The Stacks project

75.22 Producing perfect complexes

The following lemma is our main technical tool for producing perfect complexes. Later versions of this result will reduce to this by Noetherian approximation.

Lemma 75.22.1. Let $S$ be a scheme. Let $Y$ be a Noetherian algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces which is locally of finite type and quasi-separated. Let $E \in D(\mathcal{O}_ X)$ such that

  1. $E \in D^ b_{\textit{Coh}}(\mathcal{O}_ X)$,

  2. the support of $H^ i(E)$ is proper over $Y$ for all $i$,

  3. $E$ has finite tor dimension as an object of $D(f^{-1}\mathcal{O}_ Y)$.

Then $Rf_*E$ is a perfect object of $D(\mathcal{O}_ Y)$.

Proof. By Lemma 75.8.1 we see that $Rf_*E$ is an object of $D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$. Hence $Rf_*E$ is pseudo-coherent (Lemma 75.13.7). Hence it suffices to show that $Rf_*E$ has finite tor dimension, see Cohomology on Sites, Lemma 21.47.4. By Lemma 75.13.8 it suffices to check that $Rf_*(E) \otimes _{\mathcal{O}_ Y}^\mathbf {L} \mathcal{F}$ has universally bounded cohomology for all quasi-coherent sheaves $\mathcal{F}$ on $Y$. Bounded from above is clear as $Rf_*(E)$ is bounded from above. Let $T \subset |X|$ be the union of the supports of $H^ i(E)$ for all $i$. Then $T$ is proper over $Y$ by assumptions (1) and (2) and Lemma 75.7.6. In particular there exists a quasi-compact open subspace $X' \subset X$ containing $T$. Setting $f' = f|_{X'}$ we have $Rf_*(E) = Rf'_*(E|_{X'})$ because $E$ restricts to zero on $X \setminus T$. Thus we may replace $X$ by $X'$ and assume $f$ is quasi-compact. We have assumed $f$ is quasi-separated. Thus

\[ Rf_*(E) \otimes _{\mathcal{O}_ Y}^\mathbf {L} \mathcal{F} = Rf_*\left(E \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*\mathcal{F}\right) = Rf_*\left(E \otimes _{f^{-1}\mathcal{O}_ Y}^\mathbf {L} f^{-1}\mathcal{F}\right) \]

by Lemma 75.20.1 and Cohomology on Sites, Lemma 21.18.5. By assumption (3) the complex $E \otimes _{f^{-1}\mathcal{O}_ Y}^\mathbf {L} f^{-1}\mathcal{F}$ has cohomology sheaves in a given finite range, say $[a, b]$. Then $Rf_*$ of it has cohomology in the range $[a, \infty )$ and we win. $\square$

Lemma 75.22.2. Let $S$ be a scheme. Let $B$ be a Noetherian algebraic space over $S$. Let $f : X \to B$ be a morphism of algebraic spaces which is locally of finite type and quasi-separated. Let $E \in D(\mathcal{O}_ X)$ be perfect. Let $\mathcal{G}^\bullet $ be a bounded complex of coherent $\mathcal{O}_ X$-modules flat over $B$ with support proper over $B$. Then $K = Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet )$ is a perfect object of $D(\mathcal{O}_ B)$.

Proof. The object $K$ is perfect by Lemma 75.22.1. We check the lemma applies: Locally $E$ is isomorphic to a finite complex of finite free $\mathcal{O}_ X$-modules. Hence locally $E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet $ is isomorphic to a finite complex whose terms are of the form

\[ \bigoplus \nolimits _{i = a, \ldots , b} (\mathcal{G}^ i)^{\oplus r_ i} \]

for some integers $a, b, r_ a, \ldots , r_ b$. This immediately implies the cohomology sheaves $H^ i(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G})$ are coherent. The hypothesis on the tor dimension also follows as $\mathcal{G}^ i$ is flat over $f^{-1}\mathcal{O}_ Y$. $\square$

Lemma 75.22.3. Let $S$ be a scheme. Let $B$ be a Noetherian algebraic space over $S$. Let $f : X \to B$ be a morphism of algebraic spaces which is locally of finite type and quasi-separated. Let $E \in D(\mathcal{O}_ X)$ be perfect. Let $\mathcal{G}^\bullet $ be a bounded complex of coherent $\mathcal{O}_ X$-modules flat over $B$ with support proper over $B$. Then $K = Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G})$ is a perfect object of $D(\mathcal{O}_ B)$.

Proof. Since $E$ is a perfect complex there exists a dual perfect complex $E^\vee $, see Cohomology on Sites, Lemma 21.48.4. Observe that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet ) = E^\vee \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet $. Thus the perfectness of $K$ follows from Lemma 75.22.2. $\square$


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