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.
Best Answer
As Zhen Lin comments above, the converse is not true. For another example, suppose $S$ is a ring containing $R$ as a commutative subring such that $R$ is not in the center of $S$. Then $S$ is naturally an $(R,R)$-bimodule using the ring multiplication on $S$. But if $R$ is not in the center of $S$ then there exists $r \in R$, $s \in S$ such that $rs \neq sr$. So the left and right actions don't agree.
For a simple example, take $R$ to be the subring of diagonal matrices in the matrix ring $\mathbb{M}_n(k)$ for a field $k$ and $n \geq 2$.