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The usual formulation of Artin reciprocity (for example in Neukirch or Lang's book on algebraic number theory), applied to L-funcions, says that for an irreducible character $\chi$ of an abelian extension, $$L(\chi,s)=L(\tilde{\chi},s)$$ where $\tilde{\chi}$ is a Grobencharakter, and the L-series on the right is an abelian (Hecke) L-funcion.

But another formulation is that Artin reciprocity implies the equality above for all 1-dimensional characters.

How does one prove this?

I'm sure this is an easy consequence, but I don't understand it. It is easy to see that:

abelian extension $\to$ 1-dimensional character

But the converse is not true, there are plenty of 1-dimensional characters of non-abelian extensions.

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    A one-dimensional character of a finite group $G$ is a homomorphism $G \rightarrow \mathbf C^\times$, so it factors through a homomorphism $G' \rightarrow \mathbf C^\times$ where $G' = G/[G,G]$ is the abelianization of $G$. Thus any one-dim. char. of a finite group is a one-dim. character of this (maximal) abelian quotient of the group, and if $G = {\rm Gal}(E/F)$ then this quotient group is ${\rm Gal}(K/F)$ for some field $K$ between $E$ and $F$ that is in fact the maximal abelian extension of $F$ inside $E$. – 2017-01-31

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