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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