It is easy to classify conjugacy classes in $GL_n(\mathbb Q_p)$ by linear algebra. How to classify $GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class? For example, for general matrix $A \in GL_n(\mathbb Z_p)$, how many $X \in GL_n(\mathbb Z_p)$ are $GL_n(\mathbb Q_p)$ conjugated to $A$ up to $GL_n(\mathbb Z_p)$ conjugaction? In some cases it is finite, is it always finite?

Motivation: when counting invariant lattices, one comes across such problems.

Edit: could we find some numerical invariants that distinguish different $\Bbb Z_p-$ conjugacy classes?

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    $\begingroup$ One way to proceed is to multiply the ${\rm GL}(n,\mathbb{Q}_{p})$ matrix through by an element of $\mathbb{Z}_{p}$ of minimal valuation to make it integral. Then two such matrices which are conjugate in ${\rm GL}(n,\mathbb{Z}_{p})$ must have the same Smith normal form. I think people like Reiner and Zassenhaus (and maybe others before) have worked out this problem in detail. $\endgroup$ Nov 10, 2018 at 9:03
  • $\begingroup$ Note that this conjugation classification (in either sense) commutes with adding scalars, so the classification of conjugacy classes in $M_n$ boils down to that in $GL_n$. So, considering conjugacy classes in $M_n$ is not harder; on the other hand it is a nicer setting, since it's a linear action. $\endgroup$
    – YCor
    Nov 11, 2018 at 20:46
  • $\begingroup$ For n=2 you can explicitly answer your question! mathoverflow.net/questions/90370/… $\endgroup$ Nov 14, 2018 at 20:52
  • $\begingroup$ @TommasoCenteleghe Given two matrices in $\mathrm{M}_n(\mathbb{Z}_p)$, is it possible to find an explicit integer $k \geq 1$ such that the matrices are conjugate over $\mathbb{Z}_p$ if and only they are conjugate over $\mathbb{Q}_p$ and over $\mathbb{Z}/p^k$? I guess such a $k$ could be related to the $p$-adic distance between two roots of the characteristic polynomial. $\endgroup$ Nov 15, 2018 at 8:18
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    $\begingroup$ @FrançoisBrunault : Maranda had results of this nature on your question above ( mentioned in various surveys by Reiner) $\endgroup$ Nov 19, 2018 at 20:43

2 Answers 2


The answer is yes if $A \in \mathrm{GL}_n(\mathbb{Z}_p)$ is semisimple.

We may think of a matrix $A \in M_n(\mathbb{Z}_p)$ as a $\mathbb{Z}_p$-lattice of rank $n$ endowed with a $\mathbb{Z}_p$-linear endomorphism: take the standard lattice $L=\mathbb{Z}_p^n$ endowed with the endomorphism $\varphi$ associated to $A$. Clearly, two matrices $A,A' \in M_n(\mathbb{Z}_p)$ are $\mathrm{GL}_n(\mathbb{Z}_p)$-conjugated if and only if the associated pairs $(L,\varphi)$ and $(L',\varphi')$ are isomorphic.

More formally, consider the algebra $R = \mathbb{Z}_p[A]$ inside $M_n(\mathbb{Z}_p)$. Then $A$ gives rise to an $R$-module $M$ which is $\mathbb{Z}_p$-free of rank $n$. Note that if $A' \in M_n(\mathbb{Z}_p)$ is another matrix which is $\mathbb{Q}_p$-conjugated to $A$ then the algebras $\mathbb{Z}_p[A]$ and $\mathbb{Z}_p[A']$ are isomorphic, as they are both isomorphic to $R = \mathbb{Z}_p[X]/(\mu)$ where $\mu$ denotes the minimal polynomial. However, the associated $R$-modules $M$ and $M'$ need not be isomorphic. In fact, they are if and only if $A$ and $A'$ are $\mathbb{Z}_p$-conjugated.

So the problem reduces to classifying the $R$-modules which are $\mathbb{Z}_p$-free of rank $n$ up to isomorphism. In the semisimple case, the following theorem of Jordan-Zassenhaus gives a positive answer to your third question.

Theorem (Jordan-Zassenhaus). Let $K$ be a local or global field with ring of integers $\mathcal{O}_K$. Let $L$ be a (commutative) semisimple $K$-algebra, and let $R$ be an $\mathcal{O}_K$-order in $L$. Then for any integer $n \geq 1$, the set of isomorphism classes of $R$-modules which are $\mathcal{O}_K$-lattices of rank $\leq n$ is finite.

The case at hand follows by taking $K=\mathbb{Q}_p$, $L=\mathbb{Q}_p[X]/(\mu)$ and $R=\mathbb{Z}_p[X]/(\mu)$ where $\mu$ is a squarefree polynomial. A proof of the Jordan-Zassenhaus theorem is given in Reiner, Maximal orders, (26.4), p. 228. It is stated only for global fields, but you can check it also holds for local fields.

EDIT. The proof is not difficult: first one treats the case where $R$ is a maximal order, which is clear since it is a product of discrete valuation rings (hence PIDs). If $R' \subset R$ is an arbitrary order and $M'$ is an $R'$-lattice then there are only finitely many possibilities for $M=M' \otimes_{R'} R$, and since $p^N M \subset M' \subset M$ there are only finitely many possibilities for $M'$. So we get the finiteness.

Regarding the question of an explicit classification, you may be interested by Marseglia's article Computing the ideal class monoid of an order, where he gives algorithms to compute explicit representatives (in the global case).

EDIT 2. Another article giving an algorithm in the global case appeared on the arXiv today: https://arxiv.org/abs/1811.06190


Is it always finite? No.

The matrices $u_n=\begin{pmatrix}1 & p^n\\0 & 1\end{pmatrix}\in\mathrm{GL}_2(\mathbf{Z}_p)$ are all conjugate in $\mathrm{GL}_2(\mathbf{Q}_p)$, and pairwise non-conjugate in $\mathrm{GL}_2(\mathbf{Z}_p)$.

Indeed, a matrix in $\mathrm{GL}_2(\mathbf{Q}_p)$ conjugating $u_n$ to $u_{n+m}$, for $m\ge 1$, is necessarily of the form $\begin{pmatrix}p^m\lambda & p^m\lambda\mu\\0 & \lambda\end{pmatrix}$ for scalars $\lambda\neq 0$, $\mu$. If it is in $\mathrm{GL}_2(\mathbf{Z}_p)$, the determinant $p^m\lambda^2$ has to be of valuation $0$, hence $\lambda$ has valuation $-m/2<0$ and $p^m\lambda$ has valuation $m/2>0$, and hence the trace $\lambda+p^m\lambda$ also has valuation $-m/2$, contradiction since the trace should be in $\mathbf{Z}_p$.

I'd be more optimistic for semisimple conjugacy classes.

Since François proved the finiteness for semisimple conjugacy classes, let me complete the proof of the converse: namely that if the $\mathrm{GL}_d(\mathbf{Q}_p)$-conjugacy class of $A\in M_d(\mathbf{Q}_p)$ splits into finitely many $\mathrm{GL}_d(\mathbf{Z}_p)$-conjugacy classes, then $A$ is semisimple.

Indeed, first if $P$ is a nonzero polynomial, the intersection of $\mathrm{Ker}(P(A))$ with $\mathbf{Z}_p^d$ is a direct summand of the latter as $\mathbf{Z}_p$-module. Hence, up to conjugation by some element of $\mathrm{GL}_d(\mathbf{Z}_p)$, we can suppose that $A$ is block-upper triangular $$A=\begin{pmatrix} A_1 & \dots &\\ 0 & \ddots & \vdots \\ 0 & 0 &A_k\end{pmatrix}$$ such that for each diagonal block $A_i$ there is some irreducible polynomial $P_i$ over $\mathbf{Q}_p$ such that $P_i(A_i)$. Let $A'$ be the block-diagonal matrix with the same diagonal blocks $A_i$, so $A'$ is semisimple. Let $u$ be the diagonal matrix that equals $p^{k-i}$ on the $i$-th block. Then $\lim_{n\to\infty}u^iAu^{-i}=A'$.

Assume the finiteness property. Since the $\mathrm{GL}_d(\mathbf{Z}_p)$-conjugacy classes are compact and there are finitely many in the $\mathrm{GL}_d(\mathbf{Q}_p)$-conjugacy class $\mathcal{C}$ of $A$, we deduce that they are open in $\mathcal{C}$. In particular, for $n$ large enough, $u^iAu^{-i}$ belongs to the $\mathrm{GL}_d(\mathbf{Z}_p)$-conjugacy class of $A'$. This proves that $A$ is semisimple.

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    $\begingroup$ PS: there's a another argument to see that they're not conjugate: $u_n$ has a $p^n$-root in $\mathrm{GL}_2(\mathbf{Z}_p)$, but no $p^{n+1}$-root therein. Indeed, the $k$-roots of a matrix $e_{12}(x)=1+xE_{12}$, $x\neq 0$, in $\mathrm{GL}_2(\mathbf{Q}_p)$, all have the form $te_{12}(x/k)$ for some $k$-root of unity $t$. $\endgroup$
    – YCor
    Nov 12, 2018 at 0:47
  • $\begingroup$ It might be easier to note that $u_{n} -I$ is certainly not conjugate to $u_{m} -I$ when m and n are different. $\endgroup$ Nov 12, 2018 at 8:11
  • $\begingroup$ Yet another argument: if $u_n$ and $u_{n'}$ are conjugate in $\mathrm{GL}_2(\mathbf{Z}_p)$ with $n<n'$ then they are conjugate in $\mathrm{GL}_2(\mathbf{Z}/p^{n+1})$ which is clearly impossible. $\endgroup$ Nov 12, 2018 at 9:27

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