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Nov 3, 2014 at 22:41 vote accept Matthieu Romagny
Nov 3, 2014 at 22:40 comment added Matthieu Romagny OK, I get it. Thank you for this nice contribution. The particular situation that I have in mind has additional features (like quasicompactness) that your example doesn't, but it helped me anyway to understand things better. Thanks again! (And if you happen to have ideas in the quasicompact case...)
Nov 3, 2014 at 11:00 comment added user27920 @MatthieuRomagny: All local rings on $S$, and hence on $Y$, are $\mathbf{F}_2$ since everything in $A$ is idempotent.
Nov 3, 2014 at 9:51 comment added Matthieu Romagny Why it is true that $f$ is an iso on local rings at points of $Y$?
Nov 3, 2014 at 3:21 history edited user27920 CC BY-SA 3.0
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Nov 3, 2014 at 3:03 history answered user27920 CC BY-SA 3.0