MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 8 down vote accepted

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.

share|cite|improve this answer
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...

share|cite|improve this answer
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 $$U(1)\simeq \frac{E^{\times}}{F^{\times}}.$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.