Lemma 15.89.8 (0G1T)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.89: Torsion modules
Lemma 15.89.8 (cite)

Lemma 15.89.8. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $K$ be an object of $D(R)$ such that $K \otimes _ R^\mathbf {L} R/I = 0$ in $D(R)$. Then

$K \otimes _ R^\mathbf {L} R/I^ n = 0$ for all $n \geq 1$,
$K \otimes _ R^\mathbf {L} N = 0$ for any $I$-power torsion $R$-module $N$,
$K \otimes _ R^\mathbf {L} M = 0$ for any $M \in D^ b(R)$ whose cohomology modules are $I$-power torsion.

Proof. Consequence of Lemma 15.89.7 (some details omitted). $\square$

Comments (3)

Comment #8384 by Peng Du on May 08, 2023 at 12:06

Line 1 "such hat" should be "such that".

Comment #8637 by Yebo Peng on August 24, 2023 at 09:33

The proof can actually be simplified: after we've shown that $K' \otimes _{R'}^\textbf{L} R/I=K' \otimes _{R'} ^\textbf{L} R'/J=0$ , we can show directly that $K'=0$ . We do this by showing inductively that $K' \otimes _{R'} ^\textbf{L} R'/J^i=0$ for every positive integer $i$ . A method similar to that of Lemma 09AR may work, i.e. we represent $K'$ by a K-flat complex with flat terms, and then tensor it with the short exact sequence $0\rightarrow J^i/J^{i+1}\rightarrow R'/J^{i+1}\rightarrow R'/J^i\rightarrow 0$ .

Comment #8996 by Stacks project on April 17, 2024 at 11:06

@#8384: Thanks and fixed here.

@#8637: I like the proof as it stands now.

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