# Finitely generated module over an integral domain

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.

• It's not true . – Angina Seng Apr 1 '18 at 11:49
• Every projective module over a PID is free, see here. – Dietrich Burde Apr 1 '18 at 11:50
• $\;\Bbb Z/2\Bbb Z\;$ is a finitely generated $\;\Bbb Z\,-$ module, yet it is not a free $\;\Bbb Z\,-$ module... – DonAntonio Apr 1 '18 at 11:51

## 1 Answer

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: $$Y\cdot X-X\cdot Y=0.$$

• If we add some hypothesis, for example, $B$ is also an integral domain, $A$ and $B$ are both local rings, the result is still false or ir turns true? – Edude Apr 1 '18 at 12:13
• The only results I know is about integral closure of an integral domain i a finite separable extension of its field of fraction. It supposes the discriminant is a unit in $A$. In particular, $B$ is free over $A$ if $A$ is a P.I.D. – Bernard Apr 1 '18 at 12:29