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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), $i = 1,2,3$?

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    $\begingroup$ So, $S$ is defined over $\Bbb{Q}$? $\endgroup$
    – abx
    Nov 21 '17 at 10:40
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    $\begingroup$ For a variety or scheme defined over $\mathbb{C}$, say $S$, what is your definition of $S(K)$? Are you asking whether there exists a $\mathbb{C}$-point $s$ of $S$, a subfield $K\subset \mathbb{C}$, and a $K$-scheme $Y$ such that the fiber $\mathcal{X}_s$ is isomorphic to $Y\times_{\text{Spec}\ K}\text{Spec}\ \mathbb{C}$? $\endgroup$ Nov 21 '17 at 10:40
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    $\begingroup$ @Jason Starr: by writing $\mathbf{C}$ as a direct limit of finitely generated field extensions of $\mathbf{Q}$ and since $\mathcal{X}_s$ is finitely presented over $\mathbf{C}$, the answer is positive for $K$ finitely generated over $\mathbf{Q}$, but is it for $K/\mathbf{Q}$ finite, or even $K = \mathbf{Q}$? Not so clear. $\endgroup$
    – user113393
    Nov 21 '17 at 10:49
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    $\begingroup$ When I write "the answer to the question is positive", I mean the second and third questions, not the first (which is definitely not positive). I will write an answer below. $\endgroup$ Nov 21 '17 at 11:19
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    $\begingroup$ (If (a.3) has positive answer, then clearly (a.2) does too, by spreading out.) $\endgroup$
    – user95222
    Nov 21 '17 at 11:28
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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.

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