Let $R$ be a right Noetherian ring, and $S$ a multiplicative set consisting of regular elements where $1\in S$ and $0\not\in S$. Does the right ring of fractions $RS^{-1}$ exist?

This is what I know so far:

1) If $S$ was $the$ set of regular elements, then this is true by Goldie's theorem;

2) It is sufficient to show $S$ is right Ore, i.e. for all $r\in R$ $s\in S$, $rS\cap SR\neq\emptyset$.

Is it true for any set containing regular elements?


It is not true that the right ring of fractions always exists in the setting you describe. An example where it does not can be found in section 2.1.7 of McConnell and Robson's book Noncommutative Noetherian Rings.

They take the enveloping algebra $U(\mathfrak{g})$ of the $2$-dimensional $k$-Lie algebra $\mathfrak{g}=kx\oplus ky$ with $[xy]=y$ (here $k$ is some field) and show that the set of elements $S$ of $U(\mathfrak{g})$ not in the 'augmentation ideal', ie the kernel of the natural counit map $U(\mathfrak{g})\to k$, is multiplicative but does not satisfy the right Ore condition. More precisely they show that $yS\cap (x-1)R=\emptyset$.

Since every non-zero element of $U(\mathfrak{g})$ is regular and $U(\mathfrak{g})$ is both left and right Noetherian this suffices.

This example (along with many others) is developed at greater length in Jategaonkar's monograph Localisation in Noetherian Rings (see 1.3.17 for this one). This should be the starting point of any serious study of these kind of questions.

  • $\begingroup$ I just noticed that you did not say that you know that $S$ being right Ore is necessary as well as sufficient for $RS^{-1}$ to exist. However, this is true and proof can be found in both the references I mention above: Proposition 2.1.6 in McConnell and Robson or 1.1.1 in Jategaonkar. $\endgroup$ Feb 1 '17 at 14:15

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.