Algebraic function fields with continuous automorphisms Can one characterize the elements $\alpha$ of $L=\overline{\mathbb F_q(T)}\subset\Omega$, a completion of $L$ for the $\left(\frac1T\right)$-adic valuation such that for every conjugate $\beta$ of $\alpha$, there exists a continuous $\mathbb F_q\left(\left(\frac1T\right)\right)$-isomorphism of $\Omega$ with $\sigma(\alpha)=\beta$.
 A: Let $\Omega'$ be an algebraic closure of $\mathbb F_q (( \frac{1}{T}))$. Then we can embed $\Omega'$ into $\Omega$ because every extension of $\mathbb F_q (( \frac{1}{T}))$ is defined over $\mathbb F_q(T)$, and $\Omega'$ is dense in $\Omega$ because it contains $L$, so therefore $\Omega$ is the completion of $\Omega'$. Thus every continuous automorphism of $\Omega'$ extends to a continuous automorphism of its completion $\Omega$. 
Thus it's equivalent to ask that every Galois conjugate of $\alpha$ is conjugate, over $\mathbb F_q (( \frac{1}{T}))$, in $\Omega'$, to $\alpha$. Since $\Omega'$ is an algebraic closure of $\mathbb F_q (( \frac{1}{T}))$, it's equivalent to ask that the minimal polynomial of $\alpha$ over $\mathbb F_q(T)$ remains irreducible over $\mathbb F_q((1/T))$.
There are many criteria for when a polynomial over a local field is / isn't irreducible, for instance in terms of the Newton polygon, so I think this is the best way to express it.
A: Please define correctly your symbols, you meant $L$ is the algebraic closure of $\Bbb{F}_q(T)$ and $\Omega$ is the completion of $L$ for a discrete valuation above $T$. 
Take $f\in \Bbb{F}_q[T][X]$ monic irreducible such that $f\bmod T\in \Bbb{F}_q[X]$ is separable and it splits completely in $\Bbb{F}_p$ (for example $q=p=3,f=X^2-1+T$) then $f$ splits completely in $\Bbb{F}_q[[T]]\subset \Omega$ and its roots are all $\Bbb{F}_q(T)$-conjugates, of course there is no 
$\sigma\in Aut(\Omega/\Bbb{F}_q((T)))$ acting non-trivially on those roots.
A continuous automorphism of $\Omega$ restricts to an automorphism of $O_{\Omega} =\{ a\in \Omega,v(a)\ge 0\}$ sending the maximal $\mathfrak{m} = \{ a\in \Omega,v(a)> 0\}$ to itself, thus it gives an automorphism of $O_\Omega/\mathfrak{m} \cong \overline{\Bbb{F}}_q$ and since it acts trivally on $\Bbb{F}_p$ it acts trivially on the reductions on the roots thus on their unique lift to $\Bbb{F}_q[[T]]$.
