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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