# The Stacks Project

## Tag 07BS

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 an 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 33669–33674 (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 an 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}

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

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

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

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