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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.

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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".

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The only thing that Milne assumes is the Artin reciprocity law and the existence theorem. I don't see how either of them can be used to show that there's a smallest modulus. If Artin reciprocity gives us one and we know another. Then we would need to show that the gcd of them is also satisfied. I'll have to hunt down Cassels and Frolich... – pki Nov 6 '11 at 0:45

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