How natural is the reciprocity map?

Question

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.

Answer 1

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.

Answer 2

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.