Lemma 23.6.10. Let $R \to S$ be a pseudo-coherent ring map (More on Algebra, Definition 15.82.1). Then Lemma 23.6.9 holds, with the resolution $A$ of $S$ having finitely many generators in each degree.

Proof. This is proved in exactly the same way as Lemma 23.6.9. The only additional twist is that, given $A(m) \to S$ we have to show that $H_ m = H_ m(A(m))$ is a finite $R[x_1, \ldots , x_ m]$-module (so that in the next step we need only add finitely many variables). Consider the complex

$\ldots \to A(m)_{m - 1} \to A(m)_ m \to A(m)_{m - 1} \to \ldots \to A(m)_0 \to S \to 0$

Since $S$ is a pseudo-coherent $R[x_1, \ldots , x_ n]$-module and since $A(m)_ i$ is a finite free $R[x_1, \ldots , x_ n]$-module we conclude that this is a pseudo-coherent complex, see More on Algebra, Lemma 15.64.9. Since the complex is exact in (homological) degrees $> m$ we conclude that $H_ m$ is a finite $R$-module by More on Algebra, Lemma 15.64.3. $\square$

There are also:

• 2 comment(s) on Section 23.6: Tate resolutions

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).