The Stacks project

Special case of [Proposition 1.3, totaro_resolution].

Lemma 36.36.10. Let $X$ be a quasi-compact and quasi-separated scheme with the resolution property. Then $X$ has affine diagonal.

Proof. Combining Limits, Proposition 32.5.4 and Lemma 36.36.9 this reduces to the case where $X$ is Noetherian (small detail omitted). Assume $X$ is Noetherian. Recall that $X \times X$ is covered by the affine opens $U \times V$ for affine opens $U$, $V$ of $X$, see Schemes, Section 26.17. Hence to show that the diagonal $\Delta : X \to X \times X$ is affine, it suffices to show that $U \cap V = \Delta ^{-1}(U \times V)$ is affine for all affine opens $U$, $V$ of $X$, see Morphisms, Lemma 29.11.3. In particular, it suffices to show that the inclusion morphism $j : U \to X$ is affine if $U$ is an affine open of $X$. By Cohomology of Schemes, Lemma 30.3.4 it suffices to show that $R^1j_*\mathcal{G} = 0$ for any quasi-coherent $\mathcal{O}_ U$-module $\mathcal{G}$. By Proposition 36.8.3 (this is where we use that we've reduced to the Noetherian case) we can represent $Rj_*\mathcal{G}$ by a complex $\mathcal{H}^\bullet $ of quasi-coherent $\mathcal{O}_ X$-modules. Assume

\[ H^1(\mathcal{H}^\bullet ) = \mathop{\mathrm{Ker}}(\mathcal{H}^1 \to \mathcal{H}^2)/\mathop{\mathrm{Im}}(\mathcal{H}^0 \to \mathcal{H}^1) \]

is nonzero in order to get a contradiction. Then we can find a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ and a map

\[ \mathcal{F} \longrightarrow \mathop{\mathrm{Ker}}(\mathcal{H}^1 \to \mathcal{H}^2) \]

such that the composition with the projection onto $H^1(\mathcal{H}^\bullet )$ is nonzero. Namely, we can write $\mathop{\mathrm{Ker}}(\mathcal{H}^1 \to \mathcal{H}^2)$ as the filtered union of its coherent submodules by Properties, Lemma 28.22.3 and then one of these will do the job. Next, we choose a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ and a surjection $\mathcal{E} \to \mathcal{F}$ using the resolution property of $X$. This produces a map in the derived category

\[ \mathcal{E}[-1] \longrightarrow Rj_*\mathcal{G} \]

which is nonzero on cohomology sheaves and hence nonzero in $D(\mathcal{O}_ X)$. By adjunction, this is the same thing as a map

\[ j^*\mathcal{E}[-1] \to \mathcal{G} \]

nonzero in $D(\mathcal{O}_ U)$. Since $\mathcal{E}$ is finite locally free this is the same thing as a nonzero element of

\[ H^1(U, j^*\mathcal{E}^\vee \otimes _{\mathcal{O}_ U} \mathcal{G}) \]

where $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X)$ is the dual finite locally free module. However, this group is zero by Cohomology of Schemes, Lemma 30.2.2 which is the desired contradiction. (If in doubt about the step using duals, please see the more general Cohomology, Lemma 20.48.5.) $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 36.36: The resolution property

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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F8C. Beware of the difference between the letter 'O' and the digit '0'.