The Stacks project

Proof. Assume (1). In this case the morphism of algebraic spaces $\mathcal{X}_ k \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite as a base change of $f$. Hence $|\mathcal{X}_ k|$ is discrete by Morphisms of Spaces, Lemma 67.27.5. Conversely, assume (2). Pick a surjective smooth morphism $V \to \mathcal{Y}$ where $V$ is a scheme. It suffices to show that the morphism of algebraic spaces $V \times _\mathcal {Y} \mathcal{X} \to V$ is locally quasi-finite, see Properties of Stacks, Lemma 100.3.3. The morphism $V \times _\mathcal {Y} \mathcal{X} \to V$ is locally of finite type by assumption. For any morphism $\mathop{\mathrm{Spec}}(k) \to V$ where $k$ is a field

has a discrete space of points by assumption. Hence we conclude that $V \times _\mathcal {Y} \mathcal{X} \to V$ is locally quasi-finite by Morphisms of Spaces, Lemma 67.27.5. $\square$

Proof. We have seen this for quasi-DM morphisms in Lemma 101.4.4 and for locally finite type morphisms in Lemma 101.17.3. It is immediate that the condition on fibres is inherited by a base change. $\square$

Proof. By Lemma 101.17.2 we see that $V \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type. Assume, to get a contradiction, that $V \to \mathop{\mathrm{Spec}}(k)$ is not locally quasi-finite. Then there exists a nontrivial specialization $v \leadsto v'$ of points of $V$, see Morphisms, Lemma 29.20.6. In particular $\text{trdeg}_ k(\kappa (v)) > \text{trdeg}_ k(\kappa (v'))$, see Morphisms, Lemma 29.28.7. Because $|\mathcal{X}|$ is discrete we see that $|f|(v) = |f|(v')$. Consider $R = V \times _\mathcal {X} V$. Then $R$ is an algebraic space and the projections $s, t : R \to V$ are locally quasi-finite as base changes of $V \to \mathcal{X}$ (which is representable by algebraic spaces so this follows from the discussion in Properties of Stacks, Section 100.3). By Properties of Stacks, Lemma 100.4.3 we see that there exists an $r \in |R|$ such that $s(r) = v$ and $t(r) = v'$. By Morphisms of Spaces, Lemma 67.33.3 we see that the transcendence degree of $v/k$ is equal to the transcendence degree of $r/k$ is equal to the transcendence degree of $v'/k$. This contradiction proves the lemma. $\square$

Proof. We have seen this for quasi-DM morphisms in Lemma 101.4.10 and for locally finite type morphisms in Lemma 101.17.2. Let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Y} \to \mathcal{Z}$ be locally quasi-finite. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ be a morphism. It suffices to show that $|\mathcal{X}_ k|$ is discrete. By Lemma 101.23.3 the morphisms $\mathcal{X}_ k \to \mathcal{Y}_ k$ and $\mathcal{Y}_ k \to \mathop{\mathrm{Spec}}(k)$ are locally quasi-finite. In particular we see that $\mathcal{Y}_ k$ is a quasi-DM algebraic stack, see Lemma 101.4.13. By Theorem 101.21.3 we can find a scheme $V$ and a surjective, flat, locally finitely presented, locally quasi-finite morphism $V \to \mathcal{Y}_ k$. By Lemma 101.23.4 we see that $V$ is locally quasi-finite over $k$, in particular $|V|$ is discrete. The morphism $V \times _{\mathcal{Y}_ k} \mathcal{X}_ k \to \mathcal{X}_ k$ is surjective, flat, and locally of finite presentation hence $|V \times _{\mathcal{Y}_ k} \mathcal{X}_ k| \to |\mathcal{X}_ k|$ is surjective and open. Thus it suffices to show that $|V \times _{\mathcal{Y}_ k} \mathcal{X}_ k|$ is discrete. Note that $V$ is a disjoint union of spectra of Artinian local $k$-algebras $A_ i$ with residue fields $k_ i$, see Varieties, Lemma 33.20.2. Thus it suffices to show that each

is discrete, which follows from the assumption that $\mathcal{X} \to \mathcal{Y}$ is locally quasi-finite. $\square$

Proof. The implication (2) $\Rightarrow $ (3) is immediate.

Assume (1) and let $V \to \mathcal{Y}$ be as in (2). Then $\mathcal{X} \times _\mathcal {Y} V \to V$ is quasi-DM, see Lemma 101.4.4. By Lemma 101.4.3 the algebraic space $V$ is DM, hence quasi-DM. Thus $\mathcal{X} \times _\mathcal {Y} V$ is quasi-DM by Lemma 101.4.11. Hence we may apply Theorem 101.21.3 to get the morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ as in (2).

Assume (3). Let $V \to \mathcal{Y}$ and $U \to \mathcal{X} \times _\mathcal {Y} V$ be as in (3). To prove that $f$ is quasi-DM it suffices to show that $\mathcal{X} \times _\mathcal {Y} V \to V$ is quasi-DM, see Lemma 101.4.5. By Lemma 101.4.14 we see that $\mathcal{X} \times _\mathcal {Y} V$ is quasi-DM. Hence $\mathcal{X} \times _\mathcal {Y} V \to V$ is quasi-DM by Lemma 101.4.13 and (1) holds. This finishes the proof of the lemma. $\square$

Proof. Assume (1). Then $f$ is quasi-DM by assumption. Let $V \to \mathcal{Y}$ and $U \to \mathcal{X} \times _\mathcal {Y} V$ be as in (2). By Lemma 101.23.5 the composition $U \to \mathcal{X} \times _\mathcal {Y} V \to V$ is locally quasi-finite. Thus (1) implies (2).

Assume (2). Let $V \to \mathcal{Y}$ be as in (3). By Lemma 101.23.6 we can find an algebraic space $U$ and a surjective, flat, locally finitely presented, locally quasi-finite morphism $U \to \mathcal{X} \times _\mathcal {Y} V$. By (2) the composition $U \to V$ is locally quasi-finite. Thus (2) implies (3).

It is immediate that (3) implies (4).

Assume (4). We will prove (1) holds, which finishes the proof. By Lemma 101.23.6 we see that $f$ is quasi-DM. To prove that $f$ is locally of finite type it suffices to prove that $g : \mathcal{X} \times _\mathcal {Y} V \to V$ is locally of finite type, see Lemma 101.17.6. Then it suffices to check that $g$ precomposed with $h : U \to \mathcal{X} \times _\mathcal {Y} V$ is locally of finite type, see Lemma 101.17.7. Since $g \circ h : U \to V$ was assumed to be locally quasi-finite this holds, hence $f$ is locally of finite type. Finally, let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ be a morphism. Then $V \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k)$ is a nonempty algebraic space which is locally of finite presentation over $k$. Hence we can find a finite extension $k'/k$ and a morphism $\mathop{\mathrm{Spec}}(k') \to V$ such that

commutes (details omitted). Then $\mathcal{X}_{k'} \to \mathcal{X}_ k$ is representable (by schemes), surjective, and finite locally free. In particular $|\mathcal{X}_{k'}| \to |\mathcal{X}_ k|$ is surjective and open. Thus it suffices to prove that $|\mathcal{X}_{k'}|$ is discrete. Since

we see that $U \times _ V \mathop{\mathrm{Spec}}(k') \to \mathcal{X}_{k'}$ is surjective, flat, and locally of finite presentation (as a base change of $U \to \mathcal{X} \times _\mathcal {Y} V$). Hence $|U \times _ V \mathop{\mathrm{Spec}}(k')| \to |\mathcal{X}_{k'}|$ is surjective and open. Thus it suffices to show that $|U \times _ V \mathop{\mathrm{Spec}}(k')|$ is discrete. This follows from the fact that $U \to V$ is locally quasi-finite (either by our definition above or from the original definition for morphisms of algebraic spaces, via Morphisms of Spaces, Lemma 67.27.5). $\square$

Proof. Write $\mathcal{X} \to \mathcal{Y}$ as the composition

The second arrow is locally quasi-finite as a base change of $\mathcal{X} \to \mathcal{Z}$, see Lemma 101.23.3. The first arrow is locally quasi-finite by Lemma 101.4.8 as $\mathcal{Y} \to \mathcal{Z}$ is quasi-DM. Hence $\mathcal{X} \to \mathcal{Y}$ is locally quasi-finite by Lemma 101.23.5. $\square$