that the quotient algebra of ${\mathbb{C}}[e]$ modulo the ideal generated by $e-c_{\beta_1}^{-1}c_{\beta_2}$ is isomorphic to ${\mathbb{C}}$, we conclude that $$U({\mathfrak{n}})/J_c\cong(U({\mathfrak{n}})/J_0)/r(J_c)\cong(U({\widetilde}{\mathfrak{n}}_e)\otimes{\mathcal{A}}_s)/r(J_c)=(U({\widetilde}{\mathfrak{n}})/{\widetilde}J_c)\otimes{\mathbb{C}}\otimes{\mathcal{A}}_s\cong{\mathcal{A}}_t\otimes{\mathcal{A}}_s\cong{\mathcal{A}}_{t+s}.$$ Thus, $J_c$ is primitive. The proof is complete.

$\square$

Now, using this proposition we will prove our first main result, Theorem \[theo:ideal\_fd\_Bn\_Dn\]. Note that each element of $S({\mathfrak{n}})$ can be considered as a polynomial function on ${\mathfrak{n}}^*$ via the natural isomorphism $({\mathfrak{n}}^*)^*\cong{\mathfrak{n}}$.

<span style="font-variant:small-caps;">Proof of Theorem \[theo:ideal\_fd\_Bn\_Dn\]</span>. $\mathrm{(ii)}\Longrightarrow\mathrm{(iii)}$. Define $\xi$ be formula (\[formula\_xi\_via\_c\_fd\_Bn\_Dn\]), and set $f_{\xi}$ to be the corresponding Kostant form. By [@Dixmier3 6.6.9 (c)], $J(f_{\xi})$ contains ${P}_{\beta}-\xi_{\beta}(f_{\xi})$ for all $\beta\in{\mathcal{B}}$. Using explicit formulas for $\xi_{\beta}$ from Subsection \[sst:center\_fd\], one can immediately check that $\xi_{\beta}(f_{\xi})=c_{\beta}$ for all $\beta\in{\mathcal{B}}$. Thus, both $J$ and $J(f_{\xi})$ contain the centrally generated ideal $J_c$. But, thanks to Proposition \[prop:J\_c\_primitive\_Bn\_Dn\], $J_c$ is primitive (and, consequently, maximal).We conclude that $J=J_c=J(f_{\xi})$, as required.

$\mathrm{(iii)}\Longrightarrow\mathrm{(i)}$. Again by [@Dixmier3 6.6.9 (c)], $c_{\beta}=\xi_{\beta}(f_{\xi})$ for all $\beta\in{\mathcal{B}}$, hence $c_{\beta}$’s satisfy (\[formula\_xi\_via\_c\_fd\_Bn\_Dn\]). Both $J$ and $J(f_{\xi})$ contain the centrally generated ideal $J_c$, and condition (ii) is satisfied, so $J_c$ is primitive and $J=J_c=J(f_{\xi})$, as required.

$\mathrm{(i)}\Longrightarrow\mathrm{(ii)}$. Assume, to the contrary, that some scalars $c_{\beta}$, $\beta\in{\mathcal{B}}\setminus\Delta$, equal zero. Suppose that $i_1$ is the minimal number such that $c_{\beta}=0$ for $\beta\in({\mathcal{B}}\setminus\Delta)\cap{\mathcal{R}}_{i_1}$. Now, as in 