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Let $X$ be a scheme. Consider the global section functor $\Gamma : \mathrm{Qcoh}(X) \to \mathrm{Ab}$. It is well-known that $\Gamma$ preserves filtered colimits if $X$ is assumed to be quasi-compact and quasi-separated. In other words, $\mathcal{O}_X$ is a compact object. Even more, it can be shown (using the results of EGA I, 6.9) that compact objects are precisely the quasi-coherent modules which are locally of finite presentation. If we define a locally finitely presentable tensor category to be a tensor category $C$, whose underlying category is locally finitely presentable and such that a) $1_C$ is compact, b) the compact objects are closed under $\otimes_C$, then $\mathrm{Qcoh}(X)$ turns out to be such a category.

Does the converse also hold? This would be a categorical interpretation of this typical finiteness condition in algebraic geometry.

Question. When $\Gamma$ preserves filtered colimits, does it follow that $X$ is quasi-compact and quasi-separated? If not, what if we even assume that $\mathrm{Qcoh}(X)$ is a locally finitely presentable tensor category?

There are other natural conditions which might imply that $X$ is qc+qs, for example that even all cohomology functors $H^i : \mathrm{Qcoh}(X) \to \mathrm{Ab}$ preserve filtered colimits. Besides there are relative versions of this question: When $f : X \to Y$ is a morphism of schemes, then $f$ is quasi-compact and quasi-separated iff $f_\*$ or $f^\*$ has which property?

EDIT: Let us spell out what it means that $\Gamma$ preserves direct sums: Let $M_i$ be (infinitely many) quasi-coherent modules on $X$, $s_i \in \Gamma(M_i)$ global sections, and $X = \cup_j X_j$ an arbitrary open covering, then the following holds: If for every $j$ we have $s_i |_{X_j} = 0$ for almost all $i$, then $s_i = 0$ for almost all $i$. So the bounds for the $i$ with $s_i \neq 0$ on each $X_j$ are bounded. Why should this happen for a scheme which is not quasi-compact? But in order to prove this, we would have to construct appropriate quasi-coherent modules, and this can only be done my transfinite methods if $X$ is general (see for example here).

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    $\begingroup$ Even if this does not answer to your question, you might be interested to see a categorical interpretation of the property of bein quasi-compact and quasi-separated: have look at Exposé VI of SGA 4, in which you will see that quasi-compact quasi-separated schemes are precisely those for which the global section functor preserves filtered colimits of sheaves of sets for the Zariski or the étale topology (and there is also a relative version); see in particular Theorem 1.23 in loc. cit. $\endgroup$
    – D.-C. Cisinski
    Commented Mar 2, 2012 at 22:56
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    $\begingroup$ @Denis-Charles: Thanks for the reference. In this Exposé quasi-compact and quasi-separated morphisms are defined in an arbitrary topos. However it is then almost by definition the case that $X$ is qc/qs iff the final object of the corresponding topos $\mathrm{Sh}(\mathrm{Sch}/X)$ is qc/qs; see also loc. cit. 1.22. You mention Theorem 1.23, but this seems to be also a rather direct reformulation of the definition of qc objects (the proof just pulls back coverings). If we apply it to the topos associated to a scheme, I think we won't get anything new or interesting. $\endgroup$
    – Martin Brandenburg
    Commented Mar 3, 2012 at 9:24
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    $\begingroup$ The basic difference between the topos of all sheaves over $X$ and $\mathrm{Qcoh}(X)$ is that the former contains all representables. $\endgroup$
    – Martin Brandenburg
    Commented Mar 3, 2012 at 9:25
  • $\begingroup$ Martin, did you ever find a satisfactory answer including the quasi-separated condition? $\endgroup$
    – Brendan Murphy
    Commented Mar 21 at 19:00
  • $\begingroup$ No, but I also stopped working on such questions long ago. If $X$ is not qc qs, the category $\mathbf{Qcoh}(X)$ is hard to grasp. Many objects are "missing". This makes questions of this type also hard, but in any case they don't have much relevance I would say (nowadays). $\endgroup$
    – Martin Brandenburg
    Commented Aug 11 at 11:38

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Let me show that if $\Gamma$ preserves filtered colimits, then $X$ is quasicompact. (At the moment I don't know about 'quasiseparated'; but, as Martin points out, I only use the injectivity of $\varinjlim\circ \Gamma \to \Gamma\circ \varinjlim$ for filtered inductive systems).

Assume $X$ is not quasicompact. Then there is a filtered decreasing family $(Y_i)$ of nonempty closed subschemes of $X$, with empty intersection. The structure sheaves $\mathcal{O}_{Y_i}$ form an inductive system with colimit zero, but the unit section of any $Y_{i_0}$ is an element of $\varinjlim_i \Gamma(\mathcal{O}_{Y_i})$ which is nonzero because each $Y_i$ is nonempty.

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    $\begingroup$ Dear Laurent, thanks for this answer! Two minor corrections: a) "filtered colimits" instead of "colimits", b) "the family $(s_i)$ of unit sections" is an element in the limit, but in order to get an element in the colimit, we just have to specify one unit section. # Your proof only used that $\mathrm{colim}_i \Gamma(F_i) \to \Gamma(\mathrm{colim}_i F_i)$ is injective; maybe for quasi-separatedness we will need surjectivity. $\endgroup$
    – Martin Brandenburg
    Commented Mar 4, 2012 at 21:27
  • $\begingroup$ The proof can be made more direct: If $\{I_i\}$ is a filtered system of quasi-coherent ideals, then injectivity of $\mathrm{colim}_i \Gamma(\mathcal{O}_X/I_i) \to \Gamma(\mathrm{colim}_i \mathcal{O}_X/I_i) = \Gamma(\mathcal{O}_X/\sum_i I_i)$ implies: If $\sum_i I_i=\mathcal{O}_X$, then already $I_i = \mathcal{O}_X$ for some ideal (we use that $\mathcal{A}=0 \Leftrightarrow \Gamma(\mathcal{A})=0$ for algebras $\mathcal{A}$, using $1_A$). If we write $Z_i = \mathrm{supp}(\mathcal{O}_X/I_i)$, we see that $\emptyset = \cap_i Z_i$ implies $Z_i = \emptyset$ for some $i$. Hence $X$ is quasi-compact. $\endgroup$
    – Martin Brandenburg
    Commented Mar 4, 2012 at 21:44
  • $\begingroup$ @Martin: Thanks, I have edited my answer. Your second comment is indeed an alternative presentation. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 5, 2012 at 7:33

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