Lemma 70.18.8. Let $k$ be a field. Let $X$ be a proper algebraic space over $k$. Let $Z \subset X$ be a closed subspace 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 (Lemma 70.17.2) 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Your email address will not be published.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).

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.

## Comments (0)