Lemma 10.134.13 (00S7)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.134: The naive cotangent complex
Lemma 10.134.13 (cite)

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Lemma 10.134.13. Let $A \to B$ be a ring map. Let $S \subset B$ be a multiplicative subset. The canonical map $\mathop{N\! L}\nolimits _{B/A} \otimes _ B S^{-1}B \to \mathop{N\! L}\nolimits _{S^{-1}B/A}$ is a quasi-isomorphism.

Proof. We have $S^{-1}B = \mathop{\mathrm{colim}}\nolimits _{g \in S} B_ g$ where we think of $S$ as a directed set (ordering by divisibility), see Lemma 10.9.9. By Lemma 10.134.12 each of the maps $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B_ g \to \mathop{N\! L}\nolimits _{B_ g/A}$ are quasi-isomorphisms. The lemma follows from Lemma 10.134.9. $\square$

Comments (2)

Comment #2820 by Dario Weißmann on September 22, 2017 at 10:40

(Why) Isn't the canonical map in the lemma a homotopy equivalence?

Comment #2921 by Johan on October 10, 2017 at 00:16

It is indeed a homotopy equivalence. We can deduce this from a general result on two term complexes of the form $\ldots \to 0 \to L^{-1} \to L^0 \to 0 \to \ldots$ with $L^0$ projective, see Lemma 15.84.4. It would be a bit annoying (I think) to prove it here in the commutative algebra chapter and having the quasi-isomorphism statement suffices for applications, I think and hope.

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