Proposition 3

If the linearly topologized commutative ring $\mathbf{A}$ is Hausdorff and complete, the canonical homomorphism $\pi$ is a topological ring isomorphism.

For all $(n_1, \ldots, n) \in \mathbf{N}^p$ and all $3 \in \mathscr{B}$, let $\phi_{n_1 \ldots n_p}^3$ be the mapping $(A/\mathfrak{J})[X_1, \ldots, X_p] \to A/\mathfrak{J}$ which maps every polynomial to the coefficient of $X_1^{n_1} \ldots X_p^{n_p}$ in this polynomial; clearly the $\phi_{n_1 \ldots n_p}^3$ form an inverse system of $(A/\mathfrak{J})$-module homomorphisms (relative to the ordered set $\mathscr{B}$) and, as $A$ is canonically identified with $\lim_{\leftarrow} (A/\mathfrak{J})$ by hypothesis, $\phi_{n_1 \ldots n_p} = \lim_{\leftarrow} \phi_{n_1 \ldots n_p}^3$ is a continuous $A$-homomorphism from $A'$ to $A$. For every element $S = (S_\mathfrak{J})_{\mathfrak{J} \in \mathscr{B}}$ of $A'$, we shall see that the formal power series $T = \sum_{(n_i)} \phi_{n_1 \ldots n_p}(S) X_1^{n_1} \ldots X_p^{n_p}$ is restricted and satisfies $\pi(T) = S$. For all $3 \in \mathscr{B}$ and all $\mathfrak{J}' \in \mathscr{B}$ such that $\mathfrak{J}' \subset \mathfrak{J}$, the relation $\phi_{n_1 \ldots n_p}^{\mathfrak{J}'}(S_{\mathfrak{J}'}) = 0$ implies

as $S_{\mathfrak{J}}$ is a polynomial, we see that $\phi_{n_1 \ldots n_p}(S) \in \mathfrak{J}$ except for those $(n_1, \ldots, n_p)$, finite in number, such that $\phi_{n_1 \ldots n_p}^{\mathfrak{J}}(S_{\mathfrak{J}}) \neq 0$, which proves our first assertion; the second follows from the definitions. As $A$ is Hausdorff, the intersection of the $\mathbf{N}_\mathfrak{J}$ reduces to 0 and hence $\pi$ is bijective, which completes the proof, since $\pi$ is a strict morphism.