Lemma 60.12.1. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $U \to X$ be an étale morphism from a scheme to $X$. If $u, u' \in |U|$ map to the same point of $|X|$, and $u' \leadsto u$, then $u = u'$.

## 60.12 Points on decent spaces

In this section we prove some properties of points on decent algebraic spaces. The following lemma shows that specialization of points behaves well on decent algebraic spaces. Spaces, Example 57.14.9 shows that this is **not** true in general.

Lemma 60.12.2. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Then for every étale morphism $\varphi : U \to X$ from a scheme $U$ and any $u \in U$ with $\varphi (u) = x$, exists a point $u'\in U$, $u' \leadsto u$ with $\varphi (u') = x'$.

Lemma 60.12.3. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then $|X|$ is Kolmogorov (see Topology, Definition 5.8.4).

**Proof.**
Let $x_1, x_2 \in |X|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$. We have to show that $x_1 = x_2$. Pick a scheme $U$ and an étale morphism $U \to X$ such that $x_1, x_2$ are both in the image of $|U| \to |X|$. By Lemma 60.12.2 we can find a specialization $u_1 \leadsto u_2$ in $U$ mapping to $x_1 \leadsto x_2$. By Lemma 60.12.2 we can find $u_2' \leadsto u_1$ mapping to $x_2 \leadsto x_1$. This means that $u_2' \leadsto u_2$ is a specialization between points of $U$ mapping to the same point of $X$, namely $x_2$. This is not possible, unless $u_2' = u_2$, see Lemma 60.12.1. Hence also $u_1 = u_2$ as desired.
$\square$

Proposition 60.12.4. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then the topological space $|X|$ is sober (see Topology, Definition 5.8.4).

**Proof.**
We have seen in Lemma 60.12.3 that $|X|$ is Kolmogorov. Hence it remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Properties of Spaces, Lemma 58.12.4 there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. By definition this means that $Z \to X$ is a representable morphism of algebraic spaces. Hence $Z$ is a decent algebraic space by Lemma 60.5.3. By Theorem 60.10.2 we see that there exists an open dense subspace $Z' \subset Z$ which is a scheme. This means that $|Z'| \subset T$ is open dense. Hence the topological space $|Z'|$ is irreducible, which means that $Z'$ is an irreducible scheme. By Schemes, Lemma 25.11.1 we conclude that $|Z'|$ is the closure of a single point $\eta \in T$ and hence also $T = \overline{\{ \eta \} }$, and we win.
$\square$

For decent algebraic spaces dimension works as expected.

Lemma 60.12.5. Let $S$ be a scheme. Dimension as defined in Properties of Spaces, Section 58.9 behaves well on decent algebraic spaces $X$ over $S$.

If $x \in |X|$, then $\dim _ x(|X|) = \dim _ x(X)$, and

$\dim (|X|) = \dim (X)$.

**Proof.**
Proof of (1). Choose a scheme $U$ with a point $u \in U$ and an étale morphism $h : U \to X$ mapping $u$ to $x$. By definition the dimension of $X$ at $x$ is $\dim _ u(|U|)$. Thus we may pick $U$ such that $\dim _ x(X) = \dim (|U|)$. Let $d$ be an integer. If $\dim (U) \geq d$, then there exists a sequence of nontrivial specializations $u_ d \leadsto \ldots \leadsto u_0$ in $U$. Taking the image we find a corresponding sequence $h(u_ d) \leadsto \ldots \leadsto h(u_0)$ each of which is nontrivial by Lemma 60.12.1. Hence we see that the image of $|U|$ in $|X|$ has dimension at least $d$. Conversely, suppose that $x_ d \leadsto \ldots \leadsto x_0$ is a sequence of specializations in $|X|$ with $x_0$ in the image of $|U| \to |X|$. Then we can lift this to a sequence of specializations in $U$ by Lemma 60.12.2.

Part (2) is an immediate consequence of part (1), Topology, Lemma 5.10.2, and Properties of Spaces, Section 58.9. $\square$

Lemma 60.12.6. Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism of algebraic spaces over $S$. Let $x \in |X|$ with image $y \in |Y|$. Then the dimension of the local ring of $Y$ at $y$ is $\geq $ to the dimension of the local ring of $X$ at $x$.

**Proof.**
The definition of the dimension of the local ring of a point on an algebraic space is given in Properties of Spaces, Definition 58.10.2. Choose an étale morphism $(V, v) \to (Y, y)$ where $V$ is a scheme. Choose an étale morphism $U \to V \times _ Y X$ and a point $u \in U$ mapping to $x \in |X|$ and $v \in V$. Then $U \to V$ is locally quasi-finite and we have to prove that

This is Algebra, Lemma 10.124.4. $\square$

Lemma 60.12.7. Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism of algebraic spaces over $S$. Then $\dim (X) \leq \dim (Y)$.

**Proof.**
This follows from Lemma 60.12.6 and Properties of Spaces, Lemma 58.10.3.
$\square$

The following lemma is a tiny bit stronger than Properties of Spaces, Lemma 58.15.3. We will improve this lemma in Lemma 60.14.2.

Lemma 60.12.8. Let $S$ be a scheme. Let $k$ be a field. Let $X$ be an algebraic space over $S$ and assume that there exists a surjective étale morphism $\mathop{\mathrm{Spec}}(k) \to X$. If $X$ is decent, then $X \cong \mathop{\mathrm{Spec}}(k')$ where $k' \subset k$ is a finite separable extension.

**Proof.**
The assumption implies that $|X| = \{ x\} $ is a singleton. Since $X$ is decent we can find a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k') \to X$ whose image is $x$. Then the projection $U = \mathop{\mathrm{Spec}}(k') \times _ X \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism, whence $U = \mathop{\mathrm{Spec}}(k)$, see Schemes, Lemma 25.23.11. Hence the projection $\mathop{\mathrm{Spec}}(k) = U \to \mathop{\mathrm{Spec}}(k')$ is étale and we win.
$\square$

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