Let $R\_1,R\_2$ be (left and right) ore domains. Does $ Mat\_n(R\_1)\cong Mat\_m(R\_2)$ implie m=n and $q.f.(R\_1)\cong q.f.(R\_2)$?


An counter example, a proof or a reference is welcomed.


Thanks



Isomorphism of matrix ring over ore domain
In an non-commutative ring $A$, recall that an idempotent $\pi$ is an element with $\pi^2=\pi$. Two idempotents $\pi$ and $\pi'$ are called orthogonal if $\pi \pi' = \pi' \pi=0$.


**Claim** Let $D$ be a division algebra. Then $n$ is the maximal size of a set of nonzero mutually orthogonal idempotents in $M\_n(D)$.


**Proof** Clearly, there is a set of size $n$: Take $\pi\_k$ to be $1$ in position $(k,k)$ and $0$ everywhere else.


It is more convenient to work in a coordinate free manner: Let $V$ be a finite dimensional $D^{op}$ vector space, and let $E(V)$ be the $D^{op}$ endomorphism ring of $V$. We will show that there are at most $\dim\_{D^{op}} V$ mutually orthogonal idempotents in $E(V)$.


So, suppose that there are $r$ mutually orthogonal idempotents in $V$. We must show that $\dim\_{D^{\op}}(V) \geq r$. Our proof is by induction on $r$; the case $r=0$ is vacuous. 


Let $K$ be the kernel of $\pi\_r$ and let $I$ be the image. We claim that $K \cap I = \{ 0 \}$. Proof: If $w \in K \cap I$ then $w = \pi v$ for some $v \in V$ and $0=\pi w = \pi^2 v = \pi v = w$. 


So $\dim V \geq \dim K + \dim I$ and, as $\pi\_r \neq 0$, we get $\dim I \geq 1$. 


Using $\pi\_i \pi\_r = \pi\_r \pi\_i$, we see that each $\pi\_i$ takes $I$ and $K$ to themselves. So the $\pi\_i$ form orthogonal idempotents of $K$. Using $\pi\_i \pi\_r =0$, we see that $\pi\_i|\_I=0$, so the $\pi\_i$ are still nonzero when restricted to $K$. So, inductively, $\dim K \geq r-1$ and then $\dim V \geq r-1$. $\square$


That was division algebras, now for Ore domains. If $R$ is an Ore domain and $D$ its skew field of fractions, then $M\_n(R)$ embeds into $M\_n(D)$, so $M\_n(R)$ cannot have more than $n$ mutually orthogonal idempotents, and the example given in the above proof shows that $M\_n(R)$ does have that many. We have shown that $n$ can be recovered from a ring-theoretic property of $M\_n(R)$.


I don't know what q.f. means, but Andreas Thom, in a deleted answer, gives [an example of Smith](http://projecteuclid.org/download/pdf_1/euclid.ojm/1200692686) where $A \not \cong B$ but $M\_2(A) \cong M\_2(B)$. I believe that $M\_n(D)$ does determine $D$ if $D$ is a division ring, since I think that the above proof should show that $D \cong \pi M\_n(D) \pi$ where $\pi$ is part of a maximal list of mutually orthogonal idempotents, but that doesn't generalize to $M\_n(R)$ as far as I can see.


