let R be a commutative ring, and M, N two bimodules over R, such that there exists f : M -> N an isomorphism of left R-modules, and g : M -> N an isomorphism of right R-modules. Then are M and N isomorphic as R-bimodules? I don't see any reason why it should be true, but I can't find any situation where this is false. Thanks a lot!

  • $\begingroup$ This isn't really a research level question, but you might consider a situation where $R$ is an $R$-bimodule in the standard way, vs. a situation where $R$ is a left module in the standard way but a right module in a twisted way, twisted by an involution on the ring. Think of a familiar ring or field with an involution, and you should be on your way. $\endgroup$
    – user43208
    Commented Dec 11, 2014 at 1:59

1 Answer 1


As you say, there is no reason why this should be true. Consider on the one hand $R$ regarded as an $(R, R)$-bimodule in the usual way and on the other hand $R$ regarded as an $(R, R)$-bimodule where, say, the left $R$-module has been twisted by an automorphism $\varphi : R \to R$, which is to say that left multiplication now looks like

$$L_r s = \varphi(r) s.$$

Both of these bimodules are isomorphic as left modules to $R$ and as right modules to $R$, but they usually won't be isomorphic as bimodules if $\varphi$ is nontrivial. We can try to distinguish them by their centers. Recall that the center of a bimodule $_R M_R$ is

$$Z(M) = \{ m \in M : rm = mr \forall r \in R \}.$$

Then the center of the usual $R$ is the usual center $Z(R)$, but the center of the twisted $R$ is

$$\{ s \in R : \varphi(r) s = s r \forall r \in R \}$$

and this will just be different in general (as a subgroup of $R$). For example, if $R$ is commutative, then $Z(R)$ is all of $R$, but the above cannot be all of $R$ if $\varphi$ is nontrivial since, for example, it cannot include the identity.


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