The Stacks project

Lemma 33.33.5. Let $k$ be a field. Let $f : Y \to X$ be a morphism of proper schemes over $k$. Let $\mathcal{G}$ be a coherent $\mathcal{O}_ Y$-module. Then

\[ \chi (Y, \mathcal{G}) = \sum (-1)^ i \chi (X, R^ if_*\mathcal{G}) \]

Proof. The formula makes sense: the sheaves $R^ if_*\mathcal{G}$ are coherent and only a finite number of them are nonzero, see Cohomology of Schemes, Proposition 30.19.1 and Lemma 30.4.5. By Cohomology, Lemma 20.13.4 there is a spectral sequence with

\[ E_2^{p, q} = H^ p(X, R^ qf_*\mathcal{G}) \]

converging to $H^{p + q}(Y, \mathcal{G})$. By finiteness of cohomology on $X$ we see that only a finite number of $E_2^{p, q}$ are nonzero and each $E_2^{p, q}$ is a finite dimensional vector space. It follows that the same is true for $E_ r^{p, q}$ for $r \geq 2$ and that

\[ \sum (-1)^{p + q} \dim _ k E_ r^{p, q} \]

is independent of $r$. Since for $r$ large enough we have $E_ r^{p, q} = E_\infty ^{p, q}$ and since convergence means there is a filtration on $H^ n(Y, \mathcal{G})$ whose graded pieces are $E_\infty ^{p, q}$ with $p + q = n$ (this is the meaning of convergence of the spectral sequence), we conclude. Compare also with the more general Homology, Lemma 12.24.12. $\square$


Comments (2)

Comment #3214 by Junho Won on

(Typo) In the last sentence, should it be , so if is the filtration on then )?

Comment #3316 by on

Fixed this and also added a reference to a general result of this kind in the chapter on homology. See this change.


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Your email address will not be published.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 0BEK. Beware of the difference between the letter 'O' and the digit '0'.