Lemma 30.7.3. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Assume that $f$ is quasi-compact and quasi-separated and that $S$ is quasi-compact and separated. There exists a bounded below complex $\mathcal{K}^\bullet $ of quasi-coherent $\mathcal{O}_ S$-modules with the following property: For every morphism $g : S' \to S$ the complex $g^*\mathcal{K}^\bullet $ is a representative for $Rf'_*\mathcal{F}'$ with notation as in diagram (30.5.0.1).
Proof. (If $f$ is separated as well, please see Lemma 30.7.2.) The assumptions imply in particular that $X$ is quasi-compact and quasi-separated as a scheme. Let $\mathcal{B}$ be the set of affine opens of $X$. By Hypercoverings, Lemma 25.11.4 we can find a hypercovering $K = (I, \{ U_ i\} )$ such that each $I_ n$ is finite and each $U_ i$ is an affine open of $X$. By Hypercoverings, Lemma 25.5.3 there is a spectral sequence with $E_2$-page
converging to $H^{p + q}(X, \mathcal{F})$. Note that $\check{H}^ p(K, \underline{H}^ q(\mathcal{F}))$ is the $p$th cohomology group of the complex
Since each $U_ i$ is affine we see that this is zero unless $q = 0$ in which case we obtain
Thus we conclude that $R\Gamma (X, \mathcal{F})$ is computed by this complex.
For any $n$ and $i \in I_ n$ denote $f_ i : U_ i \to S$ the restriction of $f$ to $U_ i$. As $S$ is separated and $U_ i$ is affine this morphism is affine. Consider the complex of quasi-coherent sheaves
on $S$. As in Hypercoverings, Lemma 25.5.3 we obtain a map $\mathcal{K}^\bullet \to Rf_*\mathcal{F}$ in $D(\mathcal{O}_ S)$ by choosing an injective resolution of $\mathcal{F}$ (details omitted). Consider any affine scheme $V$ and a morphism $g : V \to S$. Then the base change $X_ V$ has a hypercovering $K_ V = (I, \{ U_{i, V}\} )$ obtained by base change. Moreover, $g^*f_{i, *}\mathcal{F} = f_{i, V, *}(g')^*\mathcal{F}|_{U_{i, V}}$. Thus the arguments above prove that $\Gamma (V, g^*\mathcal{K}^\bullet )$ computes $R\Gamma (X_ V, (g')^*\mathcal{F})$. This finishes the proof of the lemma as it suffices to prove the equality of complexes Zariski locally on $S'$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)