The Stacks project

Lemma 30.20.8. Let $f : X \to Y$ be a morphism of schemes. Let $y \in Y$. Assume

  1. $Y$ locally Noetherian,

  2. $f$ is proper, and

  3. $f^{-1}(\{ y\} )$ is finite.

Then for any coherent sheaf $\mathcal{F}$ on $X$ we have $(R^ pf_*\mathcal{F})_ y = 0$ for all $p > 0$.

Proof. The fibre $X_ y$ is finite, and by Morphisms, Lemma 29.20.7 it is a finite discrete space. Moreover, the underlying topological space of each infinitesimal neighbourhood $X_ n$ is the same. Hence each of the schemes $X_ n$ is affine according to Schemes, Lemma 26.11.8. Hence it follows that $H^ p(X_ n, \mathcal{F}_ n) = 0$ for all $p > 0$. Hence we see that $(R^ pf_*\mathcal{F})_ y^\wedge = 0$ by Lemma 30.20.7. Note that $R^ pf_*\mathcal{F}$ is coherent by Proposition 30.19.1 and hence $R^ pf_*\mathcal{F}_ y$ is a finite $\mathcal{O}_{Y, y}$-module. By Nakayama's lemma (Algebra, Lemma 10.20.1) if the completion of a finite module over a local ring is zero, then the module is zero. Whence $(R^ pf_*\mathcal{F})_ y = 0$. $\square$

Comments (0)

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Your email address will not be published.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 02OE. Beware of the difference between the letter 'O' and the digit '0'.