Let $k$ be an algebraically closed field and $k_0$ a sub-field. Let $X,Y$ be two projective varieties defined over $k_0$. Suppose that that there exists a dominant morphism $f$ between $X_k=X\otimes k$ and $Y_k=Y\otimes k$.

Such a dominant morphism cannot always been defined over a finite extension of $k_0$. Take by exemple an elliptic curve $X=Y$ defined over $k_0=\mathbb{Q}$, $k=\mathbb{C}$ and $f$ a translation by a point not in the algebraic closure of $\mathbb{Q}$.

But it is still natural to ask the following: With the above hypothesis on the existence of $f$, do there always exists a dominant morphism $g$ between $X$ and $Y$ defined over a finite extension $k_1$ of $k_0$ ?

According to the appendix 1 of Mumford’s book « Abelian Varieties », the answer is yes if $X$ and $Y$ are Abelian varieties. For the proof, using a translation, one reduces to the case of an homomorphism. Then the (dense set of) torsion elements of $X$ are sent to torsion elements of $Y$, therefore the homomorphism is defined over a finite extension.

The last implication is a bit obscure for me, maybe the following can help: field of definition of isogenies of abelian varieties ? Anyway I would like to understand the general case, in which one can also replace the word "dominant" by "isomorphism" or "birational morphism"...


2 Answers 2


The general technique goes as follows. Let $\{A_i\}$ be a directed system of rings with limit $A$; e.g., $A=k$ and $\{A_i\}$ the set of finitely generated $k_0$-subalgebras of $k$. Let $X$ and $Y$ be schemes of finite presentation over some $A_{i_0}$, and define $X_i = X \otimes_{A_{i_0}} A_i$ for $i \ge i_0$, $X_A = X \otimes_{A_{i_0}} A$, and similarly for $Y_i$ (with $i \ge i_0$) and $Y_A$. There is a natural map of sets $$\varinjlim {\rm{Hom}}_{A_i}(X_i,Y_i) \rightarrow {\rm{Hom}}_A(X_A,Y_A)$$ and it is bijective by EGA IV$_3$, 8.8.2(i).

So given an $A$-morphism $f:X_A \rightarrow Y_A$, there exists some $i_1$ and an $A_{i_1}$-morphism $f_{i_1}:X_{i_1} \rightarrow Y_{i_1}$ which descends $f$. But we want more: for various properties $\mathbf{P}$ of morphisms of schemes, if $f$ satisfies $\mathbf{P}$ then we want $f_i := f_{i_1} \otimes_{A_{i_1}} A_i$ to also satisfy $\mathbf{P}$ for some $i \ge i_1$. The property "isomorphism" is immediate by applying the preceding formalism to the inverse $A$-morphism too (using bijectivity in the displayed map of sets above). See IV$_3$ 8.10.5 for the tip of the iceberg on many possibilities for $\mathbf{P}$. By IV$_3$, 8.6.3, open subschemes of $X_A$ with finitely presented complement also descend to such open subschemes of some $X_i$.

For $i \ge i_1$, the (set-theoretic) image $Z_i = f_i(X_i) \subset Y_i$ is constructible and if $j \ge i$ then $Z_j$ is the preimage of $Z_i$ under $Y_j \rightarrow Y_i$ and likewise $f(X_A)$ is the preimage of $Z_i$ in $Y_A$. Define $E_i$ to be the set of of $s \in S_i := {\rm{Spec}}(A_i)$ such that $(Z_i)_s$ is dense in $(X_i)_s$, and likewise for an analogous subset $E \subset S := {\rm{Spec}}(A)$. The density or not of a constructible subset of a scheme of finite type over a field is insensitive to extension of the ground field, so for $j \ge i$ the preimage of $E_i$ under $S_j \rightarrow S_i$ is $E_j$ (due to the analogue for $Z_i \subset Y_i$ and $Z_j \subset Y_i$ under $Y_i \rightarrow Y_i$ noted above) and likewise the preimage of $E_i$ under $S \rightarrow S_i$ is $E$. But $E_i$ is constructible in $S_i$ by IV$_3$ 9.5.3 and likewise for $E$ inside $S$. Thus, if $E=S$ then $E_i = S_i$ for all large $i$ due to IV$_3$ 8.3.5.

Thus, in the initial setup of interest, if $X_k \rightarrow Y_k$ is dominant then we get a finitely generated $k_0$-subalgebra $R \subset k$ and an $R$-morphism $h:X_R \rightarrow Y_R$ which is dominant between fibers over all points of Spec($R$). In a similar manner, using the "spreading out" for open subschemes (with finitely presented complement) mentioned above, if there are dense open subschemes of $X_k$ and $Y_k$ between which $f$ restricts to an isomorphism, then by increasing $R$ further we can arrange that there exist fiberwise dense open subschemes $U \subset X_R$ and $V \subset Y_R$ between which $h$ restricts to an isomorphism. Finally, thanks to the Nullstellensatz over $k_0$, using the pullback of $h$ between fibers over any closed point of such a Spec($R$) allows us to conclude for the properties of dominance, isomorphism, or birational morphism.

This overall method is sometimes called the "Principle of finite extensions" (i.e., whatever happens after some extension of the ground field already happens over a finite extension), and complete proofs of virtually every such property you could ever imagine wanting is rigorously documented in remarkable and useful generality in EGA IV$_3$, $\S8$-$\S11$ (skip $\S10$) and IV$_4$, $\S17$-$\S18$.


The same trick, spreading out, should work for all of these. Write down all the equations defining your morphism. Look at all the coefficients. Take the algebra generated by these over $k_0$. Each point defines a morphism. Check that the set of points defining a morphism with the properties you want is constructible. Because it contains the generic point, it contains some open set, and hence contains a point defined over a finite extension of $k_0$. Win.

To check dominance is constructible, use the fact that the image of the family of maps is constructible, so the set where that image is dense is constructible as well.

  • 2
    $\begingroup$ The sentence "Each point defines a morphism" is vague, since the sentence "Write down all the equations defining your morphism" is vague. It is better to provide more precise statements/arguments or accurate literature references. $\endgroup$
    – grghxy
    Jul 19, 2015 at 17:20
  • $\begingroup$ @Will "Each point defines a morphism" do you mean each point of Spec of your algebra over $k_0$? $\endgroup$ Jul 19, 2015 at 19:01
  • $\begingroup$ @FrançoisBrunault For any field $K$ each $K$-point should define a morphism over $K$, and thus each point of the Spec should define a morphism over a residue field. $\endgroup$
    – Will Sawin
    Jul 19, 2015 at 19:03
  • $\begingroup$ @Will, thanks for the clarification. $\endgroup$ Jul 19, 2015 at 19:04
  • 1
    $\begingroup$ @WillSawin: I agree that working out some aspects of this for oneself is instructive. But actually working out all of the details (say even in the given setup over fields) can become somewhat involved if $k_0$ is not algebraically closed and the "varieties" may are not geometrically integral, when keeping track of density and open subschemes initially given to us only upstairs in such "spreading out" arguments. That is, instead of needing to reinvent the wheel every time, it is useful to alert the asker that a large body of such results has been throughly documented in the literature. $\endgroup$
    – grghxy
    Jul 19, 2015 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.