**definitions:**


A non-singular **complex** projective surface $S$ is a *ruled surface* if it is birationally equivalent to $C\times\_{\text{Spec} \mathbb C}\mathbb P^1\_{\mathbb C}$ where $C$ is a non-singular complex projective curve.


A non-singular **complex** projective surface $S$ is a *geometrically ruled surface* if there exists a surjective morphism $f:S\longrightarrow C$ over a non-singular complex projective curve $C$ such that every fiber $S\_x$ ($x\in C$) is isomorphic to $\mathbb P^1\_{\mathbb C}$.




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**Questions:**


1. Suppose that $S\cong^{\text{bir}} C \times\mathbb P^1\_{\mathbb C}$ is a ruled surface defined over $\mathbb {\overline Q}$, namely $S\cong S\_{\overline{\mathbb Q}}\times\_{\text{Spec }{\overline{\mathbb Q}}}\text{Spec }{\mathbb C}$, then can I conclude that $C$ is defined over $\overline{\mathbb Q}$?
2. Suppose that $f:S\longrightarrow C$ is a geometrically ruled surface such that the curve $C$ is defined over $\overline{\mathbb Q}$, then can I conclude that $S$ is defined over $\overline{\mathbb Q}$?


Many thanks in advance



Some questions about ruled surfaces defined over a number field
As for (1), the answer is yes. If $C$ is genus $0$, then it is clearly defined over $\overline{\mathbb{Q}}$. In higher genus, one may recover $C$ as e.g. the image of the Albanese map $S\to \text{Alb}(S)$, hence it is defined over whatever field $S$ is.


For (2), the answer is no. Let $E$ be an elliptic curve defined over $\overline{\mathbb{Q}}$, and let $x\in E(\mathbb{C})$ be a $\mathbb{C}$ point which is not an $\overline{\mathbb{Q}}$-point. Then $\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(x))$ is a counterexample. Indeed, if it was defined over $\overline{\mathbb{Q}}$, there would be a line bundle $\mathcal{L}$ on $E$ so that $\mathcal{L}\oplus \mathcal{L}(x)$ was defined over $\overline{\mathbb{Q}}$. But then both $\mathcal{L}$ and $\mathcal{L}(x)$ would be defined over $\overline{\mathbb{Q}}$, hence $\mathcal{O}(x)$ would be, which is a contradiction.


