The definition of Euclidean ring: An integral domain R is called Euclidean ring if $\exists \delta$ : $R${$0$} -> $\mathbb{N} \cup{0}$ satisfying:

(1) $\delta (a) \leqslant \delta (ab)$ if a, b $\in R${$0$}; (2) $\forall a,b \in R${$0$} => $\exists q,r \in R$ s.t. $a=bq+r$, where either $r=0$ or $\delta (r) < \delta (b)$.

How do we usually set the delta function? The norm?

Let $x=a_1+b_1\sqrt{2}, y=a_2+b_2\sqrt{2}$, and $\delta(x)=|a_1^2-2b_1^2|, \delta(y)=|a_2^2-2b_2^2|$, then, $\delta(xy)=|(a_1+b_1\sqrt{2})(a_1-b_1\sqrt{2})(a_2+b_2\sqrt{2})(a_2-b_2\sqrt{2})|$, since $\delta(a)\geq 1$, for any $a\neq 0$

Consider $y\overline{x}=x\overline{x}q_1+r_1$, $\delta(r_1)=0$ or $\delta(r_1)=\delta((y-xq_1)(\overline{x}))=\delta(y-xq_1)\delta(\overline{x})<\delta(x\overline{x})=\delta(x)\delta(\overline{x})$, $\delta(r_0)=\delta(y-xq_1)<\delta(x)$

I'm not sure what am I doing... Orz

  • 5
    $\begingroup$ The (absolute) norm is the first thing one usually tries. If it works, it's typically also the last. Have you looked at the norm here yet? $\endgroup$ – Daniel Fischer Apr 19 '15 at 12:31
  • $\begingroup$ Do you know a proof that any other quadratic ring $\mathbf Z[\sqrt{d}]$ is Euclidean, at least $d = -1$ (Gaussian integers)? $\endgroup$ – KCd Apr 19 '15 at 12:39
  • $\begingroup$ @DanielFischer, I tried and it can work. But, more than that, how about $\mathbb{Z} [i]$ or something more typically? I want to know some $\delta$ function instead of the norm. $\endgroup$ – Richard Apr 19 '15 at 13:06
  • $\begingroup$ @KCd, it is the first time I heard about "quadratic ring"... I don't know... sorry. Let me try to proof that :) $\endgroup$ – Richard Apr 19 '15 at 13:08
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    $\begingroup$ For $\mathbb{Z}[i]$, the norm works too. In cases where the norm doesn't work, I don't know of a general strategy, but I expect the algebraists have strategies that work in some families of cases. Just in case you misunderstood KCd's comment, that doesn't say that all rings $\mathbb{Z}[\sqrt{d}]$ are Euclidean (they aren't, e.g. $\mathbb{Z}[\sqrt{-5}]$ isn't a UFD), it asks whether you know the proof for some other example. $\endgroup$ – Daniel Fischer Apr 19 '15 at 13:19

In order to mark the question answered I note that your proof is finished by Kowser in this answer.


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