**Proof.**
Assume $f$ is locally quasi-finite. Let $\mathop{\mathrm{Spec}}(k) \to Y$ be as in (2). Choose a surjective étale morphism $U \to X$ where $U$ is a scheme. Then $U_ k = \mathop{\mathrm{Spec}}(k) \times _ Y U \to X_ k$ is an étale morphism of algebraic spaces by Properties of Spaces, Lemma 65.16.5. By Lemma 66.27.4 we see that $X_ k \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. By definition this means that $U_ k \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. Hence $|U_ k|$ is discrete by Morphisms, Lemma 29.20.8. Since $|U_ k| \to |X_ k|$ is surjective and open we conclude that $|X_ k|$ is discrete.

Conversely, assume (2). Choose a surjective étale morphism $V \to Y$ where $V$ is a scheme. Choose a surjective étale morphism $U \to V \times _ Y X$ where $U$ is a scheme. Note that $U \to V$ is locally of finite type as $f$ is locally of finite type. Picture

\[ \xymatrix{ U \ar[r] \ar[rd] & X \times _ Y V \ar[d] \ar[r] & V \ar[d] \\ & X \ar[r] & Y } \]

If $f$ is not locally quasi-finite then $U \to V$ is not locally quasi-finite. Hence there exists a specialization $u \leadsto u'$ for some $u, u' \in U$ lying over the same point $v \in V$, see Morphisms, Lemma 29.20.6. We claim that $u, u'$ do not have the same image in $X_ v = \mathop{\mathrm{Spec}}(\kappa (v)) \times _ Y X$ which will contradict the assumption that $|X_ v|$ is discrete as desired. Let $d = \text{trdeg}_{\kappa (v)}(\kappa (u))$ and $d' = \text{trdeg}_{\kappa (v)}(\kappa (u'))$. Then we see that $d > d'$ by Morphisms, Lemma 29.28.7. Note that $U_ v$ (the fibre of $U \to V$ over $v$) is the fibre product of $U$ and $X_ v$ over $X \times _ Y V$, hence $U_ v \to X_ v$ is étale (as a base change of the étale morphism $U \to X \times _ Y V$). If $u, u' \in U_ v$ map to the same element of $|X_ v|$ then there exists a point $r \in R_ v = U_ v \times _{X_ v} U_ v$ with $t(r) = u$ and $s(r) = u'$, see Properties of Spaces, Lemma 65.4.3. Note that $s, t : R_ v \to U_ v$ are étale morphisms of schemes over $\kappa (v)$, hence $\kappa (u) \subset \kappa (r) \supset \kappa (u')$ are finite separable extensions of fields over $\kappa (v)$ (see Morphisms, Lemma 29.36.7). We conclude that the transcendence degrees are equal. This contradiction finishes the proof.
$\square$

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