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This is something I should probably know, but it is escaping me at the moment.

Let $A$ be a commutative noetherian ring. The following corollary of Nakayama's lemma is well-known (for instance, this is Atiyah-Macdonald exercises 7.15 and 7.16)

If $M$ is a finitely generated $A$-module then $M$ is flat if and only if $\operatorname{Tor}_j^A(M,k(\mathfrak{p})) = (0)$ for all prime ideals $\mathfrak{p} \in \operatorname{Spec}(A)$ and $j\geq 1$.

Here, $k(\mathfrak{p}) = A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ is the residue field at the point $\mathfrak{p}$ of $\operatorname{Spec}(A)$. My question is the following:

Is the same true if we do not assume that $M$ is finitely generated?

Note that in the noetherian situation, flat and locally free are the same. I know that my question is not true if we replace flat by locally free statement because, for example, $\mathbf{Q}$ is flat over $\mathbf{Z}$ but not locally free. I'm willing to accept an answer that only addresses the case where $A$ is reduced, if that matters at all.

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  • $\begingroup$ Probably no. Have you considered the case $A=\mathbb{Z}$? $\endgroup$ Commented May 3, 2013 at 16:23
  • $\begingroup$ In the case of $A = \mathbf{Z}$, my question is answered in the affirmative. Indeed, if $M$ is an abelian group (not necessarily f.g) then $M$ is flat if and only if $M$ is torsion free. This follows from a different exercise in Atiyah-Macdonald (see math.stackexchange.com/questions/106793/… for a handy reference). More generally, if $A$ is a PID then my question is answered in the affirmative. $\endgroup$
    – tkr
    Commented May 3, 2013 at 16:37

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The answer to your question is yes.

Hint. Consider the set $\mathcal S=\{I\subset A: \exists\ i\ge 1 \text{ such that } \operatorname{Tor}_i(R/I,M)\neq 0\}$. If $\mathcal S$ is not empty, prove that its maximal elements are prime ideals. Then write $k(\mathfrak p)$ as a direct limit $\lim\limits_{\longrightarrow x} R/\mathfrak p$, where $x\in R-\mathfrak p$ and consider the short exact sequence $0\to R/\mathfrak p\to k(\mathfrak p)\to N\to 0$. (Here $N$ is the quotient module of $k(\mathfrak p)$ by $R/\mathfrak p$.)

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  • $\begingroup$ Thanks, I'll think about it. Do you have a reference I can cite? I actually don't need to "use" the flatness in what I am writing, but I would prefer to point out that the condition implies the flatness...it makes for cleaner exposition. $\endgroup$
    – tkr
    Commented May 4, 2013 at 21:09

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