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I'm working on a homework question, and I'm stuck. The question is:

Let $A$ and $B$ be $2n \times 2n$ rational matrices with $A^2=B^2=-I$.

The first part of the question asks to show that $A$ and $B$ are similar, and that the transition matrix is rational. I believe I've done that. However it's this second part that has me stumped:

Suppose $A$ and $B$ have integer coefficients. Can we assume that $C$ and $C^{-1}$ ($C$ is the transition matrix) have integer coefficients as well?

The hint given was to convert $\mathbb{Z}^{2n}$ into a $\mathbb{Z}[x]$-module.

While I'm sure there are other methods to doing this, I'm interested in following the direction of the hint, it seems like an interesting method, and I would like to get better at using module theory as a practical tool.

What I've Done So Far: So for starters, in order to make $\mathbb{Z}^{2n}$ into a $\mathbb{Z}[x]$-module, we need a linear map. But it seems like there is a very natural choice in this case, namely, the map $T$ underlying the similar matrices $A$ and $B$. However, I'm not sure where to go from here. The only thing I can think of it using the structure theorem for finitely -generated modules over a PID, but $\mathbb{Z}[x]$ is not a PID, so that won't work.

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First of all, note that you have two ways of making $\mathbb{Z}^{2n}$ into a $\mathbb{Z}[x]$-module: one using the action of $A$, and one using the action of $B$. The matrices $A$ and $B$ are similar over $\mathbb{Z}$ if and only if the two modules are isomorphic (exercise). The way you have written it, it seems like you are already assuming that the two definitions will give you the same module.

Now, you want to apply the theory of modules over PIDs, but $\mathbb{Z}[x]$ is not a PID. However note, that you are given the extra information that $A^2=B^2 = -I$. So when you define your $\mathbb{Z}[x]$-module by specifying that $x$ acts through $A$, say, then you know that this action factors through $\mathbb{Z}[x]/(x^2+1)$, since $x^2+1$ acts trivially. So both with the $A$-action and the $B$-action, you are actually defining $\mathbb{Z}[x]/(x^2+1)$-modules. But what is this quotient?

I will let you take it from there.

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  • $\begingroup$ So in the first part of the question (not shown here) we were asked to show that A and B were similar. The way that I did that was by noting that since $A^{2} = B^2 = -Id$, we have that $A^2 + Id = 0$ and $B^2 + Id = 0$. So $A$ and $B$ are roots of $x^2 + 1 = 0$. Then the minimal polynomial of $A$ and $B$ divide $x^2 +1$. Since $x^2 + 1$ is irreducible over $\mathbb{Q}$, the minimal polynomial has to equal $x^2 + 1$. Then $\mathbb{Q}^{2n} \cong \mathbb{Q}[x]/(x^2+1) \times \cdots \times \mathbb{Q}[x]/(x^2+1)$ (n times), so the characteristic polynomial is $(x^2 + 1)^n$. $\endgroup$
    – Samantha Y
    Apr 9, 2012 at 3:21
  • $\begingroup$ Then choosing the basis $e_1 = 1 \bmod (x^2 +1)$ and $e_2 = x \bmod (x^2 + 1)$ for each factor, the matrix for multiplication by $x$ will be a $2n \times 2n$ matrix with $2\times 2$ blocks down the diagonal of the form $\left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$. $A$ and $B$ should both be similar to this matrix, right? Then since $A$ and $B$ are similar, they represent some linear map $T: \mathbb{Q}^{2n} \rightarrow \mathbb{Q}^{2n}$ but one after a change of basis. I thought you used linear maps, not matrices, to convert vector spaces into $F[x]$-modules. $\endgroup$
    – Samantha Y
    Apr 9, 2012 at 3:30
  • $\begingroup$ You do use linear maps to define $R[x]$-modules (where $R$ is a sensible ring, and you use $R$-linear maps on $R$-free modules). But the point is that you don't know yet, whether $A$ and $B$ necessarily define the same $\mathbb{Z}$-linear map. That's what you are asked to find out. $\endgroup$
    – Alex B.
    Apr 9, 2012 at 10:23
  • $\begingroup$ So the similarity of A and B via the first half of the question (when the matrices are rational) doesn't imply the similarity of A and B for the second half of the question (when they're integral)? $\endgroup$
    – Samantha Y
    Apr 9, 2012 at 15:20
  • $\begingroup$ No, it doesn't. That's precisely the gist of the question. $\endgroup$
    – Alex B.
    Apr 9, 2012 at 16:03

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