Lemma 68.7.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \to X$ be an étale morphism from a scheme to $X$. Assume $u, u' \in |U|$ map to the same point $x$ of $|X|$, and $u' \leadsto u$. If the pair $(X, x)$ satisfies the equivalent conditions of Lemma 68.4.2 then $u = u'$.

## 68.7 Points and specializations

There exists an étale morphism of algebraic spaces $f : X \to Y$ and a nontrivial specialization between points in a fibre of $|f| : |X| \to |Y|$, see Examples, Lemma 110.50.1. If the source of the morphism is a scheme we can avoid this by imposing condition ($\alpha $) on $Y$.

**Proof.**
Assume the pair $(X, x)$ satisfies the equivalent conditions for Lemma 68.4.2. Let $U$ be a scheme, $U \to X$ étale, and let $u, u' \in |U|$ map to $x$ of $|X|$, and $u' \leadsto u$. We may and do replace $U$ by an affine neighbourhood of $u$. Let $t, s : R = U \times _ X U \to U$ be the étale projection maps.

Pick a point $r \in R$ with $t(r) = u$ and $s(r) = u'$. This is possible by Properties of Spaces, Lemma 66.4.5. Because generalizations lift along the étale morphism $t$ (Remark 68.4.1) we can find a specialization $r' \leadsto r$ with $t(r') = u'$. Set $u'' = s(r')$. Then $u'' \leadsto u'$. Thus we may repeat and find $r'' \leadsto r'$ with $t(r'') = u''$. Set $u''' = s(r'')$, and so on. Here is a picture:

In Remark 68.4.1 we have seen that there are no specializations among points in the fibres of the étale morphism $s$. Hence if $u^{(n + 1)} = u^{(n)}$ for some $n$, then also $r^{(n)} = r^{(n - 1)}$ and hence also (by taking $t$) $u^{(n)} = u^{(n - 1)}$. This then forces the whole tower to collapse, in particular $u = u'$. Thus we see that if $u \not= u'$, then all the specializations are strict and $\{ u, u', u'', \ldots \} $ is an infinite set of points in $U$ which map to the point $x$ in $|X|$. As we chose $U$ affine this contradicts the second part of Lemma 68.4.2, as desired. $\square$

Lemma 68.7.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \to X$ be an étale morphism from a scheme to $X$. Assume $u, u' \in |U|$ map to the same point $x$ of $|X|$, and $u' \leadsto u$. If $X$ is locally Noetherian, then $u = u'$.

**Proof.**
The discussion in Schemes, Section 26.13 shows that $\mathcal{O}_{U, u'}$ is a localization of the Noetherian local ring $\mathcal{O}_{U, u}$. By Properties of Spaces, Lemma 66.10.1 we have $\dim (\mathcal{O}_{U, u}) = \dim (\mathcal{O}_{U, u'})$. By dimension theory for Noetherian local rings we conclude $u = u'$.
$\square$

Lemma 68.7.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Assume the pair $(X, x')$ satisfies the equivalent conditions of Lemma 68.4.5. 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 $\varphi (u') = x'$.

**Proof.**
We may replace $U$ by an affine open neighbourhood of $u$. Hence we may assume that $U$ is affine. As $x$ is in the image of the open map $|U| \to |X|$, so is $x'$. Thus we may replace $X$ by the Zariski open subspace corresponding to the image of $|U| \to |X|$, see Properties of Spaces, Lemma 66.4.10. In other words we may assume that $U \to X$ is surjective and étale. Let $s, t : R = U \times _ X U \to U$ be the projections. By our assumption that $(X, x')$ satisfies the equivalent conditions of Lemma 68.4.5 we see that the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x'$ are finite. Say $\{ u'_1, \ldots , u'_ n\} \subset U$ and $\{ r'_1, \ldots , r'_ m\} \subset R$ form the complete inverse image of $\{ x'\} $. Consider the closed sets

Trivially we have $s(T') \subset T$. Because $R$ is an equivalence relation we also have $t(T') = s(T')$ as the set $\{ r_ j'\} $ is invariant under the inverse of $R$ by construction. Let $w \in T$ be any point. Then $u'_ i \leadsto w$ for some $i$. Choose $r \in R$ with $s(r) = w$. Since generalizations lift along $s : R \to U$, see Remark 68.4.1, we can find $r' \leadsto r$ with $s(r') = u_ i'$. Then $r' = r'_ j$ for some $j$ and we conclude that $w \in s(T')$. Hence $T = s(T') = t(T')$ is an $|R|$-invariant closed set in $|U|$. This means $T$ is the inverse image of a closed (!) subset $T'' = \varphi (T)$ of $|X|$, see Properties of Spaces, Lemmas 66.4.5 and 66.4.6. Hence $T'' = \overline{\{ x'\} }$. Thus $T$ contains some point $u_1$ mapping to $x$ as $x \in T''$. I.e., we see that for some $i$ there exists a specialization $u'_ i \leadsto u_1$ which maps to the given specialization $x' \leadsto x$.

To finish the proof, choose a point $r \in R$ such that $s(r) = u$ and $t(r) = u_1$ (using Properties of Spaces, Lemma 66.4.3). As generalizations lift along $t$, and $u'_ i \leadsto u_1$ we can find a specialization $r' \leadsto r$ such that $t(r') = u'_ i$. Set $u' = s(r')$. Then $u' \leadsto u$ and $\varphi (u') = x'$ as desired. $\square$

Lemma 68.7.4. Let $S$ be a scheme. Let $f : Y \to X$ be a flat morphism of algebraic spaces over $S$. Let $x, x' \in |X|$ and assume $x' \leadsto x$, i.e., $x$ is a specialization of $x'$. Assume the pair $(X, x')$ satisfies the equivalent conditions of Lemma 68.4.5 (for example if $X$ is decent, $X$ is quasi-separated, or $X$ is representable). Then for every $y \in |Y|$ with $f(y) = x$, there exists a point $y' \in |Y|$, $y' \leadsto y$ with $f(y') = x'$.

**Proof.**
(The parenthetical statement holds by the definition of decent spaces and the implications between the different separation conditions mentioned in Section 68.6.) Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose $v \in V$ mapping to $y$. Then we see that it suffices to prove the lemma for $V \to X$. Thus we may assume $Y$ is a scheme. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Choose $u \in U$ mapping to $x$. By Lemma 68.7.3 we may choose $u' \leadsto u$ mapping to $x'$. By Properties of Spaces, Lemma 66.4.3 we may choose $z \in U \times _ X Y$ mapping to $y$ and $u$. Thus we reduce to the case of the flat morphism of schemes $U \times _ X Y \to U$ which is Morphisms, Lemma 29.25.9.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2254 by Antoine Chambert-Loir on

Comment #2288 by Johan on