Lemma 66.12.1. 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 $u = u'$.

## 66.12 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 63.14.9 shows that this is **not** true in general.

Lemma 66.12.2. 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 66.12.3. 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 66.12.2 we can find a specialization $u_1 \leadsto u_2$ in $U$ mapping to $x_1 \leadsto x_2$. By Lemma 66.12.2 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 66.12.1. Hence also $u_1 = u_2$ as desired.
$\square$

Proposition 66.12.4. 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 66.12.3 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 64.12.3 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 66.5.3. By Theorem 66.10.2 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 26.11.1 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 66.12.5. Let $S$ be a scheme. Dimension as defined in Properties of Spaces, Section 64.9 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 66.12.1. 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 66.12.2.

Part (2) is an immediate consequence of part (1), Topology, Lemma 5.10.2, and Properties of Spaces, Section 64.9. $\square$

Lemma 66.12.6. 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 64.10.2. 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 66.12.7. 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 66.12.6 and Properties of Spaces, Lemma 64.10.3.
$\square$

The following lemma is a tiny bit stronger than Properties of Spaces, Lemma 64.15.3. We will improve this lemma in Lemma 66.14.2.

Lemma 66.12.8. 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 26.23.11. Hence the projection $\mathop{\mathrm{Spec}}(k) = U \to \mathop{\mathrm{Spec}}(k')$ is étale and we win.
$\square$

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