Solution: Since $\mathbb{Z} _5$ is a commutative ring, $\mathbb{Z} _5[x]$ is also a commutative ring. We just need to show that it does not have a zero divisor. That is, if $p(x), q(x) \in \mathbb{Z} _5[x]$ such that \[ p(x) q(x) = 0 \implies p(x) = 0 \text{ or } q(x) = 0. \] We will show that if $p(x) \neq 0$ and $q(x) \neq 0$, then $p(x) q(x) \neq 0$. Let \begin{align*} & p(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_m x^m, a_i \in \mathbb{Z} _5 \text{ and } a_m \neq 0 \\ & q(x) = b_0 + b_1 x + b_2 x^2 + \dots + b_n x^n, b_i \in \mathbb{Z} _5 \text{ and } b_n \neq 0. \end{align*} Then \begin{align*} p(x) q(x) = a_0 b_0 + \dots + a_m b_n x^{m + n}. \end{align*} Since $a_m , b_n \in \mathbb{Z} _5$, $a_m, b_n$ both are nonzero and $\mathbb{Z} _5$ is an integral domain (it is a filed), so $a_m b_n \neq 0$. Thus, $p(x) q(x) \neq 0$. Hence, $\mathbb{Z} _5[x]$ is an integral domain.