Timeline for (non)reduced stabilizer scheme

7 events

when toggle format	what		by	license	comment
Oct 9, 2023 at 10:04	answer	added	Jason Starr		timeline score: 4
Oct 9, 2023 at 2:43	comment	added	Kenta Suzuki		@JasonStarr $\mathbb A^1$ is not reductive.
Oct 9, 2023 at 0:59	comment	added	Jason Starr		Consider the action of $\mathbb{A}^1=\text{Spec}\ k[s]$ on $\mathbb{A}^2=\text{Spec}\ k[x,y]$ by $(s,(x,y))\mapsto (x,x^2s+y))$.
Oct 9, 2023 at 0:25	history	edited	Roman	CC BY-SA 4.0	added 41 characters in body
Oct 9, 2023 at 0:23	comment	added	Roman		Thanks but the question was primarily about the characteristic zero field of coefficients. Of course, a group scheme over a characteristic zero field is reduced as you say, but a (non-flat) group scheme over a base $B$ may be non-reduced even if $B$ is an algebraic variety over a field; otherwise the question about commuting pairs of elements in a reductive group had an easy answer.
Oct 8, 2023 at 12:03	comment	added	Jason Starr		All group schemes over a characteristic $0$ field are reduced (by Cartier's theorem). Let $k$ be a field of characteristic $p$. Let $G$ equal $X$ equals $\mathbb{G}_{m,k} = \text{Spec}\ k[t,t^{-1}]$. Let the action of $G$ on $X$ be $(s,t)\mapsto s^pt$. The stabilizer subgroup scheme of $G\times X$ is $\mu_p\times X$ (all products relative to $\text{Spec}\ k$).
Oct 8, 2023 at 3:15	history	asked	Roman	CC BY-SA 4.0