The Stacks project

Proposition 50.14.1. Let $X \to S$ be a morphism of schemes. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of constant rank $r$. Consider the morphism $p : P = \mathbf{P}(\mathcal{E}) \to X$. Then the map

\[ \bigoplus \nolimits _{i = 0, \ldots , r - 1} H^*_{dR}(X/S) \longrightarrow H^*_{dR}(P/S) \]

given by the rule

\[ (a_0, \ldots , a_{r - 1}) \longmapsto \sum \nolimits _{i = 0, \ldots , r - 1} c_1^{dR}(\mathcal{O}_ P(1))^ i \cup p^*(a_ i) \]

is an isomorphism.

Proof. Choose an affine open $\mathop{\mathrm{Spec}}(A) \subset X$ such that $\mathcal{E}$ restricts to the trivial locally free module $\mathcal{O}_{\mathop{\mathrm{Spec}}(A)}^{\oplus r}$. Then $P \times _ X \mathop{\mathrm{Spec}}(A) = \mathbf{P}^{r - 1}_ A$. Thus we see that $p$ is proper and smooth, see Section 50.11. Moreover, the classes $c_1^{dR}(\mathcal{O}_ P(1))^ i$, $i = 0, 1, \ldots , r - 1$ restricted to a fibre $X_ y = \mathbf{P}^{r - 1}_ y$ freely generate the de Rham cohomology $H^*_{dR}(X_ y/y)$ over $\kappa (y)$, see Lemma 50.11.4. Thus we've verified the conditions of Proposition 50.13.3 and we win. $\square$

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Your email address will not be published.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