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. $\endgroup$– user40276Commented Jun 4, 2014 at 21:18
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$\begingroup$ 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? $\endgroup$– jacobCommented Jun 4, 2014 at 21:20
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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 PetersenCommented Jun 4, 2014 at 21:23
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$\begingroup$ Oh, I've just seen your action need not be freely transitive. Then the fibers may be different. $\endgroup$– user40276Commented Jun 4, 2014 at 21:26
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1 Answer
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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.
answered Jun 5, 2014 at 2:09
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$\begingroup$ could you clarify why the observations that the map $R^G \to R$ has degree $2$ (ie $[ \operatorname{Frac}(R): \operatorname{Frac}(R^G) ]=2$) and that the fiber at $0$ has lenght $3$ (ie the ${Frac}(R)$-module $R \otimes_{R^G} \operatorname {Frac}(R)$ has length $3$ or equivalently has dimension $3$ as ${Frac}(R)$-vector space) implies that $R^G \to R$ is not flat? $\endgroup$ Commented Sep 17, 2021 at 21:03
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1$\begingroup$ Say $A\to B$ is a finite and flat homomorphism of Noetherian rings. Therefore $B$ is a finitely generated flat $A$-module and hence projective. If $A$ is a domain, $M$ a finitely generated projective $A$-module, and $P$ is a prime of $A$, let $K={\rm Frac}(A)$ and $F={\rm Frac}(A/P)$. Then $\dim_K (M\otimes_A K) = \dim_F (M\otimes_A F)$, both equal to the rank of $M$ when treated as a vector bundle on ${\rm Spec}(A)$. $\endgroup$ Commented Sep 17, 2021 at 21:16
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1$\begingroup$ P.S. The fiber at zero is the fiber over the prime $(x^2,xy,y^2)R^G = (x,y)R\cap R^G$, not over the zero ideal. $\endgroup$ Commented Sep 17, 2021 at 21:18
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