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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?

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I take it ${\bf Q}_2$ is the 2-adic completion of the rationals? – Gerry Myerson May 17 '11 at 23:56
Yes, sorry for not mentioning... unfortunately the question will become quite long if I include all the definitions. – Michalis May 17 '11 at 23:59
up vote 10 down vote accepted

I'm not sure I follow your attempted solution. However, here is how I would proceed: If $K$ is a field of characteristic not equal to $2$, then every quadratic extension of $K$ is of the form $K(\sqrt{a})$ for some $a$. (Recall that $\mathbb{Q}_2$ has characteristic zero, not two, so this is relevant.) If $\mathrm{Gal}(L/K) \cong (\mathbb{Z}/2)^3$, then $L$ must be the compositum of three disjoint quadratic extensions of $K$. So you should be trying to find $a$, $b$ and $c$ in $\mathbb{Q}_2$ such that $\mathbb{Q}_2(\sqrt{a}, \sqrt{b}, \sqrt{c})$ is a degree $8$ extension.

As a warm up, which elements in $\mathbb{Q}_2$ are squares?

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Thanks I got it. – Michalis May 18 '11 at 8:01

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.

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Thanks, the answer is $\mathbb{Q}_2(\sqrt{-1},\sqrt{5},\sqrt{2})$ – Michalis May 18 '11 at 8:03

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