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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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 $\bf Q_p$.

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.