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 $𝔭$ be a nonzero (proper) prime ideal, and choose $0\ne x\in 𝔭$. Note that $x$ is a nonunit since $𝔭$ is a proper ideal. Since $A$ is a UFD, we may write $x$ uniquely (up to units) as $x=p_1^{a_1}\mathrm{\cdots }{p}_k^{a_k}$ 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 $𝔭$ is prime and $x\in 𝔭$, it follows that some $p_i$, say $p_1$, is in $𝔭$. Then $(p_1)\subset 𝔭$. 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 $(p_1)=𝔭$ and thus $𝔭$ is principal.