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 10.98.13. Let $R \to R'$ be a ring map. Let $I \subset R$ be an ideal and $I' = IR'$. Let $M$ be an $R$-module and set $M' = M \otimes _ R R'$. The natural map $\text{Tor}_1^ R(R'/I', M) \to \text{Tor}_1^{R'}(R'/I', M')$ is surjective.

Proof. Let $F_2 \to F_1 \to F_0 \to M \to 0$ be a free resolution of $M$ over $R$. Set $F_ i' = F_ i \otimes _ R R'$. The sequence $F_2' \to F_1' \to F_0' \to M' \to 0$ may no longer be exact at $F_1'$. A free resolution of $M'$ over $R'$ therefore looks like

\[ F_2' \oplus F_2'' \to F_1' \to F_0' \to M' \to 0 \]

for a suitable free module $F_2''$ over $R'$. Next, note that $F_ i \otimes _ R R'/I' = F_ i' / IF_ i' = F_ i'/I'F_ i'$. So the complex $F_2'/I'F_2' \to F_1'/I'F_1' \to F_0'/I'F_0'$ computes $\text{Tor}_1^ R(M, R'/I')$. On the other hand $F_ i' \otimes _{R'} R'/I' = F_ i'/I'F_ i'$ and similarly for $F_2''$. Thus the complex $F_2'/I'F_2' \oplus F_2''/I'F_2'' \to F_1'/I'F_1' \to F_0'/I'F_0'$ computes $\text{Tor}_1^{R'}(M', R'/I')$. Since the vertical map on complexes

\[ \xymatrix{ F_2'/I'F_2' \ar[r] \ar[d] & F_1'/I'F_1' \ar[r] \ar[d] & F_0'/I'F_0' \ar[d] \\ F_2'/I'F_2' \oplus F_2''/I'F_2'' \ar[r] & F_1'/I'F_1' \ar[r] & F_0'/I'F_0' } \]

clearly induces a surjection on cohomology we win. $\square$


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