Lemma 15.13.2. Let $(A, I)$ be a henselian pair. The functor $B \to B/IB$ determines an equivalence between finite étale $A$-algebras and finite étale $A/I$-algebras.

Proof. Let $B, B'$ be two $A$-algebras finite étale over $A$. Then $B' \to B'' = B \otimes _ A B'$ is finite étale as well (Algebra, Lemmas 10.143.3 and 10.36.13). Now we have $1$-to-$1$ correspondences between

1. $A$-algebra maps $B \to B'$,

2. sections of $B' \to B''$, and

3. idempotents $e$ of $B''$ such that $B' \to B'' \to eB''$ is an isomorphism.

The bijection between (2) and (3) sends $\sigma : B'' \to B'$ to $e$ such that $(1 - e)$ is the idempotent that generates the kernel of $\sigma$ which exists by Algebra, Lemmas 10.143.8 and 10.143.9. There is a similar correspondence between $A/I$-algebra maps $B/IB \to B'/IB'$ and idempotents $\overline{e}$ of $B''/IB''$ such that $B'/IB' \to B''/IB'' \to \overline{e}(B''/IB'')$ is an isomorphism. However every idempotent $\overline{e}$ of $B''/IB''$ lifts uniquely to an idempotent $e$ of $B''$ (Lemma 15.11.6). Moreover, if $B'/IB' \to \overline{e}(B''/IB'')$ is an isomorphism, then $B' \to eB''$ is an isomorphism too by Nakayama's lemma (Algebra, Lemma 10.20.1). In this way we see that the functor is fully faithful.

Essential surjectivity. Let $A/I \to C$ be a finite étale map. By Algebra, Lemma 10.143.10 there exists an étale map $A \to B$ such that $B/IB \cong C$. Let $B'$ be the integral closure of $A$ in $B$. By Lemma 15.11.5 we have $B'/IB' = C \times C'$ for some ring $C'$ and $B'_ g \cong B_ g$ for some $g \in B'$ mapping to $(1, 0) \in C \times C'$. Since idempotents lift (Lemma 15.11.6) we get $B' = B'_1 \times B'_2$ with $C = B'_1/IB'_1$ and $C' = B'_2/IB'_2$. The image of $g$ in $B'_1$ is invertible. Then $B_ g = B'_ g = B'_1 \times (B_2)_ g$ and this implies that $A \to B'_1$ is étale. We conclude that $B'_1$ is finite étale over $A$ (integral étale implies finite étale by Algebra, Lemma 10.36.5 for example) and the proof is done. $\square$

Comment #2173 by JuanPablo on

I don't see why $(B'_2)_g = 0$.

But it is not necessary as $B_g=B'_g=B'_1\times(B'_2)_g$, implies $A \to B_g\to B'_1$ is étale.

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