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Let $\Omega$ be the completion of an algebraic closure of $\mathbb F_q((1/T))$ for the valuation $-\deg$. Does there exist uncountably many automorphisms of $\Omega$ that let $\mathbb F_q[T]$ stable and fixe one element of $\mathbb F_q[T]$?

Thansk in advance for any answers.

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    $\begingroup$ Isn't the Galois group of $\Omega/{\mathbb F}_q((1/T))$ already uncountable? $\endgroup$
    – user1688
    Dec 6, 2017 at 5:57
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    $\begingroup$ If $f$ is fixed in $\mathbf F_q[T]$ and is not constant then $\mathbf F_q(f)$ is fixed by all automorphisms of $\Omega$ and is a subfield of finite codimension in $\mathbf F_q(T)$. Every automorphism of $\Omega$ is an automorphism of the algebraic closure $\mathbf F_q(T)$, which equals $\overline{\mathbf F_q(f)}$, so your group can be viewed as mapping onto the absolute Galois group of $\mathbf F_q(f)$. This absolute Galois group is infinite and an infinite compact (profinite) group is uncountable. So your group is uncountable. $\endgroup$
    – KConrad
    Feb 27, 2018 at 22:24

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