# Another fix field of a certain galois group action

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 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 $m=0$?

• You mean $L=K(x \mid x^p-x-a=0)$ for some $a \in K$, right? – HeinrichD Sep 16 '16 at 17:32
• No, such $L$ would be finite over $E$, hence already $u$-adically complete. I would like to adjoin all zeroes of Artin-Schreyer Polynomials to $K$. – Louis Sep 16 '16 at 17:55
• If $K/E$ isn't Galois, then $G_E$ doesn't act on $K$. Why does it act on the completion of $L$? – znt Sep 16 '16 at 19:19
• Aren't the $L_m$ finite over $K$? – Felipe Voloch Sep 17 '16 at 21:18
• Sorry, I got my signs confused. $L_1 = K$ because if $\nu(a)>0$, then $z=-\sum_{n=0}^{\infty} a^{p^n}$ converges and $z^p-z=a$. My earlier argument shows that $L_{m}$ is finite over $L_{m+1}$ and by induction $L_m$ is finite for all $m\le 0$ (and $=K$ for $m>0$). – Felipe Voloch Sep 18 '16 at 2:48

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.

• Thank you for your answer! I indeed like this explicit argument. However, the explicit construction gives me (moral) hope, that in the case of the $L_m$ (especially $L_0$) one might obtain a positive answer. – Louis Sep 17 '16 at 19:20
• Even for $L_0$, I don’t doubt that at all. – Lubin Sep 18 '16 at 1:10
• Sorry, I'm not sure of what to make from this comment. Do you believe that $(L_0)^{G_E}=E$ or tend to $(L_0)^{G_E} \neq E$? Double negative and the stress words (even/at all) with unclear reference make it a bit confusing to me. – Louis Sep 18 '16 at 1:31
• Sorry, I am interested in the completed $\hat{L}_m$ of course (too late to edit my previous comment) – Louis Sep 18 '16 at 1:44
• (You can always delete the previous comment and rewrite.) Sorry for the unclarity, but I think your hopes are true for $L_0$: that $\widehat{L_0}\cap E^{\text{perf}}=E$ – Lubin Sep 18 '16 at 1:56