most recent 30 from http://mathoverflow.net 2013-06-20T02:34:41Z http://mathoverflow.net/feeds/question/92281 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92281/q-factorial-and-rational-singularities-on-surfaces Harry 2012-03-26T15:18:12Z 2012-03-27T20:21:37Z

<p>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?</p> <p>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.</p> <p>What about the converse?</p> <p>I know that every surface fibered over $\mathrm{Spec} \mathbf{Z}$ is $\mathbf{Q}$-factorial. Are all its singularities rational?</p> <p>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.</p>

http://mathoverflow.net/questions/92281/q-factorial-and-rational-singularities-on-surfaces/92340#92340 Karl Schwede 2012-03-27T03:39:04Z 2012-03-27T20:21:37Z

<p>Yes, this is true (at least in the excellent case). See the paper of J. Lipman <em>Rational singularities with applications to algebraic surfaces and unique factorization</em>. See in particular Proposition 17.1. In fact, Lipman proves that the divisor class group is locally finite for rational surface singularities.</p> <p>When Lipman wrote that, he hadn't yet proved resolution of singularities for excellent 2-dimensional rings. </p> <p>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).</p>

http://mathoverflow.net/questions/92281/q-factorial-and-rational-singularities-on-surfaces/92393#92393 Hailong Dao 2012-03-27T17:18:24Z 2012-03-27T17:18:24Z

<p>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 $k$. </p> <p>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 $k$! </p>