Timeline for Finite group action on quasi-projective varieties
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2016 at 7:00 | comment | added | Matthieu Romagny | Dear Ron, the best-known counterexample is due to Hironaka. It appears here and there in the literature; a good source is wikipedia's page for "Hironaka's example". | |
| Apr 18, 2016 at 5:01 | vote | accept | Ron | ||
| Apr 18, 2016 at 3:32 | answer | added | Sean Lawton | timeline score: 10 | |
| Apr 17, 2016 at 23:32 | comment | added | Ron | @MatthieuRomagny I am in characteristic zero setting. I am taking the quotient in the classical setting (not stacky quotient). As I am new to this topic could you please tell when a quotient does not exist and what conditions are necessary to ensure its existence? A reference for this will be sufficient. | |
| Apr 17, 2016 at 19:53 | comment | added | Matthieu Romagny | If a quotient exists as a variety or a scheme, then the fibres of $\alpha$ will be the same as $G$-orbits, and all contained in an open affine of $X$. But there exist group actions like in your setting where this condition fails, and hence no scheme quotient exists. Thus you must clarify what you mean by "the quotient". Another comment is that since you assume the action is free, if a quotient exists then $\alpha$ will in fact be \'etale everywhere. | |
| Apr 17, 2016 at 19:39 | comment | added | Ariyan Javanpeykar | You could have a look at Liu's book on Algebraic Geometry, and specifically Chapter 4 in which he discusses flat morphisms, etale morphisms and smooth morphisms. Of course, the same is done in Hartshorne's book. | |
| Apr 17, 2016 at 18:06 | comment | added | Ron | @AriyanJavanpeykar Thanks a lot for the answer. Could you please suggest some text to study this topic as well? | |
| Apr 17, 2016 at 17:58 | comment | added | Ariyan Javanpeykar | The morphism $X\to X/G$ is finite. Therefore, it is generically etale if and only if the extension of function fields $K(X/G) \subset K(X)$ is separable. This is the case if the base field (which you didn't specify) is of characteristic zero, because fields of characteristic zero are perfect. | |
| Apr 17, 2016 at 15:48 | history | edited | user21574 |
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| Apr 17, 2016 at 15:41 | history | asked | Ron | CC BY-SA 3.0 |