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I am reading ABELIAN VARIETIES(Gerard van der Geer and Ben Moonen) and I don't understand some part of the proof of rigidity lemma. The statement is :

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

They define a morphism $g: Y \rightarrow Z$ by $g(y)$ = $f(x_0, y)$ in the proof and would like to proof $f=g \circ pr_Y$. And so they write that "As $X \times Y$ is reduced it suffices to prove this on $k$-rational points." but I don't understand this. How can this be proven?

user816697
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Note that in the proof of the Rigidity Lemma, it is assmued WLOG that $k$ is algebraically closed. Furthermore by definition (0.4) a variety is assumed to be geometrically reduced.

Then you can simply apply the following result:

Lemma: Let $X,Y$ be two finite type $k$-schemes and assume $X$ is geometrically reduced. Given two $k$-morphisms $f,g; X\to Y$ such that $f(\bar{K})=g(\bar{K})$, we have an equality $f=g$ of morphisms as schemes.

Notone
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  • I'm sorry, I still don't understand. To use this lemma for $f,g \circ pr_Y; X \times Y \to Z$, it is necessary that $X \times Y$ is geometrically reduced. $X$ is geometrically integral, especially geometrically reduced so, since one can assume $k$ as algebraically closed, for algebraically closed field $k$ containing $k$ $X \times Spec(k)$ is reduced. Similarly $Y \times Spec(k)$ is reduced. From here can it be proof that $X \times Y$ is geometrically reduced? – user816697 Aug 11 '21 at 13:27
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    It is a general fact that the fiber product of two varieties is again a variety (i.e. the property of being geometrically reduced is preserved). See here https://math.stackexchange.com/questions/152056/is-fibre-product-of-varieties-irreducible-integral – Notone Aug 12 '21 at 09:26
  • Thank you for explaining it so courteously. $X$ and $Y$ is variety, so $X \times Y$ is variety, especially geometrically reduced. Hence one can use this lemma. The statement that "As $X \times Y$ is reduced" is properly rather the statement that "As $X \times Y$ is geometrically reduced". Now I understand. – user816697 Aug 12 '21 at 10:31