Criterion for $\mathbb{Z}/p\mathbb{Z}$ to be a field

($\Leftarrow$) Suppose $p$ is prime, and let $\bar{a} \in \mathbb{Z}/p\mathbb{Z}$ be any nonzero residue class, represented by an integer $a$ with $1 \le a \le p - 1$. Since $p$ is prime and $p \nmid a$, the greatest common divisor of $a$ and $p$ is $\gcd(a,p) = 1$.

By Bézout's identity, there exist integers $u, v \in \mathbb{Z}$ such that $au + pv = 1$. Reducing both sides modulo $p$ kills the multiple $pv$, leaving $\bar{a}\,\bar{u} = \bar{1}$ in $\mathbb{Z}/p\mathbb{Z}$. Thus $\bar{u} = \bar{a}^{-1}$ is the multiplicative inverse of $\bar{a}$, and since every nonzero element is invertible, $\mathbb{Z}/p\mathbb{Z}$ is a field.

($\Rightarrow$) Conversely, suppose $p$ is composite, so $p = bc$ for integers $1 < b, c < p$. Because $b$ and $c$ are strictly between $0$ and $p$, their residue classes $\bar{b}$ and $\bar{c}$ are both nonzero in $\mathbb{Z}/p\mathbb{Z}$, yet their product is $\bar{b}\,\bar{c} = \overline{bc} = \bar{p} = \bar{0}$.

Thus $\bar{b}$ is a zero divisor, and no zero divisor can have a multiplicative inverse: if $\bar{b}$ had an inverse $\bar{b}^{-1}$, multiplying $\bar{b}\,\bar{c} = \bar{0}$ on the left by $\bar{b}^{-1}$ would force $\bar{c} = \bar{0}$, a contradiction. Hence $\mathbb{Z}/p\mathbb{Z}$ is not a field when $p$ is composite.