# Irreducible element

In abstract algebra, a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units.

## Relationship with prime elements

Irreducible elements should not be confused with prime elements. (A non-zero non-unit element $a$ in a commutative ring $R$ is called prime if, whenever $a | bc$ for some $b$ and $c$ in $R,$ then $a|b$ or $a|c.)$ In an integral domain, every prime element is irreducible,[1][2] but the converse is not true in general. The converse is true for unique factorization domains[2] (or, more generally, GCD domains.)

Moreover, while an ideal generated by a prime element is a prime ideal, it is not true in general that an ideal generated by an irreducible element is an irreducible ideal. However, if $D$ is a GCD domain, and $x$ is an irreducible element of $D$, then the ideal generated by $x$ is a prime ideal of $D$.[3]

## Example

In the quadratic integer ring $\mathbf{Z}[\sqrt{-5}],$ it can be shown using norm arguments that the number 3 is irreducible. However, it is not a prime element in this ring since, for example,

$3 \mid \left(2 + \sqrt{-5}\right)\left(2 - \sqrt{-5}\right)=9,$

but $3$ does not divide either of the two factors.[4]

1. ^ Consider $p$ a prime that is reducible: $p=ab.$ Then $p | ab \Rightarrow p | a$ or $p | b.$ Say $p | a \Rightarrow a = pc,$ then we have $p=ab=pcb \Rightarrow p(1-cb)=0.$ Because $R$ is an integral domain we have $cb=1.$ So $b$ is a unit and $p$ is irreducible.