Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
@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

2 Answers 2

up vote 7 down vote accepted

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.

share|improve this answer
    
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 $V$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.