Irreducible elements should not be confused with prime elements. (A non-zero non-unit element ${\displaystyle a}$  in a commutative ring ${\displaystyle R}$  is called prime if, whenever ${\displaystyle a\mid bc}$  for some ${\displaystyle b}$  and ${\displaystyle c}$  in ${\displaystyle R,}$  then ${\displaystyle a\mid b}$  or ${\displaystyle a\mid c.}$ ) In an integral domain, every prime element is irreducible,^[a][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 ${\displaystyle D}$  is a GCD domain and ${\displaystyle x}$  is an irreducible element of ${\displaystyle D}$ , then as noted above ${\displaystyle x}$  is prime, and so the ideal generated by ${\displaystyle x}$  is a prime (hence irreducible) ideal of ${\displaystyle D}$ .

In the quadratic integer ring ${\displaystyle \mathbf {Z} [{\sqrt {-5}}],}$  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,

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

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

• Irreducible polynomial