The result below is from an exercise in the wonderful book Commutative Algebra by Ferretti.

Theorem

Let $A$ be a domain. If every prime ideal of $A$ is principal, then $A$ is a Principal Ideal Domain.

The proof that I have is adapted from the blog by Blum-Smith. This is an excellent proof of Cohen’s Theorem, which follows the same logic used here, though the lemma is instead applied to show that the finite generation of $I+(a)$ and $(I:a)$ implies the finite generation of $I$.

We will start with a

Lemma

Let $I \subset A$ be an ideal and $a \in A$ be arbitrary. If $I + (a)$ and $(I : a)$ are principal, then $I$ is principal.

Proof: Let $(x_1) = I + (a)$ and $(x_2) = (I : a)$. We first note that $I \subset I + (a)$ and $I \subset (I : a)$. Let $i \in I$ be arbitrary; then $i = s x_1$ and $a = t x_1$ for some $s, t \in A$. It follows that $sa = s(t x_1) = t(s x_1) = ti \in I$, so $s \in (I : a)$. Thus, $s = r x_2$ for some $r \in A$.

This gives $i = r x_1 x_2$, so we have $I \subset (x_1 x_2)$. Conversely, note that $(I + (a))(I : a) \subset I$. Since $(I + (a))(I : a) = (x_1 x_2)$, we have $(x_1 x_2) = I$, and thus $I$ is principal.

Proof of Theorem

Suppose that $A$ has some ideals that are not principal. Let $\mathcal{M}$ be the set of all such ideals, ordered by inclusion. Take some chain in this set, $P_1 \subset P_2 \subset \dots$, and let $I = \cup_i P_i$. This union of an ascending chain of ideals is itself an ideal.

If $I$ were principal, say $I = (x)$, then for some $P_i$ we would have $x \in P_i$, in which case $P_i = I$. This contradicts the assumption that the $P_i$ are not principal; thus, $I$ is not principal. By Zorn’s Lemma, this set has a maximal element; call it $I$.

By the assumption that $I$ is not principal, we know it is not prime. So, take $a, b \in A$ such that $ab \in I$ with $a, b \notin I$. By the maximality of $I$, we get that $I + (a)$ is principal. By the contrapositive of the lemma, $(I : a)$ must not be principal. Since $I \subset (I : a)$ and $I$ is maximal in $\mathcal{M}$, we must have $I = (I : a)$. But then $b \in (I : a) = I$, which is a contradiction. Thus, $I$ must be prime, so it is principal.