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
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