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 10.100.4. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

  1. $M/IM$ is flat over $R/I$, and

  2. $R' \otimes _ R M$ is flat over $R'$.

Set $I_2 = \varphi ^{-1}(\varphi (I^2)R')$. Then $M/I_2M$ is flat over $R/I_2$.

Proof. We may replace $R$, $M$, and $R'$ by $R/I_2$, $M/I_2M$, and $R'/\varphi (I)^2R'$. Then $I^2 = 0$ and $\varphi $ is injective. By Lemma 10.98.8 and the fact that $I^2 = 0$ it suffices to prove that $\text{Tor}^ R_1(R/I, M) = K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is zero. Set $M' = M \otimes _ R R'$ and $I' = IR'$. By assumption the map $I' \otimes _{R'} M' \to M'$ is injective. Hence $K$ maps to zero in

\[ I' \otimes _{R'} M' = I' \otimes _ R M = I' \otimes _{R/I} M/IM. \]

Then $I \to I'$ is an injective map of $R/I$-modules. Since $M/IM$ is flat over $R/I$ the map

\[ I \otimes _{R/I} M/IM \longrightarrow I' \otimes _{R/I} M/IM \]

is injective. This implies that $K$ is zero in $I \otimes _ R M = I \otimes _{R/I} M/IM$ as desired. $\square$


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