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Let $(X,\mathcal{O}_X)$ be a scheme (or more generally a ringed space). We know that in general the derived category of complexes of quasi-coherent modules $D(\text{Qcoh}(X))$ is not equivalent to the derived category of complexes of arbitrary $\mathcal{O}_X$-modules $D(X)$.

Nevertheless Thomason&Trobaugh's "Higher algebraic k-theory of schemes and of derived categories" B.16 and moreover SGA 6 tells us (if I understand it correctly) that

Let $X$ be either quasi-compact and semi-separated scheme, or else noetherian scheme. Then the natural functor $$ \phi: D_{+}(\text{Qcoh}(X))\rightarrow D_{+}(X) $$ is an equivalence, where $D_{+}(\bullet)$ is the derived category of complexes with bounded below cohomologies.

Now we turn to perfect complexes. For definition of perfect complex see this MO question How to prove that any perfect complex on an affine scheme is strictly perfect?

Let us denote $D_{\text{perf}}(\text{Qcoh}(X))$ the derived category of perfect complexes of quasi-coherent modules and $D_{\text{perf}}(X)$ the derived category of perfect complexes of any $\mathcal{O}_X$-modules.

By the above result in Thomason&Trobaugh, if $X$ is a quasi-compact and semi-separated scheme, or a noetherian scheme, then we have $D_{\text{perf}}(\text{Qcoh}(X))\simeq D_{\text{perf}}(X)$.

$\textbf{My question}$ is: do we have a weaker condition on $X$ which still guarantees that $D_{\text{perf}}(\text{Qcoh}(X))\simeq D_{\text{perf}}(X)$?

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    $\begingroup$ The functor you write is not an equivalence, but is rather fully faithful with essential image spanned by the complexes with quasi-coherent cohomology. That said, your claim is still true. I doubt that the assumptions can be relaxed, though. Verdier's counterexample in SGA 6 shows that this is unreasonable without something stronger than quasi-separated. While without quasi-compactness, it is not even true that perfect complexes are cohomologically bounded. What kind of generality were you hoping for? $\endgroup$ – AAK Mar 23 '15 at 9:46
  • $\begingroup$ @Adeel Yes, that's obviously my mistake. After reading your comment now I believe that maybe quasi-compact and quasi-separated is the best I can expect. $\endgroup$ – Zhaoting Wei Mar 25 '15 at 3:36

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