Sorry if the question is too long and maybe elementary.


I am reading a paper by Hirotada Naito on "Dihedral extensions of degree 8 over the rational p-adic fields". To generate dihedral extension $K(\sqrt{\epsilon},\sqrt{\epsilon^\sigma})$ in part 2-1, he said let $K=\mathbb{Q}\_2(\sqrt{m})$ for $m=2,-2, 10$ or $-10$ 
$$K^\*/K^{\*2}\cong (\langle\sqrt{m}\rangle/\langle m\rangle)\times (\mathcal{O}^\times/\langle 1+m+2\sqrt{m}, 1+ \mathfrak{p}^5 \rangle)$$ 
for $\mathfrak{p}= \langle\sqrt{m}\rangle$. 


And after that, he said "it is sufficient to examine $\epsilon$ and $\epsilon\sqrt{m}$ where $\epsilon=a+b\sqrt{m}$ for $a=1,3,5,7$ and $b=0,1,2,3$."


**My first question is "why it is enough to check these numbers?"**


In the following, he said "we take $\epsilon$ (resp. $\epsilon\sqrt{m}$) such that $\epsilon$, $\epsilon^\sigma$, $\epsilon (1+m+2\sqrt{m})$ and $\epsilon^\sigma (1+m+2\sqrt{m})$ (resp. $\epsilon$, $-\epsilon^\sigma$, $\epsilon (1+m+2\sqrt{m})$ and $-\epsilon^\sigma (1+m+2\sqrt{m})$) are different modulo $\mathfrak{p}^5$ each other." Where $\sigma$ is the generator of Galois group of $K/\mathbb{Q}\_2$.


**And in the following, for $m=2$ he took $1+\sqrt{2}$, $3+\sqrt{2}$, $\sqrt{2}$ and $3\sqrt{2}$. I couldn't get to these numbers following his method explained above, can anyone show me the calculations? And can you show me any other possible numbers other than the given ones?**


Here is the paper: [Dihedral extensions of degree 8 over the rational p-adic fields](http://projecteuclid.org/euclid.pja/1195510841).
Thank you. 



Dihedral extension of 2-adic number field
Answer to your first question: The numbers contain a set of representatives of $\mathfrak{o}^\times/(\mathfrak{o}^\times)^2$. (Naito said earlier that "we examine a representative system of $K\_i^\*/(K\_i^\*)^2$", and you copied his equation that shows representatives of this quotient group can be obtained from representatives of $\mathfrak{o}^\times/(\mathfrak{o}^\times)^2$ and their multiples by $\sqrt{m}$.)


It is easy to see that these numbers contain a set of representatives if you note that $\mathfrak{p}^5 = 4\mathfrak{p}$. (This is obvious for $m = \pm 2$. For $m = \pm 10$, note that $5$ is a unit in $\mathfrak{o}$.)


For any $x \in \mathfrak{o}^\times$, write $x^{-1} = a' + b' \sqrt{m}$, and choose $a \equiv a' \pmod{8}$ and $b \equiv b' \pmod{4}$. Then it is easy to see that $x(a + b \sqrt{m}) \in 1 + 8 \mathfrak{o} + 4\mathfrak{p} \subseteq \mathfrak{o}^{\times 2}$ (since $1 + 8 \mathfrak{o} \subseteq (\mathfrak{o}^\times)^2$).




---


Partial answer to your second question: We are supposed to check all 16 possible combinations of $a$ and $b$ (and also consider multiplying each possibility by $\sqrt{m}$). I will only do two of them (and I won't multiply by $\sqrt{m}$).


For $a = b = 1$ (with $m = 2$), we have:


1. $\epsilon = 1 + \sqrt{2}$,
2. $\epsilon^\sigma = 1 - \sqrt{2}$,
3. $\epsilon(1+m+2 \sqrt{2}) = (1 + \sqrt{2})(3+2 \sqrt{2}) = 7 + 5 \sqrt{2}$,
4. $\epsilon^\sigma(1+m+2 \sqrt{2}) = (1 - \sqrt{2})(3+2 \sqrt{2}) = -1 - \sqrt{2}$.


It is easy to check that none of these are congruent modulo $4 \sqrt{2}$:
$$x + y \sqrt{2} \equiv x' + y' \sqrt{2} \pmod{4 \sqrt{2}} \iff \text{$x \equiv x' \pmod{8}$ and $y \equiv y' \pmod{4}$} .$$
 So this gives an example, which means that $1 + 2 \sqrt{2}$ should be on Naito's list. It is indeed there.


For $a = 1$ and $b = 2$ (with $m = 2$), we have:


1. $\epsilon = 1 + 2\sqrt{2}$,
2. $\epsilon^\sigma = 1 - 2\sqrt{2}$,
3. $\epsilon(1+m+2 \sqrt{2}) = (1 + 2\sqrt{2})(3+2 \sqrt{2}) = 11 + 8 \sqrt{2}$,
4. $\epsilon^\sigma(1+m+2 \sqrt{2}) = (1 - 2\sqrt{2})(3+2 \sqrt{2}) = -1 - 2\sqrt{2}$.


Since 
 $$\epsilon - \epsilon^\sigma = (1 + 2\sqrt{2}) - (1 - 2 \sqrt{2}) = 4 \sqrt{2} \equiv 0 \pmod{4 \sqrt{2}} ,$$
this does not give an example. So $1 + 2 \sqrt{2}$ should not be on Naito's list. It is indeed not there.


By checking each of the 32 possibilities for each of the four possible values of $m$, it should be easy to determine whether his list is correct.


