Timeline for Supersingular curves over $\mathbb{F}_q$ and the splitting of $p$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29, 2021 at 19:35 | history | bounty ended | Rdrr | ||
| Jul 29, 2021 at 19:35 | vote | accept | Rdrr | ||
| Jul 29, 2021 at 14:46 | comment | added | Rdrr | Thank you so much Bjorn. This is a very insightful proof. | |
| Jul 29, 2021 at 13:37 | vote | accept | Rdrr | ||
| Jul 29, 2021 at 19:35 | |||||
| Jul 29, 2021 at 3:28 | comment | added | Bjorn Poonen | @Rdrr: Here is another way to say it. Let $\mathfrak{p}$ be the prime corresponding to $v$. The implication $v(\beta) \ge 2n \implies \beta \in (p)$ is equivalent to $\mathfrak{p}^{2n} \subset (p)$, which is the same as saying that $(p)$ divides $\mathfrak{p}^{2n}$. In this case, the factorization of the $\mathcal{O}$-ideal $(p)$ cannot involve any primes other than $\mathfrak{p}$. | |
| Jul 28, 2021 at 19:57 | comment | added | Rdrr | I must be missing a basic algebraic number theory fact. How does one get a contradiction from a high power? | |
| Jul 28, 2021 at 17:45 | comment | added | Will Sawin | @Rdrr If $v$ is not the only place above $p$, choose some element which is $0$ modulo $v$ but not zero modulo the other place above $p$, and then take a high power, for a contradiction. | |
| Jul 28, 2021 at 14:26 | comment | added | Rdrr | This is a wonderful proof. If you wouldn't mind I'd like some clarification on why $v(\beta)\geq 2n$ implies that $v$ is the only place above $p$. | |
| Jul 28, 2021 at 1:06 | history | answered | Bjorn Poonen | CC BY-SA 4.0 |