Integral Domain of Prime Order is Field

Theorem
Let $$\left({\Z_p, +_p, \times_p}\right)$$‎ be the ring of integers modulo $p$.

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

Proof

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


 * By the alternative definition of Integral Domain, $$\Z_p$$ is an integral domain iff $$\left({\Z_p^*, \times_p}\right)$$ is a semigroup.

Let $$\left({p}\right)$$ be the principal ideal of $\left({\Z, +, \times}\right)$ generated by $p$.

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

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

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 $$q_p \left({m}\right) \ne 0_p, q_p \left({n}\right) \ne 0_p$$.

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

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

Thus if $$p = m n$$, $$\left({\Z_p^*, \times_p}\right)$$ is not a semigroup.