# What are the quadratic extensions of $\mathbb{Q}_2$?

How do you classify the non-squares in $\mathbb{Q}_2$? I've tried writing down expansions for "odd" numbers in $\mathbb{Z}_2$, but unlike in $\mathbb{Z}_p$, the n$^{th}$ term in the expansion is not always uniquely determined once you know the first n-1 terms, and when you come across a non-square it's not obvious (to me at least) whether it is divisible by another non-square you've already found.

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There are seven distinct quadratic extensions of $\mathbb Q_2$. My book does not give a proof of this, but references Serre's A Course in Arithmetic. – Alex Becker Feb 25 '12 at 22:58
Aren't the squares things of the form $4^m (1 + 8n)$, for integers $m$ and 2-adic integers $n$? – Hurkyl Feb 25 '12 at 23:00
Ah, now I think I know what you were trying to describe. Try working with more digits of precision when solving. Or just use Newton's method to compute square roots. – Hurkyl Feb 25 '12 at 23:21

As for any field $K$ of characteristic different from $2$, the quadratic extensions are all of the form $K(\sqrt{d})$ for $d \in K^{\times} \setminus K^{\times 2}$; moreover $K(\sqrt{d_1}) \cong K(\sqrt{d_2}) \iff d_1 = a^2 d_2$. Thus they are parameterized by the nontrivial elements of $K^{\times}/K^{\times 2}$, it is then sufficient to understand this quotient. Note that this is an $\mathbb{F}_2$-vector space, so it's enough to determine its dimension. In what follows I will denote this $\mathbb{F}_2$-dimension simply by "$\operatorname{dim}$".

If $K$ is the fraction field of a discrete valuation ring $R$, then from $K^{\times} \cong R^{\times} \oplus \mathbb{Z}$ it is easy to see that

$\dim K^{\times}/K^{\times 2} = 1+ \dim R^{\times}/R^{\times 2}$.

So, here, you want to know the square classes in $\mathbb{Z}_2^{\times}$. I claim that an element $u \in \mathbb{Z}_2^{\times}$ is a square iff its residue modulo $8$ is a square in $\mathbb{Z}/8\mathbb{Z}$: to see this, use Hensel's Lemma. From this it follows that

$\dim \mathbb{Z}_2^{\times} / \mathbb{Z}_2^{\times 2} = \dim (\mathbb{Z}/8\mathbb{Z})^{\times} / (\mathbb{Z}/8\mathbb{Z})^{\times 2} = 2$

and thus

$\dim \mathbb{Q}_2^{\times} / \mathbb{Q}_2^{\times 2} = 3$.

This argument should give you explicit representatives as well: that is, the $2^3-1 = 7$ quadratic extensions of $\mathbb{Q}_2$ are gotten by adjoining square roots of $3,5,7,2,6,10,14$.

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Thanks. I knew Hensel's lemma was lurking in there somewhere. – Brett Frankel Feb 25 '12 at 23:30