# Automorphisms in function fields with $\mathbb F_q[T]$ globally invariant

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 $\mathbb F_q[T]$?

• Isn't the Galois group of $\Omega/{\mathbb F}_q((1/T))$ already uncountable? – user1688 Dec 6 '17 at 5:57
• 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. – KConrad Feb 27 '18 at 22:24