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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)?

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  • $\begingroup$ @user26857 yes thank you, but obviously I would like to deal with the non-noetherian case which is not immediately clear for me $\endgroup$
    – brunoh
    Jun 15 '17 at 21:08
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    $\begingroup$ 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. $\endgroup$
    – user26857
    Jun 15 '17 at 21:24
  • $\begingroup$ @user26857 why is Tor zero here ? $\endgroup$
    – brunoh
    Jun 15 '17 at 21:33
  • $\begingroup$ Did you follow the link? $\endgroup$
    – user26857
    Jun 15 '17 at 21:35
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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.

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