The center of the universal enveloping algebra of a nilpotent or solvable Lie algebra

Question

My questions are the following.

Prove that the center of the universal enveloping algebra of a nilpotent Lie algebra is generated by the center of the Lie algebra.
Give a solvable Lie algebra such that the center of its universal enveloping algebra is not generated by the center of the Lie algebra.

I heard these two are classical results, however I finally could not find the proof. Thank you.

P.S.

I know that there is a counterexample in the semi-simple Lie algebra case and I already calculated the center of universal enveloping algebras of several nilpotent Lie algebras (Heisenberg algebras, ladder algebras and so on). I want to know a general proof in the nilpotent case and I could not find such a question in the sugested.

Have a look at other posts here for this topic, e.g. here, or here. — Dietrich Burde, Jun 03 '18 at 14:19
Dietrich For the 2-dimentional non-abelian Lie algebra $L$, the center of $L$ is zero and the center of $U(L)$ is $\mathbf{C} \cdot 1$. This is not a counterexample which I am seeking. — ShyGuy, Jun 03 '18 at 14:37

ShyGuy · Accepted Answer · 2018-06-10T03:05:36.637

Excuse me but I got an answer by myself. In fact, I got a mistake but the ladder Lie algebra is a counterexample of my first question.

The ladder Lie algebra is a Lie algebra $\mathfrak{g} := \langle X_0, X_1, X_2, X_3 \rangle$ whose Lie bracket is defined by the following:

\begin{align} [X_0,X_1] = X_2, [X_0,X_2] = X_3, [X_0,X_3] = [X_1,X_2] = [X_1,X_3] = [X_2,X_3] = 0. \end{align}

In this case, the center of $\mathfrak{g}$ is $\langle X_3 \rangle$, however $X_2^2 - 2 X_1 X_3$ is contained in the center of $U(\mathfrak{g})$.

The center of the universal enveloping algebra of a nilpotent or solvable Lie algebra

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