Theorem 3

We retain the notations at the beginning of no. 7, and assume that $\mathscr{G}$ is a Chevalley order in $(\mathfrak{g},\mathfrak{h})$.

(i) $\mathscr{U}$ is a biorder in $U(\mathfrak{g})$.

(ii) Let B be a basis of R, and $(X_{\alpha})_{\alpha\in B\cup(-B)}$ a family of elements of $\mathfrak{g}$ such that $\mathscr{G}^{\alpha}=\mathbf{Z}X_{\alpha}$ for $\alpha \in B\cup (-B)$. The $\mathbf{Z}$-algebra $\mathscr{U}$ is generated by the $(^h_n)$ and the $X_{\alpha}^{(n)}(h\in \mathscr{H}, \alpha \in B\cup (-B), n\in \mathbf{N})$. If $\mathfrak{g}$ is semi-simple and $\mathscr{H}= Q(R^{\vee})$, the $\mathbf{Z}$-algebra $\mathscr{U}$ is generated by the $X_{\alpha}^{(n)}(\alpha \in B\cup (-B), n\in \mathbf{N})$.

(iii) Let B be a basis of R, $R_+$ the corresponding set of positive roots, $R_-=-R_+,\mathfrak{n}_+=\sum_{\alpha\in R_+}\mathfrak{g}^{\alpha},\mathfrak{n}_-=\sum_{\alpha\in R_-}\mathfrak{g}^{\alpha}$. Then,

Let $(h_i)_{i\in I}$ be a basis of $\mathscr{H}$. For all $\alpha \in R$, let $X_{\alpha}$ be a basis of $\mathscr{G}^{\alpha}$. Give the set $I\cup R$ a total order (we assume that $I\cap R =\emptyset$ ). For $\lambda \in I\cup R$ and $n\in \mathbf{N}$, put $e^{\langle n\rangle}_{\lambda}=(^{h_{\lambda}}_n)$ if $\lambda \in I,e^{\langle n\rangle}_{\lambda}=X_{\lambda}^{(n)}$ if $\lambda \in R$. Then the products $\prod_{\lambda\in I\cup R}e^{\langle n_{\lambda}\rangle}_{\lambda}$, where $(n_{\lambda})$ belongs to $\mathbf{N}^{I\cup R}$, form a basis of the $\mathbf{Z}$-module $\mathscr{U}$. The

products $\prod_{\lambda\in I}(^{h_{\lambda}}_{n_{\lambda}})$, where $(n_{\lambda})$ belongs to $\mathbf{N}^I$, form a basis of the $\mathbf{Z}$-module $\mathscr{U}\cap U(\mathfrak{h})$. The products $\prod_{\lambda\in R_+}X_{\lambda}^{(n_{\lambda})}$, where $(n_{\lambda})$ belongs to $\mathbf{N}^{R_+}$, form a basis of the $\mathbf{Z}$-module $\mathscr{U}\cap U(\mathfrak{n}_+)$.

Let B and $(X_{\alpha})_{\alpha\in B\cup(-B)}$ be as in (ii), and such that $[X_{-\alpha}, X_{\alpha}] =H_{\alpha}$. Let $\mathscr{U}'$ be the $\mathbf{Z}$-subalgebra of $U(\mathfrak{g})$ generated by the $(_n^h)$ and the $X_{\alpha}^{(n)}$ $(h\in \mathscr{H}, \alpha \in B\cup (-B), n\in \mathbf{N})$. We have seen in the proof of Th. 2, (i) $=\Rightarrow$ (ii), that $\mathscr{U}'$ is equal to $\mathscr{U}$ and is a biorder in $U(\mathfrak{g})$. This proves (i) and the first assertion of (ii); the second follows from Lemma 4 (ii). Assertion (iii) follows from Th. 1 (no. 3) and Prop. 3 (no. 6).