Is the number of similarity classes of integer matrices with given minimal and characteristic polynomial finite?

Question

Latimer and MacDuffee proved that there is a bijection between similarity classes of integer matrices with irreducible characteristic polynomial $\chi$ and the ideal class monoid of $\mathbb{Z}[\alpha]$, where $\alpha$ is a root of $\chi$. When $\mathbb{Z}[\alpha]$ is maximal, we can understand the similarity classes entirely by algebraic number theory. In his thesis "Similarity of Integer Matrices," David Husert generalizes this correspondence to the case of semisimple integer matrices with arbitrary characteristic polynomial by replacing ideal classes with isomorphism classes of modules over a certain order. The Jordan-Zassenhaus theorem asserts that if $A$ is a semisimple algebra of finite dimension over $\mathbb{Q}$, $\Lambda$ is a $\mathbb{Z}$-order in $A$, and $L^*$ is an $A$-module, then there are finitely many isomorphism classes of free full $\Lambda$-lattices in $L^*$. By a free full $\Lambda$-lattice in $L^*$, we mean a $\mathbb{Z}$-free $\Lambda$-module in $L^*$ whose span over $\mathbb{Q}$ is all of $L^*$. A proof of this result can be found in Section 79 of Curtis and Reiner's "Representation Theory of Finite Groups and Associative Algebras." Combining the Jordan-Zassenhaus theorem with the generalized Latimer-MacDuffee theorem demonstrates that the number of similarity classes of semisimple integer matrices with given characteristic polynomial is finite.

I am wondering if it is known whether the number of similarity classes of integer matrices with a given (not square-free) minimal polynomial and characteristic polynomial is finite. I am aware of this question: Ideal classes and integral similarity, but it only seems to address a special case of the finiteness of the number of similarity classes of semisimple integer matrices. I am not aware of any correspondence theorem for non-semisimple integer matrices that could be used to convert the problem to a number or module theoretic problem.

The Jordan-Chevalley decomposition provides a possible simplification. If $A$ is an integer matrix, we can write $A = S + N$, where $S$ is semisimple, $N$ is nilpotent, and $SN=NS$. Therefore, it suffices to show that the number of similarity classes of nilpotent integer matrices with given characteristic and minimal polynomial is finite. In Husert's thesis, he shows (proposition 4.21) that any nilpotent integer matrix is similar to a strictly block upper triangular matrix and further that each submatrix can be chosen to be in row Hermite normal form (proposition 4.25). Even with these simplifications, it is not clear to me how to prove or disprove the finiteness of the number of similarity classes.

Thanks in advance for your help!

Answer 1

No. Consider the matrices $$A_n = \begin{bmatrix} 0&n \\ 0&0 \end{bmatrix}$$ for $n$ a positive integer. They all have the same characteristic and minimal polynomials (namely, $x^2$), but no two of them are similar. To see this, note that $\mathbb{Z}^2 / A_n \mathbb{Z}^2 \cong \mathbb{Z} \times (\mathbb{Z}/n \mathbb{Z})$; similar matrices have isomorphic cokernels.

If you'd like an example with invertibles matrices, use $\begin{bmatrix} 1&n \\ 0&1 \end{bmatrix}$ instead.

I don't see how to make a counterexample with square-free characteristic polynomial.

Answer 2

David's answer fully addresses the question I asked, but after considering his examples I realized that we can resolve the finiteness question in general, so I figured I'd share. In particular, we can show that the number of similarity classes of matrices with characteristic polynomial $\chi$ and minimal polynomial $\mu$ is finite if and only if $\mu$ is square-free. As I mentioned in my question, the sufficiency of semisimplicity follows from Husert's generalized Latimer-MacDuffee theorem combined with the Jordan-Zassenahaus theorem.

We will first prove the necessity for nilpotent matrices. If $N$ is a nonzero nilpotent integer matrix, $N \sim_\mathbb{Q} kN$ for all $k \in \mathbb{N}$ by converting to Jordan form and then conjugating each block (of size $n_i$) by $\text{diag}(k, k^2, \ldots, k^{n_i})$. However, these matrices are not similar over $\mathbb{Z}$ because the GCD of their entries is different and integral similarity preserves the Smith form (one of whose invariant factors is the GCD of the entries).

We now use the Jordan-Chevalley decomposition to obtain the general result. Suppose $A \in \text{Mat}_n(\mathbb{Z})$ is not semisimple and has minimal polynomial $\mu$. Matrices similar over the integers have the same Smith form and thus the same GCD. It follows that if $kA \sim_\mathbb{Z} B$, then $B$ is divisible by $k$ and $A \sim_\mathbb{Z} \frac{1}{k}B$. Therefore, if the $\mathbb{Q}$-similarity class of $kA$ splits into $j$ integer similarity classes, then so does the $\mathbb{Q}$-similarity class of $A$. By the Jordan-Chevalley decomposition, we can uniquely write $A = S + N$, where $S$ is a semisimple rational matrix and $N$ is a nilpotent rational matrix, and $SN=NS$. Because $\mu$ is not square-free, $N$ is guaranteed to be nonzero. By choosing sufficiently large $k_1 \in \mathbb{N}$, we can write $k_1A = k_1S + k_1N$, where $S$ and $N$ are integer matrices. Because $k_1N$ is nilpotent, its minimal polynomial splits over $\mathbb{Q}$, so we can choose a rational matrix $Q$ that conjugates it to its Jordan form: $k_1QAQ^{-1} = k_1QSQ^{-1} + k_1QNQ^{-1}$. It is possible that $k_1QSQ^{-1}$ is not an integer matrix, so choose $k_2 \in \mathbb{N}$ large enough so that $k_2k_1QSQ^{-1} \in \text{Mat}_n(\mathbb{Z})$ and $k_2k_1QAQ^{-1} = k_2k_1QSQ^{-1} + k_2Jk_1QNQ^{-1}$.

Let $B(i)$ denote the $i$th Jordan block of $k_2k_1QNQ^{-1}$ and suppose that $B(i)$ has size $n_i$. Observe that if $X \in \text{Mat}_r(\mathbb{Q})$, conjugation by $\text{diag}(j, j^2, \ldots, j^r)$ acts by scaling the $m$th superdiagonal of $X$ by $j^m$ and by scaling by $j^{-m}$ on the $m$th subdiagonal. Therefore, conjugation by $\text{diag}(j, j^2, \ldots, j^{n_i})$ acts on $B(i)$ by scaling the matrix by $j$.

Let $D_j$ be the block diagonal matrix that scales $k_2k_1QNQ^{-1}$ by $j$ through conjugation. We want to conjugate by $D_j$, but we need to ensure that $k_2k_1QSQ^{-1}$ remains an integer matrix after conjugation. Therefore, let $k_3 = j^n$ and scale to obtain $k_3k_2k_1QAQ^{-1} = k_3k_2k_1QSQ^{-1} + k_3k_2k_1QNQ^{-1}$. Because conjugation by $D_j$ divides entries by at most $j^n$, our choice of $k_3$ ensures that we can safely conjugate $k_3k_2k_1QSQ^{-1}$.

Let $k = k_3k_2k_1$ and $\Lambda_j = D_jQ$. Conjugating by $D_j$, we get $k\Lambda_jA\Lambda_j^{-1} = k\Lambda_jS\Lambda_j^{-1} + k\Lambda_jN\Lambda_j^{-1}$, where both of the RHS summands are integer matrices. For any $X \in \text{Mat}_n(\mathbb{Q})$, define $X_j = k\Lambda_jX\Lambda_j^{-1}$, so $A_j = S_j + N_j$ and $S_jN_j = N_jS_j$.

Now, $D_\ell A_jD_\ell^{-1}$ is an integer matrix similar to $A_j$ over $\mathbb{Q}$, for $1\leq \ell \leq j$. However, $D_\ell A_jD_\ell^{-1}$ is not similar to $A_j$ over $\mathbb{Z}$. To see this, observe that if $D_\ell A_jD_\ell^{-1} \sim_\mathbb{Z} A_j$, then $A_j = PD_\ell S_j D_\ell^{-1}P^{-1} + PD_\ell N_j D_\ell^{-1}P^{-1}$ for some unimodular matrix $P$. By the uniqueness of the Jordan-Chevalley decomposition, we find that $PD_\ell N_j D_\ell^{-1}P^{-1} = N_j$. However, $D_\ell N_j D_\ell^{-1} = \ell N_j$, which cannot be similar to $N_j$ by the proof of the nilpotent case. Therefore, the rational similarity class of $A_j$ splits into at least $j$ integral similarity classes. By the discussion at the start of the proof, this implies that the rational similarity class of $A$ splits into at least $j$ integral similarity classes. Since $j$ was arbitrary, this implies the rational similarity class splits into infinitely many integer similarity classes.