Proposition 36.17.1. Let $X$ be a quasi-compact and quasi-separated scheme. An object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ is compact if and only if it is perfect.

Proof. If $K$ is a perfect object of $D(\mathcal{O}_ X)$ with dual $K^\vee$ (Cohomology, Lemma 20.50.5) we have

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K, M) = H^0(X, K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$

functorially in $M$. Since $K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} -$ commutes with direct sums and since $H^0(X, -)$ commutes with direct sums on $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.4.5 we conclude that $K$ is compact in $D_\mathit{QCoh}(\mathcal{O}_ X)$.

Conversely, let $K$ be a compact object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. To show that $K$ is perfect, it suffices to show that $K|_ U$ is perfect for every affine open $U \subset X$, see Cohomology, Lemma 20.49.2. Observe that $j : U \to X$ is a quasi-compact and separated morphism. Hence $Rj_* : D_\mathit{QCoh}(\mathcal{O}_ U) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ commutes with direct sums, see Lemma 36.4.5. Thus the adjointness of restriction to $U$ and $Rj_*$ implies that $K|_ U$ is a compact object of $D_\mathit{QCoh}(\mathcal{O}_ U)$. Hence we reduce to the case that $X$ is affine.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. By Lemma 36.3.5 the problem is translated into the same problem for $D(A)$. For $D(A)$ the result is More on Algebra, Proposition 15.78.3. $\square$

There are also:

• 2 comment(s) on Section 36.17: Compact and perfect objects

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).