How many blowups do we need to make a pencil base point free? Let $\Bbb{P}^2$ be the projective plane over $\Bbb{C}$.
Take a pencil of curves of degree $d$ on $\Bbb{P}^2$ given by a dominant rational map $\phi:\Bbb{P}^2\dashrightarrow \Bbb{P}^1$.
The pencil has $d^2$ base points, possibly infinitely near. If $p:S\to\Bbb{P}^2$ is the blow up at all $d^2$ base points, then $\phi\circ p:S\to\Bbb{P}^1$ is a base point free pencil on $S$.
My question is: do we always have to blow up $d^2$ times in order to get a base point free pencil?
I agree this is necessary when the $d^2$ base points are not infinitely near. But if one of the base points has many infinitely near it, then this is not clear to me.
 A: Let $\phi\colon \mathbb{P}^2\dashrightarrow \mathbb{P}^1$ be a rational map given by a pencil of curves of degree $d$. Then, denoting by $p_1,\ldots,p_r$ the base-points of the system, including infinitely near, you get $d^2=\sum_{i=1}^r m_i^2$, where $m_i$ is the multiplicity at the point $p_i$. If $r=d^2$, then $m_1=\ldots=m_r=1$, but in general you might have less than $d^2$ base-points.
To solve the indeterminacies of $\phi$ you need to blow-up ALL base-points, including infinitely near. This is in fact the definition of a base-point: a point where the rational map (or the one obtained after blowing-up firstly some other points if your point is infinitely near) is not defined.
Taking examples with tacnodes or any kind of singularities just change the position of the base-points and their multiplicities.
In the example $F=x^2z-y^3$, $G=y^2z-x^3$,you can parametrise the curve $F=0$ by choosing $x=u^3,y=vu^2,z=v^3$. This gives a birational morphism from $\mathbb{P}^1$ to the singular cubic $\{F=0\}$. Replacing in $G$ gives $u^4(v^5-u^5)$. Hence you get $6$ distinct points, namely $[0:0:1]$ and five distinct other points. The intersection multiplicity at $[0:0:1]$ is $4$, as you have $u^4$. This corresponds to the product of the multiplicities $4=2\cdot 2$, as you do not have any tangent direction. In the above notation it then gives $d=3$, $m_1=2, m_2=\cdots=m_6=1$, so you indeed get only $6$ base-points.
