## 64.4 Points of algebraic spaces

As is clear from Spaces, Example 63.14.8 a point of an algebraic space should not be defined as a monomorphism from the spectrum of a field. Instead we define them as equivalence classes of morphisms of spectra of fields exactly as explained in Schemes, Section 26.13.

Let $S$ be a scheme. Let $F$ be a presheaf on $(\mathit{Sch}/S)_{fppf}$. Let $K$ is a field. Consider a morphism

$\mathop{\mathrm{Spec}}(K) \longrightarrow F.$

By the Yoneda Lemma this is given by an element $p \in F(\mathop{\mathrm{Spec}}(K))$. We say that two such pairs $(\mathop{\mathrm{Spec}}(K), p)$ and $(\mathop{\mathrm{Spec}}(L), q)$ are equivalent if there exists a third field $\Omega$ and a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(L) \ar[d]^ q \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & F. }$

In other words, there are field extensions $K \to \Omega$ and $L \to \Omega$ such that $p$ and $q$ map to the same element of $F(\mathop{\mathrm{Spec}}(\Omega ))$. We omit the verification that this defines an equivalence relation.

Definition 64.4.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. A point of $X$ is an equivalence class of morphisms from spectra of fields into $X$. The set of points of $X$ is denoted $|X|$.

Note that if $f : X \to Y$ is a morphism of algebraic spaces over $S$, then there is an induced map $|f| : |X| \to |Y|$ which maps a representative $x : \mathop{\mathrm{Spec}}(K) \to X$ to the representative $f \circ x : \mathop{\mathrm{Spec}}(K) \to Y$.

Lemma 64.4.2. Let $S$ be a scheme. Let $X$ be a scheme over $S$. The points of $X$ as a scheme are in canonical 1-1 correspondence with the points of $X$ as an algebraic space.

Proof. This is Schemes, Lemma 26.13.3. $\square$

Lemma 64.4.3. Let $S$ be a scheme. Let

$\xymatrix{ Z \times _ Y X \ar[r] \ar[d] & X \ar[d] \\ Z \ar[r] & Y }$

be a cartesian diagram of algebraic spaces over $S$. Then the map of sets of points

$|Z \times _ Y X| \longrightarrow |Z| \times _{|Y|} |X|$

is surjective.

Proof. Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\mathrm{Spec}}(K) \to X$, $\mathop{\mathrm{Spec}}(L) \to Z$, then the assumption that they agree as elements of $|Y|$ means that there is a common extension $K \subset M$ and $L \subset M$ such that $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(K) \to X \to Y$ and $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(L) \to Z \to Y$ agree. And this is exactly the condition which says you get a morphism $\mathop{\mathrm{Spec}}(M) \to Z \times _ Y X$. $\square$

Lemma 64.4.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $f : T \to X$ be a morphism from a scheme to $X$. The following are equivalent

1. $f : T \to X$ is surjective (according to Spaces, Definition 63.5.1), and

2. $|f| : |T| \to |X|$ is surjective.

Proof. Assume (1). Let $x : \mathop{\mathrm{Spec}}(K) \to X$ be a morphism from the spectrum of a field into $X$. By assumption the morphism of schemes $\mathop{\mathrm{Spec}}(K) \times _ X T \to \mathop{\mathrm{Spec}}(K)$ is surjective. Hence there exists a field extension $K \subset K'$ and a morphism $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _ X T$ such that the left square in the diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _ X T \ar[d] \ar[r] & T \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-x & X }$

is commutative. This shows that $|f| : |T| \to |X|$ is surjective.

Assume (2). Let $Z \to X$ be a morphism where $Z$ is a scheme. We have to show that the morphism of schemes $Z \times _ X T \to T$ is surjective, i.e., that $|Z \times _ X T| \to |Z|$ is surjective. This follows from (2) and Lemma 64.4.3. $\square$

Lemma 64.4.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $X = U/R$ be a presentation of $X$, see Spaces, Definition 63.9.3. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|X|$ is the quotient of $|U|$ by this equivalence relation.

Proof. The assumption means that $U$ is a scheme, $p : U \to X$ is a surjective, étale morphism, $R = U \times _ X U$ is a scheme and defines an étale equivalence relation on $U$ such that $X = U/R$ as sheaves. By Lemma 64.4.4 we see that $|U| \to |X|$ is surjective. By Lemma 64.4.3 the map

$|R| \longrightarrow |U| \times _{|X|} |U|$

is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|X|$. Combining these two statements we get the result of the lemma. $\square$

Lemma 64.4.6. Let $S$ be a scheme. There exists a unique topology on the sets of points of algebraic spaces over $S$ with the following properties:

1. if $X$ is a scheme over $S$, then the topology on $|X|$ is the usual one (via the identification of Lemma 64.4.2),

2. for every morphism of algebraic spaces $X \to Y$ over $S$ the map $|X| \to |Y|$ is continuous, and

3. for every étale morphism $U \to X$ with $U$ a scheme the map of topological spaces $|U| \to |X|$ is continuous and open.

Proof. Let $X$ be an algebraic space over $S$. Let $p : U \to X$ be a surjective étale morphism where $U$ is a scheme over $S$. We define $W \subset |X|$ is open if and only if $|p|^{-1}(W)$ is an open subset of $|U|$. This is a topology on $|X|$ (it is the quotient topology on $|X|$, see Topology, Lemma 5.6.2).

Let us prove that the topology is independent of the choice of the presentation. To do this it suffices to show that if $U'$ is a scheme, and $U' \to X$ is an étale morphism, then the map $|U'| \to |X|$ (with topology on $|X|$ defined using $U \to X$ as above) is open and continuous; which in addition will prove that (3) holds. Set $U'' = U \times _ X U'$, so that we have the commutative diagram

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

As $U \to X$ and $U' \to X$ are étale we see that both $U'' \to U$ and $U'' \to U'$ are étale morphisms of schemes. Moreover, $U'' \to U'$ is surjective. Hence we get a commutative diagram of maps of sets

$\xymatrix{ |U''| \ar[r] \ar[d] & |U'| \ar[d] \\ |U| \ar[r] & |X| }$

The lower horizontal arrow is surjective (see Lemma 64.4.4 or Lemma 64.4.5) and continuous by definition of the topology on $|X|$. The top horizontal arrow is surjective, continuous, and open by Morphisms, Lemma 29.36.13. The left vertical arrow is continuous and open (by Morphisms, Lemma 29.36.13 again.) Hence it follows formally that the right vertical arrow is continuous and open.

To finish the proof we prove (2). Let $a : X \to Y$ be a morphism of algebraic spaces. According to Spaces, Lemma 63.11.6 we can find a diagram

$\xymatrix{ U \ar[d]_ p \ar[r]_\alpha & V \ar[d]^ q \\ X \ar[r]^ a & Y }$

where $U$ and $V$ are schemes, and $p$ and $q$ are surjective and étale. This gives rise to the diagram

$\xymatrix{ |U| \ar[d]_ p \ar[r]_\alpha & |V| \ar[d]^ q \\ |X| \ar[r]^ a & |Y| }$

where all but the lower horizontal arrows are known to be continuous and the two vertical arrows are surjective and open. It follows that the lower horizontal arrow is continuous as desired. $\square$

Definition 64.4.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The underlying topological space of $X$ is the set of points $|X|$ endowed with the topology constructed in Lemma 64.4.6.

It turns out that this topological space carries the same information as the small Zariski site $X_{Zar}$ of Spaces, Definition 63.12.6.

Lemma 64.4.8. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.

1. The rule $X' \mapsto |X'|$ defines an inclusion preserving bijection between open subspaces $X'$ (see Spaces, Definition 63.12.1) of $X$, and opens of the topological space $|X|$.

2. A family $\{ X_ i \subset X\} _{i \in I}$ of open subspaces of $X$ is a Zariski covering (see Spaces, Definition 63.12.5) if and only if $|X| = \bigcup |X_ i|$.

In other words, the small Zariski site $X_{Zar}$ of $X$ is canonically identified with a site associated to the topological space $|X|$ (see Sites, Example 7.6.4).

Proof. In order to prove (1) let us construct the inverse of the rule. Namely, suppose that $W \subset |X|$ is open. Choose a presentation $X = U/R$ corresponding to the surjective étale map $p : U \to X$ and étale maps $s, t : R \to U$. By construction we see that $|p|^{-1}(W)$ is an open of $U$. Denote $W' \subset U$ the corresponding open subscheme. It is clear that $R' = s^{-1}(W') = t^{-1}(W')$ is a Zariski open of $R$ which defines an étale equivalence relation on $W'$. By Spaces, Lemma 63.10.2 the morphism $X' = W'/R' \to X$ is an open immersion. Hence $X'$ is an algebraic space by Spaces, Lemma 63.11.3. By construction $|X'| = W$, i.e., $X'$ is a subspace of $X$ corresponding to $W$. Thus (1) is proved.

To prove (2), note that if $\{ X_ i \subset X\} _{i \in I}$ is a collection of open subspaces, then it is a Zariski covering if and only if the $U = \bigcup U \times _ X X_ i$ is an open covering. This follows from the definition of a Zariski covering and the fact that the morphism $U \to X$ is surjective as a map of presheaves on $(\mathit{Sch}/S)_{fppf}$. On the other hand, we see that $|X| = \bigcup |X_ i|$ if and only if $U = \bigcup U \times _ X X_ i$ by Lemma 64.4.5 (and the fact that the projections $U \times _ X X_ i \to X_ i$ are surjective and étale). Thus the equivalence of (2) follows. $\square$

Lemma 64.4.9. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $X' \subset X$ be an open subspace. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Then $f$ factors through $X'$ if and only if $|f| : |Y| \to |X|$ factors through $|X'| \subset |X|$.

Proof. By Spaces, Lemma 63.12.3 we see that $Y' = Y \times _ X X' \to Y$ is an open immersion. If $|f|(|Y|) \subset |X'|$, then clearly $|Y'| = |Y|$. Hence $Y' = Y$ by Lemma 64.4.8. $\square$

Lemma 64.4.10. Let $S$ be a scheme. Let $X$ be an algebraic spaces over $S$. Let $U$ be a scheme and let $f : U \to X$ be an étale morphism. Let $X' \subset X$ be the open subspace corresponding to the open $|f|(|U|) \subset |X|$ via Lemma 64.4.8. Then $f$ factors through a surjective étale morphism $f' : U \to X'$. Moreover, if $R = U \times _ X U$, then $R = U \times _{X'} U$ and $X'$ has the presentation $X' = U/R$.

Proof. The existence of the factorization follows from Lemma 64.4.9. The morphism $f'$ is surjective according to Lemma 64.4.4. To see $f'$ is étale, suppose that $T \to X'$ is a morphism where $T$ is a scheme. Then $T \times _ X U = T \times _{X'} U$ as $X' \to X$ is a monomorphism of sheaves. Thus the projection $T \times _{X'} U \to T$ is étale as we assumed $f$ étale. We have $U \times _ X U = U \times _{X'} U$ as $X' \to X$ is a monomorphism. Then $X' = U/R$ follows from Spaces, Lemma 63.9.1. $\square$

Lemma 64.4.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the map

$\{ \mathop{\mathrm{Spec}}(k) \to X \text{ monomorphism where }k\text{ is a field}\} \longrightarrow |X|$

This map is injective.

Proof. Suppose that $\varphi _ i : \mathop{\mathrm{Spec}}(k_ i) \to X$ are monomorphisms for $i = 1, 2$. If $\varphi _1$ and $\varphi _2$ define the same point of $|X|$, then we see that the scheme

$Y = \mathop{\mathrm{Spec}}(k_1) \times _{\varphi _1, X, \varphi _2} \mathop{\mathrm{Spec}}(k_2)$

is nonempty. Since the base change of a monomorphism is a monomorphism this means that the projection morphisms $Y \to \mathop{\mathrm{Spec}}(k_ i)$ are monomorphisms. Hence $\mathop{\mathrm{Spec}}(k_1) = Y = \mathop{\mathrm{Spec}}(k_2)$ as schemes over $X$, see Schemes, Lemma 26.23.11. We conclude that $\varphi _1 = \varphi _2$, which proves the lemma. $\square$

We will see in Decent Spaces, Lemma 66.11.1 that this map is a bijection when $X$ is decent.

Comment #4978 by Elyes Boughattas on

In the proof of Lemma 63.4.6 : before the first diagram, "which in addition will prove that (2) holds" should be "which in addition will prove that (3) holds". Again, at the paragraph starting before the third diagram, "To finish the proof we prove (1)" should be "To finish the proof we prove (2)".

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