Definition: Let $ E $ be a vector bundle(or locally free sheaf) of rank $ r $ over $ V. $ A projective bundle over a variety $ V $ is a map $ \pi: \mathbb{P}(E) \rightarrow V $ such that from any point $ p \in X $ there is a Zariski open neighbourhood $ U \subset X $ of $ p \in X $ with $ \pi^{-1}(U) \cong U \times \mathbb{P}^{r-1} $ as schemes over $ U; $ that is, there are commuting maps
where $ \pi_{1} $ is the first projection. The projective bundle $ \mathbb{P}(E) $ is fibred over $ V $ such that for each point $ p \in V, $ the fibre $ E_{p} $ over $ p $ is the projectivization of $ \mathbb{P}(E_{p}) $ of $ E_{p}. $ Now $ E_{p} \cong \mathbb{P}^{r-1} $ and $ \mathbb{P}(E) $ is locally isomorphic to $ U \times \mathbb{P}^{r-1} $ as mentioned. Every vector bundle $ E $ over a variety $ V $ induces a projective bundle by taking the projective space of the fibres( which are of course vector spaces themselves). This projective bundle is called the projectivation of $ E $, denoted by $ \mathbb{P}(E). $
Definition: If $ C $ be a smooth curve, then a surface $ X $ is said to be ruled over $ C, $ if there exists a smooth morphism $ \pi: X \rightarrow C $ such that each fibre of $ \pi $ is isomorphic to $ \mathbb{P}^{1}. $
Is every ruled surface a rank projective bundle over a curve?
I ask because a Hirzeburch surface $ \mathbb{F}_{a} $ is a projective bundle $ \mathbb{P}\Big(\mathcal{O}_{\mathbb{P}^{1}}(a) \oplus \mathcal{O}_{\mathbb{P}^1} \Big) $ over $ \mathbb{P}^1, $ for $ a \in \mathbb{N}, $ and I think it is a ruled surface.
Best Answer
Yes (at least if the base field is algebraically closed). One first shows that the morphism $\pi \colon X \rightarrow C$ has a section $\sigma \colon C \rightarrow X$ and then constructs a rank $2$ vector bundle $E$ by using this section. For a proof see for example Corollary $1$ at the very beginning of these notes. Let me also include the link to the theorem (Theorem $3$) that is used for the corollary.
P.S.: Hirzebruch surfaces are indeed ruled.