# Stable Conjugacy for Integer Matrices

Let $F$ be a field, and $E$ an extension field. Then two matrices in $GL_n(F)$ are conjugate if and only if they are conjugate in $GL_n(E)$. I'm curious whether the analogous fact holds for rings of integers.

Is the following true?

Two matrices in $GL_n(\mathbb Z)$ are conjugate if and only if they are conjugate in $GL_n(\mathbb A)$, where $\mathbb A$ is the ring of algebraic integers.

• Fix $A$, $B\in GL_n(\mathbb Z)$ and for each ring $R\supseteq\mathbb Z$ let $\mathcal C(R)=\{C\in GL_n(R):AC=CB\}$. If $\mathcal C(\mathbb A)\neq\emptyset$, then $\mathcal C(\overline{\mathbb Q})\neq\emptyset$ and, by the result you quote, $\mathcal C(\mathbb Q)\neq\emptyset$. So in your question you can replace $\mathbb A$ by $\mathbb Q$. – Mariano Suárez-Álvarez Apr 30 '12 at 0:37
• @Mariano: you can't replace alg. int. by rationals. Your argument shows $2 \times 2$ integral matrices with nonzero det. (not just det. $\pm 1$) that are conj. by ${\rm GL}_2({\mathbf A})$ are conjugate by ${\rm GL}_2({\mathbf Q})$, but the converse is false. For example, $A=(\begin{smallmatrix}0&4\\\ 2&0\end{smallmatrix})$ and $B=(\begin{smallmatrix}0&8\\\ 1&0\end{smallmatrix})$ are conj. by $(\begin{smallmatrix}1&0\\\ 0&1/2\end{smallmatrix})$, but if they are conj. by $(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix})$ then $a=2d$ and $b=4c$, so $ad-bc=2d^2 - 4c^2$, which (contd.) – KConrad Apr 30 '12 at 2:56
• is never a unit in the algebraic integers when $c$ and $d$ are algebraic integers. So $A$ and $B$ are not conjugate by ${\rm GL}_2({\mathbf A})$. – KConrad Apr 30 '12 at 2:56

Here is an explicit realization of the counterexample suggested by Dustin. The field $${\mathbf Q}(\sqrt{10})$$ has class number 2 and its Hilbert class field is obtained by adjoining $$\sqrt{2}$$. The ring of integers $${\mathbf Z}[\sqrt{10}]$$ has (fundamental) unit $$u:=3+\sqrt{10}$$, whose minimal polynomial over $${\mathbf Q}$$ is $$T^2 - 6T - 1$$. The two ideal classes in $${\mathbf Z}[\sqrt{10}]$$ are represented by the ideals $$(1)$$ and $$(2,\sqrt{10})$$, which have $${\mathbf Z}$$-bases $$\{1,u\}$$ and $$\{2,\sqrt{10}\}$$. Multiplication by $$u$$ on these two ideals is represented, using the indicated $$\mathbf Z$$-bases, by the respective matrices $$A = (\begin{smallmatrix}0&1\\1&6\end{smallmatrix})$$ and $$B = (\begin{smallmatrix}3&5\\2&3\end{smallmatrix})$$. These matrices are both in $${\rm GL}_2({\mathbf Z})$$, they are not conjugate in this group, but they are conjugate by the matrix $$U = (\begin{smallmatrix}\sqrt{2}&5+3\sqrt{2}\\1&3+2\sqrt{2}\end{smallmatrix})$$, which lies in $${\rm GL}_2({\mathbf Z}[\sqrt{2}])$$. That is, $$UAU^{-1} = B$$. This conjugating matrix $$U$$ has determinant $$-1$$. A matrix with determinant 1 and algebraic integer entries that satisfies $$VAV^{-1} = B$$ is $$V = (\begin{smallmatrix}2\sqrt{2}&6\sqrt{2}+5\sqrt{3}\\\ \sqrt{3}&4\sqrt{2}+3\sqrt{3}\end{smallmatrix})$$.
Quite generally, the 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 $${\mathbf Z}$$ (look at it mod 5), but we can solve it in $${\mathbf Z}[\sqrt{2}]$$ using $$a = \sqrt{2}$$ and $$c = 1$$. That is how I found $$U$$. We can solve $$2a^2 - 5c^2 = 1$$ using $$a = 2\sqrt{2}$$ and $$c = \sqrt{3}$$, which is how I found $$V$$.