Ring of Integers Modulo Prime is Field

Theorem
Let $m \in \Z: m \ge 2$.

Let $\left({\Z_m, +, \times}\right)$‎ be the ring of integers modulo $m$.

Then:
 * $m$ is prime

iff:
 * $\left({\Z_m, +, \times}\right)$ is a field.

Corollary
Let $\left({\Z_m, +, \times}\right)$‎ be the ring of integers modulo $m$.

Then:
 * $m$ is prime

iff:
 * $\left({\Z_m, +, \times}\right)$ is an integral domain.

Prime Modulus
$\left({\Z_m, +, \times}\right)$‎ is a commutative ring with unity by definition.

From Multiplicative Group of Integers Modulo m, $\left({\Z'_m, \times}\right)$ is an abelian group.

$\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$.

Now when $m$ is prime, we have, from Set of Coprime Integers:
 * $\Z'_m = \left\{{\left[\!\left[{1}\right]\!\right]_m, \left[\!\left[{2}\right]\!\right]_m, \ldots, \left[\!\left[{m-1}\right]\!\right]_m}\right\}$

That is:
 * $\Z'_m = \Z_m \setminus \left\{{\left[\!\left[{0}\right]\!\right]_m}\right\}$

where $\setminus$ denotes set difference.

Hence the result.

Composite Modulus
Now suppose $m \in \Z: m \ge 2$ is composite.

Then $\exists k, l \in \N^*: 1 < k, l < m: m = k l$.

Thus $\left[\!\left[{0}\right]\!\right]_m = \left[\!\left[{m}\right]\!\right]_m = \left[\!\left[{k l}\right]\!\right]_m = \left[\!\left[{k}\right]\!\right]_m \times \left[\!\left[{l}\right]\!\right]_m$.

Thus $\left({\Z_m, +, \times}\right)$‎ is a ring with zero divisors.

Therefore $\left({\Z_m, +, \times}\right)$ is not a division ring and therefore not a field.

Proof of Corollary
We have that a field is an integral domain.

We also have that a finite integral domain is a field.

Hence the result.