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 $S$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)