# Any commutative ring lying between a Dedekind domain and its fraction field is Dedekind?

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 ?

• 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. – Alex Wertheim Aug 16 '16 at 20:09
• @AlexWertheim When R is a PID, this is true, shown here: spot.colorado.edu/~kearnes/F09/HW/ca5p1.pdf – Quinn Greicius Aug 16 '16 at 20:53
• @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. :( – Alex Wertheim Aug 16 '16 at 20:59
• @AlexWertheim Unless I'm missing something, that link doesn't rule out your approach completely – the counterexample they use isn't a Dedekind domain. – Quinn Greicius Aug 16 '16 at 21:10

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.