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Let $X$ be a variety. Let $D^b(Coh(X))$ be the derived category of bounded complexes of coherent sheaves on $X$, and $D^b_{coh}(X)$ be the derived category of bounded complexes of sheaves of $\mathcal{O}_X$-modules with coherent sheaves as cohomologies. Similarly, we have $D^b_{qc}(X)$ and $D^b(Qco(X))$ by replacing $coherent$ sheaves to $quasi$-$coherent$ sheaves.

It is proved in "Residues and Duality" by Hartshorne (Chapt. II Corollary 7.19) that $D^b_{qc}(X)$ and $D^b(Qco(X))$ are derived equivalent. I was wondering if the same thing is still true for $D^b_{Coh}(X)$ and $D^b(coh(X))$?

Hartshorne's proof seems could not be generalized to this case because I feel that any quasi-coherent module might not be embedded to a quasi-coherent injective module.

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I think that noetherianity is necessary. Otherwise the cohomology of a complex of coherent sheaves need not be coherent. –  Fernando Muro Feb 28 at 15:59
I assume $X$ to be a variety, isn't that enough? –  Li Yutong Feb 28 at 19:39
varieties are by definition at least of finite type over a field. these are automatically noetherian, so yes. –  bananastack Mar 1 at 9:35

1 Answer 1

up vote 10 down vote accepted

For X noetherian this is still true. (Proposition 3.5 in Daniel Huybrechts' book)

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Not exactly. Huybrechts prove that $D^b(Coh(X))$ is equivalent to $D^b_{coh}(Qcoh(X))$, the derived category of quasi-coherent complexes with coherent cohomology. I do not think this is equivalent to $D^b_{coh}(X)$. –  abx Feb 28 at 15:59
I guess you are right - can we not bootstrap the fact that $D(qc(X)) = D_{qc}(Sh(X))$ to get what you want? –  bananastack Feb 28 at 16:14
I don't see how to do that. –  abx Feb 28 at 16:32
We have the embedding $D(coh(X)) \to D_{coh}(qc(X)) \to D_{coh}(Sh(X))$ and we want to show it's essentially surjective. Take a complex in $D_{coh}(Sh(X))$. By Corollary 3.4 in H this is quasi-isomorphic to a complex of quasi-coherent sheaves and by Prop 3.5 such a thing is quasi-isomorphic to a complex of coherent sheaves. In the whole process we never changed cohomology, so that ought to do it. Or am I misunderstanding something fundamental? –  bananastack Feb 28 at 16:58
Seems OK to me. I just wonder why Huybrechts did not state it that way. –  abx Feb 28 at 17:22

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