Let $R$ be an integral domain with the field of fractions $K$. Suppose $M \subseteq K$ be an $R$-module. Show that if $M$ is an invertible $R$-module, then $M^{-1}=\left\{ a \in K : am \in R \ \forall m \in M \right\}$.
I'm having a trouble showing that if $N \subseteq K$ is any $R$-module such that $MN=R$, then $M^{-1} \subseteq N$. If $M, N$ are just ideals in $R$ then it's easy ($M^{-1} = M^{-1}R=M^{-1}MN=RN=N$), but this is not useful here because $M$ is a fractional ideal.