The Stacks project

Proof. Choose a finite affine open covering. On each member of the covering we get stabilization by Algebra, Lemma 10.51.1. Hence the lemma follows. $\square$

Proof. This follows immediately from Algebra, Lemma 10.62.4 because $X$ has a finite covering by spectra of Noetherian rings. $\square$

Proof. This follows immediately from Algebra, Lemma 10.51.2 because $X$ has a finite covering by spectra of Noetherian rings. $\square$

Proof. This is a reformulation of Properties, Lemma 28.22.3 in view of the fact that a finite type quasi-coherent $\mathcal{O}_ X$-module is coherent by Lemma 30.9.1. $\square$

Proof. We first prove the second map is an isomorphism. It is injective by Properties, Lemma 28.25.3. Since $\mathcal{F}$ is the union of its coherent submodules, see Properties, Lemma 28.22.3 (and Lemma 30.9.1) we may and do assume that $\mathcal{F}$ is coherent to prove surjectivity. Let $\mathcal{F}_ n$ denote the quasi-coherent subsheaf of $\mathcal{F}$ consisting of sections annihilated by $\mathcal{I}^ n$, see Properties, Lemma 28.25.3. Since $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \ldots $ we see that $\mathcal{F}_ n = \mathcal{F}_{n + 1} = \ldots $ for some $n \geq 0$ by Lemma 30.10.1. Set $\mathcal{H} = \mathcal{F}_ n$ for this $n$. By Artin-Rees (Lemma 30.10.3) there exists an $c \geq 0$ such that $\mathcal{I}^ m\mathcal{F} \cap \mathcal{H} \subset \mathcal{I}^{m - c}\mathcal{H}$. Picking $m = n + c$ we get $\mathcal{I}^ m\mathcal{F} \cap \mathcal{H} \subset \mathcal{I}^ n\mathcal{H} = 0$. Thus if we set $\mathcal{F}' = \mathcal{I}^ m\mathcal{F}$ then we see that $\mathcal{F}' \cap \mathcal{F}_ n = 0$ and $\mathcal{F}'|_ U = \mathcal{F}|_ U$. Note in particular that the subsheaf $(\mathcal{F}')_ N$ of sections annihilated by $\mathcal{I}^ N$ is zero for all $N \geq 0$. Hence by Properties, Lemma 28.25.3 we deduce that the top horizontal arrow in the following commutative diagram is a bijection:

Since also the right vertical arrow is a bijection we conclude that the bottom horizontal arrow is surjective as desired.

Next, we prove the first arrow of the lemma is a bijection. By Lemma 30.9.1 the sheaf $\mathcal{G}$ is of finite presentation and hence the sheaf $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{F})$ is quasi-coherent, see Schemes, Section 26.24. By definition we have

Pick a $\psi $ in the right hand side of the first arrow of the lemma, i.e., $\psi \in \mathcal{H}(U)$. The result just proved applies to $\mathcal{H}$ and hence there exists an $n \geq 0$ and an $\varphi : \mathcal{I}^ n \to \mathcal{H}$ which recovers $\psi $ on restriction to $U$. By Modules, Lemma 17.22.1 $\varphi $ corresponds to a map

This is almost what we want except that the source of the arrow is the tensor product of $\mathcal{I}^ n$ and $\mathcal{G}$ and not the product. We will show that, at the cost of increasing $n$, the difference is irrelevant. Consider the short exact sequence

where $\mathcal{K}$ is defined as the kernel. Note that $\mathcal{I}^ n\mathcal{K} = 0$ (proof omitted). By Artin-Rees again we see that

for some $m$ large enough. In other words we see that

is an isomorphism. Let $\varphi '$ be the restriction of $\varphi $ to this submodule thought of as a map $\mathcal{I}^{m + n}\mathcal{G} \to \mathcal{F}$. Then $\varphi '$ gives an element of the left hand side of the first arrow of the lemma which maps to $\psi $ via the arrow. In other words we have proved surjectivity of the arrow. We omit the proof of injectivity. $\square$

Proof. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the coherent sheaf of ideals cutting out the reduced induced scheme structure on $X \setminus U$. If $X$ is Noetherian, then Lemma 30.10.5 tells us that we can take $\mathcal{F}' = \mathcal{I}^ n\mathcal{F}$ for some $n$. The general case will follow from this using Zorn's lemma.

Consider the set of triples $(U', \mathcal{F}', \varphi ')$ where $U \subset U' \subset X$ is open, $\mathcal{F}' \subset \mathcal{F}|_{U'}$ is a coherent subsheaf agreeing with $\mathcal{F}$ over $U$, and $\varphi ' : \mathcal{F}' \to \mathcal{G}|_{U'}$ restricts to $\varphi $ over $U$. We say $(U'', \mathcal{F}'', \varphi '') \geq (U', \mathcal{F}', \varphi ')$ if and only if $U'' \supset U'$, $\mathcal{F}''|_{U'} = \mathcal{F}'$, and $\varphi ''|_{U'} = \varphi '$. It is clear that if we have a totally ordered collection of triples $(U_ i, \mathcal{F}_ i, \varphi _ i)$, then we can glue the $\mathcal{F}_ i$ to a subsheaf $\mathcal{F}'$ of $\mathcal{F}$ over $U' = \bigcup U_ i$ and extend $\varphi $ to a map $\varphi ' : \mathcal{F}' \to \mathcal{G}|_{U'}$. Hence any totally ordered subset of triples has an upper bound. Finally, suppose that $(U', \mathcal{F}', \varphi ')$ is any triple but $U' \not= X$. Then we can choose an affine open $W \subset X$ which is not contained in $U'$. By the result of the first paragraph we can extend the subsheaf $\mathcal{F}'|_{W \cap U'}$ and the restriction $\varphi '|_{W \cap U'}$ to some subsheaf $\mathcal{F}'' \subset \mathcal{F}|_ W$ and map $\varphi '' : \mathcal{F}'' \to \mathcal{G}|_ W$. Of course the agreement between $(\mathcal{F}', \varphi ')$ and $(\mathcal{F}'', \varphi '')$ over $W \cap U'$ exactly means that we can extend this to a triple $(U' \cup W, \mathcal{F}''', \varphi ''')$. Hence any maximal triple $(U', \mathcal{F}', \varphi ')$ (which exist by Zorn's lemma) must have $U' = X$ and the proof is complete. $\square$