A example of a monoidal non symmetric category of $R$-bimodules

Question

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)

Edit:

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)?

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.