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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.)

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  • $\begingroup$ Do you also want the $A_i$ to have integer entries? And $S^{-1}$? $\endgroup$
    – Qiaochu Yuan
    Commented Dec 13, 2015 at 22:44
  • $\begingroup$ No I don't have the illusion that there might be such an $S$ with determinant $1$. So, well, I could have said as well "rational" instead of "integer". BTW, as I write this comment, I just saw Eric's comments, so... $\endgroup$
    – Wolfgang
    Commented Dec 14, 2015 at 7:08
  • $\begingroup$ Okay, if you're happy with rational everywhere then you can just appeal to the structure theorem for f.g. modules over a PID (I edited with this observation). $\endgroup$
    – Qiaochu Yuan
    Commented Dec 14, 2015 at 7:47
  • $\begingroup$ Well, thinking again: there is thus no use of requiring $S$ integer. But once rational $A_i$'s are found (in theory), we can obviously find for each $i$ another base in which $A_i$ has integer entries. $\endgroup$
    – Wolfgang
    Commented Dec 14, 2015 at 11:42

2 Answers 2

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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 $m = 1$.

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  • $\begingroup$ Could you explain a bit more what you mean by Galois descent? Googling for it, it looks like it refers to something much more high-powered than seems necessary here. $\endgroup$
    – Kevin Casto
    Commented Dec 13, 2015 at 19:38
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    $\begingroup$ @Kevin: I mean Galois descent for vector spaces; see math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf for details. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 13, 2015 at 19:49
  • $\begingroup$ Probably you would not expect such a matrix to exist over Z. The problem is that if you imagine the original matrix as acting on a lattice, then as Qiaochu says you can hope to decompose the lattice tensor Q into pieces, but there's no reason this decomposition might extend to Z; the obstruction is that basis vectors for the Q-vector spaces in the lattice might be congruent mod various primes (for example primes dividing the norm of the difference of two eigenvalues which aren't conjugate) and then you might find yourself in trouble. Not a proof, but an idea of how to construct a ctrex. $\endgroup$
    – eric
    Commented Dec 13, 2015 at 22:05
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    $\begingroup$ Aah, actually, there are two interpretations of the question. If you want $S$ to have integer entries but don't care about whether its inverse does, then just take Quaicho's $S$ and scale it. If on the other hand you want the inverse to have integer entries too then doesn't a matrix like $(3,1;0,1)$ kill this question immediately? Not diagonalisable mod 2 so not diagonalisable over Z in this strict sense. $\endgroup$
    – eric
    Commented Dec 13, 2015 at 22:07
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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.

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