Let $E= \mathbb{F}\_p(\!(u)\!)$, $E^s$ a separable closure of $E$ and write $G\_E= \mathrm{Gal}(E^s/E)$ for the absolute Galois group of $E$. Take a lift of the $u$-adic valuation on $E$ to $E^s$ and write $\nu \colon E^s \rightarrow \mathbb{Q}$ for the resulting valuation. Let $\hat{E^s}$ be a completion of $E^s$ with regard to this valuation. 
We obtain a $G\_E$ action on $\hat{E^s}$ by continuity. Is it true that $(\hat{E^s})^{G\_E}=E$? If not, is there an 'explicit' description for $(\hat{E^s})^{G\_E}$?


(I already asked this question on math.stackexchange, but I think it might be fit for mathoverflow)



Fix field of a certain galois group action
This question is completely answered in


J. Ax – “Zeros of polynomials over local fields—The Galois action”, J. Algebra 15 (1970), p. 417–428.


You get the completion of the perfection of E.


