Let $\mathbf{F}_q$ be a finite field with $q$ elements and let $K$ denote the local function field $\mathbf{F}_q(\!(\theta)\!)$. Let $R$ be its valuation ring $\mathbf{F}_q[\![\theta]\!]$. Let $u$ be a unit in $R$ which is not in $\mathbf{F}_q^{\times}+\theta^q R$.
Let $x\in K$ be such that the Artin-Schreier polynomial $X^q-X+u^{\ell}x$ splits in $K$ for all $\ell\geq 0$. Does this imply $x\in R$?
Some remarks:
- If $u$ would have been in $\mathbf{F}_q$, then $x=k^q-k$ for some $k\in K\setminus R$ would have provide an element $x\notin R$ such that $X^q-X+u^{\ell}x$ splits in $K$ for all $\ell\geq 0$.
- Note that if $x\in \mathfrak{m}=\theta \mathbf{F}_q[\![\theta]\!]$, then the polynomial $X^q-X+u^{\ell}x$ splits in $K$, admitting the converging series $u^{\ell}x+u^{q^{\ell}}x^q+u^{q^2\ell}x^{q^2}+...$ for root in $K$.
- As Jyrki Lahtonen pointed out, if $u$ would have been $1+\theta^{q+1}$ and $x=k-k^q$ for some $k\in K\setminus R$, the considered polynomials would also have split in $K$.
- The combination of 1.-3. explains the condition $u\notin \mathbf{F}_q^{\times}+\theta^q R$.
Many thanks in advance for your help!
Yes $x$ must be in $t\Bbb{F}_q[[t]]$.
Assume it is not.
Let $\rho(g) = g^q-g$ then $\rho(g+g_2)=\rho(g)+\rho(g_2)$.
Check that $\rho(\Bbb{F}_q[[t]]) = t \Bbb{F}_q[[t]]$.
You said that $u = b+t^m h$ with $b\in \Bbb{F}_q^\times,h\in \Bbb{F}_q[[t]]^\times $ and $m\in 1 \ldots q-1$.
As you assume that $xu^l\in \rho(\Bbb{F}_q((t))$ for all $l$, for $n$ large enough you'll have $x(u^{q^n}-b) \in t \Bbb{F}_q[[t]] \subset \rho(\Bbb{F}_q((t)))$ so $xb\in \rho(\Bbb{F}_q((t)))$, and hence $q | v(x)$.
Both $xb,xu\in \rho(\Bbb{F}_q((t))$ will give that $x t^m h \in \rho(\Bbb{F}_q((t)))$ so $q | v(x t^m h)= v(x)+m$.
As $q\nmid m$ this is a contradiction.