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After having accidentally duplicated this question, I thought I'd follow up with a related question. In an answer to the linked question, Zev Chonoles quotes the first page of Chapter 11 of Washington's Introduction to Cyclotomic Fields which states that the only $\mathbb Z[\zeta_n]$ with $\zeta_n$ a root of unity that are PIDs (with class number 1) are extensions with the following values of $n$:

 1,  3,  4,  5,  7,  8,  
 9, 11, 12, 13, 15, 16,
17, 19, 20, 21, 24, 25,
27, 28, 32, 33, 35, 36,
40, 44, 45, 48, 60, 84.

In addition, values of $n \equiv 2 \mod 4$ are also allowed, because for example, $\mathbb Z[\zeta_{30}] = \mathbb Z[\zeta_{15}]$. So my question is, which are known to be Euclidean Domains? I'm especially interested if $n = 60$ admits a form of the Euclidean division algorithm.

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    $\begingroup$ @hurkyl Good catch. I've edited the question. $\endgroup$ – hatch22 Sep 18 '17 at 17:36
  • $\begingroup$ @Hurkyl I suspect what is meant is the ring of integers of those fields. While $\mathbb Z[\zeta_n]$ is certainly contained in the ring of integers, can you confirm that the containment is actually equality? $\endgroup$ – Aaron Sep 18 '17 at 17:40
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    $\begingroup$ @Aaron mathoverflow.net/questions/17289/… $\endgroup$ – André 3000 Sep 18 '17 at 17:46
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We have the following Theorem, see here, Theorem $5.2$ on page 50:

Theorem 5.2. A cyclotomic field is Euclidean if and only if it is principal ideal domain.

So this leads again back to the duplicate question.

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  • $\begingroup$ Seems I can't win. Oh well, it's good to know that in this case these cyclotomic fields are both PIDs and Euclidean Domains. $\endgroup$ – hatch22 Sep 18 '17 at 19:01
  • $\begingroup$ The thesis you reference, Theorem 5.2, rests his conclusion on Weinberger's result which is, of course, conditional on GRH. An unconditional proof of the result that A cyclotomic field is Euclidean if and only if it is principal ideal domain appears in Harper, $\mathbb{Z}[\sqrt{14}]$ is Euclidean, cms.math.ca/10.4153/CJM-2004-003-9 $\endgroup$ – Malcolm Mar 29 '18 at 13:23
  • $\begingroup$ @Malcolm Yes, I know. I wanted to cite Harper's result, I am sorry. $\endgroup$ – Dietrich Burde Mar 29 '18 at 15:39

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