Timeline for Purely inseparable $k$-rational dominant maps between an absolutely irreducible $k$-surface and $\mathbb{P}^2$
Current License: CC BY-SA 3.0
9 events
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S Sep 11, 2016 at 17:15 | history | suggested | Edward Teach | CC BY-SA 3.0 |
If map is not regular, one should use \dashrightarrow .
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Sep 11, 2016 at 16:35 | review | Suggested edits | |||
S Sep 11, 2016 at 17:15 | |||||
Sep 10, 2016 at 14:23 | comment | added | Laurent Moret-Bailly | Then (1) means that the function field $K$ of $X$ is ($k$-isomorphic to) a purely inseparable extension of $k(x,y)$. But then $K\subset k(x^{p^{-m}},y^{p^{-m}})$ for large $m$, which implies (2) since $k(x^{p^{-m}},y^{p^{-m}})$ is $k$-isomorphic to $k(x,y)$. The converse is proved similarly. This works over any perfect $k$. | |
Sep 10, 2016 at 14:16 | comment | added | Dimitri Koshelev | Sorry. I mean by $\varphi$ and $\psi$ rational dominant maps over $k$, but not regular. | |
Sep 10, 2016 at 14:00 | comment | added | Jason Starr | @LaurentMoret-Bailly: That depends on what the OP means by "$k$-rational map". When I write that, I mean "rational transformation defined over $k$", rather than "regular morphism defined over $k$". However, based on the previous post, I suspect the OP wants the morphism to be everywhere regular. | |
Sep 10, 2016 at 12:26 | comment | added | Laurent Moret-Bailly | @JasonStarr : It is a birational problem, so how can normality be relevant? | |
Sep 10, 2016 at 12:20 | comment | added | Dimitri Koshelev | @JasonStarr, why is it true that if X is normal, then those conditions are equivalent? | |
Sep 10, 2016 at 11:18 | comment | added | Jason Starr | At least if you assume that $X$ is normal, then those conditions are equivalent. In the non-normal case, let $X$ be a surface whose seminormalization equals its normalization. If the normalization equals $\mathbb{P}^2$, then the normalization function gives $\psi$, but there need not be $\phi$. | |
Sep 10, 2016 at 8:29 | history | asked | Dimitri Koshelev | CC BY-SA 3.0 |