[x]/(x2), the corre-sponding intersection graphs of gamma sets are one and the same.

If |R| = 1 or 2, then ITΓ(R) = K1 and so hereafter we assume that |R| ≥ 3.The following lemma exhibits a relation between the ring R and the correspondingintersection graph ITΓ(R). More specifically, the relation is concerning the finitenessproperty of R and ITΓ(R).

Lemma 3.1. Let R be a commutative Artin ring with an annihilator ideal I suchthat R/I is finite. Then R is finite if and only if ITΓ(R) is finite.

Proof. Assume that ITΓ(R) is finite. If R is infinite, then I is infinite. Since oneelement from every coset of I in R form a γ-set of TΓ(R), the number of verticesin ITΓ(R) is infinite, a contradiction. The other part is trivial.

Let R be a commutative ring with an annihilator ideal J such that |R/J | isfinite and let |R/I|= min{|R/J | :J is an annihilator ideal of R}. As mentioned inRemark 2.1, if R is Artin, then we assume that Ii, for i = 1, . . . , t are the annihilatorideals of R such that |R/Ii| is finite and minimum. By Theorem 2.2, the study ofintersection graph of gamma sets in the total graph of a commutative ring naturallybreaks into two cases depending on whether R has the unique maximal annihilatorideal I with |R/I| finite and minimum (i.e. t = 1) or R has more than one suchannihilator ideal (i.e. t > 1).

The following lemma finds the cardinality of the vertex set of ITΓ(R) and thedegree of each vertex in ITΓ(R).

Lemma 3.2. Let R be a finite commutative ring and Ii (i = 1, 2, . . . , t) be anni-hilator ideals in R such that |R/Ii| is minimum. Let |Ii| = λ, |R/Ii| = µ andk = |{x ∈ R : 2x = 0}|. Then

(i)

|V (ITΓ(R))| =

2|R|−k

2 if R is an integral domain,

λµ if t = 1,

2λµ − µ!(

λ

µ

)µ

if t = 2.

Moreover if t ≥ 3, then |V (ITΓ(R))| ≥ 2λµ − µ!(λµ)µ > λµ.

(ii) If t = 1, then deg(v) =∑µ

i=1[λµ−i(λ−1)i−1]−1 for all v ∈ V (ITΓ(R)). Further

if µ = 2, then deg(v) = |R| − 2 for all v.

(iii) If t ≥ 2, then deg(v) ≥ ∑µi=1[λ

µ−i(λ − 1)i−1] − 1 + (λµ )λµ−2 for all v ∈

V (ITΓ(R)).

Proof. (i) Suppose R is an integral domain. Then by Lemma 2.3(ii), every γ-setS of TΓ(R) contains every element x ∈ R with 2x = 0 and also S contains either y

or −y whenever 2y �= 0. Hence the number of γ-sets in TΓ(R) is equal to 2|R|−k

2 .

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Suppose R is not an integral domain. Then by Theorem 2.1, any γ-set of TΓ(R)contains µ elements. Let v = {x1, x2, . . . , xµ} ∈ V (ITΓ(R)). If t = 1, then byTheorem 2.2, elements for a γ-set of TΓ(R) are from distinct cosets of I1. Fromthis, the number of choices for an element in a γ-set of TΓ(R) is λ and hence thenumber of γ-sets in TΓ(R) is λµ.Suppose t = 2. Let R ∼= R1 × R2 × · · · × Rn where each Ri is a local ring withthe maximal annihilator ideal mi = Z(Ri). By assumption on t, one can takeI1 = m1×R2×· · ·×Rn and I2 = R1×m2×· · ·×Rn. By Theorem 2.2, xj /∈ xi + I1

or xj /∈ xi + I2 for all 1 ≤ i �= j ≤ µ. As proved above, the number of γ-setsin TΓ(R) with respect to I1 as well as I2 is at the maximum 2λµ. In order toidentify the number of repetitions, let us take R/I1 = {I1, y1 + I1, . . . , yµ−1 + I1},X0 = I1 ∩ I2 = m1 ×m2 ×R3 × · · · ×Rn and Xi = (yi + m1)×m2 ×R3 × · · · ×Rn

for i = 1, . . . , µ − 1. Then I2 =⋃µ−1

i=0 Xi and so I2 can be partitioned into µ sets.Similarly each coset of I2 in R can be partitioned into µ sets and elements in each ofthese sets belong to the same coset of I1 in R. Suppose S is a repeated γ-set. Thenthe number of choices for the first element in S is µ(λ

µ), the number of choices forthe second element is (µ−1)(λ

µ) and the number of choices for the last element is λµ .

Thus the number of repeated γ-sets in TΓ(R) is µ!(λµ)µ and so the distinct number

of γ-sets in TΓ(R) is 2λµ − µ!(λµ )µ. From these facts, |V (ITΓ(R))| ≥ 2λµ − µ!(λ

µ)µ

whenever t ≥ 3.(ii) Assume that t = 1. As per the choice of elements in any γ-set of ITΓ(R), thereare λµ−1 vertices in ITΓ(R) containing x1. Similarly, there are λµ−2 vertices inITΓ(R) containing both x1 and x2 and so there are λµ−1 − λµ−2 = λµ−2(λ − 1)vertices in ITΓ(R) containing x2 but not x1. Continuing in this way, one can findall vertices in ITΓ(R) which are adjacent to v. Since ITΓ(R) is simple, v should notbe counted for deg(v) and hence we get the desired degree. If µ = 2, then λ = |R|

2

and so deg(v) = |R| − 2.(iii) Suppose t ≥ 2. Consider two annihilator ideals I1 and I2 of R with |R/I1| =|R/I2| = µ. Let R/I1 = {y1 + I1, y2 + I1, . . . , yµ + I1} and R/I2 = {z1 + I2, z2 +I2, . . . , zµ+I2}. Let v = {x1, x2, . . . , xµ} ∈ V (ITΓ(R)). By the previous case (i.e. t =1) counting degree with respect to I1, we have deg(v) ≥ ∑µ

i=1[λµ−i(λ− 1)i−1]− 1.

Without loss of generality assume that x1 ∈ y1 + I1 and x1 ∈ zj + I2. Construct aγ-set S = {a1, . . . , aµ} in TΓ(R) with a1 = x1, ai ∈ zi + I2 for i = 1, 2, . . . , µ − 1,i �= j and aµ ∈ (zµ+I2)∩I1. Since a1, aµ ∈ I1, S is not a γ-set in TΓ(R) with respectto I1. Note that each ai (i = 1, 2, . . . , µ − 1 and i �= j) has λ choices and aµ has λ

µ

choices. Thus the number of γ-sets with respect to I2 (but not with respect to I1)which contains x1 is (λ

µ)λµ−2 and so deg(v) ≥ ∑µi=1[λ

µ−i(λ− 1)i−1]− 1 + (λµ)λµ−2

for v ∈ V (ITΓ(R)).

If R is an integral domain, then by Lemma 2.3, 0 is an element in every γ-setof TΓ(R) and so from Lemma 3.2(i), we have the following corollary.

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Corollary 3.1. For any integral domain R, the graph ITΓ(R) is complete. Inparticular, if R is a finite integral domain, k = |{x ∈ R : 2x = 0}| and m = 2

|R|−k2 ,

then ITΓ(R) = Km.

Remark 3.1. Note that the Peterson graph cannot be realized as a Cayley graphof any group [18, p. 44]. Now we observe that Peterson graph cannot be realized asan intersection graph of gamma sets in TΓ(R) for any commutative Artin ring R.Note that the Peterson graph is a three regular graph with ten vertices. Supposethat the Peterson graph is a ITΓ(R) for some commutative ring R. Let I be anannihilator ideal of R with |R/I| minimum. Let |I| = λ and |R/I| = µ. Sincethe Peterson graph is not complete, R cannot be an integral domain and so byLemma 3.2(i), |V (ITΓ(R))| ≥ λµ. Therefore in our case λµ ≤ 10. This implies thateither λ = 2 with µ ≤ 3 or λ = 3 with µ = 2. Also by Lemma 3.2(ii) and (iii),deg(v) ≥ λµ−1 for all v ∈ V (ITΓ(R)) and so λµ−1 ≤ 3. If λ = 2 and µ = 3, thenλµ−1 ≥ 4, a contradiction. Thus either λ = 2 and µ = 2 or λ = 3 and µ = 2. Hencethere are four non-isomorphic rings with maximal ideals of order 2 or 3 with µ = 2,namely Z4,

Z2[x]〈x2〉 , Z2×Z2 and Z6. One can verify that none of ITΓ(Z4) or ITΓ(Z2[x]

〈x2〉 )

or ITΓ(Z2 × Z2) or ITΓ(Z6) is the Peterson graph.

Remark 3.2. Let R be a finite commutative ring which is not an integraldomain. Let I be an annihilator ideal in R with minimum |R/I| = µ and letv = {x1, . . . , xµ} ∈ V (ITΓ(R)). Since R is not an integral domain, |I| ≥ 2 andlet 0 �= z ∈ I. Thus v is adjacent to two distinct vertices {x1, . . . , xµ−1, xµ + z}and {x1 + z, x2, . . . , xµ}, and not adjacent to the vertex {x1 + z, . . . , xµ + z}. Thus2 ≤ deg(v) ≤ |V (ITΓ(R))| − 2 for v ∈ V (ITΓ(R)). Note that these bounds arealso sharp. For example if R = Z4, then deg(v) = 2 and if R = Z2 × Z2, thendeg(v) = |V (ITΓ(R))|− 2. Further, note that the lower and upper bounds are same

if R is either Z4 or Z2[x]〈x2〉 .

In the next proposition, we characterize all commutative rings R for whichdegree of every vertex in ITΓ(R) attains either the lower bound or upper boundgiven in Remark 3.2.

Proposition 3.1. Let R be a finite commutative ring which is not a field. Thenthe following holds:

(i) deg(v) = 2 for all v ∈ V (ITΓ(R)) if and only if R is either Z4 or Z2[x]〈x2〉 .

(ii) deg(v) = |V (ITΓ(R))|−2 �= 2 for all v ∈ V (ITΓ(R)) if and only if R = Z2×Z2.

Proof. If R is either Z4 or Z2[x]〈x2〉 , then as shown in Example 3.1(b), deg(v) = 2

for every v ∈ V (ITΓ(R)). If R = Z2 × Z2, then as shown in Example 3.1(b),deg(v) = |V (ITΓ(R))| − 2 for every v ∈ V (ITΓ(R)).

Conversely, assume that deg(v) = 2 or deg(v) = |V (ITΓ(R))| − 2 for everyv ∈ V (ITΓ(R)). Let I be an annihilator ideal in R with |R/I| minimum, |I| = λ

and |R/I| = µ.

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Case 1. Assume that deg(v) = 2. Suppose λ ≥ 3. By Lemma 3.2, deg(v) ≥λµ−1 ≥ 3, a contradiction. As R is not an integral domain, R contains at least twozero-divisors and so λ = 2. Note that there are only two non-isomorphic local ringswith unique maximal ideal of order 2, namely Z4 and Z2[x]

〈x2〉 . Suppose R is not local.Let Z(R) =

⋃mi=1 Ii for some m and prime ideals Ii, i = 1, . . . , m. Since Z(R) is not

an ideal of R, m ≥ 2. Since |Ii| = 2 for each i, the intersection of any two distinctIi’s is {0} and so m = 2. Due to this, Z(R) = I1 ∪ I2 for prime ideals I1 and I2

with I1 ∩ I2 = {0}. Since I1 and I2 are the only prime (maximal) ideals of R, bythe Chinese remainder theorem, we have R ∼= R/I1 × R/I2

∼= Z2 × Z2. Note thatwhen R = Z2 × Z2, by Example 3.1(b), deg(v) �= 2. Hence R ∼= Z4 or Z2[x]

〈x2〉 .

Case 2. Let deg(v) = |V (ITΓ(R))| − 2 �= 2. Suppose |I| ≥ 3 and take 0, z1, z2 ∈ I.Then v = {x1, . . . , xµ} is not adjacent to {x1 + z1, . . . , xµ + z1}, {x1 + z2, . . . , xµ +z2} ∈ V (ITΓ(R)) and so deg(v) ≤ |V (ITΓ(R))|−3, a contradiction. Hence λ = 2. ByCase 1 and Example 3.1(b), deg(v) = |V (ITΓ(R))|−2 �= 2 if and only if R = Z2×Z2.

In what follows, we give information about connectedness, diameter and girthof ITΓ(R).

Theorem 3.1. Let R be a commutative Artin ring with |R| ≥ 4 and an annihilatorideal I of R such that |R/I| is finite. Then

(i) diam(ITΓ(R)) ≤ 2. In particular, ITΓ(R) is connected.

(ii) ITΓ(R) is self-centered.

(iii) gr(ITΓ(R)) =

{4 if R ∼= Z4 or Z2[x]

〈x2〉 ,

3 otherwise.

Proof. Let Ii (1 ≤ i ≤ t) be an annihilator ideal of R with |R/Ii| finite andminimum. Let |Ii| = λ and |R/Ii| = µ.

(i) If R is an integral domain, then ITΓ(R) is a complete graph and sodiam(ITΓ(R)) = 1. On the other hand, let u = {x1, . . . , xµ} and v ={y1, . . . , yµ} be two vertices in ITΓ(R). Let u be a γ-set of TΓ(R) with respectto Ii for some 1 ≤ i ≤ t. If xj = yk for some 1 ≤ j, k ≤ µ, then u and v areadjacent. If xj �= yk for all 1 ≤ j, k ≤ µ, find a ∈ R such that y1 ∈ a+ Ii. Thenby Theorem 2.2, there exists xj ∈ u such that xj ∈ a+Ii for some j, 1 ≤ j ≤ µ.Now w = {x1, . . . , xj−1, y1, xj+1, . . . , yµ} ∈ V (ITΓ(R)) and so u − w − v is apath in ITΓ(R). Hence d(u, v) = 2.

(ii) From the proof of (i) and by Remark 3.2, we have for all v ∈ V (ITΓ(R)),

e(v) =

{1 if R is an integral domain,

2 otherwise.

Hence ITΓ(R) is self-centered.

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(iii) If R is an integral domain, then gr(ITΓ(R)) = 3. Assume that R is not anintegral domain. Let u = {x1, . . . , xµ} ∈ V (ITΓ(R)) and u is a γ-set of TΓ(R)with respect to Ii for some 1 ≤ i ≤ t. Let yj ∈ xj + Ii and yj �= xj forj = 1, . . . , µ. If µ ≥ 3, then v = {y1, x2, . . . , xµ}, w = {x1, . . . , xµ−1, yµ} aretwo distinct vertices in ITΓ(R) and u−v−w is a cycle in ITΓ(R). If λ ≥ 3, thenlet us take zj ∈ xj + Ii (different from xj and yj) for j = 1, . . . , µ. From thisv = {y1, x2, . . . , xµ}, w = {z1, x2, . . . , yµ} ∈ V (ITΓ(R)) and u−v−w is a cyclein ITΓ(R). Hence gr(ITΓ(R)) = 3 if µ ≥ 3 or λ ≥ 3. Now let us consider thecase λ = 2 and µ = 2. As mentioned in the proof of Proposition 3.1 (Case 1),R ∼= Z4 or Z2[x]

〈x2〉 or Z2 × Z2. Hence gr(ITΓ(R)) = 4 if and only if R ∼= Z4 orZ2[x]〈x2〉 .

In the next theorem, we characterize all commutative Artin rings whose ITΓ(R)belong to some known class of graphs.

Theorem 3.2. Let R be a commutative Artin ring and R has an annihilator idealI with |R/I| finite. Then

(i) ITΓ(R) is a regular graph.(ii) ITΓ(R) is a complete graph if and only if R is an integral domain.

(iii) ITΓ(R) is a bipartite graph if and only if R is either Z4 or Z2[x]〈x2〉 .

(iv) ITΓ(R) is a cycle if and only if R is either Z4 or Z2[x]〈x2〉 .

(v) ITΓ(R) is a chordal graph if and only if R is an integral domain.

Proof. Let I be an annihilator ideal of R with |R/I| finite and minimum.

(i) Follows from the fact ascertained in the proof of Lemma 3.2(ii).(ii) If R is an integral domain, then ITΓ(R) is complete. Conversely, assume that

ITΓ(R) is complete. Let v = {x1, . . . , xµ} ∈ V (ITΓ(R)). Suppose R is not anintegral domain, then |I| ≥ 2 and so let us take 0 �= y ∈ I. Clearly v is notadjacent to the vertex {x1 + y, . . . , xµ + y} ∈ V (ITΓ(R)), a contradiction.

(iii) If R ∼= Z4 or Z2[x]〈x2〉 , then ITΓ(R) = K2,2. Conversely, assume that ITΓ(R) is

bipartite. Suppose R �∼= Z4 and Z2[x]〈x2〉 , then by Theorem 3.1(ii), ITΓ(R) contains

an odd cycle, a contradiction to ITΓ(R) is bipartite.(iv) Assume that ITΓ(R) is a cycle. Then deg(v) = 2 for all v and so by Lemma 3.1

R is finite. If R has a unique annihilator ideal I of R with |R/I| finite and mini-mum, then, by Proposition 3.1, R ∼= Z4 or Z2[x]

〈x2〉 . Otherwise, by Lemma 3.2(iii),deg(v) ≥ 3 for all v, a contradiction. The other part is trivial.

(v) If R is an integral domain, then ITΓ(R) is complete and so ITΓ(R) is a chordalgraph. Conversely, ITΓ(R) is a chordal graph. Suppose R is not an integraldomain. Let 0 �= y ∈ I and {x1, . . . , xµ} be a γ-set of TΓ(R) with respect to I.Then the subgraph induced by the set {{x1, . . . , xµ}, {x1+y, x2, . . . , xµ}, {x1+y, . . . , xµ + y}, {x1, x2 + y, . . . , xµ + y}} is C4 which is a contradiction.

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Remark 3.3. From Theorem 3.2, for some prime p and m, n ∈ Z+, we have

Km (m �= pn−12 ), Cn (n �= 4), Km,n (m, n �= 2) and mK2 (m �= 2) cannot be

realized as the intersection graph of gamma sets in TΓ(R) for any R.

4. Vertex Transitivity of IT Γ(R)

In this section we are interested in vertex-transitive property of ITΓ(R). Recall that,an automorphism of a graph G is a map f of the vertex set of G with the propertythat, for any two vertices u and v, we have f(u) is adjacent to f(v) if and only if u

is adjacent to v. A graph G is called vertex-transitive if for every pair u, v ∈ V (G),there exists an automorphism f : G → G that maps u to v.

Theorem 4.1. Let R be a commutative Artin ring with the unique annihilator idealI such that |R/I| is finite and minimum. Then ITΓ(R) is vertex-transitive.

Proof. Let R/I = {z1+I = I, z2+I, . . . , zµ+I} and {x1, x2, . . . , xµ} ∈ V (ITΓ(R)).Since I is unique, all γ-sets in TΓ(R) are with respect to I only and hence one canrearrange all the vertices of ITΓ(R) such that xi ∈ zi + I for i = 1, . . . , µ. For ai ∈ I

1 ≤ i ≤ µ (not necessarily distinct), define f(a1,a2,...,aµ) : V (ITΓ(R)) → V (ITΓ(R))by f({x1, x2, . . . , xµ}) = {x1+a1, x2+a2, . . . , xµ+aµ}. Let u = {x1, x2, . . . , xµ}, v ={y1, y2, . . . , yµ} ∈ V (ITΓ(R)). If u∩v �= ∅, then clearly f(u)∩f(v) �= ∅. If u∩v = ∅,then xi + ai �= yi + ai for all i = 1, . . . , µ. Further, for all i �= j with 1 ≤ i, j ≤µ, xi /∈ yj + I and so xi + ai ∈ zi + I and yj + aj ∈ zj + I, i.e. xi + ai �=yj + aj and hence f(u) ∩ f(v) = ∅. Hence f(a1,a2,...,aµ) ∈ Aut(ITΓ(R)). Again, letu = {x1, . . . , xµ} and v = {y1, . . . , yµ} be two arbitrary vertices in ITΓ(R). Thenyi −xi ∈ I for all i = 1, . . . , µ. Therefore f(y1−x1,y2−x2,...,yµ−xµ) ∈ Aut(ITΓ(R)) andf(y1−x1,...,yµ−xµ)(u) = v. Hence ITΓ(R) is vertex-transitive.

Suppose R is a commutative ring with more than one annihilator ideal Ii

(1 ≤ i ≤ t) such that |R/Ii| is finite and minimum. Then one can consider thesubgraph of ITΓ(R) induced by the γ-sets with respect to a fixed Ii. Let us denotethe subgraph as ITΓ(Ii). By Theorem 4.1, ITΓ(Ii) is a vertex-transitive subgraphof ITΓ(R) for i = 1, . . . , t and ITΓ(R) =

⋃ti=1 ITΓ(Ii). In view of this, we suggest

the following open problem.

Open Problem 1. Let R be a commutative Artin ring with more than one anni-hilator ideal I with |R/I| finite and minimum. Then ITΓ(R) is vertex-transitive.

5. Genus of ITΓ(R)

Our next aim is to discuss about the genus of the intersection graph of gammasets in the total graph of a commutative ring and so let us recall the definitionand notation of genus. Let S� denote the sphere with � handles, where � is a non-negative integer, that is, S� is an oriented surface of genus k. The genus of a graphG, denoted g(G), is the minimal integer � such that the graph can be embedded

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in S�. A graph G with genus 0 is planar, whereas with genus 1 is toroidal. Formore details on embedding of a graph in a surface, see [27]. We list some results forfurther use.

Lemma 5.1 ([27]). Let m, n ≥ 3 be integers and, for real number x, �x� is theleast integer that is greater than or equal to x. Then

(i) g(Kn) =⌈

(n − 3)(n − 4)12

⌉.

(ii) g(Km,n) =⌈

(m − 2)(n − 2)4

⌉.

Lemma 5.2 ([28, Proposition 2.1]). If G is a graph with n vertices and genusg, then δ(G) ≤ 6 + 12g−12

n .

Maimani et al. [20] determined all finite commutative rings whose total graphis either planar or toroidal. Subsequently Tamizh Chelvam and Asir [26] character-ized all finite commutative rings whose total graph has genus two. For the sake ofreference, the relevant results in [20, 26] are given below.

Theorem 5.1 ([20, Theorem 1.5]). Let R be a commutative ring such that TΓ(R)is a planar graph. Then the following hold:

(a) If R is a local ring, then R is a field or R is isomorphic to the one of the ninefollowing rings:

Z4, Z2[x]/(x2), Z2[x]/(x3), Z2[x, y]/(x, y)2, Z4[x]/(2x, x2),

Z4[x]/(2x, x2 − 2), Z8, F4[x]/(x2), Z4[x]/(x2 + x + 1).

(b) If R is not a local ring, then R is an infinite integral domain or R is isomorphicto Z2 × Z2 or Z6.

Theorem 5.2 ([20, Theorem 1.6]). Let R be a finite ring such that TΓ(R) istoroidal. Then the following statements hold:

(a) If R is a local ring, then R is isomorphic to Z9 or Z3[X ]/(X2).(b) If R is not a local ring, then R is isomorphic to one of the following rings:

Z2 × F4, Z3 × Z3, Z2 × Z4, Z2 × Z2[x]/(x2), Z2 × Z2 × Z2.

Theorem 5.3 ([26, Theorem 4.3]). Let R be a finite commutative ring. Theng(TΓ(R)) = 2 if and only if R is isomorphic to either Z10 or Z3 × F4.

Motivated by these results, we now characterize all commutative Artin rings forwhich ITΓ(R) is either planar or toroidal or with genus 2. Let R be a commuta-tive Artin ring with an annihilator ideal I such that |R/I| is finite and minimum.Let |I| = λ and |R/I| = µ. If R is an infinite ring, then I is infinite and so byTheorem 2.2, there are infinite number of γ-sets in TΓ(R) which contains x1 andso ITΓ(R) has an infinite clique, i.e. when R is an infinite Artin ring, ITΓ(R) is

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neither planar nor toroidal nor with genus 2. Hence we take R as finite. For furtherreference, we list below all commutative rings R with |R| ≤ 7 and R is not anintegral domain.

Z4, Z2[x]/(x2), Z2 × Z2, Z6. (5.1)

Theorem 5.4. Let R be a finite commutative ring. Then

(i) ITΓ(R) is planar if and only if R is either Z3 or Z4 or Z5 or Z2[x]〈x2〉 or Z2 ×Z2

or F2n (a field with 2n elements) for some n ∈ Z+.

(ii) ITΓ(R) is toroidal if and only if R ∼= Z6.

(iii) g(ITΓ(R)) = 2 if and only if R ∼= Z7.

Proof. Let I be an annihilator ideal of R with |R/I|(= µ) minimum and let |I| = λ.

(i) We discuss the proof in two cases:Case 1. Suppose R is an integral domain.

If |R| = 2n for some n ∈ Z+, then R ∼= F2n for some n ∈ Z

+ and sok = |{x ∈ R : 2x = 0}| = |R|. By Lemma 3.2(i), ITΓ(R) = K1 and so ITΓ(R)is planar.

If |R| = pn for some prime p > 3 and n ∈ Z+, then k = |{x ∈ R : 2x =

0}| = 1. By Lemma 3.2(i), ITΓ(R) = K2

|R|−12

. By Kuratowskis’s theorem,ITΓ(R) is planar if and only if |R| ≤ 5, i.e. ITΓ(R) is planar if and only if R iseither Z3 or Z5.

Hence in this case, ITΓ(R) is planar if and only if R is either Z3 or Z5 orF2n for some n ∈ Z

+.Case 2. Suppose R is not an integral domain.

Assume that ITΓ(R) is planar. Note that by Lemma 5.2, δ(ITΓ(R)) ≤ 5.If |R| ≥ 6, then λµ ≥ 6 and so λµ−1 ≥ λ(µ − 1) ≥ 6 − λ. By Lemma 3.2(i),deg(v) ≥ λµ−1 +(λ−1)λµ−2 +(λ−1)µ−1−1 ≥ 6−λ+λ−1+λ−1−1 = λ+3.If λ ≥ 3, then deg(v) ≥ 6 for all v ∈ V (ITΓ(R)), which is contradiction toδ(ITΓ(R)) ≤ 5. Therefore λ = 2. As argued in the proof of Proposition 3.1, ifλ = 2, then |R| = 4, a contradiction to the assumption that |R| ≥ 6. Hence|R| ≤ 5. Thus by list (5.1), R ∼= Z4 or Z2[x]/(x2) or Z2 × Z2.

Conversely, by Example 3.1(b), ITΓ(Z4) and ITΓ(Z2[x]/(x2)) are planar.Also ITΓ(Z2×Z2) is planar as shown in the planar embedding of ITΓ(Z2×Z2)(Fig. 1). Note that in Fig. 1, an element (x, y) of Z2 × Z2 is denoted as xy.Hence in this case, ITΓ(R) is planar if and only if R is either Z4 or Z2[x]

〈x2〉 orZ2 × Z2.

(ii) Assume that ITΓ(R) is toroidal. If R is an integral domain, then by (i), |R| ≥ 7and |R| �= 2n for n ∈ Z

+. By Corollary 3.1, K23 ⊆ ITΓ(R) and by Lemma 5.1,g(K8) = 2. Therefore g(ITΓ(R)) ≥ 2, a contradiction. Hence R is not anintegral domain. Since genus of ITΓ(R) is one, by Lemma 5.2, δ(ITΓ(R)) ≤ 6.

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Fig. 1. Planar embedding of ITΓ(Z2 × Z2).

Fig. 2. Embedding of ITΓ(Z6) in Torus.

If |R| ≥ 7, then deg(v) ≥ λ+4. If λ ≥ 3, then deg(v) ≥ 7 for all v ∈ V (ITΓ(R)),which is contradiction to δ(ITΓ(R)) ≤ 6. Therefore λ = 2 and so |R| = 4, acontradiction. Hence |R| ≤ 6 and by (i), |R| = 6. There is only one ring oforder 6, which is not an integral domain, namely Z6. The embedding of ITΓ(Z6)in torus is given in Fig. 2.

(iii) Assume that g(ITΓ(R)) = 2. If R is an integral domain, then by (i), |R| ≥ 7 and|R| �= 2n for all n ∈ Z

+. If |R| = 7, then ITΓ(R) = K23 and so g(ITΓ(R)) = 2.If |R| ≥ 9, then K24 ⊆ ITΓ(R) and so by Lemma 5.1, g(ITΓ(R)) ≥ 13, acontradiction. Suppose R is not an integral domain. Note that by Lemma 5.2,δ(ITΓ(R)) ≤ 6+ 12

|V (ITΓ(R))| . By (i) and (ii), |R| ≥ 7. Then |V (ITΓ(R))| ≥ 7 andso δ(ITΓ(R)) ≤ 7. Since |R| ≥ 7, deg(v) ≥ λ + 5. If λ ≥ 3, then deg(v) ≥ 8 forall v ∈ V (ITΓ(R)), which is contradiction to δ(ITΓ(R)) ≤ 7. Therefore λ = 2and so |R| = 4, a contradiction. Hence g(ITΓ(R)) = 2 if and only if R ∼= Z7.

Lemma 5.3. Let R be a finite commutative ring which is not an integral domainand I be an annihilator ideal of R with |R/I| minimum. Assume that |I| = λ and|R/I| = µ. If ITΓ(R) is not planar, then g(ITΓ(R)) ≥ λ� (λµ−1−3)(λµ−1−4)

12 �.

Proof. Let I = {a1, . . . , aλ}, A1 = {v ∈ V (ITΓ(R)) : a1 ∈ v} and Ai = {v ∈V (ITΓ(R)) : ai ∈ v and v /∈ ⋃i−1

j=1 Aj} for 2 ≤ i ≤ λ. The subgraphs induced by

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each Ai (for i = 1, . . . , λ) in ITΓ(R) are vertex disjoint and each one is isomorphicto Ks. Note that by Theorem 2.2, there are at least λµ−1 number of vertices inITΓ(R) which contains x1 and so s ≥ λµ−1. Thus λKλµ−1 is a subgraph of ITΓ(R).Note that if B�’s are block of a graph G, then g(

⋃m�=1 B�) =

∑m�=1 g(B�) and

g(Kλµ−1) = � (λµ−1−3)(λµ−1−4)12 �. From this g(λKλµ−1) = λ� (λµ−1−3)(λµ−1−4)

12 �. SinceλKλµ−1 is a subgraph of ITΓ(R), we get the desired result.

Now we exhibit a relation between the total graph TΓ(R) and the correspondingintersection graph ITΓ(R).

Theorem 5.5. If R is a finite commutative ring, then g(TΓ(R)) ≤ g(ITΓ(R)).

Proof. If R is a field, then by Theorem 5.1, TΓ(R) is planar. Further, ITΓ(R) =K� for some � ∈ Z

+. This gives that g(TΓ(R)) ≤ g(ITΓ(R)). Assume that R isnot a field. Let I be an annihilator ideal of R with |R/I| = µ minimum and|I| = λ. Then by Lemma 5.1, g(TΓ(R)) ≤ � (|R|−3)(|R|−4)

12 � and by Lemma 5.3,

g(ITΓ(R)) ≥ λ� (λµ−1−3)(λµ−1−4)12 �. To conclude the proof, it is enough to prove that

λµ −3λ ≥ |R|−3 and λµ −4λ ≥ |R|−4, i.e. λµ−1−3 ≥ µ− 3λ and λµ−1−4 ≥ µ− 4

λ .If λ = 2, then R ∼= Z4 or Z2[x]/(x2) and so by Theorems 5.1 and 5.4, we getg(TΓ(R)) = g(ITΓ(R)) = 0. Hence λ ≥ 3. Now it remains to prove that λµ−1−2 ≥ µ

and λµ−1 − 3 ≥ µ. If λ ≥ 5 or µ ≥ 3, then λµ−1 − 3 ≥ µ and so consider the caseλ = 3, 4 and µ = 2. Thus |R| = 6 or 8.

If R is local, then |R| = 8. According to Corbas and Williams [15], there are5 local rings of order 8 (except F8). They are

Z2[x]/(x3), Z2[x, y]/(x, y)2, Z4[x]/(2x, x2), Z4[x]/(2x, x2 − 2), Z8

If R is not local, then R is one of the following rings:

Z6, Z2 × F4, Z2 × Z4, Z2 × Z2[x]/(x2).

Out of all these nine rings, Z6 is the only ring with λ = 3 and others are withλ = 4. Note that 0 = g(TΓ(Z6)) ≤ g(ITΓ(Z6)) = 1. In the remaining cases, byTheorems 5.1 and 5.2, TΓ(R) is either planar or toroidal, whereas by Theorem 5.4,g(ITΓ(R)) ≥ 3.

Note

We continue the study on ITΓ(R) in another paper entitled “The intersection graphof gamma sets in the total graph of a commutative ring II” in which we concentrateon Hamiltonian, coloring, perfectness and connectivity of ITΓ(R).

Acknowledgments

The work reported here is supported by the Special Assistance Programme (F510-DRS-I/2007) of University Grants Commission (UGC), India awarded to the

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Department of Mathematics, Manonmaniam Sundaranar University to the firstauthor. Also the work was supported by the INSPIRE programme (IF 110072) ofDepartment of Science and Technology, Government of India for the second author.

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