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Let $X$ be a derived fpqc stack on the $\infty$-category of connective spectral affine schemes $\mathbf{Aff}^{\mathrm{cn}}=(\mathbf{Ring}^{\mathrm{cn}}_{E_\infty})^{\mathrm{op}}$, that is to say, a functor $X:(\mathbf{Aff}^{\mathrm{cn}})^{\mathrm{op}}\to \mathcal{S}$ satisfying fpqc descent. Then we can define its $\infty$-category of quasicoherent sheaves formally by a Kan extension.

Say that a symmetric monoidal stable $\infty$-category $\mathcal{C}$ 'has enough perfect objects' if its full subcategory of dualizable objects is dense (that is to say the induced functor $\mathcal{C}\to \operatorname{Ind}(\operatorname{Perf}(\mathcal{C}))$ is fully faithful).

Are there examples of fpqc stacks $X$ as above for which $\operatorname{QCoh}(X)$ does not have enough perfect objects?

What about if we restrict our question to ((Quasi)-Geometric Stacks, Artin Stacks, Deligne-Mumford Stacks, Algebraic Spaces, Schemes)?

For certain, this is true for quasicompact quasiseparated schemes and algebraic spaces, as well as quasi-Geometric stacks $X$ such that $\operatorname{QCoh}(X)$ is compactly generated and the structure sheaf is a compact object (proven in Lurie, Spectral Algebraic Geometry).

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  • $\begingroup$ I don't know that much about this, but there's some relation to the resolution property, right? $\endgroup$ – Tim Campion May 13 at 0:00
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    $\begingroup$ Consider $X=\textrm{Spec}(k[x_1,x_2,...])$ the infinite dimensional affine space, $U=X-0$ punctured, and $Y=X\cup_U X$ with the doubled origin. The first is affine, the second not quasi-compact, the third not quasi-separated. They all have the same perfect complexes, but different QCoh. The third has extra QCoh, so fails your hypothesis. The second might be OK. @TimCampion only mildly. That is a finite question, which is automatically true in the derived setting, whereas this question is about the difference between finite and infinite. $\endgroup$ – Ben Wieland May 13 at 1:47
  • $\begingroup$ @BenWieland You should give this as an answer! Thanks! If you can also come up with cases where QCoh "doesn't have enough almost-perfects", that would also be neat, but this answer is enough if you don't feel like it or can't think of anything =). $\endgroup$ – Harry Gindi May 13 at 1:56
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    $\begingroup$ If you want to go stacks, there are serious obstructions against being generated by perfect complexes. See this nice paper of Hall, Neeman and Rydh: arxiv1405.1888 $\endgroup$ – Denis-Charles Cisinski May 14 at 18:30
  • $\begingroup$ @Denis-CharlesCisinski I was looking at that paper, a bit earlier today. I thought that their theorem is about obstructions to being generated by compact objects (which would be equivalent in the case of a quasi-geometric stack to having QCoh(X)=Ind(Perf(X))=Ind(QCoh(X)^ω). I'm asking for something a lot weaker, namely that the functor QCoh(X)→Ind(Perf(X)) is fully faithful (and with no assumption that the perfects are compact). Is there a part of the paper where they address this question too? At least from the part I read so far, they're showing that QCoh(X)^ω is often trivial. $\endgroup$ – Harry Gindi May 14 at 18:43
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Robert Thomason was the first person to draw attention to this question, before derived schemes and infinity categories. I believe that he proved that for a quasi-compact and quasi-separated scheme that $D_{qc}=\textrm{Ind}(\textrm{Perf})$. For example, see Thomason-Trobaugh section 2.3, though at first glance it appears that only proves the weaker statement that it has enough perfect complexes.

Somewhere he gives two examples to show the necessity of the two hypotheses. Consider an affine scheme with a point of infinite codimension, say, $X=\textrm{Spec}\;k[x_1,x_2,…]$. Let $U$ be the complement of the point. It is not quasi-compact. Let $Y=X\cup_U X$ be $X$ with the point doubled. It is not quasi-separated. A perfect complex is built from finitely many operations, so its support has finite codimension, so they do not notice the points of infinite codimension, so the three schemes all have the same perfect complexes. But they have different quasi-coherent complexes, such as the skyscrapers on the origins. In particular, $Y$ has two such sheaves, but they cannot be distinguished by perfect complexes. Whereas $U$ has too few sheaves, so it fails the strong hypothesis of equivalence of categories, but it still has enough perfects: $\textrm{QCoh}(U)\subset \textrm{QCoh}(X)=\textrm{Ind}(\textrm{Perf(X)})=\textrm{Ind}(\textrm{Perf(U)})$


Is there a finite dimensional example? For example, consider this non-quasi-compact scheme built from varieties. Let $Z_0=\mathbb A^2$ and $x_0=0\in Z_0$. Let $Z_{n+1}$ be the blow up of $Z_n$ at $x_n$ and let $x_{n+1}$ be a point in the exceptional fiber. Let $U_n=Z_n-\{x_n\}$. Then $U_n$ is open in $U_{n+1}$ and let $U'=\bigcup U_n$. Does it satisfy $D_{qc}=\textrm{Ind}(\textrm{Perf})$? I believe that it can be compactified by adding a 2-dimensional valuation ring. If so, we could double that point to get a non-quasi-separated scheme. Would it fail to have enough perfects?

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    $\begingroup$ For $n < \infty$, the spectrum of a valuation ring of dimension $n$ has exactly $n+1$ points (as the primes are totally ordered), so any open subset is trivially quasi-compact. $\endgroup$ – Anonymous May 14 at 16:04
  • $\begingroup$ @Anonymous thanks, edited to be more hypothetical $\endgroup$ – Ben Wieland May 14 at 17:24

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