Lemma 108.47.1. 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 $|Y|$.

## 108.47 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

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 63.14.4.

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

We conclude that $U \to X$ is an example of the promised type.

**Proof.**
See discussion above.
$\square$

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