The Stacks project

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.) $\square$


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