The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.119.6. Let $R$ be a Noetherian domain. Then $R$ is a UFD if and only if every height $1$ prime ideal is principal.

Proof. Assume $R$ is a UFD and let $\mathfrak p$ be a height 1 prime ideal. Take $x \in \mathfrak p$ nonzero and let $x = a_1 \ldots a_ n$ be a factorization into irreducibles. Since $\mathfrak p$ is prime we see that $a_ i \in \mathfrak p$ for some $i$. By Lemma 10.119.5 the ideal $(a_ i)$ is prime. Since $\mathfrak p$ has height $1$ we conclude that $(a_ i) = \mathfrak p$.

Assume every height $1$ prime is principal. Since $R$ is Noetherian every nonzero nonunit element $x$ has a factorization into irreducibles, see Lemma 10.119.3. It suffices to prove that an irreducible element $x$ is prime, see Lemma 10.119.5. Let $(x) \subset \mathfrak p$ be a prime minimal over $(x)$. Then $\mathfrak p$ has height $1$ by Lemma 10.59.10. By assumption $\mathfrak p = (y)$. Hence $x = yz$ and $z$ is a unit as $x$ is irreducible. Thus $(x) = (y)$ and we see that $x$ is prime. $\square$


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