Lemma 52.10.6. In Situation 52.10.1 the inverse systems $\{ H^ i_ T(I^ nM)\} _{n \geq 0}$ are pro-zero for $i \leq s$. Moreover, there exists an integer $m_0$ such that for all $m \geq m_0$ there exists an integer $m'(m) \geq m$ such that for $k \geq m'(m)$ the image of $H^{s + 1}_ T(I^ kM) \to H^{s + 1}_ T(I^ mM)$ maps injectively to $H^{s + 1}_ T(I^{m_0}M)$.

Proof. Fix $m$. Let $\mathfrak q \in T$. By Lemmas 52.10.3 and 52.10.4 we see that

$H^ i_\mathfrak q(M_\mathfrak q) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i_\mathfrak q(M_\mathfrak q/I^ nM_\mathfrak q)$

is an isomorphism for $i \leq s$. The inverse systems $\{ H^ i_\mathfrak q(I^ nM_\mathfrak q)\} _{n \geq 0}$ and $\{ H^ i_\mathfrak q(M/I^ nM)\} _{n \geq 0}$ satisfy the Mittag-Leffler condition for all $i$, see Lemma 52.5.2. Thus looking at the inverse system of long exact sequences

$0 \to H^0_\mathfrak q(I^ nM_\mathfrak q) \to H^0_\mathfrak q(M_\mathfrak q) \to H^0_\mathfrak q(M_\mathfrak q/I^ nM_\mathfrak q) \to H^1_\mathfrak q(I^ nM_\mathfrak q) \to H^1_\mathfrak q(M_\mathfrak q) \to \ldots$

we conclude (some details omitted) that there exists an integer $m'(m, \mathfrak q) \geq m$ such that for all $k \geq m'(m, \mathfrak q)$ the map $H^ i_\mathfrak q(I^ kM_\mathfrak q) \to H^ i_\mathfrak q(I^ mM_\mathfrak q)$ is zero for $i \leq s$ and the image of $H^{s + 1}_\mathfrak q(I^ kM_\mathfrak q) \to H^{s + 1}_\mathfrak q(I^ mM_\mathfrak q)$ is independent of $k \geq m'(m, \mathfrak q)$ and maps injectively into $H^{s + 1}_\mathfrak q(M_\mathfrak q)$.

Suppose we can show that $m'(m, \mathfrak q)$ can be chosen independently of $\mathfrak q \in T$. Then the lemma follows immediately from Local Cohomology, Lemmas 51.6.2 and 51.6.3.

Let $\omega _ A^\bullet$ be a dualizing complex. Let $\delta : \mathop{\mathrm{Spec}}(A) \to \mathbf{Z}$ be the corresponding dimension function. Recall that $\delta$ attains only a finite number of values, see Dualizing Complexes, Lemma 47.17.4. Claim: for each $d \in \mathbf{Z}$ the integer $m'(m, \mathfrak q)$ can be chosen independently of $\mathfrak q \in T$ with $\delta (\mathfrak q) = d$. Clearly the claim implies the lemma by what we said above.

Pick $\mathfrak q \in T$ with $\delta (\mathfrak q) = d$. Consider the ext modules

$E(n, j) = \text{Ext}^ j_ A(I^ nM, \omega _ A^\bullet )$

A key feature we will use is that these are finite $A$-modules. Recall that $(\omega _ A^\bullet )_\mathfrak q[-d]$ is a normalized dualizing complex for $A_\mathfrak q$ by definition of the dimension function associated to a dualizing complex, see Dualizing Complexes, Section 47.17. The local duality theorem (Dualizing Complexes, Lemma 47.18.4) tells us that the $\mathfrak qA_\mathfrak q$-adic completion of $E(n, -d - i)_\mathfrak q$ is Matlis dual to $H^ i_\mathfrak q(I^ nM_\mathfrak q)$. Thus the choice of $m'(m, \mathfrak q)$ for $i \leq s$ in the first paragraph tells us that for $k \geq m'(m, \mathfrak q)$ and $j \geq -d - s$ the map

$E(m, j)_\mathfrak q \to E(k, j)_\mathfrak q$

is zero. Since these modules are finite and nonzero only for a finite number of possible $j$ (small detail omitted), we can find an open neighbourhood $W \subset \mathop{\mathrm{Spec}}(A)$ of $\mathfrak q$ such that

$E(m, j)_{\mathfrak q'} \to E(m'(m, \mathfrak q), j)_{\mathfrak q'}$

is zero for $j \geq -d - s$ for all $\mathfrak q' \in W$. Then of course the maps $E(m, j)_{\mathfrak q'} \to E(k, j)_{\mathfrak q'}$ for $k \geq m'(m, \mathfrak q)$ are zero as well.

For $i = s + 1$ corresponding to $j = - d - s - 1$ we obtain from local duality and the results of the first paragraph that

$K_{k, \mathfrak q} = \mathop{\mathrm{Ker}}(E(m, -d - s - 1)_\mathfrak q \to E(k, -d - s - 1)_\mathfrak q)$

is independent of $k \geq m'(m, \mathfrak q)$ and that

$E(0, -d - s - 1)_\mathfrak q \to E(m, -d - s - 1)_\mathfrak q/K_{m'(m, \mathfrak q), \mathfrak q}$

is surjective. For $k \geq m'(m, \mathfrak q)$ set

$K_ k = \mathop{\mathrm{Ker}}(E(m, -d - s - 1) \to E(k, -d - s - 1))$

Since $K_ k$ is an increasing sequence of submodules of the finite module $E(m, -d - s - 1)$ we see that, at the cost of increasing $m'(m, \mathfrak q)$ a little bit, we may assume $K_{m'(m, \mathfrak q)} = K_ k$ for $k \geq m'(m, \mathfrak q)$. After shrinking $W$ further if necessary, we may also assume that

$E(0, -d - s - 1)_{\mathfrak q'} \to E(m, -d - s - 1)_{\mathfrak q'}/K_{m'(m, \mathfrak q), \mathfrak q'}$

is surjective for all $\mathfrak q' \in W$ (as before use that these modules are finite and that the map is surjective after localization at $\mathfrak q$).

Any subset, in particular $T_ d = \{ \mathfrak q \in T \text{ with }\delta (\mathfrak q) = d\}$, of the Noetherian topological space $\mathop{\mathrm{Spec}}(A)$ with the endowed topology is Noetherian and hence quasi-compact. Above we have seen that for every $\mathfrak q \in T_ d$ there is an open neighbourhood $W$ where $m'(m, \mathfrak q)$ works for all $\mathfrak q' \in T_ d \cap W$. We conclude that we can find an integer $m'(m, d)$ such that for all $\mathfrak q \in T_ d$ we have

$E(m, j)_\mathfrak q \to E(m'(m, d), j)_\mathfrak q$

is zero for $j \geq -d - s$ and with $K_{m'(m, d)} = \mathop{\mathrm{Ker}}(E(m, -d - s - 1) \to E(m'(m, d), -d - s - 1))$ we have

$K_{m'(m, d), \mathfrak q} = \mathop{\mathrm{Ker}}(E(m, -d - s - 1)_{\mathfrak q} \to E(k, -d - s - 1)_{\mathfrak q})$

for all $k \geq m'(m, d)$ and the map

$E(0, -d - s - 1)_\mathfrak q \to E(m, -d - s - 1)_\mathfrak q/K_{m'(m, d), \mathfrak q}$

is surjective. Using the local duality theorem again (in the opposite direction) we conclude that the claim is correct. This finishes the proof. $\square$

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