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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 .
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