Timeline for If the quotient of an algebraic space $X$ by a finite group is a scheme, is $X$ a scheme?
Current License: CC BY-SA 3.0
9 events
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| May 5, 2015 at 14:59 | comment | added | Dima Pasechnik | I'm voting to close this question as off-topic because it has been answered in the comments | |
| May 4, 2015 at 21:17 | comment | added | Jason Starr | Chevalley's theorem is the statement, roughly, that $Y$ is affine if and only if $X$ is affine. The easy direction is: if $Y$ is affine, then also $X$ is affine. This is the direction you are interested in: every scheme is covered by open affines, you can consider the inverse image of these open affines in $X$, and then you apply the easy direction of Chevalley's Theorem. The hard direction is the one that Rydh generalizes: if $X$ is affine, then also $Y$ is affine. | |
| May 4, 2015 at 21:06 | comment | added | cooldude99 | I googled "David Rydh Chevalley theorem." Is the Chevalley theorem you mean, "let $X$ be an affine scheme, let $Y$ be an algebraic space, and let $f:X \rightarrow Y$ be an integral and surjective morphism. Then $Y$ is affine." ? Excuse my ignorance but I don't know how to apply this. | |
| S May 4, 2015 at 20:56 | history | suggested | Rahman. M |
edit tags.
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| May 4, 2015 at 20:45 | review | Close votes | |||
| May 5, 2015 at 19:52 | |||||
| May 4, 2015 at 20:43 | comment | added | Jason Starr | This follows from the easy direction of Chevalley's theorem for algebraic spaces. For the hard direction, you can consult Knutson's "Algebraic Spaces" or the more recent proof by David Rydh in the not-necessarily-Noetherian setting. | |
| May 4, 2015 at 20:26 | review | Suggested edits | |||
| S May 4, 2015 at 20:56 | |||||
| May 4, 2015 at 20:07 | review | First posts | |||
| May 4, 2015 at 20:13 | |||||
| May 4, 2015 at 19:59 | history | asked | cooldude99 | CC BY-SA 3.0 |