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

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

S. Visweswaran

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‘₯ β‰  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‘₯ a n d 𝑦 a r e v e r t i c e s o f 𝐺 } and it is denoted by d i a m ( 𝐺 ) .

For any 𝑣 ∈ 𝑉 , the eccentricity of 𝑣 denoted by 𝑒 ( 𝑣 ) is defined as 𝑒 ( 𝑣 ) = s u p { 𝑑 ( 𝑣 , 𝑒 ) ∣ 𝑒 ∈ 𝑉 } . ( 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‘₯ ) = 2 o r 3 .

Let 𝐺 = ( 𝑉 , 𝐸 ) be a simple graph. Recall from [15, Definition, 1.1.13] that the complement of 𝐺 denoted by 𝐺 𝑐 is defined by setting 𝑉 ( 𝐺 𝑐 ) = 𝑉 and two distinct 𝑒 , 𝑣 ∈ 𝑉 are joined by an edge in 𝐺 𝑐 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 β‰  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 𝑍 𝑅 ( 𝑅 / 𝐼 ) where 𝑍 𝑅 ( 𝑅 / 𝐼 ) = { π‘₯ ∈ 𝑅 ∣ π‘₯ 𝑦 ∈ 𝐼 f o r s o m e π‘₯ ∈ 𝑅 ⧡ 𝐼 } [16]. It is well known that if { 𝑃 𝛼 } 𝛼 ∈ Ξ› is the set of all maximal 𝑁 -primes of (0) in 𝑅 , then  𝑍 ( 𝑅 ) = 𝛼 ∈ Ξ› 𝑃 𝛼 . ( 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 = ( 𝐼 ∢ 𝑅 π‘₯ ) for some π‘₯ ∈ 𝑅 [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 𝑃 1 , 𝑃 2 of ( 0 ) with 𝑃 1 ∩ 𝑃 2 β‰  ( 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 d i a m ( ( 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 d i a m ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 2 . When the condition (b) of [18, Theorem 1.1] holds, then it was shown in [18, Proposition 1.7] that d i a m ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 2 if either 𝑃 1 is not a 𝐡 -prime of ( 0 ) in 𝑅 or 𝑃 2 is not a 𝐡 -prime of ( 0 ) in 𝑅 and d i a m ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 if both 𝑃 1 and 𝑃 2 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 β‰  𝐡 , we denote it symbolically by 𝐴 βŠ‚ 𝐡 . If 𝑋 , π‘Œ are sets and if 𝑋 is not a subset of π‘Œ , we denote it symbolically 𝑋 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, d i a m ( ( 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 𝑃 1 is not a 𝐡 -prime of (0) in 𝑅 or 𝑃 2 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 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 , then it is verified that the center of ( Ξ“ ( 𝑅 ) ) 𝑐 = { π‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— ∣ 𝑃 1 β‰  ( ( 0 ) ∢ 𝑅 π‘₯ ) a n d 𝑃 2 β‰  ( ( 0 ) ∢ 𝑅 π‘₯ ) } .

Let 𝐺 = ( 𝑉 , 𝐸 ) be a graph. Recall from [15, Page 159] that the girth of 𝐺 denoted by g r ( 𝐺 ) is defined as the length of a shortest cycle in 𝐺 . If 𝐺 does not contain any cycle, then we set g r ( 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, g r ( 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 g r ( Ξ“ ( 𝑅 ) ) = 4 [5, Theorem 2.2 and Theorem 2.3], and moreover, they characterized commutative rings 𝑅 such that g r ( Ξ“ ( 𝑅 ) ) = ∞ [5, Theorem 2.4 and Theorem 2.5].

In Section 3 of this paper we study about the girth of ( Ξ“ ( 𝑅 ) ) 𝑐 where 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 g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .

Suppose that 𝑅 has only one maximal 𝑁 -prime of ( 0 ) and let it be 𝑃 . If 𝑃 2 β‰  ( 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 g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .

Suppose that 𝑅 has exactly two maximal 𝑁 -primes of ( 0 ) and let them be 𝑃 1 and 𝑃 2 . If ( Ξ“ ( 𝑅 ) ) 𝑐 is not connected, then it is shown that ( Ξ“ ( 𝑅 ) ) 𝑐 contains a cycle if and only if { 𝑃 1 ⧡ 𝑃 2 ∣ β‰₯ 3 o r | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 i f a n d o n l y i f g r ( ( 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 𝑛 β‰₯ 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 𝑛 β‰₯ 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 | n i l ( 𝑅 ) | < ∞ and n i l ( 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 n i l ( 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 𝑛 β‰₯ 2 and let 𝐾 1 , … , 𝐾 𝑛 be finite fields. Let 𝑅 = 𝐾 1 Γ— β‹― Γ— 𝐾 𝑛 . 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 = n i l ( 𝑅 ) . Moreover, if 𝑃 2 β‰  ( 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 𝑃 2 β‰  ( 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 𝐺 𝑐 is also connected. If π‘₯ is any element of 𝑉 , then 𝑒 ( π‘₯ ) β‰₯ 2 i n 𝐺 𝑐 .

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

The following lemma establishes some necessary conditions on 𝑅 in order that there exist π‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— such that 𝑑 ( π‘₯ , 𝑦 ) = 3 i n ( 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‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— with 𝑑 ( π‘₯ , 𝑦 ) = 3 i n ( 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 ) ∢ 𝑅 π‘₯ ) , ( ( 0 ) ∢ 𝑅 𝑦 ) } .

Proof. Since 𝑑 ( π‘₯ , 𝑦 ) = 3 i n ( Ξ“ ( 𝑅 ) ) 𝑐 , it follows that π‘₯ 𝑦 = 0 and for any 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— with 𝑧 βˆ‰ { π‘₯ , 𝑦 } , either 𝑧 π‘₯ = 0 or 𝑧 𝑦 = 0 . Hence we obtain that 𝑧 ∈ ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) . Since π‘₯ 𝑦 = 0 , it is clear that { 0 , π‘₯ , 𝑦 } βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) . Thus we have 𝑍 ( 𝑅 ) = ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) . Let { 𝑃 𝛼 } 𝛼 ∈ Ξ› be the set of all maximal 𝑁 -primes of ( 0 ) in 𝑅 . It is well known that ⋃ 𝑍 ( 𝑅 ) = 𝛼 ∈ Ξ› 𝑃 𝛼 , and hence we obtain that ⋃ 𝛼 ∈ Ξ› 𝑃 𝛼 = ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) . Now ( ( 0 ) ∢ 𝑅 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 ) ∢ 𝑅 π‘₯ ) βŠ† 𝑃 . Since ( ( 0 ) ∢ 𝑅 𝑦 ) ∩ ( 𝑅 ⧡ 𝑍 ( 𝑅 ) ) = βˆ… , it follows that there exists a maximal 𝑁 -prime 𝑄 of ( 0 ) in 𝑅 such that ( ( 0 ) ∢ 𝑅 𝑦 ) βŠ† 𝑄 . Now we obtain that 𝑍 ( 𝑅 ) = ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) = 𝑃 βˆͺ 𝑄 . If 𝑃 = 𝑄 , then it follows that ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) = 𝑃 is an ideal of 𝑅 . Hence it follows that either ( ( 0 ) ∢ 𝑅 π‘₯ ) βŠ† ( ( 0 ) ∢ 𝑅 𝑦 ) o r ( ( 0 ) ∢ 𝑅 𝑦 ) βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) , and so we obtain that 𝑃 is the only maximal 𝑁 -prime of ( 0 ) in 𝑅 and either 𝑃 = ( ( 0 ) ∢ 𝑅 π‘₯ ) o r 𝑃 = ( ( 0 ) ∢ 𝑅 𝑦 ) . 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 β‰  𝑄 . Now we obtain from 𝑍 ( 𝑅 ) = ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) = 𝑃 βˆͺ 𝑄 that 𝑃 and 𝑄 are the only maximal 𝑁 -primes of ( 0 ) in 𝑅 , and moreover, 𝑃 = ( ( 0 ) ∢ 𝑅 π‘₯ ) a n d 𝑄 = ( ( 0 ) ∢ 𝑅 𝑦 ) .

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 ) ∢ 𝑅 π‘₯ ) and 𝑄 = ( ( 0 ) ∢ 𝑅 𝑦 ) for some π‘₯ , 𝑦 ∈ 𝑅 . 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 𝑃 βŠ„ 𝑄 or 𝑄 βŠ„ 𝑃 . Without loss of generality, we may assume that 𝑃 βŠ„ 𝑄 . Let w ∈ 𝑃 ⧡ 𝑄 . Now 𝑀 π‘₯ = 0 ∈ 𝑄 and as 𝑀 βˆ‰ 𝑄 , it follows that π‘₯ ∈ 𝑄 = ( ( 0 ) ∢ 𝑅 𝑦 ) . 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 𝑃 1 and 𝑃 2 . If ( Ξ“ ( 𝑅 ) ) 𝑐 is connected and if 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) for some 𝑒 , 𝑣 ∈ 𝑅 , 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 𝑒 𝑣 = 0 . Thus 𝑒 , 𝑣 ∈ 𝑍 ( 𝑅 ) βˆ— and 𝑒 and 𝑣 are not adjacent in ( Ξ“ ( 𝑅 ) ) 𝑐 . Since 𝑃 1 and 𝑃 2 are the only maximal 𝑁 -primes of ( 0 ) in 𝑅 , it follows that 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑒 ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑣 ) . Thus for any 𝑀 ∈ 𝑍 ( 𝑅 ) βˆ— , either 𝑀 𝑒 = 0 o r 𝑀 𝑣 = 0 . This shows that 𝑑 ( 𝑒 , 𝑣 ) β‰₯ 3 i n ( Ξ“ ( 𝑅 ) ) 𝑐 . It is shown in the proof of [18,Proposition 1.7] that 𝑑 ( π‘₯ , 𝑦 ) ≀ 3 i n ( Ξ“ ( 𝑅 ) ) 𝑐 for any π‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— . Hence we obtain that 𝑑 ( 𝑒 , 𝑣 ) = 3 i n ( Ξ“ ( 𝑅 ) ) 𝑐 and so it follows that 𝑒 ( 𝑒 ) = 𝑒 ( 𝑣 ) = 3 i n ( 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‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑒 ( π‘₯ ) β‰₯ 2 i n ( 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‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑒 ( π‘₯ ) ≀ 2 i n ( 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 any π‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑒 ( π‘₯ ) = 2 i n ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence we obtain in the cases mentioned above that d i a m ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = π‘Ÿ ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 2 .
Assume that 𝑅 has exactly two maximal 𝑁 -primes of ( 0 ) and let them be 𝑃 1 a n d 𝑃 2 . In such a case, it was shown in the proof of [18, Proposition 1.7(i)] that there exist π‘Ž ∈ 𝑃 1 ⧡ 𝑃 2 and 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 such that π‘Ž 𝑏 β‰  0 . It follows that 𝑃 1 β‰  ( ( 0 ) ∢ 𝑅 π‘Ž ) , 𝑃 1 β‰  ( ( 0 ) ∢ 𝑅 𝑏 ) , a n d 𝑃 2 β‰  ( ( 0 ) ∢ 𝑅 π‘Ž ) , 𝑃 2 β‰  ( ( 0 ) ∢ 𝑅 𝑏 ) . Now it follows from Lemma 2.2 that 𝑑 ( π‘Ž , 𝑦 ) ≀ 2 and 𝑑 ( 𝑏 , 𝑦 ) ≀ 2 for any 𝑦 ∈ 𝑍 ( 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‰₯ 2 i n ( Ξ“ ( 𝑅 ) ) 𝑐 for any 𝑐 ∈ 𝑍 ( 𝑅 ) βˆ— , it follows that 𝑒 ( π‘Ž ) = 𝑒 ( 𝑏 ) = 2 i n ( 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 d i a m ( ( 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‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑒 ( π‘₯ ) ≀ 2 i n ( Ξ“ ( 𝑅 ) ) 𝑐 . Thus we obtain, in view of Lemma 2.1 that 𝑒 ( π‘₯ ) = 2 for any π‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— . Hence in this case also we obtain that d i a m ( ( 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 { 𝑃 1 , 𝑃 2 } 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‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— ∣ 𝑃 1 β‰  ( ( 0 ) ∢ 𝑅 π‘₯ ) a n d 𝑃 2 β‰  ( ( 0 ) ∢ 𝑅 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‘₯ ∈ 𝑀 , 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‘₯ 𝑉 ∢ 𝑉 𝑦 ) for some 𝑦 ∈ 𝑉 . Since 𝑀 β‰  𝑉 , it follows that 𝑦 βˆ‰ π‘₯ 𝑉 . As 𝑉 is a valuation domain, we obtain that π‘₯ ∈ 𝑦 𝑉 . Thus π‘₯ = 𝑦 𝑣 for some 𝑣 ∈ 𝑉 . Hence we obtain that 𝑀 = ( π‘₯ 𝑉 ∢ 𝑉 𝑦 ) = ( 𝑦 𝑣 𝑉 ∢ 𝑉 𝑦 ) = 𝑣 𝑉 . 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 d i a m ( ( 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 Γ— 𝑍 (resp. 𝑇 1 = 𝑅 Γ— 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 𝑃 1 = 𝑃 Γ— 𝑍 a n d 𝑃 2 = 𝑅 Γ— ( 0 ) (resp. 𝑄 1 = 𝑃 Γ— 𝑅 a n d 𝑄 2 = 𝑅 Γ— 𝑃 ) are the only maximal 𝑁 -primes of the zero-ideal of 𝑇 (resp. of the zero-ideal of 𝑇 1 ) and 𝑃 1 ∩ 𝑃 2 = 𝑃 Γ— ( 0 ) is not the zero-ideal of 𝑇 (resp. 𝑄 1 ∩ 𝑄 2 = 𝑃 Γ— 𝑃 is not the zero-ideal of 𝑇 1 ). Hence it follows from [18, Theorem 1.1(b)] that ( Ξ“ ( 𝑇 ) ) 𝑐 (resp. ( Ξ“ ( 𝑇 1 ) ) 𝑐 ) is connected. Since 𝑃 1 is not a 𝐡 -prime of ( 0 ) in 𝑇 (resp. 𝑄 1 and 𝑄 2 are not 𝐡 -primes of ( 0 ) in 𝑇 1 ), it follows from Remark 2.6 that d i a m ( ( Ξ“ ( 𝑇 ) ) 𝑐 ) = π‘Ÿ ( ( Ξ“ ( 𝑇 ) ) 𝑐 ) = 2 (resp. d i a m ( ( Ξ“ ( 𝑇 1 ) ) 𝑐 ) = π‘Ÿ ( ( Ξ“ ( 𝑇 1 ) ) 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 / 1 2 𝑍 . Note that 𝑃 1 = 2 𝑍 / 1 2 𝑍 a n d 𝑃 2 = 3 𝑍 / 1 2 𝑍 are the only prime ideals of the finite ring 𝑅 = 𝑍 / 1 2 𝑍 . Thus 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 . Observe that 𝑅 has exactly two maximal 𝑁 -primes of ( 0 ) and 𝑃 1 ∩ 𝑃 2 = 6 𝑍 / 1 2 𝑍 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 2 𝑍 = ( 1 2 𝑍 ∢ 𝑍 Β± 6 𝑑 ) and { Β± 6 𝑑 ∣ 𝑑 ∈ 𝑍 i s o d d a n d p o s i t i v e } is the set of all integers with the property that 2 𝑍 = ( 1 2 𝑍 ∢ 𝑍 Β± 6 𝑑 ) . Furthermore, note that { Β± 4 π‘˜ ∣ π‘˜ ∈ Z i s p o s i t i v e a n d π‘˜ ≑ 1 o r 2 ( m o d 3 ) } is the set of all integers with the property that 3 𝑍 = ( 1 2 𝑍 ∢ 𝑍 Β± 4 π‘˜ ) . Hence it follows that 𝑃 1 = ( ( 0 + 1 2 𝑍 ) ∢ 𝑅 6 + 1 2 𝑍 ) , 𝑃 2 = ( ( 0 + 1 2 Z ) ∢ 𝑅 4 + 1 2 𝑍 ) , a n d 𝑃 2 = ( ( 0 + 1 2 𝑍 ) ∢ 𝑅 8 + 1 2 𝑍 ) . Thus 𝑃 1 and 𝑃 2 are 𝐡 -primes of ( 0 ) in 𝑅 . Now it follows from [18, Proposition 1.7(b)] that d i a m ( ( 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 + 1 2 𝑍 , 1 0 + 1 2 𝑍 , 3 + 1 2 𝑍 , 9 + 1 2 𝑍 } .

Example 2.9. Let 𝑛 > 1 be such that 𝑛 admits at least three distinct prime divisors. Let { 𝑝 1 , 𝑝 2 , 𝑝 3 , … , 𝑝 𝑑 } ( 𝑑 β‰₯ 3 ) be the set of all distinct prime divisors of 𝑛 . Let 𝑅 = 𝑍 / 𝑛 𝑍 . Note that { 𝑝 1 𝑍 / 𝑛 𝑍 , 𝑝 2 𝑍 / 𝑛 𝑍 , … , 𝑝 𝑑 𝑍 / 𝑛 𝑍 } 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 d i a m ( ( 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 g r ( ( 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‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑃 βŠ„ ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ 𝑅 𝑦 .

Proof. Suppose that for some π‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑃 βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ 𝑅 𝑦 . Then either 𝑃 βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) o r 𝑃 βŠ† 𝑅 𝑦 . Since π‘₯ β‰  0 , ( ( 0 ) ∢ 𝑅 π‘₯ ) ∩ ( 𝑅 ⧡ 𝑍 ( 𝑅 ) ) = βˆ… . As 𝑦 ∈ 𝑍 ( 𝑅 ) , there exists 𝑧 ∈ 𝑅 ⧡ { 0 } such that 𝑦 𝑧 = 0 . Hence we obtain that 𝑅 𝑦 βŠ† ( ( 0 ) ∢ 𝑅 𝑧 ) . Note that ( ( 0 ) ∢ 𝑅 𝑧 ) ∩ ( 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 ) ∢ 𝑅 π‘₯ ) βŠ† 𝑄 and there exists a maximal 𝑁 -prime π‘Š of ( 0 ) in 𝑅 such that ( ( 0 ) ∢ 𝑅 𝑧 ) βŠ† π‘Š . If 𝑃 βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) , then we obtain that 𝑃 βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) βŠ† 𝑄 and hence it follows that 𝑃 = 𝑄 = ( ( 0 ) ∢ 𝑅 π‘₯ ) . This contradicts the assumption that 𝑃 is not a 𝐡 -prime of ( 0 ) in 𝑅 . If 𝑃 βŠ† 𝑅 𝑦 , then 𝑃 βŠ† ( ( 0 ) ∢ 𝑅 𝑧 ) βŠ† π‘Š and so 𝑃 = π‘Š = ( ( 0 ) ∢ 𝑅 𝑧 ) . 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‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— , 𝑃 βŠ„ ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ 𝑅 𝑦 .

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 g r ( ( 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‘₯ ∈ 𝑃 ⧡ { 0 } . By hypothesis, 𝑃 is not a 𝐡 -prime of ( 0 ) in 𝑅 and so Lemma 3.1 implies that there exists 𝑦 ∈ 𝑃 such that 𝑦 βˆ‰ 𝑅 π‘₯ and 𝑦 π‘₯ β‰  0 . We assert that π‘₯ 𝑦 βˆ‰ { π‘₯ , 𝑦 } . If π‘₯ 𝑦 = π‘₯ , then π‘₯ ( 1 βˆ’ 𝑦 ) = 0 . As 𝑦 ∈ 𝑃 , 1 βˆ’ 𝑦 βˆ‰ 𝑃 = 𝑍 ( 𝑅 ) . Hence π‘₯ ( 1 βˆ’ 𝑦 ) = 0 implies that π‘₯ = 0 . This contradicts the fact that π‘₯ β‰  0 . Similarly, it follows that π‘₯ 𝑦 β‰  𝑦 . If both π‘₯ 2 𝑦 and 𝑦 2 π‘₯ 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‘₯ 2 𝑦 = 0 or 𝑦 2 π‘₯ = 0 . Without loss of generality, we may assume that π‘₯ 2 𝑦 = 0 . As 𝑃 is not a 𝐡 -prime of ( 0 ) in 𝑅 , Lemma 3.1 implies that there exists 𝑧 ∈ 𝑃 such that 𝑧 π‘₯ 𝑦 β‰  0 and 𝑧 βˆ‰ 𝑅 𝑦 . From 𝑧 π‘₯ 𝑦 β‰  0 , it follows that 𝑧 π‘₯ β‰  0 , and 𝑧 𝑦 β‰  0 and moreover, as π‘₯ 2 𝑦 = 0 , it follows that 𝑧 β‰  π‘₯ . Furthermore, by the choice of 𝑧 , it follows that 𝑧 β‰  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 g r ( ( 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 , 𝑐 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ 𝑃 for some prime ideal 𝑃 of 𝑅 , then g r ( ( 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‘Ž 𝑏 , 𝑏 𝑐 , 𝑐 π‘Ž ∈ 𝑅 ⧡ 𝑃 and so π‘Ž 𝑏 , 𝑏 𝑐 , 𝑐 π‘Ž ∈ 𝑅 ⧡ { 0 } . Hence it follows that π‘Ž - - - 𝑏 - - - 𝑐 - - - π‘Ž is a cycle of length 3 in ( Ξ“ ( 𝑅 ) ) 𝑐 . This proves that g r ( ( 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 𝑃 1 and 𝑃 2 . Then the following hold: (i)If 𝑃 1 ∩ 𝑃 2 = ( 0 ) , then ( Ξ“ ( 𝑅 ) ) 𝑐 contains a cycle if and only if either | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 or | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 if and only if g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 . (ii)If 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , then g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 . Thus, if ( Ξ“ ( 𝑅 ) ) 𝑐 is connected, then g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .

Proof. By hypothesis, 𝑃 1 and 𝑃 2 are the only maximal 𝑁 -primes of ( 0 ) in 𝑅 . So, it follows that 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 .
(i) Assume that ( Ξ“ ( 𝑅 ) ) 𝑐 contains a cycle. Let π‘Ž 1 - - - π‘Ž 2 - - - π‘Ž 3 - - - β‹― - - - π‘Ž 𝑛 - - - π‘Ž 1 be a cycle of length 𝑛 in ( Ξ“ ( 𝑅 ) ) 𝑐 . Note that 𝑛 β‰₯ 3 and { π‘Ž 𝑖 ∣ 𝑖 = 1 , 2 , 3 , … , 𝑛 } βŠ† 𝑍 ( 𝑅 ) βˆ— with π‘Ž 1 π‘Ž 2 , π‘Ž 2 π‘Ž 3 , … , π‘Ž 𝑛 βˆ’ 1 π‘Ž 𝑛 , π‘Ž 𝑛 π‘Ž 1 ∈ 𝑅 ⧡ { 0 } . Since 𝑃 1 ∩ 𝑃 2 = ( 0 ) , it follows that either { π‘Ž 1 , π‘Ž 2 , π‘Ž 3 , … , π‘Ž 𝑛 } βŠ† 𝑃 1 ⧡ 𝑃 2 o r { π‘Ž 1 , π‘Ž 2 , π‘Ž 3 , … , π‘Ž 𝑛 } βŠ† 𝑃 2 ⧡ 𝑃 1 . Now it is clear that either | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 or | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 .
Conversely, suppose that either | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 or | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 . Since 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 and as 𝑃 1 and 𝑃 2 are prime ideals of 𝑅 , it follows from Lemma 3.3 that g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .
If g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 , then we obtain that ( Ξ“ ( 𝑅 ) ) 𝑐 contains a cycle of length 3.
(ii) Suppose that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . We know from the proof of [18, Proposition 1.7(i)] that there exist π‘Ž ∈ 𝑃 1 ⧡ 𝑃 2 and 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 such that π‘Ž 𝑏 β‰  0 . We consider two cases : Case(A): 𝑃 1 ∩ 𝑃 2 βŠ„ ( ( 0 ) ∢ 𝑅 π‘Ž ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑏 ) . Then there exists 𝑐 ∈ 𝑃 1 ∩ 𝑃 2 such that π‘Ž 𝑐 β‰  0 and 𝑏 𝑐 β‰  0 . Hence we obtain a cycle π‘Ž - - - 𝑏 - - - 𝑐 - - - π‘Ž in ( Ξ“ ( 𝑅 ) ) 𝑐 and it is of length 3. Case(B): 𝑃 1 ∩ 𝑃 2 βŠ† ( ( 0 ) ∢ 𝑅 π‘Ž ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑏 ) . Then either 𝑃 1 ∩ 𝑃 2 βŠ† ( ( 0 ) ∢ 𝑅 π‘Ž ) or 𝑃 1 ∩ 𝑃 2 βŠ† ( ( 0 ) ∢ 𝑅 𝑏 ) .
Without loss of generality, we may assume that 𝑃 1 ∩ 𝑃 2 βŠ† ( ( 0 ) ∢ 𝑅 π‘Ž ) . Let π‘₯ ∈ 𝑃 1 ∩ 𝑃 2 be such that π‘₯ β‰  0 . Since π‘Ž + 𝑏 βˆ‰ 𝑍 ( 𝑅 ) , we obtain that ( π‘Ž + 𝑏 ) π‘₯ β‰  0 . As π‘Ž π‘₯ = 0 , it follows that 𝑏 π‘₯ = ( π‘Ž + 𝑏 ) π‘₯ β‰  0 . Moreover, observe that 𝑏 + π‘₯ ∈ 𝑃 2 ⧡ 𝑃 1 , 𝑏 β‰  𝑏 + π‘₯ , and π‘Ž ( 𝑏 + π‘₯ ) = π‘Ž 𝑏 β‰  0 . As 𝑏 ( 𝑏 + π‘₯ ) ∈ 𝑃 2 ⧡ 𝑃 1 , 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 g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .
We know from [18, Theorem 1.1(b)] that ( Ξ“ ( 𝑅 ) ) 𝑐 is connected if and only if 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . Thus if 𝑅 has exactly two maximal 𝑁 -primes of ( 0 ) and if ( Ξ“ ( 𝑅 ) ) 𝑐 is connected, then g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .

We show in the next lemma that if 𝑅 has at least three maximal 𝑁 -primes of ( 0 ) , then g r ( ( 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 g r ( ( 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 𝑃 βŠ„ 𝑄 βˆͺ π‘Š , 𝑄 βŠ„ 𝑃 βˆͺ W, and π‘Š βŠ„ 𝑃 βˆͺ 𝑄 . Hence there exist elements π‘₯ ∈ 𝑃 ⧡ ( 𝑄 βˆͺ π‘Š ) , 𝑦 ∈ 𝑄 ⧡ ( 𝑃 βˆͺ π‘Š ) , and 𝑧 ∈ π‘Š ⧡ ( 𝑃 βˆͺ 𝑄 ) . Note that π‘₯ , 𝑦 , 𝑧 are distinct elements of 𝑍 ( 𝑅 ) βˆ— with π‘₯ 𝑦 , 𝑦 𝑧 , 𝑧 π‘₯ ∈ 𝑅 ⧡ { 0 } . Hence π‘₯ - - - 𝑦 - - - 𝑧 - - - π‘₯ is a cycle of length 3 in ( Ξ“ ( 𝑅 ) ) 𝑐 . This proves that g r ( ( 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 g r ( ( 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 𝑃 2 β‰  (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 g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .

Proof. Note that 𝑍 ( 𝑅 ) = 𝑃 . Assume that 𝑃 2 β‰  ( 0 ) and ( Ξ“ ( 𝑅 ) ) 𝑐 is not connected. From the assumption that 𝑃 2 β‰  ( 0 ) , it follows that there exist π‘Ž , 𝑏 ∈ 𝑃 such that π‘Ž β‰  𝑏 and π‘Ž 𝑏 β‰  0 . Moreover, by [18, Theorem 1.1(a)], 𝑃 = ( ( 0 ) ∢ 𝑅 𝑐 ) for some 𝑐 ∈ 𝑅 . It is clear that 𝑐 ∈ 𝑃 ⧡ { 0 } . Furthermore, as π‘Ž 𝑏 β‰  0 , it follows that π‘Ž β‰  𝑐 and 𝑏 β‰  𝑐 .
Suppose that ( Ξ“ ( 𝑅 ) ) 𝑐 contains a cycle. Let π‘Ž 1 - - - π‘Ž 2 - - - π‘Ž 3 - - - β‹― - - - π‘Ž 𝑛 - - - π‘Ž 1 be a cycle of length 𝑛 in ( Ξ“ ( 𝑅 ) ) 𝑐 . Note that 𝑛 β‰₯ 3 and { π‘Ž 1 , π‘Ž 2 , π‘Ž 3 , … , π‘Ž 𝑛 } βŠ† 𝑃 ⧡ { 0 } . As 𝑃 = ( ( 0 ) ∢ 𝑅 𝑐 ) and since π‘Ž 1 π‘Ž 2 , π‘Ž 2 π‘Ž 3 , … , π‘Ž 𝑛 βˆ’ 1 π‘Ž 𝑛 , a n d π‘Ž 𝑛 π‘Ž 1 ∈ 𝑅 ⧡ { 0 } , it follows that 𝑐 βˆ‰ { π‘Ž 1 , π‘Ž 2 , π‘Ž 3 , … , π‘Ž 𝑛 } . Thus { 0 , 𝑐 , π‘Ž 1 , π‘Ž 2 , π‘Ž 3 , … , π‘Ž 𝑛 } βŠ† 𝑃 and hence it follows that | 𝑃 | β‰₯ 5 .
We next show that g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 if | 𝑃 | β‰₯ 5 . Suppose that | 𝑃 | β‰₯ 5 . Observe that there exists 𝑑 ∈ 𝑃 ⧡ { 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‘Ž 𝑑 β‰  0 and 𝑏 𝑑 β‰  0 , then π‘Ž - - - 𝑑 - - - 𝑏 - - - π‘Ž is a cycle of length 3 in ( Ξ“ ( 𝑅 ) ) 𝑐 . Suppose that either π‘Ž 𝑑 = 0 or 𝑏 𝑑 = 0 .
Observe that if π‘Ž 2 = 0 and 𝑏 2 = 0 , then from ( π‘Ž + 𝑏 ) π‘Ž = π‘Ž 𝑏 = ( π‘Ž + 𝑏 ) 𝑏 = π‘Ž 𝑏 , and from the fact that π‘Ž + 𝑏 ∈ 𝑃 ⧡ { 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‘Ž 2 β‰  0 . As 𝑐 2 = 0 and π‘Ž 2 β‰  0 , it follows that π‘Ž + 𝑐 β‰  0 . If π‘Ž + 𝑐 β‰  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 𝑐 𝑑 = 0 , it follows from the assumption that either π‘Ž 𝑑 = 0 or 𝑏 𝑑 = 0 that π‘Ž 𝑑 = 𝑏 𝑑 = 0 . As 𝑐 β‰  𝑑 and π‘Ž + 𝑐 = 𝑏 , it follows that π‘Ž + 𝑑 β‰  𝑏 and it is clear that π‘Ž ( π‘Ž + 𝑑 ) = π‘Ž 2 β‰  0 and ( π‘Ž + 𝑑 ) 𝑏 = π‘Ž 𝑏 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 g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .
If g r ( ( 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 𝑃 2 β‰  ( 0 ) . Suppose that ( Ξ“ ( 𝑅 ) ) 𝑐 is not connected. Let π‘Ž , 𝑏 ∈ 𝑃 be such that π‘Ž β‰  𝑏 and π‘Ž 𝑏 β‰  0 . Let 𝑐 ∈ 𝑃 ⧡ { 0 } be such that 𝑃 = ( ( 0 ) ∢ 𝑅 𝑐 ) . Assume that g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) β‰  3 . Observe that by Proposition 3.7, g r ( ( 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† 𝑃 , it follows that 𝑃 = { 0 , π‘Ž , 𝑏 , 𝑐 } . Now π‘Ž 𝑏 ∈ 𝑃 ⧡ { 0 } = { π‘Ž , 𝑏 , 𝑐 } . We assert that π‘Ž 𝑏 = 𝑐 . If π‘Ž 𝑏 = π‘Ž , then π‘Ž ( 1 βˆ’ 𝑏 ) = 0 . This is impossible since π‘Ž β‰  0 and 1 βˆ’ 𝑏 βˆ‰ 𝑃 = 𝑍 ( 𝑅 ) . Similarly, it follows that π‘Ž 𝑏 β‰  𝑏 . Hence π‘Ž 𝑏 = 𝑐 . Now 𝑍 ( 𝑅 ) = 𝑃 = { 0 , π‘Ž , 𝑏 , 𝑐 = π‘Ž 𝑏 } 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 { 𝑍 8 , 𝑍 4 [ π‘₯ ] / ( 2 π‘₯ 𝑍 4 [ π‘₯ ] + ( π‘₯ 2 βˆ’ 2 ) 𝑍 4 [ π‘₯ ] ) , 𝑍 2 [ π‘₯ ] / π‘₯ 3 𝑍 2 [ π‘₯ ] } where 𝑍 4 [ π‘₯ ] (resp. 𝑍 2 [ π‘₯ ] ) is the polynomial ring in one variable over 𝑍 4 (resp. over 𝑍 2 ).
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 𝑄 βŠ† 𝑍 ( 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 𝑒 π‘Ž 𝑏 ∈ 𝑃 ⧡ { 0 } = { π‘Ž , 𝑏 , 𝑐 = π‘Ž 𝑏 } . We claim that 𝑒 π‘Ž 𝑏 = π‘Ž 𝑏 . If 𝑒 π‘Ž 𝑏 = π‘Ž , then π‘Ž ( 1 βˆ’ 𝑒 𝑏 ) = 0 and this is impossible. Similarly, we obtain that 𝑒 π‘Ž 𝑏 β‰  𝑏 . Hence 𝑒 π‘Ž 𝑏 = π‘Ž 𝑏 . This implies that ( 𝑒 βˆ’ 1 ) π‘Ž 𝑏 = 0 . Since π‘Ž 𝑏 β‰  0 , it follows that 𝑒 βˆ’ 1 ∈ 𝑍 ( 𝑅 ) = 𝑃 = { 0 , π‘Ž , 𝑏 , 𝑐 = π‘Ž 𝑏 } . Hence we obtain that 𝑒 ∈ { 1 , 1 + π‘Ž , 1 + 𝑏 , 1 + 𝑐 = 1 + π‘Ž 𝑏 } . Thus it is shown that 𝑅 ⧡ 𝑃 = { 1 , 1 + π‘Ž , 1 + 𝑏 , 1 + 𝑐 } . Hence 𝑅 = { 0 , π‘Ž , 𝑏 , 𝑐 = π‘Ž 𝑏 , 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 { 𝑍 8 , 𝑍 4 [ π‘₯ ] / ( 2 π‘₯ 𝑍 4 [ π‘₯ ] + ( π‘₯ 2 βˆ’ 2 ) 𝑍 4 [ π‘₯ ] ) , 𝑍 2 [ π‘₯ ] / π‘₯ 3 𝑍 2 [ π‘₯ ] } . Observe that if 𝑅 is a ring such that 𝑅 ∈ { 𝑍 8 , 𝑍 4 [ π‘₯ ] / ( 2 π‘₯ 𝑍 4 [ π‘₯ ] + ( π‘₯ 2 βˆ’ 2 ) 𝑍 4 [ π‘₯ ] ) , 𝑍 2 [ π‘₯ ] / π‘₯ 3 𝑍 2 [ π‘₯ ] } , then 𝑅 satisfies the following conditions: 𝑅 has exactly one maximal 𝑁 -prime of ( 0 ) , say 𝑃 such that 𝑃 2 β‰  ( 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 𝑃 2 β‰  ( 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 { 𝑍 8 , 𝑍 4 [ π‘₯ ] / ( 2 π‘₯ 𝑍 4 [ π‘₯ ] + ( π‘₯ 2 βˆ’ 2 ) 𝑍 4 [ π‘₯ ] ) , 𝑍 2 [ π‘₯ ] / π‘₯ 3 𝑍 2 [ 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 𝑃 1 and 𝑃 2 . Suppose that 𝑃 1 ∩ 𝑃 2 = ( 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 | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 , or | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 if and only if g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 . Suppose that | 𝑃 1 ⧡ 𝑃 2 | < 3 , | 𝑃 2 ⧡ 𝑃 1 | < 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 𝐾 1 Γ— 𝐾 2 where 𝐾 1 and 𝐾 2 are fields with either | 𝐾 1 | = | 𝐾 2 | = 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 ) = 𝑃 1 βˆͺ 𝑃 2 and 𝑃 1 ∩ 𝑃 2 = ( 0 ) , it follows that at least one between 𝑃 1 and 𝑃 2 contains exactly 3 elements. This implies that either | 𝑃 1 | = | 𝑃 2 | = 3 or exactly one between 𝑃 1 and 𝑃 2 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 𝑃 1 and 𝑃 2 are maximal ideals of 𝑅 . As 𝑃 1 ∩ 𝑃 2 = ( 0 ) , it follows from the Chinese Remainder Theorem [21, Proposition 1.10] that 𝑅 β‰ˆ ( 𝑅 / 𝑃 1 ) Γ— ( 𝑅 / 𝑃 2 ) as rings. Thus we obtain from the above discussion that R is isomorphic to the direct product of two fields 𝐾 1 and 𝐾 2 with either | 𝐾 1 | = | 𝐾 2 | = 3 or one between 𝐾 1 and 𝐾 2 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 𝐾 1 Γ— 𝐾 2 where 𝐾 1 and 𝐾 2 are fields with either | 𝐾 1 | = | 𝐾 2 | = 3 or one of them contains exactly 3 elements and the other contains exactly 2 elements. Conversely, if 𝑅 is isomorphic to 𝐾 1 × πΎ 2 where 𝐾 1 and 𝐾 2 are fields with either | 𝐾 1 | = | 𝐾 2 | = 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‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— such that π‘₯ 𝑦 β‰  0 , then g r ( ( Ξ“ ( 𝑅 ) ) 𝑐 ) = 3 .
(ii) Let 𝑅 be any ring admitting elements π‘Ž , 𝑏 ∈ 𝑍 ( 𝑅 ) βˆ— such that π‘Ž 𝑏 β‰  0 . Then g r ( ( Ξ“ ( 𝑅 [ π‘₯ ] ) ) 𝑐 ) = g r ( ( 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‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— with π‘₯ β‰  𝑦 such that π‘₯ 𝑦 β‰  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 | ≀ 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‘₯ 𝑖 ∈ 𝑍 ( 𝑅 ) ⧡ 𝑃 . Since 𝑃 is a prime ideal of R and as π‘₯ 𝑖 βˆ‰ 𝑃 for 𝑖 = 1 , 2 , 3 , … , it follows that π‘₯ 𝑖 π‘₯ 𝑗 β‰  0 for all 𝑖 , 𝑗 ∈ { 1 , 2 , 3 , … } . Observe that the subgraph of ( Ξ“ ( 𝑅 ) ) 𝑐 induced on { π‘₯ 𝑖 ∣ 𝑖 = 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 ) ⧡ 𝑃 ∣ β‰₯ 𝑛 + 1 . Let { π‘₯ 1 , … , π‘₯ 𝑛 + 1 } βŠ† 𝑍 ( 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 = { 𝑄 ∣ 𝑄 is a prime ideal of 𝑅 such that 𝑄 βŠ† 𝑍 ( 𝑅 ) but 𝑄 βŠ„ 𝑃 } . 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 𝑄 𝑖 ∈ 𝐴 for each 𝑖 = 1 , 2 , 3 , … with 𝑄 𝑖 and 𝑄 𝑗 not being comparable under inclusion for all 𝑖 , 𝑗 ∈ { 1 , 2 , 3 , … } with 𝑖 β‰  𝑗 . Now 𝑄 𝑖 βŠ„ 𝑃 for 𝑖 = 1 , 2 , 3 , … . Hence it is possible to choose π‘₯ 1 ∈ 𝑄 1 ⧡ 𝑃 . Let 𝑖 β‰₯ 2 . Now it follows from prime avoidance [21, Proposition 1.11(i)] that there exists π‘₯ 𝑖 ∈ 𝑄 𝑖 ⧡ ( 𝑃 βˆͺ 𝑄 1 βˆͺ β‹… β‹… β‹… βˆͺ 𝑄 𝑖 βˆ’ 1 ) . Observe that { π‘₯ 𝑖 ∣ 𝑖 = 1 ,