# The Stacks Project

## Tag 092Z

Theorem 15.85.25 (Olivier). Let $A \to B$ be a local homomorphism of local rings. If $A$ is strictly henselian and $A \to B$ is weakly étale, then $A = B$.

Proof. We will show that for all $\mathfrak p \subset A$ there is a unique prime $\mathfrak q \subset B$ lying over $\mathfrak p$ and $\kappa(\mathfrak p) = \kappa(\mathfrak q)$. This implies that $B \otimes_A B \to B$ is bijective on spectra as well as surjective and flat. Hence it is an isomorphism for example by the description of pure ideals in Algebra, Lemma 10.107.4. Hence $A \to B$ is a faithfully flat epimorphism of rings. We get $A = B$ by Algebra, Lemma 10.106.7.

Note that the fibre ring $B \otimes_A \kappa(\mathfrak p)$ is a colimit of étale extensions of $\kappa(\mathfrak p)$ by Lemmas 15.85.7 and 15.85.16. Hence, if there exists more than one prime lying over $\mathfrak p$ or if $\kappa(\mathfrak p) \not = \kappa(\mathfrak q)$ for some $\mathfrak q$, then $B \otimes_A L$ has a nontrivial idempotent for some (separable) algebraic field extension $L \supset \kappa(\mathfrak p)$.

Let $\kappa(\mathfrak p) \subset L$ be an algebraic field extension. Let $A' \subset L$ be the integral closure of $A/\mathfrak p$ in $L$. By Lemma 15.85.23 we see that $A'$ is a strictly henselian local ring whose residue field is a purely inseparable extension of the residue field of $A$. Thus $B \otimes_A A'$ is a local ring by Lemma 15.85.24. On the other hand, $B \otimes_A A'$ is integrally closed in $B \otimes_A L$ by Lemma 15.85.22. Since $B \otimes_A A'$ is local, it follows that the ring $B \otimes_A L$ does not have nontrivial idempotents which is what we wanted to prove. $\square$

The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 23432–23437 (see updates for more information).

\begin{theorem}[Olivier]
\label{theorem-olivier}
Let $A \to B$ be a local homomorphism of local rings.
If $A$ is strictly henselian and $A \to B$ is weakly \'etale, then
$A = B$.
\end{theorem}

\begin{proof}
We will show that for all $\mathfrak p \subset A$ there is a unique
prime $\mathfrak q \subset B$ lying over $\mathfrak p$ and
$\kappa(\mathfrak p) = \kappa(\mathfrak q)$.
This implies that $B \otimes_A B \to B$ is bijective on spectra
as well as surjective and flat. Hence it is an isomorphism
for example by the description of pure ideals in
Algebra, Lemma \ref{algebra-lemma-pure-open-closed-specializations}.
Hence $A \to B$ is a faithfully flat epimorphism of rings. We get
$A = B$ by
Algebra, Lemma \ref{algebra-lemma-faithfully-flat-epimorphism}.

\medskip\noindent
Note that the fibre ring $B \otimes_A \kappa(\mathfrak p)$
is a colimit of \'etale extensions of $\kappa(\mathfrak p)$ by
Lemmas \ref{lemma-base-change-weakly-etale} and
\ref{lemma-absolutely-flat-over-field}.
Hence, if there exists more than one prime lying over $\mathfrak p$
or if $\kappa(\mathfrak p) \not = \kappa(\mathfrak q)$ for some $\mathfrak q$,
then $B \otimes_A L$ has a nontrivial idempotent for some (separable)
algebraic field extension $L \supset \kappa(\mathfrak p)$.

\medskip\noindent
Let $\kappa(\mathfrak p) \subset L$ be an algebraic field extension.
Let $A' \subset L$ be the integral closure of $A/\mathfrak p$ in $L$.
By Lemma \ref{lemma-integral-over-henselian}
we see that $A'$ is a strictly henselian local ring
whose residue field is a purely inseparable extension of the residue
field of $A$. Thus $B \otimes_A A'$ is a local ring by
Lemma \ref{lemma-local-tensor-with-integral}.
On the other hand, $B \otimes_A A'$ is integrally closed in
$B \otimes_A L$ by Lemma \ref{lemma-normality-goes-up}.
Since $B \otimes_A A'$ is local, it follows that the ring
$B \otimes_A L$ does not have nontrivial
idempotents which is what we wanted to prove.
\end{proof}

There are no comments yet for this tag.

## Add a comment on tag 092Z

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).