Lemma 37.48.5. Let $S$ be a scheme. Let $\{ S_ i \to S\} _{i \in I}$ be an fppf covering. Then there exists a surjective integral morphism $S' \to S$ 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. Choose $S = \bigcup U_ j$, $W_ j \to U_ j$, $W_ j = \bigcup W_{j, k}$, and $T_{j, k} \to W_{j, k}$ as in Lemma 37.48.2. By Lemma 37.48.3 we can extend $W_ j \to U_ j$ to a surjective integral morphism $\overline{W}_ j \to S$. After this we can extend $T_{j, k} \to W_{j, k}$ to a surjective integral morphism $\overline{T}_{j, k} \to \overline{W}_ j$. We set $\overline{T}_ j$ equal to the product of all the schemes $\overline{T}_{j, k}$ over $\overline{W}_ j$ (Limits, Lemma 32.3.1). Then we set $S'$ equal to the product of all the schemes $\overline{T}_ j$ over $S$. If $x \in S'$, then there is a $j$ such that the image of $x$ in $S$ lies in $U_ j$. Hence there is a $k$ such that the image of $x$ under the projection $S' \to \overline{W}_ j$ lies in $W_{j, k}$. Hence under the projection $S' \to \overline{T}_ j \to \overline{T}_{j, k}$ the point $x$ ends up in $T_{j, k}$. And $T_{j, k} \to S$ factors through $S_ i$ for some $i$. Finally, the morphism $S' \to S$ is integral and surjective by Limits, Lemmas 32.3.3 and 32.3.2. $\square$
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Your email address will not be published.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)