Lemma 48.9.7. Let $X$ be a quasi-compact and quasi-separated scheme. Let $i : Z \to X$ be a pseudo-coherent closed immersion (if $X$ is Noetherian, then any closed immersion is pseudo-coherent). Let $a : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Z)$ be the right adjoint to $Ri_*$. Then there is a functorial isomorphism

$a(K) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K)$

for $K \in D_\mathit{QCoh}^+(\mathcal{O}_ X)$.

Proof. (The parenthetical statement follows from More on Morphisms, Lemma 37.60.9.) By Lemma 48.9.2 the functor $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ is a right adjoint to $Ri_* : D(\mathcal{O}_ Z) \to D(\mathcal{O}_ X)$. Moreover, by Lemma 48.9.5 and Lemma 48.3.5 both $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ and $a$ map $D_\mathit{QCoh}^+(\mathcal{O}_ X)$ into $D_\mathit{QCoh}^+(\mathcal{O}_ Z)$. Hence we obtain the isomorphism by uniqueness of adjoint functors. $\square$

There are also:

• 2 comment(s) on Section 48.9: Right adjoint of pushforward for closed immersions

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 0A9X. Beware of the difference between the letter 'O' and the digit '0'.