Skip to main content
9 events
when toggle format what by license comment
Dec 1, 2021 at 8:17 comment added Yuan Yang @JonathanLove For non-closeness of Q: if the variety base change to $\bar{Q}$ is an abelian variety, then either: it doesn’t have a rational point; or it’s an abelian variety over Q, too.
Dec 1, 2021 at 8:02 comment added Yuan Yang @JonathanLove My current status is: I have an algebraic surface in $P^1*P^1*P^1*P^1$, and I have families of elliptic curves(I actually have $P^1$ families of them), but for each family, these curves are not disjoint-they actually all intersect in one point, at one of the infinity point. But indeed, any two family intersect at only one point.
Dec 1, 2021 at 7:52 vote accept Yuan Yang
Dec 1, 2021 at 7:52 comment added Yuan Yang @JonathanLove Yes indeed. I‘ve made quite a few mistakes. It's more complicated than I thought.
Dec 1, 2021 at 1:49 comment added Jonathan Love @Yuan It sounds like your application might involve considering varieties defined over $\mathbb{Q}$? The solution afh gives here depends quite heavily on the algebraically closed assumption - there are a lot of subtleties that arise if you're trying to work over number fields (even the problem statement, as it's currently given, isn't well-posed).
Nov 30, 2021 at 2:02 comment added Yuan Yang I think we just solved the 'four-distance' problem using this result. Although I don't know whether or not it is solved yet. Can we talk by email? Please contact me at [email protected]
Nov 30, 2021 at 1:59 history edited afh CC BY-SA 4.0
added 6 characters in body
Nov 30, 2021 at 1:41 history edited afh CC BY-SA 4.0
edited body
Nov 30, 2021 at 1:32 history answered afh CC BY-SA 4.0