Proof. By [Reference Raja, Pirzada and Redmond11, Proposition 6.2], $\dim _m(\Gamma ((\mathbb {Z}_2)^6))\leq 6$ . To complete the proof, we show that $\dim _m(\Gamma ((\mathbb {Z}_2)^6)) \geq 6$ .

Let $x \in V(\Gamma ((\mathbb {Z}_2)^6))$ and let $Z(x)$ denote the number of zero components in x. Clearly, $1 \leq Z(x) \leq 5$ . Now, let $W = \{a_1, a_2, \ldots , a_h\}$ be a resolving set for $\Gamma ((\mathbb {Z}_2)^6)$ . We will prove that $h \geq 6$ . Assume, for contradiction, that $h \leq 5$ .

Define $A = \{a \in W \mid Z(a) = 3\}$ as the subset of vertices in the resolving set with exactly three zero components. We will show that $A = \emptyset $ in two steps.

Step 1: $|A| \leq 2$ . Suppose, for a contradiction, that $|A| \geq 3$ . Let $a_1, a_2, a_3$ be three distinct vertices in A. Let $B = \{x \in V(\Gamma ((\mathbb {Z}_2)^6)) \mid Z(x) = 3\}$ . We claim that $|B| - 2h \leq 3^{h - 3}$ . For each $a_i\in W\cap B$ , there are at most two vertices of B at distance at most $1$ from $a_i$ , namely $a_i$ itself and its unique neighbour in B. Hence, at most $2h$ vertices of B may have a coordinate $0$ or $1$ in their distance vectors with respect to vertices of $W\cap B$ . For every $x\in B$ outside this set, we have $(d(x,a_1),d(x,a_2),d(x,a_3))=(2,2,2)$ . Since, for the remaining $h-3$ vertices of W each distance lies in $\{1,2,3\}$ , there are at most $3^{h-3}$ distinct distance vectors. Therefore, $|B|-2h \leq 3^{h-3}$ and the claim is proved. Given that $|B| = 20$ and $h \leq 5$ , this leads to a contradiction. Therefore, $|A| \leq 2$ .

Step 2: $A = \emptyset $ . We first show that A cannot contain exactly two elements. Assume, for a contradiction, that $A = \{a_1, a_2\}$ with $Z(a_1)=Z(a_2)=3$ and $Z(a_i) \neq 3$ for $3 \leq i \leq h$ .

Define the following sets based on the zero counts of the vertices in W.

• • For $a_1, a_2 \in A$ , let $A_1 = \{ y \in V(\Gamma ) \mid d(a_1, y) = 3 \}$ , $A_2 = \{ y \in V(\Gamma ) \mid d(a_2, y) = 3 \}$ . We have $|A_1| = |A_2| = 2^{6-3} - 2 = 6$ .

• • For $a_i \in W$ ( $i \geq 3$ ) with $Z(a_i) \geq 4$ , let $A_i = \{ y \in V(\Gamma ) \mid d(a_i, y) = 3 \}$ . Then $|A_i| = 2^{6 - Z(a_i)} - 2 \leq 2^{2} - 2 = 2$ .

• • For $a_i \in W$ ( $i \geq 3$ ) with $Z(a_i) \leq 2$ , let $A_i = \{ y \in V(\Gamma ) \mid d(a_i, y) = 1 \}$ . Then $|A_i| = 2^{Z(a_i)} - 1 \leq 2^{2} - 1 = 3$ .

Now, consider any vertex $x \in V(\Gamma ) \setminus ( \bigcup _{j=1}^{h} A_j \cup W )$ . Let $D(x \mid W) = (d_1, \ldots , d_h)$ be its distance vector to the resolving set. By the construction of the sets $A_j$ , for $1 \leq j \leq h$ , the distance $d_j$ is restricted:

• • if $a_j \in A$ (that is, $Z(a_j)=3$ ), then $d_j \in \{1, 2\}$ (since $d_j \neq 3$ );

• • if $Z(a_j) \geq 4$ , then $d_j \in \{1, 2\}$ (since $d_j \neq 3$ );

• • if $Z(a_j) \leq 2$ , then $d_j \in \{2, 3\}$ (since $d_j \neq 1$ ).

Thus, each component $d_j$ of the distance vector can take at most two possible values. Consequently, there are at most $2^h$ distinct distance vectors for vertices lying outside $\bigcup _{j=1}^{h} A_j \cup W$ . This yields the inequality

$$ \begin{align*} \bigg| V(\Gamma) \setminus \bigg( \bigcup_{j=1}^{h} A_j \cup W \bigg) \bigg| \leq 2^h. \end{align*} $$

Substituting the known values $|V(\Gamma )| = 2^6 - 2 = 62$ and the upper bounds for $|A_j|$ gives

$$ \begin{align*} 62 - (6 + 6 + |A_3| + \cdots + |A_h|) - h \leq 2^h. \end{align*} $$

Since, $|A_j| \leq 3$ for $j \geq 3$ and $h \leq 5$ , this yields

$$ \begin{align*} 62 - (12 + 3(h-2)) - h \leq 2^h. \end{align*} $$

For $h = 5$ , this becomes $62 - (12 + 9) - 5 = 36 \leq 2^5 = 32$ , which is a contradiction. A similar contradiction arises for smaller h.

A similar argument shows that A cannot contain exactly one element. Assuming $A = \{a_1\}$ leads to an analogous contradiction by constructing sets $A_j$ with $|A_1|=6$ and $|A_j| \leq 3$ for $j \geq 2$ , and arriving at an inequality $62 - (6 + 3(h-1)) - h \leq 2^h$ , which fails for $h \leq 5$ .

Therefore, $A = \emptyset $ . Hence, for every $a_i \in W$ , we have $Z(a_i) \in \{1, 2, 4, 5\}$ .

To obtain the final contradiction, define for each $a_i \in W$ ,

$$ \begin{align*} C_i = \begin{cases} \{ y \in V(\Gamma) \mid d(a_i, y) = 1 \} & \text{if } Z(a_i) \in \{1,2\} \\ \{ y \in V(\Gamma) \mid d(a_i, y) = 3 \} & \text{if } Z(a_i) \in \{4,5\}. \end{cases} \end{align*} $$

Then $|C_i| \leq 3$ for all i. For any vertex x outside $\bigcup _{i=1}^{h} C_i \cup W$ , its distance vector $D(x \mid W)$ has components restricted to at most two values each (for example, $\{2,3\}$ if $Z(a_i)\leq 2$ and $\{1,2\}$ if $Z(a_i)\geq 4$ ), yielding at most $2^h$ possible vectors. This leads to

$$ \begin{align*} |V(\Gamma)| - \sum_{i=1}^{h} |C_i| - |W| \leq 2^h, \end{align*} $$

from which we derive $62 - 3h - h \leq 2^h$ , that is, $62 - 4h \leq 2^h$ . For $h \leq 5$ , the inequality fails (for example, for $h=5$ , we have $62 - 20 = 42 < 32$ ). This final contradiction completes the proof that $h \geq 6$ .

Proof. If $n \in \{5,6\}$ , then the result follows from [Reference Raja, Pirzada and Redmond11, Proposition 6.3] and Lemma 2.1, respectively. Thus, we may assume that $n \geq 7$ . By [Reference Raja, Pirzada and Redmond11, Proposition 6.2], we have $\dim _m(\Gamma ((\mathbb {Z}_2)^n)) \leq n$ . Hence, it suffices to prove the reverse inequality, $\dim _m(\Gamma ((\mathbb {Z}_2)^n)) \geq n$ .

Let $x = (x_1, x_2, \ldots , x_n) \in V(\Gamma ((\mathbb {Z}_2)^n))$ . Since, x is a zero-divisor, at least one component $x_i$ is zero. Let t denote the number of zero components in x. Clearly, $1 \leq t \leq n-1$ .

For a vertex x define the following sets:

$$ \begin{align*} V_1(x) = \{ y \mid d(x, y) = 1 \}, \quad V_2(x) = \{ y \mid d(x, y) = 2 \}, \quad V_3(x) = \{ y \mid d(x, y) = 3 \}. \end{align*} $$

By [Reference Anderson and Livingston1, Theorem 2.3], the diameter of $\Gamma (R)$ is at most 3, so

$$ \begin{align*} V\bigg(\Gamma\bigg(\prod_{i=1}^n \mathbb{Z}_2\bigg)\bigg) = V_1(x) \cup V_2(x) \cup V_3(x) \cup \{x\}. \end{align*} $$

We have the following cardinalities:

$$ \begin{align*} |V_1(x)| = 2^t - 1, \quad |V_3(x)| = 2^{n-t} - 2, \quad |V_2(x)| = 2^n - 2^t - 2^{n-t}. \end{align*} $$

For $t \neq n-1$ , it follows that $|V_2(x)| \geq |V_1(x)|$ and $|V_2(x)| \geq |V_3(x)|$ .

Let $W = \{a_1, a_2, \ldots , a_h\}$ be a resolving set for $\Gamma (\prod _{i=1}^n \mathbb {Z}_2)$ . We will prove that $h \geq n$ by considering the following cases.

Case 1: $t \neq n-1$ for all $a_i \in W$ , $1 \leq i \leq h$ . Since, $t \neq n-1$ , we have $|V_2(a_i)| \geq |V_1(a_i)|$ and $|V_2(a_i)| \geq |V_3(a_i)|$ for each $a_i \in W$ .

For each $a_i \in W$ , define

$$ \begin{align*} A(a_i) = \begin{cases} V_1(a_i) & \text{if } |V_1(a_i)| \leq |V_3(a_i)| \\ V_3(a_i) & \text{otherwise}. \end{cases} \end{align*} $$

Let $A = \bigcup _{i=1}^{h} A(a_i)$ . For every vertex $x \in V(\Gamma ((\mathbb {Z}_2)^n)) \setminus (A \cup W)$ , consider its distance vector $D(x \mid W) = (d_1, \ldots , d_h)$ . By the construction of A, for $1 \leq i \leq h$ , the distance $d_i$ is restricted to two possible values:

• • if $A(a_i) = V_1(a_i)$ , then $d_i \neq 1$ , so $d_i \in \{2, 3\}$ ;

• • if $A(a_i) = V_3(a_i)$ , then $d_i \neq 3$ , so $d_i \in \{1, 2\}$ .

This means that there are at most $2^h$ distinct distance vectors for vertices in the set $V(\Gamma ((\mathbb {Z}_2)^n)) \setminus (A \cup W)$ . This implies

$$ \begin{align*} |V(\Gamma((\mathbb{Z}_2)^n))| - |A| - |W| \leq 2^h. \end{align*} $$

Substituting $|V(\Gamma ((\mathbb {Z}_2)^n))| = 2^n - 2$ ,

$$ \begin{align*} 2^n - 2 - |A| - h \leq 2^h. \end{align*} $$

Since, $|A(a_i)| \leq \min (|V_1(a_i)|, |V_3(a_i)|) \leq 2^{\lfloor n/2 \rfloor }$ for each $a_i$ , it follows that $|A| \leq h \cdot 2^{\lfloor n/2 \rfloor }$ . Therefore,

$$ \begin{align*} 2^n - 2 - h \cdot 2^{\lfloor n/2 \rfloor} - h \leq 2^h. \end{align*} $$

We now show that this inequality forces $h \ge n$ . Assume, to the contrary, that $h \le n-1$ . Since, the function $f(h) = 2^h + h \cdot 2^{\lfloor n/2 \rfloor } + h$ is increasing in h, it suffices to check the inequality for the largest possible value $h = n-1$ . Substituting $h = n-1$ yields

$$ \begin{align*} 2^n - 2 \stackrel{?}{\leq} 2^{n-1} + (n-1)\cdot 2^{\lfloor n/2 \rfloor} + (n-1). \end{align*} $$

For $n \geq 7$ , the right-hand side is strictly less than $2^n - 2$ . For instance, for $n=7$ we have $2^7 - 2 = 126$ while the right-hand side is $2^6 + 6\cdot 2^3 + 6 = 64 + 48 + 6 = 118$ . The gap widens as n increases. This contradicts our derived inequality. Hence, our assumption $h \le n-1$ is false, and we must have $h \ge n$ .

Case 2: $t = n-1$ for some $a_i \in W$ , $1 \leq i \leq h$ . Suppose $a_i \in W$ has $t = n-1$ zero components. Then

$$ \begin{align*} |V_1(a_i)| = 2^{n-1} - 1, \quad |V_3(a_i)| = 0. \end{align*} $$

For such $a_i$ , define $A(a_i) = V_3(a_i) = \emptyset $ .

For any other $a_j \in W$ with $t \neq n-1$ , define $A(a_j)$ as in Case 1, that is,

$$ \begin{align*} A(a_j) = \begin{cases} V_1(a_j) & \text{if } |V_1(a_j)| \leq |V_3(a_j)| \\ V_3(a_j) & \text{otherwise}. \end{cases} \end{align*} $$

Let $A = \bigcup _{i=1}^{h} A(a_i)$ . The same reasoning as in Case 1 applies. For every $x \in V(\Gamma ((\mathbb {Z}_2)^n)) \setminus (A \cup W)$ , the distance vector $D(x \mid W)$ has components each confined to two values. Hence,

$$ \begin{align*} |V(\Gamma((\mathbb{Z}_2)^n))| - |A| - |W| \leq 2^h. \end{align*} $$

Vertices $a_i$ with $t = n-1$ contribute nothing to $|A|$ (since $A(a_i) = \emptyset $ ), and for the others $|A(a_j)| \leq 2^{\lfloor n/2 \rfloor }$ , so $|A| \leq h \cdot 2^{\lfloor n/2 \rfloor }$ . Thus,

$$ \begin{align*} 2^n - 2 - h \cdot 2^{\lfloor n/2 \rfloor} - h \leq 2^h, \end{align*} $$

which again forces $h \geq n$ for $n \geq 7$ .

In both cases, we conclude that $h \geq n$ , completing the proof.

Proof. Suppose that $x-a$ is an edge of ${\Gamma }(R)$ . Hence, $ax= 0$ . Since, $Rx=Ry$ , we have $\operatorname {ann}_R(x)=\operatorname {ann}_R(y)$ and so we deduce that $ay= 0$ . This means that $y-a$ is an edge of ${\Gamma }(R)$ and thus $N(x)\subseteq N(y)$ . Similarly, $N(y)\subseteq N(x)$ , as desired.

Proof. Let $R \cong \prod _{i=1}^n R_i$ , where each $(R_i, \mathfrak {m}_i)$ is a finite local ring with $\mathfrak {m}_i^2 = 0$ and $R_i$ is not a field. Let $a_i \in \mathfrak {m}_i \setminus \{0\}$ .

A key observation is that if $x, y \in Z(R)^*$ with $\operatorname {ann}(x) = \operatorname {ann}(y)$ , then $N(x) = N(y)$ . Consequently, at least one of x or y must belong to any resolving set W for $\Gamma (R)$ . Since, $|\{\operatorname {ann}(x) \mid x \in Z(R)^*\}| = 3^n - 2$ , we can take a resolving set of the form $W = Z(R)^* \setminus A$ , where

$$ \begin{align*} A = \{(x_1, \ldots, x_n) \in V(\Gamma(R)) \mid x_i \in \{0, 1, a_i\} \text{ for all } i\}. \end{align*} $$

This immediately gives the lower bound:

$$ \begin{align*} \dim_m(\Gamma(R)) \geq |Z(R)^*| - 3^n + 2. \end{align*} $$

We now prove the matching upper bound by showing that $W = Z(R)^* \setminus A$ is indeed a resolving set. Partition A into four subsets:

$$ \begin{align*} V_1 &= \{x \in A \mid x_i \in \{0,1\} \text{ for all } i\}, \\ V_2 &= \{x \in A \mid x_i \in \{0,a_i\} \text{ for all } i\}, \\ V_3 &= \{x \in A \mid x_i \neq 0 \text{ for all } i\}, \\ V_4 &= A \setminus (V_1 \cup V_2 \cup V_3). \end{align*} $$

We show that all vertices in A have distinct distance vectors to W.

Vertices within $V_1$ . The set $\{(1+a_1,0,\ldots ,0), \ldots , (0,\ldots ,0,1+a_n)\} \subseteq W$ resolves $V_1$ (compare Theorem 2.2).

Vertices within $V_2$ . For $x \neq y \in V_2$ , define $z \in W$ by $z_i = a_i$ if $x_i = a_i$ and $z_i = 1+a_i$ if $x_i = 0$ . Then $d(x,z) = 1 \neq 2 = d(y,z)$ .

Vertices within $V_3$ . For $x \neq y \in V_3$ , define $z \in W$ by $z_i = a_i$ if $x_i = 1$ and $z_i = 1+a_i$ if $x_i = a_i$ . Then $d(x,z) = 3 \neq 2 = d(y,z)$ .

Vertices within $V_4$ . For $x \neq y \in V_4$ , if $x_i = 1$ and $y_i \neq 1$ for some i, then some $z \in W$ yields $d(x,z) = 3 \neq d(y,z)$ . Otherwise, define $z \in W$ by $z_i = 0$ if $x_i = 1$ , $z_i = 1+a_i$ if $x_i = 0$ and $z_i = a_i$ if $x_i = a_i$ . Then $d(x,z) = 1 \neq 2 = d(y,z)$ .

Vertices across different $V_j$ . For $x \in V_2$ and $y \notin V_2$ , there exists $z \in W$ with $d(x,z) = 1 \neq 2 = d(y,z)$ . For $x \in V_1$ and $y \in V_3 \cup V_4$ , a suitable $z \in W$ can be constructed such that $d(x,z) \neq d(y,z)$ (for example, $d(x,z)=1$ and $d(y,z) \in \{2,3\}$ ). Other cross-type distinctions follow similarly.

Since, all pairs of distinct vertices in A are resolved by W, we conclude that

$$ \begin{align*} \dim_m(\Gamma(R)) \leq |W| = |Z(R)^*| - 3^n + 2. \end{align*} $$

Combining both inequalities yields the exact formula:

$$ \begin{align*} \dim_m(\Gamma(R)) = |Z(R)^*| - 3^n + 2. \\[-36pt] \end{align*} $$

Proof. This proof is an elaboration of the proof of Theorem 2.5. Assume that $x,y \in Z(R)^*$ such that $\operatorname {ann}(x)=\operatorname {ann}(y)$ . Then $N(x)=N(y)$ . So we can let $x\in W$ or $y\in W$ , where W is a resolving set for the graph ${\Gamma }(R)$ . This, together with $|\{\operatorname {ann}(x) \mid x\in Z(R)^*\}|=3^{n}2^{m+k}-2$ implies that we can let $Z(R)^*\setminus A\subseteq W$ , where $A=\{(x_1,\ldots , x_{n+m+k})\in V({\Gamma }(R)) \mid x_i\in \{0,1,a_i\}\,\mathrm {where}\, 0\neq a_i\in \mathfrak {m}_i\}$ . Hence, $\dim _m({\Gamma }(R)) \geq |Z(R)^*|-3^{n}2^{m+k}+2$ .

Now, we show that $\dim _m({\Gamma }(R))\leq |Z(R)^*|-3^{n}2^{m+k}+2$ . By a similar proof to that of Theorem 2.5, we can easily see that W is a resolving set, where $W=Z(R)^*\setminus A$ and hence $\dim _m({\Gamma }(R))\leq |W|=|Z(R)^*|-3^{n}2^{m+k}+2$ .