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Jan 25, 2023 at 0:12 comment added reuns If $A,B\in GL_n(\Bbb{Z}_p)$ have the same characteristic polynomial which is separable $\bmod p$ then $A=C B C^{-1}$ with $C\in GL_n(\Bbb{Z}_p)$. This is because $A\bmod p$ has a cyclic vector $\Bbb{F}_p^n = \bigoplus_{m=0}^{n-1} \Bbb{F}_pA^m v$, if $v$ is the reduction of $w$ then $\Bbb{Z}_p^n = \bigoplus_{m=0}^{n-1} \Bbb{Z}_pA^m w$ so $A$ is conjugate to its companion matrix.
Jan 24, 2023 at 17:25 history edited YCor CC BY-SA 4.0
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Aug 10, 2011 at 19:36 comment added Frieder Ladisch Afterthought: Since the answer to your first question is "No", it may also be useful to know that conjugacy mod $p^r$ implies conjugacy in $GL_n(\mathbb{Z}_p)$, when $r$ is big enough, i. e. such that $p^r$ doesn't divide $m$. This follows from the treatment of Maranda's theorem in Curtis-Reiner, Methods.
Aug 8, 2011 at 4:43 answer added Junkie timeline score: 2
Aug 7, 2011 at 10:20 history edited Tim Dokchitser CC BY-SA 3.0
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Aug 7, 2011 at 9:30 answer added Jeff Adler timeline score: 2
Aug 7, 2011 at 9:24 vote accept Tim Dokchitser
Aug 6, 2011 at 11:21 answer added Alex B. timeline score: 13
Aug 5, 2011 at 14:46 answer added Frieder Ladisch timeline score: 6
Aug 5, 2011 at 13:22 answer added Geoff Robinson timeline score: 10
Aug 5, 2011 at 13:01 comment added Geoff Robinson I think that the name of Alfredo Jones is relevant here. Try the 1962 Curtis and Reiner, and their later "Methods of Representation Theory" tome.
Aug 5, 2011 at 11:43 answer added Gjergji Zaimi timeline score: 7
Aug 5, 2011 at 9:49 history asked Tim Dokchitser CC BY-SA 3.0