Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $d>0$?


share|cite|improve this question
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.

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.

share|cite|improve this answer
Has anyone exposited the proof of the euclidean real quardatic fields in one place ? – Rene Schipperus Sep 1 '14 at 21:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.