In this section we make some remarks on points and geometric points (see Properties of Spaces, Definition 66.19.1). One way to think about a geometric point of $X$ is to consider a geometric point $\overline{s} : \mathop{\mathrm{Spec}}(k) \to S$ of $S$ and a lift of $\overline{s}$ to a morphism $\overline{x}$ into $X$. Here is a diagram
We often say “let $k$ be an algebraically closed field over $S$” to indicate that $\mathop{\mathrm{Spec}}(k)$ comes equipped with a morphism $\mathop{\mathrm{Spec}}(k) \to S$. In this situation we write
for the set of $k$-valued points of $X$. In this case the map $X(k) \to |X|$ maps into the subset $|X_ s| \subset |X|$. Here $X_ s = \mathop{\mathrm{Spec}}(\kappa (s)) \times _ S X$, where $s \in S$ is the point corresponding to $\overline{s}$. As $\mathop{\mathrm{Spec}}(\kappa (s)) \to S$ is a monomorphism, also the base change $X_ s \to X$ is a monomorphism, and $|X_ s|$ is indeed a subset of $|X|$.
Lemma 67.24.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. The following are equivalent:
$f$ is surjective, and
for every algebraically closed field $k$ over $S$ the induced map $X(k) \to Y(k)$ is surjective.
Proof. Choose a diagram
with $U$, $V$ schemes over $S$ and vertical arrows surjective and étale, see Spaces, Lemma 65.11.6. Since $f$ is locally of finite type we see that $U \to V$ is locally of finite type.
Assume (1) and let $\overline{y} \in Y(k)$. Then $U \to Y$ is surjective and locally of finite type by Lemmas 67.5.4 and 67.23.2. Let $Z = U \times _{Y, \overline{y}} \mathop{\mathrm{Spec}}(k)$. This is a scheme. The projection $Z \to \mathop{\mathrm{Spec}}(k)$ is surjective and locally of finite type by Lemmas 67.5.5 and 67.23.3. It follows from Varieties, Lemma 33.14.1 that $Z$ has a $k$ valued point $\overline{z}$. The image $\overline{x} \in X(k)$ of $\overline{z}$ maps to $\overline{y}$ as desired.
Assume (2). By Properties of Spaces, Lemma 66.4.4 it suffices to show that $|X| \to |Y|$ is surjective. Let $y \in |Y|$. Choose a $u \in U$ mapping to $y$. Let $k \supset \kappa (u)$ be an algebraic closure. Denote $\overline{u} \in U(k)$ the corresponding point and $\overline{y} \in Y(k)$ its image. By assumption there exists a $\overline{x} \in X(k)$ mapping to $\overline{y}$. Then it is clear that the image $x \in |X|$ of $\overline{x}$ maps to $y$. $\square$
In order to state the next lemma we introduce the following notation. Given a scheme $T$ we denote
In words $\lambda (T)$ is the smallest infinite cardinal bounding all the cardinalities of residue fields of $T$. Note that if $R$ is a ring then the cardinality of any residue field $\kappa (\mathfrak p)$ of $R$ is bounded by the cardinality of $R$ (details omitted). This implies that $\lambda (T) \leq \text{size}(T)$ where $\text{size}(T)$ is the size of the scheme $T$ as introduced in Sets, Section 3.9. If $L/K$ is a finitely generated field extension then $|K| \leq |L| \leq \max \{ \aleph _0, |K|\} $. It follows that if $T' \to T$ is a morphism of schemes which is locally of finite type then $\lambda (T') \leq \lambda (T)$, and if $T' \to T$ is also surjective then equality holds. Next, suppose that $S$ is a scheme and that $X$ is an algebraic space over $S$. In this case we define
where $U$ is any scheme over $S$ which has a surjective étale morphism towards $X$. The reason that this is independent of the choice of $U$ is that given a pair of such schemes $U$ and $U'$ the fibre product $U \times _ X U'$ is a scheme which admits a surjective étale morphism to both $U$ and $U'$, whence $\lambda (U) = \lambda (U \times _ X U') = \lambda (U')$ by the discussion above.
Lemma 67.24.2. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$.
As $k$ ranges over all algebraically closed fields over $S$ the collection of geometric points $\overline{y} \in Y(k)$ cover all of $|Y|$.
As $k$ ranges over all algebraically closed fields over $S$ with $|k| \geq \lambda (Y)$ and $|k| > \lambda (X)$ the geometric points $\overline{y} \in Y(k)$ cover all of $|Y|$.
For any geometric point $\overline{s} : \mathop{\mathrm{Spec}}(k) \to S$ where $k$ has cardinality $> \lambda (X)$ the map
is surjective.
Let $X \to Y$ be a morphism of algebraic spaces over $S$. For any geometric point $\overline{s} : \mathop{\mathrm{Spec}}(k) \to S$ where $k$ has cardinality $> \lambda (X)$ the map
is surjective.
Let $X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
the map $X \to Y$ is surjective,
for all algebraically closed fields $k$ over $S$ with $|k| > \lambda (X)$, and $|k| \geq \lambda (Y)$ the map $X(k) \to Y(k)$ is surjective.
\[ X(k) \longrightarrow |X_ s| \]
\[ X(k) \longrightarrow |X| \times _{|Y|} Y(k) \]
Proof. To prove part (1) choose a surjective étale morphism $V \to Y$ where $V$ is a scheme. For each $v \in V$ choose an algebraic closure $\kappa (v) \subset k_ v$. Consider the morphisms $\overline{x} : \mathop{\mathrm{Spec}}(k_ v) \to V \to Y$. By construction of $|Y|$ these cover $|Y|$.
To prove part (2) we will use the following two facts whose proofs we omit: (i) If $K$ is a field and $\overline{K}$ is algebraic closure then $|\overline{K}| \leq \max \{ \aleph _0, |K|\} $. (ii) For any algebraically closed field $k$ and any cardinal $\aleph $, $\aleph \geq |k|$ there exists an extension of algebraically closed fields $k'/k$ with $|k'| = \aleph $. Now we set $\aleph = \max \{ \lambda (X), \lambda (Y)\} ^+$. Here $\lambda ^+ > \lambda $ indicates the next bigger cardinal, see Sets, Section 3.6. Now (i) implies that the fields $k_ u$ constructed in the first paragraph of the proof all have cardinality bounded by $\lambda (X)$. Hence by (ii) we can find extensions $k_ u \subset k'_ u$ such that $|k'_ u| = \aleph $. The morphisms $\overline{x}' : \mathop{\mathrm{Spec}}(k'_ u) \to X$ cover $|X|$ as desired. To really finish the proof of (2) we need to show that the schemes $\mathop{\mathrm{Spec}}(k'_ u)$ are (isomorphic to) objects of $\mathit{Sch}_{fppf}$ because our conventions are that all schemes are objects of $\mathit{Sch}_{fppf}$; the rest of this paragraph should be skipped by anyone who is not interested in set theoretical considerations. By construction there exists an object $T$ of $\mathit{Sch}_{fppf}$ such that $\lambda (X)$ and $\lambda (Y)$ are bounded by $\text{size}(T)$. By our construction of the category $\mathit{Sch}_{fppf}$ in Topologies, Definitions 34.7.6 as the category $\mathit{Sch}_\alpha $ constructed in Sets, Lemma 3.9.2 we see that any scheme whose size is $\leq \text{size}(T)^+$ is isomorphic to an object of $\mathit{Sch}_{fppf}$. See the expression for the function $Bound$ in Sets, Equation (3.9.1.1). Since $\aleph \leq \text{size}(T)^+$ we conclude.
The notation $X_ s$ in part (3) means the fibre product $\mathop{\mathrm{Spec}}(\kappa (s)) \times _ S X$, where $s \in S$ is the point corresponding to $\overline{s}$. Hence part (2) follows from (4) with $Y = \mathop{\mathrm{Spec}}(\kappa (s))$.
Let us prove (4). Let $X \to Y$ be a morphism of algebraic spaces over $S$. Let $k$ be an algebraically closed field over $S$ of cardinality $> \lambda (X)$. Let $\overline{y} \in Y(k)$ and $x \in |X|$ which map to the same element $y$ of $|Y|$. We have to find $\overline{x} \in X(k)$ mapping to $x$ and $\overline{y}$. Choose a commutative diagram
with $U$, $V$ schemes over $S$ and vertical arrows surjective and étale, see Spaces, Lemma 65.11.6. Choose a $u \in |U|$ which maps to $x$, and denote $v \in |V|$ the image. We will think of $u = \mathop{\mathrm{Spec}}(\kappa (u))$ and $v = \mathop{\mathrm{Spec}}(\kappa (v))$ as schemes. Note that $V \times _ Y \mathop{\mathrm{Spec}}(k)$ is a scheme étale over $k$. Hence it is a disjoint union of spectra of finite separable extensions of $k$, see Morphisms, Lemma 29.36.7. As $v$ maps to $y$ we see that $v \times _ Y \mathop{\mathrm{Spec}}(k)$ is a nonempty scheme. As $v \to V$ is a monomorphism, we see that $v \times _ Y \mathop{\mathrm{Spec}}(k) \to V \times _ Y \mathop{\mathrm{Spec}}(k)$ is a monomorphism. Hence $v \times _ Y \mathop{\mathrm{Spec}}(k)$ is a disjoint union of spectra of finite separable extensions of $k$, by Schemes, Lemma 26.23.11. We conclude that the morphism $v \times _ Y \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ has a section, i.e., we can find a morphism $\overline{v} : \mathop{\mathrm{Spec}}(k) \to V$ lying over $v$ and over $\overline{y}$. Finally we consider the scheme
where $\kappa (v) \to k$ is the field map defining the morphism $\overline{v}$. Since the cardinality of $k$ is larger than the cardinality of $\kappa (u)$ by assumption we may apply Algebra, Lemma 10.35.12 to see that any maximal ideal $\mathfrak m \subset \kappa (u) \otimes _{\kappa (v)} k$ has a residue field which is algebraic over $k$ and hence equal to $k$. Such a maximal ideal will hence produce a morphism $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ and mapping to $\overline{v}$. The composition $\mathop{\mathrm{Spec}}(k) \to U \to X$ will be the desired geometric point $\overline{x} \in X(k)$. This concludes the proof of part (4).
Part (5) is a formal consequence of parts (2) and (4) and Properties of Spaces, Lemma 66.4.4. $\square$
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