A well known open question is whether the scheme of commuting pairs in a complex reductive group $G$, for example in $G=GL(n)$, is reduced. The variety of commuting pairs is a special case of a more general construction: given an action of $G$ on an algebraic variety $X$ one can form the stabilizer scheme $S$, the closed subscheme in $G\times X$ given by the equation $g(x)=x$. Are there examples when $X$ is smooth variety over a characteristic zero field but $S$ is nonreduced?
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3$\begingroup$ 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 $\text{Spec}\ k$). $\endgroup$– Jason StarrCommented Oct 8, 2023 at 12:03
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$\begingroup$ 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. $\endgroup$– RomanCommented Oct 9, 2023 at 0:23
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$\begingroup$ Consider the action of $\mathbb{A}^1=\text{Spec}\ k[s]$ on $\mathbb{A}^2=\text{Spec}\ k[x,y]$ by $(s,(x,y))\mapsto (x,x^2s+y))$. $\endgroup$– Jason StarrCommented Oct 9, 2023 at 0:59
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$\begingroup$ @JasonStarr $\mathbb A^1$ is not reductive. $\endgroup$– Kenta SuzukiCommented Oct 9, 2023 at 2:43
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1 Answer
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I am posting my comments as an answer. First of all, the hypothesis that $G$ is reductive, or even semisimple, is easily "induced" from a more general example.
Let $G$ be any linear algebraic group. Let $X$ be a smooth $k$-scheme. Let $\lambda:G\times X\to X$ be a left $G$-action. For every morphism of linear algebraic groups, $i:G\to G'$, then also there is a diagonal action of $G$ on $G'\times X$ by $$\delta:G\times (G'\times X)\to G'\times X, \ \ (g,(g',x))\mapsto (g'i(g)^{-1},\lambda(g,x)).$$ This action is free, and the quotient is a $G$-torsor (hence flat). Denote the quotient of $\delta$ by $$q:G'\times X \to X'.$$ So if $G'$ is smooth, say $\textbf{SL}_n$, then also $X'$ is smooth.
There is a natural left action of $G'$ on $G'\times X$, and this descends to a left action of $G'$ on $X'$ (so that $q$ is $G'$-equivariant). The inverse image with respect to $q$ of the stabilizer group scheme of the left $G'$-action on $X'$ is isomorphic to the inverse image of the stabilizer group scheme of the left $G$-action on $X$ with respect to the (smooth) projection, $G'\times X \to X$ (the isomorphism involves conjugation of $i(G)$ by elements of $G'$). Therefore, we can "induct" examples where $G'$ is a smooth reductive group, say $\textbf{SL}_n$, from examples where $G$ is an arbitrary linear algebraic group.
Now, let $G$ be $\mathbb{A}^1 = \text{Spec}\ k[s]$, and let $X$ be $\mathbb{A}^2=\text{Spec}\ k[x,y]$. Let the left action $\lambda$ send $(s,(x,y))$ to $(x,x^2s+y)$. The stabilizer group scheme inside $\mathbb{A}^1\times \mathbb{A}^2 = \text{Spec}\ k[s,x,y]$ is the zero scheme of $x^2y$, which is not reduced.
You can explicitly convert this to an example where $G'$ equals $\textbf{SL}_2$ by identifying $\mathbb{A}^1$ with the unipotent radical of a Borel subgroup. The scheme $X'$ will be a fourfold with a smooth morphism to $\textbf{SL}_2/\mathbb{A}^2 \cong \mathbb{A}^2\setminus\{(0,0)\}$ whose geometric fibers are each isomorphic to $\mathbb{A}^2$.
As I commented above, in characteristic $p$ you can find examples where the stabilizer group scheme is not reduced, yet it is flat over $X$ (and every geometric fiber is not reduced).