Let $X$ be a normal surface. Is any rational singularity $\mathbf{Q}$-factorial? I've seen this somewhere for surfaces over fields, but what about the general case of an integral 2-dimensional excellent normal scheme?

In this generality it might not hold so what if we assume that $X$ is fibered ( = flat projective) over a Dedekind scheme? When can we hope for such a result to hold. Probably there are some problems depending on the characteristic.

What about the converse?

I know that every surface fibered over $\mathrm{Spec} \mathbf{Z}$ is $\mathbf{Q}$-factorial. Are all its singularities rational?

I know that one has to be careful with the base scheme. Probably if the base scheme is a smooth projective curve over a field things might not work so well, but maybe if the base is $\mathrm{Spec} \mathbf{Z}$ things might become better.


2 Answers 2


Yes, this is true (at least in the excellent case). See the paper of J. Lipman Rational singularities with applications to algebraic surfaces and unique factorization. See in particular Proposition 17.1. In fact, Lipman proves that the divisor class group is locally finite for rational surface singularities.

When Lipman wrote that, he hadn't yet proved resolution of singularities for excellent 2-dimensional rings.

The converse statement is proven for Henselian local rings with algebraically closed residue fields in Theorem 17.4 (in other words, if the divisor class group locally is finite + those conditions, then the singularity is rational).

  • 1
    $\begingroup$ Just to clarify, the converse here means: "if the class group is finite locally ...", which is stronger than being $Q$-factorial. $\endgroup$ Mar 27, 2012 at 6:08
  • $\begingroup$ Karl, actually on rereading the question perhaps the OP meant the local case anyway. Sorry! $\endgroup$ Mar 27, 2012 at 17:10

About the converse: one does need all the assumptions Karl mentioned in his answer. There are $2$-dim. complete local rings which are UFD but does not have rational singularity. One such example (due to Salmon) is $k(u)[[x,y,z]]/(x^2+y^3+uz^6)$ which is factorial for any field $k$.

Removing the hensenlian condition is also a problem: $R=k[x,y,z]_{(x,y,z)}/(x^r+y^s+z^t)$ where $r,s,t$ are pairwise prime, is factorial over any field $k$!


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