The Stacks project

Proof. This lemma is a special case of Lemma 52.8.5 since condition (5)(c) is implied by condition (4) as $\delta _{max} = \delta _{min} = \delta (\mathfrak m)$. We will give the proof of this important special case as it is somewhat easier (fewer things to check).

There is no difference between $R\Gamma _\mathfrak a$ and $R\Gamma _{V(\mathfrak a)}$ in our current situation, see Dualizing Complexes, Lemma 47.10.1. Next, we observe that

by Dualizing Complexes, Lemmas 47.9.6 and 47.12.1 which explains the equality sign in the statement of the lemma.

Observe that the lemma holds for $s < 0$. This is not a trivial case because it is not a priori clear that $H^ s(R\Gamma _\mathfrak m(M)^\wedge )$ is zero for negative $s$. However, this vanishing was established in Lemma 52.5.4. We will prove the lemma by induction for $s \geq 0$.

The assumptions of Lemma 52.9.2 are satisfied by Local Cohomology, Lemma 51.4.10. The lemma for $s = 0$ follows from Lemma 52.9.2 and Dualizing Complexes, Lemma 47.12.5.

Assume $s > 0$ and the lemma holds for smaller values of $s$. Let $M' \subset M$ be the submodule of elements whose support is condained in $V(I) \cup V(J)$ for some ideal $J$ with $V(J) \cap V(I) = \{ \mathfrak m\}$. Then $M'$ is a finite $A$-module. We claim that

is an isomorphism for any choice of $J$. Namely, for any such module there is a short exact sequence $0 \to M_1 \oplus M_2 \to M' \to N \to 0$ with $M_1$ annihilated by a power of $J$, with $M_2$ annihilated by a power of $I$ and with $N$ annihilated by a power of $\mathfrak m$. In the case of $M_1$ we see that $R\Gamma _ J(M_1) = M_1$ and since $M_1$ is a finite $A$-module and $I$-adically complete we have $M_1^\wedge = M_1$. Thus the claim holds for $M_1$. In the case of $M_2$ we see that $H^ i_ J(M_2)$ is annihilated by a power of $I$ and hence derived complete. Thus the claim for $M_2$. By the same arguments the claim holds for $N$ and we conclude that the claim holds. Considering the short exact sequence $0 \to M' \to M \to M/M' \to 0$ we see that it suffices to prove the lemma for $M/M'$. This we may assume $\mathfrak p \in \text{Ass}(M)$ implies $V(\mathfrak p) \cap V(I) \not= \{ \mathfrak m\}$, i.e., $\mathfrak p$ is a prime as in (4).

Choose an ideal $J_0$ as in Lemma 52.9.2 and an integer $t > 0$ such that $(J_0I)^ t$ annihilates $H^ s_ J(M)$. Here $J$ denotes an arbitrary ideal $J \subset J_0$ with $V(J) \cap V(I) = \{ \mathfrak m\}$. The assumptions of Lemma 52.9.1 are satisfied for every $\mathfrak p \in \text{Ass}(M)$ (see previous paragraph). Thus the annihilator $\mathfrak a \subset A$ of $H^ s(R\Gamma _\mathfrak m(M)^\wedge )$ is not contained in $\mathfrak p$ for $\mathfrak p \in \text{Ass}(M)$. Thus we can find an $f \in \mathfrak a(J_0I)^ t$ not in any associated prime of $M$ which is an annihilator of both $H^ s(R\Gamma _\mathfrak m(M)^\wedge )$ and $H^ s_ J(M)$. Then $f$ is a nonzerodivisor on $M$ and we can consider the short exact sequence

Our choice of $f$ shows that we obtain

for any $J \subset J_0$ with $V(J) \cap V(I) = \{ \mathfrak m\}$. Thus if we choose $J$ such that it works for $M$ and $M/fM$ and $s - 1$ (possible by induction hypothesis), then we conclude that the lemma is true. $\square$