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\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*}

Slight generalization of [Lemme 3(a), Beauville-Laszlo]

Lemma 15.81.16. Let $(R \to R', f)$ be a glueing pair. For every $R$-module $M$, we have $\text{Tor}^ R_1(R', \mathop{\mathrm{Coker}}(M \to M_ f)) = 0$.

Proof. Set $\overline{M} = M/M[f^\infty ]$. Then $\mathop{\mathrm{Coker}}(M \to M_ f) \cong \mathop{\mathrm{Coker}}(\overline{M} \to \overline{M}_ f)$ hence we may and do assume that $f$ is a nonzerodivisor on $M$. In this case $M \subset M_ f$ and $M_ f/M = \mathop{\mathrm{colim}}\nolimits M/f^ nM$ where the transition maps are given by multiplication by $f$. Since formation of Tor groups commutes with colimits (Algebra, Lemma 10.75.2) it suffices to show that $\text{Tor}^ R_1(R', M/f^ n M) = 0$.

We first treat the case $M = R/R[f^\infty ]$. By Lemma 15.81.7 we have $M \otimes _ R R' = R'/R'[f^\infty ]$. From the short exact sequence $0 \to M \to M \to M/f^ nM \to 0$ we obtain the exact sequence

\[ \xymatrix{ \text{Tor}_1^ R(R', R/R[f^\infty ]) \ar[r] & \text{Tor}_1^ R(R', M/f^ n M) \ar[r] & R'/R'[f^\infty ] \ar[dll]_{f^ n} \\ R'/R'[f^\infty ] \ar[r] & (R'/R'[f^\infty ])/(f^ n (R'/R'[f^\infty ])) \ar[r] & 0 } \]

by Algebra, Lemma 10.74.2. Here the diagonal arrow is injective. Since the first group $\text{Tor}_1^ R(R', R/R[f^\infty ])$ is zero by Lemma 15.81.15, we deduce that $\text{Tor}_1^ R(R', M/f^ nM) = 0$ as desired.

To treat the general case, choose a surjection $F \to M$ with $F$ a free $R/R[f^\infty ]$-module, and form an exact sequence

\[ 0 \to N \to F/f^ n F \to M/f^ n M \to 0. \]

By Lemma 15.81.2 this sequence remains unchanged, and hence exact, upon tensoring with $R'$. Since $\text{Tor}^ R_1(R', F/f^ n F) = 0$ by the previous paragraph, we deduce that $\text{Tor}^ R_1(R', M/f^ n M) = 0$ as desired. $\square$


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