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I try to understand a reduction step the proof of rigidity lemma as proved in Moonen's and van der Geer's Abelian varieties (Lemma 1.11 on page 12):

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 $pr_Y : X \times_k Y → Y$.

The proof starts with reduction to $k=\bar{k}$. Why the reduction to the case $k=\bar{k}$ is allowed?

I supposed at first glance a pure category theoretical argument in mind, but it fails: Assume we can show that after base change $ - \times \bar{k}$ we prove that $(X \times Y) \times \overline{k} \to Z \times \overline{k}$ factors through the projection to $Y \times \bar{k}$, does the original morphism $f : X × Y \rightarrow Z$ factor through projection to $Y$? I don't see why this can be deduced from universal property of fiber product.

p.s.: I noticed that here was asked amongst other things the same question, but the accepted answer don't satisfy me: a "diagram chase" as suggested in an answer gives a factorisation of $f$ through $Y$ on underlying sets. There is no hint why the obtained map between sets is a morphism (in category of algebraic varieties).

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In term of $k$-algebras: Suppose $\lambda \colon A \to B \otimes C$ is a $k$-algebra homomorphism ($\lambda$ corresponds [locally] to $f$). Choose $k$-bases $b_i$ and $c_j$ for $B$ and $C$, with $1$ belonging to (both) bases, and identify $B\otimes_k C$ with $B\otimes_k C \otimes_k k$. The terms $b_i\otimes c_j$ form a $k$-basis for $B\otimes C$.

If $a \in A$, then the choice of bases determines $x_{i,j}\in k$ uniquely (almost all zero) to satisfy $$ \lambda (a) = \sum b_i\otimes c_j \otimes x_{i,j}.$$ We want, in the above expression, that the only basis elements $b_i\otimes c_j$ with non-zero coefficients $x_{i,j}$ be of the form $1\otimes c_j$.

Now, identify $A$ with its image $A\otimes 1$ in $A\otimes \overline k$, and $B\otimes_k C$ with its image in $ B\otimes_k C\otimes_k \overline k$. The elements $b_i\otimes c_j$ still form a $\overline k$ basis for $B\otimes C \otimes \overline k$. By hypothesis, we have the desired expression in $B\otimes C \otimes \overline k$: $$ \lambda (a) = (\lambda \otimes 1) (a \otimes 1 ) = \sum 1 \otimes c_j \otimes x_j,$$ with $x_j \in \overline k$. But, since the two expressions for $\lambda (a)$ are given in terms of the same basis, the expressions must be the same. Hence, the only non-zero coefficients $x_{i,j}$ correspond to $b_i = 1$.

peter a g
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  • Hi Peter, that's a great answer. Your answer bases on exploitation of vector space structure of involved objects. the only not completly solved facet of the problem is if the statement definitely not solvable only with abstract nonsense arguments (e.g. like diagram chasing) as suggested without precise explanations in the answer in https://math.stackexchange.com/questions/838434/proof-of-rigidity-lemma? I'm not sure if his argument simply wrong or I not see the allegedly simple way to see it as purposed by the answerer. –  Oct 10 '19 at 21:18
  • Just like that, (putting aside any thought to other hypotheses in the lemma) I don't think a diagram chase would do it, without another ingredient. Otherwise, it seems to me that such an argument would work with arbitrary base change, e.g., $k ={\mathbb Z}$, and $\overline k = {\mathbb F_p}$. But then, for instance, if $Z = \text{spec}\ A$, $A= {\mathbb Z}[1/p, x] \times {\mathbb F_p}$, one has $Z\times_k {\overline k}$ is a point, so there is only one map to it, and one can conclude nothing from that??? – peter a g Oct 11 '19 at 01:10