Lemma 2

Let $f(X_1, X_2, \ldots, X_n)$ be a formal power series $\neq 0$ with coefficients in a ring $E$. There exist integers $u(i) \geqslant 1$ ($1 \leqslant i \leqslant n - 1$) such that

Suppose that integers $u(i) \geqslant 1$ ($1 \leqslant i \leqslant k - 1$) are determined such that $f(X_n^{u(1)}, \ldots, X_n^{u(k-1)}, X_k, \ldots, X_n) \neq 0$. We shall determine an integer $u(k) \geqslant 1$ such that

The lemma will then be proved by induction.

Observe that the series $f(X_n^{u(1)}, \ldots, X_n^{u(k-1)}, X_k, \ldots, X)$ can be considered as a series in $X_k$ and $X_n$ with coefficients in $E[[X_{k+1}, \ldots, X_{n-1}]]$. Thus we see that it suffices to establish the lemma for $n = 2$.

Therefore let

where $f \neq 0$. Let $G \subset \mathbf{N} \times \mathbf{N}$ be the non-empty set of ordered pairs $(i, j)$ such that $e_{ij} \neq 0$. Let $\mathbf{N} \times \mathbf{N}$ be given the lexicographical ordering. Let $(c, d)$ be the least element of $G$. Choose an integer $p > d$. In the expansion of we look for the terms of degree $cp + d$. If $ip + j = cp + d, i \geq c + 1$ is impossible, for this would give nor is $i < c$ possible, for $(c, d)$ is the least element of $G$; therefore $i = c$ and then $j = d$. The term of degree $cp + d$ in $f(T^p, T)$ is therefore $e_{cd} T^{cp + d}$. Since $e_{cd} \neq 0, f(T^p, T) \neq 0$. Whence the lemma.

In the ring $E[[X_1, \ldots, X_n]]$, let $a$ be the ideal of formal power series without constant term. If $w_1, \ldots, w_n$ are elements of $a$, recall that the mapping $f(X_1, \ldots, X_n) \mapsto f(w_1, \ldots, w_n)$ is the unique endomorphism $s$ of the ring $E[[X_1, \ldots, X_n]]$ such that $s(X_i) = w_i$ for $1 \leq i \leq n$ (Chapter III, § 4, no. 5, Proposition 6).

We take $w_1 = X_1 + X_n^{u(1)}, \ldots, w_{n-1} = X_{n-1} + X_n^{u(n-1)}, w_n = X_n$, where the $u(i)$ are integers $\geq 1$. Let $s'$ be the endomorphism of $E[[X_1, \ldots, X_n]]$ which maps $X_1$ to $X_1 - X_n^{u(1)}, \ldots, X_{n-1}$ to $X_{n-1} - X_n^{u(n-1)}$ and $X_n$ to $X_n$. Then $s'(s(X_i)) = X_i$ for $1 \leq i \leq n$ and hence $s' \circ s$ is the identity automorphism; similarly for $s \circ s'$. Hence $s$ is an automorphism.