The Stacks project

Lemma 52.20.3. In Situation 52.16.1 let $(M_ n)$ be an inverse system of $A$-modules as in Lemma 52.16.2 and let $(\mathcal{F}_ n)$ be the corresponding object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $d \geq \text{cd}(A, I)$ and $s \geq 0$ be integers. With notation as above assume

  1. $A$ is local with maximal ideal $\mathfrak m = \mathfrak a$,

  2. $A$ has a dualizing complex, and

  3. $(\mathcal{F}_ n)$ satisfies the $(s, s + d)$-inequalities (Definition 52.19.1).

Let $E$ be an injective hull of the residue field of $A$. Then for $i \leq s$ there exists a finite $A$-module $N$ annihilated by a power of $I$ and for $n \gg 0$ compatible maps

\[ H^ i_\mathfrak m(M_ n) \to \mathop{\mathrm{Hom}}\nolimits _ A(N, E) \]

whose cokernels are finite length $A$-modules and whose kernels $K_ n$ form an inverse system such that $\mathop{\mathrm{Im}}(K_{n''} \to K_{n'})$ has finite length for $n'' \gg n' \gg 0$.

Proof. Let $\omega _ A^\bullet $ be a normalized dualizing complex. Then $\delta ^ Y_ Z = \delta $ is the dimension function associated with this dualizing complex. Observe that $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_ n, \omega _ A^\bullet )$ is a finite $A$-module annihilated by $I^ n$. Fix $0 \leq i \leq s$. Below we will find $n_1 > n_0 > 0$ such that if we set

\[ N = \mathop{\mathrm{Im}}(\mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_{n_0}, \omega _ A^\bullet ) \to \mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_{n_1}, \omega _ A^\bullet )) \]

then the kernels of the maps

\[ N \to \mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_ n, \omega _ A^\bullet ),\quad n \geq n_1 \]

are finite length $A$-modules and the cokernels $Q_ n$ form a system such that $\mathop{\mathrm{Im}}(Q_{n'} \to Q_{n''})$ has finite length for $n'' \gg n' \gg n_1$. This is equivalent to the statement that the system $\{ \mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_ n, \omega _ A^\bullet )\} _{n \geq 1}$ is essentially constant in the quotient of the category of finite $A$-modules modulo the Serre subcategory of finite length $A$-modules. By the local duality theorem (Dualizing Complexes, Lemma 47.18.4) and Matlis duality (Dualizing Complexes, Proposition 47.7.8) we conclude that there are maps

\[ H^ i_\mathfrak m(M_ n) \to \mathop{\mathrm{Hom}}\nolimits _ A(N, E),\quad n \geq n_1 \]

as in the statement of the lemma.

Pick $f \in \mathfrak m$. Let $B = A_ f^\wedge $ be the $I$-adic completion of the localization $A_ f$. Recall that $\omega _{A_ f}^\bullet = \omega _ A^\bullet \otimes _ A A_ f$ and $\omega _ B^\bullet = \omega _ A^\bullet \otimes _ A B$ are dualizing complexes (Dualizing Complexes, Lemma 47.15.6 and 47.22.3). Let $M$ be the finite $B$-module $\mathop{\mathrm{lim}}\nolimits M_{n, f}$ (compare with discussion in Cohomology of Schemes, Lemma 30.23.1). Then

\[ \mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_ n, \omega _ A^\bullet )_ f = \mathop{\mathrm{Ext}}\nolimits ^{-i}_{A_ f}(M_{n, f}, \omega _{A_ f}^\bullet ) = \mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M/I^ n M, \omega _ B^\bullet ) \]

Since $\mathfrak m$ can be generated by finitely many $f \in \mathfrak m$ it suffices to show that for each $f$ the system

\[ \{ \mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M/I^ n M, \omega _ B^\bullet )\} _{n \geq 1} \]

is essentially constant. Some details omitted.

Let $\mathfrak q \subset IB$ be a prime ideal. Then $\mathfrak q$ corresponds to a point $y \in U \cap Y$. Observe that $\delta (\mathfrak q) = \dim (\overline{\{ y\} })$ is also the value of the dimension function associated to $\omega _ B^\bullet $ (we omit the details; use that $\omega _ B^\bullet $ is gotten from $\omega _ A^\bullet $ by tensoring up with $B$). Assumption (3) guarantees via Lemma 52.19.2 that Lemma 52.10.4 applies to $B_\mathfrak q, IB_\mathfrak q, \mathfrak qB_\mathfrak q, M_\mathfrak q$ with $s$ replaced by $s - \delta (y)$. We obtain that

\[ H^{i - \delta (\mathfrak q)}_{\mathfrak qB_\mathfrak q}(M_\mathfrak q) = \mathop{\mathrm{lim}}\nolimits H^{i - \delta (\mathfrak q)}_{\mathfrak qB_\mathfrak q}( (M/I^ nM)_\mathfrak q) \]

and this module is annihilated by a power of $I$. By Lemma 52.5.3 we find that the inverse systems $H^{i - \delta (\mathfrak q)}_{\mathfrak qB_\mathfrak q}((M/I^ nM)_\mathfrak q)$ are essentially constant with value $H^{i - \delta (\mathfrak q)}_{\mathfrak qB_\mathfrak q}(M_\mathfrak q)$. Since $(\omega _ B^\bullet )_\mathfrak q[-\delta (\mathfrak q)]$ is a normalized dualizing complex on $B_\mathfrak q$ the local duality theorem shows that the system

\[ \mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M/I^ n M, \omega _ B^\bullet )_\mathfrak q \]

is essentially constant with value $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M, \omega _ B^\bullet )_\mathfrak q$.

To finish the proof we globalize as in the proof of Lemma 52.10.6; the argument here is easier because we know the value of our system already. Namely, consider the maps

\[ \alpha _ n : \mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M/I^ n M, \omega _ B^\bullet ) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M, \omega _ B^\bullet ) \]

for varying $n$. By the above, for every $\mathfrak q$ we can find an $n$ such that $\alpha _ n$ is surjective after localization at $\mathfrak q$. Since $B$ is Noetherian and $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M, \omega _ B^\bullet )$ a finite module, we can find an $n$ such that $\alpha _ n$ is surjective. For any $n$ such that $\alpha _ n$ is surjective, given a prime $\mathfrak q \in V(IB)$ we can find an $n' > n$ such that $\mathop{\mathrm{Ker}}(\alpha _ n)$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M/I^{n'}M, \omega _ B^\bullet )$ at least after localizing at $\mathfrak q$. Since $\mathop{\mathrm{Ker}}(\alpha _ n)$ is a finite $A$-module and since supports of sections are quasi-compact, we can find an $n'$ such that $\mathop{\mathrm{Ker}}(\alpha _ n)$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M/I^{n'}M, \omega _ B^\bullet )$. In this way we see that $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M/I^ n M, \omega _ B^\bullet )$ is essentially constant with value $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M, \omega _ B^\bullet )$. This finishes the proof. $\square$

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