Lemma 52.10.4. In Situation 52.10.1 assume $A$ is local with maximal ideal $\mathfrak m$ and $T = \{ \mathfrak m\} $. Then $H^ i_\mathfrak m(M) \to \mathop{\mathrm{lim}}\nolimits H^ i_\mathfrak m(M/I^ nM)$ is an isomorphism for $i \leq s$ and these modules are annihilated by a power of $I$.

**Proof.**
Let $A', I', \mathfrak m', M'$ be the usual $I$-adic completions of $A, I, \mathfrak m, M$. Recall that we have $H^ i_\mathfrak m(M) \otimes _ A A' = H^ i_{\mathfrak m'}(M')$ by flatness of $A \to A'$ and Dualizing Complexes, Lemma 47.9.3. Since $H^ i_\mathfrak m(M)$ is $\mathfrak m$-power torsion we have $H^ i_\mathfrak m(M) = H^ i_\mathfrak m(M) \otimes _ A A'$, see More on Algebra, Lemma 15.89.3. We conclude that $H^ i_\mathfrak m(M) = H^ i_{\mathfrak m'}(M')$. The exact same arguments will show that $H^ i_\mathfrak m(M/I^ nM) = H^ i_{\mathfrak m'}(M'/(I')^ nM')$ for all $n$ and $i$.

Lemmas 52.9.5, 52.9.2, and 52.9.4 apply to $A', \mathfrak m', I', M'$ by Lemma 52.10.3 parts (C) and (D). Thus we get an isomorphism

for $i \leq s$ where ${}^\wedge $ is derived $I'$-adic completion and these modules are annihilated by a power of $I'$. By Lemma 52.5.4 we obtain isomorphisms

for $i \leq s$. Combined with the already established comparison with local cohomology over $A$ we conclude the lemma is true. $\square$

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