# Tor and Flatness of finite type modules in local rings

We know that a finitely presented module over a local ring $(R,m)$ is free if and only if it is flat if and only if $$\text{Tor}_1(M, R/m)=0$$ We know that a finitely generated module is flat over the local ring $(R,m)$ if and only if it is free. Do we still have the equivalence with $\text{Tor}_1(M, R/m)=0$? If not, does one know a counterexample (obviously where $m$ is not of finite type)?

• @user26857 yes thank you, but obviously I would like to deal with the non-noetherian case which is not immediately clear for me Jun 15 '17 at 21:08
• Let $R$ be a valuation ring with maximal ideal $m$ satisfying $m=m^2$. Set $M=R/m$. Then $\text{Tor}_1(M, R/m)=0$; see here. On the other side $M$ is not torsion-free, hence not flat. Jun 15 '17 at 21:24
• @user26857 why is Tor zero here ? Jun 15 '17 at 21:33
Let $$R$$ be a valuation ring with maximal ideal $$\mathfrak m$$ satisfying $$\mathfrak m=\mathfrak m^2$$. Set $$M=R/\mathfrak m$$. Then $$\mathrm{Tor}_1(M,R/\mathfrak m)=0$$; see here. On the other side $$M$$ is not torsion-free, hence not flat.