Proposition 2

(i) $(^{\mathscr{V}}_{\mathbf{Z}})$ is a biorder in the bigebra $\mathbf{S}(V)$, and $(^{\mathscr{V}}_{\mathbf{Z}})\cap V =\mathscr{V}$.

(ii) The $\mathbf{Z}$-algebra $(^{\mathscr{V}}_{\mathbf{Z}})$ is generated by the $(^h_n)$ for $h\in \mathscr{V}, n\in \mathbf{N}$.

(iii) If $(h_1, . . . , h_r)$ is a basis of $\mathscr{V}$, the elements

= $\cdots$,

where $\mathbf{n}= (n_1, . . . , n_r)$ belongs to $\mathbf{N}^r$, form a basis of the $\mathbf{Z}$-module $(^{\mathscr{V}}_{\mathbf{Z}})$.

For $m\in \mathbf{N}$, put $\mathbf{S}_m(V) =\sum_{i\leq m}\mathbf{S}^i$(V), $\mathbf{S}_m(\mathscr{V}) =\sum_{i\leq m}\mathbf{S}^i(\mathscr{V})$. By Algebra, Chap. IV, §5, no. 9, Prop. 15 and Remark,

$\mathbf{S}_m(\mathscr{V})\subset \mathbf{S}_m(V)\cap (\mathscr{V}\mathbf{Z})\subset m1$!$\mathbf{S}_m(\mathscr{V})$

so $(^{\mathscr{V}}_{\mathbf{Z}})\cap V =\mathscr{V}$. Since $\mathbf{S}_m(\mathscr{V})$ is a lattice in $\mathbf{S}_m$(V), $\mathbf{S}_m(V)\cap (_{\mathbf{Z}}^{\mathscr{V}})$ is also a lattice in $\mathbf{S}_m(V)$. On the other hand, $\mathbf{S}_m(V)\cap (^{\mathscr{V}}_{\mathbf{Z}})$ is a direct factor of $\mathbf{S}_{m+1}(V)\cap (_{\mathbf{Z}}^{\mathscr{V}})$ (since the quotient is torsion-free), hence it admits a complement which is a free $\mathbf{Z}$-module. It follows that $(^{\mathscr{V}}_{\mathbf{Z}})$ is a free $\mathbf{Z}$-module. It is clear that this is a unital order in the algebra $\mathbf{S}(V)$. Let $(u_n)_{n\in\mathbf{N}}$ be a basis of the $\mathbf{Z}$-module $(^{\mathscr{V}}_{\mathbf{Z}})$. This is also a basis of the $\mathbf{Q}$-module $\mathbf{S}(V)$ and, for all

there exists a unique sequence $(v_n)$ of elements of $\mathbf{S}(V)$ such that $\varphi =\sum u_n\otimes v_n$. As above, identify $\mathbf{S}(V)$ with $A(V^*)$ and $\mathbf{S}(V)\otimes \mathbf{S}(V)$ with $A(V^*\times V^*)$. If $\varphi \in (^{\mathscr{V}\times\mathscr{V}}_{\mathbf{Z}})$, the polynomial function $x \rightarrow \varphi (x, y)$ belongs to $(^{\mathscr{V}}_{\mathbf{Z}})$ for all $y\in \mathscr{V}^*$. It follows that $v_n(y)\in \mathbf{Z}$ for all $n$ and all $y\in \mathscr{V}^*$, in other words that $v_n\in (^{\mathscr{V}}_{\mathbf{Z}})$. This proves that the coproduct maps $(^{\mathscr{V}}_{\mathbf{Z}})$ to $(_{\mathbf{Z}}^{\mathscr{V}})\otimes_{\mathbf{Z}}(_{\mathbf{Z}}^{\mathscr{V}})$. If $h\in \mathscr{V}$ and $n\in \mathbf{N}$, then $(_n^h)$ maps $u\in \mathscr{V}^*$ to the integer $(^{u(h)}_n)$, so $(^h_n)\in (_{\mathbf{Z}}^{\mathscr{V}})$. Assertion (iii) is now obtained by applying Th. 1 to the commutative Lie algebra V, and (ii) follows.