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.

  • 3
    $\begingroup$ I think that noetherianity is necessary. Otherwise the cohomology of a complex of coherent sheaves need not be coherent. $\endgroup$ – Fernando Muro Feb 28 '14 at 15:59
  • $\begingroup$ I assume $X$ to be a variety, isn't that enough? $\endgroup$ – Li Yutong Feb 28 '14 at 19:39
  • 1
    $\begingroup$ varieties are by definition at least of finite type over a field. these are automatically noetherian, so yes. $\endgroup$ – bananastack Mar 1 '14 at 9:35

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

  • $\begingroup$ 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 $D^b_{coh}(X)$. $\endgroup$ – abx Feb 28 '14 at 15:59
  • 1
    $\begingroup$ I guess you are right - can we not bootstrap the fact that $D(qc(X)) = D_{qc}(Sh(X))$ to get what you want? $\endgroup$ – bananastack Feb 28 '14 at 16:14
  • $\begingroup$ I don't see how to do that. $\endgroup$ – abx Feb 28 '14 at 16:32
  • 9
    $\begingroup$ 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? $\endgroup$ – bananastack Feb 28 '14 at 16:58
  • $\begingroup$ Seems OK to me. I just wonder why Huybrechts did not state it that way. $\endgroup$ – abx Feb 28 '14 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.