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I'm confused about admissible moduli and conductors of abelian extensions as defined in Lang's book Algebraic Number Theory. Let $k$ be a number field with idele group $J_k$, and let $\mathfrak{m}$ be a modulus of $k$. Lang defines the ray class field modulo $\mathfrak{m}$ to be the class field corresponding to the open subgroup $k^\times W_\mathfrak{m} \subset J_k$, where $W_\mathfrak{m}$ is just the product over all places $v$ of the corresponding open subgroup of $k_v^\times$ depending on the multiplicity of $v$ in $\mathfrak{m}$ (if the multiplicity is zero then the subgroup is the whole group of units in the nonarchimedean case and is the whole multiplicative group $\mathbf{R}^\times$ or $\mathbf{C}^\times$ in the archimedean case).

From the idelic statements of class field theory and basic facts about the ideles, the ray class field $K_\mathfrak{m}/K$ satisfies $I(\mathfrak{m})/P_\mathfrak{m} \cong J_{k}/k^\times W_\mathfrak{m} = J_k/k^\times N_{K_\mathfrak{m}/k}J_{K_\mathfrak{m}} \cong \mathrm{Gal}(K_\mathfrak{m}/k)$, where the last isomorphism is given by the idelic Artin map induced by the isomorphism $J_k/k^\times N_{K_\mathfrak{m}/k}J_{K_\mathfrak{m}} \cong I(\mathfrak{c})/P_\mathfrak{c}\mathfrak{N}(\mathfrak{c})$ for an admissible modulus $\mathfrak{c}$, namely one that satisfies $W_\mathfrak{c} \subseteq N_{K_\mathfrak{m}/k}J_{K_\mathfrak{m}}$ (then the global reciprocity law takes care of $I(\mathfrak{c})/P_\mathfrak{c}\mathfrak{N}(\mathfrak{c}) \cong \mathrm{Gal}(K_\mathfrak{m}/k)$). It isn't clear to me that $\mathfrak{m}$ is admissible in the sense above for $K/k$ - indeed the only thing we know is that $k^\times W_\mathfrak{m} = k^\times N_{K_\mathfrak{m}/k}J_{K_\mathfrak{m}}$, which I don't think implies $W_\mathfrak{m} \subseteq N_{K_{\mathfrak{m}}/k}J_{K_\mathfrak{m}}$. Using just the existence theorem and the class field statements in Lang, is it possible to see that the Artin map actually does give the isomorphism $I(\mathfrak{m})/P_\mathfrak{m} \to \mathrm{Gal}(K_\mathfrak{m}/k)$, even though $\mathfrak{m}$ is not necessarily admissible? I don't think it's obvious from what I've written down, since the Artin map on the idele groups in the picture are only defined in terms of idealic Artin maps for admissible moduli. The only thing that is obvious so far is that the Artin map $I(\mathfrak{m}) \to \mathrm{Gal}(K_\mathfrak{m}/k)$ exists, since $K_\mathfrak{m}/k$ is unramified outside $\mathfrak{m}$ by its definition as a class field. I just don't see why $P_\mathfrak{m}$ should be in the kernel, since it isn't clear that the big isomorphism I wrote down coincides with the Artin map.

One interesting thing is that this does become obvious when $\mathfrak{m}$ is trivial and $K_\mathfrak{m} = K$ is the Hilbert class field of $k$. Then $\mathfrak{m}$ is clearly admissible since the fact that $K/k$ is nowhere ramified means all the local units are local norms, so we can just apply the statements in Lang about admissible moduli.

This whole thing seems to reflect the fact that a modulus being admissible is much stronger than being divisible by the conductor of $K/k$. Is there a concrete way in which the conductor and the minimal admissible modulus are related?

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  • $\begingroup$ To make things more elementary you may take $F/K$ abelian with $K$ a totally complex number field (whose extensions have no ramification at infinite places) then CFT says the kernel of the Artin map contains the principal ideals $P_{a,1\bmod d}=\{(a), a \in 1+d O_K\}$ for some $d\in O_K$ contained only in ramified primes, and that for a fixed $d$ there exists some $F$ whose kernel is exactly $P_{a,1\bmod d}$ thus its subfields correspond to the quotients of the finite group $I_K / P_{a,1\bmod d}$ $\endgroup$
    – reuns
    Aug 25, 2019 at 5:52
  • $\begingroup$ The part I don't see how to prove is why there exists an $F$ whose kernel is exactly $P_{a, 1 \mod d}$. Lang only has the idelic version of the existence theorem, and the reason I asked the question was because I was trying to translate between an idealic and idelic version of it. The idelic statement of the existence theorem only guarantees the existence of an abelian $F/K$ such that the kernel of the idelic Artin map is $k^\times W_\mathfrak{m}$. This does imply the abstract isomorphism between the ray class group $I(\mathfrak{m})/P_\mathfrak{m}$ and the Galois group (see my question), but $\endgroup$
    – babu_babu
    Aug 25, 2019 at 6:20
  • $\begingroup$ (continued) does not imply that this isomorphism is actually the Artin map, since the definition of the idelic Artin map depends on an admissible modulus, which $\mathfrak{m}$ is not. $\endgroup$
    – babu_babu
    Aug 25, 2019 at 6:21

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