In 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 Inot 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 $C_2$; in english that is a $\mathbb{P^1}$-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 $X= Blowup(P)_{x_1,..., x_9}$ is the blowup of the pencil $P$ in these $ 9 $ points.

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

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 definietely not vertical and therefore I not see why it's interesection with itself should vanish.

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

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.)

In 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 Inot 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 $C_2$; in english that is a $\mathbb{P^1}$-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 $X= Blowup(P)_{x_1,..., x_9}$ is the blowup of the pencil $P$ in these $ 9 $ points.

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

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 definietely not vertical and therefore I not see why it's interesection with itself should vanish.

(I posted identical question in mse without getting any resonance. Hope that the question is not too elementary to be asked here.)

In 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 Inot 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 $C_2$; in english that is a $\mathbb{P^1}$-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 $X= Blowup(P)_{x_1,..., x_9}$ is the blowup of the pencil $P$ in these $ 9 $ points.

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

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 definietely not vertical and therefore I not see why it's interesection with itself should vanish.

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