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Jul 9, 2013 at 15:51 answer added Francesco Polizzi timeline score: 7
Jul 9, 2013 at 8:48 review First posts
Jul 9, 2013 at 8:59
Jul 9, 2013 at 8:48 comment added user36795 Hi Jason, thanks for the counterexample! I'm actually interested in the situation when the base field contains the roots of unity of order $|G|$. How do you prove the statement in that case?
Jul 9, 2013 at 8:47 history edited user36795 CC BY-SA 3.0
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Jul 9, 2013 at 8:41 comment added Jason Starr That is not true without further hypotheses. For instance, let $X$ be $\mathbb{P}^1_{\mathbb{R}}$. Let $G=\mathbb{Z}/2\mathbb{Z}$ act via the involution $[x,y]\mapsto [y,-x]$. Then the quotient $Y$ does not have any invertible sheaf $L$ of degree $-1$, a necessary condition for your assertion to hold. You either need to assume that your field contains appropriate roots of unity, or you should replace $G$ by a group scheme whose Cartier dual is split, e.g., $G=\mathbf{\mu}_n$.
Jul 9, 2013 at 8:28 history asked user36795 CC BY-SA 3.0