Lemma 15.81.2. Let $R$ be a ring and $K^\bullet $ a complex of $R$-modules. Let $m \in \mathbf{Z}$. Consider the $R$-algebra map $R[x] \to R$ which maps $x$ to zero. Then $K^\bullet $ is $m$-pseudo-coherent as a complex of $R$-modules if and only if $K^\bullet $ is $m$-pseudo-coherent as a complex of $R[x]$-modules.
Proof. This is a special case of Lemma 15.64.11. We also prove it in another way as follows.
Note that $0 \to R[x] \to R[x] \to R \to 0$ is exact. Hence $R$ is pseudo-coherent as an $R[x]$-module. Thus one implication of the lemma follows from Lemma 15.64.11. To prove the other implication, assume that $K^\bullet $ is $m$-pseudo-coherent as a complex of $R[x]$-modules. By Lemma 15.64.12 we see that $K^\bullet \otimes ^{\mathbf{L}}_{R[x]} R$ is $m$-pseudo-coherent as a complex of $R$-modules. By Lemma 15.81.1 we see that $K^\bullet \oplus K^\bullet [1]$ is $m$-pseudo-coherent as a complex of $R$-modules. Finally, we conclude that $K^\bullet $ is $m$-pseudo-coherent as a complex of $R$-modules from Lemma 15.64.8. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)