Timeline for Zariski dense $K$-points for any non-trivial finite Galois extension $K/\mathbb{Q}$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2021 at 8:58 | comment | added | Damian Rössler | @Uriya: thank you, that makes sense! | |
| May 14, 2021 at 3:43 | comment | added | R. van Dobben de Bruyn | Note also that such varieties must have points everywhere locally (i.e. over all $\mathbf Q_p$, including $\mathbf Q_\infty = \mathbf R$), so the local-global principle fails. I'm not sure if this failure could be explained by the Brauer–Manin obstruction, but my feeling is that this obstruction would still exist over some nontrivial field extension. | |
| May 13, 2021 at 19:05 | comment | added | Uriya First | In general, having a $0$-cycle of degree $1$ is the same as having finite field extensions $K_1,\dots,K_t\supseteq \mathbb{Q}$ such that $V(K_i)\neq\emptyset$ for all $i$ and $\mathrm{gcd}([K_1:{\mathbb Q}],\dots,[K_t:{\mathbb Q}])=1$. | |
| May 13, 2021 at 19:02 | comment | added | Uriya First | @DamianRössler To see that there is a $0$-cycle of degree $1$ in the sense of mathoverflow.net/questions/33774/…, take a quadratic extension $K/{Q}$ and a cubic Galois extension $L/{Q}$ with fixed embeddings into $\overline{ Q}$. Choose $x_1\in V(K)$ and $y_1\in V(L)$ and let $x_2,y_2,y_3\in V(\overline{Q})$ be their Galois conjugates (take $x_2=x_1$ if $x_1$ is fixed by $Gal(\overline{ Q}/{ Q})$ and likewise with $y_2,y_3$). Then $y_1+y_2+y_3-x_1-x_2$ is a $0$-cycle of degree $1$. | |
| May 13, 2021 at 9:24 | comment | added | Damian Rössler | @Uriya: I am probably being silly, but why does such a variety have a $0$-cycle of degree $1$? | |
| May 12, 2021 at 14:23 | comment | added | R. van Dobben de Bruyn | This is likely a very hard question, because we have very few methods for constructing rational points. Both a proof and a counterexample would have to do this. | |
| May 12, 2021 at 6:15 | comment | added | Uriya First | Varieties of the kind you consider have a $0$-cycle of degree $1$. Whether this (much weaker) condition implies the existence of a $\mathbb{Q}$-point is discussed in mathoverflow.net/questions/33774/… . It is false in general, but is true for some families of varieties. | |
| May 12, 2021 at 5:44 | comment | added | David Benjamin Lim | Remark: I think this is false if you replace $\mathbf{Q}$ with $\mathbf{R}$. | |
| May 12, 2021 at 5:34 | history | asked | commiert | CC BY-SA 4.0 |