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Suppose that F is a function field of a single variable over a finite field. The automorphism group Aut(F) acts on the places of F and permutes all places of a given degree. I have a few questions:

1) if the action an automorphism sigma on the rational places is trivial, ie sigma fixes every rational place, does it follow that sigma is the identity? (I'm assuming that F has more than one rational place).

2) is it possible for an automorphism to switch two rational places and keep the remaining rational places fixed? (I'm assuming that F has at least 3 rational places).

In the case of the Hermitian function fields, the answer is yes to 1) and no to 2).

Thanks. Hiren

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  • $\begingroup$ Have you looked at what happens for $\mathbb{F}_2(t)$? $\endgroup$
    – Daniel Loughran
    Commented Feb 11, 2015 at 21:11
  • $\begingroup$ Ah, I see that this gives an example of 2). Thanks. I am interested to know if there are examples where the function field is not a rational function field. $\endgroup$
    – Hiren
    Commented Feb 11, 2015 at 21:21

1 Answer 1

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Consider the automorphism of the curve $y^2=x^p-x$ over $\mathbb F_p$ thar sends $y$ to $-y$. This fixes every $\mathbb F_p$-point but is not trivial. Taking function fields, we get a counterexample to 1.

The same automorphism of $y^2=x^p-x^{p-1}-x+1$ fixes all the rational points but $(0,1)$ and $(0,-1)$ which it switches.

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