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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)

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    $\begingroup$ No, because the completion of the separable closure of a non-archimedean field $k$ is the same as the completed algebraic closure that is itself algebraically closed. The point is that $k_s$ is dense in $\overline{k}$ (so both have the same completion, and that of the latter is algebraically closed), essentially because slightly changing the coefficients of a monic irreducible polynomial over $k$ makes it separable. So the fixed-field you ask about contains the non-complete perfect closure of $E$, and so also its completion. $\endgroup$
    – nfdc23
    Commented Sep 1, 2016 at 13:57

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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.

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