TITLE: The conductor and minimal moduli of abelian extensions
QUESTION [6 upvotes]: Assume that $L/K$ is a finite abelian extension of global fields and $S$ the set of primes of $K$ ramifying in $L$. Then the conductor $\mathfrak{f}(L/K)$ is the smallest modulus s.t. the Artin map
$$\psi_{L/K}:I_K^S \to \operatorname{Gal}(L/K)$$
factors through the ray class group $C_{\mathfrak{f}(L/K)}$. I'm looking at Milne's notes on CFT and he seems to sort of skip this whole existence and uniqueness thing to a remark without proving anything. My question is that given two different moduli $\mathfrak{m}$ and $\mathfrak{n}$, both containing the ramifying primes of $K$, then to prove the uniqueness of the conductor, we just need to show that the Artin map factors through the ray class group of the moduli
$$\prod_{\mathfrak{p}} \mathfrak{p}^{\min\{\mathfrak{n}(\mathfrak{p}),\mathfrak{m}(\mathfrak{p})\}}.$$
Is there a slick way of proving this? I just seem to get into a mess of multiplicative congruences while trying to figure out what's in the kernel of $\psi_{L/K}$. Maybe I'm just missing the completely obvious.

REPLY [3 votes]: What are you giving yourself?  If you don't assume that Artin reciprocity law, then you don't know that there is any modulus at all.  If you assume Artin reciprocity, then a standard (but very nice, and very important) argument (explained well in Tate's article in Cassels and Frolich) allows one to pass from the ideal-theoretic to the adelic point of view, and from the latter point of view, the existence of a minimal modulus is clear.   
With regard to getting "into a mess of multiplicative congruences", this is 
precisely what the ideles are good at dealing with, so I think that the idelic approach provides at least one "slick way of proving this".