The Stacks project

Lemma 105.3.3. Let $U_1 \to \mathcal{X}$ and $U_2 \to \mathcal{X}$ be two smooth morphisms from schemes to a locally Noetherian algebraic stack $\mathcal{X}$. Let $T_1'$ and $T_2'$ be irreducible components of $|U_1|$ and $|U_2|$ respectively. Assume the closures of the images of $T_1'$ and $T_2'$ are the same irreducible component $T$ of $|\mathcal{X}|$. Then $m_{T_1', U_1} = m_{T_2', U_2}$.

Proof. Let $V_1$ and $V_2$ be dense subsets of $T_1'$ and $T'_2$, respectively, that are open in $U_1$ and $U_2$ respectively (see proof of Lemma 105.3.1). The images of $|V_1|$ and $|V_2|$ in $|\mathcal{X}|$ are non-empty open subsets of the irreducible subset $T$, and therefore have non-empty intersection. By Properties of Stacks, Lemma 98.4.3, the map $|V_1 \times _\mathcal {X} V_2| \to |V_1| \times _{|\mathcal{X}|} |V_2|$ is surjective. Consequently $V_1 \times _\mathcal {X} V_2$ is a non-empty algebraic space; we may therefore choose an ├ętale surjection $V \to V_1 \times _\mathcal {X} V_2$ whose source is a (non-empty) scheme. If we let $T'$ be any irreducible component of $V$, then Lemma 105.3.1 shows that the closure of the image of $T'$ in $U_1$ (respectively $U_2$) is equal to $T'_1$ (respectively $T'_2$).

Applying Lemma 105.3.2 twice we find that

\[ m_{T_1', U_1} = m_{T', V} = m_{T_2', U_2}, \]

as required. $\square$


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Your email address will not be published.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 0DR7. Beware of the difference between the letter 'O' and the digit '0'.