Take the tour ×  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.  share|improve
this question Even simpler: take a noncommutative ring $R$, and
let $I$ be a left ideal which is not a right ideal. –  M Turgeon
Aug 20 '12 at 18:10 I don't know either but it's a safe bet that
if you're looking for a co