The Stacks project

Lemma 66.8.1. Let $S$ be a scheme. Let $W \to X$ be a morphism of a scheme $W$ to an algebraic space $X$ which is flat, locally of finite presentation, separated, locally quasi-finite with universally bounded fibres. There exist reduced closed subspaces

\[ \emptyset = Z_{-1} \subset Z_0 \subset Z_1 \subset Z_2 \subset \ldots \subset Z_ n = X \]

such that with $X_ r = Z_ r \setminus Z_{r - 1}$ the stratification $X = \coprod _{r = 0, \ldots , n} X_ r$ is characterized by the following universal property: Given $g : T \to X$ the projection $W \times _ X T \to T$ is finite locally free of degree $r$ if and only if $g(|T|) \subset |X_ r|$.

Proof. Let $n$ be an integer bounding the degrees of the fibres of $W \to X$. Choose a scheme $U$ and a surjective ├ętale morphism $U \to X$. Apply More on Morphisms, Lemma 37.40.3 to $W \times _ X U \to U$. We obtain closed subsets

\[ \emptyset = Y_{-1} \subset Y_0 \subset Y_1 \subset Y_2 \subset \ldots \subset Y_ n = U \]

characterized by the property stated in the lemma for the morphism $W \times _ X U \to U$. Clearly, the formation of these closed subsets commutes with base change. Setting $R = U \times _ X U$ with projection maps $s, t : R \to U$ we conclude that

\[ s^{-1}(Y_ r) = t^{-1}(Y_ r) \]

as closed subsets of $R$. In other words the closed subsets $Y_ r \subset U$ are $R$-invariant. This means that $|Y_ r|$ is the inverse image of a closed subset $Z_ r \subset |X|$. Denote $Z_ r \subset X$ also the reduced induced algebraic space structure, see Properties of Spaces, Definition 64.12.6.

Let $g : T \to X$ be a morphism of algebraic spaces. Choose a scheme $V$ and a surjective ├ętale morphism $V \to T$. To prove the final assertion of the lemma it suffices to prove the assertion for the composition $V \to X$ (by our definition of finite locally free morphisms, see Morphisms of Spaces, Section 65.46). Similarly, the morphism of schemes $W \times _ X V \to V$ is finite locally free of degree $r$ if and only if the morphism of schemes

\[ W \times _ X (U \times _ X V) \longrightarrow U \times _ X V \]

is finite locally free of degree $r$ (see Descent, Lemma 35.20.30). By construction this happens if and only if $|U \times _ X V| \to |U|$ maps into $|Y_ r|$, which is true if and only if $|V| \to |X|$ maps into $|Z_ r|$. $\square$


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