Suppose that $X$ is a non-singular projective surface over $\mathbb {\overline Q}$ ( $X$ is a $\mathbb {\overline Q}$-scheme...) and suppose that there is an embedding:


$$j:\mathbb P^1\_{\mathbb C}\hookrightarrow X\times\_{\textrm{Spec }\overline{\mathbb Q}}\textrm{Spec }\mathbb C$$


Now, remembering that $\mathbb P^1\_{\mathbb C}=\mathbb P^1\_{\overline{\mathbb Q}}\times\_{\textrm{Spec }\overline{\mathbb Q}}\textrm{Spec }\mathbb C$, is it true that there exists an embedding:
$$j':\mathbb P^1\_{\overline{\mathbb Q}}\hookrightarrow X\;\textrm{?}$$




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In other words, if a projective surface defined over ${\overline{\mathbb Q}}$ contains a projective line, then does its model over ${\overline{\mathbb Q}}$ contain also a projective line?




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About $\mathbb P^1_\mathbb C$ contained in a surface
Yes, this is true, though your particular map $j$ may not come from a $\overline{\mathbb{Q}}$-map (for example, let $X=C\times \mathbb{P}^1$ where $C$ is a $\overline{\mathbb{Q}}$-curve, and let $j$ be the inclusion of a fiber over a $\mathbb{C}$-point of $C$ which is not an $\overline{\mathbb{Q}}$-point). 


Suppose $X$ is defined by some homogeneous polynomials $f\_1(\underline{x}), \cdots, f\_n(\underline{x})$ in variables $\underline{x}=(x\_0, \cdots, x\_r)$. The map $\mathbb{P}^1$ to $X$ is given by rational functions $\underline{s}=(s\_0(t), \cdots, s\_r(t))\in \mathbb{C}(t)$ so that $f\_i(\underline{s})=0$. Now the $$s\_i(t)=\frac{\sum a\_{ij}t^j}{\sum b\_{il} t^l},$$
where the $a\_{ij}\in \mathbb{C}$ (satisfying some conditions, e.g. the $b\_{il}$ are not all zero).


Consider the subvariety of $\mathbb{A}^N=\operatorname{Spec}(\overline{\mathbb{Q}}[a\_{ij}, b\_{il}]\_{i,j,l}$) defined by the equations $\{f\_r(\underline{s})=0\}$, viewed as equations in the $a\_{ij}, b\_{il}$. The assumption of the existence of $j$ shows that it is non-empty. The conditions of defining a map (e.g. the $b\_{il}$ are not all zero, etc.) which is a closed embedding are open conditions. (Showing that the space of parameters defining a closed embedding is open is non-trivial, so I'll expand on it on request). Thus we have a non-empty open set parametrizing maps of the desired form. So it contains a $\overline{\mathbb{Q}}$-point, completing the proof.


Let me remark that it is much easier to prove this if you black-box the existence of the Hilbert scheme ${\operatorname{Hilb}}(X)$, which is locally of finite type over $\overline{\mathbb{Q}}$. Then one uses the fact that $\mathbb{P}^1$ does not deform to argue that in an open neighborhood of the point of $\operatorname{Hilb}(X)$ representing $j$, we have subschemes isomorphic to $\mathbb{P}^1$. But perhaps it is illuminating to do things explicitly.


