Blow-ups in Projective Space This is in regards to a question (no solutions or comments thus far :-() I asked earlier in regards to the blow-up of an elliptic curve: Question
Let $f(x,y)=y^2-4x^3+ax+b$, where $(x,y)\in\mathbb C^2$. As described here, I need to consider the projective space $\mathbb{CP^2}$ to find the base point. Letting $x=X/Z$ and $y=Y/Z$, the homogeneous polynomial is $F(X,Y,Z)=Z^3\left(\frac{Y^2}{Z^2}-4\frac{X^3}{Z^3}+a\frac{X}{Z}+b\right)$. That is, $$F(X,Y,Z)=Y^2Z-4X^3+aXZ^2+bZ^3.$$Solving $F(X,Y,0)=0$ yields the base point$[0,1,0]$.
I want to resolve this base point through blow-ups (I believe that there is a total of 9 blow-ups). I have no idea how to perform blow-ups in projective space. Any help (hints, books, etc.) to get me started will be appreciated.
Thanks,
Jay.
 A: Edit: This is just a worked out example of a blow-up. A solution was posted by Andrew here.
I will work out the blow up for the function $f(x,y) = y^2 - x^3 - x^2$. Note that $\nabla f(0,0) = 0$, so there is a singularity at $(0,0)$. If you graph the function (over $\mathbb{R}$), you will see that it crosses over itself at $(0,0)$.
Consider $f(x,xy) = x^2(y^2 -x - 1)$. Then the blow up is $g(x,y) = y^2 - x - 1$. You can look at the graph (over $\mathbb{R}$) of $g$ and see that the singularities have been resolved. Notice that $(x,y) \mapsto (x,xy)$ is a map from $V(g) \to V(f)$. That is if I'm given a point $(x,y)$ such that $g(x,y) = 0$, then it follows that $f(x,xy) = 0$. This follows from the relation $f(x,xy) = x^2g(x,y)$. Note that $x = 0$, there is something strange going on, but this is exactly the point of the blowup. It can happen that $g(0,y) \neq 0$. This map can be seen as bring the line $x=0$ to a point (and away from zero the map is a polynomial function). 
To deal with points at infinity, one can work at the relevant portion of affine space. 
