## 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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