# Does $M_n(R_1)\cong M_n(R_2)$ imply $R_1\cong R_2$?

Let $R_1,R_2$ be two rings with identity. If for some $n\in\mathbb N$, $M_n(R_1)$ and $M_n(R_2)$ are isomorphic as rings, can we deduce that $R_1\cong R_2$? I can prove it when both $R_1,R_2$ are local or commutative.

• Multiples of the identity matrix? – jkabrg May 30 '15 at 17:21
• @user3491648 I think this method is not applicable when $R_1,R_2$ are not commutative. – Censi LI May 30 '15 at 17:26
• Since $R$ and $M_n(R)$ are Morita equivalent for every ring $R$ and $n \in \mathbb{N}$, it follows that $R_1$ and $R_2$ are Morita equivalent. According to wikipedia, this means that $R_2 \cong e M_k(R_1) e$ for some idempotent matrix $e \in M_k(R_1)$ such that $M_k(R_1) e M_k(R_1) = M_k(R_1)$. I don't know where to go from here. – Cihan Jun 10 '15 at 6:47

## 1 Answer

According to a comment by Tom Goodwillie in this MO question, there exist rings $R,S$ such that $M_2(R) \cong M_4(S) = M_2(M_2(S))$ yet $R \ncong M_2(S)$.

• Well I'd say me searching and posting a reference which answers the question is a contribution, but I've made it community wiki in case this is the norm. – Cihan Jun 10 '15 at 7:07