Lemma 10.128.3. Let $R \to S$, $M$, $\Lambda $, $R_\lambda \to S_\lambda $, $M_\lambda $ be as in Lemma 10.127.13. Assume that $M$ is flat over $R$. Then for some $\lambda \in \Lambda $ the module $M_\lambda $ is flat over $R_\lambda $.
Proof. Pick some $\lambda \in \Lambda $ and consider
\[ \text{Tor}_1^{R_\lambda }(M_\lambda , R_\lambda /\mathfrak m_\lambda ) = \mathop{\mathrm{Ker}}(\mathfrak m_\lambda \otimes _{R_\lambda } M_\lambda \to M_\lambda ). \]
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 $M$ is flat over $R$ we have that $0 = \mathop{\mathrm{Ker}}(\mathfrak m_\lambda R \otimes _ R M \to M)$. Since $\otimes $ commutes with colimits we see there exists a $\lambda ' \geq \lambda $ such that each $\xi _ i$ maps to zero in $\mathfrak m_{\lambda }R_{\lambda '} \otimes _{R_{\lambda '}} M_{\lambda '}$. Hence we see that
\[ \text{Tor}_1^{R_\lambda }(M_\lambda , R_\lambda /\mathfrak m_\lambda ) \longrightarrow \text{Tor}_1^{R_{\lambda '}}(M_{\lambda '}, R_{\lambda '}/\mathfrak m_{\lambda }R_{\lambda '}) \]
is zero. Note that $M_\lambda \otimes _{R_\lambda } R_\lambda /\mathfrak m_\lambda $ is flat over $R_\lambda /\mathfrak m_\lambda $ because this last ring is a field. Hence we may apply Lemma 10.99.14 to get that $M_{\lambda '}$ is flat over $R_{\lambda '}$. $\square$
Comments (0)