## An integral domain whose every prime ideal is principal is a principal ideal domain

Let $R$ be an integral domain, and suppose that every prime ideal in $R$ is principal.

1. Suppose $R$ contains a nonprincipal ideal. Show that $R$ contains an inclusion-maximal nonprincipal ideal
2. Let $M$ be an inclusion-maximal nonprincipal ideal, and let $a,b \in R$ such that $ab \in M$ while $a,b \notin M$. (These exist because $M$ is not prime by hypothesis.) Let $M_a = (M,a)$ and $M_b = (M,b)$, and define $N = \{r \in R \ |\ rM_a \subseteq M \}$. Prove that $M_a = (\alpha)$ and $M_b = (\beta)$ are principal, with $M \subsetneq M_b \subseteq N$ and $M_aN = (\alpha\beta) \subseteq M$.
3. If $x \in M$, show that $x = s\alpha$ for some $s \in N$. Deduce that $M = M_aN$ is principal, a contradiction.

1. Let $\mathcal{A}$ denote the set of nonprincipal ideals of $R$. Suppose $\mathcal{A}$ is nonempty. Note that $\mathcal{A}$ is partially ordered by inclusion. Let $\{C_i\}_\mathbb{N}$ be a chain in $\mathcal{A}$, and let $C = \bigcup C_i$. We know that $C$ is an ideal of $R$. Suppose now that $C$ is principal; say, $C = (\gamma)$. We have $\gamma \in C_k$ for some $k$, so that $C \subseteq C_k$, and thus $C = C_k$. But then $C_k \in \mathcal{A}$ is principal, a contradiction. Thus $C$ is not principal, and so the chain $\{C_i\}_\mathbb{N}$ has an upper bound in $\mathcal{A}$. By Zorn’s lemma, $\mathcal{A}$ contains an inclusion-maximal element; let $M$ be such an ideal.
2. Since $a \notin M$, we have $M$ properly contained in $M_a = (M,a)$. Similarly, $M$ is properly contained in $M_b$. Since $M$ is maximal with respect to nonprincipalness, both $M_a = (\alpha)$ and $M_b = (\beta)$ are principal ideals. Note moreover that $M_bM_a = (M,b)(M,a) = (M^2,Ma,Mb,ab) \subseteq M$, so that $M_b \subseteq N$. Note that $M_aN \subseteq M$ by the definition of $N$, using the fact that $R$ is commutative. Note also that $M_aN$ is principal, being the product of principal ideals.
3. Let $x \in M$. Since $M \subseteq M_a = (\alpha)$, we have $x = s\alpha$ for some $s \in R$. However, note that $sM_a = s(\alpha) = (s\alpha) = (x) \subseteq M$, so that in fact $s \in N$. Thus $M \subseteq M_aN$, and in particular $M$ is principal. This contradicts the maximal nonprincipalness of $M$, so that our original assumption that $\mathcal{A}$ is nonempty was faulty.

Thus $R$ contains no nonprincipal ideals and is (by definition) a principal ideal domain.