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Jul 28, 2019 at 7:36 comment added Carot Ok thank you! I wasn't loking for something this simple because I thought my idea worked on closed points. I will have to find a way around it then.
Jul 28, 2019 at 5:49 comment added Laurent Moret-Bailly New contributor, welcome! Take $Y=\mathbb{P}^2$, and $f=$ the bolwing-up at a rational point $y$. Then $f^{-1}(y)\cong \mathbb{P}^1$ which contains plenty of points with inseparable residue field.
Jul 27, 2019 at 18:08 history edited Martin Sleziak
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
Jul 27, 2019 at 15:45 review First posts
Jul 27, 2019 at 16:07
Jul 27, 2019 at 15:44 history asked Carot CC BY-SA 4.0