Isomorphism of matrix ring over ore domain

Question

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

Answer 1

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 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.