Suppose that $G$ is a finite group scheme over a field $k$ (we may want to assume that $k$ is perfect). How does one tell whether there exists a free action of $G$ on the function field $k(t)$ in one variable? By this I mean that there exists an action $G \times_{\mathop{\rm Spec}k}\mathop{\rm Spec}k(t) \to \mathop{\rm Spec}k(t)$, making $\mathop{\rm Spec}k(t)$ into a $G$-torsor over a scheme (necessarily of the form $\mathop{\rm Spec} k(s)$, where $s \in k(t)$, by Lüroth's theorem).

The question is a very natural one when one studies essential dimension of group schemes: see http://www.math.ubc.ca/~reichst/lens-notes6-27-8.pdf for a nice survey of the topic of essential dimension, and http://front.math.ucdavis.edu/1001.3988 for the essential dimension of group schemes. When $G$ is smooth over $k$, then it is easy to see that the action extends to an action on $\mathbb{P}^1$, so $G$ must be a subgroup of ${\rm PGL}_{2,k}$; but when $G$ is not smooth it is not all clear to us that this must happen. The sheaf of automorphisms of $k(t)$ over $k$ is enormous in positive characteristic, and we find it very hard to see what group schemes it contains.

For example, how about twisted forms of the group scheme $\mu_p$, where $p$ is the characteristic of the field? I would conjecture that most of them can't act freely on $k(t)$, but we can't prove it.

  • $\begingroup$ 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. $\endgroup$ – BCnrd Apr 5 '10 at 16:32
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    $\begingroup$ 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. $\endgroup$ – BCnrd Apr 5 '10 at 17:37
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    $\begingroup$ 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. $\endgroup$ – Angelo Apr 5 '10 at 20:19
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    $\begingroup$ 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. $\endgroup$ – BCnrd Apr 6 '10 at 5:19
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    $\begingroup$ 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. $\endgroup$ – Angelo Apr 6 '10 at 8:58

I guess by looking at it algebraically one can at least rule out the forms of $\mu_p$. Let $H$ be the function algebra of the group scheme, $H^\ast$ its dual. $H^\ast$ is a cocommutative Hopf algebra. Algebraically, you ask whether $H^\ast$ can act on $k(t)$ so that $k(t)>k(s)$ is Hopf-Galois. If I remember correctly, there is a theorem (see chapter 8 of Montogmery' Hopf Algebra actions) that this is equivalent to the semidirect product $k(t)*H^\ast$ being simple. This will necessarily require $H^\ast$ to be semisimple. Now I read your $\mu_p$ as a form of the cyclic group. Thus, your $H^\ast$ fails to be semisimple by Maschke's theorem.

Sorry, if I misunderstood or misquoted something.

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    $\begingroup$ Are you the real Bugs Bunny? I think I underestimated you. $\endgroup$ – Kevin Buzzard Apr 6 '10 at 11:26
  • $\begingroup$ Dear Bugs Bunny, thank you very much for the suggestion. However, in order to consider $\mu_p$ as a form of the cyclic group, the characteristic should be different from $p$, in which case we know a proof; we are interested exactly in the characteric $p$ case. Any form of $\mu_p$ is semisimple (in any characteristic). This said, I will take a look at Montgomery's book, there might be something in it. $\endgroup$ – Angelo Apr 6 '10 at 12:52
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    $\begingroup$ After some further thought, I realize I probably misunderstood what you were saying. On the other hand, your argument should apply to all forms of $\mu_p$; while some of them, like $\mu_p$ itself, do appear. Also, it would exclude $\alpha_p$, which also appears. $\endgroup$ – Angelo Apr 6 '10 at 14:30
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    $\begingroup$ No, Kevin, I am not real, I am p-adic :-)) Angelo, I think now that your $\mu_p$ is $p$-th roots of unity. Then, indeed, $H^*$ is semisimple and nothing I said is of any use. Is your $\alpha_p$ the Frobenius kernel of the additive group. Then it does, indeed, act via $d/dt$. I guess there is no theorem concluding that $H^*$ is semisimple or it may require a separable extension... Anyway, I will look into Childs', Taming wild extensions and see whether I can say anything intelligent. $\endgroup$ – Bugs Bunny Apr 7 '10 at 9:20

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