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?


  • $\begingroup$ 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
  • 5
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.