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I was reading some stuff on quadratic integers rings, and it seems like $R =\mathbb{Z}\left[\frac{1+\sqrt{-5}}{2}\right]$ is not supposed to be a Euclidean domain. But I got this:

If $a,b \in R$, then $\frac ab \in \mathbb{Q}\left[\frac{1+\sqrt{-5}}{2}\right]$, so $\frac ab = u+v\omega$ where $\omega = \frac{1+\sqrt{-5}}{2}$. We can find $x$ and $y$ such that $|u - x|\leq 1/2 $ and $|v - y|\leq 1/2 $. But then, as usual, we get of upper bound from the norm which is not sufficient, as it might equal 1. But if it equals 1, it means $|u - x| = 1/2$ and $|v - y| = 1/2$, then we take $x$ to be the ceil of $u$ and $y$ to be the floor of $v$ for instance. It's actually a quite similar proof as for $\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$ I came across...

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  • $\begingroup$ $-5\equiv 3\pmod{4}$, so the ring of integers is $\mathbb{Z}(\sqrt{-5})$ as Qiaochu pointed out below. I guess that is your confusion. $\endgroup$ – Bombyx mori Dec 30 '15 at 19:21
  • $\begingroup$ It is known the exact (not much, about 20) number of quadratic euclidean domains and $\mathbb{Q}(\sqrt{5})$ is not of them. $\endgroup$ – Piquito Dec 30 '15 at 19:30
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    $\begingroup$ @Piquito: it's known which imaginary quadratic fields have class number $1$, but (I guess this was a typo) $\mathbb{Q}(\sqrt{5})$ is not imaginary, and in fact it does have class number $1$. $\endgroup$ – Qiaochu Yuan Dec 30 '15 at 20:04
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This is not a ring of integers; the ring of integers of $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}[\sqrt{-5}]$, which is not a PID (so not a UFD, so not a Euclidean domain). Your argument incorrectly assumes that the norm of an element of your ring is an integer, which is false.

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    $\begingroup$ Excellent! But what's the conclusion to the question: Is $\mathbb{Z}[\frac{1+\sqrt{-5}}{2}]$ a Euclidean domain? $\endgroup$ – user26857 Dec 30 '15 at 22:09
  • $\begingroup$ @user26857: I think it's a PID, but I don't know what to say about Euclidean domains that aren't norm-Euclidean. $\endgroup$ – Qiaochu Yuan Dec 31 '15 at 1:51
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As @QiaochuYuan noted, $R={\mathbb Z}[{1+\sqrt{-5}\over 2}]$ is not the integral closure of $\mathbb Z$ in $\mathbb Q(\sqrt{-5})$, so asking whether $R$ is Euclidean is somehow a distortion of more meaningful questions that sound similar.

In fact, $R$ is the genuine ring of integers $\mathbb Z[\sqrt{-5}]$ with $1/2$ adjoined. By chance, the class number of $\mathbb Z[\sqrt{-5}]$ is $2$ and the prime $2$ is ramified $2\cdot \mathbb Z[\sqrt{-5}]=\tau^2$ for a non-principal ideal $\tau$. Thus, this localization kills the prime $\tau$, and makes the resulting ring a Dedekind ring that is a principal ideal domain, so is a unique factorization domain. (As to whether it is Euclidean, I don't know... the norm no longer takes discrete values, so there's some problem in making sense of that naive question about "Euclidean'ness".)

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  • $\begingroup$ The ring $R$ is indeed Euclidean. There's some literature on these rings, in particular on Euclidean rings of $S$-integers. $\endgroup$ – franz lemmermeyer Dec 31 '15 at 12:55
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Let $R = {\mathbb Z}[\sqrt{-5}]$ and $S = R[\frac12]$. The correct norm in $S$ is defined as usual by $N_S(A) = \# S/A$ as the cardinality of the residue class ring modulo an ideal $A$ in $S$. Since $2$ is a unit in $S$, we have $N_S(2) = 1$. In fact we can compute $N_S(A)$ by taking the usual norm $N(A)$ and then omitting the powers of $2$ that occur here. Again, an element $u \in S$ is a unit if and only if $N_S(u) = 1$.

It is now a simple exercise to show that $S$ is Euclidean with respect to the norm $N_S$.

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