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.
Tell me more
×
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.
|
|
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 $3,5,7,2,6,10,14$. |
|||
|
|
|
Ref: Stewart & Tall**, the integers of $\mathbb{Q}(\sqrt{d})$, where $d$ is square-free, are characterized according to the residue of $d \bmod 4$. N.B. If $d \equiv 1 \mod 4$ the algebraic integers of $\mathbb{Q}(d)$ are $\mathbb{Z}(\sqrt{d})$. When d is NOT congruent to 1 mod 4, the algebraic integers are of the form $1/2 + 1/2\sqrt{d}$. **Ian Stewart & David Tall (2002) Algebraic Number Theory and Fermat's Last Theorem. K. Peters Ltd. |
|||||
|
