TITLE: Image of norm map for local field 
QUESTION [6 upvotes]: Let  $F$ be a finite extension of $Q_2$ (2-adic field) or $F_2((x))$ (function field). Let $E/F$ be a separable extension of degree $2$.

What is the image of the norm map $N_{E/F}$?
In particular - is it true that the index $[F^{\star} : N_{E/F}(E^{\star})]$   depends only on the ramification $e(E|F)$?

REPLY [12 votes]: It might be worth working the example of $\mathbb{Q}_2$. Recall that $u \in \mathbb{Q}_2$ is a square if and only if $u$ is of the form $4^k (1+8 \ell)$ for $\ell \in \mathbb{Z}_2$. So $\mathbb{Q}_2^{\times} / (\mathbb{Q}_2^{\times})^2$ has order $8$, with representative elements being $1$, $3$, $5$, $7$, $2$, $6$, $10$ and $14$. This is an important computation for two reasons: 
(1) Quadratic extensions of a characteristic zero field are of the form $K(\sqrt{a})$ for some nonsquare $a$, and two different $a$'s give the same extension if there ratio is a square. So there are $7$ quadratic extensions of $\mathbb{Q}_2$. The unramified one is $\mathbb{Q}_2(\sqrt{5})$. 
(2) If $L$ is any quadratic extension, then $(\mathbb{Q}_2^\times)^2 \subset N_{L/\mathbb{Q}_2} (L^\times)$ as, for $a \in \mathbb{Q}_2$, we have $N(a)=a^2$. So we can describe the norm group by giving its image in the $8$ element group $\mathbb{Q}^{\times}_2/(\mathbb{Q}_2^{\times})^2$.
So, for example, in $\mathbb{Q}_2(\sqrt{3})$, the norms are elements of the form $a^2 - 3 b^2$. A little checking shows that the image in $\mathbb{Q}^{\times}_2/(\mathbb{Q}_2^{\times})^2$ is represented by $\{ 1, 1-3 \cdot 2^2,  3^2 -3,  1 - 3 \} \equiv \{ 1, 5, 6, 10 \}$. You can enjoy writing down the $4$ element subgroup of $\{ 1,3,5,7,2,6,10,14 \}$ corresponding to each of the $7$ quadratic extensions.