Étale ring maps lift along surjections of rings

Lemma 10.143.10. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be an étale ring map. Then there exists an étale ring map $R \to S$ such that $\overline{S} \cong S/IS$ as $R/I$-algebras.

Proof. By Lemma 10.143.2 we can write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ n)$ as in Definition 10.137.6 with $\overline{\Delta } = \det (\frac{\partial \overline{f}_ i}{\partial x_ j})_{i, j = 1, \ldots , n}$ invertible in $\overline{S}$. Just take some lifts $f_ i$ and set $S = R[x_1, \ldots , x_ n, x_{n+1}]/(f_1, \ldots , f_ n, x_{n + 1}\Delta - 1)$ where $\Delta = \det (\frac{\partial f_ i}{\partial x_ j})_{i, j = 1, \ldots , n}$ as in Example 10.137.8. This proves the lemma. $\square$

## Comments (3)

Comment #2452 by Matthieu Romagny on

Probably $c$ should be $n$ in the ideal that defines $S$ and in the indices for $\Delta$.

Comment #5376 by slogan_bot on

Suggested slogan: Étale morphisms lift along surjections of rings

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