On minimal resolution of singularities and the type of singularities - MathOverflow

On minimal resolution of singularities and the type of singularities

2010-04-11T14:16:14Z

Let $Y$ be a normal projective surface, let $X$ be a smooth projective surface and let $\pi:Y\longrightarrow X$ be a finite morphism. Why are all singularities of $Y$ cyclic quotient singularities? And what does this mean? Furthermore, why are these singularities rational? And again, what does that mean? (I edited the question. So the example of Ekedahl doesn't work anymore.)

Take a minimal resolution of singularities $\rho:Y^\prime\longrightarrow Y$. Then the above apparently shows that $R^0 \rho_\ast \mathcal{O}_{Y^\prime} = \mathcal{O}_Y$ and $R^i \rho_\ast \mathcal{O}_{Y^\prime} = 0$ for $i>0$. Is this something special for surfaces?

The reason I ask this question is the following.

If $Y$ is a normal variety (say over the field of complex numbers) with the above data, do we still have $R^0 \rho_\ast \mathcal{O}_{Y^\prime} = \mathcal{O}_Y$ and $R^i \rho_\ast \mathcal{O}_{Y^\prime} = 0$ for $i>0$.

Note. The case of a surface over the complex numbers is dealt with in Compact complex surfaces by Barth, Hulek, Peters and van de Ven. I believe they show that cyclic quotient singularities are rational in this case.

Answer by Richard Montgomery for On minimal resolution of singularities and the type of singularities

2010-06-25T05:31:03Z

Re. cyclic quotient singularity. See Kollar's book: `Resolution of Singularities' book. p. 81, item (3) and explanations that follow.

I also found Durfee. L'enseignement Math. 1979. Tome 25. fasc. 1-2. p. 131. `Fifteen characterizations of Rational Double Points' helpful in getting myself oriented with examples regarding isolated singularities for surfaces.

Answer by Sándor Kovács for On minimal resolution of singularities and the type of singularities

2010-10-03T03:56:39Z

1) You probably meant $\pi: X\to Y$ and not $\pi:Y\to X$. That way, any singularity that appears on a scheme of finite type over a field can be mapped to a smooth variety in a finite way. (This claim is implicit in VA's and Torsten Ekedahl's comments above).

2) A fairly general criterion for a singularity to be rational is given in my paper A characterization of rational singularities. In particular it covers your case in any dimension.