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?

  • 2
    $\begingroup$ 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

1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.