Suppose that $S$ is a complex projective surface defined over $\overline{\mathbb Q}$, namely there exists a surface $S\_{\overline{\mathbb Q}}$ over $\overline{\mathbb Q}$ such that:
$$S\_{\overline{\mathbb Q}}\times\_{\operatorname{Spec} \overline{\mathbb Q}}\operatorname{Spec} \mathbb C$$


Now consider a $(-1)$-curve $E\hookrightarrow S$ (namely $E\cong\mathbb P^1\_{\mathbb C}$ and $E^2=-1$), and suppose that $f:S\longrightarrow S'$ is the blow-down (contraction) of $E$ on a point, where $S'$ is another surface.


I don't understand the following statement:



> 
> There exists a surface $S'\_{\overline{\mathbb Q}}$ over $\overline{\mathbb Q}$ and a morphism
>  $$ g : S\_{\overline{\mathbb Q}} \longrightarrow S'\_{\overline{\mathbb
>  Q}} $$  
> 
>  such that $f = g \times\_{\operatorname{Spec}\overline{\mathbb Q}}\mathrm{id}\_{\operatorname{Spec}\mathbb C}$. In other words the blow-down $f$ is a morphism defined over $\overline{\mathbb Q}$.
>  (See [this question](https://math.stackexchange.com/questions/719441/what-is-a-field-of-definition-of-a-morphism) for the field of definition of a morphism.)
> 
> 
> 


The thesis should follows from the fact that, by definition, every $(-1)$-curve is defined over $\overline{\mathbb Q}$.


Many thanks in advance.



When the contraction is a morphism defined over $\overline{\mathbb Q}$
As you say, it suffices to show that any $(-1)$-curve is defined over $\overline{\mathbb{Q}}$; it is not true that this is the case "by definition." Let $C$ be a $(-1)$-curve, a priori defined over $\mathbb{C}$. Let $\mathcal{O}(1)$ be an ample divisor on $S$ and $p(t)$ the Hilbert polynomial of $C$ with respect to $\mathcal{O}(1)$; $C$ corresponds to some point $[C]$ on the Hilbert scheme $\operatorname{Hilb}^{p(t)}(S)$; by the general theory of Hilbert schemes, $\operatorname{Hilb}^{p(t)}(S)$ is finite type over $\overline{\mathbb{Q}}$. 


I claim that the point $[C]$ in $\operatorname{Hilb}^{p(t)}(S)$ is isolated; it suffices to show that it has trivial tangent space. The tangent space is given by $H^0(C, \mathcal{N}\_{C/S})$. But $$\operatorname{deg}\_C \mathcal{N}\_{C/S}=C\cdot C=-1,$$ so $H^0(C, \mathcal{N}\_{C/S})=0$. Thus $[C]$ is indeed an isolated point in $\operatorname{Hilb}^{p(t)}(S)$. 


But $\operatorname{Hilb}^{p(t)}(S)$ is finite type over $\overline{\mathbb{Q}}$, so any isolated point has residue field finite type over $\overline{\mathbb{Q}}$ (because connected components over $\mathbb{C}$ are the same as those over $\overline{\mathbb{Q}}$), hence equal to $\overline{\mathbb{Q}}$. So $C$ is defined over $\overline{\mathbb{Q}}$.


Now as Abhinav Kumar writes, Hartshorne in his proof of $V.5.7$, defines a morphism using $\mathcal{O}(N)\otimes \mathcal{O}(kE)$, where both line bundles are defined over $\overline{\mathbb{Q}}$; all the other operations he takes (e.g. normalization) also make sense over any field, so we're done.


