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Apr 18, 2016 at 17:40 comment added Dmitry Vaintrob You don't need quasi-projectiveness. Being of finite type is enough: given any (irreducible) finite-type $X$ with $G$ action, let $Y\subset X$ be a closed divisor such that the complement $X\setminus Y$ is affine. Then the complement $X\setminus G\cdot Y$ is an affine scheme, and Sean's answer applies.
Apr 18, 2016 at 17:15 history edited Sean Lawton CC BY-SA 3.0
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Apr 18, 2016 at 14:27 comment added ACL I believe you need to impose that $X$ be quasi-projective.
Apr 18, 2016 at 5:01 vote accept Ron
Apr 18, 2016 at 3:32 history answered Sean Lawton CC BY-SA 3.0