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.



Stable Conjugacy for Integer Matrices
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.


