The Stacks project

Lemma 36.21.1. Let $X$ be a quasi-compact and quasi-separated scheme. The inclusion functor $D_\mathit{QCoh}(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$ has a right adjoint $DQ_ X$.

First proof. We will use the induction principle as in Cohomology of Schemes, Lemma 30.4.1 to prove this. If $D(\mathit{QCoh}(\mathcal{O}_ X)) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ is an equivalence, then the lemma is true because the functor $RQ_ X$ of Section 36.7 is a right adjoint to the functor $D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathcal{O}_ X)$. In particular, our lemma is true for affine schemes, see Lemma 36.7.3. Thus we see that it suffices to show: if $X = U \cup V$ is a union of two quasi-compact opens and the lemma holds for $U$, $V$, and $U \cap V$, then the lemma holds for $X$.

The adjoint exists if and only if for every object $K$ of $D(\mathcal{O}_ X)$ we can find a distinguished triangle

\[ E' \to E \to K \to E'[1] \]

in $D(\mathcal{O}_ X)$ such that $E'$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and such that $\mathop{\mathrm{Hom}}\nolimits (M, K) = 0$ for all $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$. See Derived Categories, Lemma 13.39.5. Consider the distinguished triangle

\[ E \to Rj_{U, *}E|_ U \oplus Rj_{V, *}E|_ V \to Rj_{U \cap V, *}E|_{U \cap V} \to E[1] \]

in $D(\mathcal{O}_ X)$ of Cohomology, Lemma 20.33.2. By Derived Categories, Lemma 13.39.4 it suffices to construct the desired distinguished triangles for $Rj_{U, *}E|_ U$, $Rj_{V, *}E|_ V$, and $Rj_{U \cap V, *}E|_{U \cap V}$. This reduces us to the statement discussed in the next paragraph.

Let $j : U \to X$ be an open immersion corresponding with $U$ a quasi-compact open for which the lemma is true. Let $L$ be an object of $D(\mathcal{O}_ U)$. Then there exists a distinguished triangle

\[ E' \to Rj_*L \to K \to E'[1] \]

in $D(\mathcal{O}_ X)$ such that $E'$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and such that $\mathop{\mathrm{Hom}}\nolimits (M, K) = 0$ for all $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$. To see this we choose a distinguished triangle

\[ L' \to L \to Q \to L'[1] \]

in $D(\mathcal{O}_ U)$ such that $L'$ is in $D_\mathit{QCoh}(\mathcal{O}_ U)$ and such that $\mathop{\mathrm{Hom}}\nolimits (N, Q) = 0$ for all $N$ in $D_\mathit{QCoh}(\mathcal{O}_ U)$. This is possible because the statement in Derived Categories, Lemma 13.39.5 is an if and only if. We obtain a distinguished triangle

\[ Rj_*L' \to Rj_*L \to Rj_*Q \to Rj_*L'[1] \]

in $D(\mathcal{O}_ X)$. Observe that $Rj_*L'$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.4.1. On the other hand, if $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$, then

\[ \mathop{\mathrm{Hom}}\nolimits (M, Rj_*Q) = \mathop{\mathrm{Hom}}\nolimits (Lj^*M, Q) = 0 \]

because $Lj^*M$ is in $D_\mathit{QCoh}(\mathcal{O}_ U)$ by Lemma 36.3.8. This finishes the proof. $\square$

Second proof. The adjoint exists by Derived Categories, Proposition 13.38.2. The hypotheses are satisfied: First, note that $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums and direct sums commute with the inclusion functor (Lemma 36.3.1). On the other hand, $D_\mathit{QCoh}(\mathcal{O}_ X)$ is compactly generated because it has a perfect generator Theorem 36.15.3 and because perfect objects are compact by Proposition 36.17.1. $\square$


Comments (0)


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 0CR0. Beware of the difference between the letter 'O' and the digit '0'.