Local reciprocity map applied to norm

Question

This question concerns the local reciprocity homomorphism $r_L : L^\times \to G_L^\text{ab}$, where $L$ is a local field with absolute Galois group $G_L$. If $K\subset L$ is a subfield and $r_K$ is the associated reciprocity homomorphism, I know that $r_K\circ N_{L/K}=\text{res}_{L/K}\circ r_L$, where $N_{L/K}$ is the norm map and $\text{res}_{L/K}(\sigma)=\sigma|_{K^\text{ab}}$. What I want to know is, what can I say about $r_L\circ N_{L/K}$?

Answer 1

• Answering to your 1-rst question : $i_{L/K}$ . $N_{L/K}$ is just raising to the n-th power, where n is the degree of L/K, and this is it

• As for your 2-nd question, there is no functorial way (as opposed to computational way) to express the restriction of a transfer map $G^{ab} --> H^{ab}$ (where H is a subgroup of G) to a subgroup of $G^{ab}$, so I think the most natural answer would be : use local reciprocity to identify Gal($K^{ab}$/$L$) with ${L^*}$/*N*$K^{ab}{^*}$ and Gal($K^{ab}$/$K$) with K^= the completion of ${K^*}$ w.r.t. the normic topology (or the profinite topology, it's the same); then Gal($K^{ab}$/$L$)$->$ Gal($K^{ab}$/$K$)$-> Gal$($L^{ab}$/$L$) is identified with the composite of natural maps ${L^*}$/*N*$K^{ab}{^*}->$*K*^-->L^. Note that the leftmost map is induced by the norm from L to K, and *N*$K^{ab}{^*}$ goes to $1$ because the subgroup of universal norms (= intersection of all normic subgroups) of K is $1$ by local class field theory. Any more explicit description would require explicit reciprocity laws, i.e.explicit local symbols - which are explicit only in particular amenable cases.

Answer 2

The composite map in your last sentence does not make sense. I suppose that this is a mistake and that, instead of the norm $N_{L/K}$, you meant the inclusion $i_{L/K}$. Then the composite map makes sense, and is classically known to equal V. $r_K$, where V is the so called Verlagerung or transfer map. This is a natural, but non trivial map, which was used by Artin-Furtwängler to prove the famous "principal ideal theorem". The most concise and precise exposition is, I think, chapter VI (by Serre) in Cassels & Fröhlich's book "Algebraic Number Theory", Academic Press, 1967 .

Answer 3

Sorry, it's some sort of dyslexia from my part. Let's start again. On the Galois side, because of the maximality of $L^{ab}$ (= the maximal abelian extension of L), the extension $L^{ab}$/K is normal. Hence G = Gal(L/K) acts on $X_L$ = Gal($L^{ab}$/L) via inner automorphisms, in other words, $X_L$ is a module over the group algebra Z[G]. This algebra contains a specific element, the trace T = sum of all the elements of G , which induces the so called algebraic norm map $N_a$ : $X_L$ $->$ $X_L$ . On the side of the fields themselves, we also have an algebraic norm, which is $i_{L/K}$ . $N_{L/K}$, and we can immediately check that $i_{L/K}$ . $N_{L/K}$ . $r_L$ = $N_a$ . The best way to visualize this is to draw two commutative triangles with the obvious maps in the two categories (fields and Galois groups) related by the reciprocity functor .