I'm trying to prove that the complex surface $\mathbb P^1\times \mathbb P^1$ is minimal. I'd like to prove it directly, namely by showing that there are not exceptional curves.
Suppose that $E$ is a $(-1)$-curve on $\mathbb P^1\times \mathbb P^1$ and consider the projection $\pi: \mathbb P^1\times \mathbb P^1\rightarrow\mathbb P^1$. The curve $E$ can't be contained in any fibre $F$ of $\pi$ because otherwise we have $E=F$ and $E^2=0$. Therefore $\pi(E)=\mathbb P^1$ which means that $E$ is a section of $\pi$.
Now I can't continue the proof; is there a way to obtain a contradiction from this point?
$\newcommand{\Proj}{\mathbf{P}}$One approach is to note that two-dimensional integral homology of $\Proj^{1} \times \Proj^{1}$ is generated by the classes of the base $B = \Proj^{1} \times \{\text{pt}\}$ and the fibre $F = \{\text{pt}\} \times \Proj^{1}$, and $$ [B] \cdot [B] = [F] \cdot [F] = 0,\qquad [B] \cdot [F] = 1. $$ If $a$ and $b$ are integers and $[E] = a[B] + b[F]$, then $$ [E] \cdot [E] = 2ab \neq -1. $$