Lemma 37.55.2. Let $f : X \to S$ be a morphism of finite presentation between quasi-compact and quasi-separated schemes. Then there exists a $t \geq 0$ and closed subschemes
\[ S \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset \]
such that
$S_ i \to S$ is defined by a finite type ideal sheaf,
$S_0 \subset S$ is a thickening,
for each $i$ there exists a surjective finite locally free morphism $T_ i \to S_ i \setminus S_{i + 1}$,
for each $i$ there exists a $t_ i \geq 0$ and closed subschemes
\[ X_ i = X \times _ S T_ i \supset Z_{i, 0} \supset Z_{i, 1} \supset \ldots \supset Z_{i, t_ i} = \emptyset \]
such that $Z_{i, j} \to X_ i$ is defined by a finite type ideal sheaf, $Z_{i, 0} \subset X_ i$ is a thickening, and such that the morphism $Z_{i, j} \setminus Z_{i, j + 1} \to T_ i$ is smooth.
Proof.
We can find a cartesian diagram
\[ \xymatrix{ X \ar[d] \ar[r] & X_0 \ar[d] \\ S \ar[r] & S_0 } \]
such that $X_0$ and $S_0$ are of finite type over $\mathbf{Z}$. See Limits, Proposition 32.5.4 and Lemma 32.10.1. Thus we may assume $X$ and $S$ are of finite type over $\mathbf{Z}$. Namely, a solution of the problem posed by the lemma for $X_0 \to S_0$ will base change to a solution over $S$; details omitted.
Assume $X$ and $S$ are of finite type over $\mathbf{Z}$. In this case every quasi-coherent ideal is of finite type, hence we do not have to check the condition that $S_ i$ is cut out by a finite type ideal. Set $S_0 = S_{red}$ equal to the reduction of $S$. Let $\eta \in S_0$ be a generic point of an irreducible component. By Lemma 37.55.1 we can find an open subscheme $U \subset S_0$, a surjective, universally injective, finite locally free morphism $V \to U$, a $t_0 \geq 0$ and closed subschemes
\[ X \times _ S V \supset Z_{0, 0} \supset Z_{0, 1} \supset \ldots \supset Z_{0, t_0} = \emptyset \]
such that $Z_{0, i} \to X \times _ S V$ is defined by a finite type ideal sheaf, $Z_{0, 0} \subset X \times _ S V$ is a thickening, and such that the morphism $Z_{0, i} \setminus Z_{0, i + 1} \to V$ is smooth. Then we let $S_1 \subset S_0$ be the reduced induced subscheme structure on $S_0 \setminus U$. By Noetherian induction on the underlying topological space of $S$, we may assume that the lemma holds for $X \times _ S S_1 \to S_1$. This produces $t \geq 1$ and
\[ S_1 = S_1 \supset S_2 \supset \ldots \supset S_ t = \emptyset \]
and $t_ i$ and $Z_{i, j}$ as in the statement of the lemma. This proves the lemma.
$\square$
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