# fixed point scheme in caracteristic p

Let X\rightarrow A^{n} a smooth affine scheme over an affine space. Everything is defined over a field k.

Let G a finite group acting on X and suppose that his order is divisible by the caracteristic of p.

Do I have that the fixed point scheme X^{G} is flat over A^{n}?

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No. Take $X \to \mathbb A^{1} =\operatorname {Spec} k[x,y] \to \operatorname{Spec} k[x]$. Let $G=\mathbb Z/p$ act by $y \to y+x$. Then the fixed point scheme is just a copy of $\mathbb A^1$ over the point $y=0$.