= 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, $zm_\chi$ is still of <representation weight> $\chi-\rho$ and is killed by $\mathfrak n_+$, so
$$
zm_\chi=\theta_\chi(z)m_\chi.
$$
Since 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 $U(\mathfrak h)=S(\mathfrak h)$,
$$
\boxed{\theta_\chi(z)=p_z(\chi-\rho),\qquad j_z(\chi):=p_z(\chi-\rho)\in S(\mathfrak h).}
$$
This specifies the intended <polynomial> despite the printed variable mismatch. The unadorned <module> symbol in that sentence is understood as $M^\chi$.
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