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

$H^ i(K^\bullet ) \otimes _ R R' = H^ i(K^\bullet \otimes _ R R').$

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. $\square$

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