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:

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

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

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

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

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

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

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

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

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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)