# PID and UFD are equivalent in a Dedekind domain

This article shows that if $A$ is a Dedekind domain, then $A$ is a UFD if and only if it is a PID. Note that this result implies the more specific result given in the article unique factorization and ideals in ring of integers.

Since any PID is a UFD, we need only prove the other direction. So assume $A$ is a UFD, let $\mathfrak{p}$ be a nonzero (proper) prime ideal, and choose $0\neq x\in\mathfrak{p}$. Note that $x$ is a nonunit since $\mathfrak{p}$ is a proper ideal. Since $A$ is a UFD, we may write $x$ uniquely (up to units) as $x=p_{1}^{a_{1}}\cdots p_{k}^{a_{k}}$ where the $p_{i}$ are distinct irreducibles in $A$, the $a_{i}$ are positive integers, and $k>0$ since $x$ is not a unit. Since $\mathfrak{p}$ is prime and $x\in\mathfrak{p}$, it follows that some $p_{i}$, say $p_{1}$, is in $\mathfrak{p}$. Then $(p_{1})\subset\mathfrak{p}$. But $(p_{1})$ is prime since clearly in a UFD any ideal generated by an irreducible is prime. Since $A$ is Dedekind and thus has Krull dimension 1, it must be that $(p_{1})=\mathfrak{p}$ and thus $\mathfrak{p}$ is principal.

Title PID and UFD are equivalent in a Dedekind domain PIDAndUFDAreEquivalentInADedekindDomain 2013-03-22 17:53:45 2013-03-22 17:53:45 rm50 (10146) rm50 (10146) 7 rm50 (10146) Theorem msc 16D25 msc 13G05 msc 13A15 msc 11N80