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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.