Lemma 10.128.7. Let $R \to S$ be a local homomorphism of local rings. Let $I \not= R$ be an ideal in $R$. Let $M$ be an $S$-module. Assume

$S$ is essentially of finite presentation over $R$,

$M$ is of finite presentation over $S$,

$\text{Tor}_1^ R(M, R/I) = 0$, and

$M/IM$ is flat over $R/I$.

Then $M$ is flat over $R$.

**Proof.**
Let $\Lambda $, $R_\lambda \to S_\lambda $, $M_\lambda $ be as in Lemma 10.127.13. Denote $I_\lambda \subset R_\lambda $ the inverse image of $I$. In this case the system $R/I \to S/IS$, $M/IM$, $R_\lambda \to S_\lambda /I_\lambda S_\lambda $, and $M_\lambda /I_\lambda M_\lambda $ satisfies the conclusions of Lemma 10.127.13 as well. Hence by Lemma 10.128.3 we may assume (after shrinking the index set $\Lambda $) that $M_\lambda /I_\lambda M_\lambda $ is flat for all $\lambda $. Pick some $\lambda $ and consider

\[ \text{Tor}_1^{R_\lambda }(M_\lambda , R_\lambda /I_\lambda ) = \mathop{\mathrm{Ker}}(I_\lambda \otimes _{R_\lambda } M_\lambda \to M_\lambda ). \]

See Remark 10.75.9. The right hand side shows that this is a finitely generated $S_\lambda $-module (because $S_\lambda $ is Noetherian and the modules in question are finite). Let $\xi _1, \ldots , \xi _ n$ be generators. Because $\text{Tor}_1^ R(M, R/I) = 0$ and since $\otimes $ commutes with colimits we see there exists a $\lambda ' \geq \lambda $ such that each $\xi _ i$ maps to zero in $\text{Tor}_1^{R_{\lambda '}}(M_{\lambda '}, R_{\lambda '}/I_{\lambda '})$. The composition of the maps

\[ \xymatrix{ R_{\lambda '} \otimes _{R_\lambda } \text{Tor}_1^{R_\lambda }(M_\lambda , R_\lambda /I_\lambda ) \ar[d]^{\text{surjective by Lemma 00MM}} \\ \text{Tor}_1^{R_\lambda }(M_\lambda , R_{\lambda '}/I_\lambda R_{\lambda '}) \ar[d]^{\text{surjective up to localization by Lemma 00MN}} \\ \text{Tor}_1^{R_{\lambda '}}(M_{\lambda '}, R_{\lambda '}/I_\lambda R_{\lambda '}) \ar[d]^{\text{surjective by Lemma 00MM}} \\ \text{Tor}_1^{R_{\lambda '}}(M_{\lambda '}, R_{\lambda '}/I_{\lambda '}). } \]

is surjective up to a localization by the reasons indicated. The localization is necessary since $M_{\lambda '}$ is not equal to $M_\lambda \otimes _{R_\lambda } R_{\lambda '}$. Namely, it is equal to $M_\lambda \otimes _{S_\lambda } S_{\lambda '}$ and $S_{\lambda '}$ is the localization of $S_{\lambda } \otimes _{R_\lambda } R_{\lambda '}$ whence the statement up to a localization (or tensoring with $S_{\lambda '}$). Note that Lemma 10.99.12 applies to the first and third arrows because $M_\lambda /I_\lambda M_\lambda $ is flat over $R_\lambda /I_\lambda $ and because $M_{\lambda '}/I_\lambda M_{\lambda '}$ is flat over $R_{\lambda '}/I_\lambda R_{\lambda '}$ as it is a base change of the flat module $M_\lambda /I_\lambda M_\lambda $. The composition maps the generators $\xi _ i$ to zero as we explained above. We finally conclude that $\text{Tor}_1^{R_{\lambda '}}(M_{\lambda '}, R_{\lambda '}/I_{\lambda '})$ is zero. This implies that $M_{\lambda '}$ is flat over $R_{\lambda '}$ by Lemma 10.99.10.
$\square$

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