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Let $X$ be a (EDIT: non-singular, or even $\mathbf A^n$) algebraic variety over a field $k$ (alg. closed). Suppose $G$ is a finite group acting on $X$, $|G|\neq 0$ in $k$. Then $k[X]^G$ is Cohen-Macaulay. All the proofs I've seen are fairly involved.

  • Question: is there a simple proof of this fact?
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Are you assuming that $X$ was Cohen-Macaulay to begin with? – Akhil Mathew Nov 19 '11 at 16:25
Yes, I guess so. I'm even interested in a simple proof for the case $X = \mathbf A^n$. – univers Nov 19 '11 at 16:29
You can see a very simple proof as Theorem 4.3.6 in Benson's "Polynomial Invariants of Finite Groups"; here's a Google Books preview:… – Steve D Nov 19 '11 at 16:39
It's been a while since I thought about it, but my recollection is that "$X$ is non-singular" implies "$k[X]^G$ is Cohen-Macaulay" (Hochster-Roberts) and "X has rational singularities" implies "$k[X]^G$ has rational singularities" (Buto), however, it's not true that "$k[X]$ is Cohen-Macaulay" implies "$k[X]^G$ is Cohen-Macaulay", where $G$ is a reductive algebraic group. Perhaps, the situation is different for a finite group $G$? – Victor Protsak Nov 19 '11 at 17:08
It would be enough to (locally) find a regular sequence consisting of invariants as then $k[X]$ would be free over the subring generated by the sequence and so would $k[X]^G$ being a direct factor of it. It seems that for a polynomial ring the coefficients of the characteristic polynomial of a general linear polynomial might do the trick. – Torsten Ekedahl Nov 19 '11 at 17:26

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