A paper by Bondal and ven den Bergh shows that one can generate D_{QCoh}(X) by a single perfect complex if X is a quasi-compact and quasi-separated scheme.
There are two key ingredients to the proof.
The first concerns the orthogonal to a Koszul complex. Let I = (f_1, …, f_n) be a finitely generated ideal of a ring A. Let K be the Koszul complex on f_1, …, f_n over A. Set X = Spec(A) and let U = D(f_1) ∪ … ∪ D(f_n). Let j : U —> X be the inclusion morphism. Let E be a complex of A-modules, in other words an object of D_{QCoh}(X). What can we say about E if Hom(K, E[n]) = 0 for all integers n? Well, if I’m not mistaken, this means exactly that E is Rj_* of an object of D_{QCoh}(U).
Namely, one can build the Koszul complexes K^e = K(f_1^e, …, f_n^e) from the complex K(f_1, …, f_n) by taking cones. Hence the assumed vanishing gives vanishing of Hom(K^e, E[n]) also. Then we consider the distinguished triangles
I^e —> A —> K^e —> I^e[1]
where A —> K is the obvious map. Any element of H^n(U, E) comes from a map I^e —> E for some e (see for example Tag 08DD). Thus our assumption implies that RΓ(U, E) = RΓ(X, E) which is what we wanted to show.
The second ingredient is Thomason-Throghbaugh’s result that any perfect object on a quasi-compact open of a qc + qs scheme X is a direct summand of the restriction of a perfect object on X. In Bondal – van den Bergh this is proved using some abstract machinery (due to Neeman IIRC; this machinery is interesting by itself), but in essence it is elementary from the induction princple and the case of affine schemes (this is how we do it in the Stacks project).
Finally, how do we get our perfect object generating D_{QCoh}(X)? Let X = U_1 ∪ … ∪ U_n be a finite covering by affine opens with n minimal. We will find our perfect object by induction on n. Let Z be the complement of U_2 ∪ … ∪ U_n in X. Then Z is a closed subset of U_1 whose complement is quasi-compact. Writing U_1 = Spec(A_1) we can find a Koszul complex K for a finitely generated ideal I_1 in A_1 cutting out Z. This complex has no cohomology outside of Z hence is a perfect complex K on X. By the discussion above, for E in D_{QCoh}(X) being orthogonal to all shifts of K implies that E comes from U_2 ∪ … ∪ U_n. By induction we can find a perfect complex P on U_2 ∪ … ∪ U_n which generates D_{QCoh}(U_2 ∪ … ∪ U_n). By TT (see above) we can find a perfect complex Q on X whose restriction to U_2 ∪ … ∪ U_n contains P as a direct summand. Then the reader sees that K ⊕ Q is the desired generator of D_{QCoh}(X).
These results can now be found in the stacks project
- For schemes look at Section Tag 09IP, and
- for algebraic spaces look at Section Tag 09IU.
Enjoy!