TITLE: Are bimodules over a commutative ring always modules?
QUESTION [10 upvotes]: Let $R$ be a commutative ring. It is true that every module over $R$ is an $(R,R)$-bimodule. Is the converse true? In other words is it possible that there is an $R$-module where left multiplication and right multiplication do not co-incide? 
I thought perhaps a counterexample would be of the following form. If $S$ is an $R$-algebra, if we form the product $S \otimes_R S$ and think of it as an $S$-module where multiplication is given by $s(s' \otimes s'') = ss' \otimes s''$ and $(s' \otimes s'') s = s' \otimes s''s$. If we can pick the right element $s$ it is possible those two tensors are not equal. I cannot find any concrete counterexamples though. If you do this with $\mathbb{Q} \otimes_\mathbb{Z} \mathbb Q$ for example it doesn't work.
Thanks for any help.

REPLY [2 votes]: A bit more complicated but modern example: let $A$ and $B$ be curved $A_\infty$ algebras or  differential graded curved algebras and $M$ be an $A$-$B$-$A_\infty$-bimodule (dg curved algebras appear naturally in Hochschild theory and theoretical physics).
It follows that $M$ is neither a left $A_\infty$-$A$-module nor a right $A_\infty$-$B$-module due to the inclusion of the curvatures in $A$ and $B$ in the $A_\infty$ module relations. 
Even simpler, if $A$ is a dg curved algebra, then it is not a left dg $A$-module, a right dg $A$-module or a dg $A$-$A$-bimodule.