Lemma 88.14.10. 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 88.14.4 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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