The Stacks project

Proof. The existence and uniqueness of the extension is a general fact, see Categories, Lemma 4.26.2. To see that the lemma applies observe that coherent modules are of finite presentation (Modules, Lemma 17.12.2) and hence categorically compact objects of $\textit{Mod}(\mathcal{O}_ X)$ by Modules, Lemma 17.22.8. Finally, every quasi-coherent module is a filtered colimit of coherent ones for example by Properties, Lemma 28.22.3.

Assume $F$ is additive. If $\mathcal{F} = \bigoplus _{j \in J} \mathcal{H}_ j$ with $\mathcal{H}_ j$ quasi-coherent, then $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} \bigoplus _{j \in J'} \mathcal{H}_ j$. Denoting the extension of $F$ also by $F$ we obtain

Thus $F$ commutes with arbitrary direct sums.

Suppose $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to 0$ is a short exact sequence of quasi-coherent $\mathcal{O}_ X$-modules. Then we write $\mathcal{F}' = \bigcup \mathcal{F}'_ i$ as the union of its coherent submodules, see Properties, Lemma 28.22.3. Denote $\mathcal{F}''_ i \subset \mathcal{F}''$ the image of $\mathcal{F}'_ i$ and denote $\mathcal{F}_ i = \mathcal{F} \cap \mathcal{F}'_ i = \mathop{\mathrm{Ker}}(\mathcal{F}'_ i \to \mathcal{F}''_ i)$. Then it is clear that $\mathcal{F} = \bigcup \mathcal{F}_ i$ and $\mathcal{F}'' = \bigcup \mathcal{F}''_ i$ and that we have short exact sequences

Since the extension commutes with filtered colimits we have $F(\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}_ i)$, $F(\mathcal{F}') = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}'_ i)$, and $F(\mathcal{F}'') = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}''_ i)$. Since filtered colimits are exact (Modules, Lemma 17.3.2) we conclude that exactness properties of $F$ are inherited by its extension. $\square$

Proof. By Lemma 56.7.1 we obtain a unique functor $F' : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ extending $F$. The same is true for the quasi-inverse of $F$ and by the uniqueness we conclude that $F'$ is an equivalence. By Proposition 56.6.6 we find an isomorphism $f : Y \to X$ and an invertible $\mathcal{O}_ Y$-module $\mathcal{L}$ such that $F'(\mathcal{F}) = f^*\mathcal{F} \otimes \mathcal{L}$. Then $f$ and $\mathcal{L}$ work for $F$ as well. $\square$

Proof. By Zariski's main theorem we can find an open immersion $j : V \to Y$ over $X$ with $\pi : Y \to X$ finite, see More on Morphisms, Lemma 37.43.3. Since $\pi $ is affine the functor $\pi _*$ is exact and faithful on the category of coherent $\mathcal{O}_ X$-modules. Hence we see that $j_*\mathcal{K}$ is coherent and that $R^ qj_*\mathcal{K}$ is zero for $q > 0$. In other words, we reduce to the case discussed in the next paragraph.

Assume $f$ is an open immersion. We may replace $X$ by the scheme theoretic closure of $V$. Assume $X \setminus V$ is nonempty to get a contradiction. Choose a generic point $\xi \in X \setminus V$ of an irreducible component of $X \setminus V$. Looking at the situation after base change by $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }) \to X$ using flat base change and using Local Cohomology, Lemma 51.8.2 we reduce to the algebra problem discussed in the next paragraph.

Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a finite $A$-module whose support is $\mathop{\mathrm{Spec}}(A)$. Then $H^ i_\mathfrak m(M) \not= 0$ for some $i$. This is true by Dualizing Complexes, Lemma 47.11.1 and the fact that $M$ is not zero hence has finite depth. $\square$

Proof. Let $\mathcal{K}$ be as in (2). By Lemma 56.5.7 the functor $F$ given by (56.5.1.1) is exact and $k$-linear. Moreover, $F$ sends $\textit{Coh}(\mathcal{O}_ X)$ into $\textit{Coh}(\mathcal{O}_ Y)$ for example by Cohomology of Schemes, Lemma 30.26.10.

Let us construct the quasi-inverse to the construction. Let $F$ be as in (1). By Lemma 56.7.1 we can extend $F$ to a $k$-linear exact functor on the categories of quasi-coherent modules which commutes with arbitrary direct sums. By Lemma 56.5.7 the extension corresponds to a unique quasi-coherent module $\mathcal{K}$, flat over $X$, such that $R^ q\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F} \otimes _{\mathcal{O}_{X \times Y}} \mathcal{K}) = 0$ for $q > 0$ for all quasi-coherent $\mathcal{O}_ X$-modules $\mathcal{F}$. Since $F(\mathcal{O}_ X)$ is a coherent $\mathcal{O}_ Y$-module, we conclude from Lemma 56.5.11 that $\mathcal{K}$ is coherent.

For a closed point $x \in X$ denote $\mathcal{O}_ x$ the skyscraper sheaf at $x$ with value the residue field of $x$. We have

Since $x \times Y \to Y$ is finite, we see that the pushforward along this morphism is faithful. Hence if $y \in Y$ is in the image of the support of $\mathcal{K}|_{x \times Y}$, then $y$ is in the support of $F(\mathcal{O}_ x)$.

Let $Z \subset X \times Y$ be the scheme theoretic support $Z$ of $\mathcal{K}$, see Morphisms, Definition 29.5.5. We first prove that $Z \to Y$ is quasi-finite, by proving that its fibres over closed points are finite. Namely, if the fibre of $Z \to Y$ over a closed point $y \in Y$ has dimension $> 0$, then we can find infinitely many pairwise distinct closed points $x_1, x_2, \ldots $ in the image of $Z_ y \to X$. Since we have a surjection $\mathcal{O}_ X \to \bigoplus _{i = 1, \ldots , n} \mathcal{O}_{x_ i}$ we obtain a surjection

By what we said above, the point $y$ is in the support of each of the coherent modules $F(\mathcal{O}_{x_ i})$. Since $F(\mathcal{O}_ X)$ is a coherent module, this will lead to a contradiction because the stalk of $F(\mathcal{O}_ X)$ at $y$ will be generated by $< n$ elements if $n$ is large enough. Hence $Z \to Y$ is quasi-finite. Since $\text{pr}_{2, *}\mathcal{K}$ is coherent and $R^ q\text{pr}_{2, *}\mathcal{K} = 0$ for $q > 0$ we conclude that $Z \to Y$ is finite by Lemma 56.7.4. $\square$

Proof. Looking affine locally this translates into the following algebra problem. Let $A \to B$ be a ring map and let $N$ be a $B$-module which is invertible as an $A$-module. Then the annihilator $J$ of $N$ in $B$ has the property that $A \to B/J$ is an isomorphism. We omit the details. $\square$

Proof. By Lemma 56.7.6 there exists a section $s' : Y \to X$ such that $\mathcal{F} = s'_*\mathcal{L}$. Since $s'(Y)$ and $s(Y)$ have the same underlying closed subset and since both are reduced closed subschemes of $X$, they have to be equal. Hence $s = s'$ and the lemma holds. $\square$

Proof using Gabriel-Rosenberg reconstruction. This lemma is a weak form of the results discussed in Lemma 56.7.2 and Remark 56.7.3. $\square$

Proof not relying on Gabriel-Rosenberg reconstruction. By Lemma 56.7.5 we obtain a coherent $\mathcal{O}_{X \times Y}$-module $\mathcal{K}$ which is flat over $X$ with support finite over $Y$ such that $F$ is given by the restriction of the functor (56.5.1.1) to $\textit{Coh}(\mathcal{O}_ X)$. If we can show that $F(\mathcal{O}_ X)$ is an invertible $\mathcal{O}_ Y$-module, then by Lemma 56.7.6 we see that $\mathcal{K} = s_*\mathcal{L}$ for some section $s : Y \to X \times Y$ of $\text{pr}_2$ and some invertible $\mathcal{O}_ Y$-module $\mathcal{L}$. This will show that $F$ has the form indicated with $f = \text{pr}_1 \circ s$. Some details omitted.

It remains to show that $F(\mathcal{O}_ X)$ is invertible. We only sketch the proof and we omit some of the details. For a closed point $x \in X$ we denote $\mathcal{O}_ x$ in $\textit{Coh}(\mathcal{O}_ X)$ the skyscraper sheaf at $x$ with value $\kappa (x)$. First we observe that the only simple objects of the category $\textit{Coh}(\mathcal{O}_ X)$ are these skyscraper sheaves $\mathcal{O}_ x$. The same is true for $Y$. Hence for every closed point $y \in Y$ there exists a closed point $x \in X$ such that $\mathcal{O}_ y \cong F(\mathcal{O}_ x)$. Moreover, looking at endomorphisms we find that $\kappa (x) \cong \kappa (y)$ as finite extensions of $k$. Then

This implies that the stalk of the coherent $\mathcal{O}_ Y$-module $F(\mathcal{O}_ X)$ at $y \in Y$ can be generated by $1$ generator (and no less) for each closed point $y \in Y$. It follows immediately that $F(\mathcal{O}_ X)$ is locally generated by $1$ element (and no less) and since $Y$ is reduced this indeed tells us it is an invertible module. $\square$