Theorem 2

Let $\mathscr{A}$ be a generating subset of the K-vector space A. The homomorphism $\chi_{\mathscr{A}}: R_K(A)\rightarrow$ (1 + TK[[T]])$^{\mathscr{A}}$ defined by the relation $\chi_{\mathscr{A}}(x) = (\chi_a(x))_{a\in\mathscr{A}}$ is injective.

Let $x$ be an element of $R_K(A)$ such that $\chi_{\mathscr{A}}(x) = 1$. By (11), we have $\theta (x)(a) = 0$ for every $a\in \mathscr{A}$, and therefore $\theta (x) = 0$ because $\theta (x)$ is a K-linear form on A and $\mathscr{A}$ generates the K-vector space A. If the characteristic of K is zero, then this implies $x= 0$ (VIII, p. 384, Corollary of Proposition 6), and therefore the result in this case. Suppose from now on that the characteristic $p$ of K is nonzero.

Let us first treat the case when K is algebraically closed. By the above and loc. cit., we then have $x\in pR_K(A)$. Let $y\in R_K(A)$ be such that $x=py$. For every element $a$ of $\mathscr{A}$, we have $\chi_a(y)^p=\chi_a(py) =\chi_a(x) = 1$. So we have $(\chi_a(y)-1)^p= 0$, and therefore $\chi_a(y) = 1$ because the ring K[[T]] is an integral domain. So $y$ belongs to the kernel of the endomorphism $\chi_{\mathscr{A}}$. It follows by induction that $x$ belongs to $p^n R_K(A)$ for every integer $n \geq 1$. Since $R_K(A)$ is a free $\mathbf{Z}$-module, this implies $x = 0$, and therefore the injectivity of $\chi_{\mathscr{A}}$ in this case.

If $K$ is no longer supposed to be algebraically closed, then we choose an algebraic closure $\overline{K}$ of $K$ and consider the diagram of groups and group homomorphisms

where $u$ is the homomorphism deduced from the extension of scalars from $K$ to $\overline{K}$ (VIII, p. 195), $i$ is the canonical injection, and $\overline{\chi}_{\mathscr{A}}$ is the homomorphism $z \mapsto (\chi_{1 \otimes a}(z))_{a \in \mathscr{A}}$. By formula (12) of III, §9, No. 1, p. 542, diagram (12) is commutative. By the above, the homomorphism $\overline{\chi}_{\mathscr{A}}$ is injective. Since $u$ is injective (VIII, p. 195, Theorem 1), the homomorphism $\chi_{\mathscr{A}}$ is injective.