Lemma 99.7.4. In Situation 99.7.1. The functors $\text{Q}_{\mathcal{F}/X/B}$ and $\text{Q}^{fp}_{\mathcal{F}/X/B}$ satisfy the sheaf property for the fpqc topology.
Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $S$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $\mathcal{F}_ i = \mathcal{F}_{T_ i}$. Note that $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$ (Topologies on Spaces, Lemma 73.9.3) and that $X_{T_ i \times _ T T_{i'}} = X_ i \times _{X_ T} X_{i'}$. Suppose that $\mathcal{F}_ i \to \mathcal{Q}_ i$ is a collection of elements of $\text{Q}_{\mathcal{F}/X/B}(T_ i)$ such that $\mathcal{Q}_ i$ and $\mathcal{Q}_{i'}$ restrict to the same element of $\text{Q}_{\mathcal{F}/X/B}(T_ i \times _ T T_{i'})$. By Remark 99.7.3 we obtain a surjective map of quasi-coherent $\mathcal{O}_{X_ T}$-modules $\mathcal{F}_ T \to \mathcal{Q}$ whose restriction to $X_ i$ recovers the given quotients. By Morphisms of Spaces, Lemma 67.31.5 we see that $\mathcal{Q}$ is flat over $T$. Finally, in the case of $\text{Q}^{fp}_{\mathcal{F}/X/B}$, i.e., if $\mathcal{Q}_ i$ are of finite presentation, then Descent on Spaces, Lemma 74.6.2 guarantees that $\mathcal{Q}$ is of finite presentation as an $\mathcal{O}_{X_ T}$-module. $\square$
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Your email address will not be published.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)