Lemma 57.10.7. Let $k$ be a field. Let $X$ be a scheme proper and smooth over $k$. Then $D_{perf}(\mathcal{O}_ X)$ has a strong generator.

**Proof.**
Using Lemma 57.10.6 choose finite locally free $\mathcal{O}_ X$-modules $\mathcal{E}$ and $\mathcal{G}$ such that $\mathcal{O}_\Delta \in \langle \mathcal{E} \boxtimes \mathcal{G} \rangle $ in $D(\mathcal{O}_{X \times X})$. We claim that $\mathcal{G}$ is a strong generator for $D_{perf}(\mathcal{O}_ X)$. With notation as in Derived Categories, Section 13.35 choose $m, n \geq 1$ such that

This is possible by Derived Categories, Lemma 13.36.2. Let $K$ be an object of $D_{perf}(\mathcal{O}_ X)$. Since $L\text{pr}_1^*K \otimes _{\mathcal{O}_{X \times X}}^\mathbf {L} -$ is an exact functor and since

we conclude from Derived Categories, Remark 13.35.5 that

Applying the exact functor $R\text{pr}_{2, *}$ and observing that

by Derived Categories of Schemes, Lemma 36.22.1 we conclude that

The equality follows from the discussion in Example 57.9.6. Since $K$ is perfect, there exist $a \leq b$ such that $H^ i(X, K)$ is nonzero only for $i \in [a, b]$. Since $X$ is proper, each $H^ i(X, K)$ is finite dimensional. We conclude that the right hand side is contained in $smd(add(\mathcal{G}[-m + a, m + b])^{\star n})$ which is itself contained in $\langle \mathcal{G} \rangle _ n$ by one of the references given above. This finishes the proof. $\square$

## 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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)