# What is the simplest way to represent a $D_5$ singularity?

Consider this curve $f(x,y)=0$ given by

$$f(x,y) := y^3 + y^2 x + x^4 =0.$$ Is it obvious that after a change of coordinates near the origin, this curve is equivalent to

$$\hat{y}^2 \hat{x} + \hat{x}^4 = 0$$

I think, these are both $D_5$ singularities. It seems like the change of coordinates that would achieve this is of the form

$$x = \hat{x} - y + c_2 y^2 + c_3 y^3 + \ldots$$

where $x$ is an infinite power series. We can kill off the coefficients of $y^n$, for all $n$. This would give us a factor of $\hat{x}$, i.e we get something of the form $$f = \hat{x}\cdot g$$ And then we can make another change of coordinate, so that $g$ becomes $$g = \hat{y}^2 + \hat{x}^3.$$ Is there a simpler way to prove this? And aside from proving the power series converges, is there anything missing in the proof? Everything is over the complex numbers.

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Ritwik -- Maybe you should talk to Patricio. He lectured about related things in the student seminar a couple of weeks ago. – Jason Starr Oct 28 '11 at 15:14

In general, the singularities $J_{r,i}$ or $E_{r,i}$ (the notation is not uniform) are given by $y^3+y^2x^r+x^{3r+i}$ with $r \geq 1$,and $i \geq 0$. In your case $r=i=1$ and the singularity is actually $D_5$. You can recognized it because it has two branches: one smooth, and another one with an $A_2$ singularity. Those branches separates after one blowing up. (see Table A, from the latest reference), and they are the factors that you see in your calculation. This "branch behavior" is the* definition of the $D_5$ singularity, and the normal form is deduced from it. A detailed discussion is in Barth's book in compact complex surfaces, page 79.
Psd: I don't see anything missing in your argument, but it is "simpler" by using $D_5$'s resolution.