Definition 68.3.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$, and let $U$ be a scheme over $S$. Let $f : U \to X$ be a morphism over $S$. We say the *fibres of $f$ are universally bounded*^{1} if there exists an integer $n$ such that for all fields $k$ and all morphisms $\mathop{\mathrm{Spec}}(k) \to X$ the fibre product $\mathop{\mathrm{Spec}}(k) \times _ X U$ is a finite scheme over $k$ whose degree over $k$ is $\leq n$.

## 68.3 Universally bounded fibres

We briefly discuss what it means for a morphism from a scheme to an algebraic space to have universally bounded fibres. Please refer to Morphisms, Section 29.57 for similar definitions and results on morphisms of schemes.

This definition makes sense because the fibre product $\mathop{\mathrm{Spec}}(k) \times _ Y X$ is a scheme. Moreover, if $Y$ is a scheme we recover the notion of Morphisms, Definition 29.57.1 by virtue of Morphisms, Lemma 29.57.2.

Lemma 68.3.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $V \to U$ be a morphism of schemes over $S$, and let $U \to X$ be a morphism from $U$ to $X$. If the fibres of $V \to U$ and $U \to X$ are universally bounded, then so are the fibres of $V \to X$.

**Proof.**
Let $n$ be an integer which works for $V \to U$, and let $m$ be an integer which works for $U \to X$ in Definition 68.3.1. Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism, where $k$ is a field. Consider the morphisms

By assumption the scheme $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree at most $m$ over $k$, and $n$ is an integer which bounds the degree of the fibres of the first morphism. Hence by Morphisms, Lemma 29.57.4 we conclude that $\mathop{\mathrm{Spec}}(k) \times _ X V$ is finite over $k$ of degree at most $nm$. $\square$

Lemma 68.3.3. Let $S$ be a scheme. Let $Y \to X$ be a representable morphism of algebraic spaces over $S$. Let $U \to X$ be a morphism from a scheme to $X$. If the fibres of $U \to X$ are universally bounded, then the fibres of $U \times _ X Y \to Y$ are universally bounded.

**Proof.**
This is clear from the definition, and properties of fibre products. (Note that $U \times _ X Y$ is a scheme as we assumed $Y \to X$ representable, so the definition applies.)
$\square$

Lemma 68.3.4. Let $S$ be a scheme. Let $g : Y \to X$ be a representable morphism of algebraic spaces over $S$. Let $f : U \to X$ be a morphism from a scheme towards $X$. Let $f' : U \times _ X Y \to Y$ be the base change of $f$. If

and $f'$ has universally bounded fibres, then $f$ has universally bounded fibres.

**Proof.**
Let $n \geq 0$ be an integer bounding the degrees of the fibre products $\mathop{\mathrm{Spec}}(k) \times _ Y (U \times _ X Y)$ as in Definition 68.3.1 for the morphism $f'$. We claim that $n$ works for $f$ also. Namely, suppose that $x : \mathop{\mathrm{Spec}}(k) \to X$ is a morphism from the spectrum of a field. Then either $\mathop{\mathrm{Spec}}(k) \times _ X U$ is empty (and there is nothing to prove), or $x$ is in the image of $|f|$. By Properties of Spaces, Lemma 66.4.3 and the assumption of the lemma we see that this means there exists a field extension $k'/k$ and a commutative diagram

Hence we see that

Since the scheme $\mathop{\mathrm{Spec}}(k') \times _ Y (U \times _ X Y)$ is assumed finite of degree $\leq n$ over $k'$ it follows that also $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree $\leq n$ over $k$ as desired. (Some details omitted.) $\square$

Lemma 68.3.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider a commutative diagram

where $U$ and $V$ are schemes. If $g$ has universally bounded fibres, and $f$ is surjective and flat, then also $h$ has universally bounded fibres.

**Proof.**
Assume $g$ has universally bounded fibres, and $f$ is surjective and flat. Say $n \geq 0$ is an integer which bounds the degrees of the schemes $\mathop{\mathrm{Spec}}(k) \times _ X U$ as in Definition 68.3.1. We claim $n$ also works for $h$. Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism from the spectrum of a field to $X$. Consider the morphism of schemes

It is flat and surjective. By assumption the scheme on the left is finite of degree $\leq n$ over $\mathop{\mathrm{Spec}}(k)$. It follows from Morphisms, Lemma 29.57.10 that the degree of the scheme on the right is also bounded by $n$ as desired. $\square$

Lemma 68.3.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$, and let $U$ be a scheme over $S$. Let $\varphi : U \to X$ be a morphism over $S$. If the fibres of $\varphi $ are universally bounded, then there exists an integer $n$ such that each fibre of $|U| \to |X|$ has at most $n$ elements.

**Proof.**
The integer $n$ of Definition 68.3.1 works. Namely, pick $x \in |X|$. Represent $x$ by a morphism $x : \mathop{\mathrm{Spec}}(k) \to X$. Then we get a commutative diagram

which shows (via Properties of Spaces, Lemma 66.4.3) that the inverse image of $x$ in $|U|$ is the image of the top horizontal arrow. Since $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree $\leq n$ over $k$ it has at most $n$ points. $\square$

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