structure of norm one group for quadratic extension of p-adic fields Let $F$ be a p-adic field (finite extensions of $\mathbb{Q}_p$ for some prime $p$), and $E/F$ be a quadratic extension. Use $\sigma$ to denote the nontrivial element in the Galois group $Gal(E/F)$. For any $x\in E$, $N(x)=x\sigma(x)$ denotes the norm of $x$.
Now consider $U(1)=\{x\in E; N(x)=1 \}$, the group of elements of norm one in $E$.We know that $U(1)$ is a subgroup of $U_{E}$, the group of units of ring of integers in $E$, thus it is a compact subgroup (correct me if I'm wrong).
I'm wondering if there are more structural results about $U(1)$? I don't have any specific meaning about 'structure', so any answer reasonable in certain sense is welcome.
Sorry for the question is not so clear, any reference is much appreciated.
 A: The answer will depend upon whether your quadratic extension is unramified, tamely ramified, or wildly ramified.  A good place to start would be the chapter on the norm map in Fesenko-Vostokov (Ch. III in Local fields and their extensions) or the corresponding chapter in Serre's Corps locaux.  Both these accounts are based on a paper of Hasse from the early 1920s.
Addendum (2013/11/18)  Dear user1832, it so happens that I was teaching exactly this topic in my course last week, and I think my notes might be useful to you.  Here they are.
A: Just to expand a bit on Dalawat's answer, let me make a couple of observations. First, there is a very explicit description of $U(1)$, namely
$$
U(1)=\{x\in E\text{ such that }x=\frac{y}{\sigma y}\text{ for some }y\in E\}
$$
coming from Hilbert 90 which says $H^1(E/F,E^\times)=1$ coupled with the information that, since the Galois group here is cyclic$^\dagger$ there is an isomorphism
$$
H^1(E/F,E^\times)\cong \hat{H}^{-1}(E/F,E^\times)\stackrel{\text{def}}{=}U(1)/\{y/\sigma y: y\in E\}.
$$
This already tells you something but there is more to say: namely, we could try to describe $U(1)$ in terms of units only, not using non-invertible elements in $E^\times$. Then a classical result (see for instance Serre's paper in Cassels-Fröhlich) tells you that the cohomology of units in unramified extensions is trivial, so the same argument above becomes
$$
U(1)=\{x\in U_E\text{ such that }x=\frac{y}{\sigma y}\text{ for some }y\in U_E\}\quad\text{ if }E/F\text{ unramified}.
$$
If $E/F$ is ramified, there are two options: either $p=2$ or $p>2$. In the second case, the restriction-corestriction trick kills a lot of the cohomology of the units: writing $U_E=U'\times\mu_E$ with $U'$ being the principal units which are $1$ modulo the maximal ideal, we know $U'$ is a pro-$p$ group and hence has trivial cohomology. We find 
$$
H^1(E/F,\mu_E)\cong\hat{H}^{-1}(E/F,\mu_E)\stackrel{\text{res-cores}}{\cong}\hat{H}^{-1}(E/F,U_E)\stackrel{\text{def}}{=}U(1)/\{y/\sigma y: y\in U_E\}.
$$
and you see that "up to the finite group $H^1$ of roots of unity" the answer is still the same, namely that the kernel of norm is the same as elements $\sigma y/y$.
The above fails is $E/F$ is ramified and $p=2$ (the wildly ramified case Dalawat's is referring to) but in Serre's paper quoted above you will find a Lemma (I am sorry for being unable to give proper references, I do not have my Cassels-Fröhlich at hand) telling you that my argument still holds by replacing the whole $U_E$ with a smaller, but still finite-index, subgroup and hence the vague sentence that "up to a finite group the kernel of norm is the same as elements $\sigma y/y$" still holds. The exact determination of this finite group in general depends upon the extension $E/F$.
$^\dagger$: Purists would say that what I write is historically upside-down, since Hilbert first proved what I am claiming in the cyclic case and that this was only later translated in cohomological language...
A: Aside from arithmetic point of view provided by Dalawat and Edoardo, 
let me give a group theoretical point of view about this group.
If $E/F$ is a separable quadratic field extension with the Galois group $\{id, \sigma\}$ then the map
$$\phi:E^{\times}\rightarrow U(1), \ \ x\mapsto \frac{x}{\sigma(x)}$$ is surjective (as pointed out by Edoardo). We also have $\ker\phi=F^{\times}$. In other words we have a group isomorphism  $$U(1)\simeq \frac{E^{\times}}{F^{\times}}.$$
