Projective objects in the derived category of chain complexes

Question

I have been trying to understand projective objects in the derived category of chain complexes of modules over a ring.

If we stick to the category of chain complexes, the only projective objects are split exact complexes of projectives. These would all be trivial in the derived category.

What happens if we consider the derived category of chain complexes? I want to consider projectives in the infinity-categorical sense (covariant Yoneda commutes with geometric realisations). Is there a known answer? If so, could someone please include a reference?

I thought that maybe by passing to things up to homotopy, we might be able to have more projectives.

Answer 1

In a stable $\infty$-category, there are no nontrivial projectives. Of course, $0$ is always projective.

Now let $X$ be an arbitrary projective in some stable $C$, $X\simeq\Sigma \Omega X$ is a simplicial colimit of things of the form $\Omega X^n$ for some $n$'s, so that the identity $X\to X \simeq \Sigma \Omega X$ factors through some $\Omega X^n$ by projectivity. However, each of the individual maps $\Omega X^n\to \Sigma \Omega X \simeq X$ is nullhomotopic, so this implies that the identity of $X$ is nullhomotopic, so $X=0$.

What is interesting is, when $C$ has a $t$-structure, the projective objects of $C_{\geq 0}$ as those can be nontrivial. For instance, if $R$ is a connective ring spectrum and $Mod_R$ is given the usual $t$-structure, the projectives of $(Mod_R)_{\geq 0}$ are exactly what you'd expect: retracts of modules of the form $\bigoplus_I R$ - note that no shifts are allowed.

This appears in Lurie's Higher algebra, around 7.2.2.5., 7.2.2.6., 7.2.2.7..

Answer 2

If $f\colon X\to Y$ is morphism in a triangulated category, we have a distinguished triangle $X\xrightarrow{f}Y\xrightarrow{i}Cf\xrightarrow{d}\Sigma X$, so in particular $if=0$. If $f$ is an epimorphism this means that $i=0$ so $\Sigma X$ is the cofibre of $Y\xrightarrow{0}Cf$ which is $Cf\oplus\Sigma Y$. It follows (after clarifying the maps involved in the above statement) that $f$ is a split epimorphism. As all epimorphisms are split, all objects are projective. This answers your question in the triangulated category context. I don't know what to say in the context of stable $\infty$-categories.