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Proof. To explain this, we recall that $\mathbf{P}^ n_ A = \text{Proj}(A[T_0, \ldots , T_ n])$, and we write symbolically

The first arrow

in the short exact sequence above is given on each of the standard opens $D_+(T_ i) = \mathop{\mathrm{Spec}}(A[T_0/T_ i, \ldots , T_ n/T_ i])$ mentioned above by the rule

This makes sense because $1/T_ i$ is a section of $\mathcal{O}(-1)$ over $D_+(T_ i)$. The map

is given by sending $\text{d}T_ j$ to $T_ j$, more precisely, on $D_+(T_ i)$ we send the section $\sum g_ j \text{d}T_ j$ to $\sum T_ jg_ j$. We omit the verification that this produces a short exact sequence. $\square$

Proof. We are going to use the results of Cohomology of Schemes, Lemma 30.8.1 without further mention. In particular, the statements are true for $H^ q(\mathbf{P}^ n_ A, \mathcal{O}(k))$.

Proof for $p = 1$. Consider the short exact sequence

of Lemma 50.11.1. Since $\mathcal{O}(-1)$ has vanishing cohomology in all degrees, this gives that $H^ q(\mathbf{P}^ n_ A, \Omega )$ is zero except in degree $1$ where it is freely generated by the boundary of $1$ in $H^0(\mathbf{P}^ n_ A, \mathcal{O})$.

Assume $p > 1$. Let us think of the short exact sequence above as defining a $2$ step filtration on $\mathcal{O}(-1)^{\oplus n + 1}$. The induced filtration on $\wedge ^ p\mathcal{O}(-1)^{\oplus n + 1}$ looks like this

Observe that $\wedge ^ p\mathcal{O}(-1)^{\oplus n + 1}$ is isomorphic to a direct sum of $n + 1$ choose $p$ copies of $\mathcal{O}(-p)$ and hence has vanishing cohomology in all degrees. By induction hypothesis, this shows that $H^ q(\mathbf{P}^ n_ A, \Omega ^ p)$ is zero unless $q = p$ and $H^ p(\mathbf{P}^ n_ A, \Omega ^ p)$ is free of rank $1$ with generator the boundary of the generator in $H^{p - 1}(\mathbf{P}^ n_ A, \Omega ^{p - 1})$.

Let $k > 0$. Observe that $\Omega ^ n = \mathcal{O}(-n - 1)$ for example by the short exact sequence above for $p = n + 1$. Hence $\Omega ^ n(k)$ has vanishing cohomology in positive degrees. Using the short exact sequences

and descending induction on $p$ we get the vanishing of cohomology of $\Omega ^ p(k)$ in positive degrees for all $p$. $\square$

Proof. We have the vanishing and and freeness by Lemma 50.11.2. For $p = 0$ it is certainly true that $1 \in H^0(\mathbf{P}^ n_ A, \mathcal{O})$ is a generator.

Proof for $p = 1$. Consider the short exact sequence

of Lemma 50.11.1. In the proof of Lemma 50.11.2 we have seen that the generator of $H^1(\mathbf{P}^ n_ A, \Omega )$ is the boundary $\xi $ of $1 \in H^0(\mathbf{P}^ n_ A, \mathcal{O})$. As in the proof of Lemma 50.11.1 we will identify $\mathcal{O}(-1)^{\oplus n + 1}$ with $\bigoplus _{j = 0, \ldots , n} \mathcal{O}(-1)\text{d}T_ j$. Consider the open covering

We can lift the restriction of the global section $1$ of $\mathcal{O}$ to $U_ i = D_+(T_ i)$ by the section $T_ i^{-1} \text{d}T_ i$ of $\bigoplus \mathcal{O}(-1)\text{d}T_ j$ over $U_ i$. Thus the cocyle representing $\xi $ is given by

On the other hand, for each $i$ the section $T_ i$ is a trivializing section of $\mathcal{O}(1)$ over $U_ i$. Hence we see that $f_{i_0i_1} = T_{i_1}/T_{i_0} \in \mathcal{O}^*(U_{i_0i_1})$ is the cocycle representing $\mathcal{O}(1)$ in $\mathop{\mathrm{Pic}}\nolimits (\mathbf{P}^ n_ A)$, see Section 50.9. Hence $c_1^{Hodge}(\mathcal{O}(1))$ is given by the cocycle $\text{d}\log (T_{i_1}/T_{i_0})$ which agrees with what we got for $\xi $ above.

Proof for general $p$ by induction. The base cases $p = 0, 1$ were handled above. Assume $p > 1$. In the proof of Lemma 50.11.2 we have seen that the generator of $H^ p(\mathbf{P}^ n_ A, \Omega ^ p)$ is the boundary of $c_1^{Hodge}(\mathcal{O}(1))^{p - 1}$ in the long exact cohomology sequence associated to

By the calculation in Section 50.9 the cohomology class $c_1^{Hodge}(\mathcal{O}(1))^{p - 1}$ is, up to a sign, represented by the cocycle with terms

in $\Omega ^{p - 1}(U_{i_0 \ldots i_{p - 1}})$. These $\beta _{i_0 \ldots i_{p - 1}}$ can be lifted to the sections $\tilde\beta _{i_0 \ldots i_{p -1}} = T_{i_0}^{-1}\text{d}T_{i_0} \wedge \beta _{i_0 \ldots i_{p - 1}}$ of $\wedge ^ p(\bigoplus \mathcal{O}(-1) \text{d}T_ j)$ over $U_{i_0 \ldots i_{p - 1}}$. We conclude that the generator of $H^ p(\mathbf{P}^ n_ A, \Omega ^ p)$ is given by the cocycle whose components are

viewed as a section of $\Omega ^ p$ over $U_{i_0 \ldots i_ p}$. This is up to sign the same as the cocycle representing $c_1^{Hodge}(\mathcal{O}(1))^ p$ and the proof is complete. $\square$

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 $H^{2i}_{dR}(\mathbf{P}^ n_ A/A)$ 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 $0 \leq i \leq n$. $\square$