# Norm-Euclidean rings?

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.

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The norm-Euclidean quadratic fields are those with $d=2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57, 73$. OEIS. Which means $\mathbb{Z}[\sqrt{d}]$ for $d=2,3,6,7,11,19,73$, at least, are norm-euclidean. – Arturo Magidin Aug 19 '11 at 20:57
Why are the $d \equiv 1$ left out? Thanks. – Isaac Aug 19 '11 at 21:13
Because you were asking about $\mathbb{Z}[\sqrt{d}]$; the norm-Euclidean quadratic fields are those for which the ring of integers is norm Euclidean, and if $d\equiv 1\pmod{4}$, then the ring of integers is $\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$. I don't know off-hand whether $\mathbb{Z}[\sqrt{d}]$ is norm-Euclidean if $\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$ is norm-Euclidean in those cases, hence the "at least". – Arturo Magidin Aug 19 '11 at 21:19
Euclidean domains are integrally closed, so any proper subring of the full ring of integers cannot be Euclidean. – Bill Dubuque Aug 19 '11 at 21:50
... Oop @Bill D. must have been composing the same...The non-integrally-closed orders cannot be Euclidean at all, because then they'd be PIDs, which is impossible because not-integrally-closed rings cannot be even Dedekind. It's true! :) – paul garrett Aug 19 '11 at 21:51

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.