The Stacks project

Proof. The implication (2) $\Rightarrow $ (1) is trivial. The other implication holds because if the image of $\mathop{\mathrm{Spec}}(k) \to Y$ is $y$, then $X_ k = \mathop{\mathrm{Spec}}(k) \times _{\mathop{\mathrm{Spec}}(\kappa (y))} X_ y$. By definition the fibres of $f$ being universally bounded means that some $n$ exists. Finally, suppose that $X_ k = \mathop{\mathrm{Spec}}(A)$. Then $\dim _ k A = n$. Hence $A$ is Artinian, all prime ideals are maximal ideals $\mathfrak m_ i$, and $A$ is the product of the localizations at these maximal ideals. See Algebra, Lemmas 10.53.2 and 10.53.6. Then $\mathfrak m_ i$ corresponds to $x_ i$, we have $A_{\mathfrak m_ i} = \mathcal{O}_{X_ k, x_ i}$ and hence there is a surjection $A \to \bigoplus \kappa (\mathfrak m_ i) = \bigoplus \kappa (x_ i)$ which implies the inequality in the statement of the lemma by linear algebra. $\square$

Proof. This is true because any base change of $f$ is finite locally free of degree $d$ (Lemma 29.48.4) and hence the fibres of $f$ all have degree $d$. $\square$

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. $\square$

Proof. This is clear from the result of Lemma 29.57.2. $\square$

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

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.) $\square$

Proof. The integer $n = 1$ works in the definition. $\square$

Proof. This follows from Lemma 29.57.2 and the fact that the fibres $X_ y$ are disjoint unions of spectra of finite separable field extensions of $\kappa (y)$, see Lemma 29.36.7. $\square$

Proof. Since $X$ is quasi-compact, there exists a finite affine open covering $X = \bigcup _{i = 1, \ldots , n} U_ i$ and affine opens $V_ i \subset Y$, $i = 1, \ldots , n$ such that $f(U_ i) \subset V_ i$. Because of the local nature of “local quasi-finiteness” (see Lemma 29.20.6 part (4)) we see that the morphisms $f|_{U_ i} : U_ i \to V_ i$ are locally quasi-finite morphisms of affines, hence quasi-finite, see Lemma 29.20.9. For $y \in Y$ it is clear that $X_ y = \bigcup _{y \in V_ i} (U_ i)_ y$ is an open covering. Hence it suffices to prove the lemma for a quasi-finite morphism of affines (namely, if $n_ i$ works for the morphism $f|_{U_ i} : U_ i \to V_ i$, then $\sum n_ i$ works for $f$).

Assume $f : X \to Y$ is a quasi-finite morphism of affines. By Lemma 29.56.3 we can find a diagram

with $Z$ affine, $\pi $ finite and $j$ an open immersion. Since $j$ has universally bounded fibres (Lemma 29.57.7) this reduces us to showing that $\pi $ has universally bounded fibres (Lemma 29.57.4).

This reduces us to a morphism of the form $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ where $A \to B$ is finite. Say $B$ is generated by $x_1, \ldots , x_ n$ over $A$ and say $P_ i(T) \in A[T]$ is a monic polynomial of degree $d_ i$ such that $P_ i(x_ i) = 0$ in $B$ (a finite ring extension is integral, see Algebra, Lemma 10.36.3). With these notations it is clear that

is a surjective $A$-module map. Thus for any prime $\mathfrak p \subset A$ this induces a surjective map $\kappa (\mathfrak p)$-vector spaces

In other words, the integer $d_1 \ldots d_ n$ works in the definition of a morphism with universally bounded fibres. $\square$

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. $\square$