Lemma 40.12.3. Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid scheme over $S$. Let $G \to U$ be the stabilizer group scheme. Assume $s$ and $t$ are Cohen-Macaulay and locally of finite presentation. Let $u \in U$ be a finite type point of the scheme $U$, see Morphisms, Definition 29.16.3. With notation as in Situation 40.12.1 there exist an affine scheme $U'$ and a morphism $g : U' \to U$ such that

$g$ is an immersion,

$u \in U'$,

$g$ is locally of finite presentation,

the morphism $h : U' \times _{g, U, t} R \longrightarrow U$ is Cohen-Macaulay and locally of finite presentation,

the morphisms $s', t' : R' \to U'$ are Cohen-Macaulay and locally of finite presentation, and

$\dim _{e(u)}(F'_ u) = \dim (G'_ u)$.

**Proof.**
As $s$ is locally of finite presentation the scheme $F_ u$ is locally of finite type over $\kappa (u)$. Hence $\dim _{e(u)}(F_ u) < \infty $ and we may argue by induction on $\dim _{e(u)}(F_ u)$.

If $\dim _{e(u)}(F_ u) = \dim (G_ u)$ there is nothing to prove. Assume $\dim _{e(u)}(F_ u) > \dim (G_ u)$. This means that Lemma 40.12.2 applies and we find a morphism $g : U' \to U$ which has properties (1), (2), (3), instead of (6) we have $\dim _{e(u)}(F'_ u) < \dim _{e(u)}(F_ u)$, and instead of (4) and (5) we have that the composition

\[ h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \longrightarrow U \]

is Cohen-Macaulay at the point $(u, e(u))$. We apply Remark 40.6.3 and we obtain an open subscheme $U'' \subset U'$ such that $U'' \times _{g, U, t} R \subset U' \times _{g, U, t} R$ is the largest open subscheme on which $h$ is Cohen-Macaulay. Since $(u, e(u)) \in U'' \times _{g, U, t} R$ we see that $u \in U''$. Hence we may replace $U'$ by $U''$ and assume that in fact $h$ is Cohen-Macaulay everywhere! By Lemma 40.9.2 we conclude that $s', t'$ are locally of finite presentation and Cohen-Macaulay (use Morphisms, Lemma 29.21.4 and More on Morphisms, Lemma 37.22.6).

By construction $\dim _{e'(u)}(F'_ u) < \dim _{e(u)}(F_ u)$, so we may apply the induction hypothesis to $(U', R', s', t', c')$ and the point $u \in U'$. Note that $u$ is also a finite type point of $U'$ (for example you can see this using the characterization of finite type points from Morphisms, Lemma 29.16.4). Let $g' : U'' \to U'$ and $(U'', R'', s'', t'', c'')$ be the solution of the corresponding problem starting with $(U', R', s', t', c')$ and the point $u \in U'$. We claim that the composition

\[ g'' = g \circ g' : U'' \longrightarrow U \]

is a solution for the original problem. Properties (1), (2), (3), (5), and (6) are immediate. To see (4) note that the morphism

\[ h'' = s \circ \text{pr}_1 : U'' \times _{g'', U, t} R \longrightarrow U \]

is locally of finite presentation and Cohen-Macaulay by an application of Lemma 40.9.4 (use More on Morphisms, Lemma 37.22.11 to see that Cohen-Macaulay morphisms are fppf local on the target).
$\square$

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