irreducible of a UFD is prime
Proof. Let be an arbitrary irreducible element of . Thus is a non-unit. If , then with . We write as products of irreducibles:
Here, one of those first two products may me empty, i.e. it may be a unit. We have
Due to the uniqueness of prime factorization, every factor is an associate of certain of the irreducibles on the left hand side of (1). Accordingly, has to be an associate of one of the ’s or ’s. It means that either or . Thus, is a prime ideal of , and its generator must be a prime element.
|Title||irreducible of a UFD is prime|
|Date of creation||2013-03-22 18:04:35|
|Last modified on||2013-03-22 18:04:35|
|Last modified by||pahio (2872)|