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For local field, the reciprocity map establishes almost an isomorphism from the multiplicative group to the Abelian Absolute Galois group. (In global case the relationship is almost as nice). It is tempted to think that there can be no such nice accident.

Do we know any explanation which suggest that there "should be" a relationship between the multiplicative group and the Galois group?

Actually, my current belief is that the reciprocity map is half accidental. I think that there is a natural extension where we can define a natural action of the multiplicative group. In the local case this is the Lubin-Tate extension (a generalization of cyclotomic extension). The fact that this Lubin-Tate extension is the Abelian Absolute Galois group is an accident.

Do we know something that might support/ reject this view?

Please feel free to edit the question into a form that you think might be better.

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up vote 11 down vote accepted

The reciprocity map is completely natural (in the technical sense of category theory). For example, if $K$ and $L$ are two local fields, and $\sigma:K \rightarrow L$ is an isomorphism, then $\sigma$ induces an isomorphism of multiplicative groups $K^{\times} \rightarrow L^{\times}$ and also of abelian absolute Galois groups $G\_K^{ab} \rightarrow G\_L^{ab}$. The reciprocity laws for $K$ and $L$ are then compatible with these two isomorphisms induced by $\sigma$.

On the other hand the factorization $K^{\times} = {\mathbb Z} \times {\mathcal O}\_K^{\times}$ is not canonical (it depends on a choice of uniformizer), and the identification of ${\mathcal O}\_K^{\times}$ with the Galois group of a maximal totally ramified abelian extension of $K$ also depends on a choice of uniformizer (which goes into the construction of the Lubin--Tate formal group, and hence into the construction of the totally ramified extension; different choices of uniformizer will give different formal groups, and different extensions).

As others pointed out, the local reciprocity map is also a logical consequence of the global Artin map and global Artin reciprocity law (which makes no reference to local multiplicative groups, but simply to the association $\mathfrak p \mapsto Frob\_{\mathfrak p}$ of Frobenius elements to unramified prime ideals; see the beginning of Tate's article in Cassels--Frohlich for a nice explanation of this). Thus it is natural in a more colloquial sense of the word as well.

Indeed, the idelic formulation of the glocal reciprocity map and the formulation of the local reciprocity map in terms of multiplicative groups are not accidental or ad hoc inventions; they were forced on number theorists as a result of making deep investigations into the nature of global class field theory.

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Thank you for the reply. I learned a lot from your answer but still I am not convinced. I will post my objections in different comments. – abcdxyz Jan 20 '10 at 20:56
About the first part of your answer. I feel fortunate that you post this comment. I tried to find something like that for a while. In Cassels & Frohlich, Serre wrote that the change of base from $K^{\times} \rightarrow L^{\times} $ induced the transfer from $G^{ab}_K \rightarrow G^{ab}_L $. I was scared since I don't know much about group theory. Your comment above gave me a better feeling. However, this only say that if such a map exists it is natural but not yet the natural existence of such a map. – abcdxyz Jan 20 '10 at 21:06
I don't quite agree with the second paragraph. Even though, the factorization $K^{\times}= \Z \times O^{\times}_K $ is not canonical, this can be fixed by looking at the exact sequence $ O^{\times}_K \rightarrow K^{times} \rightarrow Z$ where the second map is given by the additive valuation. This non-canonical problem won't affect much. The construction of Lubin Tate formal group not only depends on the uniformizer but also on the choice of the power series. The resulting maximal totally ramified abelian extensions of K depends on the uniformizer. – abcdxyz Jan 20 '10 at 21:18
However, once these choice are made the identification of $O^{times}_K$ and the Galois group of the chosen extension is canonical. So I don't think the non-canonical fact will give a very serious problem here either. – abcdxyz Jan 20 '10 at 21:21
For the third paragraph. I am convinced that this showed that the local reciprocity map is what we must get at the end. This explanation is still not satisfying in the same manner that deriving local class field theory from global one is not very satisfying. – abcdxyz Jan 20 '10 at 21:27

The reciprocity map in the local case can be motivated by considering the unramified case. Let me try to explain this. For simplicity, let us consider the case of a finite abelian unramified extension $K_n/\bf Q_p$ of degree $n$. Denote by $k_n$ the residue field of $K_n$. The unramified condition gives rise to the isomorphism $$Gal(K_n/ {\bf Q_p}) \simeq Gal(k_n / {\bf F_p}) \simeq {\bf Z}/n.$$

Now we like to parametrize these abelian unramified extensions {$ K_n$} of $\bf Q_p$ using information from $\bf Q_p$. However, $\bf Q_p$ is not finitely generated as a module over $\bf Z_p$, let along over $\bf Z$. $\bf Q_p^\times$, on the other hand, decomposes as $$\bf Q_p^\times \simeq p^{\bf Z} \times {\bf Z}_p^\times \simeq \bf Z \times \mu_{p-1} \times Z_p,$$ which, if anything, at least contains a copy of $\bf Z$ (depending on the choice of a uniformizer, say $p$).

It then seems somewhat natural to consider the map $${\bf Q_p^\times} \xrightarrow{Art} Gal({\bf Q^{ab, un}_p}/{\bf Q_p}) \qquad p \mapsto Frob$$

where $Frob$ is a choice of a topological generator for $Gal({\bf Q_p^{ab, un}}/{\bf Q_p})$, the Galois group of the maximal abelian unramified extension ${\bf Q_p^{ab, un}}$ of $\bf Q_p$, which is naturally isomorphic to $Gal(\bar{\bf F_p} / \bf F_p)$ (which is itself non-canonically isomorphic to $\hat {\bf Z}$).

Now compose the Artin map $Art$ with the restriction map $Gal({\bf Q_p^{ab, un}}/{\bf Q_p}) \to Gal(K_n/ {\bf Q_p})$, we obtain a map $${\bf Q_p}^\times \xrightarrow{Art_n} Gal(K_n/ {\bf Q_p})$$ whose kernel is $$p^{n {\bf Z}} \times \bf Z_p^\times,$$ which coincidentally is also the image of the norm of $K_n^\times$ in $\bf Q_p^\times$.

This also sheds some light on the global situation, where you have Frobenius at all but finitely many primes. Don't quote me on this, but I recall the Artin reciprocity map is uniquely determined by its action on the Frobenii (I assume a Chebatorev density argument will show this, and you can prove Chebatorev Density Theorem independent of class field theory if I remember correctly.)

Lastly, we saw that the (local) reciprocity map depends on the choice of a Frobenius as well as that of a uniformizer.

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Thank you for the reply. I think that this explains the part related to unramified extensions. However, it appears to me that we still don't have the part about totally ramified extensions yet. – abcdxyz Jan 17 '10 at 19:13
Yes, one can prove the Chebotarev Density Theorem without class field theory (in fact, this is what Chebotarev himself did). Lenstra and Stevenhagen wrote an article on the theorem for the Intelligencer and included a short proof not requiring class field theory as an appendix. It is available here: – Ben Linowitz Jan 18 '10 at 0:15
I just read your answer again. It appears to me the fact that $Q_P^{\times}$ contains $Z$ is not compelling enough to define the Artin reciprocity map as above. Is there a good reason for the domain of the Artin reciprocity map to be $Q_p^{\times} $ and not $Z \times Z_p^{\times} $? – abcdxyz Jan 18 '10 at 1:29

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