Suppose that $S$ is a nonsingular complex projective surface that is defined over $\overline{\mathbb Q}$, namely $S\cong\text{Proj}\frac{\mathbb C[T_1,T_2,\ldots,T_n]}{(f_1,\ldots,f_n)}$ where $f_i$ has coefficients in $\overline{\mathbb Q}$ for every $i=1,\ldots,n$. If $S'$ is a minimal model of $S$, then can we conclude that also $S'$ is defined over $\overline{\mathbb Q}$? Does the blowdown preserve the field of definition (for algebraically closed fields)?
1 Answer
I think so, yes.
1) If $V_{/\overline{\mathbb{Q}}}$ be a smooth projective variety. Then on NeronSeveri groups we have $NS(V) = NS(V_{/\mathbb{C}})$. I believe this follows from the existence of the Picard scheme: the component group of a group scheme does not change under extension from one algebraically closed field to another.
2) Let $S_{\overline{\mathbb{Q}}}$ be a smooth, projective surface. By Castelnuovo's Criterion, minimal models are obtained precisely by a finite sequence of contractions of $1$curves. By part 1), every $(1)$curve in the NeronSeveri group of $S_{\mathbb{C}}$ is, up to algebraic equivalence, defined over $\overline{\mathbb{Q}}$, so the contractions can be done over $\mathbb{\overline{Q}}$.
Added: In the above there is an implicit assumption that the isomorphism class of the blowdown depends only on the NeronSeveri class of the $(1)$curve. I hope that's true; at the moment, I don't quite remember. Alternatively, the above argument certainly produces at least one $\mathbb{C}$minimal model which is defined over $\overline{\mathbb{Q}}$. If the surface is irrational, minimal models are unique. If the surface is rational, then the minimal models are all known: see e.g. here. They can all be defined over $\overline{\mathbb{Q}}$.

$\begingroup$ Beautiful answer but I have still a doubt when you at the end write "...so the contractions can be done over $\overline{\mathbb Q}$". I formally don't see this. Why the contraction of a $(1)$curve defined over $\overline {\mathbb Q}$ is a morphism defined over $\overline {\mathbb Q}$? $\endgroup$– DubiousCommented Oct 6, 2014 at 13:49

1$\begingroup$ Hartshorne, Theorem V.5.7, proves Castelnuovo's Criterion over any algebraically closed field. $\endgroup$ Commented Oct 6, 2014 at 16:01