Is it always possible to "separate" the eigenvalues of an integer matrix? Call a square matrix Galois-irreducible if all its eigenvalues are Galois conjugates of each other.  

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 $A_i$ ?

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:  

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 $?  

(It is clear that for non-trivial Jordan blocks the latter isn't possible.)
 A: 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 $m = 1$. 
A: If a matrix $M$ is not Galois-irreducile, then its minimal annihilating polynomial $\mu_M(x)$ factors as a product of two coprime non-constant polynomials, $\mu_M(x)=p_1(x)p_2(x)$. Then $\mathbb Q^n=\mathop{\rm Ker} p_1(M)\oplus \mathop{\rm Ker} p_2(M)$, both subspaces being $M$-invariant, which leads to the required representation.
