Lemma 33.45.8. Let $k$ be a field. Let $X$ be proper over $k$. Let $Z \subset X$ be a closed subscheme of dimension $d$. Let $\mathcal{L}_1, \ldots , \mathcal{L}_ d$ be invertible $\mathcal{O}_ X$-modules. Assume there exists an effective Cartier divisor $D \subset Z$ such that $\mathcal{L}_1|_ Z \cong \mathcal{O}_ Z(D)$. Then
Proof. We may replace $X$ by $Z$ and $\mathcal{L}_ i$ by $\mathcal{L}_ i|_ Z$. Thus we may assume $X = Z$ and $\mathcal{L}_1 = \mathcal{O}_ X(D)$. Then $\mathcal{L}_1^{-1}$ is the ideal sheaf of $D$ and we can consider the short exact sequence
Set $P(n_1, \ldots , n_ d) = \chi (X, \mathcal{L}_1^{\otimes n_1} \otimes \ldots \otimes \mathcal{L}_ d^{\otimes n_ d})$ and $Q(n_1, \ldots , n_ d) = \chi (D, \mathcal{L}_1^{\otimes n_1} \otimes \ldots \otimes \mathcal{L}_ d^{\otimes n_ d}|_ D)$. We conclude from additivity that
Because the total degree of $P$ is at most $d$, we see that the coefficient of $n_1 \ldots n_ d$ in $P$ is equal to the coefficient of $n_2 \ldots n_ d$ in $Q$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)