Lemma 15.64.3. Let $R$ be a ring. Let $K^\bullet$ be a complex of $R$-modules. Let $m \in \mathbf{Z}$.

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

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

Proof. Proof of (1). Choose a bounded complex $E^\bullet$ of finite projective $R$-modules and a map $\alpha : E^\bullet \to K^\bullet$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. It is clear that it suffices to prove the result for $E^\bullet$. Let $n$ be the largest integer such that $E^ n \not= 0$. If $n = m$, then the result is clear. If $n > m$, then $E^{n - 1} \to E^ n$ is surjective as $H^ n(E^\bullet ) = 0$. As $E^ n$ is finite projective we see that $E^{n - 1} = E' \oplus E^ n$. Hence it suffices to prove the result for the complex $(E')^\bullet$ which is the same as $E^\bullet$ except has $E'$ in degree $n - 1$ and $0$ in degree $n$. We win by induction on $n$.

Proof of (2). Choose a bounded complex $E^\bullet$ of finite projective $R$-modules and a map $\alpha : E^\bullet \to K^\bullet$ 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 $E^ i = 0$ for $i > m + 1$. Then we see that $H^{m + 1}(K^\bullet ) \cong H^{m + 1}(E^\bullet ) = \mathop{\mathrm{Coker}}(E^ m \to E^{m + 1})$ which is of finite presentation. $\square$

There are also:

• 7 comment(s) on Section 15.64: Pseudo-coherent modules, I

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).