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 ${\displaystyle a}$ in a commutative ring ${\displaystyle R}$ is called prime if, whenever ${\displaystyle a|bc}$ for some ${\displaystyle b}$ and ${\displaystyle c}$ in ${\displaystyle R,}$ then ${\displaystyle a|b}$ or ${\displaystyle a|c.)}$ 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 ${\displaystyle D}$ 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 ideal of ${\displaystyle D}$.[3]

Example

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.[4]

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