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I try to understand the proof of an extension theorem proved in Moonen's and van der Geer's draft Abelian varieties(in online accessible notes Theorem 1.18 on page 14):

(1.18) Theorem.

Let $X$ be an abelian variety over a field $k$. If $V$ is a smooth $k$-variety then any rational map $f: V \dashrightarrow X$ extends to a morphism $V \to X$.

Proof:

We may assume that $k = \bar{k}$, for if a morphism $V_{\overline{k}}= V \times_k \overline{k} \to X_{\overline{k}}$ is defined over $k$ on some dense open subset of $V_{\overline{k}}$, then it is defined over $k$. Let $U \subset V$ be the maximal open subset on which $f$ is defined. Our goal is to show that $U = V$. [...]

Questions: the assumption $k = \bar{k}$ I not understand. if we succeed to prove that $V_{\overline{k}} \to X_{\overline{k}}$ extends to a morphism, why $f$ is also extendable? Afterwards, if we proof $V_{\overline{k}} \to X_{\overline{k}}$ extends to a morphism, why if it is defined over $k$ on some dense open subset of $V_{\overline{k}}$ (=preimage of a dense open $U \subset V$ under projection $V_{\overline{k}} \to V$), then it is defined over $k$?

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    $\begingroup$ Suppose we know what we want over $\overline{k}$. Then we can descend this to a finite Galois extension $L/k$, i.e., $U_L\to X_L$ extends to a morphism $V_L\to X_L$. Now, we consider the action of $G= Gal(L/k)$ on the morphism $V_L\to X_L$. Note that the restriction morphism $U_L\to X_L$ is fixed. Since $U_L\subset V_L$ is dense open, this implies that $V_L\to X_L$ is in fact fixed by $G$, so it descends to a morphism $V\to X$. For another proof of this extension property: see Lemma 3.5 of arxiv.org/abs/1807.03665 and use the fact that abelian varieties have no rational curves. $\endgroup$ – Ariyan Javanpeykar Oct 10 at 17:39
  • $\begingroup$ @AriyanJavanpeykar: This is a nice idea, thank you a lot. a little remark: lastly I asked a similar question in a slightly other context about factorization problem reduced to algebraically closed ground field: math.stackexchange.com/questions/3385478/…. although I got a good answer I'm keenly curious if this reduction to $k=\bar{k}$ can also be solved with a descent argument similar to that you used above? My first suspicion was with Stein factorization $\endgroup$ – katalaveino Oct 11 at 12:51
  • $\begingroup$ I can assume that fibers are geometrically connected and could "somehow go down" using properness (especially closedness of the map, but from there I could not proceed ahead. $\endgroup$ – katalaveino Oct 11 at 12:51

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