Timeline for Is there a purely inseparable covering $\mathbb{A}^2 \to K$ of a Kummer surface $K$ over $\mathbb{F}_{p^2}$?
Current License: CC BY-SA 4.0
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| May 29, 2018 at 13:34 | history | bounty ended | CommunityBot | ||
| May 25, 2018 at 20:15 | comment | added | Felipe Voloch | @DimaKoshelev The extensions that you will get if you bother to do the calculations for $p \equiv 3 \pmod{4}, p \equiv 2 \pmod{3}$ are like that. So maybe you should try to do them yourself, because I am not going to do it for you. | |
| May 25, 2018 at 19:26 | comment | added | Dimitri Koshelev | This is obvious. Could you clarify please, what do you mean? | |
| May 25, 2018 at 16:24 | comment | added | Felipe Voloch | @DimaKoshelev I am not sure what kind of general statement you need but if $y=x^m, (m,p)=1$, then $y^{1/p} = x^{m/p}$. | |
| May 25, 2018 at 14:21 | comment | added | Dimitri Koshelev | Is the extension $\mathbb{F}_{p^2}(\mathbb{A}^2) \supset \mathbb{F}_{p^2}(K)$ normal? Why does the lift of the Galois action on $X$ to $\mathbb{A}^2$ exist? | |
| May 24, 2018 at 6:59 | comment | added | Felipe Voloch | @DimaKoshelev Yes, a priori, only over the algebraic closure. I believe the cases of $p \equiv 2 \pmod{3}, p \equiv 3 \pmod{4}$ already mentioned will be quite explicit and can be checked directly. Shioda's paper Math. Ann. 211 (1974), 233–236, does the Fermat surface of degree $p+1$ explicitly and shows it's a Zariski surface over $\mathbb{F}_{p^2}$. | |
| May 24, 2018 at 6:16 | comment | added | Dimitri Koshelev | However this quotient is a priori rational only over the algebraic closure. Am I right? | |
| May 23, 2018 at 20:45 | comment | added | Noam D. Elkies | Likewise for $p \equiv 3 \bmod 4$ by sending $y^2 = x^{p+1} - 1$ to $y^2 = X^4 - 1$ where $X = x^{(p+1)/4}$. | |
| May 23, 2018 at 19:53 | comment | added | Felipe Voloch | @DimaKoshelev I missed the fact you wanted purely inseparable. But if $\mathbb{A}^2 \to X \to K$ and $X \to K$ is Galois, you can try to lift the Galois action to $\mathbb{A}^2$ and take the quotient, which will also be a rational surface. | |
| May 23, 2018 at 19:26 | comment | added | Dimitri Koshelev | How can we explicitly find a purely inseparable part $S \to K$ of the map $\mathbb{A}^2 \to K$, where $S$ is some "intermediate" surface? | |
| May 23, 2018 at 18:54 | history | edited | Felipe Voloch | CC BY-SA 4.0 |
added 433 characters in body
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| May 23, 2018 at 18:32 | history | answered | Felipe Voloch | CC BY-SA 4.0 |