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 $r$. $\square$

Comment #1540 by jojo on

I think that the conclusion of this lemma should be that $s,t : R_r \to U_r$ are finite locally free of rank $r$ not $1$.

Comment #1541 by jojo on

You explain that $pr_{1}(z) \to \text{Spec}(\kappa(z))$ is a base change of

$$s^{-1}(u) \to \text{Spec}(\kappa(u)) \;\;\text{ by }\;\; $\text{Spec}(\kappa(z)) \to \text{Spec}(\kappa(u))$ and of$$ s^{-1}(u') \to \text{Spec}(\kappa(u')) \;\;\text{ by }\;\; $\text{Spec}(\kappa(z)) \to \text{Spec}(\kappa(u'))$$Instead of concluding directly that it means that $U_r$ is $R$-stable I think that it might be nice to add the following short argument : The first base change implies that $pr_1$ is of rank $r$ over $z$ because $s$ is of rank $r$ above $u$. Now the second base change implies that $s$ is of rank $r$ over $u'$ because $s$ is faithfully flat and affine (because surjective flat and finite). So this means that $u'$ is in $U_r$ which in turn implies that $U_r$ is $R$-stable. Comment #1568 by on Thanks for the typo. I do not agree entirely with your second comment because I think faithfully flat descent isn't needed here because we already have the opens $U_r$. In fact, in some cases one can obtain invariant loci like this without assuming that $s$ and $t$ are flat; take a look at the (unfortunately rather hard to parse) Section 40.6. For the edits corresponding to your comments, see here. Thanks very much! There are also: • 6 comment(s) on Section 39.23: Finite flat groupoids, affine case ## Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi\$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

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