The anticanonical divisor of geometric ruled surface $X$ is not numerical positive except $\mathbb{F}_i, i =0, 1$ ?
Obviously, $\mathbb{F}_0$ and $\mathbb{F}_1$ is del Pezzo surface.
Let $X = \mathbb{P}_C(E)$ where $E$ is rank two locally free sheaf on curve $C$ with genus $g$.
And we know that $[K_X] = -2h + (\operatorname{deg}(E) + 2g -2)f$, where $h, f$ are the class of $\mathcal{O}_X(1)$ and the fibre of structure map $p: X \to C$, respectively. And also $h$ is section of this map, with $h^2 = \operatorname{deg}(E)$.
$(-K_X \cdot h) = \operatorname{deg}(E) + 2 - 2g \geq 0$
I confused here, we can tensor a negative degree line bundle $L$ with $E$ and then decrease the degree of $E$, then we can make the intersection number above always negative. But it seems impossible?
And if $E$ is decomposable and normalized (c.f.: Hartshorne P373) then $\operatorname{deg}(E) \leq 0$, therefore $\operatorname{deg}(E) = 0/-1, g = 0$, i.e. $X = \mathbb{F}_0$ or $\mathbb{F}_1$.
But for $E$ is indecomposable, I have no idea to prove this argument. Can anyone give me a hint or idea?