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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.

  • 0
    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](http://oeis.org/A048981). Which means $\mathbb{Z}[\sqrt{d}]$ for $d=2,3,6,7,11,19,73$, at least, are norm-euclidean.2011-08-19
  • 0
    Why are the $d \equiv 1$ left out? Thanks.2011-08-19
  • 2
    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".2011-08-19
  • 4
    Euclidean domains are integrally closed, so any proper subring of the full ring of integers cannot be Euclidean.2011-08-19
  • 0
    ... 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! :)2011-08-19
  • 0
    @Bill: Duh to me; I knew that, but it flew out of my head.2011-08-20

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