# Question about the modulus of a number field

Milne defines the conductor of an abelian extension $L/K$ to be the smallest modulus $\mathfrak{m}$ s.t. the Artin map factors as

$$\psi_{L/K}:I_K^{\mathfrak{m}}\to \textrm{Cl}_\mathfrak{m}(K)\to \textrm{Gal}(L/K)$$

I'm trying to understand the class field theoretic proof of the Kronecker-Weber theorem. For the proof to work, one needs to show that given some conductor $\mathfrak{m}$ and its ray class field $L_\mathfrak{m}$, then the abelian extensions of $K$ contained in $L_\mathfrak{m}$ are precisely the fields s.t. their conductor divides $\mathfrak{m}$. Hence, one needs to prove (denoting the conductor of $L/K$ by $\mathfrak{f}(L/K)$) that

$$\mathfrak{f}(L/K)\mid \mathfrak{m}\Rightarrow L\subset L_\mathfrak{m}$$

Anyone know how to prove this? I've been looking at a few other books, but they seem to define things differently, so the proofs did not translate to Milne's definition, which is what I'm currently trying to understand.

EDIT: This question is about trying to fill in the details in Remark 3.8 on page 155 in his notes: http://www.jmilne.org/math/CourseNotes/CFT.pdf

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One way to argue is as follows: If the conductor $\mathfrak f$ of $L$ over $K$ divides $\mathfrak m$, then all the primes that split completely in $L_{\mathfrak m}$ also split completely in $L$. Consider now the compositum of $L$ and $L_{\mathfrak m}$; one sees that the primes that split completely in this extension are precisely the primes that split completely in $L_{\mathfrak m}$.
But a standard consequence of Cebotarev density is: if $E_1$ and $E_2$ are two Galois extensions of the number field $K$ (in some fixed algebraic closure $\overline{K}$) and the sets of primes that splits completely in each of them coincide (perhaps away from a finite number of bad primes), then $E_1$ and $E_2$ are equal.
This implies that $L_{\mathfrak m} L = L_{\mathfrak m}$, i.e. that $L \subset L_{\mathfrak m}$, as desired.
[In the context of abelian extensions there are other proofs that avoid Cebotarev; e.g. once you have the whole class field theory apparatus, the inclusion you ask about follows immediately. But the above argument is a stanard and important one, which gets at the heart of how splitting behaviour can pin down a field --- after all, $L_{\mathfrak m}$ is defined a priori by the splitting behaviour of primes, and it is the above result that shows that this splitting behaviour suffices to pin down $L_{\mathfrak m}$ uniquely.]
Thanks. The fact that $(p,L/K)=1\Rightarrow (p,L_\mathfrak{m}/K)=1$ is an obvious consequence of the definition, yet I had never thought about how useful this is. I don't mind that the proof uses Chebotarev as the Chebotarev part is a very standard argument. –  pki Nov 6 '11 at 0:25