Lemma 29.56.4. A composition of morphisms with universally bounded fibres is a morphism with universally bounded fibres. More precisely, assume that $n$ bounds the degrees of the fibres of $f : X \to Y$ and $m$ bounds the degrees of $g : Y \to Z$. Then $nm$ bounds the degrees of the fibres of $g \circ f : X \to Z$.

**Proof.**
Let $f : X \to Y$ and $g : Y \to Z$ have universally bounded fibres. Say that $\deg (X_ y/\kappa (y)) \leq n$ for all $y \in Y$, and that $\deg (Y_ z/\kappa (z)) \leq m$ for all $z \in Z$. Let $z \in Z$ be a point. By assumption the scheme $Y_ z$ is finite over $\mathop{\mathrm{Spec}}(\kappa (z))$. In particular, the underlying topological space of $Y_ z$ is a finite discrete set. The fibres of the morphism $f_ z : X_ z \to Y_ z$ are the fibres of $f$ at the corresponding points of $Y$, which are finite discrete sets by the reasoning above. Hence we conclude that the underlying topological space of $X_ z$ is a finite discrete set as well. Thus $X_ z$ 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 $X_ z$). Write $X_ z = \mathop{\mathrm{Spec}}(A)$, $Y_ z = \mathop{\mathrm{Spec}}(B)$, and $k = \kappa (z)$. Then $k \to B \to A$ and we know that (a) $\dim _ k(B) \leq m$, and (b) for every maximal ideal $\mathfrak m \subset B$ we have $\dim _{\kappa (\mathfrak m)}(A/\mathfrak mA) \leq n$. We claim this implies that $\dim _ k(A) \leq nm$. Note that $B$ is the product of its localizations $B_{\mathfrak m}$, for example because $Y_ z$ is a disjoint union of $1$-point schemes, or by Algebra, Lemmas 10.53.2 and 10.53.6. So we see that $\dim _ k(B) = \sum _{\mathfrak m} \dim _ k(B_{\mathfrak m})$ and $\dim _ k(A) = \sum _{\mathfrak m} \dim _ k(A_{\mathfrak m})$ where in both cases $\mathfrak m$ runs over the maximal ideals of $B$ (not of $A$). By the above, and Nakayama's Lemma (Algebra, Lemma 10.20.1) we see that each $A_{\mathfrak m}$ is a quotient of $B_{\mathfrak m}^{\oplus n}$ as a $B_{\mathfrak m}$-module. Hence $\dim _ k(A_{\mathfrak m}) \leq n \dim _ k(B_{\mathfrak m})$. Putting everything together we see that

as desired. $\square$

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