## 65.13 Coherent sheaves on Noetherian spaces

In this section we mention some properties of coherent sheaves on Noetherian algebraic spaces.

Lemma 65.13.1. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. The ascending chain condition holds for quasi-coherent submodules of $\mathcal{F}$. In other words, given any sequence

$\mathcal{F}_1 \subset \mathcal{F}_2 \subset \ldots \subset \mathcal{F}$

of quasi-coherent submodules, then $\mathcal{F}_ n = \mathcal{F}_{n + 1} = \ldots$ for some $n \geq 0$.

Proof. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$ (see Properties of Spaces, Lemma 62.6.3). Then $U$ is a Noetherian scheme (by Morphisms of Spaces, Lemma 63.23.5). If $\mathcal{F}_ n|_ U = \mathcal{F}_{n + 1}|_ U = \ldots$ then $\mathcal{F}_ n = \mathcal{F}_{n + 1} = \ldots$. Hence the result follows from the case of schemes, see Cohomology of Schemes, Lemma 29.10.1. $\square$

Lemma 65.13.2. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{F}$ be a coherent sheaf on $X$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals corresponding to a closed subspace $Z \subset X$. Then there is some $n \geq 0$ such that $\mathcal{I}^ n\mathcal{F} = 0$ if and only if $\text{Supp}(\mathcal{F}) \subset Z$ (set theoretically).

Proof. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$ (see Properties of Spaces, Lemma 62.6.3). Then $U$ is a Noetherian scheme (by Morphisms of Spaces, Lemma 63.23.5). Note that $\mathcal{I}^ n\mathcal{F}|_ U = 0$ if and only if $\mathcal{I}^ n\mathcal{F} = 0$ and similarly for the condition on the support. Hence the result follows from the case of schemes, see Cohomology of Schemes, Lemma 29.10.2. $\square$

Lemma 65.13.3 (Artin-Rees). Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{F}$ be a coherent sheaf on $X$. Let $\mathcal{G} \subset \mathcal{F}$ be a quasi-coherent subsheaf. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Then there exists a $c \geq 0$ such that for all $n \geq c$ we have

$\mathcal{I}^{n - c}(\mathcal{I}^ c\mathcal{F} \cap \mathcal{G}) = \mathcal{I}^ n\mathcal{F}$

Proof. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$ (see Properties of Spaces, Lemma 62.6.3). Then $U$ is a Noetherian scheme (by Morphisms of Spaces, Lemma 63.23.5). The equality of the lemma holds if and only if it holds after restricting to $U$. Hence the result follows from the case of schemes, see Cohomology of Schemes, Lemma 29.10.3. $\square$

Lemma 65.13.4. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Denote $Z \subset X$ the corresponding closed subspace and set $U = X \setminus Z$. There is a canonical isomorphism

$\mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{I}^ n\mathcal{G}, \mathcal{F}) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{G}|_ U, \mathcal{F}|_ U).$

In particular we have an isomorphism

$\mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{I}^ n, \mathcal{F}) \longrightarrow \Gamma (U, \mathcal{F}).$

Proof. Let $W$ be an affine scheme and let $W \to X$ be a surjective étale morphism (see Properties of Spaces, Lemma 62.6.3). Set $R = W \times _ X W$. Then $W$ and $R$ are Noetherian schemes, see Morphisms of Spaces, Lemma 63.23.5. Hence the result hold for the restrictions of $\mathcal{F}$, $\mathcal{G}$, and $\mathcal{I}$, $U$, $Z$ to $W$ and $R$ by Cohomology of Schemes, Lemma 29.10.4. It follows formally that the result holds over $X$. $\square$

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).