16D25-PIDAndUFDAreEquivalentInADedekindDomain.tex

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\pmtitle{PID and UFD are equivalent in a Dedekind domain}
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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.

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