The Stacks project

Lemma 20.43.10. In the situation of Lemma 20.43.9 if $\mathcal{F}^\bullet $ is K-flat, then $\mathcal{H}^\bullet $ is K-flat.

Proof. Observe that $\mathcal{H}^\bullet $ is simply the hom complex $\mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ since the boundedness of the strictly prefect complex $\mathcal{E}^\bullet $ insures that the products in the definition of the hom complex turn into direct sums. Let $\mathcal{K}^\bullet $ be an acyclic complex of $\mathcal{O}_ X$-modules. Consider the map

\[ \gamma : \text{Tot}(\mathcal{K}^\bullet \otimes \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \mathcal{F}^\bullet )) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \text{Tot}(\mathcal{K}^\bullet \otimes \mathcal{F}^\bullet )) \]

of Lemma 20.38.3. Since $\mathcal{F}^\bullet $ is K-flat, the complex $\text{Tot}(\mathcal{K}^\bullet \otimes \mathcal{F}^\bullet )$ is acyclic, and hence by Lemma 20.43.8 (or Lemma 20.43.9 if you like) the target of $\gamma $ is acyclic too. Hence to prove the lemma it suffices to show that $\gamma $ is an isomorphism of complexes. To see this, we may argue by induction on the length of the complex $\mathcal{E}^\bullet $. If the length is $\leq 1$ then the $\mathcal{E}^\bullet $ is a direct summand of $\mathcal{O}_ X^{\oplus n}[k]$ for some $n \geq 0$ and $k \in \mathbf{Z}$ and in this case the result follows by inspection. If the length is $> 1$, then we reduce to smaller length by considering the termwise split short exact sequence of complexes

\[ 0 \to \sigma _{\geq a + 1} \mathcal{E}^\bullet \to \mathcal{E}^\bullet \to \sigma _{\leq a} \mathcal{E}^\bullet \to 0 \]

for a suitable $a \in \mathbf{Z}$, see Homology, Section 12.15. Then $\gamma $ fits into a morphism of termwise split short exact sequences of complexes. By induction $\gamma $ is an isomorphism for $\sigma _{\geq a + 1} \mathcal{E}^\bullet $ and $\sigma _{\leq a} \mathcal{E}^\bullet $ and hence the result for $\mathcal{E}^\bullet $ follows. Some details omitted. $\square$


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Your email address will not be published.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 0GM5. Beware of the difference between the letter 'O' and the digit '0'.