how to calculate the norm of the above fields $K^{\times}$? Let $K$ be the quadratic extension of p-adic field $\mathbb{Q}_2$, then $K=\mathbb{Q}_2(\sqrt n)$, where $n=-1, \pm 2, \pm 3, \pm 6$. 
Then I have seen a table in page $34$ of  the book $\text{Arithmetic of quadratic forms}$ by Goro Shimura as follows:
\begin{array}{c  | c}
K & N_{K/\mathbb{Q}_2}(K^{\times}) \\ 
\hline
\mathbb{Q}_2(\sqrt{-3}) & 4^{\mathbb{Z}} \mathbb{Z}_2^{\times} \\ \hline
\mathbb{Q}_2(\sqrt{-1}) & 2^{\mathbb{Z}} \cdot \{ x \in \mathbb{Z}_2 |x-1 \in 4 \mathbb{Z}_2\} \\ \hline
\mathbb{Q}_2(\sqrt{3}) & (-2)^{\mathbb{Z}} \cdot \{ x \in \mathbb{Z}_2|x-1 \in 4 \mathbb{Z}_2 \} \\ \hline 
\mathbb{Q}_2(\sqrt{2}) & 2^{\mathbb{Z}} \cdot \{ x \in \mathbb{Z}_2 |x\pm 1 \in 8 \mathbb{Z}_2\} \\ \hline
 \mathbb{Q}_2(\sqrt{-2})& 2^{\mathbb{Z}} \cdot \{ x \in \mathbb{Z}_2 |x- 1 \in 8 \mathbb{Z}_2 \ \text{or} \ x-3 \in 8 \mathbb{Z}_2\}\\ \hline
\mathbb{Q}_2(\sqrt{6}) &  (-2)^{\mathbb{Z}} \cdot \{ x \in \mathbb{Z}_2 |x- 1 \in 8 \mathbb{Z}_2 \ \text{or} \ x-3 \in 8 \mathbb{Z}_2\}\\ \hline
\mathbb{Q}_2(\sqrt{-6}) & 6^{\mathbb{Z}} \cdot \{ x \in \mathbb{Z}_2 |x\pm 1 \in 8 \mathbb{Z}_2 \}
\\ \hline 
\end{array}
where $N_{K/\mathbb{Q}_2}$ denotes the field norm.
My question is-
I know to calculate field norm at a point of a field in few ways like by multiplication map, by using Galois group etc.
(i)But how to calculate the norm of the above fields $K^{\times}$ ?
(ii) Does it mean $N(K^{\times})=\{N(a)  \ | \  \forall a \in K^{\times} \}$?
(iii) Can you please help me in case $-1, -2$ in the above cases ?
 A: The starting points are:


*

*The norm map $N:\Bbb{Q}_2(\sqrt n)\to\Bbb{Q}_2$ looks like $N(a+b\sqrt n)=a^2-nb^2$.

*The multiplicative group of the smaller field is a direct product $\Bbb{Q}_2^*=2^{\Bbb{Z}}\times \Bbb{Z}_2^*$. Of these the latter factor is equal to $1+2\Bbb{Z}_2$, and often it is prudent to separate the torsion subgroup by writing it as $\Bbb{Z}_2^*=\langle -1\rangle\times(1+4\Bbb{Z}_2)$.

*The squares of elements of $\Bbb{Q}_2^*$ form the subgroup $4^{\Bbb{Z}}\times(1+8\Bbb{Z}_2)$. The modulo $8$ part is a consequence of the extended Hensel's lemma. A trick solution that works for all 2-adic integers. This subgroup is contained in all the images of the norm maps (set $b=0$), so the real question is to determine what else is in the image of the norm.


As $\mathrm{Im}(N)$ is a union of cosets of the subgroup of squares $S=\{q^2\mid q\in\Bbb{Q}_2^*\}$ we record as a consequence of the last bullet the following

$\Bbb{Q}_2^*$ is the union of the eight cosets of $S$ of the form $2^\epsilon u S$ with $\epsilon\in\{0,1\}$ and $u\in\{\pm1,\pm3\}$. The game we play is to determine which cosets belong to the image of $N$ for each choice of $n$. Observe that the choices for $u$ are the odd integers, pairwise non-congruent modulo $8$.

You asked specifically about $n=-1$. In that case we need to observe that:


*

*$N(1+\sqrt{-1})=1^2+1^2=2$, so the coset $2S$ is contained in the image.

*$N(2+\sqrt{-1})=5\equiv-3\pmod8$, so the coset $5S=(-3)S$ is contained in the image. Because the image is a subgroup, it obviously contains the cosets $S$ and $(-6)S$ as well.

*We have seen that $\mathrm{Im}(N)$ contains the listed four cosets of $S$. Because it is a subgroup, by Lagrange and the correspondence theorem, either the image is exactly the union of those four cosets, or the image is all of $\Bbb{Q}_2^*$.

*But $N(a+b\sqrt{-1})=a^2+b^2$ can never be congruent to $3$ modulo $4$ (the proof familiar from the ring $\Bbb{Z}$ survives after you observe that for $N(a+b\sqrt{-1})$ to be a $2$-adic integer it is necessary for both $a,b$ to be in $\Bbb{Z}_2$). Therefore $N$ is not surjective.


So we can conclude that

when $n=-1$, the image of the norm is
  $$\mathrm{Im}(N)=S\cup 2S\cup (-3)S\cup (-6)S,$$ and this matches with the description in your source.

