Let $\phi: X \dashrightarrow Y$ be a rational map
of two $k$-varieties. A "variety over field $k$" is here a
separated $k$-scheme of finite type which is geometrically integral.
Assume that if we base change to algebraic closure $\bar{k}$ of $k$
the induced map $\bar{\phi}: X_{\bar{k}} \to Y_{\bar{k}}$ becomes
a morphism of $\bar{k}$-varieties.
How can it be shown that this assumption implies that the rational map
$\phi: X \dashrightarrow Y$ was already a morphism?
This problem emerges in (Abelian varities, Theorem
1.17) as part of a reduction step.
what I tried before:
This is a local problem therefore we can assume
$X, Y$ to be affine with coordinate rings $R$ and $A$.
The map $\phi$ is rational means this is defned
on a dense open subset $U \subset X$ which
we can see as localization of $A$ and $R$. Let
$f \in A$ and assume the ring map
$\phi_f: A_s \to R_{s}$ exists.
We also assumed that
$\bar{\phi}: X_{\bar{k}} \to Y_{\bar{k}}$ is a
morphism and therefore there exist a map
$\bar{\phi}: A \otimes \bar{k} \to
R \otimes \bar{k}$. Then the image of
$A$ is contained in $R_{s} \cap B \otimes \bar{k}$.
But there seems no reason that
$R_{s} \cap R \otimes \bar{k}=R$.
Best Answer
First, we can replace $\overline{k}$ with a finite extension $K$ of $k$.
We know that $\mathcal{F}=\mathrm{Hom}_k(-,Y)$ is a fpqc-sheaf. We know that there is some $\phi’ \in \mathcal{F}(X_K)$ (which is the composition of $\overline{\phi}:X_K \rightarrow Y_K$ and $Y_K \rightarrow Y$).
As $X_K \rightarrow X$ is a fpqc cover, $\phi’$ comes from a morphism $\phi_b \in \mathcal{F}(X)$ iff $\phi’ \circ p_1=\phi’ \circ p_2$, where the $p_i$ are the projections $X_K \times_X X_K \rightarrow X$. It is then easy to see that such a $\phi_b$ is a morphism whose base change to $K$ is $\overline{\phi}$, and thus extends $\phi$.
Let $f_i=\phi’ \circ p_i$, and we want to show that $f_1=f_2$. Let $U \subset X$ be an open subset such that $\phi$ is defined on $U$. Then $\overline{\phi}_{|U_K}$ is a base change of $\phi$, so that $f_1$ and $f_2$ have the same restriction to $U_K \times_U U_K$. If $Z \in \{U,X\}$, $Z_K \times_Z Z_K \cong Z \times_k (K \otimes_k K)$ (functorially in $Z$), and the spectrum of $K \otimes_k K$ is a finite set of isolated points, so that $U_K \times_U U_K$ is an everywhere dense open subset of $X_K \times_X X_K$.
As $Y_K$ is separated over $k$, it follows that $f_1=f_2$ topologically and thus that, given any local section $s$ of $Y$, $f_1^{\sharp}s -f_2^{\sharp}s$ is a nilpotent section vanishing on $U_K \times_U U_K$.
To conclude, it is thus enough to show that if $A$ is a geometrically integral $k$-algebra (the ring of sections of any open affine subset $V$ of $X$) and $B$ is a localization of it (the ring of sections of some principal open subset of $V \cap U$), then the base change $A \otimes_k C \rightarrow B \otimes_k C$ is injective (where $C=K \otimes_k K$). But this is true because $C$ is flat over $k$.