A Dedekind UFD is a PID

Remarks

Let $P$ be a nonzero prime ideal and let $a \in P$ be nonzero. The element $a$ is not a unit, and one of its irreducible factors $\pi$ lies in $P$. In a UFD, $\pi$ is a prime element, so $(\pi)$ is a nonzero prime ideal inside $P$; as nonzero primes of a Dedekind domain are maximal, $(\pi) = P$. So every nonzero prime is principal, and by the unique factorization of ideals every nonzero ideal is a product of principal ideals, hence principal.

Example: $\Z[\sqrt{-5}]$ is a Dedekind domain, but $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ gives two different factorizations into irreducibles, so it is not a UFD; consistent with this, $(2, 1 + \sqrt{-5})$ is not principal. The Dedekind hypothesis cannot be dropped: $k[x, y]$ for a field $k$ is a UFD, but the ideal $(x, y)$ is not principal.

Going deeper

For a student

In terms of the ideal class group, a Dedekind domain is a UFD exactly when its class group is trivial, and this is the same as being a PID. The class group thus measures the failure of unique factorization of elements. For example, the class group of $\Z[\sqrt{-5}]$ has order $2$. The proof uses only that nonzero primes are maximal and that ideals factor into primes, so it is a good first exercise with the ideal factorization theorem.