Base field is $\mathbb{C}$. Let $C_1:F_1(X,Y,Z)=0, C_2:F_2(X,Y,Z)=0$ be two non-degenerate cubic curves in $\mathbb{P^2}$ intersecting at 9 distinct points. Consider the pencil $$C_{(\lambda:\mu)}:{\lambda}F_1+{\mu}F_2=0, (\lambda:\mu)\in\mathbb{P^1}.$$ I want to figure out the number of degenerate cubic curves in this pencil, but I do not know how to do it.
In the conic case the answer is 3, so I guess the answer is 4 here. The problem is I do not know a way to check whether a cubic curve is degenerate or not. For a conic curve, I can write out its matrix and compute the determinant, so the degenerate conic curves are just the solutions to the equation $\mathbb{det}(\lambda A_1+\mu A_2)=0$, which has exactly 3 solutions (with multiplicities). Can I do the similar things here in the cubic case and how?
Best Answer
There are a number of ways to do this. Over $\mathbb C$, this is my favorite: a general point in $\mathbb P^2$ lies on a unique member of this pencil, which induces a rational map $\mathbb P^2 \dashrightarrow \mathbb P^1$ whose base locus is the $9$ points of $C_1 \cap C_2$. Blowing up the base points, we get a family of elliptic curves $\pi: X \to \mathbb P^1$. Now we apply a principal sometimes called the "topological Hurwitz formula." The basic idea is that while a smooth cubic is topologically a torus T, a nodal cubic (this is the only singularity occurring in a general pencil) is topologically a $2$-sphere with two points identified, and we can detect the difference via the topological Euler characteristic: $\chi(T) = 0$, but the nodal cubic has $\chi = 1$.
Suppose we had a pencil with no singular fibers. Then its structure map is a topological fiber bundle, and its Euler characteristic is $\chi(\mathbb P^1) \times \chi(T) = 2 \times 0 = 0$. Since $\chi(\mathbb P^2) = 3$ and blowing up a smooth point increases $\chi$ by $1$, $\chi(X) = 12$. Since this differs from the Euler characteristic of a family with no singular fibers by $12$, we conclude that there are $12$ singular cubics.