Let $\Bbb{P}^2$ be the projective plane over $\Bbb{C}$.
Let $\mathcal{L}$ be a pencil of curves with degree $d$ on $\Bbb{P}^2$ with no fixed components.
Does $\mathcal{L}$ have necessarily $d^2$ base points (counted with multiplicity)?
Here's my attempt to prove this:
Take two distinct curves on $\mathcal{L}$ defined by homogeneous polynomials $F,G\in\Bbb{C}[x,y,z]$. The set $S:=\{sF+tG=0\}_{(s:t)\in\Bbb{P}^1}$ defines a pencil without fixed components and with $F\cdot G=d^2$ base points, counted with multiplicity. I'm tempted to say that $\mathcal{L}=S$, but I don't know how to formalize this.
Best Answer
A pencil (or linear system in general) has a number of equivalent definitions, one of which is that the parametrizing $\mathbb P^1$ is $\mathbb P(V)$ for $V \subset H^0(\mathbb P^2, \mathcal O(d))$ a two-dimensional subspace. Any two non-proportional sections $F,G \in V$ span, and we can take their scheme-theoretic intersection as the definition of base locus, and its length (i.e. the number of base points with multiplicity) is $d^2$.