Call a square matrix **Galois-irreducible** if all its eigenvalues are Galois conjugates of each other. 



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> Let $M$ be an integer $n\times n$ matrix which is not Galois-irreducible. Is it always possible to find an integer matrix $S$ such that $S^{-1}MS=diag(A\_1,...,A\_k)$ is a block diagonal matrix with Galois-irreducible matrices $A\_i$ ?
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Intuitively this should be true, but I have no idea how to construct such a matrix $S$.


Note that this is not true when replacing everywhere "Galois-irreducibility" by "irreducibility of the characteristic polynomial" because any non-trivial Jordan blocks like $\begin{pmatrix}\lambda&1\\ 0&\lambda\\ \end{pmatrix}$ for, say, $\lambda\in\mathbb Z$, would yield a counterexample. 


In a similar vein: 



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> Suppose $M$ is this time Galois-irreducible, but with each eigenvalue of multiplicity $k$ and trivial Jordan blocks. Does there exist $S$ as above, maybe even such that $A\_1=\cdots=A\_k $? 
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(It is clear that for non-trivial Jordan blocks the latter isn't possible.)



Is it always possible to "separate" the eigenvalues of an integer matrix?
Think of $M$ first as a linear operator acting on $V = \mathbb{Q}^n$. Pass to a splitting field $K$ and consider the induced action on $V \otimes K$. This splits up into a direct sum of generalized eigenspaces of $M$, which are also then permuted by the Galois action of $G = \text{Gal}(K/\mathbb{Q})$ into orbits. By Galois descent we can find $G$-invariant bases of these sums over each Galois orbit of generalized eigenspaces. With respect to these bases, $M$ acts on each orbit by a Galois-irreducible rational matrix. If $M$ has no nontrivial Jordan blocks then you can replace "generalized eigenspaces" with "eigenspaces" and I think with a little more fiddling you get your second statement as well. 


So that's over $\mathbb{Q}$. I'm less sure what to do over $\mathbb{Z}$. Presumably you should look at the induced action on $\mathbb{Z}^n \otimes \mathcal{O}\_K$. 


**Edit:** Again over $\mathbb{Q}$, you can avoid appealing to Galois descent as follows: the structure theorem for f.g. modules over a PID implies that $V$ above, as a $\mathbb{Q}[x]$-module (where $x$ acts by $M$), breaks up as a direct sum of modules of the form $\mathbb{Q}[x]/f(x)^m$ where $f$ is irreducible. Multiplication by $x$ always acts by a Galois-irreducible matrix on such submodules. Trivial Jordan blocks means we can take $m = 1$. 


