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EDIT: Since the original question was to vague I will pose some stronger conditions:

Given $n \in \mathbb{N}$ with $n \geq 3$ I want to know if there is a ring $R$ with the following properties:

  • $\mathbb{Z}[\sqrt{-n}] \subset R$
  • $R$ is a UFD but not a field (in particular I want that there are only finitely many integers $k \in \mathbb{Z}$ that are invertible in $R$)
  • For any $\alpha \in R$ there is a unit $\eta \in R^*$ such that $\alpha \eta \in \mathbb{Z}[\sqrt{-n}]$

If $n=3$, one can take $R$ to be the ring of integers of $\mathbb{Q}(\sqrt{-n})$. This does not work for $n \geq 4$, since then the ring of integers of $\mathbb{Q}(\sqrt{-n})$ has only two units. Are there other rings $R$ that could work?

Original question: Let $n \in \mathbb{N}, n \geq 3$. Then $\mathbb{Z}[\sqrt{-n}]$ is not a UFD. I want to know which extensions $\mathbb{Z}[\sqrt{-n}]$ are (or can be) a UFD. Two examples would be:

  • For some $n$, the ring of integers of $\mathbb{Q}[\sqrt{-n}]$ is a UFD and this is an extension of $\mathbb{Z}[\sqrt{-n}]$.
  • Every field extension of $\mathbb{Z}[\sqrt{-n}]$ (for example $\mathbb{Q}(\sqrt{-n})$) is a UFD

Are there any other extensions of $\mathbb{Z}[\sqrt{-n}]$ (for example using localization, adjoining other algebraic numbers, other concepts, or combining some of these) that are UFD?

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    $\begingroup$ What do you mean by "extension"? Just a ring which contains $\mathbb Z[\sqrt{-n}]$? $\endgroup$
    – Wojowu
    Sep 12, 2017 at 12:57
  • $\begingroup$ @Wojowu yes (preferably "small"), but first I won't pose any condition other then it should be a ring containing $\mathbb{Z}[\sqrt{-n}]$ $\endgroup$
    – Martin
    Sep 12, 2017 at 13:20
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    $\begingroup$ This question is waaaay too vague to have a useful answer. Every field and DVR (e.g., localizing the ring of integers of $\mathbf Q(\sqrt{-n})$ at a prime ideal) is a UFD. A field in turn has a ton of its own field extensions, and all are UFDs. It does not seem very interesting. Be more focused on explaining what you are really looking for and why. $\endgroup$
    – KConrad
    Sep 12, 2017 at 13:48
  • $\begingroup$ @KConrad I edited the question $\endgroup$
    – Martin
    Sep 12, 2017 at 14:32
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    $\begingroup$ Staying within the ambient field, if you localize by allowing denominators in finitely many prime ideals representing the ideal classes, the resulting slightly-localized ring is always a PID, for example. This is an old, standard result. $\endgroup$
    – paul garrett
    Sep 12, 2017 at 14:39

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