The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EFX. Beware of the difference between the letter 'O' and the digit '0'.