# When is the enveloping algebra finitely generated over its center as a module?

Let $$g$$ be a Lie algebra with enveloping algebra $$U(g)$$ and $$Z(g)$$ the center of $$U(g)$$.

Question 1: When is $$Z(g)$$ noetherian?

Question 2: When is $$U(g)$$ a finitely generated $$Z(g)$$-module? Is this true when $$g$$ is finite dimensional?

For Question 2, I think the answer is "very rarely" in characteristic 0 (unless $$\mathfrak{g}$$ is abelian).

For example, let $$\mathfrak{g} = \langle x, y, z\rangle_{\mathbb{C}}$$ with relation $$[x,y] = z$$ and $$[x,z] = [y,z] = 0$$. It follows that $$Z(\mathfrak{g}) = \mathbb{C}[z]$$. Indeed, let $$a = \sum_{m, n, l\in \mathbb{N}} c_{m, n, l}x^my^nz^l\in Z(\mathfrak{g}),\quad c_{m,n,l}\in \mathbb{C}.$$ Then we have $$[x,a] = \sum_{\substack{m, n, l\in \mathbb{N} \\ n \ge 1}}n c_{m, n, l}x^my^{n-1}z^{l+1}$$ so $$c_{m,n,l} = 0$$ unless $$n = 0$$. Similarly, $$c_{m,n,l} = 0$$ unless $$m = 0$$.

However, in characteristic $$p > 0$$, for an arbitrary Lie algebra $$\mathfrak{g}$$ we have $$U(\mathfrak{g}^p) = \mathbb{k}[x^p\;;\;x\in \mathfrak{g}]\subset Z(\mathfrak{g})$$ so $$U(\mathfrak{g})$$ is a finite $$Z(\mathfrak{g})$$-module provided that $$\mathfrak{g}$$ is finite dimensional.

• Thanks, how is the situation when $g$ is semisimple in characteristic 0?
– Mare
Nov 7, 2020 at 8:06
• en.wikipedia.org/wiki/Harish-Chandra_isomorphism Nov 7, 2020 at 8:07
• Thanks, so it is module-finite when g is semisimple and the characteristic is 0?! (it seems wikipedia doesnt say it explicitly)
– Mare
Nov 7, 2020 at 8:09
• Ah no, sorry. I said something idiot. In the semisimple case the centre $\mathfrak{g}$ is just a ring of symmetric polynomial of dimension equal to the rank of $\mathfrak{g}$, which is smaller than the dimension of $\mathfrak{g}$. It's not module finite. Nov 7, 2020 at 11:03
• Thanks, is the universal enveloping algebra for a semisimple g module-finite over some other (canonical) commutative noetherian ring?
– Mare
Nov 7, 2020 at 13:29