Theorem 1

Let $\mathscr{U}$ be a biorder in the bigebra $U(\mathfrak{g})$. Let $\mathscr{G}=\mathscr{U}\cap \mathfrak{g}$, which is an order in the Lie algebra $\mathfrak{g}$. Let $(x_i)_{i\in I}$ be a basis of $\mathscr{G}$. Give I a total order, and assume that we are given, for all $\mathbf{n}\in \mathbf{N}^I$, an element $[\mathbf{n}]$ of $\mathscr{U}$ such that $[\mathbf{n}]-\mathbf{x}^{(\mathbf{n})}$ has filtration $<|\mathbf{n}|$ in $U(\mathfrak{g})$. Then, the family of the $[\mathbf{n}]$ for $\mathbf{n}\in \mathbf{N}^I$ is a basis of the $\mathbf{Z}$-module $\mathscr{U}$.

For $p\in \mathbf{N}$, let $U_p(\mathfrak{g})$ be the set of elements of $U(\mathfrak{g})$ of filtration $\leq p$; then the images in $U_p(\mathfrak{g})/U_{p-1}(\mathfrak{g})$ of the $\mathbf{x}^{(\mathbf{n})}$ such that $|\mathbf{n}|=p$ form a basis of this $\mathbf{Q}$-vector space (Chap. I, §2, no. 7, Th. 1); hence the $[\mathbf{n}]$ form a basis of the $\mathbf{Q}$-vector space $U(\mathfrak{g})$. It remains to prove the following assertion (in which we put $M =\mathbf{N}^I$)$:$

(*) if $u\in \mathscr{U}, (a_{\mathbf{n}})\in \mathbf{Z}^{(M)}$, and $d\in$ N --$\{0\}$ are such that

then $d$ divides each $a_{\mathbf{n}}$.

For each integer $r\geq 0$, introduce the iterated coproduct

by definition, $c_0$ is the counit of $\mathscr{U},c_1=$ Id$_{\mathscr{U}},c_2=c$ (the coproduct of $\mathscr{U}$), and, for $r\geq 2,c_{r+1}$ is defined as the composite $p\circ (c_r\otimes 1)\circ c:$

where $p$ is defined by using the multiplication in the algebra $\mathbf{T}(\mathscr{U})$. Further, consider the canonical projection $\pi$ of $\mathscr{U}$ onto $\mathscr{U}^+=$ Ker $c_0$, and the composite