Lemma 10.99.11. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Assume
$R$ is a Noetherian ring,
$S$ is a Noetherian ring,
$M$ is a finite $S$-module, and
for each $n \geq 1$ the module $M/I^ n M$ is flat over $R/I^ n$.
Then for every $\mathfrak q \in V(IS)$ the localization $M_{\mathfrak q}$ is flat over $R$. In particular, if $S$ is local and $IS$ is contained in its maximal ideal, then $M$ is flat over $R$.
Proof. We are going to use Lemma 10.99.10. By assumption $M/IM$ is flat over $R/I$. Hence it suffices to check that $\text{Tor}_1^ R(M, R/I)$ is zero on localization at $\mathfrak q$. By Remark 10.75.9 this Tor group is equal to $K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$. We know for each $n \geq 1$ that the kernel $\mathop{\mathrm{Ker}}(I/I^ n \otimes _{R/I^ n} M/I^ nM \to M/I^ nM)$ is zero. Since there is a module map $I/I^ n \otimes _{R/I^ n} M/I^ nM \to (I \otimes _ R M)/I^{n - 1}(I \otimes _ R M)$ we conclude that $K \subset I^{n - 1}(I \otimes _ R M)$ for each $n$. By the Artin-Rees lemma, and more precisely Lemma 10.51.5 we conclude that $K_{\mathfrak q} = 0$, as desired. $\square$
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Comment #8476 by Et on