The Stacks project

Proof. We will use the criterion of Homology, Lemma 12.10.2. Moreover, we will use that if $0 \to (\mathcal{G}_ n) \to (\mathcal{F}_ n) \to (\mathcal{H}_ n) \to 0$ is a short exact sequence of $\textit{Coh}(X, \mathcal{I})$, then (a) $\mathcal{G}_ n \to \mathcal{F}_ n \to \mathcal{H}_ n \to 0$ is exact for all $n \geq 1$ and (b) $\mathcal{G}_ n$ is a quotient of $\mathop{\mathrm{Ker}}(\mathcal{F}_ m \to \mathcal{H}_ m)$ for some $m \geq n$. See proof of Lemma 30.23.2.

Proof of (1). Let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(X, \mathcal{I})$. Then $\text{Supp}(\mathcal{F}_ n) = \text{Supp}(\mathcal{F}_1)$ for all $n \geq 1$. Hence by remarks (a) and (b) above we see that for any short exact sequence $0 \to (\mathcal{G}_ n) \to (\mathcal{F}_ n) \to (\mathcal{H}_ n) \to 0$ of $\textit{Coh}(X, \mathcal{I})$ we have $\text{Supp}(\mathcal{G}_1) \cup \text{Supp}(\mathcal{H}_1) = \text{Supp}(\mathcal{F}_1)$. This proves that the category defined in (1) is a Serre subcategory of $\textit{Coh}(X, \mathcal{I})$.

Proof of (2). Here we argue the same way. Let $0 \to (\mathcal{G}_ n) \to (\mathcal{F}_ n) \to (\mathcal{H}_ n) \to 0$ be a short exact sequence of $\textit{Coh}(X, \mathcal{I})$. If $Z \subset X$ is a closed subscheme and $\mathcal{I}_ Z$ annihilates $\mathcal{F}_ n$ for all $n$, then $\mathcal{I}_ Z$ annihilates $\mathcal{G}_ n$ and $\mathcal{H}_ n$ for all $n$ by (a) and (b) above. Hence if $Z \to S$ is proper, then we conclude that the category defined in (2) is closed under taking sub and quotient objects inside of $\textit{Coh}(X, \mathcal{I})$. Finally, suppose that $Z \subset X$ and $Y \subset X$ are closed subschemes proper over $S$ such that $\mathcal{I}_ Z \mathcal{G}_ n = 0$ and $\mathcal{I}_ Y \mathcal{H}_ n = 0$ for all $n \geq 1$. Then it follows from (a) above that $\mathcal{I}_{Z \cup Y} = \mathcal{I}_ Z \cdot \mathcal{I}_ Y$ annihilates $\mathcal{F}_ n$ for all $n$. By Lemma 30.26.6 (and via Definition 30.26.2 which tells us we may choose an arbitrary scheme structure used on the union) we see that $Z \cup Y \to S$ is proper and the proof is complete. $\square$