# Derived functor of global sections of coherent sheaves

Hi,

I want to consider a scheme $X$ which is proper over a field $k$. With $Qcoh(X)$ resp. $Coh(X)$ I mean the abelian category of quasicoherent resp. coherent sheaves on $X$. With $Vec(k)$ resp. $Vec_{f}(k)$ I mean the category of k-vector spaces resp. finitedimensional k-vector-spaces.

I consider the left exact functor

$\Gamma: Qcoh(X) \rightarrow Vec(k)$

Then, as $Qcoh(X)$ has enough Injectives one has a right derived functor

$R\Gamma: D^{+}(Qcoh(X)) \rightarrow D^{+}(Vec(k))$.

My question: how can I get the right derived functor of

$\Gamma: Coh(X) \rightarrow Vec_{f}(k)$?

I know that it exists, it is constantly used e.g. in Huybrechts book about Fourier-Mukai. But I don't see how you get it.

Of course one first thinks of just composing

$D^{+}(Coh(X)) \rightarrow D^{+}(Qcoh(X)) \rightarrow D^{+}(Vec(k))$,

but who tells me that this will satisfy the universal property for derived functors? (That you land in the finitedimensional vecs is not that important for me.)

As a stimulus see discussion after Theorem 3.21 in Huybrechts' book about Fourier-Mukai. As I say below in a comment I can't cope with his treatment at that point.

Thanks a lot!

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I believe the standard way to handle such problems if you are happy to assume the scheme is Noetherian, is to rather than consider D^+(Coh(X)) we consider for example the equivalent category D^+_{Coh}(QCoh(X)) or even D^+_{Coh}(Mod(X))(this might require to assume quasi-compact and seperated). That is the derived category of quasicoherent sheaves with coherent cohomology. This category has injectives as usual... – Daniel Pomerleano Aug 19 '11 at 13:58
Huybrechts, in that very book, actually states this quite carefully and examines it pretty thoroughly. See Proposition 2.42 and the surrounding material. – Matt Aug 19 '11 at 15:56
I know the book and my problem actually arose from the discussion in that book, especially what he says after Theorem 3.21. I don't think that it suffices what he is saying. In fact, he just takes the composition of the functors as I mentioned it in my question, then shows that you land in the finitedimensional vector spaces and that you can take the bounded categories, but definitely not why this really is the derived functor. You don't a priori know that it exists in the sense of its universal property. – Descartes Aug 19 '11 at 16:07