Lemma 79.9.10. In Situation 79.9.1 assume $G$ locally of finite type. For all $g \in |G|$
$\dim (G) = \dim _ g(G)$,
if the transcendence degree of $g$ over $k$ is $0$, then $\dim (G) = \dim (\mathcal{O}_{G, \overline{g}})$.
Proof. Immediate from Lemma 79.9.9 via (79.9.2.1). $\square$
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