Consider $x^2+y+\alpha=0$ (made-up example), where $(x,y)\in\mathbb{C}^2$ and $\alpha\in\mathbb C$ is a free parameter. The equation defines a family of curves.
The curves have no common points in the Cartesian plane therefore we consider the complex projective plane $\mathbb{CP}^2$. Homogenizing by using $x=X/Z$ and $y=Y/Z$ yields $X^2+YZ+\alpha Z^2=0$.
The only base point is $[X,Y,Z]=[0,1,0]$ (corresponds to line at infinity). We need to blow-up this base point.
This is the part I do not understand:
In the chart $Y=1$, we have $X^2+Z+\alpha Z^2=0$, which has the base point $(X,Z)=(0,0)$. We then proceed to blow-up this base point in the $XZ$ coordinates.
Why do we consider the chart $Y=1$. Say, if the base point was $[X,Y,Z]=[0,1,-1]$, which chart do I consider?
Edit: Please ignore the following. This has been resolved.
I was thinking about another example: Consider $y-mx=0$, where $m$ is a free parameter. The common point in the Cartesian plane is $(x,y)=(0,0)$. Suppose, we embed the equation in $\mathbb{CP}^2$. Then, $x=X/Z$ and $y=Y/Z$ yields $Y-mX=0$. If $Z=1$ the base points are $[0,0,1]$ and $[1,1,1]$. If $Z=0$ the base point is $[1,0,0]$. But if I apply Bezout's Theorem, then there should be only 1 base points. Why am I getting a total of 3? What is my mistake?
Thanks, Jay.
Your pencil of curves $x^2+y+\alpha=0$ in $\mathbb{C}^2$ has equation $x^2+yz+\alpha z^2$ in $\mathbb{P}^2$. But then, as you observed, there is a base-point which is $[x:y:z]=[0:1:0]$. This point is on the chart $y=1$, that is why you want to choose this chart to blow-up the point, because blow-up a point of $\mathbb{C}^2$ is easy. But you can also blow-up directly this point of $\mathbb{P}^2$ if you prefer.
In general, you have always one of the three charts $x=1$, $y=1$ or $z=1$ which contains your point and you can make change of coordinates (translations) so that it is the origin.