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The Stacks project

64.13 Cohomology of nice complexes

The following is a special case of a more general result about compactly supported cohomology of objects of D_{ctf}(X, \Lambda ).

Proposition 64.13.1. Let X be a projective curve over a field k, \Lambda a finite ring and K\in D_{ctf}(X, \Lambda ). Then R\Gamma (X_{\bar k}, K)\in D_{perf}(\Lambda ).

Sketch of proof.. The first step is to show:

  1. The cohomology of R\Gamma (X_{\bar k}, K) is bounded.

Consider the spectral sequence

H^ i(X_{\bar k}, \underline H^ j(K)) \Rightarrow H^{i+j} (R\Gamma (X_{\bar k}, K)).

Since K is bounded and \Lambda is finite, the sheaves \underline H^ j(K) are torsion. Moreover, X_{\bar k} has finite cohomological dimension, so the left-hand side is nonzero for finitely many i and j only. Therefore, so is the right-hand side.

  1. The cohomology groups H^{i+j} (R\Gamma (X_{\bar k}, K)) are finite.

Since the sheaves \underline H^ j(K) are constructible, the groups H^ i(X_{\bar k}, \underline H^ j(K)) are finite (Étale Cohomology, Section 59.83) so it follows by the spectral sequence again.

  1. R\Gamma (X_{\bar k}, K) has finite \text{Tor}-dimension.

Let N be a right \Lambda -module (in fact, since \Lambda is finite, it suffices to assume that N is finite). By the projection formula (change of module),

N \otimes ^\mathbf {L}_\Lambda R \Gamma (X_{\bar k}, K) = R\Gamma (X_{\bar k}, N \otimes ^\mathbf {L}_\Lambda K).

Therefore,

H^ i (N \otimes ^\mathbf {L}_\Lambda R\Gamma (X_{\bar k}, K)) = H^ i(R\Gamma (X_{\bar k}, N \otimes _{\Lambda }^\mathbf {L} K)).

Now consider the spectral sequence

H^ i (X_{\bar k}, \underline H^ j (N \otimes _{\Lambda }^\mathbf {L} K)) \Rightarrow H^{i+j}(R\Gamma (X_{\bar k}, N \otimes _{\Lambda }^\mathbf {L} K)).

Since K has finite \text{Tor}-dimension, \underline H^ j (N \otimes _{\Lambda }^\mathbf {L} K) vanishes universally for j small enough, and the left-hand side vanishes whenever i < 0. Therefore R\Gamma (X_{\bar k}, K) has finite \text{Tor}-dimension, as claimed. So it is a perfect complex by Lemma 64.12.2. \square


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