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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*}

Lemma 15.81.11. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. The sequence (15.81.10.1) is

  1. exact on the right,

  2. exact on the left if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is injective, and

  3. exact in the middle if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective.

Thus $M$ is glueable for $(R \to R', f)$ if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is bijective. If $(R \to R', f)$ is a glueing pair, then $M$ is glueable for $(R \to R', f)$ if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is injective. For example, if $(R, f)$ is a glueing pair, then $M$ is glueable if and only if $M[f^\infty ] \to (M \otimes _ R R^\wedge )[f^\infty ]$ is injective.

Proof. We will use the results of Lemma 15.81.7 without further mention. The functor $M \otimes _ R -$ is right exact (Algebra, Lemma 10.11.10) hence we get (1).

The kernel of $M \to M \otimes _ R R_ f = M_ f$ is $M[f^\infty ]$. Thus (2) follows.

If the sequence is exact in the middle, then elements of the form $(x, 0)$ with $x \in (M \otimes _ R R')[f^\infty ]$ are in the image of the first arrow. This implies that $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective. Conversely, assume that $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective. Let $(x, y)$ be an element in the middle which maps to zero on the right. Write $y = y'/f^ n$ for some $y' \in M$. Then we see that $f^ n x - y'$ is annihilated by some power of $f$ in $M \otimes _ R R'$. By assumption we can write $f^ nx - y' = z$ for some $z \in M[f^\infty ]$. Then $y = y''/f^ n$ where $y'' = y' + z$ is in the kernel of $M \to M/f^ nM$. Hence we see that $y$ can be represented as $y'''/1$ for some $y''' \in M$. Then $x - y'''$ is in $(M \otimes _ R R')[f^\infty ]$. Thus $x - y''' = z' \in M[f^\infty ]$. Then $(x, y'''/1) = (y''' + z', (y''' + z')/1)$ as desired.

If $(R \to R', f)$ is a glueing pair, then (15.81.10.1) is exact in the middle for any $M$ by Algebra, Lemma 10.11.10. This gives the penultimate statement of the lemma. The final statement of the lemma follows from this and the fact that $(R, f)$ is a glueing pair if and only if $(R \to R^\wedge , f)$ is a glueing pair. $\square$


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