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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)