Lemma 110.50.1. There exists an étale morphism of algebraic spaces $f : X \to Y$ and a nontrivial specialization of points $x \leadsto x'$ in $|X|$ with $f(x) = f(x')$ in $|Y|$.
110.50 Specializations between points in fibre étale morphism
If $f : X \to Y$ is an étale, or more generally a locally quasi-finite morphism of schemes, then there are no specializations between points of fibres, see Morphisms, Lemma 29.20.8. However, for morphisms of algebraic spaces this doesn't hold in general.
To give an example, let $k$ be a field. Set
\[ P = k[u, u^{-1}, y, \{ x_ n\} _{n \in \mathbf{Z}}]. \]
Consider the action of $\mathbf{Z}$ on $P$ by $k$-algebra maps generated by the automorphism $\tau $ given by the rules $\tau (u) = u$, $\tau (y) = uy$, and $\tau (x_ n) = x_{n + 1}$. For $d \geq 1$ set $I_ d = ((1 - u^ d)y, x_ n - x_{n + d}, n \in \mathbf{Z})$. Then $V(I_ d) \subset \mathop{\mathrm{Spec}}(P)$ is the fix point locus of $\tau ^ d$. Let $S \subset P$ be the multiplicative subset generated by $y$ and all $1 - u^ d$, $d \in \mathbf{N}$. Then we see that $\mathbf{Z}$ acts freely on $U = \mathop{\mathrm{Spec}}(S^{-1}P)$. Let $X = U/\mathbf{Z}$ be the quotient algebraic space, see Spaces, Definition 65.14.4.
Consider the prime ideals $\mathfrak p_ n = (x_ n, x_{n + 1}, \ldots )$ in $S^{-1}P$. Note that $\tau (\mathfrak p_ n) = \mathfrak p_{n + 1}$. Hence each of these define point $\xi _ n \in U$ whose image in $X$ is the same point $x$ of $X$. Moreover we have the specializations
\[ \ldots \leadsto \xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \]
We conclude that $U \to X$ is an example of the promised type.
Proof. See discussion above. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)