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?

  • 2
    $\begingroup$ 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 '15 at 22:30
  • $\begingroup$ Which are the four domains? $\endgroup$ Nov 8 '15 at 16:20
  • 5
    $\begingroup$ 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 '15 at 3:58
  • 1
    $\begingroup$ 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 '17 at 15:55
  • 1
    $\begingroup$ 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 '17 at 20:00

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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