Timeline for What finite group schemes can act freely on a rational function field in one variable?
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| Jun 22, 2022 at 7:16 | history | edited | CommunityBot |
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| Apr 6, 2010 at 10:11 | answer | added | Bugs Bunny | timeline score: 3 | |
| Apr 6, 2010 at 8:58 | comment | added | Angelo | Indeed, this would be logical. The structure of the group is complicated, but it has a large unipotent part, which should not interfere with a form of $\mu_p$. I am fairly sure that the 1-dimensional torus is maximal. Thanks, Brian. | |
| Apr 6, 2010 at 5:19 | comment | added | BCnrd | You're right, and this error of mine is especially ironic since not more than a day ago I explained to a colleague why the automorphism scheme of $F(a^{1/p})$ over $F$ has positive dimension. Passing to a geometric point over $F$, this becomes the automorphism scheme of $F[y]/(y^p)$ as an $F$-algebra, which is parameterized by the possible images of $y$, namely $c_ 0 + c_ 1 y + ...$ with $c_ 1$ a unit and $c^p _0 = 0$. So it has dimension $p-1$. So one should first work out the structure of this group, especially if the evident 1-dimensional torus is maximal in its "smooth" part. | |
| Apr 5, 2010 at 20:19 | comment | added | Angelo | I am not convinced that $t^{1/p}$ must go to $ut^{1/p}$; you can also add nilpotents. This is what makes it complicated. For example, $\alpha_p$ can add freely by translations. In fact, I think that the automorphism group scheme of $k(t^{1/p})$ over $k(t)$ isn't even finite. | |
| Apr 5, 2010 at 20:08 | comment | added | BCnrd | The covering map for quotient is infinitesimal and quotient is a smooth curve, hence open $U$ in the projective line, and by shrinking and changing coordinate downstairs we can arrange that the cover map is $A := k[t]_f \rightarrow k[t]_f[t^{1/p}]$. For any $k$-algebra $R$, an $R$-algebra automorphism of $A_R[t^{1/p}]$ must carry $t^{1/p}$ to $u t^{1/p}$ where $u^p = 1$ (since $t^{1/p}$ is not a zero divisor, even after scalar extension by any $R$). So in this coordinatization the action is given by the usual one of $\mu_ p$ composed with an automorphism of $\mu_ p$. Does it look ok? | |
| Apr 5, 2010 at 19:29 | comment | added | Angelo | I tried that, it was certainly concrete but also complicated. But I may have missed something, if you say it works I'll try again. | |
| Apr 5, 2010 at 19:14 | comment | added | BCnrd | There are no others, since can take quotient by action of finite flat group scheme to get smooth quotient and over an alg. closed field the only extension field of $k(t)$ of degree $p$ that isn't \'etale is $k(t^{1/p})$ so can work out Aut-schemes of the possible degree-$p$ covers quite concretely. | |
| Apr 5, 2010 at 17:44 | comment | added | Angelo | Yes, but what I meant to say is that I don't know any other example of a form of $\mu_p$ that acts on $k(t)$. I would imagine that there are none, but I can't prove it. | |
| Apr 5, 2010 at 17:37 | comment | added | BCnrd | Any elt. of Lie alg. with ss adjoint action is tangent to a torus, and for infinitesimal gps of height $\le 1$ a homomorphism to another $k$-gp of finite type is same as map on $p$-Lie algebras. So for any form $\mu$ of $\mu_p$, nontrivial action on proj. line factors through embedding of $\mu$ into $k$-torus, which in turn is 1-dim'l. Hence, the possible $\mu$ are the $p$-torsion in $k$-tori of ${\rm{PGL}}_ 2$, which correspond to maximal $k$-tori in ${\rm{GL}}_ 2$, which correspond to deg-2 etale comm. algebras (i.e., split or separable quad. field). So no examples beyond what you know. | |
| Apr 5, 2010 at 17:14 | comment | added | Angelo | I mean free action. In any case, I think that if a form of $\mu_p$ acts faithfully, it also acts freely. There are some forms of $\mu_p$ that can act: $\mu_p$ itself, of course, and also those obtained from the involution $x \mapsto x^{-1}$ via a quadratic extension. I don't know any other example. | |
| Apr 5, 2010 at 16:32 | comment | added | BCnrd | Do you mean free action or faithful action? And $\mu_p$ can act faithfully since ${\rm{PGL}}_ 2$ contains a nontrivial split torus (e.g., $\zeta.[x,y] = [\zeta x, y]$). Anyway, by viewing the projective line over the finite base $G$, by looking at geometric fibers over $G$ your setup amounts to an action of $G$ on a dense open of the projective line (as for ordinary finite groups). But then it perhaps get complicated, since automorphism functor of such opens is generally not representable. | |
| Apr 5, 2010 at 14:46 | history | asked | Angelo | CC BY-SA 2.5 |