maximal abelian extension of exponent $q-1$ of $\mathbb F_q((t))$ I would like to find the maximal abelian extension of exponent $q-1$ of $K=\mathbb F_q((t))$ and find its Galois group. Due to Kummer theory this extension is $K(\sqrt[q-1]{K^*})$ and it's Galois group is $Hom(K^*/K^{*(q-1)},\mathbb F_q^*)$.But I can't understand what is this group $K^*/K^{*(q-1)}$. So I can't find Galois group of this extension, basis of this extension or even degree.
 A: $K^{*(q-1)}$ is the group of $(q-1)$th powers of $K^*$.
$K^*/K^{*(q-1)}$ is the quotient obtained from $K^*$ when two element are considered equal when their quotient is a $(q-1)$th power.
Let $R$ be the (multiplicative) group of formal power series with constant coefficient $1$ ($R = 1+t\Bbb F_q[[t]]$). We have an isomorphism of groups $\Bbb Z \times \Bbb F_q^* \times R \to \Bbb F_q((t))^*$ given by $(d,c,1+p(t)) \mapsto c t^d (1+p(t))$
We can then study the $(q-1)$th power endomorphism, and the quotient $K^*/K^{*(q-1)}$ piece by piece.
On the $\Bbb Z$ component, it is multiplication by $q-1$, which gives us a $\Bbb Z/(q-1)\Bbb Z$ piece.
On the $\Bbb F_q^*$ component, it is the trivial endomorphism sending everyone to $1$, which gives a $\Bbb F_q^*$ piece.
On the $R$ component, it is an isomorphism (everything is a $(q-1)$th power by Hensel's lemma, and there is exactly one $(q-1)$th root starting with a $1$).
Hence $K^*/K^{*(q-1)} \simeq (\Bbb Z/(q-1)\Bbb Z) \times \Bbb F_q^*$
Let $G$ be the Galois group. If $\sigma \in G$ and  $p \in K^*$, if $r^{q-1} = p$ then $\sigma(r)/r \in \Bbb F_q^*$ doesn't depend on the choice of $r(t)$, so we have a map $G \to \hom(K^* , \Bbb F_q^*)$. If $p$ was a $q-1$th power then $p \in K^*$, then $\sigma(r)/r = r/r = 1$ so this induces a map $G \to \hom(K^*/K^{*(q-1)} , \Bbb F_q^*)$.
If $\mu$ is a generator of $\Bbb F_q^*$ then it is enough to know the images of $t$ and $\mu$, so we have an isomorphism  $G \to \Bbb F_q^{*2}$ given by $\sigma \mapsto ( \sigma(\sqrt[q-1]{t})/\sqrt[q-1]{t} \;,\; \sigma(\sqrt[q-1]\mu)/\sqrt[q-1]\mu))$.
And finally, $K(\sqrt[q-1]{K^*}) = K(\sqrt[q-1]t, \sqrt[q-1]\mu) = \Bbb F_{q^{q-1}}((\sqrt[q-1]t))$ 
