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Local class field theory concerns with norm subgroups, in particular they correspond to abelian extension of the base field. Giving that in a galois extension L/K can never be that: $N^L_K(L^\times) = K^\times$, as this would imply the abelianized of such group to be trivial. This cannot happen since galois of local fields are solvable,, by ramification filtration.

I cannot prove for L/K not necesserly Galois, my attempt:

consider L/K and take F the galois clousure of L over K, denote $L = F^H$, then we have the diagram given by local reciprocity map:

I cannot upload it, but there's the inclusion $H^{ab}\to G^{ab}$ on the right and on the left there are $N^L_K: L^\times/N^F_L(F^\times)\to K^\times/N^F_K(F^\times)$.

Where the map on the left, $N^L_K: L^\times/N^F_L(F^\times)\to K^\times/N^F_K(F^\times)$ is surjective, obtained from $H^{ab}\to G^{ab}$ by local reciprocity map. Does this give a contradiction?

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You can have ${\rm N}_{L/K}(L^\times) = K^\times$ in the Galois case: use $L = K$. Of course that is a trivial situation.

And for non-Galois extensions you can have ${\rm N}_{L/K}(L^\times) = K^\times$ when $[L:K] > 1$! In fact, we always have ${\rm N}_{L/K}(L^\times) = {\rm N}_{M/K}(M^\times)$ where $M$ is the maximal abelian extension of $K$ in $L$. This is called the norm limitation theorem. Thus ${\rm N}_{L/K}(L^\times) = K^\times$ if and only if the maximal abelian extension of $K$ in $L$ is $K$ itself.

For example, ${\rm N}_{L/K}(L^\times) = K^\times$ when $K = \mathbf Q_2$ and $L = \mathbf Q_2(\sqrt[3]{2})$.

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  • $\begingroup$ right, thanks! I was getting confused. $\endgroup$ Commented Jul 19 at 16:42
  • $\begingroup$ In this case it does in fact happen that the normal clousure has Galois D_3 and L is corresponding to the reflection. (call such group H) Proving that it does in fact happen that $H^{ab} = H $ is equal to $G^{ab} = D_3/<r> = H$ $\endgroup$ Commented Jul 19 at 16:44

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