the blowing up of a plane curve playing me tricks. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:54:13Z http://mathoverflow.net/feeds/question/74132 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74132/the-blowing-up-of-a-plane-curve-playing-me-tricks the blowing up of a plane curve playing me tricks. pmath 2011-08-31T03:18:47Z 2011-08-31T05:53:05Z <p>Sorry for the easy question but this is driving me crazy. Consider the blowing up of the curve \$(y^2-x^3)^2+y^5\$ at the origin. </p> <p>On the first blowing up, on the chart that intersects the exceptional divisor I have: \$x^4(y^4-2xy^2+x^2+xy^5)\$. </p> <p>The second blowing up, on the chart that intersects the exceptional divisor: \$y^2(y^4x+y^2-2xy+x^2)\$</p> <p>On the third blowing up, on the chart that intersects the exceptional divisor: \$x^2(x^3y^4+y^2-2y+1)\$. </p> <p>So the last strict transform is smooth at \$(0,0)\$ (the same for the other chart). I naively thought that this is the end, and I solved the singularity. </p> <p>However, the multiplicity sequence for the resolution is (4,2,2,2,1,1) , and the dual graph of the resolution is like a T, and it has six exceptional divisors. (the dual graph is here <a href="http://www.freeimagehosting.net/c2d80" rel="nofollow">link text</a>) So what am I missing? I tried several examples, and I am running into trouble specially when there is a node on my way. How can I keep going with the resolution all the way until the end?</p> <p>Ps: Calculations like the multiplicity sequence were done in Singular for avoiding trivial mistakes.</p> http://mathoverflow.net/questions/74132/the-blowing-up-of-a-plane-curve-playing-me-tricks/74138#74138 Answer by Michael Joyce for the blowing up of a plane curve playing me tricks. Michael Joyce 2011-08-31T05:53:05Z 2011-08-31T05:53:05Z <p>On your third blowup, you need to check for singularities at all points that lie above your singular point, not just the point with local coordinates (0,0). Fortunately, the equations you get from setting the partials to zero are not bad ... (or alternatively, some inspection should reveal the point \$P\$ where the equation will be in \$\mathfrak{m}_P^2\$).</p>