Let $f'$ be the unique unital algebra homomorphism of $A(X)$ into B extending $f$ (Algebra, Chapter III, § 2, no. 7, Proposition 7). We show that $f'$ is continuous: $f'(A^n(X)) \subset B_{n\lambda}$ whence $f'(\hat{A}_n(X) \cap A(X)) \subset B_{n\lambda}$. Therefore $f'$ can be extended in one and only one way by continuity to a homomorphism $\hat{f} : \hat{A}(X) \to B$.

We preserve the hypotheses and notation of Proposition 1 and let $u \in \hat{A}(X)$. The element $\hat{f}(u)$ is denoted by $u((f(x))_{x \in X})$ and called the result of substituting the f(x) for the x in u. In particular, $u((x)_{x \in X}) = u$. Now let $u = (u_y)_{y \in Y}$ be a family of elements of $\hat{A}_1(X)$ and let $v \in \hat{A}(Y)$. The above allows us to define the element $v((u_y)_{y \in Y}) \in \hat{A}(X)$. It is denoted by $v \circ u$. As $u_y((f(x))) \in B_\lambda$, the elements $u_y((f(x)))$ can be substituted for the y in v. The mappings $v \mapsto (v \circ u)((f(x)))$ and $v \mapsto v((u_y((f(x))))))$ are then two continuous homomorphisms of unital algebras of $\hat{A}(X)$ into B taking the same value $u_y((f(x)))$ at the element $y \in Y$. Therefore (Proposition 1)

for all $v \in \hat{A}(Y)$.

If $a$ is invertible in $\hat{\mathbf{A}}(\mathbf{X})$, $\varepsilon(a)$ is invertible in $K$. Conversely, if $\varepsilon(a)$ is invertible in $K$, there exists $u \in \hat{\mathbf{A}}_1(\mathbf{X})$ such that $a = \varepsilon(a)(1-u)$; we write $b = \left( \sum_{n \geq 0} u^n \right) \varepsilon(a)^{-1}$. Then $ab = ba = 1$ and $a$ is invertible.

The set of elements of $\hat{\mathbf{A}}(\mathbf{X})$ of constant term 1 is therefore a subgroup of the multiplicative monoid $\hat{\mathbf{A}}(\mathbf{X})$, called the Magnus group constructed on $\mathbf{X}$ (relative to $K$). In this chapter it will be denoted by $\Gamma(\mathbf{X})$ or simply $\Gamma$. For every integer $n \geq 1$, we denote by $\Gamma_n$ the set of $a \in \Gamma$ such that $\omega(a-1) \geq n$. By Proposition 2 of § 4, no. 5, the sequence $(\Gamma_n)_{n \geq 1}$ is an integral central filtration on $\Gamma$.

We first prove three lemmas.

Let $m$ be a maximal ideal of $K$ and $k$ the field $K/m$; let $p : \hat{\mathbf{A}}_K(\mathbf{X}) \to \hat{\mathbf{A}}_k(\mathbf{X})$ be the unique continuous homomorphism of unital $K$-algebras such that $p(x) = x$ for $x \in \mathbf{X}$ (no. 1, Proposition 1). It suffices to prove that $p\left( \prod_{i=1}^s (1+x_i)^{n_i} \right) \neq 1$ and the problem is reduced to the case where $K$ is a field.

In the notation of Lemma 2:

By Lemma 2, there exist integers $a_i > 0$ such that $c_{a_i, n_i} \neq 0$ ($1 \leq i \leq s$). By Algebra, Chapter I, § 7, no. 4, Proposition 6, no monomial $x_1^{b_1} \cdots x_s^{b_s}$ such that $b_i \geq 0$ and $(b_1, \ldots, b_s) \neq (a_1, \ldots, a_s)$ can be equal to $x_1^{a_1} \ldots x_s^{a_s}$. It follows that the coefficient of $x_1^{a_1} \ldots x_s^{a_s}$ in $\prod_{i=1}^s (1 + x_i)^{n_i}$ is $c_{a_1, n_1} \ldots c_{a_s, n_s} \neq 0$, which implies the result.

$\sigma(x) \equiv x \mod \hat{A}_2(X)$ 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.

Loosely speaking, the graded Lie $\mathbf{Z}$-algebra associated with the free group $F(X)$ (with the lower central series) is the free Lie $\mathbf{Z}$-algebra $L_\mathbf{Z}(X)$.

We write $F(X) = F, \Gamma(X) = \Gamma, \hat{A}(X) = \hat{A}, \hat{A}_\mathbf{Z}(X) = \hat{A}_\mathbf{Z}, C^n F(X) = C^n, \Gamma_n = 1 + \hat{A}_n(X)$ and let $\alpha : L_\mathbf{Z}(X) \to g$ be the homomorphism introduced in the statement of Theorem 3.

(A) Preliminary reductions.

Let $\gamma$ denote the homomorphism of $F$ into $\Gamma$ defined by $\gamma(x) = 1 + x$ for $x \in X$. By Lemma 4 there exists an automorphism $\sigma$ of the algebra $\hat{A}$ compatible with the filtration on $\hat{A}$ and such that $\sigma(1 + x) = g(x)$ for all $x \in X$; then $\sigma(\Gamma_n) = \Gamma_n$ for all $n$. As the homomorphisms $g$ and $\sigma \circ \gamma$ of $F$ into $\Gamma$ coincide on $X, g = \sigma \circ \gamma$ and hence $g^{-1}(\Gamma_n) = \gamma^{-1}(\Gamma_n)$. Under the hypotheses of Theorem 2, $\mathbf{Z}$ can be identified with a subring of $K$; the Magnus algebra $\hat{A}_\mathbf{Z}$ is therefore identified with a subring of $\hat{A}$ and the filtration on $\hat{A}_\mathbf{Z}$ is induced by that on $\hat{A}$. As $\gamma$ maps $F$ into $\hat{A}_\mathbf{Z}$, we see that it suffices to prove Theorems 2 and 3 under the supplementary hypotheses $K = \mathbf{Z}, r = 0$ and hence $g = \gamma$, hypotheses which we shall henceforth make.

(B) Surjectivity of $\alpha$.

As $X$ generates the group $F = C^1$, the set $c(X)$ generates the $\mathbf{Z}$-module $g^1 = C^1/C^2$. But $g^1$ generates the Lie $\mathbf{Z}$-algebra $g$ (§ 4, no. 6, Proposition 5) and hence $c(X)$ generates $g$, which proves that $\alpha$ is surjective.

(C) We identify the graded algebra $\mathrm{gr}(\hat{A})$ with $A(X)$ under the canonical isomorphisms $A^n(X) \to \hat{A}_n/\hat{A}_{n+1}$. For every integer $n \geq 1$, we write $F^n = \gamma^{-1}(\Gamma_n)$; we know (§ 4, no. 5) that $(F^n)_{n \geq 1}$ is an integral central filtration on $F$. Let $g'$ denote the associated graded Lie $\mathbf{Z}$-algebra (§ 4, no. 4). Let $f$ be the Lie algebra homomorphism of $g'$ into $A(X)$ associated with $\gamma$ (§ 4, no. 5, Proposition 3). Now $C^n \subset F^n$ for every integer $n \geq 1$ (§ 4, no. 6, Proposition 4) and hence there is a canonical homomorphism $\varepsilon$ of $g = \bigoplus_{n \geq 1} C^n/C^{n+1}$ into $g' = \bigoplus_{n \geq 1} F^n/F^{n+1}$

We write $\beta = f \circ \varepsilon$; we give $\beta$ explicitly as follows: if $u$ is the class modulo $C^{n+1}$ of an element $w$ of $C^n$, then $\gamma(w) - 1$ is of order $\geq n$ in $\hat{A}$ and $\beta(u)$ is the homogeneous component of $\gamma(w) - 1$ of degree $n$. In particular,

(D) Proof of Theorems 2 and 3.

The Lie algebra homomorphism $\beta \circ \alpha : L_\mathbf{Z}(X) \to A(X)$ restricted to $X$ is the identity by (3) and hence is the canonical injection (§ 3, no. 1). Therefore $\alpha$ is injective and hence bijective by (B); this proves Theorem 3. As $\beta \circ \alpha = f \circ \varepsilon \circ \alpha$ is injective and $\alpha$ is bijective, $\varepsilon$ is injective. For all $n \geq 1$,

is injective and hence

C^1 = F = F^1; if C^n = F^n, then C^n \cap F^{n+1} = F^{n+1} whence C^{n+1} = F^{n+1} which proves Theorem 2 by induction on n \geqslant 1.

Applying Theorem 2 with K = \mathbf{Z} and r = 0,

Arguing by induction on $n$, it suffices to prove that $F_n^{(p)}(X)/F_{n+1}^{(p)}(X)$ is a finite commutative $p$-group for all $n \geqslant 1$. For all $w \in F_n^{(p)}(X)$, the element $\gamma(w) - 1$ of $\hat{A}(X)$ is of order $\geqslant n$; we denote by $\delta_n(w)$ the homogeneous component of $\gamma(w) - 1$ of degree $n$. The mapping $\delta_n : F_n^{(p)}(X) \to A^n(X)$ is a homomorphism with kernel $F_{n+1}^{(p)}(X)$ (§ 4, no. 5, Proposition 3) and hence $F_n^{(p)}(X)/F_{n+1}^{(p)}(X)$ is isomorphic to a subgroup of $A^n(X)$. Since $X$ is finite, $A^n(X)$ is a finite-dimensional vector space over $\mathbf{F}_p$ and hence a finite commutative $p$-group and so is $F_n^{(p)}(X)/F_{n+1}^{(p)}(X)$.

There exist elements $x_1, \ldots, x_r$ of $X$ and integers $n_1, \ldots, n_r$ such that $w = x_1^{n_1} \ldots x_r^{n_r}$. Let $Y = \{x_1, \ldots, x_r\}$. The canonical injection of $Y$ into $X$ extends to a homomorphism $\alpha : F(Y) \to F(X)$; on the other hand, let $\beta$ be the homomorphism of $F(X)$ into $F(Y)$ whose restriction to $Y$ is the identity and which maps $X - Y$ to $\{1\}$. Then $\beta(\alpha(y)) = y$ for $y \in Y$ and hence $\beta \circ \alpha$ is the identity automorphism of $F(Y)$. There obviously exists $w'$ in $F(Y)$ such that $w = \alpha(w')$; then $\beta(w) = w' \neq 1$; now $\bigcap_{n \geqslant 1} F_n^{(p)}(Y) = \{1\}$ and there therefore exists an integer $n \geqslant 1$ such that $\beta(w) \notin F_n^{(p)}(Y)$. By Proposition 2, the group $G = F(Y)/F_n^{(p)}(Y)$ is a finite $p$-group. If $f$ is the composition of $\beta$ and the canonical homomorphism of $F(Y)$ onto $G$, then $f(w) \neq 1$.