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

$A$ is $f$-adically complete,

$H^ i_\mathfrak a(M)$ 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.**
By induction on $s$. If $s = 0$, the assertion is empty. If $s = 1$, then the result is Lemma 52.12.7. Assume $s > 1$. By induction it suffices to prove the result for $i = s - 1 \geq 1$. We may apply Lemma 52.3.2 to $U$ and $\mathcal{F} = \widetilde{M}|_ U$ because $\mathcal{F}$ is a Noetherian object in the category of coherent $\mathcal{O}_ U$-modules. Observe that $H^ j(U, \mathcal{F}) = H^{j + 1}_\mathfrak a(M)$ for all $j$ by Local Cohomology, Lemma 51.8.2. Thus for $j = s = (s - 1) + 1$ this is annihilated by a power of $f$ by assumption. Thus it follows from Lemma 52.3.2 that $\mathop{\mathrm{lim}}\nolimits H^{s - 1}(U, \mathcal{F}/f^ n\mathcal{F})$ is the usual $f$-adic completion of $H^{s - 1}(U, \mathcal{F})$. Then again using that this module is killed by a power of $f$ we see that the completion is simply equal to $H^{s - 1}(U, \mathcal{F})$ as desired.
$\square$

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