# Why is the blow-up of 9 points an elliptic surface?

One example of elliptic fibration is obtained as follows:

Let $$Z(F),Z(G)\subset\Bbb{P}^2$$ be two non-singular cubics intersecting in distinct points $$P_1,...,P_9$$ and take the rational map \begin{align*} \varphi:\Bbb{P}^2&\to\Bbb{P}^1\\ P&\mapsto (F(P):G(P)) \end{align*}

If $$p:X\to \Bbb{P}^2$$ is the blow-up of $$\Bbb{P}^2$$ in $$P_1,...,P_9$$, then $$\pi:=\varphi\circ p:X\to\Bbb{P}^1$$ defines an elliptic fibration.

I'm trying to understand why almost all fibers of $$\pi$$ are elliptic curves.

Here's where I'm at: If $$(a:b)\in\Bbb{P}^1$$, we see that $$\varphi^{-1}(a:b)=C\setminus\{P_1,...,P_9\}$$ where $$C:=Z(bF-aG)$$.

I'm not sure how to prove this, but intuitively I'm convinced that $$C$$ is irreducible for almost all $$(a:b)$$.

Now if $$\widetilde{C}\subset X$$ the strict transform of $$C$$, we have $$\pi^{-1}(a:b)=\widetilde{C}$$, and we should be able to prove that $$g(\widetilde{C})=1$$. If $$m_i$$ is the multiplicity of $$C$$ in $$P_i$$, then $$m_i\cdot m_F(P_i)\leq I(P_i, C\cap F)=I(P_i, F\cap G)=1$$, so $$m_i=1$$ for all $$i=1,...,9$$. Therefore: $$\widetilde{C}^2=C^2-(m_1^2+...+m_9^2)=9-(1+...+1)=0$$

This seems relevant, but I don't know how to conclude that $$g(\widetilde{C})=1$$.

• For the genus: use the adjunction formula $\operatorname{deg} K_C = (K_X+C)\cdot C$ and the fact that $K_X=-C$ (where I have dropped the tilde from your notation). – Lazzaro Campeotti Apr 21 at 21:19
• @LazzaroCampeotti, how do we prove that $K_X=-\widetilde{C}$? Doesn't it matter which $C$ we chose? – rmdmc89 Sep 1 at 21:34
• Maybe I was too sloppy in not distinguishing between curves and linear equivalence clases. If I am understanding your notation correctly, $\tilde{C}$ is the proper transform of a cubic curve that passes through all the 9 points. Standard computations that you can find e.g. in Shafarevich Vol 1 Chapter 3 (IIRC) show that any such $\tilde{C}$ has linear equivalence class $3\tilde{l} -E_1-\ldots-E_9=-K_X$. It doesn't matter which curve you choose. – Lazzaro Campeotti Sep 1 at 22:40
• By the way, a simpler answer to the original question is the following: the curves $C$ and $\tilde{C}$ are isomorphic. Since $C$ is a plane cubic, it has genus 1. – Lazzaro Campeotti Sep 1 at 22:44
• Yes, the geometric genus of a singular plane cubic is zero (although the arithmetic genus is still 1). To see that that is the case, you can use the discriminant: a plane curve is nonsingular if and only if its discriminant is nonzero. The discriminant of $bF-aG$ is a homogeneous polynomial $\Delta(a,b)$ in $a$ and $b$ (the actual formula doesn't matter for your purposes). Since by assumption both $F$ and $G$ are nonsingular, this means $\Delta$ is not identically zero, hence there are only finitely many $(a:b) \in \mathbf P^1$ where it is zero. – Lazzaro Campeotti Sep 2 at 8:41