Some results on the total zero-divisor graph of a commutative ring

Visweswaran, Subramanian

doi:10.1108/AJMS-10-2023-0039

Purpose

The purpose of this paper is to characterize a commutative ring R with identity which is not an integral domain such that ZT(R), the total zero-divisor graph of R is connected and to determine the diameter and radius of ZT(R) whenever ZT(R) is connected. Also, the purpose is to generalize some of the known results proved by Duric et al. on the total zero-divisor graph of R.

Design/methodology/approach

We use the methods from commutative ring theory on primary decomposition and strong primary decomposition of ideals in commutative rings. The structure of ideals, the primary ideals, the prime ideals, the set of zero-divisors of the finite direct product of commutative rings is used in this paper. The notion of maximal Nagata prime of the zero-ideal of a commutative ring is also used in our discussion.

Findings

For a commutative ring R with identity, ZT(R) is the intersection of the zero-divisor graph of R and the total graph of R induced by the set of all non-zero zero-divisors of R. The zero-divisor graph of R and the total graph of R induced by the set of all non-zero zero-divisors of R are well studied. Hence, we determine necessary and sufficient condition so that ZT(R) agrees with the zero-divisor graph of R (respectively, agrees with the total graph induced by the set of non-zero zero-divisors of R). If Z(R) is an ideal of R, then it is noted that ZT(R) agrees with the zero-divisor graph of R. Hence, we focus on rings R such that Z(R) is not an ideal of R. We are able to characterize R such that ZT(R) is connected under the assumptions that the zero ideal of R admits a strong primary decomposition and Z(R) is not an ideal of R. With the above assumptions, we are able to determine the domination number of ZT(R).

Research limitations/implications

Duric et al. characterized Artinian rings R such that ZT(R) is connected. In this paper, we extend their result to rings R such that the zero ideal of R admits a strong primary decomposition and Z(R) is not an ideal of R. As an Artinian ring is isomorphic to the direct product of a finite number of Artinian local rings, we try to characterize R such that ZT(R) is connected under the assumption that R is ta finite direct product of rings R1, R2, … Rn with Z(Ri) is an ideal of Ri for each i between 1 to n. Their result on domination number of ZT(R) is also generalized in this paper. We provide several examples to illustrate our results proved.

Practical implications

The implication of this paper is that the existing result of Duric et al. is applicable to large class of commutative rings thereby yielding more examples. Moreover, the results proved in this paper make us to understand the structure of commutative rings better. It also helps us to learn the interplay between the ring-theoretic properties and the graph-theoretic properties of the graph associated with it.

Originality/value

The results proved in this paper are original and they provide more insight into the structure of total zero-divisor graph of a commutative ring. This paper provides several examples. Not much work done in the area of total zero-divisor graph of a commutative ring. This paper is a contribution to the area of graphs and rings and may inspire other researchers to study the total zero-divisor graph in further detail.

1. Introduction

The rings considered in this paper are commutative with identity and unless otherwise specified, they are not integral domains. Throughout this paper, we use R to denote a ring. We denote the set of all zero-divisors of R by Z(R) and the set Z(R)\{0} by Z(R)*. Motivated by the results proved on the zero-divisor graphs of commutative rings (for example, refer [1–4]) and the interesting survey article on zero-divisor graphs in commutative rings in Ref. [5] and the note-worthy theorems proved on total graphs of commutative rings in Refs. [6–8], Ðurić et al. in Ref. [9] introduced an undirected graph called the total zero-divisor graph of R denoted by ZT(R) and investigated its graph-theoretic properties. The graphs considered in this paper are undirected and simple. For a graph G, we denote the vertex set of G by V(G) and the edge set of G by E(G). Let R be such that Z(R)* ≠ ∅. Recall that the total zero-divisor graph of R denoted by ZT(R) is an undirected graph with V(ZT(R)) = Z(R)* and distinct vertices x and y are adjacent in ZT(R) if and only if xy = 0 and x + y ∈ Z(R) [9]. It is useful to recall that the zero-divisor graph of R denoted by Γ(R) is an undirected graph with V(Γ(R)) = Z(R)* and distinct vertices x and y are adjacent in Γ(R) if and only if xy = 0 [1]. Recall that the total graph of R denoted by T(Γ(R)) is an undirected graph with V(T(Γ(R))) = R and distinct vertices x and y are adjacent in T(Γ(R)) if and only if x + y ∈ Z(R) [6]. We denote the subgraph of T(Γ(R)) induced by Z(R)* by $T_{Z {(R)}^{*}} (Γ (R))$ ⁠. A subgraph H of a graph G is said to be a spanning subgraph of G if V(H) = V(G) [10]. In such a case, we say that G is a spanning supergraph of H. It is clear from the definition of ZT(R) that both Γ(R) and $T_{Z {(R)}^{*}} (Γ (R))$ are spanning supergraphs of ZT(R) and $Z T (R) = Γ (R) \cap T_{Z {(R)}^{*}} (Γ (R))$ ⁠.

Let R be such that Z(R)* ≠ ∅. In Ref. [9], Ðurić et al. proved several interesting results on ZT(R). It was observed in Ref. [9] that even if R is Artinian, ZT(R) can fail to be connected. It was proved in [[9], Theorem 3.1] that for an Artinian ring R, ZT(R) is connected if and only if $p \cap A n n (p) \neq (0)$ for each $p \in A s s (R)$ ⁠. Let R be Artinian such that ZT(R) is connected. Then the diameter of ZT(R) was determined in [[9], Theorem 3.2]. For any $m \in N$ with m ≥ 2, we denote the ring of integers modulo m by $Z_{m}$ ⁠. In Section 4 of [9], Ðurić et al. discussed several graph-theoretic properties of $Z T (Z_{m})$ ⁠.

Let R be a ring. We denote the set of all prime ideals of R by Spec(R), the set of all maximal ideals of R by Max(R), and the set of all minimal prime ideals of R by Min(R). We denote the nilradical of R by nil(R) and R is said to be reduced if nil(R) = (0). Let I be an ideal of R with I ≠ R. Recall that $p \in S p e c (R)$ is said to be a maximal N-prime of I if $p$ is maximal with respect to the property of being contained in $Z_{R} (\frac{R}{I}) = {r \in R ∣ r x \in I for some x \in R \ I}$ [11]. Thus $p \in S p e c (R)$ is a maximal N-prime of (0) if $p$ is maximal with respect to the property of being contained in Z(R). For convenience, we denote the set of all maximal N-primes of (0) in R by MNP(R). It is clear that S = R\Z(R) is a multiplicatively closed subset of R. If I is an ideal of R with I ∩ S = ∅, then it follows from Zorn’s lemma and [[12], Theorem 1] that there exists $p \in M N P (R)$ with $I \subseteq p$ ⁠. In particular, if x ∈ Z(R), then Rx ∩ S = ∅ and so, there exists $p \in M N P (R)$ such that $x \in p$ ⁠. It follows from the above argument that if $M N P (R) = {p_{α}}_{α \in Λ}$ ⁠, then $Z (R) = ⋃_{α \in Λ} p_{α}$ ⁠. We denote the cardinality of a set A by |A|. Observe that Z(R) is an ideal of R if and only if |MNP(R)| = 1. The ring S⁻¹R, where S = R\Z(R) is called the total quotient ring of R and is denoted by Tot(R). Let I be an ideal of R with I ≠ R. Recall that $p \in S p e c (R)$ is said to be an associated prime of I in the sense of Bourbaki if $p = (I :_{R} x)$ for some x ∈ R [13]. In such a case, we say that $p$ is a B-prime of I. For a subset E of R, let us denote the set ${p \in S p e c (R) ∣ p \supseteq E}$ by V(E). If E = {a} for some a ∈ R, then we denote V(E) by V(a). We say that R is quasi-local if |Max(R)| = 1. A Noetherian quasi-local ring is called a local ring. If a set A is a subset of a set B and A ≠ B, then we denote it by A ⊂ B. The Krull dimension of R is referred to as the dimension of R and is denoted by dim R. An element e of R is said to be idempotent if e = e². An idempotent e of R is said to be non-trivial if e∉{0, 1}. We denote the group of units of R by U(R) and the set of all non-units of R by NU(R). For a connected graph G, we denote the diameter of G by diam(G) and the radius of G by r(G).

Let R be such that Z(R)* ≠ ∅. This paper aims to state and prove some results on ZT(R) that were not considered in Ref. [9] and to generalize some results that were proved in Ref. [9]. This paper consists of four sections including the introduction. As $Z T (R) = Γ (R) \cap T_{Z {(R)}^{*}} (Γ (R))$ ⁠, Γ(R) and T(Γ(R)) were investigated by many researchers in the literature, in Section 2 of this article, among other basic results on ZT(R), we try to answer when the total zero-divisor graph of R is equal to Γ(R) and when it is equal to $T_{Z {(R)}^{*}} (Γ (R))$ ⁠. The following results are proved in Section 2.

ZT(R) = Γ(R) if and only if Tot(R) has no non-trivial idempotent (see Theorem 2.5).
If $M N P (R) = {p}$ ⁠, then ZT(R) = Γ(R) (see Corollary 2.7). And $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ if and only if $p^{2} = (0)$ (see Lemma 2.12).
If |MNP(R)|≥ 2, then $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ if and only if $R ≅ Z_{2} \times Z_{2}$ as rings (see Theorem 2.17).

For a ring R with |MNP(R)|≥ 2, in Section 3 of this paper, we determine some necessary (respectively, sufficient) conditions so that ZT(R) is connected. The following results are proved in Section 3.

If ZT(R) is connected, then for any a ∈ Z(R)* with MNP(R)⊈V(a), there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅ (see Lemma 3.1).
If ZT(R) is connected, then for any a ∈ Z(R)* with |V(a) ∩ MNP(R)| = 1, there exists x ∈ Z(R)*\{a} such that ax = 0 and x belongs to each member of MNP(R) (see Corollary 3.2).
Let $n \in N \ {1}$ ⁠. If $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ and if ZT(R) is connected, then $⋂_{i = 1}^{n} p_{i} \neq (0)$ (see Corollary 3.3).
If $n, p_{i} (1 \leq i \leq n)$ are as in the statement of Corollary 3.3, then the following statements are equivalent: (1) ZT(R) is connected; $(2) ⋂_{i = 1}^{n} p_{i} \neq (0)$ and for any a ∈ Z(R)* with MNP(R)⊈V(a), there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅. Moreover, if either (1) or (2) holds, then diam(ZT(R)) = 3 and if R is not reduced, then r(ZT(R)) = 2 (see Theorem 3.8).
If the zero ideal of R admits a strong primary decomposition, then ZT(R) is connected if and only if $p \cap A n n (p) \neq (0)$ for each $p \in M N P (R)$ ⁠. Moreover, if ZT(R) is connected, then diam(ZT(R)) = 3 and r(ZT(R)) = 2 (see Theorem 3.9).
If R is reduced with 2 ≤ |Min(R)| < ∞, then ZT(R) is not connected (see Corollary 3.11).
Let $n \in N \ {1}$ and let R_i be a ring with Z(R_i) is an ideal of R for each i ∈ {1, 2, …, n}. If R = R₁ × R₂ × . . . × R_n, then the following statements are equivalent: (1) ZT(R) is connected; (2) Z(R_i) ≠ (0) for each i ∈ {1, 2, …, n}. Moreover, if either (1) or (2) holds, then diam(ZT(R)) = 3, and if R is not reduced, then r(ZT(R)) = 2 (see Proposition 3.15).
For each $n \in N$ ⁠, let R_n be a ring with Z(R_n) is an ideal of R_n. Let $R = \prod_{n \in N} R_{n}$ ⁠. Then the following statements are equivalent: (1) ZT(R) is connected; (2) Z(R_n) ≠ (0) for each $n \in N$ ⁠. Moreover, if either (1) or (2) holds, then diam(ZT(R)) = 3 and if R is not reduced, then r(ZT(R)) = 2 (see Proposition 3.16).
If R is von Neumann regular, then the following statements are equivalent: (1) ZT(R) is connected; (2) No maximal ideal of R is principal. Moreover, if either (1) or (2) holds, then diam(ZT(R)) = 3 = r(ZT(R)) (see Theorem 3.18).
There exists a von Neumann regular ring R such that no maximal ideal of R is principal (see Example 3.19).

For a ring R with Z(R)* ≠ ∅, motivated by [[9], Theorem 4.4], in Section 4 of this paper, we discuss some results on the dominating sets and the domination number of ZT(R). For a graph G, we denote the domination number of G by γ(G). The following results are proved in Section 4.

If R is such that |Z(R)*|≥ 2, then the following statements are equivalent: (1) γ(ZT(R)) = 1; (2) ZT(R) is connected and r(ZT(R)) = 1; (3) |MNP(R)| = 1 and the unique member of MNP(R) is a B-prime of (0) in R (see Proposition 4.1).
There exists a ring R with |MNP(R)| = 1 such that ZT(R) does not admit any finite dominating set (see Example 4.2).
If (0) admits a strong primary decomposition in R with |MNP(R)|≥ 2 and if ZT(R) is connected, then γ(ZT(R)) = |MNP(R)| (see Proposition 4.7).
There exists a ring R such that (0) admits a strong primary decomposition in R and |MNP(R)| = 2 but γ(ZT(R)) = 3 (see Example 4.10).
There exists a ring T such that |MNP(T)| = 2 but ZT(T) does not admit any finite dominating set (see Example 4.11).

2. Some basic properties of ZT(R)

Let R be a ring. This section aims to discuss some basic properties of ZT(R). Let f: R → Tot(R) denote the usual homomorphism of rings defined by $f (r) = \frac{r}{1}$ ⁠. Observe that f is injective. As mentioned in the introduction, unless otherwise specified, the rings considered in this paper are commutative with identity which admit at least one non-zero zero-divisor. It is noted in Section 1 that $Z T (R) = Γ (R) \cap T_{Z {(R)}^{*}} (Γ (R))$ ⁠. Hence, we first try to characterize R such that ZT(R) = Γ(R) (respectively, $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠). We prove in Theorem 2.5 that ZT(R) = Γ(R) if and only if Tot(R) has no non-trivial idempotent. We first state and prove some lemmas which are needed for its proof.

Lemma 2.1.

The following statements are equivalent:

Proof. For a ring T, as Γ(T) and $T_{Z {(T)}^{*}} (Γ (T))$ are spanning supergraphs of ZT(T) with $Z T (T) = Γ (T) \cap T_{Z {(T)}^{*}} (Γ (T))$ ⁠, it follows that ZT(T) = Γ(T) if and only if Γ(T) is a subgraph of $T_{Z {(T)}^{*}} (Γ (T))$ ⁠.

(1) ⇒ (2) Assume that ZT(R) = Γ(R). Let z₁, z₂ ∈ Z(Tot(R))* be such that z₁ and z₂ are adjacent in Γ(Tot(R)). Then z₁ ≠ z₂ and $z_{1} z_{2} = \frac{0}{1}$ ⁠. Observe that there exist x₁, x₂ ∈ Z(R)* and s ∈ R\Z(R) such that $z_{i} = \frac{x_{i}}{s}$ for each i ∈ {1, 2}. From z₁ ≠ z₂, it follows that x₁ ≠ x₂ and from $z_{1} z_{2} = \frac{0}{1}$ ⁠, we get that x₁x₂ = 0. Therefore, x₁ and x₂ are adjacent in Γ(R). As ZT(R) = Γ(R) by assumption, it follows that x₁ and x₂ are adjacent in $T_{Z {(R)}^{*}} (Γ (R))$ ⁠. Hence, x₁ + x₂ ∈ Z(R) and so, z₁ + z₂ ∈ Z(Tot(R)). This shows that Γ(Tot(R)) is a subgraph of $T_{Z {(T o t (R))}^{*}} (Γ (T o t (R)))$ and therefore, ZT(Tot(R)) = Γ(Tot(R)).

(2) ⇒ (1) Assume that ZT(Tot(R)) = Γ(Tot(R)). Let x₁, x₂ ∈ Z(R)* be such that x₁ and x₂ are adjacent in Γ(R). Then x₁ ≠ x₂ and x₁x₂ = 0. Let $z_{i} = \frac{x_{i}}{1}$ for each i ∈ {1, 2}. It is clear that z₁ ≠ z₂ and $z_{1} z_{2} = \frac{0}{1}$ ⁠. Therefore, z₁ and z₂ are adjacent in Γ(Tot(R)). As ZT(Tot(R)) = Γ(Tot(R)) by assumption, it follows that z₁ and z₂ are adjacent in $T_{Z {(T o t (R))}^{*}} (Γ (T o t (R)))$ ⁠. Hence, z₁ + z₂ ∈ Z(Tot(R)) and so, x₁ + x₂ ∈ Z(R). This shows that Γ(R) is a subgraph of $T_{Z {(R)}^{*}} (Γ (R))$ and therefore, ZT(R) = Γ(R). □

ZT(R) = Γ(R).
ZT(Tot(R)) = Γ(Tot(R)).

Remark 2.2.

Using arguments similar to those that are used in the proof of Lemma 2.1, it can be shown that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ if and only if $Z T (T o t (R)) = T_{Z {(T o t (R))}^{*}} (Γ (T o t (R)))$ ⁠.

Lemma 2.3.

Let R = R₁ × R₂ be the direct product of rings R₁ and R₂. Then ZT(R) ≠ Γ(R).

Proof. Let x = (1, 0) and let y = (0, 1). It is clear that x, y ∈ Z(R)*, x ≠ y, and xy = (0, 0). Thus x and y are adjacent in Γ(R). Note that x + y = (1, 1) ∉ Z(R). Therefore, x and y are not adjacent in ZT(R) and so, ZT(R) ≠ Γ(R). □

Lemma 2.4.

If there exist x, y ∈ NU(R) such that xy = 0 and x + y ∈ U(R), then R admits at least two non-trivial idempotent elements.

Proof. From xy = 0, x, y ∈ NU(R), and x + y ∈ U(R), it follows that x ≠ 0, y ≠ 0. As x + y ∈ U(R), there exists u ∈ U(R) such that (x + y)u = 1. Let us denote xu by x₁ and yu by y₁. Note that x₁ + y₁ = 1 and from xy = 0, it follows that x₁y₁ = 0. It is clear that $x_{1} = 1 x_{1} = (x_{1} + y_{1}) x_{1} = x_{1}^{2}$ and $y_{1} = 1 y_{1} = (x_{1} + y_{1}) y_{1} = y_{1}^{2}$ ⁠. Observe that x₁ ≠ y₁ and x₁, y₁∉{0, 1}. Therefore, x₁ and y₁ are non-trivial idempotent elements of R. □

Theorem 2.5.

The following statements are equivalent:

Proof. (1) ⇒ (2) Assume that ZT(R) = Γ(R). Note that ZT(Tot(R)) = Γ(Tot(R)) by (1) ⇒ (2) of Lemma 2.1. Suppose that Tot(R) admits a non-trivial idempotent. Let $e \in T o t (R) \ {\frac{0}{1}, \frac{1}{1}}$ be such that e = e². Let T₁ = Tot(R)e and let $T_{2} = T o t (R) (\frac{1}{1} - e)$ ⁠. Observe that the mapping f: Tot(R) → T₁ × T₂ given by $f (t) = (t e, t (\frac{1}{1} - e))$ is an isomorphism of rings. Let us denote the ring T₁ × T₂ by T. We know from Lemma 2.3 that ZT(T) ≠ Γ(T). Since Tot(R) ≅ T as rings, it follows that ZT(Tot(R)) ≠ Γ(Tot(R)). This is a contradiction, since ZT(Tot(R)) = Γ(Tot(R)). Therefore, Tot(R) has no non-trivial idempotent.

(2) ⇒ (1) Assume that Tot(R) has no non-trivial idempotent. Let z₁, z₂ ∈ Z(Tot(R))* be such that z₁ and z₂ are adjacent in Γ(Tot(R)). Then $z_{1} z_{2} = \frac{0}{1}$ ⁠. Since Tot(R) has no non-trivial idempotent by assumption, it follows from Lemma 2.4 that z₁ + z₂ ∈ NU(Tot(R)). Hence, z₁ + z₂ ∈ Z(Tot(R)), as U(Tot(R)) = Tot(R)\Z(Tot(R)). This shows that Γ(Tot(R)) is a subgraph of $T_{Z {(T o t (R))}^{*}} (Γ (T o t (R)))$ and so, ZT(Tot(R)) = Γ(Tot(R)). Hence, ZT(R) = Γ(R) by (2) ⇒ (1) of Lemma 2.1. □

ZT(R) = Γ(R).
Tot(R) has no non-trivial idempotent.

Recall that a simple graph G = (V, E) is said to be complete if each pair of distinct vertices of G are adjacent in G [[10], Definition 1.2.11]. Let $n \in N$ ⁠. A complete graph with n vertices is denoted by K_n.

If R is a ring with |MNP(R)| = 1, then we prove in Corollary 2.7 that ZT(R) = Γ(R). We use the following lemma in its proof.

Lemma 2.6.

Let G₁, G₂ be spanning supergraphs of a graph g with g = G₁ ∩ G₂. If G₂ is complete, then g = G₁.

Proof. Observe that g is a spanning subgraph of both G₁ and G₂. Let x, y ∈ V(G₁) = V(G₂) = V(g) be such that x and y are adjacent in G₁. As G₂ is complete by hypothesis, x and y are adjacent in G₂. Therefore, x and y are adjacent in G₁ ∩ G₂ = g. This proves that G₁ is a spanning subgraph of g and so, g = G₁. □

Corollary 2.7.

If |MNP(R)| = 1, then ZT(R) = Γ(R).

Proof. Note that Γ(R) and $T_{Z {(R)}^{*}} (Γ (R))$ are spanning supergraphs of ZT(R) and $Z T (R) = Γ (R) \cap T_{Z {(R)}^{*}} (Γ (R))$ ⁠. By hypothesis, |MNP(R)| = 1. Let $M N P (R) = {p}$ ⁠. Observe that $Z (R) = p$ ⁠. Let x, y ∈ Z(R)* be such that x ≠ y. Then x + y ∈ Z(R). Hence, x and y are adjacent in $T_{Z {(R)}^{*}} (Γ (R))$ ⁠. This shows that $T_{Z {(R)}^{*}} (Γ (R))$ is complete. Therefore, ZT(R) = Γ(R) by Lemma 2.6. □

Remark 2.8.

Let R be such that |MNP(R)| = 1. In this remark, we give another proof of Corollary 2.7. Let $M N P (R) = {p}$ ⁠. Note that $Z (R) = p$ and $T o t (R) = R_{p}$ is quasi-local with $p R_{p}$ as its unique maximal ideal. Hence, Tot(R) has no non-trivial idempotent [[14], Exercise 12, p.11]. Therefore, ZT(R) = Γ(R) by (2) ⇒ (1) of Theorem 2.5.

Let G = (V, E) be a graph. Let a, b ∈ V with a ≠ b. Suppose that there exists a path in G between a and b. Recall that the distance between a and b denoted by d(a, b) is defined as the length of the shortest path in G between a and b. We set d(a, b) = ∞ if there exists no path in G between a and b. We set d(a, a) = 0 [[10], Definition 1.5.5]. Recall that a graph G = (V, E) is said to be connected if there exists at least one path in G between a and b for each a, b ∈ V with a ≠ b [[10], Definition 1.5.4]. Let G = (V, E) be a connected graph. Then the diameter of G denoted by diam(G) is defined as diam(G) = sup{d(a, b)∣a, b ∈ V} [[10], Definition 4.3.1(1)].

Corollary 2.9.

Let R be such that |MNP(R)| = 1. Then ZT(R) is connected and diam(ZT(R)) ≤ 3.

Proof. For any ring T with Z(T)* ≠ ∅, it is known that Γ(T) is connected and diam(Γ(T)) ≤ 3 [[1], Theorem 2.3]. Note that ZT(R) = Γ(R) by Corollary 2.7. Therefore, ZT(R) is connected and diam(ZT(R)) ≤ 3. □

We denote the polynomial ring in one variable X over R by R[X]. In the following remark, we recall [[4], Theorem 2.6].

Remark 2.10.

Let R be a ring which admits $p$ as the unique member of MNP(R). As ZT(R) = Γ(R), it follows from [[4], Theorem 2.6] that

diam(ZT(R)) = 0 if and only if either $R ≅ Z_{4}$ or $R ≅ \frac{Z_{2} [X]}{X^{2} Z_{2} [X]}$ as rings.
diam(ZT(R)) = 1 if and only if $p^{2} = (0)$ and $| p | \geq 3$ ⁠.
diam(ZT(R)) = 2 if and only if $p^{2} \neq (0)$ and for any x, y ∈ Z(R) with x ≠ y, there exists a non-zero r ∈ R such that rx = ry = 0.
diam(ZT(R)) = 3 if and only if there are x, y ∈ Z(R) with x ≠ y such that ((0):_RRx + Ry) = (0). (Since Z(R) is an ideal of R, in the case R is reduced, it is not hard to verify that R has an infinite number of minimal prime ideals.)

Let G = (V, E) be a connected graph. Let a ∈ V. Recall that the eccentricity of a denoted by e(a) is defined as e(a) = sup{d(a, b)∣b ∈ V} [[10], Definition 4.3.1(2)]. The radius of G denoted by r(G) is defined as r(G) = min{e(a)∣a ∈ V} [[10], Definition 4.3.1(3)]. Let R be such that |MNP(R)| = 1. We next discuss some results regarding r(ZT(R)).

Let G = (V, E) be a simple graph. Recall that the complement of G denoted by G^c is defined by setting V(G^c) = V and distinct vertices a and b are joined by an edge in G^c if and only if there exists no edge joining a and b in G [[10], Definition 1.2.13].

Lemma 2.11.

Let G = (V, E) be a graph such that |V|≥ 2. If both G and G^c are connected, then r(G^c) ≥ 2 and r(G) ≥ 2.

Proof. Assume that G and G^c are connected. Let x ∈ V. Since |V|≥ 2 by hypothesis, there exists v ∈ V such that d(x, v) = 1 in G. Hence, d(x, v) ≥ 2 in G^c. Therefore, e(x) ≥ 2 in G^c and so, r(G^c) ≥ 2. Similarly, it follows that r(G) ≥ 2. □

Lemma 2.12.

Let R be such that $M N P (R) = {p}$ ⁠. Suppose that |Z(R)*|≥ 2. Then the following statements are equivalent:

Proof. Note that ZT(R) = Γ(R) by Corollary 2.7 and so, (ZT(R))^c = (Γ(R))^c. As |Z(R)*|≥ 2 by hypothesis and ZT(R) is connected by Corollary 2.9, it follows that r(ZT(R)) ≥ 1.

(1) ⇒ (2) Assume that r(ZT(R)) = 1. Suppose that $p$ is not a B-prime of (0) in R. Then (Γ(R))^c is connected by [[15], Proposition 1.2(i)]. Thus both ZT(R) and (ZT(R))^c are connected. Hence, r(ZT(R)) ≥ 2 by Lemma 2.11 and this contradicts the assumption r(ZT(R)) = 1. Therefore, $p$ is a B-prime of (0) in R.

(2) ⇒ (1) Assume that $p$ is a B-prime of (0) in R. Hence, there exists $x \in Z {(R)}^{*} = p \ {0}$ such that $p = ((0) :_{R} x)$ ⁠. Let y ∈ Z(R)* be such that y ≠ x. (It is possible to find y ∈ Z(R)* with y ≠ x, since by hypothesis, |Z(R)*|≥ 2.) Note that yx = 0 and so, d(x, y) = 1 in Γ(R) = ZT(R). This shows that e(x) = 1 in ZT(R) and so, r(ZT(R)) = 1. □

(1) r(ZT(R)) = 1.
$(2) p$ is a B-prime of (0) in R.

Lemma 2.13.

Let R be a non-reduced ring. Then r(Γ(R)) ≤ 2.

Proof. By hypothesis, R is not a reduced ring. Let x ∈ nil(R)\{0}. Let y ∈ Z(R)* be such that y ≠ x. If xy = 0, then d(x, y) = 1 in Γ(R). Suppose that xy ≠ 0. As y ∈ Z(R), there exists r ∈ R\{0} such that yr = 0. If xr = 0, then x − r − y is a path of length two between x and y in Γ(R). Suppose that rx ≠ 0. Since x is nilpotent, it is possible to find n ≥ 2 least with the property that rxⁿ = 0. Then rxⁿ⁻¹ ≠ 0. It is clear that x − rxⁿ⁻¹ − y is a path of length 2 between x and y in Γ(R). Thus for any y ∈ Z(R)* with y ≠ x, d(x, y) ≤ 2 in Γ(R). This shows that e(x) ≤ 2 in Γ(R) and so, r(Γ(R)) ≤ 2. □

The following example illustrates Lemmas 2.12 and 2.13.

Example 2.14.

Proof.

Let p be an odd prime and let $R = Z_{p^{2}}$ ⁠. Then diam(ZT(R)) = r(ZT(R)) = 1.
Let $R = Z_{8}$ ⁠. Then diam(ZT(R)) = 2 and r(ZT(R)) = 1.
Let V be a rank one valuation domain that is not discrete and let $m$ be its unique maximal ideal. Let $m \in m \ {0}$ ⁠. Let $R = \frac{V}{V m}$ ⁠. Then diam(ZT(R)) = r(ZT(R)) = 2.

Note that Z(R) = pR is an ideal of R and (pR)² = (0). Observe that pR is the unique member of MNP(R) with (pR)² = (0). Since p is an odd prime, it follows that the p, 2p ∈ Z(R)* are distinct. Hence, we obtain from Remark 2.10 (2) that diam(ZT(R)) = 1 and so, r(ZT(R)) = 1.
Note that MNP(R) = 2R and (2R)² = 4R ≠ (0). It is clear that 2R = ((0):_R4) is a B-prime of (0) in R. It now follows from Remark 2.10(3) that diam(ZT(R)) = 2 and r(ZT(R)) = 1 by (2) ⇒ (1) of Lemma 2.12.
It was already noted in the proof of [[15], Example 3.1(ii)] that $Z (R) = \frac{m}{V m}$ and $\frac{m}{V m}$ is not a B-prime of the zero ideal in R. Let us denote $\frac{m}{V m}$ by $p$ ⁠. Thus $M N P (R) = {p}$ and $p$ is not a B-prime of the zero ideal in R. Hence, $p^{2}$ is not the zero ideal of R. Let x, y ∈ Z(R)*. Since any two ideals of R are comparable under inclusion, it follows that either Rx + Ry = Rx or Rx + Ry = Ry. Hence, there exists r ∈ R\{0 + Vm} such that xr = yr = 0 + Vm. It now follows from Remark 2.10(3) that diam(ZT(R)) = 2. Since $p$ is not a B-prime of the zero ideal in R, we obtain from (1) ⇒ (2) of Lemma 2.12 that r(ZT(R)) ≥ 2. From diam(ZT(R)) = 2, we get that r(ZT(R)) ≤ 2 and so, diam(ZT(R)) = r(ZT(R)) = 2. □

Remark 2.15.

As Γ(R) and $T_{Z {(R)}^{*}} (Γ (R))$ are spanning supergraphs of ZT(R) and $Z T (R) = Γ (R) \cap T_{Z {(R)}^{*}} (Γ (R))$ ⁠, we obtain that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ if and only if $T_{Z {(R)}^{*}} (Γ (R))$ is a subgraph of Γ(R).

If |MNP(R)| = 1, then in the following proposition, we determine necessary and sufficient conditions so that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠.

Proposition 2.16.

If R is such that $M N P (R) = {p}$ ⁠, then the following statements are equivalent:

Proof. Note that $Z (R) = p$ ⁠.

(1) ⇒ (2) Assume that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠. Suppose that $| p | = 2$ ⁠. Let $p = {0, p}$ ⁠. As $1 - p \notin p$ ⁠, it follows that p² ≠ p and so, p² = 0. Hence, $p^{2} = (0)$ ⁠. In the case, $| p | = 2$ ⁠, it is well-known that either $R ≅ Z_{4}$ or $R ≅ \frac{Z_{2} [X]}{X^{2} Z_{2} [X]}$ as rings (see [[1], Example 2.1(a)]). Suppose that $| p | \geq 3$ ⁠. Let x, y ∈ Z(R)* be such that x ≠ y. Then $x + y \in p = Z (R)$ ⁠. Therefore, x and y are adjacent in $T_{Z {(R)}^{*}} (Γ (R))$ ⁠. As $T_{Z {(R)}^{*}} (Γ (R)) = Z T (R)$ by assumption, we obtain from Remark 2.15 that x and y are adjacent in Γ(R). This shows that Γ(R) is complete and so, $p^{2} = (0)$ by [[1], Theorem 2.8].

(2) ⇒ (3) Let x, y ∈ Z(R)* be such that x ≠ y. Then $x + y \in p = Z (R)$ ⁠. Therefore, $T_{Z {(R)}^{*}} (Γ (R))$ is complete. From $p^{2} = (0)$ ⁠, it follows that xy = 0 and so, Γ(R) is complete. Since $T_{Z {(R)}^{*}} (Γ (R))$ and Γ(R) are spanning supergraphs of ZT(R) with $Z T (R) = Γ (R) \cap T_{Z {(R)}^{*}} (Γ (R))$ ⁠, we obtain that $Z T (R) = Γ (R) = T_{Z {(R)}^{*}} (Γ (R))$ is complete.

(3) ⇒ (1) This is clear. □

$(1) Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠.
$(2) p^{2} = (0)$ ⁠.
$(3) Z T (R) = Γ (R) = T_{Z {(R)}^{*}} (Γ (R))$ is complete.

For a ring R with |MNP(R)|≥ 2, the following theorem characterizes R such that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠.

Theorem 2.17.

Let R be such that |MNP(R)|≥ 2. The following statements are equivalent:

Proof. By hypothesis, |MNP(R)|≥ 2. Let $p_{1}$ and $p_{2}$ be distinct members of MNP(R).

(1) ⇒ (2) Assume that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠. Since $p_{1}$ and $p_{2}$ are distinct members of MNP(R), we get that $p_{1} ⊈ p_{2}$ and $p_{2} ⊈ p_{1}$ ⁠. Let $x \in p_{1} \ p_{2}$ and let $y \in p_{2} \ p_{1}$ ⁠. We claim that $p_{1} \ p_{2} = {x}$ and $p_{2} \ p_{1} = {y}$ ⁠. Suppose that there exists $x^{'} \in p_{1} \ p_{2}$ with x′ ≠ x. Then $x + x^{'} \in p_{1} \subset Z (R)$ ⁠. This implies that x and x′ are adjacent in $T_{Z {(R)}^{*}} (Γ (R)) = Z T (R)$ and so, xx′ = 0. This is impossible, since $x x^{'} \in p_{1} \ p_{2}$ ⁠. Hence, we obtain that $p_{1} \ p_{2} = {x}$ and similarly, it follows that $p_{2} \ p_{1} = {y}$ ⁠. Let $z \in ⋂_{i = 1}^{2} p_{i}$ ⁠. Then $x + z \in p_{1} \ p_{2}$ and so, x + z = x. Hence, z = 0 and this shows that $⋂_{i = 1}^{2} p_{i} = (0)$ ⁠. We next verify that $| \frac{R}{p_{i}} | = 2$ for each i ∈ {1, 2}. Let r ∈ R be such that $r \notin p_{1}$ ⁠. Then $r y \in p_{2} \ p_{1}$ and so, ry = y. This implies that $(r - 1) y = 0 \in p_{1}$ and so, $r - 1 \in p_{1}$ ⁠. This shows that $\frac{R}{p_{1}} = {0 + p_{1}, 1 + p_{1}}$ ⁠. Similarly, it can be shown that $\frac{R}{p_{2}} = {0 + p_{2}, 1 + p_{2}}$ ⁠. This proves that $| \frac{R}{p_{i}} | = 2$ for each i ∈ {1, 2}. Hence, $p_{i} \in M a x (R)$ for each i ∈ {1, 2}. Now, $p_{1} + p_{2} = R$ and $⋂_{i = 1}^{2} p_{i} = (0)$ and so, we obtain from [[14], Proposition 1.10(ii) and (iii)] that $R ≅ \prod_{i = 1}^{2} \frac{R}{p_{i}}$ as rings. Since $| \frac{R}{p_{i}} | = 2$ for each i ∈ {1, 2}, we obtain that $R ≅ Z_{2} \times Z_{2}$ as rings.

(2) ⇒ (1) Assume that $R ≅ Z_{2} \times Z_{2}$ as rings. Let us denote $Z_{2} \times Z_{2}$ by T. Note that Z(T)* = {(0, 1), (1, 0)}. Observe that $T_{Z {(T)}^{*}} (Γ (T))$ has no edges and so, $T_{Z {(T)}^{*}} (Γ (T))$ is a subgraph of Γ(T). Hence, we obtain from Remark 2.15 that $Z T (T) = T_{Z {(T)}^{*}} (Γ (T))$ ⁠. Since R ≅ T as rings, we get that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠. □

$(1) Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠.
$(2) R ≅ Z_{2} \times Z_{2}$ as rings.

The following corollary characterizes R such that $Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠.

Corollary 2.18.

The following statements are equivalent:

Proof. The proof of this corollary follows immediately from Proposition 2.16 and Theorem 2.17. □

$(1) Z T (R) = T_{Z {(R)}^{*}} (Γ (R))$ ⁠.
(2) Either |MNP(R)| = 1 in which case, $p^{2} = (0)$ ⁠, where $M N P (R) = {p}$ or |MNP(R)|≥ 2 in which case, $R ≅ Z_{2} \times Z_{2}$ as rings.

3. On the connectedness of ZT(R), where 2 ≤ |MNP(R)| < ∞

Let R be a ring such that 2 ≤ |MNP(R)| < ∞. Let |MNP(R)| = n. Note that $n \in N$ with n ≥ 2. Let $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. This section aims to determine a necessary and sufficient condition so that ZT(R) is connected and to determine the diameter of ZT(R) in the case when it is connected. In Corollary 3.3, we provide a necessary condition so that ZT(R) is connected. We use the following lemma and Corollary 3.2 in its proof.

For any ring R and a ∈ R, we denote Spec(R)\V(a) by D(a).

Lemma 3.1.

Let R be a ring. Let a ∈ Z(R)* be such that MNP(R)⊈V(a). If ZT(R) is connected, then there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅.

Proof. Assume that ZT(R) is connected and a ∈ Z(R)* with MNP(R)⊈V(a). As a ∈ Z(R)*, there exists $p \in M N P (R)$ such that $a \in p$ ⁠. By assumption, MNP(R)⊈V(a). Hence, there exists $p^{'} \in M N P (R)$ such that $a \notin p^{'}$ ⁠. Let $b \in p^{'} \ {0}$ ⁠. Then b ∈ Z(R)* and b ≠ a. Thus |Z(R)*|≥ 2. Let us denote V(a) ∩ MNP(R) by A and D(a) ∩ MNP(R) by B. Observe that $p \in A$ and $p^{'} \in B$ and so, A and B are non-empty. As V(ZT(R)) = Z(R)*, |Z(R)*|≥ 2, and ZT(R) is connected by assumption, there exists x ∈ Z(R)*\{a} such that a and x are adjacent in ZT(R). Hence, ax = 0 and a + x ∈ Z(R). From ax = 0, it follows that x belongs to each member of B. If $M N P (R) = {p_{α}}_{α \in Λ}$ ⁠, then $Z (R) = ⋃_{α \in Λ} p_{α}$ ⁠. It is clear that MNP(R) = A ∪ B. From a + x ∈ Z(R), it follows that $a + x \in p_{α}$ for some α ∈ Λ. Since x is in every member of B and a is not in any member of B, we get that $p_{α} \in A$ ⁠. As $a \in p_{α}$ ⁠, we obtain that $x \in p_{α}$ ⁠. This proves that given a ∈ Z(R)* with MNP(R)⊈V(a), there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅. □

Corollary 3.2.

Let R be such that |MNP(R)|≥ 2. Let a ∈ Z(R)* be such that $V (a) \cap M N P (R) = {p}$ ⁠. If ZT(R) is connected, then there exists x ∈ Z(R)*\{a} such that ax = 0 and x belongs to each member of MNP(R).

Proof. By hypothesis, |MNP(R)|≥ 2. Assume a ∈ Z(R)* is such that $V (a) \cap M N P (R) = {p}$ and ZT(R) is connected. Therefore, MNP(R)⊈V(a). Since ZT(R) is connected by assumption, there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅ by Lemma 3.1. Hence, $x \in p$ ⁠. From ax = 0, it follows that $x \in q$ for each $q \in D (a) \cap M N P (R)$ ⁠. From $M N P (R) = {p} \cup (D (a) \cap M N P (R))$ ⁠, we get that x belongs to each member of MNP(R). □

Corollary 3.3.

Let $n \in N \ {1}$ ⁠. Let $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. If ZT(R) is connected, then $⋂_{i = 1}^{n} p_{i} \neq (0)$ ⁠.

Proof. As $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠, it follows that $Z (R) = ⋃_{i = 1}^{n} p_{i}$ ⁠. Since distinct members of MNP(R) are not comparable under inclusion, $p_{1} ⊈ ⋃_{j = 2}^{n} p_{j}$ by [[14], Proposition 1.11(i)]. Hence, there exists $a \in p_{1} \ (⋃_{j = 2}^{n} p_{j})$ and so, $V (a) \cap M N P (R) = {p_{1}}$ ⁠. As |MNP(R)|≥ 2 by hypothesis and ZT(R) is connected by assumption, it follows from Corollary 3.2 that there exists x ∈ Z(R)* such that ax = 0 and $x \in ⋂_{i = 1}^{n} p_{i}$ ⁠. This shows that $⋂_{i = 1}^{n} p_{i} \neq (0)$ ⁠. □

For a ring R with 2 ≤ |MNP(R)| < ∞, in Theorem 3.8, we provide a necessary and sufficient condition so that ZT(R) is connected. Moreover, if ZT(R) is connected, then we determine diam(ZT(R)). Lemmas 3.4 to 3.7 are needed for its proof.

Lemma 3.4.

Let R be such that |MNP(R)|≥ 2 and $⋂_{p \in M N P (R)} p \neq (0)$ ⁠. Let $a \in (⋂_{p \in M N P (R)} p) \ {0}$ ⁠. Then for any x ∈ Z(R)*, a + x ∈ Z(R).

Proof. Let $p \in M N P (R)$ be such that $x \in p$ ⁠. Since a is in each member of MNP(R), we get that $a + x \in p \subset Z (R)$ ⁠. Hence, a + x ∈ Z(R). □

Lemma 3.5.

Let R be such that |MNP(R)|≥ 2 and $⋂_{p \in M N P (R)} p \neq (0)$ ⁠. Let $a, b \in (⋂_{p \in M N P (R)} p) \ {0}$ be distinct. Then there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R).

Proof. It is clear that a, b ∈ Z(R)*. Note that $a + b \in ⋂_{p \in M N P (R)} p$ and so, a + b ∈ Z(R). If ab = 0, then a and b are adjacent in ZT(R). Hence, d(a, b) = 1 in ZT(R). Suppose that ab ≠ 0. It follows that either d(a, b) = 2 in Γ(R) or d(a, b) = 3 in Γ(R), since Γ(R) is connected and diam(Γ(R)) ≤ 3 by [[1], Theorem 2.3]. Suppose that d(a, b) = 2 in Γ(R). Let x ∈ Z(R)* be such that a − x − b is a path in Γ(R) between a and b. As both a and b belong to each member of MNP(R), we obtain from Lemma 3.4 that a + x, b + x ∈ Z(R). Since ax = 0 = bx, we obtain that a − x − b is a path in ZT(R) between a and b and so, d(a, b) = 2 in ZT(R). Suppose that d(a, b) = 3 in Γ(R). Let x, y ∈ Z(R)* be such that a − x − y − b is a path in Γ(R) between a and b. Hence, ax = xy = by = 0. As both a and b belong to each member of MNP(R), we obtain from Lemma 3.4 that a + x, y + b ∈ Z(R). Observe that ab ≠ 0 and (x + y)ab = 0 and so, x + y ∈ Z(R). This shows that a − x − y − b is a path in ZT(R) between a and b. Since d(a, b) = 3 in Γ(R) and Γ(R) is a spanning supergraph of ZT(R), we obtain that d(a, b) = 3 in ZT(R). This proves that there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R). □

Lemma 3.6.

Let R be such that |MNP(R)|≥ 2 and $⋂_{p \in M N P (R)} p \neq (0)$ ⁠. Let a ∈ Z(R)* be such that $a \notin ⋂_{p \in M N P (R)} p$ ⁠. Suppose that there exists x ∈ Z(R)*\{a} with ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅. Then for any $b \in (⋂_{p \in M N P (R)} p) \ {0}$ ⁠, there exists a path in ZT(R) between a and b such that 1 ≤ d(a, b) ≤ 3 in ZT(R).

Proof. It is clear from the given hypotheses that a and x are adjacent in ZT(R) and hence, we can assume that b ≠ x. As b belongs to each member of MNP(R), we obtain from Lemma 3.4 that a + b, x + b ∈ Z(R). If ab = 0, then a and b are adjacent in ZT(R) and so, d(a, b) = 1 in ZT(R). Hence, we can assume that ab ≠ 0. If bx = 0, then a − x − b is a path of length two between a and b in ZT(R) and so, d(a, b) = 2 in ZT(R). Hence, we can assume that bx ≠ 0. Note that b ∈ Z(R)*. As |Z(R)*|≥ 2, there exists y ∈ Z(R)*\{b} such that by = 0 by [[1], Theorem 2.3]. It is clear that y∉{a, x}. As (a + xy)bx = 0, it follows that a + xy ∈ Z(R), and from Lemma 3.4, we get that xy + b ∈ Z(R). Therefore, if xy ≠ 0, then a − xy − b is a path in ZT(R) between a and b and so, d(a, b) = 2 in ZT(R). Suppose that xy = 0. From (x + y)ab = 0, it follows that x + y ∈ Z(R). Since a + x, x + y, b + y ∈ Z(R), we get that a − x − y − b is a path of length three in ZT(R) between a and b. This proves that there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R). □

Lemma 3.7.

Let R be such that |MNP(R)|≥ 2. Let a, b ∈ Z(R)* be distinct. Suppose there exist x, y ∈ Z(R)* such that a ≠ x, ax = 0, V(a) ∩ V(x) ∩ MNP(R) ≠ ∅ (respectively, b ≠ y, by = 0, V(b) ∩ V(y) ∩ MNP(R) ≠ ∅). Then there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R).

Proof. Let $p \in V (a) \cap V (x) \cap M N P (R)$ and let $p^{'} \in V (b) \cap V (y) \cap M N P (R)$ ⁠. Observe that $a + x y, a + b x \in p \subset Z (R)$ and $b + x y, b + a y \in p^{'} \subset Z (R)$ ⁠. We consider the following cases.

If xy ≠ 0, then from ax = 0 = by, it follows that a − xy − b is a path in ZT(R) between a and b and so, d(a, b) = 2 in ZT(R). Suppose that xy = 0. As a + b ∉ Z(R) and ax = by = 0, we get that bx ≠ 0 and ay ≠ 0. Observe that $b x + a y \in p \subset Z (R)$ ⁠. If bx = ay, then we obtain that a − bx = ay − b is a path in ZT(R) between a and b and so, d(a, b) = 2 in ZT(R). Suppose that bx ≠ ay. It is clear that bx, ay∉{a, b}. Observe that a + bx, b + ay ∈ Z(R), and a − bx − ay − b is a path in ZT(R) between a and b.

If ab = 0, then a and b are adjacent in ZT(R) and so, d(a, b) = 1 in ZT(R). Suppose that ab ≠ 0. If xy ≠ 0, then a − xy − b is a path in ZT(R) between a and b and so, d(a, b) = 2 in ZT(R). Suppose that xy = 0. From (x + y)ab = 0, it follows that x + y ∈ Z(R). If x = y, then a − x − b is a path in ZT(R) between a and b, and hence, d(a, b) = 2 in ZT(R). Hence, we can assume that x ≠ y. It is clear that x, y∉{a, b}. Observe that a − x − y − b is a path in ZT(R) between a and b.

Case(1). a + b ∉ Z(R).

Case(2). a + b ∈ Z(R).

It is clear from the above discussion that there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R). □

Theorem 3.8.

Let R be such that 2 ≤ n = |MNP(R)| < ∞. Let $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. The following statements are equivalent:

Moreover, if (2) holds, then diam(ZT(R)) = 3 and if R is not reduced, then r(ZT(R)) = 2.

Proof. (1) ⇒ (2) Assume that ZT(R) is connected. Hence, $⋂_{i = 1}^{n} p_{i} \neq (0)$ by Corollary 3.3. Let a ∈ Z(R)* be such that MNP(R)⊈V(a). It follows from Lemma 3.1 that there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅.

(2) ⇒ (1) Assume that the statement (2) holds. Let a, b ∈ Z(R)* be distinct. We consider the following cases.

In this case, by Lemma 3.5, there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R).

Without loss of generality, we can assume that $b \in ⋂_{i = 1}^{n} p_{i}$ ⁠. In this case, by Lemma 3.6, there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R).

In this case, by Lemma 3.7, there exists a path in ZT(R) between a and b with 1 ≤ d(a, b) ≤ 3 in ZT(R).

(1) ZT(R) is connected.
$(2) ⋂_{i = 1}^{n} p_{i} \neq (0)$ and for given a ∈ Z(R)* with MNP(R)⊈V(a), there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅.

Case(1). $a, b \in ⋂_{i = 1}^{n} p_{i}$ ⁠.

Case(2). Exactly one between a and b belongs to $⋂_{i = 1}^{n} p_{i}$ ⁠.

Case(3). $a, b \notin ⋂_{i = 1}^{n} p_{i}$ ⁠.

It is clear from the above discussion that ZT(R) is connected and 1 ≤ diam(ZT(R)) ≤ 3.

Assume that (2) holds. As $p_{1}$ and $p_{2}$ are not comparable under inclusion, it follows from Zorn’s lemma and [[12], Theorem 1] that $p_{1} + p_{2} ⊈ Z (R)$ ⁠. Hence, there exist $a \in p_{1}, b \in p_{2}$ such that a + b ∉ Z(R). Observe that a, b ∈ Z(R)* and a ≠ b. As a + b ∉ Z(R), it follows that d(a, b) ≥ 2 in ZT(R). We claim that d(a, b) ≠ 2 in ZT(R). If d(a, b) = 2 in ZT(R), then there exists z ∈ Z(R)* such that a − z − b is a path of length two in ZT(R). This implies that az = bz = 0 and so, (a + b)z = 0. This is impossible, since a + b ∉ Z(R). Therefore, d(a, b) ≥ 3 in ZT(R). It is already verified in the proof of (2) ⇒ (1) of this theorem that diam(ZT(R)) ≤ 3. Hence, d(a, b) = 3 in ZT(R) and so, diam(ZT(R)) = 3.

Assume that (2) holds and R is not reduced. It is shown in the previous paragraph that diam(ZT(R)) = 3. Hence, e(z) ≥ 2 in ZT(R) for each z ∈ Z(R)* and so, r(ZT(R)) ≥ 2. By assumption, R is not reduced. Hence, it is possible to find x ∈ nil(R)\{0} such that x² = 0. Let a ∈ Z(R)*\{x}. Note that x + a ∈ Z(R) by [[4], Lemma 2.3]. Thus if ax = 0, then d(x, a) = 1 in ZT(R). Suppose that ax ≠ 0. Since ZT(R) is connected and |Z(R)*|≥ 2, there exists b ∈ Z(R)*\{a} such that a and b are adjacent in ZT(R). Since x ∈ nil(R), x + b ∈ Z(R) by [[4], Lemma 2.3]. If bx = 0, then x − b − a is a path of length two between x and a in ZT(R). Suppose that bx ≠ 0. As x² = 0, it follows that x − bx − a is a path of length two in ZT(R) between x and a. This shows that d(x, a) ≤ 2 in ZT(R) for each a ∈ Z(R)* with a ≠ x. Hence, e(x) ≤ 2 in ZT(R) and so, e(x) = 2 in ZT(R). This proves that r(ZT(R)) ≤ 2 and therefore, r(ZT(R)) = 2. □

Let R be a ring. Recall that R is said to be Laskerian (respectively, strongly Laskerian) if each ideal I of R with I ≠ R admits a primary decomposition (respectively, strong primary decomposition) [16]. It is well-known that any Noetherian ring is strongly Laskerian (see [[14], Theorem 7.13 and Proposition 7.14]). In Ref. [16], Heinzer and Lantz proved that Laskerian rings possess several important properties of Noetherian rings and many non-trivial examples of non-Noetherian Laskerian rings are known in the literature (see for example [16–18]). As any Artinian ring is Noetherian by [[14], Theorem 8.5], it follows that any Artinian ring is strongly Laskerian.

Let R be such that (0) admits a strong primary decomposition. Let $(0) = ⋂_{i = 1}^{t} q_{i}$ be an irredundant strong primary decomposition of (0) in R. For each i ∈ {1, …, t}, let $q_{i}$ be a $p_{i}$ -primary ideal of R with $p_{i}^{m_{i}} \subseteq q_{i}$ for some $m_{i} \in N$ ⁠. Let $n \in N$ with n ≤ t be such that there are exactly n maximal members among $p_{1}, \dots, p_{t}$ ⁠. After a suitable rearrangement of 1, …, t, we can assume without loss of generality that $p_{1}, \dots, p_{n}$ are maximal members among $p_{1}, \dots, p_{t}$ ⁠. As $Z (R) = ⋃_{i = 1}^{t} p_{i}$ by [[14], Proposition 4.7], it follows that $Z (R) = = ⋃_{j = 1}^{n} p_{j}$ ⁠. Observe that $M N P (R) = {p_{j} ∣ j \in {1, \dots, n}}$ ⁠. Suppose that |MNP(R)|≥ 2. Hence, n ≥ 2. Let i ∈ {1, 2, …, t}. Since $q_{i}$ is a strongly primary ideal of R, proceeding as in the proof of [[14], Proposition 7.17], it can be shown that there exists x_i ∈ R\{0} such that $p_{i} = ((0) :_{R} x_{i})$ ⁠. Theorem 3.9 provided below is inspired by [[9], Theorem 3.1] and it generalizes [[9], Theorem 3.1].

Let R be a ring. If S is a non-empty subset of R, then the annihilator of S in R denoted by Ann(S) is defined as Ann(S) = {r ∈ R∣rS = (0)}. Note that Ann(S) is an ideal of R.

Theorem 3.9.

Let R be a ring such that |MNP(R)|≥ 2. Suppose that (0) admits a strong primary decomposition. Let $(0) = ⋂_{i = 1}^{t} q_{i}$ be an irredundant strong primary decomposition of (0) in R and let $p_{i}$ (i ∈ {1, 2, …, t}), $p_{j}$ (j ∈ {1, 2, …, n}), and x_i (i ∈ {1, 2, …, t}) be as described in the paragraph which appears just preceding the statement of this theorem. The following statements are equivalent:

Moreover, if (3) holds, then diam(ZT(R)) = 3 and r(ZT(R)) = 2.

Proof. (1) ⇒ (2) Assume that ZT(R) is connected. We first verify that $x_{1} \in ⋂_{i = 1}^{t} p_{i}$ ⁠. Since $p_{1} x_{1} = (0)$ and $p_{1} ⊈ p_{k}$ for each k ∈ {2, …, t}, it follows that $x_{1} \in ⋂_{k = 2}^{t} p_{k}$ ⁠. Observe that $p_{1} ⊈ ⋃_{k = 2}^{t} p_{k}$ by [[14], Proposition 1.11(i)]. Let $a \in p_{1} \ (⋃_{k = 2}^{t} p_{k})$ ⁠. It is clear that MNP(R)⊈V(a). Since ZT(R) is connected by assumption, we obtain from Lemma 3.1 that there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅. It is clear from the choice of a that $x \in p_{1}$ ⁠. As x ≠ 0, it follows that $x \notin q_{i}$ for some i ∈ {1, 2, …, t}. From $a x = 0 \in q_{i}$ ⁠, we get that $a \in p_{i}$ ⁠. From the choice of a, it follows that i = 1. Therefore, $x \notin q_{1}$ ⁠. As $p_{1} = ((0) :_{R} x_{1})$ and $x \in p_{1}$ ⁠, we get that $x x_{1} = 0 \in q_{1}$ ⁠. Since $x \notin q_{1}$ ⁠, it follows that $x_{1} \in p_{1}$ ⁠. This shows that $x_{1} \in ⋂_{i = 1}^{t} p_{i}$ ⁠. Similarly, it can be shown that $x_{j} \in ⋂_{i = 1}^{t} p_{i}$ for each j ∈ {2, …, n}.

(2) ⇒ (3) Let j ∈ {1, 2, …, n}. As $x_{j} \in p_{j}$ and $p_{j} = ((0) :_{R} x_{j})$ ⁠, it follows that $x_{j} \in p_{j} \cap A n n (p_{j})$ ⁠. Therefore, $p_{j} \cap A n n (p_{j}) \neq (0)$ ⁠.

(3) ⇒ (1) Assume that $p_{j} \cap A n n (p_{j}) \neq (0)$ for each j ∈ {1, 2, …, n}. Hence, for each j ∈ {1, 2, …, n}, we can choose y_j ∈ R\{0} such that $y_{j} \in p_{j} \cap A n n (p_{j})$ ⁠. Let j ∈ {1, 2, …, n}. Note that $y_{j} \in p_{j}$ and $p_{j} y_{j} = (0)$ ⁠. As $p_{j} ⊈ ⋃_{k \in {1,2, \dots, t} \ {j}} p_{k}$ ⁠, we get that $y_{j} \in ⋂_{k \in {1,2, \dots, t} \ {j}} p_{k}$ ⁠. Therefore, $y_{j} \in ⋂_{i = 1}^{t} p_{i}$ ⁠. Hence, $⋂_{j = 1}^{n} p_{j} \neq (0)$ ⁠. Let a ∈ Z(R)* be such that MNP(R)⊈V(a). Note that $a \in p_{j}$ for some j ∈ {1, 2, …, n}. As $p_{j} y_{j} = (0)$ ⁠, it follows that ay_j = 0. As $y_{j} \in ⋂_{i = 1}^{t} p_{i}$ but $a \notin p_{k}$ for at least one k ∈ {1, 2, …, n}, it follows that a ≠ y_j. Note that $p_{j} \in V (a) \cap V (y_{j})$ and so, V(a) ∩ V(y_j) ∩ MNP(R) ≠ ∅. Hence, the statement (2) of Theorem 3.8 holds. Therefore, ZT(R) is connected by (2) ⇒ (1) of Theorem 3.8.

(1) ZT(R) is connected.
$(2) x_{j} \in ⋂_{i = 1}^{t} p_{i}$ for each j ∈ {1, 2, …, n}.
$(3) p_{j} \cap A n n (p_{j}) \neq (0)$ for each j ∈ {1, 2, …, n}.

Assume that the statement (3) of this theorem holds. Then it is noted in the proof of (3) ⇒ (1) of this theorem that the statement (2) of Theorem 3.8 holds. Hence, we obtain from the moreover part of Theorem 3.8 that diam(ZT(R)) = 3. Therefore, e(a) ≥ 2 in ZT(R) for each a ∈ Z(R)* and so, r(ZT(R)) ≥ 2. Let j ∈ {1, 2, …, n}. Let y_j be as in the proof of (3) ⇒ (1) of this theorem. As $p_{j} y_{j} = (0)$ and $y_{j} \in p_{j}$ ⁠, we get that y_j ≠ 0 and $y_{j}^{2} = 0$ ⁠. Hence, R is not reduced and we obtain from the moreover part of Theorem 3.8 that r(ZT(R)) = 2. Indeed, we know from the proof of the moreover part of Theorem 3.8 that e(y_j) = 2 in ZT(R). □

Remark 3.10.

In this remark, we give another proof of (2) ⇒ (1) of the previous theorem. Let a, b ∈ Z(R)* be distinct. We claim that there exists a path in ZT(R) between a and b. First, we show that if either a or b belongs to {x_j∣j ∈ {1, 2, …, n}}, then there is a path in ZT(R) between a and b. Let j ∈ {1, 2, …, n}. Let z ∈ Z(R)*\{x_j}. We claim that there exists a path of length at most two between z and x_j in ZT(R). As $x_{j}^{2} = 0$ ⁠, it follows that z + x_j ∈ Z(R) by [[4], Lemma 2.3]. Suppose that z and x_j are not adjacent in ZT(R). Then zx_j ≠ 0 and so, $z \in p_{i}$ for some i ∈ {1, 2, …, n}\{j}. Hence, zx_i = 0. Since x_ix_j = 0, it follows that z ≠ x_i. From $x_{i}^{2} = 0$ ⁠, we get that z + x_i ∈ Z(R) by [[4], Lemma 2.3]. From x_ix_j = 0 and x_i + x_j ∈ Z(R), it follows that z − x_i − x_j is a path between z and x_j in ZT(R). Hence, we can assume that a, b∉{x_j∣j ∈ {1, 2, …, n}}. Assume that a and b are not adjacent in ZT(R). Observe that $a \in p_{j}, b \in p_{k}$ for some j, k ∈ {1, 2, …, n}. If j = k, then ax_j = bx_j = 0 and $a + x_{j}, b + x_{j} \in p_{j} \subset Z (R)$ ⁠. Hence, a − x_j − b is a path of length two between a and b in ZT(R). Suppose that j ≠ k. Then ax_j = bx_k = 0, $a + x_{j} \in p_{j} \subset Z (R), b + x_{k} \in p_{k} \subset Z (R), x_{j} x_{k} = 0$ ⁠, and $x_{j} + x_{k} \in ⋂_{i = 1}^{t} p_{i} \subset Z (R)$ ⁠. Therefore, a − x_j − x_k − b is a path of length three between a and b in ZT(R). This shows that ZT(R) is connected and diam(ZT(R)) ≤ 3. It is possible to find $a \in p_{1}, b \in p_{2}$ such that a + b ∉ Z(R). Then it is already observed in the proof of the moreover part of Theorem 3.8 that d(a, b) ≥ 3 in ZT(R) and so, diam(ZT(R)) = 3.

For a reduced ring R with 2 ≤ |Min(R)| < ∞, we deduce in the following corollary that ZT(R) is not connected.

Corollary 3.11.

Let $n \in N \ {1}$ ⁠. If R is a reduced ring with |Min(R)| = n, then ZT(R) is not connected.

Proof. Let $M i n (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. As nil(R) = (0), it follows from [[14], Proposition 1.8] and [[12], Theorem 10] that $(0) = ⋂_{i = 1}^{n} p_{i}$ ⁠. Observe that $(0) = ⋂_{i = 1}^{n} p_{i}$ is an irredundant strong primary decomposition of (0) in R and $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. Since $⋂_{i = 1}^{n} p_{i} = (0)$ ⁠, it follows from (1) ⇒ (2) of Theorem 3.8 that ZT(R) is not connected. □

Let R be a ring such that the zero ideal of R admits a strong primary decomposition. Let |MNP(R)|≥ 2. Let $t, n, q_{i}, p_{i}, x_{i}$ (for i ∈ {1, 2, …, t}) be as in the statement of Theorem 3.9. If ZT(R) is connected, then it is shown in the proof of (1) ⇒ (2) of Theorem 3.9 that $x_{j} \in ⋂_{i = 1}^{t} p_{i}$ for each j ∈ {1, 2, …, n} with the assumption that $M N P (R) = {p_{j} ∣ j \in {1,2, \dots, n}}$ ⁠. In the following example, we provide a ring R such that the zero ideal of R admits a strong primary decomposition, ${P_{i} ∣ i \in {1,2,3,4}}$ is the set of prime ideals of R belonging to the zero ideal of R, $M N P (R) = {P_{1}, P_{2}}$ ⁠, ZT(R) is connected, but there exists no $z_{j} \in ⋂_{i = 1}^{4} P_{i}$ such that $P_{j} = ((0) :_{R} z_{j})$ for each j ∈ {3, 4}.

Example 3.12.

Let S = K[X, Y] be the polynomial ring in two variables X, Y over a field K. Let I = SX² + SXY. Let $T = \frac{S}{I}$ and let R = T × T. Then (0 + I) × (0 + I), the zero ideal of R admits a strong primary decomposition with the set of prime ideals of R belonging to the zero ideal of R equals ${P_{i} ∣ i \in {1,2,3,4}}$ ⁠, $M N P (R) = {P_{1}, P_{2}}$ and for each j ∈ {3, 4}, there exists no $z_{j} \in ⋂_{i = 1}^{4} P_{i}$ such that $P_{j} = ((0 + I) \times (0 + I) :_{R} z_{j})$ ⁠.

Proof. It is clear that I = (SX² + SXY + SY²) ∩ SX is an irredundant strong primary decomposition of I in S with SX² + SXY + SY² is a SX + SY-primary ideal of S and SX ∈ Spec(S). Hence, $(0 + I) = \frac{S X^{2} + S X Y + S Y^{2}}{I} \cap \frac{S X}{I}$ is an irredundant strong primary decomposition of the zero ideal in T with $\frac{S X^{2} + S X Y + S Y^{2}}{I}$ is a $\frac{S X + S Y}{I}$ -primary ideal of T and $\frac{S X}{I} \in S p e c (T)$ ⁠. We denote $\frac{S X + S Y}{I}$ by $m$ and $\frac{S X}{I}$ by $p$ ⁠. Observe that $m \in M a x (T)$ ⁠, $p \subset m$ ⁠, and $Z (T) = m$ by [[14], Proposition 4.7]. Note that (0 + I) × (0 + I) is the zero ideal of R. Let us denote $\frac{S X^{2} + S X Y + S Y^{2}}{I}$ by $q$ ⁠. Hence, $(0 + I) = q \cap p$ is an irredundant strong primary decomposition of (0 + I) in T. From (0 + I) × (0 + I) = ((0 + I) × T) ∩ (T × (0 + I)), it follows that $(0 + I) \times (0 + I) = (q \times T) \cap (p \times T) \cap (T \times q) \cap (T \times p)$ ⁠. Let us denote $m \times T$ by $P_{1}$ ⁠, $T \times m$ by $P_{2}$ ⁠, $p \times T$ by $P_{3}$ ⁠, and $T \times p$ by $P_{4}$ ⁠. Note that $P_{i} \in M a x (R)$ for each i ∈ {1, 2}, $P_{j} \in S p e c (R)$ for each j ∈ {3, 4}, $q \times T$ is a $P_{1}$ -primary ideal of R, and $T \times q$ is a $P_{2}$ -primary ideal of R. It is clear that $(0 + I) \times (0 + I) = (q \times T) \cap (T \times q) \cap (⋂_{j = 3}^{4} P_{j})$ is an irredundant strong primary decomposition of the zero ideal in R. Moreover, $P_{3} \subset P_{1}$ and $P_{4} \subset P_{2}$ ⁠. By [[14], Theorem 4.5], it follows that ${P_{i} ∣ i \in {1,2,3,4}}$ is the set of prime ideals belonging to the zero ideal of R and it is clear that $M N P (R) = {P_{i} ∣ i \in {1,2}}$ ⁠. From $m = ((0 + I) :_{T} X + I)$ and $p = ((0 + I) :_{T} X + Y + I)$ ⁠, it follows that $P_{1} = ((0 + I) \times (0 + I) :_{R} (X + I, 0 + I))$ ⁠, $P_{2} = ((0 + I) \times (0 + I) :_{R} (0 + I, X + I))$ ⁠, $P_{3} = ((0 + I) \times (0 + I) :_{R} (X + Y + I, 0 + I))$ ⁠, and $P_{4} = ((0 + I) \times (0 + I) :_{R} (0 + I, X + Y + I))$ ⁠. As $P_{3} \subset P_{1}$ and $P_{4} \subset P_{2}$ ⁠, we get that $⋂_{i = 1}^{4} P_{i} = P_{3} \cap P_{4}$ ⁠. Note that $(X + I, 0 + I), (0 + I, X + I) \in ⋂_{j = 3}^{4} P_{j}$ ⁠. Hence, ZT(R) is connected by (2) ⇒ (1) of Theorem 3.9. We claim that there exists no $z_{j} \in ⋂_{i = 1}^{4} P_{i}$ such that $P_{j} = ((0 + I) \times (0 + I) :_{R} z_{j})$ for each j ∈ {3, 4}. Observe that $⋂_{i = 1}^{4} P_{i} = ⋂_{j = 3}^{4} P_{j} = p \times p$ ⁠. Suppose that there exists $z_{3} \in ⋂_{j = 3}^{4} P_{j}$ such that $P_{3} = ((0 + I) \times (0 + I) :_{R} z_{3})$ ⁠. Then z₃ ≠ (0 + I, 0 + I) and thus there exists s₃ ∈ SX\I such that $p = ((0 + I) :_{T} s_{3} + I)$ ⁠. This implies that SX = (I:_Ss₃). As XY ∈ I, it follows that Y ∈ (I:_Ts₃). This is impossible, since Y ∉ SX. Therefore, there exists no $z_{3} \in ⋂_{j = 3}^{4} P_{j}$ such that $P_{3} = ((0 + I) \times (0 + I) :_{R} z_{3})$ ⁠. Similarly, it can be shown that there exists no $z_{4} \in ⋂_{j = 3}^{4} P_{j}$ such that $P_{4} = ((0 + I) \times (0 + I) :_{R} z_{4})$ ⁠. □

Let R be an Artinian ring. Then there exists $n \in N$ such that $R ≅ \prod_{i = 1}^{n} R_{i}$ as rings, where R_i is an Artinian local ring for each i ∈ {1, …, n} by [[14], Theorem 8.7]. Ðurić et al. determined a necessary and sufficient condition so that ZT(R) is connected (see [[9], Theorem 3.1]).

Let $n \in N \ {1}$ ⁠. Let R_i be a ring for each i ∈ {1, 2, …, n} and let R = R₁ × R₂ × . . . × R_n be their direct product. Motivated by [[9], Theorem 3.1], we want to determine a necessary and sufficient condition so that ZT(R) is connected. In Corollary 3.14, we prove that R_i is not an integral domain for each i ∈ {1, 2, …, n} is a necessary condition for ZT(R) to be connected. In Proposition 3.15, we prove that this necessary condition is also sufficient if Z(R_i) is an ideal of R_i for each i ∈ {1, 2, …, n}. We use the following lemma in the proof of Corollary 3.14.

Lemma 3.13.

Let S, T be rings, and let R = S × T be their direct product. If ZT(R) is connected, then S and T are not integral domains.

Proof. Assume that ZT(R) is connected. Let e₁ = (1, 0) and let e₂ = (0, 1). Note that e₁, e₂ ∈ Z(R)* and e₁ + e₂ = (1, 1) ∉ Z(R). Hence, e₁ and e₂ are not adjacent in ZT(R). As ZT(R) is connected by assumption, there exists (s, t) ∈ Z(R)* such that e₂ and (s, t) are adjacent in ZT(R). Therefore, e₂(s, t) = (0, 0) and e₂ + (s, t) ∈ Z(R). From e₂(s, t) = (0, 0), we get that t = 0 and so, s ≠ 0. Note that (s, 1) = (s, 0) + e₂ ∈ Z(R). Hence, s ∈ Z(S)*. This shows that S is not an integral domain. Similarly, there exists (s′, t′) ∈ Z(R)* such that e₁ and (s′, t′) are adjacent in ZT(R) and this implies that s′ = 0 and t′ ∈ Z(T)*. Therefore, T is not an integral domain. □

Corollary 3.14.

Let $n \in N \ {1}$ ⁠. Let R_i be a ring for each i ∈ {1, 2, …, n} and let R = R₁ × R₂ × . . . × R_n be their direct product. If ZT(R) is connected, then R_i is not an integral domain for each i ∈ {1, 2, …, n}.

Proof. Assume that ZT(R) is connected. Let i ∈ {1, 2, …, n}. Let us denote the ring ∏_{j∈{1,2,…,n}\{i}}R_j by T_i. Then R ≅ R_i × T_i as rings. Hence, ZT(R_i × T_i) is connected. Therefore, R_i is not an integral domain by Lemma 3.13. □

Proposition 3.15.

Let $n \in N \ {1}$ ⁠. Let i ∈ {1, 2, …, n} and let R_i be a ring such that Z(R_i) is an ideal of R_i. Let R = R₁ × R₂ × . . . × R_n be their direct product. Then the following statements are equivalent:

Moreover, if (1) holds, then diam(ZT(R)) = 3 and if R is not reduced, then r(ZT(R)) = 2.

Proof. Let i ∈ {1, 2, …, n}. It is convenient to denote Z(R_i) by $p_{i}$ ⁠. Let us denote the ideal I₁ × I₂× . . . × I_n of R with $I_{i} = p_{i}$ and I_j = R_j for each j ∈ {1, 2, …, n}\{i} by $P_{i}$ ⁠. It is clear that $P_{i} \in S p e c (R)$ and $Z (R) = ⋃_{i = 1}^{n} P_{i}$ ⁠. Since $P_{i}$ and $P_{j}$ are not comparable under inclusion for all distinct i, j ∈ {1, 2, …, n}, it follows that $M N P (R) = {P_{i} ∣ i \in {1,2, \dots, n}}$ by [[14], Proposition 1.11(i)]. Thus |MNP(R)| = n.

(1) ⇒ (2) Assume that ZT(R) is connected. It follows from Corollary 3.14 that Z(R_i) ≠ (0) for each i ∈ {1, 2, …, n}. (This part of the proof does not need the assumption that Z(R_i) is an ideal of R for each i ∈ {1, 2, …, n}.)

(2) ⇒ (1) Observe that $M N P (R) = {P_{i} ∣ i \in {1,2, \dots, n}}$ and $⋂_{i = 1}^{n} P_{i} = p_{1} \times p_{2} \times ... \times p_{n} \neq (0) \times (0) \times . . . \times (0)$ ⁠. Let a ∈ Z(R)* be such that MNP(R)⊈V(a). Let a = (a₁, a₂, …, a_n). From $Z (R) = ⋃_{i = 1}^{n} P_{i}$ ⁠, it follows that there exists at least one i ∈ {1, 2, …, n} such that $a \in P_{i}$ ⁠. Hence, $a_{i} \in p_{i}$ ⁠. As MNP(R)⊈V(a), it follows that $a_{j} \notin p_{j}$ for at least one j ∈ {1, 2, …, n}\{i}. As a_i ∈ Z(R_i), there exists b_i ∈ R_i\{0} such that a_ib_i = 0. We can assume that $b_{i} \in p_{i}$ ⁠. If a_i = 0, then we can take b_i to be any non-zero element of $p_{i}$ ⁠. If a_i ≠ 0, then the element b_i ∈ R_i\{0} with a_ib_i = 0 satisfies $b_{i} \in p_{i}$ ⁠. Let b ∈ R be defined as follows: the ith coordinate of b equals b_i and the kth coordinate of b equals 0 for all k ∈ {1, 2, …, n}\{i}. Observe that a_j ≠ 0 but the jth coordinate of b equals 0. It is clear that b ∈ Z(R)*, a ≠ b, ab = (0, 0, …, 0), and $P_{i} \in V (a) \cap V (b)$ ⁠. Thus the statement (2) of Theorem 3.8 holds. Hence, ZT(R) is connected by (2) ⇒ (1) of Theorem 3.8.

ZT(R) is connected.
Z(R_i) ≠ (0) for each i ∈ {1, 2, …, n}.

Assume that (1) holds. Then the statement (1) of Theorem 3.8 holds and so, the statement (2) of Theorem 3.8 also holds. Hence, it follows from the moreover part of Theorem 3.8 that diam(ZT(R)) = 3, and if R is not reduced, then r(ZT(R)) = 2. □

Let R be such that MNP(R) is infinite. Now, Lemmas 3.1, 3.5, 3.6, and 3.7 imply that ZT(R) is connected if and only if given a ∈ Z(R)* with MNP(R)⊈V(a), there exists x ∈ Z(R)*\{a} such that ax = 0 and V(a) ∩ V(x) ∩ MNP(R) ≠ ∅. Let $n \in N$ and let R_n be a ring such that Z(R_n) is an ideal of R_n. Let R = R₁ × R₂ × R₃ × . . . be their direct product. In the following proposition, we verify that MNP(R) is infinite and show that ZT(R) is connected if and only if Z(R_n) ≠ (0) for each $n \in N$ ⁠.

Proposition 3.16.

Let $n \in N$ and let R_n be a ring such that Z(R_n) is an ideal of R_n. Let R = R₁ × R₂ × R₃ × . . . be their direct product. Then MNP(R) is infinite and the following statements are equivalent:

Moreover, if either (1) or (2) holds, then diam(ZT(R)) = 3, and if R is not reduced, then r(ZT(R)) = 2.

Proof. Let $n \in N$ ⁠. It is convenient to denote Z(R_n) by $p_{n}$ ⁠. Let $P_{n}$ be the ideal I₁ × I₂ × I₃ × . . . of R with $I_{n} = p_{n}$ and I_j = R_j for all $j \in N \ {n}$ ⁠. Since $p_{n} \in S p e c (R_{n})$ ⁠, it follows that $P_{n} \in S p e c (R)$ ⁠. As $Z (R_{n}) = p_{n}$ ⁠, we get that $P_{n} \subseteq Z (R)$ ⁠. Hence, $⋃_{n = 1}^{\infty} P_{n} \subseteq Z (R)$ ⁠. Let r = (r₁, r₂, r₃, …) ∈ Z(R). Then there exists s = (s₁, s₂, s₃, …) ∈ R\{(0, 0, 0, …)} such that r_is_i = 0 for each $i \in N$ ⁠. Note that s_n ≠ 0 for some $n \in N$ and hence, from r_ns_n = 0, we get that $r_{n} \in p_{n}$ and so, $r \in P_{n}$ ⁠. This proves that $Z (R) \subseteq ⋃_{n = 1}^{\infty} P_{n}$ and therefore, $Z (R) = ⋃_{n = 1}^{\infty} P_{n}$ ⁠. It is clear that $P_{n} \neq P_{m}$ for all distinct $n, m \in N$ ⁠. For each $n \in N$ ⁠, let e_n be the element of R whose nth coordinate equals 1 and jth coordinate equals zero for all $j \in N$ with j ≠ n. Let $n, m \in N$ with n ≠ m. Observe that $(1,1,1, \dots) - e_{n} \in P_{n}$ and $e_{n} \in P_{m}$ and so, $(1,1,1, \dots) \in P_{n} + P_{m}$ ⁠. Hence, $P_{n} + P_{m} = R$ ⁠. Let $n \in N$ ⁠. We next verify that $P_{n} \in M N P (R)$ ⁠. Let $P \in S p e c (R)$ be such that $P \subseteq Z (R)$ and $P_{n} \subseteq P$ ⁠. Let $r = (r_{1}, r_{2}, r_{3}, \dots) \in P$ ⁠. Then $e_{n} r \in P$ ⁠. As $(1,1,1, \dots) - e_{n} \in P_{n} \subseteq P$ ⁠, it follows that $s = (1,1,1, \dots) - e_{n} + e_{n} r \in P$ ⁠. Observe that the nth coordinate of s equals r_n and the jth coordinate of s equals 1 for all $j \in N \ {n}$ ⁠. As s ∈ Z(R), we get that $r_{n} \in p_{n}$ ⁠. Hence, $r \in P_{n}$ ⁠. This shows that $P \subseteq P_{n}$ and so, $P = P_{n}$ ⁠. This proves that $P_{n} \in M N P (R)$ ⁠. Hence, we get that ${P_{n} ∣ n \in N} \subseteq M N P (R)$ and so, MNP(R) is infinite. Let $P \in M N P (R)$ ⁠. If $(1,1,1, \dots) - e_{n} \in P$ for some $n \in N$ ⁠, then we claim that $P = P_{n}$ ⁠. Let $r = (r_{1}, r_{2}, r_{3}, \dots) \in P$ ⁠. Note that $(1,1,1, \dots) - e_{n} + e_{n} r \in P$ ⁠, and as shown above, we get that $r_{n} \in p_{n}$ and so, $r \in P_{n}$ ⁠. Therefore, $P \subseteq P_{n}$ and so, $P = P_{n}$ ⁠.

(1) ⇒ (2) Assume that ZT(R) is connected. Let $n \in N$ and let us denote the ring $\prod_{m \in N \ {n}} R_{m}$ by T_n. Note that R ≅ R_n × T_n as rings. Since ZT(R_n × T_n) is connected, Z(R_n) ≠ (0) by Lemma 3.13.

(2) ⇒ (1) Assume that Z(R_n) ≠ (0) for each $n \in N$ ⁠. Let a = (a₁, a₂, a₃…) ∈ Z(R)* be such that MNP(R)⊈V(a). Observe that $a \in P_{n}$ for some $n \in N$ ⁠. Hence, $a_{n} \in p_{n}$ ⁠. Thus $P_{n} \in V (a)$ ⁠. As MNP(R)⊈V(a) by assumption, there exists $P \in M N P (R) \ {P_{n}}$ such that $a \notin P$ ⁠. Note that $(1,1,1, \dots) - e_{n} \notin P$ (see the first paragraph of this proof). Therefore, $e_{n} \in P$ ⁠, since e_n((1, 1, 1, …) − e_n) = (0, 0, 0, …). If a_j = 0 for each $j \in N \ {n}$ ⁠, then $a = a e_{n} \in P$ ⁠. This is impossible and so, a_j ≠ 0 for at least one $j \in N \ {n}$ ⁠. Since $Z (R_{n}) = p_{n}$ ⁠, it follows as in the proof of Proposition 3.15 that there exists $b_{n} \in p_{n}, b_{n} \neq 0$ such that a_nb_n = 0. Let b ∈ Z(R)* be defined as follows: the nth coordinate of b equals b_n and the jth coordinate of b equals 0 for all $j \in N \ {n}$ ⁠. Observe that a ≠ b, ab = (0, 0, 0, …), and $P_{n} \in V (a) \cap V (b)$ ⁠.

ZT(R) is connected.
Z(R_n) ≠ (0) for each $n \in N$ ⁠.

Let x, y ∈ Z(R)* with x ≠ y. It follows from Lemmas 3.5, 3.6, and 3.7 that there exists a path in ZT(R) between x and y with 1 ≤ d(x, y) ≤ 3 in ZT(R). Therefore, we obtain that ZT(R) is connected with 1 ≤ diam(ZT(R)) ≤ 3.

Assume that (2) holds. It is verified in the proof of (2) ⇒ (1) of this proposition that ZT(R) is connected and diam(ZT(R)) ≤ 3. We can find $x \in P_{1}, y \in P_{2}$ such that x + y = (1, 1, 1, …) ∉ Z(R). Hence, it follows as in the proof of the moreover part of Theorem 3.8 that d(x, y) = 3 in ZT(R) and so, diam(ZT(R)) = 3. If R is not reduced, then it can be shown as in the proof of the moreover part of Theorem 3.8 that r(ZT(R)) = 2. □

We provide Example 3.17 to illustrate Theorems 3.8, 3.9, and Propositions 3.15, 3.16.

Example 3.17.

Proof.

Let R be as in Example 2.14(3). Let T = R × R. Then |MNP(T)| = 2, the zero ideal of T admits a primary decomposition but does not admit any strong primary decomposition and ZT(T) is connected with diam(ZT(T)) = 3 and r(ZT(T)) = 2.
Let ${X_{i}}_{i \in N}$ be a set of indeterminates. Let $D = ⋃_{n = 1}^{\infty} K [[X_{1}, \dots, X_{n}]]$ ⁠, where K[[X₁, …, X_n]] is the power series ring in X₁, …, X_n over a field K. Let I be the ideal of D generated by ${X_{i} X_{j} ∣ i, j \in N, i \neq j}$ ⁠. Let $R = \frac{D}{I}$ ⁠. If T = R × R, then T is a reduced ring, |MNP(T)| = 2, the zero ideal of T does not admit primary decomposition, and ZT(T) is connected with diam(ZT(T)) = 3 and r(ZT(T)) = 2.
Let $R = Z_{4} \times Z_{4}$ ⁠. Then |MNP(R)| = 2, R is a finite ring, and ZT(R) is connected with diam(ZT(R)) = 3 and r(ZT(R)) = 2.
Let $R = Z_{4} \times F$ ⁠, where F is a field. Then R is Artinian, |MNP(R)| = 2 but ZT(R) is not connected.
Let R be as in Example 2.14(3). Let R_n = R for each $n \in N$ and let T = R₁ × R₂ × R₃ × . . .. Then MNP(T) is infinite, ZT(T) is connected with diam(ZT(T)) = 3 and r(ZT(T)) = 2.

In the notation of Example 2.14(3), $M N P (R) = {p = \frac{m}{V m}}$ ⁠. It follows as in the proof of Proposition 3.15 that $M N P (T) = {p \times R, R \times p}$ and so, |MNP(T)| = 2. Note that dim R = 0 and $S p e c (R) = M a x (R) = {p}$ and so, $n i l (R) = p$ by [[14], Proposition 1.8]. As $p \neq (0 + V m)$ ⁠, we get that R is not reduced and so, T is not reduced. As (0 + Vm) is the zero ideal of R with $\sqrt{(0 + V m)} = p$ ⁠, we get that (0 + Vm) is a $p$ -primary ideal of R by [[14], Proposition 4.2]. It is clear that (0 + Vm) × (0 + Vm) is the zero ideal of T and (0 + Vm) × (0 + Vm) = ((0 + Vm) × R) ∩ (R × (0 + Vm)). Let us denote the ideal $p \times R$ by $P_{1}$ and $R \times p$ by $P_{2}$ ⁠. Since $p \in M a x (R)$ ⁠, it follows that $P_{i} \in M a x (T)$ for each i ∈ {1, 2}. Let us denote the ideal (0 + Vm) × R by $Q_{1}$ and the ideal R × (0 + Vm) by $Q_{2}$ ⁠. Let i ∈ {1, 2}. As $\sqrt{Q_{i}} = P_{i} \in M a x (T)$ ⁠, we obtain that $Q_{i}$ is a $P_{i}$ -primary ideal of T by [[14], Proposition 4.2]. From $(0 + V m) \times (0 + V m) = ⋂_{i = 1}^{2} Q_{i}$ ⁠, it follows that the zero ideal of T admits a primary decomposition. Note that $S p e c (T) = M a x (T) = {P_{i} ∣ i \in {1,2}}$ ⁠. Since V is a rank one non-discrete valuation domain, it follows that $m = m^{2} = m^{n}$ for all n ≥ 2 and so, $p = p^{n}$ for all n ≥ 2. Hence, $P_{i} = P_{i}^{n}$ for all n ≥ 2 and for each i ∈ {1, 2}. Observe that $p \neq (0 + V m)$ and so, $P_{i}^{n} ⊈ Q_{i}$ for all $n \in N$ and for each i ∈ {1, 2}. Therefore, $(0 + V m) \times (0 + V m) = ⋂_{i = 1}^{2} Q_{i}$ is not a strong primary decomposition of the zero ideal in T. We claim that $(0 + V m) \times (0 + V m) = ⋂_{i = 1}^{2} Q_{i}$ is the only irredundant primary decomposition of the zero ideal in T. Let $(0 + V m) \times (0 + V m) = ⋂_{j = 1}^{t} Q_{i}^{'}$ be an irredundant primary decomposition of (0 + Vm) × (0 + Vm) in T. Then t = 2 by [[14], Theorem 4.5] and without loss of generality, we can assume that $Q_{1}^{'} = q_{1} \times R$ for some $p$ -primary ideal $q_{1}$ of R. Then $Q_{2}^{'}$ must be of the form $R \times q_{2}$ for some $p$ -primary ideal $q_{2}$ of R. Hence, it follows from $(0 + V m) \times (0 + V m) = ⋂_{i = 1}^{2} Q_{i}^{'}$ that $q_{1} = (0 + V m) = q_{2}$ ⁠. Therefore, $Q_{i}^{'} = Q_{i}$ for each i ∈ {1, 2}. This shows that the zero ideal of T admits a primary decomposition but it does not admit any strong primary decomposition. Note that T = R × R and $Z (R) = p$ is an ideal of R with $p \neq (0 + V m)$ ⁠. Hence, ZT(T) is connected by (2) ⇒ (1) of Proposition 3.15. It follows from the moreover part of Proposition 3.15 that diam(ZT(T)) = 3 and r(ZT(T)) = 2, since T is not reduced.
The ring R mentioned in (2) is due to Gilmer and Heinzer and it appeared in [[19], Example, p. 16]. Let $i \in N$ ⁠. It is convenient to denote X_i + I by x_i. Observe that x_i ≠ 0 + I. It was noted in [[19], Example, p. 16] that R is a quasi-local reduced ring with $m = \sum_{n = 1}^{\infty} R x_{n}$ as its unique maximal ideal. It is clear that $Z (R) \subseteq m$ ⁠. Let $r \in m$ ⁠. Then there exist $n \in N$ and r_i ∈ R for each i ∈ {1, …, n} such that $r = \sum_{i = 1}^{n} r_{i} x_{i}$ ⁠. Note that rx_n+1 = 0 + I and so, r ∈ Z(R). Hence, $m \subseteq Z (R)$ and therefore, $Z (R) = m$ ⁠. As R is reduced, it follows that T = R × R is reduced. It follows as in the proof of Proposition 3.15 that $M N P (T) = {m \times R, R \times m}$ ⁠. Thus |MNP(T)| = 2. It is convenient to denote $m \times R$ by $M_{1}$ and $R \times m$ by $M_{2}$ ⁠. It is clear that $M_{i} \in M a x (T)$ for each i ∈ {1, 2}. It was already noted in [[19], Example, p. 16] that Min(R) is infinite, and indeed, $M i n (R) = {p_{i} ∣ i \in N}$ ⁠, where for each $i \in N$ ⁠, $p_{i}$ is the ideal of R generated by ${x_{j} ∣ j \in N \ {i}}$ ⁠. Hence, $M i n (T) = {p_{i} \times R ∣ i \in N} \cup {R \times p_{i} ∣ i \in N}$ and so, Min(T) is infinite. Note that (0 + I) × (0 + I) is the zero ideal of T. As Min(T) is infinite, we obtain from [[14], Proposition 4.6] that the zero ideal of T does not admit primary decomposition. Observe that $Z (R) = m$ is an ideal of R with $m \neq (0 + I)$ ⁠. As T = R × R, we obtain that ZT(T) is connected by (2) ⇒ (1) of Proposition 3.15 and from the moreover part of Proposition 3.15, we get that diam(ZT(T)) = 3. From diam(ZT(T)) = 3, it follows that e(t) ≥ 2 in ZT(T) for each t ∈ Z(T)* and so, r(ZT(T)) ≥ 2. Let $m \in m$ with m ≠ 0 + I. Let t = (m, 0 + I). We claim that e(t) = 2 in ZT(T). Let t′ ∈ Z(T)*\{t}. Observe that either t′ = (m′, r) for some $m^{'} \in m$ and r ∈ R or t′ = (a, m″) for some a ∈ R and $m^{″} \in m$ ⁠. Suppose that t′ = (m′, r). As $m + m^{'} \in m = Z (R)$ ⁠, it follows that t + t′ = (m + m′, r) ∈ Z(T). If tt′ = (0 + I, 0 + I), then t and t′ are adjacent in ZT(T) and so, d(t, t′) = 1 in ZT(T). Suppose that tt′ ≠ (0 + I, 0 + I). Note that Rm + Rm′ is a finitely generated ideal of R with $R m + R m^{'} \subset m$ and so, there exists $n \in N$ such that $R m + R m^{'} \subseteq \sum_{i = 1}^{n} R x_{i}$ ⁠. Hence, mx_n+1 = m′x_n+1 = 0 + I. Let t″ = (x_n+1, 0 + I). Then t″ ∈ Z(T)* and t − t″ − t′ is a path of length two between t and t′ in ZT(T). Suppose that t′ = (a, m″) for some a ∈ R and $m^{″} \in m$ ⁠. It is clear that t + t′ ∈ Z(T). If tt′ = (0 + I, 0 + I), then d(t, t′) = 1 in ZT(T). Suppose that d(t, t′) > 1 in ZT(T). It is possible to find $j \in N$ such that m″x_j = 0 + I. Let t″ = (0 + I, x_j). Then t″ ∈ Z(T)* is such that t − t″ − t′ is a path of length two between t and t′ in ZT(T). This shows that d(t, t′) ≤ 2 in ZT(T) for any t′ ∈ Z(T)* with t′ ≠ t. Therefore, e(t) ≤ 2 in ZT(T) and so, e(t) = 2 in ZT(T). Hence, it follows that r(ZT(T)) ≤ 2 and so, r(ZT(T)) = 2.
Since $Z_{4}$ is not reduced, we get that $R = Z_{4} \times Z_{4}$ is not reduced. It is clear that |R| = 16. Since $Z (Z_{4}) = 2 Z_{4}$ ⁠, it follows as in the proof of Proposition 3.15 that $M N P (R) = {2 Z_{4} \times Z_{4}, Z_{4} \times 2 Z_{4}}$ ⁠. Note that $Z_{4}$ is local with $m = 2 Z_{4}$ as its unique maximal ideal. It is convenient to denote $m \times Z_{4}$ by $M_{1}$ and $Z_{4} \times m$ by $M_{2}$ ⁠. Thus $M N P (R) = {M_{i} ∣ i \in {1,2}}$ ⁠. Therefore, |MNP(R)| = 2. Since (0) of $Z_{4}$ is a $m$ -primary ideal of $Z_{4}$ ⁠, we get that $Q_{1} = (0) \times Z_{4}$ is a $M_{1}$ -primary ideal of R and $Q_{2} = Z_{4} \times (0)$ is a $M_{2}$ -primary ideal of R. Observe that $M_{i}^{2} = Q_{i}$ for each i ∈ {1, 2} and $(0) \times (0) = ⋂_{i = 1}^{2} Q_{i}$ is an irredundant strong primary decomposition of the zero ideal in R. Note that $(2,0), (0,2) \in ⋂_{i = 1}^{2} M_{i}$ ⁠, $M_{1} = ((0) \times (0) :_{R} (2,0))$ and $M_{2} = ((0) \times (0) :_{R} (0,2))$ ⁠. Hence, ZT(R) is connected by (2) ⇒ (1) of Theorem 3.9. It follows from the moreover part of Theorem 3.9 that diam(ZT(R)) = 3 and r(ZT(R)) = 2.
Since F is a field, it is clear that $R = Z_{4} \times F$ is Artinian. Proceeding as in (3), it can be shown that $M N P (R) = {M_{1} = 2 Z_{4} \times F, M_{2} = Z_{4} \times (0)}$ ⁠. Therefore, |MNP(R)| = 2. Observe that $Q_{1} = (0) \times F$ is a $M_{1}$ -primary ideal of R with $M_{1}^{2} = Q_{1}$ and $(0) \times (0) = Q_{1} \cap M_{2}$ is an irredundant strong primary decomposition of the zero ideal in R. Let $r \in M_{2} \ {(0,0)}$ ⁠. Then r = (n, 0) for some $n \in Z_{4} \ {0}$ ⁠. Note that $(1,0) \in M_{2}$ and (1, 0)(n, 0) ≠ (0, 0). Therefore, there exists no r ∈ R\{(0, 0)} with $r \in M_{2}$ such that $M_{2} = ((0,0) :_{R} r)$ ⁠. Hence, ZT(R) is not connected by (1) ⇒ (2) of Theorem 3.9. One can also apply the following argument to conclude that ZT(R) is not connected. As $R = Z_{4} \times F$ with Z(F) = (0), it follows from (1) ⇒ (2) of Proposition 3.15 that ZT(R) is not connected.
In the notation of Example 2.14(3), $Z (R) = \frac{m}{V m}$ is a non-zero ideal of R. As R_n = R by assumption for each $n \in N$ ⁠, we get that Z(R_n) is a non-zero ideal of R_n. As T = R₁ × R₂ × R₃ × . . ., we obtain from Proposition 3.16 that MNP(T) is infinite. It follows from (2) ⇒ (1) of Proposition 3.16 that ZT(T) is connected. From the moreover part of Proposition 3.16, we get that diam(ZT(T)) = 3 and as R is not reduced, it follows that T is not reduced and so, r(ZT(T)) = 2. □

Recall that a ring R is said to be von Neumann regular if given a ∈ R, there exists b ∈ R such that a = a²b [[20], Exercise 16, p.111]. We know from (a) ⇔ (d) of [[20], Exercise 16, p.111] that a ring R is von Neumann regular if and only if dim R = 0 and R is reduced. Thus if a ring R is von Neumann regular, then Spec(R) = Max(R) = Min(R) and R is reduced, and so, $⋂_{m \in M a x (R)} m = (0)$ by [[14], Proposition 1.8]. Moreover, $m \subseteq Z (R)$ for each $m \in M a x (R)$ by [[12], Theorem 84] and so, Max(R) = MNP(R). Let R be von Neumann regular and let a ∈ R. Note that a = ue for some u ∈ U(R) and e ∈ R is idempotent by (1) ⇒ (3) of [[20], Exercise 29, p.113]. For a von Neumann regular ring R which is not a field, in the following theorem, we provide a necessary and sufficient condition so that ZT(R) is connected.

Theorem 3.18.

Let R be a von Neumann regular ring which is not a field. The following statements are equivalent:

Moreover, if either (1) or (2) holds, then diam(ZT(R)) = r(ZT(R)) = 3.

Proof. As R is a von Neumann regular ring which is not a field, |Max(R)|≥ 2. It is already noted that Spec(R) = Max(R) = Min(R) = MNP(R) and $⋂_{m \in M N P (R)} m = (0)$ ⁠.

(1) ⇒ (2) Assume that ZT(R) is connected. Since $⋂_{m \in M N P (R)} m = (0)$ ⁠, there exists no a ∈ Z(R)* such that |V(a) ∩ MNP(R)| = 1 by Corollary 3.2. Let $m \in M a x (R) = M N P (R)$ ⁠. We claim that $m$ is not principal. Suppose that $m$ is principal. Let $m \in m$ be such that $m = R m$ ⁠. Then m ∈ Z(R)* and $V (m) = {m}$ ⁠. Since Spec(R) = Max(R) = MNP(R), we obtain that $V (m) \cap M N P (R) = V (m) = {m}$ ⁠. Hence, m ∈ Z(R)* is such that |V(m) ∩ MNP(R)| = 1. This is a contradiction. Therefore, no maximal ideal of R is principal.

(2) ⇒ (1) Assume that no maximal ideal of R is principal. It is already noted that |Max(R)|≥ 2. Let a ∈ Z(R)*. As $⋂_{m \in M N P (R)} m = (0)$ ⁠, it follows that there exists at least one $m \in M N P (R)$ such that $a \notin m$ ⁠. Therefore, MNP(R)⊈V(a). Let a = ue for some u ∈ U(R) and e ∈ R\{0, 1} is idempotent. Let $m_{1} \in M N P (R)$ be such that $a \in m_{1}$ ⁠. Then $e \in m_{1}$ ⁠. Since $m_{1}$ is not principal, it is possible to find $b \in m_{1}$ such that b ∉ Re. Observe that b(1 − e) ∈ Z(R)*\{a}, ab(1 − e) = 0, and $m_{1} \in V (a) \cap V (b (1 - e)) \cap M N P (R)$ ⁠. Let x, y ∈ Z(R)* be such that x ≠ y. It now follows from the above argument that MNP(R)⊈V(x), MNP(R)⊈V(y), there exist z, w ∈ Z(R)* such that x ≠ z, xz = 0, and V(x) ∩ V(z) ∩ MNP(R) ≠ ∅ and y ≠ w, yw = 0, and V(y) ∩ V(w) ∩ MNP(R) ≠ ∅. Hence, there exists a path in ZT(R) between x and y such that 1 ≤ d(x, y) ≤ 3 in ZT(R) by Lemma 3.7. This proves that ZT(R) is connected and diam(ZT(R)) ≤ 3.

ZT(R) is connected.
No maximal ideal of R is principal.

Assume that (2) holds. Then it is verified in the proof of (2) ⇒ (1) of this theorem that ZT(R) is connected and 1 ≤ diam(ZT(R)) ≤ 3. Hence, r(ZT(R)) ≤ 3. Let a ∈ Z(R)*. Observe that a = ue for some u ∈ U(R) and e ∈ R is a non-trivial idempotent. It is clear that 1 − e ∈ R is a non-trivial idempotent. As any maximal ideal of R contains exactly one between e and 1 − e, it follows that $a + 1 - e \notin m$ for each $m \in M a x (R)$ ⁠. Hence, a + 1 − e ∈ U(R) by [[14], Corollary 1.5]. Hence, a + 1 − e ∉ Z(R). Therefore, it follows as in the proof of the moreover part of Theorem 3.8 that d(a, 1 − e) ≥ 3 in ZT(R) and so, d(a, 1 − e) = 3 in ZT(R). This proves that the eccentricity of a in ZT(R) equals 3 for each a ∈ Z(R)*. Therefore, diam(ZT(R)) = r(ZT(R)) = 3. □

The following example illustrates Theorem 3.18.

Example 3.19.

Let L be the field of algebraic numbers (that is, L is the algebraic closure of $Q$ ⁠). Let A be the ring of all algebraic integers. Let $R = \frac{A}{\sqrt{2 A}}$ ⁠. Then R is a von Neumann regular ring and ZT(R) is connected with diam(ZT(R)) = r(ZT(R)) = 3.

Proof. It was verified in the proof of [[21], Example 2.20(3)] that R is a von Neumann regular ring. With the help of [[20], Proposition 42.8(1)], it was noted in the proof of [[21], Example 2.20(3)] that Max(R) is uncountable, and moreover, it was verified in the proof of [[21], Example 2.20(3)] that no maximal ideal of R is principal. It now follows from (2) ⇒ (1) of Theorem 3.18 that ZT(R) is connected. Also, we obtain from the moreover part of Theorem 3.18 that diam(ZT(R)) = r(ZT(R)) = 3. □

4. On the domination number of ZT(R)

Let R be a ring such that Z(R)* ≠ ∅. This section aims to discuss some results on the dominating sets and the domination number of ZT(R). Let G = (V, E) be a graph. Recall that a set S ⊆ V is called a dominating set of G if every vertex u ∈ V\S has a neighbor v ∈ S. Equivalently, every vertex of G is either in S or in the neighbor set N(S) = ⋃_v ∈ SN(v) of S in G [[10], Definition 10.2.1]. A γ-set of G is a minimum dominating set of G, that is, a dominating set of G whose cardinality is minimum. A dominating set S of G is minimal if S properly contains no dominating set S′ of G [[10], Definition 10.2.2]. The domination number of G is the cardinality of a minimum dominating set (that is, γ-set) of G; it is denoted by γ(G) [[10], Definition 10.2.3]. In Ref. [22], Rad et al. investigated on the dominating sets and the domination number of Γ(R). If |Z(R)*|≥ 2, then in the following proposition, we determine necessary and sufficient conditions so that γ(ZT(R)) to be equal to 1.

Proposition 4.1.

Let R be such that |Z(R)*|≥ 2. The following statements are equivalent:

Proof. (1) ⇒ (2) Assume that γ(ZT(R)) = 1. Let x ∈ Z(R)* be such that {x} is a dominating set of ZT(R). By hypothesis, |Z(R)*|≥ 2. For any y ∈ Z(R)* with y ≠ x, x and y are adjacent in ZT(R). Hence, for any distinct a, b ∈ Z(R)*, either a and b are adjacent in ZT(R) or a − x − b is a path of length two in ZT(R) between a and b. This proves that ZT(R) is connected and r(ZT(R)) = 1.

(2) ⇒ (3) Assume that ZT(R) is connected and r(ZT(R)) = 1. Let x ∈ Z(R)* be such that e(x) = 1 in ZT(R). Let y ∈ Z(R)* be such that y ≠ x. From e(x) = 1 in ZT(R), we get that x and y are adjacent in ZT(R). Therefore, xy = 0 and x + y ∈ Z(R). Thus Z(R)\{x}⊆ ((0):_Rx). As |Z(R)*|≥ 2 by hypothesis, there exists y ∈ Z(R)* with y ≠ x. Note that x + y ∈ Z(R)\{x}. Hence, (x + y)x = xy = 0 and so, x² = 0. This shows that Z(R) = ((0):_Rx). Therefore, Z(R) is an ideal of R. Let us denote Z(R) by $p$ ⁠. As $M N P (R) = {p}$ ⁠, we get that |MNP(R)| = 1 and it is already verified that $p = ((0) :_{R} x)$ is a B-prime of (0) in R.

(3) ⇒ (1) By (3), R admits a unique member of MNP(R) and the unique member is a B-prime of (0) in R. Let $M N P (R) = {p}$ and let $x \in p \ {0}$ be such that $p = ((0) :_{R} x)$ ⁠. Then it is clear that {x} is a dominating set of ZT(R) and so, γ(ZT(R)) = 1. □

γ(ZT(R)) = 1.
ZT(R) is connected and r(ZT(R)) = 1.
|MNP(R)| = 1 and the unique member of MNP(R) is a B-prime of (0) in R.

In the following example, we provide a ring R with |MNP(R)| = 1 such that ZT(R) does not admit any finite dominating set. Recall that a ring R is called a chained ring if the set of ideals of R is linearly ordered by inclusion [[20], Exercise 8, p. 184].

Example 4.2.

Let R be as in Example 2.14(3). Then |MNP(R)| = 1 and ZT(R) = Γ(R) does not admit any finite dominating set.

Proof. In the notation of Example 2.14(3), $R = \frac{V}{V m}$ and $M N P (R) = {p = \frac{m}{V m}}$ ⁠. Therefore, |MNP(R)| = 1 and hence, ZT(R) = Γ(R) by Corollary 2.7. It is already noted in the proof of Example 2.14(3) that $p$ is not a B-prime of zero ideal in R. Let {m₁ + Vm, …, m_k + Vm} be any finite non-empty subset of Z(R)*. Since R is a chained ring, after a suitable rearrangement of 1, …, k, we can assume without loss of generality that R(m₁ + Vm) ⊆ . . . ⊆ R(m_k + Vm). Observe that R(m₁ + Vm) is a non-zero ideal of R and m_k + Vm ∈ Z(R)*. Since $p$ is not a B-prime of the zero ideal in R, it follows that $p ⊈ R (m_{k} + V m)$ and $p ⊈ ((0 + V m) :_{R} m_{1} + V m)$ ⁠. Hence, $p ⊈ R (m_{k} + V m) \cup ((0 + V m) :_{R} m_{1} + V m)$ ⁠. Therefore, there exists $m^{'} \in m$ such that m′ + Vm ∉ R(m_k + Vm) and m′m₁∉Vm. As R(m₁ + Vm) ⊆ . . . ⊆ R(m_k + Vm), it is clear from the choice of m′ + Vm that m′ + Vm ∈ Z(R)*\{m₁ + Vm, …, m_k + Vm}, (m′ + Vm)(m_i + Vm) ≠ 0 + Vm for each i ∈ {1, …, k}. Thus m′ + Vm ∈ Z(R)* is such that m′ + Vm∉{m_i + Vm∣i ∈ {1, …, k}} and m′ + Vm is not adjacent to any element from {m_i + Vm∣i ∈ {1, …, k}} in Γ(R). This proves that ZT(R) = Γ(R) does not admit any finite dominating set. □

Let R be such that 2 ≤ |MNP(R)| < ∞. Let |MNP(R)| = n and let $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. If ZT(R) is connected, then we verify in the following proposition that γ(ZT(R)) ≥ n.

Proposition 4.3.

Let R be a ring such that 2 ≤ |MNP(R)| = n < ∞. Let $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. If ZT(R) is connected, then γ(ZT(R)) ≥ n.

Proof. Let D ⊆ Z(R)* be any dominating set of ZT(R). By hypothesis, |MNP(R)| = n ≥ 2 and $M N P (R) = {p_{i} ∣ i \in {1,2, \dots, n}}$ ⁠. We know from the proof of [[23], Proposition 3.10] that it is possible to find elements a₁, a₂, …, a_n ∈ Z(R)* and r₁, r₂…, r_n ∈ R such that r_ia_i + r_ja_j ∉ Z(R) for all distinct i, j ∈ {1, 2, …, n}. Let i ∈ {1, 2, …, n}. Note that a_i ∈ Z(R)* and MNP(R)⊈V(a_i). Since |Z(R)*|≥ 2 and ZT(R) is connected by assumption, there exists x_i ∈ Z(R)* such that a_i and x_i are adjacent in ZT(R). Any such x_i satisfies a_ix_i = 0 and a_i + x_i ∈ Z(R). From a_ix_i = 0 and a_i + x_i ∈ Z(R), it follows that $a_{i} + x_{i} \in p_{k}$ for some $p_{k} \in M N P (R) \cap V (a_{i}) \cap V (x_{i})$ ⁠. Therefore, Ra_i + Rx_i ⊆ Z(R). For any i ∈ {1, 2, …, n}, let us denote {a_i} by A_i and let B_i = {x_i ∈ Z(R)*∣ a_i and x_i are adjacent in ZT(R)}. It is clear that either a_i ∈ D or D ∩ B_i ≠ ∅. Thus for each i ∈ {1, 2, …, n}, |D ∩ (A_i ∪ B_i)|≥ 1. Let i, j ∈ {1, 2, …, n} with i ≠ j. Note that a_i ≠ a_j. As Ra_i + Rx_i ⊆ Z(R) for any x_i ∈ B_i, Ra_j + Rx_j ⊆ Z(R) for any x_j ∈ B_j, whereas Ra_i + Ra_j⊈Z(R), it follows that a_i ∉ B_j, and a_j ∉ B_i. Thus A_i ∩ A_j = A_i ∩ B_j = A_j ∩ B_i = ∅. As a_ix_i = 0 for any x_i ∈ B_i, a_jx_j = 0 for any x_j ∈ B_j, Ra_i + Ra_j⊈Z(R), we get that B_i ∩ B_j = ∅. It follows that (A_i ∪ B_i) ∩ (A_j ∪ B_j) = ∅. Therefore, (D ∩ (A_i ∪ B_i)) ∩ (D ∩ (A_j ∪ B_j)) = ∅. It is clear that $⋃_{i = 1}^{n} (D \cap (A_{i} \cup B_{i})) \subseteq D$ and so, $| D | \geq | ⋃_{i = 1}^{n} (D \cap (A_{i} \cup B_{i})) | = \sum_{i = 1}^{n} | D \cap (A_{i} \cup B_{i}) | \geq n$ ⁠. Therefore, γ(ZT(R)) ≥ n. □

Remark 4.4.

Let $n \in N$ be such that n ≥ 2. Let R be such that |MNP(R)| = n. Note that Γ(R) is connected by [[1], Theorem 2.3]. In this remark, we provide an example of a ring R with |MNP(R)| = 2 to illustrate that the conclusion of Proposition 4.3 can fail to hold for Γ(R). Let $R = Z_{2} \times Z_{2}$ ⁠. Observe that $M N P (R) = {(0) \times Z_{2}, Z_{2} \times (0)}$ and so, |MNP(R)| = 2. It is clear that Z(R)* = {(0, 1), (1, 0)} and as (0, 1)(1, 0) = (0, 0), it follows that {(0, 1)} is a dominating set of Γ(R) and so, γ(Γ(R)) = 1. This example also illustrates that (1) ⇒ (3) of Proposition 4.1 can fail to hold for the zero-divisor graph.

Lemma 4.5.

Let n ≥ 2 and let R_i be a ring for each i ∈ {1, 2, …, n}. Suppose that Z(R_i) is an ideal of R_i for each i ∈ {1, 2, …, n} with Z(R_i) ≠ (0). Let R = R₁ × R₂ × . . . × R_n. Then |MNP(R)| = n, ZT(R) is connected, γ(ZT(R)) ≥ n, and γ(Γ(R)) ≥ n.

Proof. We know from the proof of Proposition 3.15 that |MNP(R)| = n. As Z(R_i) ≠ (0) for each i ∈ {1, 2, …, n} by assumption, it follows from (2) ⇒ (1) of Proposition 3.15 that ZT(R) is connected and so, γ(ZT(R)) ≥ n by Proposition 4.3. For each i ∈ {1, 2, …, n}, let e_i be the element of R whose ith coordinate equals 1 and whose jth coordinate equals 0 for each j ∈ {1, 2, …, n}\{i}, and let $x_{i} \in Z (R_{i}) *$ ⁠. For each i ∈ {1, 2, …, n}, let a_i ∈ R be such that the ith coordinate of a_i equals x_i and the jth coordinate of a_i equals 1 for all j ∈ {1, 2, …, n}\{i}. Since |Z(R)*|≥ 2 and Γ(R) is connected by [[1], Theorem 2.3], it is possible to find b_i ∈ Z(R)* such that a_i and b_i are adjacent in Γ(R). Observe that any such b_i must be such that the jth coordinate of b_i equals 0 for all j ∈ {1, 2, …, n}\{i} and the ith coordinate of b_i equals y_i for some $y_{i} \in Z {(R_{i})}^{*}$ with x_iy_i = 0. Let i ∈ {1, 2, …, n}. Let A_i = {a_i} and let B_i = {b_i ∈ Z(R)*∣ a_i and b_i are adjacent in Γ(R)}. Let i, j ∈ {1, 2, …, n} be distinct. It is clear that a_i ≠ a_j, a_i ∉ B_j, a_j ∉ B_i, and B_i ∩ B_j = ∅. Therefore, (A_i ∪ B_i) ∩ (A_j ∪ B_j) = ∅ for all distinct i, j ∈ {1, 2, …, n}. Let D be any dominating set of Γ(R). Let i ∈ {1, 2, …, n}. Note that either a_i ∈ D or D ∩ B_i ≠ ∅, and so, D ∩ (A_i ∪ B_i) ≠ ∅. Thus |D ∩ (A_i ∪ B_i)|≥ 1. Since (D ∩ (A_i ∪ B_i)) ∩ (D ∩ (A_j ∪ B_j)) = ∅ for all distinct i, j ∈ {1, 2, …, n}, we get that $| D | \geq | ⋃_{i = 1}^{n} (D \cap (A_{i} \cup B_{i})) | = \sum_{i = 1}^{n} | D \cap (A_{i} \cup B_{i}) | \geq n$ ⁠. Therefore, we obtain that γ(Γ(R)) ≥ n. □

Remark 4.6.

Let n ≥ 2 and let R_i be as in the statement of Lemma 4.5. Let R = R₁ × R₂ × . . . × R_n. It is noted in the proof of Lemma 4.5 that ZT(R) is connected and it is concluded with the help of Proposition 4.3 that γ(ZT(R)) ≥ n. In this remark, we give another argument for the conclusion that γ(ZT(R)) ≥ n. Let D be any dominating set of ZT(R). As Γ(R) is a spanning supergraph of ZT(R), it follows that D is also a dominating set of Γ(R). Hence, |D|≥ n by Lemma 4.5 and so, γ(ZT(R)) ≥ n.

Suppose R is such that (0) admits a strong primary decomposition with |MNP(R)| = n ≥ 2. If ZT(R) is connected, then in the following proposition, we show that γ(ZT(R)) = n.

Proposition 4.7.

Assume R is such that the zero ideal of R admits a strong primary decomposition. Let |MNP(R)| = n ≥ 2. If ZT(R) is connected, then γ(ZT(R)) = n.

Proof. Let $t, n, q_{i}, p_{i}, x_{i} (i \in {1,2, \dots, t}), p_{j} (j \in {1,2, \dots, n})$ be as in the statement of Theorem 3.9. Assume that ZT(R) is connected. Hence, we obtain from (1) ⇒ (2) of Theorem 3.9 that $x_{j} \in ⋂_{i = 1}^{t} p_{i}$ for each j ∈ {1, 2, …, n}. Observe that $Z (R) = ⋃_{i = 1}^{t} p_{i} = ⋃_{j = 1}^{n} p_{j}$ ⁠. Since $p_{i} = ((0) :_{R} x_{i})$ for each i ∈ {1, 2, …, t} and $p_{i} \neq p_{j}$ for all distinct i, j ∈ {1, 2, …, t}, we get that x_i ≠ x_j. We claim that D = {x_j∣j ∈ {1, 2, …, n}} is a dominating set of ZT(R). Let a ∈ Z(R)*\D. Note that $a \in p_{j}$ for some j ∈ {1, 2, …, n}. Hence, ax_j = 0 and $a + x_{j} \in p_{j} \subset Z (R)$ ⁠. Therefore, a and x_j are adjacent in ZT(R). This shows that D = {x_j∣j ∈ {1, 2, …, n}} is a dominating set of ZT(R). Therefore, γ(ZT(R)) ≤ n. Since |MNP(R)| = n and ZT(R) is connected by assumption, we obtain from Proposition 4.3 that γ(ZT(R)) ≥ n and so, γ(ZT(R)) = n. □

Remark 4.8.

Suppose R is such that its zero ideal admits a strong primary decomposition. Let $t, n, q_{i}, p_{i}, x_{i} (i \in {1,2, \dots, t}), p_{j} (j \in {1,2, \dots, n})$ be as in the statement of Theorem 3.9. We verify in this remark that γ(Γ(R)) ≤ n. Note that $Z (R) = ⋃_{j = 1}^{n} p_{j}$ ⁠. Let a ∈ Z(R)*\{x_j∣j ∈ {1, 2, …, n}}. Now, $a \in p_{j} = ((0) :_{R} x_{j})$ for some j ∈ {1, 2, …, n} and so, ax_j = 0. Hence, a and x_j are adjacent in Γ(R). This shows that {x_j∣j ∈ {1, 2, …, n}} is a dominating set of Γ(R). Therefore, γ(Γ(R)) ≤ n.

Remark 4.9.

Let R be a Noetherian ring such that Z(R)* ≠ ∅. We know from [[14], Theorem 7.13 and Proposition 7.14] that each proper ideal of R admits a strong primary decomposition. Let |MNP(R)| = k. If k = 1, then $Z (R) = p$ is an ideal of R and ZT(R) = Γ(R). As $p$ is a B-prime of (0) in R, it follows that γ(ZT(R)) = 1. Suppose that k ≥ 2. If ZT(R) is connected, then γ(ZT(R)) = k by Proposition 4.7. We know from Remark 4.8 that γ(Γ(R)) ≤ k. We know from [[14], Theorem 8.5] that any Artinian ring is Noetherian. Let R be an Artinian ring with Z(R)* ≠ ∅. From [[14], Proposition 8.1], it follows that Spec(R) = Max(R) = Min(R). As any minimal prime ideal of a ring is contained in its set of zero-divisors by [[12], Theorem 84], we get that MNP(R) = Max(R). Thus if |Max(R)| = k and if ZT(R) is connected, then it follows that γ(ZT(R)) = k. Since R is Artinian with |Max(R)| = k, it follows from [[14], Theorem 8.7] that there exist Artinian local rings $(R_{i}, m_{i})$ for each i ∈ {1, 2, …, k} such that R ≅ R₁ × R₂ × … × R_k as rings. Observe that $Z (R_{i}) = m_{i}$ is an ideal of R_i for each i ∈ {1, 2, …, k}. As ZT(R) is connected by assumption, it follows from (1) ⇒ (2) of Proposition 3.15 that Z(R_i) ≠ (0) for each i ∈ {1, 2, …, k}. Therefore, we obtain from Lemma 4.5 that γ(Γ(R)) ≥ k. Since any Artinian ring is Noetherian, it is already noted in this remark that γ(Γ(R)) ≤ k and so, γ(Γ(R)) = k. Thus if R is any finite ring with Z(R)* ≠ ∅, |Max(R)| = n, and if ZT(R) is connected, then γ(ZT(R)) = n and this generalizes [[9], Proposition 4.4]. Also, for such a finite ring R, γ(Γ(R)) = n.

The following example illustrates that the conclusion of Proposition 4.7 can fail to hold if the hypothesis ZT(R) is connected is omitted in the statement of Proposition 4.7.

Example 4.10.

Let $R = Z_{4} \times Z_{2}$ ⁠. Then the zero ideal of R admits a strong primary decomposition, |MNP(R)| = 2 but γ(ZT(R)) = 3.

Proof. It is already verified in the proof of Example 3.17(4) that the zero ideal of R admits a strong primary decomposition and $M N P (R) = {2 Z_{4} \times Z_{2}, Z_{4} \times (0)}$ ⁠. Therefore, |MNP(R)| = 2. It is also verified in the proof of Example 3.17(4) that ZT(R) is not connected. Observe that Z(R)* = {(1, 0), (2, 0), (3, 0), (0, 1), (2, 1)}. It is not hard to verify that {(1, 0), (3, 0)} is the set of isolated vertices of ZT(R). Thus if D is any dominating set of ZT(R), then D ⊇{(1, 0), (3, 0)}. Note that (0, 1)∉{(1, 0), (3, 0)} and (0, 1) is neither adjacent to (1, 0) nor adjacent to (3, 0) in ZT(R). Hence, {(1, 0), (3, 0)} is not a dominating set of ZT(R). Hence, we obtain that γ(ZT(R)) ≥ 3. It is clear that (2, 0)(0, 1) = (0, 0), (2, 0) + (0, 1) = (2, 1) ∈ Z(R) and (2, 0)(2, 1) = (0, 0), (2, 0) + (2, 1) = (0, 1) ∈ Z(R). Thus it follows that {(1, 0), (3, 0), (2, 0)} is a dominating set of ZT(R). Hence, γ(ZT(R)) ≤ 3 and so, we get that γ(ZT(R)) = 3. Note that Z(R)* = {(1, 0), (2, 0), (3, 0), (0, 1), (2, 1)}. It is clear that no two members from {(1, 0), (2, 0), (3, 0)} are adjacent in Γ(R) and (0, 1) and (2, 1) are not adjacent in Γ(R). Hence, it follows that γ(Γ(R)) ≥ 2. It can be easily verified that {(2, 0), (0, 1)} is a dominating set of Γ(R). Hence, γ(Γ(R)) ≤ 2 and so, γ(Γ(R)) = 2. □

In the following example, we provide a ring T with |MNP(T)| = 2 such that Γ(T) (and hence, ZT(T)) does not admit any finite dominating set.

Example 4.11.

Let R be as in Example 2.14(3). Let T = R × R. Then |MNP(T)| = 2 and Γ(T) does not admit any finite dominating set and hence, ZT(T) does not admit any finite dominating set.

Proof. In the notation of Example 2.14(3), $Z (R) = p = \frac{m}{V m}$ ⁠. It is already noted in the proof of Example 3.17(1) that|MNP(T)| = 2 and in the notation of Example 3.17(1), $M N P (T) = {P_{1} = p \times R, P_{2} = R \times p}$ ⁠. Note that $Z (T) = ⋃_{i = 1}^{2} P_{i}$ ⁠. Suppose there exists a finite non-empty subset D of Z(T)* such that D is a dominating set of Γ(T). We first verify that $D \cap (⋂_{i = 1}^{2} P_{i}) \neq \emptyset$ ⁠. Since V is a rank one non-discrete valuation domain, $m$ is not finitely generated. Let $x \in m \ V m$ ⁠. Then $(x + V m, x + V m) \in ⋂_{i = 1}^{2} P_{i} \subset Z {(T)}^{*}$ ⁠. If (x + Vm, x + Vm) ∈ D, then it follows that $D \cap (⋂_{i = 1}^{2} P_{i}) \neq \emptyset$ ⁠. If (x + Vm, x + Vm) ∉ D, then D being a dominating set of Γ(T) by assumption, there exists d ∈ D such that d and (x + Vm, x + Vm) are adjacent in Γ(T). Either d = (z + Vm, v + Vm) for some $z \in m, v \in V$ or d = (u + Vm, w + Vm) for some $u \in V, w \in m$ ⁠. If d = (z + Vm, v + Vm), then from d(x + Vm, x + Vm) = (0 + Vm, 0 + Vm), it follows that vx ∈ Vm. As x ∉ Vm, we get that $v \in m$ and so, $d = (z + V m, v + V m) \in ⋂_{i = 1}^{2} P_{i}$ ⁠. Similarly, if d = (u + Vm, w + Vm), then from ux ∈ Vm, it follows that $u \in m$ and so, $d = (u + V m, w + V m) \in ⋂_{i = 1}^{2} P_{i}$ ⁠. This shows that $D \cap (⋂_{i = 1}^{2} P_{i}) \neq \emptyset$ ⁠. Indeed, the above arguments show that for any $x \in m \ V m$ ⁠, either (x + Vm, x + Vm) ∈ D or (x + Vm, x + Vm) is adjacent in Γ(T) to an element in D of the form (v₁ + Vm, v₂ + Vm) for some $v_{1}, v_{2} \in m$ ⁠. Let $D \cap (⋂_{i = 1}^{2} P_{i}) = {(z_{1} + V m, w_{1} + V m), \dots, (z_{k} + V m, w_{k} + V m)}$ ⁠. It is clear that $z_{i}, w_{i} \in m$ for each i ∈ {1, …, k}. Observe that for each i ∈ {1, …, k}, either z_i ∉ Vm or w_i ∉ Vm. Let us denote the set of all distinct non-zero elements among z₁ + Vm, w₁ + Vm, …, z_k + Vm, w_k + Vm by A. Let z + Vm ∈ Z(R)*. Suppose that z + Vm ∉ A. Then (z + Vm, z + Vm) ∉ D and so, (z + Vm, z + Vm) is adjacent in Γ(T) to (z_i + Vm, w_i + Vm) for some i ∈ {1, …, k}. Hence, either z + Vm ∈ A or z + Vm is adjacent in Γ(R) to an element from A. This shows that A is a dominating set of Γ(R). This is impossible since we know from Example 4.2 that Γ(R) does not admit any finite dominating set. Therefore, we obtain that Γ(T) does not admit any finite dominating set. Since Γ(T) is a spanning supergraph of ZT(T), it follows that ZT(T) does not admit any finite dominating set. It is already noted in the proof of Example 3.17(1) that the zero ideal of T admits a primary decomposition but it does not admit any strong primary decomposition and it is also noted in the proof of Example 3.17(1) that ZT(T) is connected. Thus this example also illustrates that the conclusion of Proposition 4.7 can fail to hold if the hypothesis (0) admits a strong primary decomposition is omitted in the statement of Proposition 4.7. □

I am very much thankful to the reviewers for their helpful suggestions. I am also very much thankful to Dr Malik Talbi for his support.

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Subramanian Visweswaran

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this license may be seen at http://creativecommons.org/licences/by/4.0/legalcode

Some results on the total zero-divisor graph of a commutative ring

1. Introduction

2. Some basic properties of ZT(R)

3. On the connectedness of ZT(R), where 2 ≤ |MNP(R)| < ∞

4. On the domination number of ZT(R)

References

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Some results on the total zero-divisor graph of a commutative ring

1. Introduction

2. Some basic properties of ZT(R)

3. On the connectedness of ZT(R), where 2 ≤ |MNP(R)| < ∞

4. On the domination number of ZT(R)

References

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