Timeline for Projection from closure of locally closed subscheme is Etale

5 events

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Aug 10, 2020 at 14:53	history	edited	user267839	CC BY-SA 4.0	edited body
Aug 10, 2020 at 14:48	comment	added	user267839		The natural choice for $R$ is $R:=\overline{\Delta(U)}$ since if $S$ is not separated, the image $\Delta(U)$ isn't closed. That is we need etaleness of projection maps $p_i: \overline{\Delta(U)} \to U$. Do you know an argument why that's true?
Aug 10, 2020 at 14:48	comment	added	user267839		So the attempt to generalize the issue was too optimistic. But in more concrete situation when $f$ is the diagonal map $\Delta: S \to S \times S$. Why is the restriction $p_1 \vert _{\overline{\Delta(S)}}:\overline{\Delta(S)} \subset S \times S \to S$ of the projection $p_1: S \times S \to S$ nevertheless etale? The issue that I wanted to understand here mathoverflow.net/questions/365735/… that a usual scheme $S$ is an algebraic space. So we need data $(U,R)$ with $U:=S$ and $R \subset U \times U$ closed (!).
Jul 16, 2020 at 14:19	comment	added	Francesco Polizzi		Take a degree $d \geq 3$ covering of smooth projective curves $f \colon X \to Y$, branched at a single point $y \in Y$ and such that $$f^{*}(y)=kx_0 + x_1 + \ldots + x_{d-k},$$ with $k <d$. Then the restriction $$f \colon X-\{x_0 \} \to Y$$ is surjective and unramified, hence étale, but it extension to $X$ clearly is not.
Jul 16, 2020 at 10:35	history	asked	user267839	CC BY-SA 4.0