Let $X$ be a scheme. Let $A$ be a sheaf of associative algebras with 1 on $X$, containing $\mathcal O_X$, which is quasi-coherent as a left and right module over it. Under what conditions is it true that the derived category of quasi-coherent $A$-modules is equal to the derived category of $A$-modules with quasi-coherent cohomology? Do we need to work with bounded below derived category for this? Are there any necessary assumptions on $X$ or $A$ (noetherian?). Either a proof or a reference will be greatly appreciated.

  • $\begingroup$ A relevant reference is Thomason, Trobaugh "Higher algebraic K-theory of schemes and of derived categories", Appendix B, Proposition B.16 and Section B.17. It explains that one needs to assume either that $X$ is quasi-compact and semi-separated, or otherwise that $X$ is Noetherian (which implies quasi-compact and quasi-separated, but not necessarily semi-separated). The weaker "locally Noetherian" assumption may be enough in some assertions of this kind (needs to be checked). $\endgroup$ – Leonid Positselski Apr 16 at 16:28
  • $\begingroup$ This covers the case when $A=\mathcal O_X$. The case with $A$ a quasi-coherent algebra over $X$ with $X$ quasi-compact and semi-separated should be provable similarly. For $X$ Noetherian, you want $A$ to be Noetherian as well (see Efimov, Positselski "Coherent analogues of matrix factorizations and relative singularity categories", Section A.3 in Appendix A, for a relevant result about injective quasi-coherent modules over quasi-coherent Noetherian algebras). $\endgroup$ – Leonid Positselski Apr 16 at 16:36
  • $\begingroup$ Finally, the case when $\mathcal O_X$ is a sheaf of subrings in $A$ which is not central. The assumptions that $A$ is quasi-coherent both as a left and as a right $\mathcal O_X$-module sound weak to me. If they are not enough to make the arguments work, adding the assumption that $A$ is a differential $\mathcal O_X$-$\mathcal O_X$-bimodule (i.e., concentrated set-theoretically on the diagonal in $X\times X$) may help. $\endgroup$ – Leonid Positselski Apr 16 at 16:41
  • $\begingroup$ Thanks a lot! How do you think one can use the assumption that $A$ is concentrated on the diagonal? $\endgroup$ – Alexander Braverman Apr 17 at 1:21

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