`ISRN AlgebraVolumeΒ 2012Β (2012), Article IDΒ 282054, 13 pageshttp://dx.doi.org/10.5402/2012/282054`
Research Article

## When Is the Complement of the Zero-Divisor Graph of a Commutative Ring Complemented?

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

Received 12 March 2012; Accepted 3 April 2012

Academic Editors: D.Β Anderson, A. V.Β Kelarev, and C.Β Munuera

Copyright Β© 2012 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 be a commutative ring with identity which has at least two nonzero zero-divisors. Suppose that the complement of the zero-divisor graph of has at least one edge. Under the above assumptions on , it is shown in this paper that the complement of the zero-divisor graph of is complemented if and only if is isomorphic to as rings. Moreover, if is not isomorphic to as rings, then, it is shown that in the complement of the zero-divisor graph of , either no vertex admits a complement or there are exactly two vertices which admit a complement.

#### 1. Introduction

The rings considered in this paper are commutative rings with identity satisfying the further condition that there exist two distinct zero-divisors whose product is nonzero. Let be a commutative ring with identity. Recall from [1] that the zero-divisor graph of denoted by is an undirected graph whose vertex set is the set of all nonzero zero-divisors of and two distinct nonzero zero-divisors , of are joined by an edge in this graph if and only if . Several researchers investigated the properties of zero-divisor graphs of commutative rings with identity. The following survey article [2] gives a very clear account of the problems solved in the area of zero-divisor graphs of commutative rings along with necessary history of the problems attempted in this area.

All graphs considered in this paper are undirected graphs. Let be a graph. Recall from [3, 4] that two distinct vertices , of are said to be orthogonal, written if and are adjacent in and there is no vertex of which is adjacent to both and in ; that is, the edge of is not a part of any triangle in . Let be a vertex of . Recall from [3] that a vertex of is said to be a complement of if . Moreover, recall from [3] that is complemented if each vertex of admits a complement in . Furthermore, is said to be uniquely complemented if is complemented and whenever the vertices , , of are such that and , then are not adjacent in and a vertex of is adjacent to in if and only if is adjacent to in . In Section  3 of [3], the authors characterized commutative rings such that is complemented (resp., is uniquely complemented).

Let be a simple graph. Recall from [5, Definition ] that the complement of denoted by is defined as a graph with and two distinct elements , are joined by an edge in if and only if there is no edge of joining and .

Let be a commutative ring with identity. We denote by the set of all nonzero zero-divisors of . Suppose that . With this assumption on , in [6, 7], we investigated the relationship between some graph-theoretic properties of and the ring-theoretic properties of . Motivated by the interesting theorems proved in [3], in this paper, we discuss the question of when is complemented.

Before we proceed further, let us recall the following definitions and results from commutative ring theory. Let be a commutative ring with identity. Let be an ideal of . Recall from [8] that a prime ideal of is said to be a maximal -prime of if is maximal with respect to the property of being contained in for some . Thus a prime ideal of is a maximal -prime of (0) if is maximal with respect to the property of being contained in . Note that is a multiplicatively closed subset of . If , then . Hence it follows from Zorn’s lemma and [9, Theorem 1.1] that there exists a maximal -prime of (0) in such that . Thus if denotes the set of all maximal -primes of (0) in , then .

Let be an ideal of a ring . Recall from [10] that a prime ideal of is said to be an associated prime of in the sense of Bourbaki if for some . In this case we say that is a -prime of .

Let be a commutative ring with identity and suppose that . In Section 2 of this paper we assume that has exactly one maximal -prime of (0). Let be the unique maximal -prime of (0) in . If is not a -prime of (0) in , then it is proved in Proposition 2.1 that no vertex of admits a complement in . Suppose that is a -prime of (0) in and if , then it is shown in Proposition 2.2 that no vertex of admits a complement in . If and if admits at least one edge (i.e., equivalently, if there exist distinct with ), then it is observed in Remark 2.3 that , moreover, we note that, except for the unique isolated vertex of , the other two vertices are complements of each other, and furthermore, with the help of results from [11], it is deduced that is isomorphic to exactly one of the rings from the collection where (resp., ) denotes the polynomial ring in one variable over (resp., over ). Throughout this paper unless otherwise specified we denote by , the polynomial ring in one variable over for any .

In Section 3 we consider commutative rings with identity such that has exactly two maximal -primes of (0). Let denote the set of all maximal -primes of (0) in . The main results proved in Section 3 are Theorem 3.9 and Proposition 3.11. If , then it is proved in Theorem 3.9 that has vertices which admit a complement in if and only if either is isomorphic to or is isomorphic to as rings if and only if has exactly two vertices which admit a complement in . If and if admits at least one edge, then it is shown in Proposition 3.11 that has vertices which admit a complement in if and only if is isomorphic to , where is an integral domain, and it is noted that is complemented if and only if is isomorphic to as rings, and in the case, when is not isomorphic to , then has exactly two vertices which admit a complement in .

In Section 4 we consider commutative rings with identity such that has at least three maximal -primes of (0) and it is shown in Proposition 4.1 that, for such rings, no vertex of admits a complement in .

#### 2. π Has Exactly One Maximal π -Prime of (0)

Let be a commutative ring with identity such that and has exactly one maximal -prime of (0). Let be the unique maximal -prime of (0) in . Observe that either is not a -prime of (0) in or is a -prime of (0) in ; that is, equivalently in terms of graph theoretic terms, either is connected or is not connected [6, Theorem 1.1(a)]. In this section, we prove that is not complemented if is not a -prime of (0) in . Indeed, if is not a -prime of (0) in , then we prove in Proposition 2.1 that no vertex of admits a complement in . If is a -prime of (0) in and if , then we prove in Proposition 2.2 that no vertex of admits a complement in . Moreover, if is a -prime of (0) in and if , then with the help of [11, Theorem 3.2] we describe, in Remark 2.3, rings such that has vertices which admit a complement in . We often make use of the following fact (which is an immediate consequence of the definition of a complement of a vertex in a graph) that, in a graph , no vertex admits a complement in if and only if each edge of is an edge of a triangle in . We begin with the following proposition.

Proposition 2.1. Let be a commutative ring with identity. Let , and suppose that has exactly one maximal -prime of (0). Let be the unique maximal -prime of (0) in . If is not a -prime of (0) in , then no vertex of admits a complement in .

Proof. In view of the hypothesis that is the only maximal -prime of (0) in , it follows that . Let be any edge of . We prove that the edge is an edge of a triangle in ; that is, there exists such that and . Proceeding as in the proof of [7, Lemma 3.2], it can be shown that there exists such that and . This shows that any edge of is an edge of a triangle in . Thus if is any element of , then does not admit a complement in .

With the assumption that is a -prime of (0) in , the following proposition provides a sufficient condition under which no vertex of admits a complement in .

Proposition 2.2. Let be a commutative ring with identity. Suppose that has exactly one maximal -prime of (0), and let it be . If is a -prime of (0) in and if , then no vertex of admits a complement in .

Proof. As is a -prime of (0) in , there exists such that . Let be any edge of . We now verify that there exists such that , , and . Proceeding as in the proof of [7, Proposition 3.7], it can be shown that there exists such that , , and . This proves that any edge of is an edge of a triangle in . Hence we obtain that no vertex of admits a complement in .

The following remark characterizes commutative rings with identity satisfying the following conditions: has exactly one maximal -prime of (0), the unique maximal -prime of (0) in is a -prime of (0) in , and has at least one edge and has at least one vertex which admits a complement in .

Remark 2.3. First observe that is one of the rings from the collection , then has properties , , and mentioned above. We show in this remark with the help of [11, Theorem 3.2] that if is a ring with the above three properties, then is isomorphic to exactly one of the rings given in the above collection. Let , be as mentioned in the beginning of this section. Suppose that is a -prime of (0) in and moreover, contains at least one edge that is, there exist distinct such that . Let be such that . We want to characterize such that has at least one vertex which admits a complement in . It follows from Proposition 2.2 that . Note that , and hence it follows that . Now proceeding as in [7, Remark 3.8], it follows using [11, Theorem 3.2] that must be isomorphic to one of the rings given in the above collection. Moreover, note that has exactly three vertices with one isolated vertex and the other two vertices being complements of each other in .

#### 3. π Has Exactly Two Maximal π -Primes of (0)

Let be a commutative ring with identity. Suppose that has exactly two maximal -primes of (0). Let denote the set of all maximal -primes of (0) in .

Suppose that . It is useful to remark here that if and only if is connected [6, Theorem 1.1.(b)]. We prove in Theorem 3.9 that has at least one vertex which admits a complement in if and only if either is isomorphic to or is isomorphic to where denotes the polynomial ring in one variable over .

Let , , be as mentioned in the beginning of this section. Suppose that . In Proposition 3.11, we determine up to isomorphism of rings, rings such that has at least one vertex which admits a complement in . It is indeed proved in Proposition 3.11 that has vertices which admit a complement in if and only if is isomorphic to , where is an integral domain. Moreover, it is noted in Proposition 3.11 that is complemented if and only if is isomorphic to as rings. Furthermore, if is not isomorphic to , then it is observed that has exactly two vertices which admit a complement in .

We first state and prove several lemmas that are needed for proving Theorem 3.9. We start with the following.

Lemma 3.1. Let , , be as mentioned in the beginning of this section. If one between and is not a -prime of (0) in , then no vertex of admits a complement in .

Proof. Without loss of generality we may assume that is not a -prime of (0) in . Observe that, to prove the lemma, it is enough to prove the following: if is any edge of , then there exists such that and . We consider the following cases.
Case 1. Both and belong to .
If , then as , it follows from [9, Theorem 81] that . Hence there exists such that and . Since , whereas , it is clear that .
Suppose that , then it follows that either or . Without loss of generality we may assume that . Since , it follows that . Thus . Since is not a -prime of (0) in , and . Now it follows from [9, Theorem 81] that . Hence there exists such that . Since , it follows that . As , we obtain that and so .
Case 2. Both and belong to .
Since is not a -prime of (0) in , it follows that and . Hence we obtain from [9, Theorem 81] that . So there exists such that and . Since , whereas , it is clear that .
Case 3. .
Now , are such that . Proceeding as in the proof of [7, Lemma 3.4(ii)], we obtain that there exists such that and .
Thus it is shown that any edge of is an edge of a triangle in . Hence we obtain that no vertex of admits a complement in .

Suppose that both and are -primes of (0) in . The following lemma gives a sufficient condition under which no vertex of admits a complement in .

Lemma 3.2. Let be as mentioned in the beginning of this section. If both and are -primes of (0) in and if , then no vertex of admits a complement in .

Proof. By assumption both and are -primes of (0) in . Hence there exist such that and . We know from [12, Lemma 3.6] that . We now proceed to show that no vertex of admits a complement in . As is remarked in the introduction, it is enough to show that any edge of is an edge of a triangle in . Let be any edge of . We want to show that there exists such that and . We consider the following cases.
Case 1. Both and belong to .
It now follows as in the proof of Lemma 3.1 Case 1 that we may assume without loss of generality that . Since , we obtain that . Thus . Note that for any , and so . Hence . Moreover, since , we obtain that . By hypothesis, . Hence there exist distinct . It is clear that and . As , it follows that either or . We may assume without loss of generality that . Now is such that , , and .
Case 2. Both and belong to .
The proof of the fact that there exists such that and is similar to the proof of Case 1 of this lemma.
Case 3. , .
Now . It follows as in the proof of [7, Lemma 3.4(ii)] that there exists such that and .
This proves that any edge of is an edge of a triangle in , and so we obtain that no vertex of admits a complement in .

If both and are -primes of (0) in , we provide in the following lemma some conditions under which no vertex of admits a complement in .

Lemma 3.3. Let , , be as mentioned in the beginning of this section. Suppose that both and are -primes of (0) in . If and , then no vertex of admits a complement in .

Proof. Let be any edge of . We prove that there exists such that and . We consider the following cases.Case 1. Both and belong to .
Proceeding as in the proof of Lemma 3.1 Case 1, we may assume without loss of generality that . Since , it follows that . Thus . If , then . Hence . By hypothesis . Hence there exists . Now it is clear that and . Suppose that . Then and so . Note that is such that and . Moreover, , , and so .
Case 2. Both and belong to .
The hypotheses regarding and are symmetric. Hence it follows as in the proof of Case 1 of this lemma that there exists such that and .
Case 3. , .
Now it follows as in the proof of [7, Lemma 3.4(ii)] that there exists such that and .
This shows that any edge of is an edge of a triangle in , and so we obtain that no vertex of admits a complement in .

If , then we prove in Lemma 3.5 that is not complemented; that is, there exists at least one vertex of which does not admit a complement in . We make use of the following lemma in the proof of Lemma 3.5.

Lemma 3.4. Let , , be as mentioned in the beginning of this section. Suppose that both and are -primes of (0) in . Let be such that and . If , then either or .

Proof. Since and are the only maximal -primes of (0) in , it follows that . By assumption, and . Now for any , from , it follows that . Similarly for any , from , we obtain that . By hypothesis, . Let . Now and so . Hence . Thus either or . If , then as and , we obtain that . If , then it follows that since and . This proves that either or .

We next have the following lemma which shows that is not complemented if .

Lemma 3.5. Let be mentioned in the beginnig of this section. If , then there exists at least one vertex of which does not admit a complement in .

Proof. If at least one between and is not a -prime of (0) in , then it is proved in Lemma 3.1 that no vertex of admits a complement in . Hence we may assume that both and are -primes of (0) in . Let be such that and . We know from Lemma 3.4 that either or . Without loss of generality we may assume that .
We assert that does not admit a complement in . It is enough to prove the following: if is any edge of , then there exists such that and . Since , it follows that . Note that is such that , , , and .
This proves that has at least one vertex which does not admit a complement in .

With the assumption that , we next attempt to characterize rings such that has vertices which admit a complement in . Towards that goal, we begin with the following lemma.

Lemma 3.6. Let , , be as mentioned in the beginning of this section. If , then has vertices which admit a complement in if and only if one of the following holds:(i) and for some .(ii) and for some .

Proof. Suppose that has vertices which admit a complement in . Then it follows from Lemma 3.1 that there exist such that and . Moreover, it follows from Lemma 3.2 that . Furthermore, we know from Lemma 3.3 that either or .
Let . Note that, for any , . Hence . Similarly it follows that . Suppose that . Then we obtain that . It is shown in the proof of Lemma 3.4 that and . If , then we arrive at and with . Hence holds. Suppose that . Then as , it follows that . Hence . We assert that . Suppose that it does not hold. Then . Let be such that admits a complement in . It is shown in the proof of Lemma 3.5 that does not admit a complement in . Hence it follows that . As , it follows that . If one between and is in and the other is in , then it follows as in the proof of [7, Lemma 3.4(ii)] that there exists which is adjacent to both and in . This is not possible since is a complement of in . Thus either both and belong to or both and belong to . If both and belong to , then is such that and . Hence is adjacent to both and in . This is impossible. If both and belong to , then, since we are assuming that , any satisfies and . This cannot happen as is a complement of in . Hence . This together with the fact that and implies that holds. Thus if has vertices which admit a complement in , then either or holds.
Conversely assume that either or holds. Suppose that holds. Now for some , and . Let . Since , it follows that and hence . Moreover, if , then must be in and so . Hence there exists no vertex of which is adjacent to both and in . Therefore, is a complement of in . This shows that has vertices which admit a complement in . Similarly if holds and if , then it follows that and are complements of each other in . This proves that either or holds, then has vertices which admit a complement in .
This completes the proof of Lemma 3.6.

Suppose that . In Theorem 3.9 we characterize rings such that has vertices which admit a complement in . We need the following lemma for proving Theorem 3.9.

Lemma 3.7. Let , , be as mentioned in the beginning of this section. Suppose that , and if has vertices which admit a complement in , then is isomorphic to as rings where is a ring with and is an integral domain.

Proof. We are assuming that there are vertices of which admit a complement in . Hence we obtain from Lemma 3.1 that there exist such that and . Now it follows from Lemma 3.2 that . Let . We claim that . Suppose that . Then there exists a maximal ideal of such that . Let and . Now , and since , it follows from the choice of the elements and that . Let . Note that , and as , we obtain that . Since , it follows that . Thus , and so . This is impossible since , whereas . Hence we obtain that . Observe that , and as , it follows that . Hence . As , it follows that . Now we obtain from the Chinese remainder theorem [13, Proposition 1.10(ii)] that is isomorphic to as rings. Indeed, it follows from the Chinese remainder theorem that the mapping given by for any is an isomorphism of rings. The isomorphism maps onto . Hence . Thus either or . We may assume without loss of generality that and . Let and . Thus is isomorphic to as rings. Since , it follows that is an integral domain. We next verify that . Note that . As and , it follows that . Since , we obtain that .
Thus if and if has vertices which admit a complement in , then is isomorphic to as rings with and is an integral domain.

Let be a commutative ring with identity. It is well known [1, Example 2.1(a)] that (i.e., equivalently, is a graph on a single vertex) if and only if is either isomorphic to or is isomorphic to as rings. Let , where either or and is an integral domain. In the following lemma, we determine when has vertices which admit a complement in .

Lemma 3.8. Let , be as in the previous paragraph. Let be an integral domain. Let , where either or . Then the following statements are equivalent.(i) has vertices which admit a complement in .(ii) is isomorphic to as rings.(iii) has exactly two vertices which admit a complement in .

Proof. Let where is an integral domain. Note that is the union of two prime ideals and . Moreover, and are the only maximal -primes of the zero ideal in . Furthermore, , and hence is not the zero ideal of . Furthermore, and . We now show that the statements to are equivalent.
Suppose that has vertices which admit a complement in . We claim that for any . Suppose that for some . Note that either or . Observe that is such that . This shows that for any . Hence condition of Lemma 3.6 does not hold. Therefore condition of Lemma 3.6 must hold. Thus and for some . We now show that . Suppose that . Let with . Note that , and this implies that which contradicts the fact that . This proves that , and so is isomorphic to as rings.
Now . Observe that . Observe that the edge of is not an edge of any triangle in . Indeed, is a pendant vertex of . Thus the vertices and are complements of each other in . It can be easily verified that each of the other edges of is an edge of a triangle in . Hence has exactly two vertices which admit a complement in .
This is clear.
If where and is an integral domain, then the proof of the fact that the statements , , and are equivalent is exactly similar and hence is omitted.

Suppose that . The following theorem characterizes rings such that has vertices which admit a complement in .

Theorem 3.9. Let , , be as mentioned in the beginning of this section. Suppose that . The following statements are equivalent.(i) has vertices which admit a complement in .(ii) Either is isomorphic to or is isomorphic to as rings.(iii) has exactly two vertices which admit a complement in .

Proof. We know from Lemma 3.7 that is isomorphic to as rings with and is an integral domain. Since , it follows from [1, Example 2.1(a)] that either is isomorphic to or is isomorphic to as rings. Let and . Since either is isomorphic to or is isomorphic to , implies that either has vertices which admit a complement in or has vertices which admit a complement in . Now it follows from of Lemma 3.8 that is isomorphic to as rings. Hence we obtain that either is isomorphic to or is isomorphic to as rings.
It follows from of Lemma 3.8 that has exactly two vertices which admit a complement in .
This is clear.
This completes the proof of Theorem 3.9.

Suppose that ; that is, equivalently, by [6, Theorem 1.1(b)], is not connected. Assume that admits at least one edge. Our next aim is to determine when no vertex of admits a complement in and to determine rings such that has vertices which admit a complement in .

Lemma 3.10. Let , , be as mentioned in the beginning of this section. Suppose that . Then the following hold.(i) If , then no element of admits a complement in .(ii) If , then no element of admits a complement in .(iii) If and , then no vertex of admits a complement in .

Proof. (i) Let . Suppose that admits a complement in . Let be a complement of in . Then and . Since , , and , it follows that . By hypothesis, . Now for any , and . This is impossible since is a complement of in . This proves that if , then no element of admits a complement in .
(ii) The proof of is similar to the proof of .
(iii) Since and , it follows that . Note that is the vertex set of . Thus if and , then it follows from and that no vertex of admits a complement in .
This completes the proof of Lemma 3.10.

The following proposition describes rings such that , has at least one edge, and moreover, has vertices which admit a complement in .

Proposition 3.11. Let , , be as mentioned in the beginning of this section. If and if has at least one edge, then the following statements are equivalent.(i) has vertices which admit a complement in .(ii) is isomorphic to as rings, where is an integral domain.(iii) Exactly one of the following holds.(a)   is isomorphic to as rings, and in this case is a graph on three vertices and it admits exactly one isolated vertex and the two vertices are complements of each other in .(b)   is isomorphic to as rings, and in this case is a graph on four vertices and is complemented.(c)   is isomorphic to as rings, where is an integral domain with , and in this case is a graph on more than four vertices and it has exactly two vertices which admit a complement in .

Proof. Suppose that has vertices which admit a complement in . Then it follows from Lemma 3.10 that either or . Without loss of generality we may assume that . Then either or .
Suppose that . Let . Since , it follows that . Moreover, . We assert that . Since we are assuming that has at least one edge, we obtain that . Suppose that , then it follows from Lemma 3.10 that no element of admits a complement in . Observe that is an isolated vertex in and hence it does not admit a complement in . Hence we obtain that no vertex of admits a complement in . This is in contradiction to the hypothesis that has vertices which admit a complement in . Therefore, we obtain that . This shows that . Thus . Now is a finite set, and hence it follows from [14, Theorem 1] that is finite. Since any prime ideal of a finite ring is a maximal ideal, it follows that and are maximal ideals of . Thus . Since , it follows from the Chinese remainder theorem [13, Proposition 1.10(ii)] that the mapping given by for any is an isomorphism of rings. Note that and . Hence and . So we obtain that and as rings. Thus as rings. Thus with , we obtain that .
Suppose that . Let . Note that . We claim that . Suppose that . Then there exists a maximal ideal of such that . Let . Now , and as , it follows from the choice of the elements and that . Let . Note that