TITLE: Galois extension of $\mathbb{Q}_2$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$
QUESTION [8 upvotes]: I am trying to solve this exercise:
Prove that $\mathbb{Q}_2$ has a unique Galois extension F with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$. Compute it's ramification groups.
Here is what I have done so far: 
Im will use the following lemma: If F has residue class degree f, then it contains $K=\mathbb{Q}_2(\zeta_{q^f-1})$ and F/K is totally ramified, while $K/\mathbb{Q}_2$ is unramified. The extension is of degree 8 and so it has residue class degree f = 1,2,4 or 8.
f=8: This can't happen since unramified extensions of $\mathbb{Q}_2$ have cyclic Galois group.
f=4: This can't happen either, since the extension has an unramified part of order 4, which is $\mathbb{Q}_2(\zeta_{15})/\mathbb{Q}_2$ and the Galois group of this extension is cyclic of order 4. (F has no element of order 4)
f=2: This works for example for $F=\mathbb{Q}_2(\zeta_3,\zeta_8)$, so this must be the only solution... Im not sure how to prove it though. By the lemma I've been using it must contain $\zeta_3$ and the other extension is an Eisenstein polynomial of order 4.
f=1: F is a totally ramified extension, so it comes from adjoining a root of an Eisensteinpolynomial. How do I get a contradiction here?

REPLY [9 votes]: You can prove that $\mathbb{Q}_2$ has exactly three quadratic extensions such that any one of them is disjoint from the compositum of the other two. More precisely, you can prove that $\mathbb{Q}_2^\times/(\mathbb{Q}_2^\times)^2\cong\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. That will be enough because, as David said in his answer, coset representatives of $\mathbb{Q}_2^\times/(\mathbb{Q}_2^\times)^2$ are in bijection with quadratic extensions of $\mathbb{Q}_2$.
To prove the claimed isomorphism, write down everything you know about the group structure of $\mathbb{Q}_2^\times$.
As an exercise, you might want to repeat for $\mathbb{Q}_p$, $p$ odd.