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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.

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  • $\begingroup$ There are many more normal surface singularities than cyclic quotient singularities or for that matter rational singularities (which is defined by the condition of the second paragraph). You must have misinterpreted something. To take just one example: The cone point of the affine cone of an elliptic curve (embedded in the projective plane say) is not a rational singularity. $\endgroup$
    – Torsten Ekedahl
    Commented Apr 11, 2010 at 14:29
  • $\begingroup$ @Torsten: My thought is that Amira has something crossed from the ADE singularities. Not familiar enough with the general theory though to be sure of what. $\endgroup$
    – Charles Siegel
    Commented Apr 11, 2010 at 14:44
  • $\begingroup$ Ok. My question is too general. I edited it. (I'm really confused here.) Could somebody maybe give a reference to where all this material on "singularity types" is written down carefully? I'd like to understand how to show that cyclic quotient singularities are rational in my case. $\endgroup$
    – Amira
    Commented Apr 11, 2010 at 15:44
  • $\begingroup$ This is still a nonsense question. Take $Y$ = the cone over an elliptic curve, and $f:Y\to X$ a projection to $Y=\mathbb P^2$. $Y$ is normal, the singularity is not cyclic quotient. $\endgroup$
    – VA.
    Commented Apr 11, 2010 at 16:02
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    $\begingroup$ I think you may be thinking of the case of a finite morphism $\pi\colon X \to Y$ where $X$ is smooth. In that case the singularities of $Y$ are indeed quotient singularities. Not necessarily cyclic quotient singularities however; an $E_8$-singularity for instance is the quotient of $\mathbb C^2$ by the icosahedral group which is not cyclic. $\endgroup$
    – Torsten Ekedahl
    Commented Apr 11, 2010 at 18:30

2 Answers 2

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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.

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  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.

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