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, or equivalently, if every factoring of such element contains at least one unit.

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 $a\mid bc$ for some $b$ and $c$ in $R,$ then $a\mid b$ or $a\mid c.$ ) 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 as noted above $x$ is prime, and so the ideal generated by $x$ is a prime (hence irreducible) 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]

References[]

^ Consider $p$ a prime element of $R$ and suppose $p=ab.$ Then $p\mid ab\Rightarrow p\mid a$ or $p\mid b.$ Say $p\mid a\Rightarrow a=pc,$ 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.
^ ^a ^b Sharpe (1987) p.54
^ "Archived copy". Archived from the original on 2010-06-20. Retrieved 2009-03-18.{{cite web}}: CS1 maint: archived copy as title (link)
^ William W. Adams and Larry Joel Goldstein (1976), Introduction to Number Theory, p. 250, Prentice-Hall, Inc., ISBN 0-13-491282-9