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. 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
\[ (\mathbf{P}^1_ k)^ n \xrightarrow {projection} (\mathbf{P}^1_ k)^ m \xrightarrow {finite} Z. \]
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. 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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