irreducible of a UFD is prime
Any irreducible element of a factorial ring D is a prime element
of D.
Proof. Let p be an arbitrary irreducible element of D. Thus p is a non-unit. If ab∈(p)∖{0}, then ab=cp with c∈D. We write a,b,c as products of irreducibles:
a=p1⋯pl,b=q1⋯qm,c=r1⋯rn |
Here, one of those first two products may me empty, i.e. it may be a unit. We have
p1⋯plq1⋯qm=r1⋯rnp. | (1) |
Due to the uniqueness of prime factorization, every factor rk is an associate
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 ideal
of D, and its generator
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 |