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Let $R$ be a commutative ring with unity, $D$ be a Dedekind domain, $K$ be its fraction field such that $D \subseteq R \subseteq K$. Then is it true that $R$ is Dedekind ?

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  • $\begingroup$ A nice question, one which I've regrettably been able to make little progress on. Any localization of a Dedekind domain is again a Dedekind domain, so if you can prove that any ring sitting between an integral domain $R$ and its field of fractions $F$ is a localization of $R$, then the claim follows. However, I'm unable to actually prove this - maybe you might consider posting it as a separate question. $\endgroup$ – Alex Wertheim Aug 16 '16 at 20:09
  • $\begingroup$ @AlexWertheim When R is a PID, this is true, shown here: spot.colorado.edu/~kearnes/F09/HW/ca5p1.pdf $\endgroup$ – Quinn Greicius Aug 16 '16 at 20:53
  • $\begingroup$ @QuinnGreicius: thanks! That question had been eating at me and I was unable to find a reference. Pity that the idea doesn't help here. :( $\endgroup$ – Alex Wertheim Aug 16 '16 at 20:59
  • $\begingroup$ @AlexWertheim Unless I'm missing something, that link doesn't rule out your approach completely – the counterexample they use isn't a Dedekind domain. $\endgroup$ – Quinn Greicius Aug 16 '16 at 21:10
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Let $p$ be a prime of $R$, and $q$ it's restriction to $D$. We have $D_q\subseteq R_p \subseteq K$ with $D_q$ a DVR. It is immediate to see that there are no proper intermediate rings between a DVR and its fraction field, hence if $p$ is nonzero then $D_q=R_p\neq K$, if $p=(0)$ then $D_q=R_p= K$. Hence all localization at prime ideals coincide with those of $R$, in particular localizations at maximals are DVR. To conclude, it's enough to show that $R$ is noetherian, i.e. it satisfies the ascending chain condition on ideals.

But this is easy: we already know that $D$ is noetherian, and hence it is enough to show that the map sending an ideal $I\subseteq R$ to $I\cap D\subseteq D$ is injective. But this can be checked at level of primes: two ideals are equal if and only if their localizations at every prime are equal, and since $D_q=R_p$ for every prime $p\subseteq R$, we conclude.

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The answer is positive; this is exactly the statement of Proposition 0.63 from the book Finite Rank Torsion Free Modules Over Dedekind Domains by E. L. Lady. (Here's the direct link to the PDF file of Chapter 0.)

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