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.


2 Answers 2


This is treated (in much greater generality) in EGA IV$_3$, Théorème 8.8.2. The existence of $X_0$ and $Y_0$ follows from part (ii), whereas the existence of $f_0$ is part (i).


[EGA IV$_3$] A. Grothendieck, Éléments de géométrie algébrique. IV: Étude locale des schémas et des morphismes de schémas. (Troisième partie). Publ. Math., Inst. Hautes Étud. Sci. 28, 1-255 (1966). ZBL0144.19904. DOI: 10.1007/BF02684343

  • 3
    $\begingroup$ If one prefers English, it is also in [Görtz-Wedhorn], Chapter 10, "Schemes over inductive limits of rings". $\endgroup$
    – user19475
    Oct 7, 2018 at 14:52

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.

  • 1
    $\begingroup$ Thank you, @anon! I have included your proof in my paper with a reference to "anonymous Mathoverflow user". $\endgroup$ Oct 6, 2018 at 13:57
  • 1
    $\begingroup$ Ex ungue leonem! $\endgroup$ Oct 6, 2018 at 13:59
  • $\begingroup$ Dear Mikhail: and who might this modern Newton be? $\endgroup$ Oct 29, 2018 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .