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Let $X$ be a scheme. The sheaf cohomology on $X$ is given by the derived functors of the global sections functor $$ \Gamma(X, -): \mathfrak{Mod}(X) \longrightarrow \mathfrak{Ab}, $$ which we denote by $H^{i}(X, -)$. There is a canonical inclusion functor of the subcategory of quasicoherent sheaves on $X$, $$ \iota: \mathfrak{qcoh}(X) \hookrightarrow \mathfrak{Mod}(X). $$ Then we can consider two functors from $\mathfrak{qcoh}(X)$, the derived functors of the composition, $$ \Gamma(X, -) \circ \iota: \mathfrak{qcoh}(X) \longrightarrow \mathfrak{Ab} $$ and the restriction of the sheaf cohomology functors, $$ H^{i}(X, -)|_{\mathfrak{qcoh}(X)}: \mathfrak{qcoh}(X) \longrightarrow \mathfrak{Ab}, $$ When do these agree?

In other words, when can we compute the sheaf cohomology of a quasi-coherent sheaf by taking injective resolutions in the category of quasicoherent sheaves? I am fairly certain this is related to the fact that for a noetherian scheme, quasicoherent sheaves can be embedded in a flasque quasicoherent sheaf. But I am not sure how to use that fact.

Luke
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    The condition you want is that the resolution of any sheaf by quasi-coherent injective objects is again a resolution by acyclic objects of the whole category of abelian sheaves. This is true when $X$ is noetherian (there exists a resolution by injective sheaves which are quasi-coherent) and is explained in my answer to your question from earlier this year. Are you asking for the converse, that is, this condition for all quasi-coherent sheaves implies $X$ noetherian? – KReiser Nov 10 '20 at 00:38
  • @KReiser Thanks for your reply. I was going back over your answer I realised I don't understand. You claim "Even if quasicoherent $\mathcal{O}{X}$-modules didn't have enough injectives, Hartshorne has proven that one can compute cohomology and higher direct image using just the category of quasicoherent modules and get the same answer as one would in the category of all $\mathcal{O}{X}$-modules". I don't see how this works if the category of quasicoherent sheaves didn't have enough injectives. Surely the argument depends on the derived functors existing, which requires enough injectives? – Luke Nov 10 '20 at 00:45
  • And I know that the category of quasicoherent modules DOES have enough injectives, but the fact that I can't see why it is independent of that fact is making me concerned I am misunderstanding something. And moreover wouldn't it need those objects which are injective in the category of quasicoherent sheaves to remain injective when viewed as objects in the category of sheaves of modules? – Luke Nov 10 '20 at 00:46
  • Derived functors can be computed on acyclic resolutions, assuming there is a class of acyclic objects closed under sums in to which every object embeds and satisfying a restricted 2-out-of-3 property - my first comment here should have said "there exists a resolution by acyclic sheaves which are quasi-coherent". Here's a question about why a nice collection of acyclic objects satsifying those 3 further conditions suffice to define the functors you want. – KReiser Nov 10 '20 at 00:52
  • "Derived functors can be computed on acyclic resolutions"

    But this is only true if the derived functor exists in the first place. I'm sorry if I am being dense, but the statement seems to presuppose that the derived functors you are computing actually exist. And that existence requires injective resolutions. Once you know they exist then you can appeal to acyclic resolutions.

     – Luke Nov 10 '20 at 01:07
  • You can define the cohomology functors out of the category of q-coh sheaves on a noetherian scheme before knowing it has enough injectives by using the resolution technique from Hartshorne corollary III.3.6 and then showing that since they agree with the functors from $Ab(X)$, everything works. But this is getting rather far afield from your posted question - if you'd like to make an edit to include this in the question, I'd be happy to take a look at writing an answer. My original comment still stands: are you asking for an iff characterization in your question, or is "here's a yes" enough? – KReiser Nov 10 '20 at 02:06

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