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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?

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    $\begingroup$ In the case where $A$ is Dedekind every ideal is a product of prime ideals and the product of principal ideals is principal. $\endgroup$ – Antoine Giard May 15 '18 at 21:05
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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.

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