Integral Domain of Prime Order is a 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.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 24$: Theorem $24.6$