nt.number theory - How natural is the reciprocity map?

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.