Lemma 15.88.7 (0G1T)—The Stacks project

Lemma 15.88.7. 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. Proof of (2). We can write $N = \bigcup N[I^ n]$. We have $K \otimes _ R^\mathbf {L} N = \text{hocolim}_ n K \otimes _ R^\mathbf {L} N[I^ n]$ as tensor products commute with colimits (details omitted; hint: represent $K$ by a K-flat complex and compute directly). Hence we may assume $N$ is annihilated by $I^ n$. Consider the $R$-algebra $R' = R/I^ n \oplus N$ where $N$ is an ideal of square zero. It suffices to show that $K' = K \otimes _ R^\mathbf {L} R'$ is $0$ in $D(R')$. We have a surjection $R' \to R/I$ of $R$-algebras whose kernel $J$ is nilpotent (any product of $n$ elements in the kernel is zero). We have

\[ 0 = K \otimes _ R^\mathbf {L} R/I = (K \otimes _ R^\mathbf {L} R') \otimes _{R'}^\mathbf {L} R/I = K' \otimes _{R'}^\mathbf {L} R/I \]

by Lemma 15.60.5. Hence by Lemma 15.78.4 we find that $K'$ is a perfect complex of $R'$-modules. In particular $K'$ is bounded above and if $H^ b(K')$ is the right-most nonvanishing cohomology module (if it exists), then $H^ b(K')$ is a finite $R'$-module (use Lemmas 15.74.2 and 15.64.3) with $H^ b(K') \otimes _{R'} R'/J = H^ b(K')/JH^ b(K') = 0$ (because $K' \otimes _{R'}^\mathbf {L} R'/J = 0$). By Nakayama's lemma (Algebra, Lemma 10.20.1) we find $H^ b(K') = 0$, i.e., $K' = 0$ as desired.

Part (1) follows trivially from part (2). Part (3) follows from part (2), induction on the number of nonzero cohomology modules of $M$, and the distinguished triangles of truncation from Derived Categories, Remark 13.12.4. 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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