Norm-Euclidean rings?

Question

For which integer $d$ is the ring $\mathbb{Z}[\sqrt{d}]$ norm-Euclidean?

Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a,b \in \mathbb{Z}\}$, not the ring of integers of $\mathbb{Q}[\sqrt{d}]$.

For $d < 0$, it is easy to show that only $d = -1, -2$ suffice; but what about $d>0$?

Thanks.

Answer 1

The ring of integers of the real quadratic number field $\rm\:\mathbb Q(\sqrt{d})\:$ is norm-Euclidean iff $\rm\:d = 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57, 73\:.\:$ For this result and much more of interest see Franz Lemmermeyer's excellent survey The Euclidean Algorithm in Algebraic Number Fields.

Regarding the edited question: since a Euclidean domain is integrally closed, any proper subring of the full ring of integers, being not integrally closed, is not Euclidean. That Euclidean domains are integrally closed is nothing more than the standard simple proof of the Rational Root Test.