Hello all,


Does anyone have an example in mind of a ring $R$ for which $R^n\cong R^m$ *as $R,R$ bimodules* for some positive integers $n\neq m$?


I would be a little surprised if someone showed no such thing could exist, but that would also be a welcome answer. 


Thanks!


P.S.: Naturally such a ring could not have [IBN](http://en.wikipedia.org/wiki/Invariant_basis_number). I don't recall deciding whether or not the "easiest" ring without IBN (the endomorphism ring of an $\aleph\_0$-dimensional vector space $V$) precluded this, so that is a starting point.



Bimodule version of IBN
No. For a bimodule $M$ let $Z(M) = \{ m : rm = mr \forall r \in R \}$. Then $Z(R^n) \cong Z(R)^n$, so if $R^n \cong R^m$ as $(R, R)$-bimodules then $Z(R)^n \cong Z(R)^m$ as $Z(R)$-modules, and commutative rings satisfy IBN. 


