 Solution

Use a triangular decomposition of a Lie algebra $\mathfrak{g}={\mathfrak{n}}_{-}\oplus \mathfrak{h}\oplus {\mathfrak{n}}_{+}$. The highest weight space of $M^{\chi }$ has dimension one. If $z$ is central, $z{m}_{\chi }$ is still of representation weight $\chi -\rho$ and is killed by ${\mathfrak{n}}_{+}$, soSince the module is generated by $m_{\chi }$ and $z$ commutes with $U(\mathfrak{g})$, the same scalar acts on every vector. Addition, multiplication and the unit show that $z\mapsto {\theta }_{\chi }(z)$ is an algebra homomorphism to $\mathbb{C}$.

For the polynomial assertion, the relevant variable for an element of $S(\mathfrak{h})$ is $\chi$. The dependence on $z$ itself is linear, as already follows from the scalar action. Apply the Poincaré-Birkhoff-Witt theorem ordering $U({\mathfrak{n}}_{-})U(\mathfrak{h})U({\mathfrak{n}}_{+})$. Because $z$ has adjoint representation weight zero, its surviving action on the highest vector is precisely the pure $U(\mathfrak{h})$ component $p_z$; terms with a positive root factor kill that vector, and a term with only a nontrivial negative root factor could not have representation weight zero. Since $U(\mathfrak{h})=S(\mathfrak{h})$,This specifies the intended polynomial despite the printed variable mismatch. The unadorned module symbol in that sentence is understood as $M^{\chi }$.