The Stacks project

Lemma 37.13.11. Let $f : X \to Y$ and $Y \to Z$ be morphisms of schemes. Assume $X \to Y$ is a complete intersection morphism. Then there is a canonical distinguished triangle

\[ f^*\mathop{N\! L}\nolimits _{Y/Z} \to \mathop{N\! L}\nolimits _{X/Z} \to \mathop{N\! L}\nolimits _{X/Y} \to f^*\mathop{N\! L}\nolimits _{Y/Z}[1] \]

in $D(\mathcal{O}_ X)$ which recovers the $6$-term exact sequence of Lemma 37.13.10.

Proof. It suffices to show the canonical map

\[ f^*\mathop{N\! L}\nolimits _{Y/Z} \to \text{Cone}(\mathop{N\! L}\nolimits _{X/Z} \to \mathop{N\! L}\nolimits _{X/Y})[-1] \]

of Modules, Lemma 17.31.7 is an isomorphism in $D(\mathcal{O}_ X)$. In order to show this, it suffices to show that the $6$-term sequence has a zero on the left, i.e., that $H^{-1}(f^*\mathop{N\! L}\nolimits _{Y/Z}) \to H^{-1}(\mathop{N\! L}\nolimits _{X/Z})$ is injective. Affine locally this follows from the corresponding algebra result in More on Algebra, Lemma 15.33.6. To translate into algebra use Lemma 37.13.2. $\square$

Comments (2)

Comment #8873 by Eoin on

In the proof, in the cone of the first equation, the terms and are swapped.

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Your email address will not be published.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 0FV3. Beware of the difference between the letter 'O' and the digit '0'.