The Stacks project

Proof. Conditions (4) and (5) are equivalent for any algebraic space by definition, see Properties of Spaces, Definition 66.10.2. Observe that any $Y$ as in (2) and (3) is decent by Lemma 68.6.6. Thus it suffices to prove the equivalence of (1) and (4) as then the equivalence with (2) and (3) follows since the dimension of the local ring of $Y$ at $y$ is equal to the dimension of the local ring of $X$ at $x$. Let $f : U \to X$ be an étale morphism from an affine scheme and let $u \in U$ be a point mapping to $x$.

Assume (1). Let $u' \leadsto u$ be a specialization in $U$. Then $f(u') = f(u) = x$. By Lemma 68.12.1 we see that $u' = u$. Hence $u$ is a generic point of an irreducible component of $U$. Thus $\dim (\mathcal{O}_{U, u}) = 0$ and we see that (4) holds.

Assume (4). The point $x$ is contained in an irreducible component $T \subset |X|$. Since $|X|$ is sober (Proposition 68.12.4) we $T$ has a generic point $x'$. Of course $x' \leadsto x$. Then we can lift this specialization to $u' \leadsto u$ in $U$ (Lemma 68.12.2). This contradicts the assumption that $\dim (\mathcal{O}_{U, u}) = 0$ unless $u' = u$, i.e., $x' = x$. $\square$

Proof. Choose a scheme $U$, a point $u \in U$, and an étale morphism $U \to X$ sending $u$ to $\xi $. Then any sequence of nontrivial specializations $\xi _ e \leadsto \ldots \leadsto \xi _0 = \xi $ can be lifted to a sequence $u_ e \leadsto \ldots \leadsto u_0 = u$ in $U$ by Lemma 68.12.2. Conversely, any sequence of nontrivial specializations $u_ e \leadsto \ldots \leadsto u_0 = u$ in $U$ maps to a sequence of nontrivial specializations $\xi _ e \leadsto \ldots \leadsto \xi _0 = \xi $ by Lemma 68.12.1. Because $|X|$ and $U$ are sober topological spaces we conclude that the codimension of $T$ in $|X|$ and of $\overline{\{ u\} }$ in $U$ are the same. In this way the lemma reduces to the schemes case which is Properties, Lemma 28.10.3. $\square$

Proof. Let $U$ be an affine scheme and let $a : U \to X$ be an étale morphism. We have to show that the fibres of $a$ are universally bounded. By assumption (1) the scheme $U$ has finitely many irreducible components. Let $u_1, \ldots , u_ n \in U$ be the generic points of these irreducible components. Let $\{ x_1, \ldots , x_ m\} \subset |X|$ be the image of $\{ u_1, \ldots , u_ n\} $. Each $x_ j$ is a point of codimension $0$. By assumption (2) we may choose a monomorphism $\mathop{\mathrm{Spec}}(k_ j) \to X$ representing $x_ j$. By Properties of Spaces, Lemma 66.11.1 we have

This is a scheme finite over $\mathop{\mathrm{Spec}}(k_ j)$ of degree $d_ j = \sum _{a(u_ i) = x_ j} [\kappa (u_ i) : k_ j]$. Set $n = \max d_ j$.

Observe that $a$ is separated (Properties of Spaces, Lemma 66.6.4). Consider the stratification

associated to $U \to X$ in Lemma 68.8.2. By our choice of $n$ above we conclude that $X_{n + 1}$ is empty. Namely, if not, then $a^{-1}(X_{n + 1})$ is a nonempty open of $U$ and hence would contain one of the $x_ i$. This would mean that $X_{n + 1}$ contains $x_ j = a(u_ i)$ which is impossible. Hence we see that the fibres of $U \to X$ are universally bounded (in fact by the integer $n$). $\square$

Proof. In the proof we use Properties of Spaces, Lemma 66.11.1 without further mention. Assume (1). Then $X$ has a dense open subscheme $X'$ by Theorem 68.10.2. Since the closure of an irreducible component of $|X'|$ is an irreducible component of $|X|$, we see that $|X'|$ has finitely many irreducible components. Thus (3) holds.

Assume $X' \subset X$ is as in (3). Let $\{ x'_1, \ldots , x'_ m\} $ be the generic points of the irreducible components of $X'$. Let $a : U \to X$ be an étale morphism with $U$ a quasi-compact scheme. To prove (2) it suffices to show that $U$ has finitely many irreducible components whose generic points lie over $\{ x'_1, \ldots , x'_ m\} $. It suffices to prove this for the members of a finite affine open cover of $U$, hence we may and do assume $U$ is affine. Note that $U' = a^{-1}(X') \subset U$ is a dense open. Since $U' \to X'$ is an étale morphism of schemes, we see the generic points of irreducible components of $U'$ are the points lying over $\{ x'_1, \ldots , x'_ m\} $. Since $x'_ j \to X$ is quasi-compact there are finitely many points of $U$ lying over $x'_ j$ (Lemma 68.4.5). Hence $U'$ has finitely many irreducible components, which implies that the closures of these irreducible components are the irreducible components of $U$. Thus (2) holds.

Assume (2). This implies (1) and the final statement by Lemma 68.20.3. (We also use that a reasonable algebraic space is decent, see discussion following Definition 68.6.1.) $\square$