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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. 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.

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    $\begingroup$ $R=0$ is an example. This also shows that Qiaochu's proof misses a tiny detail ;-). $\endgroup$
    – Martin Brandenburg
    Commented May 3, 2012 at 10:12
  • $\begingroup$ @Martin: right. Nonzero commutative rings have IBN... $\endgroup$
    – Qiaochu Yuan
    Commented May 3, 2012 at 15:58
  • $\begingroup$ planetmath.org/… $\endgroup$
    – Bhaskar Vashishth
    Commented Jun 24, 2015 at 7:02
  • $\begingroup$ @BhaskarVashishth That link is not relevant since the isomorphism is not a bimodule morphism. $\endgroup$
    – rschwieb
    Commented Jun 24, 2015 at 9:55

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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.

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  • $\begingroup$ Nice! I should have thought about it longer! The "center" of a bimodule - I think I had one-sided module blinders on. $\endgroup$
    – rschwieb
    Commented May 2, 2012 at 15:59
  • $\begingroup$ This is of course the 0-Hochschild cohomology. $\endgroup$
    – Benjamin Steinberg
    Commented May 2, 2012 at 16:03
  • $\begingroup$ @BenjaminSteinberg Thank you for the connection. To date I have not had the opportunity to learn anything about Hochschild cohomology. $\endgroup$
    – rschwieb
    Commented May 2, 2012 at 17:16

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