The Stacks project

Lemma 37.47.6. Let $S$ be a quasi-compact and quasi-separated scheme. Let $\{ S_ i \to S\} _{i \in I}$ be an fppf covering. Then there exists a surjective finite morphism $S' \to S$ of finite presentation and an open covering $S' = \bigcup U'_\alpha $ such that for each $\alpha $ the morphism $U'_\alpha \to S$ factors through $S_ i \to S$ for some $i$.

Proof. Let $Y \to X$ be the integral surjective morphism found in Lemma 37.47.5. Choose a finite affine open covering $Y = \bigcup V_ j$ such that $V_ j \to X$ factors through $S_{i(j)}$. We can write $Y = \mathop{\mathrm{lim}}\nolimits Y_\lambda $ with $Y_\lambda \to X$ finite and of finite presentation, see Limits, Lemma 32.7.3. For large enough $\lambda $ we can find affine opens $V_{\lambda , j} \subset Y_\lambda $ whose inverse image in $Y$ recovers $V_ j$, see Limits, Lemma 32.4.11. For even larger $\lambda $ the morphisms $V_ j \to S_{i(j)}$ over $X$ come from morphisms $V_{\lambda , j} \to S_{i(j)}$ over $X$, see Limits, Proposition 32.6.1. Setting $S' = Y_\lambda $ for this $\lambda $ finishes the proof. $\square$


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Your email address will not be published.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 0CP0. Beware of the difference between the letter 'O' and the digit '0'.