# 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.

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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-Alvarez 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

Well, I think the answer is "no". Here's a construction: let R be the ring of integers of a real quadratic field K of class number > 1, let M be an invertible R-module of rank one which is not isomorphic to R, and let x be a fundamental unit in R. Then the action of x on M (viewed as a Z-module) determines a well-defined conjugacy class C_M in GL_2(Z), and similarly the action of x on R determines a conjugacy class C_R. I claim that these conjugacy classes are distinct, but become equal in GL_2(A).

They are distinct: indeed, M is recovered up to isomorphism from C_M since R identifies with the commutant algebra of x acting on the Z-module M.

They become equal in GL_2(A): in fact they become equal in GL_2 of the ring of integers in the Hilbert class field of K, since M and R become isomorphic there.

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Thanks this is helpful. –  Steven Spallone May 1 '12 at 1:29

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 $V$.

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In case of extreme distress, you can always wrap the whole LaTeX thing in backticks ` and sometimes it'll get rid of the problem. –  Mariano Suárez-Alvarez Apr 30 '12 at 2:22
Mariano: Thanks. –  KConrad Apr 30 '12 at 2:35
Thanks, Keith -- nice to see things worked out! –  Dustin Clausen Apr 30 '12 at 4:04
Thank you very much this is good to know. –  Steven Spallone May 1 '12 at 1:29