## Tag `04D1`

Chapter 10: Commutative Algebra > Section 10.136: Étale ring maps

Lemma 10.136.11. 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 10.136.2 we can write $\overline{S} = (R/I)[x_1, \ldots, x_n]/(\overline{f}_1, \ldots, \overline{f}_n)$ as in Definition 10.130.6 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 $S = R[x_1, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_c, x_{n + 1}\Delta - 1)$ where $\Delta = \det(\frac{\partial f_i}{\partial x_j})_{i, j = 1, \ldots, c}$ as in Example 10.130.8. This proves the lemma. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 35062–35068 (see updates for more information).

```
\begin{lemma}
\label{lemma-lift-etale}
Let $R$ be a ring and let $I \subset R$ be an ideal.
Let $R/I \to \overline{S}$ be an \'etale ring map.
Then there exists an \'etale ring map
$R \to S$ such that $\overline{S} \cong S/IS$ as $R/I$-algebras.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-etale-standard-smooth} we can write
$\overline{S} =
(R/I)[x_1, \ldots, x_n]/(\overline{f}_1, \ldots, \overline{f}_n)$
as in Definition \ref{definition-standard-smooth} 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
$S = R[x_1, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_c, x_{n + 1}\Delta - 1)$
where $\Delta = \det(\frac{\partial f_i}{\partial x_j})_{i, j = 1, \ldots, c}$
as in Example \ref{example-make-standard-smooth}.
This proves the lemma.
\end{proof}
```

## Comments (0)

## Add a comment on tag `04D1`

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

There are no comments yet for this tag.