Lemma 15.66.3 (0654)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.66: Tor dimension
Lemma 15.66.3 (cite)

Lemma 15.66.3. Let $R$ be a ring. Let $K^\bullet $ be an object of $D(R)$. Let $a, b \in \mathbf{Z}$. The following are equivalent

$K^\bullet $ has tor-amplitude in $[a, b]$.
$K^\bullet $ is quasi-isomorphic to a complex $E^\bullet $ of flat $R$-modules with $E^ i = 0$ for $i \not\in [a, b]$.

Proof. If (2) holds, then we may compute $K^\bullet \otimes _ R^\mathbf {L} M = E^\bullet \otimes _ R M$ and it is clear that (1) holds. Assume that (1) holds. We may replace $K^\bullet $ by a projective resolution with $K^ i = 0$ for $i > b$. See Derived Categories, Lemma 13.19.3. Set $E^\bullet = \tau _{\geq a}K^\bullet $. Everything is clear except that $E^ a$ is flat which follows immediately from Lemma 15.66.2 and the definitions. $\square$

Comments (2)

Comment #4357 by Manuel Hoff on August 25, 2019 at 05:56

I find the proof somewhat confusing. Instead of first choosing a projective resolution for $K^\bullet$ and then chopping of the higher terms one could also just directly choose a projective resolution with vanishing higher terms as $H^i(K^\bullet ) = 0$ for $i > b$ , see Lemma 05T6 (or its dual version).

Comment #4501 by Johan on September 02, 2019 at 12:56

A rare instance where the proof gets shorter. Thanks. Fixed here.

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