Finitely generated module over an integral domain

Question

I am trying to prove that if $B$ is a finitely generated $A$-module and $A$ is an integral domain, then $B$ is a free module. I don't know if there exists some result about this problem or is just straigthforward.

Answer 1

It is simply false: $\mathbf Z/n\mathbf Z$ is finitely generated (cyclic, generated by $1+n\mathbf Z$) but certainly not free, since $\;n\cdot(1+n\mathbf Z)=0+n\mathbf Z$.

However the assertion is true for finitely generated torsion-free modules over P.I.D.s.

For other integral domains, a counter-example is the ring $A=K[X,Y]$, where $K$ is a field: the ideal $(X,Y)$ is minimally generated by $X$ and $Y$, it is torsion-free, but there's a linear relation between its generators: $$Y\cdot X-X\cdot Y=0.$$