Not sure of a better title (and I'm open to suggestions). But when following Laffey's notes on Integer Matrices, found here (and in particular, starting on page 13), I came across an alternative proof that ring of integers of the cyclotomic number field $\mathbb Q(\zeta_n)$ is $\mathbb Z[\zeta_n]$ using the sufficient condition: any matrix with characteristic polynomial $\Phi_n$ is non derogatory when reduced over any prime field. The application to cyclotomic fields used an inductive argument, starting first with prime power cyclotomic polynomials (which is straightforward), before working with the general case for $n = p^r n_0$ with $p\nmid n_0$ and $r \ge 1$. I follow the general argument all the way up to the penultimate step.

A basic outline of this argument is as follows: for $A$ with characteristic polynomial $\Phi_n$, assuming first for a contradiction that $A \pmod p$ is derogatory for some $p$ (and so necessarily $p \mid n$, hence write $n = p^r n_0$), then noting there exist matrices $B,C$ with respective minimal polynomials $\Phi_{p^r}$ and $\Phi_{n_0}$ such that $A = BC = CB$ (take $B = A^{n_0}$ and $C = A^{1 - n_0}$, for instance), we then have (by induction and the prime power case) $B$ is similar to a block diagonal matrix with identical $s \times s$ full Jordan blocks $J$ (for $s = \varphi(p^r)$) with eigenvalue 1 and $C$ is similar (now working over $\mathbb F_p(\zeta_{n_0})$) to a diagonal matrix that can be written as a block diagonal matrix with blocks $\lambda_i I$ for $I$ the $s \times s$ identity, and $\lambda_i$ a (distinct) root of $\Phi_{n_0}$. The final step then states (paraphrasing):

...since $A = BC = CB$, then $A$ is similar to the block diagonal matrix with blocks $\lambda_i J$...

which is similar to the block diagonal matrix with corresponding Jordan blocks per $\lambda_i$, which then yields $A \pmod p$ is non-derogatory, a contradiction, and hence we have $A \pmod q$ is non-derogatory for all $q$.

So that is my issue: how can we properly conclude $A$ is similar to that block diagonal matrix just by the fact $B,C$ commute? Is there some nonsingular $U$ that simultaneously puts $B,C$ into the above Jordan forms (which obviously would do the job)?

I feel like I want to focus on eigenspaces (and I don't need to work explicitly with a choice of $B,C$ for my starting $A$, but maybe it's "obvious" from their construction...?). Since $B,C$ commute, they preserve respective eigenspaces. In particular, $B$ preserves each $s$-dimensional eigenspace $V_{\lambda_i}$ of $C$. I know commuting matrices can be simultaneously upper triangularized, but I don't recall this having to do with any particular JNF. So, I wasn't entirely sure where else to go.

Edit 1: A quick reread, and Laffey does explicitly use the implication that $C$ being diagonalized would simultaneously block diagonalize $B$, with each block being similar to the full Jordan block, from which the result does follow. But I'll let my question still stand as is, since I don't fully follow why this holds.

Edit 2: And now actually writing out an attempt, it does follows as directed. If $U$ diagonalizes $C$, say to $Q = \lambda_1I\oplus \cdots\oplus \lambda_t I$, then $A^U = B^U Q = QB^U$, so now $B^U$ must preserve the eigenspaces of $Q$, and hence must be block diagonal $X_1\oplus\cdots\oplus X_t$ of the same size, with blocks each being similar to the full Jordan block with eigenvalue 1. Now it does follow and so $A^U$ is block diagonal $\lambda_1 X_1\oplus \cdots \oplus \lambda_t X_t$, etc.


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