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Why do they have to be units? $\mathbb{Q}_{2}[\sqrt{2}]$ is a quadratic extension but $|2|_{2}=\frac{1}{2}\neq 1$.

Where do I need them to be units below?

$\mathbb{Q}_{2}[\sqrt{a}]=\mathbb{Q}_{2}[\sqrt{b}]$ iff $a=bk^{2}$ i.e. $a\equiv b~ mod (\mathbb{Q}_{2})^{2}$ because:

if $\sqrt{a}=c+d\sqrt{b}\Rightarrow a=c^2+d^2b+2\sqrt{b}cd\Rightarrow a\equiv b ~mod (\mathbb{Q}_{2})^{2}$

if $a=bk^{2}\Rightarrow \sqrt{a}=c\sqrt{b}$

Jyrki Lahtonen
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Thomas Kojar
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    $2$ is invertible in $\Bbb Q_2$ so it's an element of $\Bbb Q_2^*$ (the subgroup of invertible elements of $\Bbb Q_2$). – mercio Apr 22 '14 at 22:19

1 Answers1

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A tacit assumption in that argument surely is that neither $a$ nor $b$ has a square root in $\Bbb{Q}_2$. So once you reached the equation $$ a=c^2+d^2b+2\sqrt{b}cd $$ you can conclude that $cd=0$ for otherwise you can solve for $\sqrt b$ and see that it would be in $\Bbb{Q}_2$. If here $d=0$, then you are left with $\sqrt a=c$ contradicting the assumption that $a$ has no square root. Thus $c=0$ and we are left with $$ \sqrt a=d\sqrt b $$ for some non-zero $d$. So $d\in\Bbb{Q}_2^*$ and $a=bd^2$. Or $a\in b(\Bbb{Q}_2^*)^2$.

Or did you confuse the group of units of the field $\Bbb{Q}_2$ with the group of units of $\Bbb{Z}_2$? In that case all the above was probably clear to you. Sorry.

Jyrki Lahtonen
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