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.

On the first blowing up, on the chart that intersects the exceptional divisor I have: $x^4(y^4-2xy^2+x^2+xy^5)$.

The second blowing up, on the chart that intersects the exceptional divisor: $y^2(y^4x+y^2-2xy+x^2)$

On the third blowing up, on the chart that intersects the exceptional divisor: $x^2(x^3y^4+y^2-2y+1)$.

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.

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 link text) 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?

Ps: Calculations like the multiplicity sequence were done in Singular for avoiding trivial mistakes.

  • $\begingroup$ You might be missing some singularities by looking at only one chart since there is always a point on the exceptional divisor which is not contained in a given chart. $\endgroup$
    – naf
    Aug 31 '11 at 6:24
  • $\begingroup$ This is completely irrelevant to the question, but here is the curve $\left(y^2-x^3\right)^2+y^5=0$: <hr /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; !Cusp <hr /> $\endgroup$ Apr 5 '15 at 13:58

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 $\mathfrak{m}_P^2$).

  • $\begingroup$ Thanks for the answer. In the third blowup, the strict transform intersects the exceptional divisor at (0,0) and (0,1). So the point (0,0) is non-singular, but (0,1) is singular!!. If I move that point to the origin, I obtain $x^3(y+1)^4+y^2 = 0$. So, you are saying that I must follow the resolution by blowing at it?. $\endgroup$
    – eventually
    Aug 31 '11 at 6:37

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