The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.127.3. Let $R \to S$, $M$, $\Lambda $, $R_\lambda \to S_\lambda $, $M_\lambda $ be as in Lemma 10.126.13. 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 10.74.9. 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 10.98.14 to get that $M_{\lambda '}$ is flat over $R_{\lambda '}$. $\square$


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