irreducible of a UFD is prime

Proof.  Let p be an arbitrary irreducible element of D.  Thus p is a non-unit.  If  ab(p){0},  then  ab=cp  with  cD.  We write a,b,c as products of irreducibles:


Here, one of those first two products may me empty, i.e. it may be a unit.  We have

p1plq1qm=r1rnp. (1)

Due to the uniqueness of prime factorizationMathworldPlanetmath, every factor rk is an associateMathworldPlanetmath of certain of the l+m irreducibles on the left hand side of (1).  Accordingly, p has to be an associate of one of the pi’s or qj’s.  It means that either  a(p)  or  b(p).  Thus, (p) is a prime idealMathworldPlanetmathPlanetmath of D, and its generatorPlanetmathPlanetmathPlanetmath must be a prime element.

Title irreducible of a UFD is prime
Canonical name IrreducibleOfAUFDIsPrime
Date of creation 2013-03-22 18:04:35
Last modified on 2013-03-22 18:04:35
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 7
Author pahio (2872)
Entry type Theorem
Classification msc 13G05
Classification msc 13F15
Related topic PrimeElementIsIrreducibleInIntegralDomain