Lemma 29.56.6. Let $f : X \to Y$ be a morphism of schemes. Let $Y' \to Y$ be a morphism of schemes, and let $f' : X' = X_{Y'} \to Y'$ be the base change of $f$. If $Y' \to Y$ is surjective and $f'$ has universally bounded fibres, then $f$ has universally bounded fibres. More precisely, if $n$ bounds the degree of the fibres of $f'$, then also $n$ bounds the degrees of the fibres of $f$.

Proof. Let $n \geq 0$ be an integer bounding the degrees of the fibres of $f'$. We claim that $n$ works for $f$ also. Namely, if $y \in Y$ is a point, then choose a point $y' \in Y'$ lying over $y$ and observe that

$X'_{y'} = \mathop{\mathrm{Spec}}(\kappa (y')) \times _{\mathop{\mathrm{Spec}}(\kappa (y))} X_ y.$

Since $X'_{y'}$ is assumed finite of degree $\leq n$ over $\kappa (y')$ it follows that also $X_ y$ is finite of degree $\leq n$ over $\kappa (y)$. (Some details omitted.) $\square$

## 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Your email address will not be published.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 03J8. Beware of the difference between the letter 'O' and the digit '0'.