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Let $R$ be a (Noetherian) ring, and $G$ a finite group acting on $R$. Consider the subring $R^G$. Is the map $R^G\rightarrow R$ faithfully flat?

If not, does this become true if we restrict to varieties?

Thanks!

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  • $\begingroup$ Apparently yes, since it's just the categorial quotient. The extension is integral and the fibers are isomorphic. $\endgroup$ – user40276 Jun 4 '14 at 21:18
  • $\begingroup$ Sorry, I'm afraid I don't follow. Do you mind adding more details? I see that the extension is integral, but in what sense are the fibers isomorphic? $\endgroup$ – jacob Jun 4 '14 at 21:20
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    $\begingroup$ Consider two affine spaces (of dim $\geq 2$) glued together along a single point, with the $\mathbf Z/2$-action switching the two spaces. This is an affine variety, and the quotient map is not flat. See mathoverflow.net/a/85713/1310 $\endgroup$ – Dan Petersen Jun 4 '14 at 21:23
  • $\begingroup$ Oh, I've just seen your action need not be freely transitive. Then the fibers may be different. $\endgroup$ – user40276 Jun 4 '14 at 21:26
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The answer is no even for $G=\mathbb{Z}/2$ acting on $R=k[x,y]$ by swapping $x$ with $−x$ and $y$ with $−y$. In this case $R$ is finite, but not flat, over $R^G=k[x^2,xy,y^2]$, for example because the length of the fiber at 0 is 3, while the map has degree 2.

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