# Question about unbounded derived categories of quasicoherent sheaves

This is a bit of a strange question since I more or less want to ask the MO crowd whether I've understood the situation correctly. If you have an unbounded complex of quasicoherent injective sheaves $I$ over a variety over a field k where all local rings are regular (the property that I want to isolate is that every quasicoherent sheaf have a finite injective resolution), is it K-injective in the sense of Spaltenstein? My source for this material is Krause's "The Stable Derived Category of a Noetherian Scheme".

Since the topic of this question maybe of somewhat broad interest(I had mostly ignored unbounded complexes until today and I assume many people have the done same), I'll give a little bit of background. Recall that for a bounded complex of sheaves, one can compute right derived functors by replacing your complex $K$ with a quasi-isomorphic complex of injectives $I$ and apply your functor to this complex. One needs a more nuanced approach. A complex $KI$ is called K-injective if $Hom^{*}(A,KI)=0$ is an acyclic if A is acyclic. Every bounded complex of injectives is K-injective, but this is false for unbounded complexes. A very simple example is given for the dual projective question in the beginning of Spaltenstein's paper. It is a theorem of Spaltenstein that such resolutions always exist.

My question is, in the case of schemes as above, is a complex of injectives K-injective? My reasoning is pretty simple. In the paper above, Krause explains that the inclusion from the homotopy category of injectives $K(Inj)\mapsto K(QCoh)$ has a left adjoint L. Now take an acylic object A viewed as an object in $K(QCoh)$, then we have $Hom_{K(QCoh)}(A, I)= Hom_{K(Inj)}(L(A),I)=0$ because Krause proves that all acyclic injective complexes are zero in $K(Inj)$ when all objects have finite injective resolutions. Edit: We know that L(A) is still acyclic since Krause proves that if $Q_I$ denotes $K(Inj) \mapsto D(QCoh)$(the derived category) and $Q$ denotes the map $K(QCoh) \mapsto D(QCoh)$ we know that $Q$ is isomorphic to $Q_I \circ L$. Now shift A to get the result. Am I missing something? If this breaks down for some reason, does it at least hold when your complex is two periodic? If one only cares about computing derived functors of $\Gamma$, the global sections functor.

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Yes, you are right. In any abelian category of finite homological dimension with enough injectives, any complex of injectives is K-injective. Your reasoning is correct, one just has to also use the fact that the functor L preserves acyclicity. But this is true. –  Leonid Positselski Oct 10 '10 at 11:15
What I think is not true is the assertion that K-injectivity of arbitrary complexes of injectives fails in general because the inverse limits are not exact. These seem to be unrelated phenomenons. Rather, it is the simple construction of K-injective complexes of modules that fails for sheaves because infinite products of sheaves are not exact. –  Leonid Positselski Oct 10 '10 at 11:15
Thanks Professor Positselski for confirming this! I'll edit it to make it better. –  Daniel Pomerleano Oct 10 '10 at 16:56