The Stacks project

Proof. To prove the final assertion of the theorem we apply Local Cohomology, Proposition 51.10.1 with $T \subset V(I) \subset \mathop{\mathrm{Spec}}(A)$. Namely, suppose that $\mathfrak p \not\in V(I)$, $\mathfrak q \in T$ with $\mathfrak p \subset \mathfrak q$. Then either there exists a prime $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ with $\mathfrak r \in V(I) \setminus T$ and we get

by (4) in Situation 52.10.1 or there does not exist an $\mathfrak r$ and we get $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) > s$ by Lemma 52.10.2. In all three cases we see that $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > s$. Thus Local Cohomology, Proposition 51.10.1 (2) holds and we find that a power of $I$ annihilates $H^ i_ T(M)$ for $i \leq s$.

We already know the other two assertions of the theorem hold for $i < s$ by Lemma 52.10.7 and for the module $I^{m_0}M$ for $i = s$ and $m_0$ large enough. To finish of the proof we will show that in fact these assertions for $i = s$ holds for $M$.

Let $M' = H^0_ I(M)$ and $M'' = M/M'$ so that we have a short exact sequence

and $M''$ has $H^0_ I(M') = 0$ by Dualizing Complexes, Lemma 47.11.6. By Artin-Rees (Algebra, Lemma 10.51.2) we get short exact sequences

for $n$ large enough. Consider the long exact sequences

Now it is a simple matter to see that if we have Mittag-Leffler for the inverse system $\{ H^ s_ T(M''/I^ nM'')\} _{n \geq 0}$ then we have Mittag-Leffler for the inverse system $\{ H^ s_ T(M/I^ nM)\} _{n \geq 0}$. (Note that the ML condition for an inverse system of groups $G_ n$ only depends on the values of the inverse system for sufficiently large $n$.) Moreover the sequence

is exact because we have ML in the required spots, see Homology, Lemma 12.31.4. Hence, if $H^ s_ T(M'') \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M''/I^ nM'')$ is an isomorphism, then $H^ s_ T(M) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM)$ is an isomorphism too by the five lemma (Homology, Lemma 12.5.20). This reduces us to the case discussed in the next paragraph.

Assume that $H^0_ I(M) = 0$. Choose generators $f_1, \ldots , f_ r$ of $I^{m_0}$ where $m_0$ is the integer found for $M$ in Lemma 52.10.7. Then we consider the exact sequence

defining $Q$. Some observations: the first map is injective exactly because $H^0_ I(M) = 0$. The cokernel $Q$ of this injection is a finite $A$-module such that for every $1 \leq j \leq r$ we have $Q_{f_ j} \cong (M_{f_ j})^{\oplus r - 1}$. In particular, for a prime $\mathfrak p \subset A$ with $\mathfrak p \not\in V(I)$ we have $Q_\mathfrak p \cong (M_\mathfrak p)^{\oplus r - 1}$. Similarly, given $\mathfrak q \in T$ and $\mathfrak p' \subset A' = (A_\mathfrak q)^\wedge $ not contained in $V(IA')$, we have $Q'_{\mathfrak p'} \cong (M'_{\mathfrak p'})^{\oplus r - 1}$ where $Q' = (Q_\mathfrak q)^\wedge $ and $M' = (M_\mathfrak q)^\wedge $. Thus the conditions in Situation 52.10.1 hold for $A, I, T, Q$. (Observe that $Q$ may have nonvanishing $H^0_ I(Q)$ but this won't matter.)

For any $n \geq 0$ we set $F^ nM = M \cap I^ n(I^{m_0}M)^{\oplus r}$ so that we get short exact sequences

By Artin-Rees (Algebra, Lemma 10.51.2) there exists a $c \geq 0$ such that $I^ n M \subset F^ nM \subset I^{n - c}M$ for all $n \geq c$. Let $m_0$ be the integer and let $m'(m)$ be the function defined for $m \geq m_0$ found in Lemma 52.10.6 applied to $M$. Note that the integer $m_0$ is the same as our integer $m_0$ chosen above (you don't need to check this: you can just take the maximum of the two integers if you like). Finally, by Lemma 52.10.6 applied to $Q$ for every integer $m$ there exists an integer $m''(m) \geq m$ such that $H^ s_ T(I^ kQ) \to H^ s_ T(I^ mQ)$ is zero for all $k \geq m''(m)$.

Fix $m \geq m_0$. Choose $k \geq m'(m''(m + c))$. Choose $\xi \in H^{s + 1}_ T(I^ kM)$ which maps to zero in $H^{s + 1}_ T(M)$. We want to show that $\xi $ maps to zero in $H^{s + 1}_ T(I^ mM)$. Namely, this will show that $\{ H^ s_ T(M/I^ nM)\} _{n \geq 0}$ is Mittag-Leffler exactly as in the proof of Lemma 52.10.7. Picture to help visualize the argument:

The image of $\xi $ in $H^{s + 1}_ T(I^ k(I^{m_0}M)^{\oplus r})$ maps to zero in $H^{s + 1}_ T((I^{m_0}M)^{\oplus r})$ and hence maps to zero in $H^{s + 1}_ T(I^{m''(m + c)}(I^{m_0}M)^{\oplus r})$ by choice of $m'(-)$. Thus the image $\xi ' \in H^{s + 1}_ T(F^{m''(m + c)}M)$ maps to zero in $H^{s + 1}_ T(I^{m''(m + c)}(I^{m_0}M)^{\oplus r})$ and hence $\xi ' = \delta (\eta )$ for some $\eta \in H^ s_ T(I^{m''(m + c)}Q)$. By our choice of $m''(-)$ we find that $\eta $ maps to zero in $H^ s_ T(I^{m + c}Q)$. This in turn means that $\xi '$ maps to zero in $H^{s + 1}_ T(F^{m + c}M)$. Since $F^{m + c}M \subset I^ mM$ we conclude.

Finally, we prove the statement on limits. Consider the short exact sequences

We have $\mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM) = \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/F^ nM)$ as these inverse systems are pro-isomorphic. We obtain a commutative diagram

The right column is exact because we have ML in the required spots, see Homology, Lemma 12.31.4. The lowest horizontal arrow is injective (!) by part (5) of Lemma 52.10.7. The horizontal arrow above it is bijective by part (4) of Lemma 52.10.7. The arrows in cohomological degrees $\leq s - 1$ are isomorphisms. Thus we conclude $H^ s_ T(M) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM)$ is an isomorphism by the five lemma (Homology, Lemma 12.5.20). This finishes the proof of the theorem. $\square$