Let $p$ be a prime number, $K$ a finite extension of $\mathbb{Q}_p$, $\mathfrak{o}$ its ring of integers, $\mathfrak{p}$ the unique maximal ideal of $\mathfrak{o}$, $k=\mathfrak{o}/\mathfrak{p}$ the residue field, and $q=\operatorname{Card} k$.

Recall that a polynomial $\varphi=T^n+c_{n-1}T^{n-1}+\cdots+c_1T+c_0$ ($n>0$) in $K[T]$ is said to be **Eisenstein** if $c_i\in\mathfrak{p}$ for $i\in[0,n[$ and if $c_0\notin\mathfrak{p}^2$.

**Question.** When is the extension $L_\varphi$ defined by $\varphi$ galoisian (resp. abelian, resp. cyclic) over $K$ ?

**Background.** Every Eisenstein polymonial $\varphi$ is irreducible, the extension $L_\varphi=K[T]/\varphi K[T]$ is totally ramified over $K$, and every root of $\varphi$ in $L_\varphi$ is a uniformiser of $L_\varphi$. There is a converse.

If the degree $n$ of $\varphi$ is prime to $p$, then the extension $L_\varphi|K$ is tamely ramified and can be defined by the polynomial $T^n-\pi$ for some uniformiser $\pi$ of $K$. Thus $L_\varphi|K$ is galoisian if and only if $n|q-1$, and, when such is the case, $L_\varphi|K$ is actually cyclic.

**Real question**. Is there a similar criterion, in case $n=p^m$ is a power of $p$, for deciding if $L_\varphi|K$ is galoisian (resp. abelian, resp. cyclic) ?

`$SACRED_BOOK`

says that one should not define $\mathbb N$ as the free semigroup on one generator? :) – Mariano Suárez-Alvarez♦ Mar 6 '10 at 16:57