MarketIrreducible element
Company Profile

Irreducible element

In algebra, an irreducible element of an integral domain is a non-zero element that is not invertible, and is not the product of two non-invertible elements.

Definition in an integral domain
Let R be an integral domain. An element a \in R is irreducible if it is not a unit and whenever a = bc, either b or c is a unit. == Definition in rings with zero divisors ==
Definition in rings with zero divisors
Anderson and Valdes-Leon in 1996 defined irreducible elements in arbitrary commutative rings (potentially with zero divisors): they define elements to be very strongly irreducible, m-irreducible, strongly irreducible, and irreducible (in decreasing order of strength) based on different conditions on b and c (Theorem 2.13). All definitions require a to be not a unit. Their very strongly irreducible corresponds to the definition above. The condition m-irreducible is that whenever a = bc, (b) = (1) or (b) = (a). The condition strongly irreducible is that whenever a = bc, a is equivalent to either b or c up to multiplication by a unit. Finally their irreducible is the condition that, whenever a = bc, either (a) = (b) or (a) = (c). == Relationship with prime 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 ==
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. == See also ==
tickerdossier.comtickerdossier.substack.com