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Tag 07EN

Chapter 15: More on Algebra > Section 15.73: Pseudo-coherent and perfect ring maps

Lemma 15.73.6. A local complete intersection homomorphism is perfect.

Proof. Let $A \to B$ he a local complete intersection homomorphism. By Definition 15.30.2 this means that $B = A[x_1, \ldots, x_n]/I$ where $I$ is a Koszul ideal in $A[x_1, \ldots, x_n]$. By Lemmas 15.73.2 and 15.67.3 it suffices to show that $I$ is a perfect module over $A[x_1, \ldots, x_n]$. By Lemma 15.67.11 this is a local question. Hence we may assume that $I$ is generated by a Koszul-regular sequence (by Definition 15.29.1). Of course this means that $I$ has a finite free resolution and we win. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 18513–18516 (see updates for more information).

    \begin{lemma}
    \label{lemma-lci-perfect}
    A local complete intersection homomorphism is perfect.
    \end{lemma}
    
    \begin{proof}
    Let $A \to B$ he a local complete intersection homomorphism.
    By Definition \ref{definition-local-complete-intersection} this
    means that $B = A[x_1, \ldots, x_n]/I$ where $I$ is a Koszul ideal
    in $A[x_1, \ldots, x_n]$. 
    By Lemmas \ref{lemma-perfect-ring-map} and \ref{lemma-perfect-module}
    it suffices to show that $I$ is a perfect module over $A[x_1, \ldots, x_n]$.
    By Lemma \ref{lemma-glue-perfect} this is a local question. Hence we
    may assume that $I$ is generated by a Koszul-regular sequence (by
    Definition \ref{definition-regular-ideal}).
    Of course this means that $I$ has a finite free resolution and we win.
    \end{proof}

    Comments (1)

    Comment #2954 by Ko Aoki on October 11, 2017 a 1:05 pm UTC

    Typo in the proof: "Let $A \to B$ he a ..." should be replaced by "Let $A \to B" be a ...".

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