Lemma 50.11.4. For $0 \leq i \leq n$ the de Rham cohomology $H^{2i}_{dR}(\mathbf{P}^ n_ A/A)$ is a free $A$-module of rank $1$ with basis element $c_1^{dR}(\mathcal{O}(1))^ i$. In all other degrees the de Rham cohomology of $\mathbf{P}^ n_ A$ over $A$ is zero.

**Proof.**
Consider the Hodge-to-de Rham spectral sequence of Section 50.6. By the computation of the Hodge cohomology of $\mathbf{P}^ n_ A$ over $A$ done in Lemma 50.11.3 we see that the spectral sequence degenerates on the $E_1$ page. In this way we see that $H^ i_{dR}(\mathbf{P}^ n_ A/A)$ is a free $A$-module of rank $1$ for $0 \leq i \leq n$ and zero else. Observe that $c_1^{dR}(\mathcal{O}(1))^ i \in H^{2i}_{dR}(\mathbf{P}^ n_ A/A)$ for $i = 0, \ldots , n$ and that for $i = n$ this element is the image of $c_1^{Hodge}(\mathcal{L})^ n$ by the map of complexes

This follows for example from the discussion in Remark 50.9.2 or from the explicit description of cocycles representing these classes in Section 50.9. The spectral sequence shows that the induced map

is an isomorphism and since $c_1^{Hodge}(\mathcal{L})^ n$ is a generator of of the source (Lemma 50.11.3), we conclude that $c_1^{dR}(\mathcal{L})^ n$ is a generator of the target. By the $A$-bilinearity of the cup products, it follows that also $c_1^{dR}(\mathcal{L})^ i$ is a generator of $H^{2i}_{dR}(\mathbf{P}^ n_ A/A)$ for $0 \leq i \leq n$. $\square$

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## Comments (1)

Comment #4884 by Matt Larson on