Let $E=\mathbb{F}\_p(\!(u)\!)$, the Laurent series field over $\mathbb{F}\_p$. Let $K/E$ be a finite normal separable extension. Consider the field $L=K(x \mid x^p-x-a=0 \text{ for some } a \in K)$. Let $\hat{L}$ be the $u$-adic completion of $L$. We write $G\_E=\mathrm{Gal}(E^s/E)$ for the absolute Galois group of $E$. By continuity $G\_E$ acts on $\hat{L}$. What is $(\hat{L})^{G\_E}$?


In particular, is $(\hat{L})^{G\_E}= E$? One has $\hat{L} \subseteq \widehat{E^s}$ and [following the answer to one of my last questions](https://mathoverflow.net/questions/248819/fix-field-of-a-certain-galois-group-action) one also has $(\widehat{E^s})^{G\_E}=\widehat{E^{\mathrm{perf}}}$. Therefore, we are left with the question if $\hat{L} \cap \widehat{E^{\mathrm{perf}}}=E$?


Edit: and what if we consider for some $m \in \mathbb{Z}$ the field $\hat{L}\_m$ where $L\_m= K(x \mid x^p-x-a=0 \text{ for some } a \in K \text{ with } \nu(a) \geq m)$? What in the special case where $m=0$?



Another fix field of a certain galois group action
The right way to do this would be to give a short, efficient, abstract argument. I, however, will give a fairly messy example to show that $u^{1/p}\in\hat L$.


I’ll use $K=E=\Bbb F\_p((u))$, and set $a\_n=u^{1-pn}$ for $n>0$, and set $f\_n(x)=x^p-x-a\_n$, an irreducible and separable polynomial over $E$. Its roots are of valuation $-n+1/p$, and I’ll multiply these all by $u^n$ to form the polynomial $g\_n(x)=x^{pn}f(X/u^n)=x^p-u^{(p-1)n}x-u$. This is Eisenstein, so a root $\alpha\_n$ of $g\_n$ will be a generator of a field $K\_n$, which by its construction is one of the fields whose compositum defines $L$.


Now I claim that $\lim\_n\alpha\_n=u^{1/p}$. For this, I need $v(\alpha\_n-u^{1/p})$, where $v$ is the $u$-adic valuation normalized so that $v(u)=1$, and I’ll exhibit this as $\frac1{p^2}v\bigl(\mathbf N^{K\_n(t^{1/p})}\_E(\alpha\_n-t^{1/p})\bigr)$.


Since $\mathrm{Irr}(\alpha\_n,E)=g\_n(x)=x^p-u^{(p-1)n}x-u$, this is also $\mathrm{Irr}\bigl(\alpha\_n,E(u^{1/p})\bigr)$, and so $\mathrm{Irr}\bigl(\alpha\_n-u^{1/p},E(u^{1/p})\bigr)=g(x+u^{1/p})=x^p-u^{(p-1)n}x-u^{(p-1)n+1/p}$. To get the minimal polynomial for $\alpha\_n-u^{1/p}$ over $E$, just raise to the $p$-power. We get:
$$\mathrm{Irr}(\alpha\_n-u^{1/p},E\,) = x^{p^2}-u^{p(p-1)n}x^p-u^{p(p-1)n+1}\,.
$$
Thus $v(\alpha\_n-u^{1/p})=\frac1{p^2}\bigl(p(p-1)n+1\bigr)>n/2$, which establishes the claim.


