Let $\mathfrak{g}$ be a Lie algebra. Suppose $\mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$. Then let $P$ be the left ideal in $U(\mathfrak{g})$ generated by $\mathfrak{g}$.
I am trying to show that $S= U(\mathfrak{g}) \setminus P$ is multiplicatively closed but not left localisable.
I am pretty sure that the multiplicatively closed part is easy as elements in $P$ are those of word length at least $1$. However I am not sure how to use the condition to show that it is not left localisable.
I know that a multiplicatively closed set is left localisable iff it is a left Ore left reversible set. So I have tried now using the Left Ore condition: for any $a \in A, s \in S, As\cap Sa \ne \emptyset$.
Letting $\{x_1, ..., x_n\}$ be a basis for $\mathfrak{g}$. I took $s= x_n +1, a= x_n$. Then if $\exists \tilde a \in A, \tilde s \in S$ with $ \tilde a (x_n + 1) = \tilde s x_n$ then $(\tilde a - \tilde s)x_n + \tilde a = 0$. But then Poincaré-Birkhoff-Witt says that $\{x_1^{a_1}...x_n^{a_n} : a \in \Bbb{N}^n \}$ is a basis for $U(\mathfrak{g})$.
So $\tilde a=0$ and then $\tilde s = 0$ which contradicts $\tilde s \in S$.
However this hasn’t used the fact that that $[\mathfrak{g}, \mathfrak{g}] = \mathfrak{g}$ anywhere so I am sure I have gone wrong somewhere. Thanks in advance for any help!!