I not understand the following proof in the paper Abelian varieties on 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 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?

do I not understand the following proof in the paper Abelian varieties on 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 not understand the following proof in the paper Abelian varieties on 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 not understand folowing proof in this paper on abelian varieties by Moonen & van der Geert:

(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 $z ∈ Z(k)$ then $f$ factors through the projection $\text{pr}_Y : X \times Y \to Y$.

Proof. We may assume that $k = k$. 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 a little lovely diagram chase be involved but I not know how finally to construct the morphism $ Y \to Z$.

Finally there is the maschinery of fppf-descent which justify 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?

do I not understand the following proof in the paper Abelian varieties on 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?