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

Authors: S. Visweswaran
Journal: ISRN Algebra
 (2011)

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## Abstract

LetR 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.


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## Body

## 1. Introduction

All rings considered in this note are nonzero commutative rings with identity.  Unless otherwise specified, we consider ringsR such that R admits at least two nonzero zero-divisors.LetR 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 [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)* 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 x≠y, 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)∣xandyare vertices ofG} and it is denoted by diam(G).For anyv∈V, the eccentricity of v denoted by e(v) is defined as(1.1)e(v)=sup⁡{d(v,u)∣u∈V}.The set of vertices ofG 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 ringR 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.LetG=(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,v∈V 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. LetI be an ideal of a ring R,I≠R. A prime ideal P of R is said to be a maximal N-prime of I in R if P is maximal with respect to the property of being contained in ZR(R/I) where ZR(R/I)={x∈R∣xy∈Ifor somex∈R∖I} [16].   It is well known that if {Pα}α∈Λ is the set of all maximal N-primes of (0) in R, then(1.2)Z(R)=⋃α∈ΛPα.LetI 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:Rx) for some x∈R [17].  In this case, we say that P is a B-prime of I.LetR 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 P1∩P2≠(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 setA, we denote by |A|, the cardinality of A. Whenever a set A is a subset of a set B and A≠B, we denote it symbolically by A⊂B. If X,Y are sets and if X is not a subset of Y, we denote it symbolically X⊄Y.LetR 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={x∈Z(R)*∣P1≠((0):Rx)andP2≠((0):Rx)}.LetG=(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].LetR 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 Section3 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 thatR 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 thatR 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 {P1∖P2∣≥3or|P2∖P1|≥3if and only ifgr((Γ(R))c)=3.LetG=(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 n≥1 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 n≥1.LetR 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].LetR 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 n≥2   and let  K1,…,Kn be finite fields. Let R=K1×⋯×Kn. We describe a method of determining ω((Γ(R))c).LetR 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).LetR 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 ringR 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

LetR 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.
LetG=(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.
Letx∈V. Since |V|≥2 and G is connected, it follows that there exists y∈V 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 onR in order that there exist x,y∈Z(R)*  such that d(x,y)=3in(Γ(R))c.Lemma 2.2.
LetR be a ring with |Z(R)*|≥2.  Suppose that (Γ(R))c is connected.  If there exist x,y∈Z(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.
Sinced(x,y)=3in(Γ(R))c, it follows that xy=0 and for any z∈Z(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}α∈Λ be the set of all maximal N-primes of (0) in R. It is well known that Z(R)=⋃α∈ΛPα, and hence we obtain that ⋃α∈ΛPα=((0):Rx)∪((0):Ry). Now ((0):Rx)∩(R∖Z(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)∩(R∖Z(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)=P∪Q. 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 P≠Q. Now we obtain from Z(R)=((0):Rx)∪((0):Ry)=P∪Q 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.
LetR 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,y∈R. Then xy=0.Proof.
For the sake of completeness, we give below an argument for the fact thatxy = 0. Since P≠Q. either P⊄Q or Q⊄P. Without loss of generality, we may assume that P⊄Q.  Let w∈P∖Q. Now wx=0∈Q and as w∉Q, it follows that x∈Q=((0):Ry). Hence xy=0.We provide in the next lemma some sufficient conditions onR in order that (Γ(R))c admits vertices x such that e(x)=3 in (Γ(R))c.Lemma 2.4.
LetR 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,v∈R, 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 Lemma2.3 that uv=0.  Thus u,v∈Z(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)=P1∪P2=((0):Ru)∪((0):Rv). Thus for any w∈Z(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,y∈Z(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.
LetR 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 thatΓ(R) is connected [1, Theorem 2.3].  Hence it follows from Lemma 2.1 that for any x∈Z(R)*,e(x)≥2in(Γ(R))c.
IfR 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 x∈Z(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 anyx∈Z(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 thatR 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 exista∈P1∖P2 and b∈P2∖P1 such that ab≠0.  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 y∈Z(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 c∈Z(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.
LetR 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.
IfR 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 x∈Z(R)*,e(x)≤2in(Γ(R))c. Thus we obtain, in view of Lemma 2.1 that e(x)=2for anyx∈Z(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 thatR 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={x∈Z(R)*∣P1≠((0):Rx)andP2≠((0):Rx)}.We next present some examples to illustrate the results proved in this section.Example 2.7.
LetV be a rank 1 valuation domain which is not discrete. Let M denote the unique maximal ideal of V.  Let x∈M,x≠0. 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=(xV:Vy) for some y∈V.  Since M≠V,  it follows that y∉xV. As V is a valuation domain, we obtain that x∈yV.  Thus x=yv for some v∈V.  Hence we obtain that M=(xV:Vy)=(yvV:Vy)=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) LetR 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 P1∩P2=P×(0) is not the zero-ideal of T (resp. Q1∩Q2=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) LetR=Z/12Z.  Note that P1=2Z/12ZandP2=3Z/12Z are the only prime ideals of the finite ring R=Z/12Z. Thus Z(R)=P1∪P2.   Observe that R has exactly two maximal N-primes of (0) and P1∩P2=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=(12Z:Z±6t) and {±6t∣t∈Zis odd and positive} is the set of all integers with the property that 2Z=(12Z:Z±6t).  Furthermore, note that {±4k∣k∈Zis positive andk≡1or2(mod3)} is the set of all integers with the property that 3Z=(12Z:Z±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.
Letn>1 be such that n admits at least three distinct prime divisors.  Let {p1,p2,p3,…,pt}(t≥3) 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

LetR 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 Lemma3.2.Lemma 3.1.
LetR 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,y∈Z(R)*, P⊄((0):Rx)∪Ry.Proof.
Suppose that for somex,y∈Z(R)*, P⊆((0):Rx)∪Ry.  Then either P⊆((0):Rx)orP⊆Ry.  Since x≠0,  ((0):Rx)∩(R∖Z(R))=∅.  As y∈Z(R), there exists z∈R∖{0} such that yz=0. Hence we obtain that Ry⊆((0):Rz).  Note that ((0):Rz)∩(R∖Z(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 P⊆Ry, 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,y∈Z(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.
LetR 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.
LetP 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 x∈P∖{0}. By hypothesis, P is not a B-prime of (0) in R and so Lemma 3.1 implies that there exists y∈P such that y∉Rx and yx≠0.  We assert that xy∉{x,y}. If xy=x, then x(1-y)=0. As y∈P, 1-y∉P=Z(R). Hence x(1-y)=0 implies that x=0. This contradicts the fact that x≠0.  Similarly, it follows that xy≠y.  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 z∈P such that zxy≠0 and z∉Ry.  From zxy≠0, it follows that zx≠0, and zy≠0 and moreover, as x2y=0, it follows that z≠x.  Furthermore, by the choice of z, it follows that z≠y.  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.
LetR be a commutative ring with identity. If there exist distinct elements a,b,c∈Z(R)*∖P for some prime ideal P of R, then gr((Γ(R))c)=3.Proof.
AsP is a prime ideal of R and a,b,c are elements of R which are not in P, we obtain that ab,bc,ca∈R∖P and so ab,bc,ca∈R∖{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.
LetR 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)
IfP1∩P2=(0), then (Γ(R))c contains a cycle if and only if either |P1∖P2|≥3 or |P2∖P1|≥3 if and only if gr((Γ(R))c)=3.(ii)
IfP1∩P2≠(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)=P1∪P2.
(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 n≥3 and  {ai∣i=1,2,3,…,n}⊆Z(R)* with a1a2,a2a3,…,an-1an,ana1∈R∖{0}.  Since P1∩P2=(0), it follows that either {a1,a2,a3,…,an}⊆P1∖P2or{a1,a2,a3,…,an}⊆P2∖P1.  Now it is clear that either |P1∖P2|≥3 or |P2∖P1|≥3.
Conversely, suppose that either|P1∖P2|≥3 or |P2∖P1|≥3.  Since Z(R)=P1∪P2 and as P1 and P2 are prime ideals of R, it follows from Lemma 3.3 that gr((Γ(R))c)=3.
Ifgr((Γ(R))c)=3, then we obtain that (Γ(R))c contains a cycle of length 3.
(ii) Suppose thatP1∩P2≠(0).  We know from the proof of [18, Proposition 1.7(i)] that there exist a∈P1∖P2 and b∈P2∖P1 such that ab≠0.  We consider two cases : Case(A): P1∩P2⊄((0):Ra)∪((0):Rb). Then there exists c∈P1∩P2 such that ac≠0 and bc≠0.  Hence we obtain a cycle  a—b—c—a  in (Γ(R))c and it is of length 3. Case(B): P1∩P2⊆((0):Ra)∪((0):Rb).  Then either P1∩P2⊆((0):Ra) or P1∩P2⊆((0):Rb).
Without loss of generality, we may assume thatP1∩P2⊆((0):Ra). Let x∈P1∩P2 be such that x≠0.  Since a+b∉Z(R), we obtain that (a+b)x≠0. As ax=0, it follows that bx=(a+b)x≠0.  Moreover, observe that b+x∈P2∖P1,b≠b+x, and a(b+x)=ab≠0. As b(b+x)∈P2∖P1, 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 P1∩P2≠(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 ifR has at least three maximal N-primes of (0), then gr((Γ(R))c)=3.Lemma 3.5.
LetR 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 thatR 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 P⊄Q∪W, Q⊄P∪W, and W⊄P∪Q.  Hence there exist elements x∈P∖(Q∪W), y∈Q∖(P∪W), and z∈W∖(P∪Q).  Note that x,y,z are distinct elements of Z(R)* with xy,yz,zx∈R∖{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.
LetR 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 ringsR 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.
LetR 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 thatZ(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,b∈P such that a≠b and ab≠0. Moreover, by [18, Theorem 1.1(a)], P=((0):Rc) for some c∈R.  It is clear that c∈P∖{0}. Furthermore, as ab≠0, it follows that a≠c and b≠c.
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 n≥3 and {a1,a2,a3,…,an}⊆P∖{0}.  As P=((0):Rc) and since a1a2,a2a3,…,an-1an,andana1∈R∖{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 thatgr((Γ(R))c)=3  if |P|≥5.  Suppose that |P|≥5.  Observe that there exists d∈P∖{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 ad≠0 and bd≠0, 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 ifa2=0 and b2=0, then from (a+b)a=ab=(a+b)b=ab, and from the fact that a+b∈P∖{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 a2≠0.  As c2=0 and a2≠0, it follows that a+c≠0.  If a+c≠b, 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 c≠d and a+c=b, it follows that a+d≠b and it is clear that a(a+d)=a2≠ 0 and (a+d)b=ab≠0. 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.
Ifgr((Γ(R))c)=3, then (Γ(R))c contains a cycle of length 3.The following remark characterizes ringsR 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.
LetR 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,b∈P be such that a≠b and ab≠0.  Let c∈P∖{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 ab∈P∖{0}={a,b,c}. We assert that ab=c. If ab=a, then a(1-b)=0. This is impossible since a≠0 and 1-b∉P=Z(R).  Similarly, it follows that ab≠b. 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]+(x2-2)Z4[x]),Z2[x]/x3Z2[x]} where Z4[x] (resp. Z2[x]) is the polynomial ring in one variable over Z4 (resp. over Z2).
SinceR 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⊆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 uab∈P∖{0}={a,b,c=ab}. We claim that uab=ab.  If uab=a, then a(1-ub)=0and this is impossible. Similarly, we obtain that uab≠b.  Hence uab=ab.  This implies that (u-1)ab=0. Since ab≠0, it follows that u-1∈Z(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 R∖P={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]+(x2-2)Z4[x]),Z2[x]/x3Z2[x]}. Observe that if R is a ring such that R∈{Z8,Z4[x]/(2xZ4[x]+(x2-2)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.
LetR 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]+(x2-2)Z4[x]),Z2[x]/x3Z2[x]}.We determine in the following remark ringsR 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.
LetR be a ring admitting exactly two maximal N-primes of (0). Let them be P1 and P2.  Suppose that P1∩P2=(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 |P1∖P2|≥3, or |P2∖P1|≥3 if and only if gr((Γ(R))c)=3. Suppose that |P1∖P2|<3, |P2∖P1|<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)=P1∪P2 and P1∩P2=(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 P1∩P2=(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)  LetR be an infinite ring.  If there exist x,y∈Z(R)* such that xy≠0, then gr((Γ(R))c)=3.
(ii)  LetR be any ring admitting elements a,b∈Z(R)* such that ab≠0. 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

LetR 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,y∈Z(R)* with x≠y such that xy≠0), 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.
LetR 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)
Ifω((Γ(R))c) is finite, then Z(R)∖P is finite and indeed, |Z(R)∖P|≤ω((Γ(R))c).Proof.
(i) Suppose thatZ(R)∖P is infinite. Then we can choose an infinite sequence of distinct elements xi∈Z(R)∖P.  Since P is a prime ideal of R and as xi∉P for i=1,2,3,…, it follows that xixj≠0 for all  i,j∈{1,2,3,…}.  Observe that the subgraph of (Γ(R))c induced on {xi∣i=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) Letω((Γ(R))c)=n. We assert that |Z(R)∖P|≤n. Suppose that {Z(R)∖P∣≥n+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 Lemma4.1 and prime avoidance [21, Proposition 1.11(i)], we obtain the following result.Lemma 4.2.
LetR 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⊆Z(R) but Q⊄P}.  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)
Ifω((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 Qi∈A for each i=1,2,3,… with Qi and Qj not being comparable under inclusion for all i,j∈{1,2,3,…} with i≠j. Now Qi⊄P for i=1,2,3,…. Hence it is possible to choose x1∈Q1∖P.  Let i≥2.  Now it follows from prime avoidance [21, Proposition 1.11(i)] that there exists xi∈Qi∖(P∪Q1∪···∪Qi-1).  Observe that {xi∣i=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) Letω((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 i≠j.  Let x1∈Q1∖P.  Let i∈{2,…,n+1}.  As in (i), we can choose xi∈Qi∖(P∪Q1∪···∪Qi-1). Observe that {xi∣i=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 Lemma4.2, 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.
LetR 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)
Ifω((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 k≥3, then k≤ω((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 k≥3, then k≤ω((Γ(R))c).Proof.
(i)(a) LetP be a maximal N-prime of (0) in R.  Let A={Q∣Q is a maximal N-prime of (0) in R and Q≠P}.  Since any maximal N-prime of (0) in R is a subset of Z(R) and as distinct maximal N-primes of (0) in R are not comparable under inclusion, it follows from Lemma 4.2(i) that A is finite. Observe that the set of all maximal N-primes of (0) in R=A∪{P}.  It is now clear that R can admit only a finite number of maximal N-primes of (0).
(i)(b) IfP is any minimal prime ideal of R, then P⊆Z(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) LetP, 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 thatR admits exactly k maximal N-primes of (0) with k≥3.  Let {P1,P2,P3,…,Pk}  be the set of all maximal N-primes of  (0) in R. Note that Z(R)=∪i=1kPi.  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},  ∃xi∈Pi∖(⋃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  k≥3, 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 xixj≠0.  Thus we obtain that the subgraph of (Γ(R))c induced on {x1,x2,x3,…,xk} is a clique and so k≤ω((Γ(R))c).
(ii)(b) This can be proved using similar arguments as in the proof of (ii)(a) and using  Lemma4.2(ii).The following proposition is one among the main results in this section.  We show in this proposition that if a ringR 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.
LetR 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 m≤n+1 and by the hypothesis, m≥2.  Observe that
(4.1)Z(R)=P1∪P2∪⋯∪Pm=(P1∩P2∩⋯∩Pm)∪(Z(R)∖P1)∪(Z(R)∖P2)∪⋯∪(Z(R)∖Pm).
We know from Lemma 4.1(ii) that
(4.2)|Z(R)∖Pi|≤nfori=1,2,…,m.
We now verify that |P1∩P2∩⋯∩Pm|≤n.  Let x∈P1∖P2. Note that {x+y∣y∈P1∩P2∩⋯∩Pm}⊆Z(R)∖P2.   Hence it follows that
(4.3)|P1∩P2∩⋯∩Pm|≤n.
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.
LetR 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.LetR 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)=P1∩···∩Pn.  As R is not an integral domain, it follows that n≥2.  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 n≥2 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.
LetR be a commutative ring with identity which is not an integral domain.  For any a,b∈Z(R)*, define a~b 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 elementa∈Z(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 a∈R there exists b∈R such that a=a2b.The following fact is important, and we make use of it in the proof of Lemma4.8.Fact 4.7.
LetR 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 a∈Z(R)*, there exists a unique idempotent element e∈Z(R)* such that [a]=[e].Proof.
The Fact4.7 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,b∈Z(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 a∈Z(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 a∈Z(R)*, it follows that e∈Z(R)*.  This proves that a~e and so [a]=[e].
Note that ife is any idempotent element in a ring R (R not necessarily von Neumann regular), then ((0):Re)=R(1-e). 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(1-e)=R(1-f).  Hence 1-e=1-f and so e=f.
Now it follows from the above two preceding paragraphs that given anya∈Z(R)* where R is a von Neumann regular ring, then there exists a unique idempotent e∈Z(R)* such that [a]=[e].LetR 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.
LetR 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 eiej≠0 for all i,j∈{1,…,m}.  Let A=[e1]∪···∪[em].  Then the subgraph of (Γ(R))c induced on A is a clique.Proof.
Leti∈{1,…,m}.  Let a,b∈[ei] with a≠b.   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=uvei≠0.  This proves that any two distinct elements of [ei] are adjacent in (Γ(R))c for each i∈{1,…,m}.
Leti,j∈{1,…,m} with i≠j. Let x∈[ei], y∈[ej].  Note that x=uei and y=vej   for some units u,v in R. Hence xy=uveiej≠0, since eiej≠0 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 ab≠0, then we assert that for any x∈[a] and y∈[b], xy≠0. 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 xy≠0.
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.
Letn≥2.  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 2n-2 and the number of equivalence classes determined by the equivalence relation ~ defined on Z(R)* equals 2n-2.Proof.
The only idempotent elements in a field are 1 and 0. Using this observation, it follows thatR=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 2n-2.
Let{ei∣i=1,2,…,2n-2} 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,…,2n-2}.Letn≥2. 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.
Letn≥2.  Let Ki be a field for i=1,2,…,n. Let R=K1×⋯×Kn. Let A⊆Z(R)* be such that the subgraph of (Γ(R))c induced on A is a clique.  Let E={ei∣i=1,2,…,2n-2} be the set of all idempotents in Z(R)*.  Then there exists a nonempty subset B of E such that b1b2≠0 for all b1,  b2∈Band A⊆⋃b∈B[b].Proof.
We know from the proof of Lemma4.9 that {[ei]∣i=1,2,…,2n-2} is the set of all equivalence classes determined by ~. Thus we obtain that Z(R)*=⋃e∈E[e]. Now A being a subset of Z(R)*, it follows that A=⋃e∈E(A∩[e]).  Let B⊆E be such that A∩[b] is non-empty for each b∈B.  Since A is non-empty, it follows that B is non-empty. We now verify that for any   b1,b2∈B,b1b2≠0.   This is clear if b1=b2 since any element of B is a nonzero idempotent in R. Suppose that b1≠b2.
Leta1∈A∩[b1] and a2∈A∩[b2].  Since [b1]∩[b2]=∅, it follows that a1≠a2. As the subgraph of (Γ(R))c induced on A is a clique, we obtain that a1a2≠0.  Note that Rai=Rbi for i=1,2.  Hence we obtain that Rb1b2=Ra1a2≠(0) and so b1b2≠0. This proves that there exists a non-empty subset B of E such that b1b2≠ 0 for allb1, b1∈B and moreover, A=⋃b∈B(A∩[b])⊆⋃b∈B[b].LetR 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.
Letn≥2 and let Ki(i=1,2,…,n) be finite fields.  Let R=K1×K2×···×Kn. Let E={ei∣i=1,2,…,2n-2} be the set of all idempotents in Z(R)*.  Then ω((Γ(R))c)=max{(∑b∈B|[b]|)∣B   varies over all non-empty subsets of E satisfying the property that b1b2≠ 0 for any b1,b2∈B}.Proof.
LetX={B∣B is a non-empty subset of E satisfying the property that for any b1,b2∈B,b1b2≠0}. Let B∈X.  Let A=⋃b∈B[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|=∑b∈B|[b]|. Hence we obtain that
(4.4)ω((Γ(R))c)≥max⁡{(∑b∈B|[b]|)∣B∈X}.
LetA⊆Z(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 B∈X such that A⊆⋃b∈B[b].  Hence we obtain that |A|≤∑b∈B|[b]|.  This implies that
(4.5)ω((Γ(R))c)≤max⁡{(∑b∈B|[b]|)∣B∈X}.
From (4.4) and (4.5), it follows that ω((Γ(R))c)=max{(∑b∈B|[b]|)∣B∈X}.We make use of the following useful remark in Example4.13(i) and (ii).Remark 4.12.
LetR 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 e∈Z(R)*,  [e]={ue∣uisaunitinR}.
Letn≥2. Let K1,K2,…,Kn be finite fields and let R=K1×K2×⋯×Kn. Let e∈Z(R)* be an idempotent.  Let C={i∈{1,2,…,n}∣theith component ofeis1}.  Observe that |[e]|=∏i∈C(|Ki|-1).We next have the following example.Example 4.13.
(i) LetK1,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)}.LetR 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.
LetR 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 ofR is a zero-divisor of R and since Z(R)=P, it follows that nil(R)⊆Z(R)=P.  Let x∈P. We assert that x is nilpotent. Suppose that x is not nilpotent. Then for any i,j∈{1,2,3,…},  xixj≠0, and moreover, for all distinct i,j∈{1,2,3,…},xi≠xj.  Hence  {xk∣k=1,2,3,…}⊆Z(R)* is such that the subgraph of (Γ(R))c induced on {xk∣k=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 P⊆nil(R) and so P=nil(R).
(ii) Suppose thatR/P is infinite.  We first assert that x2=0 for each x∈P.  Suppose that there exists x∈P such that x2≠0.  Since we are assuming that R/P is infinite, it is possible to find an infinite sequence of elements {rk∈R∖Pfork=1,2,3,…} such that ri-rj∉P and ri+rj∉P for all distinct i,j∈{1,2,3,…}.  Note that for all i,j∈{1,2,3,…}, rirj∉P=Z(R), and as x2≠0, it follows that  x2rirj≠0.  Moreover, as for all distinct i,j∈{1,2,3,…}, ri-rj∉P, it follows that xri≠xrj. Hence we obtain that the subgraph of (Γ(R))c induced on {xrk∣k=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 x∈P. As P2≠(0), there exist a,b∈P such that a≠b and ab≠0.  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,xk∈P and axk=abrk≠0, since ab≠0 and rk∉P=Z(R).  Moreover, for all distinct i,j∈{1,2,3,…},xixj=ab(ri+rj)≠0, since ab≠0 and ri+rj∉P=Z(R).  Furthermore, as b≠0 and ri-rj∉P, for all distinct i,j∈{1,2,3,…}, it follows that xi≠xj for all distinct i,j∈{1,2,3,…}.  Hence we obtain that the subgraph of (Γ(R))c induced on {xk∣k=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 thatP is a B-prime of (0) in R.  We consider two cases. Case (A): P is finitely generated.
By (i), we haveP=nil(R) and hence we obtain that P is a nilpotent ideal of R. Let m≥2 be least with the property that Pm=(0).  Now for any x∈Pm-1∖{0},P⊆((0):Rx)⊆Z(R)=P,and soP=((0):Rx) is a B-prime of (0) in R. Case (B): P is not finitely generated.
We haveP2≠(0), by assumption. Hence there exist a1,a2∈P such that a1≠a2 anda1a2≠0.  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+Ra2⊆P and since P is not finitely generated, it follows that P⊄Ra1+Ra2.  Hence P⊄(Ra1+Ra2)∪((0):Ra1a2).  Hence there existsa3∈P∖((Ra1+Ra2)∪((0):Ra1a2)).   Thus a1,a2,a3∈P are distinct and a1a2a3≠0.  Let k be any positive integer with k≥3.  Assume that there exists a subset {a1,a2,a3,…,ak} of P with   a1a2a3⋯ak≠0.  Observe that P⊄Ra1+Ra2+Ra3+⋯+Rak and P⊄((0):Ra1a2a3⋯ak).  Hence there exists ak+1∈P∖((Ra1+⋯+Rak)∪((0):Ra1a2a3⋯ak)). 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 {ai∣i=1,2,3,…} of P such that the product a1 · · ·  ak≠0 for k=1,2,3,….  This implies that the subgraph of (Γ(R))c induced on {ai∣i=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) thatR/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 xi∈P for i=1,2,3,… such that Rx1⊃Rx2⊃Rx3⊃⋯. Note that there exist ai∈P for i=1,2,3,… such that xi+1=aixi. Hence(I)xk+1=(ak⋯a1)x1,∀k≥1.
Since xk≠0 for k=1,2,3,…, it follows that  the  elements ak∈P(k=1,2,3,…) satisfy  aiaj≠0 for all  distinct  i,j∈{1,2,3,…}.  As each element of P is nilpotent, it follows from (I) and the fact that xk≠0 for k=1,2,3,… that there exist positive integers k1<k2<k3<⋯ such that for all distinct i,j∈{k1,k2,…},ai≠aj. Let A={ki∣i=1,2,3,…}.  Observe that the subgraph of (Γ(R))c induced on {ai∣i∈A} 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.LetR 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 thatR 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 n≥1. Since P2≠(0), it follows that n≥3.  Observe that for each integer k,0≤k≤n-1,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,…,n-1. Now Pn-1,Pn-2/Pn-1 are finite, and hence it follows that Pn-2 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 ringR 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).LetR 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.
LetR be a commutative ring with identity which is not an integral domain. Suppose that R is a chained ring and moreover, there exist x,y∈Z(R)* with x≠y such that xy≠0. 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)={x∈P∣x2=0}.  It is known that for any x,y∈P∖N(R), xy≠0 [6, Lemma 4.2(3)]. Since (Γ(R))c does not admit any infinite clique, it follows that P∖N(R)is finite.  Now by the assumption that R is a chained ring and there exist x,y∈Z(R)* with x≠y such that xy≠0, it follows that Z(R)∖N(R) is non-empty.  Let P∖N(R)={x1,…,xm}.  Since each element of P is nilpotent, it follows that there exist t>2 such that pt=0 for each p∈P.  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 P≠P2. Now R is a chained ring with P as its unique maximal ideal satisfying the further condition that P≠P2.  In such a case it is well known that P=Rp for anyp∈P∖P2.  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 determinesω((Γ(R))c) for any finite chained ring which is not an integral domain.Remark 4.17.
LetR 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 n≥3  be least with the property that Pn=(0).  Then the following hold.(i)
ω((Γ(R))c)=|P∖Pn/2|+1if n is even.(ii)
ω((Γ(R))c)=|P∖P(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 anyp∈P∖P2.
(i) Suppose thatn is even and let n=2m.  By the choice of n,pn-1≠0.  Ifa∈P∖Pm,b∈P∖Pm, then a=upt for some t,1≤t<m and a unit u in R and b=vps for some s,1≤s<m and a unit v in R.  Hence ab=uvpt+s≠0, since t+s<2m=n and pn-1≠0. Moreover, observe that pmx≠0 for any x∈P∖Pm. Hence the subgraph of (Γ(R))c induced on (P∖Pm)∪{pm} is a clique. This implies that ω((Γ(R))c)≥|P∖Pm|+1.  Let A⊆Z(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 a∈A such that Ra⊆Rb for each b∈A.  Hence Rab⊆Rb2. Since ab≠0 for eachb∈A∖{a}, we obtain that b2≠0 for eachb∈A∖{a}. As Pn=P2m=(0), it follows that A∖{a}⊆P∖Pm. Thus A={a}∪(A∖{a})⊆{a}∪(P∖Pm) and so |A|≤|P∖Pm|+1. This proves that ω((Γ(R))c)≤|P∖Pm|+1.  Hence ω((Γ(R))c)=|P∖Pm|+1.
(ii) Suppose thatn is odd and let n=2k+1 for some k≥1.  In this case, we verify that ω((Γ(R))c)=|P∖Pk+1| where k=(n-1)/2.  It is clear that for any x,y∈P∖Pk+1,xy≠0.  This shows that the subgraph of (Γ(R))c induced on P∖Pk+1 is a clique. Hence we obtain that ω((Γ(R))c)≥|P∖Pk+1|.  Let A⊆Z(R)*=P∖{0} be such that the subgraph of (Γ(R))c induced on A is a clique.  We assert that |A|≤|P∖Pk+1|.  Let a∈A be such that Ra⊆Rb for each b∈A.  Now, it follows as in the proof of (i) that b2≠0 for each b∈A∖{a}. Since P2k+1=(0), it follows that A∖{a}⊆P∖Pk+1.  If there exists at least one b∈A∖{a} such that b∈Pk, then, as ab≠0, it follows that a∉Pk+1 and so we obtain that A⊆P∖Pk+1.  Hence it follows that |A|≤|P∖Pk+1|. Otherwise, we obtain that A∖{a}⊆P∖Pk. Note that A={a}∪(A∖{a})⊆{a}∪(P∖Pk) and so |A|≤1+|P∖Pk|≤|Pk∖Pk+1|+|P∖Pk|=|P∖Pk+1|. This proves that ω((Γ(R))c)≤|P∖Pk+1|.  Hence we obtain that
(4.6)ω((Γ(R))c)=|P∖Pk+1|=|P∖P(n+1)/2|.We next mention an example to illustrate that in Remark4.17, one cannot drop the assumption that R is a chained ring.Example 4.18.
LetF=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+1∈F[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 α∈K be a root of p(x). It is clear that K=F(α) and [K:F]=2.  Indeed, {1,α} is a basis of K as a vector space over F.  Moreover, K={0,1,α,α2=-α-1=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,e∈F}.  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 N∖N2={x+I,αx+I, α2x+I,(x+αx2)+I, (αx+αx2)+I, (α2x+αx2)+I}. Thus |N∖N2|=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
(4.7)N∖N2.
Asα∉F,αx2∉I=x2S,andα2x2=(1+α)x2∉I.  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
(4.8)(x+I)((x+αx2)+I)=x2+I=0+I.
We have α3=αα2=α(1+α)=α+α2=α+(1+α)=1. Note that
(4.9)(αx+I)(α2x+I)=x2+I=0+I.
Moreover,
(4.10)((αx+αx2)+I)((α2x+αx2)+I)=x2+I=0+I.
It is clear from  (4.7), (4.8), (4.9), and (4.10) that if A⊆Z(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<|N∖N2|=6.
LetT 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 a∈A be such that a∈(N1)2.  Note that a=(βx2+γx3)+J for some β,γ∈K.  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.
β≠0. Observe that β=u+vα for some u,v∈F 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+J∈N1∖(N1)2. The element (αx2+γx3)+J annihilates α2x+J∈N1∖(N1)2 and the element (α2x2+γx3)+J annihilates αx+J∈N1∖(N1)2.This shows that ifA 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, b∈N1∖(N1)2.  Note that a=(β1x+β2x2+β3x3)+J and b=(γ1x+γ2x2+γ3x3)+J for some βi,γi∈K (fori=1,2,3) with β1,γ1∈K∖{0}.  Note that ab = β1γ1x2+w+J  for some w∈x3K[[x]].  Since, J=x3S⊆x3K[[x]] and as β1γ1x2∉x3K[[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|.LetR 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.
LetR 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 x∈P,xm=0 where m=ω((Γ(R))c)+1.Proof.
By hypothesis,ω((Γ(R))c) is finite. Let x∈P.  We assert that xm=0 where  m=ω((Γ(R))c)+1. Suppose that xm≠0.  For each k∈{1,2,…,m}, let yk=x+···+xk. Note that Z(R)=P and r1=1∉P=Z(R) and for 2≤k≤m,rk=1+···+xk-1∉P=Z(R) and moreover,  yk=xrkfork=1,2,…,m.  Since x2≠0 and rirj∉Z(R), it follows that yiyj≠0 for all i,j∈{1,2,…,m}.   Let i,j∈{1,2,…,m} with i≠j.  We claim that yi≠yj.  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, xm≠0.  Hence yi≠yj for all distinct i,j∈{1,2,…,m} and moreover, yiyj≠0.  Hence we obtain that the subgraph of (Γ(R))c induced on {yk∣k=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 x∈P.The next remark provides examples of ringsR for which ω((Γ(R))c)=∞.Remark 4.20.
We remark here that Lemma4.19 is motivated by [3, Theorem 3.4].  Let R be a commutative ring with identity  which is not reduced.  Let x∈nil(R).  Recall that by the index of nilpotence of x, we mean the least positive integer n such that xn=0.  Suppose that x∈nil(R) with x≠0.  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+···+xk∣k=1,…,n-1} is a clique in (Γ(R))c. Hence it follows that ω((Γ(R))c)≥n-1.  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)=∞.LetR 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.
LetR 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.
Letm=ω((Γ(R))c)+1. We claim that Pn=(0) with n=(m-1)2+1=(ω((Γ(R))c))2+1.   Suppose that Pn≠(0).  Then there exist xk∈P for k=1,2,…,n such that x1x2···xn≠0. 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}  = ⋃j=1sAj,  and moreover, Ai∩Aj=∅ for all distinct i,j∈{1,…,s}.  Hence we obtain that n=∑j=1s|Aj|=n1+···+ns. Since x1x2···xn≠0, it follows that x1···xs≠0, 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 x∈P.  Hence it follows that nj≤m-1 for j=1,…,s.  Furthermore, observe that the subgraph of (Γ(R))c induced on {x1,…,xs} is a clique and so s≤ω((Γ(R))c)=m-1.  Thus we obtain that (m-1)2+1=n=∑j=1snj≤s(m-1)≤(m-1)(m-1).   This is impossible. Hence it follows that Pn=(0) with n=(m-1)2+1.We conclude this note with the following example of an infinite ringR 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.
LetT=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 {xk2∣k=1,2,3,…}∪{x1xk∣k=3,4,5,…}∪{xixj∣i,j∈N,2≤i<j}.  Let R=T/I.  Let M be the ideal of T generated by {xk∣k=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 A⊆Z(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,w∈A with z≠w, zw≠0+I.
Letz,w∈Z(R)*. Observe that z=f1(x1,x2,…,xm)+I, w=f2(x1,x2,…,xm)+I, for some m≥2 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
(4.11)fi(x1,…,xm)fori=1,2.
We are assuming thatA contains at least 4 elements. Let {z1,z2,z3,z4}⊆A.  Let zk=gk(x1,x2,…,xm)+I, for some m≥2 and for some gk(x1,x2,…,xm)∈Z2x1+Z2x2+⋯+Z2xm+Z2x1x2fork=1,2,3,4.  Sincezizj≠0+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,ak12∈Z2.  Since z1z2≠0+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 ifA⊆Z(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.



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*Source: 591041-2011-10-04.xml*