Timeline for On the arithmetic of powers of subseries of the exponential series
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29 at 15:32 | answer | added | Ira Gessel | timeline score: 1 | |
| Jul 5, 2021 at 13:50 | vote | accept | user313592 | ||
| Jul 4, 2021 at 12:23 | answer | added | Will Sawin | timeline score: 2 | |
| Jul 4, 2021 at 11:35 | comment | added | user313592 | @WillSawin That is a great observation! Thanks so much! | |
| Jul 4, 2021 at 4:07 | comment | added | user313592 | Right, my notation there is a little abusive. The ring you get by reducing modulo p the coefficients of the lattice generated the $q$th roots of unity is actually a quotient of $\mathbf{F}_p[\zeta]/(\zeta^q-1)$. I'm just pointing out the result about the Frobenius already holds in that bigger ring. | |
| Jul 4, 2021 at 3:59 | comment | added | markvs | If $\zeta^q=1$ (by the definition of $\zeta$) then the ideal $(\zeta^q-1)$ is $0$? | |
| Jul 4, 2021 at 3:53 | comment | added | user313592 | @MarkSapir "Frobenius is the identity" means that for any element $x$ of $\mathbf{F}_p[\zeta]/(\zeta^q-1)$, $x^p=x$. | |
| Jul 4, 2021 at 2:46 | comment | added | Will Sawin | Can't the coefficient of $z^q$ be handled by going back to the original formula for $f$, which has only one term of degree at most $q$? | |
| Jul 4, 2021 at 2:41 | comment | added | markvs | "Frobenius is the identity" What does it mean? | |
| Jul 4, 2021 at 2:32 | review | First posts | |||
| Jul 4, 2021 at 9:07 | |||||
| Jul 4, 2021 at 2:28 | history | asked | user313592 | CC BY-SA 4.0 |