So, I feel like I'm missing something obvious, but I have the following situation:

Let $X\to Y$ be a finite group quotient of schemes (in fact, varieties) by the finite group $G$. Let $\tilde{Y}\to Y$ be any resolution of singularities. Then we have a natural map $X\times_Y \tilde{Y}\to X$ which is birational, and $X\times_Y \tilde{Y}\to \tilde{Y}$ is a quotient by $G$. I want to be able to say that $X\times_Y \tilde{Y}$ is a resolution of $X$, and if $G$ acts freely, I can see how. But what if there's a locus (let's say a divisor) where $G$'s action has stabilizers?

EDIT: I'm leaving the above, though that situation is no good, thanks to comments below and abx's answer. But perhaps this is more reasonable:

Given $X\to Y$ a quotient by $G$ a finite group and $\tilde{Y}\to Y$ a resolution of $Y$, can I always find a resolution $\tilde{X}\to X$ such that $G$ acts on $\tilde{X}$ and $\tilde{X}/G\cong \tilde{Y}$?

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    $\begingroup$ The fiber product you write is typically singular. $\endgroup$ – Lev Borisov Jun 9 '14 at 11:17
  • $\begingroup$ Is there any reasonable to check criterion for when it's smooth? $\endgroup$ – Charles Siegel Jun 9 '14 at 12:10
  • $\begingroup$ Why do you believe that $X \times_Y \widetilde{Y} \rightarrow \widetilde{Y}$ is the quotient by the natural $G$-action on the source (say assuming $X$ and $Y$ are quasi-projective varieties over an algebraically closed field)? This entails some delicate algebraic conditions, not just topological ones, and the formation of such quotients generally does not commute with non-flat base change (beyond the case of a free $G$-action). $\endgroup$ – user76758 Jun 9 '14 at 12:53
  • $\begingroup$ User76758: Good point, I am working in a specific situation and can check it there. Though between your comment, Lev's comment, and abx's answer, I think I've realized that I didn't quite ask the question that I wanted to, but a much stronger statement. I'm editing the question to rephrase. $\endgroup$ – Charles Siegel Jun 9 '14 at 13:01

I suspect that you will have trouble with trying to "simultaneously" resolve $X$ and $X/G$. We actually run into this issue in the joint paper with Anatoly Libgober arXiv:math/0206241 (although it may not be evident from the paper) and consequently had to settle for working with $\hat X\to\hat Y$ which was $G$-equivariant morphism of smooth varieties which is not necessarily a quotient map. It didn't matter for us a whole lot, since we were working with the formalism of log-terminal pairs.

However, it may not be completely impossible to find this "simultaneous resolution", especially if you know something about $X$ (say it is of small dimension). You can start by using abelianization trick of Batyrev (arXiv:math.AG/0009043) to reduce to the situation when the stabilizers of the action of $G$ are abelian. But then you get into a bit of trouble.

Let me illustrate this trouble by the simplest example, that of $A_2$ singularity. Let $G=\mathbb Z_3$ act on $X=\mathbb C^2$ by $(x,y)\mapsto (xw,yw^{-1})$ where $w=\exp(2\pi I/3)$.

One may view the quotient in terms of toric geometry: one starts with the toric variety that corresponds to the cone $C=(\mathbb Q_{\geq 0})^2$ in the lattice $N=\mathbb Z^2$. Taking the quotient by the above group has the effect of enlarging the lattice $N$ to a coindex three suplattice $N_1=\mathbb Z^2+ {\mathbb Z}(\frac 13,-\frac 13)$. The minimal resolution of singularities of the quotient corresponds to subdividing the cone $C$ into three cones $$ \mathbb Q_{\geq 0}(1,0)+\mathbb Q_{\geq 0}(\frac 23,\frac 13),~~ \mathbb Q_{\geq 0}(\frac 23,\frac 13)+\mathbb Q_{\geq 0}(\frac 13,\frac 23),~~ \mathbb Q_{\geq 0}(\frac 13,\frac 23)+\mathbb Q_{\geq 0}(0,1) $$ which are generated by lattice elements of $N_1$ (the two additional rays correspond to the two exceptional divisors of the minimal resolution).

Taking the obvious $\hat X$, which is the normalization of resolution of $Y$ in the field of functions of $X$ corresponds to considering the same subdivision, but now in the original lattice $N$. The same cones can be written now in terms of minimum generators as $$ \mathbb Q_{\geq 0}(1,0)+\mathbb Q_{\geq 0}(2,1),~~ \mathbb Q_{\geq 0}(2,1)+\mathbb Q_{\geq 0}(1,2),~~ \mathbb Q_{\geq 0}(1,2)+\mathbb Q_{\geq 0}(0,1). $$ The middle cone is now not generated by a basis of the lattice (determinant is three) and thus gives a singular point.

If one is looking for a toric "simultaneous" resolution of this singularity, then one needs to make a subdivision in such a way that for each of the smaller cones taking minimum generators in either lattice $N$ or $N_1$ gives a basis of the said lattice. This requires that in each cone exactly one of minimum generators of $N$ coincides with that of $N_1$. It is possible to do it in this example by splitting the middle cone into $$ \mathbb Q_{\geq 0}(2,1)+\mathbb Q_{\geq 0}(1,1),~~ \mathbb Q_{\geq 0}(1,1)+\mathbb Q_{\geq 0}(1,2) $$ which becomes $$ \mathbb Q_{\geq 0}(\frac 23,\frac 13)+\mathbb Q_{\geq 0}(1,1),~~ \mathbb Q_{\geq 0}( 1,1)+\mathbb Q_{\geq 0}(\frac 13,\frac 23) $$ in $N_1$.

So in this example such $\hat X$ exists. You can try playing with other surface singularities, say $A_n$ for starters. If I were to bet on this, I would say that if you can handle $A_4$, then you should be able to get all of $A_n$.

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  • $\begingroup$ Thanks a lot, looks like I'll need to find a workaround in my situation. $\endgroup$ – Charles Siegel Jun 11 '14 at 0:06

It is a good exercise to make the computation in the simplest possible case: $X=\mathbb{C}^2$, $G=\mathbb{Z}/2$ acting by the antipodal involution, so that $Y$ is a quadratic cone, $\tilde{Y}$ obtained by blowing up the vertex of the cone. Locally $\tilde{X} :=X\times _Y\tilde{Y} $ is given in $\mathbb{C}^3$ by the equations $x(y-tx)=y(y-tx)=0$. That is, $\tilde{X} $ is the plane $y=tx$, with an embedded line along $x=y=0$ (corresponding to the exceptional divisor).

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  • $\begingroup$ This is a nice example, but there is something I am missing in it. If you consider the blow-up $Z \rightarrow X$ of $X$ along $0$, then $Z$ maps surjectively to $\tilde{X}$. But $Z$ being integral, I think this implies that $\tilde{X}$ is integral (because the map $Z \rightarrow \tilde{X}$ is surjective). $\endgroup$ – Libli Jul 8 '14 at 22:22
  • $\begingroup$ That's not true, think of a non-reduced scheme $X$ and the map $X_{red}\rightarrow X$. $\endgroup$ – abx Jul 9 '14 at 4:56
  • $\begingroup$ Well the map you mention is not surjective at the scheme-theoretic level but I understand what you mean. It seems that the map from $Z$ to $\tilde{X}$ is only surjective at the set-theoretic level and not at the scheme-theoretic level. $\endgroup$ – Libli Jul 9 '14 at 17:33

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