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Let $(R,\mathfrak m)$ be a local noetherian ring, and $M$ an arbitrary $R$-module. Suppose that $\mathrm{Tor}_1(M,R/\mathfrak m)=0$. Does it follow that $M$ is flat?

The answer is positive when $M$ is finitely generated, or when $\mathfrak m$ is nilpotent (that is, $R$ is artinian). The same can be said if there is a local homomorphism of local rings $R\to S$ and $M$ is a finitely generated $S$-module (local criterion for flatness).

If one assumes that $\mathrm{Tor}_1(M,R/\mathfrak p)=0$ for every prime ideal $\mathfrak p\subset R$, then $M$ is flat.

There are reasonable doubts that the answer is positive in general, but I don't have a counterexample.

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  • $\begingroup$ I'm sorry for my stupid hint. There is a counterexample. Consider $R=\mathbb{Z}_{(p)}[x]_{(x)}$ and $M=\mathbb{Q}$ where $x$ is acting on $M$ by identity. Then we can apply stacks.math.columbia.edu/tag/00MD to prove $\mathrm{Tor}_1(M,R/\mathfrak{m})$ but $M$ is not flat over $R$. $\endgroup$
    – user196717
    Commented Apr 26, 2021 at 11:06
  • $\begingroup$ I have just considered local ring of Krull dim. 2 and a non-flat module but first Tor can be vanishing... $\mathbb{Q}$ is a $R=\mathbb{Z}_{(p)}[x]_{(x)}$-module bacause of the composition of $$ \mathbb{Z}_{(p)}[x]_{(x)}\to \mathbb{Q}[x]_{(x)}\to \mathbb{Q}$$ ... oh I have wrongly described the action of $x$ on $\mathbb{Q}$ sorry. The action of $x$ is $0$. Then we can compute the Tor functor. by $$ 0\to \mathbb{Q}\to \mathbb{Q}(x)\to \mathbb{Q}[x]/\mathbb{Q}\to 0$$ we have $\mathrm{Tor}_1(\mathbb{Q},\mathbb{Z}/(p))=\mathrm{Tor}_2(\mathbb{Q}(x)/\mathbb{Q},\mathbb{Z}/(p))$ and by $\endgroup$
    – user196717
    Commented Apr 26, 2021 at 14:54
  • $\begingroup$ $$ 0\to \mathbb{Z}_{(p)}\to \mathbb{Z}_{(p)}\to \mathbb{Z}/(p)\to 0$$ also we have $$ \mathrm{Tor}_2(\mathbb{Q}(x)/\mathbb{Q},\mathbb{Z}/p)=\mathrm{Ker}(\mathrm{Tor}_1(\mathbb{Q}(x)/\mathbb{Q},\mathbb{Z})\to \mathrm{Tor}_1(\mathbb{Q}(x)/\mathbb{Q},\mathbb{Z}))$$ and $$ \mathrm{Tor}_1(\mathbb{Q}(x)/\mathbb{Q},\mathbb{Z})=\mathrm{Ker}(\mathbb{Q}\otimes \mathbb{Z}\to \mathbb{Q}(x)\otimes \mathbb{Z})$$ and any element of that module is $p$-divisible. So $\mathrm{Tor}_1(\mathbb{Q},\mathbb{Z}/p)=0$ but $\mathbb{Q}$ is not flat over $R$. $\endgroup$
    – user196717
    Commented Apr 26, 2021 at 15:02
  • $\begingroup$ The problem is that for arbitrary $M$, we must check for any prime ideal not just the maximal ideal. In my example, $\mathrm{Tor}_1(\mathbb{Q},\mathbb{Z})$ does not vanish and using $$ 0\to \mathbb{Z}[x]_{(x)}\to \mathbb{Z}[x]_{(x)}\to \mathbb{Z}\to 0$$ $\endgroup$
    – user196717
    Commented Apr 26, 2021 at 15:10
  • $\begingroup$ we have $\mathrm{Tor}_1(\mathbb{Q},\mathbb{Z})=\mathbb{Q}$. This is somewhat small but not $0$. I think the following is true. Theorem. Let $R$ be a noetherian ring, $M$ be a $R$-module. Then if $\mathrm{Tor}_1(M,R/\mathfrak{p})=0$ for any prime ideal $\mathfrak{p}$, then $M$ is flat over $R$. $\endgroup$
    – user196717
    Commented Apr 26, 2021 at 15:15

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