Lemma 52.13.2. Let $A$ be a Noetherian ring. Let $f \in \mathfrak a \subset A$ be an element of an ideal of $A$. Let $M$ be a finite $A$-module. Let $s \geq 0$. Assume

1. $A$ is $f$-adically complete,

2. $H^ i_\mathfrak a(M)$ is annihilated by a power of $f$ for $i \leq s + 1$.

Then with $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$ the map

$H^ i(U, \widetilde{M}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i(U, \widetilde{M/f^ nM})$

is an isomorphism for $i < s$.

Proof. The proof is the same as the proof of Lemma 52.12.6. We may apply Lemma 52.3.6 to $U$ and $\mathcal{F} = \widetilde{M}|_ U$ because $\mathcal{F}$ is a Noetherian object in the category of coherent $\mathcal{O}_ U$-modules. Since $H^ i(U, \mathcal{F}) = H^{i + 1}_\mathfrak a(M)$ (Local Cohomology, Lemma 51.8.2) is annihilated by a power of $f$ for $i \leq s$, we see that its $f$-adic Tate module is zero. Hence the lemma shows $\mathop{\mathrm{lim}}\nolimits H^{i - 1}(U, \mathcal{F}/f^ n \mathcal{F})$ is the $0$th cohomology group of the derived $f$-adic completion of $H^{i - 1}(U, \mathcal{F})$. However, if $s \geq i > 1$, then this equal to the $f$-power torsion module $H^ i_\mathfrak a(M)$ and hence equal to its own (derived) completion. For $i = 0$, we refer to Lemma 52.12.6. $\square$

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