Timeline for Is the set of hyperelliptic curves with a K-point closed?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2020 at 20:41 | vote | accept | Erik Walsberg | ||
| Oct 29, 2020 at 9:32 | answer | added | Laurent Moret-Bailly | timeline score: 7 | |
| Oct 29, 2020 at 6:33 | comment | added | Asvin | I think there will always exist some points for which stuff like this happens. It's related to the aut group of the curve being non trivial, I am not sure how "non generic" a non trivial automorphism group is. | |
| Oct 29, 2020 at 3:10 | comment | added | Erik Walsberg | Does the same thing always happen for the coarse moduli space of genus $g$ curves? If that's the case then I probably need to work with something that is either more sophisticated or less sophisticated. | |
| Oct 29, 2020 at 2:58 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Oct 29, 2020 at 2:52 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Oct 29, 2020 at 1:30 | comment | added | Joe Silverman | @Asvin But there's always a twist with a rational point, just take your favorite $x_0,y_0\in K$ and set $d=f(x_0)/y_0^2$. Maybe one could instead work on the moduli stack, instead of the moduli space? | |
| Oct 29, 2020 at 1:00 | comment | added | Asvin | I don't understand this very well: I think on the coarse moduli space, all the twists of a hyperelliptic curve (so of the form $dy^2 = f(x)$ for varying d) correspond to the same point and having $K$ point depends on which twist you take (?). So the question doesn't seem to be well defined as is. Perhaps you want to say that some twist has a $K$ point? | |
| Oct 28, 2020 at 23:54 | history | asked | Erik Walsberg | CC BY-SA 4.0 |