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The ring $\mathbb{Z}[2\cos(\frac{\pi}{k})]$ is known to be a Euclidean domain for $k=3,4,5$ and $6$, because in those cases $2\cos(\frac{\pi}{k}) = 1, \sqrt{2},$ the golden ratio $\phi$, and $\sqrt{3}$ respectively, and $\mathbb{Z}, \mathbb{Z}[\sqrt{2}], \mathbb{Z}[\phi]$, and $\mathbb{Z}[\sqrt{3}]$ are all known to be Euclidean domains.

I haven't been able to find a reference that will tell me whether $\mathbb{Z}[2\cos(\frac{\pi}{k})]$ is (or isn't) a Euclidean domain for integers $k$ higher than $6$, though. Is this an open problem?

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This is the ring of integers of the real subfield of the $k$-th cyclotomic field and it's known not to be a PID, let alone Euclidean, for infinitely many $k$ (maybe even all but finitely many). –  Felipe Voloch May 9 '12 at 23:13
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Thanks so much- that's a very good piece of information to know! Do you know some of the infinitely many values of $k$ for which it isn't a PID? Or whether there are any values of $k$ higher than $6$ for which it is a Euclidean domain? –  Collin Hazlett May 9 '12 at 23:40
    
This thesis, by M.A. Simachew, www.algant.eu/documents/theses/simachew.pdf has a comprehensive survey of what was known in 2009. –  Victor Miller May 10 '12 at 2:10
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up vote 7 down vote accepted

There's certainly no reason to expect that it will always be a Euclidean domain, since (as Voloch points out) to be a Euclidean domain it must necessarily be a PID. On the other hand, the converse is (essentially) true in this case. Namely, a theorem of Weinberger implies that if $K/\mathbb{Q}$ is Galois, totally real, and has degree $\ge 4$, then $\mathcal{O}_K$ is a Euclidean domain if and only if it is a PID. So you are really asking about the invariant known as $h^{+}$, that is, the class numbers of totally real subfields of cyclotomic fields. If one supposes that $p$ is prime, then the smallest example which is provably not a PID is $p = 163$, since there exists a totally real $A_4$ extension $L/\mathbb{Q}$ which is unramified over the degree three extension $L \cap \mathbb{Q}(\zeta_{163})$. (Indeed, assuming GRH, this is even the first example. There's a paper of Schoof where he talks about computing these numbers.) The problem of proving that there are infinitely many such fields with class number one seems as least as hard as proving the result for real quadratic fields of prime conductor, which is totally open.

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Thank you very much! –  Collin Hazlett May 10 '12 at 0:56
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Weinberger's Theorem assumes the validity of GRH. Removing this assumption is not easy, and was achieved by Murty's students (Clarke, Harper, ...) for certain fields with sufficiently large unit rank. –  Franz Lemmermeyer May 10 '12 at 4:14
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