TITLE: Totally ramified extensions of $\mathbb{Q}_p$
QUESTION [10 upvotes]: There's a theorem that gives you all unramified extensions of $\mathbb{Q}_p$: they correspond to the finite extensions of the residue field $\mathbb{F}_p$. Is there a similar result for totally ramified extensions?
I have a more specific question - I'm trying to find all degree three extensions of $\mathbb{Q}_2$. The ramification index is either 1 (in which case the extension is unramified), or 3 (in which case it's totally ramified). I know exactly how to construct a unramified extension of degree 3 - it's $\mathbb{Q}(\zeta_{2^3-1}) = \mathbb{Q}(\zeta_7)$. But I'm not sure what to do with the totally ramified ones.
I know that $\mathbb{Q}_{\zeta_{2^m}}$ are totally ramified extensions of $\mathbb{Q}_2$, but none of such field has degree 3 because $[\mathbb{Q}_{\zeta_{2^m}}:\mathbb{Q}_2]=2^{m-1}$. 
I know that $\mathbb{Q}_2$ adjoint a root of an Eisenstein polynomial would be totally ramified, so I can adjoint a root of $X^3-2$,  which gives me a totally ramified degree 3 extension.
Is that all of it? 
Edit: I also know that all totally ramified extensions are obtained by adjoinint a root of an Eisenstein polynomial, but that still doesn't tell me how to find all of them.

REPLY [4 votes]: The OP asks for the determination of all the ramified cubic extensions of $\mathbf Q_2$, but I think that Lubin's (impeccable) proof gives more : $\mathbf Q_2 (\omega, \sqrt [3] 2)$ is the only ramified dihedral extension $N/\mathbf Q_2$ of degree $6$ (i.e. $N/\mathbf Q_2$ is normal, with Galois group $\cong D_6 \cong S_3)$ and ramification index $\ge 3$. Indeed :
1) From  $\mathbf {Q^*_2}\cong \mathbf Z \times \mathbf Z_2 \times \mathbf Z/2$, it follows that $\mathbf {Q^*_2}/\mathbf {Q^*_2}^3 \cong \mathbf Z/3$, and local CFT tells us that $\mathbf Q_2$ admits a single normal cubic extension, which is naturally the unramified one. So our dihedral $N$ must be the normal closure of a cubic non normal ramified extension $K/\mathbf Q_2$ .
2) One knows that such a $K$ is of the form $\mathbf Q_2 (\alpha)$, where $\alpha$ is a root of a monic Eisenstein polynomial $f\in \mathbf Z_2 [X]$, and  the normal closure $N$ of $K$ is the splitting field of $f$. Because $S_3$ has a unique subgroup of index $2$ (necessarily normal), $N$ contains a unique quadratic subfield which is no other than $\mathbf Q_2 (\sqrt \Delta)$
, where $\Delta$ is the dicriminant of $f$. By Kummer's theory, and slightly abusing language, the quadratic extensions of $\mathbf Q_2$ are obtained by taking square roots of representatives of classes of $\mathbf {Q^*_2}/\mathbf {Q^*_2}^2$. A complete set of such representatives is {$\pm 1, \pm 2, \pm 3, \pm 6$} (see e.g. Serre's "A course in arithmetic", end of chap. II), so $\Delta$ must be (multiplicatively) congruent mod $\mathbf {Q^*_2}^2$ to one and only one of these (excluding $1$). But it has been shown by Lubin that $\Delta \equiv -3$ mod $8$, hence $-3.\Delta \equiv 1$ mod $8$. Since any $2$-adic unit $\equiv 1$ mod $8$ is a square (Serre, op. cit.), one concludes that $\mathbf Q_2 (\sqrt \Delta)=\mathbf Q_2 (\sqrt {-3})=\mathbf Q_2 (\omega)$, hence $N$ has the desired form. The three ramified cubic extensions of $\mathbf Q_2$ are then the subfields of $N$ fixed by the three transpositions in $S_3$ .
NB : One could avoid computing the discriminant, starting from the fact that the ramification condition on $3$ implies that $3 \mid \Delta$ .