Timeline for Variant of Hilbert 90 for Galois extensions
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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May 28, 2014 at 8:09 | vote | accept | joaopa | ||
May 28, 2014 at 7:31 | answer | added | user76758 | timeline score: 11 | |
May 27, 2014 at 21:04 | comment | added | Laurent Moret-Bailly | @Daniel Loughran: If $g=1$, $\mathrm{Aut}(K)$ is an extension of the finite group of (pointed) automorphisms of an elliptic curve by the group of translations of the same, which is also finite since the ground field is. So, it seems that $\mathrm{Aut}(K)$ is finite in all cases. | |
May 27, 2014 at 11:38 | history | edited | Daniel Loughran |
edited tags
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May 27, 2014 at 11:38 | comment | added | Daniel Loughran | Let $g$ denote the genus of $C$. If $g=0$ then one gets a copy of $\mathrm{PGL}_2$ plus perhaps something else finite, so you can study explicitly what happens in this case. If $g > 1$ then $Aut(K)$ should be finite, hence one might guess that the usual proof of Hilbert theorem 90 should show that $H^1(Aut(K),K^*)=1$. For $g=1$ one obtains an elliptic curve, where things are more complicated and I'm not sure what should happen. I hope this helps. | |
May 27, 2014 at 11:38 | comment | added | Daniel Loughran | I think you might gain something by translating this into a geometric problem. Namely, let $C$ be the unique smooth projective algebraic curve with function field K. Then $Aut(K)$ is the automorphism group of $C$ over $\mathbf{F}_q$. | |
May 27, 2014 at 11:37 | comment | added | Peter Mueller | Did you check the other extreme case, $K=\mathbb F_q(x)$, so $\text{Aut}(K)=\text{PGL}_2(\mathbb F_q)$? | |
S May 27, 2014 at 6:28 | history | suggested | jmc | CC BY-SA 3.0 |
Corrects formatting and typo's
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May 27, 2014 at 6:20 | review | Suggested edits | |||
S May 27, 2014 at 6:28 | |||||
May 27, 2014 at 5:22 | history | edited | joaopa | CC BY-SA 3.0 |
added 52 characters in body; edited title
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May 27, 2014 at 5:16 | history | asked | joaopa | CC BY-SA 3.0 |