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.

cyclicquotient singularities however; an $E_8$-singularity for instance is the quotient of $\mathbb C^2$ by the icosahedral group which is not cyclic. – Torsten Ekedahl Apr 11 '10 at 18:30