The Stacks project

Proof. As usual, the hardest part is to show descent for objects. To see this let $\{ U_ i \to U\}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Let $\xi _ i = (U_ i, b_ i, \mathcal{L}_ i)$ be an object of $\mathcal{P}\! \mathit{ic}_{X/B}$ lying over $U_ i$, and let $\varphi _{ij} : \text{pr}_0^*\xi _ i \to \text{pr}_1^*\xi _ j$ be a descent datum. This implies in particular that the morphisms $b_ i$ are the restrictions of a morphism $b : U \to B$. Write $X_ U = U \times _{b, B} X$ and $X_ i = U_ i \times _{b_ i, B} X = U_ i \times _ U U \times _{b, B} X = U_ i \times _ U X_ U$. Observe that $\mathcal{L}_ i$ is an invertible $\mathcal{O}_{X_ i}$-module. Note that $\{ X_ i \to X_ U\}$ forms an fppf covering as well. Moreover, the descent datum $\varphi _{ij}$ translates into a descent datum on the invertible sheaves $\mathcal{L}_ i$ relative to the fppf covering $\{ X_ i \to X_ U\}$. Hence by Descent on Spaces, Proposition 74.4.1 we obtain a unique invertible sheaf $\mathcal{L}$ on $X_ U$ which recovers $\mathcal{L}_ i$ and the descent data over $X_ i$. The triple $(U, b, \mathcal{L})$ is therefore the object of $\mathcal{P}\! \mathit{ic}_{X/B}$ over $U$ we were looking for. Details omitted. $\square$