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Tag 07BS

Chapter 10: Commutative Algebra > Section 10.132: The naive cotangent complex

Lemma 10.132.11. Let $S \subset A$ is a multiplicative subset of $A$. Let $S^{-1}A \to B$ be a ring map. Then $\mathop{N\!L}\nolimits_{B/A} \to \mathop{N\!L}\nolimits_{B/S^{-1}A}$ is a homotopy equivalence.

Proof. Choose a presentation $\alpha : P \to B$ of $B$ over $A$. Then $\beta : S^{-1}P \to B$ is a presentation of $B$ over $S^{-1}A$. A direct computation shows that we have $\mathop{N\!L}\nolimits(\alpha) = \mathop{N\!L}\nolimits(\beta)$ which proves the lemma as the naive cotangent complex is well defined up to homotopy by Lemma 10.132.2. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 33541–33546 (see updates for more information).

    \begin{lemma}
    \label{lemma-NL-localize-bottom}
    Let $S \subset A$ is a multiplicative subset of $A$.
    Let $S^{-1}A \to B$ be a ring map.
    Then $\NL_{B/A} \to \NL_{B/S^{-1}A}$ is a homotopy equivalence.
    \end{lemma}
    
    \begin{proof}
    Choose a presentation $\alpha : P \to B$ of $B$ over $A$.
    Then $\beta : S^{-1}P \to B$ is a presentation of $B$ over $S^{-1}A$.
    A direct computation shows that we have $\NL(\alpha) = \NL(\beta)$
    which proves the lemma as the naive cotangent complex is well defined
    up to homotopy by Lemma \ref{lemma-NL-homotopy}.
    \end{proof}

    Comments (2)

    Comment #2797 by Dario WeiƟmann on September 5, 2017 a 4:58 pm UTC

    The second sentence of the statement should be: ... is a homotopy equivalence.

    Comment #2900 by Johan (site) on October 7, 2017 a 3:24 pm UTC

    THanks, fixed here.

    There are also 7 comments on Section 10.132: Commutative Algebra.

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