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Tag 05YT

Chapter 10: Commutative Algebra > Section 10.141: Étale ring maps

Lemma 10.141.11. Consider a commutative diagram $$ \xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } $$ with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps whose kernels are ideals of square zero. If $A \to B$ is étale, and $J = I \otimes_A B$, then $A' \to B'$ is étale.

Proof. By Lemma 10.141.10 there exists an étale ring map $A' \to C$ such that $C/IC = B$. Then $A' \to C$ is formally smooth (by Proposition 10.136.13) hence we get an $A'$-algebra map $\varphi : C \to B'$. Since $A' \to C$ is flat we have $I \otimes_A B = I \otimes_A C/IC = IC$. Hence the assumption that $J = I \otimes_A B$ implies that $\varphi$ induces an isomorphism $IC \to J$ and an isomorphism $C/IC \to B'/IB'$, whence $\varphi$ is an isomorphism. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 37569–37587 (see updates for more information).

    \begin{lemma}
    \label{lemma-lift-etale-infinitesimal}
    Consider a commutative diagram
    $$
    \xymatrix{
    0 \ar[r] &
    J \ar[r] &
    B' \ar[r] &
    B \ar[r] & 0 \\
    0 \ar[r] &
    I \ar[r] \ar[u] &
    A' \ar[r] \ar[u] &
    A \ar[r] \ar[u] & 0
    }
    $$
    with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps
    whose kernels are ideals of square zero. If $A \to B$ is \'etale,
    and $J = I \otimes_A B$, then $A' \to B'$ is \'etale.
    \end{lemma}
    
    \begin{proof}
    By
    Lemma \ref{lemma-lift-etale}
    there exists an \'etale ring map $A' \to C$ such that $C/IC = B$.
    Then $A' \to C$ is formally smooth (by
    Proposition \ref{proposition-smooth-formally-smooth})
    hence we get an $A'$-algebra map $\varphi : C \to B'$.
    Since $A' \to C$ is flat we have $I \otimes_A B = I \otimes_A C/IC = IC$.
    Hence the assumption that $J = I \otimes_A B$ implies that
    $\varphi$ induces an isomorphism $IC \to J$ and an isomorphism
    $C/IC \to B'/IB'$, whence $\varphi$ is an isomorphism.
    \end{proof}

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