Image of norm map for local field  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 $2$.


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*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 $e(E|F)$?
 A: As KConrad points out, you perhaps mean to say that $F$ is a finite extension of $\mathbf{Q}_2$ or of $\mathbf{F}_2((x))$, and that the quadratic extesnions $E|F$ is separable (and hence galoisian) in the second case.
With this interpretation of the question, $N_{E|F}(E^\times)$ is a closed subgroup of index $2$ in $F^\times$, and every closed subgroup of index $2$ in $F^\times$ is of this form.  In particular, the ramification index $e_{E|F}$ does not determine the subgroup in question.
For more on this, see the relevant chapter in Serre's Corps locaux (=Local fields) or the book by Fesenko and Vostokov, among many other places.
Addendum.  In the same vein as David Speyer's example, it might also be instructive to work out the case $F=\mathbf{F}_2((x))$,by using "Artin-Schreier   theory" instead of "Kummer theory".  Cyclic quadratic extensions of $F$ correspond to $\mathbf{F}_2$-lines $D\subset F/\wp(F)$, where $\wp$ is the endomorhphism $t\mapsto t^2-t$ of the additive group of $F$.  The quotient $F/\wp(F)$ carries a natural filtration, coming from the filtration on the additive group $F$, so every line $D$ has a level in this filtration.  The ramification index of the cyclic quadratic extension $E_D=F(\wp^{-1}(D))$ of $F$ corresponding to $D$ depends only on this level.
It would be interesting to work out the norm subgroup $N_{E_D|F}(E_D^\times)\subset F^\times$ for each line $D$.
By the way, the Second Extended Edition of Fesenko-Vostokov is available online.
Addendum 2.  See this answer for a generalisation from $p=2$ to arbitrary primes $p$.
A: 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.
