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. $\square$

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