Lemma 52.10.9. Let $I \subset \mathfrak a \subset A$ be ideals of a Noetherian ring $A$ and let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. Suppose that

1. $A, I, V(\mathfrak a), M$ satisfy the assumptions of Situation 52.10.1 for $s$ and $d$, and

2. $A, I, \mathfrak a, M$ satisfy the conditions of Lemma 52.8.5 for $s + 1$ and $d$ with $J = \mathfrak a$.

Then there exists an ideal $J_0 \subset \mathfrak a$ with $V(J_0) \cap V(I) = V(\mathfrak a)$ such that for any $J \subset J_0$ with $V(J) \cap V(I) = V(\mathfrak a)$ the map

$H^{s + 1}_ J(M) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM)$

is an isomorphism.

Proof. Namely, we have the existence of $J_0$ and the isomorphism $H^{s + 1}_ J(M) = H^{s + 1}(R\Gamma _\mathfrak a(M)^\wedge )$ by Lemma 52.8.5, we have a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM) \to H^{s + 1}(R\Gamma _\mathfrak a(M)^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM) \to 0$

by Dualizing Complexes, Lemma 47.12.4, and the module $R^1\mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM)$ is zero because $\{ H^ s_\mathfrak a(M/I^ nM)\} _{n \geq 0}$ has Mittag-Leffler by Theorem 52.10.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).