Problem: Let $R$ be a commutative ring and $Z(R)$ denotes the zero divisor of $R$, that is,
\[
Z(R) \coloneqq \{ r \in R: \exists ~ s (\neq 0) \in R \text{ such that } rs = 0 \}.
\]
Define a graph $\Gamma (R)$ whose vertex set is nonzero zero divisors of $R$ and there exists an edge between $r_1$ and $r_2$ if and only if $r_1 \neq r_2$ and $r_1 r_2 = 0$. For example, the graph for $\mathbb{Z} _8$ will be linear with vertex set $2,4,6$ and there is an edge between $2,4$ and $4,6$. Prove that if $R = \mathbb{Z} _{p^2},~ p $ is a prime, then $\Gamma (R)$ is complete, that is between any two vertices, there is an edge.
Solution: Note that for $R = \mathbb{Z} _{p^2}$ the zero divisors are
\[
Z(R) = \left\{ kp: 0 \leq k \leq p - 1 \right\}.
\]
Therefore the vertex set will be
\[
V = \left\{ kp : 1 \leq k \leq p-1 \right\}.
\]
Take any two distinct vertices $k_1 p$ and $k_2 p$. We have
\[
\left( k_1 p \right) \left( k_2 p \right) \equiv 0 \left( \text{ mod } p^2 \right).
\]
Thus, between any two vertices, there is an edge and hence the corresponding graph is complete.