A Dedekind ring is a UFD if and only if it is a PID, but not all UFDs are PIDs, so it must be the case that there are UFDs that are not Dedekind. But on the other hand, since unique factorization of ideals into prime ideals seems like a much weaker condition than unique factorization of elements into primes, shouldn't all UFDs be Dedekind and therefore PIDs?
What are the errors in my logic, and (if they do exist), what's an example of a UFD that is not Dedekind?
On
The best example (a UFD with Krull dimension $> 1$) has already been given in one answer as well as the comments. However, I thought I might point out that another criterion for being Dedekind can also fail for a UFD, namely being noetherian.
For an example of a non-noetherian UFD, consider $R = \mathbb{Z}[x_{1}, x_{2}, \ldots]$. It is not hard to see that $R$ is a UFD. However, $R$ is not noetherian, as the ideal $\langle x_{1}, x_{2}, \ldots \rangle$ is not finitely generated. (Alternatively, the ascending chain of ideals $\langle x_{1} \rangle \subset \langle x_{1}, x_{2} \rangle \subset \langle x_{1}, x_{2}, x_{3} \rangle \subset \cdots$ does not stabilize, so $R$ is not noetherian by an equivalent characterization.)
Note that $\mathbb{Q}[x,y]$ is a UFD, but it is not Dedekind domain since $(x)$ is a non-zero prime ideal, which is not maximal.