1.18 $\mathbb{Z}_m$ is a field $\iff$ $m$ is a prime

Proof:
$(\implies)$ Assume $\mathbb{Z}_m$ is a field. Then $\forall z \in \mathbb{Z}_m z \neq 0 \implies \exists e \in \mathbb{Z}_m, ze = 1 \: \text{(mod m)} \implies \exists t \in \mathbb{Z}, ze = 1 + mt \implies gcd(z, m) = 1 \implies \text{m is prime}$

$(\Longleftarrow)$ Assume $m$ is a prime. Then $\forall a \in \mathbb{Z}_{0<a<m}, gcd(m, a) = 1 \implies \exists x, y \in \mathbb{Z}, ax+my = 1 \: \text{(by bezout's identity)} \implies ax = 1 \: \text{(mod m)} \implies \text{x is the multiplicative inverse of a}$