# Tautological quotient bundle of ruled surface

I am reading the book Complex algebraic surfaces by Beauville, the chapter on Ruled surfaces. I have some doubts.

Let $C$ be a smooth curve. $E$ be a rank 2 vector bundle on $C$. Let $X=\mathbb{P}(E)$ the ruled surface associated to $E$. $p:X\rightarrow C$ is the projection morphism.

We have the canonical short exact sequence on $X$, $0\rightarrow N\rightarrow p^*E\rightarrow O_X(1)\rightarrow 0$.

Here $N$ is the line bundle whose fiber over a point associated to a one dimensional subspace of the corresponding fiber of $E$ is the subspace itself. $O_X(1)$ is the quotient.

1) why is $O_X(1)$ a line bundle.

2) Also it says if $F$ is the class of a fiber of $p$ in $Pic\,X$ then $F.O_X(1)=1$. Why is this so?

3) What are the sections of $O_X(1)$? That is there a divisor $D$ on $X$ such that $O_X(1)=O_X(D)$? Is this like bundle ample? Or nef, big?

1) Because the map $N \to p^*E$ is an embedding at every point of $X$.
2) Because, when you restrict the sequence to a fiber, you get the Euler's sequence $$0 \to O_F(-1) \to O_F \oplus O_F \to O_F(1) \to 0.$$
3) By pushing forward to $C$ one gets $H^0(X,O_X(1)) = H^0(C,E)$, so it has sections if and only if $E$ has. Furthermore, $O_X(1)$ is ample if and only if $E$ is ample on $C$.
• what do you mean by ampleness of vector bundle $E$ in (3). – user52991 Oct 13 '16 at 15:42
• sorry in part 3 you say $O_X(1)$ is ample if and only if $E$ is ample. But by definition $E$ is ample if $O_X(1)$ is ample. I am confused because the argument seems cyclic.. – user52991 Oct 13 '16 at 16:55
• i thought your statement meant that if I want to check if $O_X(1)$ is ample, then I need to check that $E$ is ample. Which by definition I should check if $O_X(1)$ is ample. So ampleness of $E$ is just terminology? – user52991 Oct 13 '16 at 18:10