Given a matrix $A \in M_d(\mathbb{Z}_p)$ (with nonzero determinant), viewed as a map $\mathbb{Z}_p^d \to \mathbb{Z}_p^d$, I am interested in the sequence of abelian $p$-groups $\{coker(A^n)\}_{n \geq 1}$ (equivalently, invariant factors/Smith normal forms of all powers). For certain matrices, it is quite clear that this sequence of groups follows a pattern, e.g. if $A$ is diagonal we only need to know the cokernel of $A$ and this determines all the others. In general, I am hoping there is some nice, finite combinatorial object that packages all of the information of this sequence, but any way to specify any such sequence via, say, a finite number of integers would be a very good start.

Note: Someone asking more or less the same question but over any PID made a stronger conjecture which people in the comments were optimistic could be proven, but this proof never appeared. Such a proof would immediately solve my problem.

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    $\begingroup$ After conjugating by a suitable matrix (in $M_n(Q_p)$) we can get a diagonal block decomposition in which every block has all its eigenvalues of the same norm ($\le 1$ by your hypothesis). Hence, up to a bounded error (due to the conjugation possibly involving denominators - I'm not sure if this can be done without denominators) one boils down to a matrix with all roots of the same norm. $\endgroup$
    – YCor
    Commented Jan 29, 2018 at 2:39
  • $\begingroup$ Thanks for the idea. If I understand this right, you're saying I should conjugate my matrix $A$ by some matrix $B \in GL_d(\mathbb{Q}_p)$ to put my matrix in rational canonical form over $\mathbb{Q}_p$ (which I can certainly do), raise that to a power, and then conjugate by $B^{-1}$ to get back a matrix in $M_d(\mathbb{Z}_p)$. I'm confused what you mean by these blocks having all eigenvalues of the same norm--shouldn't each block just have a single eigenvalue with an eigenvector and some generalized eigenvectors corresponding to it? I'm also confused what you mean by the "bounded error". $\endgroup$ Commented Jan 29, 2018 at 22:03
  • $\begingroup$ No because eigenvalues belong to extensions of $\mathbf{Q}_p$ and the conjugation would involve a matrix in such an extension, which would destroy too much information. $\endgroup$
    – YCor
    Commented Jan 29, 2018 at 22:28
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    $\begingroup$ Bounded error: define a graph where the vertices are abelian groups (up to isomorphy), and draw an edge of length $\log k$ between two vertices if one passes from one to another as a subgroup of index $k$, or by taking the quotient by a finite subgroup of of order $k$. Consider the distance it yields. For instance, the distance between finite groups $F,F'$ of cardinal $d\le d'$ belongs to $[\log(d')-\log(d),\log(d')+\log(d)]$. A subgroup with bounded distance to $C_p^n$ ($C_p=Z/pZ$) should have the form $C_p^{n+m_n}\oplus F_n$ with $m_n\in\mathbf{Z}$ and group $F_n$ of bounded order (wrt $n$). $\endgroup$
    – YCor
    Commented Jan 29, 2018 at 22:34
  • $\begingroup$ Ok that helps to clarify a bit, thanks! $\endgroup$ Commented Jan 30, 2018 at 18:16


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