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.

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.