If $X$ is reduced scheme then same morphism on k-rational points is the same.

Question

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?

Answer 1

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.