Are bimodules over a commutative ring always modules? 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.
 A: 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 $n \geq 2$.
A: Supplementing other answers and comments: first, I would argue against thinking that "left" and "right" have genuine content, but I would argue in favor of thinking of these as notational artifacts of our left-to-right writing system, etc. (I am reminded again of Herstein's advocacy of writing functions on the right of their arguments, at least in English.)
E.g., an $R,S$-bimodule's main property is that the $R$-action and $S$-action commute with each other, and that the order-of-multiplication in $S$ is "backward", so (in English, with left-to-right conventions) an $R,S$-bimodule is equivalently a ("left") $R\otimes S^{\rm opp}$-module, with the "opposite" ring. 
It is true that sometimes the notational left and right are mnemonically helpful, but their content should not be over-estimated.
"Even" in looking at $Hom_?(M,N)$ with non-commutative $R$, $S,R$-bimodule $M$, and $R,T$-bimodule $N$, the "left/right" structures might better be called pre-composition and/or post-composition and such, refering to the more basic convention of order of composition of functions: those closest the argument are applied first. Luckily, the $M\otimes_R N$ structures are more directly correctly suggested by left-right conventions, but, still, can be converted to other expressions, as noted above.
Edit: forgot to emphasize that associativity is, or should, if done right, built into all these set-ups. So the $(rm)s=r(ms)$ principle ought not be something one is worrying about in the face of all the other issues.
A: 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.
A: For a commutative ring $R$, it is not true that every $R$ module is a bimodule. (Are you possibly thinking that modules which are left and right $R$ modules are called "bimodules"? This is not the case...)
Mariano Suárez-Alvarez gave an example here: An $R$ module and $S$ module that cannot be an $R$-$S$ bimodule of a left $R$ right $S$ module that is not an $R-S$ bimodule. In his example he made his $R\neq S$, but I don't see why that is necessary, because he only used dimensionality in his arguments. So, I think you can replace $S$ in the example with another copy of $R$, and still have a counterexample.
It will also show that the left and right actions of $R$ are not the same.
Over any ring it is trivial that any $R-R$ bimodule is both a left module and a right module over $R$. It's just the restriction of the bimodule action to one side.

I misled myself with the first statement, and didn't see the simple claim you were making about $R$ modules. Yes, every $R$ module has the "naive" $R-R$ bimodule structure. Sorry for the distraction! I think the rest of my answer pertains to your question though! It produces separate module actions which do not commute.
