Exercise A0.V.2.10

Proposition

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

Proof

The forward direction was proven in 2.9. We will now prove the converse.

Let $R$ be a Noetherian domain such that every prime ideal of height $1$ in $R$ is principal.

Suppose $r \in R$ is irreducible. By Hauptidealsatz there exists a prime ideal $I$ of height $1$ such that $r \in I.$ By our main hypothesis, $I$ is principal. Let $I = (i).$ In that case, $r = iv$ for some $v \in R.$ Since $r$ is irreducible, either $i$ is a unit or $v$ is a unit. But $(i)$ is prime and this precludes $i$ from being a unit. Therefore, $v$ is a unit. We conclude that $r$ and $i$ are associate, that is, $(i) = (r).$ This proves that $r$ is prime. Since $r$ was generic, every irreducible of $R$ is prime.

Let $I_1 \subseteq I_2 \subseteq I_3 \subseteq \ldots$ be an ascending chain of principal ideals. By Noetherianity of $R$ this chain stabilizes.

Thus we can conclude that $R$ is a UFD.

$\blacksquare$