Lemma 4

*Let $\sigma$ be the continuous endomorphism of $\hat{A}(X)$ such that $\sigma(x) = x + r(x)$ for $x \in X$ (no. 1, Proposition 1). Then $\sigma$ is an automorphism and $\sigma(\hat{A}_m(X)) = \hat{A}_m(X)$ for all $m \in \mathbf{N}$.

$\sigma(x) \equiv x \mod \hat{A}_2(X)$ for $x \in X$, whence, for $n \geq 1$ and $x_1, \ldots, x_n$ in $X$,

it follows by linearity that $\sigma(a) \equiv a$ modulo $\hat{A}_{n+1}(X)$ for all $a \in A^n(X)$ and in particular $\sigma(A^n(X)) \subset \hat{A}_n(X)$. It follows that $\sigma(A^n(X)) \subset \hat{A}_n(X)$ for $m \geq n$, whence $\sigma(\hat{A}_n(X)) \subset \hat{A}_n(X)$. In other words, $\sigma$ is compatible with the filtration $(\hat{A}_m(X))$ on $A(X)$ and its restriction to the associated graded ring is the identity. Hence $\sigma$ is bijective (Commutative Algebra, Chapter III, § 2, no. 8, Corollary 3 to Theorem 1).

Finally we prove Theorem 1. Let $w \neq 1$ be an element of $F(X)$. By Algebra, Chapter I, § 7, no. 5, Proposition 7, there exist $x_1, \ldots, x_s$ in $X$ and non-zero rational integers $n_1, \ldots, n_s$ such that $s \geq 1$, $x_i \neq x_{i+1}$ ($1 \leq i \leq s-1$) and

In the notation of Lemma 4, hence $g(w) \neq 1$ by Lemmas 3 and 4.