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
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