Lemma 76.11.3. 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

where on the right hand side we use multiplication in the group. By Lemma 76.11.2 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 76.9.10 and the fact that $G$ is locally of finite type over $k$. $\square$

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