Noetherian domain whose fraction field is such that some specific proper submodules are projective 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 $K$ ?    
 A: 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 $k[[T]]$.
