# Equivalence of Definitions of Irreducible Element of Ring

## Theorem

The following definitions of the concept of Irreducible Element of Ring are equivalent:

### Definition 1

$x$ is defined as irreducible if and only if it has no non-trivial factorization in $D$.

That is, if and only if $x$ cannot be written as a product of two non-units.

### Definition 2

$x$ is defined as irreducible if and only if the only divisors of $x$ are its associates and the units of $D$.

That is, if and only if $x$ has no proper divisors.

## Proof

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.

### $(1)$ implies $(2)$

By definition:

$x$ has no non-trivial factorization in $D$.

Let $x = y \circ z$ for some $y, z \in D$.

By definition, it cannot be the case that neither $y$ nor $z$ are units of $D$.

So either $y$ or $z$ is a unit of $D$.

Without loss of generality, suppose $y$ is a unit of $D$.

Then by definition $z$ is an associate of $x$.

Contrariwise, suppose $z$ is a unit of $D$.

Then by definition $y$ is an associate of $x$.

Thus both $y$ and $z$ are either a unit of $D$ or an associate of $x$.

$x = y \circ z$ is an arbitrary factorization of $x$ in $D$.

It follows that the only divisors of $x$ are its associates and the units of $D$.

$\Box$

### $(2)$ implies $(1)$

Then by definition:

the only divisors of $x$ are its associates and the units of $D$.

Let $x = y \circ z$.

Then either:

$y$ is an associate of $x$ and $z$ is a unit of $D$

or:

$z$ is an associate of $x$ and $y$ is a unit of $D$.

In either case, $y \circ z$ is a trivial factorization of $x$.

$\blacksquare$