When the contraction is a morphism defined over $\overline{\mathbb Q}$

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 for the field of definition of a morphism.)

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

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.
If the exceptional curve $E$ is defined over $\overline{\bf Q}$, then the morphism can be defined over that field too. See the proof of Castelnuovo's contractibility criterion (for example, in Beauville's algebraic surfaces book), and follow the steps there to see that you can do everything over $\overline{\bf Q}$.
• I have seen the proof of the Castelnuovo's contraction theorem on Beauville's and Hartshorne's books, but I don't see what you say. In particular, the contraction comes from a morphism $f:X\longrightarrow P^N_{\mathbb C}$, but I don't understand why this map is defined over $\overline{\mathbb Q}$. If you have enough time, I'd like more details about the argument that solves my trouble. Many thanks in advance. – Dubious Oct 9 '14 at 7:48
• Why is it a map $f : X \to \mathbb P^N_{\mathbb C}$? The map comes from some explicit line bundle Hartshorne cooks up that gives a map to projective space. When you're working over a field, line bundles give maps to projective space over that same field. Is there somewhere along the way that you have to starting working over $\mathbb C$ instead of $\bar{\mathbb Q}$? (At least in Hartshorne, I think not.) – user47305 Oct 9 '14 at 13:21