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 10.97.3. Let $A$ be a graded ring. Let $I \subset A_+$ be a homogeneous ideal. Denote $A' = \mathop{\mathrm{lim}}\nolimits A/I^ n$. Let $(G_ n)$ be an inverse system of graded $A$-modules with $G_ n$ annihilated by $I^ n$. Let $M$ be a graded $A$-module and let $\varphi _ n : M \to G_ n$ be a compatible system of graded $A$-module maps. If the induced map

\[ \varphi : M \otimes _ A A' \longrightarrow \mathop{\mathrm{lim}}\nolimits G_ n \]

is an isomorphism, then $M_ d \to \mathop{\mathrm{lim}}\nolimits G_{n, d}$ is an isomorphism for all $d \in \mathbf{Z}$.

Proof. By convention graded rings are in degrees $\geq 0$ and graded modules may have nonzero parts of any degree, see Section 10.55. The map $\varphi $ exists because $\mathop{\mathrm{lim}}\nolimits G_ n$ is a module over $A'$ as $G_ n$ is annihilated by $I^ n$. Another useful thing to keep in mind is that we have

\[ \bigoplus \nolimits _{d \in \mathbf{Z}} \mathop{\mathrm{lim}}\nolimits G_{n, d} \subset \mathop{\mathrm{lim}}\nolimits G_ n \subset \prod \nolimits _{d \in \mathbf{Z}} \mathop{\mathrm{lim}}\nolimits G_{n, d} \]

where a subscript ${\ }_ d$ indicates the $d$th graded part.

Injective. Let $x \in M_ d$. If $x \mapsto 0$ in $\mathop{\mathrm{lim}}\nolimits G_{n, d}$ then $x \otimes 1 = 0$ in $M \otimes _ A A'$. Then we can find a finitely generated submodule $M' \subset M$ with $x \in M'$ such that $x \otimes 1$ is zero in $M' \otimes _ A A'$. Say $M'$ is generated by homogeneous elements sitting in degrees $d_1, \ldots , d_ r$. Let $n = d - \min (d_ i) + 1$. Since $A'$ has a map to $A/I^ n$ and since $A \to A/I^ n$ is an isomorphism in degrees $\leq n - 1$ we see that $M' \to M' \otimes _ A A'$ is injective in degrees $\leq n - 1$. Thus $x = 0$ as desired.

Surjective. Let $y \in \mathop{\mathrm{lim}}\nolimits G_{n, d}$. Choose a finite sum $\sum x_ i \otimes f'_ i$ in $M \otimes _ A A'$ mapping to $y$. We may assume $x_ i$ is homogeneous, say of degree $d_ i$. Observe that although $A'$ is not a graded ring, it is a limit of the graded rings $A/I^ nA$ and moreover, in any given degree the transition maps eventually become isomorphisms (see above). This gives

\[ A = \bigoplus \nolimits _{d \geq 0} A_ d \subset A' \subset \prod \nolimits _{d \geq 0} A_ d \]

Thus we can write

\[ f'_ i = \sum \nolimits _{j = 0, \ldots , d - d_ i - 1} f_{i, j} + f_ i + g'_ i \]

with $f_{i, j} \in A_ j$, $f_ i \in A_{d - d_ i}$, and $g'_ i \in A'$ mapping to zero in $\prod _{j \leq d - d_ i} A_ j$. Now if we compute $\varphi _ n(\sum _{i, j} f_{i, j}x_ i) \in G_ n$, then we get a sum of homogeneous elements of degree $< d$. Hence $\varphi (\sum x_ i \otimes f_{i, j})$ maps to zero in $\mathop{\mathrm{lim}}\nolimits G_{n, d}$. Similarly, a computation shows the element $\varphi (\sum x_ i \otimes g'_ i)$ maps to zero in $\prod _{d' \leq d} \mathop{\mathrm{lim}}\nolimits G_{n, d'}$. Since we know that $\varphi (\sum x_ i \otimes f'_ i)$ is $y$, we conclude that $\sum f_ ix_ i \in M_ d$ maps to $y$ as desired. $\square$


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