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 that the kernel of $I/I^ n \otimes _{R/I^ n} M/I^ nM \to M/I^ nM$ is zero for all $n \geq 1$. Hence an element of $K$ maps to zero in $I/I^ n \otimes _{R/I^ n} M/I^ nM$. Since

\[ I/I^ n \otimes _{R/I^ n} M/I^ nM = I/I^ n \otimes _ R M = (I \otimes _ R M)/I^{n - 1}(I \otimes _ R M) \]

we conclude that $K \subset I^{n - 1}(I \otimes _ R M)$ for all $n \geq 1$. By the Artin-Rees lemma, and more precisely Lemma 10.51.5 we conclude that $K_{\mathfrak q} = 0$, as desired.
$\square$

## Comments (2)

Comment #8476 by Et on

Comment #9092 by Stacks project on