The title says it all.

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.
$\square$

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