Timeline for Conjugacy for p-adic matrices of finite order II
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 18, 2011 at 2:06 | answer | added | Florian Eisele | timeline score: 3 | |
| Aug 11, 2011 at 17:19 | answer | added | Alex B. | timeline score: 1 | |
| Aug 11, 2011 at 9:30 | answer | added | Geoff Robinson | timeline score: 5 | |
| Aug 10, 2011 at 7:32 | comment | added | Tim Dokchitser | @Junkie: This is indeed the direction that I want! But can you expand your argument a bit (and possibly post it as an answer)? I think you are using that every ${\mathbb Q}_p$-conjugate of your matrix in $GL_n({\mathbb Z}_p)$ has the same ${\bar J}$. I cannot see why this must be true. But if it is, I agree, this proves the statement. | |
| Aug 10, 2011 at 7:20 | history | edited | Tim Dokchitser | CC BY-SA 3.0 |
Corrected the question
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| Aug 10, 2011 at 4:40 | comment | added | Junkie | I think my answer, for the other question (I misread) is now correct for the direction of $F_p$ conjugacy implying $Q_p$ conjugacy. Namely, given a general Jordan block $J$ of dimension $d$ in the decomposition for $A$, this $J$ is a companion matrix (ones above the diagonal), which remains the case upon mod $p$ reduction, and so the minimal polynomial of the reduced block $\bar J$ is also of degree $d$, so that this block does not split. This means (referring to the argument there) that $\bar J$ determines $J$, so two distinct $J$ and $J'$ cannot reduce the same mod $p$. | |
| Aug 9, 2011 at 20:25 | answer | added | user91132 | timeline score: 4 | |
| Aug 9, 2011 at 18:53 | comment | added | Igor Rivin | This seems relevant: mathoverflow.net/questions/69578/… | |
| Aug 9, 2011 at 18:44 | history | asked | Tim Dokchitser | CC BY-SA 3.0 |