 Solution

The height of a prime ideal $P$ is

The Krull principal ideal theorem states that if $R$ is Noetherian and $P$ is minimal among the primes containing a proper principal ideal $(a)$, thenSuppose otherwise that $P_0⊊Q⊊P$. Quotient by $P_0$ and localize at $P$; it is enough to consider a Noetherian local domain whose maximal ideal $P$ is the only prime containing $(a)$ and which has $0⊊Q⊊P$.

For $n\ge 1$, putThis is a $Q$-primary ideal. Since $R/(a)$ is a zero-dimensional Noetherian ring, it is Artinian, and the descending chain $(Q^{(n)}+(a))/(a)$ eventually stabilizes. Thus, for all sufficiently large $n$, every $q_n\in Q^{(n)}$ can be writtenNow $ra\in Q^{(n)}$, while $a\notin Q$; $Q$-primaryness gives $r\in Q^{(n)}$. HenceThe Nakayama lemma applied to $Q^{(n)}/Q^{(n+1)}$ gives $Q^{(n)}=Q^{(n+1)}$. Localizing at $Q$ makes all sufficiently large powers of the nonzero maximal ideal $Q{R}_Q$ equal. A nonzero element of the stable power then belongs to ${\bigcap }_n{Q}^n{R}_Q$, contradicting the Krull intersection theorem. This proves the theorem.

Now let $R$ be a Noetherian integral domain. If $R$ is a unique factorization domain and $P$ has height one, choose $0\ne x\in P$ and an irreducible factor $p$ of $x$. In a UFD, $p$ is prime, soHeight one forces $P=(p)$.

Conversely, suppose every height-one prime is principal. Noetherianity makes $R$ an atomic domain. Given an irreducible element $p$, choose a prime $P$ minimal over $(p)$. The principal ideal theorem gives $\mathrm{ht}P=1$, so $P=(q)$ by hypothesis. Since $p=qr$ and $p$ is irreducible, $r$ is a unit; hence $(p)=P$ is prime. Thus every irreducible is a prime element, and an atomic domain with this property is a UFD. Therefore