Ideals of $U(\mathrm{gl}(n,\mathbb{C}))$ and their intersection with center Does every non-null two-sided ideal of $U(\mathrm{gl}(n,\mathbb{C}))$ have nonzero intersection with the center of $U$?
 A: The answer is yes. Let's say that a domain $R$ has the intersection property if every non-zero two-sided ideal $I$ of $R$ has non-zero intersection with the centre of $R$. The result follows from the Theorem and Lemma below.

Theorem Let $\mathfrak{g}$ be a semisimple Lie algebra over a field $k$ of characteristic zero. Then $U(\mathfrak{g})$ has the intersection property.

This is Proposition 4.2.2 of Dixmier's book Enveloping algebras. 
First one shows that the intersection of the annihilators in $U(\mathfrak{g})$ of all finite-dimensional representations of $\mathfrak{g}$ is zero; this is a non-trivial result due to Harish-Chandra. 
Now if $J$ is a non-zero two-sided ideal of $U(\mathfrak{g})$, then we can find some finite-dimensional representation $V$ of $\mathfrak{g}$ such that $J \nsubseteq I := \mathrm{Ann}(V)$. Since $\mathfrak{g}$ is semisimple, by Weyl's theorem on complete reducibility we can replace $V$ by one of its irreducible direct summands and thereby assume that $V$ is itself irreducible. Therefore $\mathrm{End}_k(V)$ is a simple ring, and it follows that
$$ \frac{J}{J \cap I} \cong \frac{I + J}{I} = \frac{U(\mathfrak{g})}{I} \cong \mathrm{End}_k(V) $$
because $I + J$ is a two-sided ideal in $U(\mathfrak{g})$ properly containing $I$. This is an isomorphism of $\mathfrak{g}$-bimodules.
Next, the adjoint representation of $\mathfrak{g}$ in $U(\mathfrak{g})$ is locally finite (i.e. is the union of its finite-dimensional submodules). Since $\mathfrak{g}$ is semisimple, the short exact sequence of $\mathfrak{g}$-modules
$$ 0 \to J \cap I \to J \to \mathrm{End}_k(V) \to 0 $$
splits, so we can find a $\mathfrak{g}$-module complement $W$ for $J \cap I$ inside $J$. Now the centre $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ is just the set of $\mathfrak{g}$-invariants in $U(\mathfrak{g})$ under the adjoint representation, so that
$$ W^{\mathfrak{g}} = W \cap Z(\mathfrak{g}) \subseteq J \cap Z(\mathfrak{g}).$$
However $W \cong \mathrm{End}_k(V)$ as a $\mathfrak{g}$-module, and 
$$ \mathrm{End}_k(V)^\mathfrak{g} = \mathrm{End}_{\mathfrak{g}}(V) \neq 0$$
because, for example, the identity map $\mathrm{id} : V \to V$ is a non-zero $\mathfrak{g}$-invariant. Hence $J \cap Z(\mathfrak{g})$ is non-zero, so $U(\mathfrak{g})$ has the intersection property.

Lemma Let $R$ be a domain. If $R$ has the intersection property, then so does the polynomial ring $R[x]$.

Let $Z$ be the centre of $R$, let $S := Z \backslash \{0\}$, and let $Q := S^{-1}R$ be a partial localisation of $R$. Then $Q$ is a simple ring: if $J$ is a non-zero two-sided ideal of $Q$, then $J \cap R$ is a non-zero two-sided ideal of $R$, so by the intersection property $J$ contains a non-zero element of $Z$, which is a unit in $Q$ by construction. [In fact the converse is also true, but we will not need this.]
Let $I$ be a non-zero two-sided ideal of $R[x]$, and consider the two-sided ideal $S^{-1}I$ of $S^{-1} R[x] = Q[x]$. Choose a non-zero element $\alpha$ in $S^{-1} I$ of least possible degree:
$$\alpha = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in Q[x].$$
Then $a_n \neq 0$, so the ideal $Q a_n Q$ of $Q$ is equal to $Q$ by the simplicity of $Q$. Hence there exist elements $c_1,\ldots, c_m, d_1,\ldots, d_m \in Q$ such that $\sum_{i=1}^m c_i a_n d_i = 1$. The element
$$\alpha' := \sum_{i=1}^m c_i \alpha d_i = x^n + a_{n-1}' x^{n-1}  + \cdots + a_1'x + a_0'$$
lies in $S^{-1}I$, and is now monic. Let $u \in R$ and consider the commutator
$$[u,\alpha'] = [u, a_{n-1}']x^{n-1} + \cdots + [u,a_1'] x + [u,a_0'].$$
This is still an element of our ideal $S^{-1}I$ of degree stricly smaller than $n$, so the minimality of $n$ forces it to be zero. So $[u, a_i'] = 0$ for all $u \in R$ and all $i < n$. Because $Q = S^{-1}R$, it follows that each $a_i'$ is a central element of $Q$. Now if $a = rz^{-1} \in Q$ is central with $r \in R$ and $z \in Z$, then $r = az$ is central in $R$. So $a$ is actually an element of $F$, the field of fractions of $Z$.
Thus we have found a non-zero element $\alpha' \in S^{-1} I \cap F[x]$. Hence there is some $s \in S$ such that $s \alpha' \in I \cap F[x]$. This element is non-zero, and by clearing denominators, we see that $I \cap Z[x]$ is also non-zero. 
