Lemma 79.11.2. Let k be a field. Let n \geq 1 and let (\mathbf{P}^1_ k)^ n be the n-fold self product over \mathop{\mathrm{Spec}}(k). Let f : (\mathbf{P}^1_ k)^ n \to Z be a morphism of algebraic spaces over k. If Z is separated of finite type over k, then f factors as
A morphism from a nonempty product of projective lines over a field to a separated finite type algebraic space over a field factors as a finite morphism after a projection to a product of projective lines.
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
Comments (1)
Comment #887 by Konrad Voelkel on