The Stacks project

Proof. By Properties of Spaces, Lemma 66.6.3 we can choose an affine scheme $U$ and a surjective étale morphism $U \to X$. For $p \geq 0$ set

where the fibre product has $p$ factors. Since $U$ is separated, the morphism $U \to X$ is separated and all fibre products $U \times _ X \ldots \times _ X U$ are separated schemes. Since $U \to X$ is separated the diagonal $U \to U \times _ X U$ is a closed immersion. Since $U \to X$ is étale the diagonal $U \to U \times _ X U$ is an open immersion, see Morphisms of Spaces, Lemmas 67.39.10 and 67.38.9. Similarly, all the diagonal morphisms are open and closed immersions and $W_ p$ is an open and closed subscheme of $U \times _ X \ldots \times _ X U$. Moreover, the morphism

is locally quasi-finite and separated (Morphisms of Spaces, Lemma 67.4.5) and its target is an affine scheme. Hence every finite set of points of $U \times _ X \ldots \times _ X U$ is contained in an affine open, see More on Morphisms, Lemma 37.45.1. Therefore, the same is true for $W_ p$. There is a free action of the symmetric group $S_ p$ on $W_ p$ over $X$ (because we threw out the fix point locus from $U \times _ X \ldots \times _ X U$). By the above and Properties of Spaces, Proposition 66.14.1 the quotient $V_ p = W_ p/S_ p$ is a scheme. Since the action of $S_ p$ on $W_ p$ was over $X$, there is a morphism $V_ p \to X$. Since $W_ p \to X$ is étale and since $W_ p \to V_ p$ is surjective étale, it follows that also $V_ p \to X$ is étale, see Properties of Spaces, Lemma 66.16.3. Observe that $V_ p$ is a separated scheme by Properties of Spaces, Lemma 66.14.3.

We let $U_ p \subset X$ be the open subspace which is the image of $V_ p \to X$. By construction a morphism $\mathop{\mathrm{Spec}}(k) \to X$ with $k$ algebraically closed, factors through $U_ p$ if and only if $U \times _ X \mathop{\mathrm{Spec}}(k)$ has $\geq p$ points; as usual observe that $U \times _ X \mathop{\mathrm{Spec}}(k)$ is scheme theoretically a disjoint union of (possibly infinitely many) copies of $\mathop{\mathrm{Spec}}(k)$, see Remark 68.4.1. It follows that the $U_ p$ give a filtration of $X$ as stated in the lemma. Moreover, our morphism $\mathop{\mathrm{Spec}}(k) \to X$ factors through $T_ p$ if and only if $U \times _ X \mathop{\mathrm{Spec}}(k)$ has exactly $p$ points. In this case we see that $V_ p \times _ X \mathop{\mathrm{Spec}}(k)$ has exactly one point. Set $Z_ p = f_ p^{-1}(T_ p) \subset V_ p$. This is a closed subscheme of $V_ p$. Then $Z_ p \to T_ p$ is an étale morphism between algebraic spaces which induces a bijection on $k$-valued points for any algebraically closed field $k$. To be sure this implies that $Z_ p \to T_ p$ is universally injective, whence an open immersion by Morphisms of Spaces, Lemma 67.51.2 hence an isomorphism and (1) has been proved.

Let $x : \mathop{\mathrm{Spec}}(k) \to X$ be a quasi-compact morphism where $k$ is a field. Then the composition $\mathop{\mathrm{Spec}}(\overline{k}) \to \mathop{\mathrm{Spec}}(k) \to X$ is quasi-compact as well (Morphisms of Spaces, Lemma 67.8.5). In this case the scheme $U \times _ X \mathop{\mathrm{Spec}}(\overline{k})$ is quasi-compact. In view of the fact (seen above) that it is a disjoint union of copies of $\mathop{\mathrm{Spec}}(\overline{k})$ we find that it has finitely many points. If the number of points is $p$, then we see that indeed $x \in T_ p$ and the proof is finished. $\square$