Lemma 29.57.10. Consider a commutative diagram of morphisms of schemes

$\xymatrix{ X \ar[rd]_ g \ar[rr]_ f & & Y \ar[ld]^ h \\ & Z & }$

If $g$ has universally bounded fibres, and $f$ is surjective and flat, then also $h$ has universally bounded fibres. More precisely, if $n$ bounds the degree of the fibres of $g$, then also $n$ bounds the degree of the fibres of $h$.

Proof. Assume $g$ has universally bounded fibres, and $f$ is surjective and flat. Say the degree of the fibres of $g$ is bounded by $n \in \mathbf{N}$. We claim $n$ also works for $h$. Let $z \in Z$. Consider the morphism of schemes $X_ z \to Y_ z$. It is flat and surjective. By assumption $X_ z$ is a finite scheme over $\kappa (z)$, in particular it is the spectrum of an Artinian ring (by Algebra, Lemma 10.53.2). By Lemma 29.11.13 the morphism $X_ z \to Y_ z$ is affine in particular quasi-compact. It follows from Lemma 29.25.12 that $Y_ z$ is a finite discrete as this holds for $X_ z$. Hence $Y_ z$ is an affine scheme (this is a nice exercise; it also follows for example from Properties, Lemma 28.29.1 applied to the set of all points of $Y_ z$). Write $Y_ z = \mathop{\mathrm{Spec}}(B)$ and $X_ z = \mathop{\mathrm{Spec}}(A)$. Then $A$ is faithfully flat over $B$, so $B \subset A$. Hence $\dim _ k(B) \leq \dim _ k(A) \leq n$ as desired. $\square$

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).