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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like
$\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.