Irreducible element

In

invertible (that is, is not a unit

), and is not the product of two non-invertible elements.

The irreducible elements are the terminal elements of a

number fields are not unique factorization domains, and, therefore, that some irreducible elements can appear in some factorization of an element and not in other factorizations of the same element. The ignorance of this fact is the main error in many of the wrong proofs of Fermat's Last Theorem that were given during the three centuries between Fermat's statement and Wiles's proof of Fermat's Last Theorem

.

If $R$ is an integral domain, then $a$ is an irreducible element of $R$ if and only if, for all $b,c\in R$ , the equation $a=bc$ implies that the ideal generated by $a$ is equal to the ideal generated by $b$ or equal to the ideal generated by $c$ . This equivalence does not hold for general commutative rings, which is why the assumption of the ring having no nonzero zero divisors is commonly made in the definition of irreducible elements. It results also that there are several ways to extend the definition of an irreducible element to an arbitrary commutative ring.^[1]

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,^[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 $D$ 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 $D$ .

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.^[3]

Notes

^ 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.