An example of nonzero ring $R$ with $R^m\cong R^n$, $m\neq n$

Question

I'd like to find an example of a nonzero ring $R$ and positive integers $m,n$ with $m\neq n$ such that $R^m\cong R^n$ as $R$-modules.

Answer 1

For commutative rings with one this does not happen. For a non-commutative example take $R = \text{End}_K(K[X])$ for a field $K$. Then $R$ is a free $R$-module with basis $\{\text{id}_{K[X]}\}$. Now consider $$f_1(X^n) := \begin{cases} X^{n/2} &\mbox{if $n$ is even}, \\ 0 &\mbox{otherwise}\end{cases}$$ and $$f_2(X^n) := \begin{cases} 0 &\mbox{if $n$ is even}, \\ X^{(n-1)/2} &\mbox{otherwise}\end{cases}.$$ Then $\{f_1, f_2\}$ is also a basis of $R$, i.e. $R \simeq R^2$. Inductively we see $R \simeq R^n$.