# smooth quotient out of a singular variety?

If $X$ is a smooth quasi-projective variety over $\mathbb{C}$ and $G$ is a finite group acting faithfully on $X$, then the Shepard-Todd theorem gives us some criterion for $X/G$ to be smooth.

My question goes a little bit backward. Assume that $X$ is a variety (irreducible, reduced and separated scheme of finite typer over $\mathbb{C}$) with a faithful action of finite group $G$ on it.

Assume that the quotient $X/G$ is smooth. Can we say something on $X$ (is it Cohen-Macaulay?) Or on the quotient morphism $p : X \rightarrow X/G$ (is it flat?).

Thanks a lot!

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Here's an example: Take $\mathbb C^2$ minus the origin and identify the points $(1,1)$ and $(-1,-1)$ and let this be $X$. Then $G=\mathbb Z/2$ acts by $(x,y) \mapsto (-x,-y)$ and the quotient is smooth, but $X$ is not Cohen-Macauley (two planes intersecting in a point is not Cohen-Macaulay). The quotient map is not flat either, since a finite morphism with smooth target is flat iff the source is CM.
This seems to be a nice answer, but could you explain why the quotient is smooth at the image of $(1,1) = (-1,-1)$? –  Libli Jul 8 at 9:43
Is it also clear that your $X$ is quasi-projective? –  Libli Jul 8 at 10:18
If you consider $\mathbb C^2 \setminus \{0,0\}$ and divide by $\pm 1$, the quotient is just $\mathbb C^2 \setminus \{0,0\}$. The same is true for $X / \pm 1$; informally, $(1,1)$ and $(-1,-1)$ were going to be identified in the quotient anyway, so it does not matter if we identify them before quotienting. Yes, it should be clear that it is quasi-projective, but a direct argument escapes me at the moment. You can prove it by a nuke: contracting a closed subscheme of an affine variety to a point produces a new affine variety, so $\mathbb C^2$ with two points identified is affine... –  Dan Petersen Jul 8 at 13:19
Yes, but it seems to me that in Schwede's paper you are quoting it is claimed that contracting a subscheme of an affine scheme to a point one obtains a scheme, not necessarily an affine scheme. For instance, he shows that contracting a line in $\mathbb{A}^2$ the resulting scheme is not noetherian, hence it cannot be $\rm{Spec}\, A$ where $A$ is an noetherian ring. For this reason, I'm still not convinced that this argument shows the quasi-projectivity of $X$. –  Francesco Polizzi Jul 8 at 14:07