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Let $K$ be a local field, for example the $p$-adic numbers. In Neukirch's book "Algebraic number theory", there is the statement: if $K$ contains the $n$-th roots of unity and if the characteristic of $K$ does not divide $n$, and we set $L=K(\sqrt[n]{K^{\times}})$, then one has $N_{L/K}(L^{\times})=K^{\times n}$.

My questions are the following, for a (finite) Galois extension $L$ of $K$:\

(1) What happens if the characteristic of $K$ divides $n$? Can one obtain an explicit form of the image of the norm of $L^\times$?\

(2) If we don't add the condition that $K$ contains the $n$-th roots of unity, what is the image of the norm operator? If $L = K(x)$, can one write it in terms of the primitive element $x$ and $K^{\times n}$?\

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In general the image of the norm map between local fields is equal to the image of the norm map from the largest abelian extension of $K$ inside $L$. When $L/K$ is abelian, the image of the norm map in general can't be written as concretely as the special case worked out by Neukirch. That Neukirch gave that special example was precisely because in that case it can be worked out very explicitly, unlike in general. –  KConrad Mar 21 '11 at 20:47
For some other special cases where the image of the norm map can be written down explicitly (using local class field theory), see my answer below. –  Chandan Singh Dalawat Mar 22 '11 at 5:30
To KConrad: Thank you for rewrite the question. –  user13726 Mar 22 '11 at 7:15

3 Answers 3

up vote 10 down vote accepted

Let $L/K$ be a finite abelian extension of local fields. Although, there is no generic form for the image of the norm map, $N^{L}_K$, in practice one can follow the following procedure to determine its image.

Choose a uniformizer $\pi_L$ in $\mathcal{O}_L.$ Then $L^{\times}$ is equal to the group generated by $\pi_L$ and $\mathcal{O}_L^{\times}.$ It follows that $N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L) \rangle N^{L}_K \mathcal{O}_L^{\times},$ Hence to determine $N^{L}_K L^{\times}$ it is enough to establish the image of $\mathcal{O}_L^{\times}$ under the norm mapping.

The group $N^L_K \mathcal{O}_L^{\times}$ is a subgroup of $\mathcal{O}_K^{\times}$ and by a group cohomology argument it can be shown that

$[\mathcal{O}_K^{\times}:N^{L}_K\mathcal{O}_L^{\times}] = e(L|K) = [L^{\times} : K^{\times}\mathcal{O}_L^{\times}].$

In particular, if $L/K$ is unramified, $\mathcal{O}_K^{\times} =N^{L}_K\mathcal{O}_L^{\times}$ and hence $N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L) \rangle \mathcal{O}_K^{\times}.$

The norm group of a tamely ramified extension is similarly easy to deduce. Write

$$\mathcal{O}_L^{\times} = \langle \zeta_{q_L - 1} \rangle U_L$$

where $q_L$ is the residue field characteristic of $L$ and $\zeta_n$ denotes a primitive n-th root of unity. Denote the residue field of $L$ by $l$ and that of $K$ by $k,$ then

$$N^L_K(\zeta_{q_L -1}) = N^l_k(\zeta_{q_L -1})^{e(L|K)} = \zeta_{q_K -1}^{e(L|K)}.$$

As $U_K$ is a pro-$p$ group and contains the image of $U_L$ under $N^L_K,$ we have in the tamely ramified case that $U_K = N^L_K(U_L)$ else $\mathcal{O}_K^{\times}/N^{L}_K\mathcal{O}_L^{\times}$ would contain an element of $p$-power order contradicting the equality

$$[\mathcal{O}_K^{\times}:N^{L}_K\mathcal{O}_L^{\times}] = e(L|K)$$

and the fact that $L/K$ was assumed tamely ramfied. It follows in the tamely ramified case that $$N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L), \zeta_{q_K -1}^{e(L|K)}\rangle U_K$$

The case of wild ramification is more difficult. But two facts are helpful. First, in the case $K$ is a p-adic field $U_K^{ap^n} \supset 1 +\mathcal{M}_K^{2ne(k|\mathbb{Q}_p) + 1}$ where $a$ and $p$ are relatively prime. Hence, it is enough to determine the image of the norm mapping in the units of $\mathcal{O}_K \mod \mathcal{M}_K^{2ne(k|\mathbb{Q}_p) + 1},$ a finite set.

Another technique is to determine the higher ramification filtration of $Gal(L/K)^{ab}.$ In practice this can be done by examining the derivatives of the irreducible polynomial of $\pi_L.$ These groups map under the inverse of the artin map to the higher unit groups. As the domain of the inverse of the artin map is $K^{\times}/N^{L}_K L^{\times}$ their sizes reveal norm indexes within the higher unit groups. For a good exposition see Serre's book Local Fields.

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I completely agree with you. –  user13726 Mar 22 '11 at 18:17

Here is another interesting case where the image of the norm map can be written down explicitly. Let $F$ be a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root $\zeta$ of $1$, and denote the filtration on the multiplicative group $F^\times$ by $$ \ldots\subset U_2\subset U_1\subset\mathfrak{o}^\times\subset F^\times. $$ We thus get the extensions $L_n=F(\root p\of{U_n})$. It can be checked that $L_{pe_1+1}=F$, where $e_1$ is the absolute ramification index of $F$ divided by $p-1$, and that $L_{pe_1}$ is the unramified degree-$p$ extension of $F$, so the image of the norm map $L_{pe_1}^\times\to F^\times$ is $\mathfrak{o}^\times F^{\times p}$.

What is the image of the norm map $L_{n}^\times\to F^\times$ for other $n$ ? Local class field theory and a certain orthogonality relation for the Kummer pairing (see for example the last section of arXiv:0711.3878) can be used to answer this question. Basically, for $n\in[1,pe_1]$, $N(L_n^\times)=U_mF^{\times p}$, where $m=pe_1+1-n$.

There are similar results for elementary abelian $p$-extensions of finite extensions of $\mathbf{F}_p((t))$. See for example the last section of arXiv:0909.2541.

These two papers have appeared in J. Ramanujan Math. Soc. 25 (2010), no. 1, 25–80, and 25 (2010), no. 4, 393–417.

There are other instances where the image of the norm map can be computed explicitly. This happens for the cyclotomic extension $K_m$ of $\mathbf{Q}_p$ obtained by adjoining $\root{p^m}\of1$. It can be shown that $p\in N(K_m^\times)$, and that the image of the units of $K_m$ under the norm map down to $\mathbf{Q}_p$ is $1+p^m\mathbf{Z}_p$. See for example Artin, Algebraic numbers and algebraic functions, p. 208, or Neukirch, Class Field Theory, p. 45.

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Thanks a lot! the example you given is very interesting for me, the research papers sometimes are very complicated. –  user13726 Mar 22 '11 at 18:44

(1): You want inseparable extensions? It seems quite easy to handle the inseparable part of the extension.

(2): Since the norm map is functorial, the norm group you ask about is the n-th power of $Im N:K(\zeta_n)^{\times} \to K$. On the other hand, I don't know of a nice way to describe the latter group.

Upd. My (2) is wrong since we don't add all $n$-th power roots of elements of $K(\zeta_n)$. One can only say that the group in question contains $Im(N)^n$ and is contained in $Im(N)$.

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Mikhail Bondarko, thank you. I think the question (1) finally involves the case wild ramification. Chandan has indicated some recent papers. –  user13726 Mar 22 '11 at 18:52

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