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This is an elementary question which did not get answered on math.stackexchange. I would like to know the answer for expository purposes.

I need either a reference or a counter-example to the following statement. Let $A$ be a noetherian ring which is Jacobson (i.e. every prime ideal $\mathfrak{p} \subset A$ is an intersection of maximal ideals). Suppose that $M$ is an $A$-module.

Is it true that $M$ is flat if and only if $\operatorname{Tor}_1^A(M,A/\mathfrak{m}) = (0)$ for all maximal ideals $\mathfrak{m} \subset A$?

Note hat I have not assumed that $M$ be finitely generated. If $M$ is finitely generated then the answer is in the affirmative, as is well-known (and requires just the noetherian hypothesis). Note as well that under just the noetherian hypothesis, the same statement is true with maximal replaced by prime. See, for example, Lemma 2.1 of these notes. I would be happy if the Jacobson property was superfluous here as well.

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  • $\begingroup$ No. Take $A = k[x,y]$, and as $M$ the quotient field of $k[x] = k[x,y]/(y)$. $\endgroup$
    – Angelo
    May 14, 2013 at 21:51

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