Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

1 Answer

up vote 4 down vote accepted

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 $K^{\times}/K^{\times 2}$. 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

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

and thus

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

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 $3,5,7,2,6,10,14$.

share|improve this answer
    
Thanks. I knew Hensel's lemma was lurking in there somewhere. –  Brett Frankel Feb 25 '12 at 23:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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