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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 $g(\widetilde{C})=1$.

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    $\begingroup$ 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). $\endgroup$ – Lazzaro Campeotti Apr 21 at 21:19
  • $\begingroup$ @LazzaroCampeotti, how do we prove that $K_X=-\widetilde{C}$? Doesn't it matter which $C$ we chose? $\endgroup$ – rmdmc89 Sep 1 at 21:34
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    $\begingroup$ 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. $\endgroup$ – Lazzaro Campeotti Sep 1 at 22:40
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    $\begingroup$ 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. $\endgroup$ – Lazzaro Campeotti Sep 1 at 22:44
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    $\begingroup$ 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. $\endgroup$ – Lazzaro Campeotti Sep 2 at 8:41

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