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Let $R$ be a Noetherian domain (which is not a field) with fraction field $K$. Suppose every proper $R$-submodule of $K$ of the form $R[1/a]$, where $a\in R$, is projective as an $R$-module. Then, I can show that $R$ has exactly one non-zero prime ideal.

My question is : Does it follow that $R$ is normal i.e. integrally closed in $K$ ?

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  • $\begingroup$ When you write $R[1/a]$, I assume $a\neq 0$ and it is just the localization, that is just inverting $a$. What is your example where your hypothesis is satisfied and $R$ is not a field? $\endgroup$ – Mohan Dec 1 '18 at 23:24
  • $\begingroup$ @Mohan: Any DVR ... and my aim is to prove that they can only be DVR ... $\endgroup$ – user521337 Dec 1 '18 at 23:47
  • $\begingroup$ if $R$ is a dvr an $a\neq o$ and not unit, then $R[1/a]$ is the fraction field and it is not projective over $R$. $\endgroup$ – Mohan Dec 2 '18 at 2:05
  • $\begingroup$ @Mohan: yes, in that case there are no proper submodules of the form $R[1/a]$ and my condition is trivially satisfied ... $\endgroup$ – user521337 Dec 2 '18 at 2:33
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In any $1$-dimensional local (not necessarily Noetherian) domain, $K=R[1/a]$ for every nonunit $a \not= 0$. This is a special case of the following fact:

Let $R$ be a domain with fraction field $K$, $R \subsetneq K$. Then there exists some $a \not=0$ in the intersection of the (nonzero) minimal primes of $R$ (sometimes called the pseudoradical of $R$) iff $K = R[1/a]$ for some $a \in R$.

Proof of fact: Suppose $a$ is in every minimal prime. Let $S$ be the multiplicatively closed set $S = \{a^n\}_{n=0}^{\infty}$. The saturation of $S$ must be $R \setminus\{0\}$, otherwise the complement of $S$ would contain a prime and therefore absurdly contain $a$. Conversely, if $K = R[1/a]$, pick an element $p \in P$ prime. Noting $\frac{1}{p} = \frac{b}{a^n}$ for some $b \in R$, it is clear that $a \in P$.

Therefore any $1$-dimensional Noetherian local domain that is not normal provides a counterexample by trivializing your additional assumption about proper $R$-submodules of $K$.

An easy example is.....

$R = k[[T^2, T^3]] \cong k[[X,Y]]/ (X^2 - Y^3)$.

$R$ is 1-dimensional, Noetherian, and local with maximal ideal $(T^2, T^3)$. Clearly $R$ is not normal (indeed its integral closure is $k[[T]]$.

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