# 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.

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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.

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Thanks a lot for your references very much! – user1832 Oct 28 '13 at 0:10
Much appreciated for the new notes! – user1832 Nov 18 '13 at 11:55

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...

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Thanks a lot for this detailed explanation. – user1832 Oct 28 '13 at 0:11
A trivial typo: "non-invertible elements in $E^\times$" should be "non-invertible elements in $U_E$". – L Spice Feb 24 '15 at 4:39
By definition, $U_E$ are the units, so all its elements are invertible... – Filippo Alberto Edoardo Feb 24 '15 at 6:20
Oh, sorry, I thought it was the ring of integers (I don't know why). Well, "non-invertible elements in the ring of integers of $E$" (however you like to denote that). – L Spice Feb 25 '15 at 23:32

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 $$U(1)\simeq \frac{E^{\times}}{F^{\times}}.$$

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nice observation, thanks. – user1832 Oct 29 '13 at 0:09
However, it is worth noting that this isomorphism does not fit into the canonical sequence $1 \to U(1) \to E^\times \to F^\times \to 1$; that is, that the induced map $E^\times/F^\times \cong U(1) \to E^\times \to E^\times/F^\times$ is not the identity. – L Spice Feb 24 '15 at 4:38