Resolution of "nice" and zero-dimensional singularities on a surface - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:48:26Z http://mathoverflow.net/feeds/question/71588 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71588/resolution-of-nice-and-zero-dimensional-singularities-on-a-surface Resolution of "nice" and zero-dimensional singularities on a surface Jesko Hüttenhain 2011-07-29T16:30:31Z 2011-07-30T00:33:16Z <p>Assume I have a singular algebraic surface \$X\$ over an algebraically closed field (characteristic zero if you want) which is singular in a <em>finite set of points</em>. I am looking for a condition as to the nature of these singularities which will guarantee that after blowing up \$X\$ in each of the singular points <b>once</b>, I will get a smooth surface.</p> <p>In my fever dreams, you find a reference for such a statement from a text book with a very laid-out, comprehensible proof. Any reference, however, is welcome. Thanks!</p> http://mathoverflow.net/questions/71588/resolution-of-nice-and-zero-dimensional-singularities-on-a-surface/71591#71591 Answer by Francesco Polizzi for Resolution of "nice" and zero-dimensional singularities on a surface Francesco Polizzi 2011-07-29T16:50:56Z 2011-07-29T18:14:32Z <p>The condition you are looking for has a name: <em>absolute isolatedness</em>. </p> <p>In fact, a surface singularity is called <em>absolutely isolated</em> if it can be resolved by using only quadratic transformations centered at reduced points, that is, no normalizations will be required.</p> <p>In general, isolated surface singularities are not absolutely isolated. But, for instance, <em>rational singularities</em> are so.</p> <p>Googling "absolutely isolated \$2\$-dimensional singularities" you can find a lot of references. For example, Tyurina's paper <a href="http://www.springerlink.com/content/n3l1t82035550886/" rel="nofollow">Absolute isolatedness of rational singularities and triple rational points </a> can be surely useful.</p>