The Stacks project

Lemma 48.9.5. Let $i : Z \to X$ be a pseudo-coherent closed immersion of schemes (any closed immersion if $X$ is locally Noetherian). Then

  1. $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ maps $D^+_\mathit{QCoh}(\mathcal{O}_ X)$ into $D^+_\mathit{QCoh}(\mathcal{O}_ Z)$, and

  2. if $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(B)$, then the diagram

    \[ \xymatrix{ D^+(B) \ar[r] & D_\mathit{QCoh}^+(\mathcal{O}_ Z) \\ D^+(A) \ar[r] \ar[u]^{R\mathop{\mathrm{Hom}}\nolimits (B, -)} & D_\mathit{QCoh}^+(\mathcal{O}_ X) \ar[u]_{R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)} } \]

    is commutative.

Proof. To explain the parenthetical remark, if $X$ is locally Noetherian, then $i$ is pseudo-coherent by More on Morphisms, Lemma 37.60.9.

Let $K$ be an object of $D^+_\mathit{QCoh}(\mathcal{O}_ X)$. To prove (1), by Morphisms, Lemma 29.4.1 it suffices to show that $i_*$ applied to $H^ n(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K))$ produces a quasi-coherent module on $X$. By Lemma 48.9.3 this means we have to show that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Z, K)$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$. Since $i$ is pseudo-coherent the sheaf $\mathcal{O}_ Z$ is a pseudo-coherent $\mathcal{O}_ X$-module. Hence the result follows from Derived Categories of Schemes, Lemma 36.10.8.

Assume $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(B)$ as in (2). Let $I^\bullet $ be a bounded below complex of injective $A$-modules representing an object $K$ of $D^+(A)$. Then we know that $R\mathop{\mathrm{Hom}}\nolimits (B, K) = \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ viewed as a complex of $B$-modules. Choose a quasi-isomorphism

\[ \widetilde{I^\bullet } \longrightarrow \mathcal{I}^\bullet \]

where $\mathcal{I}^\bullet $ is a bounded below complex of injective $\mathcal{O}_ X$-modules. It follows from the description of the functor $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ in Lemma 48.9.1 that there is a map

\[ \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet ) \longrightarrow \Gamma (Z, \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \mathcal{I}^\bullet )) \]

Observe that $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \mathcal{I}^\bullet )$ represents $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \widetilde{K})$. Applying the universal property of the $\widetilde{\ }$ functor we obtain a map

\[ \widetilde{\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \widetilde{K}) \]

in $D(\mathcal{O}_ Z)$. We may check that this map is an isomorphism in $D(\mathcal{O}_ Z)$ after applying $i_*$. However, once we apply $i_*$ we obtain the isomorphism of Derived Categories of Schemes, Lemma 36.10.8 via the identification of Lemma 48.9.3. $\square$


Comments (0)

There are also:

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

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 0A78. Beware of the difference between the letter 'O' and the digit '0'.