Lemma 15.64.15. Let $R$ be a ring. Let $m \in \mathbf{Z}$. Let $K^\bullet $ be a complex of $R$-modules. Let $R \to R'$ be a faithfully flat ring map. If the complex $K^\bullet \otimes _ R R'$ is $m$-pseudo-coherent (resp. pseudo-coherent), then $K^\bullet $ is $m$-pseudo-coherent (resp. pseudo-coherent).

**Proof.**
We will use without further mention that $- \otimes _ R R'$ is an exact functor and that therefore

Assume $K^\bullet \otimes _ R R'$ is $m$-pseudo-coherent. Let $n \in \mathbf{Z}$ be the largest integer such that $H^ n(K^\bullet )$ is nonzero; then $n$ is also the largest integer such that $H^ n(K^\bullet \otimes _ R R')$ is nonzero. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma 15.64.7. If $n \geq m$, then $H^ n(K^\bullet ) \otimes _ R R'$ is a finite $R'$-module, see Lemma 15.64.3. Hence $H^ n(K^\bullet )$ is a finite $R$-module, see Algebra, Lemma 10.83.2. Choose a finite free $R$-module $E$ and a surjection $E \to H^ n(K^\bullet )$. As $E$ is projective we can lift this to a map of complexes $\alpha : E[-n] \to K^\bullet $. Then the cone $C(\alpha )^\bullet $ has vanishing cohomology in degrees $\geq n$. On the other hand, the complex $C(\alpha )^\bullet \otimes _ R R'$ is $m$-pseudo-coherent, see Lemma 15.64.2. Hence by induction we see that $C(\alpha )^\bullet $ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.64.2 once more we conclude. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: