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Let $S$ be a projective surface over an algebraically closed field. Suppose that $q(S)=h^1(\mathcal O_S)=0$ and $P_2(S)=h^0(\mathcal O_S(2K_S))=0$. If $S$ is smooth, Castelnuovo's rationality criterion (proved by Zariski in characteristic p) tells us that $S$ is rational.

Does this extend to singular surfaces if the characteristic is 0? I'm happy if it extends only to 'some' singular surfaces (e.g. canonical).

I have looked for this in the literature but I haven't found anything to prove it or disprove it. I assume someone has asked this question before (although not in MO, it seems).

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It does not hold in general: a cone over a smooth plane cubic satisfies $q=P_2=0$ but is not rational.

On the other hand if $S$ has canonical singularities and $\tilde{S} \rightarrow S$ is any resolution, one has $P_2(S)=P_2(\tilde{S} )$ and $q(S)=q(\tilde{S} )$, hence $\tilde{S} $ and therefore $S$ are rational.

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    $\begingroup$ If $q$ were to be defined as the dimension of the Albanese of $S$ (for a reasonable definition of Albanese in the singular case), then this example would not work, as the Albanese should be the cubic. What happens with this definition? I suppose if one uses birational invariants, then it doesn't matter if the surface is smooth or not. $\endgroup$
    – Felipe Voloch
    Commented Jun 9, 2014 at 23:18
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    $\begingroup$ The OP gave the definitions he is using, namely $q(S)=h^1(\mathcal O_S)$ and $P_2(S)=h^0(\mathcal O_S(2K_S))$. This is what I use in the examples. $\endgroup$
    – abx
    Commented Jun 10, 2014 at 6:26
  • $\begingroup$ Ooops, the du Val case is so obvious, you are right. Glad to see that it is sharp for singularities. Thanks. $\endgroup$
    – Jesus Martinez Garcia
    Commented Jun 13, 2014 at 17:33

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