Let $A$ be a Dedekind domain. PID implies UFD. So for the other direction assume $A$ is an UFD. In this proof the author only considers prime ideals instead of any proper ideal. Why is this sufficient?
2026-04-12 10:18:00.1775989080
UFD iff PID in Dedekind domain.
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In general, an integral domain in which every prime ideal is principal is a PID.
In the case of Dedekind domains, the story is much simpler. Every ideal factorises (uniquely) as a product of prime ideals. Since a product of principal ideals is principal, it is sufficient to show that prime ideals are principal.