Skip to main content
6 events
when toggle format what by license comment
Feb 17, 2021 at 22:48 comment added Ben Wieland The fundamental group does not respect products. That requires either completeness or coprime characteristic, so it's ok here, but it's not true in general. If it were true, then the fundamental group of a group scheme would be abelian. But it's not for $\mathbb A^1$. @Faris
Feb 17, 2021 at 18:03 comment added R. van Dobben de Bruyn See for example Tag 0BTX combined with the computation of $\pi_1(\mathbf P^n_{\bar k})$. (See here for a fun Riemann–Hurwitz-free proof when $n=1$.)
Feb 17, 2021 at 14:09 comment added Faris The fundamental group of $\mathbb{P}^n_{\overline{\mathbb{F}_q}}$ can be computed by noting that the fundamental group $\mathbb{P}^1_{\overline{\mathbb{F}_q}}$ is trivial, that the fundamental group of a direct product is the direct product of the fundamental groups and that fundamental group is a birational invariant.
Feb 17, 2021 at 10:57 review Close votes
Feb 18, 2021 at 21:25
Feb 17, 2021 at 10:38 comment added abx This is just the Galois group $\operatorname{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q) $.
Feb 17, 2021 at 9:56 history asked hennlu CC BY-SA 4.0