The Stacks project

Proof. We may assume $s > 0$ as the case $s = 0$ was done in Proposition 52.12.3.

Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction to $U$ of the coherent module associated to $M$, see Local Cohomology, Lemma 51.8.2. Set $d = \text{cd}(A, I)$. Let $\mathfrak p$ be a prime of $A$ not contained in $V(I)$ and let $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$. Then either $\text{depth}(M_\mathfrak p) \geq s + 1 > s$ or we have $\dim ((A/\mathfrak p)_\mathfrak q) > d + s + 1$ by (2). By Lemma 52.10.5 we conclude that the assumptions of Situation 52.10.1 are satisfied for $A, I, V(\mathfrak a), M, s, d$. On the other hand, the hypotheses of Lemma 52.8.5 are satisfied for $s + 1$ and $d$; this is where condition (3) is used.

Applying Lemma 52.8.5 we find 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 maps

is an isomorphism for $i \leq s + 1$.

For $i \leq s$ the map $H^ i_\mathfrak a(M) \to H^ i_ J(M)$ is an isomorphism by Lemmas 52.10.3 and 52.8.4. Using the comparison of cohomology and local cohomology (Local Cohomology, Lemma 51.2.2) we deduce $H^ i(U, \mathcal{F}) \to H^ i(V,\mathcal{F})$ is an isomorphism for $V = \mathop{\mathrm{Spec}}(A) \setminus V(J)$ and $i < s$.

By Theorem 52.10.8 we have $H^ i_\mathfrak a(M) = \mathop{\mathrm{lim}}\nolimits H^ i_\mathfrak a(M/I^ nM)$ for $i \leq s$. By Lemma 52.10.9 we have $H^{s + 1}_\mathfrak a(M) = \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM)$.

The isomorphism $H^0(U, \mathcal{F}) = H^0(V, \mathcal{F}) = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ follows from the above and Proposition 52.12.3. For $0 < i < s$ we get the desired isomorphisms $H^ i(U, \mathcal{F}) = H^ i(V, \mathcal{F}) = \mathop{\mathrm{lim}}\nolimits H^ i(U, \mathcal{F}/I^ n\mathcal{F})$ in the same manner using the relation between local cohomology and cohomology; it is easier than the case $i = 0$ because for $i > 0$ we have

Similarly for the final statement. $\square$