Lemma 15.75.8. 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

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

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

3. $K$ is pseudo-coherent, and

4. $(R, I)$ is a henselian pair.

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

Proof. We apply Lemma 15.75.2 using the class $\mathcal{P}$ of all finite 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 finite projective $R/I$-modules to finite projective $R$-modules, see Lemma 15.13.1. Property (5) holds because a pseudo-coherent complex can be represented by a bounded above complex of finite free $R$-modules. Thus Lemma 15.75.2 applies and we find $P^\bullet$ as desired. The final assertion of the lemma follows from Lemma 15.3.5. $\square$

Comment #3475 by Ravi Vakil on

Dumb question: what is part (5) in the proof?\ref{BCE}

Comment #3504 by on

Dear Ravi, fair question. It is the 5th condition in Lemma 15.75.2. I have edited the proof to clarify. See here.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).