TITLE: Is the center of the universal enveloping algebra generated by the center of the lie algebra?
QUESTION [14 upvotes]: Let $\mathfrak{g}$ be a Lie algebra over a field $k$, and let $U(\mathfrak{g})$ be its universal enveloping algebra.  $\mathfrak{g}$ is canonically embedded in $U(\mathfrak{g})$; identify it with its image there. Fixing (totally standard) notation, let
$$Z(\mathfrak{g}) = \{x\in \mathfrak{g}\mid [x,y]=0,\forall y\in\mathfrak{g}\}$$
be the center of $\mathfrak{g}$ (as a lie algebra) and let
$$Z(U(\mathfrak{g})) = \{f\in U(\mathfrak{g}) \mid fg = gf,\forall g\in U(\mathfrak{g})\}$$
be the center of $U(\mathfrak{g})$ (as an associative algebra). The canonical embedding of $\mathfrak{g}$ in $U(\mathfrak{g})$ also embeds $Z(\mathfrak{g})$ in $Z(U(\mathfrak{g}))$.  Here is my question:

Is it the case that $Z(U(\mathfrak{g}))$ is generated as an associative $k$-algebra by $Z(\mathfrak{g})$? What's the proof?

Intuitively I believe that it should, and I have verified this by hand with an ad-hoc calculation in the special cases that $\mathfrak{g}$ is (1) the nonabelian 2-dimensional lie algebra over $\mathbb{C}$, and (2) the Heisenberg algebra.  While my calculation "feels" a little general, I haven't been able to eke the general argument out of it.
Thanks in advance for your thoughts.

REPLY [2 votes]: In general no. However, there exists an important class of Lie algebras $\mathfrak g$ called square integrable (see p. 134 https://www.sciencedirect.com/science/article/pii/S0021869303001467), introduced by Ooms, which verify that $Z(U(\mathfrak g))=U(Z(\mathfrak g))$ when the characteristic of the field is zero. Examples of such Lie algebras include the
Heisenberg Lie algebras. 
I belive that in prime characteristic the equality $Z(U(\mathfrak g))=U(Z(\mathfrak g))$ is possible only if $Z(\mathfrak g)=\mathfrak g$. Because, if $Z(\mathfrak g)$ is proper then by Lemma 5 from Chapter VI, of the Jacobson's book: Lie Algebras,  implies that there exist a $p$-polynomial expression in $Z(U(\mathfrak g))$ that is not contained in $U(Z(\mathfrak g))$. That is $Z(U(\mathfrak g))\not\subseteq U(Z(\mathfrak g))$.