# Genus of the desingularization of a plane curve through meromorphic function

Could someone help me? Consider, in the complex projective plane, the curve $C$ given by the the points $[X,Y,Z]$ for which $F(X,Y,Z)=X^2Y^3+YZ^4+Z^5=0$. I have to desingularize the curve and consider the compact Riemann surface $S$ associated to $C$. Then I have to use the function $f=X/Y$ from $S$ to the extended complex space (the Aleksandrov one point compactification of the complexes) to calculate the genus of $X$. I'm new to this and I have some problems. The curve has $2$ singular points: $P=[0,1,0]$ and $Q=[1,0,0]$. I started blowing-up the point $P$, but to find a regular point I have to do it twice. I don't really get what this process leads to and I can't understand which are the actual points that "replace" the point $P$ in $X$. In more simple but similar exercises (where the curve is not singular or where I have to blow-up the points just once) I would check $f$ also on these "new" points in order to use Hurwitz formula to calculate the genus. I'm sorry for the inaccuracy of my question, I'm confused!

• Write everything locally. In each chart you'll see how the equation of the curve and the map $f$ transform under the blowups. Another thing that may help is to write $f$ as a map to the projective line, $f = (X:Y)$. – Alan Muniz Aug 24 '18 at 11:46

In the chart $Y=1$, $F= X^2 +Z^4 +Z^5$. Blowing up we have that the strict transform of $C$ is given in the chart $(t,Z)$, $X=tZ$, by $F_1= t^2 + Z^2 + Z^3$. (The other chart is not important)
Then we take another blowup $Zu = tw$. In the chart $(t,w)$ we have that the strict transform of the curve is given by $F_2 = 1+w^2 + tw^3$ which is smooth and intersects the exceptional divisor in $(0, \pm i)$.
Now for the map $f = X/Y$. In the chart $Y=1$ we have $f(X,Z) = X$. In the first blowup we have $f(t,Z) = tZ$ and in the second blowup it transforms to $f(t,w) = t^2w$.