What invariants of a matrix determine the Smith Normal Form (SNF) of all the powers of a matrix?
The question makes sense over any PID $R$. If we let $M = M_n(R)$ and $G=Gl_n(R)$, then SNF is a parameterization of the double coset space $G\backslash M/G$ and I am asking about the image of a sequence $(y^i : i \geq 1)$ under the projection $M \longrightarrow G \backslash M/G$. Clearly this factors through the quotient of $M$ under conjugation by $G$.
It is easy to produce examples to show that the characteristic polynomial is insufficient, even for 2x2 matrices.
The reason I care is that, in computing local cohomology groups for graded one dimensional rings, one often comes across a ring $S$, free as a module over a PID $R$, and an element $y \in S$ for which one wants to know all the quotients $S/(y^i)$ explicitly as an $R$-module.
If we let $A_i$ be the cokernel of $y^i$, then we have short exact sequences $ 0 \longrightarrow A_i \longrightarrow A_{i+j} \longrightarrow A_j \longrightarrow 0$ relating these quotients (at least when $det(y) \neq 0$).
I conjecture that for any $y \in M$ there exist an integer $d > 0$ and a diagonal matrix $D$ such that $SNF(y^{i+d}) = D*SNF(y^i).$
The work on the possible values of SNF(AB), given SNF(A) and SNF(B), masterfully recounted in Fulton's "Eigenvalues, invariant factors, highest weights, and Schubert calculus", Bull. AMS 37 (2000), no. 3, 209–249, is probably relevant, though $SNF(A^i)$ is in some sense the worst case, since $SNF(AB)$ is most constrained when $det(A)$ and $det(B)$ share few factors. If I understood that work, perhaps I would know that the answer is already known.
I should note that nothing crucial depends on this question: I am simply curious.
