Integral Domain of Prime Order is Field

Theorem
Let $\struct {\Z_p, +_p, \times_p}$‎ be the ring of integers modulo $p$.

The following statements are equivalent:
 * $(1): \quad p$ is a prime.
 * $(2): \quad \struct {\Z_p, +_p, \times_p}$ is an integral domain.
 * $(3): \quad \struct {\Z_p, +_p, \times_p}$ is a field.

Proof
By Prime Number iff Generates Principal Maximal Ideal and Maximal Ideal iff Quotient Ring is Field, $(1)$ implies $(3)$, and from Field is Integral Domain, $(3)$ implies $(2)$.

By definition of Integral Domain, $\Z_p$ is an integral domain $\struct {\Z_p^*, \times_p}$ is a semigroup.

Let $\ideal p$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.

From the Canonical Epimorphism from Integers by Principal Ideal, $\struct {\Z_p, +_p, \times_p}$ is isomorphic to $\struct {\Z, +, \times} / \ideal p$.

So, we can let $\map {q_p} m: \Z \to \Z_p$ be the quotient mapping from $\struct {\Z, +, \times}$ to $\struct {\Z_p, +_p, \times_p}$.

Let $0_p$ denote the zero of $\Z_p$.

Suppose $p = m n$ where $1 < m < p, 1 < n < p$.

Then in the ring $\Z_p$ we have $\map {q_p} m \ne 0_p, \map {q_p} n \ne 0_p$.

But as $q_p$ is an epimorphism and therefore obeys the morphism property, $\map {q_p} m \times_p \map {q_p} n = \map {q_p} {m n} = \map {q_p} p = 0_p$.

But by definition, $0_p \notin \Z_p^*$.

Thus if $p = m n$, then $\struct {\Z_p^*, \times_p}$ is not a semigroup.