Lemma 4

Let $x, y, h\in A$ be such that

(i) For $m, n\in \mathbf{N}$, we have

$xy$ = $yx$. (16)

$p\geq 0$

(ii) Let $A'$ be the $\mathbf{Z}$-subalgebra of A generated by the $x^{(m)}$ and the $y^{(m)}$ for $m\in \mathbf{N}$. Then $(^h_n)\in A'$ for all $n\in \mathbf{N}$.

Formula (16) can be written in the equivalent form

(ad $x$)$y$ = $yx.(17_m)$

$p\geq 1$

This is trivial for $m= 0$. We argue by induction on $m$. From $(17_m)$, we obtain

$(m+$ 1)(ad $x^{(n)}$)$y^{(m+1)}=$ (ad $x^{(n)}$)$y^{(m)}.y+y^{(m)}$.(ad $x^{(n)}$)$y$ (18)

$p\geq 1$

(§1, no. 1, Lemma 1). Now, applying the same lemma, and then Lemma 3,

we have

= $(yx+ (n-p-1-h)x^{(np1)})$

+ $(n-p-1-h)x^{(np1)}$.

Inserting this into (18), we obtain

$(m+$ 1)(ad $x^{(n)}$)$y^{(m+1)}$

$p\geq 1$ $+\sum y^{(mp)}(n-p-1-h)x^{(np1)}$ $p\geq 1$ $=\sum(m-p+ 1)y^{(mp+1)}x^{(np)}$ $p\geq 1$

$+\sum y(n-p-1-h)x^{(np1)}$.

$p\geq 0$ Changing $p$ to $p-1$ in the second sum, and regrouping the terms, we obtain

$(m+$ 1)(ad $x^{(n)}$)$y^{(m+1)}=\sum_{p\geq 1}y^{(m-p+1)}A_px^{(n-p)}$ (19)

with

$A_p= (m-p+ 1)+ (n-p-h)$.

Putting $z=m+n-p-h$, this can also be written as

$A_p=p1$!$(m-p+ 1)(z+ 1)z(z-1). . .(z-p+ 2)$

+ 1 $(z-m)z(z-1). . .(z-p+ 2)$

$(p-$ 1)!

$=p1$! $z(z-1). . .(z-p+ 2)[(m-p+ 1)(z+ 1) +p(z-m)]$

Inserting this into (19), we obtain $(17_{m+1})$, hence (i).

Assume that $(^h_p)\in A'$ for $p < n$. Then, for all $P\in \mathbf{Q}[T]$ of degree $< n$ such that $P(\mathbf{Z})\subset \mathbf{Z}$, we have $P(h)\in A'$ (no. 4, Cor. of Prop. 2). Hence, in view of (16) with $m=n$,

$(-1)^n$ =

$=-x^{(n)}y^{(n)}+\sum y^{(np)}x^{(np)}\in A$ ;

$p=0$

hence (ii) by induction on $n$.