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Given integers $m$ and $n$ and $d_1, \ldots, d_m \in \mathbb{Z}/n \mathbb{Z}$, consider the set of all lower-triangular matrices of dimension $m$ with diagonal elements equal to $d_i$. What can be said about the distribution of the Smith Normal Form of such matrices? What kind of Smith Normal Form can arise?

EDIT: As an example, let $n = p^2$, $m = 2$, and $d_1 = d_2 = p$. Then we are considering the set of all matrices of form

\begin{bmatrix} p & 0 \\ x & p \\ \end{bmatrix}

for $x \in \{0, 1, \ldots, p^2-1\}$. In this case we can show that the SNF takes form \begin{bmatrix} \gcd(x,p) & 0 \\ 0 & p^2/\gcd(x,p) \\ \end{bmatrix} . So the SNF can take any value given the constraint that the determinant is $p^2$. Is this true for all dimensions?

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  • $\begingroup$ (Over the integers Z,) Primarily, the square free part of the determinant appears at the end of the diagonal of the SNF, multiplied with (some powers of) all the primes belonging to the square full part. Since most determinants are not powerful numbers, this implies that the SNF diagonal often starts as a sequence of ones. Gerhard "And What Were You Expecting?" Paseman, 2019.11.16. $\endgroup$
    – Gerhard Paseman
    Commented Nov 16, 2019 at 15:46
  • $\begingroup$ @GerhardPaseman thanks for the comment but I was looking for a more fine-grained answer: just "often starts as a sequence of ones" is not good enough for me. $\endgroup$
    – hao chen
    Commented Nov 18, 2019 at 1:32

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