The Stacks project

Lemma 79.9.11. In Situation 79.9.2 assume $s, t$ are locally of finite type. Let $G = \mathop{\mathrm{Spec}}(k) \times _{\Delta , \mathop{\mathrm{Spec}}(k) \times _ B \mathop{\mathrm{Spec}}(k), t \times s} R$ be the stabilizer group algebraic space. Then we have $\dim (R) = \dim (G)$.

Proof. Since $G$ and $R$ are equidimensional (see Lemmas 79.9.9 and 79.9.10) it suffices to prove that $\dim _ e(R) = \dim _ e(G)$. Let $V$ be an affine scheme, $v \in V$, and let $\varphi : V \to R$ be an ├ętale morphism of schemes such that $\varphi (v) = e$. Note that $V$ is a Noetherian scheme as $s \circ \varphi $ is locally of finite type as a composition of morphisms locally of finite type and as $V$ is quasi-compact (use Morphisms of Spaces, Lemmas 67.23.2, 67.39.8, and 67.28.5 and Morphisms, Lemma 29.15.6). Hence $V$ is locally connected (see Properties, Lemma 28.5.5 and Topology, Lemma 5.9.6). Thus we may replace $V$ by the connected component containing $v$ (it is still affine as it is an open and closed subscheme of $V$). Set $T = V_{red}$ equal to the reduction of $V$. Consider the two morphisms $a, b : T \to \mathop{\mathrm{Spec}}(k)$ given by $a = s \circ \varphi |_ T$ and $b = t \circ \varphi |_ T$. Note that $a, b$ induce the same field map $k \to \kappa (v)$ because $\varphi (v) = e$! Let $k_ a \subset \Gamma (T, \mathcal{O}_ T)$ be the integral closure of $a^\sharp (k) \subset \Gamma (T, \mathcal{O}_ T)$. Similarly, let $k_ b \subset \Gamma (T, \mathcal{O}_ T)$ be the integral closure of $b^\sharp (k) \subset \Gamma (T, \mathcal{O}_ T)$. By Varieties, Proposition 33.31.1 we see that $k_ a = k_ b$. Thus we obtain the following commutative diagram

\[ \xymatrix{ k \ar[rd]^ a \ar[rrrd] \\ & k_ a = k_ b \ar[r] & \Gamma (T, \mathcal{O}_ T) \ar[r] & \kappa (v) \\ k \ar[ru]_ b \ar[rrru] } \]

As discussed above the long arrows are equal. Since $k_ a = k_ b \to \kappa (v)$ is injective we conclude that the two morphisms $a$ and $b$ agree. Hence $T \to R$ factors through $G$. It follows that $R_{red} = G_{red}$ in an open neighbourhood of $e$ which certainly implies that $\dim _ e(R) = \dim _ e(G)$. $\square$

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