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ISRN Algebra
VolumeΒ 2011Β (2011), Article IDΒ 591041, 24 pages
http://dx.doi.org/10.5402/2011/591041
Research Article

Some Properties of the Complement of the Zero-Divisor Graph of a Commutative Ring

Department of Mathematics, Saurashtra University, Rajkot 360 005, India

Received 19 April 2011; Accepted 17 May 2011

Academic Editors: D.Β Anderson, V.Β Drensky, and D.Β Herbera

Copyright Β© 2011 S. Visweswaran. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let R be a commutative ring with identity admitting at least two nonzero zero-divisors. Let (Ξ“(𝑅))𝑐 denote the complement of the zero-divisor graph Ξ“(𝑅) of 𝑅. It is shown that if (Ξ“(𝑅))𝑐 is connected, then its radius is equal to 2 and we also determine the center of (Ξ“(𝑅))𝑐. It is proved that if (Ξ“(𝑅))𝑐 is connected, then its girth is equal to 3, and we also discuss about its girth in the case when (Ξ“(𝑅))𝑐 is not connected. We discuss about the cliques in (Ξ“(𝑅))𝑐.

1. Introduction

All rings considered in this note are nonzero commutative rings with identity. Unless otherwise specified, we consider rings 𝑅 such that 𝑅 admits at least two nonzero zero-divisors.

Let 𝑅 be a commutative ring with identity which is not an integral domain. Recall from [1] that the zero-divisor graph of 𝑅, denoted by Ξ“(𝑅), is the graph whose vertex set is the set of all nonzero zero-divisors of 𝑅 and distinct vertices π‘₯, 𝑦 are joined by an edge in this graph if and only if π‘₯𝑦=0. Several researchers studied the zero-divisor graphs of commutative rings and proved several interesting and inspiring theorems in this area [1–14]. The research paper of Beck [9], the research paper of Anderson and Naseer [2], and the research paper of Anderson and Livingston [1] are first among several research papers that inspired a lot of work in the area of zero-divisor graphs. We denote by 𝑍(𝑅) the set of all zero-divisors of 𝑅, and by 𝑍(𝑅)βˆ— the set of all nonzero zero-divisors of 𝑅.

Before we describe the results that are proved in this note, it is useful to recall the following definitions from [15]. Let 𝐺=(𝑉,𝐸) be a connected graph. For any vertices π‘₯, 𝑦 of 𝐺 with π‘₯≠𝑦, 𝑑(π‘₯,𝑦) is the length of a shortest path in 𝐺 from π‘₯ to 𝑦 and 𝑑(π‘₯,π‘₯)=0 and the diameter of 𝐺 is defined as sup{𝑑(π‘₯,𝑦)∣π‘₯and𝑦areverticesof𝐺} and it is denoted by diam(𝐺).

For any π‘£βˆˆπ‘‰, the eccentricity of 𝑣 denoted by 𝑒(𝑣) is defined as𝑒(𝑣)=sup{𝑑(𝑣,𝑒)βˆ£π‘’βˆˆπ‘‰}.(1.1)

The set of vertices of 𝐺 with minimal eccentricity is called the center of the graph, and the minimum eccentric value is called the radius of 𝐺 and is denoted by π‘Ÿ(𝐺).

It is known that for any commutative ring 𝑅 with identity which is not an integral domain, Ξ“(𝑅) is connected and diam(Ξ“(𝑅))≀3 [1, Theorem 2.3]. In [13, Theorem 2.3], Redmond proved that for any Noetherian ring 𝑅 with identity which is not an integral domain, π‘Ÿ(Ξ“(𝑅))≀2. Moreover, in Section 3 of [13] Redmond determined the center of Ξ“(𝑅) for any Artinian ring 𝑅. It is known that there are rings 𝑅 for which π‘Ÿ(Ξ“(𝑅))=3 [8, Corollary 1.6]. In [14, Theorem 2.4], Karim Samei characterized vertices π‘₯ of Ξ“(𝑅) such that 𝑒(π‘₯)=1 where 𝑅 is a reduced ring. Furthermore, in the same theorem under some additional hypotheses on 𝑅, he described vertices π‘₯ of Ξ“(𝑅) such that 𝑒(π‘₯)=2or3.

Let 𝐺=(𝑉,𝐸) be a simple graph. Recall from [15, Definition, 1.1.13] that the complement of 𝐺 denoted by 𝐺𝑐 is defined by setting 𝑉(𝐺𝑐)=𝑉 and two distinct 𝑒,π‘£βˆˆπ‘‰ are joined by an edge in 𝐺𝑐 if and only if there exists no edge in 𝐺 joining 𝑒,𝑣.

It is useful to recall the following definitions from commutative ring theory before we proceed further. Let 𝐼 be an ideal of a ring 𝑅,𝐼≠𝑅. A prime ideal 𝑃 of 𝑅 is said to be a maximal 𝑁-prime of 𝐼 in 𝑅 if 𝑃 is maximal with respect to the property of being contained in 𝑍𝑅(𝑅/𝐼) where 𝑍𝑅(𝑅/𝐼)={π‘₯βˆˆπ‘…βˆ£π‘₯π‘¦βˆˆπΌforsomeπ‘₯βˆˆπ‘…β§΅πΌ} [16]. It is well known that if {𝑃𝛼}π›ΌβˆˆΞ› is the set of all maximal 𝑁-primes of (0) in 𝑅, thenξšπ‘(𝑅)=π›ΌβˆˆΞ›π‘ƒπ›Ό.(1.2)

Let 𝐼 be an ideal of a ring 𝑅. A prime ideal 𝑃 of 𝑅 is said to be an associated prime of 𝐼 in the sense of Bourbaki if 𝑃=(πΌβˆΆπ‘…π‘₯) for some π‘₯βˆˆπ‘… [17]. In this case, we say that 𝑃 is a 𝐡-prime of 𝐼.

Let 𝑅 be a commutative ring with identity admitting at least two nonzero zero-divisors. In [18, Theorem 1.1], it was shown that (Ξ“(𝑅))𝑐 is connected if and only if one of the following conditions holds.(a)𝑅 has exactly one maximal 𝑁-prime 𝑃 of (0) such that 𝑃 is not a 𝐡-prime of (0) in 𝑅.(b)𝑅 has exactly two maximal 𝑁-primes 𝑃1, 𝑃2 of (0) with 𝑃1βˆ©π‘ƒ2β‰ (0).(c)𝑅 has more than two maximal 𝑁-primes of (0).

Moreover, it was shown in [18] that if (Ξ“(𝑅))𝑐 is connected, then diam((Ξ“(𝑅))𝑐)≀3. In fact, it was shown in [18] that if either the condition (a) or the condition (c) of [18, Theorem 1.1] holds, then diam((Ξ“(𝑅))𝑐)=2. When the condition (b) of [18, Theorem 1.1] holds, then it was shown in [18, Proposition 1.7] that diam((Ξ“(𝑅))𝑐)=2 if either 𝑃1 is not a 𝐡-prime of (0) in 𝑅 or 𝑃2 is not a 𝐡-prime of (0) in 𝑅 and diam((Ξ“(𝑅))𝑐)=3 if both 𝑃1 and 𝑃2 are 𝐡-primes of (0) in 𝑅.

For any set 𝐴, we denote by |𝐴|, the cardinality of 𝐴. Whenever a set 𝐴 is a subset of a set 𝐡 and 𝐴≠𝐡, we denote it symbolically by π΄βŠ‚π΅. If 𝑋,π‘Œ are sets and if 𝑋 is not a subset of π‘Œ, we denote it symbolically π‘‹βŠ„π‘Œ.

Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)βˆ—|β‰₯2. If (Ξ“(𝑅))𝑐 is connected, then we prove in Section 2 of this note that the radius of (Ξ“(𝑅))𝑐 is equal to 2. Moreover, we observe that except in the case when condition (b) of [18, Theorem 1.1] holds, diam((Ξ“(𝑅))𝑐)=π‘Ÿ((Ξ“(𝑅))𝑐)=2 and so every vertex of (Ξ“(𝑅))𝑐 is in the center of (Ξ“(𝑅))𝑐. Furthermore, if condition (b) of [18,Theorem 1.1] holds and if (i) 𝑅 satisfies the further condition that either 𝑃1 is not a 𝐡-prime of (0) in 𝑅 or 𝑃2 is not a 𝐡-prime of (0) in 𝑅, then it is noted that each vertex of (Ξ“(𝑅))𝑐 is in the center of (Ξ“(𝑅))𝑐 and if (ii) both 𝑃1 and 𝑃2 are 𝐡-primes of (0) in 𝑅, then it is verified that the center of (Ξ“(𝑅))𝑐={π‘₯βˆˆπ‘(𝑅)βˆ—βˆ£π‘ƒ1β‰ ((0)βˆΆπ‘…π‘₯)and𝑃2β‰ ((0)βˆΆπ‘…π‘₯)}.

Let 𝐺=(𝑉,𝐸) be a graph. Recall from [15, Page 159] that the girth of 𝐺 denoted by gr(𝐺) is defined as the length of a shortest cycle in 𝐺. If 𝐺 does not contain any cycle, then we set gr(𝐺)=∞ [5].

Let 𝑅 be a commutative ring with identity which is not an integral domain. Several results are known about the girth of Ξ“(𝑅) [5, 7]. Indeed, it is known that for any commutative ring with identity which is not an integral domain, gr(Ξ“(𝑅))≀4 if Ξ“(𝑅) contains a cycle [7, Proposition 2.2] and [12, 1.4]. In [5], Anderson and Mulay characterized commutative rings 𝑅 such that gr(Ξ“(𝑅))=4 [5, Theorem 2.2 and Theorem 2.3], and moreover, they characterized commutative rings 𝑅 such that gr(Ξ“(𝑅))=∞ [5, Theorem 2.4 and Theorem 2.5].

In Section 3 of this paper we study about the girth of (Ξ“(𝑅))𝑐where 𝑅 is a commutative ring with identity satisfying the further condition that |𝑍(𝑅)βˆ—|β‰₯2. If (Ξ“(𝑅))𝑐 is connected, then it is shown that gr((Ξ“(𝑅))𝑐)=3.

Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝑃2β‰ (0) and if (Ξ“(𝑅))𝑐 is not connected, then it is proved that (Ξ“(𝑅))𝑐 contains a cycle if and only if |𝑃|β‰₯5 if and only if gr((Ξ“(𝑅))𝑐)=3.

Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and let them be 𝑃1 and 𝑃2. If (Ξ“(𝑅))𝑐 is not connected, then it is shown that (Ξ“(𝑅))𝑐 contains a cycle if and only if {𝑃1⧡𝑃2∣β‰₯3or|𝑃2⧡𝑃1|β‰₯3ifandonlyifgr((Ξ“(𝑅))𝑐)=3.

Let 𝐺=(𝑉,𝐸) be a graph. Recall from [15, Definition 1.2.2] that a clique of 𝐺 is a complete subgraph of 𝐺. Moreover, it is useful to recall the definition of the clique number of 𝐺. Let 𝐺=(𝑉,𝐸) be a simple graph. The clique number of 𝐺 denoted by πœ”(𝐺) is defined as the largest integer 𝑛β‰₯1 such that 𝐺 contains a clique on 𝑛 vertices [15, Definition, Page 185]. We set πœ”(𝐺)=∞ if 𝐺 contains a clique on 𝑛 vertices for all 𝑛β‰₯1.

Let 𝑅 be a commutative ring with identity which is not an integral domain. It is known that (i) πœ”(Ξ“(𝑅))=∞ if and only if Ξ“(𝑅) contains an infinite clique, (ii) πœ”(Ξ“(𝑅))<∞ if and only if |nil(𝑅)|<∞ and nil(𝑅) is a finite intersection of prime ideals of 𝑅 (i.e., the set of all minimal prime ideals of 𝑅 is finite) [9, Theorem 3.9] where nil(𝑅) is the nilradical of 𝑅. More interesting theorems were proved on πœ”(Ξ“(𝑅)) in [3, Section 3].

Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)βˆ—|β‰₯2. In Section 4 of this paper we observe that if 𝑅 has at least two maximal 𝑁-primes of (0), then (Ξ“(𝑅))𝑐 does not contain any infinite clique if and only if 𝑅 is finite if and only if πœ”((Ξ“(𝑅))𝑐) is finite. Let 𝑛β‰₯2 and let 𝐾1,…,𝐾𝑛 be finite fields. Let 𝑅=𝐾1×⋯×𝐾𝑛. We describe a method of determining πœ”((Ξ“(𝑅))𝑐).

Let 𝑅 be a ring admitting exactly one maximal 𝑁-prime of (0). Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. Suppose that (Ξ“(𝑅))𝑐 does not contain any infinite clique. Then it is verified in Section 4 that 𝑃=nil(𝑅). Moreover, if 𝑃2β‰ (0), then it is shown in Section 4 that 𝑃 is a 𝐡-prime of (0) in 𝑅 and furthermore, 𝑅/𝑃 is a finite field and 𝑅 satisfies d. c. c. on principal ideals. As a corollary, we deduce that if 𝑅 is either a Noetherian ring or a chained ring, then (Ξ“(𝑅))𝑐 does not contain any infinite clique if and only if 𝑅 is finite. Let 𝑅 be a finite chained ring with 𝑃 as its unique maximal 𝑁-prime of (0). If 𝑃2β‰ (0), then we provide a formula for computing πœ”((Ξ“(𝑅))𝑐).

Let 𝑅 be a ring with |𝑍(𝑅)βˆ—|β‰₯2. Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If πœ”((Ξ“(𝑅))𝑐) is finite, then it is shown in Section 4 that 𝑃 is nilpotent.

We end this note with an example of an infinite ring 𝑅 such that 𝑅 has exactly one maximal 𝑁-prime of (0) with the property that πœ”((Ξ“(𝑅))𝑐)=3.

2. The Radius of (Ξ“(𝑅))𝑐

Let 𝑅 be a ring with |𝑍(𝑅)βˆ—|β‰₯2. We assume that (Ξ“(𝑅))𝑐 is connected. The aim of this section is to show that the radius of (Ξ“(𝑅))𝑐 is equal to 2. We make use of the following lemmas for proving the result that π‘Ÿ((Ξ“(𝑅))𝑐)=2.

Lemma 2.1. Let 𝐺=(𝑉,𝐸) be a simple and connected graph with |𝑉|β‰₯2 satisfying the further condition that 𝐺𝑐 is also connected. If π‘₯ is any element of 𝑉,then 𝑒(π‘₯)β‰₯2in𝐺𝑐.

Proof. Let π‘₯βˆˆπ‘‰. Since |𝑉|β‰₯2 and 𝐺 is connected, it follows that there exists π‘¦βˆˆπ‘‰ such that π‘₯ is adjacent to 𝑦 in 𝐺. Thus π‘₯ is not adjacent to 𝑦 in 𝐺𝑐. Hence 𝑑(π‘₯,𝑦)β‰₯2in𝐺𝑐. This proves that 𝑒(π‘₯)β‰₯2in𝐺𝑐.

The following lemma establishes some necessary conditions on 𝑅 in order that there exist π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ— such that 𝑑(π‘₯,𝑦)=3in(Ξ“(𝑅))𝑐.

Lemma 2.2. Let 𝑅 be a ring with |𝑍(𝑅)βˆ—|β‰₯2. Suppose that (Ξ“(𝑅))𝑐 is connected. If there exist π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ— with 𝑑(π‘₯,𝑦)=3in(Ξ“(𝑅))𝑐, then 𝑅 has exactly two maximal 𝑁-primes of (0) and moreover, both of them are 𝐡-primes of (0) in 𝑅. Indeed, if 𝑃 and 𝑄 are the maximal 𝑁-primes of (0) in 𝑅, then {𝑃,𝑄}={(0)βˆΆπ‘…π‘₯),((0)βˆΆπ‘…π‘¦)}.

Proof. Since 𝑑(π‘₯,𝑦)=3in(Ξ“(𝑅))𝑐, it follows that π‘₯𝑦=0 and for any π‘§βˆˆπ‘(𝑅)βˆ— with π‘§βˆ‰{π‘₯,𝑦}, either 𝑧π‘₯=0 or 𝑧𝑦=0. Hence we obtain that π‘§βˆˆ((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦). Since π‘₯𝑦=0, it is clear that {0,π‘₯,𝑦}βŠ†((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦). Thus we have 𝑍(𝑅)=((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦). Let {𝑃𝛼}π›ΌβˆˆΞ› be the set of all maximal 𝑁-primes of (0) in 𝑅. It is well known that ⋃𝑍(𝑅)=π›ΌβˆˆΞ›π‘ƒπ›Ό, and hence we obtain that β‹ƒπ›ΌβˆˆΞ›π‘ƒπ›Ό=((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦). Now ((0)βˆΆπ‘…π‘₯)∩(𝑅⧡𝑍(𝑅))=βˆ…. Hence by [19, Theorem 2.2, Page 378], we obtain that there exists a maximal 𝑁-prime 𝑃 of (0) in 𝑅 such that ((0)βˆΆπ‘…π‘₯)βŠ†π‘ƒ. Since ((0)βˆΆπ‘…π‘¦)∩(𝑅⧡𝑍(𝑅))=βˆ…, it follows that there exists a maximal 𝑁-prime 𝑄 of (0) in 𝑅 such that ((0)βˆΆπ‘…π‘¦)βŠ†π‘„. Now we obtain that 𝑍(𝑅)=((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦)=𝑃βˆͺ𝑄. If 𝑃=𝑄, then it follows that ((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦)=𝑃 is an ideal of 𝑅. Hence it follows that either ((0)βˆΆπ‘…π‘₯)βŠ†((0)βˆΆπ‘…π‘¦)or((0)βˆΆπ‘…π‘¦)βŠ†((0)βˆΆπ‘…π‘₯), and so we obtain that 𝑃 is the only maximal 𝑁-prime of (0) in 𝑅 and either 𝑃=((0)βˆΆπ‘…π‘₯)or𝑃=((0)βˆΆπ‘…π‘¦). Now by hypothesis (Ξ“(𝑅))𝑐 is connected. Hence it follows from [18, Theorem 1.1 (a)] that 𝑃 is not a 𝐡-prime of (0) in 𝑅 and so 𝑃≠𝑄. Now we obtain from 𝑍(𝑅)=((0)βˆΆπ‘…π‘₯)βˆͺ((0)βˆΆπ‘…π‘¦)=𝑃βˆͺ𝑄 that 𝑃 and 𝑄 are the only maximal 𝑁-primes of (0) in 𝑅, and moreover, 𝑃=((0)βˆΆπ‘…π‘₯)and𝑄=((0)βˆΆπ‘…π‘¦).

The next lemma is [9, Lemma 3.6]. We make use of it in the proof of Lemma 2.4.

Lemma 2.3. Let 𝑅 be a ring. Let 𝑃,𝑄 be distinct 𝐡-prime ideals of (0) in 𝑅 with 𝑃=((0)βˆΆπ‘…π‘₯) and 𝑄=((0)βˆΆπ‘…π‘¦) for some π‘₯,π‘¦βˆˆπ‘…. Then π‘₯𝑦=0.

Proof. For the sake of completeness, we give below an argument for the fact that xy = 0. Since P β‰ Q. either π‘ƒβŠ„π‘„ or π‘„βŠ„π‘ƒ. Without loss of generality, we may assume that π‘ƒβŠ„π‘„. Let w βˆˆπ‘ƒβ§΅π‘„. Now 𝑀π‘₯=0βˆˆπ‘„ and as π‘€βˆ‰π‘„, it follows that π‘₯βˆˆπ‘„=((0)βˆΆπ‘…π‘¦). Hence π‘₯𝑦=0.

We provide in the next lemma some sufficient conditions on 𝑅 in order that (Ξ“(𝑅))𝑐 admits vertices x such that 𝑒(π‘₯)=3 in (Ξ“(𝑅))𝑐.

Lemma 2.4. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and let them be 𝑃1 and 𝑃2. If (Ξ“(𝑅))𝑐 is connected and if 𝑃1=((0)βˆΆπ‘…π‘’) and 𝑃2=((0)βˆΆπ‘…π‘£) for some 𝑒,π‘£βˆˆπ‘…, then 𝑑(𝑒,𝑣)=3 in (Ξ“(𝑅))𝑐 and so 𝑒(𝑒)=𝑒(𝑣)=3 in (Ξ“(𝑅))𝑐.

Proof. The proof of this lemma is contained in the proof of [18, Proposition 1.7], though it was not stated there in the above form. For the sake of completeness, we include a proof of it here.
We know from Lemma 2.3 that 𝑒𝑣=0. Thus 𝑒,π‘£βˆˆπ‘(𝑅)βˆ— and 𝑒 and 𝑣 are not adjacent in (Ξ“(𝑅))𝑐. Since 𝑃1 and 𝑃2 are the only maximal 𝑁-primes of (0) in 𝑅, it follows that 𝑍(𝑅)=𝑃1βˆͺ𝑃2=((0)βˆΆπ‘…π‘’)βˆͺ((0)βˆΆπ‘…π‘£). Thus for any π‘€βˆˆπ‘(𝑅)βˆ—, either 𝑀𝑒=0or𝑀𝑣=0. This shows that 𝑑(𝑒,𝑣)β‰₯3in(Ξ“(𝑅))𝑐. It is shown in the proof of [18,Proposition 1.7] that 𝑑(π‘₯,𝑦)≀3in(Ξ“(𝑅))𝑐 for any π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ—. Hence we obtain that𝑑(𝑒,𝑣)=3in(Ξ“(𝑅))𝑐 and so it follows that 𝑒(𝑒)=𝑒(𝑣)=3in(Ξ“(𝑅))𝑐.

We next state and prove the main theorem of this section.

Theorem 2.5. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. If (Ξ“(𝑅))𝑐 is connected, then the radius of (Ξ“(𝑅))𝑐 is equal to 2.

Proof. It is well known that Ξ“(𝑅) is connected [1, Theorem 2.3]. Hence it follows from Lemma 2.1 that for any π‘₯βˆˆπ‘(𝑅)βˆ—,𝑒(π‘₯)β‰₯2in(Ξ“(𝑅))𝑐.
If 𝑅 has either exactly one maximal 𝑁-prime of (0) or more than two maximal 𝑁-primes of (0), then it follows from Lemma 2.2. that for any π‘₯βˆˆπ‘(𝑅)βˆ—,𝑒(π‘₯)≀2in(Ξ“(𝑅))𝑐. Thus we obtain that if 𝑅 has either exactly one maximal 𝑁-prime of (0) or more than two maximal 𝑁-primes of (0), then for anyπ‘₯βˆˆπ‘(𝑅)βˆ—,𝑒(π‘₯)=2in(Ξ“(𝑅))𝑐. Hence we obtain in the cases mentioned above that diam((Ξ“(𝑅))𝑐)=π‘Ÿ((Ξ“(𝑅))𝑐)=2.
Assume that 𝑅 has exactly two maximal 𝑁-primes of (0) and let them be 𝑃1and𝑃2. In such a case, it was shown in the proof of [18, Proposition 1.7(i)] that there existπ‘Žβˆˆπ‘ƒ1⧡𝑃2 and π‘βˆˆπ‘ƒ2⧡𝑃1 such that π‘Žπ‘β‰ 0. It follows that 𝑃1β‰ ((0)βˆΆπ‘…π‘Ž),𝑃1β‰ ((0)βˆΆπ‘…π‘),and𝑃2β‰ ((0)βˆΆπ‘…π‘Ž),𝑃2β‰ ((0)βˆΆπ‘…π‘). Now it follows from Lemma 2.2 that 𝑑(π‘Ž,𝑦)≀2 and 𝑑(𝑏,𝑦)≀2 for any π‘¦βˆˆπ‘(𝑅)βˆ— in (Ξ“(𝑅))𝑐. Hence we obtain that 𝑒(π‘Ž)≀2 and 𝑒(𝑏)≀2 in (Ξ“(𝑅))𝑐. Since 𝑒(𝑐)β‰₯2in(Ξ“(𝑅))𝑐 for any π‘βˆˆπ‘(𝑅)βˆ—, it follows that 𝑒(π‘Ž)=𝑒(𝑏)=2in(Ξ“(𝑅))𝑐. Thus in this case also, we arrive at the conclusion that the radius of (Ξ“(𝑅))𝑐 is equal to 2.

The following remark determines the center of (Ξ“(𝑅))𝑐.

Remark 2.6. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. Assume that (Ξ“(𝑅))𝑐 is connected. In this remark, we determine the center of (Ξ“(𝑅))𝑐.
If 𝑅 has either exactly one maximal 𝑁-prime of (0) or more than two maximal 𝑁-primes of (0), then it was shown in the proof of Theorem 2.5 that diam((Ξ“(𝑅))𝑐)=π‘Ÿ((Ξ“(𝑅))𝑐)=2. Hence the center of (Ξ“(𝑅))𝑐 is the set of all vertices of (Ξ“(𝑅))𝑐. Moreover, if 𝑅 has exactly two maximal 𝑁-primes of (0) and if at least one of them is not a 𝐡-prime of (0) in 𝑅, then it follows from Lemma 2.2 that for any π‘₯βˆˆπ‘(𝑅)βˆ—,𝑒(π‘₯)≀2in(Ξ“(𝑅))𝑐. Thus we obtain, in view of Lemma 2.1 that 𝑒(π‘₯)=2for anyπ‘₯βˆˆπ‘(𝑅)βˆ—. Hence in this case also we obtain that diam((Ξ“(𝑅))𝑐)=π‘Ÿ((Ξ“(𝑅))𝑐)=2 and so each vertex of (Ξ“(𝑅))𝑐 is in the center of (Ξ“(𝑅))𝑐.
Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and both are 𝐡-primes of (0) in 𝑅. Let {𝑃1,𝑃2} be the set of all maximal 𝑁-primes of (0) in 𝑅. Then it follows from Lemmas 2.1, 2.2, and 2.4 that the center of (Ξ“(𝑅))𝑐={π‘₯βˆˆπ‘(𝑅)βˆ—βˆ£π‘ƒ1β‰ ((0)βˆΆπ‘…π‘₯)and𝑃2β‰ ((0)βˆΆπ‘…π‘₯)}.

We next present some examples to illustrate the results proved in this section.

Example 2.7. Let 𝑉 be a rank 1 valuation domain which is not discrete. Let 𝑀 denote the unique maximal ideal of 𝑉. Let π‘₯βˆˆπ‘€,π‘₯β‰ 0. Let 𝑅=𝑉/π‘₯𝑉. Observe that 𝑀/π‘₯𝑉 is the only prime ideal of 𝑅 and 𝑍(𝑅)=𝑀/π‘₯𝑉. Let us denote 𝑀/π‘₯𝑉 by 𝑃. We assert that 𝑃 is not a 𝐡-prime of (0) in 𝑅. Suppose that 𝑃 is a 𝐡-prime of (0) in 𝑅. Then it can be easily verified that 𝑀=(π‘₯π‘‰βˆΆπ‘‰π‘¦) for some π‘¦βˆˆπ‘‰. Since 𝑀≠𝑉, it follows that π‘¦βˆ‰π‘₯𝑉. As 𝑉 is a valuation domain, we obtain that π‘₯βˆˆπ‘¦π‘‰. Thus π‘₯=𝑦𝑣 for some π‘£βˆˆπ‘‰. Hence we obtain that 𝑀=(π‘₯π‘‰βˆΆπ‘‰π‘¦)=(π‘¦π‘£π‘‰βˆΆπ‘‰π‘¦)=𝑣𝑉. This is impossible since 𝑀 is not finitely generated. Thus 𝑃 is not a 𝐡-prime of (0) in 𝑅. Now it follows from [18, Theorem 1.1 (a)] that (Ξ“(𝑅))𝑐 is connected. We obtain from Theorem 2.5 that diam((Ξ“(𝑅))𝑐)=π‘Ÿ((Ξ“(𝑅))𝑐)=2 and each vertex of (Ξ“(𝑅))𝑐 is in the center of (Ξ“(𝑅))𝑐.

Example 2.8. (i) Let 𝑅 be as in Example 2.7 and let 𝑇=𝑅×𝑍 (resp. 𝑇1=𝑅×𝑅) be the direct product of 𝑅 and the ring of integers (resp. 𝑅 and 𝑅). We know from Example 2.7 that 𝑃 is the unique maximal 𝑁-prime of (0) in 𝑅 and 𝑃 is not a 𝐡-prime of (0) in 𝑅. Note that 𝑃1=𝑃×𝑍and𝑃2=𝑅×(0) (resp. 𝑄1=𝑃×𝑅and𝑄2=𝑅×𝑃) are the only maximal 𝑁-primes of the zero-ideal of 𝑇 (resp. of the zero-ideal of 𝑇1) and 𝑃1βˆ©π‘ƒ2=𝑃×(0) is not the zero-ideal of 𝑇 (resp. 𝑄1βˆ©π‘„2=𝑃×𝑃 is not the zero-ideal of 𝑇1). Hence it follows from [18, Theorem 1.1(b)] that (Ξ“(𝑇))𝑐 (resp. (Ξ“(𝑇1))𝑐) is connected. Since 𝑃1 is not a 𝐡-prime of (0) in 𝑇 (resp. 𝑄1 and 𝑄2 are not 𝐡-primes of (0) in 𝑇1), it follows from Remark 2.6 that diam((Ξ“(𝑇))𝑐)=π‘Ÿ((Ξ“(𝑇))𝑐)=2 (resp. diam((Ξ“(𝑇1))𝑐)=π‘Ÿ((Ξ“(𝑇1))𝑐)=2) and each vertex of (Ξ“(𝑇))𝑐 is in the center of (Ξ“(𝑇))𝑐 (resp. each vertex of (Ξ“(T1))c is in the center of (Ξ“(T1))c).
(ii) Let 𝑅=𝑍/12𝑍. Note that 𝑃1=2𝑍/12𝑍and𝑃2=3𝑍/12𝑍 are the only prime ideals of the finite ring 𝑅=𝑍/12𝑍. Thus 𝑍(𝑅)=𝑃1βˆͺ𝑃2. Observe that 𝑅 has exactly two maximal 𝑁-primes of (0) and 𝑃1βˆ©π‘ƒ2=6𝑍/12𝑍 is not the zero-ideal of 𝑅. Hence it follows from [18, Theorem 1.1(b)] that (Ξ“(𝑅))𝑐 is connected. Moreover, observe that 2𝑍=(12π‘βˆΆπ‘Β±6𝑑) and {Β±6π‘‘βˆ£π‘‘βˆˆπ‘isoddandpositive} is the set of all integers with the property that 2𝑍=(12π‘βˆΆπ‘Β±6𝑑). Furthermore, note that {Β±4π‘˜βˆ£π‘˜βˆˆZispositiveandπ‘˜β‰‘1or2(mod3)} is the set of all integers with the property that 3𝑍=(12π‘βˆΆπ‘Β±4π‘˜). Hence it follows that 𝑃1=((0+12𝑍)βˆΆπ‘…6+12𝑍), 𝑃2=((0+12Z)βˆΆπ‘…4+12𝑍),and𝑃2=((0+12𝑍)βˆΆπ‘…8+12𝑍). Thus 𝑃1 and 𝑃2 are 𝐡-primes of (0) in 𝑅. Now it follows from [18, Proposition 1.7(b)] that diam((Ξ“(𝑅))𝑐)=3, and we know from Theorem 2.5 that π‘Ÿ((Ξ“(𝑅))𝑐)=2. Moreover, we obtain from Remark 2.6 and from the above discussion that the set of centers of (Ξ“(𝑅))𝑐={2+12𝑍,10+12𝑍,3+12𝑍,9+12𝑍}.

Example 2.9. Let 𝑛>1 be such that 𝑛 admits at least three distinct prime divisors. Let {𝑝1,𝑝2,𝑝3,…,𝑝𝑑}(𝑑β‰₯3) be the set of all distinct prime divisors of 𝑛. Let 𝑅=𝑍/𝑛𝑍. Note that {𝑝1𝑍/𝑛𝑍,𝑝2𝑍/𝑛𝑍,…,𝑝𝑑𝑍/𝑛𝑍} is the set of all maximal 𝑁-primes of the zero-ideal of 𝑅. We know from [18, Theorem 1.1(c)] that (Ξ“(𝑅))𝑐 is connected and moreover, it is known from Remark 2.6 that diam((Ξ“(𝑅))𝑐)=π‘Ÿ((Ξ“(𝑅))𝑐)=2 and each vertex of (Ξ“(𝑅))𝑐 is in the center of (Ξ“(𝑅))𝑐.

3. The Girth of (Ξ“(𝑅))𝑐

Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)βˆ—|β‰₯2. The aim of this section is to study about the girth of (Ξ“(𝑅))𝑐. If (Ξ“(𝑅))𝑐 is connected, then we prove in this section that gr((Ξ“(𝑅))𝑐)=3. Moreover, we also discuss about the girth of (Ξ“(𝑅))𝑐 in the case when (Ξ“(𝑅))𝑐 is not connected.

For the sake of convenience we split the results proved in this section into several lemmas. We begin with the following lemma. We make use of this lemma in the proof of Lemma 3.2.

Lemma 3.1. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. Let 𝑃 be a maximal 𝑁-prime of (0) in 𝑅. If 𝑃 is not a 𝐡-prime of (0) in 𝑅, then for any π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ—, π‘ƒβŠ„((0)βˆΆπ‘…π‘₯)βˆͺ𝑅𝑦.

Proof. Suppose that for some π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ—, π‘ƒβŠ†((0)βˆΆπ‘…π‘₯)βˆͺ𝑅𝑦. Then either π‘ƒβŠ†((0)βˆΆπ‘…π‘₯)orπ‘ƒβŠ†π‘…π‘¦. Since π‘₯β‰ 0, ((0)βˆΆπ‘…π‘₯)∩(𝑅⧡𝑍(𝑅))=βˆ…. As π‘¦βˆˆπ‘(𝑅), there exists π‘§βˆˆπ‘…β§΅{0} such that 𝑦𝑧=0. Hence we obtain that π‘…π‘¦βŠ†((0)βˆΆπ‘…π‘§). Note that ((0)βˆΆπ‘…π‘§)∩(𝑅⧡𝑍(𝑅))=βˆ…. Now it follows from [19, Theorem 2.2, Page 378] that there exists a maximal N-prime Q of (0) in R such that ((0)βˆΆπ‘…π‘₯)βŠ†π‘„ and there exists a maximal 𝑁-prime π‘Š of (0) in 𝑅 such that ((0)βˆΆπ‘…π‘§)βŠ†π‘Š. If π‘ƒβŠ†((0)βˆΆπ‘…π‘₯), then we obtain that π‘ƒβŠ†((0)βˆΆπ‘…π‘₯)βŠ†π‘„ and hence it follows that 𝑃=𝑄=((0)βˆΆπ‘…π‘₯). This contradicts the assumption that 𝑃 is not a 𝐡-prime of (0) in 𝑅. If π‘ƒβŠ†π‘…π‘¦, then π‘ƒβŠ†((0)βˆΆπ‘…π‘§)βŠ†π‘Š and so 𝑃=π‘Š=((0)βˆΆπ‘…π‘§). This is also impossible since 𝑃 is not a 𝐡-prime of (0) in 𝑅. This proves that if a maximal 𝑁-prime 𝑃 of (0) in 𝑅 is not a 𝐡-prime of (0) in 𝑅, then for any π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ—, π‘ƒβŠ„((0)βˆΆπ‘…π‘₯)βˆͺ𝑅𝑦.

In the following lemma, we determine the girth of (Ξ“(𝑅))𝑐 under the assumptions that 𝑅 has exactly one maximal 𝑁-prime of (0) and (Ξ“(𝑅))𝑐 is connected.

Lemma 3.2. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. Suppose that 𝑅 has only one maximal 𝑁-prime of (0). If (Ξ“(𝑅))𝑐 is connected, then gr((Ξ“(𝑅))𝑐)=3.

Proof. Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. Since (Ξ“(𝑅))𝑐 is connected, we obtain from [18, Theorem 1.1(a)] that 𝑃 is not a 𝐡-prime of (0) in 𝑅. Note that 𝑍(𝑅)=𝑃. Let π‘₯βˆˆπ‘ƒβ§΅{0}. By hypothesis, 𝑃 is not a 𝐡-prime of (0) in 𝑅 and so Lemma 3.1 implies that there exists π‘¦βˆˆπ‘ƒ such that π‘¦βˆ‰π‘…π‘₯ and 𝑦π‘₯β‰ 0. We assert that π‘₯π‘¦βˆ‰{π‘₯,𝑦}. If π‘₯𝑦=π‘₯, then π‘₯(1βˆ’π‘¦)=0. As π‘¦βˆˆπ‘ƒ, 1βˆ’π‘¦βˆ‰π‘ƒ=𝑍(𝑅). Hence π‘₯(1βˆ’π‘¦)=0 implies that π‘₯=0. This contradicts the fact that π‘₯β‰ 0. Similarly, it follows that π‘₯𝑦≠𝑦. If both π‘₯2𝑦 and 𝑦2π‘₯ are nonzero, then we obtain that π‘₯---π‘₯𝑦---𝑦---π‘₯ is a cycle of length 3 in (Ξ“(𝑅))𝑐. Suppose that either π‘₯2𝑦=0 or 𝑦2π‘₯=0. Without loss of generality, we may assume that π‘₯2𝑦=0. As 𝑃 is not a 𝐡-prime of (0) in 𝑅, Lemma 3.1 implies that there exists π‘§βˆˆπ‘ƒ such that 𝑧π‘₯𝑦≠0 and π‘§βˆ‰π‘…π‘¦. From 𝑧π‘₯𝑦≠0, it follows that 𝑧π‘₯β‰ 0, and 𝑧𝑦≠0 and moreover, as π‘₯2𝑦=0, it follows that 𝑧≠π‘₯. Furthermore, by the choice of 𝑧, it follows that 𝑧≠𝑦. Now π‘₯---𝑧---𝑦---π‘₯ is a cycle of length 3 in (Ξ“(𝑅))𝑐. This shows that if 𝑅 has only one maximal 𝑁-prime of (0) and if (Ξ“(𝑅))𝑐 is connected, then there exists a cycle of length 3 in (Ξ“(𝑅))𝑐 and hence gr((Ξ“(𝑅))𝑐)=3.

Though the following lemma is elementary, we include it for the sake of future reference.

Lemma 3.3. Let 𝑅 be a commutative ring with identity. If there exist distinct elements π‘Ž,𝑏,π‘βˆˆπ‘(𝑅)βˆ—β§΅π‘ƒ for some prime ideal 𝑃 of 𝑅, then gr((Ξ“(𝑅))𝑐)=3.

Proof. As 𝑃 is a prime ideal of 𝑅 and π‘Ž,𝑏,𝑐 are elements of 𝑅 which are not in 𝑃, we obtain that π‘Žπ‘,𝑏𝑐,π‘π‘Žβˆˆπ‘…β§΅π‘ƒ and so π‘Žπ‘,𝑏𝑐,π‘π‘Žβˆˆπ‘…β§΅{0}. Hence it follows that π‘Ž---𝑏---𝑐---π‘Ž is a cycle of length 3 in (Ξ“(𝑅))𝑐. This proves that gr((Ξ“(𝑅))𝑐)=3.

The next lemma discusses the girth of (Ξ“(𝑅))𝑐 where 𝑅 is a ring with (0) of 𝑅 admitting exactly two maximal 𝑁-primes.

Lemma 3.4. Let 𝑅 be a commutative ring with identity. Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and let them be 𝑃1 and 𝑃2. Then the following hold: (i)If 𝑃1βˆ©π‘ƒ2=(0), then (Ξ“(𝑅))𝑐 contains a cycle if and only if either |𝑃1⧡𝑃2|β‰₯3 or |𝑃2⧡𝑃1|β‰₯3 if and only if gr((Ξ“(𝑅))𝑐)=3. (ii)If 𝑃1βˆ©π‘ƒ2β‰ (0), then gr((Ξ“(𝑅))𝑐)=3. Thus, if (Ξ“(𝑅))𝑐 is connected, then gr((Ξ“(𝑅))𝑐)=3.

Proof. By hypothesis, 𝑃1 and 𝑃2 are the only maximal 𝑁-primes of (0) in 𝑅. So, it follows that 𝑍(𝑅)=𝑃1βˆͺ𝑃2.
(i) Assume that (Ξ“(𝑅))𝑐 contains a cycle. Let π‘Ž1---π‘Ž2---π‘Ž3---β‹―---π‘Žπ‘›---π‘Ž1 be a cycle of length 𝑛 in (Ξ“(𝑅))𝑐. Note that 𝑛β‰₯3 and {π‘Žπ‘–βˆ£π‘–=1,2,3,…,𝑛}βŠ†π‘(𝑅)βˆ— with π‘Ž1π‘Ž2,π‘Ž2π‘Ž3,…,π‘Žπ‘›βˆ’1π‘Žπ‘›,π‘Žπ‘›π‘Ž1βˆˆπ‘…β§΅{0}. Since 𝑃1βˆ©π‘ƒ2=(0), it follows that either {π‘Ž1,π‘Ž2,π‘Ž3,…,π‘Žπ‘›}βŠ†π‘ƒ1⧡𝑃2or{π‘Ž1,π‘Ž2,π‘Ž3,…,π‘Žπ‘›}βŠ†π‘ƒ2⧡𝑃1. Now it is clear that either |𝑃1⧡𝑃2|β‰₯3 or |𝑃2⧡𝑃1|β‰₯3.
Conversely, suppose that either |𝑃1⧡𝑃2|β‰₯3 or |𝑃2⧡𝑃1|β‰₯3. Since 𝑍(𝑅)=𝑃1βˆͺ𝑃2 and as 𝑃1 and 𝑃2 are prime ideals of 𝑅, it follows from Lemma 3.3 that gr((Ξ“(𝑅))𝑐)=3.
If gr((Ξ“(𝑅))𝑐)=3, then we obtain that (Ξ“(𝑅))𝑐 contains a cycle of length 3.
(ii) Suppose that 𝑃1βˆ©π‘ƒ2β‰ (0). We know from the proof of [18, Proposition 1.7(i)] that there exist π‘Žβˆˆπ‘ƒ1⧡𝑃2 and π‘βˆˆπ‘ƒ2⧡𝑃1 such that π‘Žπ‘β‰ 0. We consider two cases : Case(A): 𝑃1βˆ©π‘ƒ2βŠ„((0)βˆΆπ‘…π‘Ž)βˆͺ((0)βˆΆπ‘…π‘). Then there exists π‘βˆˆπ‘ƒ1βˆ©π‘ƒ2 such that π‘Žπ‘β‰ 0 and 𝑏𝑐≠0. Hence we obtain a cycle π‘Ž---𝑏---𝑐---π‘Ž in (Ξ“(𝑅))𝑐 and it is of length 3. Case(B): 𝑃1βˆ©π‘ƒ2βŠ†((0)βˆΆπ‘…π‘Ž)βˆͺ((0)βˆΆπ‘…π‘). Then either 𝑃1βˆ©π‘ƒ2βŠ†((0)βˆΆπ‘…π‘Ž) or 𝑃1βˆ©π‘ƒ2βŠ†((0)βˆΆπ‘…π‘).
Without loss of generality, we may assume that 𝑃1βˆ©π‘ƒ2βŠ†((0)βˆΆπ‘…π‘Ž). Let π‘₯βˆˆπ‘ƒ1βˆ©π‘ƒ2 be such that π‘₯β‰ 0. Since π‘Ž+π‘βˆ‰π‘(𝑅), we obtain that (π‘Ž+𝑏)π‘₯β‰ 0. As π‘Žπ‘₯=0, it follows that 𝑏π‘₯=(π‘Ž+𝑏)π‘₯β‰ 0. Moreover, observe that 𝑏+π‘₯βˆˆπ‘ƒ2⧡𝑃1,𝑏≠𝑏+π‘₯, and π‘Ž(𝑏+π‘₯)=π‘Žπ‘β‰ 0. As 𝑏(𝑏+π‘₯)βˆˆπ‘ƒ2⧡𝑃1, it follows that 𝑏(𝑏+π‘₯)β‰ 0. Note that π‘Ž---𝑏---𝑏+π‘₯---π‘Ž is a cycle of length 3 in (Ξ“(𝑅))𝑐.
Thus in both the cases, (Ξ“(𝑅))𝑐 admits a cycle of length 3. Hence we obtain that gr((Ξ“(𝑅))𝑐)=3.
We know from [18, Theorem 1.1(b)] that (Ξ“(𝑅))𝑐 is connected if and only if 𝑃1βˆ©π‘ƒ2β‰ (0). Thus if 𝑅 has exactly two maximal 𝑁-primes of (0) and if (Ξ“(𝑅))𝑐 is connected, then gr((Ξ“(𝑅))𝑐)=3.

We show in the next lemma that if 𝑅 has at least three maximal 𝑁-primes of (0), then gr((Ξ“(𝑅))𝑐)=3.

Lemma 3.5. Let 𝑅 be a ring. Suppose that 𝑅 admits more than two maximal 𝑁-primes of (0). Then gr((Ξ“(𝑅))𝑐)=3.

Proof. Since by the assumption that 𝑅 admits more than two maximal 𝑁-primes of (0), we can find at least three maximal 𝑁-primes of (0) in 𝑅. Let {𝑃,𝑄,π‘Š} be a subset of the set of all maximal 𝑁-primes of (0) in 𝑅. It is clear that π‘ƒβŠ„π‘„βˆͺπ‘Š, π‘„βŠ„π‘ƒβˆͺW, and π‘ŠβŠ„π‘ƒβˆͺ𝑄. Hence there exist elements π‘₯βˆˆπ‘ƒβ§΅(𝑄βˆͺπ‘Š), π‘¦βˆˆπ‘„β§΅(𝑃βˆͺπ‘Š), and π‘§βˆˆπ‘Šβ§΅(𝑃βˆͺ𝑄). Note that π‘₯,𝑦,𝑧 are distinct elements of 𝑍(𝑅)βˆ— with π‘₯𝑦,𝑦𝑧,𝑧π‘₯βˆˆπ‘…β§΅{0}. Hence π‘₯---𝑦---𝑧---π‘₯ is a cycle of length 3 in (Ξ“(𝑅))𝑐. This proves that gr((Ξ“(𝑅))𝑐)=3.

With the help of the above lemmas, we obtain the main theorem of this section.

Theorem 3.6. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. If (Ξ“(𝑅))𝑐 is connected, then gr((Ξ“(𝑅))𝑐)=3.

Proof. The proof of this theorem follows immediately from [18, Theorem 1.1] and Lemmas 3.2, 3.4(ii), and 3.5.

We next proceed to consider rings 𝑅 such that (Ξ“(𝑅))𝑐 is not connected and discuss about the girth of (Ξ“(𝑅))𝑐. In this direction, we first have the following proposition.

Proposition 3.7. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝑃2β‰  (0) and if (Ξ“(𝑅))𝑐 is not connected, then (Ξ“(𝑅))𝑐 contains a cycle if and only if |𝑃|β‰₯5 if and only if gr((Ξ“(𝑅))𝑐)=3.

Proof. Note that 𝑍(𝑅)=𝑃. Assume that 𝑃2β‰ (0) and (Ξ“(𝑅))𝑐 is not connected. From the assumption that 𝑃2β‰ (0), it follows that there exist π‘Ž,π‘βˆˆπ‘ƒ such that π‘Žβ‰ π‘ and π‘Žπ‘β‰ 0. Moreover, by [18, Theorem 1.1(a)], 𝑃=((0)βˆΆπ‘…π‘) for some π‘βˆˆπ‘…. It is clear that π‘βˆˆπ‘ƒβ§΅{0}. Furthermore, as π‘Žπ‘β‰ 0, it follows that π‘Žβ‰ π‘ and 𝑏≠𝑐.
Suppose that (Ξ“(𝑅))𝑐 contains a cycle. Let π‘Ž1---π‘Ž2---π‘Ž3---β‹―---π‘Žπ‘›---π‘Ž1 be a cycle of length 𝑛 in (Ξ“(𝑅))𝑐. Note that 𝑛β‰₯3 and {π‘Ž1,π‘Ž2,π‘Ž3,…,π‘Žπ‘›}βŠ†π‘ƒβ§΅{0}. As 𝑃=((0)βˆΆπ‘…π‘) and since π‘Ž1π‘Ž2,π‘Ž2π‘Ž3,…,π‘Žπ‘›βˆ’1π‘Žπ‘›,andπ‘Žπ‘›π‘Ž1βˆˆπ‘…β§΅{0}, it follows that π‘βˆ‰{π‘Ž1,π‘Ž2,π‘Ž3,…,π‘Žπ‘›}. Thus {0,𝑐,π‘Ž1,π‘Ž2,π‘Ž3,…,π‘Žπ‘›}βŠ†π‘ƒ and hence it follows that |𝑃|β‰₯5.
We next show that gr((Ξ“(𝑅))𝑐)=3 if |𝑃|β‰₯5. Suppose that |𝑃|β‰₯5. Observe that there exists π‘‘βˆˆπ‘ƒβ§΅{0,π‘Ž,𝑏,𝑐} where π‘Ž,𝑏,𝑐 are as in the first paragraph of this proof. We claim that there exists a cycle of length 3 in (Ξ“(𝑅))𝑐. If π‘Žπ‘‘β‰ 0 and 𝑏𝑑≠0, then π‘Ž---𝑑---𝑏---π‘Ž is a cycle of length 3 in (Ξ“(𝑅))𝑐. Suppose that either π‘Žπ‘‘=0 or 𝑏𝑑=0.
Observe that if π‘Ž2=0 and 𝑏2=0, then from (π‘Ž+𝑏)π‘Ž=π‘Žπ‘=(π‘Ž+𝑏)𝑏=π‘Žπ‘, and from the fact that π‘Ž+π‘βˆˆπ‘ƒβ§΅{0,π‘Ž,𝑏}, it follows that π‘Ž---π‘Ž+𝑏---𝑏---π‘Ž is a cycle of length 3 in (Ξ“(𝑅))𝑐. Now let us suppose without loss of generality that π‘Ž2β‰ 0. As 𝑐2=0 and π‘Ž2β‰ 0, it follows that π‘Ž+𝑐≠0. If π‘Ž+𝑐≠𝑏, then π‘Ž---π‘Ž+𝑐---𝑏---π‘Ž is a cycle of length 3 in (Ξ“(𝑅))𝑐. Suppose that π‘Ž+𝑐=𝑏. Then, as 𝑐𝑑=0, it follows from the assumption that either π‘Žπ‘‘=0 or 𝑏𝑑=0 that π‘Žπ‘‘=𝑏𝑑=0. As 𝑐≠𝑑 and π‘Ž+𝑐=𝑏, it follows that π‘Ž+𝑑≠𝑏 and it is clear that π‘Ž(π‘Ž+𝑑)=π‘Ž2β‰  0 and (π‘Ž+𝑑)𝑏=π‘Žπ‘β‰ 0. Hence π‘Ž---π‘Ž+𝑑---𝑏---π‘Ž is a cycle of length 3 in (Ξ“(𝑅))𝑐.
Thus if |𝑃|β‰₯5, then it is shown that there exists a cycle of length 3 in (Ξ“(𝑅))𝑐. Hence gr((Ξ“(𝑅))𝑐)=3.
If gr((Ξ“(𝑅))𝑐)=3, then (Ξ“(𝑅))𝑐 contains a cycle of length 3.

The following remark characterizes rings 𝑅 satisfying the following conditions: (i) 𝑅 has exactly one maximal 𝑁-prime of (0), (ii) (Ξ“(𝑅))𝑐 contains at least one edge, (iii) (Ξ“(𝑅))𝑐 is not connected, and (iv) (Ξ“(𝑅))𝑐 does not contain any cycle.

Remark 3.8. Let 𝑅 be a ring and let |𝑍(𝑅)βˆ—|β‰₯2. Suppose that 𝑃 is the only maximal 𝑁-prime of (0) in 𝑅. Assume that 𝑃2β‰ (0). Suppose that (Ξ“(𝑅))𝑐 is not connected. Let π‘Ž,π‘βˆˆπ‘ƒ be such that π‘Žβ‰ π‘ and π‘Žπ‘β‰ 0. Let π‘βˆˆπ‘ƒβ§΅{0} be such that 𝑃=((0)βˆΆπ‘…π‘). Assume that gr((Ξ“(𝑅))𝑐)β‰ 3. Observe that by Proposition 3.7, gr((Ξ“(𝑅))𝑐)β‰ 3 if and only if (Ξ“(𝑅))𝑐 does not contain any cycle. Now it follows from Proposition 3.7 that |𝑃|≀4 and as {0,π‘Ž,𝑏,𝑐}βŠ†π‘ƒ, it follows that 𝑃={0,π‘Ž,𝑏,𝑐}. Now π‘Žπ‘βˆˆπ‘ƒβ§΅{0}={π‘Ž,𝑏,𝑐}. We assert that π‘Žπ‘=𝑐. If π‘Žπ‘=π‘Ž, then π‘Ž(1βˆ’π‘)=0. This is impossible since π‘Žβ‰ 0 and 1βˆ’π‘βˆ‰π‘ƒ=𝑍(𝑅). Similarly, it follows that π‘Žπ‘β‰ π‘. Hence π‘Žπ‘=𝑐. Now 𝑍(𝑅)=𝑃={0,π‘Ž,𝑏,𝑐=π‘Žπ‘} is finite and hence we obtain from [20, Theorem 1] that R is a finite ring. We verify in this remark that |𝑅|=8. Moreover, we observe with the help of [3, Theorem 3.2] that 𝑅 is isomorphic to exactly one of the rings from the set {𝑍8,𝑍4[π‘₯]/(2π‘₯𝑍4[π‘₯]+(π‘₯2βˆ’2)𝑍4[π‘₯]),𝑍2[π‘₯]/π‘₯3𝑍2[π‘₯]} where 𝑍4[π‘₯] (resp. 𝑍2[π‘₯]) is the polynomial ring in one variable over 𝑍4 (resp. over 𝑍2).
Since 𝑅 is a finite ring, any prime ideal of 𝑅 is a maximal ideal of 𝑅, and moreover, if 𝑄 is any prime ideal of 𝑅, then π‘„βŠ†π‘(𝑅). Since 𝑍(𝑅)=𝑃, it follows that 𝑃 is the only prime ideal of 𝑅. Now 𝑅 is a local ring with unique maximal ideal 𝑃. Hence we obtain that 𝑅⧡𝑃 is the set of all units in 𝑅. Let u be a unit in 𝑅. Then π‘’π‘Žπ‘βˆˆπ‘ƒβ§΅{0}={π‘Ž,𝑏,𝑐=π‘Žπ‘}. We claim that π‘’π‘Žπ‘=π‘Žπ‘. If π‘’π‘Žπ‘=π‘Ž, then π‘Ž(1βˆ’π‘’π‘)=0and this is impossible. Similarly, we obtain that π‘’π‘Žπ‘β‰ π‘. Hence π‘’π‘Žπ‘=π‘Žπ‘. This implies that (π‘’βˆ’1)π‘Žπ‘=0. Since π‘Žπ‘β‰ 0, it follows that π‘’βˆ’1βˆˆπ‘(𝑅)=𝑃={0,π‘Ž,𝑏,𝑐=π‘Žπ‘}. Hence we obtain that π‘’βˆˆ{1,1+π‘Ž,1+𝑏,1+𝑐=1+π‘Žπ‘}. Thus it is shown that 𝑅⧡𝑃={1,1+π‘Ž,1+𝑏,1+𝑐}. Hence 𝑅={0,π‘Ž,𝑏,𝑐=π‘Žπ‘,1,1+π‘Ž,1+𝑏,1+𝑐} is a ring containing exactly 8 elements. Note that πœ”(Ξ“(𝑅))=2. Now [3, Theorem 3.2] implies that 𝑅 is isomorphic to exactly one of the rings from the set {𝑍8,𝑍4[π‘₯]/(2π‘₯𝑍4[π‘₯]+(π‘₯2βˆ’2)𝑍4[π‘₯]),𝑍2[π‘₯]/π‘₯3𝑍2[π‘₯]}. Observe that if 𝑅 is a ring such that π‘…βˆˆ{𝑍8,𝑍4[π‘₯]/(2π‘₯𝑍4[π‘₯]+(π‘₯2βˆ’2)𝑍4[π‘₯]),𝑍2[π‘₯]/π‘₯3𝑍2[π‘₯]}, then 𝑅 satisfies the following conditions: 𝑅 has exactly one maximal 𝑁-prime of (0), say 𝑃 such that 𝑃2β‰ (0), (Ξ“(𝑅))𝑐 is not connected, and (Ξ“(𝑅))𝑐 does not contain any cycle.
Let 𝑅 be a ring such that 𝑅 has only one maximal 𝑁-prime of (0), say 𝑃, and 𝑅 satisfies the further conditions that 𝑃2β‰ (0) and (Ξ“(𝑅))𝑐 is not connected. The above discussion implies that (Ξ“(𝑅))𝑐 does not contain any cycle if and only if 𝑅 is isomorphic to exactly one of the rings from the set {𝑍8,𝑍4[π‘₯]/(2π‘₯𝑍4[π‘₯]+(π‘₯2βˆ’2)𝑍4[π‘₯]),𝑍2[π‘₯]/π‘₯3𝑍2[π‘₯]}.

We determine in the following remark rings 𝑅 satisfying the following conditions: (i) 𝑅 admits exactly two maximal 𝑁-primes of (0), (ii) (Ξ“(𝑅))𝑐 contains at least one edge, (iii) (Ξ“(𝑅))𝑐 is not connected, and (iv) (Ξ“(𝑅))𝑐 does not contain any cycle.

Remark 3.9. Let 𝑅 be a ring admitting exactly two maximal 𝑁-primes of (0). Let them be 𝑃1 and 𝑃2. Suppose that 𝑃1βˆ©π‘ƒ2=(0) (that is, equivalently (Ξ“(𝑅))𝑐 is not connected). In such a case, it is shown in Lemma 3.4(i) that (Ξ“(𝑅))𝑐 contains a cycle if and only if either |𝑃1⧡𝑃2|β‰₯3, or |𝑃2⧡𝑃1|β‰₯3 if and only if gr((Ξ“(𝑅))𝑐)=3. Suppose that |𝑃1⧡𝑃2|<3, |𝑃2⧡𝑃1|<3 and that (Ξ“(𝑅))𝑐 contains at least one edge. We verify in this remark that either |𝑅|=9 or |𝑅|=6. Moreover, we verify that 𝑅 is isomorphic to 𝐾1×𝐾2 where 𝐾1 and 𝐾2 are fields with either |𝐾1|=|𝐾2|=3 or one of them contains exactly 3 elements and the other contains exactly 2 elements.
We are assuming that (Ξ“(𝑅))𝑐 contains at least one edge. Since 𝑍(𝑅)=𝑃1βˆͺ𝑃2 and 𝑃1βˆ©π‘ƒ2=(0), it follows that at least one between 𝑃1 and 𝑃2 contains exactly 3 elements. This implies that either |𝑃1|=|𝑃2|=3 or exactly one between 𝑃1 and 𝑃2 contains exactly 3 elements and the other contains exactly 2 elements. Thus either |𝑍(𝑅)|=5 or |𝑍(𝑅)|=4. As 𝑍(𝑅) is a finite set, it follows from [20, Theorem 1] that 𝑅 is a finite ring. Since any prime ideal of a finite ring is a maximal ideal, it follows that 𝑃1 and 𝑃2 are maximal ideals of 𝑅. As 𝑃1βˆ©π‘ƒ2=(0), it follows from the Chinese Remainder Theorem [21, Proposition 1.10] that π‘…β‰ˆ(𝑅/𝑃1)Γ—(𝑅/𝑃2) as rings. Thus we obtain from the above discussion that R is isomorphic to the direct product of two fields 𝐾1 and 𝐾2 with either |𝐾1|=|𝐾2|=3 or one between 𝐾1 and 𝐾2 contains exactly 3 elements and the other contains exactly 2 elements. Hence we obtain that either |𝑅|=9 or |𝑅|=6 and 𝑅 is isomorphic to 𝐾1×𝐾2 where 𝐾1 and 𝐾2 are fields with either |𝐾1|=|𝐾2|=3 or one of them contains exactly 3 elements and the other contains exactly 2 elements. Conversely, if 𝑅 is isomorphic to 𝐾1×𝐾2 where 𝐾1 and 𝐾2 are fields with either |𝐾1|=|𝐾2|=3 or one of them contains exactly 3 elements and the other contains exactly 2 elements, then it is clear that 𝑅 has the following properties: 𝑅 admits exactly two maximal 𝑁-primes of (0), (Ξ“(𝑅))𝑐 contains at least one edge, (Ξ“(𝑅))𝑐 is not connected and it does not contain any cycle.

We next have the following corollary, the proof of which is immediate from the results proved in this section.

Corollary 3.10. (i) Let 𝑅 be an infinite ring. If there exist π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ— such that π‘₯𝑦≠0, then gr((Ξ“(𝑅))𝑐)=3.
(ii) Let 𝑅 be any ring admitting elements π‘Ž,π‘βˆˆπ‘(𝑅)βˆ— such that π‘Žπ‘β‰ 0. Then gr((Ξ“(𝑅[π‘₯]))𝑐)=gr((Ξ“(𝑅[[π‘₯]]))𝑐)=3 where 𝑅[π‘₯] (resp. 𝑅[[π‘₯]]) is the polynomial (resp.the power series) ring in one variable over 𝑅.

4. Cliques in (Ξ“(𝑅))𝑐

Let 𝑅 be a commutative ring with identity which is not an integral domain. In this section, we prove that if a ring 𝑅 admits more than one maximal 𝑁-prime of (0), then the clique number of (Ξ“(𝑅))𝑐 is finite if and only if (Ξ“(𝑅))𝑐 does not contain any infinite clique if and only if 𝑅 is finite. Moreover, if a ring 𝑅 is such that R has only one maximal 𝑁-prime of (0), we obtain some necessary conditions in order that the clique number of (Ξ“(𝑅))𝑐 is finite. Furthermore, if 𝑅 is either a Noetherian ring or a chained ring and if (Ξ“(𝑅))𝑐 admits at least one edge (that is, there exist π‘₯,π‘¦βˆˆπ‘(𝑅)βˆ— with π‘₯≠𝑦 such that π‘₯𝑦≠0), then it is shown that the clique number of (Ξ“(𝑅))𝑐 is finite if and only if (Ξ“(𝑅))𝑐 does not contain any infinite clique if and only if 𝑅 is finite.

We first prove some elementary lemmas which are of interest in their own right and which are useful in proving the main results of this section. We begin with the following lemma.

Lemma 4.1. Let 𝑅 be a commutative ring with identity which is not an integral domain. Let 𝑃 be any prime ideal of 𝑅. Then the following hold.(i)If (Ξ“(𝑅))𝑐 does not contain any infinite clique, then 𝑍(𝑅)⧡𝑃 is finite. (ii)If πœ”((Ξ“(𝑅))𝑐) is finite, then 𝑍(𝑅)⧡𝑃 is finite and indeed, |𝑍(𝑅)⧡𝑃|β‰€πœ”((Ξ“(𝑅))𝑐).

Proof. (i) Suppose that 𝑍(𝑅)⧡𝑃 is infinite. Then we can choose an infinite sequence of distinct elements π‘₯π‘–βˆˆπ‘(𝑅)⧡𝑃. Since 𝑃 is a prime ideal of R and as π‘₯π‘–βˆ‰π‘ƒ for 𝑖=1,2,3,…, it follows that π‘₯𝑖π‘₯𝑗≠0 for all 𝑖,π‘—βˆˆ{1,2,3,…}. Observe that the subgraph of (Ξ“(𝑅))𝑐 induced on {π‘₯π‘–βˆ£π‘–=1,2,3,…} is an infinite clique. This contradicts the assumption that (Ξ“(𝑅))𝑐 does not contain any infinite clique. Hence we obtain that 𝑍(𝑅)⧡𝑃 is finite.
(ii) Let πœ”((Ξ“(𝑅))𝑐)=𝑛. We assert that |𝑍(𝑅)⧡𝑃|≀𝑛. Suppose that {𝑍(𝑅)β§΅π‘ƒβˆ£β‰₯𝑛+1. Let {π‘₯1,…,π‘₯𝑛+1}βŠ†π‘(𝑅)⧡𝑃. Then it is clear that the subgraph of (Ξ“(𝑅))𝑐 induced on {π‘₯1,…,π‘₯𝑛+1} is a clique. This is impossible since πœ”((Ξ“(𝑅))𝑐)=𝑛. This shows that |𝑍(𝑅)⧡𝑃|≀𝑛=πœ”((Ξ“(𝑅))𝑐).

Using Lemma 4.1 and prime avoidance [21, Proposition 1.11(i)], we obtain the following result.

Lemma 4.2. Let 𝑅 be a commutative ring with identity which is not an integral domain. Let 𝑃 be any prime ideal of 𝑅. Let 𝐴={π‘„βˆ£π‘„ is a prime ideal of 𝑅 such that π‘„βŠ†π‘(𝑅) but π‘„βŠ„π‘ƒ}. Then the following hold.(i)If (Ξ“(𝑅))𝑐 does not contain any infinite clique, then 𝐴 can admit only a finite number of elements which are pairwise incomparable under inclusion. (ii)If πœ”((Ξ“(𝑅))𝑐) is finite, then 𝐴 can admit at most πœ”((Ξ“(𝑅))𝑐) elements which are pairwise incomparable under inclusion.

Proof. (i) Suppose that (Ξ“(𝑅))𝑐 does not contain any infinite clique. We want to verify that 𝐴 can admit only a finite number of elements which are pairwise incomparable under inclusion. Suppose that there exist infinitely many elements in 𝐴 which are pairwise incomparable under inclusion. Hence there exist π‘„π‘–βˆˆπ΄ for each 𝑖=1,2,3,… with 𝑄𝑖 and 𝑄𝑗 not being comparable under inclusion for all 𝑖,π‘—βˆˆ{1,2,3,…} with 𝑖≠𝑗. Now π‘„π‘–βŠ„π‘ƒ for 𝑖=1,2,3,…. Hence it is possible to choose π‘₯1βˆˆπ‘„1⧡𝑃. Let 𝑖β‰₯2. Now it follows from prime avoidance [21, Proposition 1.11(i)] that there exists π‘₯π‘–βˆˆπ‘„π‘–β§΅(𝑃βˆͺ𝑄1βˆͺβ‹…β‹…β‹…βˆͺπ‘„π‘–βˆ’1). Observe that {π‘₯π‘–βˆ£π‘–=1,2,3,…}βŠ†π‘(𝑅)⧡𝑃. Hence we obtain that 𝑍(𝑅)⧡𝑃 is infinite. This contradicts Lemma 4.1(i). This proves that if (Ξ“(𝑅))𝑐 does not contain any infinite clique, then 𝐴 can admit only a finite number of elements which are pairwise incomparable under inclusion.
(ii) Let πœ”((Ξ“(𝑅))𝑐)=𝑛. Suppose that 𝐴 admits more than 𝑛 elements which are pairwise incomparable under inclusion. Let {𝑄1,…,𝑄𝑛+1}βŠ†π΄ be such that 𝑄𝑖 and 𝑄𝑗 are not comparable for all 𝑖,π‘—βˆˆ{1,2,…,𝑛+1} with 𝑖≠𝑗. Let π‘₯1βˆˆπ‘„1⧡𝑃. Let π‘–βˆˆ{2,…,𝑛+1}. As in (i), we can choose π‘₯π‘–βˆˆπ‘„π‘–β§΅(𝑃βˆͺ𝑄1βˆͺβ‹…β‹…β‹…βˆͺπ‘„π‘–βˆ’1). Observe that {π‘₯π‘–βˆ£π‘–=1,2,…,𝑛+1}βŠ†π‘(𝑅)⧡𝑃. This implies that |𝑍(𝑅)⧡𝑃|β‰₯𝑛+1>πœ”((Ξ“(𝑅))𝑐). This contradicts Lemma 4.1 (ii). Hence we obtain that 𝐴 can admit at most 𝑛=πœ”((Ξ“(𝑅))𝑐) elements which are pairwise incomparable under inclusion.

We next study in the following corollary to Lemma 4.2, the effect of the nature of the cliques of (Ξ“(𝑅))𝑐 on the set of maximal 𝑁-primes of (0), and the set of minimal prime ideals of 𝑅.

Corollary 4.3. Let 𝑅 be a commutative ring with identity which is not an integral domain. Then the following hold. (i)If (Ξ“(𝑅))𝑐 does not contain any infinite clique then (π‘Ž) the set of maximal 𝑁-primes of (0) in 𝑅 is finite (𝑏) the set of minimal prime ideals of 𝑅 is finite. (ii)If πœ”((Ξ“(𝑅))𝑐) is finite, then (π‘Ž)𝑅 can admit at most πœ”((Ξ“(𝑅))𝑐)+1 maximal 𝑁-primes of (0) and if 𝑅 admits exactly π‘˜ maximal 𝑁-primes of (0) with π‘˜β‰₯3, then π‘˜β‰€πœ”((Ξ“(𝑅))𝑐)(b)𝑅 can admit at most πœ”((Ξ“(𝑅))𝑐)+1 minimal prime ideals, and if π‘˜ is the number of minimal prime ideals of 𝑅 with π‘˜β‰₯3, then π‘˜β‰€πœ”((Ξ“(𝑅))𝑐).

Proof. (i)(a) Let 𝑃 be a maximal 𝑁-prime of (0) in 𝑅. Let 𝐴={π‘„βˆ£π‘„ is a maximal 𝑁-prime of (0) in 𝑅 and 𝑄≠𝑃}. Since any maximal 𝑁-prime of (0) in 𝑅 is a subset of 𝑍(𝑅) and as distinct maximal 𝑁-primes of (0) in 𝑅 are not comparable under inclusion, it follows from Lemma 4.2(i) that 𝐴 is finite. Observe that the set of all maximal 𝑁-primes of (0) in 𝑅=𝐴βˆͺ{𝑃}. It is now clear that 𝑅 can admit only a finite number of maximal 𝑁-primes of (0).
(i)(b) If 𝑃 is any minimal prime ideal of 𝑅, then π‘ƒβŠ†π‘(𝑅) [22, Theorem 84]. Since distinct minimal prime ideals of 𝑅 are not comparable under inclusion, it follows using the same arguments as in the proof of (i)(a) that 𝑅 can admit only a finite number of minimal prime ideals.
(ii)(a) Let 𝑃, 𝐴 be as in the proof of (i)(a). Let πœ”((Ξ“(𝑅))𝑐)=𝑛. Now using the same arguments as in the proof of (i)(a), it follows from Lemma 4.2(ii) that |𝐴|≀𝑛. Since the set of all maximal 𝑁-primes of (0) in 𝑅=𝐴βˆͺ{𝑃}, we obtain that 𝑅 can admit at most 𝑛+1 maximal 𝑁-primes of (0).
Suppose that 𝑅 admits exactly π‘˜ maximal 𝑁-primes of (0) with π‘˜β‰₯3. Let {𝑃1,𝑃2,𝑃3,…,π‘ƒπ‘˜} be the set of all maximal 𝑁-primes of (0) in 𝑅. Note that 𝑍(𝑅)=βˆͺπ‘˜π‘–=1𝑃𝑖. Since distinct maximal 𝑁-primes of (0) in 𝑅 are not comparable under the inclusion relation, it follows from [21, Proposition 1.11(i)] that for each π‘–βˆˆ{1,2,3,…,π‘˜}, βˆƒπ‘₯π‘–βˆˆπ‘ƒπ‘–β‹ƒβ§΅(π‘—βˆˆ{1,…,π‘˜}⧡{𝑖}𝑃𝑗). Then it is clear that for any distinct 𝑖,π‘—βˆˆ{1,2,3,…,π‘˜}, π‘₯𝑖 and π‘₯𝑗 are distinct nonzero zero-divisors of 𝑅, and as π‘˜β‰₯3, it follows that there exists at least one π‘‘βˆˆ{1,2,3,…,π‘˜} such that both π‘₯𝑖 and π‘₯𝑗 are not in 𝑃𝑑 and hence π‘₯𝑖π‘₯𝑗≠0. Thus we obtain that the subgraph of (Ξ“(𝑅))𝑐 induced on {π‘₯1,π‘₯2,π‘₯3,…,π‘₯π‘˜} is a clique and so π‘˜β‰€πœ”((Ξ“(𝑅))𝑐).
(ii)(b) This can be proved using similar arguments as in the proof of (ii)(a) and using Lemma 4.2(ii).

The following proposition is one among the main results in this section. We show in this proposition that if a ring 𝑅 admits at least two maximal 𝑁-primes of (0), then πœ”((Ξ“(𝑅))𝑐) is finite if and only if 𝑅 is finite.

Proposition 4.4. Let 𝑅 be a commutative ring with identity. Suppose that 𝑅 has at least two maximal 𝑁-primes of (0). Then the following conditions are equivalent:
(i)β€‰β€‰πœ”((Ξ“(𝑅))𝑐) is finite.
(ii)𝑅 is finite.
(iii)(Ξ“(𝑅))𝑐 does not contain any infinite clique.

Proof. (i) β‡’ (ii) Let πœ”((Ξ“(𝑅))𝑐)=𝑛. Now it follows from Corollary 4.3(ii)(a), that 𝑅 can admit at most 𝑛+1 maximal 𝑁<