Skip to main content
11 events
when toggle format what by license comment
Oct 10, 2021 at 17:55 comment added Donu Arapura @R.vanDobbendeBruyn I guess I missed that.
Oct 10, 2021 at 17:43 history became hot network question
Oct 10, 2021 at 17:05 comment added R. van Dobben de Bruyn @DonuArapura the post states "over $\operatorname{Spec} R$", i.e. as $R$-schemes.
Oct 10, 2021 at 16:55 comment added Donu Arapura If by a "map" you mean a morphism of schemes, and not necessarily $R$-schemes, then there other possibilities. Simply take an etale morphism $R\to R$ and base change to $\mathbb{P}^1_R$
Oct 10, 2021 at 12:45 answer added R. van Dobben de Bruyn timeline score: 4
Oct 10, 2021 at 11:52 comment added R. van Dobben de Bruyn @Jef since $\mathbf P^1_R$ is proper over $R$, the map $f$ is proper as well. Since everything is of finite type, $f$ is quasi-finite, and therefore finite (Tag 02LS).
Oct 10, 2021 at 10:50 answer added Martin Skilleter timeline score: 2
Oct 10, 2021 at 10:29 comment added Jef Do you assume that $f$ is finite?
Oct 10, 2021 at 9:48 history edited Alex M. CC BY-SA 4.0
added 70 characters in body; edited tags
S Oct 10, 2021 at 9:42 review First questions
Oct 10, 2021 at 9:48
S Oct 10, 2021 at 9:42 history asked FiniteEtaleP1 CC BY-SA 4.0