## 36.5 Affine morphisms

In this section we collect some information about pushforward along an affine morphism of schemes.

Lemma 36.5.1. Let $f : X \to S$ be an affine morphism of schemes. Let $\mathcal{F}^\bullet$ be a complex of quasi-coherent $\mathcal{O}_ X$-modules. Then $f_*\mathcal{F}^\bullet = Rf_*\mathcal{F}^\bullet$.

Proof. Combine Lemma 36.4.2 with Cohomology of Schemes, Lemma 30.2.3. An alternative proof is to work affine locally on $S$ and use Lemma 36.3.7. $\square$

Lemma 36.5.2. Let $f : X \to S$ be an affine morphism of schemes. Then $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ S)$ reflects isomorphisms.

Proof. The statement means that a morphism $\alpha : E \to F$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ is an isomorphism if $Rf_*\alpha$ is an isomorphism. We may check this on cohomology sheaves. In particular, the question is local on $S$. Hence we may assume $S$ and therefore $X$ is affine. In this case the statement is clear from the description of the derived categories $D_\mathit{QCoh}(\mathcal{O}_ X)$ and $D_\mathit{QCoh}(\mathcal{O}_ S)$ given in Lemma 36.3.5. Some details omitted. $\square$

Lemma 36.5.3. Let $f : X \to S$ be an affine morphism of schemes. For $E$ in $D_\mathit{QCoh}(\mathcal{O}_ S)$ we have $Rf_* Lf^* E = E \otimes ^\mathbf {L}_{\mathcal{O}_ S} f_*\mathcal{O}_ X$.

Proof. Since $f$ is affine the map $f_*\mathcal{O}_ X \to Rf_*\mathcal{O}_ X$ is an isomorphism (Cohomology of Schemes, Lemma 30.2.3). There is a canonical map $E \otimes ^\mathbf {L} f_*\mathcal{O}_ X = E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X \to Rf_* Lf^* E$ adjoint to the map

$Lf^*(E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X) = Lf^*E \otimes ^\mathbf {L} Lf^*Rf_*\mathcal{O}_ X \longrightarrow Lf^* E \otimes ^\mathbf {L} \mathcal{O}_ X = Lf^* E$

coming from $1 : Lf^*E \to Lf^*E$ and the canonical map $Lf^*Rf_*\mathcal{O}_ X \to \mathcal{O}_ X$. To check the map so constructed is an isomorphism we may work locally on $S$. Hence we may assume $S$ and therefore $X$ is affine. In this case the statement is clear from the description of the derived categories $D_\mathit{QCoh}(\mathcal{O}_ X)$ and $D_\mathit{QCoh}(\mathcal{O}_ S)$ and the functor $Lf^*$ given in Lemmas 36.3.5 and 36.3.8. Some details omitted. $\square$

Let $Y$ be a scheme. Let $\mathcal{A}$ be a sheaf of $\mathcal{O}_ Y$-algebras. We will denote $D_\mathit{QCoh}(\mathcal{A})$ the inverse image of $D_\mathit{QCoh}(\mathcal{O}_ X)$ under the restriction functor $D(\mathcal{A}) \to D(\mathcal{O}_ X)$. In other words, $K \in D(\mathcal{A})$ is in $D_\mathit{QCoh}(\mathcal{A})$ if and only if its cohomology sheaves are quasi-coherent as $\mathcal{O}_ X$-modules. If $\mathcal{A}$ is quasi-coherent itself this is the same as asking the cohomology sheaves to be quasi-coherent as $\mathcal{A}$-modules, see Morphisms, Lemma 29.11.6.

Lemma 36.5.4. Let $f : X \to Y$ be an affine morphism of schemes. Then $f_*$ induces an equivalence

$\Phi : D_\mathit{QCoh}(\mathcal{O}_ X) \longrightarrow D_\mathit{QCoh}(f_*\mathcal{O}_ X)$

whose composition with $D_\mathit{QCoh}(f_*\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ is $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. Recall that $Rf_*$ is computed on an object $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ by choosing a K-injective complex $\mathcal{I}^\bullet$ of $\mathcal{O}_ X$-modules representing $K$ and taking $f_*\mathcal{I}^\bullet$. Thus we let $\Phi (K)$ be the complex $f_*\mathcal{I}^\bullet$ viewed as a complex of $f_*\mathcal{O}_ X$-modules. Denote $g : (X, \mathcal{O}_ X) \to (Y, f_*\mathcal{O}_ X)$ the obvious morphism of ringed spaces. Then $g$ is a flat morphism of ringed spaces (see below for a description of the stalks) and $\Phi$ is the restriction of $Rg_*$ to $D_\mathit{QCoh}(\mathcal{O}_ X)$. We claim that $Lg^*$ is a quasi-inverse. First, observe that $Lg^*$ sends $D_\mathit{QCoh}(f_*\mathcal{O}_ X)$ into $D_\mathit{QCoh}(\mathcal{O}_ X)$ because $g^*$ transforms quasi-coherent modules into quasi-coherent modules (Modules, Lemma 17.10.4). To finish the proof it suffices to show that the adjunction mappings

$Lg^*\Phi (K) = Lg^*Rg_*K \to K \quad \text{and}\quad M \to Rg_*Lg^*M = \Phi (Lg^*M)$

are isomorphisms for $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $M \in D_\mathit{QCoh}(f_*\mathcal{O}_ X)$. This is a local question, hence we may assume $Y$ and therefore $X$ are affine.

Assume $Y = \mathop{\mathrm{Spec}}(B)$ and $X = \mathop{\mathrm{Spec}}(A)$. Let $\mathfrak p = x \in \mathop{\mathrm{Spec}}(A) = X$ be a point mapping to $\mathfrak q = y \in \mathop{\mathrm{Spec}}(B) = Y$. Then $(f_*\mathcal{O}_ X)_ y = A_\mathfrak q$ and $\mathcal{O}_{X, x} = A_\mathfrak p$ hence $g$ is flat. Hence $g^*$ is exact and $H^ i(Lg^*M) = g^*H^ i(M)$ for any $M$ in $D(f_*\mathcal{O}_ X)$. For $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ we see that

$H^ i(\Phi (K)) = H^ i(Rf_*K) = f_*H^ i(K)$

by the vanishing of higher direct images (Cohomology of Schemes, Lemma 30.2.3) and Lemma 36.3.4 (small detail omitted). Thus it suffice to show that

$g^*g_*\mathcal{F} \to \mathcal{F} \quad \text{and}\quad \mathcal{G} \to g_*g^*\mathcal{F}$

are isomorphisms where $\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ X$-module and $\mathcal{G}$ is a quasi-coherent $f_*\mathcal{O}_ X$-module. This follows from Morphisms, Lemma 29.11.6. $\square$

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).