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:

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

Consider the spectral sequence

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.

*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.

*$R\Gamma (X_{\bar k}, K)$ 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),

Therefore,

Now consider the spectral sequence

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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