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