Prime Element of Integral Domain is Irreducible

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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 $\struct {D, +, \circ}$.

Aiming for a contradiction, suppose:

$p = a \circ b$

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 $p \divides b$


Without loss of generality, suppose $p \divides a$.

That is:

$a = p \circ c$

for some $c \in D$.

\(\ds p \circ 1_D\) \(=\) \(\ds p\) Definition of Unity of Ring
\(\ds \) \(=\) \(\ds a \circ b\)
\(\ds \) \(=\) \(\ds p \circ c \circ b\) $a = p \circ c$
\(\ds \leadsto \ \ \) \(\ds 1_D\) \(=\) \(\ds c \circ b\) Cancellation Law for Ring Product of Integral Domain

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 $\struct {D, +, \circ}$.

$\blacksquare$