Let $X$ be a Noetherian scheme. Is the obvious functor $D(\operatorname{Coh}(X))\to D(\operatorname{QCoh}(X))$ fully faithful?

If this is true then $D(\operatorname{Coh}(X))$ is equivalent to the full subcategory of $D(\operatorname{QCoh}(X))$ consisting of those complexes whose cohomology sheaves are coherent. The bounded above version of the latter statement is here: https://stacks.math.columbia.edu/tag/0FDA

Are any counterexamples known? If yes, how does one "usually" define the unbounded derived category of coherent sheaves?

P.S. It appears that the "modification" of $D(\operatorname{Coh}(X))$ consisting of those quasi-coherent complexes whose cohomology sheaves are coherent is more "useful". This is the version that is relevant for Grothendieck duality; isn't it?


1 Answer 1


No, not always.


Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, Article ID 106722, 23 p. (2021). ZBL1464.18015.

the authors show in Theorem 3.1 that the natural functor $D(\operatorname{mod}R)\to D(\operatorname{Mod}R)$ is not fully faithful when $R$ is a quasi-Frobenius ring of infinite global dimension (such as $R=\mathbb{Z}/4\mathbb{Z}$).

They also have some positive results. For example, Corollary 5.12 states that $D(\operatorname{coh}X)\to D(\operatorname{Coh}X)$ is fully faithful if $X$ is a regular Noetherian scheme of finite Krull dimension.


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