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Let $X$ be an absolutely irreducible surface over an algebraically closed or finite field $k$ of characteristic $p$. Is it true that the following conditions are equivalent?

  1. There is a purely inseparable $k$-rational dominant map $\varphi\!: X \dashrightarrow \mathbb{P}^2$.

  2. There is a purely inseparable $k$-rational dominant map $\psi\!: \mathbb{P}^2 \dashrightarrow X$, i.e., $X$ is a generalized Zariski surface.

Are degrees of $\varphi$ and $\psi$ equal?

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  • $\begingroup$ 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 $\phi$. $\endgroup$
    – Jason Starr
    Commented Sep 10, 2016 at 11:18
  • $\begingroup$ @JasonStarr, why is it true that if X is normal, then those conditions are equivalent? $\endgroup$
    – Dimitri Koshelev
    Commented Sep 10, 2016 at 12:20
  • $\begingroup$ @JasonStarr : It is a birational problem, so how can normality be relevant? $\endgroup$
    – Laurent Moret-Bailly
    Commented Sep 10, 2016 at 12:26
  • $\begingroup$ @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. $\endgroup$
    – Jason Starr
    Commented Sep 10, 2016 at 14:00
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    $\begingroup$ 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 $k$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Sep 10, 2016 at 14:23

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