## Tag `08XE`

Chapter 34: Descent > Section 34.4: Descent for universally injective morphisms

Theorem 34.4.26. If $A \otimes_R S$ has one of the following properties as an $S$-algebra

- (a) of finite type;
- (b) of finite presentation;
- (c) formally unramified;
- (d) unramified;
- (e) étale;
then so does $A$ as an $R$-algebra (and of course conversely).

Proof.To prove (a), choose a finite set $\{x_i\}$ of generators of $A \otimes_R S$ over $S$. Write each $x_i$ as $\sum_j y_{ij} \otimes s_{ij}$ with $y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending $e_{ij}$ to $y_{ij}$. Then $F \otimes_R S\to A \otimes_R S$ is surjective, so $\text{Coker}(F \to A) \otimes_R S$ is zero and hence $\text{Coker}(F \to A)$ is zero. This proves (a).To see (b) assume $A \otimes_R S$ is a finitely presented $S$-algebra. Then $A$ is finite type over $R$ by (a). Choose a surjection $R[x_1, \ldots, x_n] \to A$ with kernel $I$. Then $I \otimes_R S \to S[x_1, \ldots, x_n] \to A \otimes_R S \to 0$ is exact. By Algebra, Lemma 10.6.3 the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$ is a finitely generated ideal. Thus we can find finitely many elements $y_1, \ldots, y_t \in I$ such that the images of $y_i \otimes 1$ in $S[x_1, \ldots, x_n]$ generate the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$. Let $I' \subset I$ be the ideal generated by $y_1, \ldots, y_t$. Then $A' = R[x_1, \ldots, x_n]/I'$ is a finitely presented $R$-algebra with a morphism $A' \to A$ such that $A' \otimes_R S \to A \otimes_R S$ is an isomorphism. Thus $A' \cong A$ as desired.

To prove (c), recall that $A$ is formally unramified over $R$ if and only if the module of relative differentials $\Omega_{A/R}$ vanishes, see Algebra, Lemma 10.144.2 or [EGA4, Proposition 17.2.1]. Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$, the vanishing descends by Theorem 34.4.22.

To deduce (d) from the previous cases, recall that $A$ is unramified over $R$ if and only if $A$ is formally unramified and of finite type over $R$, see Algebra, Lemma 10.147.2.

To prove (e), recall that by Algebra, Lemma 10.147.8 or [EGA4, Théorème 17.6.1] the algebra $A$ is étale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$. $\square$

The code snippet corresponding to this tag is a part of the file `descent.tex` and is located in lines 1436–1452 (see updates for more information).

```
\begin{theorem}
\label{theorem-descend-algebra-properties}
If $A \otimes_R S$ has one of the following properties as an $S$-algebra
\begin{enumerate}
\item[(a)]
of finite type;
\item[(b)]
of finite presentation;
\item[(c)]
formally unramified;
\item[(d)]
unramified;
\item[(e)]
\'etale;
\end{enumerate}
then so does $A$ as an $R$-algebra (and of course conversely).
\end{theorem}
\begin{proof}
To prove (a), choose a finite set $\{x_i\}$ of generators of $A \otimes_R S$
over $S$. Write each $x_i$ as $\sum_j y_{ij} \otimes s_{ij}$ with
$y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra
on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending
$e_{ij}$ to $y_{ij}$. Then $F \otimes_R S\to A \otimes_R S$ is surjective, so
$\Coker(F \to A) \otimes_R S$ is zero and hence $\Coker(F \to A)$
is zero. This proves (a).
\medskip\noindent
To see (b) assume $A \otimes_R S$ is a finitely presented $S$-algebra.
Then $A$ is finite type over $R$ by (a). Choose a surjection
$R[x_1, \ldots, x_n] \to A$ with kernel $I$.
Then $I \otimes_R S \to S[x_1, \ldots, x_n] \to A \otimes_R S \to 0$ is exact.
By Algebra, Lemma \ref{algebra-lemma-finite-presentation-independent}
the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$
is a finitely generated ideal. Thus we can find finitely many elements
$y_1, \ldots, y_t \in I$ such that the images of $y_i \otimes 1$ in
$S[x_1, \ldots, x_n]$ generate the kernel of
$S[x_1, \ldots, x_n] \to A \otimes_R S$.
Let $I' \subset I$ be the ideal generated by $y_1, \ldots, y_t$.
Then $A' = R[x_1, \ldots, x_n]/I'$ is a finitely presented $R$-algebra
with a morphism $A' \to A$ such that $A' \otimes_R S \to A \otimes_R S$
is an isomorphism. Thus $A' \cong A$ as desired.
\medskip\noindent
To prove (c), recall that $A$ is formally unramified over $R$ if and only
if the module of relative differentials $\Omega_{A/R}$ vanishes, see
Algebra, Lemma \ref{algebra-lemma-characterize-formally-unramified} or
\cite[Proposition~17.2.1]{EGA4}.
Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$,
the vanishing descends by Theorem \ref{theorem-descent}.
\medskip\noindent
To deduce (d) from the previous cases, recall that $A$ is unramified
over $R$ if and only if $A$ is formally unramified and of finite type
over $R$, see
Algebra, Lemma \ref{algebra-lemma-formally-unramified-unramified}.
\medskip\noindent
To prove (e), recall that by
Algebra, Lemma \ref{algebra-lemma-etale-flat-unramified-finite-presentation}
or \cite[Th\'eor\`eme~17.6.1]{EGA4} the algebra
$A$ is \'etale over $R$ if and only if
$A$ is flat, unramified, and of finite presentation over $R$.
\end{proof}
```

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