The Stacks project

Lemma 50.23.3. Let $X \to S$ be a morphism of schemes. Let $Z \to X$ be a closed immersion of finite presentation whose conormal sheaf $\mathcal{C}_{Z/X}$ is locally free of rank $c$. Then there is a canonical map

\[ \gamma ^ p : \Omega ^ p_{Z/S} \to \mathcal{H}^ c_ Z(\Omega ^{p + c}_{X/S}) \]

which is locally given by the maps $\gamma ^ p_{f_1, \ldots , f_ c}$ of Remark 50.23.1.

Proof. The assumptions imply that given $x \in Z \subset X$ there exists an open neighbourhood $U$ of $x$ such that $Z$ is cut out by $c$ elements $f_1, \ldots , f_ c \in \mathcal{O}_ X(U)$. Thus it suffices to show that given $f_1, \ldots , f_ c$ and $g_1, \ldots , g_ c$ in $\mathcal{O}_ X(U)$ cutting out $Z \cap U$, the maps $\gamma ^ p_{f_1, \ldots , f_ c}$ and $\gamma ^ p_{g_1, \ldots , g_ c}$ are the same. To do this, after shrinking $U$ we may assume $g_ j = \sum a_{ji} f_ i$ for some $a_{ji} \in \mathcal{O}_ X(U)$. Then we have $c_{f_1, \ldots , f_ c} = \det (a_{ji}) c_{g_1, \ldots , g_ c}$ by Derived Categories of Schemes, Lemma 36.6.12. On the other hand we have

\[ \text{d}(g_1) \wedge \ldots \wedge \text{d}(g_ c) \equiv \det (a_{ji}) \text{d}(f_1) \wedge \ldots \wedge \text{d}(f_ c) \bmod (f_1, \ldots , f_ c)\Omega ^ c_{X/S} \]

Combining these relations, a straightforward calculation gives the desired equality. $\square$


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Your email address will not be published.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 0G86. Beware of the difference between the letter 'O' and the digit '0'.