Lemma 79.11.3. 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).
No complete rational curves on groups.
Proof. Assume f is nonconstant. Consider the morphisms
\mathbf{P}^1_ k \times _{\mathop{\mathrm{Spec}}(k)} \ldots \times _{\mathop{\mathrm{Spec}}(k)} \mathbf{P}^1_ k \longrightarrow G, \quad (t_1, \ldots , t_ n) \longmapsto f(g_1) \ldots f(g_ n)
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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