Timeline for Conjugacy for $p$-adic matrices of finite order
Current License: CC BY-SA 4.0
14 events
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| Jan 24, 2023 at 17:44 | history | edited | LSpice | CC BY-SA 4.0 |
Links, while this is on the front page
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| Aug 8, 2011 at 10:36 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
Added proof of last statement
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| Aug 6, 2011 at 13:28 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
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| Aug 6, 2011 at 13:17 | comment | added | Frieder Ladisch | @vytas: I agree that for fixed $n$, the statement is true for almost all primes $p$. I think for $p$ fixed, there is a counterexample of dimension $2p^3$ and order $p^3$, but didn't check carefully. | |
| Aug 6, 2011 at 12:47 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
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| Aug 6, 2011 at 12:47 | comment | added | Frieder Ladisch | @at all: thanks for pointing out the wrong statement. Sorry for the mistake and the late reaction, I wasn't online the last 20 hours or so. I think I fixed it, but the fix turns out to be nearly the same argument as Alex's. | |
| Aug 6, 2011 at 12:39 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
corrected wrong statements
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| Aug 6, 2011 at 4:54 | comment | added | Junkie | With my above comment, it seems that the companion matrix of $\Psi_{p^3}$ when reduced mod $p$ has a different (Jordan) form than $p$ copies of $\Psi_{p^2}$ when resp. reduced, so this direction in cyclotomic polys is murky. I am back to looking at Alex Bartel's comment, as to why the lifting step should occur. | |
| Aug 6, 2011 at 4:22 | comment | added | Junkie | Put another way I think, $\Psi_{p^3}\equiv\Psi_{p^2}^p$ modulo $p$ for the cyclotomic polynomials $\Psi$ and from this the desired matrices can appear with companions. EDIT: The companions $A$ and $B^{\oplus p}$ do not have the order ($p^3$ resp. $p^2$), but summing each to $A$ will. However, I am still unsure it all works, for the final matrices are not conjugate in $F_p$ when I calculate. | |
| Aug 5, 2011 at 18:12 | comment | added | vytas | @Geoff Robinson. I am saying that if $p-1> n$ then $\GL_n(\mathbb Z_p)$ does not contain an element of order $p$. Given that, and the last sentence in F.Ladisch's answer, we obtain: "Yes, if $p-1>n$". Ah! I have uncofused myself, Ladisch is saying for every prime $p$ there exists an $n$, such that the statement is false, I am saying for a fixed $n$ the statement is true for almost all primes $p$. | |
| Aug 5, 2011 at 17:43 | comment | added | vytas | @F.Ladisch: I am confused. Isn't it true that if $p>n$ then a pro-$p$ Sylow of $GL_n(\mathbb Z_p)$ does not contain an element of finite order. If I recall this correctly this has been remarked by Lazard in his IHES paper on p-adic analytic groups. So if $p> n$ then all the elements of finite order in $GL_n(\mathbb Z_p)$ have order prime to $p$. So your answer should say "Yes, for almost all primes". | |
| Aug 5, 2011 at 16:23 | comment | added | Alex B. | Could you briefly explain why you can always lift the modular matrices to integral matrices of order dividing $p^3$? | |
| Aug 5, 2011 at 15:18 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
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| Aug 5, 2011 at 14:46 | history | answered | Frieder Ladisch | CC BY-SA 3.0 |