Reduction step to $k=\bar{k}$ in the proof of rigidity lemma

Question

I do not understand the following proof in the paper Abelian varieties by Edixhoven, van der Geert, and Moonen:

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

Proof. We may assume that $k = \bar k$. Choose a point $x_0 \in X(k)$, and define a morphism $g: Y \to Z$ by $ g(y) = f(x_0,y)$ (here is the passage to to algebraic closed field involved, since otherwise we may not find such point $x_0$…).

Question: Why $ k$ can be assumed to be algebraically closed? Assuming the lemma has been proved for fibre products $X_{\bar{k}}$, $Y_{\bar{k}}$, $Z_{\bar{k}}$, how can we derive the statement for schemes over not algebraically closed $k$? It appears that there should be a little lovely diagram chase involved but I do not know how finally to construct the morphism $ Y \to Z$.

Finally there is the machinery of fppf-descent which justifies immediately the reduction to $k=\overline{k}$. But I would like to know if this can be also showed with elemenary methods — presumably a (tricky?) diagram chase. It looks rather similar to the problem Proof of rigidity lemma I posted recently, and there it turned out that the reduction to $k=\overline{k}$ can be justified by simple diagram chase. Is it here also possible to argue in similar way without ‘deep’ methods?

I apologise for my edit mathoverflow.net/revisions/440411/2 in which a misplaced keystroke led me to put "do" at the beginning of the sentence rather than as the 2nd word. I see you deleted it, which is quite the right action for where I had wrongly put it. Since there was another typo (a missing right parenthesis), I restored it to its proper place and fixed that other typo. — LSpice
– LSpice, Commented Feb 15, 2023 at 2:00

R. van Dobben de Bruyn · Accepted Answer · 2023-02-14 23:55:28Z

One way to argue is as follows: given a morphism $f \colon X \times Y \to Z$, consider its graph $\Gamma_f \subseteq X \times Y \times Z$. Let $W$ be the scheme-theoretic image of $\Gamma_f$ under the projection $\pi_{Y \times Z} \colon X \times Y \times Z \to Y \times Z$, and let $X \times W \subseteq X \times Y \times Z$ be the preimage of $W$ under $\pi_{Y \times Z}$.

Then we have $\Gamma_f \subseteq X \times W$, and the statement that $f$ factors via $\pi_Y \colon X \times Y \to Y$ is equivalent to the statement that $\Gamma_f = X \times W$. Indeed, if $f$ factors through a morphism $g \colon Y \to Z$, then $W = \Gamma_g$ and $\Gamma_f = X \times W$. Conversely, if $\Gamma_f = X \times W$, then the projection $\pi_Y \colon W \to Y$ is an isomorphism since the same holds after multiplying with $X$ (this is a form of fppf descent, I suppose ― if you want you can choose a closed point in $X$ to reduce to the case of base change along a finite field extension). Thus $W$ is the graph of a morphism (namely $Y \underset{\pi_Y}{\stackrel\sim\leftarrow} W \underset{\pi_Z}\to Z$).

All of the above constructions are defined over $k$. Formation of scheme-theoretic image commutes with flat base change (on the level of sheaves, it is given by image factorisation $\mathcal O_{Y \times Z} \twoheadrightarrow \mathcal O_W \hookrightarrow \pi_{Y \times Z,*} \mathcal O_{X \times Y \times Z}$), and checking whether an inclusion of subschemes is an equality can be done over $\bar k$. $\square$

There is probably also a commutative algebra argument, but your suggestion of a diagrammatic proof led me to the one presented above. (The local computation is that checking whether $C \to A \otimes B$ lands in $k \otimes B$ can be done over $\bar k$, which should be clear by considering the map of $k$-vector spaces $C \to (A/k) \otimes B$. But reducing to the affine case is a bit annoying, because the inverse image of an affine open in $Z$ need not be of the form $U \times V$ for $U \subseteq X$ and $V \subseteq Y$ open.) — R. van Dobben de Bruyn
– R. van Dobben de Bruyn, Commented Feb 15, 2023 at 0:38

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