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Tag 092Z

Chapter 15: More on Algebra > Section 15.85: Weakly étale ring maps

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}

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