This question is about extending modules of fractions to bimodules of fractions. I would not be surprised if the result is known, I have tried looking in Goodearl and Warfield, but may have missed the point as I am not an expert in algebra. The motivation lies in looking at invertible bimodules in non commutative algebraic geometry.

Given sufficient conditions, for a ring $R$ and a multiplicative set of regular elements $X$ we can form the ring of fractions $RX^{-1}$. Now for an right $R$ module $E$, we can form the right $RX^{-1}$ module $E \otimes_R RX^{-1}$.

However suppose that $E$ is actually a $R$-bimodule. When is $E \otimes_R RX^{-1}$ then a $RX^{-1}$ bimodule?

As I see it, this question can be phrased in the same sort of terms as the Ore condition on the ring, that is given $x\in X$ and $e\in E$ is there $y\in X$ and $f\in E$ so that $ey=xf$. Is this the right approach? If so, are there general conditions saying when this construction works? Where can I find out about it?

Apologies for my lack of knowledge here, I hope that this does not sound too trivial to the experts. The idea is to apply it to localisation of bimodules over quantum groups and their homogenous spaces.

Edit: I should say that the reason why I want $E \otimes_R RX^{-1}$ rather than something potentially bigger to be the bimodule is that otherwise taking fractions will not commute with tensor product, or rather we will likely not get a monoiodal functor between bimodule categories.


1 Answer 1


Suppose $R,S,T$ are rings. Let $_RM_S$ and $_SN_T$ be bimodules. The tensor product $M\otimes_S N$ is naturally an $R$-$T$-bimodule. The "middle" $S$-structure is gone.

In your case, $E$ is an $R$-$R$-bimodule, and you are viewing $RX^{-1}$ as an $R$-$RX^{-1}$-bimodule. That makes $E\otimes_R RX^{-1}$ naturally into an $R$-$RX^{-1}$-bimodule. The left $R$-module structure is unaffected, and there is no natural way to make $E\otimes_R RX^{-1}$ have a left $RX^{-1}$ structure. Of course, the double tensor $RX^{-1}\otimes_R E\otimes_R RX^{-1}$ does have such a structure.

To justify my "no natural" claim, take $R=E=\mathbb{Z}[x]$. Define a right $R$-module structure on $E$ in the obvious way. Define a left $R$-module structure on $E$ by letting $\mathbb{Z}$ act normally but left multiplication by $x$ acts like multiplying by $0$. This makes $E$ an $R$-$R$-bimodule. Now taking $X$ to be the set of all regular elements of $R$, we have $S:=RX^{-1}=\mathbb{Q}(x)$ and $E\otimes_R S\cong R\otimes_R S\cong S=\mathbb{Q}(x)$. This is now an $R$-$S$-bimodule, with $x$ acting as $0$ from the left. If we try to invert the $0$-multiplication, we get the zero module. There is no way to make the non-zero module $E\otimes_R S$ have a left $RX^{-1}$ structure, which extends its given left $R$-module structure.


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