Lemma 39.23.3. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $s, t : R \to U$ finite locally free. Then

\[ U = \coprod \nolimits _{r \geq 1} U_ r \]

is a disjoint union of $R$-invariant opens such that the restriction $R_ r$ of $R$ to $U_ r$ has the property that $s, t : R_ r \to U_ r$ are finite locally free of rank $r$.

**Proof.**
By Morphisms, Lemma 29.48.5 there exists a decomposition $U = \coprod \nolimits _{r \geq 0} U_ r$ such that $s : s^{-1}(U_ r) \to U_ r$ is finite locally free of rank $r$. As $s$ is surjective we see that $U_0 = \emptyset $. Note that $u \in U_ r \Leftrightarrow $ if and only if the scheme theoretic fibre $s^{-1}(u)$ has degree $r$ over $\kappa (u)$. Now, if $z \in R$ with $s(z) = u$ and $t(z) = u'$ then using notation as in Lemma 39.13.4

\[ \text{pr}_1^{-1}(z) \to \mathop{\mathrm{Spec}}(\kappa (z)) \]

is the base change of both $s^{-1}(u) \to \mathop{\mathrm{Spec}}(\kappa (u))$ and $s^{-1}(u') \to \mathop{\mathrm{Spec}}(\kappa (u'))$ by the lemma cited. Hence $u \in U_ r \Leftrightarrow u' \in U_ r$, in other words, the open subsets $U_ r$ are $R$-invariant. In particular the restriction of $R$ to $U_ r$ is just $s^{-1}(U_ r)$ and $s : R_ r \to U_ r$ is finite locally free of rank $r$. As $t : R_ r \to U_ r$ is isomorphic to $s$ by the inverse of $R_ r$ we see that it has also rank $r$.
$\square$

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