## 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AVV. Beware of the difference between the letter 'O' and the digit '0'.