The Stacks project

Proof. Recall that (3) is the definition of (2), see Algebraic Stacks, Definition 94.12.2. The implication (3) $\Rightarrow$ (1) is Lemma 101.4.14. Assume (1). Let $x \in |\mathcal{X}|$ be a finite type point. We will produce a scheme over $\mathcal{X}$ which “works” in a neighbourhood of $x$. At the end of the proof we will take the disjoint union of all of these to conclude.

By Lemma 101.18.7 the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ exists and $\mathcal{Z}_ x \to \mathcal{X}$ is locally of finite type. By Lemma 101.4.16 the algebraic stack $\mathcal{Z}_ x$ is DM. By Lemma 101.21.4 there exists a field $k$ and a surjective étale morphism $z : \mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$. In particular the composition $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is locally of finite type (by Morphisms of Spaces, Lemmas 67.23.2 and 67.39.9).

Pick a scheme $U$ and a smooth morphism $U \to \mathcal{X}$ such that $x$ is in the image of $|U| \to |\mathcal{X}|$. Consider the following fibre square

in other words $F = U \times _{\mathcal{X}, x} \mathop{\mathrm{Spec}}(k)$. By Properties of Stacks, Lemma 100.4.3 we see that $F$ is nonempty. As $\mathcal{Z}_ x \to \mathcal{X}$ is a monomorphism we have

with étale projection maps to $\mathop{\mathrm{Spec}}(k)$ by construction of $z$. Since

we see that the projections maps $F \times _ U F \to F$ are étale as well. It follows that $\Delta _{F/U} : F \to F \times _ U F$ is étale (see Morphisms of Spaces, Lemma 67.39.11). By Morphisms of Spaces, Lemma 67.51.2 this implies that $\Delta _{F/U}$ is an open immersion, which finally implies by Morphisms of Spaces, Lemma 67.38.9 that $F \to U$ is unramified.

Pick a nonempty affine scheme $V$ and an étale morphism $V \to F$. (This could be avoided by working directly with $F$, but it seems easier to explain what's going on by doing so.) Picture

Then $V \to \mathop{\mathrm{Spec}}(k)$ is a smooth morphism of schemes and $V \to U$ is an unramified morphism of schemes (see Morphisms of Spaces, Lemmas 67.37.2 and 67.38.3). Pick a closed point $v \in V$ with $k \subset \kappa (v)$ finite separable, see Varieties, Lemma 33.25.6. Let $u \in U$ be the image point. The local ring $\mathcal{O}_{V, v}$ is regular (see Varieties, Lemma 33.25.3) and the local ring homomorphism

coming from the morphism $V \to U$ is such that $\varphi (\mathfrak m_ u)\mathcal{O}_{V, v} = \mathfrak m_ v$, see Morphisms, Lemma 29.35.14. Hence we can find $f_1, \ldots , f_ d \in \mathcal{O}_{U, u}$ such that the images $\varphi (f_1), \ldots , \varphi (f_ d)$ form a basis for $\mathfrak m_ v/\mathfrak m_ v^2$ over $\kappa (v)$. Since $\mathcal{O}_{V, v}$ is a regular local ring this implies that $\varphi (f_1), \ldots , \varphi (f_ d)$ form a regular sequence in $\mathcal{O}_{V, v}$ (see Algebra, Lemma 10.106.3). After replacing $U$ by an open neighbourhood of $u$ we may assume $f_1, \ldots , f_ d \in \Gamma (U, \mathcal{O}_ U)$. After replacing $U$ by a possibly even smaller open neighbourhood of $u$ we may assume that $V(f_1, \ldots , f_ d) \to \mathcal{X}$ is flat and locally of finite presentation, see Lemma 101.21.1. By construction

is étale and $V(f_1, \ldots , f_ d) \times _ U V$ is the closed subscheme $T \subset V$ cut out by $f_1|_ V, \ldots , f_ d|_ V$. Hence by construction $v \in T$ and

a finite separable extension of $k$. It follows that $T \to \mathop{\mathrm{Spec}}(k)$ is unramified at $v$, see Morphisms, Lemma 29.35.14. By definition of an unramified morphism of algebraic spaces this means that $V(f_1, \ldots , f_ d) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is unramified at the image of $v$ in $V(f_1, \ldots , f_ d) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$. Applying Lemma 101.21.5 we see that on shrinking $U$ to yet another open neighbourhood of $u$ the morphism $V(f_1, \ldots , f_ d) \to \mathcal{X}$ is étale.

We conclude that for every finite type point $x$ of $\mathcal{X}$ there exists an étale morphism $f_ x : W_ x \to \mathcal{X}$ with $x$ in the image of $|f_ x|$. Set $W = \coprod _ x W_ x$ and $f = \coprod f_ x$. Then $f$ is étale. In particular the image of $|f|$ is open, see Properties of Stacks, Lemma 100.4.7. By construction the image contains all finite type points of $\mathcal{X}$, hence $f$ is surjective by Lemma 101.18.6 (and Properties of Stacks, Lemma 100.4.4). $\square$