Inalgebra, anirreducible element of anintegral domain is a non-zero element that is notinvertible (that is, is not aunit), and is not the product of two non-invertible elements.

The irreducible elements are the terminal elements of afactorization process; that is, they are the factors that cannot be further factorized. If the irreducible factors of every non-zero non-unit element are uniquely defined,up to the multiplication by a unit, then the integral domain is called aunique factorization domain, but this does not need to happen in general for every integral domain. It was discovered in the 19th century that therings of integers of somenumber 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 ofFermat's Last Theorem that were given during the three centuries between Fermat's statement andWiles's proof of Fermat's Last Theorem.

If${\displaystyle R}$ is an integral domain, then${\displaystyle a}$ is an irreducible element of${\displaystyle R}$ if and only if, for all${\displaystyle b,c\in R}$, the equation${\displaystyle a=bc}$ implies that the ideal generated by${\displaystyle a}$ is equal to the ideal generated by${\displaystyle b}$ or equal to the ideal generated by${\displaystyle c}$. 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 arbitrarycommutative ring.^[1]

Let${\displaystyle R}$ be an integral domain. An element${\displaystyle a\in R}$ isirreducible if it is not a unit and whenever${\displaystyle a=bc}$, either${\displaystyle b}$ or${\displaystyle c}$ is a unit.

Anderson and Valdes-Leon in 1996 defined irreducible elements in arbitrarycommutative rings (potentially withzero divisors): they define elements to bevery strongly irreducible,m-irreducible,strongly irreducible, andirreducible (in decreasing order of strength) based on different conditions on${\displaystyle b}$ and${\displaystyle c}$ (Theorem 2.13).^[1] All definitions require${\displaystyle a}$ to be not a unit.^[1] Theirvery strongly irreducible corresponds to the definition above. The conditionm-irreducible is that whenever${\displaystyle a=bc}$,${\displaystyle (b)=(1)}$ or${\displaystyle (b)=(a)}$. The conditionstrongly irreducible is that whenever${\displaystyle a=bc}$,${\displaystyle a}$ is equivalent to either${\displaystyle b}$ or${\displaystyle c}$ up to multiplication by a unit. Finally theirirreducible is the condition that, whenever${\displaystyle a=bc}$, either${\displaystyle (a)=(b)}$ or${\displaystyle (a)=(c)}$.^[1]

Irreducible elements should not be confused withprime elements. (A non-zero non-unit element${\displaystyle a}$ in acommutative 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 anintegral domain, every prime element is irreducible,^[a][2] but the converse is not true in general. The converse is true forunique factorization domains^[2] (or, more generally,GCD domains).

Moreover, while an ideal generated by a prime element is aprime ideal, it is not true in general that an ideal generated by an irreducible element is anirreducible 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 thequadratic integer ring${\displaystyle \mathbf{Z}[\sqrt{-5}],}$ it can be shown usingnorm 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

• Ring theory
• Algebraic properties of elements

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