The Stacks project

Lemma 99.11.5. In Situation 99.11.1 let $\sigma : B \to X$ be a section. Then $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$ as defined above is a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$.

Proof. We already know that $\mathcal{P}\! \mathit{ic}_{X/B}$ is a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$ by Examples of Stacks, Lemma 95.16.1. Let us show descent for objects for $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$. Let $\{ T_ i \to T\} $ be an fppf covering and let $\xi _ i = (T_ i, h_ i, \mathcal{L}_ i, \alpha _ i)$ be an object of $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$ lying over $T_ i$, and let $\varphi _{ij} : \text{pr}_0^*\xi _ i \to \text{pr}_1^*\xi _ j$ be a descent datum. Applying the result for $\mathcal{P}\! \mathit{ic}_{X/B}$ we see that we may assume we have an object $(T, h, \mathcal{L})$ of $\mathcal{P}\! \mathit{ic}_{X/B}$ over $T$ which pulls back to $\xi _ i$ for all $i$. Then we get

\[ \alpha _ i : \mathcal{O}_{T_ i} \to \sigma _{T_ i}^*\mathcal{L}_ i = (T_ i \to T)^*\sigma _ T^*\mathcal{L} \]

Since the maps $\varphi _{ij}$ are compatible with the $\alpha _ i$ we see that $\alpha _ i$ and $\alpha _ j$ pullback to the same map on $T_ i \times _ T T_ j$. By descent of quasi-coherent sheaves (Descent, Proposition 35.5.2, we see that the $\alpha _ i$ are the restriction of a single map $\alpha : \mathcal{O}_ T \to \sigma _ T^*\mathcal{L}$ as desired. We omit the proof of descent for morphisms. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0D29. Beware of the difference between the letter 'O' and the digit '0'.