Prime Element of Integral Domain is Irreducible

Theorem
Let $D$ be an integral domain whose unity is $1_D$.

Let $p$ be a prime element of $D$.

Then $p$ is an irreducible element of $D$.

Proof
By definition of prime element, $p$ is neither zero nor a unit of $D$.

Let $p = a b$ for some $a, b \in D$.

$a$ and $b$ are not units of $D$.

From Element of Integral Domain is Divisor of Itself, $p \divides a b$.

By definition of prime element, $p \divides a$ or $p \divides b$.

, suppose $p \divides a$.

That is, $a = p c$ for some $c \in D$.

Thus $b$ is a unit of $D$.

This contradicts $a$ and $b$ are not units of $D$.

Thus $p$ is an irreducible element of $D$.