Lemma 86.8.7. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be a finite type $A$-algebra such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Let $C$ be a Noetherian $A$-algebra. Then any $A$-algebra map $B^\wedge \to C^\wedge$ of $I$-adic completions comes from a unique $A$-algebra map

$B \longrightarrow C^ h$

where $C^ h$ is the henselization of the pair $(C, IC)$ as in More on Algebra, Lemma 15.12.1. Moreover, any $A$-algebra homomorphism $B \to C^ h$ factors through some étale $C$-algebra $C'$ such that $C/IC \to C'/IC'$ is an isomorphism.

Proof. Uniqueness follows from the fact that $C^ h$ is a subring of $C^\wedge$, see for example More on Algebra, Lemma 15.12.4. The final assertion follows from the fact that $C^ h$ is the filtered colimit of these $C$-algebras $C'$, see proof of More on Algebra, Lemma 15.12.1. Having said this we now turn to the proof of existence.

Let $\varphi : B^\wedge \to C^\wedge$ be the given map. This defines a section

$\sigma : (B \otimes _ A C)^\wedge \longrightarrow C^\wedge$

of the completion of the map $C \to B \otimes _ A C$. We may replace $(A, I, B, C, \varphi )$ by $(C, IC, B \otimes _ A C, C, \sigma )$. In this way we see that we may assume that $A = C$.

Proof of existence in the case $A = C$. In this case the map $\varphi : B^\wedge \to A^\wedge$ is necessarily surjective. By Lemmas 86.8.4 and 86.3.5 we see that the cohomology groups of $\mathop{N\! L}\nolimits _{A^\wedge /\! _\varphi B^\wedge }^\wedge$ are annihilated by a power of $I$. Since $\varphi$ is surjective, this implies that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Ker}}(\varphi )^2$ is annihilated by a power of $I$. Hence $\varphi : B^\wedge \to A^\wedge$ is the completion of a finite type $B$-algebra $B \to D$, see More on Algebra, Lemma 15.105.4. Hence $A \to D$ is a finite type algebra map which induces an isomorphism $A^\wedge \to D^\wedge$. By Lemma 86.8.4 we may replace $D$ by a localization and assume that $A \to D$ is étale away from $V(I)$. Since $A^\wedge \to D^\wedge$ is an isomorphism, we see that $\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(A)$ is also étale in a neighbourhood of $V(ID)$ (for example by More on Morphisms, Lemma 37.12.3). Thus $\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(A)$ is étale. Therefore $D$ maps to $A^ h$ and the lemma is proved. $\square$

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