**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

\[ 0 \to \Omega \to \mathcal{O}(-1)^{\oplus n + 1} \to \mathcal{O} \to 0 \]

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

\[ 0 \to \Omega ^ p \to \wedge ^ p\left(\mathcal{O}(-1)^{\oplus n + 1}\right) \to \Omega ^{p - 1} \to 0 \]

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

\[ 0 \to \Omega ^ p(k) \to \wedge ^ p\left(\mathcal{O}(-1)^{\oplus n + 1}\right)(k) \to \Omega ^{p - 1}(k) \to 0 \]

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

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