Lemma 88.6.4. Let $A$ be a Noetherian G-ring. Let $I \subset A$ be an ideal. Let $B, C$ be finite type $A$-algebras. For any $A$-algebra map $\varphi : B^\wedge \to C^\wedge $ of $I$-adic completions and any $N \geq 1$ there exist
an étale ring map $C \to C'$ which induces an isomorphism $C/IC \to C'/IC'$,
an $A$-algebra map $\varphi : B \to C'$
such that $\varphi $ and $\psi $ agree modulo $I^ N$ into $C^\wedge = (C')^\wedge $.
Proof. The statement of the lemma makes sense as $C \to C'$ is flat (Algebra, Lemma 10.143.3) hence induces an isomorphism $C/I^ nC \to C'/I^ nC'$ for all $n$ (More on Algebra, Lemma 15.89.2) and hence an isomorphism on completions. Let $C^ h$ be the henselization of the pair $(C, IC)$, see More on Algebra, Lemma 15.12.1. Then $C^ h$ is the filtered colimit of the algebras $C'$ and the maps $C \to C' \to C^ h$ induce isomorphism on completions (More on Algebra, Lemma 15.12.4). Thus it suffices to prove there exists an $A$-algebra map $B \to C^ h$ which is congruent to $\psi $ modulo $I^ N$. Write $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. The ring map $\psi $ corresponds to elements $\hat c_1, \ldots , \hat c_ n \in C^\wedge $ with $f_ j(\hat c_1, \ldots , \hat c_ n) = 0$ for $j = 1, \ldots , m$. Namely, as $A$ is a Noetherian G-ring, so is $C$, see More on Algebra, Proposition 15.50.10. Thus Smoothing Ring Maps, Lemma 16.14.1 applies to give elements $c_1, \ldots , c_ n \in C^ h$ such that $f_ j(c_1, \ldots , c_ n) = 0$ for $j = 1, \ldots , m$ and such that $\hat c_ i - c_ i \in I^ NC^ h$. This determines the map $B \to C^ h$ as desired. $\square$
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