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In algebra, an **irreducible element** of an 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 was 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 were given during the three centuries between Fermat's statement and Wiles's proof of Fermat's Last Theorem.

If is an integral domain, then is an irreducible element of if and only if, for all , the equation implies that the ideal generated by is equal to the ideal generated by or equal to the ideal generated by . 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]}

Irreducible elements should not be confused with prime elements. (A non-zero non-unit element in a commutative ring is called prime if, whenever for some and in then or ) 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 is a GCD domain and is an irreducible element of , then as noted above is prime, and so the ideal generated by is a prime (hence irreducible) ideal of .

In the quadratic integer ring 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,

but 3 does not divide either of the two factors.^{[3]}

**^**Consider a prime element of and suppose Then or Say then we have Because is an integral domain we have So is a unit and is irreducible.

**^**Anderson, D. D.; Valdes-Leon, Silvia (1996-06-01). "Factorization in Commutative Rings with Zero Divisors".*Rocky Mountain Journal of Mathematics*.**26**(2): 439–480. doi:10.1216/rmjm/1181072068. ISSN 0035-7596.- ^
^{a}^{b}Sharpe, David (1987).*Rings and factorization*. Cambridge University Press. p. 54. ISBN 0-521-33718-6. Zbl 0674.13008. **^**William W. Adams and Larry Joel Goldstein (1976),*Introduction to Number Theory*, p. 250, Prentice-Hall, Inc., ISBN 0-13-491282-9