Lemma 86.14.10. Let $A$ be a Noetherian adic topological ring. Denote $A\{ x_1, \ldots , x_ n\}$ the restricted power series over $A$. Let $\mathfrak q \subset A\{ x_1, \ldots , x_ n\}$ be a prime ideal. Set $\mathfrak q' = A[x_1, \ldots , x_ n] \cap \mathfrak q$ and $\mathfrak p = A \cap \mathfrak q$. If $\mathfrak q$ and $\mathfrak p$ are rig-closed, then the map

$A[x_1, \ldots , x_ n]_{\mathfrak q'} \to A\{ x_1, \ldots , x_ n\} _\mathfrak q$

defines an isomorphism on completions with respect to their maximal ideals.

Proof. By Lemma 86.14.4 the ring map $A/\mathfrak p \to A\{ x_1, \ldots , x_ n\} /\mathfrak q$ is finite. For every $m \geq 1$ the module $\mathfrak q^ m/\mathfrak q^{m + 1}$ is finite over $A$ as it is a finite $A\{ x_1, \ldots , x_ n\} /\mathfrak q$-module. Hence $A\{ x_1, \ldots x_ n\} /\mathfrak q^ m$ is a finite $A$-module. Hence $A[x_1, \ldots , x_ n] \to A\{ x_1, \ldots , x_ n\} /\mathfrak q^ m$ is surjective (as the image is dense and an $A$-submodule). It follows in a straightforward manner that $A[x_1, \ldots , x_ n]/(\mathfrak q')^ m \to A\{ x_1, \ldots , x_ n\} /\mathfrak q^ m$ is an isomorphism for all $m$. From this the lemma easily follows. Hint: Pick a topologically nilpotent $g \in A$ which is not contained in $\mathfrak p$. Then the map of completions is the map

$\mathop{\mathrm{lim}}\nolimits _ m \left(A[x_1, \ldots , x_ n]/(\mathfrak q')^ m\right)_ g \longrightarrow \left(A\{ x_1, \ldots , x_ n\} /\mathfrak q^ m\right)_ g$

Some details omitted. $\square$

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