In this section we prove that there are no nonconstant morphisms from \mathbf{P}^1 to a group algebraic space locally of finite type over a field.
Lemma 79.11.1. Let S be a scheme. Let B be an algebraic space over S. Let f : X \to Y and g : X \to Z be morphisms of algebraic spaces over B. Assume
Y \to B is separated,
g is surjective, flat, and locally of finite presentation,
there is a scheme theoretically dense open V \subset Z such that f|_{g^{-1}(V)} : g^{-1}(V) \to Y factors through V.
Then f factors through g.
Proof. Set R = X \times _ Z X. By (2) we see that Z = X/R as sheaves. Also (2) implies that the inverse image of V in R is scheme theoretically dense in R (Morphisms of Spaces, Lemma 67.30.11). The we see that the two compositions R \to X \to Y are equal by Morphisms of Spaces, Lemma 67.17.8. The lemma follows. \square
Lemma 79.11.2.slogan 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
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
Lemma 79.11.3.slogan Let k be a field. Let G be a separated group algebraic space locally of finite type over k. There does not exist a nonconstant morphism f : \mathbf{P}^1_ k \to G over \mathop{\mathrm{Spec}}(k).
Proof. Assume f is nonconstant. Consider the morphisms
where on the right hand side we use multiplication in the group. By Lemma 79.11.2 and the assumption that f is nonconstant this morphism is finite onto its image. Hence \dim (G) \geq n for all n, which is impossible by Lemma 79.9.10 and the fact that G is locally of finite type over k. \square
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