Timeline for Number of solutions to polynomial congruences
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29, 2022 at 19:30 | comment | added | Victor Wang | I think Serre's Théorème 8 says something a little different; see the distinction in "Exemple. hypersurfaces algébriques" (several paragraphs after Théorème 8) between $Y_n$ vs. $\tilde{Y}_n$ (reduction mod p^n of a p-adic zero set vs. mod p^n points of a p-adic scheme), especially around (57). | |
| Mar 3, 2018 at 7:32 | comment | added | Daniel Loughran | I don't think my proof completely works for prime powers, so I deleted my answer with a view to possibly fixing it later when I have more time. | |
| Mar 3, 2018 at 4:11 | comment | added | Joël | I just realized that everything I said was already said in Daniel's deleted answer (only visible to high-rep users I suppose). Why was this fine answer deleted? | |
| Mar 3, 2018 at 4:05 | comment | added | Joël | As mentioned in my comment under Igor's answer, even knowing that $C_p$ is independent on $p$ would not immediately give the result for all integers $q$. Looking at Serre's proof, it was not immediately clear to me that you can take the same $C_p$ for all $p$. Of course, having this fr almost all $p$ would be enough, so you can throw away bad $p$'s and assume your $p$-adic space $V$ is smooth, in which case Serre's proof is much simpler (in the non-smooth case, it uses Hironaka's resolution of singularities). But even in the smooth case, it is not clear to me how to control $C_p$. | |
| Mar 3, 2018 at 3:56 | history | answered | Joël | CC BY-SA 3.0 |