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Intuitively, it's not hard to believe that for a ring $R$, the matrix ring $M_{mn}(R)$ is isomorphic to $M_m(M_n(R))$. Taking a matrix in $M_{mn}(R)$ and turning the $n\times n$ blocks into single entries would give a matrix in $M_m(M_n(R))$, and taking a matrix in $M_m(M_n(R))$ and "erasing" the brackets of each entry would give something that looks like it belongs to $M_{mn}(R)$.

This is embarrassingly informal though, what's the proper way to show they are isomorphic?

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    You've described a map (and its inverse) that gives an isomorphism. I guess the only thing that isn't immediate from this is checking that the multiplication is preserved.2012-06-12
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    It's the "right" map to pick. To prove it is multiplicative, you have to convince yourself that "blockwise" matrix multiplication works. If you really wanted to be formal you could actually work out an argument using isomorphisms of $Hom(R^{mn},R^{mn})$ and induction.2012-06-12

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