The Stacks project

Proof. We may view $X$ as an algebraic space over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Spaces, Definition 65.16.2 and Properties of Spaces, Definition 66.3.1. Thus we may apply Proposition 70.8.1. It follows that we can find an affine morphism $X \to X_0$ with $X_0$ of finite presentation over $\mathbf{Z}$. If we can prove the lemma for $X_0$, then we can pull back the stratification and the morphisms to $X$ and get the result for $X$; some details omitted. This reduces us to the case discussed in the next paragraph.

Assume $X$ is of finite presentation over $\mathbf{Z}$. Then $X$ is Noetherian and $|X|$ is a Noetherian topological space (with finitely many irreducible components) of finite dimension. Hence we may use induction on $\dim (|X|)$. Any finite morphism towards $X$ is of finite presentation, so we can ignore that requirement in the rest of the proof. By Lemma 70.16.3 there exists a surjective finite morphism $Y \to X$ which is finite étale over a dense open $U \subset X$. Set $Z_0 = X$ and let $Z_1 \subset X$ be the reduced closed subspace with $|Z_1| = |X| \setminus |U|$. By induction we find an integer $t \geq 0$ and a filtration

by closed subspaces, where $Z_{1, 0} \to Z_1$ is a thickening and there exist finite surjective morphisms $Y_{1, i} \to Z_{1, i}$ which are finite étale over $Z_{1, i} \setminus Z_{1, i + 1}$. Since $Z_1$ is reduced, we have $Z_1 = Z_{1, 0}$. Hence we can set $Z_ i = Z_{1, i - 1}$ and $Y_ i = Y_{1, i - 1}$ for $i \geq 1$ and the lemma is proved. $\square$