# Localising a right Noetherian ring at a set of regular elements

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?

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.
• 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. – Simon Wadsley Feb 1 '17 at 14:15