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For me it is easy to see $\mathbb{Z}[\sqrt{5}]$ is not Dedekind domain since it is not integrally closed nor a UFD because $4=(\sqrt{5}-1)(\sqrt{5}+1)=2\times2$ (not even a GDC domain).

I have some problems for this ring $\mathbb{Z}[\frac{\sqrt{5}+1}{2}]$. It is the ring of integer of $\mathbb{Q}(\sqrt{5})$, hence, a Dedekind domain. But I cannot see whether it is a PID or UFD directly (indirectly apply Minkovski bound and compute the class number). For example, usually, we will prove it is a Euclidean domain, like $\mathbb{Q}(i)$, $\mathbb{Q}(\omega)$, $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, we could simply define the absolute value of the norm to be the Euclidean functions. But here, I am not sure. So my question is are they Euclidean domain.

Upc
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    $\mathbb{Z}[\frac{\sqrt{5}+1}{2}]$ is Euclidean – J. W. Tanner Jan 08 '20 at 14:51
  • @J.W.Tanner Thank you that helps a lot! Is it possible for $\mathbb{Q}(\sqrt{p})$ to have Euclidean but not norm-Euclidean with $p$ a prime? In the example, $d=14, 69$ are not prime. – Upc Jan 08 '20 at 15:05
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    Weinberger showed under GRH that class number 1 implies Euclidean for a ring of integers with an infinite unit group, such as a real quadratic field. It is expected, but still unproved, that infinitely many real quadratic fields have class number 1. So there should be infinitely many Euclidean real quadratic fields. It is known that there are only finitely many norm-Euclidean real quadratic fields: see the Wikipedia page on norm-Euclidean fields. The complete list of primes $p$ for which $\mathbf Q(\sqrt{p})$ is norm-Euclidean are 2, 3, 5, 7, 11, 13, 17, 19, 29, 37, 41, and 73. – user1728 Jan 08 '20 at 15:28
  • Taking $p$ to be any of the other primes below $100$, except $79$, the ring of integers of $\mathbf Q(\sqrt{p})$ is definitely not norm-Euclidean and you can check (by yourself of see tables) that it has class number 1, so under GRH the ring of integers of $\mathbf Q(\sqrt{p})$ should be Euclidean. I don't believe the Euclidean nature of any of those rings has been proved without using GRH. For example, ${\mathbf Z}[\sqrt{23}]$ is a PID and is not norm-Euclidean, GRH implies it is Euclidean, but I don't think it's been proved Euclidean without using GRH. – user1728 Jan 08 '20 at 15:40
  • @user1728 I see. So no one constructed (explicitly) a Euclid function that is not the norm on the ring of integer for some $\mathbb{Q}(\sqrt{p})$ where $p$ is a positive prime? – Upc Jan 12 '20 at 00:01

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As you say, the rings of integers $\mathcal{O}_K$ of $K=\Bbb Q(\sqrt{d})$ satisfying the Minkowski bound $B_k<2$ automatically have class number $1$, i.e., they are factorial rings and PIDs. This is certainly true for $$ d=-7,-3,-2,-1,2,3,5,13. $$ In particular, for $d=5$, the ring of integers $\mathcal{O}_5=\Bbb Z[(1+\sqrt{5})/2]$ is factorial and a PID. But is indeed a Euclidean ring. A direct proof for being norm-Euclidean can be found on this site, too, for example here for $d=2,3,7,11$ etc. It should be possible to do this for $d=5$ as well.

References:

Proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain

Proof that $\mathbb Z[\sqrt{3}]$ is a Euclidean Domain

Factoring 1001 in $\Bbb Z[\sqrt 7]$

$\mathbb{Z}[\sqrt{11}]$ is norm-euclidean

Dietrich Burde
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