Timeline for IMO 2017/6 via arithmetic geometry
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 26, 2017 at 11:31 | comment | added | R. van Dobben de Bruyn | This produces an $f$ with $f(x,y) \in \mathbb Z^{\times} = \{\pm 1\}$ for all $(x,y) \in S$. To get $f(x,y) = 1$, take the square. | |
| Jul 26, 2017 at 7:04 | comment | added | js21 | Oh, right! Thank you for the clarification. | |
| Jul 26, 2017 at 6:56 | comment | added | dhy | @js21: Not so! Every irreducible component of $S$ is a copy of $\operatorname{Spec}{\mathbb{Z}}$, but they may be joined at some finite primes. As I see it, this is the essential source of difficulty for the elementary solutions to this problem. | |
| Jul 26, 2017 at 6:35 | comment | added | js21 | Here the subscheme $S$ is a finite coproduct of copies of $\mathrm{Spec}(\mathbb{Z})$, so its Picard group is actually trivial (so lemma $7.3$ is not needed here). | |
| Jul 25, 2017 at 21:23 | vote | accept | Evan Chen | ||
| Jul 25, 2017 at 20:37 | history | answered | dhy | CC BY-SA 3.0 |