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Let $R$ be a Noetherian commutative ring. A complex of $R$-modules $P^{\bullet}$ is K-projective if for any acyclic complex $A^{\bullet}$, the complex of abelian groups $ Hom(P^{\bullet}, A^{\bullet})$ is acyclic. K-projective complexes were defined by Spaltenstein:

http://www.numdam.org/article/CM_1988__65_2_121_0.pdf.

Suppose now that $R$ has finite homological dimension. I've heard it stated that a (possibly unbounded) complex of projective modules $P^\bullet$ is K-projective. Is there a (preferably elementary) reference for this or proof?

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    $\begingroup$ It would be good to make this self-contained by including the relevant definition in the post. $\endgroup$ Sep 29, 2021 at 14:41
  • $\begingroup$ @StevenLandsburg I included the definition. Thanks for your comment. $\endgroup$ Sep 29, 2021 at 16:18
  • $\begingroup$ Thank you...... $\endgroup$ Sep 29, 2021 at 16:45

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There's a nice, short proof in

Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, Article ID 106722, 23 p. (2021). ZBL1464.18015.

where this is Proposition 4.1(b).

Suppose $R$ has global dimension $d<\infty$. Let $P^\bullet$ be a complex of projectives, and let $\alpha:P_K^\bullet\to P^\bullet$ be a $K$-projective resolution. To prove that $P^\bullet$ is $K$-projective, it suffices to prove that $\alpha$ is a homotopy equivalence, or equivalently that the mapping cone of $\alpha$ is contractible.

So we just need to prove that an acyclic complex of projectives is contractible.

Let $Q^\bullet$ be an acyclic complex of projectives. Then the truncation $$\dots\to Q^{-2}\to Q^{-1}\to Q^0\to 0\to\dots$$ is a projective resolution of some module, which has projective dimension at most $d$ by the assumption on the global dimension of $R$. Therefore the image of the differential $Q^{-d}\to Q^{-d+1}$ is projective.

Applying the same argument to shifts of $Q^\bullet$, every differential of $Q^\bullet$ has projective image, which (together with the fact that $Q^\bullet$ is acyclic) implies that $Q^\bullet$ is contractible.

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  • $\begingroup$ Thanks for this help! $\endgroup$ Sep 30, 2021 at 14:52

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