I saw a proof somewhere (can't find it at the moment) that shows $\textbf{Z}[\sqrt{-2}]$ is norm-Euclidean because for any pair of nonzero numbers $a, b$ it's possible to find a remainder $r$ such that $$N(r) \leq \frac{3}{4} N(b).$$ In Bolker's error-prone book, a proof covering $\textbf{Z}[\sqrt{-2}]$, $\textbf{Z}[i]$, $\textbf{Z}[\sqrt{2}]$ shows that in each of these domains it's always possible to find $$N(r) \leq \frac{1}{2} N(b).$$ Maybe I'm recalling incorrectly, because it seems to me that if $\frac{1}{2}$ does it for these four domains, then $\frac{3}{4}$ is too much for $\textbf{Z}[\sqrt{-2}]$. Basic logic suggests that at the most general level, $1$ is the absolute maximum, with strict inequality, of course.

Do any of the norm-Euclidean quadratic rings require $1$ for the proof interval? Or is $\frac{1}{2}$ always sufficient? Or is there more variety than this?

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    $\begingroup$ Perhaps you're referring to dpmms.cam.ac.uk/~par31/notes/ed.pdf ? The penultimate line asserts $$N(r) \leq \frac{3}{4} N(c + d \sqrt{-2}).$$ $\endgroup$
    – Mr. Brooks
    Nov 5, 2015 at 22:30
  • $\begingroup$ Which are the four domains? $\endgroup$ Nov 8, 2015 at 16:20
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    $\begingroup$ As I think more about it, I think you've misread Bolker. My understanding is that the interval of proof is $$\frac{1}{4} + \frac{|d|}{4}$$ for $d = -2, -1,$ 2 or 3. For the remaining negative $d$ corresponding to norm-Euclidean domains, Bolker gives $$\frac{1}{4} + \frac{|d|}{16}$$ in Theorem 33.5, giving $$\frac{15}{16}, \frac{11}{16}, \frac{7}{16}$$ and for $d = 5$ or 13, $$\frac{5}{16}, \frac{13}{16}.$$ $\endgroup$ Nov 11, 2015 at 3:58
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    $\begingroup$ I apologize for all the errors. You can find a list at cs.umb.edu/~eb/numberTheory . I think @RobertSoupe 's comment confirms my argument. Of course that doesn't answer your more general question. $\endgroup$ May 8, 2017 at 15:55
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    $\begingroup$ That reminds me, @Ethan, I have to travel back in time and make you make those mistakes. Without my meddling, your text would have had less than a page's worth of errata. I know, I know, it's one of those pesky predestination paradoxes. $\endgroup$ May 8, 2017 at 20:00

1 Answer 1


The infimum of all values of $c$ such that $f(r) \le c f(b)$ for all pairs of $r$ and $b$ is called the Euclidean minimum for the (Euclidean) function $f$. If $f$ is the absolute value of the norm, the classical name is "inhomogeneous minimum of the norm form".

If $f$ is the minimal Euclidean function on a domain as defined by Motzkin, then $c = 1$. If $f$ is the absolute value of the norm, then we must have $c < 1$ if the ring is norm-Euclidean, as was shown by Barnes and Swinnerton-Dyer in the quadratic and by Cerri in the case of general number fields. If the ring contains a principal ideal of norm $a > 1$, then we always have $c > 1/a$. I do not know whether every rational value of $c > 0$ does occur as a Euclidean minimum for the norm function.

For more information, see my survey.


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