I'm trying to understand the proof of proposition III.18 in Beauville's book on complex, algebraic surfaces.
Let $E$ be a rank 2 vector bundle on a curve C, and let $X=\mathbb P(E)$ with projection $\pi:X\to C$. The tautological bundle $\mathscr O_X(1)$ is defined via the exact sequence
$$0\to N\to\pi^*E\to\mathscr O_X(1)\to0$$
where $N$ is the bundle whose fiber above a line $\ell\in\mathbb P(E)$ is the line $\ell$ itself. Beauville claims that $\mathscr O_X(1)\cdot\pi^*\bigwedge^2E=\mathrm{deg}(E)$, but I do not see why this is the case.
As far as I can tell, you want to be able to say something like $\mathscr O_X(1)=\mathscr O_X(S)$ where $S$ is (the image of) a section of $\pi$. This would give the claim, but I haven't been able to see how to show this. A comment in this question
https://mathoverflow.net/questions/75105/two-basic-questions-concerning-geometrically-ruled-surfaces
seems to suggest that (something like) this should follow from the fact that $\mathscr O_X(1)\cdot F=1$ for $F$ a fiber, but I don't see the connection.
I think I see the connection now.
Let $D\in\operatorname{Div}(X)$ be any divisor such that $DF=1$. We'll show that $D\sim S+nF$ for some section $S$ and integer $n\in\mathbb Z$. For each $n\in\mathbb Z$, let $D_n=D+nF$, and let $K$ be a canonical divisor on $X$. Since $F$ is a fiber, $F^2=0$ and the adjunction formula then shows that $KF=-2$. Given this, we have the equalities \begin{align*} D_n^2=D^2+2n && D_nF = 1 && D_nK=DK-2n \end{align*} Using Riemann-Roch, we now see that $$h^0(D_n)+h^0(K-D_n)\ge\frac12D_n(D_n-K)+\chi(\mathscr O_X)=\chi(\mathscr O_X(D))+2n.$$ Furthermore, we have $(K-D_n)F=-3<0$, so $h^0(K-D_n)=0$. Hence, for $n$ large enough, we have $h^0(D_n)>0$ and so there's some effective divisor $E\sim D_n$. Now, because $E$ is effective and because $EF=1$, it must be of the form $E=S+mF$ where $S$ is a section and $m\ge0$. Hence $D\sim S+(m-n)F$ as desired.
The above shows that there's some divisor of the form $D=S+nF$ for which $\mathscr O_X(D)\simeq\mathscr O_X(1)$. Hence, $$\mathscr O_X(1)\cdot\pi^*\bigwedge^2E=\deg(\pi^*E\vert_S)+n\pi^*\left(\bigwedge^2E\cdot c\right)=\deg(\pi^*E\vert_S)=\deg(E),$$ where $c\in C$ is just some point. The second to last equality comes from the fact that, as a cohomology class, $\bigwedge^2E\cdot c$ lives in $\operatorname H^4(C,\mathbb Z)=0$, and the last equality comes from the fact that $\pi\vert_S:S\to C$ is an isomorphism sending $\pi^*E$ to $E$.
Edit: The above argument is essentially correct, but contains a slight error. I am treating all fibers as equivalent divisors, but this is not quite the case. All fibers (or rather, their first Chern classes) are equivalent in $H^2(X;\mathbb Z)$ (which is all that matters of computing intersection numbers), but they are not necessarily equivalent in $\mathrm{Pic}(X)$. The only change this has on the argument is that you fix a particular fiber $F$, consider these same $D_n$'s, and then when you get an effective $E\sim D_n$ (for $n\gg0$), you cannot conclude that $E$ is of the form $S+nF$; instead, you know it is of the form $S+\pi^*P$ where $P\in\mathrm{Div}(C)$ is some divisor on $C$. This is still good enough for the short argument in the last paragraph.