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Let $\mathcal{O}$ be a Dedekind ring which is also a Euclidean domain, and let $\omega$ be a root of unity. Is it true that $\mathcal{O}[\omega]$ is a principal ideal domain if and only if $\mathcal{O}[\omega]$ is a Euclidean domain?

The statement is true for $\mathcal{O} = \mathbb{Z}$, since there are only finitely many roots of unity $\omega$ with the property that $\mathbb{Z}[\omega]$ is a principal ideal domain (see When is $\Bbb{Z[\zeta_n]}$ a PID?), and all of them turn out to be Euclidean domains (see http://www.rzuser.uni-heidelberg.de/~hb3/publ/survey.pdf). The question whether the equivalence holds more generally came up while hoping there would be a proof without checking all the cases.

Levi
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Lenstra showed that ${\mathbb Z}[\zeta_{32}]$ is not norm-Euclidean, so the answer to your question is "no". Another counterexample is the field ${\mathbb Q}(\sqrt{-1},\sqrt{11})$, whose ring of integers is not norm-Euclidean although both ${\mathbb Q}(\sqrt{11})$ and ${\mathbb Q}(\sqrt{-11})$ are.