Timeline for A property of varieties between unirational and retract rational
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2020 at 6:40 | comment | added | Arno Fehm | As Daniel Loughran explained to me by email, the universal torsors of these Châtelet surfaces are in fact of dimension 8, so this example does show that $(3)\not\Rightarrow(5)$, but it does not answer question Q2. | |
| May 12, 2020 at 18:26 | vote | accept | Arno Fehm | ||
| May 22, 2020 at 6:40 | |||||
| May 12, 2020 at 11:28 | history | edited | Daniel Loughran | CC BY-SA 4.0 |
added 894 characters in body
|
| May 12, 2020 at 11:07 | comment | added | Daniel Loughran | Hi. Yes you are right - to prove stable rationality one needs some genericity assumptions, e.g. the polynomial $f$ is irreducible with Galois group $S_3$. Anyway, you seem happy with my answer so I will upgrade it and add more details. | |
| May 10, 2020 at 7:34 | comment | added | Arno Fehm | @OlivierBenoist: Thanks a lot. So if I understand correctly, the example would indeed answer the question. The only thing that is not clear to me yet is why a universal torsor with a real point necessarily exists. Is that a general fact? (A reference would be appreciated - I have not yet been able to access CT-S' La descente sur les variétés rationneles where apparently they were defined.) | |
| May 9, 2020 at 14:43 | comment | added | Olivier Benoist | I think you are both right. The (smooth projective model of) $X$ does not satisfy (4) or (5) because it has non-connected real locus. It satisfies (3) because a universal torsor with a real point is stably rational (by Colliot-Thélène, Sansuc and Swinnerton-Dyer, Intersections of two quadrics and Châtelet surfaces, II, Theorem 8.1). | |
| May 8, 2020 at 20:41 | comment | added | Arno Fehm | Thanks, I will have a closer look at that paper. I am a bit confused though: $X(\mathbb{R})$ is a real manifold with two connected components, hence so is $(X\times\mathbb{A}^n)(\mathbb{R})$. Shouldn't this imply that $X\times\mathbb{A}^n$ and $\mathbb{A}^m$ are not birationally equivalent (over $\mathbb{R}$)? | |
| May 8, 2020 at 18:51 | history | answered | Daniel Loughran | CC BY-SA 4.0 |