Lemma 99.11.6. In Situation 99.11.1 let $\sigma : B \to X$ be a section. The morphism $\mathcal{P}\! \mathit{ic}_{X/B, \sigma } \to \mathcal{P}\! \mathit{ic}_{X/B}$ is representable, surjective, and smooth.
Proof. Let $T$ be a scheme and let $(\mathit{Sch}/T)_{fppf} \to \mathcal{P}\! \mathit{ic}_{X/B}$ be given by the object $\xi = (T, h, \mathcal{L})$ of $\mathcal{P}\! \mathit{ic}_{X/B}$ over $T$. We have to show that
is representable by a scheme $V$ and that the corresponding morphism $V \to T$ is surjective and smooth. See Algebraic Stacks, Sections 94.6, 94.9, and 94.10. The forgetful functor $\mathcal{P}\! \mathit{ic}_{X/B, \sigma } \to \mathcal{P}\! \mathit{ic}_{X/B}$ is faithful on fibre categories and for $T'/T$ the set of isomorphism classes is the set of isomorphisms
See Algebraic Stacks, Lemma 94.9.2. We know this functor is representable by an affine scheme $U$ of finite presentation over $T$ by Proposition 99.4.3 (applied to $\text{id} : T \to T$ and $\mathcal{O}_ T$ and $\sigma ^*\mathcal{L}$). Working Zariski locally on $T$ we may assume that $\sigma _ T^*\mathcal{L}$ is isomorphic to $\mathcal{O}_ T$ and then we see that our functor is representable by $\mathbf{G}_ m \times T$ over $T$. Hence $U \to T$ Zariski locally on $T$ looks like the projection $\mathbf{G}_ m \times T \to T$ which is indeed smooth and surjective. $\square$
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Your email address will not be published.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)