Lemma 15.75.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $E^\bullet $ be a complex of $R/I$-modules. Let $K$ be an object of $D(R)$. Assume that

$E^\bullet $ is a bounded above complex of projective $R/I$-modules,

$K \otimes _ R^\mathbf {L} R/I$ is represented by $E^\bullet $ in $D(R/I)$, and

$I$ is a nilpotent ideal.

Then there exists a bounded above complex $P^\bullet $ of projective $R$-modules representing $K$ in $D(R)$ such that $P^\bullet \otimes _ R R/I$ is isomorphic to $E^\bullet $.

**Proof.**
We apply Lemma 15.75.2 using the class $\mathcal{P}$ of all projective $R$-modules. Properties (1) and (2) of the lemma are immediate. Property (3) follows from Nakayama's lemma (Algebra, Lemma 10.20.1). Property (4) follows from the fact that we can lift projective $R/I$-modules to projective $R$-modules, see Algebra, Lemma 10.77.5. To see that (5) holds it suffices to show that $K$ is in $D^{-}(R)$. We are given that $K \otimes _ R^\mathbf {L} R/I$ is in $D^{-}(R/I)$ (because $E^\bullet $ is bounded above). We will show by induction on $n$ that $K \otimes _ R^\mathbf {L} R/I^ n$ is in $D^{-}(R/I^ n)$. This will finish the proof because $I$ being nilpotent exactly means that $I^ n = 0$ for some $n$. We may represent $K$ by a K-flat complex $K^\bullet $ with flat terms (Lemma 15.59.10). Then derived tensor products are represented by usual tensor products. Thus we consider the exact sequence

\[ 0 \to K^\bullet \otimes _ R I^ n/I^{n + 1} \to K^\bullet \otimes _ R R/I^{n + 1} \to K^\bullet \otimes _ R R/I^ n \to 0 \]

Thus the cohomology of $K \otimes _ R^\mathbf {L} R/I^{n + 1}$ sits in a long exact sequence with the cohomology of $K \otimes _ R^\mathbf {L} R/I^ n$ and the cohomology of

\[ K \otimes _ R^\mathbf {L} I^ n/I^{n + 1} = K \otimes _ R^\mathbf {L} R/I \otimes _{R/I}^\mathbf {L} I^ n/I^{n + 1} \]

The first cohomologies vanish above a certain degree by induction assumption and the second cohomologies vanish above a certain degree because $K^\bullet \otimes _ R^\mathbf {L} R/I$ is bounded above and $I^ n/I^{n + 1}$ is in degree $0$.
$\square$

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