Is the center of the universal enveloping algebra generated by the center of the lie algebra? 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.
 A: No, the universal enveloping algebra can have a nontrivial center even if the Lie algebra itself has only $0$ as center.  The (or at least a) concept you want to look up is "Casimir element; see for example http://en.wikipedia.org/wiki/Casimir_element .
A: No, it's bigger in general. This is already the case for $\mathfrak{sl}_2$. More generally see Harish-Chandra isomorphism. 
A: 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 $Z(U(\mathfrak g))\not\subseteq U(Z(\mathfrak g))$. 
A: If the center of $\mathfrak{g}$ is 0, this only means that $U(\mathfrak{g})$ has no central elements of degree 1, but there could be central elements of higher degree.  There is a natural $\mathfrak{g}$-module map $$\mathfrak{g} \otimes \mathfrak{g} \to U(\mathfrak{g}).$$ Now if, say, $\mathfrak{g}$ is semisimple so that it has a nondegenerate Killing form (a bilinear symmetric invariant form), then there are $\mathfrak{g}$-module isomorphisms $$\mathfrak{g} \otimes \mathfrak{g} \cong \mathfrak{g} \otimes \mathfrak{g}^* \cong \text{End}(\mathfrak{g})$$ Clearly $\text{End}(\mathfrak{g})$ has nonzero central elements, namely scalars, and if you take the corresponding element $\sum e_i \otimes f_i \in \mathfrak{g} \otimes \mathfrak{g}$ and its image $\sum e_i f_i \in U(\mathfrak{g})$, you get a central element of $U(\mathfrak{g})$ called the Casimir element. 
(We are not quite done; we still have to show that the element we've constructed is $\ne 0$.  This follows from the Poincare-Birkhoff-Witt theorem.)
