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Angelo
  • Member for 14 years, 9 months
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What finite group schemes can act freely on a rational function field in one variable?
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.
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What finite group schemes can act freely on a rational function field in one variable?
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.
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What finite group schemes can act freely on a rational function field in one variable?
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.
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What finite group schemes can act freely on a rational function field in one variable?
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.
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What finite group schemes can act freely on a rational function field in one variable?
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.
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What finite group schemes can act freely on a rational function field in one variable?
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.
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What finite group schemes can act freely on a rational function field in one variable?
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.
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what mistakes did the Italian algebraic geometers actually make?
I understand, and most certainly I was not offended. My point is that the general opinion that the old Italian algebraic geometers made mistakes because they did not have the proper foundations may be roughly right, but also simplistic. It it true that the Italian school went slowly astray, as discussed in Mumford's very interesting email message; but how much of it was due to personalities of the leaders of the school (particularly Severi) and how much to lack of proper foundations, I'll leave to others more competent than me to answer.
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