Lemma 67.24.2. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$.

1. As $k$ ranges over all algebraically closed fields over $S$ the collection of geometric points $\overline{y} \in Y(k)$ cover all of $|Y|$.

2. 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 $|Y|$.

3. For any geometric point $\overline{s} : \mathop{\mathrm{Spec}}(k) \to S$ where $k$ has cardinality $> \lambda (X)$ the map

$X(k) \longrightarrow |X_ s|$

is surjective.

4. 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

$X(k) \longrightarrow |X| \times _{|Y|} Y(k)$

is surjective.

5. Let $X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

1. the map $X \to Y$ is surjective,

2. 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.

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

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

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

$u \times _{V, \overline{v}} \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(\kappa (u) \otimes _{\kappa (v)} k)$

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0488. Beware of the difference between the letter 'O' and the digit '0'.