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Timeline for smooth morphism on schemes

Current License: CC BY-SA 3.0

6 events
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Nov 13, 2012 at 23:20 answer added Qing Liu timeline score: 2
Nov 13, 2012 at 15:13 comment added Karl Schwede Matthieu, you are right. I missed the integral implies connected...
Nov 13, 2012 at 13:27 comment added Matthieu Romagny You write that geometric fibres of $C\to S$ are integral, hence connected ; thus $S'=S$ does the job, doesn't it ?
Nov 13, 2012 at 13:14 answer added Rob timeline score: -1
Nov 13, 2012 at 12:30 comment added Karl Schwede Probably you want $S' \to S$ to be a finite morphism, not just a map of finite type.
Nov 13, 2012 at 11:25 history asked Frank CC BY-SA 3.0