This is related to a question I asked earlier: Link
So, the Hesse pencil is given by $\lambda (x^3+y^3+z^3)+\mu xyz=0$, where $[\lambda,\mu]\in\mathbb{CP^1}$ and $[x,y,z]\in\mathbb{CP^2}$. I can compute the base-points as follows:
Let $x=0$ such that $xyz=0$ (corresponds to the line at infinity). Then, we have $\lambda(y^3+z^3)=0$. Let $\lambda\neq 0$; rescaling yields $\lambda=1$. So, $y^3+z^3=0$. Let $y=1$ such that $z^3=-1$. So, $z=\exp\left(\left(\frac{\pi+2\pi n}{3}\right)i\right)$, $n=0,1,2$. This yields $z=-1,\exp\left(\frac{\pi}{3}i\right),\exp\left(-\frac{\pi}{3}i\right)$. We can introduce $a=\exp\left(\frac{2\pi}{3}i\right)$ and rewrite the last two roots as $-a^2$ and $-a$, respectively.
So, the base points are $[0,1,-1]$, $[0,1,-a^2]$ and $[0,1,-a]$. I can repeat this for $y=0$ and $z=0$.
My question is how do I resolve these base-points through blow-ups. For example, what procedure (basically, how to get the ball rolling) do I need to follow to resolve $[0,1,-1]$?
Edit As far as the procedure is concerned, an example within this forum is Example. For this example, base point $[u,v,w]=[0,1,0]$ is transformed to (x,y) coordinates by setting $v=1$ in the homogenized equation and letting $(u,w)=(x,y)$. Subsequently, blow-up of the base-point (x,y)=(0,0) is discussed.
But I do not know how this procedure is applicable to my case and what to do (start-off) with my base-point $[0,1,−1]$. Do I basically consider the chart $x=0$ and consider the curve $y^3+z^3$ with the base-point $(y,z)=(1,−1)$ and proceed to blow this up?
Thanks, Radz.