Lemma 21.43.7. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K$ be an object of $D(\mathcal{O})$. Let $m \in \mathbf{Z}$.

1. If $K$ is $m$-pseudo-coherent and $H^ i(K) = 0$ for $i > m$, then $H^ m(K)$ is a finite type $\mathcal{O}$-module.

2. If $K$ is $m$-pseudo-coherent and $H^ i(K) = 0$ for $i > m + 1$, then $H^{m + 1}(K)$ is a finitely presented $\mathcal{O}$-module.

Proof. Proof of (1). Let $U$ be an object of $\mathcal{C}$. We have to show that $H^ m(K)$ is can be generated by finitely many sections over the members of a covering of $U$ (see Modules on Sites, Definition 18.23.1). Thus during the proof we may (finitely often) choose a covering $\{ U_ i \to U\}$ and replace $\mathcal{C}$ by $\mathcal{C}/U_ i$ and $U$ by $U_ i$. In particular, by our definitions we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet$ and a map $\alpha : \mathcal{E}^\bullet \to K$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. It suffices to prove the result for $\mathcal{E}^\bullet$. Let $n$ be the largest integer such that $\mathcal{E}^ n \not= 0$. If $n = m$, then $H^ m(\mathcal{E}^\bullet )$ is a quotient of $\mathcal{E}^ n$ and the result is clear. If $n > m$, then $\mathcal{E}^{n - 1} \to \mathcal{E}^ n$ is surjective as $H^ n(E^\bullet ) = 0$. By Lemma 21.42.5 we can (after replacing $U$ by the members of a covering) find a section of this surjection and write $\mathcal{E}^{n - 1} = \mathcal{E}' \oplus \mathcal{E}^ n$. Hence it suffices to prove the result for the complex $(\mathcal{E}')^\bullet$ which is the same as $\mathcal{E}^\bullet$ except has $\mathcal{E}'$ in degree $n - 1$ and $0$ in degree $n$. We win by induction on $n$.

Proof of (2). Pick an object $U$ of $\mathcal{C}$. As in the proof of (1) we may work locally on $U$. Hence we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet$ and a map $\alpha : \mathcal{E}^\bullet \to K$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. As in the proof of (1) we can reduce to the case that $\mathcal{E}^ i = 0$ for $i > m + 1$. Then we see that $H^{m + 1}(K) \cong H^{m + 1}(\mathcal{E}^\bullet ) = \mathop{\mathrm{Coker}}(\mathcal{E}^ m \to \mathcal{E}^{m + 1})$ which is of finite presentation. $\square$

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