11

If I am reading this post correctly, any smooth projective $\mathbb{C}$-morphism of schemes $X\rightarrow \mathbb{P}^1$ admits a section. I am afraid of the topological argument presented there. Is there an algebraic proof of this fact? What happens over fields other than $\mathbb{C}$?

  • 2
    If $k$ is an algebraically closed field of characteristic zero and $f:X\to \mathbb{P}^1$ is a smooth projective morphism then $f$ will have a section. You can prove this by "reducing" to the case that $k=\mathbb{C}$. This is an example of the so-called "Lefschetz principle". To get a feeling of how this goes, assume for simplicity that $k\subset \mathbb{C}$. Now, consider a section $\sigma:\mathbb{P}^1_{\mathbb{C}}\to X_{\mathbb{C}}$ of $f_{\mathbb{C}}$. – Ariyan Javanpeykar Jul 12 '19 at 15:58
  • 2
    Now, use "spreading out" (of your section) over a variety over $k$, and then specialize the spread-out section to get a section of $f$. – Ariyan Javanpeykar Jul 12 '19 at 15:59
  • It is not so clear to me what one should expect when $k=\overline{\mathbb{F}_p}$. – Ariyan Javanpeykar Jul 12 '19 at 16:10
  • Since any smooth morphism admits a section locally in the étale topology, I am wondering if one can not prove this using: i) the simply connectedness of $\mathbb{A}^1$ ii) the fact that any rational map between projective curves can be extended to an actual morphism? – Libli Jul 15 '19 at 08:29

0 Answers0