Are two bimodules isomomorphic as left and right modules also isomorphic as bimodules? 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!
 A: 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. 
