Lemma 68.3.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $V \to U$ be a morphism of schemes over $S$, and let $U \to X$ be a morphism from $U$ to $X$. If the fibres of $V \to U$ and $U \to X$ are universally bounded, then so are the fibres of $V \to X$.

Proof. Let $n$ be an integer which works for $V \to U$, and let $m$ be an integer which works for $U \to X$ in Definition 68.3.1. Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism, where $k$ is a field. Consider the morphisms

$\mathop{\mathrm{Spec}}(k) \times _ X V \longrightarrow \mathop{\mathrm{Spec}}(k) \times _ X U \longrightarrow \mathop{\mathrm{Spec}}(k).$

By assumption the scheme $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree at most $m$ over $k$, and $n$ is an integer which bounds the degree of the fibres of the first morphism. Hence by Morphisms, Lemma 29.57.4 we conclude that $\mathop{\mathrm{Spec}}(k) \times _ X V$ is finite over $k$ of degree at most $nm$. $\square$

## 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Your email address will not be published.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.

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 03JM. Beware of the difference between the letter 'O' and the digit '0'.