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I am looking for an illuminating example of a UFD that is not a PID. The standard one is $\mathbb{Z}[x]$, but since I am studying algebraic number theory, I am wondering if there is any subring of $\overline{\mathbb{Q}}$ (field of algebraic numbers) that is a UFD but not a PID.

I know that algebraic number rings are Dedekind domains and for those PID and UFD are equivalent, so naturally the example can not be a number ring. My first guess would therefore be $\mathbb{Z}[\sqrt{ m}]$ for $m \equiv 1$ $(\mod 4)$. But are any of those even UFDs?

EinStone
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    Any finitely generated ring of algebraic integers that is a UFD is normal, thus is Dedekind and a UFD, therefore is a PID (it is enough to show that prime ideals are principal, but nonzero prime ideals have height $1$ and contain the ideal generated by some irreducible element, which is prime of positive height as well, so these two ideals are equal). – Aphelli Jun 04 '20 at 09:34
  • OK makes sense, so no finitely generated examples can exist. What about infinite ones? – EinStone Jun 04 '20 at 09:37
  • From what I understand that would either be not a UFD at all, or already be a PID @DietrichBurde ? – EinStone Jun 04 '20 at 09:40
  • Ah, I see. This MO-post is helpful. – Dietrich Burde Jun 04 '20 at 09:45
  • According to the answer you linked here, this question was asked and answered with yes. – EinStone Jun 04 '20 at 09:45
  • Yes, exactly. So it is known (again a duplicate), or not? – Dietrich Burde Jun 04 '20 at 09:49
  • It is known that for $\mathbb{Z}[\sqrt{ m}]$ and, according to the comment of @Mindlack, that for all finitely generated rings consisting of algebraic integers UFD is equivalent to PID. However, my original question didn't include finitely generated, so it is still open as to whether there exist non finitely generated examples. – EinStone Jun 04 '20 at 09:58
  • This would be a good new question. I think, the present one "But are any of those even UFDs?" is answered. Yes, there are many $\Bbb Z[\sqrt{m}]$ which are indeed UFDs, but they are also PIDs. For example, $m=14$. – Dietrich Burde Jun 04 '20 at 10:58
  • The question "But are any of those even UFDs?" was just a hint of what my thought process was to find an example. The real question I have is to find an example, and this also exactly what I wrote in the title. Do you want me to recreate the post but leaving the second paragraph out? – EinStone Jun 04 '20 at 11:19
  • You could improve and work out the first part too, summarising what we know about finitely generated and asking for not finitely generated. Also I am not sure what you find "illuminating" and what exactly you are looking for. Clearly there are enough examples of UFDs which are not PIDs. You should ask for more than this in the title. – Dietrich Burde Jun 04 '20 at 16:39

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