Lemma 40.10.10. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(k)$ with $k$ a field. Assume $s, t$ are locally of finite type. Then $\dim (R) = \dim (G)$ where $G$ is the stabilizer group scheme of $R$.

Proof. Let $Z \subset R$ be the irreducible component passing through $e$ (see Lemma 40.10.8) thought of as an integral closed subscheme of $R$. Let $k'_ s$, resp. $k'_ t$ be the integral closure of $s(k)$, resp. $t(k)$ in $\Gamma (Z, \mathcal{O}_ Z)$. Recall that $k'_ s$ and $k'_ t$ are fields, see Varieties, Lemma 33.28.4. By Varieties, Proposition 33.31.1 we have $k'_ s = k'_ t$ as subrings of $\Gamma (Z, \mathcal{O}_ Z)$. As $e$ factors through $Z$ we obtain a commutative diagram

$\xymatrix{ k \ar[rd]_ t \ar[rrd]^1 \\ & \Gamma (Z, \mathcal{O}_ Z) \ar[r]^ e & k \\ k \ar[ru]^ s \ar[rru]_1 }$

This on the one hand shows that $k'_ s = s(k)$, $k'_ t = t(k)$, so $s(k) = t(k)$, which combined with the diagram above implies that $s = t$! In other words, we conclude that $Z$ is a closed subscheme of $G = R \times _{(t, s), U \times _ S U, \Delta } U$. The lemma follows as both $G$ and $R$ are equidimensional, see Lemma 40.10.9 and Groupoids, Lemma 39.8.1. $\square$

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