Consider a Lie algebra ${\frak g}$, and denote its universal enveloping algebra with ${\cal U}({\frak g})$.
I've read, e.g. in these notes (Link to pdf), Remark 3.1.3 at page 19, that the two-sided ideal $\mathcal U_+$ is generated by the image of $\mathfrak g$ in $\mathcal U(\mathfrak g)$, where here $\mathcal U_+$ denotes the subspace of $\mathcal U(\mathfrak g)$ of elements of degree more than $0$.
I don't understand how this is possible. I've read e.g. in the relevant Wikipedia page that $\mathcal U(\mathfrak g)$ is generated by products of elements of $\mathfrak g$ of the form $x_1^{k_1}\cdots x_n^{k_n}$ (when $\mathfrak g$ is finite-dimensional). While nontrivial, I can understand this being the case. However, this sounds different than saying that $\mathcal U_+$ is "generated by the image of $\mathfrak g$ in the UEA". I understand the mapping $\mathfrak g\to\mathcal U(\mathfrak g)$ as sending $g\in\mathfrak g$ into the corresponding rank-1 element in $\mathcal U(\mathfrak g)$. If $\mathfrak g$ is finite-dimensional, how can the image of this mapping not be also finite-dimensional? This seems at odds with the fact that $\mathcal U(\mathfrak g)$ is supposed to always be infinite-dimensional, as stated e.g. in the Wikipedia page.
