# 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 $(2)$

Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 1.

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 $D$.

Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 2.

$\Box$

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

Let $x$ be an irreducible element of $\struct {D, +, \circ}$ by definition 2.

Then by definition:

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

Let $x = y \circ z$.

Then either:

or:

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

Thus $x$ is an irreducible element of $\struct {D, +, \circ}$ by definition 1.

$\blacksquare$