# K-projectivity for rings of finite homological dimension

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:

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?

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

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.

• Thanks for this help! Sep 30, 2021 at 14:52