The Stacks project

Lemma 15.62.15. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ be elements which generate the unit ideal. Let $m \in \mathbf{Z}$. Let $K^\bullet $ be a complex of $R$-modules. If for each $i$ the complex $K^\bullet \otimes _ R R_{f_ i}$ 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 $- \otimes _ R R_{f_ i}$ is an exact functor and that therefore

\[ H^ i(K^\bullet )_{f_ i} = H^ i(K^\bullet ) \otimes _ R R_{f_ i} = H^ i(K^\bullet \otimes _ R R_{f_ i}). \]

Assume $K^\bullet \otimes _ R R_{f_ i}$ is $m$-pseudo-coherent for $i = 1, \ldots , r$. Let $n \in \mathbf{Z}$ be the largest integer such that $H^ n(K^\bullet \otimes _ R R_{f_ i})$ is nonzero for some $i$. This implies in particular that $H^ i(K^\bullet ) = 0$ for $i > n$ (and that $H^ n(K^\bullet ) \not= 0$) see Algebra, Lemma 10.22.2. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma 15.62.8. If $n \geq m$, then $H^ n(K^\bullet )_{f_ i}$ is a finite $R_{f_ i}$-module for each $i$, see Lemma 15.62.3. Hence $H^ n(K^\bullet )$ is a finite $R$-module, see Algebra, Lemma 10.22.2. 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 complexes $C(\alpha )^\bullet \otimes _ R R_{f_ i}$ are $m$-pseudo-coherent for each $i$, see Lemma 15.62.2. Hence by induction we see that $C(\alpha )^\bullet $ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.62.2 once more we conclude. $\square$


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