Lemma 39.8.1. Let $k$ be a field. Let $G$ be a locally algebraic group scheme over $k$. Then $G$ is equidimensional and $\dim (G) = \dim _ g(G)$ for all $g \in G$. For any closed point $g \in G$ we have $\dim (G) = \dim (\mathcal{O}_{G, g})$.

**Proof.**
Let us first prove that $\dim _ g(G) = \dim _{g'}(G)$ for any pair of points $g, g' \in G$. By Morphisms, Lemma 29.28.3 we may extend the ground field at will. Hence we may assume that both $g$ and $g'$ are defined over $k$. Hence there exists an automorphism of $G$ mapping $g$ to $g'$, whence the equality. By Morphisms, Lemma 29.28.1 we have $\dim _ g(G) = \dim (\mathcal{O}_{G, g}) + \text{trdeg}_ k(\kappa (g))$. On the other hand, the dimension of $G$ (or any open subset of $G$) is the supremum of the dimensions of the local rings of $G$, see Properties, Lemma 28.10.3. Clearly this is maximal for closed points $g$ in which case $\text{trdeg}_ k(\kappa (g)) = 0$ (by the Hilbert Nullstellensatz, see Morphisms, Section 29.16). Hence the lemma follows.
$\square$

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