An infinite family of Artin-Schreier polynomials which all split in $\mathbf{F}_q(\!(\theta)\!)$

Solution 1:

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.