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Let $K$ be an arbitrary number field. I know that $\mathcal{O}_K$ is a Dedekind domain. I know that it need not be a PID. However, despite a long search, I am having a hard time finding either an affirmative or negative answer to the question of whether $\mathcal{O}_K$ is necessarily a GCD domain. If anyone knows the answer to this question, then please let me know.

I also have a related question: Let $S$ be any finite set of places of $K$ that contains all the Archimedian places. Let $\mathcal{O}_{K, S}$ be the ring of $S$-integers of $K$; that is, the set of all $x \in K$ such that for each place $v \notin S,$ we have $|x|_v \leq 1.$ Is $\mathcal{O}_{K, S}$ necessarily a GCD domain?

Thank you very much for your attention.

Mishel Skenderi
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    Being a localization of $O_K$, $O_{K,S}$ is a Dedekind domain and its ideal class group is the quotient of $C_K$ by the classes of the primes in $S$. Iff $S$ contains enough primes to generate $C_K$ then $O_{K,S}$ is a PID – reuns Jun 23 '19 at 03:36
  • @reuns Thank you! – Mishel Skenderi Jun 23 '19 at 03:53
  • @reuns Sorry, I do not understand your second comment. Also, could you please tell me by which multiplicatively closed set the ring of S-integers is a localization of the ring of integers? Thanks! – Mishel Skenderi Jun 23 '19 at 13:54
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    See also this Answer to a previous Question with a related but somewhat different focus. – hardmath Jun 23 '19 at 14:08
  • @hardmath Thanks! – Mishel Skenderi Jun 23 '19 at 14:21
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    My comment holds at least if $S$ contains all or none of the prime ideals above $p$ : if $S$ contains all the prime ideals above $p \in \Bbb{Z}$ then $O_{K,S}$ contains $p^{-1}$ so it is clear every ideal of $O_K$ of norm a power of $p$ becomes trivial in $O_{K,S}$. If $S$ contains only one prime ideal above $p$ and that it is not principal then it is more complicated – reuns Jun 23 '19 at 23:20
  • @reuns I see; thank you. Also, is it a localization by powers of a single element, as in the case of K = \Q ? – Mishel Skenderi Jun 24 '19 at 13:23

1 Answers1

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A Dedekind domain $D$ is Noetherian so atomic, i.e. nonunits $\neq 0$ factor into atoms = irreducibles.

Thus $D$ is a GCD domain $\iff D$ is a UFD, by the same proof as in $\,\Bbb Z,\,$ i.e. gcds exist therefore Euclid's Lemma holds, which implies atoms are prime.

$D$ Dedekind $\Rightarrow D$ is $1$-dimensional (prime ideals $\!\neq\! 0$ are max), thus $D$ is a UFD $\!\iff\! D$ is a PID.

Hence a Dedekind domain $D$ is a GCD domain $\iff D$ is a UFD $\iff D$ is a PID.

Bill Dubuque
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