Prime Element of Integral Domain is Irreducible

Theorem
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.

Let $p$ be a prime element of $\struct {D, +, \circ}$.

Then $p$ is an irreducible element of $\struct {D, +, \circ}$.

Proof
By definition of prime element, $p$ is neither zero nor a unit of $\struct {D, +, \circ}$.


 * $p = a \circ b$
 * $p = a \circ b$

for some non-units $a, b \in D$.

From Element of Integral Domain is Divisor of Itself:


 * $p \divides a \circ b$

By definition of prime element:


 * $p \divides a$ or $p \divides b$

, suppose $p \divides a$.

That is:


 * $a = p \circ c$

for some $c \in D$.

Thus $b$ is a unit of $\struct {D, +, \circ}$.

This contradicts the assertion that $a$ and $b$ are not units of $\struct {D, +, \circ}$.

Thus by Proof by Contradiction $p$ is an irreducible element of $\struct {D, +, \circ}$.