TITLE: Defining abstract varieties and their morphisms over a finitely generated subfield of the base field
QUESTION [12 upvotes]: Let $k$ be an algebraically closed field.
By a finitely generated subfield of $k$ I mean a subfield $k_0\subset k$ that is finitely generated over the prime subfield of $k$ (that is, over $\mathbb Q$ or $\mathbb F_p$).
I need a reference or a proof for the following (well-known? evident?) proposition:

Proposition.
Let
  $$f\colon X\to V$$
  be a morphism of $k$-varieties.
  Then the triple  $(X,Y,f)$ can be defined over a finitely generated subfield of $k$.
  In other words, there exists a finitely generated subfield $k_0\subset k$
  and a morphism of $k_0$-varieties
  $$f_0\colon X_0\to Y_0$$
  such that
  $(X_0,Y_0,f_0)\times_{k_0} k$ is isomorphic to $(X,Y,f)$.

A referee asked me to prove this assertion. I know how to prove this for affine and projective varieties, but not for abstract varieties.

REPLY [10 votes]: A variety $V$ is a finite union of open affine varieties $V_i$, and because $V$ is separated (usually part of the definition of variety) the intersections $V_i\cap V_j$ are also affine. Now $V$ can be reconstructed from the affine varieties $V_i,V_i\cap V_j$ and the maps of affine varieties $V_i\cap V_j\to V_i$. Obviously, this system is defined over a finitely generated field.
Hence $X$ and $V$ are defined over a finitely generated field. The graph of $f$ is a closed subvariety of $X\times V$, and so is defined by a coherent sheaf of ideals in the structure sheaf of $X\times V$, which is obviously defined over a finitely generated field.