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ISRN Algebra
VolumeΒ 2012Β (2012), Article IDΒ 282054, 13 pages
http://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 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 / 3 𝐙 Γ— 𝐙 / 3 𝐙 as rings. Moreover, if 𝑅 is not isomorphic to 𝐙 / 3 𝐙 Γ— 𝐙 / 3 𝐙 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 Ξ“ ( 𝑅 ) is an undirected graph whose vertex set is the set of all nonzero zero-divisors of 𝑅 and two distinct nonzero zero-divisors π‘₯ , 𝑦 of 𝑅 are joined by an edge in this graph if and only if π‘₯ 𝑦 = 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 π‘Ž , 𝑏 of 𝐺 are said to be orthogonal, written π‘Ž βŸ‚ 𝑏 if π‘Ž and 𝑏 are adjacent in 𝐺 and there is no vertex 𝑐 of 𝐺 which is adjacent to both π‘Ž and 𝑏 in 𝐺 ; that is, the edge π‘Ž βˆ’ 𝑏 of 𝐺 is not a part of any triangle in 𝐺 . Let π‘Ž be a vertex of 𝐺 . Recall from [3] that a vertex 𝑏 of 𝐺 is said to be a complement of π‘Ž if π‘Ž βŸ‚ 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 π‘Ž , 𝑏 , 𝑐 of 𝐺 are such that π‘Ž βŸ‚ 𝑏 and π‘Ž βŸ‚ 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 Ξ“ ( 𝑅 ) is complemented (resp., Ξ“ ( 𝑅 ) is uniquely complemented).

Let 𝐺 = ( 𝑉 , 𝐸 ) be a simple graph. Recall from [5, Definition 1 . 1 . 1 3 ] 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 π‘₯ , 𝑦 ∈ 𝑉 are joined by an edge in 𝐺 𝑐 if and only if there is no edge of 𝐺 joining π‘₯ and 𝑦 .

Let 𝑅 be a commutative ring with identity. We denote by 𝑍 ( 𝑅 ) βˆ— the set of all nonzero zero-divisors of 𝑅 . Suppose that | 𝑍 ( 𝑅 ) βˆ— | β‰₯ 2 . With this assumption on 𝑅 , in [6, 7], we investigated the relationship between some graph-theoretic properties of ( Ξ“ ( 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 ( Ξ“ ( 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 ) = { π‘Ÿ ∈ 𝑅 ∢ π‘Ÿ 𝑠 ∈ 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 π‘₯ ∈ 𝑍 ( 𝑅 ) , then 𝑅 π‘₯ ∩ 𝑆 = βˆ… . 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 π‘₯ ∈ 𝑃 . Thus if { 𝑃 𝛼 } 𝛼 ∈ Ξ› denotes the set of all maximal 𝑁 -primes of (0) in 𝑅 , then ⋃ 𝑍 ( 𝑅 ) = 𝛼 ∈ Ξ› 𝑃 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 ∢ 𝑅 π‘₯ ) for some π‘₯ ∈ 𝑅 . In this case we say that 𝑃 is a 𝐡 -prime of 𝐼 .

Let 𝑅 be a commutative ring with identity and suppose that | 𝑍 ( 𝑅 ) βˆ— | β‰₯ 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 𝐡 -prime of (0) in 𝑅 , then it is proved in Proposition 2.1 that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Suppose that 𝑃 is a 𝐡 -prime of (0) in 𝑅 and if | 𝑃 | β‰₯ 5 , then it is shown in Proposition 2.2 that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . If | 𝑃 | < 5 and if ( Ξ“ ( 𝑅 ) ) 𝑐 admits at least one edge (i.e., equivalently, if there exist distinct π‘₯ , 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— with π‘₯ 𝑦 β‰  0 ), then it is observed in Remark 2.3 that | 𝑃 | = 4 , moreover, we note that, except for the unique isolated vertex of ( Ξ“ ( 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 / 8 𝐙 , ( 𝐙 / 4 𝐙 ) [ π‘₯ ] / ( ( 2 + 4 𝐙 ) π‘₯ ( 𝐙 / 4 𝐙 ) [ π‘₯ ] + ( π‘₯ 2 βˆ’ ( 2 + 4 𝐙 ) ) ( 𝐙 / 4 𝐙 ) [ π‘₯ ] ) , ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 3 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] } where ( 𝐙 / 4 𝐙 ) [ π‘₯ ] (resp., ( 𝐙 / 2 𝐙 ) [ π‘₯ ] ) denotes the polynomial ring in one variable over 𝐙 / 4 𝐙 (resp., over 𝐙 / 2 𝐙 ). Throughout this paper unless otherwise specified we denote by ( 𝐙 / 𝑛 𝐙 ) [ π‘₯ ] , the polynomial ring in one variable over 𝐙 / 𝑛 𝐙 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 { 𝑃 1 , 𝑃 2 } 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 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , then it is proved in Theorem 3.9 that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 if and only if either 𝑅 is isomorphic to 𝐙 / 4 𝐙 Γ— 𝐙 / 2 𝐙 or 𝑅 is isomorphic to ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] Γ— 𝐙 / 2 𝐙 as rings if and only if ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . If 𝑃 1 ∩ 𝑃 2 = ( 0 ) and if ( Ξ“ ( 𝑅 ) ) 𝑐 admits at least one edge, then it is shown in Proposition 3.11 that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 if and only if 𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝑇 , where 𝑇 is an integral domain, and it is noted that ( Ξ“ ( 𝑅 ) ) 𝑐 is complemented if and only if 𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝐙 / 3 𝐙 as rings, and in the case, when 𝑇 is not isomorphic to 𝐙 / 3 𝐙 , then ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 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 ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

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

Let 𝑅 be a commutative ring with identity such that | 𝑍 ( 𝑅 ) βˆ— | β‰₯ 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 𝐡 -prime of (0) in 𝑅 or 𝑃 is a 𝐡 -prime of (0) in 𝑅 ; that is, equivalently in terms of graph theoretic terms, either ( Ξ“ ( 𝑅 ) ) 𝑐 is connected or ( Ξ“ ( 𝑅 ) ) 𝑐 is not connected [6, Theorem 1.1(a)]. In this section, we prove that ( Ξ“ ( 𝑅 ) ) 𝑐 is not complemented if 𝑃 is not a 𝐡 -prime of (0) in 𝑅 . Indeed, if 𝑃 is not a 𝐡 -prime of (0) in 𝑅 , then we prove in Proposition 2.1 that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . If 𝑃 is a 𝐡 -prime of (0) in 𝑅 and if | 𝑃 | β‰₯ 5 , then we prove in Proposition 2.2 that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Moreover, if 𝑃 is a 𝐡 -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 ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 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 ) βˆ— | β‰₯ 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 𝐡 -prime of (0) in 𝑅 , then no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 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 π‘₯ βˆ’ 𝑦 be any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 . We prove that the edge π‘₯ βˆ’ 𝑦 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 ; that is, there exists 𝑧 ∈ 𝑃 ⧡ { π‘₯ , 𝑦 } such that π‘₯ 𝑧 β‰  0 and 𝑦 𝑧 β‰  0 . Proceeding as in the proof of [7, Lemma 3.2], it can be shown that there exists 𝑧 ∈ 𝑃 ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . This shows that any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 . Thus if π‘₯ is any element of 𝑍 ( 𝑅 ) βˆ— , then π‘₯ does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

With the assumption that 𝑃 is a 𝐡 -prime of (0) in 𝑅 , the following proposition provides a sufficient condition under which no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 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 𝐡 -prime of (0) in 𝑅 and if | 𝑃 | β‰₯ 5 , then no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. As 𝑃 is a 𝐡 -prime of (0) in 𝑅 , there exists 𝑐 ∈ 𝑃 ⧡ { 0 } such that 𝑃 = ( ( 0 ) ∢ 𝑅 𝑐 ) . Let π‘₯ βˆ’ 𝑦 be any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 . We now verify that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— = 𝑃 ⧡ { 0 } such that 𝑧 βˆ‰ { π‘₯ , 𝑦 } , 𝑧 π‘₯ β‰  0 , and 𝑧 𝑦 β‰  0 . Proceeding as in the proof of [7, Proposition 3.7], it can be shown that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— such that 𝑧 βˆ‰ { π‘₯ , 𝑦 } , 𝑧 π‘₯ β‰  0 , and 𝑧 𝑦 β‰  0 . This proves that any edge π‘₯ βˆ’ 𝑦 of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence we obtain that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 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), ( i i ) the unique maximal 𝑁 -prime of (0) in 𝑅 is a 𝐡 -prime of (0) in 𝑅 , and ( i i i ) ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one edge and has at least one vertex which admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Remark 2.3. First observe that 𝑅 is one of the rings from the collection { 𝐙 / 8 𝐙 , ( 𝐙 / 4 𝐙 ) [ π‘₯ ] / ( ( 2 + 4 𝐙 ) π‘₯ ( 𝐙 / 4 𝐙 ) [ π‘₯ ] + ( π‘₯ 2 βˆ’ ( 2 + 4 𝐙 ) ) ( 𝐙 / 4 𝐙 ) [ π‘₯ ] ) , ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 3 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] } , then 𝑅 has properties ( i ) , ( i i ) , and ( i i i ) 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 𝐡 -prime of (0) in 𝑅 and moreover, ( Ξ“ ( 𝑅 ) ) 𝑐 contains at least one edge that is, there exist distinct π‘₯ , 𝑦 ∈ 𝑃 ⧡ { 0 } such that π‘₯ 𝑦 β‰  0 . Let 𝑐 ∈ 𝑃 ⧡ { 0 } be such that 𝑃 = ( ( 0 ) ∢ 𝑅 𝑐 ) . We want to characterize 𝑅 such that ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one vertex which admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . It follows from Proposition 2.2 that | 𝑃 | < 5 . Note that { 0 , π‘₯ , 𝑦 , 𝑐 } βŠ† 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 ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly three vertices with one isolated vertex and the other two vertices being complements of each other in ( Ξ“ ( 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 { 𝑃 1 , 𝑃 2 } denote the set of all maximal 𝑁 -primes of (0) in 𝑅 .

Suppose that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . It is useful to remark here that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) if and only if ( Ξ“ ( 𝑅 ) ) 𝑐 is connected [6, Theorem 1.1.(b)]. We prove in Theorem 3.9 that ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one vertex which admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 if and only if either 𝑅 is isomorphic to 𝐙 / 4 𝐙 Γ— 𝐙 / 2 𝐙 or 𝑅 is isomorphic to ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] Γ— 𝐙 / 2 𝐙 where ( 𝐙 / 2 𝐙 ) [ π‘₯ ] denotes the polynomial ring in one variable over 𝐙 / 2 𝐙 .

Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. Suppose that 𝑃 1 ∩ 𝑃 2 = ( 0 ) . In Proposition 3.11, we determine up to isomorphism of rings, rings 𝑅 such that ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one vertex which admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . It is indeed proved in Proposition 3.11 that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 if and only if 𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝑇 , where 𝑇 is an integral domain. Moreover, it is noted in Proposition 3.11 that ( Ξ“ ( 𝑅 ) ) 𝑐 is complemented if and only if 𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝐙 / 3 𝐙 as rings. Furthermore, if 𝑇 is not isomorphic to 𝐙 / 3 𝐙 , then it is observed that ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 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 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. If one between 𝑃 1 and 𝑃 2 is not a 𝐡 -prime of (0) in 𝑅 , then no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. Without loss of generality we may assume that 𝑃 1 is not a 𝐡 -prime of (0) in 𝑅 . Observe that, to prove the lemma, it is enough to prove the following: if π‘₯ βˆ’ 𝑦 is any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 , then there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . We consider the following cases.
Case 1. Both π‘₯ and 𝑦 belong to 𝑃 1 .
If 𝑃 2 ΜΈ βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) , then as 𝑃 2 ΜΈ βŠ† 𝑃 1 , it follows from [9, Theorem 81] that 𝑃 2 ΜΈ βŠ† 𝑃 1 βˆͺ ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) . Hence there exists 𝑧 ∈ 𝑃 2 ⧡ 𝑃 1 such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . Since π‘₯ , 𝑦 ∈ 𝑃 1 , whereas 𝑧 βˆ‰ 𝑃 1 , it is clear that 𝑧 βˆ‰ { π‘₯ , 𝑦 } .
Suppose that 𝑃 2 βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) , then it follows that either 𝑃 2 = ( 0 ) ∢ 𝑅 π‘₯ ) or 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑦 ) . Without loss of generality we may assume that 𝑃 2 = ( ( 0 ) ∢ 𝑅 π‘₯ ) . Since π‘₯ 𝑦 β‰  0 , it follows that 𝑦 βˆ‰ 𝑃 2 . Thus 𝑦 ∈ 𝑃 1 ⧡ 𝑃 2 . Since 𝑃 1 is not a 𝐡 -prime of (0) in 𝑅 , 𝑃 1 ΜΈ βŠ† 𝑅 π‘₯ and 𝑃 1 ΜΈ βŠ† 𝑅 𝑦 . Now it follows from [9, Theorem 81] that 𝑃 1 ΜΈ βŠ† 𝑅 π‘₯ βˆͺ 𝑅 𝑦 βˆͺ 𝑃 2 . Hence there exists 𝑧 ∈ 𝑃 1 ⧡ { 𝑣 , 𝑦 } such that 𝑧 βˆ‰ 𝑃 2 . Since 𝑃 2 = ( ( 0 ) ∢ 𝑅 π‘₯ ) , it follows that 𝑧 π‘₯ β‰  0 . As 𝑦 , 𝑧 ∈ 𝑃 1 ⧡ 𝑃 2 , we obtain that 𝑦 𝑧 ∈ 𝑃 1 ⧡ 𝑃 2 and so 𝑧 𝑦 β‰  0 .
Case 2. Both π‘₯ and 𝑦 belong to 𝑃 2 .
Since 𝑃 1 is not a 𝐡 -prime of (0) in 𝑅 , it follows that 𝑃 1 ΜΈ βŠ† ( ( 0 ) ∢ 𝑅 π‘₯ ) and 𝑃 1 ΜΈ βŠ† ( ( 0 ) ∢ 𝑅 𝑦 ) . Hence we obtain from [9, Theorem 81] that 𝑃 1 ΜΈ βŠ† 𝑃 2 βˆͺ ( ( 0 ) ∢ 𝑅 π‘₯ ) βˆͺ ( ( 0 ) ∢ 𝑅 𝑦 ) . So there exists 𝑧 ∈ 𝑃 1 ⧡ 𝑃 2 such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . Since π‘₯ , 𝑦 ∈ 𝑃 2 , whereas 𝑧 βˆ‰ 𝑃 2 , it is clear that 𝑧 βˆ‰ { π‘₯ , 𝑦 } .
Case 3. π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 , 𝑦 ∈ 𝑃 2 ⧡ 𝑃 1 .
Now π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 , 𝑦 ∈ 𝑃 2 ⧡ 𝑃 1 are such that π‘₯ 𝑦 β‰  0 . Proceeding as in the proof of [7, Lemma 3.4(ii)], we obtain that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 .
Thus it is shown that any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence we obtain that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Suppose that both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 . The following lemma gives a sufficient condition under which no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Lemma 3.2. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. If both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 and if | 𝑃 1 ∩ 𝑃 2 | β‰₯ 3 , then no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. By assumption both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 . Hence there exist 𝑒 , 𝑣 ∈ 𝑅 such that 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) . We know from [12, Lemma 3.6] that 𝑒 𝑣 = 0 . We now proceed to show that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . As is remarked in the introduction, it is enough to show that any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 . Let π‘₯ βˆ’ 𝑦 be any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 . We want to show that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . We consider the following cases.
Case 1. Both π‘₯ and 𝑦 belong to 𝑃 1 .
It now follows as in the proof of Lemma 3.1 Case 1 that we may assume without loss of generality that 𝑃 2 = ( ( 0 ) ∢ 𝑅 π‘₯ ) . Since π‘₯ 𝑦 β‰  0 , we obtain that 𝑦 βˆ‰ 𝑃 2 . Thus 𝑦 ∈ 𝑃 1 ⧡ 𝑃 2 . Note that for any 𝑀 ∈ 𝑃 1 ∩ 𝑃 2 , 𝑦 + 𝑀 ∈ 𝑃 1 ⧡ 𝑃 2 and so ( 𝑦 + 𝑀 ) 𝑦 ∈ 𝑃 1 ⧡ 𝑃 2 . Hence ( 𝑦 + 𝑀 ) 𝑦 β‰  0 . Moreover, since π‘₯ 𝑀 = 0 , we obtain that π‘₯ ( 𝑦 + 𝑀 ) = π‘₯ 𝑦 β‰  0 . By hypothesis, | 𝑃 1 ∩ 𝑃 2 | β‰₯ 3 . Hence there exist distinct 𝑀 1 , 𝑀 2 ∈ ( 𝑃 1 ∩ 𝑃 2 ) ⧡ { 0 } . It is clear that 𝑦 + 𝑀 1 β‰  𝑦 and 𝑦 + 𝑀 2 β‰  𝑦 . As 𝑀 1 β‰  𝑀 2 , it follows that either 𝑦 + 𝑀 1 β‰  π‘₯ or 𝑦 + 𝑀 2 β‰  π‘₯ . We may assume without loss of generality that 𝑦 + 𝑀 1 β‰  π‘₯ . Now 𝑧 = 𝑦 + 𝑀 1 is such that 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } , 𝑧 π‘₯ β‰  0 , and 𝑧 𝑦 β‰  0 .
Case 2. Both π‘₯ and 𝑦 belong to 𝑃 2 .
The proof of the fact that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 is similar to the proof of Case 1 of this lemma.
Case 3. π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 , 𝑦 ∈ 𝑃 2 ⧡ 𝑃 1 .
Now π‘₯ 𝑦 β‰  0 . It follows as in the proof of [7, Lemma 3.4(ii)] that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 .
This proves that any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 , and so we obtain that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

If both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 , we provide in the following lemma some conditions under which no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Lemma 3.3. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. Suppose that both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 . If | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 and | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 , then no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. Let π‘₯ βˆ’ 𝑦 be any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 . We prove that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . We consider the following cases.Case 1. Both π‘₯ and 𝑦 belong to 𝑃 1 .
Proceeding as in the proof of Lemma 3.1 Case 1, we may assume without loss of generality that 𝑃 2 = ( ( 0 ) ∢ 𝑅 π‘₯ ) . Since π‘₯ 𝑦 β‰  0 , it follows that 𝑦 βˆ‰ 𝑃 2 . Thus 𝑦 ∈ 𝑃 1 ⧡ 𝑃 2 . If π‘₯ 2 β‰  0 , then π‘₯ βˆ‰ 𝑃 2 . Hence π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 . By hypothesis | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 . Hence there exists 𝑧 ∈ ( 𝑃 1 ⧡ 𝑃 2 ) ⧡ { π‘₯ , 𝑦 } . Now it is clear that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . Suppose that π‘₯ 2 = 0 . Then π‘₯ ∈ 𝑃 2 and so π‘₯ ∈ 𝑃 1 ∩ 𝑃 2 . Note that 𝑧 = π‘₯ + 𝑦 is such that 𝑧 ∈ 𝑃 1 ⧡ 𝑃 2 and 𝑧 βˆ‰ { π‘₯ , 𝑦 } . Moreover, 𝑧 π‘₯ = ( π‘₯ + 𝑦 ) π‘₯ = π‘₯ 𝑦 β‰  0 , 𝑧 𝑦 ∈ 𝑃 1 ⧡ 𝑃 2 , and so 𝑧 𝑦 β‰  0 .
Case 2. Both π‘₯ and 𝑦 belong to 𝑃 2 .
The hypotheses regarding 𝑃 1 and 𝑃 2 are symmetric. Hence it follows as in the proof of Case 1 of this lemma that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 .
Case 3. π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 , 𝑦 ∈ 𝑃 2 ⧡ 𝑃 1 .
Now it follows as in the proof of [7, Lemma 3.4(ii)] that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { π‘₯ , 𝑦 } such that 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 .
This shows that any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 , and so we obtain that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

If 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , then we prove in Lemma 3.5 that ( Ξ“ ( 𝑅 ) ) 𝑐 is not complemented; that is, there exists at least one vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 which does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . We make use of the following lemma in the proof of Lemma 3.5.

Lemma 3.4. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. Suppose that both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 . Let 𝑒 , 𝑣 ∈ 𝑅 be such that 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) . If 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , then either 𝑒 ∈ 𝑃 1 ∩ 𝑃 2 or 𝑣 ∈ 𝑃 1 ∩ 𝑃 2 .

Proof. Since 𝑃 1 and 𝑃 2 are the only maximal 𝑁 -primes of (0) in 𝑅 , it follows that 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 . By assumption, 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) . Now for any π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 , from π‘₯ 𝑒 = 0 , it follows that 𝑒 ∈ 𝑃 2 . Similarly for any 𝑦 ∈ 𝑃 2 ⧡ 𝑃 1 , from 𝑦 𝑣 = 0 , we obtain that 𝑣 ∈ 𝑃 1 . By hypothesis, 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . Let 𝑧 ∈ ( 𝑃 1 ∩ 𝑃 2 ) ⧡ { 0 } . Now 𝑧 𝑒 = 𝑧 𝑣 = 0 and so ( 𝑒 + 𝑣 ) 𝑧 = 0 . Hence 𝑒 + 𝑣 ∈ 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 . Thus either 𝑒 + 𝑣 ∈ 𝑃 1 or 𝑒 + 𝑣 ∈ 𝑃 2 . If 𝑒 + 𝑣 ∈ 𝑃 1 , then as 𝑣 ∈ 𝑃 1 and 𝑒 ∈ 𝑃 2 , we obtain that 𝑒 ∈ 𝑃 1 ∩ 𝑃 2 . If 𝑒 + 𝑣 ∈ 𝑃 2 , then it follows that 𝑣 ∈ 𝑃 1 ∩ 𝑃 2 since 𝑒 ∈ 𝑃 2 and 𝑣 ∈ 𝑃 1 . This proves that either 𝑒 ∈ 𝑃 1 ∩ 𝑃 2 or 𝑣 ∈ 𝑃 1 ∩ 𝑃 2 .

We next have the following lemma which shows that ( Ξ“ ( 𝑅 ) ) 𝑐 is not complemented if 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) .

Lemma 3.5. Let 𝑅 , 𝑃 1 , 𝑃 2 be mentioned in the beginnig of this section. If 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , then there exists at least one vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 which does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. If at least one between 𝑃 1 and 𝑃 2 is not a 𝐡 -prime of (0) in 𝑅 , then it is proved in Lemma 3.1 that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence we may assume that both 𝑃 1 and 𝑃 2 are 𝐡 -primes of (0) in 𝑅 . Let 𝑒 , 𝑣 ∈ 𝑅 be such that 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) . We know from Lemma 3.4 that either 𝑒 ∈ 𝑃 1 ∩ 𝑃 2 or 𝑣 ∈ 𝑃 1 ∩ 𝑃 2 . Without loss of generality we may assume that 𝑒 ∈ 𝑃 1 ∩ 𝑃 2 .
We assert that 𝑒 does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . It is enough to prove the following: if 𝑒 βˆ’ 𝑀 is any edge of ( Ξ“ ( 𝑅 ) ) 𝑐 , then there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { 𝑒 , 𝑀 } such that 𝑧 𝑒 β‰  0 and 𝑧 𝑀 β‰  0 . Since 𝑒 𝑀 β‰  0 , it follows that 𝑀 ∈ 𝑃 2 ⧡ 𝑃 1 . Note that 𝑧 = 𝑀 + 𝑒 is such that 𝑧 ∈ 𝑃 2 ⧡ 𝑃 1 , 𝑧 βˆ‰ { 𝑒 , 𝑀 } , 𝑧 𝑒 β‰  0 , and 𝑧 𝑀 β‰  0 .
This proves that ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one vertex which does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

With the assumption that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , we next attempt to characterize rings 𝑅 such that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Towards that goal, we begin with the following lemma.

Lemma 3.6. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. If 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , then ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 if and only if one of the following holds:(i) | 𝑃 2 ⧡ 𝑃 1 | = 2 and 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑏 ) for some 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 .(ii) | 𝑃 1 ⧡ 𝑃 2 | = 2 and 𝑃 2 = ( ( 0 ) ∢ 𝑅 π‘Ž ) for some π‘Ž ∈ 𝑃 1 ⧡ 𝑃 2 .

Proof. Suppose that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Then it follows from Lemma 3.1 that there exist 𝑒 , 𝑣 ∈ 𝑅 such that 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) . Moreover, it follows from Lemma 3.2 that | 𝑃 1 ∩ 𝑃 2 | = 2 . Furthermore, we know from Lemma 3.3 that either | 𝑃 1 ⧡ 𝑃 2 | < 3 or | 𝑃 2 ⧡ 𝑃 1 | < 3 .
Let 𝑃 1 ∩ 𝑃 2 = { 0 , 𝑀 } . Note that, for any π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 , π‘₯ + 𝑀 ∈ ( 𝑃 1 ⧡ 𝑃 2 ) ⧡ { π‘₯ } . Hence | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 2 . Similarly it follows that | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 2 . Suppose that | 𝑃 2 ⧡ 𝑃 1 | < 3 . Then we obtain that | 𝑃 2 ⧡ 𝑃 1 | = 2 . It is shown in the proof of Lemma 3.4 that 𝑒 ∈ 𝑃 2 and 𝑣 ∈ 𝑃 1 . If 𝑒 βˆ‰ 𝑃 1 , then we arrive at | 𝑃 2 ⧡ 𝑃 1 | = 2 and 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) with 𝑒 ∈ 𝑃 2 ⧡ 𝑃 1 . Hence ( i ) holds. Suppose that 𝑒 ∈ 𝑃 1 ∩ 𝑃 2 . Then as | 𝑃 1 ∩ 𝑃 2 | = 2 , it follows that 𝑣 βˆ‰ 𝑃 1 ∩ 𝑃 2 . Hence 𝑣 ∈ 𝑃 1 ⧡ 𝑃 2 . We assert that | 𝑃 1 ⧡ 𝑃 2 | = 2 . Suppose that it does not hold. Then | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 . Let π‘₯ ∈ 𝑍 ( 𝑅 ) βˆ— be such that π‘₯ admits a complement 𝑦 in ( Ξ“ ( 𝑅 ) ) 𝑐 . It is shown in the proof of Lemma 3.5 that 𝑒 does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence it follows that π‘₯ , 𝑦 βˆ‰ { 𝑒 } . As 𝑃 1 ∩ 𝑃 2 = { 0 , 𝑒 } , it follows that π‘₯ , 𝑦 βˆ‰ 𝑃 1 ∩ 𝑃 2 . If one between π‘₯ and 𝑦 is in 𝑃 1 ⧡ 𝑃 2 and the other is in 𝑃 2 ⧡ 𝑃 1 , then it follows as in the proof of [7, Lemma 3.4(ii)] that there exists 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— which is adjacent to both π‘₯ and 𝑦 in ( Ξ“ ( 𝑅 ) ) 𝑐 . This is not possible since 𝑦 is a complement of π‘₯ in ( Ξ“ ( 𝑅 ) ) 𝑐 . Thus either both π‘₯ and 𝑦 belong to 𝑃 2 ⧡ 𝑃 1 or both π‘₯ and 𝑦 belong to 𝑃 1 ⧡ 𝑃 2 . If both π‘₯ and 𝑦 belong to 𝑃 2 ⧡ 𝑃 1 , then 𝑒 ∈ 𝑍 ( 𝑅 ) βˆ— is such that π‘₯ 𝑒 β‰  0 and 𝑦 𝑒 β‰  0 . Hence 𝑒 is adjacent to both π‘₯ and 𝑦 in ( Ξ“ ( 𝑅 ) ) 𝑐 . This is impossible. If both π‘₯ and 𝑦 belong to 𝑃 1 ⧡ 𝑃 2 , then, since we are assuming that | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 , any 𝑧 ∈ ( 𝑃 1 ⧡ 𝑃 2 ) ⧡ { π‘₯ , 𝑦 } satisfies 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . This cannot happen as 𝑦 is a complement of π‘₯ in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence | 𝑃 1 ⧡ 𝑃 2 | = 2 . This together with the fact that 𝑣 ∈ 𝑃 1 ⧡ 𝑃 2 and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) implies that ( i i ) holds. Thus if ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 , then either ( i ) or ( i i ) holds.
Conversely assume that either ( i ) or ( i i ) holds. Suppose that ( i ) holds. Now 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑏 ) for some 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 , and | 𝑃 2 ⧡ 𝑃 1 | = 2 . Let 𝑃 2 ⧡ 𝑃 1 = { 𝑏 , 𝑐 } . Since 𝑏 , 𝑐 ∈ 𝑃 2 ⧡ 𝑃 1 , it follows that 𝑏 𝑐 ∈ 𝑃 2 ⧡ 𝑃 1 and hence 𝑏 𝑐 β‰  0 . Moreover, if 𝑧 ∈ 𝑍 ( 𝑅 ) βˆ— ⧡ { 𝑏 , 𝑐 } , then 𝑧 must be in 𝑃 1 and so 𝑧 𝑏 = 0 . Hence there exists no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 which is adjacent to both 𝑏 and 𝑐 in ( Ξ“ ( 𝑅 ) ) 𝑐 . Therefore, 𝑐 is a complement of 𝑏 in ( Ξ“ ( 𝑅 ) ) 𝑐 . This shows that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Similarly if ( i i ) holds and if 𝑃 1 ⧡ 𝑃 2 = { π‘Ž , 𝑑 } , then it follows that π‘Ž and 𝑑 are complements of each other in ( Ξ“ ( 𝑅 ) ) 𝑐 . This proves that either ( i ) or ( i i ) holds, then ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .
This completes the proof of Lemma 3.6.

Suppose that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . In Theorem 3.9 we characterize rings 𝑅 such that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . We need the following lemma for proving Theorem 3.9.

Lemma 3.7. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. Suppose that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) , and if ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 , then 𝑅 is isomorphic to 𝑇 1 Γ— 𝑇 2 as rings where 𝑇 1 is a ring with | 𝑍 ( 𝑇 1 ) βˆ— | = 1 and 𝑇 2 is an integral domain.

Proof. We are assuming that there are vertices of ( Ξ“ ( 𝑅 ) ) 𝑐 which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence we obtain from Lemma 3.1 that there exist 𝑒 , 𝑣 ∈ 𝑅 such that 𝑃 1 = ( ( 0 ) ∢ 𝑅 𝑒 ) and 𝑃 2 = ( ( 0 ) ∢ 𝑅 𝑣 ) . Now it follows from Lemma 3.2 that | 𝑃 1 ∩ 𝑃 2 | = 2 . Let 𝑃 1 ∩ 𝑃 2 = { 0 , 𝑀 } . We claim that 𝑃 1 + 𝑃 2 = 𝑅 . Suppose that 𝑃 1 + 𝑃 2 β‰  𝑅 . Then there exists a maximal ideal 𝑀 of 𝑅 such that 𝑃 1 + 𝑃 2 βŠ† 𝑀 . Let π‘Ž ∈ 𝑃 1 ⧡ 𝑃 2 and 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 . Now π‘Ž + 𝑏 ∈ 𝑃 1 + 𝑃 2 βŠ† 𝑀 , and since 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 , it follows from the choice of the elements π‘Ž and 𝑏 that π‘Ž + 𝑏 βˆ‰ 𝑍 ( 𝑅 ) . Let 𝑧 = π‘Ž + 𝑏 . Note that 𝑧 ∈ 𝑀 ⧡ 𝑍 ( 𝑅 ) , and as 𝑀 β‰  0 , we obtain that 𝑧 𝑀 β‰  0 . Since 𝑃 1 ∩ 𝑃 2 = { 0 , 𝑀 } , it follows that 𝑧 𝑀 = 𝑀 . Thus 𝑀 ( 1 βˆ’ 𝑧 ) = 0 , and so 1 βˆ’ 𝑧 ∈ 𝑍 ( 𝑅 ) . This is impossible since 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 βŠ† 𝑀 , whereas 1 βˆ’ 𝑧 βˆ‰ 𝑀 . Hence we obtain that 𝑃 1 + 𝑃 2 = 𝑅 . Observe that 𝑀 2 ∈ 𝑃 1 ∩ 𝑃 2 = { 0 , 𝑀 } , and as 1 βˆ’ 𝑀 βˆ‰ 𝑍 ( 𝑅 ) , it follows that 𝑀 2 = 0 . Hence ( 𝑃 1 ∩ 𝑃 2 ) 2 = ( 0 ) . As 𝑃 1 + 𝑃 2 = 𝑅 , it follows that ( 0 ) = ( 𝑃 1 ∩ 𝑃 2 ) 2 = 𝑃 2 1 ∩ 𝑃 2 2 . Now we obtain from the Chinese remainder theorem [13, Proposition 1.10(ii)] that 𝑅 is isomorphic to 𝑅 / 𝑃 2 1 Γ— 𝑅 / 𝑃 2 2 as rings. Indeed, it follows from the Chinese remainder theorem that the mapping 𝑓 ∢ 𝑅 β†’ 𝑅 / 𝑃 2 1 Γ— 𝑅 / 𝑃 2 2 given by 𝑓 ( π‘Ÿ ) = ( π‘Ÿ + 𝑃 2 1 , π‘Ÿ + 𝑃 2 2 ) for any π‘Ÿ ∈ 𝑅 is an isomorphism of rings. The isomorphism 𝑓 maps 𝑃 1 ∩ 𝑃 2 onto 𝑃 1 / 𝑃 2 1 Γ— 𝑃 2 / 𝑃 2 2 . Hence 2 = | 𝑃 1 ∩ 𝑃 2 | = | 𝑓 ( 𝑃 1 ∩ 𝑃 2 ) | = | 𝑃 1 / 𝑃 2 1 Γ— 𝑃 2 / 𝑃 2 2 | = | 𝑃 1 / 𝑃 2 1 β€– 𝑃 2 / 𝑃 2 2 | . Thus either | 𝑃 1 / 𝑃 2 1 | = 2 , | 𝑃 2 / 𝑃 2 2 | = 1 or | 𝑃 1 / 𝑃 2 1 | = 1 , | 𝑃 2 / 𝑃 2 2 | = 2 . We may assume without loss of generality that | 𝑃 1 / 𝑃 2 1 | = 2 and | 𝑃 2 / 𝑃 2 2 | = 1 . Let 𝑇 1 = 𝑅 / 𝑃 2 1 and 𝑇 2 = 𝑅 / 𝑃 2 2 . Thus 𝑅 is isomorphic to 𝑇 1 Γ— 𝑇 2 as rings. Since 𝑃 2 = 𝑃 2 2 , it follows that 𝑇 2 = 𝑅 / 𝑃 2 is an integral domain. We next verify that | 𝑍 ( 𝑇 1 ) βˆ— | = 1 . Note that 𝑓 ( 𝑍 ( 𝑅 ) ) = 𝑍 ( 𝑇 1 Γ— 𝑇 2 ) = ( 𝑍 ( 𝑇 1 ) Γ— 𝑇 2 ) βˆͺ ( 𝑇 1 Γ— 𝑍 ( 𝑇 2 ) ) . As 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 and 𝑓 ( 𝑃 1 βˆͺ 𝑃 2 ) = ( 𝑃 1 / 𝑃 2 1 Γ— 𝑇 2 ) βˆͺ ( 𝑇 1 Γ— 𝑃 2 / 𝑃 2 2 ) , it follows that 𝑍 ( 𝑇 1 ) = 𝑃 1 / 𝑃 2 1 . Since | 𝑃 1 / 𝑃 2 1 | = 2 , we obtain that | 𝑍 ( 𝑇 1 ) βˆ— | = 1 .
Thus if 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) and if ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 , then 𝑅 is isomorphic to 𝑇 1 Γ— 𝑇 2 as rings with | 𝑍 ( 𝑇 1 ) βˆ— | = 1 and 𝑇 2 is an integral domain.

Let 𝑇 be a commutative ring with identity. It is well known [1, Example 2.1(a)] that | 𝑍 ( 𝑇 ) βˆ— | = 1 (i.e., equivalently, Ξ“ ( 𝑇 ) is a graph on a single vertex) if and only if 𝑇 is either isomorphic to 𝑇 1 1 = 𝐙 / 4 𝐙 or 𝑇 is isomorphic to 𝑇 1 2 = ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] as rings. Let 𝑅 = 𝑇 1 Γ— 𝑇 2 , where either 𝑇 1 = 𝑇 1 1 or 𝑇 1 = 𝑇 1 2 and 𝑇 2 is an integral domain. In the following lemma, we determine when ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Lemma 3.8. Let 𝑇 1 1 , 𝑇 1 2 be as in the previous paragraph. Let 𝑇 2 be an integral domain. Let 𝑅 = 𝑇 1 Γ— 𝑇 2 , where either 𝑇 1 = 𝑇 1 1 or 𝑇 1 = 𝑇 1 2 . Then the following statements are equivalent.(i) ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .(ii) 𝑇 2 is isomorphic to 𝐙 / 2 𝐙 as rings.(iii) ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. Let 𝑅 = 𝑇 1 1 Γ— 𝑇 2 = 𝐙 / 4 𝐙 Γ— 𝑇 2 where 𝑇 2 is an integral domain. Note that 𝑍 ( 𝑅 ) is the union of two prime ideals 𝑃 1 = ( 2 + 4 𝐙 ) 𝑇 1 1 Γ— 𝑇 2 and 𝑃 2 = 𝑇 1 1 Γ— ( 0 ) . Moreover, 𝑃 1 and 𝑃 2 are the only maximal 𝑁 -primes of the zero ideal in 𝑅 . Furthermore, ( 2 + 4 𝐙 , 0 ) ∈ 𝑃 1 ∩ 𝑃 2 , and hence 𝑃 1 ∩ 𝑃 2 is not the zero ideal of 𝑅 . Furthermore, 𝑃 1 = ( ( 0 + 4 𝐙 , 0 ) ∢ 𝑅 ( 2 + 4 𝐙 , 0 ) ) and 𝑃 2 = ( ( 0 + 4 𝐙 , 0 ) ∢ 𝑅 ( 0 + 4 𝐙 , 1 ) ) . We now show that the statements ( i ) to ( i i i ) are equivalent.
( i ) β‡’ ( i i ) Suppose that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . We claim that 𝑃 1 β‰  ( ( 0 + 4 𝐙 , 0 ) ∢ 𝑅 𝑏 ) for any 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 . Suppose that 𝑃 1 = ( ( 0 + 4 𝐙 , 0 ) ∢ 𝑅 𝑏 ) for some 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 . Note that either 𝑏 = ( 1 + 4 𝐙 , 0 ) or 𝑏 = ( 3 + 4 𝐙 , 0 ) . Observe that π‘Ž = ( 2 + 4 𝐙 , 0 ) ∈ 𝑃 1 is such that π‘Ž ( 1 + 4 𝐙 , 0 ) = π‘Ž = π‘Ž ( 3 + 4 𝐙 , 0 ) β‰  ( 0 + 4 𝐙 , 0 ) . This shows that 𝑃 1 β‰  ( ( 0 + 4 𝐙 , 0 ) ∢ 𝑅 𝑏 ) for any 𝑏 ∈ 𝑃 2 ⧡ 𝑃 1 . Hence condition ( 𝑖 ) of Lemma 3.6 does not hold. Therefore condition ( i i ) of Lemma 3.6 must hold. Thus | 𝑃 1 ⧡ 𝑃 2 | = 2 and 𝑃 2 = ( ( 0 + 4 𝐙 , 0 ) ∢ 𝑅 𝑐 ) for some 𝑐 ∈ 𝑃 1 ⧡ 𝑃 2 . We now show that | 𝑇 2 | = 2 . Suppose that | 𝑇 2 | β‰₯ 3 . Let 𝑦 , 𝑧 ∈ 𝑇 2 ⧡ { 0 } with 𝑦 β‰  𝑧 . Note that { ( 0 + 4 𝐙 , 𝑦 ) , ( 0 + 4 𝐙 , 𝑧 ) , ( 2 + 4 𝐙 , 𝑦 ) } βŠ† 𝑃 1 ⧡ 𝑃 2 , and this implies that | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 which contradicts the fact that | 𝑃 1 ⧡ 𝑃 2 | = 2 . This proves that | 𝑇 2 | = 2 , and so 𝑇 2 is isomorphic to 𝐙 / 2 𝐙 as rings.
( i i ) β‡’ ( i i i ) Now 𝑇 2 = { 0 , 1 } . Observe that 𝑍 ( 𝑅 ) βˆ— = { ( 0 + 4 𝐙 , 1 ) , ( 2 + 4 𝐙 , 0 ) , ( 2 + 4 𝐙 , 1 ) , ( 1 + 4 𝐙 , 0 ) , ( 3 + 4 𝐙 , 0 ) } . Observe that the edge ( 0 + 4 𝐙 , 1 ) βˆ’ ( 2 + 4 𝐙 , 1 ) of ( Ξ“ ( 𝑅 ) ) 𝑐 is not an edge of any triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 . Indeed, ( 0 + 4 𝐙 , 1 ) is a pendant vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 . Thus the vertices ( 0 + 4 𝐙 , 1 ) and ( 2 + 4 𝐙 , 1 ) are complements of each other in ( Ξ“ ( 𝑅 ) ) 𝑐 . It can be easily verified that each of the other edges of ( Ξ“ ( 𝑅 ) ) 𝑐 is an edge of a triangle in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .
( i i i ) β‡’ ( i ) This is clear.
If 𝑅 = 𝑇 1 2 Γ— 𝑇 2 where 𝑇 1 2 = ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] and 𝑇 2 is an integral domain, then the proof of the fact that the statements ( i ) , ( i i ) , and ( i i i ) are equivalent is exactly similar and hence is omitted.

Suppose that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . The following theorem characterizes rings 𝑅 such that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Theorem 3.9. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. Suppose that 𝑃 1 ∩ 𝑃 2 β‰  ( 0 ) . The following statements are equivalent.(i) ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .(ii) Either 𝑅 is isomorphic to 𝐙 / 4 𝐙 Γ— 𝐙 / 2 𝐙 or 𝑅 is isomorphic to ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] Γ— 𝐙 / 2 𝐙 as rings.(iii) ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. ( i ) β‡’ ( i i ) We know from Lemma 3.7 that 𝑅 is isomorphic to 𝑇 1 Γ— 𝑇 2 as rings with | 𝑍 ( 𝑇 1 ) βˆ— | = 1 and 𝑇 2 is an integral domain. Since | 𝑍 ( 𝑇 1 ) βˆ— | = 1 , it follows from [1, Example 2.1(a)] that either 𝑇 1 is isomorphic to 𝐙 / 4 𝐙 or 𝑇 1 is isomorphic to ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] as rings. Let 𝑇 1 1 = 𝐙 / 4 𝐙 and 𝑇 1 2 = ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] . Since either 𝑅 is isomorphic to 𝑇 1 1 Γ— 𝑇 2 or 𝑅 is isomorphic to 𝑇 1 2 Γ— 𝑇 2 , ( 𝑖 ) implies that either ( Ξ“ ( 𝑇 1 1 Γ— 𝑇 2 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑇 1 1 Γ— 𝑇 2 ) ) 𝑐 or ( Ξ“ ( 𝑇 1 2 Γ— 𝑇 2 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑇 1 2 Γ— 𝑇 2 ) ) 𝑐 . Now it follows from ( i ) β‡’ ( i i ) of Lemma 3.8 that 𝑇 2 is isomorphic to 𝐙 / 2 𝐙 as rings. Hence we obtain that either 𝑅 is isomorphic to 𝐙 / 4 𝐙 Γ— 𝐙 / 2 𝐙 or 𝑅 is isomorphic to ( 𝐙 / 2 𝐙 ) [ π‘₯ ] / π‘₯ 2 ( 𝐙 / 2 𝐙 ) [ π‘₯ ] Γ— 𝐙 / 2 𝐙 as rings.
( i i ) β‡’ ( i i i ) It follows from ( i i ) β‡’ ( i i i ) of Lemma 3.8 that ( Ξ“ ( 𝑅 ) ) 𝑐 has exactly two vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .
( i i i ) β‡’ ( i ) This is clear.
This completes the proof of Theorem 3.9.

Suppose that 𝑃 1 ∩ 𝑃 2 = ( 0 ) ; that is, equivalently, by [6, Theorem 1.1(b)], ( Ξ“ ( 𝑅 ) ) 𝑐 is not connected. Assume that ( Ξ“ ( 𝑅 ) ) 𝑐 admits at least one edge. Our next aim is to determine when no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 and to determine rings 𝑅 such that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Lemma 3.10. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. Suppose that 𝑃 1 ∩ 𝑃 2 = ( 0 ) . Then the following hold.(i) If | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 , then no element of 𝑃 1 ⧡ 𝑃 2 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .(ii) If | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 , then no element of 𝑃 2 ⧡ 𝑃 1 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .(iii) If | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 and | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 , then no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. (i) Let π‘₯ ∈ 𝑃 1 ⧡ 𝑃 2 . Suppose that π‘₯ admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Let 𝑦 ∈ 𝑍 ( 𝑅 ) βˆ— be a complement of π‘₯ in ( Ξ“ ( 𝑅 ) ) 𝑐 . Then π‘₯ β‰  𝑦 and π‘₯ 𝑦 β‰  0 . Since 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 , 𝑃 1 ∩ 𝑃 2 = ( 0 ) , and π‘₯ 𝑦 β‰  ( 0 ) , it follows that 𝑦 ∈ 𝑃 1 ⧡ 𝑃 2 . By hypothesis, | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 . Now for any 𝑧 ∈ ( 𝑃 1 ⧡ 𝑃 2 ) ⧡ { π‘₯ , 𝑦 } , 𝑧 π‘₯ β‰  0 and 𝑧 𝑦 β‰  0 . This is impossible since 𝑦 is a complement of π‘₯ in ( Ξ“ ( 𝑅 ) ) 𝑐 . This proves that if | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 , then no element of 𝑃 1 ⧡ 𝑃 2 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .
(ii) The proof of ( i i ) is similar to the proof of ( i ) .
(iii) Since 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 and 𝑃 1 ∩ 𝑃 2 = ( 0 ) , it follows that 𝑍 ( 𝑅 ) βˆ— = ( 𝑃 1 ⧡ 𝑃 2 ) βˆͺ ( 𝑃 2 ⧡ 𝑃 1 ) . Note that 𝑍 ( 𝑅 ) βˆ— is the vertex set of ( Ξ“ ( 𝑅 ) ) 𝑐 . Thus if | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 and | 𝑃 2 ⧡ 𝑃 1 | β‰₯ 3 , then it follows from ( i ) and ( i i ) that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .
This completes the proof of Lemma 3.10.

The following proposition describes rings 𝑅 such that 𝑃 1 ∩ 𝑃 2 = ( 0 ) , ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one edge, and moreover, ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proposition 3.11. Let 𝑅 , 𝑃 1 , 𝑃 2 be as mentioned in the beginning of this section. If 𝑃 1 ∩ 𝑃 2 = ( 0 ) and if ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one edge, then the following statements are equivalent.(i) ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .(ii) 𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝑇 as rings, where 𝑇 is an integral domain.(iii) Exactly one of the following holds.(a)   𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝐙 / 2 𝐙 as rings, and in this case ( Ξ“ ( 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 ( Ξ“ ( 𝑅 ) ) 𝑐 .(b)   𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝐙 / 3 𝐙 as rings, and in this case ( Ξ“ ( 𝑅 ) ) 𝑐 is a graph on four vertices and is complemented.(c)   𝑅 is isomorphic to 𝐙 / 3 𝐙 Γ— 𝑇 as rings, where 𝑇 is an integral domain with | 𝑇 | β‰₯ 4 , and in this case ( Ξ“ ( 𝑅 ) ) 𝑐 is a graph on more than four vertices and it has exactly two vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 .

Proof. ( i ) β‡’ ( i i ) Suppose that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Then it follows from Lemma 3.10 ( i i i ) that either | 𝑃 1 ⧡ 𝑃 2 | < 3 or | 𝑃 2 ⧡ 𝑃 1 | < 3 . Without loss of generality we may assume that | 𝑃 2 ⧡ 𝑃 1 | < 3 . Then either | 𝑃 2 ⧡ 𝑃 1 | = 1 or | 𝑃 2 ⧡ 𝑃 1 | = 2 .
Suppose that | 𝑃 2 ⧡ 𝑃 1 | = 1 . Let 𝑃 2 ⧡ 𝑃 1 = { 𝑏 } . Since 𝑃 1 ∩ 𝑃 2 = ( 0 ) , it follows that 𝑃 2 = { 0 , 𝑏 } . Moreover, 𝑍 ( 𝑅 ) βˆ— = ( 𝑃 1 ⧡ 𝑃 2 ) βˆͺ { 𝑏 } . We assert that | 𝑃 1 ⧡ 𝑃 2 | = 2 . Since we are assuming that ( Ξ“ ( 𝑅 ) ) 𝑐 has at least one edge, we obtain that | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 2 . Suppose that | 𝑃 1 ⧡ 𝑃 2 | β‰₯ 3 , then it follows from Lemma 3.10 ( i ) that no element of 𝑃 1 ⧡ 𝑃 2 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Observe that 𝑏 is an isolated vertex in ( Ξ“ ( 𝑅 ) ) 𝑐 and hence it does not admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Hence we obtain that no vertex of ( Ξ“ ( 𝑅 ) ) 𝑐 admits a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . This is in contradiction to the hypothesis that ( Ξ“ ( 𝑅 ) ) 𝑐 has vertices which admit a complement in ( Ξ“ ( 𝑅 ) ) 𝑐 . Therefore, we obtain that | 𝑃 1 ⧡ 𝑃 2 | ≀ 2 . This shows that | 𝑃 1 ⧡ 𝑃 2 | = 2 . Thus | 𝑍 ( 𝑅 ) | = | 𝑃 1 βˆͺ 𝑃 2 | = 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 𝑃 1 and 𝑃 2 are maximal ideals of 𝑅 . Thus 𝑃 1 + 𝑃 2 = 𝑅 . Since 𝑃 1 ∩ 𝑃 2 = ( 0 ) , it follows from the Chinese remainder theorem [13, Proposition 1.10(ii)] that the mapping 𝑓 ∢ 𝑅 β†’ 𝑅 / 𝑃 1 Γ— 𝑅 / 𝑃 2 given by 𝑓 ( π‘Ÿ ) = ( π‘Ÿ + 𝑃 1 , π‘Ÿ + 𝑃 2 ) for any π‘Ÿ ∈ 𝑅 is an isomorphism of rings. Note that 𝑓 ( 𝑃 1 ) = 𝑃 1 / 𝑃 1 Γ— 𝑅 / 𝑃 2 and 𝑓 ( 𝑃 2 ) = 𝑅 / 𝑃 1 Γ— 𝑃 2 / 𝑃 2 . Hence 3 = | 𝑃 1 | = | 𝑓 ( 𝑃 1 ) | = | 𝑅 / 𝑃 2 | and 2 = | 𝑃 2 | = | 𝑓 ( 𝑃 2 ) | = | 𝑅 / 𝑃 1 | . So we obtain that 𝑅 / 𝑃 1 β‰… 𝐙 / 2 𝐙 and 𝑅 / 𝑃 2 β‰… 𝐙 / 3 𝐙 as rings. Thus 𝑅 β‰… 𝑅 / 𝑃 1 Γ— 𝑅 / 𝑃 2 β‰… 𝐙 / 2 𝐙 Γ— 𝐙 / 3 𝐙 β‰… 𝐙 / 3 𝐙 Γ— 𝐙 / 2 𝐙 as rings. Thus with 𝑇 = 𝐙 / 2 𝐙 , we obtain that 𝑅 β‰… 𝐙 / 3 𝐙 Γ— 𝑇 .
Suppose that | 𝑃 2 ⧡ 𝑃 1 | = 2 . Let 𝑃 2 ⧡ 𝑃 1 = { 𝑏 1 , 𝑏 2 } . Note that 𝑃 2 = { 0 , 𝑏 1 , 𝑏 2 } . We claim that 𝑃 1 + 𝑃 2 = 𝑅 . Suppose that 𝑃 1 + 𝑃 2 β‰  𝑅 . Then there exists a maximal ideal 𝑀 of 𝑅 such that 𝑃 1 + 𝑃 2 βŠ† 𝑀 . Let π‘Ž ∈ 𝑃 1 ⧡ 𝑃 2 . Now π‘Ž + 𝑏 1 ∈ 𝑃 1 + 𝑃 2 βŠ† 𝑀 , and as 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 , it follows from the choice of the elements π‘Ž and 𝑏 1 that π‘Ž + 𝑏 1 βˆ‰ 𝑍 ( 𝑅 ) . Let 𝑧 = π‘Ž + 𝑏 1 . Note that 𝑧 𝑏 1 ∈ 𝑃 2 ⧡ { 0 } = { 𝑏 1 , 𝑏 2 } . Observe that 𝑧 𝑏 1 β‰  𝑏 1 . If 𝑧 𝑏 1 = 𝑏 1 , then ( 1 βˆ’ 𝑧 ) 𝑏 1 = 0 , and this implies that 1 βˆ’ 𝑧 ∈ 𝑍 ( 𝑅 ) = 𝑃 1 βˆͺ 𝑃 2 βŠ† 𝑀 . Thus both