Timeline for Symmetric power of an etale map of curves
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
S Jan 29, 2015 at 9:07 | history | bounty ended | CommunityBot | ||
S Jan 29, 2015 at 9:07 | history | notice removed | CommunityBot | ||
S Jan 21, 2015 at 7:57 | history | bounty started | Lisa S. | ||
S Jan 21, 2015 at 7:57 | history | notice added | Lisa S. | Improve details | |
Jan 19, 2015 at 17:18 | comment | added | Lisa S. | @abx: I have already tried to do this along the lines you indicate but I don't find it easy. That is why I am asking this here. Your comments do not answer my question. | |
Jan 19, 2015 at 17:04 | comment | added | abx | This is an easy exercise, you should be able to do it by yourself. First reduce to the case of a point $(x,\ldots ,x)$ in $X^{[n]}$, then choose a map $u:X\rightarrow \mathbb{A}^1$ étale at $x$, and show that $u^{[n]}$ is étale at $(x,\ldots ,x)$. | |
Jan 19, 2015 at 16:57 | comment | added | Lisa S. | @abx: The "hence" part of your comment is precisely what I am asking to explain. I understand the general strategy that you mention, but my question is: how to make it precise? | |
Jan 19, 2015 at 16:52 | comment | added | abx | Answer to the EDIT question: locally for the étale topology, $X$ is isomorphic to $\mathbb{A}^1$, hence $X^{[n]}$ is isomorphic to $(\mathbb{A}^1)^{[n]}\cong \mathbb{A}^n$, and therefore smooth. | |
Jan 19, 2015 at 16:31 | history | edited | Lisa S. | CC BY-SA 3.0 |
added 258 characters in body
|
Jan 19, 2015 at 7:46 | comment | added | Charles Siegel | As abx said, this is false. A slightly expanded explanation is that it is true for product, but not symmetric product. That's because the symmetric group doesn't act freely. | |
Jan 19, 2015 at 7:14 | comment | added | abx | This is obviously false. Say $f$ has degree 2, and $f^{-1}(y)=\{x_1,x_2\} $; then the fiber of $f^{[2]}$ above $(y,y)$ consists of 3 points $(x_1,x_1)$, $(x_2,x_2)$ and $(x_1,x_2)$, while $f^{[2]}$ has degree 4. | |
Jan 19, 2015 at 6:50 | history | asked | Lisa S. | CC BY-SA 3.0 |