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
Comments (0)