The Stacks project

Lemma 52.7.3. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $X$ be a Noetherian scheme over $A$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Assume that $H^ p(X, \mathcal{F})$ is a finite $A$-module for all $p$. Then there are short exact sequences

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}/I^ n\mathcal{F}) \to H^ p(X, \mathcal{F})^\wedge \to \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}/I^ n\mathcal{F}) \to 0 \]

of $A$-modules where $H^ p(X, \mathcal{F})^\wedge $ is the usual $I$-adic completion. If $f$ is proper, then the $R^1\mathop{\mathrm{lim}}\nolimits $ term is zero.

Proof. Consider the two spectral sequences of Lemma 52.6.20. The first degenerates by More on Algebra, Lemma 15.93.4. We obtain $H^ p(X, \mathcal{F})^\wedge $ in degree $p$. This is where we use the assumption that $H^ p(X, \mathcal{F})$ is a finite $A$-module. The second degenerates because

\[ \mathcal{F}^\wedge = \mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F} = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F} \]

is a sheaf by Lemma 52.7.2. We obtain $H^ p(X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F})$ in degree $p$. Since $R\Gamma (X, -)$ commutes with derived limits (Injectives, Lemma 19.13.6) we also get

\[ R\Gamma (X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F}) = R\Gamma (X, R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F}) = R\mathop{\mathrm{lim}}\nolimits R\Gamma (X, \mathcal{F}/I^ n\mathcal{F}) \]

By More on Algebra, Remark 15.86.6 we obtain exact sequences

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}/I^ n\mathcal{F}) \to H^ p(X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F}) \to \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}/I^ n\mathcal{F}) \to 0 \]

of $A$-modules. Combining the above we get the first statement of the lemma. The vanishing of the $R^1\mathop{\mathrm{lim}}\nolimits $ term follows from Cohomology of Schemes, Lemma 30.20.4. $\square$

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