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?



K-projectivity for rings of finite homological dimension
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?**](http://dx.doi.org/10.1016/j.jpaa.2021.106722), J. Pure Appl. Algebra 225, No. 11, Article ID 106722, 23 p. (2021). [ZBL1464.18015](https://zbmath.org/?q=an:1464.18015).


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.


