Example 66.18.1. If $X, Y, Z$ are schemes, then the set $F_{x, z}$ is equal to the spectrum of $\kappa (x) \otimes _{\kappa (y)} \kappa (z)$ (Schemes, Lemma 26.17.5). Thus we obtain a finite set if either $\kappa (y) \subset \kappa (x)$ is finite or if $\kappa (y) \subset \kappa (z)$ is finite. In particular, this is always the case if $g$ is quasi-finite at $z$ (Morphisms, Lemma 29.20.5).

## 66.18 Points of fibres

Let $S$ be a scheme. Consider a cartesian diagram

of algebraic spaces over $S$. Let $x \in |X|$ and $z \in |Z|$ be points mapping to the same point $y \in |Y|$. We may ask: When is the set

finite?

Example 66.18.2. Let $K$ be a characteristic $0$ field endowed with an automorphism $\sigma $ of infinite order. Set $Y = \mathop{\mathrm{Spec}}(K)/\mathbf{Z}$ and $X = \mathbf{A}^1_ K/\mathbf{Z}$ where $\mathbf{Z}$ acts on $K$ via $\sigma $ and on $\mathbf{A}^1_ K = \mathop{\mathrm{Spec}}(K[t])$ via $t \mapsto t + 1$. Let $Z = \mathop{\mathrm{Spec}}(K)$. Then $W = \mathbf{A}^1_ K$. Picture

Take $x$ corresponding to $t = 0$ and $z$ the unique point of $\mathop{\mathrm{Spec}}(K)$. Then we see that $F_{x, z} = \mathbf{Z}$ as a set.

Lemma 66.18.3. In the situation of (66.18.0.1) if $Z' \to Z$ is a morphism and $z' \in |Z'|$ maps to $z$, then the induced map $F_{x, z'} \to F_{x, z}$ is surjective.

**Proof.**
Set $W' = X \times _ Y Z' = W \times _ Z Z'$. Then $|W'| \to |W| \times _{|Z|} |Z'|$ is surjective by Properties of Spaces, Lemma 64.4.3. Hence the surjectivity of $F_{x, z'} \to F_{x, z}$.
$\square$

Lemma 66.18.4. In diagram (66.18.0.1) the set (66.18.0.2) is finite if $f$ is of finite type and $f$ is quasi-finite at $x$.

**Proof.**
The morphism $q$ is quasi-finite at every $w \in F_{x, z}$, see Morphisms of Spaces, Lemma 65.27.2. Hence the lemma follows from Morphisms of Spaces, Lemma 65.27.9.
$\square$

Lemma 66.18.5. In diagram (66.18.0.1) the set (66.18.0.2) is finite if $y$ can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to Y$ where $k$ is a field and $g$ is quasi-finite at $z$. (Special case: $Y$ is decent and $g$ is étale.)

**Proof.**
By Lemma 66.18.3 applied twice we may replace $Z$ by $Z_ k = \mathop{\mathrm{Spec}}(k) \times _ Y Z$ and $X$ by $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$. We may and do replace $Y$ by $\mathop{\mathrm{Spec}}(k)$ as well. Note that $Z_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at $z$ by Morphisms of Spaces, Lemma 65.27.2. Choose a scheme $V$, a point $v \in V$, and an étale morphism $V \to Z_ k$ mapping $v$ to $z$. Choose a scheme $U$, a point $u \in U$, and an étale morphism $U \to X_ k$ mapping $u$ to $x$. Again by Lemma 66.18.3 it suffices to show $F_{u, v}$ is finite for the diagram

The morphism $V \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at $v$ (follows from the general discussion in Morphisms of Spaces, Section 65.22 and the definition of being quasi-finite at a point). At this point the finiteness follows from Example 66.18.1. The parenthetical remark of the statement of the lemma follows from the fact that on decent spaces points are represented by monomorphisms from fields and from the fact that an étale morphism of algebraic spaces is locally quasi-finite. $\square$

Lemma 66.18.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $y \in |Y|$ and assume that $y$ is represented by a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to Y$. Then $|X_ k| \to |X|$ is a homeomorphism onto $f^{-1}(\{ y\} ) \subset |X|$ with induced topology.

**Proof.**
We will use Properties of Spaces, Lemma 64.16.7 and Morphisms of Spaces, Lemma 65.10.9 without further mention. Let $V \to Y$ be an étale morphism with $V$ affine such that there exists a $v \in V$ mapping to $y$. Since $\mathop{\mathrm{Spec}}(k) \to Y$ is quasi-compact there are a finite number of points of $V$ mapping to $y$ (Lemma 66.4.5). After shrinking $V$ we may assume $v$ is the only one. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Consider the commutative diagram

Since $U_ v \to U_ V$ identifies $U_ v$ with a subset of $U_ V$ with the induced topology (Schemes, Lemma 26.18.5), and since $|U_ V| \to |X_ V|$ and $|U_ v| \to |X_ v|$ are surjective and open, we see that $|X_ v| \to |X_ V|$ is a homeomorphism onto its image (with induced topology). On the other hand, the inverse image of $f^{-1}(\{ y\} )$ under the open map $|X_ V| \to |X|$ is equal to $|X_ v|$. We conclude that $|X_ v| \to f^{-1}(\{ y\} )$ is open. The morphism $X_ v \to X$ factors through $X_ k$ and $|X_ k| \to |X|$ is injective with image $f^{-1}(\{ y\} )$ by Properties of Spaces, Lemma 64.4.3. Using $|X_ v| \to |X_ k| \to f^{-1}(\{ y\} )$ the lemma follows because $X_ v \to X_ k$ is surjective. $\square$

Lemma 66.18.7. Let $X$ be an algebraic space locally of finite type over a field $k$. Let $x \in |X|$. Consider the conditions

$\dim _ x(|X|) = 0$,

$x$ is closed in $|X|$ and if $x' \leadsto x$ in $|X|$ then $x' = x$,

$x$ is an isolated point of $|X|$,

$\dim _ x(X) = 0$,

$X \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at $x$.

Then (2), (3), (4), and (5) are equivalent. If $X$ is decent, then (1) is equivalent to the others.

**Proof.**
Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemmas 65.34.7 and 65.34.8.

Let $U \to X$ be an étale morphism where $U$ is an affine scheme and let $u \in U$ be a point mapping to $x$. Moreover, if $x$ is a closed point, e.g., in case (2) or (3), then we may and do assume that $u$ is a closed point. Observe that $\dim _ u(U) = \dim _ x(X)$ by definition and that this is equal to $\dim (\mathcal{O}_{U, u})$ if $u$ is a closed point, see Algebra, Lemma 10.113.6.

If $\dim _ x(X) > 0$ and $u$ is closed, by the arguments above we can choose a nontrivial specialization $u' \leadsto u$ in $U$. Then the transcendence degree of $\kappa (u')$ over $k$ exceeds the transcendence degree of $\kappa (u)$ over $k$. It follows that the images $x$ and $x'$ in $X$ are distinct, because the transcendence degree of $x/k$ and $x'/k$ are well defined, see Morphisms of Spaces, Definition 65.33.1. This applies in particular in cases (2) and (3) and we conclude that (2) and (3) imply (4).

Conversely, if $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite at $x$, then $U \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite at $u$, hence $u$ is an isolated point of $U$ (Morphisms, Lemma 29.20.6). It follows that (5) implies (2) and (3) as $|U| \to |X|$ is continuous and open.

Assume $X$ is decent and (1) holds. Then $\dim _ x(X) = \dim _ x(|X|)$ by Lemma 66.12.5 and the proof is complete. $\square$

Lemma 66.18.8. Let $X$ be an algebraic space locally of finite type over a field $k$. Consider the conditions

$|X|$ is a finite set,

$|X|$ is a discrete space,

$\dim (|X|) = 0$,

$\dim (X) = 0$,

$X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite,

Then (2), (3), (4), and (5) are equivalent. If $X$ is decent, then (1) implies the others.

**Proof.**
Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemma 65.34.7.

Let $U \to X$ be a surjective étale morphism where $U$ is a scheme.

If $\dim (U) > 0$, then choose a nontrivial specialization $u \leadsto u'$ in $U$ and the transcendence degree of $\kappa (u)$ over $k$ exceeds the transcendence degree of $\kappa (u')$ over $k$. It follows that the images $x$ and $x'$ in $X$ are distinct, because the transcendence degree of $x/k$ and $x'/k$ is well defined, see Morphisms of Spaces, Definition 65.33.1. We conclude that (2) and (3) imply (4).

Conversely, if $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite, then $U$ is locally Noetherian (Morphisms, Lemma 29.15.6) of dimension $0$ (Morphisms, Lemma 29.29.5) and hence is a disjoint union of spectra of Artinian local rings (Properties, Lemma 28.10.5). Hence $U$ is a discrete topological space, and since $|U| \to |X|$ is continuous and open, the same is true for $|X|$. In other words, (4) implies (2) and (3).

Assume $X$ is decent and (1) holds. Then we may choose $U$ above to be affine. The fibres of $|U| \to |X|$ are finite (this is a part of the defining property of decent spaces). Hence $U$ is a finite type scheme over $k$ with finitely many points. Hence $U$ is quasi-finite over $k$ (Morphisms, Lemma 29.20.7) which by definition means that $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. $\square$

Lemma 66.18.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $x \in |X|$ with image $y \in |Y|$. Let $F = f^{-1}(\{ y\} )$ with induced topology from $|X|$. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Y$ be in the equivalence class defining $y$. Set $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$. Let $\tilde x \in |X_ k|$ map to $x \in |X|$. Consider the following conditions

$\dim _ x(F) = 0$,

$x$ is isolated in $F$,

$x$ is closed in $F$ and if $x' \leadsto x$ in $F$, then $x = x'$,

$\dim _{\tilde x}(|X_ k|) = 0$,

$\tilde x$ is isolated in $|X_ k|$,

$\tilde x$ is closed in $|X_ k|$ and if $\tilde x' \leadsto \tilde x$ in $|X_ k|$, then $\tilde x = \tilde x'$,

$\dim _{\tilde x}(X_ k) = 0$,

$f$ is quasi-finite at $x$.

Then we have

If $Y$ is decent, then conditions (2) and (3) are equivalent to each other and to conditions (5), (6), (7), and (8). If $Y$ and $X$ are decent, then all conditions are equivalent.

**Proof.**
By Lemma 66.18.7 conditions (5), (6), and (7) are equivalent to each other and to the condition that $X_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at $\tilde x$. Thus by Morphisms of Spaces, Lemma 65.27.2 they are also equivalent to (8). If $f$ is decent, then $X_ k$ is a decent algebraic space and Lemma 66.18.7 shows that (4) implies (5).

If $Y$ is decent, then we can pick a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k') \to Y$ in the equivalence class of $y$. In this case Lemma 66.18.6 tells us that $|X_{k'}| \to F$ is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted. $\square$

Lemma 66.18.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $y \in |Y|$. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Y$ be in the equivalence class defining $y$. Set $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ and let $F = f^{-1}(\{ y\} )$ with the induced topology from $|X|$. Consider the following conditions

$F$ is finite,

$F$ is a discrete topological space,

$\dim (F) = 0$,

$|X_ k|$ is a finite set,

$|X_ k|$ is a discrete space,

$\dim (|X_ k|) = 0$,

$\dim (X_ k) = 0$,

$f$ is quasi-finite at all points of $|X|$ lying over $y$.

Then we have

If $Y$ is decent, then conditions (2) and (3) are equivalent to each other and to conditions (5), (6), (7), and (8). If $Y$ and $X$ are decent, then (1) implies all the other conditions.

**Proof.**
By Lemma 66.18.8 conditions (5), (6), and (7) are equivalent to each other and to the condition that $X_ k \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. Thus by Morphisms of Spaces, Lemma 65.27.2 they are also equivalent to (8). If $f$ is decent, then $X_ k$ is a decent algebraic space and Lemma 66.18.8 shows that (4) implies (5).

The map $|X_ k| \to F$ is surjective by Properties of Spaces, Lemma 64.4.3 and we see (4) $\Rightarrow $ (1).

If $Y$ is decent, then we can pick a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k') \to Y$ in the equivalence class of $y$. In this case Lemma 66.18.6 tells us that $|X_{k'}| \to F$ is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted. $\square$

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