local class field theory (Norm map)

Question

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 $K^{\times n}$?\

Answer 1

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.

Answer 2

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.

Answer 3

(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 $Im(N)$.