Prove that for local rings $R,S$, $M_m(R)\cong M_n(S)\implies R\cong S$ and $m=n$

Question

Prove that for local rings (not necessarily commutative, and local means the nonunits form a two-sided ideal) $R,S$, $M_m(R)\cong M_n(S)\implies R\cong S$ and $m=n$.

My first attempt was to consider the natural embeddings $R\to M_m(R),\ S\to M_n(S)$ that sends, e.g. $r\in R$ to the scalar matrix $rI$ and $s\in S$ to $sI$, and then maybe restrict the isomorphism $M_m(R)\cong M_n(S)$ to the scalar matrices. The problem is I can't prove that this isomorphism maps scalar matrices to scalar matrices.

I am sensing there is something wrong with my attempt. After all, I didn't even use $R,S$ being local. Am I on the right track? How should I use the assumption that they are local rings?

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Related:

The same question without assuming $R,S$ local.

Commutative case

Answer 1

If you know a little Morita theory, the proof is pretty short.

1. $M_m(R)\cong M_n(S)$ would imply that $S$ is Morita equivalent to $R$, which amounts to $S\cong End(P_R)$ where $P$ is a finitely generated projective $R$ module.

2. Since $R$ is a local ring, a finitely generated projective module is isomorphic to $R^k$ for some $k$, and that makes $End(P_R)\cong M_k(R)$.

3. $M_k(R)$ has distinct maximal right ideals if $k>1$, so in order to be local it has to be that $k=1$, i.e. $S\cong R$.

Finally, suppose $M_n(R)\cong M_m(R)$ for some $m$ and $n$. If $\mathfrak m$ is the unique maximal ideal of $R$, we know $R/\mathfrak m$ is a division ring. By ideal correspondence, it has to be that $M_n(R/\mathfrak m)\cong M_m(R/\mathfrak m)$. By the Artin-Wedderburn theorem, we know that $m=n$ necessarily.