The Stacks project

84.8 Right adjoint of pushforward for proper flat morphisms

For proper, flat, and finitely presented morphisms of quasi-compact and quasi-separated algebraic spaces the right adjoint of pushforward enjoys some remarkable properties.

Lemma 84.8.1. Let $S$ be a scheme. Let $Y$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces which is proper, flat, and of finite presentation. Let $a$ be the right adjoint for $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 84.3.1. Then $a$ commutes with direct sums.

Proof. Let $P$ be a perfect object of $D(\mathcal{O}_ X)$. By Derived Categories of Spaces, Lemma 73.25.4 the complex $Rf_*P$ is perfect on $Y$. Let $K_ i$ be a family of objects of $D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P, a(\bigoplus K_ i)) & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*P, \bigoplus K_ i) \\ & = \bigoplus \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*P, K_ i) \\ & = \bigoplus \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P, a(K_ i)) \end{align*}

because a perfect object is compact (Derived Categories of Spaces, Proposition 73.16.1). Since $D_\mathit{QCoh}(\mathcal{O}_ X)$ has a perfect generator (Derived Categories of Spaces, Theorem 73.15.4) we conclude that the map $\bigoplus a(K_ i) \to a(\bigoplus K_ i)$ is an isomorphism, i.e., $a$ commutes with direct sums. $\square$

Lemma 84.8.2. Let $S$ be a scheme. Let $Y$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces which is proper, flat, and of finite presentation. The map (84.7.0.1) is an isomorphism for every object $K$ of $D_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. By Lemma 84.8.1 we know that $a$ commutes with direct sums. Hence the collection of objects of $D_\mathit{QCoh}(\mathcal{O}_ Y)$ for which (84.7.0.1) is an isomorphism is a strictly full, saturated, triangulated subcategory of $D_\mathit{QCoh}(\mathcal{O}_ Y)$ which is moreover preserved under taking direct sums. Since $D_\mathit{QCoh}(\mathcal{O}_ Y)$ is a module category (Derived Categories of Spaces, Theorem 73.17.3) generated by a single perfect object (Derived Categories of Spaces, Theorem 73.15.4) we can argue as in More on Algebra, Remark 15.57.13 to see that it suffices to prove (84.7.0.1) is an isomorphism for a single perfect object. However, the result holds for perfect objects, see Lemma 84.7.1. $\square$

Lemma 84.8.3. Let $Y$ be an affine scheme. Let $f : X \to Y$ be a morphism of algebraic spaces which is proper, flat, and of finite presentation. Let $a$ be the right adjoint for $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 84.3.1. Then

  1. $a(\mathcal{O}_ Y)$ is a $Y$-perfect object of $D(\mathcal{O}_ X)$,

  2. $Rf_*a(\mathcal{O}_ Y)$ has vanishing cohomology sheaves in positive degrees,

  3. $\mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a(\mathcal{O}_ Y), a(\mathcal{O}_ Y))$ is an isomorphism.

Proof. We will repeatedly use that $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$, see Lemma 84.3.3. Let $E$ be a perfect object of $D(\mathcal{O}_ X)$ with dual $E^\vee $, see Cohomology on Sites, Lemma 21.46.4. Then

\[ Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y)) = Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , a(\mathcal{O}_ Y)) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y) \]

By Derived Categories of Spaces, Lemma 73.25.4 the complex $Rf_*E^\vee $ is perfect. Hence the dual $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y)$ is perfect as well. We conclude that $a(\mathcal{O}_ Y)$ is pseudo-coherent by Derived Categories of Spaces, Lemma 73.25.7 amd More on Morphisms of Spaces, Lemma 74.51.4.

Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ Y$-module. By Lemma 84.8.2 we have

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

Second equality by Cohomology on Sites, Lemma 21.18.5. By Lemma 84.3.2 there exists an integer $N$ such that $H^ i(a(\mathcal{F})) = 0$ for $i \leq -N$. Looking at stalks we conclude that $a(\mathcal{O}_ Y)$ has finite tor dimension (details omitted; hint: for $y \in Y$ any $\mathcal{O}_{Y, y}$-module occurs as $\mathcal{F}_ y$ for some quasi-coherent module on the affine scheme $Y$).

Combining the results of the previous two paragraphs we find that $a(\mathcal{O}_ Y)$ is $Y$-perfect, see More on Morphisms of Spaces, Definition 74.52.1. This proves (1).

Let $M$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*a(\mathcal{O}_ Y)) & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M, a(\mathcal{O}_ Y)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*Lf^*M, \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*\mathcal{O}_ Y, \mathcal{O}_ Y) \end{align*}

The first equality holds by Cohomology on Sites, Lemma 21.19.1. The second equality by construction of $a$. The third equality by Derived Categories of Spaces, Lemma 73.20.1. Recall $Rf_*\mathcal{O}_ X$ is perfect of tor amplitude in $[0, N]$ for some $N$, see Derived Categories of Spaces, Lemma 73.25.4. Thus we can represent $Rf_*\mathcal{O}_ X$ by a complex of finite projective modules sitting in degrees $[0, N]$ (using More on Algebra, Lemma 15.70.2 and the fact that $Y$ is affine). Hence if $M = \mathcal{O}_ Y[-i]$ for some $i > 0$, then the last group is zero. Since $Y$ is affine we conclude that $H^ i(Rf_*a(\mathcal{O}_ Y)) = 0$ for $i > 0$. This proves (2).

Let $E$ be a perfect object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Then we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ X(E, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a(\mathcal{O}_ Y), a(\mathcal{O}_ Y)) & = \mathop{\mathrm{Hom}}\nolimits _ X(E \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y), a(\mathcal{O}_ Y)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y)), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , a(\mathcal{O}_ Y))), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y), \mathcal{O}_ Y) \\ & = R\Gamma (Y, Rf_*E^\vee ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(E, \mathcal{O}_ X) \end{align*}

The first equality holds by Cohomology on Sites, Lemma 21.34.2. The second equality is the definition of $a$. The third equality comes from the construction of the dual perfect complex $E^\vee $, see Cohomology on Sites, Lemma 21.46.4. The fourth equality is given in the first line of the proof. The fifth equality holds by double duality for perfect complexes (Cohomology on Sites, Lemma 21.46.4) and the fact that $Rf_*E$ is perfect by Derived Categories of Spaces, Lemma 73.25.4 The last equality is Leray for $f$. This string of equalities essentially shows (3) holds by the Yoneda lemma. Namely, the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (a(\mathcal{O}_ Y), a(\mathcal{O}_ Y))$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Spaces, Lemma 73.13.10. Taking $E = \mathcal{O}_ X$ in the above we get a map $\alpha : \mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a(\mathcal{O}_ Y), a(\mathcal{O}_ Y))$ corresponding to $\text{id}_{\mathcal{O}_ X} \in \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, \mathcal{O}_ X)$. Since all the isomorphisms above are functorial in $E$ we see that the cone on $\alpha $ is an object $C$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ such that $\mathop{\mathrm{Hom}}\nolimits (E, C) = 0$ for all perfect $E$. Since the perfect objects generate (Derived Categories of Spaces, Theorem 73.15.4) we conclude that $\alpha $ is an isomorphism. $\square$


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