It is well know that if $R$ is a commutative ring, then the category of modules $_R\operatorname{Mod}$ is monoidal symmetric. (Edit. I wrote the category of $R$-bimodules, as F. Muro explained this is wrong)

A quoted result is that for (some) $R$ noncommutative the category of $R$-bimodules is a example of a monoidal non symmetrical (biclosed) category

this is quoted in :

G.M. Kelly "Basic concepts of enriched category theory" pg. 8, find this book in TAC:

"A non-symmetric example is the category of bimodules over a non-commutative ring $R$, with $\otimes_R$ as $\otimes$"

Now for to have a symmetry isn't enough a natural isomorphism, but also coherence conditions.

then I ask:

1) Exist some example of a noncommutative ring $R$ and two $R$-bimodules $M$ and $N$ such that $M\otimes_RN$ isn't isomorphic to $N\otimes_R M$? (as $R$-bimodule)

2) If this natural isomorphism exist, is $_R\operatorname{Mod}_R$ braided, or other else? (the article cited under give a deep and beautiful answer)


As showed in http://arxiv.org/pdf/1108.2575v3.pdf the question is more interesting of what should seems:

there could be braids on $A$-bimodules category and these are all symmetries, then the question become : find a ring $A$ without any canonical R-matrices (this concept and the meaning of $R$ is explained in the article).

In the commutative case we have a braid (that is a symmetry) if and only if the morphisms $Z\to A$ is a ephimorphism of rings (in the article they consider $K\to A$ where $A$ is a $K$-algebra, $K$ a commutative ring).

How about qestion (1)?

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    $\begingroup$ The category of bimodules is not symmetric even for R commutative since the action may not be the same on both sides. I'd say this question is too elementary for this forum. $\endgroup$
    – Fernando Muro
    Jun 30, 2013 at 10:01
  • $\begingroup$ Since you want something explicit, take a field k with two non commuting automorphismsand twist the right action on k by these automorphisms the two tensor products of these two k-bimodules are not isomorphic. $\endgroup$
    – Fernando Muro
    Jun 30, 2013 at 10:13
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    $\begingroup$ It's easy to fall into the trap of thinking that a non-commutative ring is a monoidal category enriched in $Ab$, where the monoidal structure is given by multiplication in the ring (whence one could use Day convolution for the modules), but: it isn't monoidal because the interchange law fails (i.e., the monoidal product will not actually be functorial). Of course, if $R$ is a bialgebra, then you can endow the category of modules with a monoidal structure. $\endgroup$
    – user43208
    Jun 30, 2013 at 12:55
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    $\begingroup$ Buschi Sergio, it seems that in your second edit you are thinking that bialgebras naturally come with bimodules. This is not so, the "bi" refers to something different in both cases. $\endgroup$ Jun 30, 2013 at 15:46
  • 2
    $\begingroup$ What? The category of $R$-bimodules is always monoidal. And don't you mean $M \otimes_R N$ is not isomorphic to $N \otimes_R M$? $\endgroup$ Jun 30, 2013 at 19:35

1 Answer 1


Let $R$ be any ring with two non-commuting automorphisms $\sigma,\tau\colon R\cong R$, $\sigma\tau\neq\tau\sigma$, where all units are contained in the center, e.g. a number field with non-commutative Galois group, or the free non-commutative algebra $\mathbb{Q}\langle x,y,z\rangle$ if you insist in looking for a non-commutative $R$.

Let $M={}_1R{}_\sigma$ be the abelian group $R$ with the following $R$-bimodule structure $a\cdot b\cdot c=ab\sigma(c)$, $a,b,c\in R$. The notation $\cdot$ indicates the bimodule structure, while juxtaposition is the ring product in $R$. Similarly, let $N={}_1R{}_\tau$.

We have an $R$-bimodule isomorphism

$$\varphi\colon M\otimes_RN\cong {}_1R{}_{\sigma\tau},\quad \varphi(a\otimes b)=a\sigma(b).$$ Similarly $N\otimes_RM\cong {}_1R{}_{\tau\sigma}$. But ${}_1R{}_{\sigma\tau}\ncong {}_1R{}_{\tau\sigma}$. Indeed, let $\psi\colon {}_1R{}_{\sigma\tau}\rightarrow {}_1R{}_{\tau\sigma}$ be an $R$-bimodule isomorphism. Since it is a left $R$-module isomorphism, it is given by $\psi(a)=ab$ for a certain unit $b\in R$ and any $a\in R$. Since it is a right module homomorphism it must satisfy $ab=b\tau\sigma\tau^{-1}\sigma^{-1}(a)$ for any $a\in R$. Since $b$ is a unit, it is in the center, and we deduce from the previous equation that $a=\tau\sigma\tau^{-1}\sigma^{-1}(a)$ for any $a\in R$, but this is false since $\sigma\tau\neq\tau\sigma$.

This will answer your first question once you correct the misprint.

Question 2) is impossible to answer since it asks about the possible existence of some a priori unespecified isomorphism.

  • $\begingroup$ Why is $b$ in the center of $R$? $\endgroup$
    – user44400
    Jun 17, 2015 at 7:30

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