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Let $K/\mathbb F_q(x)$ be a finite Galois extension with Galois group $G$. Let $Aut(K)$ be the group of $\mathbb F_q$-automorphisms of $K$. Obviously, $G\subseteq Aut(K)$. It is well known that $H^1(G,K^*) = 1$ [Hilbert 90]. But does the following hold: $H^1(Aut(K), K^*)=1$?

Thanks in advance.

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    $\begingroup$ Did you check the other extreme case, $K=\mathbb F_q(x)$, so $\text{Aut}(K)=\text{PGL}_2(\mathbb F_q)$? $\endgroup$
    – Peter Mueller
    Commented May 27, 2014 at 11:37
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    $\begingroup$ 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 $\mathbf{F}_q$. $\endgroup$
    – Daniel Loughran
    Commented May 27, 2014 at 11:38
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    $\begingroup$ 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. $\endgroup$
    – Daniel Loughran
    Commented May 27, 2014 at 11:38
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    $\begingroup$ @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. $\endgroup$
    – Laurent Moret-Bailly
    Commented May 27, 2014 at 21:04

1 Answer 1

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This is an application of Artin's Lemma to reduce to Hilbert 90. Namely, let $K$ be any field at all, and $G$ any finite subgroup of ${\rm{Aut}}(K)$ (as noted in this comments, such finiteness holds for the $K$ in the question with $G = {\rm{Aut}}(K)$). Then by Artin's Lemma the field $K$ is finite Galois over its subfield $K^G$ of $G$-invariants with $G \rightarrow {\rm{Gal}}(K/K^G)$ an isomorphism. Hence, ${\rm{H}}^1(G,K)$ vanishes by Hilbert 90.

(In the geometric setting of the question, $K^G$ is the function field of the quotient $C/G$ where $C$ is connected regular proper $\mathbf{F}_p$-scheme of dimension 1 with function field $K$.)

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