Conjugacy in $\text{GL}(2, \mathbb{Z})$ vs $\text{GL}(2, \mathbb{Q})$

Question

When are two elements $x,y\in\text{GL}(2, \mathbb{Z})$ conjugate in $\text{GL}(2, \mathbb{Q})$, but not in $\text{GL}(2, \mathbb{Z})$? Does this ever happen? I feel that it should sometimes be the case, but cannot come up with any concrete examples.

EDIT: so it is possible; but is it possible if we require the conjugating element in $\text{GL}(2, \mathbb{Q})$ to have determinant $\pm 1$?

Answer 1

For $n \times n$ matrices, not just $2 \times 2$ matrices, if the characteristic polynomial of the two matrices is irreducible over $\mathbf Q$ then the distinction between their conjugacy as rational matrices and as integral matrices is directly related to ideal classes in algebraic number theory. See here. For instance, the example given by Dietrich Burde is two matrices sharing irreducible characteristic polynomial $x^2-6x-1$, whose roots are $3\pm\sqrt{10}$, and the ring generated by either root is $\mathbf Z[\sqrt{10}]$, which has two ideal classes: see Example 3.2 at the link.

Answer 2

Take $A = (\begin{smallmatrix}0&1\\1&6\end{smallmatrix})$ and $B = (\begin{smallmatrix}3&5\\2&3\end{smallmatrix})$. These matrices are both in ${\rm GL}_2({\Bbb Z})$, but they are not conjugate in this group. Indeed, any matrix $M = (\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ satisfies $MA = BM$ if and only if $b=3a+5c$ and $d = 2a+3c$, and then $\det M = 2a^2 - 5c^2$. We can't solve $2a^2 - 5c^2 = \pm 1$ in ${\Bbb Z}$ - look at it modulo $5$. Note that any matrix $M$ in $GL_2(\Bbb Z)$ has determinant $\pm 1$. However, over $\Bbb Q$ the determinant can be every nonzero rational number. So it is easy to see, that they are conjugated in $GL_2(\Bbb Q)$. For example, take $M=(\begin{smallmatrix}1&8\\1&5\end{smallmatrix})$.

Answer 3

The matrices of order 2 $s=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$ and $d=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$ are conjugate in $\mathrm{GL}_2(\mathbf{Q})$ (these are two reflections) but not over $\mathbf{Z}$ (since their reductions modulo 2 are not conjugate).

Moreover the conjugating matrix can be chosen to have determinant one: if $s=udu^{-1}$, choose $v=\begin{pmatrix}\det(u) & 0\\0 & 1\end{pmatrix}$ and then $s=(uv^{-1})d(uv^{-1})^{-1}$.