Let R and S be commutative rings with a 1 different from zero. Let m and n be positive integers. Assume the ring of mbym matrices over R is isomorphic to the ring of nbyn matrices over S. Does it follow that R is isomorphic to S? Does it follow that m = n? Does either of those follow from the other? I'm interested in both where R,S are finite and where R,S are infinite. (although the second question is trivial in the former case)
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Yes and yes. Let $T=M_m(R)=M_n(S)$. The center of $T$ is isomorphic to both $R$ and $S$. The $1\times m$ matrices over $R$ form an $(R,T)$bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$bimodule. Tensor these over $T$ to get an $(R,S)$bimodule. As an $S$module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $R$module, $n$ divides $m$. 

