# Totally ramified extensions of $\mathbb{Q}_p$

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.

This is an exercise I’ve never done, but it should be a lot of fun. What is the general Eisenstein polynomial in this case? it’ll be $$X^3 + 2aX^2+2bX+2(1+2c)\,,$$ where $a$, $b$, and $c$ can be any $2$-adic integers. Notice that the constant term has to be indivisible by any higher power of $2$, so of form $2$ times a unit, and the units of $\Bbb Z_2$ are exactly the things of form $1+2c$. So your parameter space is $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$, pleasingly compact, and a general result of Krasner says that if you jiggle the coefficients a little, the extension doesn’t change. You might be able to use all this to construct your (finitely many) fields.

Not much of an answer, I know, but it was too long for a comment. It’s a nice question, though, and I think I’m going to worry it over a little.

EDIT — Expansion:

I told no lies above, but that’s not the way to look at this problem. As I reached the solution, I realized that there are really two questions here. Consider the simplest case, which you mentioned, the Eisenstein polynomial $X^3-2$. If you think of it abstractly, there’s only the one extension of $\Bbb Q_2$ here, but if you think of the subfields of some algebraically closed containing field, there are three fields, generated by $\lambda$, $\omega\lambda$, and $\omega^2\lambda$, where $\lambda$ is a chosen cube root of $2$ and $\omega$ is a primitive cube root of unity.

As usual, if you take the displayed cubic above and make a substitution $X'=X-2a/3$, you’ll get a new Eisenstein polynomial, but without a quadratic term. Now, if you calculate the discriminant of $X^3+2bX+2(1+2c)$, you’ll get $\Delta=-32b^3-27(1+4c+4c^2)$; and since $c$ and $c^2$ have same parity, we get $\Delta\equiv-3\pmod8$, definitely not a square, indeed $\sqrt\Delta\in\Bbb Q(\omega)$, hardly a surprise, I suppose. And the splitting field of our polynomial will be a cubic extension of $k=\Bbb Q_2(\omega)$, all of which we know. We need only calculate the group $k^*/(k^*)^3$, and its cyclic subgroups (of order $3$) will tell us the cubic extensions of $k$. That’s Kummer Theory, as I’m sure you know.

Let’s call $\Bbb Z_2[\omega]=\mathfrak o$, that’s the ring of integers of $k$. To know $k^*/(k^*)^3$ we have to look at the groups $1+2\mathfrak o\subset \mathfrak o^*\subset k^*$. Now the principal units $1+2\mathfrak o$ are uniquely $3$-divisible, so no contribution to $k^*/(k^*)^3$; the next layer, $\mathfrak o^*/(1+2\mathfrak o)$ is cyclic of order $3$, generated by $\omega$, and $k^*/\mathfrak o^*$ is infinite cyclic, that’s the value group. So $k^*/(k^*)^3$ is of dimension two as an $\Bbb F_3$-vector space, and has only four one-dimensional subspaces. One is spanned by $\omega$, and its cube roots generate an unramified extension, so is not of interest to us. The other three are spanned by $2$, $2\omega$, and $2\omega^2$. ( ! )

And that’s it. Contrary to my expectation and perhaps yours, the only cubic ramified extensions of $\Bbb Q_2$ within an algebraic closure are the three I mentioned in the first paragraph of this Edit.

• Didn't see your (updated) answer till now... this is great, thanks! – Aaron Johnson Feb 13 '18 at 4:24

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 $3 \mid \Delta$ .

If one considers any ramified extension I would say that we cannot expect a general theory as simple as the unramified case. Check this question.

Nonetheless, there is a pleasant description of the abelian extensions of a local field $K$ given by the theory of Lubin-Tate formal groups, local class field theory. For example one can describe all abelian ramified extensions of a local field. This helps in your case to prove that every cubic (totally) ramified extension of $\mathbb Q_2$ is not Galois. Indeed, if such a thing exists it is an abelian extension. Local class field theory builts a group homomorphism $$\mathbb Q_2^\times \longrightarrow Gal(\mathbb Q_2^{ab}/\mathbb Q_2)$$ with some cool properties. For instance index-n subgroups of $\mathbb Q_2^\times$ correspond bijectively to abelian degree-n extensions of $\mathbb Q_2$.

The group $\mathbb Q_2^\times \simeq \mathbb Z\times \mathbb Z_2\times \{\pm1\}$ has one index-3 subgroup, $3\mathbb Z\times \mathbb Z_2\times \{\pm1\}$. By unicity $\mathbb Q_2(\zeta_7)$ is the only Galois cubic extension of $\mathbb Q_2$.

So, cubic ramified extensions of $\mathbb Q_2$ have dihedral Galois closure. Maybe next step is to study $D_3$- Galois extensions of $\mathbb Q_2$. Elliptic curves provide some examples I guess.