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Can the following cases happen?

  1. A proper subring of $\Bbb Q$ with finitely many prime ideals
  2. A proper subring of $\Bbb Q$ which is not a UFD

For case 1 I was thinking about some localisation of $\Bbb Z$ which would be a UFD and have unique prime = maximal ideal.

Sumanta
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Angry_Math_Person
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    Your thoughts on #1 are good. For #2, note that every subring of $\mathbb{Q}$ must contain $\mathbb{Z}$ (in a canonical way). You might try showing the more general fact: let $R$ be a PID, and let $F$ be its field of fractions. Then any subring of $F$ containing $R$ is also a PID. (I encourage you to try to do this yourself, but if you get stuck, the answers to following question provide solutions: https://math.stackexchange.com/q/137876/73817.) – Alex Wertheim Dec 06 '19 at 09:09
  • In fact every subring will also be a ED right? – Angry_Math_Person Dec 06 '19 at 09:17
  • If $R$ is a Euclidean domain (e.g., if $R=\mathbb{Z}$, as in your case), then yes. The linked post shows that any subring between a PID $R$ and its field of fractions $F$ is a localization of $R$. Any (nonzero) localization of a Euclidean domain is again a Euclidean domain (though this is nonobvious!), which proves the claim. – Alex Wertheim Dec 06 '19 at 09:31
  • So what is the final answer to my questions?
    1. is possible
    2. is not
     – Angry_Math_Person Dec 06 '19 at 09:34
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    Yes, that's right. I would encourage you to write an answer to your own question to ensure that you understand the details of what is happening. (For instance: you should make precise what kind of localization of $\mathbb{Z}$ you are talking about in #1.) – Alex Wertheim Dec 06 '19 at 09:35

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