The Stacks project

Lemma 52.2.4. Let $X$ be a scheme. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Let

\[ \ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1 \]

be an inverse system of quasi-coherent $\mathcal{O}_ X$-modules such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/\mathcal{I}^ n\mathcal{F}_{n + 1}$. Set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Then

  1. $\mathcal{F} = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$,

  2. for any affine open $U \subset X$ we have $H^ p(U, \mathcal{F}) = 0$ for $p > 0$, and

  3. for each $p$ there is a short exact sequence $0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}_ n) \to H^ p(X, \mathcal{F}) \to \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}_ n) \to 0$.

If moreover $\mathcal{I}$ is of finite type, then

  1. $\mathcal{F}_ n = \mathcal{F}/\mathcal{I}^ n\mathcal{F}$, and

  2. $\mathcal{I}^ n \mathcal{F} = \mathop{\mathrm{lim}}\nolimits _{m \geq n} \mathcal{I}^ n\mathcal{F}_ m$.

Proof. Parts (1), (2), and (3) are general facts about inverse systems of quasi-coherent modules with surjective transition maps, see Derived Categories of Schemes, Lemma 36.3.2 and Cohomology, Lemma 20.36.1. Next, assume $\mathcal{I}$ is of finite type. Let $U \subset X$ be affine open. Say $U = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{I}|_ U$ corresponds to $I \subset A$. Observe that $I$ is a finitely generated ideal. By the equivalence of categories between quasi-coherent $\mathcal{O}_ U$-modules and $A$-modules (Schemes, Lemma 26.7.5) we find that $M_ n = \mathcal{F}_ n(U)$ is an inverse system of $A$-modules with $M_ n = M_{n + 1}/I^ nM_{n + 1}$. Thus

\[ M = \mathcal{F}(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U) = \mathop{\mathrm{lim}}\nolimits M_ n \]

is an $I$-adically complete module with $M/I^ nM = M_ n$ by Algebra, Lemma 10.98.2. This proves (4). Part (5) translates into the statement that $\mathop{\mathrm{lim}}\nolimits _{m \geq n} I^ nM/I^ mM = I^ nM$. Since $I^ mM = I^{m - n} \cdot I^ nM$ this is just the statement that $I^ mM$ is $I$-adically complete. This follows from Algebra, Lemma 10.96.3 and the fact that $M$ is complete. $\square$


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