The Stacks project

66.3 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 29.55 for similar definitions and results on morphisms of schemes.

Definition 66.3.1. 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 bounded1 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 $\leq n$.

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 29.55.1 by virtue of Morphisms, Lemma 29.55.2.

Lemma 66.3.2. 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 66.3.1. Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism, where $k$ is a field. Consider the morphisms

\[ \mathop{\mathrm{Spec}}(k) \times _ X V \longrightarrow \mathop{\mathrm{Spec}}(k) \times _ X U \longrightarrow \mathop{\mathrm{Spec}}(k). \]

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 29.55.5 we conclude that $\mathop{\mathrm{Spec}}(k) \times _ X V$ is finite over $k$ of degree at most $nm$. $\square$

Lemma 66.3.3. 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 66.3.4. 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

\[ \mathop{\mathrm{Im}}(|f| : |U| \to |X|) \subset \mathop{\mathrm{Im}}(|g| : |Y| \to |X|) \]

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 66.3.1 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 64.4.3 and the assumption of the lemma we see that this means there exists a field extension $k \subset k'$ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r] \ar[d] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & X } \]

Hence we see that

\[ \mathop{\mathrm{Spec}}(k') \times _ Y (U \times _ X Y) = \mathop{\mathrm{Spec}}(k') \times _{\mathop{\mathrm{Spec}}(k)} (\mathop{\mathrm{Spec}}(k) \times _ X U) \]

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

\[ \xymatrix{ U \ar[rd]_ g \ar[rr]_ f & & V \ar[ld]^ h \\ & X & } \]

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 66.3.1. 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

\[ \mathop{\mathrm{Spec}}(k) \times _ X V \longrightarrow \mathop{\mathrm{Spec}}(k) \times _ X U \]

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 29.55.11 that the degree of the scheme on the right is also bounded by $n$ as desired. $\square$

Lemma 66.3.6. 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 66.3.1 works. Namely, pick $x \in |X|$. Represent $x$ by a morphism $x : \mathop{\mathrm{Spec}}(k) \to X$. Then we get a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k) \times _ X U \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r]^ x & X } \]

which shows (via Properties of Spaces, Lemma 64.4.3) 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. $\square$

[1] This is probably nonstandard notation.

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