Picard group of a ruled surface

Question

Consider the ruled surface $F$ over $\Bbb P^1$ obtained by projectivizing $\mathcal{O}\oplus \mathcal{O}(-4)$. Let $s$ be denote the zero section $(0:1)$. Note that $\text{Pic}(F)=\Bbb Zs \oplus \Bbb Zf$ where $f$ is a fiber of $F$. Since $\mathcal{O}_F(1), \mathcal{O}_F(1)\otimes \mathcal{O}_{\Bbb P^1}(1) \in \text{Pic}(F)$, they are linear combinations of $s$ and $f$. But is there a way to determine the coefficients?

Answer 1

Actually the basis you chose is not the one considered in Beauvilles book (he choses $\{f,h\}$ where $h$ is the class of the tautological bundle). Hence, I think it is sufficiently non-standard to write an answer. I will work with the vector bundle $E = \mathcal{O} \oplus \mathcal{O}(n)$ to make it more general

In Proposition 3.18 in Beauville's book here are two equations concerning the class of the tautological bundle, which we call $h$, which essentially follow from the definition of ruled surface. Namely: $$1). h.f = 1 , $$ and $$2). h^2 = \deg(E) = n.$$

Now we consider the basis for the Picard group you, chose and try to write $h$ in terms of this. Unless I am mistaken you have chosen the section with negative self-intersection $-n$. So putting this together with the fact that the tautolical bundle restricts to $\mathcal{O}(1)$ on each fibre, gives that $s^2 = -n$, $s.f=1$ and $f^2=0$. Writing $h = as+ b f$ we get from 1). above $$ a=1$$ and from 2.) $$ n = (s+bf)^2 = -n + 2b ,$$ hence $b=n$, i.e. $$h = s + nf.$$ Everything else can be worked out via direct calculation.

One final remark. If $s'$ is the other section with $s'^2=n$, then note that we have $s'-s = nf$, hence we have have $h = s'$ as classes in the Picard group.