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

Authors: S. Visweswaran
Journal: ISRN Algebra
 (2012)

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## Abstract

LetR 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.


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## Body

## 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. LetR 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. LetG=(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).LetG=(V,E) be a simple graph. Recall from [5, Definition 1.1.13] that the complement of G denoted by Gc is defined as a graph with V(Gc)=V(G)=V and two distinct elements x, y∈V are joined by an edge in Gc if and only if there is no edge of G joining x and y.LetR 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. LetR 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 ZR(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 {Pα}α∈Λ denotes the set of all maximal N-primes of (0) in R, then Z(R)=⋃α∈ΛPα.LetI 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.LetR 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 Section3 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 Section4 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)

LetR 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.
LetR 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 thatP 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 thatP 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.
LetR 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.
AsP 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 ringsR 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 thatR 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)

LetR 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 thatP1∩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.LetR, 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 Theorem3.9. We start with the following.Lemma 3.1.
LetR, 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 thatP1 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.
Bothx and y belong to P1.
IfP2⊈((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 thatP2⊆((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.
Bothx and y belong to P2.
SinceP1 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.
Nowx∈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 bothP1 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.
LetR,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 bothP1 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.
Bothx and y belong to P1.
It now follows as in the proof of Lemma3.1 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.
Bothx and y belong to P2.
The proof of the fact that there existsz∈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.
Nowxy≠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 bothP1 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.
LetR, 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.
Letx-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.
Bothx and y belong to P1.
Proceeding as in the proof of Lemma3.1 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.
Bothx and y belong to P2.
The hypotheses regardingP1 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.IfP1∩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.
LetR, 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.
SinceP1 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.
LetR,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 betweenP1 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 thatu 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 thatP1∩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.
LetR, 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.
LetP1∩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 Lemma3.6.Suppose thatP1∩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.
LetR, 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=P12∩P22. Now we obtain from the Chinese remainder theorem [13, Proposition 1.10(ii)] that R is isomorphic to R/P12×R/P22 as rings. Indeed, it follows from the Chinese remainder theorem that the mapping f:R→R/P12×R/P22 given by f(r)=(r+P12,r+P22) for any r∈R is an isomorphism of rings. The isomorphism f maps P1∩P2 onto P1/P12×P2/P22. Hence 2=|P1∩P2|=|f(P1∩P2)|=|P1/P12×P2/P22|=|P1/P12∥P2/P22|. Thus either |P1/P12|=2,|P2/P22|=1 or |P1/P12|=1,|P2/P22|=2. We may assume without loss of generality that |P1/P12|=2 and |P2/P22|=1. Let T1=R/P12 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/P12×T2)∪(T1×P2/P22), it follows that Z(T1)=P1/P12. Since |P1/P12|=2, we obtain that |Z(T1)*|=1.
Thus ifP1∩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.LetT 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.
LetT11, 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.
LetR=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.
IfR=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 thatP1∩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.
LetR, 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)
EitherR 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 Theorem3.9.Suppose thatP1∩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.
LetR, 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) Letx∈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) SinceZ(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 Lemma3.10.The following proposition describes ringsR 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.
LetR, 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 Proposition3.11.


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

LetR 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.
LetR 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 {Pα}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∈ΛPα. We consider the following cases.Case 1.
There existsα∈Λ such that both x and y belong to Pα. Let β,γ∈Λ∖{α} be such that β≠γ. Since Pβ≠Pγ and xy≠0, it follows that either Pβ⊈((0):Rxy) or Pγ⊈((0):Rxy). We may assume without loss of generality that Pβ⊈((0):Rxy). Note that Pβ⊈Pα∪((0):Rxy). Hence there exists z∈Pβ such that z∉Pα and zxy≠0. Since both x and y belong to Pα, whereas z∉Pα, it is clear that z∉{x,y}. As z∈Pβ⊆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.
Letα, β be distinct elements of Λ such that x∈Pα∖Pβ and y∈Pβ∖Pα. 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∈Pα∖Pβ, it follows that ((0):Rx)⊆Pβ. Similarly since y∈Pβ∖Pα, it follows that ((0):Ry)⊆Pα. By hypothesis |Λ|≥3. Let γ∈Λ∖{α,β}. Note that Pγ⊈Pα∪Pβ. Hence there exists z∈Pγ such that z∉Pα∪Pβ. It is now clear that z∈Z(R)∖{x,y}. Since ((0):Rx)⊆Pβ and z∉Pβ, it follows that zx≠0. Similarly since ((0):Ry)⊆Pα and z∉Pα, we obtain that zy≠0.
This shows that no vertex of(Γ(R))c admits a complement in (Γ(R))c.



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*Source: 282054-2012-06-16.xml*