# Local-globalism for similar matrices?

My background on number theory is very weak, so please bear with me...

Given two matrices $A$ and $B$ in $\mathbb{Z}^{n\times n}$. Assume that for every prime $p$, the images of $A$ and $B$ in $\mathbb{F}_p^{n\times n}$ are similar to each other. Does this yield the existence of a matrix $X\in\mathrm{SL}_n\left(\mathbb{Z}\right)$ satisfying $AX=XB$ ? What if we additionally assume $A$ and $B$ to be similar in $\mathbb{Q}^{n\times n}$ ?

-

Your question reminds me of a classical theorem of Latimer and MacDuffee (Annals of Math. 1933). To be sure, the theorem does not answer your question, but it seems relevant.

A nice contemporary treatment of this theorem (with references) can be found at

[At the moment that I write this, math.uconn.edu seems to be down. I presume this is only temporary!]

-
Many thanks! Example 12 from this PDF (which Google has in its cache - accessible through the preview link even when the site is down) negatively answers my question. –  darij grinberg Dec 12 '09 at 23:51
Ah. Well, that was fortunate. Thank goodness for Keith Conrad and his many wonderful blurbs. –  Pete L. Clark Dec 12 '09 at 23:59
Darij, that is a nice observation. I added it to the end of Example 12 and reposted it. –  KConrad Jan 18 '10 at 0:41

The answer is no. Here is a counter-example $$\left( \begin{matrix} 0 & -5 \\\\ 1 & 0 \end{matrix} \right) \quad \mbox{and} \quad \left( \begin{matrix} -1 & -3 \\\\ 2 & 1 \end{matrix} \right).$$

Both of these matrices have characteristic polynomial $x^2+5$. For $p \neq 2$, $5$, this polynomial has no repeated factors so any matrices with this polynomial are similar. By the same argument, they are similar over $\mathbb{Q}$. By brute force computation, they are also similar at $2$ and $5$.

However, they are not similar over $\mathbb{Z}$. Consider $\mathbb{Z}^2$ as a module for $\mathbb{Z}[t]$ where $t$ acts by one of the two matrices above. Both of these matrices square to $-5$, so these are in fact $\mathbb{Z}[\sqrt{-5}]$-modules. If the matrices were similar, the similarity would give an isomorphism of $\mathbb{Z}[\sqrt{-5}]$-modules. But these are not isomorphic: the former is free on one generator while the latter is isomorphic to the ideal $\langle 2, 1+ \sqrt{-5} \rangle$.

In general, the way to classify similarity of matrices over $\mathbb{Z}$ is the following: If the matrices do not have the same characteristic polynomial over $\mathbb{Q}$, they are not similar. If they do, let $f$ be the characteristic polynomial and let $R=\mathbb{Z}[t]/f(t)$. Then your matrices give $R$-modules, and the matrices are similar if and only if the $R$-modules are isomorphic. If $R$ is the ring of integers of a number field, then $R$-modules are classified by the ideal class group. In general, they are related to the ideal class group, but there are various correction factors related to how $R$ fails to be the ring of integers of its fraction field (or how it fails to be a domain at all). I don't know the details here.

-
By the way, the details are explain in detail in the link given by Pete. –  Mariano Suárez-Alvarez Dec 13 '09 at 16:51
Concerning David's second to last sentence, the conjugacy over Z is based on ALL ideals classes for R (including non-invertible ones if R is a non-maximal order). So this problem is a natural and interesting one where the "ideal class monoid" of all R-fractional ideals classes is relevant. If someone ever asks you for a concrete reason that noninvertible ideal classes (for a non-maximal order) could be useful, this conjugacy problem over Z is the simplest example I can think of. –  KConrad Jan 18 '10 at 0:47

It is a well known fact that over any field a square matrix and its transpose are conjugated. You should look for a matrix $A$ over the integers such that $A$ and $A^t$ are non conjugated. If I did not messed up my calculations the matrix with entries $A_{1,1}=1, A_{1,2}=2, A_{1,3}=3, A_{1,4}=4$ has the property that $A$ and $A^t$ are non-conjugated by any element of $GL_2(\mathbb{Z})$. On the other hand they are conjugated over any field(you can always think of them as elements in $M_{2 \times 2}(F)$ for any field $F$).

-
I can't check your example (it leads to the question whether $3d^2-2c^2-3cd=\pm 3$ is solvable over $\mathbb Z$, what seems to require Pell equation theory) but Example 12 in the article referenced by Pete gives another example of a matrix which is not similar to its transpose over $\mathbb Z$, so I consider my question answered. Thanks anyway. –  darij grinberg Dec 13 '09 at 12:25
Your example is wrong. The matrix is A = [1,2;3,4] and if you seek an integral matrix U with determinant +/-1 satisfying UA = AU (it's easier to do the algebra for that than for UAU^(-1) = A then we find a solution: U = [0,1;1,1]. So A and its transpose are conjugate in M_2(Z). The reason I knew this in advance is that A has char. poly. X^2 - 5X - 2, whose root 2 + (1 + sqrt(33))/2 generates the ring of integers of its fraction field Q(sqrt(33)), whose class number is ONE. So any matrices in M_2(Z) with that char. polynomial are conjugate over Z (see link to my handout in Pete's answer). –  KConrad Jan 18 '10 at 0:39
@KConrad: I just read the handout and I enjoyed very much, I did not know about this cool result. I was trying to find the best possible candidates for a counterexample, i.e. $A$ and $B=A^{t}$ for some matrix $A$. To be honest after that I just plugged some numbers that seemed to work, but I was not careful about it. –  Guillermo Mantilla Feb 1 '10 at 4:03