ISRN AlgebraVolume 2011 (2011), Article ID 591041, 24 pagesdoi: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. Anderson, V. Drensky, and D. Herbera

Copyright © 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 denote the complement of the zero-divisor graph of . It is shown that if is connected, then its radius is equal to 2 and we also determine the center of . It is proved that if is connected, then its girth is equal to 3, and we also discuss about its girth in the case when is not connected. We discuss about the cliques in .

1. Introduction

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

Let be a commutative ring with identity which is not an integral domain. Recall from [1] that the zero-divisor graph of , denoted by , is the graph whose vertex set is the set of all nonzero zero-divisors of and distinct vertices , are joined by an edge in this graph if and only if . 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 the set of all zero-divisors of , and by the set of all nonzero zero-divisors of .

Before we describe the results that are proved in this note, it is useful to recall the following definitions from [15]. Let be a connected graph. For any vertices , of with , is the length of a shortest path in from to and and the diameter of is defined as sup and it is denoted by .

For any , the eccentricity of denoted by is defined as

The set of vertices of with minimal eccentricity is called the center of the graph, and the minimum eccentric value is called the radius of and is denoted by .

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

Let be a simple graph. Recall from [15, Definition, 1.1.13] that the complement of denoted by is defined by setting and two distinct are joined by an edge in if and only if there exists no edge in joining .

It is useful to recall the following definitions from commutative ring theory before we proceed further. Let be an ideal of a ring . A prime ideal of is said to be a maximal -prime of in if is maximal with respect to the property of being contained in where [16]. It is well known that if is the set of all maximal -primes of (0) in , then

Let be an ideal of a ring . A prime ideal of is said to be an associated prime of in the sense of Bourbaki if for some [17]. In this case, we say that is a -prime of .

Let be a commutative ring with identity admitting at least two nonzero zero-divisors. In [18, Theorem 1.1], it was shown that is connected if and only if one of the following conditions holds.(a) has exactly one maximal -prime of (0) such that is not a -prime of (0) in .(b) has exactly two maximal -primes , of with .(c) has more than two maximal -primes of .

Moreover, it was shown in [18] that if is connected, then . In fact, it was shown in [18] that if either the condition (a) or the condition (c) of [18, Theorem 1.1] holds, then . When the condition (b) of [18, Theorem 1.1] holds, then it was shown in [18, Proposition 1.7] that if either is not a -prime of in or is not a -prime of in and if both and are -primes of in .

For any set , we denote by , the cardinality of . Whenever a set is a subset of a set and , we denote it symbolically by . If are sets and if is not a subset of , we denote it symbolically .

Let be a commutative ring with identity and let . If is connected, then we prove in Section 2 of this note that the radius of is equal to 2. Moreover, we observe that except in the case when condition (b) of [18, Theorem 1.1] holds, and so every vertex of is in the center of . Furthermore, if condition (b) of [18,Theorem 1.1] holds and if (i) satisfies the further condition that either is not a -prime of (0) in or is not a -prime of in , then it is noted that each vertex of is in the center of and if (ii) both and are -primes of (0) in , then it is verified that the center of .

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

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

In Section 3 of this paper we study about the girth of where is a commutative ring with identity satisfying the further condition that . If is connected, then it is shown that .

Suppose that has only one maximal -prime of and let it be . If and if is not connected, then it is proved that contains a cycle if and only if if and only if .

Suppose that has exactly two maximal -primes of and let them be and . If is not connected, then it is shown that contains a cycle if and only if .

Let be a graph. Recall from [15, Definition 1.2.2] that a clique of is a complete subgraph of . Moreover, it is useful to recall the definition of the clique number of . Let be a simple graph. The clique number of denoted by is defined as the largest integer such that contains a clique on vertices [15, Definition, Page 185]. We set if contains a clique on vertices for all .

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

Let be a commutative ring with identity and let . In Section 4 of this paper we observe that if has at least two maximal -primes of , then does not contain any infinite clique if and only if is finite if and only if is finite. Let and let be finite fields. Let . We describe a method of determining .

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

Let be a ring with . Suppose that has only one maximal -prime of and let it be . If is finite, then it is shown in Section 4 that is nilpotent.

We end this note with an example of an infinite ring such that has exactly one maximal -prime of with the property that .

2. The Radius of

Let be a ring with . We assume that is connected. The aim of this section is to show that the radius of is equal to 2. We make use of the following lemmas for proving the result that .

Lemma 2.1. Let be a simple and connected graph with satisfying the further condition that is also connected. If is any element of then .

Proof. Let . Since and is connected, it follows that there exists such that is adjacent to in . Thus is not adjacent to in . Hence . This proves that .

The following lemma establishes some necessary conditions on in order that there exist such that .

Lemma 2.2. Let be a ring with . Suppose that is connected. If there exist with , then has exactly two maximal -primes of and moreover, both of them are -primes of in . Indeed, if and are the maximal -primes of in , then .

Proof. Since , it follows that and for any with , either or . Hence we obtain that . Since , it is clear that . Thus we have . Let be the set of all maximal -primes of in . It is well known that , and hence we obtain that . Now . Hence by [19, Theorem 2.2, Page 378], we obtain that there exists a maximal -prime of in such that . Since , it follows that there exists a maximal -prime of in such that . Now we obtain that . If , then it follows that is an ideal of . Hence it follows that either , and so we obtain that is the only maximal -prime of in and either . Now by hypothesis is connected. Hence it follows from [18, Theorem 1.1 (a)] that is not a -prime of in and so . Now we obtain from that and are the only maximal -primes of in , and moreover, .

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

Lemma 2.3. Let be a ring. Let be distinct -prime ideals of in with and for some . Then .

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

We provide in the next lemma some sufficient conditions on in order that admits vertices x such that in .

Lemma 2.4. Let be a ring and let . Suppose that has exactly two maximal -primes of and let them be and . If is connected and if and for some , then in and so in .

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 . Thus and and are not adjacent in . Since and are the only maximal -primes of in , it follows that . Thus for any , either . This shows that . It is shown in the proof of [18,Proposition 1.7] that for any . Hence we obtain that and so it follows that .

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

Theorem 2.5. Let be a ring and let . If is connected, then the radius of is equal to 2.

Proof. It is well known that is connected [1, Theorem 2.3]. Hence it follows from Lemma 2.1 that for any .If has either exactly one maximal -prime of or more than two maximal -primes of , then it follows from Lemma 2.2. that for any ,. Thus we obtain that if has either exactly one maximal -prime of or more than two maximal -primes of , then for any. Hence we obtain in the cases mentioned above that .Assume that has exactly two maximal -primes of and let them be . In such a case, it was shown in the proof of [18, Proposition 1.7(i)] that there exist and such that . It follows that . Now it follows from Lemma 2.2 that and for any in . Hence we obtain that and in . Since for any , it follows that . Thus in this case also, we arrive at the conclusion that the radius of is equal to 2.

The following remark determines the center of .

Remark 2.6. Let be a ring and let . Assume that is connected. In this remark, we determine the center of . If has either exactly one maximal -prime of or more than two maximal -primes of , then it was shown in the proof of Theorem 2.5 that . Hence the center of is the set of all vertices of . Moreover, if has exactly two maximal -primes of and if at least one of them is not a -prime of in , then it follows from Lemma 2.2 that for any . Thus we obtain, in view of Lemma 2.1 that for any. Hence in this case also we obtain that and so each vertex of is in the center of .Suppose that has exactly two maximal -primes of and both are -primes of in . Let be the set of all maximal -primes of in . Then it follows from Lemmas 2.1, 2.2, and 2.4 that the center of .

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

Example 2.7. Let be a rank 1 valuation domain which is not discrete. Let denote the unique maximal ideal of . Let . Let . Observe that is the only prime ideal of and . Let us denote by . We assert that is not a -prime of in . Suppose that is a -prime of in . Then it can be easily verified that for some . Since , it follows that . As is a valuation domain, we obtain that . Thus for some . Hence we obtain that . This is impossible since is not finitely generated. Thus is not a -prime of in . Now it follows from [18, Theorem 1.1 (a)] that is connected. We obtain from Theorem 2.5 that and each vertex of is in the center of .

Example 2.8. (i) Let be as in Example 2.7 and let (resp. ) be the direct product of and the ring of integers (resp. and ). We know from Example 2.7 that is the unique maximal -prime of in and is not a -prime of in . Note that (resp. ) are the only maximal -primes of the zero-ideal of (resp. of the zero-ideal of ) and is not the zero-ideal of (resp. is not the zero-ideal of ). Hence it follows from [18, Theorem 1.1(b)] that (resp. ) is connected. Since is not a -prime of in (resp. and are not -primes of in ), it follows from Remark 2.6 that (resp. ) and each vertex of is in the center of (resp. each vertex of ((T₁))c is in the center of ((T₁))c). (ii) Let . Note that are the only prime ideals of the finite ring . Thus . Observe that has exactly two maximal -primes of and is not the zero-ideal of . Hence it follows from [18, Theorem 1.1(b)] that is connected. Moreover, observe that and is the set of all integers with the property that . Furthermore, note that is the set of all integers with the property that . Hence it follows that , . Thus and are -primes of in . Now it follows from [18, Proposition 1.7(b)] that , and we know from Theorem 2.5 that . Moreover, we obtain from Remark 2.6 and from the above discussion that the set of centers of .

Example 2.9. Let be such that admits at least three distinct prime divisors. Let be the set of all distinct prime divisors of . Let . Note that is the set of all maximal -primes of the zero-ideal of . We know from [18, Theorem 1.1(c)] that is connected and moreover, it is known from Remark 2.6 that and each vertex of is in the center of .

3. The Girth of

Let be a commutative ring with identity and let . The aim of this section is to study about the girth of . If is connected, then we prove in this section that . Moreover, we also discuss about the girth of in the case when 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 be a ring and let . Let be a maximal -prime of in . If is not a -prime of in , then for any , .

Proof. Suppose that for some , . Then either . Since , . As , there exists such that . Hence we obtain that . Note that . Now it follows from [19, Theorem 2.2, Page 378] that there exists a maximal N-prime Q of (0) in R such that and there exists a maximal -prime of in such that . If , then we obtain that and hence it follows that . This contradicts the assumption that is not a -prime of in . If , then and so . This is also impossible since is not a -prime of in . This proves that if a maximal -prime of in is not a -prime of in , then for any , .

In the following lemma, we determine the girth of under the assumptions that has exactly one maximal -prime of and is connected.

Lemma 3.2. Let be a ring and let . Suppose that has only one maximal -prime of . If is connected, then .

Proof. Let be the unique maximal -prime of in . Since is connected, we obtain from [18, Theorem 1.1(a)] that is not a -prime of in . Note that . Let . By hypothesis, is not a -prime of in and so Lemma 3.1 implies that there exists such that and . We assert that . If , then . As , . Hence implies that . This contradicts the fact that . Similarly, it follows that . If both and are nonzero, then we obtain that is a cycle of length 3 in . Suppose that either or . Without loss of generality, we may assume that . As is not a -prime of in , Lemma 3.1 implies that there exists such that and . From , it follows that , and and moreover, as , it follows that . Furthermore, by the choice of , it follows that . Now is a cycle of length 3 in . This shows that if has only one maximal -prime of and if is connected, then there exists a cycle of length 3 in and hence .

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

Lemma 3.3. Let be a commutative ring with identity. If there exist distinct elements for some prime ideal of , then .

Proof. As is a prime ideal of and are elements of which are not in , we obtain that and so . Hence it follows that is a cycle of length 3 in . This proves that .

The next lemma discusses the girth of where is a ring with of admitting exactly two maximal -primes.

Lemma 3.4. Let be a commutative ring with identity. Suppose that has exactly two maximal -primes of and let them be and . Then the following hold: (i)If , then contains a cycle if and only if either or if and only if . (ii)If , then . Thus, if is connected, then .

Proof. By hypothesis, and are the only maximal -primes of in . So, it follows that . (i) Assume that contains a cycle. Let be a cycle of length in . Note that and with . Since , it follows that either . Now it is clear that either or .Conversely, suppose that either or . Since and as and are prime ideals of , it follows from Lemma 3.3 that .If , then we obtain that contains a cycle of length 3.(ii) Suppose that . We know from the proof of [18, Proposition 1.7(i)] that there exist and such that . We consider two cases : Case(A): . Then there exists such that and . Hence we obtain a cycle in and it is of length 3. Case(B): . Then either or .Without loss of generality, we may assume that . Let be such that . Since , we obtain that . As , it follows that . Moreover, observe that , and . As , it follows that . Note that is a cycle of length 3 in .Thus in both the cases, admits a cycle of length 3. Hence we obtain that .We know from [18, Theorem 1.1(b)] that is connected if and only if . Thus if has exactly two maximal -primes of and if is connected, then .

We show in the next lemma that if has at least three maximal -primes of , then .

Lemma 3.5. Let be a ring. Suppose that admits more than two maximal -primes of . Then .

Proof. Since by the assumption that admits more than two maximal -primes of , we can find at least three maximal -primes of in . Let be a subset of the set of all maximal -primes of in . It is clear that , W, and . Hence there exist elements , , and . Note that are distinct elements of with . Hence is a cycle of length 3 in . This proves that .

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

Theorem 3.6. Let be a ring and let . If is connected, then .

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 such that is not connected and discuss about the girth of . In this direction, we first have the following proposition.

Proposition 3.7. Let be a ring and let . Suppose that has only one maximal -prime of and let it be . If (0) and if is not connected, then contains a cycle if and only if if and only if .

Proof. Note that . Assume that and is not connected. From the assumption that , it follows that there exist such that and . Moreover, by [18, Theorem 1.1(a)], for some . It is clear that . Furthermore, as , it follows that and . Suppose that contains a cycle. Let be a cycle of length in . Note that and . As and since , it follows that . Thus and hence it follows that .We next show that if . Suppose that . Observe that there exists where are as in the first paragraph of this proof. We claim that there exists a cycle of length 3 in . If and , then is a cycle of length 3 in . Suppose that either or .Observe that if and , then from , and from the fact that , it follows that is a cycle of length 3 in . Now let us suppose without loss of generality that . As and , it follows that . If , then is a cycle of length 3 in . Suppose that . Then, as , it follows from the assumption that either or that . As and , it follows that and it is clear that 0 and . Hence is a cycle of length 3 in .Thus if , then it is shown that there exists a cycle of length 3 in . Hence .If , then contains a cycle of length 3.

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

Remark 3.8. Let be a ring and let . Suppose that is the only maximal -prime of (0) in . Assume that . Suppose that is not connected. Let be such that and . Let be such that . Assume that . Observe that by Proposition 3.7, if and only if does not contain any cycle. Now it follows from Proposition 3.7 that and as , it follows that . Now . We assert that . If , then . This is impossible since and . Similarly, it follows that . Hence . Now is finite and hence we obtain from [20, Theorem 1] that R is a finite ring. We verify in this remark that . Moreover, we observe with the help of [3, Theorem 3.2] that is isomorphic to exactly one of the rings from the set where (resp. ) is the polynomial ring in one variable over (resp. over ). Since is a finite ring, any prime ideal of is a maximal ideal of , and moreover, if is any prime ideal of , then . Since , it follows that is the only prime ideal of . Now is a local ring with unique maximal ideal . Hence we obtain that is the set of all units in . Let u be a unit in . Then . We claim that . If , then and this is impossible. Similarly, we obtain that . Hence . This implies that . Since , it follows that . Hence we obtain that . Thus it is shown that . Hence is a ring containing exactly 8 elements. Note that . Now [3, Theorem 3.2] implies that is isomorphic to exactly one of the rings from the set . Observe that if is a ring such that , then satisfies the following conditions: has exactly one maximal -prime of , say such that , is not connected, and does not contain any cycle.Let be a ring such that has only one maximal -prime of , say , and satisfies the further conditions that and is not connected. The above discussion implies that does not contain any cycle if and only if is isomorphic to exactly one of the rings from the set .

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

Remark 3.9. Let be a ring admitting exactly two maximal -primes of . Let them be and . Suppose that (that is, equivalently is not connected). In such a case, it is shown in Lemma 3.4(i) that contains a cycle if and only if either , or if and only if . Suppose that , and that contains at least one edge. We verify in this remark that either or . Moreover, we verify that is isomorphic to where and are fields with either or one of them contains exactly 3 elements and the other contains exactly 2 elements. We are assuming that contains at least one edge. Since and , it follows that at least one between and contains exactly 3 elements. This implies that either or exactly one between and contains exactly 3 elements and the other contains exactly 2 elements. Thus either or . As is a finite set, it follows from [20, Theorem 1] that is a finite ring. Since any prime ideal of a finite ring is a maximal ideal, it follows that and are maximal ideals of . As , it follows from the Chinese Remainder Theorem [21, Proposition 1.10] that as rings. Thus we obtain from the above discussion that R is isomorphic to the direct product of two fields and with either or one between and contains exactly 3 elements and the other contains exactly 2 elements. Hence we obtain that either or and is isomorphic to where and are fields with either or one of them contains exactly 3 elements and the other contains exactly 2 elements. Conversely, if is isomorphic to × where and are fields with either or one of them contains exactly 3 elements and the other contains exactly 2 elements, then it is clear that has the following properties: admits exactly two maximal -primes of , contains at least one edge, 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 be an infinite ring. If there exist such that , then . (ii) Let be any ring admitting elements such that . Then where (resp. ) is the polynomial (resp.the power series) ring in one variable over .

4. Cliques in

Let be a commutative ring with identity which is not an integral domain. In this section, we prove that if a ring admits more than one maximal -prime of , then the clique number of is finite if and only if does not contain any infinite clique if and only if is finite. Moreover, if a ring is such that R has only one maximal -prime of , we obtain some necessary conditions in order that the clique number of is finite. Furthermore, if is either a Noetherian ring or a chained ring and if admits at least one edge (that is, there exist with such that ), then it is shown that the clique number of is finite if and only if does not contain any infinite clique if and only if 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 be a commutative ring with identity which is not an integral domain. Let be any prime ideal of . Then the following hold. (i)If does not contain any infinite clique, then is finite. (ii)If is finite, then is finite and indeed, .

Proof. (i) Suppose that is infinite. Then we can choose an infinite sequence of distinct elements . Since is a prime ideal of R and as for , it follows that 0 for all . Observe that the subgraph of induced on is an infinite clique. This contradicts the assumption that does not contain any infinite clique. Hence we obtain that is finite.(ii) Let . We assert that . Suppose that . Let . Then it is clear that the subgraph of induced on is a clique. This is impossible since . This shows that .

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

Lemma 4.2. Let be a commutative ring with identity which is not an integral domain. Let be any prime ideal of . Let is a prime ideal of such that but . Then the following hold. (i)If does not contain any infinite clique, then can admit only a finite number of elements which are pairwise incomparable under inclusion. (ii)If is finite, then can admit at most elements which are pairwise incomparable under inclusion.

Proof. (i) Suppose that does not contain any infinite clique. We want to verify that can admit only a finite number of elements which are pairwise incomparable under inclusion. Suppose that there exist infinitely many elements in which are pairwise incomparable under inclusion. Hence there exist for each with and not being comparable under inclusion for all with . Now for . Hence it is possible to choose . Let . Now it follows from prime avoidance [21, Proposition 1.11(i)] that there exists . Observe that . Hence we obtain that is infinite. This contradicts Lemma 4.1(i). This proves that if does not contain any infinite clique, then can admit only a finite number of elements which are pairwise incomparable under inclusion.(ii) Let . Suppose that admits more than elements which are pairwise incomparable under inclusion. Let be such that and are not comparable for all with . Let . Let . As in (i), we can choose . Observe that . This implies that . This contradicts Lemma 4.1 (ii). Hence we obtain that can admit at most elements which are pairwise incomparable under inclusion.

We next study in the following corollary to Lemma 4.2, the effect of the nature of the cliques of on the set of maximal -primes of , and the set of minimal prime ideals of .

Corollary 4.3. Let be a commutative ring with identity which is not an integral domain. Then the following hold. (i)If does not contain any infinite clique then the set of maximal -primes of in is finite the set of minimal prime ideals of is finite. (ii)If is finite, then can admit at most maximal -primes of and if admits exactly maximal -primes of with , then can admit at most minimal prime ideals, and if is the number of minimal prime ideals of with , then .

Proof. (i)(a) Let be a maximal -prime of in . Let is a maximal -prime of in and . Since any maximal -prime of in is a subset of and as distinct maximal -primes of in are not comparable under inclusion, it follows from Lemma 4.2(i) that is finite. Observe that the set of all maximal -primes of in . It is now clear that can admit only a finite number of maximal -primes of .(i)(b) If is any minimal prime ideal of , then [22, Theorem 84]. Since distinct minimal prime ideals of are not comparable under inclusion, it follows using the same arguments as in the proof of (i)(a) that can admit only a finite number of minimal prime ideals.(ii)(a) Let , be as in the proof of (i)(a). Let . Now using the same arguments as in the proof of (i), it follows from Lemma 4.2(ii) that . Since the set of all maximal -primes of in , we obtain that can admit at most maximal -primes of .Suppose that admits exactly maximal -primes of with . Let be the set of all maximal primes of in . Note that . Since distinct maximal -primes of in are not comparable under the inclusion relation, it follows from [21, Proposition 1.11(i)] that for each , . Then it is clear that for any distinct , and are distinct nonzero zero-divisors of , and as , it follows that there exists at least one such that both and are not in and hence . Thus we obtain that the subgraph of induced on is a clique and so .(ii)(b) This can be proved using similar arguments as in the proof of (ii)(a) and using Lemma 4.2(ii).

The following proposition is one among the main results in this section. We show in this proposition that if a ring admits at least two maximal -primes of , then is finite if and only if is finite.

Proposition 4.4. Let be a commutative ring with identity. Suppose that has at least two maximal -primes of . Then the following conditions are equivalent: (i)   is finite. (ii) is finite. (iii) does not contain any infinite clique.

Proof. (i) ⇒ (ii) Let . Now it follows from Corollary 4.3(ii)(a), that can admit at most maximal -primes of . Let be the number of maximal -primes of in and let be the set of all maximal -primes of in . Note that and by the hypothesis, . Observe that We know from Lemma 4.1(ii) that We now verify that . Let . Note that . Hence it follows that 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 is finite. It follows from Corollary 4.3(i)(a) that can admit only a finite number of maximal -primes of . Now one can proceed as in the proof of (i) ⇒ (ii), and use Lemma 4.1(i) to obtain that is finite. It is now clear that is finite.

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

Let be a finite commutative reduced ring with identity which is not an integral domain. Since is a finite ring, any prime ideal of is maximal. Let be the set of all prime ideals of . Since is reduced, . As is not an integral domain, it follows that . Moreover, by the Chinese Remainder Theorem [21, Proposition 1.10], it follows that is isomorphic to . Thus is isomorphic to a finite direct product of finite fields. Let and let be finite fields. Let . We now proceed to describe . We make use of some of the techniques from [4].

We first recall the following facts from [4].

Fact 4.6. Let be a commutative ring with identity which is not an integral domain. For any , define if and only if . Then ~ is an equivalence relation on .

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

For an element , we denote by , the equivalence class determined by ~ containing “a”.

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

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

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

Proof. The Fact 4.7 can be proved easily with the help of the ideas contained in the proof of [4, Lemma 3.1] and [10, Lemma 2.11]. In fact, [4, Lemma 3.1] and [10, Lemma 2.11] assert that for any von Neumann regular ring and for , if and only if . 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 . Now there exists a unit in and an idempotent element in such that . Then it is clear that . Since , it follows that . This proves that and so .Note that if is any idempotent element in a ring ( not necessarily von Neumann regular), then . Moreover, it is easy to check that for idempotent elements in a ring if and only if . If , are idempotent elements in a ring such that , then it follows that . Hence and so .Now it follows from the above two preceding paragraphs that given any where is a von Neumann regular ring, then there exists a unique idempotent such that .

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

Lemma 4.8. Let be a von Neumann regular ring which is not an integral domain. Let be a set of idempotents in such that for all . Let . Then the subgraph of induced on is a clique.

Proof. Let . Let with . Note that . Now it follows from the proof of Fact 4.7 that there exist units in such that and . Hence . This proves that any two distinct elements of are adjacent in for each . Let with . Let , . Note that and for some units in . Hence , since by the hypothesis.Indeed, it is true that if and are equivalence classes determined by ~ defined on where is a commutative ring with identity which is not an integral domain ( need not be von Neumann regular) and if , then we assert that for any and , . Note that if , then . Hence . This implies that . Hence and this is a contradiction. Thus we obtain that .This proves that the subgraph of induced on A = is a clique.

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

Lemma 4.9. Let . Let be a field for . Let be their direct product. Then the number of idempotents in (that is, the number of nontrivial idempotents in ) equals and the number of equivalence classes determined by the equivalence relation ~ defined on equals .

Proof. The only idempotent elements in a field are 1 and 0. Using this observation, it follows that has exactly idempotents. Among them except and , the rest of them are in . Thus we obtain that the number of idempotents in (that is, the number of nontrivial idempotents in ) equals . Let be the set of all idempotent elements in . Since is von Neumann regular, it follows from Fact 4.7 that the set of all equivalence classes determined by ~ is .

Let . Let where is a field for . The following lemma describes the cliques of .

Lemma 4.10. Let . Let be a field for . Let . Let be such that the subgraph of induced on is a clique. Let be the set of all idempotents in . Then there exists a nonempty subset of such that for all , and .

Proof. We know from the proof of Lemma 4.9 that is the set of all equivalence classes determined by ~. Thus we obtain that . Now being a subset of , it follows that . Let be such that is non-empty for each . Since is non-empty, it follows that is non-empty. We now verify that for any . This is clear if since any element of is a nonzero idempotent in . Suppose that .Let and . Since , it follows that . As the subgraph of induced on is a clique, we obtain that . Note that for . Hence we obtain that and so . This proves that there exists a non-empty subset of such that 0 for all, and moreover, .

Let be as in Lemma 4.10 with the further assumption that is finite for . We determine in the following proposition.

Proposition 4.11. Let and let be finite fields. Let . Let be the set of all idempotents in . Then varies over all non-empty subsets of satisfying the property that 0 for any .

Proof. Let is a non-empty subset of satisfying the property that for any . Let . Let . Now it follows from Lemma 4.8 that the subgraph of induced on is a clique. Thus . Hence we obtain that Let be such that the subgraph of induced on is a clique. We know from Lemma 4.10 that there exists such that . Hence we obtain that . This implies that From (4.4) and (4.5), it follows that .

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

Remark 4.12. Let 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 , . Let . Let be finite fields and let . Let be an idempotent. Let . Observe that .

We next have the following example.

Example 4.13. (i) Let be finite fields. Let . Note that is the set of all idempotents in . Now it follows from Proposition 4.11 and Remark 4.12 that . Note that this fact can also be verified directly. (ii) Let , , be finite fields. Let . Note that is the set of all idempotents in . Now it follows from Proposition 4.11 and Remark 4.12 that ,.

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

Lemma 4.14. Let be a commutative ring with identity and let . Suppose that has exactly one maximal -prime of and let it be . If does not contain any infinite clique, then the nilradical of and if furthermore, , then is finite, is a -prime of in satisfies descending chain condition (d. c. c.) on principal ideals.

Proof. (i) As any nilpotent element of is a zero-divisor of and since , it follows that . Let . We assert that is nilpotent. Suppose that is not nilpotent. Then for any , , and moreover, for all distinct ,. Hence is such that the subgraph of induced on is an infinite clique. This is in contradiction to the assumption that does not contain any infinite clique. Hence is nilpotent. This shows that and so .(ii) Suppose that is infinite. We first assert that for each . Suppose that there exists such that . Since we are assuming that is infinite, it is possible to find an infinite sequence of elements such that and for all distinct . Note that for all , , and as , it follows that . Moreover, as for all distinct , , it follows that . Hence we obtain that the subgraph of induced on is an infinite clique. This contradicts the assumption that does not contain any infinite clique. Hence for each . As , there exist such that and . Note that we have . For any positive integer , let . It is clear that for any positive integer and , since and . Moreover, for all distinct , since and . Furthermore, as and , for all distinct , it follows that for all distinct . Hence we obtain that the subgraph of induced on is an infinite clique. This contradicts the assumption that does not contain any infinite clique. Hence it follows that is finite.(iii) We now verify that is a -prime of in . We consider two cases. Case (A): is finitely generated. By (i), we have and hence we obtain that is a nilpotent ideal of . Let be least with the property that . Now for any , is a -prime of in . Case (B): is not finitely generated.We have , by assumption. Hence there exist such that and. Suppose that is not a -prime of in . Then, as , it follows that . As and since is not finitely generated, it follows that . Hence . Hence there exists. Thus are distinct and . Let k be any positive integer with . Assume that there exists a subset of with . Observe that and . Hence there exists . This shows that if is not finitely generated and if is not a -prime of in , then there exists an infinite subset of such that the product · · · for . This implies that the subgraph of induced on is an infinite clique.This contradicts the hypothesis that does not contain any infinite clique. Hence is a -prime of in . (iv) We obtain from (ii) that is finite. Since any finite integral domain is a field, it follows that is a maximal ideal of . By (i), . Hence we obtain that 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 for such that . Note that there exist for such that . Hence Since for , it follows that the elements satisfy for all distinct . As each element of is nilpotent, it follows from (I) and the fact that for that there exist positive integers such that for all distinct . Let . Observe that the subgraph of induced on is an infinite clique. This contradicts the assumption that does not contain any infinite clique. Hence it follows that satisfies d. c. c. on principal ideals.

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

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

Proof. (i) ⇒ (ii) This is clear.(ii) ⇒ (iii) We know from Lemma 4.14 that the nilradical of and is finite. Now by hypothesis, is a Noetherian ring. Hence is finitely generated. Therefore, for some . Since , it follows that . Observe that for each integer is a finite-dimensional vector space over the finite field . Hence it follows that is finite for . Now are finite, and hence it follows that is finite. Proceeding in this way, we obtain that is finite. Thus and are finite. Hence we obtain that is finite.(iii) ⇒ (i) This is clear.

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

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

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

Proof. (i) ⇒ (ii) Since the ideals of are linearly ordered under inclusion, it follows that admits exactly one maximal -prime of . Let be the unique maximal -prime of in . We are assuming that does not admit any infinite clique. So, we obtain from Lemma 4.14(i) that the nilradical of . Note that . Let . It is known that for any , [6, Lemma 4.2(3)]. Since does not admit any infinite clique, it follows that is finite. Now by the assumption that is a chained ring and there exist with such that , it follows that is non-empty. Let . Since each element of is nilpotent, it follows that there exist such that for each . As is a chained ring, we obtain that . Moreover, it follows from Lemma 4.14(ii) that is finite. Hence is a maximal ideal of and since is a chained ring, it follows that is quasilocal with as its unique maximal ideal. As is nilpotent and , it follows that . Now is a chained ring with as its unique maximal ideal satisfying the further condition that . In such a case it is well known that for any. Using the same reasoning as in the proof of (ii) ⇒ (iii) of Proposition 4.15, it now follows that is finite. (ii) ⇒ (iii) and (iii) ⇒ (i) are clear.

The following remark determines for any finite chained ring which is not an integral domain.

Remark 4.17. Let be a finite chained ring which is not an integral domain. Let denote the unique maximal ideal of . Suppose that . Let be least with the property that . Then the following hold.(i)if is even. (ii) if is odd.

Proof. As is already observed in the proof of (i) ⇒ (ii) of Proposition 4.16, for any.(i) Suppose that is even and let . By the choice of . If, then for some and unit in and for some and unit in . Hence , since and . Moreover, observe that for any . Hence the subgraph of induced on is a clique. This implies that . Let be such that the subgraph of induced on is a clique. Let . Since is a chained ring, it is possible to find such that for each . Hence . Since for each, we obtain that for each. As , it follows that . Thus and so . This proves that . Hence .(ii) Suppose that is odd and let for some . In this case, we verify that where . It is clear that for any . This shows that the subgraph of ((R))c induced on is a clique. Hence we obtain that . Let be such that the subgraph of induced on is a clique. We assert that . Let be such that for each . Now, it follows as in the proof of (i) that for each . Since , it follows that . If there exists at least one such that , then, as , it follows that and so we obtain that . Hence it follows that . Otherwise, we obtain that . Note that and so . This proves that . Hence we obtain that

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

Example 4.18. Let be the field containing exactly two elements. It is convenient to denote simply by . Let where is the polynomial ring in one variable over . Note that is irreducible over . Let be the splitting field of over . Let be a root of . It is clear that and . Indeed, is a basis of as a vector space over . Moreover, . Let be the power series ring in one variable over . Note that where . Let . Consider the ring . Let us denote by I. Observe that . Note that is a finite local ring with unique maximal ideal satisfying the zero-ideal of , but is nonzero. We have and thus , , , . Observe that . Now , , , . Thus . We assert that . Since the zero-ideal of , it follows that the vertex set of any clique in must be a subset of As . Hence the subgraph of induced on is a clique. This shows that . Observe that We have . Note that Moreover, It is clear from (4.7), (4.8), (4.9), and (4.10) that if is such that the subgraph of induced on is a clique, then can contain at most 3 elements. Hence we obtain that . Thus .Let and be as above. Let . Note that is a finite local ring with as its unique maximal ideal. Thus . Moreover, note that is the zero- ideal of but is nonzero. It is convenient to denote the ideal of by . Note that . We assert that . Since , it follows that if is any subset of such that the subgraph of induced on is a clique, then can admit at most one element of . We claim that . This is clear if does not contain any element of . Suppose that has an element of . In such a case, we verify that cannot contain all the elements of . Let be such that . Note that for some . We consider two cases.

Case i. . In such a case, the element annihilates each element of and hence it annihilates each element of .

Case ii. . Observe that for some with at least one of is different from 0. Thus . Note that annihilates the element . The element annihilates and the element annihilates .

This shows that if contains an element of , then that element annihilates at least one element of . Since the subgraph of induced on is a clique, it follows that A cannot contain all the elements of . This proves that and hence we obtain that.

We next claim that the subgraph of induced on is a clique. Let , . Note that and for some (for) with . Note that ab = for some . Since, and as , it follows that ab is a nonzero element of . This proves that the subgraph of induced on is a clique. Hence we obtain that and so .

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

Lemma 4.19. Let be a commutative ring with identity which is not an integral domain. Suppose that is the only maximal -prime of in . If is finite, then for any where .

Proof. By hypothesis, is finite. Let . We assert that where . Suppose that . For each , let . Note that and and for and moreover, . Since and , it follows that for all . Let with . We claim that . Suppose that . We may assume without loss of generality that . Then implies that . As , it follows that , and this is not possible since by assumption, . Hence for all distinct and moreover, . Hence we obtain that the subgraph of induced on is a clique. This implies that . This is impossible since . Hence we obtain that for any .

The next remark provides examples of rings for which .

Remark 4.20. We remark here that Lemma 4.19 is motivated by [3, Theorem 3.4]. Let be a commutative ring with identity which is not reduced. Let . Recall that by the index of nilpotence of , we mean the least positive integer such that . Suppose that with . Let be the index of nilpotence of . 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 is a clique in . Hence it follows that . Thus if a commutative ring with identity is such that there is no bound on the index of nilpotence of nilpotent elements of , then it follows that .

Let and be as in Lemma 4.19. If is finite, then with the help of Lemma 4.19, we prove in the following proposition that is nilpotent.

Proposition 4.21. Let be a commutative ring with identity which is not an integral domain. Suppose that has only one maximal -prime of and let it be . If is finite, then is nilpotent.

Proof. Let . We claim that with . Suppose that . Then there exist for such that . Let s be the number of distinct elements among . Note that we may assume without loss of generality that are the distinct elements among . Let , and let . Let for . Note that = , and moreover, for all distinct . Hence we obtain that . Since , it follows that , and moreover, for each , the product of factors of is different from 0. We know from Lemma 4.19 that for any . Hence it follows that for . Furthermore, observe that the subgraph of induced on is a clique and so . Thus we obtain that . This is impossible. Hence it follows that with .

We conclude this note with the following example of an infinite ring such that has exactly one maximal -prime of satisfying the property that , thereby illustrating that Proposition 4.15 need not hold for non-Noetherian rings.

Example 4.22. Let be the polynomial ring in an infinite number of variables over . Let be the ideal of generated by . Let . Let be the ideal of generated by . Observe that is the only prime ideal of . Thus . It is clear that is infinite. Note that is a cycle of length 3 in and hence we obtain that . We next verify that . Let be such that the subgraph of induced on is a clique. We assert that can contain at most 3 elements. Suppose that contains more than 3 elements. Note that for any with , . Let . Observe that , , for some and for some , . It is clear that if the coefficient of is 0 in We are assuming that contains at least 4 elements. Let . Let , for some and for some . Since for all distinct , it follows from (4.11) that the coefficient of must be 1 for all except possibly one value of . We may assume without loss of generality that where . Since , it follows that exactly one among must be 1. We may assume without loss of generality that and . Now either or . If , then we arrive at, which is impossible. If , then we obtain that , and this is also impossible.This proves that if is such that the subgraph of induced on is a clique, then . Hence we obtain that . This proves that.

Acknowledgments

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

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