Let $X$ be a smooth projective algebraic variety over the complex numbers.


(a) Do there exist:


a smooth proper map $\pi : \mathcal{X}\to S$ of algebraic varieties over the complex numbers, such that $\mathcal{X}$ and $S$ are smooth, $S$ is connected, $X$ is isomorphic to the fiber of $\pi$ over some $\mathbf{C}$-point of $S$, and such that there exists $s\in S(\mathbf{C})$, a $K$-scheme $X\_0$, and an isomorphism of $\mathbf{C}$-schemes:
$$(X\_0)\_{\mathbf{C}} \simeq \mathcal{X}\_s$$


with $K$ either:


(a.1) $K = \mathbf{Q}$, (a.2) $K/\mathbf{Q}$ a finite extension, (a.3) $K = \bar{\mathbf{Q}}$?


(b) What can be reasonably called an "obstruction" to (a.i), $i = 1,2,3$?



Smooth proper fibration of complex projective varieties
The answers to Questions (2) and (3) are positive, and the answer to Question (1) is negative. The positive answers follow by a version of the "Lefschetz principle", often also called "spreading out". 


Let $X\subset \mathbb{P}^n\_{\mathbb{C}}$ be a closed subscheme. Denote by $\mathcal{I}$ the corresponding ideal sheaf. By ampleness / Serre vanishing, there exists an integer $d\geq 0$ such that $\mathcal{I}(d)$ is globally generated. By finiteness of cohomology of coherent sheaves on proper schemes, the $\mathbb{C}$-vector space $H^0(\mathbb{P}\_{\mathbb{C}}^n,\mathcal{I}(d))$ has finite dimension. 


Let $f\_1,\dots,f\_m$ be a basis for this finite-dimensional $\mathbb{C}$-vector space. For a choice of homogeneous coordinates, i.e., for a basis $(x\_0,\dots,x\_n)$ for $H^0(\mathbb{P}^n\_{\mathbb{C}},\mathcal{O}(1))$, each $f\_i$ is a linear combination of monomials, $$f\_i =\sum\_{(e\_0,\dots,e\_n)} c\_{i;e\_0,\dots,e\_n} x\_0^{e\_0}\cdots x\_n^{e\_n}, $$ where the indexing set is all $(n+1)$-tuples of nonnegative integers that sum to $d$. 


Let $R\subset \mathbb{C}$ be the finitely generated $\mathbb{Z}$-algebra generated by all of the coefficients $c\_{i;e\_0,\dots,e\_n}$. This is an integral domain that contains $\mathbb{Z}$. Since the coefficients are in $R$, every $f\_i$ is a well-defined element in $H^0(\mathbb{P}^n\_R,\mathcal{O}(d))$. Let $X\_R$ denote the zero scheme of $f\_1,\dots,f\_m$ inside $\mathbb{P}^n\_R$. Then $X\_R$ is a finite type scheme over $\text{Spec}\ R$. The base change of $X\_R$ by $R\hookrightarrow \mathbb{C}$ equals $X$.


By Grothendieck's Generic Freeness result, after replacing $R$ by a dense, Zariski open affine subscheme, the $R$-scheme $X\_R$ is $R$-flat. Since the base change to $\mathbb{C}$ is smooth, after shrinking further, also $X\_R$ is $R$-smooth. 


Since $R$ is a finitely generated integral domain over $\mathbb{Z}$ that contains $\mathbb{Z}$, also $R\otimes\_{\mathbb{Z}}\mathbb{Q}$ is a finitely generated $\mathbb{Q}$-algebra that is an integral domain that contains $\mathbb{Q}$. By Zorn's Lemma, there exists a maximal ideal $\mathfrak{m}$ in this nonzero ring. By the Nullstellensatz, for every maximal ideal $\mathfrak{m}\subset R\otimes\_{\mathbb{Z}}\mathbb{Q},$ the corresponding residue field $K$ is a finite field extension of $\mathbb{Q}$. Thus, there exists a finite field extension $K/\mathbb{Q}$ and there exists a homomorphism of commutative rings with $1$, $f:R\to K$. Denote by $X\_K$ the base change $K$-scheme, $$X\_K = X\_R\times\_{\text{Spec}\ R} \text{Spec}\ \mathbb{C}.$$


Now define $R\_{\mathbb{C}}$ to be $R\otimes\_{\mathbb{Z}} \mathbb{C}$, define $S$ to be $\text{Spec}\ R\_{\mathbb{C}}$, and define $\mathcal{X}\to S$ to be the base change of $X\_R\to \text{Spec}\ R$ by $S\to \text{Spec}\ R$. The ring homomorphism $f$ induces a ring homomorphism, $$f\_{\mathbb{C}}:R\otimes\_{\mathbb{Z}} \mathbb{C} \to K\otimes\_{\mathbb{Z}} \mathbb{C}.$$ By the Fundamental Theorem of Algebra, $K\otimes\_{\mathbb{Z}}\mathbb{C}$ is a direct product of finitely many copies of $\mathbb{C}$. For any of these factors, say $$p:K\otimes\_{\mathbb{Z}}\mathbb{C} \twoheadrightarrow \mathbb{C},$$ the composite $p\circ f\_{\mathbb{C}}$ gives a $\mathbb{C}$-point of $S$ such that the fiber of $\mathcal{X}$ over that point equals the base change of $X\_K$ by the field homomorphism, $$K\hookrightarrow K\otimes\_{\mathbb{Z}}\mathbb{C} \xrightarrow{p} \mathbb{C}.$$ This gives a positive answer to Question (2) and to Question (3).


The answer to Question (1) is negative. There are Abelian varieties $A$ of dimension $2$ defined over a number field $K$ that have a finite set of endomorphisms that do not (simultaneously) admit any infinitesimal deformations to nearby Abelian varieties. The blowing up $X\_K$ of $A\times\_{\text{Spec}\ K}A$ along the graphs of these endomorphisms gives a $K$-scheme with no nontrivial infinitesimal deformations. So if $X\_K$ is not isomorphic to the base change of $\mathbb{Q}$-scheme, then there is also no family $\mathcal{X}\to S$ as in the question. 


