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

Abstract

Let 𝑅 be a commutative ring with identity which has at least two nonzero zero-divisors. Suppose that the complement of the zero-divisor graph of 𝑅 has at least one edge. Under the above assumptions on 𝑅, it is shown in this paper that the complement of the zero-divisor graph of 𝑅 is complemented if and only if 𝑅 is isomorphic to 𝐙/3𝐙×𝐙/3𝐙 as rings. Moreover, if 𝑅 is not isomorphic to 𝐙/3𝐙×𝐙/3𝐙 as rings, then, it is shown that in the complement of the zero-divisor graph of 𝑅, 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 𝑅 be a commutative ring with identity. Recall from [1] that the zero-divisor graph of 𝑅 denoted by Γ(𝑅) is an undirected graph whose vertex set is the set of all nonzero zero-divisors of 𝑅 and two distinct nonzero zero-divisors 𝑥, 𝑦 of 𝑅 are joined by an edge in this graph if and only if 𝑥𝑦=0. 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 𝐺=(𝑉,𝐸) be a graph. Recall from [3, 4] that two distinct vertices 𝑎, 𝑏 of 𝐺 are said to be orthogonal, written 𝑎⟂𝑏 if 𝑎 and 𝑏 are adjacent in 𝐺 and there is no vertex 𝑐 of 𝐺 which is adjacent to both 𝑎 and 𝑏 in 𝐺; that is, the edge 𝑎−𝑏 of 𝐺 is not a part of any triangle in 𝐺. Let 𝑎 be a vertex of 𝐺. Recall from [3] that a vertex 𝑏 of 𝐺 is said to be a complement of 𝑎 if 𝑎⟂𝑏. Moreover, recall from [3] that 𝐺 is complemented if each vertex of 𝐺 admits a complement in 𝐺. Furthermore, 𝐺 is said to be uniquely complemented if 𝐺 is complemented and whenever the vertices 𝑎, 𝑏, 𝑐 of 𝐺 are such that 𝑎⟂𝑏 and 𝑎⟂𝑐, then 𝑏,𝑐 are not adjacent in 𝐺 and a vertex 𝑣 of 𝐺 is adjacent to 𝑏 in 𝐺 if and only if 𝑣 is adjacent to 𝑐 in 𝐺. In Section  3 of [3], the authors characterized commutative rings 𝑅 such that Γ(𝑅) is complemented (resp., Γ(𝑅) is uniquely complemented).

Let 𝐺=(𝑉,𝐸) be a simple graph. Recall from [5, Definition 1.1.13] that the complement of 𝐺 denoted by 𝐺^𝑐 is defined as a graph with 𝑉(𝐺^𝑐)=𝑉(𝐺)=𝑉 and two distinct elements 𝑥, 𝑦∈𝑉 are joined by an edge in 𝐺^𝑐 if and only if there is no edge of 𝐺 joining 𝑥 and 𝑦.

Let 𝑅 be a commutative ring with identity. We denote by 𝑍(𝑅)^∗ the set of all nonzero zero-divisors of 𝑅. Suppose that |𝑍(𝑅)^∗|≥2. With this assumption on 𝑅, in [6, 7], we investigated the relationship between some graph-theoretic properties of (Γ(𝑅))^𝑐 and the ring-theoretic properties of 𝑅. Motivated by the interesting theorems proved in [3], in this paper, we discuss the question of when (Γ(𝑅))^𝑐 is complemented.

Before we proceed further, let us recall the following definitions and results from commutative ring theory. Let 𝑅 be a commutative ring with identity. Let 𝐼 be an ideal of 𝑅. Recall from [8] that a prime ideal 𝑃 of 𝑅 is said to be a maximal 𝑁-prime of 𝐼 if 𝑃 is maximal with respect to the property of being contained in 𝑍_𝑅(𝑅/𝐼)={𝑟∈𝑅∶𝑟𝑠∈𝐼 for some 𝑠∈𝑅⧵𝐼}. Thus a prime ideal 𝑃 of 𝑅 is a maximal 𝑁-prime of (0) if 𝑃 is maximal with respect to the property of being contained in 𝑍(𝑅). Note that 𝑆=𝑅⧵𝑍(𝑅) is a multiplicatively closed subset of 𝑅. If 𝑥∈𝑍(𝑅), then 𝑅𝑥∩𝑆=∅. Hence it follows from Zorn’s lemma and [9, Theorem 1.1] that there exists a maximal 𝑁-prime 𝑃 of (0) in 𝑅 such that 𝑥∈𝑃. Thus if {𝑃_𝛼}_𝛼∈Λ denotes the set of all maximal 𝑁-primes of (0) in 𝑅, then 𝑍(𝑅)=⋃_𝛼∈Λ𝑃_𝛼.

Let 𝐼 be an ideal of a ring 𝑅. Recall from [10] that 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. In this case we say that 𝑃 is a 𝐵-prime of 𝐼.

Let 𝑅 be a commutative ring with identity and suppose that |𝑍(𝑅)^∗|≥2. In Section 2 of this paper we assume that 𝑅 has exactly one maximal 𝑁-prime of (0). Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. If 𝑃 is not a 𝐵-prime of (0) in 𝑅, then it is proved in Proposition 2.1 that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. Suppose that 𝑃 is a 𝐵-prime of (0) in 𝑅 and if |𝑃|≥5, then it is shown in Proposition 2.2 that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. If |𝑃|<5 and if (Γ(𝑅))^𝑐 admits at least one edge (i.e., equivalently, if there exist distinct 𝑥,𝑦∈𝑍(𝑅)^∗ with 𝑥𝑦≠0), then it is observed in Remark 2.3 that |𝑃|=4, moreover, we note that, except for the unique isolated vertex of (Γ(𝑅))^𝑐, the other two vertices are complements of each other, and furthermore, with the help of results from [11], it is deduced that 𝑅 is isomorphic to exactly one of the rings from the collection {𝐙/8𝐙,(𝐙/4𝐙)[𝑥]/((2+4𝐙)𝑥(𝐙/4𝐙)[𝑥]+(𝑥²−(2+4𝐙))(𝐙/4𝐙)[𝑥]),(𝐙/2𝐙)[𝑥]/𝑥³(𝐙/2𝐙)[𝑥]} where (𝐙/4𝐙)[𝑥] (resp., (𝐙/2𝐙)[𝑥]) denotes the polynomial ring in one variable over 𝐙/4𝐙 (resp., over 𝐙/2𝐙). Throughout this paper unless otherwise specified we denote by (𝐙/𝑛𝐙)[𝑥], the polynomial ring in one variable over 𝐙/𝑛𝐙 for any 𝑛>1.

In Section 3 we consider commutative rings 𝑅 with identity such that 𝑅 has exactly two maximal 𝑁-primes of (0). Let {𝑃₁,𝑃₂} denote the set of all maximal 𝑁-primes of (0) in 𝑅. The main results proved in Section 3 are Theorem 3.9 and Proposition 3.11. If 𝑃₁∩𝑃₂≠(0), then it is proved in Theorem 3.9 that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐 if and only if either 𝑅 is isomorphic to 𝐙/4𝐙×𝐙/2𝐙 or 𝑅 is isomorphic to (𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥]×𝐙/2𝐙 as rings if and only if (Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐. If 𝑃₁∩𝑃₂=(0) and if (Γ(𝑅))^𝑐 admits at least one edge, then it is shown in Proposition 3.11 that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐 if and only if 𝑅 is isomorphic to 𝐙/3𝐙×𝑇, where 𝑇 is an integral domain, and it is noted that (Γ(𝑅))^𝑐 is complemented if and only if 𝑅 is isomorphic to 𝐙/3𝐙×𝐙/3𝐙 as rings, and in the case, when 𝑇 is not isomorphic to 𝐙/3𝐙, then (Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.

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

2. 𝑅 Has Exactly One Maximal 𝑁-Prime of (0)

Let 𝑅 be a commutative ring with identity such that |𝑍(𝑅)^∗|≥2 and 𝑅 has exactly one maximal 𝑁-prime of (0). Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. Observe that either 𝑃 is not a 𝐵-prime of (0) in 𝑅 or 𝑃 is a 𝐵-prime of (0) in 𝑅; that is, equivalently in terms of graph theoretic terms, either (Γ(𝑅))^𝑐 is connected or (Γ(𝑅))^𝑐 is not connected [6, Theorem 1.1(a)]. In this section, we prove that (Γ(𝑅))^𝑐 is not complemented if 𝑃 is not a 𝐵-prime of (0) in 𝑅. Indeed, if 𝑃 is not a 𝐵-prime of (0) in 𝑅, then we prove in Proposition 2.1 that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. If 𝑃 is a 𝐵-prime of (0) in 𝑅 and if |𝑃|≥5, then we prove in Proposition 2.2 that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. Moreover, if 𝑃 is a 𝐵-prime of (0) in 𝑅 and if |𝑃|<5, then with the help of [11, Theorem 3.2] we describe, in Remark 2.3, rings 𝑅 such that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. 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 𝐺=(𝑉,𝐸), no vertex admits a complement in 𝐺 if and only if each edge of 𝐺 is an edge of a triangle in 𝐺. We begin with the following proposition.

Proposition 2.1. Let 𝑅 be a commutative ring with identity. Let |𝑍(𝑅)^∗|≥2, and suppose that 𝑅 has exactly one maximal 𝑁-prime of (0). Let 𝑃 be the unique maximal 𝑁-prime of (0) in 𝑅. If 𝑃 is not a 𝐵-prime of (0) in 𝑅, then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. In view of the hypothesis that 𝑃 is the only maximal 𝑁-prime of (0) in 𝑅, it follows that 𝑍(𝑅)=𝑃. Let 𝑥−𝑦 be any edge of (Γ(𝑅))^𝑐. We prove that the edge 𝑥−𝑦 is an edge of a triangle in (Γ(𝑅))^𝑐; that is, there exists 𝑧∈𝑃⧵{𝑥,𝑦} such that 𝑥𝑧≠0 and 𝑦𝑧≠0. Proceeding as in the proof of [7, Lemma 3.2], it can be shown that there exists 𝑧∈𝑃⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. This shows that any edge of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐. Thus if 𝑥 is any element of 𝑍(𝑅)^∗, then 𝑥 does not admit a complement in (Γ(𝑅))^𝑐.

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

Proposition 2.2. Let 𝑅 be a commutative ring with identity. Suppose that 𝑅 has exactly one maximal 𝑁-prime of (0), and let it be 𝑃. If 𝑃 is a 𝐵-prime of (0) in 𝑅 and if |𝑃|≥5, then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. As 𝑃 is a 𝐵-prime of (0) in 𝑅, there exists 𝑐∈𝑃⧵{0} such that 𝑃=((0)∶_𝑅𝑐). Let 𝑥−𝑦 be any edge of (Γ(𝑅))^𝑐. We now verify that there exists 𝑧∈𝑍(𝑅)^∗=𝑃⧵{0} such that 𝑧∉{𝑥,𝑦}, 𝑧𝑥≠0, and 𝑧𝑦≠0. Proceeding as in the proof of [7, Proposition 3.7], it can be shown that there exists 𝑧∈𝑍(𝑅)^∗ such that 𝑧∉{𝑥,𝑦}, 𝑧𝑥≠0, and 𝑧𝑦≠0. This proves that any edge 𝑥−𝑦 of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐. Hence we obtain that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

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

Remark 2.3. First observe that 𝑅 is one of the rings from the collection {𝐙/8𝐙,(𝐙/4𝐙)[𝑥]/((2+4𝐙)𝑥(𝐙/4𝐙)[𝑥]+(𝑥²−(2+4𝐙))(𝐙/4𝐙)[𝑥]),(𝐙/2𝐙)[𝑥]/𝑥³(𝐙/2𝐙)[𝑥]}, then 𝑅 has properties (i), (ii), and (iii) mentioned above. We show in this remark with the help of [11, Theorem 3.2] that if 𝑅 is a ring with the above three properties, then 𝑅 is isomorphic to exactly one of the rings given in the above collection. Let 𝑅, 𝑃 be as mentioned in the beginning of this section. Suppose that 𝑃 is a 𝐵-prime of (0) in 𝑅 and moreover, (Γ(𝑅))^𝑐 contains at least one edge that is, there exist distinct 𝑥,𝑦∈𝑃⧵{0} such that 𝑥𝑦≠0. Let 𝑐∈𝑃⧵{0} be such that 𝑃=((0)∶_𝑅𝑐). We want to characterize 𝑅 such that (Γ(𝑅))^𝑐 has at least one vertex which admits a complement in (Γ(𝑅))^𝑐. It follows from Proposition 2.2 that |𝑃|<5. Note that {0,𝑥,𝑦,𝑐}⊆𝑃, and hence it follows that |𝑃|=4. Now proceeding as in [7, Remark 3.8], it follows using [11, Theorem 3.2] that 𝑅 must be isomorphic to one of the rings given in the above collection. Moreover, note that (Γ(𝑅))^𝑐 has exactly three vertices with one isolated vertex and the other two vertices being complements of each other in (Γ(𝑅))^𝑐.

3. 𝑅 Has Exactly Two Maximal 𝑁-Primes of (0)

Let 𝑅 be a commutative ring with identity. Suppose that 𝑅 has exactly two maximal 𝑁-primes of (0). Let {𝑃₁,𝑃₂} denote the set of all maximal 𝑁-primes of (0) in 𝑅.

Suppose that 𝑃₁∩𝑃₂≠(0). It is useful to remark here that 𝑃₁∩𝑃₂≠(0) if and only if (Γ(𝑅))^𝑐 is connected [6, Theorem 1.1.(b)]. We prove in Theorem 3.9 that (Γ(𝑅))^𝑐 has at least one vertex which admits a complement in (Γ(𝑅))^𝑐 if and only if either 𝑅 is isomorphic to 𝐙/4𝐙×𝐙/2𝐙 or 𝑅 is isomorphic to (𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥]×𝐙/2𝐙 where (𝐙/2𝐙)[𝑥] denotes the polynomial ring in one variable over 𝐙/2𝐙.

Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. Suppose that 𝑃₁∩𝑃₂=(0). In Proposition 3.11, we determine up to isomorphism of rings, rings 𝑅 such that (Γ(𝑅))^𝑐 has at least one vertex which admits a complement in (Γ(𝑅))^𝑐. It is indeed proved in Proposition 3.11 that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐 if and only if 𝑅 is isomorphic to 𝐙/3𝐙×𝑇, where 𝑇 is an integral domain. Moreover, it is noted in Proposition 3.11 that (Γ(𝑅))^𝑐 is complemented if and only if 𝑅 is isomorphic to 𝐙/3𝐙×𝐙/3𝐙 as rings. Furthermore, if 𝑇 is not isomorphic to 𝐙/3𝐙, then it is observed that (Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.

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

Lemma 3.1. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. If one between 𝑃₁ and 𝑃₂ is not a 𝐵-prime of (0) in 𝑅, then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. Without loss of generality we may assume that 𝑃₁ is not a 𝐵-prime of (0) in 𝑅. Observe that, to prove the lemma, it is enough to prove the following: if 𝑥−𝑦 is any edge of (Γ(𝑅))^𝑐, then there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. We consider the following cases.
Case 1. Both 𝑥 and 𝑦 belong to 𝑃₁.
If 𝑃₂⊈((0)∶_𝑅𝑥)∪((0)∶_𝑅𝑦), then as 𝑃₂⊈𝑃₁, it follows from [9, Theorem 81] that 𝑃₂⊈𝑃₁∪((0)∶_𝑅𝑥)∪((0)∶_𝑅𝑦). Hence there exists 𝑧∈𝑃₂⧵𝑃₁ such that 𝑧𝑥≠0 and 𝑧𝑦≠0. Since 𝑥,𝑦∈𝑃₁, whereas 𝑧∉𝑃₁, it is clear that 𝑧∉{𝑥,𝑦}.
Suppose that 𝑃₂⊆((0)∶_𝑅𝑥)∪((0)∶_𝑅𝑦), then it follows that either 𝑃₂=(0)∶_𝑅𝑥) or 𝑃₂=((0)∶_𝑅𝑦). Without loss of generality we may assume that 𝑃₂=((0)∶_𝑅𝑥). Since 𝑥𝑦≠0, it follows that 𝑦∉𝑃₂. Thus 𝑦∈𝑃₁⧵𝑃₂. Since 𝑃₁ is not a 𝐵-prime of (0) in 𝑅, 𝑃₁⊈𝑅𝑥 and 𝑃₁⊈𝑅𝑦. Now it follows from [9, Theorem 81] that 𝑃₁⊈𝑅𝑥∪𝑅𝑦∪𝑃₂. Hence there exists 𝑧∈𝑃₁⧵{𝑣,𝑦} such that 𝑧∉𝑃₂. Since 𝑃₂=((0)∶_𝑅𝑥), it follows that 𝑧𝑥≠0. As 𝑦,𝑧∈𝑃₁⧵𝑃₂, we obtain that 𝑦𝑧∈𝑃₁⧵𝑃₂ and so 𝑧𝑦≠0.
Case 2. Both 𝑥 and 𝑦 belong to 𝑃₂.
Since 𝑃₁ is not a 𝐵-prime of (0) in 𝑅, it follows that 𝑃₁⊈((0)∶_𝑅𝑥) and 𝑃₁⊈((0)∶_𝑅𝑦). Hence we obtain from [9, Theorem 81] that 𝑃₁⊈𝑃₂∪((0)∶_𝑅𝑥)∪((0)∶_𝑅𝑦). So there exists 𝑧∈𝑃₁⧵𝑃₂ such that 𝑧𝑥≠0 and 𝑧𝑦≠0. Since 𝑥,𝑦∈𝑃₂, whereas 𝑧∉𝑃₂, it is clear that 𝑧∉{𝑥,𝑦}.
Case 3. 𝑥∈𝑃₁⧵𝑃₂,𝑦∈𝑃₂⧵𝑃₁.
Now 𝑥∈𝑃₁⧵𝑃₂, 𝑦∈𝑃₂⧵𝑃₁ are such that 𝑥𝑦≠0. Proceeding as in the proof of [7, Lemma 3.4(ii)], we obtain that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0.
Thus it is shown that any edge of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐. Hence we obtain that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Suppose that both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅. The following lemma gives a sufficient condition under which no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Lemma 3.2. Let 𝑅,𝑃₁,𝑃₂ be as mentioned in the beginning of this section. If both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅 and if |𝑃₁∩𝑃₂|≥3, then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. By assumption both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅. Hence there exist 𝑢,𝑣∈𝑅 such that 𝑃₁=((0)∶_𝑅𝑢) and 𝑃₂=((0)∶_𝑅𝑣). We know from [12, Lemma 3.6] that 𝑢𝑣=0. We now proceed to show that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. As is remarked in the introduction, it is enough to show that any edge of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐. Let 𝑥−𝑦 be any edge of (Γ(𝑅))^𝑐. We want to show that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. We consider the following cases.
Case 1. Both 𝑥 and 𝑦 belong to 𝑃₁.
It now follows as in the proof of Lemma 3.1 Case 1 that we may assume without loss of generality that 𝑃₂=((0)∶_𝑅𝑥). Since 𝑥𝑦≠0, we obtain that 𝑦∉𝑃₂. Thus 𝑦∈𝑃₁⧵𝑃₂. Note that for any 𝑤∈𝑃₁∩𝑃₂, 𝑦+𝑤∈𝑃₁⧵𝑃₂ and so (𝑦+𝑤)𝑦∈𝑃₁⧵𝑃₂. Hence (𝑦+𝑤)𝑦≠0. Moreover, since 𝑥𝑤=0, we obtain that 𝑥(𝑦+𝑤)=𝑥𝑦≠0. By hypothesis, |𝑃₁∩𝑃₂|≥3. Hence there exist distinct 𝑤₁,𝑤₂∈(𝑃₁∩𝑃₂)⧵{0}. It is clear that 𝑦+𝑤₁≠𝑦 and 𝑦+𝑤₂≠𝑦. As 𝑤₁≠𝑤₂, it follows that either 𝑦+𝑤₁≠𝑥 or 𝑦+𝑤₂≠𝑥. We may assume without loss of generality that 𝑦+𝑤₁≠𝑥. Now 𝑧=𝑦+𝑤₁ is such that 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦}, 𝑧𝑥≠0, and 𝑧𝑦≠0.
Case 2. Both 𝑥 and 𝑦 belong to 𝑃₂.
The proof of the fact that there exists 𝑧∈𝑍(𝑅))^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0 is similar to the proof of Case 1 of this lemma.
Case 3. 𝑥∈𝑃₁⧵𝑃₂, 𝑦∈𝑃₂⧵𝑃₁.
Now 𝑥𝑦≠0. It follows as in the proof of [7, Lemma 3.4(ii)] that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0.
This proves that any edge of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐, and so we obtain that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

If both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅, we provide in the following lemma some conditions under which no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Lemma 3.3. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. Suppose that both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅. If |𝑃₁⧵𝑃₂|≥3 and |𝑃₂⧵𝑃₁|≥3, then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. Let 𝑥−𝑦 be any edge of (Γ(𝑅))^𝑐. We prove that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. We consider the following cases.Case 1. Both 𝑥 and 𝑦 belong to 𝑃₁.
Proceeding as in the proof of Lemma 3.1 Case 1, we may assume without loss of generality that 𝑃₂=((0)∶_𝑅𝑥). Since 𝑥𝑦≠0, it follows that 𝑦∉𝑃₂. Thus 𝑦∈𝑃₁⧵𝑃₂. If 𝑥²≠0, then 𝑥∉𝑃₂. Hence 𝑥∈𝑃₁⧵𝑃₂. By hypothesis |𝑃₁⧵𝑃₂|≥3. Hence there exists 𝑧∈(𝑃₁⧵𝑃₂)⧵{𝑥,𝑦}. Now it is clear that 𝑧𝑥≠0 and 𝑧𝑦≠0. Suppose that 𝑥²=0. Then 𝑥∈𝑃₂ and so 𝑥∈𝑃₁∩𝑃₂. Note that 𝑧=𝑥+𝑦 is such that 𝑧∈𝑃₁⧵𝑃₂ and 𝑧∉{𝑥,𝑦}. Moreover, 𝑧𝑥=(𝑥+𝑦)𝑥=𝑥𝑦≠0, 𝑧𝑦∈𝑃₁⧵𝑃₂, and so 𝑧𝑦≠0.
Case 2. Both 𝑥 and 𝑦 belong to 𝑃₂.
The hypotheses regarding 𝑃₁ and 𝑃₂ are symmetric. Hence it follows as in the proof of Case 1 of this lemma that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0.
Case 3. 𝑥∈𝑃₁⧵𝑃₂, 𝑦∈𝑃₂⧵𝑃₁.
Now it follows as in the proof of [7, Lemma 3.4(ii)] that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0.
This shows that any edge of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐, and so we obtain that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

If 𝑃₁∩𝑃₂≠(0), then we prove in Lemma 3.5 that (Γ(𝑅))^𝑐 is not complemented; that is, there exists at least one vertex of (Γ(𝑅))^𝑐 which does not admit a complement in (Γ(𝑅))^𝑐. We make use of the following lemma in the proof of Lemma 3.5.

Lemma 3.4. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. Suppose that both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅. Let 𝑢,𝑣∈𝑅 be such that 𝑃₁=((0)∶_𝑅𝑢) and 𝑃₂=((0)∶_𝑅𝑣). If 𝑃₁∩𝑃₂≠(0), then either 𝑢∈𝑃₁∩𝑃₂ or 𝑣∈𝑃₁∩𝑃₂.

Proof. Since 𝑃₁ and 𝑃₂ are the only maximal 𝑁-primes of (0) in 𝑅, it follows that 𝑍(𝑅)=𝑃₁∪𝑃₂. By assumption, 𝑃₁=((0)∶_𝑅𝑢) and 𝑃₂=((0)∶_𝑅𝑣). Now for any 𝑥∈𝑃₁⧵𝑃₂, from 𝑥𝑢=0, it follows that 𝑢∈𝑃₂. Similarly for any 𝑦∈𝑃₂⧵𝑃₁, from 𝑦𝑣=0, we obtain that 𝑣∈𝑃₁. By hypothesis, 𝑃₁∩𝑃₂≠(0). Let 𝑧∈(𝑃₁∩𝑃₂)⧵{0}. Now 𝑧𝑢=𝑧𝑣=0 and so (𝑢+𝑣)𝑧=0. Hence 𝑢+𝑣∈𝑍(𝑅)=𝑃₁∪𝑃₂. Thus either 𝑢+𝑣∈𝑃₁ or 𝑢+𝑣∈𝑃₂. If 𝑢+𝑣∈𝑃₁, then as 𝑣∈𝑃₁ and 𝑢∈𝑃₂, we obtain that 𝑢∈𝑃₁∩𝑃₂. If 𝑢+𝑣∈𝑃₂, then it follows that 𝑣∈𝑃₁∩𝑃₂ since 𝑢∈𝑃₂ and 𝑣∈𝑃₁. This proves that either 𝑢∈𝑃₁∩𝑃₂ or 𝑣∈𝑃₁∩𝑃₂.

We next have the following lemma which shows that (Γ(𝑅))^𝑐 is not complemented if 𝑃₁∩𝑃₂≠(0).

Lemma 3.5. Let 𝑅,𝑃₁,𝑃₂ be mentioned in the beginnig of this section. If 𝑃₁∩𝑃₂≠(0), then there exists at least one vertex of (Γ(𝑅))^𝑐 which does not admit a complement in (Γ(𝑅))^𝑐.

Proof. If at least one between 𝑃₁ and 𝑃₂ is not a 𝐵-prime of (0) in 𝑅, then it is proved in Lemma 3.1 that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. Hence we may assume that both 𝑃₁ and 𝑃₂ are 𝐵-primes of (0) in 𝑅. Let 𝑢,𝑣∈𝑅 be such that 𝑃₁=((0)∶_𝑅𝑢) and 𝑃₂=((0)∶_𝑅𝑣). We know from Lemma 3.4 that either 𝑢∈𝑃₁∩𝑃₂ or 𝑣∈𝑃₁∩𝑃₂. Without loss of generality we may assume that 𝑢∈𝑃₁∩𝑃₂.
We assert that 𝑢 does not admit a complement in (Γ(𝑅))^𝑐. It is enough to prove the following: if 𝑢−𝑤 is any edge of (Γ(𝑅))^𝑐, then there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑢,𝑤} such that 𝑧𝑢≠0 and 𝑧𝑤≠0. Since 𝑢𝑤≠0, it follows that 𝑤∈𝑃₂⧵𝑃₁. Note that 𝑧=𝑤+𝑢 is such that 𝑧∈𝑃₂⧵𝑃₁, 𝑧∉{𝑢,𝑤}, 𝑧𝑢≠0, and 𝑧𝑤≠0.
This proves that (Γ(𝑅))^𝑐 has at least one vertex which does not admit a complement in (Γ(𝑅))^𝑐.

With the assumption that 𝑃₁∩𝑃₂≠(0), we next attempt to characterize rings 𝑅 such that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. Towards that goal, we begin with the following lemma.

Lemma 3.6. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. If 𝑃₁∩𝑃₂≠(0), then (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐 if and only if one of the following holds:(i)|𝑃₂⧵𝑃₁|=2 and 𝑃₁=((0)∶_𝑅𝑏) for some 𝑏∈𝑃₂⧵𝑃₁.(ii)|𝑃₁⧵𝑃₂|=2 and 𝑃₂=((0)∶_𝑅𝑎) for some 𝑎∈𝑃₁⧵𝑃₂.

Proof. Suppose that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. Then it follows from Lemma 3.1 that there exist 𝑢,𝑣∈𝑅 such that 𝑃₁=((0)∶_𝑅𝑢) and 𝑃₂=((0)∶_𝑅𝑣). Moreover, it follows from Lemma 3.2 that |𝑃₁∩𝑃₂|=2. Furthermore, we know from Lemma 3.3 that either |𝑃₁⧵𝑃₂|<3 or |𝑃₂⧵𝑃₁|<3.
Let 𝑃₁∩𝑃₂={0,𝑤}. Note that, for any 𝑥∈𝑃₁⧵𝑃₂, 𝑥+𝑤∈(𝑃₁⧵𝑃₂)⧵{𝑥}. Hence |𝑃₁⧵𝑃₂|≥2. Similarly it follows that |𝑃₂⧵𝑃₁|≥2. Suppose that |𝑃₂⧵𝑃₁|<3. Then we obtain that |𝑃₂⧵𝑃₁|=2. It is shown in the proof of Lemma 3.4 that 𝑢∈𝑃₂ and 𝑣∈𝑃₁. If 𝑢∉𝑃₁, then we arrive at |𝑃₂⧵𝑃₁|=2 and 𝑃₁=((0)∶_𝑅𝑢) with 𝑢∈𝑃₂⧵𝑃₁. Hence (i) holds. Suppose that 𝑢∈𝑃₁∩𝑃₂. Then as |𝑃₁∩𝑃₂|=2, it follows that 𝑣∉𝑃₁∩𝑃₂. Hence 𝑣∈𝑃₁⧵𝑃₂. We assert that |𝑃₁⧵𝑃₂|=2. Suppose that it does not hold. Then |𝑃₁⧵𝑃₂|≥3. Let 𝑥∈𝑍(𝑅)^∗ be such that 𝑥 admits a complement 𝑦 in (Γ(𝑅))^𝑐. It is shown in the proof of Lemma 3.5 that 𝑢 does not admit a complement in (Γ(𝑅))^𝑐. Hence it follows that 𝑥,𝑦∉{𝑢}. As 𝑃₁∩𝑃₂={0,𝑢}, it follows that 𝑥,𝑦∉𝑃₁∩𝑃₂. If one between 𝑥 and 𝑦 is in 𝑃₁⧵𝑃₂ and the other is in 𝑃₂⧵𝑃₁, then it follows as in the proof of [7, Lemma 3.4(ii)] that there exists 𝑧∈𝑍(𝑅)^∗ which is adjacent to both 𝑥 and 𝑦 in (Γ(𝑅))^𝑐. This is not possible since 𝑦 is a complement of 𝑥 in (Γ(𝑅))^𝑐. Thus either both 𝑥 and 𝑦 belong to 𝑃₂⧵𝑃₁ or both 𝑥 and 𝑦 belong to 𝑃₁⧵𝑃₂. If both 𝑥 and 𝑦 belong to 𝑃₂⧵𝑃₁, then 𝑢∈𝑍(𝑅)^∗ is such that 𝑥𝑢≠0 and 𝑦𝑢≠0. Hence 𝑢 is adjacent to both 𝑥 and 𝑦 in (Γ(𝑅))^𝑐. This is impossible. If both 𝑥 and 𝑦 belong to 𝑃₁⧵𝑃₂, then, since we are assuming that |𝑃₁⧵𝑃₂|≥3, any 𝑧∈(𝑃₁⧵𝑃₂)⧵{𝑥,𝑦} satisfies 𝑧𝑥≠0 and 𝑧𝑦≠0. This cannot happen as 𝑦 is a complement of 𝑥 in (Γ(𝑅))^𝑐. Hence |𝑃₁⧵𝑃₂|=2. This together with the fact that 𝑣∈𝑃₁⧵𝑃₂ and 𝑃₂=((0)∶_𝑅𝑣) implies that (ii) holds. Thus if (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐, then either (i) or (ii) holds.
Conversely assume that either (i) or (ii) holds. Suppose that (i) holds. Now 𝑃₁=((0)∶_𝑅𝑏) for some 𝑏∈𝑃₂⧵𝑃₁, and |𝑃₂⧵𝑃₁|=2. Let 𝑃₂⧵𝑃₁={𝑏,𝑐}. Since 𝑏,𝑐∈𝑃₂⧵𝑃₁, it follows that 𝑏𝑐∈𝑃₂⧵𝑃₁ and hence 𝑏𝑐≠0. Moreover, if 𝑧∈𝑍(𝑅)^∗⧵{𝑏,𝑐}, then 𝑧 must be in 𝑃₁ and so 𝑧𝑏=0. Hence there exists no vertex of (Γ(𝑅))^𝑐 which is adjacent to both 𝑏 and 𝑐 in (Γ(𝑅))^𝑐. Therefore, 𝑐 is a complement of 𝑏 in (Γ(𝑅))^𝑐. This shows that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. Similarly if (ii) holds and if 𝑃₁⧵𝑃₂={𝑎,𝑑}, then it follows that 𝑎 and 𝑑 are complements of each other in (Γ(𝑅))^𝑐. This proves that either (i) or (ii) holds, then (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.
This completes the proof of Lemma 3.6.

Suppose that 𝑃₁∩𝑃₂≠(0). In Theorem 3.9 we characterize rings 𝑅 such that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. We need the following lemma for proving Theorem 3.9.

Lemma 3.7. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. Suppose that 𝑃₁∩𝑃₂≠(0), and if (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐, then 𝑅 is isomorphic to 𝑇₁×𝑇₂ as rings where 𝑇₁ is a ring with |𝑍(𝑇₁)^∗|=1 and 𝑇₂ is an integral domain.

Proof. We are assuming that there are vertices of (Γ(𝑅))^𝑐 which admit a complement in (Γ(𝑅))^𝑐. Hence we obtain from Lemma 3.1 that there exist 𝑢,𝑣∈𝑅 such that 𝑃₁=((0)∶_𝑅𝑢) and 𝑃₂=((0)∶_𝑅𝑣). Now it follows from Lemma 3.2 that |𝑃₁∩𝑃₂|=2. Let 𝑃₁∩𝑃₂={0,𝑤}. We claim that 𝑃₁+𝑃₂=𝑅. Suppose that 𝑃₁+𝑃₂≠𝑅. Then there exists a maximal ideal 𝑀 of 𝑅 such that 𝑃₁+𝑃₂⊆𝑀. Let 𝑎∈𝑃₁⧵𝑃₂ and 𝑏∈𝑃₂⧵𝑃₁. Now 𝑎+𝑏∈𝑃₁+𝑃₂⊆𝑀, and since 𝑍(𝑅)=𝑃₁∪𝑃₂, it follows from the choice of the elements 𝑎 and 𝑏 that 𝑎+𝑏∉𝑍(𝑅). Let 𝑧=𝑎+𝑏. Note that 𝑧∈𝑀⧵𝑍(𝑅), and as 𝑤≠0, we obtain that 𝑧𝑤≠0. Since 𝑃₁∩𝑃₂={0,𝑤}, it follows that 𝑧𝑤=𝑤. Thus 𝑤(1−𝑧)=0, and so 1−𝑧∈𝑍(𝑅). This is impossible since 𝑍(𝑅)=𝑃₁∪𝑃₂⊆𝑀, whereas 1−𝑧∉𝑀. Hence we obtain that 𝑃₁+𝑃₂=𝑅. Observe that 𝑤²∈𝑃₁∩𝑃₂={0,𝑤}, and as 1−𝑤∉𝑍(𝑅), it follows that 𝑤²=0. Hence (𝑃₁∩𝑃₂)²=(0). As 𝑃₁+𝑃₂=𝑅, it follows that (0)=(𝑃₁∩𝑃₂)²=𝑃²¹∩𝑃²². Now we obtain from the Chinese remainder theorem [13, Proposition 1.10(ii)] that 𝑅 is isomorphic to 𝑅/𝑃²¹×𝑅/𝑃²² as rings. Indeed, it follows from the Chinese remainder theorem that the mapping 𝑓∶𝑅→𝑅/𝑃²¹×𝑅/𝑃²² given by 𝑓(𝑟)=(𝑟+𝑃²¹,𝑟+𝑃²²) for any 𝑟∈𝑅 is an isomorphism of rings. The isomorphism 𝑓 maps 𝑃₁∩𝑃₂ onto 𝑃₁/𝑃²¹×𝑃₂/𝑃²². Hence 2=|𝑃₁∩𝑃₂|=|𝑓(𝑃₁∩𝑃₂)|=|𝑃₁/𝑃²¹×𝑃₂/𝑃²²|=|𝑃₁/𝑃²¹‖𝑃₂/𝑃²²|. Thus either |𝑃₁/𝑃²¹|=2,|𝑃₂/𝑃²²|=1 or |𝑃₁/𝑃²¹|=1,|𝑃₂/𝑃²²|=2. We may assume without loss of generality that |𝑃₁/𝑃²¹|=2 and |𝑃₂/𝑃²²|=1. Let 𝑇₁=𝑅/𝑃²¹ and 𝑇₂=𝑅/𝑃²². Thus 𝑅 is isomorphic to 𝑇₁×𝑇₂ as rings. Since 𝑃₂=𝑃²², it follows that 𝑇₂=𝑅/𝑃₂ is an integral domain. We next verify that |𝑍(𝑇₁)^∗|=1. Note that 𝑓(𝑍(𝑅))=𝑍(𝑇₁×𝑇₂)=(𝑍(𝑇₁)×𝑇₂)∪(𝑇₁×𝑍(𝑇₂)). As 𝑍(𝑅)=𝑃₁∪𝑃₂ and 𝑓(𝑃₁∪𝑃₂)=(𝑃₁/𝑃²¹×𝑇₂)∪(𝑇₁×𝑃₂/𝑃²²), it follows that 𝑍(𝑇₁)=𝑃₁/𝑃²¹. Since |𝑃₁/𝑃²¹|=2, we obtain that |𝑍(𝑇₁)^∗|=1.
Thus if 𝑃₁∩𝑃₂≠(0) and if (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐, then 𝑅 is isomorphic to 𝑇₁×𝑇₂ as rings with |𝑍(𝑇₁)^∗|=1 and 𝑇₂ is an integral domain.

Let 𝑇 be a commutative ring with identity. It is well known [1, Example 2.1(a)] that |𝑍(𝑇)^∗|=1 (i.e., equivalently, Γ(𝑇) is a graph on a single vertex) if and only if 𝑇 is either isomorphic to 𝑇₁₁=𝐙/4𝐙 or 𝑇 is isomorphic to 𝑇₁₂=(𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥] as rings. Let 𝑅=𝑇₁×𝑇₂, where either 𝑇₁=𝑇₁₁ or 𝑇₁=𝑇₁₂ and 𝑇₂ is an integral domain. In the following lemma, we determine when (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.

Lemma 3.8. Let 𝑇₁₁, 𝑇₁₂ be as in the previous paragraph. Let 𝑇₂ be an integral domain. Let 𝑅=𝑇₁×𝑇₂, where either 𝑇₁=𝑇₁₁ or 𝑇₁=𝑇₁₂. Then the following statements are equivalent.(i)(Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.(ii)𝑇₂ is isomorphic to 𝐙/2𝐙 as rings.(iii)(Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.

Proof. Let 𝑅=𝑇₁₁×𝑇₂=𝐙/4𝐙×𝑇₂ where 𝑇₂ is an integral domain. Note that 𝑍(𝑅) is the union of two prime ideals 𝑃₁=(2+4𝐙)𝑇₁₁×𝑇₂ and 𝑃₂=𝑇₁₁×(0). Moreover, 𝑃₁ and 𝑃₂ are the only maximal 𝑁-primes of the zero ideal in 𝑅. Furthermore, (2+4𝐙,0)∈𝑃₁∩𝑃₂, and hence 𝑃₁∩𝑃₂ is not the zero ideal of 𝑅. Furthermore, 𝑃₁=((0+4𝐙,0)∶_𝑅(2+4𝐙,0)) and 𝑃₂=((0+4𝐙,0)∶_𝑅(0+4𝐙,1)). We now show that the statements (i) to (iii) are equivalent.
(i)⇒(ii) Suppose that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. We claim that 𝑃₁≠((0+4𝐙,0)∶_𝑅𝑏) for any 𝑏∈𝑃₂⧵𝑃₁. Suppose that 𝑃₁=((0+4𝐙,0)∶_𝑅𝑏) for some 𝑏∈𝑃₂⧵𝑃₁. Note that either 𝑏=(1+4𝐙,0) or 𝑏=(3+4𝐙,0). Observe that 𝑎=(2+4𝐙,0)∈𝑃₁ is such that 𝑎(1+4𝐙,0)=𝑎=𝑎(3+4𝐙,0)≠(0+4𝐙,0). This shows that 𝑃₁≠((0+4𝐙,0)∶_𝑅𝑏) for any 𝑏∈𝑃₂⧵𝑃₁. Hence condition (𝑖) of Lemma 3.6 does not hold. Therefore condition (ii) of Lemma 3.6 must hold. Thus |𝑃₁⧵𝑃₂|=2 and 𝑃₂=((0+4𝐙,0)∶_𝑅𝑐) for some 𝑐∈𝑃₁⧵𝑃₂. We now show that |𝑇₂|=2. Suppose that |𝑇₂|≥3. Let 𝑦,𝑧∈𝑇₂⧵{0} with 𝑦≠𝑧. Note that {(0+4𝐙,𝑦),(0+4𝐙,𝑧),(2+4𝐙,𝑦)}⊆𝑃₁⧵𝑃₂, and this implies that |𝑃₁⧵𝑃₂|≥3 which contradicts the fact that |𝑃₁⧵𝑃₂|=2. This proves that |𝑇₂|=2, and so 𝑇₂ is isomorphic to 𝐙/2𝐙 as rings.
(ii)⇒(iii) Now 𝑇₂={0,1}. Observe that 𝑍(𝑅)^∗={(0+4𝐙,1),(2+4𝐙,0),(2+4𝐙,1),(1+4𝐙,0),(3+4𝐙,0)}. Observe that the edge (0+4𝐙,1)−(2+4𝐙,1) of (Γ(𝑅))^𝑐 is not an edge of any triangle in (Γ(𝑅))^𝑐. Indeed, (0+4𝐙,1) is a pendant vertex of (Γ(𝑅))^𝑐. Thus the vertices (0+4𝐙,1) and (2+4𝐙,1) are complements of each other in (Γ(𝑅))^𝑐. It can be easily verified that each of the other edges of (Γ(𝑅))^𝑐 is an edge of a triangle in (Γ(𝑅))^𝑐. Hence (Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.
(iii)⇒(i) This is clear.
If 𝑅=𝑇₁₂×𝑇₂ where 𝑇₁₂=(𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥] and 𝑇₂ is an integral domain, then the proof of the fact that the statements (i), (ii), and (iii) are equivalent is exactly similar and hence is omitted.

Suppose that 𝑃₁∩𝑃₂≠(0). The following theorem characterizes rings 𝑅 such that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.

Theorem 3.9. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. Suppose that 𝑃₁∩𝑃₂≠(0). The following statements are equivalent.(i)(Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.(ii) Either 𝑅 is isomorphic to 𝐙/4𝐙×𝐙/2𝐙 or 𝑅 is isomorphic to (𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥]×𝐙/2𝐙 as rings.(iii)(Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.

Proof. (i)⇒(ii) We know from Lemma 3.7 that 𝑅 is isomorphic to 𝑇₁×𝑇₂ as rings with |𝑍(𝑇₁)^∗|=1 and 𝑇₂ is an integral domain. Since |𝑍(𝑇₁)^∗|=1, it follows from [1, Example 2.1(a)] that either 𝑇₁ is isomorphic to 𝐙/4𝐙 or 𝑇₁ is isomorphic to (𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥] as rings. Let 𝑇₁₁=𝐙/4𝐙 and 𝑇₁₂=(𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥]. Since either 𝑅 is isomorphic to 𝑇₁₁×𝑇₂ or 𝑅 is isomorphic to 𝑇₁₂×𝑇₂, (𝑖) implies that either (Γ(𝑇₁₁×𝑇₂))^𝑐 has vertices which admit a complement in (Γ(𝑇₁₁×𝑇₂))^𝑐 or (Γ(𝑇₁₂×𝑇₂))^𝑐 has vertices which admit a complement in (Γ(𝑇₁₂×𝑇₂))^𝑐. Now it follows from (i)⇒(ii) of Lemma 3.8 that 𝑇₂ is isomorphic to 𝐙/2𝐙 as rings. Hence we obtain that either 𝑅 is isomorphic to 𝐙/4𝐙×𝐙/2𝐙 or 𝑅 is isomorphic to (𝐙/2𝐙)[𝑥]/𝑥²(𝐙/2𝐙)[𝑥]×𝐙/2𝐙 as rings.
(ii)⇒(iii) It follows from (ii)⇒(iii) of Lemma 3.8 that (Γ(𝑅))^𝑐 has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.
(iii)⇒(i) This is clear.
This completes the proof of Theorem 3.9.

Suppose that 𝑃₁∩𝑃₂=(0); that is, equivalently, by [6, Theorem 1.1(b)], (Γ(𝑅))^𝑐 is not connected. Assume that (Γ(𝑅))^𝑐 admits at least one edge. Our next aim is to determine when no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐 and to determine rings 𝑅 such that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.

Lemma 3.10. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. Suppose that 𝑃₁∩𝑃₂=(0). Then the following hold.(i) If |𝑃₁⧵𝑃₂|≥3, then no element of 𝑃₁⧵𝑃₂ admits a complement in (Γ(𝑅))^𝑐.(ii) If |𝑃₂⧵𝑃₁|≥3, then no element of 𝑃₂⧵𝑃₁ admits a complement in (Γ(𝑅))^𝑐.(iii) If |𝑃₁⧵𝑃₂|≥3 and |𝑃₂⧵𝑃₁|≥3, then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. (i) Let 𝑥∈𝑃₁⧵𝑃₂. Suppose that 𝑥 admits a complement in (Γ(𝑅))^𝑐. Let 𝑦∈𝑍(𝑅)^∗ be a complement of 𝑥 in (Γ(𝑅))^𝑐. Then 𝑥≠𝑦 and 𝑥𝑦≠0. Since 𝑍(𝑅)=𝑃₁∪𝑃₂, 𝑃₁∩𝑃₂=(0), and 𝑥𝑦≠(0), it follows that 𝑦∈𝑃₁⧵𝑃₂. By hypothesis, |𝑃₁⧵𝑃₂|≥3. Now for any 𝑧∈(𝑃₁⧵𝑃₂)⧵{𝑥,𝑦}, 𝑧𝑥≠0 and 𝑧𝑦≠0. This is impossible since 𝑦 is a complement of 𝑥 in (Γ(𝑅))^𝑐. This proves that if |𝑃₁⧵𝑃₂|≥3, then no element of 𝑃₁⧵𝑃₂ admits a complement in (Γ(𝑅))^𝑐.
(ii) The proof of (ii) is similar to the proof of (i).
(iii) Since 𝑍(𝑅)=𝑃₁∪𝑃₂ and 𝑃₁∩𝑃₂=(0), it follows that 𝑍(𝑅)^∗=(𝑃₁⧵𝑃₂)∪(𝑃₂⧵𝑃₁). Note that 𝑍(𝑅)^∗ is the vertex set of (Γ(𝑅))^𝑐. Thus if |𝑃₁⧵𝑃₂|≥3 and |𝑃₂⧵𝑃₁|≥3, then it follows from (i) and (ii) that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.
This completes the proof of Lemma 3.10.

The following proposition describes rings 𝑅 such that 𝑃₁∩𝑃₂=(0), (Γ(𝑅))^𝑐 has at least one edge, and moreover, (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.

Proposition 3.11. Let 𝑅, 𝑃₁, 𝑃₂ be as mentioned in the beginning of this section. If 𝑃₁∩𝑃₂=(0) and if (Γ(𝑅))^𝑐 has at least one edge, then the following statements are equivalent.(i)(Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐.(ii)𝑅 is isomorphic to 𝐙/3𝐙×𝑇 as rings, where 𝑇 is an integral domain.(iii) Exactly one of the following holds.(a)  𝑅 is isomorphic to 𝐙/3𝐙×𝐙/2𝐙 as rings, and in this case (Γ(𝑅))^𝑐 is a graph on three vertices and it admits exactly one isolated vertex and the two vertices are complements of each other in (Γ(𝑅))^𝑐.(b)  𝑅 is isomorphic to 𝐙/3𝐙×𝐙/3𝐙 as rings, and in this case (Γ(𝑅))^𝑐 is a graph on four vertices and is complemented.(c)  𝑅 is isomorphic to 𝐙/3𝐙×𝑇 as rings, where 𝑇 is an integral domain with |𝑇|≥4, and in this case (Γ(𝑅))^𝑐 is a graph on more than four vertices and it has exactly two vertices which admit a complement in (Γ(𝑅))^𝑐.

Proof. (i)⇒(ii) Suppose that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. Then it follows from Lemma 3.10(iii) that either |𝑃₁⧵𝑃₂|<3 or |𝑃₂⧵𝑃₁|<3. Without loss of generality we may assume that |𝑃₂⧵𝑃₁|<3. Then either |𝑃₂⧵𝑃₁|=1 or |𝑃₂⧵𝑃₁|=2.
Suppose that |𝑃₂⧵𝑃₁|=1. Let 𝑃₂⧵𝑃₁={𝑏}. Since 𝑃₁∩𝑃₂=(0), it follows that 𝑃₂={0,𝑏}. Moreover, 𝑍(𝑅)^∗=(𝑃₁⧵𝑃₂)∪{𝑏}. We assert that |𝑃₁⧵𝑃₂|=2. Since we are assuming that (Γ(𝑅))^𝑐 has at least one edge, we obtain that |𝑃₁⧵𝑃₂|≥2. Suppose that |𝑃₁⧵𝑃₂|≥3, then it follows from Lemma 3.10(i) that no element of 𝑃₁⧵𝑃₂ admits a complement in (Γ(𝑅))^𝑐. Observe that 𝑏 is an isolated vertex in (Γ(𝑅))^𝑐 and hence it does not admit a complement in (Γ(𝑅))^𝑐. Hence we obtain that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐. This is in contradiction to the hypothesis that (Γ(𝑅))^𝑐 has vertices which admit a complement in (Γ(𝑅))^𝑐. Therefore, we obtain that |𝑃₁⧵𝑃₂|≤2. This shows that |𝑃₁⧵𝑃₂|=2. Thus |𝑍(𝑅)|=|𝑃₁∪𝑃₂|=4. Now 𝑍(𝑅) is a finite set, and hence it follows from [14, Theorem 1] that 𝑅 is finite. Since any prime ideal of a finite ring is a maximal ideal, it follows that 𝑃₁ and 𝑃₂ are maximal ideals of 𝑅. Thus 𝑃₁+𝑃₂=𝑅. Since 𝑃₁∩𝑃₂=(0), it follows from the Chinese remainder theorem [13, Proposition 1.10(ii)] that the mapping 𝑓∶𝑅→𝑅/𝑃₁×𝑅/𝑃₂ given by 𝑓(𝑟)=(𝑟+𝑃₁,𝑟+𝑃₂) for any 𝑟∈𝑅 is an isomorphism of rings. Note that 𝑓(𝑃₁)=𝑃₁/𝑃₁×𝑅/𝑃₂ and 𝑓(𝑃₂)=𝑅/𝑃₁×𝑃₂/𝑃₂. Hence 3=|𝑃₁|=|𝑓(𝑃₁)|=|𝑅/𝑃₂| and 2=|𝑃₂|=|𝑓(𝑃₂)|=|𝑅/𝑃₁|. So we obtain that 𝑅/𝑃₁≅𝐙/2𝐙 and 𝑅/𝑃₂≅𝐙/3𝐙 as rings. Thus 𝑅≅𝑅/𝑃₁×𝑅/𝑃₂≅𝐙/2𝐙×𝐙/3𝐙≅𝐙/3𝐙×𝐙/2𝐙 as rings. Thus with 𝑇=𝐙/2𝐙, we obtain that 𝑅≅𝐙/3𝐙×𝑇.
Suppose that |𝑃₂⧵𝑃₁|=2. Let 𝑃₂⧵𝑃₁={𝑏₁,𝑏₂}. Note that 𝑃₂={0,𝑏₁,𝑏₂}. We claim that 𝑃₁+𝑃₂=𝑅. Suppose that 𝑃₁+𝑃₂≠𝑅. Then there exists a maximal ideal 𝑀 of 𝑅 such that 𝑃₁+𝑃₂⊆𝑀. Let 𝑎∈𝑃₁⧵𝑃₂. Now 𝑎+𝑏₁∈𝑃₁+𝑃₂⊆𝑀, and as 𝑍(𝑅)=𝑃₁∪𝑃₂, it follows from the choice of the elements 𝑎 and 𝑏₁ that 𝑎+𝑏₁∉𝑍(𝑅). Let 𝑧=𝑎+𝑏₁. Note that 𝑧𝑏₁∈𝑃₂⧵{0}={𝑏₁,𝑏₂}. Observe that 𝑧𝑏₁≠𝑏₁. If 𝑧𝑏₁=𝑏₁, then (1−𝑧)𝑏₁=0, and this implies that 1−𝑧∈𝑍(𝑅)=𝑃₁∪𝑃₂⊆𝑀. Thus both 𝑧 and 1−𝑧 belong to 𝑀. Hence 1=𝑧+1−𝑧∈𝑀, and this is impossible since 𝑀≠𝑅. This shows that 𝑧𝑏₁=𝑏₂. Similarly it follows that 𝑧𝑏₂=𝑏₁. Hence we obtain that 𝑧²𝑏₁𝑏₂=𝑏₁𝑏₂. Hence (1−𝑧²)𝑏₁𝑏₂=0. As 𝑏₁𝑏₂∉𝑃₁, it follows that 1−𝑧²∈𝑃₁⊆𝑀. This is impossible since 𝑧²∈𝑀. This proves that 𝑃₁+𝑃₂=𝑅. Since 𝑃₁∩𝑃₂=(0), we obtain from the Chinese remainder theorem that the mapping 𝑓∶𝑅→𝑅/𝑃₁×𝑅/𝑃₂ given by 𝑓(𝑟)=(𝑟+𝑃₁,𝑟+𝑃₂) is an isomorphism of rings. As |𝑃₂|=3 and as 𝑓(𝑃₂)=𝑅/𝑃₁×𝑃₂/𝑃₂, it follows that |𝑅/𝑃₁|=3. Hence 𝑅/𝑃₁≅𝐙/3𝐙 as rings. Let 𝑇=𝑅/𝑃₂. Then 𝑇 is an integral domain and 𝑅≅𝑅/𝑃₁×𝑅/𝑃₂≅𝐙/3𝐙×𝑇 as rings.
This proves (i)⇒(ii).
(ii)⇒(iii) Now by hypothesis 𝑅 is isomorphic to 𝐙/3𝐙×𝑇 as rings where 𝑇 is an integral domain. If |𝑇|=2, then 𝑅≅𝐙/3𝐙×𝐙/2𝐙 as rings. Consider the ring 𝑆₁=𝐙/3𝐙×𝐙/2𝐙. Note that 𝑍(𝑆₁)^∗={(0+3𝐙,1+2𝐙),(1+3𝐙,0+2𝐙),(2+3𝐙,0+2𝐙)}. Note that (Γ(𝑆₁))^𝑐 is a graph on three vertices and it admits exactly one isolated vertex and the other two vertices are complements of each other in (Γ(𝑆₁))^𝑐. Since 𝑅≅𝑆₁ as rings, it follows that (Γ(𝑅))^𝑐 also has the same properties. Hence (a) holds.
Suppose that |𝑇|=3. Then 𝑇≅𝐙/3𝐙 as rings. Hence 𝑅 is isomorphic to 𝐙/3𝐙×𝐙/3𝐙 as rings. Let 𝑆₂=𝐙/3𝐙×𝐙/3𝐙. Note that 𝑍(𝑆₂)^∗={(0+3𝐙,1+3𝐙),(0+3𝐙,2+3𝐙),(1+3𝐙,0+3𝐙),(2+3𝐙,0+3𝐙)}. Note that (Γ(𝑆₂))^𝑐 is a graph on four vertices, (0+3𝐙,1+3𝐙), (0+3𝐙,2+3𝐙) are complements of each other in (Γ(𝑆₂))^𝑐, and (1+3𝐙,0+3𝐙), (2+3𝐙,0+3𝐙) are complements of each other in (Γ(𝑆₂))^𝑐. Thus (Γ(𝑆₂))^𝑐 is complemented. Since 𝑅≅𝑆₂ as rings, it follows that (Γ(𝑅))^𝑐 is a graph on four vertices and is complemented. Hence (b) holds.
Suppose that |𝑇|≥4. Consider the ring 𝑆₃=𝐙/3𝐙×𝑇. Note that 𝑍(𝑆₃)^∗={(0+3𝐙,𝑡),(1+3𝐙,0),(2+3𝐙,0)|𝑡∈𝑇⧵{0}}. Since |𝑇|≥4, it is clear that (Γ(𝑆₃))^𝑐 is a graph on more than four vertices. The vertex (2+3𝐙,0) is a complement of (1+3𝐙,0) in (Γ(𝑆₃))^𝑐. Since |𝑇⧵{0}|≥3, it is easy to verify that no vertex of the form (0+3𝐙,𝑡), where 𝑡∈𝑇⧵{0}, admits a complement in (Γ(𝑆₃))^𝑐. Since 𝑅≅𝑆₃ as rings, it follows that (Γ(𝑅))^𝑐 is a graph on more than four vertices, and moreover, there are exactly two vertices which admit a complement in (Γ(𝑅))^𝑐. Hence (c) holds.
(iii)⇒(i) This is clear.
This completes the proof of Proposition 3.11.

4. 𝑅 Has More Than Two Maximal 𝑁-Primes of (0)

Let 𝑅 be a commutative ring with identity. Suppose that 𝑅 has more than two maximal 𝑁-primes of (0). In the following proposition, we prove that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proposition 4.1. Let 𝑅 be a commutative ring with identity. If 𝑅 has more than two maximal 𝑁-primes of (0), then no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Proof. Note that to prove that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐, it is enough to prove the following: if 𝑥−𝑦 is any edge of (Γ(𝑅))^𝑐, then there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. Let {𝑃_𝛼}_𝛼∈Λ denote the set of all maximal 𝑁-primes of (0) in 𝑅. By hypothesis 𝑅 has more than two maximal 𝑁-primes of (0). Hence |Λ|≥3. Note that 𝑍(𝑅)=⋃_𝛼∈Λ𝑃_𝛼. We consider the following cases.Case 1. There exists 𝛼∈Λ such that both 𝑥 and 𝑦 belong to 𝑃_𝛼. Let 𝛽,𝛾∈Λ⧵{𝛼} be such that 𝛽≠𝛾. Since 𝑃_𝛽≠𝑃_𝛾 and 𝑥𝑦≠0, it follows that either 𝑃_𝛽⊈((0)∶_𝑅𝑥𝑦) or 𝑃_𝛾⊈((0)∶_𝑅𝑥𝑦). We may assume without loss of generality that 𝑃_𝛽⊈((0)∶_𝑅𝑥𝑦). Note that 𝑃_𝛽⊈𝑃_𝛼∪((0)∶_𝑅𝑥𝑦). Hence there exists 𝑧∈𝑃_𝛽 such that 𝑧∉𝑃_𝛼 and 𝑧𝑥𝑦≠0. Since both 𝑥 and 𝑦 belong to 𝑃_𝛼, whereas 𝑧∉𝑃_𝛼, it is clear that 𝑧∉{𝑥,𝑦}. As 𝑧∈𝑃_𝛽⊆𝑍(𝑅), we obtain that 𝑧∈𝑍(𝑅). From 𝑧𝑥𝑦≠0, it follows that 𝑧𝑥≠0 and 𝑧𝑦≠0.Case 2. 𝑥 and 𝑦 do not belong to the same maximal 𝑁-prime of (0) in 𝑅.
Let 𝛼, 𝛽 be distinct elements of Λ such that 𝑥∈𝑃_𝛼⧵𝑃_𝛽 and 𝑦∈𝑃_𝛽⧵𝑃_𝛼. Now 𝑥𝑦≠0. Proceeding as in the proof of [7, Lemma 3.4(ii)], it can be shown that there exists 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. We give another argument to exhibit an element 𝑧∈𝑍(𝑅)^∗⧵{𝑥,𝑦} such that 𝑧𝑥≠0 and 𝑧𝑦≠0. Since 𝑥∈𝑃_𝛼⧵𝑃_𝛽, it follows that ((0)∶_𝑅𝑥)⊆𝑃_𝛽. Similarly since 𝑦∈𝑃_𝛽⧵𝑃_𝛼, it follows that ((0)∶_𝑅𝑦)⊆𝑃_𝛼. By hypothesis |Λ|≥3. Let 𝛾∈Λ⧵{𝛼,𝛽}. Note that 𝑃_𝛾⊈𝑃_𝛼∪𝑃_𝛽. Hence there exists 𝑧∈𝑃_𝛾 such that 𝑧∉𝑃_𝛼∪𝑃_𝛽. It is now clear that 𝑧∈𝑍(𝑅)⧵{𝑥,𝑦}. Since ((0)∶_𝑅𝑥)⊆𝑃_𝛽 and 𝑧∉𝑃_𝛽, it follows that 𝑧𝑥≠0. Similarly since ((0)∶_𝑅𝑦)⊆𝑃_𝛼 and 𝑧∉𝑃_𝛼, we obtain that 𝑧𝑦≠0.
This shows that no vertex of (Γ(𝑅))^𝑐 admits a complement in (Γ(𝑅))^𝑐.

Acknowledgments

The author is very much thankful to the Academic Editors Professor D. F. Anderson, Professor A. V. Kelarev, and Professor C. Munuera for their valuable suggestions and for their support. Moreover, he is very much thankful to Professor A. V. Kelarev for bringing to his attention two useful and valuable books pertaining to the investigation of graphs associated to algebraic structures [15, 16].