Flatness of module over field of fractions

Question

This is from Liu 1.2.9.

Let $A$ be an integral domain, and $K$ its field of fractions. Let $M$ be a finitely generated sub-$A$-module of $K$. Why do $M$ is flat if and only if $M_{\mathfrak p}$ is free of rank $1$ over $A_\mathfrak p$ for every prime ideal $\mathfrak p$ of $A$?

Answer 1

For a finitely generated module over a local ring being flat is the same as being free. Can the fraction field of a domain have two elements linearly independent over the domain? Try this for $\mathbf{Z}$, for example.