**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

Comment #8637 by Yebo Peng on

Comment #8996 by Stacks project on