Abstract

Let R be a commutative ring with identity admitting at least two nonzero zero-divisors. Let (Γ(𝑅))𝑐 denote the complement of the zero-divisor graph Γ(𝑅) of 𝑅. It is shown that if (Γ(𝑅))𝑐 is connected, then its radius is equal to 2 and we also determine the center of (Γ(𝑅))𝑐. It is proved that if (Γ(𝑅))𝑐 is connected, then its girth is equal to 3, and we also discuss about its girth in the case when (Γ(𝑅))𝑐 is not connected. We discuss about the cliques in (Γ(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 Γ(𝑅), is the graph whose vertex set is the set of all nonzero zero-divisors of 𝑅 and distinct vertices 𝑥, 𝑦 are joined by an edge in this graph if and only if 𝑥𝑦=0. Several researchers studied the zero-divisor graphs of commutative rings and proved several interesting and inspiring theorems in this area [114]. 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) 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 𝑥, 𝑦 of 𝐺 with 𝑥𝑦, 𝑑(𝑥,𝑦) is the length of a shortest path in 𝐺 from 𝑥 to 𝑦 and 𝑑(𝑥,𝑥)=0 and the diameter of 𝐺 is defined as sup{𝑑(𝑥,𝑦)𝑥and𝑦areverticesof𝐺} and it is denoted by diam(𝐺).

For any 𝑣𝑉, the eccentricity of 𝑣 denoted by 𝑒(𝑣) is defined as𝑒(𝑣)=sup{𝑑(𝑣,𝑢)𝑢𝑉}.(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 𝑟(𝐺).

It is known that for any commutative ring 𝑅 with identity which is not an integral domain, Γ(𝑅) is connected and diam(Γ(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, 𝑟(Γ(𝑅))2. Moreover, in Section 3 of [13] Redmond determined the center of Γ(𝑅) for any Artinian ring 𝑅. It is known that there are rings 𝑅 for which 𝑟(Γ(𝑅))=3 [8, Corollary 1.6]. In [14, Theorem 2.4], Karim Samei characterized vertices 𝑥 of Γ(𝑅) such that 𝑒(𝑥)=1 where 𝑅 is a reduced ring. Furthermore, in the same theorem under some additional hypotheses on 𝑅, he described vertices 𝑥 of Γ(𝑅) such that 𝑒(𝑥)=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 𝑢,𝑣𝑉 are joined by an edge in 𝐺𝑐 if and only if there exists no edge in 𝐺 joining 𝑢,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,𝐼𝑅. 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)={𝑥𝑅𝑥𝑦𝐼forsome𝑥𝑅𝐼} [16]. It is well known that if {𝑃𝛼}𝛼Λ is the set of all maximal 𝑁-primes of (0) in 𝑅, then𝑍(𝑅)=𝛼Λ𝑃𝛼.(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=(𝐼𝑅𝑥) for some 𝑥𝑅 [17]. In this case, we say that 𝑃 is a 𝐵-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 (Γ(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 𝐵-prime of (0) in 𝑅.(b)𝑅 has exactly two maximal 𝑁-primes 𝑃1, 𝑃2 of (0) with 𝑃1𝑃2(0).(c)𝑅 has more than two maximal 𝑁-primes of (0).

Moreover, it was shown in [18] that if (Γ(𝑅))𝑐 is connected, then diam((Γ(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((Γ(𝑅))𝑐)=2. When the condition (b) of [18, Theorem 1.1] holds, then it was shown in [18, Proposition 1.7] that diam((Γ(𝑅))𝑐)=2 if either 𝑃1 is not a 𝐵-prime of (0) in 𝑅 or 𝑃2 is not a 𝐵-prime of (0) in 𝑅 and diam((Γ(𝑅))𝑐)=3 if both 𝑃1 and 𝑃2 are 𝐵-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 𝐵 and 𝐴𝐵, we denote it symbolically by 𝐴𝐵. If 𝑋,𝑌 are sets and if 𝑋 is not a subset of 𝑌, we denote it symbolically 𝑋𝑌.

Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)|2. If (Γ(𝑅))𝑐 is connected, then we prove in Section 2 of this note that the radius of (Γ(𝑅))𝑐 is equal to 2. Moreover, we observe that except in the case when condition (b) of [18, Theorem 1.1] holds, diam((Γ(𝑅))𝑐)=𝑟((Γ(𝑅))𝑐)=2 and so every vertex of (Γ(𝑅))𝑐 is in the center of (Γ(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 𝐵-prime of (0) in 𝑅 or 𝑃2 is not a 𝐵-prime of (0) in 𝑅, then it is noted that each vertex of (Γ(𝑅))𝑐 is in the center of (Γ(𝑅))𝑐 and if (ii) both 𝑃1 and 𝑃2 are 𝐵-primes of (0) in 𝑅, then it is verified that the center of (Γ(𝑅))𝑐={𝑥𝑍(𝑅)𝑃1((0)𝑅𝑥)and𝑃2((0)𝑅𝑥)}.

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 Γ(𝑅) [5, 7]. Indeed, it is known that for any commutative ring with identity which is not an integral domain, gr(Γ(𝑅))4 if Γ(𝑅) contains a cycle [7, Proposition 2.2] and [12, 1.4]. In [5], Anderson and Mulay characterized commutative rings 𝑅 such that gr(Γ(𝑅))=4 [5, Theorem 2.2 and Theorem 2.3], and moreover, they characterized commutative rings 𝑅 such that gr(Γ(𝑅))= [5, Theorem 2.4 and Theorem 2.5].

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

Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝑃2(0) and if (Γ(𝑅))𝑐 is not connected, then it is proved that (Γ(𝑅))𝑐 contains a cycle if and only if |𝑃|5 if and only if gr((Γ(𝑅))𝑐)=3.

Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and let them be 𝑃1 and 𝑃2. If (Γ(𝑅))𝑐 is not connected, then it is shown that (Γ(𝑅))𝑐 contains a cycle if and only if {𝑃1𝑃23or|𝑃2𝑃1|3ifandonlyifgr((Γ(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 𝜔(𝐺) is defined as the largest integer 𝑛1 such that 𝐺 contains a clique on 𝑛 vertices [15, Definition, Page 185]. We set 𝜔(𝐺)= if 𝐺 contains a clique on 𝑛 vertices for all 𝑛1.

Let 𝑅 be a commutative ring with identity which is not an integral domain. It is known that (i) 𝜔(Γ(𝑅))= if and only if Γ(𝑅) contains an infinite clique, (ii) 𝜔(Γ(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 𝜔(Γ(𝑅)) in [3, Section 3].

Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)|2. In Section 4 of this paper we observe that if 𝑅 has at least two maximal 𝑁-primes of (0), then (Γ(𝑅))𝑐 does not contain any infinite clique if and only if 𝑅 is finite if and only if 𝜔((Γ(𝑅))𝑐) is finite. Let 𝑛2 and let 𝐾1,,𝐾𝑛 be finite fields. Let 𝑅=𝐾1××𝐾𝑛. We describe a method of determining 𝜔((Γ(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 (Γ(𝑅))𝑐 does not contain any infinite clique. Then it is verified in Section 4 that 𝑃=nil(𝑅). Moreover, if 𝑃2(0), then it is shown in Section 4 that 𝑃 is a 𝐵-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 (Γ(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 𝑃2(0), then we provide a formula for computing 𝜔((Γ(𝑅))𝑐).

Let 𝑅 be a ring with |𝑍(𝑅)|2. Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝜔((Γ(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 𝜔((Γ(𝑅))𝑐)=3.

2. The Radius of (Γ(𝑅))𝑐

Let 𝑅 be a ring with |𝑍(𝑅)|2. We assume that (Γ(𝑅))𝑐 is connected. The aim of this section is to show that the radius of (Γ(𝑅))𝑐 is equal to 2. We make use of the following lemmas for proving the result that 𝑟((Γ(𝑅))𝑐)=2.

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

Proof. Let 𝑥𝑉. Since |𝑉|2 and 𝐺 is connected, it follows that there exists 𝑦𝑉 such that 𝑥 is adjacent to 𝑦 in 𝐺. Thus 𝑥 is not adjacent to 𝑦 in 𝐺𝑐. Hence 𝑑(𝑥,𝑦)2in𝐺𝑐. This proves that 𝑒(𝑥)2in𝐺𝑐.

The following lemma establishes some necessary conditions on 𝑅 in order that there exist 𝑥,𝑦𝑍(𝑅) such that 𝑑(𝑥,𝑦)=3in(Γ(𝑅))𝑐.

Lemma 2.2. Let 𝑅 be a ring with |𝑍(𝑅)|2. Suppose that (Γ(𝑅))𝑐 is connected. If there exist 𝑥,𝑦𝑍(𝑅) with 𝑑(𝑥,𝑦)=3in(Γ(𝑅))𝑐, then 𝑅 has exactly two maximal 𝑁-primes of (0) and moreover, both of them are 𝐵-primes of (0) in 𝑅. Indeed, if 𝑃 and 𝑄 are the maximal 𝑁-primes of (0) in 𝑅, then {𝑃,𝑄}={(0)𝑅𝑥),((0)𝑅𝑦)}.

Proof. Since 𝑑(𝑥,𝑦)=3in(Γ(𝑅))𝑐, it follows that 𝑥𝑦=0 and for any 𝑧𝑍(𝑅) with 𝑧{𝑥,𝑦}, either 𝑧𝑥=0 or 𝑧𝑦=0. Hence we obtain that 𝑧((0)𝑅𝑥)((0)𝑅𝑦). Since 𝑥𝑦=0, it is clear that {0,𝑥,𝑦}((0)𝑅𝑥)((0)𝑅𝑦). Thus we have 𝑍(𝑅)=((0)𝑅𝑥)((0)𝑅𝑦). Let {𝑃𝛼}𝛼Λ be the set of all maximal 𝑁-primes of (0) in 𝑅. It is well known that 𝑍(𝑅)=𝛼Λ𝑃𝛼, and hence we obtain that 𝛼Λ𝑃𝛼=((0)𝑅𝑥)((0)𝑅𝑦). Now ((0)𝑅𝑥)(𝑅𝑍(𝑅))=. 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)𝑅𝑥)𝑃. Since ((0)𝑅𝑦)(𝑅𝑍(𝑅))=, it follows that there exists a maximal 𝑁-prime 𝑄 of (0) in 𝑅 such that ((0)𝑅𝑦)𝑄. Now we obtain that 𝑍(𝑅)=((0)𝑅𝑥)((0)𝑅𝑦)=𝑃𝑄. If 𝑃=𝑄, then it follows that ((0)𝑅𝑥)((0)𝑅𝑦)=𝑃 is an ideal of 𝑅. Hence it follows that either ((0)𝑅𝑥)((0)𝑅𝑦)or((0)𝑅𝑦)((0)𝑅𝑥), and so we obtain that 𝑃 is the only maximal 𝑁-prime of (0) in 𝑅 and either 𝑃=((0)𝑅𝑥)or𝑃=((0)𝑅𝑦). Now by hypothesis (Γ(𝑅))𝑐 is connected. Hence it follows from [18, Theorem 1.1 (a)] that 𝑃 is not a 𝐵-prime of (0) in 𝑅 and so 𝑃𝑄. Now we obtain from 𝑍(𝑅)=((0)𝑅𝑥)((0)𝑅𝑦)=𝑃𝑄 that 𝑃 and 𝑄 are the only maximal 𝑁-primes of (0) in 𝑅, and moreover, 𝑃=((0)𝑅𝑥)and𝑄=((0)𝑅𝑦).

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 𝐵-prime ideals of (0) in 𝑅 with 𝑃=((0)𝑅𝑥) and 𝑄=((0)𝑅𝑦) for some 𝑥,𝑦𝑅. Then 𝑥𝑦=0.

Proof. For the sake of completeness, we give below an argument for the fact that xy = 0. Since P Q. either 𝑃𝑄 or 𝑄𝑃. Without loss of generality, we may assume that 𝑃𝑄. Let w 𝑃𝑄. Now 𝑤𝑥=0𝑄 and as 𝑤𝑄, it follows that 𝑥𝑄=((0)𝑅𝑦). Hence 𝑥𝑦=0.

We provide in the next lemma some sufficient conditions on 𝑅 in order that (Γ(𝑅))𝑐 admits vertices x such that 𝑒(𝑥)=3 in (Γ(𝑅))𝑐.

Lemma 2.4. Let 𝑅 be a ring and let |𝑍(𝑅)|2. Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and let them be 𝑃1 and 𝑃2. If (Γ(𝑅))𝑐 is connected and if 𝑃1=((0)𝑅𝑢) and 𝑃2=((0)𝑅𝑣) for some 𝑢,𝑣𝑅, then 𝑑(𝑢,𝑣)=3 in (Γ(𝑅))𝑐 and so 𝑒(𝑢)=𝑒(𝑣)=3 in (Γ(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 𝑢𝑣=0. Thus 𝑢,𝑣𝑍(𝑅) and 𝑢 and 𝑣 are not adjacent in (Γ(𝑅))𝑐. Since 𝑃1 and 𝑃2 are the only maximal 𝑁-primes of (0) in 𝑅, it follows that 𝑍(𝑅)=𝑃1𝑃2=((0)𝑅𝑢)((0)𝑅𝑣). Thus for any 𝑤𝑍(𝑅), either 𝑤𝑢=0or𝑤𝑣=0. This shows that 𝑑(𝑢,𝑣)3in(Γ(𝑅))𝑐. It is shown in the proof of [18,Proposition 1.7] that 𝑑(𝑥,𝑦)3in(Γ(𝑅))𝑐 for any 𝑥,𝑦𝑍(𝑅). Hence we obtain that𝑑(𝑢,𝑣)=3in(Γ(𝑅))𝑐 and so it follows that 𝑒(𝑢)=𝑒(𝑣)=3in(Γ(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)|2. If (Γ(𝑅))𝑐 is connected, then the radius of (Γ(𝑅))𝑐 is equal to 2.

Proof. It is well known that Γ(𝑅) is connected [1, Theorem 2.3]. Hence it follows from Lemma 2.1 that for any 𝑥𝑍(𝑅),𝑒(𝑥)2in(Γ(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 𝑥𝑍(𝑅),𝑒(𝑥)2in(Γ(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 any𝑥𝑍(𝑅),𝑒(𝑥)=2in(Γ(𝑅))𝑐. Hence we obtain in the cases mentioned above that diam((Γ(𝑅))𝑐)=𝑟((Γ(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 exist𝑎𝑃1𝑃2 and 𝑏𝑃2𝑃1 such that 𝑎𝑏0. It follows that 𝑃1((0)𝑅𝑎),𝑃1((0)𝑅𝑏),and𝑃2((0)𝑅𝑎),𝑃2((0)𝑅𝑏). Now it follows from Lemma 2.2 that 𝑑(𝑎,𝑦)2 and 𝑑(𝑏,𝑦)2 for any 𝑦𝑍(𝑅) in (Γ(𝑅))𝑐. Hence we obtain that 𝑒(𝑎)2 and 𝑒(𝑏)2 in (Γ(𝑅))𝑐. Since 𝑒(𝑐)2in(Γ(𝑅))𝑐 for any 𝑐𝑍(𝑅), it follows that 𝑒(𝑎)=𝑒(𝑏)=2in(Γ(𝑅))𝑐. Thus in this case also, we arrive at the conclusion that the radius of (Γ(𝑅))𝑐 is equal to 2.

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

Remark 2.6. Let 𝑅 be a ring and let |𝑍(𝑅)|2. Assume that (Γ(𝑅))𝑐 is connected. In this remark, we determine the center of (Γ(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((Γ(𝑅))𝑐)=𝑟((Γ(𝑅))𝑐)=2. Hence the center of (Γ(𝑅))𝑐 is the set of all vertices of (Γ(𝑅))𝑐. Moreover, if 𝑅 has exactly two maximal 𝑁-primes of (0) and if at least one of them is not a 𝐵-prime of (0) in 𝑅, then it follows from Lemma 2.2 that for any 𝑥𝑍(𝑅),𝑒(𝑥)2in(Γ(𝑅))𝑐. Thus we obtain, in view of Lemma 2.1 that 𝑒(𝑥)=2for any𝑥𝑍(𝑅). Hence in this case also we obtain that diam((Γ(𝑅))𝑐)=𝑟((Γ(𝑅))𝑐)=2 and so each vertex of (Γ(𝑅))𝑐 is in the center of (Γ(𝑅))𝑐.
Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0) and both are 𝐵-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 (Γ(𝑅))𝑐={𝑥𝑍(𝑅)𝑃1((0)𝑅𝑥)and𝑃2((0)𝑅𝑥)}.

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 𝑥𝑀,𝑥0. Let 𝑅=𝑉/𝑥𝑉. Observe that 𝑀/𝑥𝑉 is the only prime ideal of 𝑅 and 𝑍(𝑅)=𝑀/𝑥𝑉. Let us denote 𝑀/𝑥𝑉 by 𝑃. We assert that 𝑃 is not a 𝐵-prime of (0) in 𝑅. Suppose that 𝑃 is a 𝐵-prime of (0) in 𝑅. Then it can be easily verified that 𝑀=(𝑥𝑉𝑉𝑦) for some 𝑦𝑉. Since 𝑀𝑉, it follows that 𝑦𝑥𝑉. As 𝑉 is a valuation domain, we obtain that 𝑥𝑦𝑉. Thus 𝑥=𝑦𝑣 for some 𝑣𝑉. Hence we obtain that 𝑀=(𝑥𝑉𝑉𝑦)=(𝑦𝑣𝑉𝑉𝑦)=𝑣𝑉. This is impossible since 𝑀 is not finitely generated. Thus 𝑃 is not a 𝐵-prime of (0) in 𝑅. Now it follows from [18, Theorem 1.1 (a)] that (Γ(𝑅))𝑐 is connected. We obtain from Theorem 2.5 that diam((Γ(𝑅))𝑐)=𝑟((Γ(𝑅))𝑐)=2 and each vertex of (Γ(𝑅))𝑐 is in the center of (Γ(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 𝐵-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 𝑃1𝑃2=𝑃×(0) is not the zero-ideal of 𝑇 (resp. 𝑄1𝑄2=𝑃×𝑃 is not the zero-ideal of 𝑇1). Hence it follows from [18, Theorem 1.1(b)] that (Γ(𝑇))𝑐 (resp. (Γ(𝑇1))𝑐) is connected. Since 𝑃1 is not a 𝐵-prime of (0) in 𝑇 (resp. 𝑄1 and 𝑄2 are not 𝐵-primes of (0) in 𝑇1), it follows from Remark 2.6 that diam((Γ(𝑇))𝑐)=𝑟((Γ(𝑇))𝑐)=2 (resp. diam((Γ(𝑇1))𝑐)=𝑟((Γ(𝑇1))𝑐)=2) and each vertex of (Γ(𝑇))𝑐 is in the center of (Γ(𝑇))𝑐 (resp. each vertex of (Γ(T1))c is in the center of (Γ(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)=𝑃1𝑃2. Observe that 𝑅 has exactly two maximal 𝑁-primes of (0) and 𝑃1𝑃2=6𝑍/12𝑍 is not the zero-ideal of 𝑅. Hence it follows from [18, Theorem 1.1(b)] that (Γ(𝑅))𝑐 is connected. Moreover, observe that 2𝑍=(12𝑍𝑍±6𝑡) and {±6𝑡𝑡𝑍isoddandpositive} is the set of all integers with the property that 2𝑍=(12𝑍𝑍±6𝑡). Furthermore, note that {±4𝑘𝑘Zispositiveand𝑘1or2(mod3)} is the set of all integers with the property that 3𝑍=(12𝑍𝑍±4𝑘). Hence it follows that 𝑃1=((0+12𝑍)𝑅6+12𝑍), 𝑃2=((0+12Z)𝑅4+12𝑍),and𝑃2=((0+12𝑍)𝑅8+12𝑍). Thus 𝑃1 and 𝑃2 are 𝐵-primes of (0) in 𝑅. Now it follows from [18, Proposition 1.7(b)] that diam((Γ(𝑅))𝑐)=3, and we know from Theorem 2.5 that 𝑟((Γ(𝑅))𝑐)=2. Moreover, we obtain from Remark 2.6 and from the above discussion that the set of centers of (Γ(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,,𝑝𝑡}(𝑡3) be the set of all distinct prime divisors of 𝑛. Let 𝑅=𝑍/𝑛𝑍. Note that {𝑝1𝑍/𝑛𝑍,𝑝2𝑍/𝑛𝑍,,𝑝𝑡𝑍/𝑛𝑍} is the set of all maximal 𝑁-primes of the zero-ideal of 𝑅. We know from [18, Theorem 1.1(c)] that (Γ(𝑅))𝑐 is connected and moreover, it is known from Remark 2.6 that diam((Γ(𝑅))𝑐)=𝑟((Γ(𝑅))𝑐)=2 and each vertex of (Γ(𝑅))𝑐 is in the center of (Γ(𝑅))𝑐.

3. The Girth of (Γ(𝑅))𝑐

Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)|2. The aim of this section is to study about the girth of (Γ(𝑅))𝑐. If (Γ(𝑅))𝑐 is connected, then we prove in this section that gr((Γ(𝑅))𝑐)=3. Moreover, we also discuss about the girth of (Γ(𝑅))𝑐 in the case when (Γ(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)|2. Let 𝑃 be a maximal 𝑁-prime of (0) in 𝑅. If 𝑃 is not a 𝐵-prime of (0) in 𝑅, then for any 𝑥,𝑦𝑍(𝑅), 𝑃((0)𝑅𝑥)𝑅𝑦.

Proof. Suppose that for some 𝑥,𝑦𝑍(𝑅), 𝑃((0)𝑅𝑥)𝑅𝑦. Then either 𝑃((0)𝑅𝑥)or𝑃𝑅𝑦. Since 𝑥0, ((0)𝑅𝑥)(𝑅𝑍(𝑅))=. As 𝑦𝑍(𝑅), there exists 𝑧𝑅{0} such that 𝑦𝑧=0. Hence we obtain that 𝑅𝑦((0)𝑅𝑧). Note that ((0)𝑅𝑧)(𝑅𝑍(𝑅))=. 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)𝑅𝑥)𝑄 and there exists a maximal 𝑁-prime 𝑊 of (0) in 𝑅 such that ((0)𝑅𝑧)𝑊. If 𝑃((0)𝑅𝑥), then we obtain that 𝑃((0)𝑅𝑥)𝑄 and hence it follows that 𝑃=𝑄=((0)𝑅𝑥). This contradicts the assumption that 𝑃 is not a 𝐵-prime of (0) in 𝑅. If 𝑃𝑅𝑦, then 𝑃((0)𝑅𝑧)𝑊 and so 𝑃=𝑊=((0)𝑅𝑧). This is also impossible since 𝑃 is not a 𝐵-prime of (0) in 𝑅. This proves that if a maximal 𝑁-prime 𝑃 of (0) in 𝑅 is not a 𝐵-prime of (0) in 𝑅, then for any 𝑥,𝑦𝑍(𝑅), 𝑃((0)𝑅𝑥)𝑅𝑦.

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

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

Proof. Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. Since (Γ(𝑅))𝑐 is connected, we obtain from [18, Theorem 1.1(a)] that 𝑃 is not a 𝐵-prime of (0) in 𝑅. Note that 𝑍(𝑅)=𝑃. Let 𝑥𝑃{0}. By hypothesis, 𝑃 is not a 𝐵-prime of (0) in 𝑅 and so Lemma 3.1 implies that there exists 𝑦𝑃 such that 𝑦𝑅𝑥 and 𝑦𝑥0. We assert that 𝑥𝑦{𝑥,𝑦}. If 𝑥𝑦=𝑥, then 𝑥(1𝑦)=0. As 𝑦𝑃, 1𝑦𝑃=𝑍(𝑅). Hence 𝑥(1𝑦)=0 implies that 𝑥=0. This contradicts the fact that 𝑥0. Similarly, it follows that 𝑥𝑦𝑦. If both 𝑥2𝑦 and 𝑦2𝑥 are nonzero, then we obtain that 𝑥---𝑥𝑦---𝑦---𝑥 is a cycle of length 3 in (Γ(𝑅))𝑐. Suppose that either 𝑥2𝑦=0 or 𝑦2𝑥=0. Without loss of generality, we may assume that 𝑥2𝑦=0. As 𝑃 is not a 𝐵-prime of (0) in 𝑅, Lemma 3.1 implies that there exists 𝑧𝑃 such that 𝑧𝑥𝑦0 and 𝑧𝑅𝑦. From 𝑧𝑥𝑦0, it follows that 𝑧𝑥0, and 𝑧𝑦0 and moreover, as 𝑥2𝑦=0, it follows that 𝑧𝑥. Furthermore, by the choice of 𝑧, it follows that 𝑧𝑦. Now 𝑥---𝑧---𝑦---𝑥 is a cycle of length 3 in (Γ(𝑅))𝑐. This shows that if 𝑅 has only one maximal 𝑁-prime of (0) and if (Γ(𝑅))𝑐 is connected, then there exists a cycle of length 3 in (Γ(𝑅))𝑐 and hence gr((Γ(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 𝑎,𝑏,𝑐𝑍(𝑅)𝑃 for some prime ideal 𝑃 of 𝑅, then gr((Γ(𝑅))𝑐)=3.

Proof. As 𝑃 is a prime ideal of 𝑅 and 𝑎,𝑏,𝑐 are elements of 𝑅 which are not in 𝑃, we obtain that 𝑎𝑏,𝑏𝑐,𝑐𝑎𝑅𝑃 and so 𝑎𝑏,𝑏𝑐,𝑐𝑎𝑅{0}. Hence it follows that 𝑎---𝑏---𝑐---𝑎 is a cycle of length 3 in (Γ(𝑅))𝑐. This proves that gr((Γ(𝑅))𝑐)=3.

The next lemma discusses the girth of (Γ(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 𝑃1𝑃2=(0), then (Γ(𝑅))𝑐 contains a cycle if and only if either |𝑃1𝑃2|3 or |𝑃2𝑃1|3 if and only if gr((Γ(𝑅))𝑐)=3. (ii)If 𝑃1𝑃2(0), then gr((Γ(𝑅))𝑐)=3. Thus, if (Γ(𝑅))𝑐 is connected, then gr((Γ(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)=𝑃1𝑃2.
(i) Assume that (Γ(𝑅))𝑐 contains a cycle. Let 𝑎1---𝑎2---𝑎3------𝑎𝑛---𝑎1 be a cycle of length 𝑛 in (Γ(𝑅))𝑐. Note that 𝑛3 and {𝑎𝑖𝑖=1,2,3,,𝑛}𝑍(𝑅) with 𝑎1𝑎2,𝑎2𝑎3,,𝑎𝑛1𝑎𝑛,𝑎𝑛𝑎1𝑅{0}. Since 𝑃1𝑃2=(0), it follows that either {𝑎1,𝑎2,𝑎3,,𝑎𝑛}𝑃1𝑃2or{𝑎1,𝑎2,𝑎3,,𝑎𝑛}𝑃2𝑃1. Now it is clear that either |𝑃1𝑃2|3 or |𝑃2𝑃1|3.
Conversely, suppose that either |𝑃1𝑃2|3 or |𝑃2𝑃1|3. Since 𝑍(𝑅)=𝑃1𝑃2 and as 𝑃1 and 𝑃2 are prime ideals of 𝑅, it follows from Lemma 3.3 that gr((Γ(𝑅))𝑐)=3.
If gr((Γ(𝑅))𝑐)=3, then we obtain that (Γ(𝑅))𝑐 contains a cycle of length 3.
(ii) Suppose that 𝑃1𝑃2(0). We know from the proof of [18, Proposition 1.7(i)] that there exist 𝑎𝑃1𝑃2 and 𝑏𝑃2𝑃1 such that 𝑎𝑏0. We consider two cases : Case(A): 𝑃1𝑃2((0)𝑅𝑎)((0)𝑅𝑏). Then there exists 𝑐𝑃1𝑃2 such that 𝑎𝑐0 and 𝑏𝑐0. Hence we obtain a cycle 𝑎---𝑏---𝑐---𝑎 in (Γ(𝑅))𝑐 and it is of length 3. Case(B): 𝑃1𝑃2((0)𝑅𝑎)((0)𝑅𝑏). Then either 𝑃1𝑃2((0)𝑅𝑎) or 𝑃1𝑃2((0)𝑅𝑏).
Without loss of generality, we may assume that 𝑃1𝑃2((0)𝑅𝑎). Let 𝑥𝑃1𝑃2 be such that 𝑥0. Since 𝑎+𝑏𝑍(𝑅), we obtain that (𝑎+𝑏)𝑥0. As 𝑎𝑥=0, it follows that 𝑏𝑥=(𝑎+𝑏)𝑥0. Moreover, observe that 𝑏+𝑥𝑃2𝑃1,𝑏𝑏+𝑥, and 𝑎(𝑏+𝑥)=𝑎𝑏0. As 𝑏(𝑏+𝑥)𝑃2𝑃1, it follows that 𝑏(𝑏+𝑥)0. Note that 𝑎---𝑏---𝑏+𝑥---𝑎 is a cycle of length 3 in (Γ(𝑅))𝑐.
Thus in both the cases, (Γ(𝑅))𝑐 admits a cycle of length 3. Hence we obtain that gr((Γ(𝑅))𝑐)=3.
We know from [18, Theorem 1.1(b)] that (Γ(𝑅))𝑐 is connected if and only if 𝑃1𝑃2(0). Thus if 𝑅 has exactly two maximal 𝑁-primes of (0) and if (Γ(𝑅))𝑐 is connected, then gr((Γ(𝑅))𝑐)=3.

We show in the next lemma that if 𝑅 has at least three maximal 𝑁-primes of (0), then gr((Γ(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((Γ(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,𝑊} be a subset of the set of all maximal 𝑁-primes of (0) in 𝑅. It is clear that 𝑃𝑄𝑊, 𝑄𝑃W, and 𝑊𝑃𝑄. Hence there exist elements 𝑥𝑃(𝑄𝑊), 𝑦𝑄(𝑃𝑊), and 𝑧𝑊(𝑃𝑄). Note that 𝑥,𝑦,𝑧 are distinct elements of 𝑍(𝑅) with 𝑥𝑦,𝑦𝑧,𝑧𝑥𝑅{0}. Hence 𝑥---𝑦---𝑧---𝑥 is a cycle of length 3 in (Γ(𝑅))𝑐. This proves that gr((Γ(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)|2. If (Γ(𝑅))𝑐 is connected, then gr((Γ(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 (Γ(𝑅))𝑐 is not connected and discuss about the girth of (Γ(𝑅))𝑐. In this direction, we first have the following proposition.

Proposition 3.7. Let 𝑅 be a ring and let |𝑍(𝑅)|2. Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝑃2 (0) and if (Γ(𝑅))𝑐 is not connected, then (Γ(𝑅))𝑐 contains a cycle if and only if |𝑃|5 if and only if gr((Γ(𝑅))𝑐)=3.

Proof. Note that 𝑍(𝑅)=𝑃. Assume that 𝑃2(0) and (Γ(𝑅))𝑐 is not connected. From the assumption that 𝑃2(0), it follows that there exist 𝑎,𝑏𝑃 such that 𝑎𝑏 and 𝑎𝑏0. Moreover, by [18, Theorem 1.1(a)], 𝑃=((0)𝑅𝑐) for some 𝑐𝑅. It is clear that 𝑐𝑃{0}. Furthermore, as 𝑎𝑏0, it follows that 𝑎𝑐 and 𝑏𝑐.
Suppose that (Γ(𝑅))𝑐 contains a cycle. Let 𝑎1---𝑎2---𝑎3------𝑎𝑛---𝑎1 be a cycle of length 𝑛 in (Γ(𝑅))𝑐. Note that 𝑛3 and {𝑎1,𝑎2,𝑎3,,𝑎𝑛}𝑃{0}. As 𝑃=((0)𝑅𝑐) and since 𝑎1𝑎2,𝑎2𝑎3,,𝑎𝑛1𝑎𝑛,and𝑎𝑛𝑎1𝑅{0}, it follows that 𝑐{𝑎1,𝑎2,𝑎3,,𝑎𝑛}. Thus {0,𝑐,𝑎1,𝑎2,𝑎3,,𝑎𝑛}𝑃 and hence it follows that |𝑃|5.
We next show that gr((Γ(𝑅))𝑐)=3 if |𝑃|5. Suppose that |𝑃|5. Observe that there exists 𝑑𝑃{0,𝑎,𝑏,𝑐} where 𝑎,𝑏,𝑐 are as in the first paragraph of this proof. We claim that there exists a cycle of length 3 in (Γ(𝑅))𝑐. If 𝑎𝑑0 and 𝑏𝑑0, then 𝑎---𝑑---𝑏---𝑎 is a cycle of length 3 in (Γ(𝑅))𝑐. Suppose that either 𝑎𝑑=0 or 𝑏𝑑=0.
Observe that if 𝑎2=0 and 𝑏2=0, then from (𝑎+𝑏)𝑎=𝑎𝑏=(𝑎+𝑏)𝑏=𝑎𝑏, and from the fact that 𝑎+𝑏𝑃{0,𝑎,𝑏}, it follows that 𝑎---𝑎+𝑏---𝑏---𝑎 is a cycle of length 3 in (Γ(𝑅))𝑐. Now let us suppose without loss of generality that 𝑎20. As 𝑐2=0 and 𝑎20, it follows that 𝑎+𝑐0. If 𝑎+𝑐𝑏, then 𝑎---𝑎+𝑐---𝑏---𝑎 is a cycle of length 3 in (Γ(𝑅))𝑐. Suppose that 𝑎+𝑐=𝑏. Then, as 𝑐𝑑=0, it follows from the assumption that either 𝑎𝑑=0 or 𝑏𝑑=0 that 𝑎𝑑=𝑏𝑑=0. As 𝑐𝑑 and 𝑎+𝑐=𝑏, it follows that 𝑎+𝑑𝑏 and it is clear that 𝑎(𝑎+𝑑)=𝑎2 0 and (𝑎+𝑑)𝑏=𝑎𝑏0. Hence 𝑎---𝑎+𝑑---𝑏---𝑎 is a cycle of length 3 in (Γ(𝑅))𝑐.
Thus if |𝑃|5, then it is shown that there exists a cycle of length 3 in (Γ(𝑅))𝑐. Hence gr((Γ(𝑅))𝑐)=3.
If gr((Γ(𝑅))𝑐)=3, then (Γ(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) (Γ(𝑅))𝑐 contains at least one edge, (iii) (Γ(𝑅))𝑐 is not connected, and (iv) (Γ(𝑅))𝑐 does not contain any cycle.

Remark 3.8. Let 𝑅 be a ring and let |𝑍(𝑅)|2. Suppose that 𝑃 is the only maximal 𝑁-prime of (0) in 𝑅. Assume that 𝑃2(0). Suppose that (Γ(𝑅))𝑐 is not connected. Let 𝑎,𝑏𝑃 be such that 𝑎𝑏 and 𝑎𝑏0. Let 𝑐𝑃{0} be such that 𝑃=((0)𝑅𝑐). Assume that gr((Γ(𝑅))𝑐)3. Observe that by Proposition 3.7, gr((Γ(𝑅))𝑐)3 if and only if (Γ(𝑅))𝑐 does not contain any cycle. Now it follows from Proposition 3.7 that |𝑃|4 and as {0,𝑎,𝑏,𝑐}𝑃, it follows that 𝑃={0,𝑎,𝑏,𝑐}. Now 𝑎𝑏𝑃{0}={𝑎,𝑏,𝑐}. We assert that 𝑎𝑏=𝑐. If 𝑎𝑏=𝑎, then 𝑎(1𝑏)=0. This is impossible since 𝑎0 and 1𝑏𝑃=𝑍(𝑅). Similarly, it follows that 𝑎𝑏𝑏. Hence 𝑎𝑏=𝑐. Now 𝑍(𝑅)=𝑃={0,𝑎,𝑏,𝑐=𝑎𝑏} 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[𝑥]/(2𝑥𝑍4[𝑥]+(𝑥22)𝑍4[𝑥]),𝑍2[𝑥]/𝑥3𝑍2[𝑥]} where 𝑍4[𝑥] (resp. 𝑍2[𝑥]) 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 𝑄𝑍(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 𝑅𝑃 is the set of all units in 𝑅. Let u be a unit in 𝑅. Then 𝑢𝑎𝑏𝑃{0}={𝑎,𝑏,𝑐=𝑎𝑏}. We claim that 𝑢𝑎𝑏=𝑎𝑏. If 𝑢𝑎𝑏=𝑎, then 𝑎(1𝑢𝑏)=0and this is impossible. Similarly, we obtain that 𝑢𝑎𝑏𝑏. Hence 𝑢𝑎𝑏=𝑎𝑏. This implies that (𝑢1)𝑎𝑏=0. Since 𝑎𝑏0, it follows that 𝑢1𝑍(𝑅)=𝑃={0,𝑎,𝑏,𝑐=𝑎𝑏}. Hence we obtain that 𝑢{1,1+𝑎,1+𝑏,1+𝑐=1+𝑎𝑏}. Thus it is shown that 𝑅𝑃={1,1+𝑎,1+𝑏,1+𝑐}. Hence 𝑅={0,𝑎,𝑏,𝑐=𝑎𝑏,1,1+𝑎,1+𝑏,1+𝑐} is a ring containing exactly 8 elements. Note that 𝜔(Γ(𝑅))=2. Now [3, Theorem 3.2] implies that 𝑅 is isomorphic to exactly one of the rings from the set {𝑍8,𝑍4[𝑥]/(2𝑥𝑍4[𝑥]+(𝑥22)𝑍4[𝑥]),𝑍2[𝑥]/𝑥3𝑍2[𝑥]}. Observe that if 𝑅 is a ring such that 𝑅{𝑍8,𝑍4[𝑥]/(2𝑥𝑍4[𝑥]+(𝑥22)𝑍4[𝑥]),𝑍2[𝑥]/𝑥3𝑍2[𝑥]}, then 𝑅 satisfies the following conditions: 𝑅 has exactly one maximal 𝑁-prime of (0), say 𝑃 such that 𝑃2(0), (Γ(𝑅))𝑐 is not connected, and (Γ(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 𝑃2(0) and (Γ(𝑅))𝑐 is not connected. The above discussion implies that (Γ(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[𝑥]/(2𝑥𝑍4[𝑥]+(𝑥22)𝑍4[𝑥]),𝑍2[𝑥]/𝑥3𝑍2[𝑥]}.

We determine in the following remark rings 𝑅 satisfying the following conditions: (i) 𝑅 admits exactly two maximal 𝑁-primes of (0), (ii) (Γ(𝑅))𝑐 contains at least one edge, (iii) (Γ(𝑅))𝑐 is not connected, and (iv) (Γ(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 𝑃1𝑃2=(0) (that is, equivalently (Γ(𝑅))𝑐 is not connected). In such a case, it is shown in Lemma 3.4(i) that (Γ(𝑅))𝑐 contains a cycle if and only if either |𝑃1𝑃2|3, or |𝑃2𝑃1|3 if and only if gr((Γ(𝑅))𝑐)=3. Suppose that |𝑃1𝑃2|<3, |𝑃2𝑃1|<3 and that (Γ(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 (Γ(𝑅))𝑐 contains at least one edge. Since 𝑍(𝑅)=𝑃1𝑃2 and 𝑃1𝑃2=(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 𝑃1𝑃2=(0), it follows from the Chinese Remainder Theorem [21, Proposition 1.10] that 𝑅(𝑅/𝑃1)×(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), (Γ(𝑅))𝑐 contains at least one edge, (Γ(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 𝑥,𝑦𝑍(𝑅) such that 𝑥𝑦0, then gr((Γ(𝑅))𝑐)=3.
(ii) Let 𝑅 be any ring admitting elements 𝑎,𝑏𝑍(𝑅) such that 𝑎𝑏0. Then gr((Γ(𝑅[𝑥]))𝑐)=gr((Γ(𝑅[[𝑥]]))𝑐)=3 where 𝑅[𝑥] (resp. 𝑅[[𝑥]]) is the polynomial (resp.the power series) ring in one variable over 𝑅.

4. Cliques in (Γ(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 (Γ(𝑅))𝑐 is finite if and only if (Γ(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 (Γ(𝑅))𝑐 is finite. Furthermore, if 𝑅 is either a Noetherian ring or a chained ring and if (Γ(𝑅))𝑐 admits at least one edge (that is, there exist 𝑥,𝑦𝑍(𝑅) with 𝑥𝑦 such that 𝑥𝑦0), then it is shown that the clique number of (Γ(𝑅))𝑐 is finite if and only if (Γ(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 (Γ(𝑅))𝑐 does not contain any infinite clique, then 𝑍(𝑅)𝑃 is finite. (ii)If 𝜔((Γ(𝑅))𝑐) is finite, then 𝑍(𝑅)𝑃 is finite and indeed, |𝑍(𝑅)𝑃|𝜔((Γ(𝑅))𝑐).

Proof. (i) Suppose that 𝑍(𝑅)𝑃 is infinite. Then we can choose an infinite sequence of distinct elements 𝑥𝑖𝑍(𝑅)𝑃. Since 𝑃 is a prime ideal of R and as 𝑥𝑖𝑃 for 𝑖=1,2,3,, it follows that 𝑥𝑖𝑥𝑗0 for all 𝑖,𝑗{1,2,3,}. Observe that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥𝑖𝑖=1,2,3,} is an infinite clique. This contradicts the assumption that (Γ(𝑅))𝑐 does not contain any infinite clique. Hence we obtain that 𝑍(𝑅)𝑃 is finite.
(ii) Let 𝜔((Γ(𝑅))𝑐)=𝑛. We assert that |𝑍(𝑅)𝑃|𝑛. Suppose that {𝑍(𝑅)𝑃𝑛+1. Let {𝑥1,,𝑥𝑛+1}𝑍(𝑅)𝑃. Then it is clear that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥1,,𝑥𝑛+1} is a clique. This is impossible since 𝜔((Γ(𝑅))𝑐)=𝑛. This shows that |𝑍(𝑅)𝑃|𝑛=𝜔((Γ(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={𝑄𝑄 is a prime ideal of 𝑅 such that 𝑄𝑍(𝑅) but 𝑄𝑃}. Then the following hold.(i)If (Γ(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 𝜔((Γ(𝑅))𝑐) is finite, then 𝐴 can admit at most 𝜔((Γ(𝑅))𝑐) elements which are pairwise incomparable under inclusion.

Proof. (i) Suppose that (Γ(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 𝑄𝑖𝐴 for each 𝑖=1,2,3, with 𝑄𝑖 and 𝑄𝑗 not being comparable under inclusion for all 𝑖,𝑗{1,2,3,} with 𝑖𝑗. Now 𝑄𝑖𝑃 for 𝑖=1,2,3,. Hence it is possible to choose 𝑥1𝑄1𝑃. Let 𝑖2. Now it follows from prime avoidance [21, Proposition 1.11(i)] that there exists 𝑥𝑖𝑄𝑖(𝑃𝑄1𝑄𝑖1). Observe that {𝑥𝑖𝑖=1,2,3,}𝑍(𝑅)𝑃. Hence we obtain that 𝑍(𝑅)𝑃 is infinite. This contradicts Lemma 4.1(i). This proves that if (Γ(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 𝜔((Γ(𝑅))𝑐)=𝑛. Suppose that 𝐴 admits more than 𝑛 elements which are pairwise incomparable under inclusion. Let {𝑄1,,𝑄𝑛+1}𝐴 be such that 𝑄𝑖 and 𝑄𝑗 are not comparable for all 𝑖,𝑗{1,2,,𝑛+1} with 𝑖𝑗. Let 𝑥1𝑄1𝑃. Let 𝑖{2,,𝑛+1}. As in (i), we can choose 𝑥𝑖𝑄𝑖(𝑃𝑄1𝑄𝑖1). Observe that {𝑥𝑖𝑖=1,2,,𝑛+1}𝑍(𝑅)𝑃. This implies that |𝑍(𝑅)𝑃|𝑛+1>𝜔((Γ(𝑅))𝑐). This contradicts Lemma 4.1 (ii). Hence we obtain that 𝐴 can admit at most 𝑛=𝜔((Γ(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 (Γ(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 (Γ(𝑅))𝑐 does not contain any infinite clique then (𝑎) 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 𝜔((Γ(𝑅))𝑐) is finite, then (𝑎)𝑅 can admit at most 𝜔((Γ(𝑅))𝑐)+1 maximal 𝑁-primes of (0) and if 𝑅 admits exactly 𝑘 maximal 𝑁-primes of (0) with 𝑘3, then 𝑘𝜔((Γ(𝑅))𝑐)(b)𝑅 can admit at most 𝜔((Γ(𝑅))𝑐)+1 minimal prime ideals, and if 𝑘 is the number of minimal prime ideals of 𝑅 with 𝑘3, then 𝑘𝜔((Γ(𝑅))𝑐).

Proof. (i)(a) Let 𝑃 be a maximal 𝑁-prime of (0) in 𝑅. Let 𝐴={𝑄𝑄 is a maximal 𝑁-prime of (0) in 𝑅 and 𝑄𝑃}. 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=𝐴{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 𝑃𝑍(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 𝜔((Γ(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=𝐴{𝑃}, we obtain that 𝑅 can admit at most 𝑛+1 maximal 𝑁-primes of (0).
Suppose that 𝑅 admits exactly 𝑘 maximal 𝑁-primes of (0) with 𝑘3. Let {𝑃1,𝑃2,𝑃3,,𝑃𝑘} be the set of all maximal 𝑁-primes of (0) in 𝑅. Note that 𝑍(𝑅)=𝑘𝑖=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 𝑖{1,2,3,,𝑘}, 𝑥𝑖𝑃𝑖(𝑗{1,,𝑘}{𝑖}𝑃𝑗). Then it is clear that for any distinct 𝑖,𝑗{1,2,3,,𝑘}, 𝑥𝑖 and 𝑥𝑗 are distinct nonzero zero-divisors of 𝑅, and as 𝑘3, it follows that there exists at least one 𝑡{1,2,3,,𝑘} such that both 𝑥𝑖 and 𝑥𝑗 are not in 𝑃𝑡 and hence 𝑥𝑖𝑥𝑗0. Thus we obtain that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥1,𝑥2,𝑥3,,𝑥𝑘} is a clique and so 𝑘𝜔((Γ(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 𝜔((Γ(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)  𝜔((Γ(𝑅))𝑐) is finite.
(ii)𝑅 is finite.
(iii)(Γ(𝑅))𝑐 does not contain any infinite clique.

Proof. (i) (ii) Let 𝜔((Γ(𝑅))𝑐)=𝑛. Now it follows from Corollary 4.3(ii)(a), that 𝑅 can admit at most 𝑛+1 maximal 𝑁-primes of (0). Let 𝑚 be the number of maximal 𝑁-primes of (0) in 𝑅 and let {𝑃1,,𝑃𝑚} be the set of all maximal 𝑁-primes of (0) in 𝑅. Note that 𝑚𝑛+1 and by the hypothesis, 𝑚2. Observe that 𝑍(𝑅)=𝑃1𝑃2𝑃𝑚=𝑃1𝑃2𝑃𝑚𝑍(𝑅)𝑃1𝑍(𝑅)𝑃2𝑍(𝑅)𝑃𝑚.(4.1) We know from Lemma 4.1(ii) that ||𝑍(𝑅)𝑃𝑖||𝑛for𝑖=1,2,,𝑚.(4.2) We now verify that |𝑃1𝑃2𝑃𝑚|𝑛. Let 𝑥𝑃1𝑃2. Note that {𝑥+𝑦𝑦𝑃1𝑃2𝑃𝑚}𝑍(𝑅)𝑃2. Hence it follows that ||𝑃1𝑃2𝑃𝑚||𝑛.(4.3) Now it follows from (4.1), (4.2), and (4.3) that Z(R) is finite. Hence it follows from [20, Theorem  1] that R is finite.
(ii) (iii) This is obvious.
(iii) (i) We first show that 𝑅 is finite. It follows from Corollary 4.3(i)(a) that 𝑅 can admit only a finite number of maximal 𝑁-primes of (0). Now one can proceed as in the proof of (i) (ii), and use Lemma 4.1(i) to obtain that 𝑅 is finite. It is now clear that 𝜔((Γ(𝑅))𝑐) is finite.

Remark 4.5. Let 𝑅 be an infinite ring and let 𝑅 admit at least two maximal 𝑁-primes of (0). It follows from Proposition 4.4 that 𝜔((Γ(𝑅))𝑐)=. Motivated by the interesting theorems proved on cliques in Γ(𝑅) in [3], and in particular, [3, Theorems 3.7 and 3.8], we attempt to determine 𝜔((Γ(𝑅))𝑐) for a finite commutative ring 𝑅 with identity which admits at least two maximal 𝑁-primes of (0). We are able to describe 𝜔((Γ(𝑅))𝑐) in the case when 𝑅 is a finite reduced ring which is not an integral domain.

Let 𝑅 be a finite commutative reduced ring with identity which is not an integral domain. Since 𝑅 is a finite ring, any prime ideal of 𝑅 is maximal. Let {𝑃1,,𝑃𝑛} be the set of all prime ideals of 𝑅. Since 𝑅 is reduced, (0)=nil(𝑅)=𝑃1𝑃𝑛. As 𝑅 is not an integral domain, it follows that 𝑛2. Moreover, by the Chinese Remainder Theorem [21, Proposition 1.10], it follows that 𝑅 is isomorphic to 𝑅/𝑃1××𝑅/𝑃𝑛. Thus 𝑅 is isomorphic to a finite direct product of finite fields. Let 𝑛2 and let 𝐾1,𝐾𝑛 be finite fields. Let 𝑅=𝐾1××𝐾𝑛. We now proceed to describe 𝜔((Γ(𝑅))𝑐). We make use of some of the techniques from [4].

We first recall the following facts from [4].

Fact 4.6. Let 𝑅 be a commutative ring with identity which is not an integral domain. For any 𝑎,𝑏𝑍(𝑅), define 𝑎𝑏 if and only if ((0)𝑅𝑎)=((0)𝑅𝑏). Then ~ is an equivalence relation on 𝑍(𝑅).

Proof. This fact is easy to check. The relation ~ can be defined on the whole of 𝑅. As our interest is on 𝑍(𝑅), we consider this relation defined on 𝑍(𝑅).

For an element 𝑎𝑍(𝑅), we denote by [𝑎], the equivalence class determined by ~ containing “a”.

Recall from [23] that a commutative ring 𝑅 with identity is said to be von Neumann regular if for each 𝑎𝑅 there exists 𝑏𝑅 such that 𝑎=𝑎2𝑏.

The following fact is important, and we make use of it in the proof of Lemma 4.8.

Fact 4.7. Let 𝑅 be a von Neumann regular ring which is not an integral domain (equivalently, which is not a field). Let ~ be the relation defined on 𝑍(𝑅) as in Fact 4.6. Then for any 𝑎𝑍(𝑅), there exists a unique idempotent element 𝑒𝑍(𝑅) such that [𝑎]=[𝑒].

Proof. The Fact 4.7 can be proved easily with the help of the ideas contained in the proof of [4, Lemma 3.1] and [10, Lemma 2.11]. In fact, [4, Lemma 3.1] and [10, Lemma 2.11] assert that for any von Neumann regular ring 𝑅 and for 𝑎,𝑏𝑍(𝑅), [𝑎]=[𝑏] if and only if 𝑅𝑎=𝑅𝑏. Yet for the sake of completeness, we present a proof of Fact 4.7.
It is well known that any element of a von Neumann regular ring can be expressed as the product of a unit and an idempotent [24, Lemma 2.5]. Let 𝑎𝑍(𝑅). Now there exists a unit 𝑢 in 𝑅 and an idempotent element 𝑒 in 𝑅 such that 𝑎=𝑢𝑒. Then it is clear that ((0)𝑅𝑎)=((0)𝑅𝑒). Since 𝑎𝑍(𝑅), it follows that 𝑒𝑍(𝑅). This proves that 𝑎𝑒 and so [𝑎]=[𝑒].
Note that if 𝑒 is any idempotent element in a ring 𝑅 (𝑅 not necessarily von Neumann regular), then ((0)𝑅𝑒)=𝑅(1𝑒). Moreover, it is easy to check that for idempotent elements 𝑒1,𝑒2 in a ring 𝑅,𝑅𝑒1=𝑅𝑒2 if and only if 𝑒1=𝑒2. If 𝑒, 𝑓 are idempotent elements in a ring 𝑅 such that ((0)𝑅𝑒)=((0)𝑅𝑓), then it follows that 𝑅(1𝑒)=𝑅(1𝑓). Hence 1𝑒=1𝑓 and so 𝑒=𝑓.
Now it follows from the above two preceding paragraphs that given any 𝑎𝑍(𝑅) where 𝑅 is a von Neumann regular ring, then there exists a unique idempotent 𝑒𝑍(𝑅) such that [𝑎]=[𝑒].

Let 𝑅 be a von Neumann regular ring which is not an integral domain. In the following lemma, we exhibit some cliques of (Γ(𝑅))𝑐.

Lemma 4.8. Let 𝑅 be a von Neumann regular ring which is not an integral domain. Let {𝑒1,,𝑒𝑚}𝑅{0,1} be a set of idempotents in 𝑅 such that 𝑒𝑖𝑒𝑗0 for all 𝑖,𝑗{1,,𝑚}. Let 𝐴=[𝑒1][𝑒𝑚]. Then the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique.

Proof. Let 𝑖{1,,𝑚}. Let 𝑎,𝑏[𝑒𝑖] with 𝑎𝑏. Note that [𝑎]=[𝑏]=[𝑒𝑖]. Now it follows from the proof of Fact 4.7 that there exist units 𝑢,𝑣 in 𝑅 such that 𝑎=𝑢𝑒𝑖 and 𝑏=𝑣𝑒𝑖. Hence 𝑎𝑏=𝑢𝑣𝑒𝑖0. This proves that any two distinct elements of [𝑒𝑖] are adjacent in (Γ(𝑅))𝑐 for each 𝑖{1,,𝑚}.
Let 𝑖,𝑗{1,,𝑚} with 𝑖𝑗. Let 𝑥[𝑒𝑖], 𝑦[𝑒𝑗]. Note that 𝑥=𝑢𝑒𝑖 and 𝑦=𝑣𝑒𝑗 for some units 𝑢,𝑣 in 𝑅. Hence 𝑥𝑦=𝑢𝑣𝑒𝑖𝑒𝑗0, since 𝑒𝑖𝑒𝑗0 by the hypothesis.
Indeed, it is true that if [𝑎] and [𝑏] are equivalence classes determined by ~ defined on 𝑍(𝑅) where 𝑅 is a commutative ring with identity which is not an integral domain (𝑅 need not be von Neumann regular) and if 𝑎𝑏0, then we assert that for any 𝑥[𝑎] and 𝑦[𝑏], 𝑥𝑦0. Note that if 𝑥𝑦=0, then 𝑦((0)𝑅𝑥)=((0)𝑅𝑎). Hence 𝑎𝑦=0. This implies that 𝑎((0)𝑅𝑦)=((0)𝑅𝑏). Hence 𝑎𝑏=0 and this is a contradiction. Thus we obtain that 𝑥𝑦0.
This proves that the subgraph of(Γ(𝑅))𝑐 induced on A = [𝑒1][𝑒𝑚] is a clique.

We include the following simple lemma for the sake of completeness.

Lemma 4.9. Let 𝑛2. Let 𝐾𝑖 be a field for 𝑖=1,2,,𝑛. Let 𝑅=𝐾1××𝐾𝑛 be their direct product. Then the number of idempotents in 𝑍(𝑅) (that is, the number of nontrivial idempotents in 𝑅) equals 2𝑛2 and the number of equivalence classes determined by the equivalence relation ~ defined on 𝑍(𝑅) equals 2𝑛2.

Proof. The only idempotent elements in a field are 1 and 0. Using this observation, it follows that 𝑅=𝐾1××𝐾𝑛 has exactly 2𝑛 idempotents. Among them except (0,,0) and (1,,1), the rest of them are in 𝑍(𝑅). Thus we obtain that the number of idempotents in 𝑍(𝑅) (that is, the number of nontrivial idempotents in 𝑅) equals 2𝑛2.
Let {𝑒𝑖𝑖=1,2,,2𝑛2} be the set of all idempotent elements in 𝑍(𝑅). Since 𝑅=𝐾1××𝐾𝑛 is von Neumann regular, it follows from Fact 4.7 that the set of all equivalence classes determined by ~ is {[𝑒𝑖]𝑖=1,2,,2𝑛2}.

Let 𝑛2. Let 𝑅=𝐾1××𝐾𝑛 where 𝐾𝑖 is a field for 𝑖=1,2,,𝑛. The following lemma describes the cliques of (Γ(𝑅))𝑐.

Lemma 4.10. Let 𝑛2. Let 𝐾𝑖 be a field for 𝑖=1,2,,𝑛. Let 𝑅=𝐾1××𝐾𝑛. Let 𝐴𝑍(𝑅) be such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique. Let 𝐸={𝑒𝑖𝑖=1,2,,2𝑛2} be the set of all idempotents in 𝑍(𝑅). Then there exists a nonempty subset 𝐵 of 𝐸 such that 𝑏1𝑏20 for all 𝑏1, 𝑏2𝐵and 𝐴𝑏𝐵[𝑏].

Proof. We know from the proof of Lemma 4.9 that {[𝑒𝑖]𝑖=1,2,,2𝑛2} is the set of all equivalence classes determined by ~. Thus we obtain that 𝑍(𝑅)=𝑒𝐸[𝑒]. Now 𝐴 being a subset of 𝑍(𝑅), it follows that 𝐴=𝑒𝐸(𝐴[𝑒]). Let 𝐵𝐸 be such that 𝐴[𝑏] is non-empty for each 𝑏𝐵. Since 𝐴 is non-empty, it follows that 𝐵 is non-empty. We now verify that for any 𝑏1,𝑏2𝐵,𝑏1𝑏20. This is clear if 𝑏1=𝑏2 since any element of 𝐵 is a nonzero idempotent in 𝑅. Suppose that 𝑏1𝑏2.
Let 𝑎1𝐴[𝑏1] and 𝑎2𝐴[𝑏2]. Since [𝑏1][𝑏2]=, it follows that 𝑎1𝑎2. As the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique, we obtain that 𝑎1𝑎20. Note that 𝑅𝑎𝑖=𝑅𝑏𝑖 for 𝑖=1,2. Hence we obtain that 𝑅𝑏1𝑏2=𝑅𝑎1𝑎2(0) and so 𝑏1𝑏20. This proves that there exists a non-empty subset 𝐵 of 𝐸 such that 𝑏1𝑏2 0 for all𝑏1, 𝑏1𝐵 and moreover, 𝐴=𝑏𝐵(𝐴[𝑏])𝑏𝐵[𝑏].

Let 𝑅 be as in Lemma 4.10 with the further assumption that 𝐾𝑖 is finite for 𝑖=1,2,,𝑛. We determine 𝜔((Γ(𝑅))𝑐) in the following proposition.

Proposition 4.11. Let 𝑛2 and let 𝐾𝑖(𝑖=1,2,,𝑛) be finite fields. Let 𝑅=𝐾1×𝐾2××𝐾𝑛. Let 𝐸={𝑒𝑖𝑖=1,2,,2𝑛2} be the set of all idempotents in 𝑍(𝑅). Then 𝜔((Γ(𝑅))𝑐)=max{(𝑏𝐵|[𝑏]|)𝐵 varies over all non-empty subsets of 𝐸 satisfying the property that 𝑏1𝑏2 0 for any 𝑏1,𝑏2𝐵}.

Proof. Let 𝑋={𝐵𝐵 is a non-empty subset of 𝐸 satisfying the property that for any 𝑏1,𝑏2𝐵,𝑏1𝑏20}. Let 𝐵𝑋. Let 𝐴=𝑏𝐵[𝑏]. Now it follows from Lemma 4.8 that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique. Thus 𝜔((Γ(𝑅))𝑐)|𝐴|=𝑏𝐵|[𝑏]|. Hence we obtain that 𝜔((Γ(𝑅))𝑐)max𝑏𝐵||[𝑏]||𝐵𝑋.(4.4)
Let 𝐴𝑍(𝑅) be such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique. We know from Lemma 4.10 that there exists 𝐵𝑋 such that 𝐴𝑏𝐵[𝑏]. Hence we obtain that |𝐴|𝑏𝐵|[𝑏]|. This implies that 𝜔((Γ(𝑅))𝑐)max𝑏𝐵||[𝑏]||𝐵𝑋.(4.5)
From (4.4) and (4.5), it follows that 𝜔((Γ(𝑅))𝑐)=max{(𝑏𝐵|[𝑏]|)𝐵𝑋}.

We make use of the following useful remark in Example 4.13(i) and (ii).

Remark 4.12. Let 𝑅 be a von Neumann regular ring which is not a field. Let ~ be the equivalence relation which was considered in Fact 4.7. Observe that for any idempotent element 𝑒𝑍(𝑅), [𝑒]={𝑢𝑒𝑢isaunitin𝑅}.
Let 𝑛2. Let 𝐾1,𝐾2,,𝐾𝑛 be finite fields and let 𝑅=𝐾1×𝐾2××𝐾𝑛. Let 𝑒𝑍(𝑅) be an idempotent. Let 𝐶={𝑖{1,2,,𝑛}the𝑖thcomponentof𝑒is1}. Observe that |[𝑒]|=𝑖𝐶(|𝐾𝑖|1).

We next have the following example.

Example 4.13. (i) Let 𝐾1,𝐾2 be finite fields. Let 𝑅=𝐾1×𝐾2. Note that {𝑒1=(1,0),𝑒2=(0,1)} is the set of all idempotents in 𝑍(𝑅). Now it follows from Proposition 4.11 and Remark 4.12 that 𝜔((Γ(𝑅))𝑐)=max{|𝐾1|1,|𝐾2|1}. Note that this fact can also be verified directly. (ii) Let 𝐾1, 𝐾2, 𝐾3 be finite fields. Let 𝑅=𝐾1×𝐾2×𝐾3. Note that {𝑒1=(1,0,0),𝑒2=(0,1,0),𝑒3=(0,0,1),𝑒4=(1,1,0),𝑒5=(1,0,1),𝑒6=(0,1,1)} is the set of all idempotents in 𝑍(𝑅). Now it follows from Proposition 4.11 and Remark 4.12 that 𝜔((Γ(𝑅))𝑐)=max{(|𝐾1|1)+(|𝐾1|1)(|𝐾2|1)+(|𝐾1|1)(|𝐾3|1),(|𝐾2|1)+(|𝐾1|1)(|𝐾2|1)+(|𝐾2|1)(|𝐾3|1),(|𝐾3|1)+(|𝐾1|1)(|𝐾3|1)+(|𝐾2|1)(|𝐾3|1),(|𝐾1|1)(|𝐾2|1)+(|𝐾2|1)(|𝐾3|1)+(|𝐾1|1)(|𝐾3|1)}.

Let 𝑅 be a ring with |𝑍(𝑅)|2. Suppose that 𝑅 has only one maximal 𝑁-prime of (0). We, in the following result, determine some necessary conditions on 𝑅 in order that (Γ(𝑅))𝑐 does not contain any infinite clique.

Lemma 4.14. Let 𝑅 be a commutative ring with identity and let |𝑍(𝑅)|2. Suppose that 𝑅 has exactly one maximal 𝑁-prime of (0) and let it be 𝑃. If (Γ(𝑅))𝑐 does not contain any infinite clique, then (𝑖)𝑃= the nilradical of 𝑅 and if furthermore, 𝑃2(0), then (𝑖𝑖)𝑅/𝑃 is finite, (𝑖𝑖𝑖)𝑃 is a 𝐵-prime of (0) in 𝑅,and(𝑖𝑣)𝑅 satisfies descending chain condition (d. c. c.) on principal ideals.

Proof. (i) As any nilpotent element of 𝑅 is a zero-divisor of 𝑅 and since 𝑍(𝑅)=𝑃, it follows that nil(𝑅)𝑍(𝑅)=𝑃. Let 𝑥𝑃. We assert that 𝑥 is nilpotent. Suppose that 𝑥 is not nilpotent. Then for any 𝑖,𝑗{1,2,3,}, 𝑥𝑖𝑥𝑗0, and moreover, for all distinct 𝑖,𝑗{1,2,3,},𝑥𝑖𝑥𝑗. Hence {𝑥𝑘𝑘=1,2,3,}𝑍(𝑅) is such that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥𝑘𝑘=1,2,3,} is an infinite clique. This is in contradiction to the assumption that (Γ(𝑅))𝑐 does not contain any infinite clique. Hence 𝑥 is nilpotent. This shows that 𝑃nil(𝑅) and so 𝑃=nil(𝑅).
(ii) Suppose that 𝑅/𝑃 is infinite. We first assert that 𝑥2=0 for each 𝑥𝑃. Suppose that there exists 𝑥𝑃 such that 𝑥20. Since we are assuming that 𝑅/𝑃 is infinite, it is possible to find an infinite sequence of elements {𝑟𝑘𝑅𝑃for𝑘=1,2,3,} such that 𝑟𝑖𝑟𝑗𝑃 and 𝑟𝑖+𝑟𝑗𝑃 for all distinct 𝑖,𝑗{1,2,3,}. Note that for all 𝑖,𝑗{1,2,3,}, 𝑟𝑖𝑟𝑗𝑃=𝑍(𝑅), and as 𝑥20, it follows that 𝑥2𝑟𝑖𝑟𝑗0. Moreover, as for all distinct 𝑖,𝑗{1,2,3,}, 𝑟𝑖𝑟𝑗𝑃, it follows that 𝑥𝑟𝑖𝑥𝑟𝑗. Hence we obtain that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥𝑟𝑘𝑘=1,2,3,} is an infinite clique. This contradicts the assumption that (Γ(𝑅))𝑐 does not contain any infinite clique. Hence 𝑥2=0 for each 𝑥𝑃. As 𝑃2(0), there exist 𝑎,𝑏𝑃 such that 𝑎𝑏 and 𝑎𝑏0. Note that we have 𝑎2=𝑏2=0. For any positive integer 𝑘, let 𝑥𝑘=𝑎+𝑏𝑟𝑘. It is clear that for any positive integer 𝑘,𝑥𝑘𝑃 and 𝑎𝑥𝑘=𝑎𝑏𝑟𝑘0, since 𝑎𝑏0 and 𝑟𝑘𝑃=𝑍(𝑅). Moreover, for all distinct 𝑖,𝑗{1,2,3,},𝑥𝑖𝑥𝑗=𝑎𝑏(𝑟𝑖+𝑟𝑗)0, since 𝑎𝑏0 and 𝑟𝑖+𝑟𝑗𝑃=𝑍(𝑅). Furthermore, as 𝑏0 and 𝑟𝑖𝑟𝑗𝑃, for all distinct 𝑖,𝑗{1,2,3,}, it follows that 𝑥𝑖𝑥𝑗 for all distinct 𝑖,𝑗{1,2,3,}. Hence we obtain that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥𝑘𝑘=1,2,3,} is an infinite clique. This contradicts the assumption that (Γ(𝑅))𝑐 does not contain any infinite clique. Hence it follows that 𝑅/𝑃 is finite.
(iii) We now verify that 𝑃 is a 𝐵-prime of (0) in 𝑅. We consider two cases. Case (A): 𝑃 is finitely generated.
By (i), we have 𝑃=nil(𝑅) and hence we obtain that 𝑃 is a nilpotent ideal of 𝑅. Let 𝑚2 be least with the property that 𝑃𝑚=(0). Now for any 𝑥𝑃𝑚1{0},𝑃((0)𝑅𝑥)𝑍(𝑅)=𝑃,andso𝑃=((0)𝑅𝑥) is a 𝐵-prime of (0) in 𝑅. Case (B): 𝑃 is not finitely generated.
We have 𝑃2(0), by assumption. Hence there exist 𝑎1,𝑎2𝑃 such that 𝑎1𝑎2 and𝑎1𝑎20. Suppose that 𝑃 is not a 𝐵-prime of (0) in 𝑅. Then, as ((0)𝑅𝑎1𝑎2)𝑍(𝑅)=𝑃, it follows that 𝑃((0)𝑅𝑎1𝑎2). As 𝑅𝑎1+𝑅𝑎2𝑃 and since 𝑃 is not finitely generated, it follows that 𝑃𝑅𝑎1+𝑅𝑎2. Hence 𝑃(𝑅𝑎1+𝑅𝑎2)((0)𝑅𝑎1𝑎2). Hence there exists𝑎3𝑃((𝑅𝑎1+𝑅𝑎2)((0)𝑅𝑎1𝑎2)). Thus 𝑎1,𝑎2,𝑎3𝑃 are distinct and 𝑎1𝑎2𝑎30. Let k be any positive integer with 𝑘3. Assume that there exists a subset {𝑎1,𝑎2,𝑎3,,𝑎𝑘} of 𝑃 with 𝑎1𝑎2𝑎3𝑎𝑘0. Observe that 𝑃𝑅𝑎1+𝑅𝑎2+𝑅𝑎3++𝑅𝑎𝑘 and 𝑃((0)𝑅𝑎1𝑎2𝑎3𝑎𝑘). Hence there exists 𝑎𝑘+1𝑃((𝑅𝑎1++𝑅𝑎𝑘)((0)𝑅𝑎1𝑎2𝑎3𝑎𝑘)). This shows that if 𝑃 is not finitely generated and if 𝑃 is not a 𝐵-prime of (0) in 𝑅, then there exists an infinite subset {𝑎𝑖𝑖=1,2,3,} of 𝑃 such that the product 𝑎1 · · · 𝑎𝑘0 for 𝑘=1,2,3,. This implies that the subgraph of (Γ(𝑅))𝑐 induced on {𝑎𝑖𝑖=1,2,3,} is an infinite clique.
This contradicts the hypothesis that (Γ(𝑅))𝑐 does not contain any infinite clique. Hence 𝑃 is a 𝐵-prime of (0) in 𝑅.
(iv) We obtain from (ii) that 𝑅/𝑃 is finite. Since any finite integral domain is a field, it follows that 𝑃 is a maximal ideal of 𝑅. By (i), 𝑃=nil(𝑅). Hence we obtain that 𝑃 is the only prime ideal of R. We now verify that R satisfies d. c. c. on principal ideals. Suppose that R does not satisfy d. c. c. on principal ideals. Then there exist nonzero elements 𝑥𝑖𝑃 for 𝑖=1,2,3, such that 𝑅𝑥1𝑅𝑥2𝑅𝑥3. Note that there exist 𝑎𝑖𝑃 for 𝑖=1,2,3, such that 𝑥𝑖+1=𝑎𝑖𝑥𝑖. Hence𝑥𝑘+1=𝑎𝑘𝑎1𝑥1,𝑘1.(I) Since 𝑥𝑘0 for 𝑘=1,2,3,, it follows that the elements 𝑎𝑘𝑃(𝑘=1,2,3,) satisfy 𝑎𝑖𝑎𝑗0 for all distinct 𝑖,𝑗{1,2,3,}. As each element of 𝑃 is nilpotent, it follows from (I) and the fact that 𝑥𝑘0 for 𝑘=1,2,3, that there exist positive integers 𝑘1<𝑘2<𝑘3< such that for all distinct 𝑖,𝑗{𝑘1,𝑘2,},𝑎𝑖𝑎𝑗. Let 𝐴={𝑘𝑖𝑖=1,2,3,}. Observe that the subgraph of (Γ(𝑅))𝑐 induced on {𝑎𝑖𝑖𝐴} is an infinite clique. This contradicts the assumption that (Γ(𝑅))𝑐 does not contain any infinite clique. Hence it follows that 𝑅 satisfies d. c. c. on principal ideals.

Let 𝑅 and 𝑃 be as in Lemma 4.14. Suppose that 𝑃2(0). I do not know any necessary and sufficient condition in order that (Γ(𝑅))𝑐 does not contain any infinite clique. However, the following proposition shows that if the ring 𝑅 is Noetherian, then (Γ(𝑅))𝑐 does not contain any infinite clique if and only if 𝑅 is finite.

Proposition 4.15. Let R be a Noetherian ring which is not an integral domain. Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝑃2(0), then the following conditions are equivalent.(i)𝜔((Γ(𝑅))𝑐) is finite.(ii)(Γ(𝑅))𝑐 does not contain any infinite clique.(iii)𝑅 is finite.

Proof. (i) (ii) This is clear.
(ii) (iii) We know from Lemma 4.14 that 𝑃= the nilradical of 𝑅 and 𝑅/𝑃 is finite. Now by hypothesis, 𝑅 is a Noetherian ring. Hence 𝑃 is finitely generated. Therefore, 𝑃𝑛=(0) for some 𝑛1. Since 𝑃2(0), it follows that 𝑛3. Observe that for each integer 𝑘,0𝑘𝑛1,𝑃𝑘/𝑃𝑘+1 is a finite-dimensional vector space over the finite field 𝑅/𝑃. Hence it follows that 𝑃𝑘/𝑃𝑘+1 is finite for 𝑘=0,1,2,,𝑛1. Now 𝑃𝑛1,𝑃𝑛2/𝑃𝑛1 are finite, and hence it follows that 𝑃𝑛2 is finite. Proceeding in this way, we obtain that 𝑃 is finite. Thus 𝑃 and 𝑅/𝑃 are finite. Hence we obtain that 𝑅 is finite.
(iii) (i) This is clear.

Recall that a commutative ring 𝑅 with identity is said to be a chained ring if the principal ideals of 𝑅 are linearly ordered under inclusion (equivalently, the ideals of 𝑅 are linearly ordered under inclusion).

Let 𝑅 be a chained ring which is not an integral domain. Then, it is clear that 𝑅 must have exactly one maximal 𝑁-prime of (0). If 𝑃 is the only maximal 𝑁-prime of (0) and if 𝑃2(0), then the following proposition characterizes when (Γ(𝑅))𝑐 can admit infinite cliques.

Proposition 4.16. Let 𝑅 be a commutative ring with identity which is not an integral domain. Suppose that 𝑅 is a chained ring and moreover, there exist 𝑥,𝑦𝑍(𝑅) with 𝑥𝑦 such that 𝑥𝑦0. Then the following conditions are equivalent.(i)(Γ(𝑅))𝑐 does not admit any infinite clique.(ii)𝑅 is finite.(iii)𝜔((Γ(𝑅))𝑐) is finite.

Proof. (i) (ii) Since the ideals of 𝑅 are linearly ordered under inclusion, it follows that 𝑅 admits exactly one maximal 𝑁-prime of (0). Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. We are assuming that (Γ(𝑅))𝑐 does not admit any infinite clique. So, we obtain from Lemma 4.14(i) that 𝑃= the nilradical of 𝑅. Note that 𝑍(𝑅)=𝑃. Let 𝑁(𝑅)={𝑥𝑃𝑥2=0}. It is known that for any 𝑥,𝑦𝑃𝑁(𝑅), 𝑥𝑦0 [6, Lemma 4.2(3)]. Since (Γ(𝑅))𝑐 does not admit any infinite clique, it follows that 𝑃𝑁(𝑅)is finite. Now by the assumption that 𝑅 is a chained ring and there exist 𝑥,𝑦𝑍(𝑅) with 𝑥𝑦 such that 𝑥𝑦0, it follows that 𝑍(𝑅)𝑁(𝑅) is non-empty. Let 𝑃𝑁(𝑅)={𝑥1,,𝑥𝑚}. Since each element of 𝑃 is nilpotent, it follows that there exist 𝑡>2 such that 𝑝𝑡=0 for each 𝑝𝑃. As 𝑅 is a chained ring, we obtain that 𝑃𝑡=(0). Moreover, it follows from Lemma 4.14(ii) that 𝑅/𝑃 is finite. Hence 𝑃 is a maximal ideal of 𝑅 and since 𝑅 is a chained ring, it follows that 𝑅 is quasilocal with 𝑃 as its unique maximal ideal. As 𝑃 is nilpotent and 𝑃(0), it follows that 𝑃𝑃2. Now 𝑅 is a chained ring with 𝑃 as its unique maximal ideal satisfying the further condition that 𝑃𝑃2. In such a case it is well known that 𝑃=𝑅𝑝 for any𝑝𝑃𝑃2. Using the same reasoning as in the proof of (ii) (iii) of Proposition 4.15, it now follows that 𝑅 is finite.
(ii) (iii) and (iii) (i) are clear.

The following remark determines 𝜔((Γ(𝑅))𝑐) for any finite chained ring which is not an integral domain.

Remark 4.17. Let 𝑅 be a finite chained ring which is not an integral domain. Let 𝑃 denote the unique maximal ideal of 𝑅. Suppose that 𝑃2(0). Let 𝑛3 be least with the property that 𝑃𝑛=(0). Then the following hold.(i)𝜔((Γ(𝑅))𝑐)=|𝑃𝑃𝑛/2|+1if 𝑛 is even. (ii)𝜔((Γ(𝑅))𝑐)=|𝑃𝑃(𝑛+1)/2| if 𝑛 is odd.

Proof. As is already observed in the proof of (i) (ii) of Proposition 4.16, 𝑃=𝑅𝑝 for any𝑝𝑃𝑃2.
(i) Suppose that 𝑛 is even and let 𝑛=2𝑚. By the choice of 𝑛,𝑝𝑛10. If𝑎𝑃𝑃𝑚,𝑏𝑃𝑃𝑚, then 𝑎=𝑢𝑝𝑡 for some 𝑡,1𝑡<𝑚 and 𝑎 unit 𝑢 in 𝑅 and 𝑏=𝑣𝑝𝑠 for some 𝑠,1𝑠<𝑚 and 𝑎 unit 𝑣 in 𝑅. Hence 𝑎𝑏=𝑢𝑣𝑝𝑡+𝑠0, since 𝑡+𝑠<2𝑚=𝑛 and 𝑝𝑛10. Moreover, observe that 𝑝𝑚𝑥0 for any 𝑥𝑃𝑃𝑚. Hence the subgraph of (Γ(𝑅))𝑐 induced on (𝑃𝑃𝑚){𝑝𝑚} is a clique. This implies that 𝜔((Γ(𝑅))𝑐)|𝑃𝑃𝑚|+1. Let 𝐴𝑍(𝑅)=𝑃{0} be such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique. Let |𝐴|=𝑡. Since 𝑅 is a chained ring, it is possible to find 𝑎𝐴 such that 𝑅𝑎𝑅𝑏 for each 𝑏𝐴. Hence 𝑅𝑎𝑏𝑅𝑏2. Since 𝑎𝑏0 for each𝑏𝐴{𝑎}, we obtain that 𝑏20 for each𝑏𝐴{𝑎}. As 𝑃𝑛=𝑃2𝑚=(0), it follows that 𝐴{𝑎}𝑃𝑃𝑚. Thus 𝐴={𝑎}(𝐴{𝑎}){𝑎}(𝑃𝑃𝑚) and so |𝐴||𝑃𝑃𝑚|+1. This proves that 𝜔((Γ(𝑅))𝑐)|𝑃𝑃𝑚|+1. Hence 𝜔((Γ(𝑅))𝑐)=|𝑃𝑃𝑚|+1.
(ii) Suppose that 𝑛 is odd and let 𝑛=2𝑘+1 for some 𝑘1. In this case, we verify that 𝜔((Γ(𝑅))𝑐)=|𝑃𝑃𝑘+1| where 𝑘=(𝑛1)/2. It is clear that for any 𝑥,𝑦𝑃𝑃𝑘+1,𝑥𝑦0. This shows that the subgraph of (Γ(R))c induced on 𝑃𝑃𝑘+1 is a clique. Hence we obtain that 𝜔((Γ(𝑅))𝑐)|𝑃𝑃𝑘+1|. Let 𝐴𝑍(𝑅)=𝑃{0} be such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique. We assert that |𝐴||𝑃𝑃𝑘+1|. Let 𝑎𝐴 be such that 𝑅𝑎𝑅𝑏 for each 𝑏𝐴. Now, it follows as in the proof of (i) that 𝑏20 for each 𝑏𝐴{𝑎}. Since 𝑃2𝑘+1=(0), it follows that 𝐴{𝑎}𝑃𝑃𝑘+1. If there exists at least one 𝑏𝐴{𝑎} such that 𝑏𝑃𝑘, then, as 𝑎𝑏0, it follows that 𝑎𝑃𝑘+1 and so we obtain that 𝐴𝑃𝑃𝑘+1. Hence it follows that |𝐴||𝑃𝑃𝑘+1|. Otherwise, we obtain that 𝐴{𝑎}𝑃𝑃𝑘. Note that 𝐴={𝑎}(𝐴{𝑎}){𝑎}(𝑃𝑃𝑘) and so |𝐴|1+|𝑃𝑃𝑘||𝑃𝑘𝑃𝑘+1|+|𝑃𝑃𝑘|=|𝑃𝑃𝑘+1|. This proves that 𝜔((Γ(𝑅))𝑐)|𝑃𝑃𝑘+1|. Hence we obtain that 𝜔((Γ(𝑅))𝑐||)=𝑃𝑃𝑘+1||=||𝑃𝑃(𝑛+1)/2||.(4.6)

We next mention an example to illustrate that in Remark 4.17, one cannot drop the assumption that 𝑅 is a chained ring.

Example 4.18. Let 𝐹=𝑍/2𝑍 be the field containing exactly two elements. It is convenient to denote 𝐹 simply by 𝐹={0,1}. Let 𝑝(𝑥)=𝑥2+𝑥+1𝐹[𝑥] where 𝐹[𝑥] is the polynomial ring in one variable over 𝐹. Note that 𝑝(𝑥) is irreducible over 𝐹. Let 𝐾 be the splitting field of 𝑝(𝑥) over 𝐹. Let 𝛼𝐾 be a root of 𝑝(𝑥). It is clear that 𝐾=𝐹(𝛼) and [𝐾𝐹]=2. Indeed, {1,𝛼} is a basis of 𝐾 as a vector space over 𝐹. Moreover, 𝐾={0,1,𝛼,𝛼2=𝛼1=𝛼+1}. Let 𝑇=𝐾[[𝑥]] be the power series ring in one variable over 𝐾. Note that 𝑇=𝐾+𝑀 where 𝑀=𝑥𝐾[[𝑥]]=𝑥𝑇. Let 𝑆=𝐹+𝑀. Consider the ring 𝑅=𝑆/𝑥2𝑆. Let us denote 𝑥2𝑆 by I. Observe that 𝑅={(𝑎+(𝑏+𝑐𝛼)𝑥+(𝑑+𝑒𝛼)𝑥2)+𝐼|𝑎,𝑏,𝑐,𝑑,𝑒𝐹}. Note that 𝑅 is a finite local ring with unique maximal ideal 𝑁=𝑀/𝐼satisfying 𝑁3= the zero-ideal of 𝑅, but 𝑁2 is nonzero. We have 𝛼2=1+𝛼 and thus 𝑁={0+𝐼,𝑥+𝐼,𝛼𝑥+𝐼,𝛼2𝑥+𝐼,𝛼𝑥2+𝐼, (𝑥+𝛼𝑥2)+𝐼, (𝛼𝑥+𝛼𝑥2)+𝐼, (𝛼2𝑥+𝛼𝑥2)+𝐼}. Observe that 𝑍(𝑅)=𝑁. Now 𝑁𝑁2={𝑥+𝐼,𝛼𝑥+𝐼, 𝛼2𝑥+𝐼,(𝑥+𝛼𝑥2)+𝐼, (𝛼𝑥+𝛼𝑥2)+𝐼, (𝛼2𝑥+𝛼𝑥2)+𝐼}. Thus |𝑁𝑁2|=6. We assert that 𝜔((Γ(𝑅))𝑐)=3. Since 𝑁3= the zero-ideal of 𝑅, it follows that the vertex set of any clique in (Γ(𝑅))𝑐 must be a subset of 𝑁𝑁2.(4.7)
As 𝛼𝐹,𝛼𝑥2𝐼=𝑥2𝑆,and𝛼2𝑥2=(1+𝛼)𝑥2𝐼. Hence the subgraph of (Γ(𝑅))𝑐 induced on {𝑥+𝐼,𝛼𝑥+𝐼,(𝛼𝑥+𝛼𝑥2)+𝐼} is a clique. This shows that 𝜔((Γ(𝑅))𝑐)3. Observe that (𝑥+𝐼)𝑥+𝛼𝑥2+𝐼=𝑥2+𝐼=0+𝐼.(4.8) We have 𝛼3=𝛼𝛼2=𝛼(1+𝛼)=𝛼+𝛼2=𝛼+(1+𝛼)=1. Note that 𝛼(𝛼𝑥+𝐼)2𝑥+𝐼=𝑥2+𝐼=0+𝐼.(4.9) Moreover, 𝛼𝑥+𝛼𝑥2𝛼+𝐼2𝑥+𝛼𝑥2+𝐼=𝑥2+𝐼=0+𝐼.(4.10) It is clear from (4.7), (4.8), (4.9), and (4.10) that if 𝐴𝑍(𝑅) is such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique, then 𝐴 can contain at most 3 elements. Hence we obtain that 𝜔((Γ(𝑅))𝑐)3. Thus 𝜔((Γ(𝑅))𝑐)=3<|𝑁𝑁2|=6.
Let 𝑇 and 𝑆 be as above. Let 𝑅1=𝑆/𝑥3𝑆. Note that 𝑅1 is a finite local ring with 𝑁1=𝑀/𝑥3𝑆 as its unique maximal ideal. Thus 𝑍(𝑅1)=𝑁1. Moreover, note that (𝑁1)4 is the zero- ideal of 𝑅1 but (𝑁1)3 is nonzero. It is convenient to denote the ideal 𝑥3𝑆 of 𝑆 by 𝐽. Note that 𝛼2=𝛼+1. We assert that 𝜔((Γ(𝑅1))𝑐)=|𝑁1(𝑁1)2|. Since (𝑁1)4=(0), it follows that if 𝐴 is any subset of 𝑍(𝑅1)=𝑁1{0+𝐽} such that the subgraph of (Γ(𝑅1))𝑐 induced on 𝐴 is a clique, then 𝐴 can admit at most one element of (𝑁1)2. We claim that |𝐴||𝑁1(𝑁1)2|. This is clear if 𝐴 does not contain any element of (𝑁1)2. Suppose that 𝐴 has an element of (𝑁1)2. In such a case, we verify that 𝐴 cannot contain all the elements of 𝑁1(𝑁1)2. Let 𝑎𝐴 be such that 𝑎(𝑁1)2. Note that 𝑎=(𝛽𝑥2+𝛾𝑥3)+𝐽 for some 𝛽,𝛾𝐾. We consider two cases.

Case i. 𝛽=0. In such a case, the element 𝑎=𝛾𝑥3+𝐽 annihilates each element of 𝑁1 and hence it annihilates each element of 𝑁1(𝑁1)2.

Case ii. 𝛽0. Observe that 𝛽=𝑢+𝑣𝛼 for some 𝑢,𝑣𝐹 with at least one of 𝑢,𝑣 is different from 0. Thus 𝑎{(𝑥2+𝛾𝑥3)+𝐽,(𝛼𝑥2+𝛾𝑥3)+𝐽,(𝛼2𝑥2+𝛾𝑥3)+𝐽}. Note that(𝑥2+𝛾𝑥3)+𝐽 annihilates the element 𝑥+𝐽𝑁1(𝑁1)2. The element (𝛼𝑥2+𝛾𝑥3)+𝐽 annihilates 𝛼2𝑥+𝐽𝑁1(𝑁1)2 and the element (𝛼2𝑥2+𝛾𝑥3)+𝐽 annihilates 𝛼𝑥+𝐽𝑁1(𝑁1)2.

This shows that if 𝐴 contains an element of (𝑁1)2, then that element annihilates at least one element of 𝑁1(𝑁1)2. Since the subgraph of (Γ(𝑅1))𝑐 induced on 𝐴 is a clique, it follows that A cannot contain all the elements of 𝑁1(𝑁1)2. This proves that |𝐴||𝑁1(𝑁1)2| and hence we obtain that𝜔((Γ(𝑅1))𝑐)|𝑁1(𝑁1)2|.

We next claim that the subgraph of (Γ(𝑅1))𝑐 induced on 𝑁1(𝑁1)2 is a clique. Let 𝑎, 𝑏𝑁1(𝑁1)2. Note that 𝑎=(𝛽1𝑥+𝛽2𝑥2+𝛽3𝑥3)+𝐽 and 𝑏=(𝛾1𝑥+𝛾2𝑥2+𝛾3𝑥3)+𝐽 for some 𝛽𝑖,𝛾𝑖𝐾 (for𝑖=1,2,3) with 𝛽1,𝛾1𝐾{0}. Note that ab = 𝛽1𝛾1𝑥2+𝑤+𝐽 for some 𝑤𝑥3𝐾[[𝑥]]. Since, 𝐽=𝑥3𝑆𝑥3𝐾[[𝑥]] and as 𝛽1𝛾1𝑥2𝑥3𝐾[[𝑥]], it follows that ab is a nonzero element of 𝑅1. This proves that the subgraph of (Γ(𝑅1))𝑐 induced on 𝑁1(𝑁1)2 is a clique. Hence we obtain that 𝜔((Γ(𝑅1))𝑐)|𝑁1(𝑁1)2| and so 𝜔((Γ(𝑅1))𝑐)=|𝑁1(𝑁1)2|.

Let 𝑅 be a commutative ring with identity which is not an integral domain. Suppose that 𝑅 has exactly one maximal 𝑁-prime of (0) and let it be 𝑃. If (Γ(𝑅))𝑐 does not admit any infinite clique, then it was shown in Lemma 4.14 that each element of 𝑃 is nilpotent. The following lemma describes the elements of 𝑃 under the hypothesis that 𝜔((Γ(𝑅))𝑐) is finite.

Lemma 4.19. Let 𝑅 be a commutative ring with identity which is not an integral domain. Suppose that 𝑃 is the only maximal 𝑁-prime of (0) in 𝑅. If 𝜔((Γ(𝑅))𝑐) is finite, then for any 𝑥𝑃,𝑥𝑚=0 where 𝑚=𝜔((Γ(𝑅))𝑐)+1.

Proof. By hypothesis, 𝜔((Γ(𝑅))𝑐) is finite. Let 𝑥𝑃. We assert that 𝑥𝑚=0 where 𝑚=𝜔((Γ(𝑅))𝑐)+1. Suppose that 𝑥𝑚0. For each 𝑘{1,2,,𝑚}, let 𝑦𝑘=𝑥++𝑥𝑘. Note that 𝑍(𝑅)=𝑃 and 𝑟1=1𝑃=𝑍(𝑅) and for 2𝑘𝑚,𝑟𝑘=1++𝑥𝑘1𝑃=𝑍(𝑅) and moreover, 𝑦𝑘=𝑥𝑟𝑘for𝑘=1,2,,𝑚. Since 𝑥20 and 𝑟𝑖𝑟𝑗𝑍(𝑅), it follows that 𝑦𝑖𝑦𝑗0 for all 𝑖,𝑗{1,2,,𝑚}. Let 𝑖,𝑗{1,2,,𝑚} with 𝑖𝑗. We claim that 𝑦𝑖𝑦𝑗. Suppose that 𝑦𝑖=𝑦𝑗. We may assume without loss of generality that 𝑖<𝑗. Then 𝑦𝑖=𝑦𝑗 implies that 𝑥𝑖+1(1++𝑥𝑗(𝑖+1))=0. As 1++𝑥𝑗(𝑖+1)𝑍(𝑅), it follows that 𝑥𝑖+1=0, and this is not possible since by assumption, 𝑥𝑚0. Hence 𝑦𝑖𝑦𝑗 for all distinct 𝑖,𝑗{1,2,,𝑚} and moreover, 𝑦𝑖𝑦𝑗0. Hence we obtain that the subgraph of (Γ(𝑅))𝑐 induced on {𝑦𝑘𝑘=1,2,,𝑚} is a clique. This implies that 𝜔((Γ(𝑅))𝑐)𝑚. This is impossible since 𝑚=𝜔((Γ(𝑅))𝑐)+1. Hence we obtain that 𝑥𝑚=0 for any 𝑥𝑃.

The next remark provides examples of rings 𝑅 for which 𝜔((Γ(𝑅))𝑐)=.

Remark 4.20. We remark here that Lemma 4.19 is motivated by [3, Theorem 3.4]. Let 𝑅 be a commutative ring with identity which is not reduced. Let 𝑥nil(𝑅). Recall that by the index of nilpotence of 𝑥, we mean the least positive integer 𝑛 such that 𝑥𝑛=0. Suppose that 𝑥nil(𝑅) with 𝑥0. Let 𝑛 be the index of nilpotence of 𝑥. Using the fact that the sum of a nilpotent element and a unit in any ring is a unit, it can be shown as in the proof of Lemma 4.19 that {𝑦𝑘=𝑥++𝑥𝑘𝑘=1,,𝑛1} is a clique in (Γ(𝑅))𝑐. Hence it follows that 𝜔((Γ(𝑅))𝑐)𝑛1. Thus if a commutative ring 𝑅 with identity is such that there is no bound on the index of nilpotence of nilpotent elements of 𝑅, then it follows that 𝜔((Γ(𝑅))𝑐)=.

Let 𝑅 and 𝑃 be as in Lemma 4.19. If 𝜔((Γ(𝑅))𝑐) is finite, then with the help of Lemma 4.19, we prove in the following proposition that 𝑃 is nilpotent.

Proposition 4.21. Let 𝑅 be a commutative ring with identity which is not an integral domain. Suppose that 𝑅 has only one maximal 𝑁-prime of (0) and let it be 𝑃. If 𝜔((Γ(𝑅))𝑐) is finite, then 𝑃 is nilpotent.

Proof. Let 𝑚=𝜔((Γ(𝑅))𝑐)+1. We claim that 𝑃𝑛=(0) with 𝑛=(𝑚1)2+1=(𝜔((Γ(𝑅))𝑐))2+1. Suppose that 𝑃𝑛(0). Then there exist 𝑥𝑘𝑃 for 𝑘=1,2,,𝑛 such that 𝑥1𝑥2𝑥𝑛0. Let s be the number of distinct elements among 𝑥1,𝑥2,,𝑥𝑛. Note that we may assume without loss of generality that 𝑥1,,𝑥𝑠 are the distinct elements among 𝑥1,𝑥2,,𝑥𝑛. Let 𝑗{1,,𝑠}, and let 𝐴𝑗={𝑘{1,2,,𝑛}𝑥𝑘=𝑥𝑗}. Let |𝐴𝑗|=𝑛𝑗 for 𝑗=1,,𝑠. Note that {1,2,,𝑛} = sj=1𝐴𝑗, and moreover, 𝐴𝑖𝐴𝑗= for all distinct 𝑖,𝑗{1,,𝑠}. Hence we obtain that 𝑛=𝑠𝑗=1|𝐴𝑗|=𝑛1++𝑛𝑠. Since 𝑥1𝑥2𝑥𝑛0, it follows that 𝑥1𝑥𝑠0, and moreover, for each 𝑗=1,,𝑠, the product of 𝑛𝑗 factors of 𝑥𝑗is different from 0. We know from Lemma 4.19 that 𝑥𝑚=0 for any 𝑥𝑃. Hence it follows that 𝑛𝑗𝑚1 for 𝑗=1,,𝑠. Furthermore, observe that the subgraph of (Γ(𝑅))𝑐 induced on {𝑥1,,𝑥𝑠} is a clique and so 𝑠𝜔((Γ(𝑅))𝑐)=𝑚1. Thus we obtain that (𝑚1)2+1=𝑛=𝑠𝑗=1𝑛𝑗𝑠(𝑚1)(𝑚1)(𝑚1). This is impossible. Hence it follows that 𝑃𝑛=(0) with 𝑛=(𝑚1)2+1.

We conclude this note with the following example of an infinite ring 𝑅 such that 𝑅 has exactly one maximal 𝑁-prime of (0) satisfying the property that 𝜔((Γ(𝑅))𝑐)=3, thereby illustrating that Proposition 4.15 need not hold for non-Noetherian rings.

Example 4.22. Let 𝑇=𝑍2[𝑥1,𝑥2,𝑥3,] be the polynomial ring in an infinite number of variables over 𝑍2. Let 𝐼 be the ideal of 𝑇 generated by {𝑥2𝑘𝑘=1,2,3,}{𝑥1𝑥𝑘𝑘=3,4,5,}{𝑥𝑖𝑥𝑗𝑖,𝑗𝐍,2𝑖<𝑗}. Let 𝑅=𝑇/𝐼. Let 𝑀 be the ideal of 𝑇 generated by {𝑥𝑘𝑘=1,2,3,}. Observe that 𝑃=𝑀/𝐼 is the only prime ideal of 𝑅. Thus 𝑍(𝑅)=𝑃. It is clear that 𝑅 is infinite. Note that 𝑥1+𝐼---(𝑥1+𝑥2)+𝐼---𝑥2+𝐼---𝑥1+𝐼 is a cycle of length 3 in (Γ(𝑅))𝑐 and hence we obtain that 𝜔((Γ(𝑅))𝑐)3. We next verify that 𝜔((Γ(𝑅))𝑐)3. Let 𝐴𝑍(𝑅) be such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique. We assert that 𝐴 can contain at most 3 elements. Suppose that 𝐴 contains more than 3 elements. Note that for any 𝑧,𝑤𝐴 with 𝑧𝑤, 𝑧𝑤0+𝐼.
Let 𝑧,𝑤𝑍(𝑅). Observe that 𝑧=𝑓1(𝑥1,𝑥2,,𝑥𝑚)+𝐼, 𝑤=𝑓2(𝑥1,𝑥2,,𝑥𝑚)+𝐼, for some 𝑚2 and for some 𝑓1(𝑥1,,𝑥𝑚), 𝑓2(𝑥1,,𝑥𝑚)Z2𝑥1+Z2𝑥2++Z2𝑥𝑚+Z2𝑥1𝑥2. It is clear that 𝑧𝑤=0+𝐼 if the coefficient of 𝑥2 is 0 in 𝑓𝑖𝑥1,,𝑥𝑚for𝑖=1,2.(4.11)
We are assuming that 𝐴 contains at least 4 elements. Let {𝑧1,𝑧2,𝑧3,𝑧4}𝐴. Let 𝑧𝑘=𝑔𝑘(𝑥1,𝑥2,,𝑥𝑚)+𝐼, for some 𝑚2 and for some 𝑔𝑘(𝑥1,𝑥2,,𝑥𝑚)Z2𝑥1+Z2𝑥2++Z2𝑥𝑚+Z2𝑥1𝑥2for𝑘=1,2,3,4. Since𝑧𝑖𝑧𝑗0+𝐼 for all distinct 𝑖,𝑗{1,2,3,4}, it follows from (4.11) that the coefficient of 𝑥2 must be 1 for all 𝑔𝑘(𝑥1,𝑥2,,𝑥𝑚)(𝑘=1,2,3,4) except possibly one value of 𝑘. We may assume without loss of generality that 𝑔𝑘(𝑥1,𝑥2,,𝑥𝑚)=𝑎𝑘1𝑥1+𝑥2++𝑎𝑘𝑚𝑥𝑚+𝑎𝑘12𝑥1𝑥2for𝑘=1,2,3 where 𝑎𝑘1,𝑎𝑘3,,𝑎𝑘𝑚,𝑎𝑘12Z2. Since 𝑧1𝑧20+𝐼, it follows that exactly one among 𝑎11,𝑎21 must be 1. We may assume without loss of generality that 𝑎11=1 and 𝑎21=0. Now either 𝑎31=0 or 𝑎31=1. If 𝑎31=0, then we arrive at𝑧2𝑧3=0+𝐼, which is impossible. If 𝑎31=1, then we obtain that 𝑧1𝑧3=0+𝐼, and this is also impossible.
This proves that if 𝐴𝑍(𝑅) is such that the subgraph of (Γ(𝑅))𝑐 induced on 𝐴 is a clique, then |𝐴|3. Hence we obtain that 𝜔((Γ(𝑅))𝑐)3. This proves that𝜔((Γ(𝑅))𝑐)=3.

Acknowledgments

The author is very much thankful to the academic editors Professor David F. Anderson, Professor Vesselin Drensky, and Professor Dolors Herbera for their useful and valuable suggestions.