The Stacks project

Proof. See Categories, Definition 4.26.1 for our notion of categorically compact objects in a category. If $\mathcal{F}$ is of finite presentation then it is categorically compact by Modules, Lemma 17.22.8. Conversely, any quasi-coherent module $\mathcal{F}$ can be written as a filtered colimit $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ of finitely presented (hence quasi-coherent) $\mathcal{O}_ X$-modules, see Properties, Lemma 28.22.7. If $\mathcal{F}$ is categorically compact, then we find some $i$ and a morphism $\mathcal{F} \to \mathcal{F}_ i$ which is a right inverse to the given map $\mathcal{F}_ i \to \mathcal{F}$. We conclude that $\mathcal{F}$ is a direct summand of a finitely presented module, and hence finitely presented itself. $\square$

Proof. Let us translate the statement into algebra. Let $A$ be a ring. Let $M$ be a finitely presented $A$-module. Let $N$ be a nonzero $A$-module. Assume $\text{Supp}(N) \subset \text{Supp}(M)$. To show: $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)$ is nonzero. We may assume $N = A/I$ is cyclic (replace $N$ by any nonzero cyclic submodule). Choose a presentation

Recall that $\text{Supp}(M)$ is cut out by $\text{Fit}_0(M)$ which is the ideal generated by the $n \times n$ minors of the matrix $T$. See More on Algebra, Lemma 15.8.4. The assumption $\text{Supp}(N) \subset \text{Supp}(M)$ now means that the elements of $\text{Fit}_0(M)$ are nilpotent in $A/I$. Consider the exact sequence

We have to show that $T^ t$ cannot be injective; we urge the reader to find their own proof of this using the nilpotency of elements of $\text{Fit}_0(M)$ in $A/I$. Here is our proof. Since $\text{Fit}_0(M)$ is finitely generated, the nilpotency means that the annihilator $J \subset A/I$ of $\text{Fit}_0(M)$ in $A/I$ is nonzero. To show the non-injectivity of $T^ t$ we may localize at a prime. Choosing a suitable prime we may assume $A$ is local and $J$ is still nonzero. Then $T^ t$ has a nonzero kernel by More on Algebra, Lemma 15.15.6. $\square$

Proof. Observe that the statement makes sense as finitely presented $\mathcal{O}_ X$-modules are quasi-coherent. Since $\mathcal{A}$ is a Serre subcategory closed under extensions and direct sums and since $\mathcal{F}$ is an object of $\mathcal{A}$ we see that $\mathcal{B} \subset \mathcal{A}$. Thus it remains to show that $\mathcal{A}$ is contained in $\mathcal{B}$.

Let $\mathcal{E}$ be an object of $\mathcal{A}$. There exists a maximal submodule $\mathcal{E}' \subset \mathcal{E}$ which is in $\mathcal{B}$. Namely, suppose $\mathcal{E}_ i \subset \mathcal{E}$, $i \in I$ is the set of subobjects which are objects of $\mathcal{B}$. Then $\bigoplus \mathcal{E}_ i$ is in $\mathcal{B}$ and so is

This is clearly the maximal submodule we were looking for.

Now suppose that we have a nonzero map $\mathcal{G} \to \mathcal{E}/\mathcal{E}'$ with $\mathcal{G}$ in $\mathcal{B}$. Then $\mathcal{G}' = \mathcal{E} \times _{\mathcal{E}/\mathcal{E}'} \mathcal{G}$ is in $\mathcal{B}$ as an extension of $\mathcal{E}'$ and $\mathcal{G}$. Then the image $\mathcal{G}' \to \mathcal{E}$ would be strictly bigger than $\mathcal{E}'$, contradicting the maximality of $\mathcal{E}'$. Thus it suffices to show the claim in the following paragraph.

Let $\mathcal{E}$ be an nonzero object of $\mathcal{A}$. We claim that there is a nonzero map $\mathcal{G} \to \mathcal{E}$ with $\mathcal{G}$ in $\mathcal{B}$. We will prove this by induction on the minimal number $n$ of affine opens $U_ i$ of $X$ such that $\text{Supp}(\mathcal{E}) \subset U_1 \cup \ldots \cup U_ n$. Set $U = U_ n$ and denote $j : U \to X$ the inclusion morphism. Denote $\mathcal{E}' = \mathop{\mathrm{Im}}(\mathcal{E} \to j_*\mathcal{E}|_ U)$. Then the kernel $\mathcal{E}''$ of the surjection $\mathcal{E} \to \mathcal{E}'$ has support contained in $U_1 \cup \ldots \cup U_{n - 1}$. Thus if $\mathcal{E}''$ is nonzero, then we win. In other words, we may assume that $\mathcal{E} \subset j_*\mathcal{E}|_ U$. In particular, we see that $\mathcal{E}|_ U$ is nonzero. By Lemma 56.6.2 there exists a nonzero map $\mathcal{F}|_ U \to \mathcal{E}|_ U$. This corresponds to a map

whose restriction to $U$ is nonzero. Setting $\mathcal{G} = \varphi ^{-1}(\mathcal{E})$ we conclude. $\square$

Proof. By the universal property of the quotient construction (Homology, Lemma 12.10.6) we certainly obtain an induced functor $\mathit{QCoh}(\mathcal{O}_ X)/\mathcal{A} \cong \mathit{QCoh}(\mathcal{O}_ U)$. Denote $j : U \to X$ the inclusion morphism. Since $j$ is quasi-compact and quasi-separated we obtain a functor $j_* : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_ X)$. The reader shows that this defines a quasi-inverse; details omitted. $\square$

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

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. $\square$

Proof. Of course $F$ is additive, exact, commutes with all limits, commutes with all colimits, commutes with direct sums, etc. Let $U \subset X$ be an affine open subscheme. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a finite type quasi-coherent sheaf of ideals such that $Z = V(\mathcal{I})$ is the complement of $U$ in $X$, see Properties, Lemma 28.24.1. Then $\mathcal{O}_ X/\mathcal{I}$ is a finitely presented $\mathcal{O}_ X$-module. Hence $\mathcal{G} = F(\mathcal{O}_ X/\mathcal{I})$ is a finitely presented $\mathcal{O}_ Y$-module by Lemma 56.6.1. Denote $T \subset Y$ the support of $\mathcal{G}$ and set $V = Y \setminus T$. Since $\mathcal{G}$ is of finite presentation, the scheme $V$ is a quasi-compact open of $Y$. By Lemma 56.6.3 we see that $F$ induces an equivalence between

1. the full subcategory of $\mathit{QCoh}(\mathcal{O}_ X)$ consisting of modules supported on $Z$, and

2. the full subcategory of $\mathit{QCoh}(\mathcal{O}_ Y)$ consisting of modules supported on $T$.

By Lemma 56.6.4 we obtain a commutative diagram

where the vertical arrows are the restruction functors and the horizontal arrows are equivalences. By Lemma 56.6.5 we conclude that $V$ is affine. For the affine case we have Lemma 56.3.8. Thus we find that there is an isomorphism $f_ U : V \to U$ and an invertible $\mathcal{O}_ V$-module $\mathcal{L}_ U$ such that $F_ U$ is the functor $\mathcal{F} \mapsto f_ U^*\mathcal{F} \otimes \mathcal{L}_ U$.

The proof can be finished by noticing that the diagrams above satisfy an obvious compatibility with regards to inclusions of affine open subschemes of $X$. Thus the morphisms $f_ U$ and the invertible modules $\mathcal{L}_ U$ glue. We omit the details. $\square$