# When is the extension defined by an Eisenstein polynomial galoisian or abelian or cyclic ?

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) ?

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Do you mean [0,n-1]? –  S. Carnahan Feb 3 '10 at 6:25
Yes. $[0,n[=[0,n-1]$, as intervals in $\mathbb{N}$. –  Chandan Singh Dalawat Feb 3 '10 at 8:06
Definition. $\mathbb{N}$ is the free monoid on one generator, namely 1. Corollary. $0\in\mathbb{N}$. –  Chandan Singh Dalawat Feb 3 '10 at 9:50
@Chandan, my understanding is that 0 tends to be in N in several European countries, but not in America. I have been told to use the notation Z_{\ge 0} to avoid ambiguity. (Personally I agree with you, but for a slightly different reason - the non-negative integers are precisely the set of cardinalities of finite sets.) –  Qiaochu Yuan Mar 6 '10 at 16:20

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