Lemma 59.29.5. Let $S$ be a scheme. Let $\overline{s}$ be a geometric point of $S$. Let $(U, \overline{u})$ be an étale neighborhood of $\overline{s}$. Let $\mathcal{U} = \{ \varphi _ i : U_ i \to U \} _{i\in I}$ be an étale covering. Then there exist $i \in I$ and $\overline{u}_ i : \overline{s} \to U_ i$ such that $\varphi _ i : (U_ i, \overline{u}_ i) \to (U, \overline{u})$ is a morphism of étale neighborhoods.
Proof. As $U = \bigcup _{i\in I} \varphi _ i(U_ i)$, the fibre product $\overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i$ is not empty for some $i$. Then look at the cartesian diagram
\[ \xymatrix{ \overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i \ar[d]^{\text{pr}_1} \ar[r]_-{\text{pr}_2} & U_ i \ar[d]^{\varphi _ i} \\ \mathop{\mathrm{Spec}}(k) = \overline{s} \ar@/^1pc/[u]^\sigma \ar[r]^-{\overline{u}} & U } \]
The projection $\text{pr}_1$ is the base change of an étale morphisms so it is étale, see Proposition 59.26.2. Therefore, $\overline{s} \times _{\overline{u}, U, \varphi _ i} U_ i$ is a disjoint union of finite separable extensions of $k$, by Proposition 59.26.2. Here $\overline{s} = \mathop{\mathrm{Spec}}(k)$. But $k$ is algebraically closed, so all these extensions are trivial, and there exists a section $\sigma $ of $\text{pr}_1$. The composition $\text{pr}_2 \circ \sigma $ gives a map compatible with $\overline{u}$. $\square$
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Your email address will not be published.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)
There are also: