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 15.81.7. 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 (15.81.6.1) is

  1. exact on the right,

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

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

In particular, $(R \to R', f)$ is a glueing pair if and only if $R[f^\infty ] \to R'[f^\infty ]$ is bijective. For example, $(R, f)$ is a glueing pair if and only if $R[f^\infty ] \to R^\wedge [f^\infty ]$ is bijective.

Proof. Let $x \in R'_ f$. Write $x = x'/f^ n$ with $x' \in R'$. Write $x' = x'' + f^ n y$ with $x'' \in R$ and $y \in R'$. Then we see that $(y, -x''/f^ n)$ maps to $x$. Thus (1) holds.

Part (2) follows from the fact that $\mathop{\mathrm{Ker}}(R \to R_ f) = R[f^\infty ]$.

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

The last statement of the lemma is a special case of the penultimate statement by Lemma 15.81.1. $\square$


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