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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.


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

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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Nice! I should have thought about it longer! The "center" of a bimodule - I think I had one-sided module blinders on. – rschwieb May 2 '12 at 15:59
This is of course the 0-Hochschild cohomology. – Benjamin Steinberg May 2 '12 at 16:03
@BenjaminSteinberg Thank you for the connection. To date I have not had the opportunity to learn anything about Hochschild cohomology. – rschwieb May 2 '12 at 17:16

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