Lemma 27.21.5. Let $S$ be a scheme. Let $n \geq 0$. Then $\mathbf{P}^ n_ S$ is a projective bundle over $S$.

Proof. Note that

$\mathbf{P}^ n_{\mathbf{Z}} = \text{Proj}(\mathbf{Z}[T_0, \ldots , T_ n]) = \underline{\text{Proj}}_{\mathop{\mathrm{Spec}}(\mathbf{Z})} \left(\widetilde{\mathbf{Z}[T_0, \ldots , T_ n]}\right)$

where the grading on the ring $\mathbf{Z}[T_0, \ldots , T_ n]$ is given by $\deg (T_ i) = 1$ and the elements of $\mathbf{Z}$ are in degree $0$. Recall that $\mathbf{P}^ n_ S$ is defined as $\mathbf{P}^ n_{\mathbf{Z}} \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} S$. Moreover, forming the relative homogeneous spectrum commutes with base change, see Lemma 27.16.10. For any scheme $g : S \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ we have $g^*\mathcal{O}_{\mathop{\mathrm{Spec}}(\mathbf{Z})}[T_0, \ldots , T_ n] = \mathcal{O}_ S[T_0, \ldots , T_ n]$. Combining the above we see that

$\mathbf{P}^ n_ S = \underline{\text{Proj}}_ S(\mathcal{O}_ S[T_0, \ldots , T_ n]).$

Finally, note that $\mathcal{O}_ S[T_0, \ldots , T_ n] = \text{Sym}(\mathcal{O}_ S^{\oplus n + 1})$. Hence we see that $\mathbf{P}^ n_ S$ is a projective bundle over $S$. $\square$

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