Timeline for Conjugacy for $p$-adic matrices of finite order
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8, 2011 at 13:24 | history | edited | Junkie | CC BY-SA 3.0 |
added 296 characters in body
|
| Aug 8, 2011 at 10:35 | comment | added | Junkie | For the last difficulty, do you need to change the $\phi$-function definition over $Q_p(\mu_p)$, so that the factor of $(p-1)$ disappears into the units? In other words, there is additional splitting over $Q_p(\mu_p)$ viz. $Q_p$, but do you start at $F_p(\mu_p)$ now also? I would say, my statement "this classifies it" corresponds to the Jordan form part of the argument, not the rest of it where the reduction is done. That part uses that $Q_p$ and $F_p$ splitting are the same, up to ramification, but $Q_p(\mu_p)$ can have more. | |
| Aug 8, 2011 at 8:39 | comment | added | Tim Dokchitser | This sounds like the right approach, but I am worried about the "Furthernote,..." step. Why is this true? I can only see that in every $Q_p$-conjugacy class there is a matrix for which the claim holds. E.g, in particular, this step implies immediately that a matrix of order $m>1$ can't reduce to identity. This is indeed true (when $p \ne 2$), but I think this is a non-trivial statement. Or am I wrong? Also, "This classifies it over a field" is worrisome, because the statement is false over $Q_p[\mu_p]$. | |
| Aug 8, 2011 at 4:48 | history | edited | Junkie | CC BY-SA 3.0 |
added 93 characters in body
|
| Aug 8, 2011 at 4:43 | history | answered | Junkie | CC BY-SA 3.0 |