## 48.12 Right adjoint of pushforward for proper flat morphisms

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

Lemma 48.12.1. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes 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 48.3.1. Then $a$ commutes with direct sums.

Proof. Let $P$ be a perfect object of $D(\mathcal{O}_ X)$. By Derived Categories of Schemes, Lemma 36.30.4 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 Schemes, Proposition 36.17.1). Since $D_\mathit{QCoh}(\mathcal{O}_ X)$ has a perfect generator (Derived Categories of Schemes, Theorem 36.15.3) 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 48.12.2. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes 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 48.3.1. Then

1. for every closed $T \subset Y$ if $Q \in D_\mathit{QCoh}(Y)$ is supported on $T$, then $a(Q)$ is supported on $f^{-1}(T)$,

2. for every open $V \subset Y$ and any $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ the map (48.4.1.1) is an isomorphism, and

Lemma 48.12.3. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes which is proper, flat, and of finite presentation. The map (48.8.0.1) is an isomorphism for every object $K$ of $D_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. By Lemma 48.12.1 we know that $a$ commutes with direct sums. Hence the collection of objects of $D_\mathit{QCoh}(\mathcal{O}_ Y)$ for which (48.8.0.1) 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 Schemes, Theorem 36.18.3) generated by a single perfect object (Derived Categories of Schemes, Theorem 36.15.3) we can argue as in More on Algebra, Remark 15.59.11 to see that it suffices to prove (48.8.0.1) is an isomorphism for a single perfect object. However, the result holds for perfect objects, see Lemma 48.8.1. $\square$

The following lemma shows that the base change map (48.5.0.1) is an isomorphism for proper, flat morphisms of finite presentation. We will see in Example 48.15.2 that this does not remain true for perfect proper morphisms; in that case one has to make a tor independence condition.

Lemma 48.12.4. Let $g : Y' \to Y$ be a morphism of quasi-compact and quasi-separated schemes. Let $f : X \to Y$ be a proper, flat morphism of finite presentation. Then the base change map (48.5.0.1) is an isomorphism for all $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. By Lemma 48.12.2 formation of the functors $a$ and $a'$ commutes with restriction to opens of $Y$ and $Y'$. Hence we may assume $Y' \to Y$ is a morphism of affine schemes, see Remark 48.6.1. In this case the statement follows from Lemma 48.6.2. $\square$

Remark 48.12.5. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a proper, flat morphism of finite presentation. Let $a$ be the adjoint of Lemma 48.3.1 for $f$. In this situation, $\omega _{X/Y}^\bullet = a(\mathcal{O}_ Y)$ is sometimes called the relative dualizing complex. By Lemma 48.12.3 there is a functorial isomorphism $a(K) = Lf^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet$ for $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. Moreover, the trace map

$\text{Tr}_{f, \mathcal{O}_ Y} : Rf_*\omega _{X/Y}^\bullet \to \mathcal{O}_ Y$

of Section 48.7 induces the trace map for all $K$ in $D_\mathit{QCoh}(\mathcal{O}_ Y)$. More precisely the diagram

$\xymatrix{ Rf_*a(K) \ar[rrr]_{\text{Tr}_{f, K}} \ar@{=}[d] & & & K \ar@{=}[d] \\ Rf_*(Lf^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet ) \ar@{=}[r] & K \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*\omega _{X/Y}^\bullet \ar[rr]^-{\text{id}_ K \otimes \text{Tr}_{f, \mathcal{O}_ Y}} & & K }$

where the equality on the lower right is Derived Categories of Schemes, Lemma 36.22.1. If $g : Y' \to Y$ is a morphism of quasi-compact and quasi-separated schemes and $X' = Y' \times _ Y X$, then by Lemma 48.12.4 we have $\omega _{X'/Y'}^\bullet = L(g')^*\omega _{X/Y}^\bullet$ where $g' : X' \to X$ is the projection and by Lemma 48.7.1 the trace map

$\text{Tr}_{f', \mathcal{O}_{Y'}} : Rf'_*\omega _{X'/Y'}^\bullet \to \mathcal{O}_{Y'}$

for $f' : X' \to Y'$ is the base change of $\text{Tr}_{f, \mathcal{O}_ Y}$ via the base change isomorphism.

Remark 48.12.6. Let $f : X \to Y$, $\omega ^\bullet _{X/Y}$, and $\text{Tr}_{f, \mathcal{O}_ Y}$ be as in Remark 48.12.5. Let $K$ and $M$ be in $D_\mathit{QCoh}(\mathcal{O}_ X)$ with $M$ pseudo-coherent (for example perfect). Suppose given a map $K \otimes _{\mathcal{O}_ X}^\mathbf {L} M \to \omega ^\bullet _{X/Y}$ which corresponds to an isomorphism $K \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(M, \omega ^\bullet _{X/Y})$ via Cohomology, Equation (20.40.0.1). Then the relative cup product (Cohomology, Remark 20.28.7)

$Rf_*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*M \to Rf_*(K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \to Rf_*\omega ^\bullet _{X/Y} \xrightarrow {\text{Tr}_{f, \mathcal{O}_ Y}} \mathcal{O}_ Y$

determines an isomorphism $Rf_*K \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*M, \mathcal{O}_ Y)$. Namely, since $\omega ^\bullet _{X/Y} = a(\mathcal{O}_ Y)$ the canonical map (48.3.5.1)

$Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(M, \omega ^\bullet _{X/Y}) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*M, \mathcal{O}_ Y)$

is an isomorphism by Lemma 48.3.6 and Remark 48.3.8 and the fact that $M$ and $Rf_*M$ are pseudo-coherent, see Derived Categories of Schemes, Lemma 36.30.5. To see that the relative cup product induces this isomorphism use the commutativity of the diagram in Cohomology, Remark 20.40.12.

Lemma 48.12.7. Let $Y$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes which is proper, flat, and of finite presentation with relative dualizing complex $\omega _{X/Y}^\bullet$ (Remark 48.12.5). Then

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

2. $Rf_*\omega _{X/Y}^\bullet$ has vanishing cohomology sheaves in positive degrees,

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

Proof. In view of the fact that formation of $\omega _{X/Y}^\bullet$ commutes with base change (see Remark 48.12.5), we may and do assume that $Y$ is affine. For a perfect object $E$ of $D(\mathcal{O}_ X)$ we have

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

For the first equality, see Cohomology, Lemma 20.48.5. For the second equality, see Lemma 48.3.6, Remark 48.3.8, and Derived Categories of Schemes, Lemma 36.30.4. The third equality is the definition of the dual. In particular these references also show that the outcome is a perfect object of $D(\mathcal{O}_ Y)$. We conclude that $\omega _{X/Y}^\bullet$ is $Y$-perfect by More on Morphisms, Lemma 37.67.6. 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_*\omega _{X/Y}^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M, \omega _{X/Y}^\bullet ) \\ & = \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}_ X, \mathcal{O}_ Y) \end{align*}

The first equality holds by Cohomology, Lemma 20.28.1. The second equality by construction of $a$. The third equality by Derived Categories of Schemes, Lemma 36.22.1. Recall $Rf_*\mathcal{O}_ X$ is perfect of tor amplitude in $[0, N]$ for some $N$, see Derived Categories of Schemes, Lemma 36.30.4. 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 15.74.2 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 $H^ i(Rf_*\omega _{X/Y}^\bullet ) = 0$ 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}(\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _ X(E \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet ), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet )), \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, Lemma 20.40.2. The second equality is the definition of $\omega _{X/Y}^\bullet$. The third equality comes from the construction of the dual perfect complex $E^\vee$, see Cohomology, Lemma 20.48.5. The fourth equality follows from the equality $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet ) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y)$ shown in the first paragraph of the proof. The fifth equality holds by double duality for perfect complexes (Cohomology, Lemma 20.48.5) and the fact that $Rf_*E$ is perfect by Derived Categories of Schemes, Lemma 36.30.4. The last equality is Leray for $f$. This string of equalities essentially shows (3) holds by the Yoneda lemma. Namely, the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet )$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.10.8. Taking $E = \mathcal{O}_ X$ in the above we get a map $\alpha : \mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _{X/Y}^\bullet , \omega _{X/Y}^\bullet )$ 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 Schemes, Theorem 36.15.3) we conclude that $\alpha$ is an isomorphism. $\square$

Lemma 48.12.8 (Rigidity). Let $Y$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a proper, flat morphism of finite presentation with relative dualizing complex $\omega _{X/Y}^\bullet$ (Remark 48.12.5). There is a canonical isomorphism

48.12.8.1
$$\label{duality-equation-pre-rigid} \mathcal{O}_ X = c(L\text{pr}_1^*\omega _{X/Y}^\bullet ) = c(L\text{pr}_2^*\omega _{X/Y}^\bullet )$$

and a canonical isomorphism

48.12.8.2
$$\label{duality-equation-rigid} \omega _{X/Y}^\bullet = c\left(L\text{pr}_1^*\omega _{X/Y}^\bullet \otimes _{\mathcal{O}_{X \times _ Y X}}^\mathbf {L} L\text{pr}_2^*\omega _{X/Y}^\bullet \right)$$

where $c$ is the right adjoint of Lemma 48.3.1 for the diagonal $\Delta : X \to X \times _ Y X$.

Proof. Let $a$ be the right adjoint to $Rf_*$ as in Lemma 48.3.1. Consider the cartesian square

$\xymatrix{ X \times _ Y X \ar[r]_ q \ar[d]_ p & X \ar[d]_ f \\ X \ar[r]^ f & Y }$

Let $b$ be the right adjoint for $p$ as in Lemma 48.3.1. Then

\begin{align*} \omega _{X/Y}^\bullet & = c(b(\omega _{X/Y}^\bullet )) \\ & = c(Lp^*\omega _{X/Y}^\bullet \otimes _{\mathcal{O}_{X \times _ Y X}}^\mathbf {L} b(\mathcal{O}_ X)) \\ & = c(Lp^*\omega _{X/Y}^\bullet \otimes _{\mathcal{O}_{X \times _ Y X}}^\mathbf {L} Lq^*a(\mathcal{O}_ Y)) \\ & = c(Lp^*\omega _{X/Y}^\bullet \otimes _{\mathcal{O}_{X \times _ Y X}}^\mathbf {L} Lq^*\omega _{X/Y}^\bullet ) \end{align*}

as in (48.12.8.2). Explanation as follows:

1. The first equality holds as $\text{id} = c \circ b$ because $\text{id}_ X = p \circ \Delta$.

2. The second equality holds by Lemma 48.12.3.

3. The third holds by Lemma 48.12.4 and the fact that $\mathcal{O}_ X = Lf^*\mathcal{O}_ Y$.

4. The fourth holds because $\omega _{X/Y}^\bullet = a(\mathcal{O}_ Y)$.

Equation (48.12.8.1) is proved in exactly the same way. $\square$

Remark 48.12.9. Lemma 48.12.8 means our relative dualizing complex is rigid in a sense analogous to the notion introduced in . Namely, since the functor on the right of (48.12.8.2) is “quadratic” in $\omega _{X/Y}^\bullet$ and the functor on the left of (48.12.8.2) is “linear” this “pins down” the complex $\omega _{X/Y}^\bullet$ to some extent. There is an approach to duality theory using “rigid” (relative) dualizing complexes, see for example , , and . We will return to this in Section 48.28.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).