The Stacks project

65.24 Points and geometric points

In this section we make some remarks on points and geometric points (see Properties of Spaces, Definition 64.19.1). 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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k) \ar[r]_-{\overline{x}} \ar[rd]_{\overline{s}} & X \ar[d] \\ & S. } \]

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

\[ X(k) = \mathop{\mathrm{Mor}}\nolimits _ S(\mathop{\mathrm{Spec}}(k), X) = \{ \overline{x} \in X\text{ lying over }\overline{s}\} \]

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 65.24.1. 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:

  1. $f$ is surjective, and

  2. for every algebraically closed field $k$ over $S$ the induced map $X(k) \to Y(k)$ is surjective.

Proof. Choose a 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 63.11.6. 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 65.5.4 and 65.23.2. 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 65.5.5 and 65.23.3. 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 64.4.4 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

\[ \lambda (T) = \sup \{ \aleph _0, |\kappa (t)| ; t \in T\} . \]

In words $\lambda (T)$ is the smallest infinite cardinal bounding all the cardinalities of residue fields ot $T$. 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 $K \subset L$ 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

\[ \lambda (X) := \lambda (U) \]

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 65.24.2. 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 $k \subset k'$ 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 63.11.6. 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 64.4.4. $\square$


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