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.

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    $\begingroup$ Even simpler: take a noncommutative ring $R$, and let $I$ be a left ideal which is not a right ideal. $\endgroup$ – M Turgeon Aug 20 '12 at 18:10
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    $\begingroup$ I don't know either but it's a safe bet that if you're looking for a counterexample you're going to have to pick something less well-behaved than $\mathbb{Z}$ and $\mathbb{Q}$. $\endgroup$ – Paul Z Aug 20 '12 at 18:10
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    $\begingroup$ Take $\mathbb{R} \otimes_\mathbb{Z} \mathbb{R}$ and let $\mathbb{R}$ act on each side separately. $\endgroup$ – Zhen Lin Aug 20 '12 at 18:12
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    $\begingroup$ You could also just take $\mathbb{C}$ with the following two module structures: One is given by multiplication and the other one is given by multiplication with the conjugate. $\endgroup$ – HenrikRueping Aug 20 '12 at 18:34
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    $\begingroup$ Related question $\endgroup$ – Pierre-Yves Gaillard Aug 21 '12 at 7:36

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 $n \geq 2$.


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 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.


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.

  • $\begingroup$ Every module is an $(R,R)$-bimodule by definition of an $R$-module. There is no distinction between left and right multiplication in a module so you automatically get that it is a bimodule $\endgroup$ – Paul Slevin Aug 20 '12 at 18:51
  • $\begingroup$ @PaulSlevin Again, $R-R$ bimodule does not mean "Is a left and right $R$ module". There is an additional requirement that the two actions commute with each other. It may happen in a left-right $R$ module that $(rm)s\neq r(ms)$, and that would prevent it from being a bimodule. $\endgroup$ – rschwieb Aug 20 '12 at 18:54
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    $\begingroup$ They do! for a bimodule you need (rm)s = r(ms) which are both just equal to (rs)m for modules. Over a commutative ring. Just to remind you I am talking about commutative rings here $\endgroup$ – Paul Slevin Aug 20 '12 at 18:56
  • $\begingroup$ @PaulSlevin I think there is some confusion of module operations here. There are two actions which we can write on the right side. Using different symbols for the actions, we need to show $(m\ast x)\cdot y=(m\cdot y)\ast x$. The rule $(m\cdot r)\cdot s$ doesn't cover that (right?!). Said another way: If $f:R\rightarrow End(M)$ and $g:R\rightarrow End(M)$ are the module actions, how do you prove that $Im(g)$ and $Im(f)$ commute with each other? $\endgroup$ – rschwieb Aug 20 '12 at 19:30
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    $\begingroup$ @PaulSlevin Aha, wait, now I see what you meant :) Adding to my answer. I should have seen what you meant earlier. (What I've written here isn't wrong though, by the way.) $\endgroup$ – rschwieb Aug 20 '12 at 19:34

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