In $GL(n, \mathbb{Q}\_p)$, what are the orbits under conjugation of $GL(n, \mathbb{Z}\_p)$?



Conjugation in GL(n) (p-adic setting) 
The problem can be reduced to that of classifying $GL(n,\mathbf{Z}\_p)$ conjugacy classes in $M(n,\mathbf{Z}\_p)$. The situation for general $n$ is complicated, but for $n=2$ the problem is settled by the following.


Let $F\in M(2,\mathbf{Z}\_p)$ be any matrix, let $f(x)$ be its characteristic polynomial, and
let


$n(F)=\sup\_i(i\in\mathbf{Z}\_{\ge 0},$ $F$ mod $p^i$ is multiplication by a scalar $)$.


Then the $GL(2,\mathbf{Z}\_p)$ conjugacy class of $F$ is uniquely determined by $f(x)$ and $n(F)$.


If $n(F)$ is infinite, then $F$ is a scalar matrix and therefore central.


If $n(F)$ is finite then there exists a unique integer $\lambda\in\mathbf{Z}$, with $0\le\lambda\le p^{n(F)}-1$ such that $F$ is conjugate to
$\begin{pmatrix}
\lambda&0\\
0&\lambda
\end{pmatrix}+p^{n(F)}\begin{pmatrix}
0&-a\_0\\
1&-a\_1
\end{pmatrix}$.


Here $a\_0$ and $a\_1$ are the constant and linear term of the polynomial
$f\_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$, which has coefficients in $\mathbf{Z}\_p$.


We have that $p^{n(F)}$ is the index of the ring $\mathbf{Z}\_p[F]$ inside the ring $R\_F:=\mathbf{Q}\_p[F]\cap M(2, \mathbf{Z}\_p)$.


All rings are viewed as subrings of $M(2,\mathbf{Q}\_p)$. We will sometime think of $F$ and of elements of $R\_F$ as endomorphisms of the standard lattice $\mathbf{Z}\_p^2$ inside $\mathbf{Q}\_p^2$.


Proof: if $n(F)$ is infinite then there is not much to prove, therefore we assume $n(F)$ finite, and $F$ not central. The ring $R\_F$ as defined above contains $\mathbf{Z}\_p[F]$ with finite index, since they both are finite free $\mathbf{Z}\_p$-modules of rank two. This is clear for $\mathbf{Z}\_p[F]$, since $f(x)$ is the minimal polynomial of $F$, since $F$ is not central. For $R\_F$ it follows from the fact that $\mathbf{Q}\_p[F]\cap M(2, \mathbf{Z}\_p)$ is open, compact, and non-empty in $\mathbf{Q}\_p[F]$, which has rank two over $\mathbf{Q}\_p$, since $F$ is not central.


The ring $R\_F$ has a $\mathbf{Z}\_p$-basis of the form $(1, F')$ where $F'=(a+bF)/p^{h}$, for some $a,b\in\mathbf{Z}\_p$ not both divisible by $p$, and where $p^h$, with $h\ge 0$, is the index of $\mathbf{Z}\_p[F]$ in $R\_F$. The $p$-adic integer $b$ is a unit, for otherwise $p^{h-1}F'-b'F=a/p$, with $b'=b/p\in\mathbf{Z}\_p$, would belong to $R\_F$, which is not possible since $a/p$ is not a $p$-adic integer. This shows that the natural action of $F$ on $\mathbf{Z}\_p^2/(p^h)$ is multiplication by $-ab^{-1}$ mod $p^h$. Therefore $h\leq n(F)$.


On the other hand, if $\lambda$ is any integer such that $F-\lambda$ is zero mod $p^{n(F)}$, then $F'':=(F-\lambda)/p^{n(F)}$ is an element of $R\_F$, since $F-\lambda$ commutes with $F$ and it is divisible by $p^{n(F)}$ in $M(2,\mathbf{Z}\_p)$. Therefore $n(F)\leq h$. Thus $n(F)=p^h$ and $(1, F'')$ is a $\mathbf{Z}\_p$-basis of $R\_F$ (since it spans a lattice of the correct index).


Now, by the maximality of $n(F)$ we see that $F''$ does not act via scalar multiplication on $\mathbf{Z}\_p^2/p$. This implies that there is a $\mathbf{Z}\_p$-basis $(e\_1, e\_2)$ of $\mathbf{Z}\_p^2$ such that
$F''(\mathbf{Z}\_p\cdot e\_1)\not\equiv \mathbf{Z}\_p e\_1$ mod $p$. It follows that $(e\_1, F''(e\_1))$ is also a $\mathbf{Z}\_p$-basis of $\mathbf{Z}\_p^2$.


With respect to this basis the action of $F''$ is given by a matrix of the form
$\begin{pmatrix}
0&-a\_0\\
1&-a\_1
\end{pmatrix}$, where $a\_0$ and $a\_1$ are the constant and the linear term of the characteristic polynomial of $F''$, which is $f\_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$ and has coefficients in $\mathbf{Z}\_p$. By picking $\lambda$ in the range $0,\ldots,p^{n(F)}-1$, we see that the action of $F$ with respect to the basis $(e\_1, F''(e\_1))$ is that given by the statement.




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Notice that this shows that $\mathbf{Z}\_p^2$ is a free $R\_F$-module of rank one, and classifying the action of $F$ on $\mathbf{Z}\_p^2$ is roughly equivalent to finding $R\_F$. I was interested exactly in this in the context of Tate modules of elliptic curves over finite fields ($F=$Frobenius). Probably there is a more conceptual/simpler proof. I would be interested to hear what you get in higher dimension. It won't be that easy, I expect. What makes this case simple is that orders of $\mathbf{Q}\_p[F]$ containing $F$ are classified by the index with which $\mathbf{Z}\_p[F]$ sits in them.


