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I've read that quotient singularities (that is, spectra of invariant subrings of finite groups acting linearly on polynomial rings) have rational singularities. Is there an elementary proof of this fact in dimension two? I have one proof, but it uses big guns like Grothendieck spectral sequences etc.

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5 Answers 5

up vote 6 down vote accepted

Let me add a quick proof, also working in characteristic zero, that doesn't rely on classification (this proof is essentially originally due to S\'andor, see his Duke paper on rational singularities). It does rely on canonical modules and a certain functoriality for them. I will use the following characterization of rational singularities though (which I believe is originally due to Kempf), $Z$ has rational singularities if and only if it is Cohen-Macaulay and for a resolution $\pi : Z' \to Z$, we have $$\pi_* \omega_Z' \cong \omega_Z.$$

Suppose that $R \subseteq S$ is the invariant subring of some finite group action. We know $f : R \to S$ splits as a map of $R$-module (trace map splits everything).

Take a resolution $\pi : X \to \text{Spec} R$ and another proper birational map $\pi' : Y \to \text{Spec} S$ such that $f \circ \pi'$ factors through $\pi$ (we can always do this, we can even assume $Y$ is also smooth if we feel like it). So $f \circ \pi' = \pi \circ \eta$ and $\eta : Y \to X$ is some proper map.

Now, we have the following map

$$\pi_* \eta_* \omega_Y \to \omega_R$$

which can either be factored as $$\pi_* \eta_* \omega_Y \to \pi_* \omega_X \to \omega_R$$ or $$f_* \pi'_* \omega_Y \to f_* \omega_S \to \omega_R.$$

In the second factorization, the first map is surjective because $S$ is regular and the second map $f_* \omega_S \to \omega_R$ is surjective because $R \to S$ splits (that map can be obtained by applying $Hom_R( \cdot , \omega_R)$ for instance). Thus the original map is surjective and so is $\pi_* \omega_X \to \omega_R$.

If we want to prove rational singularities, we now just need to prove Cohen-Macaulayness. But that is obvious for normal surfaces. Of course, the Cohen-Macaulayness of a summand of a regular ring is also pretty easy ($H^i_{\mathfrak{m}}(R) \to H^i_{\mathfrak{m}}(S)$ injects).

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Thanks, Karl. This is the sort of proof I knew had to exist. –  Graham Leuschke Feb 14 '11 at 15:09

For dim 2 and field $\mathbb{C}$ the calculations are elementary. First one proves by elementary group theory that the only finite subgroups of $SL(2)$ are the Cyclic, Dihedral, Tetrahedral, Octahedral and the Icosahedral groups. Now for these one can by hand write down the equations for the quotients. Look at the nice article by Slodowy titled Platonic solids, Kelinian singularities and Lie groups. Also isolated rational singularities on surfaces were classified by Artin and include the above. Look at the paper of Artin "Isolated rational singularities on surfaces".

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In positive characteristic, quotient singularities need not be rational. For an example, see Artin's paper "Wildly ramified Z/2-actions in dimension two".

In characteristic zero, quotient singularities are log terminal, and log terminal singularities are rational.

The first fact is a simple exercise on discrepancies (which I think first appeared in a paper of Shokurov), basically a straightforward application of Riemann-Hurwitz ramification formula.

For the second one, an elementary proof is contained in the book of Koll\'ar and Mori.

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Well, Avan's answer covers crepant quotient singularities. There are also non-crepant ones coming finite subgroups in $GL_2$. You can handle those in a similar way. The part of the finite group contained in the center of $GL_2$ acts torically and so you can check that the quotient by this cenral piece has rational singularities (as does every toric variety) by simply writing a toric resolution. After that you have to pass to a quotient by the image of your finite group in $PGL_2$ and there you can run again Avan's prescription.

Also, I may be missing something here but let me mention that the fact that the singularities of affine quotients by reductive groups are rational (in any dimension) is a well known theorem of Boutot and Hochster-Roberts:

J.-F. BOUTOT, Singularites rationnelles et quotients par les groupes reductifs, Invent. Math. 88 (1987), 65-68.

M. HOCHSTER AND J. L. ROBERTS, Rings of invariants of reductive groups acting on regular rings are Cohen-Macaulay, Adv. Math. 13 (1974), 115-175.

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Let's assume the characteristic is $0$.

Let $R$ be a normal $N$-graded ring of dimension 2 with homogenous maximal ideal $m$. Then $R$ has rational singularity if and only if the non-negative degrees part of the graded local cohomology module $H_m(R)$ vanish. That is $H_m(R)_{(i)}=0$ for $i\geq 0$. This is due to Watanabe, see Theorem 2.2 in: http://www.ams.org/tran/2003-355-03/S0002-9947-02-03186-0/home.html

It is safe to work in affine situation, so let $S=k[x,y]$ and $R=S^G$. Then $R$ is normal and generated by forms of positive degree. Since $H_{(x,y)}(S)_{(i)}=0$ for $i\geq 0$, it follows that $H_m(R)_{(i)}=0$ for $i\geq 0$ (one can compute local cohomology in $S$ by using a system of parameters which are elements in $R$).

In characteristic $p$ one probably has to use Frobenius. Note that Boutot's theorem fails in this case (I think it is still true for finite group though).

A truly easy proof is probably not easy to find unless one has a truly elementary definition of rationality.

EDIT: There are other ways to see this:

II) Again, assume $k$ is algebraically closed of characteristic $0$.Let $S=k[[x,y]]$ and $R=S^G$. Then the following 2 facts will suffice (using same notation as above):

1) There are only finitely many indecomposable reflexive modules over $R$. (Proof not hard, they have to be summands of $S$). In particular, the class group of $R$ is finite.

2) Since $R$ is complete, $R$ has rational singularity is equivalent to the class group of $R$ is finite. This is Theorem 17.4 in Lipman paper on rational singularity.

III) Finally, one can quote Prop 5.15 of Kollar-Mori book on birational geometry. It gave the exact statement, but the proof uses general machinery, and probably close to what you already knew.

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