Lemma 39.8.6. Let $k$ be an algebraically closed field. Let $G$ be a locally algebraic group scheme over $k$. Let $g_1, \ldots , g_ n \in G(k)$ be $k$-rational points. Then there exists an affine open $U \subset G$ containing $g_1, \ldots , g_ n$.
Proof. We first argue by induction on $n$ that we may assume all $g_ i$ are on the same connected component of $G$. Namely, if not, then we can find a decomposition $G = W_1 \amalg W_2$ with $W_ i$ open in $G$ and (after possibly renumbering) $g_1, \ldots , g_ r \in W_1$ and $g_{r + 1}, \ldots , g_ n \in W_2$ for some $0 < r < n$. By induction we can find affine opens $U_1$ and $U_2$ of $G$ with $g_1, \ldots , g_ r \in U_1$ and $g_{r + 1}, \ldots , g_ n \in U_2$. Then
\[ g_1, \ldots , g_ n \in (U_1 \cap W_1) \cup (U_2 \cap W_2) \]
is a solution to the problem. Thus we may assume $g_1, \ldots , g_ n$ are all on the same connected component of $G$. Translating by $g_1^{-1}$ we may assume $g_1, \ldots , g_ n \in G^0$ where $G^0 \subset G$ is as in Proposition 39.7.11. Choose an affine open neighbourhood $U$ of $e$, in particular $U \cap G^0$ is nonempty. Since $G^0$ is irreducible we see that
\[ G^0 \cap (Ug_1^{-1} \cap \ldots \cap Ug_ n^{-1}) \]
is nonempty. Since $G \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type, also $G^0 \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type, hence any nonempty open has a $k$-rational point. Thus we can pick $g \in G^0(k)$ with $g \in Ug_ i^{-1}$ for all $i$. Then $g_ i \in g^{-1}U$ for all $i$ and $g^{-1}U$ is the affine open we were looking for. $\square$
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