The Stacks project

Lemma 71.6.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $D \subset X$ be a closed subspace. The following are equivalent:

  1. The subspace $D$ is an effective Cartier divisor on $X$.

  2. For some scheme $U$ and surjective étale morphism $U \to X$ the inverse image $D \times _ X U$ is an effective Cartier divisor on $U$.

  3. For every scheme $U$ and every étale morphism $U \to X$ the inverse image $D \times _ X U$ is an effective Cartier divisor on $U$.

  4. For every $x \in |D|$ there exists an étale morphism $(U, u) \to (X, x)$ of pointed algebraic spaces such that $U = \mathop{\mathrm{Spec}}(A)$ and $D \times _ X U = \mathop{\mathrm{Spec}}(A/(f))$ with $f \in A$ not a zerodivisor.

Proof. The equivalence of (1) – (3) follows from Definition 71.6.1 and the references preceding it. Assume (1) and let $x \in |D|$. Choose a scheme $W$ and a surjective étale morphism $W \to X$. Choose $w \in D \times _ X W$ mapping to $x$. By (3) $D \times _ X W$ is an effective Cartier divisor on $W$. Hence we can find affine étale neighbourhood $U$ by choosing an affine open neighbourhood of $w$ in $W$ as in Divisors, Lemma 31.13.2.

Assume (4). Then we see that $\mathcal{I}_ D|_ U$ is invertible by Divisors, Lemma 31.13.2. Since we can find an étale covering of $X$ by the collection of all such $U$ and $X \setminus D$, we conclude that $\mathcal{I}_ D$ is an invertible $\mathcal{O}_ X$-module. $\square$


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Your email address will not be published.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 083C. Beware of the difference between the letter 'O' and the digit '0'.