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I not understand an assumption done at the beginning of the proof of Rigidity lemma in Moonens and van der Geers book about Abelian variaties (page 12). Here is it:

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Question: Why the assumption $k= \overline{k}$ is legit? That is, if we take our varieties $X,Y,Z$ over $k$, build the fiber bundles $X \times_k \overline{k}, Y \times_k \overline{k}, Z \times_k \overline{k}$. Recall, by abuse of notation we mean by $X \times_k \overline{k}$ formally more correctly $X \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}$.

assume we have proved the claim for $X \times_k \overline{k}, Y \times_k \overline{k}, Z \times_k \overline{k}$. How we can descent the claim for initial varieties $X,Y,Z$? I think that the author's had previously a Galoi-descent argument in mind but up to now I failed in working it out correctly.

More precisely, the essence of descent theory is that if we want to verify a property of a morphism, we can also do base change to $\overline{k}$ and verify the same property of the resulted new morphism, see eg here. Unfortunately, here we have a factorization problem and thus we haven't from the starting point a morphism of $k$-varieties which we can pullback to a morphism over $\overline{k}$-varieties. On the other hand, if we simply pullback $f$ and build a factorization over $\overline{k}$ I don't see how this factorization can be descent down through $f$.

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    $\begingroup$ There is a candidate morphism. You can pick a k rational point x in X (e.g. identity when X is ab.var) and use this to consider k-morphism $Y\rightarrow X\times Y\rightarrow Z$. $\endgroup$
    – GTA
    Commented Feb 26, 2020 at 18:29
  • $\begingroup$ @GTA: The lemma not assumes that $X$ is Abelian variety, thus I'm not sure why we can assume that $X$ has a rational point if $k$ is not algebraically closed (e.g. $x^2+y^2+1$ over $\mathbb{R}$). $\endgroup$
    – user267839
    Commented Feb 26, 2020 at 20:13
  • $\begingroup$ If the map factors through the projection, it factors uniquely (the projection is dominant) . So the factorization is automatically galois invariant. In other words factoring is a property, not extra data, so it can be checked after base change. $\endgroup$
    – Phil Tosteson
    Commented Feb 27, 2020 at 1:55
  • $\begingroup$ @user7391733 I was trying to give the simplest reason why it should hold in your context without trying to invoke actual Galois descent. Having a candidate object over $k$ is too simple to be called Galois descent. It is usually rather about whether an object over $\overline{k}$ satisfying compatibility condition actually descends down to an object (not defined a priori) over $k$. This case is the case of descent of morphism of quasicoherent sheaves so it is always effective. $\endgroup$
    – GTA
    Commented Feb 27, 2020 at 2:09
  • $\begingroup$ @PhilTosteson: one remark on what you mean by Galois-action on morphisms: When you talking about Galois invariant morphism $h$(in our case the factorization map) between $\bar{k}$-schemes $A,B$, which action you concretly mean? ie how the Galois group "act" here on a morphism $h: A \to B$? Do we have in this context "the action" (ie a canonical one)? The only one with which I'm familar works as follows: Since $h$ incuces a map between $\bar{k}$-valued points $h(\bar{k}): A(\bar{k}) \to B(\bar{k})$ defined by composing $h$ with morphisms $\operatorname{Spec} \ \bar{k} \to A$. $\endgroup$
    – user267839
    Commented Feb 27, 2020 at 2:59

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