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Let $R,S,T$ be (associative, with a 1) rings, and $m,n,r,s$ be non-negative integers.


If

(a) $R$ is isomorphic to $M_{m\times m}(T)$ and $S$ is isomorphic to $M_{n\times n}(T)$ and $m\cdot r = n\cdot s$
or
(b) $R$ and $S$ both have exactly one element

then $M_{r\times r}(R)$ is isomorphic to $M_{s\times s}(S)$.



Based on the last comment on Tom Goodwillie's answer to my earlier question, (a) and (b) apparently do not exhaust the cases in which $M_{r\times r}(R)$ is isomorphic to $M_{s\times s}(S)$.


What are examples to show that? Can they be finite?

Is there a relatively simple exhaustion of when $M_{r\times r}(R)$ is isomorphic to $M_{s\times s}(S)$ in terms of $R,S,r,s$? What if $R$ and $S$ are assumed to be finite?

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1  
You can take $R$ to be the column finite, row finite matrices. Then all of its matrix rings are isomorphic to it. –  Mariano Suárez-Alvarez Jan 9 '11 at 4:03
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Off-topic, but since your profile has no contact details: you may or may not want to respond to remarks following your comment at mathoverflow.net/questions/52032 –  Yemon Choi Jan 16 '11 at 4:24
    
Thanks, I just did. (I'd completely forgotten about that question.) –  Ricky Demer Jan 17 '11 at 5:31
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