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.

  • 1
    $\begingroup$ Didn't see your (updated) answer till now... this is great, thanks! $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.