Proposition 6

With the above notation, suppose that $a_0 = A$ and let $h$ denote the mapping $P \mapsto (\alpha_{e,P})_{e \in \mathbf{N}^s}$. Then $h$ is an isomorphism of the additive topological group $A$ onto the additive topological group $A^{\mathbf{N}^s}$. The polynomial ring $A[X_1, \ldots, X_s]$ is dense in $A'$; if $A$ is complete, so is $A'$.

Clearly $h$ is bijective; $V_n = h(a'_n)$ is the set of $(a_e) \in A^{\mathbf{N}^s}$ such that $a_e \in a_{n-|e|}$ for all $e \in \mathbf{N}^s$ such that $|e| \leq n$; as these elements $e$ are finite in number, $V_n$ is a neighbourhood of 0 in $A^{\mathbf{N}^s}$. Conversely, if $V$ is a neighbourhood of 0 in $A^{\mathbf{N}^s}$, there is a finite subset $E$ of $\mathbf{N}^s$ and an integer $v$ such that the conditions $a_e \in a_v$ for all $e \in E$ imply $(a_e) \in V$; if then $n$ is the greatest of the integers $v + |e|$ for $e \in E$, then $h(a'_n) \subset V$, which proves the first assertion of Proposition 6.

Moreover, with $n$ and $E$ defined as above, $h(P - \sum_{e \in E} \alpha_{e,P}X^e) \in V$ for all $P \in A'$, which shows that $A[X_1, \ldots, X_s]$ is dense in $A'$. The last assertion follows from the first and the fact that a product of complete spaces is complete.

Let $m$ be an ideal of $A$ and suppose that $(a_i)$ is the $m$-adic filtration; then, if $n$ is the ideal of $A'$ generated by $m$ and the $X_i$ ($1 \leq i \leq s$), the filtration $(a_i)$ is the $n$-adic filtration. For clearly, for all $k \geq 0$, $n^k$ is generated by the elements $aX^e$ such that $a \in m^{k-|e|}$ for all $e \in \mathbf{N}^s$ such that $|e| \leq k$, whence $n^k \subset a \&$. Let us prove conversely that $a'_k \subset n^k$. For all $P \in a'_k$, $P = P' + P''$, where $P' = \sum_{|e| < k} \alpha_{e,P}X^e$, $P'' = \sum_{|e| \geq k} \alpha_{e,P}X^e$. Clearly it is possible to write $P'' = \sum_{|e|=k} X^e Q_e$, where the $Q_e$ are elements of $A'$, whence $P'' \in n^k$; on the other hand, clearly $\alpha_{e,P}X^e \in n^k$ for all $e \in \mathbf{N}^s$, whence $P' \in n^k$. Then $n^k = a'_k$.