The Stacks project

Proof. We may assume $k$ is algebraically closed (details omitted); we only do this so we may argue using rational points, but the reader can work around this if she/he so desires. In the proof products are over $k$. The automorphism group algebraic space of $(\mathbf{P}^1_ k)^ n$ contains $G = (\text{GL}_{2, k})^ n$. If $C \subset (\mathbf{P}^1_ k)^ n$ is a closed subvariety (in particular irreducible over $k$) which is mapped to a point, then we can apply More on Morphisms of Spaces, Lemma 76.35.3 to the morphism

over $G$. Hence $g(C)$ is mapped to a point for $g \in G(k)$ lying in a Zariski open $U \subset G$. Suppose $x = (x_1, \ldots , x_ n)$, $y = (y_1, \ldots , y_ n)$ are $k$-valued points of $(\mathbf{P}^1_ k)^ n$. Let $I \subset \{ 1, \ldots , n\} $ be the set of indices $i$ such that $x_ i = y_ i$. Then

is Zariski dense in the fibre of the projection $\pi _ I : (\mathbf{P}^1_ k)^ n \to \prod _{i \in I} \mathbf{P}^1_ k$ (exercise). Hence if $x, y \in C(k)$ are distinct, we conclude that $f$ maps the whole fibre of $\pi _ I$ containing $x, y$ to a single point. Moreover, the $U(k)$-orbit of $C$ meets a Zariski open set of fibres of $\pi _ I$. By Lemma 79.11.1 the morphism $f$ factors through $\pi _ I$. After repeating this process finitely many times we reach the stage where all fibres of $f$ over $k$ points are finite. In this case $f$ is finite by More on Morphisms of Spaces, Lemma 76.35.2 and the fact that $k$ points are dense in $Z$ (Spaces over Fields, Lemma 72.16.2). $\square$