# Rigidity lemma over non-algebraically closed field

I would like to extend the rigidity lemma (as in Mumford's "Abelian varieties") to the case in which the base field $k$ is not algebraically closed.

I found a suitable proof in the draft of "Abelian varieties" by van der Geer and Moonen, and now I would like to undestand this proof in detail. It can be found here http://staff.science.uva.nl/~bmoonen/boek/DefBasEx.pdf [Lemma 1.11, p. 12].

The statement in this reference is

Lemma Let $X$, $Y$ and $Z$ be algebraic varieties over a field $k$. Suppose that $X$ is complete. If $f : X × Y \rightarrow Z$ is a morphism with the property that, for some $y \in Y (k)$, the fibre $X \times_k \{y\}$ is mapped to a point $z \in Z(k)$ then $f$ factors through the projection $pr_Y : X \times_k Y → Y$.

I have the following questions concernig the proof. (For questions 4 and 5 I am following the notation in the proof in the link).

1. Why we can reduce to the case $k=\bar{k}$?
2. Why can we choose $x_0\in X(k)$, that is, why $X(k)$ is non-empty?
3. Which result allow me to say that since $X\times_k Y$ is reduced it sufficies to prove the statement for $k$-rational points?
4. Why $f$ has to be constant on $X\times_k \{P\}$?
5. At the end they say: $f=g \circ pr_Y$ on the non-empy open set $X\times_k (Y-V)$. So $f=g \circ pr_Y$ everywhere because $X\times_k Y$ is irreducible. Why?

For question (5) I would argue like this: since $X\times_k Y$ is irreducible, $X\times_k (Y-V)$ is a dense subset. Then by continuity $f=g \circ pr_Y$ everywhere. Is this argument correct?

Any help is appreciated, thank you.

Note I posted a question almost identical to this on MSE, and even putting a bounty on it I did not get any answer. I think that the level of the question is reasonably high to propose it here. Please, if I am wrong let me notice. (Here is the link to the original question http://math.stackexchange.com/questions/838434/proof-of-rigidity-lemma).

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1. This is a bit of an advanced technique for this application, but the morphism $\operatorname{spec}\overline{k} \rightarrow \operatorname{spec}k$ is a fpqc-covering(faithfully flat and quasi-compact), and so as a very simple case of fpqc-descent, the "base change map" $\operatorname{Hom}(A,B) \rightarrow \operatorname{Hom}(A_{\overline{k}},B_{\overline{k}})$ is injective. This should be enough, and you probably don't have to appeal to fpqc-descent.

2. You can choose a $k$-rational point because you assumed $\overline{k}=k$, but keep in mind we want to apply this to abelian varieties which by definition have at least one rational point.

3. Some searching found this: http://math.stackexchange.com/questions/630465/morphisms-of-k-schemes-who-agree-on-overlinek-points

4. $f$ is mapping $X \times P \cong X$ into an affine variety. A proper variety has no non-constant maps to any affine space. This is an exercise in Hartshorne, presumably in the section about separatedness and properness.

5. I think that's fine for showing the morphisms agree topologically, but one must also worry about the structure sheaf. This is really addressed in the MSE link above.

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Dear @Joe B, How does (1) show that it is enough to prove the result over $\overline{k}$? The injectivity of the map $\mathrm{Hom}(A,B)\to\mathrm{Hom}(A_{\overline{k}},B_{\overline{k}})$ is the easy part of fpqc descent, but the issue is with whether or not the map you produce over $\overline{k}$ is in the image of this map. Descent theory says that it is if it respects the descent data on the $A_{\overline{k}}$ and $B_{\overline{k}}$. Using limit arguments one can go from $\overline{k}$ to a finite Galois $K$ extension of $k$, but then one must prove $\mathrm{Gal}(K/k)$-invariance of the –  Keenan Kidwell Jul 2 '14 at 18:33
morphism over $K$ to deduce that it descends all the way to $k$. –  Keenan Kidwell Jul 2 '14 at 18:33
My thought was much simpler, if $f_{\overline{k}}$ factors through the projection $p=\operatorname{pr}_{Y_{\overline{k}}}$ then I should be able to factor $f$ itself through $\operatorname{pr}_Y$. I thought it was fairly straightforward but now I see it's not entirely. If $f_{\overline{k}}= g \circ p$, then since $p$ descends(and is thus Galois-invariant) we should get that $g_{\sigma} \circ p = g_{\tau} \circ p$, but $p$ is epic so that $g$ is Galois invariant. Probably there is a nicer argument, but this is just what I came up with. –  Joe B Jul 3 '14 at 0:41