# Irreducible element

In algebra, an irreducible element of a integral domain is a non-zero element that is not 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 factorization process; that is, they are the factors that cannot be further factorized. The irreducible factors of an element are uniquely defined, up to the multiplication by a unit, if the integral domain is a unique factorization domain. It has been discovered in the 19th century that the rings of integers of some 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 have been given during the three centuries between Fermat's statement and Wiles's proof of Fermat's Last Theorem.

The definition can be and is usually extended verbatim to the elements of an arbitrary commutative ring. For a general ring $R$ , an element $a$ of $R$ is called irreducible if it is neither left-invertible nor right-invertible, and if there exists no left-invertible element $b\in R$ together with a right-invertible element $c\in R$ such that $a=bc$ .

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 zero divisors is commonly made in the definition of irreducible elements.

## 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, but the converse is not true in general. The converse is true for unique factorization domains (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.