Subring of $R$ with fraction field $=$ Frac $R$ Let $R$ be an integral domain with fraction field $K$. Of course all the overrings of $R$  share the same fraction field $K$ (by an overring I mean a subring $S\subset K$ containing $R$ as a subring). Do you have a reference for a criterion saying when a subring of $R$ has fraction field $K$?
 A: Allow me to rephrase slightly Simon Markett's excellent solution.
In his notation, the condition is $x S \cap S \ne \{ 0 \}$ for all $0 \ne x \in R$. This seems to me to be equivalent to saying that $R/S$ is a torsion $S$-module.
A: Not a reference but an answer:
Let $S\subset R$ be a subring with $\text{Frac }S=\text{Frac }R=K$. In other words, the natural map $\text{Frac }S\to\text{Frac }R=K$ needs to be an isomorphism. It is clearly injective so it remains to show surjectivity.
For any $x,y\in R$ ($y\neq 0$) we need to find $s,t\in S$ ($t\neq 0$) such that $\frac{x}{y}=\frac{s}{t}\in K$ ($\ast$).
Necessary and sufficient condition: For all $x\in R$ exists a $0\neq t\in S$ such that $tx\in S$.
It is indeed necessary since we need equation $(\ast)$ to hold for $y=1$, i.e. we need $s,t\in S$ ($t\neq0$) such that $tx=s$. (Note that $R$ is a domain).
Now it is also sufficient: Let $x,y\in R$ ($y\neq 0$) and $0\neq s,t\in S$ such that $tx,sy\in S$. Then
$$\frac xy=\frac{stx}{sty}\in \text{Frac }S$$
