Lemma 40.14.1. Let $(U, R, s, t, c)$ be a groupoid scheme over a scheme $S$. Assume $s, t$ are finite. There exists a sequence of $R$-invariant closed subschemes
such that $\bigcap Z_ r = \emptyset $ and such that $s^{-1}(Z_{r - 1}) \setminus s^{-1}(Z_ r) \to Z_{r - 1} \setminus Z_ r$ is finite locally free of rank $r$.
Proof. Let $\{ Z_ r\} $ be the stratification of $U$ given by the Fitting ideals of the finite type quasi-coherent modules $s_*\mathcal{O}_ R$. See Divisors, Lemma 31.9.6. Since the identity $e : U \to R$ is a section to $s$ we see that $s_*\mathcal{O}_ R$ contains $\mathcal{O}_ S$ as a direct summand. Hence $U = Z_{-1} = Z_0$ (details omitted). Since formation of Fitting ideals commutes with base change (More on Algebra, Lemma 15.8.4) we find that $s^{-1}(Z_ r)$ corresponds to the $r$th Fitting ideal of $\text{pr}_{1, *}\mathcal{O}_{R \times _{s, U, t} R}$ because the lower right square of diagram (40.3.0.1) is cartesian. Using the fact that the lower left square is also cartesian we conclude that $s^{-1}(Z_ r) = t^{-1}(Z_ r)$, in other words $Z_ r$ is $R$-invariant. The morphism $s^{-1}(Z_{r - 1}) \setminus s^{-1}(Z_ r) \to Z_{r - 1} \setminus Z_ r$ is finite locally free of rank $r$ because the module $s_*\mathcal{O}_ R$ pulls back to a finite locally free module of rank $r$ on $Z_{r - 1} \setminus Z_ r$ by Divisors, Lemma 31.9.6. $\square$
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