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Rick Miranda's "The basic theory of elliptic surfaces" the Example (I.5.1) see page 7 on a pencil of plane curves contains an argument that I do not understand yet.

Let $C_1$ be a smooth cubic curve in $\mathbb{P^2}$ and let $C_2$ be any other cubic. By intersection theory and Bezout's theorem the intersection number $C_1 \cdot C_2$ is $9$. We form a pencil $P \subset \mathbb{P^2} $ generated by $C_1$ and $C_2$; it is the $\mathbb{P^1}$-family of curves (or more generally divisors) $[ \lambda C_1 + \mu C_2 ]$, which has $9$ base points $x_1,..., x_9$ . This gives only a rational map to $\mathbb{P^1}$. After blowing them up the fundamental locus of this rational map is resolved and we obtain a honest morphism $\pi: X \to \mathbb{P^1}$ where $X= \operatorname{Bl}(\mathbb{P}^2)_{x_1,..., x_9}$ is the blowup of the plane at these $ 9 $ points.

Then it is claimed that the canonical class of $X$ is $-C_1$ and that this implies that $K_X^2= (-C_1)^2 =0$.

Question. How to verify that the canonical class of $X$ equals $-C_1$ and why does it have as consequence self-intersection number zero? The divisor $-C_1$ is definitely not vertical and therefore I not see why its intersection with itself should vanish.

(I posted identical question a week ago on MSE without getting any resonance. Hope that the question is not too elementary to be asked here.)

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    $\begingroup$ There is a standard formula for the canonical divisor of a blow up, which gives $K_X\equiv\pi ^*(-3L)+\sum E_i$, where $L$ is a line and $E_i$ the exceptional divisors. By construction $C_1\equiv 3\pi^*L-\sum E_i$. $\endgroup$
    – abx
    Commented Oct 24, 2022 at 4:17

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As explained in abx's comment, the canonical divisor of your surface is given by $$K_X=-3 \pi^*L + \sum E_i,$$ and this is precisely the class of $-\widetilde{C}_1$ (note that $C_1$ is a curve in $\mathbb{P}^2$, so I put the tilde to specify that we are considering its strict transform in $X$).

Now, $\widetilde{C}_1$ is the strict transform of a smooth cubic curve: since the self-intersection number of a cubic is $9$ and you are blowing up nine points on it, the self-intersection of the strict transform is $(C_1)^2=9-9=0,$ as claimed.

Finally, the strict transform of your pencil of cubics is a base-point free pencil of elliptic curves in $X$, providing an elliptic fibration $\pi \colon X \to \mathbb{P}^1$, endowed with nine sections corresponding to the nine exceptional divisors of the blow-up. By construction, the curve $\widetilde{C}_1$ is a fibre of $\pi$, hence it is a "vertical divisor", after all.

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