Finitely generated $R$-module in the product of quotient field. Let $R$ be an integral domain with the quotient field $K$. 

Let $M$ be a finitely generated $R$-submodule in $K^n$. 

Is it true that $M$ is free $R$-module?
Non, it is almost never true. E.g. an finitely generated ideal $M$ in $R$ is free only if it is generated by one element because two elements $x_1, x_2\in M$ are always related by the linear relation $x_1.x_2+(-x_2).x_1=0$, thus any basis of $M$ can only have one element. 

Example: $R=\mathbb C[X, Y]$, $M=XR+YR$. If $M$ was free, it would be generated by one element $P(X,Y)$. Then $P(X,Y)$ divides $X$ and $Y$,so $P(X,Y)$ is constant and generates the unit ideal $R$. But $M\ne R$.
Let $\,M=\langle\,m_1,...,m_s\,\rangle_R\leq K^n\,$ be a finitely generated $\,R-\,$submodule of $\,K^n\,$. Assuming  the above is a minimal generating set for $\,M\,$, we thus clearly have $(1)\,\,\,s\leq n\,$ (why?) and also $(2)\,\,\,\{m_1,...,m_s\}\,$ are $\,K-\,$ linearly independent (why?). Thus

$$r_1m_1+...+r_sm_s=0\,\,,\,\,r_i\in R\Longrightarrow r_i=0\,\,\forall\,i$$

since otherwise we get a contradiction to (2) above.