TITLE: Subrings of fraction fields
QUESTION [11 upvotes]: Let $R$ be an integral domain and let $S$ be a ring with $R \le S \le \text{Frac}(R)$ (fraction field). 
Question: Is there a multiplicatively closed subset $U \subseteq R\setminus \{0\}$ such that $S=R[U^{-1}]$ ?

REPLY [3 votes]: Another class of counterexamples (to which Martin's example belongs to) is: 

If $R \lneqq S \le \text{Frac}(R)$ is finitely generated as $R$-module, then $S$ is no localization of $R$.  

Proof: Suppose $S=R[U^{-1}]$ is generated by $s_i \in S\;(i=1,\ldots,n)$. By taking a common denominator, we can write $s_i=r_i/u$ with $r_i \in R, u \in U$. Thus $S$ is generated by $1/u$ over $R$. Therefore we find $r \in R$ such that $1/u^2=r/u$, i.e. $1/u=r$ is a unit in $R$. This yields $S=R(1/u) = Rr=R$. q.e.d. 
In combination with BenjaLim's comment we obtain the corollary:

If $R$ is a PID, then no ring $R \lneqq S \le \text{Frac}(R)$ is finitely generated over $R$.