I know that, in general, rational singularities are not necessarily $\mathbb{Q}$-Gorenstein. So I ask:
is there any positive result in this direction known for surfaces?
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The answer is yes. In fact, there is the following
Proposition 1. Every $2$-dimensional rational singularity $(X, \, x)$ is analytically $\mathbb{Q}$-factorial, i.e. there exists an analytic neighborhood $V$ of $x$ such that every Weil divisor on $V$ is a $\mathbb{Q}$-Cartier divisor.
In particular, every $2$-dimensional rational singularity is $\mathbb{Q}$-Gorenstein.
Concerning the particular case of rational Gorenstein singularities, we have
Proposition 2. For a $2$-dimensional normal singularity $(X, \, x)$, the following are equivalent:
- $(X, \, x)$ is a rational double point (i.e., a Du Val singularity);
- $(X, \, x)$ is a rational hypersurface singularity;
- $(X, \, x)$ is a rational Gorenstein singularity;
- $(X, \, x)$ is a canonical singularity.
See S. Ishii, Introduction to Singularities, Theorem 7.3.2 and Theorem 7.5.1.
answered Jun 6, 2016 at 17:44
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$\begingroup$ Thank you. I'm wondering also if there is a special case in which they are even Gorenstein? $\endgroup$ Commented Jun 7, 2016 at 16:32
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$\begingroup$ Yes, they are classified. I have edited the answer. $\endgroup$ Commented Jun 8, 2016 at 12:41
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$\begingroup$ It might be worth pointing out that the local divisor class group of a rational surface singularity is even a finite group, see for instance section 17 of numdam.org/item?id=PMIHES_1969__36__195_0 $\endgroup$ Commented Jun 8, 2016 at 13:42
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$\begingroup$ The converse also holds (for singularities over algebraically closed fields too). Finite divisor class group implies rational singularity for surfaces. $\endgroup$ Commented Jun 8, 2016 at 13:54