every PID is a UFD - alternative proof

Proposition. If $R$ is a principal ideal domain, then $R$ is a unique factorization domain.

Proof. Recall, that due to Kaplansky Theorem (see this article (http://planetmath.org/EquivalentDefinitionsForUFD) for details) it is enough to show that every nonzero prime ideal in $R$ contains a prime element.

On the other hand, recall that an element $p\in R$ is prime if and only if an ideal $(p)$ generated by $p$ is nonzero and prime.

Thus, if $P$ is a nonzero prime ideal in $R$, then (since $R$ is a PID) there exists $p\in R$ such that $P=(p)$. This completes the proof. $\mathrm{\square }$

Title	every PID is a UFD - alternative proof
Canonical name	EveryPIDIsAUFDAlternativeProof
Date of creation	2013-03-22 19:04:26
Last modified on	2013-03-22 19:04:26
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Theorem
Classification	msc 13F07
Classification	msc 16D25
Classification	msc 13G05
Classification	msc 11N80
Classification	msc 13A15