Flatness versus vanishing of Tor-groups for a non-finitely generated module

Question

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.

Answer 1

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 $R/\mathfrak p$.)