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Jun 11, 2014 at 19:35 comment added user27920 @Qfwfq: The case of finite groups with order invertible on the base is in Edixhoven's 1992 paper "Neron models and tame ramification" in Compositio 81 (see 3.1--3.4), and the general case is Prop. A.8.11 in the book "Pseudo-reductive groups" (which gives a proof in a more general setting over rings that generalizes Edixhoven's Prop. 3.4; various aspects simplify when working over a field).
Jun 11, 2014 at 19:03 comment added Qfwfq @user52824: could you please also give a reference?
Jun 11, 2014 at 12:27 comment added user27920 If $H$ is a linearly reductive affine group scheme over any field $k$ (e.g., any affine algebraic group with reductive identity component in char. 0, which includes any finite group, as well as any group of multiplicative type in any characteristic) and $Y$ is a separated $k$-scheme of equipped with an $H$-action then the functor $Y^H$ of $H$-fixed points is represented by a closed subscheme of $Y$ (that much doesn't use the linear reductivity) which moreover is smooth whenever $Y$ is smooth. In particular, irreducible components of $Y^H$ are always pairwise disjoint.
Jun 10, 2014 at 19:17 vote accept Qfwfq
Jun 10, 2014 at 19:10 answer added abx timeline score: 9
Jun 10, 2014 at 19:08 comment added Francesco Polizzi I think this is a consequence of Cartan Lemma. In a neighborhood of a fixed point $p$ the action can be linearized, so locally analytically we may assume $X=\mathbb{C}^n$, $p=0$ and $G \subset \rm{GL}(n, \, \mathbb{C})$. Then in a neighborhood of $0$ the fixed locus is given by a union of linear subspaces and the claim follows.
Jun 10, 2014 at 18:55 history edited Qfwfq CC BY-SA 3.0
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Jun 10, 2014 at 18:49 history asked Qfwfq CC BY-SA 3.0