# Smith Normal Form of powers of a matrix

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.

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The conjecture sounds right; I will write down a proof when I have time. The point is that one can reformulate this (and many other algebraic questions about SNF) in the language of affine buildings and then solve it using geometry of such buildings. – Misha May 22 '13 at 15:39

We can split the question into a question local at each prime of $R$, because one can recover a finitely-generated module from its localizations. So we can work over a complete DVR. Over a complete DVR, we can sometimes decompose a matrix, up to conjugacy, into two blocks separated by $0$s, so we can reduce to the indecomposable case.
I don't see why you can reduce to the case of a single Jordan block; the problem is only invariant under $GL(R)$ conjugacy, not $GL(K)$. – David Speyer May 22 '13 at 12:38
I was thinking about $GL(R/p)$, not $GL(K)$. But my argument for it is oversimplified and wrong, and the conclusion is not quite true as I stated it. I'll need to think about it. – Will Sawin May 22 '13 at 13:38