basis $\{e_{\alpha},~\alpha\in\Phi^+\}$ of ${\mathfrak{n}}$.

**Definition ..** [To a map $\xi\colon{\mathcal{B}}\to{\mathbb{C}}$ we assign the linear form $f_{\xi}=\sum_{\beta\in{\mathcal{B}}}\xi(\beta)e_{\beta}^*\in{\mathfrak{n}}^*$. We call a form $f_{\xi}$ a *Kostant form* if $\xi(\beta)\neq0$ for any $\beta\in{\mathcal{B}}\setminus\Delta$.]{}

Let $V$ be a simple ${\mathfrak{n}}$-module and $J={\mathrm{Ann}\,}_{U({\mathfrak{n}})}{V}$ be the corresponding primitive ideal of $U({\mathfrak{n}})$. By a version of Schur’s Lemma [@Dixmier4], each central element of $U({\mathfrak{n}})$ acts on $V$ as a scalar operator. For $A_{n-1}$ and $C_n$ (respectively, for $B_n$ and $D_n$), let $c_{\beta}$ be the scalar corresponding to $\Delta_{\beta}$ (respectively, to $P_{\beta}$) for $\beta\in{\mathcal{B}}$, and $J_c$ be the ideal of $U({\mathfrak{n}})$ generated by all $\Delta_{\beta}-c_\beta$ (respectively, by all $P_{\beta}-c_{\beta}$), $\beta\in{\mathcal{B}}$. Clearly, $J_c\subseteq J$. Further, since $Z({\mathfrak{n}})$ is a polynomial ring and the center of $U({\mathfrak{n}})/J$ is trivial, $J$ is centrally generated if and only if $J=J_c$.

The following result was proved in [@IgnatyevPenkov1 Theorem 3.1].

**Theorem ..** [*Suppose $\Phi$ is of type $A_{n-1}$ or $C_n$. The following conditions on a primitive ideal $J\subset U({\mathfrak{n}})$ are equivalent $$\begin{split}
&\text{\textup{i)} $J$ is centrally generated \textup{(}or\textup{,} equivalently\textup{,} $J=J_c$\textup{)}};\\
&\text{\textup{ii)} the scalars $c_{\beta}$\textup, $\beta\in{\mathcal{B}}\setminus\Delta$\textup, are nonzero};\\
&\text{\textup{iii)} $J=J(f_{\xi})$ for a Kostant form $f_{\xi}\in{\mathfrak{n}}^*$}.\\
\end{split}$$ If these\[theo:ideal\_fd\_An\_Cn\] conditions are satisfied then the map $\xi$ is reconstructed by $J$ $$\xi(\beta_k)=(-1)^{k+1}\label{formula_xi_via_c_fd_An_Cn}c_kc_{k-1}^{-1},$$ where $c_0=1$ and $c_k=c_{\beta_k}$ for 