For blow ups, I have worked only in $\mathbb{CP}^2$. Once I locate the base-point, say $[x,y,z]=[0,1,0]$, I go back to $\mathbb{C}^2$ by considering the chart $y=1$.
I then proceed to blow up $(x,z)=(0,0)$ in $\mathbb{C}^2$. For example, in my first chart I use the transformation (define new coordinates) $x=x_{1}$ and $z=x_{1}z_{1}$, that is, $x_{1}=x$ and $z_{1}=z/x$, where the exceptional divisor is $x_{1}=0$. A similar approach holds for the second chart.
However, I am totally clueless when it comes to higher dimensions. For example, I have the base point $[x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8]=[0,0,0,0,0,0,1,0]$ in $\mathbb{CP}^7$.
What do I do (transformations) for my first blow up? How many charts are possible? Is there any explicit example (in higher dimensions) that I could go over?
I have added my attempt at solving this problem as a solution for suggestions and comments.
Thanks, Radz.
Maybe I just need to extend what I know for $\mathbb{CP}^2$ and $\mathbb{C}^2$. Please do let me know if this approach is correct (or incorrect).
For the base-point $[x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8]=[0,0,0,0,0,0,1,0]\in\mathbb{CP}^7$, we consider the chart $x_7=1$. Then, let $[x_1,x_2,x_3,x_4,x_5,x_6,x_8]=[y_{10},y_{20},y_{30},y_{40},y_{50},y_{60},y_{70}]\in\mathbb{C}^7$. The base-point in $\mathbb{C^7}$ is $[0,0,0,0,0,0,0]$.
In chart 1 of the first blow-up, the transformations are: $y_{10}=y_{11}$, $y_{20}=y_{11}y_{21}$, $y_{30}=y_{11}y_{31}$, $y_{40}=y_{11}y_{41}$, $y_{50}=y_{11}y_{51}$, $y_{60}=y_{11}y_{61}$ and $y_{70}=y_{11}y_{71}$.
That is, $$y_{11}=y_{10}$$ and $$y_{i1}=\frac{y_{i0}}{y_{10}},\quad\quad i=2,3,\ldots,7.$$
The exceptional divisor in this chart is defined by $y_{11}=0$.
We apply a similar procedure for the remaining 6 charts.
Thanks, Radz.