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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