Lemma 52.12.7. 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. Assume
$A$ is $f$-adically complete,
$H^1_\mathfrak a(M)$ and $H^2_\mathfrak a(M)$ are annihilated by a power of $f$.
Then with $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$ the map
is an isomorphism.
Proof. 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. Since $H^1(U, \mathcal{F}) = H^2_\mathfrak a(M)$ (Local Cohomology, Lemma 51.8.2) is annihilated by a power of $f$, we see that its $f$-adic Tate module is zero. Hence the lemma shows $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/f^ n \mathcal{F})$ is equal to the usual $f$-adic completion of $H^0(U, \mathcal{F})$. Consider the short exact sequence
of Local Cohomology, Lemma 51.8.2. Since $M/H^0_\mathfrak a(M)$ is a finite $A$-module, it is complete, see Algebra, Lemma 10.97.1. Since $H^1_\mathfrak a(M)$ is killed by a power of $f$, we conclude from Algebra, Lemma 10.96.4 that $H^0(U, \mathcal{F})$ is complete as well. This finishes the proof. $\square$
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