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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 $2$.

  1. What is the image of the norm map $N_{E/F}$?

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

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  • $\begingroup$ What does "some extension" mean? If you meant it to be a finite extension or an algebraic extension, please say so. Also, is the quadratic extension in the function field case supposed to be Galois or not? $\endgroup$
    – KConrad
    Commented Feb 14, 2011 at 11:09
  • $\begingroup$ Thanks for you comment KConard. I have edited the question. $\endgroup$
    – Pooja Singla
    Commented Feb 14, 2011 at 12:55
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    $\begingroup$ Just a remark: while (as the answers below show) it is not true that $[F^\times : N_{E/F} E^\times]$ has anything to do with ramification, it is true for any finite abelian extension of local fields that $[\mathcal{O}_F^\times : N_{E/F} \mathcal{O}_E^\times] = e_{E/F}$. $\endgroup$
    – David Loeffler
    Commented Feb 14, 2011 at 15:49
  • $\begingroup$ Thank you very much for your comment David. It is very useful for me. Would you suggest a reference for this result? I am also trying to understand the image of groups $1 + \pi^t \mathcal{O}_E$, $t \in \mathbb{Z}$ under the norm map in above situation. Do you have any comments for these groups? $\endgroup$
    – Pooja Singla
    Commented Feb 15, 2011 at 8:24
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    $\begingroup$ Serre's "Local Fields" is the bible here. Milne's notes on local class field theory (from his website) are also very good. As for the subgroups you mention: if t is positive and reasonably large, the p-adic logarithm map will give an isomorphism between $(1 + \pi^t \mathcal{O}_E, \times)$ and $(\pi^t \mathcal{O}_E, +)$, and the norm on the left-hand side corresponds to the trace on the right, which reduces it to a much easier question. For t positive but small there might be some messier behaviour. I don't know what the question means for $t < 0$. $\endgroup$
    – David Loeffler
    Commented Feb 15, 2011 at 14:01

2 Answers 2

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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 $p$.

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  • $\begingroup$ Thank you very much for your reply. I can not read the fourth line of your reply, would you please edit it a little bit? Thanks. $\endgroup$
    – Pooja Singla
    Commented Feb 14, 2011 at 13:01
  • $\begingroup$ Is it better now ? $\endgroup$
    – Chandan Singh Dalawat
    Commented Feb 14, 2011 at 13:04
  • $\begingroup$ Yes, surely it is. Thank you very much for your reply. $\endgroup$
    – Pooja Singla
    Commented Feb 15, 2011 at 7:57
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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.

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