Lemma 48.13.1. Let $f : X \to Y$ be a perfect proper morphism of Noetherian schemes. 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 More on Morphisms, Lemma 37.58.13 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. $\square$

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.

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 0A9R. Beware of the difference between the letter 'O' and the digit '0'.