Lemma 56.6.5. Let $X$ be a quasi-compact and quasi-separated scheme. If $\mathit{QCoh}(\mathcal{O}_ X)$ is equivalent to the category of modules over a ring, then $X$ is affine.

Proof. Say $F : \text{Mod}_ R \to \mathit{QCoh}(\mathcal{O}_ X)$ is an equivalence. Then $\mathcal{F} = F(R)$ has the following properties:

1. it is a finitely presented $\mathcal{O}_ X$-module (Lemma 56.6.1),

2. $\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, -)$ is exact,

3. $\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, \mathcal{F})$ is a commutative ring,

4. every object of $\mathit{QCoh}(\mathcal{O}_ X)$ is a quotient of a direct sum of copies of $\mathcal{F}$.

Let $x \in X$ be a closed point. Consider the surjection

$\mathcal{O}_ X \to i_*\kappa (x)$

where the target is the pushforward of $\kappa (x)$ by the inclusion morphism $i : x \to X$. We have

$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, i_*\kappa (x)) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, \kappa (x))$

This first by (4) implies that $\mathcal{F}_ x$ is nonzero. From (2) we deduce that every map $\mathcal{F}_ x \to \kappa (x)$ lifts to a map $\mathcal{F}_ x \to \mathcal{O}_{X, x}$ (as it even lifts to a global map $\mathcal{F} \to \mathcal{O}_ X$). Since $\mathcal{F}_ x$ is a finite $\mathcal{O}_{X, x}$-module, this implies that $\mathcal{F}_ x$ is a (nonzero) finite free $\mathcal{O}_{X, x}$-module. Then since $\mathcal{F}$ is of finite presentation, this implies that $\mathcal{F}$ is finite free of positive rank in an open neighbourhood of $x$ (Modules, Lemma 17.11.6). Since every closed subset of $X$ contains a closed point (Topology, Lemma 5.12.8) this implies that $\mathcal{F}$ is finite locally free of positive rank. Similarly, the map

$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, \mathcal{F}) \to \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, i_*i^*\mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _{\kappa (x)}(\mathcal{F}_ x/\mathfrak m_ x \mathcal{F}_ x, \mathcal{F}_ x/\mathfrak m_ x \mathcal{F}_ x)$

is surjective. By property (3) we conclude that the rank $\mathcal{F}_ x$ must be $1$. Hence $\mathcal{F}$ is an invertible $\mathcal{O}_ X$-module. But then we conclude that the functor

$\mathcal{H} \longmapsto \Gamma (X, \mathcal{H}) = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, \mathcal{H}) = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, \mathcal{H} \otimes _{\mathcal{O}_ X} \mathcal{F})$

on $\mathit{QCoh}(\mathcal{O}_ X)$ is exact too. This implies that the first $\mathop{\mathrm{Ext}}\nolimits$ group

$\mathop{\mathrm{Ext}}\nolimits ^1_{\mathit{QCoh}(\mathcal{O}_ X)}(\mathcal{O}_ X, \mathcal{H}) = 0$

computed in the abelian category $\mathit{QCoh}(\mathcal{O}_ X)$ vanishes for all $\mathcal{H}$ in $\mathit{QCoh}(\mathcal{O}_ X)$. However, since $\mathit{QCoh}(\mathcal{O}_ X) \subset \textit{Mod}(\mathcal{O}_ X)$ is closed under extensions (Schemes, Section 26.24) we see that $\mathop{\mathrm{Ext}}\nolimits ^1$ between quasi-coherent modules computed in $\mathit{QCoh}(\mathcal{O}_ X)$ is the same as computed in $\textit{Mod}(\mathcal{O}_ X)$. Hence we conclude that

$H^1(X, \mathcal{H}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\textit{Mod}(\mathcal{O}_ X)}(\mathcal{O}_ X, \mathcal{H}) = 0$

for all $\mathcal{H}$ in $\mathit{QCoh}(\mathcal{O}_ X)$. This implies that $X$ is affine for example by Cohomology of Schemes, Lemma 30.3.1. $\square$

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