The Stacks project

Lemma 75.13.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $K, L, M$ be objects of $D_\mathit{QCoh}(\mathcal{O}_ X)$. The map

\[ K \otimes _{\mathcal{O}_ X}^\mathbf {L} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, L) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} L) \]

of Cohomology on Sites, Lemma 21.35.7 is an isomorphism in the following cases

  1. $M$ perfect, or

  2. $K$ is perfect, or

  3. $M$ is pseudo-coherent, $L \in D^+(\mathcal{O}_ X)$, and $K$ has finite tor dimension.

Proof. Checking whether or not the map is an isomorphism can be done ├ętale locally hence we may assume $X$ is an affine scheme. Then we can write $K, L, M$ as $\epsilon ^*K_0, \epsilon ^*L_0, \epsilon ^*M_0$ for some $K_0, L_0, M_0$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 75.4.2. Then we see that Lemma 75.13.9 reduces cases (1) and (3) to the case of schemes which is Derived Categories of Schemes, Lemma 36.10.9. If $K$ is perfect but no other assumptions are made, then we do not know that either side of the arrow is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ but the result is still true because $K$ will be represented (after localizing further) by a finite complex of finite free modules in which case it is clear. $\square$


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Your email address will not be published.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 0E4Q. Beware of the difference between the letter 'O' and the digit '0'.