Lemma 10.120.6. 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 10.120.5 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 10.120.3. It suffices to prove that an irreducible element $x$ is prime, see Lemma 10.120.5. Let $(x) \subset \mathfrak p$ be a prime minimal over $(x)$. Then $\mathfrak p$ has height $1$ by Lemma 10.60.11. 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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